EverythingScience Grade10
EverythingScience Grade10
EverythingScience Grade10
1 IA
1
2.2
H
3
2 IIA
1,01
0.98 4
1.57
No
13 IIIA
5
EN
14 IVA
2.04 6
15 VA
2.55 7
16 VIA
3.04 8
He
17 VIIA
3.44 9
4,00
3.98 10
Li
Be
Element
6,94
9,01
AMU
10,8
12,0
14,0
16,0
19,0
11
0.93 12
13
1.31
Na
Mg
23,0
24,3
19
0.93 20
3 IIIB
1.00 21
4 IVB
1.36 22
5 VB
1.54 23
6 VIB
1.63 24
7 VIIB
1.66 25
8 VII
1.55 26
9 VII
1.33 27
10 VII
1.88 28
11 IB
1.91 29
1.61 14
2.19 16
2.58 17
20,2
3.16 18
Al
Si
Cl
Ar
27,0
12 IIB
1.90 30
1.90 15
Ne
28,1
31,0
32,1
35,45
39,9
1.65 31
1.81 32
2.01 33
2.18 34
2.55 35
2.96 36
Ca
Sc
Ti
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
39,1
40,1
45,0
47,9
50,9
52,0
54,9
55,8
58,9
58,7
63,5
65,4
69,7
72,6
74,9
79,0
79,9
83,8
37
0.82 38
0.95 39
1.22 40
1.33 41
2.16 43
1.60 42
1.90 44
2.20 45
2.28 46
2.20 47
1.93 48
1.69 49
1.78 50
1.96 51
2.05 52
2.10 53
2.66 54
Rb
Sr
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
Xe
85,5
87,6
88,9
91,2
92,9
95,9
(98)
101,1
102,9
106,4
107,9
112,4
114,8
118,7
121,8
127,6
126,9
131,3
55
0.79 56
0.89
57-71
72
1.30 73
1.50 74
2.36 75
1.90 76
2.20 77
2.20 78
2.28 79
2.54 80
2.00 81
1.62 82
2.02 84
2.33 83
2.00 85
2.20 86
Cs
Ba
La-Lu
Hf
Ta
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
132,9
137,3
Lanthanides
178,5
180,9
183,8
186,2
190,2
192,2
195,1
197,0
200,6
204,4
207,2
209,0
(209)
(210)
(222)
87
0.7 88
0.9 89-103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
Fr
Ra
Ac-Lr
Rf
Db
Sg
Bh
Hs
Mt
Ds
Rg
Cn
Uut
Uuq
Uup
Uuh
Uus
Uuo
(223)
226,0
Actinides
(261)
(262)
(263)
(262)
(265)
(266)
(269)
(272)
(277)
(284)
(289)
(288)
(293)
(282)
(282)
Transition Metal
57
Metal
Actinide
89
1.12 59
1.10 90
1.14 61
1.13 60
Pr
140,1
140,9
1.30 91
62
1.17 63
64
Nd
Ce
138,9
Noble Gas
Lanthanide
1.10 58
La
Metalloid
Non-metal
Pm
Sm
Eu
144,2
(145)
150,4
152,0
1.50 92
1.38 93
1.36 94
1.28 95
1.20 65
66
1.22 67
1.23 68
1.24 69
Gd
1.30 96
Tb
Dy
Ho
Er
157,3
158,9
162,5
164,9
167,3
1.30 97
1.30 98
1.30 99
1.30 100
1.25 70
71
1.27
Tm
Yb
Lu
168,9
173,0
175,0
1.30 101
1.30 102
1.30 103
Ac
Th
Pa
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
227,0
232,0
231,0
238,0
237,0
(244)
(243)
(247)
(247)
(251)
(252)
(257)
(258)
(258)
(260)
Everything Science
Grade 10 Physical Science
Version 1 CAPS
Copyright notice
Authors List
This book is based upon the original Free High School Science Text which was
entirely written by volunteer academics, educators and industry professionals.
Their vision was to see a curriculum aligned set of mathematics and physical
science textbooks which are freely available to anybody and exist under an
open copyright license.
Siyavula core team
Mark Horner; Heather Williams; Ren Toerien; Jayanthi SK Maharaj (Veena); Morongwa Masemula; Elize Jones;
Kevin Reddy; Marius Diergaardt; Wetsie Visser
Moon; Calvin Moore; Bhavani Morarjee; Kholofelo Moyaba; Nina Gitau Muchunu; Kate Murphy; Emmanuel
Musonza; Tom Mutabazi; David Myburgh; Kamie Naidu; Nolene Naidu; Gokul Nair; Vafa Naraghi; Bridget
Nash; Tyrone Negus; Huw Newton-Hill; Buntu Ngcebetsha; Dr. Markus Oldenburg; Thomas ODonnell; Dr.
Jaynie Padayachee; Poveshen Padayachee; Masimba Paradza; Dave Pawson; Justin Pead; Nicolette Pekeur; Sirika
Pillay; Jacques Plaut; Barry Povey; Andrea Prinsloo; Joseph Raimondo; Sanya Rajani; Prof. Sergey Rakityansky;
Alastair Ramlakan; Dr. Matina J. Rassias; Dr. Jocelyn Read; Jonathan Reader; Jane Reddick; Dr. Matthew
Reece; Razvan Remsing; Laura Richter; Max Richter; Sean Riddle; Dr. David Roberts; Christopher Roberts; Helen
Robertson; Evan Robinson; Raoul Rontsch; Dr. Andrew Rose; Katie Ross; Jeanne-Mari Roux; Mark Roux; Bianca
Ruddy; Nitin Rughoonauth; Katie Russell; Steven Sam; Dr. Carl Schefer; Cho Hee Shrader; Nathaniel Schwartz;
Duncan Scott; Helen Seals; Relebohile Sefako; Prof. Sergey Rakityansky; Sandra Serumaga-Zake; Paul Shangase;
Cameron Sharp; Ian Sherratt; Dr. James Short; Roger Sieloff; Brandon Sim; Bonga Skozana; Clare Slotow; Bradley
Smith; Greg Solomon; Nicholas Spaull; Dr. Andrew Stacey; Dr. Jim Stasheff; Mike Stay; Mike Stringer; Masixole
Swartbooi; Tshenolo Tau; Tim Teatro; Ben Thompson; Shen Tian; Xolani Timbile; Robert Torregrosa; Jimmy
Tseng; Tim van Beek; Neels van der Westhuizen; Frans van Eeden; Pierre van Heerden; Dr. Marco van Leeuwen;
Marina van Zyl; Pieter Vergeer; Rizmari Versfeld; Mfundo Vezi; Mpilonhle Vilakazi; Ingrid von Glehn; Tamara
von Glehn; Kosma von Maltitz; Helen Waugh; Leandra Webb; Dr. Dawn Webber; Michelle Wen; Neels van der
Westhuizen; Dr. Alexander Wetzler; Dr. Spencer Wheaton; Vivian White; Dr. Gerald Wigger; Harry Wiggins;
Heather Williams; Wendy Williams; Julie Wilson; Timothy Wilson; Andrew Wood; Emma Wormauld; Dr. Sahal
Yacoob; Jean Youssef; Ewald Zietsman;
iv
Everything Science
When we look outside at everything in nature, look around us at everything manufactured or look up at everything
in space we cannot but be struck by the incredible diversity and complexity of life; so many things, that look so
different, operating in such unique ways. The physical universe really contains incredible complexity.
Yet, what is even more remarkable than this seeming complexity is the fact that things in the physical universe
are knowable. We can investigate them, analyse them and understand them. It is this ability to understand the
physical universe that allows us to transform elements and make technological progress possible.
If we look back at some of the things that developed over the last century space travel, advances in medicine,
wireless communication (from television to mobile phones) and materials a thousand times stronger than steel
we see they are not the consequence of magic or some inexplicable phenomena. They were all developed
through the study and systematic application of the physical sciences. So as we look forward at the 21st century
and some of the problems of poverty, disease and pollution that face us, it is partly to the physical sciences we
need to turn.
For however great these challenges seem, we know that the physical universe is knowable and that the dedicated
study thereof can lead to the most remarkable advances. There can hardly be a more exciting challenge than
laying bare the seeming complexity of the physical universe and working with the incredible diversity therein to
develop products and services that add real quality to peoples lives.
Physical sciences is far more wonderful, exciting and beautiful than magic! It is everywhere. See introductory
video by Dr. Mark Horner:
VPsfk at www.everythingscience.co.za
Everything Science is not just a Science textbook. It has everything you expect from your regular printed school
textbook, but comes with a whole lot more. For a start, you can download or read it on-line on your mobile
phone, computer or iPad, which means you have the convenience of accessing it wherever you are.
We know that some things are hard to explain in words. That is why every chapter comes with video lessons and
explanations which help bring the ideas and concepts to life. Summary presentations at the end of every chapter
offer an overview of the content covered, with key points highlighted for easy revision.
All the exercises inside the book link to a service where you can get more practice, see the full solution or test
your skills level on mobile and PC.
We are interested in what you think, wonder about or struggle with as you read through the book and attempt the
exercises. That is why we made it possible for you to use your mobile phone or computer to digitally pin your
question to a page and see what questions and answers other readers pinned up.
Go directly to a section
(V123)
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To watch the videos on-line, practise your skills or post a question, go to the Everything Science website at
www.everythingscience.co.za on your mobile or PC and enter the short-code in the navigation box.
Video lessons
Look out for the video icons inside the book. These will take you to video lessons that help bring the ideas and
concepts on the page to life. Get extra insight, detailed explanations and worked examples. See the concepts in
action and hear real people talk about how they use maths and science in their work.
See video explanation
(Video: V123)
Video exercises
Wherever there are exercises in the book you will see icons and short-codes for video solutions, practice and
help. These short-codes will take you to video solutions of select exercises to show you step-by-step how to solve
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CONTENTS
CONTENTS
Contents
1 Skills for science
1.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1
1.2
Mathematical skills . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3
Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4
1.5
Hazard signs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2 Classication of matter
23
2.1
Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
2.2
2.3
2.4
2.5
2.6
2.7
2.8
Mixtures . . . . . . . . . . . . . . . . . . . . . . . .
Pure substances . . . . . . . . . . . . . . . . . . . .
Names and formulae of substances . . . . . . . . . .
Metals, Metalloids and Non-metals . . . . . . . . . .
Electrical conductors, semi-conductors and insulators
Thermal Conductors and Insulators . . . . . . . . . .
Magnetic and Non-magnetic Materials . . . . . . . .
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63
63
64
67
70
76
79
95
6 Chemical bonding
6.1
6.2
6.3
55
States of matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
The kinetic molecular theory . . . . . . . . . . . . . . . . . . . . . . . . . 59
4 The atom
4.1
Introduction . . . . . . . .
4.2
Models of the atom . . . .
4.3
Atomic mass and diameter
4.4
Structure of the atom . . .
4.5
Isotopes . . . . . . . . . .
4.6
Electronic conguration . .
5.1
5.2
25
30
35
42
45
47
49
105
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
Lewis structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
Covalent Bonding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
i
CONTENTS
6.4
6.5
6.6
CONTENTS
7 Transverse pulses
7.1
7.2
7.3
124
8 Transverse waves
8.1
Introduction . . . . . . . . . .
8.2
What is a transverse wave? . .
8.3
Crests and troughs . . . . . . .
8.4
Amplitude . . . . . . . . . . .
8.5
Points in phase . . . . . . . .
8.6
Period and frequency . . . . .
8.7
Speed of a transverse wave . .
139
. . . . . . . . . . . . . . . . . . . . . . . . 139
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139
141
142
146
147
149
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157
158
159
160
161
10 Sound
10.1 Introduction . . . . . . . . . . .
10.2 Speed of sound . . . . . . . . .
10.3 Characteristics of a sound wave .
10.4 Ultrasound . . . . . . . . . . . .
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166
166
168
172
177
9 Longitudinal waves
9.1
9.2
9.3
9.4
9.5
157
11 Electromagnetic radiation
186
11.1
11.2
11.3
11.4
11.5
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187
190
193
196
203
213
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
Conservation of atoms and mass in reactions . . . . . . . . . . . . . . . . 221
Law of constant composition . . . . . . . . . . . . . . . . . . . . . . . . . 225
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230
CONTENTS
14.1
14.2
CONTENTS
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230
Balancing chemical equations . . . . . . . . . . . . . . . . . . . . . . . . 232
15 Magnetism
245
15.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245
15.2 The compass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253
16 Electrostatics
16.1 Introduction and key concepts
16.2 Two kinds of charge . . . . . .
16.3 Conservation of charge . . . .
16.4 Quantisation of charge . . . .
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17 Electric circuits
17.1
17.2
17.3
17.4
17.5
280
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259
259
259
265
267
280
282
287
290
296
314
Introduction . . . . . . . . . . . . . . .
Ions in aqueous solution . . . . . . . .
Electrolytes, ionisation and conductivity
Precipitation reactions . . . . . . . . .
Other types of reactions . . . . . . . . .
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.
.
.
.
.
.
.
314
314
317
320
325
Composition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342
19.3
19.4
364
20.1
20.2
20.3
. . . . . . . . . . . . . . . . . . . . . . . . 375
21.3
21.4
Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402
21.5
21.6
21.7
iii
CONTENTS
CONTENTS
22 Mechanical energy
443
22.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443
22.2
22.3
22.4
22.5
23 The hydrosphere
470
23.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470
23.2
23.3
23.4
23.5
23.6
481
. . . . . . . . . . . . . . . . . . . . . . . . . 482
484
25.1
25.2
25.3
25.4
25.5
25.6
25.7
25.8
25.9
iv
Introduction
ESAA
In the physical sciences there are many skills that you need to learn. These include working
with units, basic mathematics skills and laboratory skills. In this chapter we will revise some
of these skills that you should know before starting to study physical science. This chapter
is intended as a reference guide to assist you in your journey of studying physical science.
See introductory video: (
Mathematical skills
ESAB
You should be comfortable with scientic notation and how to write scientic notation.
You should also be able to easily convert between different units and change the subject of
a formula. In addition, concepts such as rate, direct and indirect proportion, fractions and
ratios and the use of constants in equations are important.
Rounding off
ESAC
Certain numbers may take an innite amount of paper and ink to write out. Not only is
that impossible, but writing numbers out to a high precision (many decimal places) is very
inconvenient and rarely gives better answers. For this reason we often estimate the number
to a certain number of decimal places.
Rounding off a decimal number to a given number of decimal places is the quickest way
to approximate a number. For example, if you wanted to round-off 2, 6525272 to three
decimal places then you would rst count three places after the decimal. Next you mark
this point with a |: 2, 652|5272. All numbers to the right of | are ignored after you determine
whether the number in the third decimal place must be rounded up or rounded down. You
round up the nal digit (make the digit one more) if the rst digit after the | is greater than
1
1.2
or equal to 5 and round down (leave the digit alone) otherwise. So, since the rst digit
after the | is a 5, we must round up the digit in the third decimal place to a 3 and the nal
ESAD
Scientic notation
In science one often needs to work with very large or very small numbers. These can be
written more easily (and more compactly) in scientic notation, in the general form:
N 10n
where N is a decimal number between 0 and 10 that is rounded off to a few decimal
places. n is known as the exponent and is an integer. If n > 0 it represents how many
times the decimal place in N should be moved to the right. If n < 0, then it represents how
many times the decimal place in N should be moved to the left. For example 3, 24 103
represents 3 240 (the decimal moved three places to the right) and 3, 24 103 represents
0, 00324 (the decimal moved three places to the left).
If a number must be converted into scientic notation, we need to work out how many
times the number must be multiplied or divided by 10 to make it into a number between
1 and 10 (i.e. the value of n) and what this number between 1 and 10 is (the value of N ).
We do this by counting the number of decimal places the decimal comma must move.
For example, write the speed of light (299 792 458 m s1 ) in scientic notation, to two
decimal places. First, we nd where the decimal comma must go for two decimal places (to
nd N ) and then count how many places there are after the decimal comma to determine
n.
In this example, the decimal comma must go after the rst 2, but since the number after
the 9 is 7, N = 3, 00. n = 8 because there are 8 digits left after the decimal comma. So the
speed of light in scientic notation, to two decimal places is 3, 00 108 m s1 .
We can also perform addition, subtraction, multiplication and division with scientic notation. The following two worked examples show how to do this:
1.2
SOLUTION
Step 1 : Make all the exponents the same
To add or subtract numbers in scientic notation we must make all the
exponents the same:
1, 99 1026 = 0, 199 1025 and
1, 67 1027 = 0, 0167 1025
Note that we follow the same process if the exponents are positive. For example 5, 1
103 + 4, 2 104 = 4, 71 104 .
Example 2:
notation
QUESTION
1, 6 1019 3, 2 1019 5 1021
SOLUTION
1.3
Note that we follow the same process if the exponents are positive. For example: 5, 1
103 4, 2 104 = 21, 42 107 = 2, 142 108
Units
ESAE
Imagine you had to make curtains and needed to buy fabric. The shop assistant would
need to know how much fabric you needed. Telling her you need fabric 2 wide and 6 long
would be insufcient you have to specify the unit (i.e. 2 metres wide and 6 metres long).
Without the unit the information is incomplete and the shop assistant would have to guess.
If you were making curtains for a dolls house the dimensions might be 2 centimetres wide
and 6 centimetres long!
It is not just lengths that have units, all physical quantities have units (e.g. time, temperature, distance, etc.).
There are many different systems of units. The main systems of units are:
SI units
c.g.s units
Imperial units
Natural units
1.3
SI Units
ESAF
We will be using the SI units in this course. SI units are the internationally agreed upon
units.
DEFINITION: SI Units
The name SI units comes from the French Systme International dUnits,
which means international system of units.
There are seven base SI units. These are listed in table 1.1. All physical quantities have
units which can be built from these seven base units. So, it is possible to create a different
set of units by dening a different set of base units.
These seven units are called base units because none of them can be expressed as combinations of the other six. These base units are like the 26 letters of the alphabet for English.
Many different words can be formed by using these letters.
Base quantity
Name
Symbol
length
metre
mass
kilogram
kg
time
second
electric current
ampere
temperature
kelvin
amount of substance
mole
mol
luminous intensity
candela
cd
ESAG
The SI Units are not the only units available, but they are most widely used. In Science
there are three other sets of units that can also be used. These are mentioned here for
5
1.3
interest only.
c.g.s. Units
In the c.g.s. system, the metre is replaced by the centimetre and the kilogram is
replaced by the gram. This is a simple change but it means that all units derived from
these two are changed. For example, the units of force and work are different. These
units are used most often in astrophysics and atomic physics.
Imperial Units
Imperial units arose when kings and queens decided the measures that were to be
used in the land. All the imperial base units, except for the measure of time, are
different to those of SI units. This is the unit system you are most likely to encounter
if SI units are not used. Examples of imperial units are pounds, miles, gallons and
yards. These units are used by the Americans and British. As you can imagine,
having different units in use from place to place makes scientic communication
very difcult. This was the motivation for adopting a set of internationally agreed
upon units.
Natural Units
This is the most sophisticated choice of units. Here the most fundamental discovered
quantities (such as the speed of light) are set equal to 1. The argument for this choice
is that all other quantities should be built from these fundamental units. This system
of units is used in high energy physics and quantum mechanics.
ESAH
To make working with units easier, some combinations of the base units are given special
names, but it is always correct to reduce everything to the base units. Table 1.2 lists some
examples of combinations of SI base units that are assigned special names. Do not be
concerned if the formulae look unfamiliar at this stage - we will deal with each in detail in
the chapters ahead (as well as many others)!
It is very important that you are able to recognise the units correctly. For example, the
newton (N) is another name for the kilogram metre per second squared (kgms2 ), while
the kilogram metre squared per second squared (kgm2 s2 ) is called the joule (J).
1.3
Quantity
Formula
Name of Combination
Force
ma
kg m s2
N (newton)
Frequency
1
T
s1
Hz (hertz)
Work
Fs
kg m2 s2
J (joule)
Table 1.2: Some examples of combinations of SI base units assigned special names
ESAI
Tip
Now that you know how to write numbers in scientic notation, another important aspect
of units is the prexes that are used with the units. In the case of units, the prexes have
a special use. The kilogram (kg) is a simple example. 1 kg is equal to 1 000 g or 1 103
g. Grouping the 103 and the g together we can replace the 103 with the prex k (kilo).
Therefore the k takes the place of the 103 . The kilogram is unique in that it is the only SI
base unit containing a prex.
In science, all the prexes used with units are some power of 10. Table 1.3 lists some of
these prexes. You will not use most of these prexes, but those prexes listed in bold
should be learnt. The case of the prex symbol is very important. Where a letter features
twice in the table, it is written in uppercase for exponents bigger than one and in lowercase
for exponents less than one. For example M means mega (106 ) and m means milli (103 ).
1.3
Prex
Symbol
Exponent
Prex
Symbol
Exponent
yotta
1024
yocto
1024
zetta
1021
zepto
1021
exa
1018
atto
1018
peta
1015
femto
1015
tera
1012
pico
1012
giga
109
nano
109
mega
106
micro
106
kilo
103
milli
103
hecto
102
centi
102
deca
da
101
deci
101
Tip
Exercise 1 - 1
1.3
d. 250 nm
e. 0, 00035 hg
3. Write the following using the prexes in Table 1.3.
a. 1, 602 1019 C
b. 1, 992 106 J
c. 5, 98 104 N
d. 25 104 A
e. 0, 0075 106 m
More practice
(1.) 02u4
(2.) 01v7
video solutions
or help at www.everythingscience.co.za
(3.) 01v8
ESAJ
Without units much of our work as scientists would be meaningless. We need to express
our thoughts clearly and units give meaning to the numbers we measure and calculate.
Depending on which units we use, the numbers are different. For example if you have
12 water, it means nothing. You could have 12 ml of water, 12 litres of water, or even 12
bottles of water. Units are an essential part of the language we use. Units must be specied
when expressing physical quantities. Imagine that you are baking a cake, but the units, like
grams and millilitres, for the our, milk, sugar and baking powder are not specied!
1.3
The Mars Climate Orbiter, a key craft in the space agencys exploration of the
red planet, vanished on 23 September after a 10 month journey. It is believed
that the craft came dangerously close to the atmosphere of Mars, where it presumably burned and broke into pieces.
An investigation board concluded that NASA engineers failed to convert English measures of rocket thrusts to newton, a metric system measuring rocket
force. One English pound of force equals 4,45 newtons. A small difference
between the two values caused the spacecraft to approach Mars at too low an
altitude and the craft is thought to have smashed into the planets atmosphere
and was destroyed.
The spacecraft was to be a key part of the exploration of the planet. From its
station about the red planet, the Mars Climate Orbiter was to relay signals from
the Mars Polar Lander, which is scheduled to touch down on Mars next month.
The root cause of the loss of the spacecraft was a failed translation of English
units into metric units and a segment of ground-based, navigation-related mission software, said Arthur Stephenson, chairman of the investigation board.
Questions:
1. Why did the Mars Climate Orbiter crash? Answer in your own words.
2. How could this have been avoided?
3. Why was the Mars Orbiter sent to Mars?
4. Do you think space exploration is important? Explain your answer.
ESAK
It is very important that you are aware that different systems of units exist. Furthermore, you
must be able to convert between units. Being able to change between units (for example,
converting from millimetres to metres) is a useful skill in Science.
The following conversion diagrams will help you change from one unit to another.
mm
1000
1000
1000
km
1000
1.3
change volumes:
m
cm3
1000
1000
dm3
1000
k
m3
1000
Example 3: Conversion 1
QUESTION
Express 3 800 mm in metres.
SOLUTION
Example 4: Conversion 2
11
1.3
QUESTION
Convert 4,56 kg to g.
SOLUTION
Step 1 : Find the two units on the conversion diagram.
Use Figure 1.2. Kilogram is the same as kilometre and gram is the same
as metre.
Step 2 : Decide whether you are moving to the left or to the right.
You need to go from kg to g, so it is from right to left.
Step 3 : Read from the diagram what you must do and nd the answer.
4, 56 kg 1 000 = 4 560 g
12
1.3
ESAL
Very often in science you will have to change the subject of a formula. We will look at two
examples. (Do not worry if you do not yet know what the terms and symbols mean, these
formulae will be covered later in the book.)
1. Moles
m
The equation to calculate moles from molar mass is: n = M , where n is the number
of moles, m is the mass and M is the molar mass. As it is written we can easily nd
the number of moles of a substance. But what if we have the number of moles and
want to nd the molar mass? We note that we can simply multiply both sides of the
equation by the molar mass and then divide both sides by the number of moles:
n
nM
m
M
m
m
n
Plancks constant, c is the speed of light and is the wavelength. To get c we can do
the following:
E
hc
E
h
h
hc
E
E
c .
ESAM
In science we often want to know how a quantity relates to another quantity or how something changes over a period of time. To do this we need to know about rate, proportion
and ratios.
Rate:
13
1.3
The rate at which something happens is the number of times that it happens over a period
of time. The rate is always a change per time unit. So we can get rate of change of velocity
per unit time ( v ) or the rate of change in concentration per unit time (or C ). (Note that
t
t
represents a change in).
Ratios and fractions:
A fraction is a number which represents a part of a whole and is written as a , where a is
b
the numerator and b is the denominator. A ratio tells us the relative size of one quantity
(e.g. number of moles of reactants) compared to another quantity (e.g. number of moles of
product): 2 : 1, 4 : 3, etc. Ratios can also be written as fractions as percentages (fractions
with a denominator of 100).
Proportion:
Proportion is a way of describing relationships between values or between constants. We
can say that x is directly proportional to y (x y) or that a is inversely proportional to
b (a 1 ). It is important to understand the difference between directly and inversely
b
proportional.
Directly proportional
Two values or constants are directly proportional when a change in one leads to the
same change in the other. This is a more-more relationship. We can represent this as
y x or y = kx where k is the proportionality constant. We have to include k since
we do not know by how much x changes when y changes. x could change by 2 for
every change of 1 in y. If we plot two directly proportional variables on a graph, then
we get a straight line graph that goes through the origin (0; 0):
Inversely proportional
Two values or constants are inversely proportional when a change in one leads to the
k
opposite change in the other. We can represent this as y = x . This is a more-less
relationship. If we plot two inversely proportional variables we get a curve that never
14
1.3
Constants in equations
ESAN
A constant in an equation always has the same value. For example the speed of light
(c = 2, 99108 m s1 ), Plancks constant (h) and Avogadros number (NA ) are all examples
of constants that are use in science. The following table lists all the constants that you will
encounter in this book.
Constant
Symbol
SI Units
1, 67 1024 g
1, 67 1027 kg
Charge on an electron
1, 6 1019 C
1, 6 1019 s A
344m s1
Speed of light
3 108 m s1
Plancks constant
6, 626 1034 J s
Avogadros number
NA
6, 022 1023
Gravitational acceleration
9, 8 m s1
Trigonometry
6, 626 1034 kg m2 s1
ESAO
se
nu
te
po
hy
opposite
Trigonometry is the relationship between the angles and sides of right angled triangles.
Trigonometrical relationships are ratios and therefore have no units. You should know the
following trigonometric ratios:
A
adjacent
Sine
This is dened as sinA =
Cosine
opposite
hypotenuse
adjacent
hypotenuse
15
1.3
Tangent
This is dened as tanA =
opposite
adjacent
Exercise 1 - 2
More practice
(1.) 01v9
16
(2.) 01va
video solutions
(3.) 01vb
(4.) 01vc
or help at www.everythingscience.co.za
(5.) 01vd
(6.) 01ve
1.4
ESAP
To carry out experiments in the laboratory you need to know how to properly present
your experimental results, you also need to know how to read instruments and how to
interpret your data. A laboratory (be it for physics, chemistry or other sciences) can be a
very dangerous and daunting place. However, if you follow a few simple guidelines you
can safely carry out experiments in the laboratory without endangering yourself or others
around you.
Experiments
ESAQ
This process is known as the scientic method. In the work that you will do you will be
given the rst three items and be required to determine the last four items. For verifying
results you should see what your classmates obtained for their experiment.
In science the recording of practical work follows a specic layout. You should always
present your work using this layout, as it will help any other person be able to understand
and repeat your experiment.
Aim: A brief sentence describing the purpose of the experiment.
Apparatus: A sketch of the apparatus and a list of the apparatus
17
1.4
Laboratory apparatus
ESAR
Listed here are some of the common pieces of apparatus that you will be working with in
the laboratory. You should be able to name all the apparatus listed here as well as make a
simple sketch of it.
18
Item
Photo
1.4
Sketch
Beaker
Flask
Test tubes
Bunsen burner
Measuring cylinder
Pipette
Watch glass
Thermometer
Funnel
19
1.4
The following image shows the correct setup for heating liquids on a Bunsen burner:
When reading any instrument (such as a measuring cylinder, a pipette, etc.) always make
sure that the instrument is level and that your eye is at the level of the top of the liquid.
ESAS
The following are some of the general guidelines and rules that you should always observe
when working in a laboratory.
1. You are responsible for your own safety as well as the safety of others in the laboratory.
2. Do not eat or drink in the laboratory. Do not use laboratory glassware to eat or drink
from.
3. Always behave responsibly in the laboratory. Do not run around or play practical
jokes.
4. In case of accidents or chemical spills call your teacher at once.
5. Always check with your teacher how to dispose of waste. Chemicals should not be
disposed of down the sink.
6. Only perform the experiments that your teacher instructs you to. Never mix chemicals for fun.
7. Never perform experiments alone.
8. Always check the safety data of any chemicals you are going to use.
9. Follow the given instructions exactly. Do not mix up steps or try things in a different
order.
10. Be alert and careful when handling chemicals, hot glassware, etc.
11. Ensure all Bunsen burners are turned off at the end of the practical and all chemical
containers are sealed.
12. Never add water to acid. Always add the acid to water.
13. Never heat thick glassware as it will break. (i.e. do not heat measuring cylinders).
14. When you are smelling chemicals, place the container on a laboratory bench and
20
1.5
use your hand to gently waft (fan) the vapours towards you.
15. Do not take chemicals from the laboratory.
16. Always work in a well ventilated room. Whenever you perform experiments, you
should open the windows.
17. Do not leave Bunsen burners and ames unattended.
18. Never smell, taste or touch chemicals unless instructed to do so.
19. Never point test tubes at people or yourself. When heating chemicals, always point
the mouth of the test tube away from you and your classmates.
Hazard signs
ESAT
The table below lists some of the common hazards signs that you may encounter. You
should know what all of these mean.
Sign
Symbol
C
Meaning
Corrosive. Chemicals with this label can burn your
skin and eyes and
burn holes in your
clothes. An example is hydrochloric
acid.
Sign
Symbol
N
Meaning
Environmentally
harmful. Chemicals
with this label are
damaging to the
environment.
An
example is CFCs.
21
1.5
Sign
Symbol
E
Xn
Meaning
Sign
Symbol
Meaning
Harmful.
Xi
Irritant.
Chemi-
Chemi-
this
gener-
cause irritation to
ally considered to
be
An example is hy-
are
damaging
to
humans.
O
Oxidising.
drogen peroxide.
Chemi-
Toxic.
Chemicals
contain
that
may
oxygen
cause
ample is mercury.
other materials to
combust.
An ex-
ample is potassium
dichromate.
ESAU
You can nd safety data sheets at http://www.msds.com/. You should always look at these
data sheets anytime you work with a new chemical. These data sheets contain information
about how to work with chemicals and what dangers the chemicals pose to you and the
environment. You should always try dispose of chemicals correctly and safely. Many
chemicals cannot simply be washed down the sink.
22
Classication of matter
Materials
ESAV
All the objects that we see in the world around us, are made of matter. Matter makes up
the air we breathe, the ground we walk on, the food we eat and the animals and plants that
live around us. Even our own human bodies are made of matter!
Different objects can be made of different
types of materials (the matter from which objects are made). For example, a cupboard (an
Cupboard
Materials that conduct heat (e.g. metals) are called thermal conductors. Materials
that conduct electricity (e.g. copper wire) are electrical conductors.
Brittle materials break easily (e.g. plastic). Materials that are malleable can be easily
formed into different shapes (e.g. clay, dough). Ductile materials are able to be
formed into long wires (e.g. copper).
Density is the mass per unit volume. Examples of dense materials include concrete
and stones.
The boiling and melting points of substances tells us the temperature at which the
substance will boil or melt. This helps us to classify substances as solids, liquids or
gases at a specic temperature.
The diagram below shows one way in which matter can be classied (grouped) according
Chemistry: Matter and Materials
23
2.1
to its different properties. As you read further in this chapter, you will see that there are
also other ways of classifying materials, for example according to whether or not they are
good electrical conductors.
MATTER
MIXTURES
Homogeneous
PURE SUBSTANCES
Heterogeneous
Metals
Magnetic
Compounds
Elements
Non-metals
Non-magnetic
Activity:
items
24
Activity:
2.2
Classifying materials
Picture by
owcomm on
Flickr.com
Mixtures
ESAW
We see mixtures all the time in our everyday lives. A stew, for example, is a mixture of
different foods such as meat and vegetables; sea water is a mixture of water, salt and other
substances, and air is a mixture of gases such as carbon dioxide, oxygen and nitrogen.
DEFINITION: Mixture
A mixture is a combination of two or more substances, where these substances are not bonded (or joined) to each other and no chemical reaction
occurs between the substances.
Imagine, for example, that you have 250 ml of water and you add sand to the water.
It doesnt matter whether you add 20 g, 40 g, 100 g or any other mass of sand to the
In the example we used of sand and water, neither of these substances has changed
in any way when they are mixed together. The sand is still sand and the water is still
water.
To separate something by mechanical means, means that there is no chemical process involved. In our sand and water example, it is possible to separate the mixture by
25
2.2
simply pouring the water through a lter. Something physical is done to the mixture,
rather than something chemical.
We can group mixtures further by dividing them into those that are heterogeneous and
those that are homogeneous.
Heterogeneous mixtures
ESAX
A heterogeneous mixture does not have a denite composition. Cereal in milk is an example of a heterogeneous mixture. Soil is another example. Soil has pebbles, plant matter
and sand in it. Although you may add one substance to the other, they will stay separate
in the mixture. We say that these heterogeneous mixtures are non-uniform, in other words
they are not exactly the same throughout.
Cereal
26
2.2
Phases of matter
Name of mixture
Example
liquid-liquid
emulsion
oil in water
solid-liquid
suspension
muddy water
gas-liquid
aerosol
zzy drinks
gas-solid
smoke
smog
Homogeneous mixtures
ESAY
A homogeneous mixture has a denite composition, and specic properties. In a homogeneous mixture, the different parts cannot be seen. A solution of salt dissolved in water is
an example of a homogeneous mixture. When the salt dissolves, it spreads evenly through
the water so that all parts of the solution are the same, and you can no longer see the salt
as being separate from the water. Think also of coffee without milk. The air we breathe is
another example of a homogeneous mixture since it is made up of different gases which
are in a constant ratio, and which cant be visually distinguished from each other (i.e. you
cant see the different components).
See video: VPabz at www.everythingscience.co.za
Coffee
FACT
An alloy is a homogeneous mixture of two
or more elements, at
where
ties.
For example
27
2.2
Example 1: Mixtures
QUESTION
For each of the following mixtures state whether it is a homogeneous or a heterogeneous mixture:
a. sugar dissolved in water
b. our and iron lings (small pieces of iron)
SOLUTION
Step 1 : Look at the denition
We rst look at the denition of a heterogeneous and homogeneous
mixture.
Step 2 : Decide whether or not you can see the components
a. We cannot see the sugar in the water.
b. We are able to make out the pieces of iron in the our.
Step 3 : Decide whether or not the components are mixed uniformly
a. The two components are mixed uniformly.
b. In this mixture there may be places where there are a lot of iron lings and
places where there is more our, so it is not uniformly mixed.
Step 4 : Give the nal answer
a. Homogeneous mixture.
b. Heterogeneous mixture.
Activity:
Making mixtures
Make mixtures of sand and water, potassium dichromate and water, iodine and
ethanol, iodine and water. Classify these as heterogeneous or homogeneous. Give
reasons for your choice.
28
2.2
water
stones
cereal
salt
sugar
choice.
iodine (bottom)
Exercise 2 - 1
Non-mixture or
Heterogeneous
Homogeneous
mixture
mixture
mixture
tap water
brass (an alloy of copper and zinc)
concrete
aluminium foil (tinfoil)
Coca Cola
soapy water
black tea
sugar water
baby milk formula
More practice
video solutions
or help at www.everythingscience.co.za
(1.) 0000
29
2.3
Pure substances
ESAZ
Any material that is not a mixture, is called a pure substance. Pure substances include
elements and compounds. It is much more difcult to break down pure substances into
their parts, and complex chemical methods are needed to do this.
See video: VPacc at www.everythingscience.co.za
We can use melting and boiling points and chromatography to test for pure substances.
Pure substances have a sharply dened (one temperature) melting or boiling point. Impure
substances have a temperature range over which they melt or boil. Chromatography is
the process of separating substances into their individual components. If a substance is
pure then chromatography will only produce one substance at the end of the process. If a
substance is impure then several substances will be seen at the end of the process.
Activity:
an eye dropper.
Place a smartie in the centre of a piece
of lter paper. Carefully drop a few drops
of water onto the smartie, until the smartie is quite wet and there is a ring of water on the lter paper. After some time
you should see a coloured ring on the paper around the smartie. This is because
the food colouring that is used to make
the smartie colourful dissolves in the water and is carried through the paper away
from the smartie.
30
Smartie chromatography
2.3
ESAAA
Elements
An element is a chemical substance that cant be divided or changed into other chemical substances by any ordinary chemical means. The smallest unit of an element is the
atom.
DEFINITION: Element
FACT
Recently
it
was
to
the
list
There are 112 ofcially named elements and about 118 known elements. Most of these are
of ofcially named
natural, but some are man-made. The elements we know are represented in the periodic
table, where each element is abbreviated to a chemical symbol. Table 2.3 gives the rst
20 elements and some of the common transition metals.
elements
number
The
name
Element name
Element symbol
Element name
Element symbol
Hydrogen
Phosphorus
element 116 it is
Helium
He
Sulphur
moscovium.
This
brings
total
Lithium
Li
Chlorine
Cl
number of ofcially
Beryllium
Be
Argon
Ar
Boron
Potassium
Carbon
Calcium
Ca
Nitrogen
Iron
Fe
Oxygen
Nickel
Ni
Fluorine
Copper
Cu
Neon
Ne
Zinc
Zn
Sodium
Na
Silver
Ag
Magnesium
Mg
Platinum
Pt
Aluminium
Al
Gold
Au
Silicon
Si
Mercury
Hg
erovium
the
for
named elements to
114.
Table 2.2: List of the rst 20 elements and some common transition metals
Chemistry: Matter and Materials
and
31
2.3
Compounds
ESAAB
A compound is a chemical substance that forms when two or more different elements
combine in a xed ratio. Water (H2 O), for example, is a compound that is made up of
two hydrogen atoms for every one oxygen atom. Sodium chloride (NaCl) is a compound
made up of one sodium atom for every chlorine atom. An important characteristic of a
compound is that it has a chemical formula, which describes the ratio in which the atoms
of each element in the compound occur.
DEFINITION: Compound
A substance made up of two or more different elements that are joined together in a xed ratio.
ture and compound. Iron (Fe) and sulphur (S) are two elements. When they are added
together, they form a mixture of iron and sulphur. The iron and sulphur are not joined
together. However, if the mixture is heated, a new compound is formed, which is called
iron sulphide (FeS).
In a submiS
croscopic representa-
Fe
Fe
To show a
Fe
Fe
compound, we draw
Fe
Fe
An atom
of the element sulphur (S)
An atom
of the element iron
(Fe)
We can also use symbols to represent elements, mixtures and compounds. The symbols
for the elements are all found on the periodic table. Compounds are shown as two or more
element names written right next to each other. Subscripts may be used to show that there
ture.
is more than one atom of a particular element. (e.g. H2 O or NH3 ). Mixtures are written as:
a mixture of element (or compound) A and element (or compound) B. (e.g. a mixture of Fe
and S).
32
2.3
SOLUTION
Activity:
33
2.3
Exercise 2 - 2
1. In the following table, tick whether each of the substances listed is a mixture or a pure substance. If it is a mixture, also say whether it is a homogeneous or heterogeneous mixture.
Substance
Mixture or pure
zzy colddrink
steel
oxygen
iron lings
smoke
limestone (CaCO3 )
More practice
34
video solutions
or help at www.everythingscience.co.za
(1.) 0001
2.4
(2.) 0002
ESAAC
Think about what you call your friends. Some of your friends might have full names (long
names) and a nickname (short name). These are the words we use to tell others who or
what we are referring to. Their full name is like the substances name and their nickname
is like the substances formulae. Without these names your friends would have no idea
which of them you are referring to. Chemical substances have names, just like people have
names. This helps scientists to communicate efciently.
It is easy to describe elements and mixtures. We simply use the names that we nd on
the periodic table for elements and we use words to describe mixtures. But how are compounds named? In the example of iron sulphide that was used earlier, the compound name
is a combination of the names of the elements but slightly changed.
See video: VPadm at www.everythingscience.co.za
The following are some guidelines for naming compounds:
1. The compound name will always include the names of the elements that are part of
it.
A compound of iron (Fe) and sulphur (S) is iron sulph ide (FeS)
2. In a compound, the element that is on the left of the Periodic Table, is used rst
when naming the compound. In the example of NaCl, sodium is a group 1 element
on the left hand side of the table, while chlorine is in group 7 on the right of the
table. Sodium therefore comes rst in the compound name. The same is true for FeS
and KBr.
3. The symbols of the elements can be used to represent compounds e.g. FeS, NaCl,
KBr and H2 O. These are called chemical formulae. In the rst three examples, the
ratio of the elements in each compound is 1:1. So, for FeS, there is one atom of iron
for every atom of sulphur in the compound. In the last example (H2 O) there are two
atoms of hydrogen for every atom of oxygen in the compound.
4. A compound may contain ions (an ion is an atom that has lost or gained electrons).
These ions can either be simple (consist of only one element) or compound (consist
of several elements). Some of the more common ions and their formulae are given
in Table 2.3 and in Table 2.4. You should know all these ions.
Chemistry: Matter and Materials
35
2.4
Compound ion
Formula
Compound ion
Formula
Compound ion
Formula
Hydrogen
H+
Lithium
Li+
Sodium
Na+
Potassium
K+
Silver
Ag+
Mercury (I)
Hg+
Copper (I)
Cu+
Ammonium
NH+
4
Beryllium
Be2+
Magnesium
Mg2+
Calcium
Ca2+
Barium
Ba2+
Tin (II)
Sn2+
Lead (II)
Pb
Chromium (II)
Cr2+
Manganese (II)
Mn2+
Iron (II)
Fe2+
Cobalt (II)
Co2+
Nickel
Ni2+
Copper (II)
Cu2+
Zinc
Zn2+
Aluminium
Al
3+
Chromium (III)
Cr3+
Iron (III)
Fe3+
Cobalt (III)
Co3+
Chromium (VI)
Cr6+
Manganese (VII)
Mn7+
2+
Compound ion
Formula
Fluoride
Chloride
Cl
Bromide
Br
Compound ion
Formula
Oxide
O2
Peroxide
O2
2
CO2
3
Carbonate
Sulphide
S2
Hydroxide
OH
Sulphite
SO2
3
Nitrite
NO
2
Sulphate
SO2
4
Nitrate
NO
3
Thiosulphate
S2 O2
3
Iodide
Hydrogen carbonate
HCO
3
Chromate
CrO2
4
Hydrogen sulphite
HSO
3
Dichromate
Cr2 O2
7
Hydrogen sulphate
HSO
4
Manganate
MnO2
4
Dihydrogen phosphate
H2 PO
4
Oxalate
Hydrogen phosphate
ClO3
Nitride
MnO
4
Phosphate
PO3
4
CH3 COO
Phosphide
P3
Hypochlorite
ClO
Chlorate
Permanganate
Acetate (ethanoate)
(COO)2 /C2 O2
2
4
HPO2
4
N3
2.4
This is used for non-metals. For metals, we add a roman number (I, II, III, IV) in
brackets after the metal ion to indicate the ratio. You should know the following
prexes: mono (one), di (two) and tri (three).
CO (carbon monoxide) - There is one atom of oxygen for every one atom of
carbon
NO2 (nitrogen dioxide) - There are two atoms of oxygen for every one atom of
nitrogen
SO3 (sulphur trioxide) - There are three atoms of oxygen for every one atom of
sulphur
Tip
When numbers are
written as subscripts
in compounds (i.e.
they are written below and to the right of
the element symbol),
this tells us how many
atoms of that element
there are in relation
The above guidelines also help us to work out the formula of a compound from the name
of the compound. The following worked examples will look at names and formulae in
detail.
We can use these rules to help us name both ionic compounds and covalent compounds.
However, covalent compounds are often given other names by scientists to simplify the
name (or because the molecule was named long before its formula was discovered). For
example, if we have 2 hydrogen atoms and one oxygen atom the above naming rules would
tell us that the substance is dihydrogen monoxide. But this compound is better known as
water!
Some common covalent compounds are given in table 2.4
to other elements in
the compound.
For
example in nitrogen
dioxide (NO2 ) there
are two oxygen atoms
for every one atom
of nitrogen.
Later,
Name
Formula
Name
Formula
water
H2 O
hydrochloric acid
HCl
sulphuric acid
H2 SO4
methane
CH4
ethane
C2 H6
ammonia
NH3
nitric acid
HNO3
SOLUTION
Step 1 : List the ions involved:
We have the sodium ion (Na+ ) and the uoride ion (F ). (You can look
these up on the tables of cations and anions.)
37
2.4
SOLUTION
Step 1 : List the ions involved
Mg2+ and Cl
Mg2+
Cl
Draw a cross as above, and then you can see that Mg 1 and
Cl 2.
38
2.4
39
2.4
SOLUTION
SOLUTION
40
Cu2+
NO
3
2.4
2
1
Tip
Cu(NO3 )2
Activity:
Your teacher will assign each of you a different ion (written on a piece of card).
Stick this to yourself. You will also get cards with the numbers 1 - 5 on them. Now
walk around the class and try to work out who you can pair up with and in what
ratio. Once you have found a partner, indicate your ratio using the numbered cards.
Check your results with your classmates or your teacher.
Exercise 2 - 3
f. Na2 SO4
g. Fe(NO3 )3
c. KMnO4
h. PbSO3
d. NO2
e. NH4 OH
i. Cu(HCO3 )2
41
2.5
a. potassium nitrate
b. sodium oxide
c. barium sulphate
d. aluminium chloride
More practice
(1.) 0003
(2.) 0004
e. magnesium phosphate
f. tin(II) bromide
g. manganese(II) phosphide
video solutions
or help at www.everythingscience.co.za
(3.) 0005
ESAAD
The elements in the periodic table can also be divided according to whether they are
metals, metalloids or non-metals. The zigzag line separates all the elements that are metals
from those that are non-metals. Metals are found on the left of the line, and non-metals
are those on the right. Along the line you nd the metalloids. You should notice that there
are more metals then non-metals. Metals, metalloids and non-metals all have their own
specic properties.
See video: VPaec at www.everythingscience.co.za
H
Metalloids
Non-metals
Metals
Metalloids
42
2.5
ESAAE
Metals
Examples of metals include copper (Cu), zinc
Copper wire
Metals are good conductors of heat and are therefore used in cooking utensils such
Metals are good conductors of electricity, and are therefore used in electrical con-
ducting wires.
Shiny metallic lustre
Metals have a characteristic shiny appearance and are often used to make jewellery.
Malleable and ductile
This means that they can be bent into shape without breaking (malleable) and can be
stretched into thin wires (ductile) such as copper.
Melting point
Metals usually have a high melting point and can therefore be used to make cooking
pots and other equipment that needs to become very hot, without being damaged.
Density
Metals have a high density.
Magnetic properties
Only three main metals (iron, cobalt and nickel) are magnetic, the others are nonmagnetic.
You can see how the properties of metals make them very useful in certain applications.
43
2.5
Activity:
safety pins
cooking pots
jewellery
scissors
cutlery (knives, forks, spoons)
ESAAF
Non-metals
In contrast to metals, non-metals are poor
thermal conductors, good electrical insulators (meaning that they do not conduct electrical charge) and are neither malleable nor
ductile. The non-metals include elements
such as sulphur (S), phosphorus (P), nitrogen
(N) and oxygen (O).
Metalloids
Sulphur
ESAAG
44
2.6
Silicon chips
portant in digital electronics, such as computers. The metalloids include elements such as
silicon (Si) and germanium (Ge).
Electrical conductors,
semi-conductors and
insulators
ESAAH
Electrical conductors are usually metals. Copper is one of the best electrical conductors,
and this is why it is used to make conducting wire. In reality, silver actually has an even
higher electrical conductivity than copper, but silver is too expensive to use.
See video: VPaex at www.everythingscience.co.za
In the overhead power lines that we see
above us, aluminium is used. The aluminium
usually surrounds a steel core which adds
makes it stronger so that it doesnt break
when it is stretched across distances. Sometimes gold is used to make wire because it
is very resistant to surface corrosion. Corrosion is when a material starts to deteriorate
because of its reactions with oxygen and water in the air.
Power lines
DEFINITION: Insulators
An insulator is a non-conducting material that does not carry any charge.
45
2.6
Examples of insulators are plastic and wood. Semi-conductors behave like insulators when
they are cold, and like conductors when they are hot. The elements silicon and germanium
are examples of semi-conductors.
Aim:
To investigate the electrical conductivity of a number of substances
Apparatus:
cells
wire leads
light bulb
light bulb
crocodile clips
test substance
X
crocodile clip
Method:
1. Set up the circuit as shown above, so that the test substance is held between
the two crocodile clips. The wire leads should be connected to the cells and
the light bulb should also be connected into the circuit.
2. Place the test substances one by one between the crocodile clips and see
what happens to the light bulb. If the light bulb shines it means that current
is owing and the substance you are testing is an electrical conductor.
Results:
Record your results in the table below:
46
Test
sub-
stance
2.7
Metal/non-
Does
the
Conductor
metal
light
bulb
or insulator
glow?
Conclusions:
In the substances that were tested, the metals were able to conduct electricity and
the non-metals were not. Metals are good electrical conductors and non-metals are
not.
See simulation: (
Thermal Conductors
and Insulators
ESAAI
A thermal conductor is a material that allows energy in the form of heat, to be transferred
within the material, without any movement of the material itself. An easy way to understand
this concept is through a simple demonstration.
See video: VPafb at www.everythingscience.co.za
Aim:
To demonstrate the ability of different substances to conduct heat.
Apparatus:
47
2.7
plastic spoon
metal spoon
boiling water
boiling water
a metal spoon
a plastic spoon.
Method:
Pour boiling water into the two cups so that they are about half full.
Place a metal spoon into one cup and a plastic spoon in the other.
Note which spoon heats up more quickly
Warning:
FACT
Be careful when working with boiling water and when you touch the
Well-insulated build-
heating
than
Two
Results:
materials
The metal spoon heats up faster than the plastic spoon. In other words, the metal
and
more
worldwide,
are
mineral
and
wool
polystyrene.
Min-
An insulator is a material that does not allow a transfer of electricity or energy. Materials
that are poor thermal conductors can also be described as being good thermal insulators.
Since air is a
number of different materials, and then answer the questions that follow.
The higher the number in the second column, the better the material is
at conducting heat (i.e. it is a good thermal conductor). Remember that
a material that conducts heat efciently, will also lose heat more quickly
than an insulating material.
It has
48
Material
2.8
Thermal Conductivity
(W m1 K1 )
Silver
429
Stainless steel
16
Standard glass
1.05
Concrete
0.9 - 2
Red brick
0.69
Water
0.58
Polyethylene (plastic)
0.42 - 0.51
Wood
0.04 - 0.12
Polystyrene
0.03
Air
0.024
Magnetic and
Non-magnetic Materials
ESAAJ
We have now looked at a number of ways in which matter can be grouped, such as into
metals, semi-metals and non-metals; electrical conductors and insulators, and thermal conductors and insulators. One way in which we can further group metals, is to divide them
into those that are magnetic and those that are non-magnetic.
See video: VPaga at www.everythingscience.co.za
49
2.8
DEFINITION: Magnetism
Magnetism is a force that certain kinds of objects, which are called magnetic objects, can exert on each other without physically touching. A magnetic object is surrounded by a magnetic eld that gets weaker as one
moves further away from the object.
Investigation: Magnetism
50
Object
2.8
Magnetic
or
non-
magnetic
Chapter 2 | Summary
See the summary presentation (
All the objects and substances that we see in the world are made of matter.
The main characteristics of mixtures are that the substances that make them up are
not in a xed ratio, these substances keep their physical properties and these substances can be separated from each other using mechanical means.
Chemistry: Matter and Materials
51
2.8
A heterogeneous mixture is one that consists of two or more substances. It is nonuniform and the different components of the mixture can be seen. An example would
be a mixture of sand and water.
A homogeneous mixture is one that is uniform, and where the different components
An element is a substance that cannot be broken down into other substances through
chemical means.
All the elements are found on the periodic table. Each element has its own chemical
symbol. Examples are iron (Fe), sulphur (S), calcium (Ca), magnesium (Mg) and
uorine (F).
A compound is a A substance made up of two or more different elements that are
joined together in a xed ratio. Examples of compounds are sodium chloride (NaCl),
iron sulphide (FeS), calcium carbonate (CaCO3 ) and water (H2 O).
elements will combine in the compound and where the elements are in the periodic
table. A number of rules can then be followed to name the compound.
Another way of classifying matter is into metals (e.g. iron, gold, copper), metalloids
(e.g. silicon and germanium) and non-metals (e.g. sulphur, phosphorus and nitrogen).
Metals are good electrical and thermal conductors, they have a shiny lustre, they
are malleable and ductile, and they have a high melting point. Metals also have a
high density. These properties make metals very useful in electrical wires, cooking
utensils, jewellery and many other applications.
Matter can also be classied into electrical conductors, semi-conductors and insula-
tors.
An electrical conductor allows an electrical current to pass through it. Most metals
are good electrical conductors.
An electrical insulator is a non-conducting material that does not carry any charge.
Examples are plastic, wood, cotton material and ceramic.
Materials may also be classied as thermal conductors or thermal insulators depending on whether or not they are able to conduct heat.
Materials may also be magnetic or non-magnetic. Magnetism is a force that certain
kinds of objects, which are called magnetic objects, can exert on each other without
physically touching. A magnetic object is surrounded by a magnetic eld that gets
weaker as one moves further away from the object.
52
Chapter 2
2.8
Column B
1. iron
2. H2 S
B. a heterogeneous mixture
3. sugar solution
C. a metal alloy
D. an element
5. steel
E. a homogeneous mixture
5. You are given a test tube that contains a mixture of iron lings and sulphur.
You are asked to weigh the amount of iron in the sample.
a. Suggest one method that you could use to separate the iron lings
from the sulphur.
b. What property of metals allows you to do this?
Chemistry: Matter and Materials
53
2.8
6. Given the following descriptions, write the chemical formula for each of
the following substances:
a. silver metal
b. a compound that contains only potassium and bromine
c. a gas that contains the elements carbon and oxygen in a ratio of 1:2
7. Give the names of each of the following compounds:
a. NaBr
b. Ba(NO2 )2
c. SO2
d. H2 SO4
8. Give the formula for each of the following compounds:
a. iron (II) sulphate
b. boron triuoride
c. potassium permanganate
d. zinc chloride
9. For each of the following materials, say what properties of the material
make it important in carrying out its particular function.
a. tar on roads
b. iron burglar bars
c. plastic furniture
d. metal jewellery
e. clay for building
f. cotton clothing
More practice
(1.) 0006
(2.) 0007
video solutions
(3.) 0008
(4.) 0009
or help at www.everythingscience.co.za
(5.) 000a
(6.) 000b
(7.) 000c
(9.) 000e
54
(8.) 000d
States of matter
ESAAK
In this chapter we will explore the states of matter and then look at the kinetic molecular
theory. Matter exists in three states: solid, liquid and gas. We will also examine how the
kinetic theory of matter helps explain boiling and melting points as well as other properties
of matter.
See introductory video: (
All matter is made up of particles. We can see this when we look at diffusion.
DEFINITION: Diffusion
Diffusion is the movement of particles from a high concentration to a low
concentration.
Diffusion is a result of the constant thermal motion of particles. In 1828 Robert Brown
observed that pollen grains suspended in water moved about in a rapid, irregular motion.
This motion has since become known as Brownian motion. Brownian motion is essentially
diffusion of many particles. Brownian motion can also be seen as the random to and fro
movement of particles.
55
3.1
Matter exists in one of three states, namely solid, liquid and gas. A solid has a xed shape
and volume. A liquid takes on the shape of the container that it is in. A gas completely lls
the container that it is in. Matter can change between these states by either adding heat
or removing heat. This is known as a change of state. As we heat an object (e.g. water) it
goes from a solid to a liquid to a gas. As we cool an object it goes from a gas to a liquid to
a solid. The changes of state that you should know are:
Melting
Freezing
Evaporation
Evaporation is the process of going from a liquid to a gas. Evaporation from a liquids
surface can happen at a wide range of temperatures. If more energy is added then
bubbles of gas appear inside the liquid and this is known as boiling.
If we know the melting and boiling point of a substance then we can say what state (solid,
liquid or gas) it will be in at any temperature.
The gure 3.1 summarises these processes:
56
Condensation
Liquid
Gas
3.1
Solid
Melting
Sublimation
Bunsen burner
thermometer
water
Method:
Thermometer
Thermometer
Beaker
Ice
Beaker
Water
Warning:
Be careful when handling the beaker of hot water. Do not touch the
beaker with your hands, you will burn yourself.
Results:
1. Record your results in the following table:
57
3.1
Heating of ice
Time (min)
Temperature (in C)
etc.
Temperature (in C)
etc.
2. Plot a graph of time (independent variable, x-axis) against temperature (dependent variable, y-axis) for the ice melting and the boiling
water cooling.
Discussion and conclusion: You should nd that the temperature of the
ice increases until the rst drops of liquid appear and then the temperature remains the same, until all the ice is melted. You should also nd
that when you cool water down from boiling, the temperature remains
constant for a while, then starts decreasing.
In the above experiment, you investigated the heating and cooling curves of water. We can
draw heating and cooling curves for any substance. A heating curve of a substance gives
the changes in temperature as we move from a solid to a liquid to a gas. A cooling curve
gives the changes in temperature as we move from gas to liquid to solid. An important
observation is that as a substance melts or boils, the temperature remains constant until
the substance has changed state. This is because all the heat energy goes into breaking or
forming the bonds between the molecules.
Temperature (C)
Temperature (C)
The following diagram gives an example of what heating and cooling curves look like:
Time (min)
58
Time (min)
3.2
ESAAL
The kinetic theory of matter helps us to explain why matter exists in different phases (i.e.
solid, liquid and gas), and how matter can change from one phase to the next. The kinetic
theory of matter also helps us to understand other properties of matter. It is important to
realise that what we will go on to describe is only a theory. It cannot be proved beyond
doubt, but the fact that it helps us to explain our observations of changes in phase, and
other properties of matter, suggests that it probably is more than just a theory.
Broadly, the kinetic theory of matter says that all matter is composed of particles which
have a certain amount of energy which allows them to move at different speeds depending
on the temperature (energy). There are spaces between the particles and also attractive
forces between particles when they come close together.
solid
liquid
gas
Table 3.2 summarises the characteristics of the particles that are in each phase of matter.
Taking copper as an example we nd that in the solid phase the copper atoms have little
energy. They vibrate in xed positions. The atoms are held closely together in a regular
pattern called a lattice. If the copper is heated, the energy of the atoms increases. This
means that some of the copper atoms are able to overcome the forces that are holding
them together, and they move away from each other to form liquid copper. This is why
liquid copper is able to ow, because the atoms are more free to move than when they
were in the solid lattice. If the liquid is heated further, it will become a gas. Gas particles
have lots of energy and are far away from each other. That is why it is difcult to keep a
gas in a specic area! The attractive forces between the particles are very weak. Gas atoms
will ll the container they are in. Figure 3.1 shows the changes in phase that may occur in
matter, and the names that describe these processes.
Activity:
59
3.2
Property of matter
Solid
Liquid
Gas
Particles
Atoms or molecules
Atoms or molecules
Atoms or molecules
Spaces
particles
between
Weaker
forces
than in solids, but
stronger forces than
in gases.
Changes in phase.
A liquid becomes
a gas if its temperature is increased.
A liquid becomes a
solid if its temperature decreases.
In general a gas
becomes a liquid
or solid when it is
cooled.
Particles
have less energy
and therefore move
closer together so
that the attractive
forces
become
stronger, and the
gas becomes a
liquid or a solid.
Chapter 3 | Summary
See the summary presentation (
60
3.2
The kinetic theory of matter says that all matter is composed of particles which have
a certain amount of energy which allows them to move at different speeds depending on the temperature (energy). There are spaces between the particles and also
attractive forces between particles when they come close together.
Chapter 3
61
3.2
Element
Melting point ( C)
Boiling point ( C)
copper
1083
2567
magnesium
650
1107
oxygen
-218,4
-183
carbon
3500
4827
helium
-272
-268,6
sulphur
112,8
444,6
a. What state of matter (i.e. solid, liquid or gas) will each of these elements be in at room temperature (25 C)?
b. Which of these elements has the strongest forces between its atoms?
Give a reason for your answer.
c. Which of these elements has the weakest forces between its atoms?
Give a reason for your answer.
5. Complete the following submicroscopic diagrams to show what magnesium will look like in the solid, liquid and gas phase.
solid
More practice
(1.) 000f
62
(2.) 000g
liquid
video solutions
(3.) 000h
(4.) 000i
gas
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(5.) 000j
The atom
Introduction
ESAAM
We have now looked at many examples of the types of matter and materials that exist
around us and we have investigated some of the ways that materials are classied. But what
is it that makes up these materials? And what makes one material different from another? In
order to understand this, we need to take a closer look at the building blocks of matter - the
atom. Atoms are the basis of all the structures and organisms in the universe. The planets,
sun, grass, trees, air we breathe and people are all made up of different combinations of
atoms.
See introductory video: (
63
4.2
ESAAN
It is important to realise that a lot of what we know about the structure of atoms has been
developed over a long period of time. This is often how scientic knowledge develops,
with one person building on the ideas of someone else. We are going to look at how our
modern understanding of the atom has evolved over time.
See video: VPakv at www.everythingscience.co.za
The idea of atoms was invented by two Greek philosophers, Democritus and Leucippus
in the fth century BC. The Greek word oo (atom) means indivisible because they
believed that atoms could not be broken into smaller pieces.
Nowadays, we know that atoms are made up of a positively charged nucleus in the centre
surrounded by negatively charged electrons. However, in the past, before the structure
of the atom was properly understood, scientists came up with lots of different models or
pictures to describe what atoms look like.
DEFINITION: Model
A model is a representation of a system in the real world. Models help us to
understand systems and their properties.
For example, an atomic model represents what the structure of an atom could look like,
based on what we know about how atoms behave. It is not necessarily a true picture of the
exact structure of an atom.
Models are often simplied. The small toy cars that you may have played with as a child are
models. They give you a good idea of what a real car looks like, but they are much smaller
and much simpler. A model cannot always be absolutely accurate and it is important that
we realise this, so that we do not build up an incorrect idea about something.
64
4.2
ESAAO
ESAAP
FACT
Two
-
models
electrons
and
the
Saturnian
soup of
positive charge
were
lie
imagined
at
the
to
corners
of a cube.
In the
Saturnian
model,
the
electrons
imagined
to
a very big,
nucleus.
The discovery of radiation was the next step along the path to building an accurate picture
of atomic structure. In the early twentieth century, Marie and Pierre Curie, discovered that
some elements (the radioactive elements) emit particles, which are able to pass through
matter in a similar way to X-rays (read more about this in Grade 11). It was Ernest Rutherford
who, in 1911, used this discovery to revise the model of the atom.
other
65
were
orbit
heavy
4.2
ESAAQ
nucleus
the atom.
ESAAR
electron orbit
nucleus
James Chadwick
Figure 4.4:
atom
ESAAS
Rutherford predicted (in 1920) that another kind of particle must be present in the nucleus
along with the proton. He predicted this because if there were only positively charged
protons in the nucleus, then it should break into bits because of the repulsive forces between the like-charged protons! To make sure that the atom stays electrically neutral, this
particle would have to be neutral itself. In 1932 James Chadwick discovered the neutron
66
4.3
ESAAT
Although the most commonly used model of the atom is the Bohr model, scientists are
still developing new and improved theories on what the atom looks like. One of the most
important contributions to atomic theory (the eld of science that looks at atoms) was the
development of quantum theory. Schrodinger, Heisenberg, Born and many others have
had a role in developing quantum theory.
Exercise 4 - 1
Column B
A. Niels Bohr
2. Arrangement of electrons
D. JJ Thomson
5. Discovery of radiation
E. Rutherford
More practice
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(1.) 000k
ESAAU
It is difcult sometimes to imagine the size of an atom, or its mass, because we cannot see
an atom and also because we are not used to working with such small measurements.
Chemistry: Matter and Materials
67
4.3
ESAAV
It is possible to determine the mass of a single atom in kilograms. But to do this, you would
need special instruments and the values you would get would be very clumsy and difcult
to work with. The mass of a carbon atom, for example, is about 1, 99 1026 kg, while
the mass of an atom of hydrogen is about 1, 67 1027 kg. Looking at these very small
numbers makes it difcult to compare how much bigger the mass of one atom is when
compared to another.
To make the situation simpler, scientists use a different unit of mass when they are describing the mass of an atom. This unit is called the atomic mass unit (amu). We can abbreviate
(shorten) this unit to just u. Scientists use the carbon standard to determine amu. The
carbon standard gives carbon an atomic mass of 12, 0 u. Compared to carbon the mass of
a hydrogen atom will be 1 u. Atomic mass units are therefore not giving us the actual mass
of an atom, but rather its mass relative to the mass of one (carefully chosen) atom in the
periodic table. In other words it is only a number in comparison to another number. The
atomic masses of some elements are shown in table 4.1.
Element
Carbon (C)
12, 0
Nitrogen (N)
14, 0
Bromine (Br)
79, 9
Magnesium (Mg)
24, 3
Potassium (K)
39, 1
Calcium (Ca)
40, 1
Oxygen (O)
16, 0
68
4.3
ESAAW
Radioactive elements emit different types of particles. Some of these are positively charged
alpha () particles. Rutherford wanted to nd out where the positive charge in an atom is.
He carried out a series of experiments where he bombarded sheets of gold foil with alpha
particles (since these would be repelled by the positive nucleus). A simplied diagram of
his experiment is shown in gure 4.5.
C
B
A
radioactive
substance
A
particles
particles
gold sheet
nucleus of
gold atom
C
(a)
B
C
(b)
Figure 4.5: Rutherfords gold foil experiment. Figure (a) shows the path of the particles
after they hit the gold sheet. Figure (b) shows the arrangement of atoms in the gold sheets
and the path of the particles in relation to this.
69
4.4
ESAAX
A stadium
ESAAY
The relative atomic mass of an element is the average mass of all the naturally occurring isotopes of that element. The units for relative atomic mass
are atomic mass units.
The relative atomic mass of an element is the number you will nd on the periodic table.
ESAAZ
As a result of the work done by previous scientists on atomic models, scientists now have
a good idea of what an atom looks like. This knowledge is important because it helps us
to understand why materials have different properties and why some materials bond with
others. Let us now take a closer look at the microscopic structure of the atom (what the
atom looks like inside).
70
4.4
ESABA
The Electron
The electron is a very tiny particle. It has a mass of 9, 11 1031 kg. The electron carries
one unit of negative electric charge (i.e. 1, 6 1019 C).
ESABB
The Nucleus
Unlike the electron, the nucleus can be broken up into smaller building blocks called
protons and neutrons. Together, the protons and neutrons are called nucleons.
FACT
Scientists believe that
the electron can be
The Proton
The electron carries one unit of negative electric charge (i.e. 1, 610
19
C, C is Coulombs).
Each proton carries one unit of positive electric charge (i.e. +1, 6 10
C). Since we
know that atoms are electrically neutral, i.e. do not carry any extra charge, then the num19
ber of protons in an atom has to be the same as the number of electrons to balance out the
positive and negative charge to zero. The total positive charge of a nucleus is equal to the
number of protons in the nucleus. The proton is much heavier than the electron (10 000
times heavier!) and has a mass of 1, 6726 1027 kg. When we talk about the atomic mass
of an atom, we are mostly referring to the combined mass of the protons and neutrons, i.e.
the nucleons.
The Neutron
The neutron is electrically neutral i.e. it carries no charge at all. Like the proton, it is
much heavier than the electron and its mass is 1, 6749 1027 kg (slightly heavier than the
proton).
71
4.4
proton
neutron
electron
Mass (kg)
1, 6726 1027
1, 6749 1027
9, 11 1031
Units of charge
+1
Charge (C)
1, 6 1019
1, 6 1019
ESABC
The chemical properties of an element are determined by the charge of its nucleus, i.e. by
the number of protons. This number is called the atomic number and is denoted by the
letter Z.
FACT
Currently
element
being
Scientists
that
after
formed.
You can nd the atomic number on the periodic table (see periodic table at front of book).
The atomic number is an integer and ranges from 1 to about 118.
The mass of an atom depends on how many nucleons its nucleus contains. The number of
nucleons, i.e. the total number of protons plus neutrons, is called the atomic mass number
and is denoted by the letter A.
believe
element
which
higher
elements
atomic
The atomic number (Z) and the mass number (A) are indicated using a standard notation,
for example carbon will look like this: 12 C
6
Standard notation shows the chemical symbol, the atomic mass number and the atomic
number of an element as follows:
72
4.4
number of nucleons
FACT
A
Z
A nuclide is a distinct
chemical symbol
number of protons
cleus characterised by
the number of protons
and neutrons in the
atom.
For example, the iron nucleus which has 26 protons and 30 neutrons, is denoted as:
To be abso-
56
26 Fe
where the atomic number is Z = 26 and the mass number A = 56. The number of neutrons
nuclides.
Tip
For a neutral atom the number of electrons is the same as the number of protons, since the
charge on the atom must balance. But what happens if an atom gains or loses electrons?
Does it mean that the atom will still be part of the same element? A change in the number
of electrons of an atom does not change the type of atom that it is. However, the charge
of the atom will change. The neutrality of the atom has changed. If electrons are added,
then the atom will become more negative. If electrons are taken away then the atom will
we
have
this
informa-
On
become more positive. The atom that is formed in either of these two cases is called an
the
ion. An ion is a charged atom. For example: a neutral sodium atom can lose one electron
to become a positively charged sodium atom (Na+ ). A neutral chlorine atom can gain one
periodic
usually
table,
appears
electron to become a negatively charged chlorine ion (Cl ). Another example is Li which
has lost one electron and now has only 2 electrons, instead of 3. Or consider F which has
gained one electron and now has 10 electrons instead of 9.
or immediately above
top
in
the
left-hand
QUESTION
Use standard notation to represent sodium and give the number of protons, neu-
in "Isotopes".
example of iron is
shown below.
26
SOLUTION
Fe
55.85
11 Na
73
The
4.4
23
11 Na.
Exercise 4 - 2
Atomic
mass
units
Atomic
number
Mg
24
12
Number
protons
of
Number of
electrons
Number
neutrons
8
17
Ni
28
40
20
Zn
0
C
12
3+
Al
O2
74
6
13
18
Chemistry: Matter and Materials
of
4.4
13
6 C
7
3 Li
15
7 N
6. In each of the following cases, give the number or the element symbol
represented by X.
a.
b.
c.
40
18 X
x
20 Ca
31
x P
235
92 U
238
92 U
More practice
(1.) 000m
(2.) 000n
video solutions
(3.) 000p
(4.) 000q
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(5.) 000r
(6.) 000s
(7.) 000t
75
4.5
Isotopes
ESABD
The chemical properties of an element depend on the number of protons and electrons
inside the atom. So if a neutron or two is added or removed from the nucleus, then the
FACT
In
Greek,
same
place
reads
as
o o o (isos
topos).
atoms
This is why
which
have
chemical properties will not change. This means that such an atom would remain in the
same place in the periodic table. For example, no matter how many neutrons we add or
subtract from a nucleus with 6 protons, that element will always be called carbon and have
the element symbol C (see the periodic table). Atoms which have the same number of
protons (i.e. same atomic number Z), but a different number of neutrons (i.e. different N
and therefore different mass number A), are called isotopes.
DEFINITION: Isotope
Isotopes of an element have the same number of protons (same Z), but a
different number of neutrons (different N ).
table!
The chemical properties of the different isotopes of an element are the same, but they might
vary in how stable their nucleus is. We can also write elements as E - A where the E is the
element symbol and the A is the atomic mass of that element. For example Cl-35 has an
atomic mass of 35 u (17 protons and 18 neutrons), while Cl-37 has an atomic mass of 37 u
(17 protons and 20 neutrons).
In nature the different isotopes occur in different percentages. For example Cl-35 might
make up 75% of all chlorine atoms on Earth, and Cl-37 makes up the remaining 25%. The
following worked example will show you how to calculate the average atomic mass for
these two isotopes:
76
4.5
SOLUTION
Step 3 : Add the two values to arrive at the average relative atomic mass of
chlorine
Relative atomic mass of chlorine = 26, 25 u + 9, 25 u = 35, 5 u
If you look on the periodic table (see front of book), the average relative atomic mass for
chlorine is 35, 5 u. See simulation: (
Exercise 4 - 3
1. Atom A has 5 protons and 5 neutrons, and atom B has 6 protons and 5
neutrons. These atoms are:
a. allotropes
b. isotopes
c. isomers
d. atoms of different elements
2. For the sulphur isotopes, 32 S and 34 S, give the number of:
16
16
a. protons
b. nucleons
c. electrons
d. neutrons
3. Which of the following are isotopes of 35 Cl?
17
a.
b.
c.
17
35 Cl
35
17 Cl
37
17 Cl
77
4.5
238
92 E
238
90 E
235
92 E
Protons
Neutrons
Electrons
Carbon-12
Carbon-14
Iron-54
Iron-56
Iron-57
6. If a sample contains 19, 9% boron-10 and 80, 1% boron-11, calculate the
relative atomic mass of an atom of boron in that sample.
7. If a sample contains 79% Mg-24, 10% Mg-25 and 11% Mg-26, calculate
the relative atomic mass of an atom of magnesium in that sample.
8. For the element 234 U (uranium), use standard notation to describe:
92
a. the isotope with 2 fewer neutrons
b. the isotope with 4 more neutrons
9. Which of the following are isotopes of 40 Ca?
20
a.
b.
c.
40
19 K
42
20 Ca
40
18 Ar
More practice
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(3.) 000z
(4.) 0010
(7.) 0013
78
(2.) 000y
(8.) 0014
(9.) 0015
or help at www.everythingscience.co.za
(10.) 0016
4.6
Electronic conguration
The energy of electrons
ESABE
ESABF
The electrons of an atom all have the same charge and the same mass, but each electron
has a different amount of energy. Electrons that have the lowest energy are found closest
to the nucleus (where the attractive force of the positively charged nucleus is the greatest)
and the electrons that have higher energy (and are able to overcome the attractive force of
the nucleus) are found further away.
See video: VPama at www.everythingscience.co.za
Electron arrangement
ESABG
We will start with a very simple view of the arrangement or conguration of electrons
around an atom. This view simply states that electrons are arranged in energy levels (or
shells) around the nucleus of an atom. These energy levels are numbered 1, 2, 3, etc.
Electrons that are in the rst energy level (energy level 1) are closest to the nucleus and will
have the lowest energy. Electrons further away from the nucleus will have a higher energy.
In the following examples, the energy levels are shown as concentric circles around the
central nucleus. The important thing to know for these diagrams is that the rst energy
level can hold 2 electrons, the second energy level can hold 8 electrons and the third
energy level can hold 8 electrons.
1. Lithium
Lithium (Li) has an atomic number
of 3, meaning that in a neutral atom,
electrons
rst energy level
79
4.6
2. Fluorine
Fluorine (F) has an atomic number
of 9, meaning that a neutral atom
also has 9 electrons. The rst 2 electrons are found in the rst energy
level, while the other 7 are found in
the second energy level (Figure 4.7).
3. Neon
But the situation is slightly more complicated than this. Within each energy level, the
electrons move in orbitals. An orbital denes the spaces or regions where electrons move.
This
The rst energy level contains only one s orbital, the second energy level contains one s
orbital and three p orbitals and the third energy level contains one s orbital and three p
orbitals (as well as ve d orbitals). Within each energy level, the s orbital is at a lower
energy than the p orbitals. This arrangement is shown in Figure 4.9.
This diagram also helps us when we are working out the electron conguration of an
element. The electron conguration of an element is the arrangement of the electrons in the
shells and subshells. There are a few guidelines for working out the electron conguration.
These are:
Each orbital can only hold two electrons. Electrons that occur together in an orbital
are called an electron pair.
An electron will always try to enter an orbital with the lowest possible energy.
80
4.6
4s
3p
Third main
energy level
2p
Second main
energy level
3s
E
N
E
R
G
Y
2s
First main
energy level
1s
Figure 4.9: The positions of the rst ten orbitals of an atom on an energy diagram.
An electron will occupy an orbital on its own, rather than share an orbital with
another electron. An electron would also rather occupy a lower energy orbital with
another electron, before occupying a higher energy orbital. In other words, within
one energy level, electrons will ll an s orbital before starting to ll p orbitals.
Tip
When there are two
electrons in an or-
bital,
the electrons
The way that the electrons are arranged in an atom is called its electron conguration.
If the orbital
tron.
posite directions.
Aufbau diagrams
ESABH
Electron pairs
81
4.6
4. Put one electron in each of the three p orbitals in the second energy level (the 2p
orbitals) and then if there are still electrons remaining, go back and place a second
Scientists
You can think of Aufbau diagrams as being similar to people getting on a bus or a train.
People will rst sit in empty seats with empty seats between them and the other people
(unless they know the people and then they will sit next to them). This is the lowest energy.
When all the seats are lled like this, any more people that get on will be forced to sit next
to someone. This is higher in energy. As the bus or train lls even more the people have to
stand to t on. This is the highest energy.
ESABI
A special type of notation is used to show an atoms electron conguration. The notation
describes the energy levels, orbitals and the number of electrons in each. For example, the
electron conguration of lithium is 1s2 2s1 . The number and letter describe the energy level
and orbital and the number above the orbital shows how many electrons are in that orbital.
Aufbau diagrams for the elements uorine and argon are shown in Figures 4.10 and 4.11 respectively. Using spectroscopic notation, the electron conguration of uorine is 1s2 2s2 2p5
and the electron conguration of argon is 1s2 2s2 2p5 3s2 3p6 .
82
4.6
3p
3p
3s
3s
2p
2p
2s
2s
1s
1s
Tip
The
spectroscopic
electron
congura-
form is written as
[noble gas]electrons,
where the noble gas
is the nearest one
that
occurs
before
the element.
For
example, magnesium
can be represented as
[Ne]3s2 and carbon
as [H]2s2 2p2 .
is
QUESTION
known
condensed
as
SOLUTION
Step 1 : Give the number of electrons
Nitrogen has seven electrons.
Step 2 : Place two electrons in the 1s orbital
We start by placing two electrons in the 1s orbital: 1s2 .
1s
83
the
electron
conguration.
Give the electron conguration for nitrogen (N) and draw an Aufbau diagram.
This
4.6
2s
1s
2p
2p
2s
2s
2s
1s
1s
1s
anion. For example chlorine will gain one electron and become Cl or oxygen will gain
two electrons and become O2 .
Aufbau diagrams and electron congurations can be done for cations and anions as well.
The following worked example will show you how.
84
4.6
SOLUTION
Step 1 : Give the number of electrons
Oxygen has eight electrons. The oxygen anion has gained two electrons
and so the total number of electrons is ten.
Step 2 : Place two electrons in the 1s orbital
We start by placing two electrons in the 1s orbital: 1s2 .
1s
1s
85
4.6
2p
2s
FACT
1s
we
shape
draw
that
a
a
closed
has
boundary
(i.e.
shape).
from
nucleus
in
we
95%
are
the
which
sure
In reality
the electrons of an
Orbital shapes
ESABJ
distance
away
Each of the orbitals has a different shape. The s orbitals are spherical and the p orbitals are
dumbbell shaped.
Figure 4.12: The orbital shapes. From left to right: an s orbital, a p orbital, the three p
orbitals
ESABK
Electrons in the outermost energy level of an atom are called valence electrons. The
electrons that are in the energy shells closer to the nucleus are called core electrons. Core
electrons are all the electrons in an atom, excluding the valence electrons. An element that
has its valence energy level full is more stable and less likely to react than other elements
with a valence energy level that is not full.
86
4.6
Exercise 4 - 4
Element or Ion
Electron conguration
Core electrons
Valence electrons
Potassium (K)
Helium (He)
Oxygen ion (O2 )
Magnesium ion (Mg2+ )
More practice
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(1.) 0017
ESABL
By this stage, you may well be wondering why it is important for you to understand how
electrons are arranged around the nucleus of an atom. Remember that during chemical
Chemistry: Matter and Materials
87
4.6
reactions, when atoms come into contact with one another, it is the electrons of these
atoms that will interact rst. More specically, it is the valence electrons of the atoms that
will determine how they react with one another.
To take this a step further, an atom is at its most stable (and therefore unreactive) when all
its orbitals are full. On the other hand, an atom is least stable (and therefore most reactive)
when its valence electron orbitals are not full. This will make more sense when we go on
to look at chemical bonding in a later chapter. To put it simply, the valence electrons are
largely responsible for an elements chemical behaviour and elements that have the same
number of valence electrons often have similar chemical properties.
The most stable congurations are the ones that have full energy levels. These congurations occur in the noble gases. The noble gases are very stable elements that do not react
easily (if at all) with any other elements. This is due to the full energy levels. All elements
would like to reach the most stable electron congurations, i.e. all elements want to be
noble gases. This principle of stability is sometimes referred to as the octet rule. An octet
is a set of 8, and the number of electrons in a full energy level is 8.
See video: VPamk at www.everythingscience.co.za
Aim: To determine what colour a metal cation will cause a ame to be.
Apparatus:
Watch glass
Bunsen burner
methanol
Warning:
Be careful when working with Bunsen burners as you can easily burn
yourself. Make sure all scarves/loose clothing are securely tucked in and
long hair is tied back. Ensure that you work in a well-ventilated space and
that there is nothing ammable near the open ame.
88
4.6
The above experiment on ame tests relates to the line emission spectra of the metals.
These line emission spectra are a direct result of the arrangement of the electrons in metals.
Each metal salt has a uniquely coloured ame.
Exercise 4 - 5
1. Draw Aufbau diagrams to show the electron conguration of each of the
following elements:
a. magnesium
b. potassium
d. neon
e. nitrogen
c. sulphur
2. Use the Aufbau diagrams you drew to help you complete the following
table:
89
4.6
Element
No. of energy
No. of electrons
levels
Electron conguration
(standard notation)
Mg
K
S
Ne
N
Ca2+
Cl
3. Rank the elements used above in order of increasing reactivity. Give reasons for the order you give.
More practice
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(1.-3.) 0018
Earlier in this chapter, we talked about different models of the atom. In science, one of
the uses of models is that they can help us to understand the structure of something that we
cant see. In the case of the atom, models help us to build a picture in our heads of what
the atom looks like.
4.6
Build an atom
See simulation: (
See simulation: (
Chapter 4 | Summary
See the summary presentation (
Some of the scientists who have contributed to the theory of the atom include J.J.Thomson
(discovery of the electron, which led to the Plum Pudding Model of the atom),
Marie and Pierre Curie (work on radiation), Ernest Rutherford (discovery that positive charge is concentrated in the centre of the atom) and Niels Bohr (the arrangement
of electrons around the nucleus in energy levels).
Because of the very small mass of atoms, their mass is measured in atomic mass units
(u). 1 u = 1, 67 1024 g.
The relative atomic mass of an element is the average mass of all the naturally occurring isotopes of that element. The units for relative atomic mass are atomic mass
units. The relative atomic mass is written under the elements symbol on the periodic
table.
An atom is made up of a central nucleus (containing protons and neutrons), surrounded by electrons. Most of the atom is empty space.
The atomic number (Z) is the number of protons in an atom.
The atomic mass number (A) is the number of protons and neutrons in the nucleus
of an atom.
91
4.6
The standard notation that is used to write an element, is A X, where X is the element
Z
symbol, A is the atomic mass number and Z is the atomic number.
The isotope of a particular element is made up of atoms which have the same number
of protons as the atoms in the original element, but a different number of neutrons.
This means that not all atoms of an element will have the same atomic mass.
Within each energy level, an electron may move within a particular shape of orbital.
An orbital denes the space in which an electron is most likely to be found.
The electron conguration is the arrangement of electrons in an atom, molecule or
other physical structure.
Energy diagrams such as Aufbau diagrams are used to show the electron conguration of atoms.
The electron conguration of an atom can be given using spectroscopic notation.
Different orbitals have different shapes: s orbitals are spherically shaped and p orbitals are dumbbell shaped.
The electrons in the outermost energy level are called valence electrons.
The electrons in an atom that are not valence electrons are called core electrons.
Atoms whose outermost energy level is full, are less chemically reactive and therefore
more stable, than those atoms whose outermost energy level is not full.
Chapter 4
1. Write down only the word/term for each of the following descriptions.
a. The sum of the number of protons and neutrons in an atom
b. The dened space around an atoms nucleus, where an electron is
most likely to be found
2. For each of the following, say whether the statement is true or false. If it
is false, re-write the statement correctly.
a. 20 Ne and 22 Ne each have 10 protons, 12 electrons and 12 neutrons.
10
10
b. The atomic mass of any atom of a particular element is always the
same.
c. It is safer to use helium gas rather than hydrogen gas in balloons.
d. Group 1 elements readily form negative ions.
3. The three basic components of an atom are:
a. protons, neutrons, and ions
b. protons, neutrons, and electrons
c. protons, neutrinos, and ions
d. protium, deuterium, and tritium
4. The charge of an atom is:
a. positive
92
4.6
b. neutral
c. negative
d. none of the above
5. If Rutherford had used neutrons instead of alpha particles in his scattering
experiment, the neutrons would:
a. not deect because they have no charge
b. have deected more often
c. have been attracted to the nucleus easily
d. have given the same results
6. Consider the isotope 234 U. Which of the following statements is true?
92
a. The element is an isotope of 234 Pu
94
b. The element contains 234 neutrons
c. The element has the same electron conguration as 238 U
92
d. The element has an atomic mass number of 92
7. The electron conguration of an atom of chlorine can be represented using the following notation:
a. 1s2 2s8 3s7
b. 1s2 2s2 2p6 3s2 3p5
c. 1s2 2s2 2p6 3s2 3p6
d. 1s2 2s2 2p5
8. Give the standard notation for the following elements:
a. beryllium
b. carbon-12
c. titanium-48
d. uorine
9. Give the electron congurations and Aufbau diagrams for the following
elements:
a. aluminium
b. phosphorus
c. carbon
d. oxygen ion
e. calcium ion
10. For each of the following elements give the number of protons, neutrons
and electrons in the element:
a.
b.
c.
d.
e.
195
78 Pt
40
18 Ar
59
27 Co
7
3 Li
11
5 B
11. For each of the following elements give the element or number represented by x:
a.
b.
103
45 X
35
x Cl
93
4.6
c.
x
4 Be
12
25 Mg
26
12 Mg
24
13 Al
Electron conguration
Core electrons
Valence electrons
Boron (B)
Calcium (Ca)
Silicon (Si)
Lithium ion (Li+ )
Neon (Ne)
More practice
video solutions
or help at www.everythingscience.co.za
(1.) 0019
(3.) 001b
(4.) 001c
(5.) 001d
(6.) 001e
(7.) 001f
(8.) 001g
(9.) 001h
(10.) 001i
(11.) 001j
(12.) 001k
(13.) 001m
94
(2.) 001a
(14.) 001n
ESABM
The periodic table of the elements is a method of showing the chemical elements in a
table with the elements arranged in order of increasing atomic number. Most of the work
that was done to arrive at the periodic table that we know can be attributed to a Russian
chemist named Dmitri Mendeleev. Mendeleev designed the table in 1869 in such a way
that recurring ("periodic") trends or patterns in the properties of the elements could be
shown. Using the trends he observed, he left gaps for those elements that he thought were
missing. He also predicted the properties that he thought the missing elements would
have when they were discovered. Many of these elements were indeed discovered and
Mendeleevs predictions were proved to be correct.
See introductory video: ( Video: VParg at www.everythingscience.co.za)
To show the recurring properties that he had observed, Mendeleev began new rows in his
table so that elements with similar properties were in the same vertical columns, called
groups. Each row was referred to as a period. Figure 5.3 shows a simplied version of the
periodic table. The full periodic table is reproduced at the front of this book. You can view
an online periodic table at http://periodictable.com/.
group number
1
18
Period
13
14
15
16
Li
Be
Ne
Na Mg
Al
Si
Cl
Ar
Ni Cu Zn Ga Ge As
Se
Br
Kr
Ca
Sc
Ti
Cr Mn Fe
Co
17 He
Group
Figure 5.1: A simplied diagram showing part of the periodic table. Metals are given in
gray, metalloids in light blue and non-metals in turquoise.
95
5.1
ESABN
Before we can talk about the trends in the periodic table, we rst need to dene some terms
that are used:
Atomic radius
Ionisation energy
The rst ionisation energy is the energy needed to remove one electron from an atom
in the gas phase. The ionisation energy is different for each element. We can also
dene second, third, fourth, etc. ionisation energies. These are the energies needed
to remove the second, third, or fourth electron respectively.
Electron afnity
Electronegativity
A group is a vertical column in the periodic table and is considered to be the most
important way of classifying the elements. If you look at a periodic table, you will
see the groups numbered at the top of each column. The groups are numbered from
left to right starting with 1 and ending with 18. This is the convention that we will
use in this book. On some periodic tables you may see that the groups are numbered
from left to right as follows: 1, 2, then an open space which contains the transition
elements, followed by groups 3 to 8. Another way to label the groups is using Roman
numerals.
A period is a horizontal row in the periodic table of the elements. The periods are
labelled from top to bottom, starting with 1 and ending with 7.
For each element on the periodic table we can give its period number and its group number.
For example, B is in period 2 and group 13. We can also determine the electronic structure
of an element from its position on the periodic table. In chapter 4 you worked out the
electronic conguration of various elements. Using the periodic table we can easily give
the electronic congurations of any element. To see how this works look at the following:
96
5.1
group number
1
18
13 14 15 16 17
p-block
s-block
d-block
We also note that the period number gives the energy level that is being lled. For example,
phosphorus (P) is in the third period and group 15. Looking at the gure above, we see
that the p-orbital is being lled. Also the third energy level is being lled. So its electron
conguration is: [Ne]3s2 3p3 . (Phosphorus is in the third group in the p-block, so it must
have 3 electrons in the p shell.)
ESABO
The following diagram illustrates some of the key trends in the periods:
period number
1
2
3
4
5
6
7
Electronegativity
Ionization energy
Atomic radius
97
5.1
Element
23
11 Na
24
12 Mg
27
13 Al
28
14 Si
31
15 P
Chlorides
NaCl
MgCl2
AlCl3
SiCl4
PCl5
PCl3
Oxides
Na2 O
MgO
Al2 O3
SiO2
P4 O6 or
P4 O10
SO3
SO4
Valence elec-
3s1
3s2
3s2 3p1
3s2 3p2
3s2 3p3
3s2 3p4
32
16 S
or
35
17 Cl
S2 Cl2
no chlorides
or
Cl2 O7 or
Cl2 O
3s2 3p5
trons
Atomic
radius
Electro-
negativity
Melting and
boiling point
Electrical
conductivity
Note that we have left argon (40 Ar) out. Argon is a noble gas with electron conguration:
18
[Ne]3s2 3p6 . Argon does not form any compounds with oxygen or chlorine.
Exercise 5 - 1
1. Use Table 5.1 and Figure 5.2 to help you produce a similar table for the
elements in period 2.
2. Refer to the data table below which gives the ionisation energy (in kJ
1
mol ) and atomic number (Z) for a number of elements in the periodic
table:
98
Name of element
5.2
Ionization energy
Name of element
Ionization energy
1310
10
2072
2360
11
494
517
12
734
895
13
575
797
14
783
1087
15
1051
1397
16
994
1307
17
1250
1673
18
1540
More practice
(1.) 001p
video solutions
or help at www.everythingscience.co.za
(2.) 001q
Chemical properties of
the groups
ESABP
In some groups, the elements display very similar chemical properties and some of the
groups are even given special names to identify them. The characteristics of each group are
mostly determined by the electron conguration of the atoms of the elements in the group.
The names of the groups are summarised in Figure 5.3
99
Halogens
Noble gases
Transition metals
Group 13
Group 14
Group 15
Group 16
Alkali metals
Alkali earth metals
5.2
The halogens and the alkali earth metals are very reactive groups.
Atomic
radius
Table 5.2 summarises the patterns or trends in the properties of the elements in group 1.
Similar trends are observed for the elements in the other groups of the periodic table. We
can use the information in 5.2 to predict the chemical properties of unfamiliar elements.
For example, given the element Francium (Fr) we can say that its electronic structure will
be [Rn]7s1 , it will have a lower rst ionisation energy than caesium (Cs) and its melting and
boiling point will also be lower than caesium.
You should also recall from chapter 2 that the metals are found on the left of the periodic
table, non-metals are on the right and metalloids are found on the zig-zag line that starts at
boron.
100
Element
7
3 Li
Electron structure
Group 1 chlorides
5.2
7
3 Na
7
3 Rb
7
3 Cs
[Kr]4s1
[Xe]5s1
LiCl
RbCl
CsCl
NaCl
7
3K
KCl
Group 1 oxides
Na2 O
K2 O
Rb2 O
Cs2 O
Atomic radius
Electronegativity
Density
Exercise 5 - 2
1. Use Table 5.2 and Figure 5.4 to help you produce similar tables for group
2 and group 17.
2. The following two elements are given. Compare these elements in terms
of the following properties. Explain the differences in each case. 24 Mg
12
and 40 Ca.
20
a. Size of the atom (atomic radius)
b. Electronegativity
c. First ionisation energy
d. Boiling point
3. Study the following graph and explain the trend in electronegativity of the
group 2 elements.
1.5
1.0
0.5
0
Mg
Be
Ca
Sr
Ba
4. Refer to the elements listed below:
101
5.2
Oxygen (O)
Lithium (Li)
Chlorine (Cl)
Calcium (Ca)
Carbon (C)
Magnesium (Mg)
Neon (Ne)
More practice
(1.) 001r
Activity:
(2.) 001s
video solutions
(3.) 001t
(4.) 001u
102
or help at www.everythingscience.co.za
Chapter 5 | Summary
See the summary presentation (
5.2
Elements are arranged in periods and groups on the periodic table. The elements are
arranged according to increasing atomic number.
A group is a column on the periodic table containing elements with similar properties. A period is a row on the periodic table.
Across a period the ionisation energy and electronegativity increase. The atomic
radius decreases across a period.
The groups on the periodic table are labelled from 1 to 18. Group 1 is known as the
alkali metals, group 2 is known as the alkali earth metals, group 17 is known as the
halogens and the group 18 is known as the noble gases. The elements in a group
have similar properties.
The atomic radius and the density both increase down a group. The ionisation energy, electronegativity, and melting and boiling points all decrease down a group.
Chapter 5
1. For the following questions state whether they are true or false. If they are
false, correct the statement
a. The group 1 elements are sometimes known as the alkali earth metals.
b. The group 8 elements are known as the noble gases.
c. Group 7 elements are very unreactive.
d. The transition elements are found between groups 3 and 4.
2. Give one word or term for each of the following:
a. The energy that is needed to remove one electron from an atom
b. A horizontal row on the periodic table
c. A very reactive group of elements that is missing just one electron
from their outer shells.
3. Given
80
35 Br
and
35
17 Cl.
103
5.2
Na
Mg
Al
Si
Cl
Ar
Atomic number
11
12
13
14
15
16
17
18
Density (g cm3 )
0,97
1,74
2,70
2,33
1,82
2,08
3,17
1,78
Melting point ( C)
370,9
923,0
933,5
1687
317,3
388,4
171,6
83,8
Boiling point ( C)
1156
1363
2792
3538
550
717,8
239,1
87,3
Electronegativity
0.93
1.31
1.61
1.90
2.19
2.58
3.16
Na
Mg
Al
Si
Cl
Ar
More practice
(1.) 001v
104
(2.) 001w
video solutions
(3.) 001x
(4.) 001y
or help at www.everythingscience.co.za
(5.) 001z
Chemical bonding
Introduction
ESABQ
When you look at everything around you and what it is made of, you will realise that atoms
seldom exist on their own. More often, the things around us are made up of different
atoms that have been joined together. This is called chemical bonding. Chemical bonding
is one of the most important processes in chemistry because it allows all sorts of different
molecules and combinations of atoms to form, which then make up the objects in the
complex world around us.
See introductory video: (
ESABR
A chemical bond is formed when atoms are held together by attractive forces. This attraction occurs when electrons are shared between atoms, or when electrons are exchanged
between the atoms that are involved in the bond. The sharing or exchange of electrons
takes place so that the outer energy levels of the atoms involved are lled, making the
atoms are more stable. If an electron is shared, it means that it will spend its time moving
in the electron orbitals around both atoms. If an electron is exchanged it means that it is
transferred from one atom to another. In other words one atom gains an electron while the
other loses an electron.
105
6.2
bonding.
You need to remember that it is the valence electrons (those in the outermost level) that are
involved in bonding and that atoms will try to ll their outer energy levels so that they are
more stable. The noble gases have completely full outer energy levels, so are very stable
and do not react easily with other atoms.
Lewis structures
ESABS
Lewis notation uses dots and crosses to represent the valence electrons on different atoms.
Tip
The chemical symbol of the element is used to represent the nucleus and the inner electrons of the atom. To determine which are the valence electrons we look at the last energy
level in the atoms electronic structure (chapter 4). For example, chlorines electronic struc-
ture can be written as:1s2 2s2 2p6 3s2 3p5 or [Ne]3s2 3p5 . The last energy level is the third one
Cl
Cl
The dot and cross in between the two atoms, represent the pair of electrons that are shared
in the covalent bond.
Table 6.1 gives some further examples of Lewis diagrams.
HCN
Hydrogen cyanide
CO2
Carbon dioxide
H2 O
H O
Water
I2
Iodine
lence electrons.
For example:
guration, then we
106
6.2
For carbon dioxide, you can see how we represent a double bond in Lewis notation. As
there are two bonds between each oxygen atom and the carbon atom, two pairs of valence
electrons link them. Similarly, hydrogen cyanide shows how to represent a triple bond.
Exercise 6 - 1
1. Represent each of the following atoms using Lewis notation:
a. beryllium
b. calcium
c. lithium
2. Represent each of the following molecules using Lewis notation:
a. bromine gas (Br2 )
b. carbon dioxide (CO2 )
Which of these two molecules contains a double bond?
3. Two chemical reactions are described below.
nitrogen and hydrogen react to form NH3
More practice
(1.) 0020
(2.) 0021
video solutions
(3.) 0022
or help at www.everythingscience.co.za
(4.) 0023
107
6.3
Covalent Bonding
ESABT
ESABU
Covalent bonding occurs between the atoms of non-metals. The outermost orbitals of the
atoms overlap so that unpaired electrons in each of the bonding atoms can be shared. By
overlapping orbitals, the outer energy shells of all the bonding atoms are lled. The shared
electrons move in the orbitals around both atoms. As they move, there is an attraction between these negatively charged electrons and the positively charged nuclei. This attractive
force holds the atoms together in a covalent bond.
See video: VPaxf at www.everythingscience.co.za
Tip
There is a relationship
odic table.
For the
elements in groups 1
and 2, the valency
is the group number.
ments in groups 13
You will have noticed in table 6.1 that the number of electrons that are involved in bonding
varies between atoms. We can say the following:
In
A single covalent bond is formed when two electrons are shared between the same
two atoms, one electron from each atom.
A double covalent bond is formed when four electrons are shared between the same
two atoms, two electrons from each atom.
A triple covalent bond is formed when six electrons are shared between the same
two atoms, three electrons from each atom.
You should also have noticed that compounds can have a mixture of single, double and
triple bonds and that an atom can have several bonds. In other words, an atom does
not need to share all its valence electrons with one other atom, but can share its valence
electrons with several different atoms.
We say that the valency of the atoms is different.
108
6.3
DEFINITION: Valency
The number of electrons in the outer shell of an atom which are able to be
used to form bonds with other atoms.
Below are a few examples. Remember that it is only the valence electrons that are involved
in bonding and so when diagrams are drawn to show what is happening during bonding,
it is only these electrons that are shown.
SOLUTION
Step 1 : Determine the electron conguration of each of the bonding atoms.
A chlorine atom has 17 electrons and an electron conguration of
[Ne]3s2 3p5 . A hydrogen atom has only one electron and an electron
conguration of 1s1 .
Step 2 : Determine how many of the electrons are paired or unpaired.
Chlorine has seven valence electrons. One of these electrons is unpaired. Hydrogen has one valence electron and it is unpaired.
Step 3 : Work out how the electrons are shared
The hydrogen atom needs one more electron to complete its outermost
energy level. The chlorine atom also needs one more electron to complete its outermost energy level. Therefore one pair of electrons must be
shared between the two atoms. A single covalent bond will be formed.
xx
H
Cl
xx
xx
x
x
Cl
x
x
xx
109
6.3
SOLUTION
xx
3
N
x
110
xx
x
x
H
6.3
SOLUTION
xx
x
x
xx
x+
x
x
x
x
ESABV
Covalent compounds have several properties that distinguish them from ionic compounds
and metals. These properties are:
111
6.3
1. The melting and boiling points of covalent compounds are generally lower than those
of ionic compounds.
2. Covalent compounds are generally more exible than ionic compounds. The molecules
in covalent compounds are able to move around to some extent and can sometimes
slide over each other (as is the case with graphite, which is why the lead in your
pencil feels slightly slippery). In ionic compounds, all the ions are tightly held in
place.
3. Covalent compounds generally are not very soluble in water, for example plastics are
covalent compounds and many plastics are water resistant.
4. Covalent compounds generally do not conduct electricity when dissolved in water,
for example iodine dissolved in pure water does not conduct electricity.
Exercise 6 - 2
1. Explain the difference between the valence electrons and the valency of
an element.
2. Complete the table below by lling in the number of valence electrons for
each of the elements shown:
Element
Group number
No. of electrons
needed to ll outer
shell
He
Li
B
C
F
Ne
Na
Al
P
S
Ca
Kr
112
6.4
3. Draw simple diagrams to show how electrons are arranged in the following covalent molecules:
a. hydrogen sulphide (H2 S)
b. chlorine (Cl2 )
c. nitrogen (N2 )
d. carbon monoxide (CO)
More practice
(1.) 0025
(2.) 0026
video solutions
or help at www.everythingscience.co.za
(3.) 0027
Ionic bonding
The nature of the ionic bond
ESABW
ESABX
When electrons are transferred from one atom to another it is called ionic bonding.
Electronegativity is a property of an atom, describing how strongly it attracts or holds onto
electrons. Ionic bonding takes place when the difference in electronegativity between the
two atoms is more than 1.7. This usually happens when a metal atom bonds with a nonmetal atom. When the difference in electronegativity is large, one atom will attract the
shared electron pair much more strongly than the other, causing electrons to be transferred
to the atom with higher electronegativity. When ionic bonds form, a metal donates one or
more electrons, due to having a low electronegativity, to form a positive ion or cation. The
non-metal atom has a high electronegativity, and therefore readily gains electrons to form a
negative ion or anion. The two ions are then attracted to each other by electrostatic forces.
See video: VPaxl at www.everythingscience.co.za
113
6.4
Example 1:
In the case of NaCl, the difference in electronegativity between Na (0,93) and Cl (3,16)
is 2, 1. Sodium has only one valence electron, while chlorine has seven. Because the
electronegativity of chlorine is higher than the electronegativity of sodium, chlorine will
Note
attract the valence electron of the sodium atom very strongly. This electron from sodium is
Chlorine is a diatomic
Sodium is
Cl
it to take part in
+
Na + electron
Na
[Cl ]
+ electron
+
Na + [ Cl ]
[Na] [ Cl ]
2Mg + O2 2MgO
Because oxygen is a diatomic molecule, two magnesium atoms will be needed to combine with one oxygen molecule (which has two oxygen atoms) to produce two units of
magnesium oxide (MgO).
114
6.4
ESABY
Ionic substances are actually a combination of lots of ions bonded together into a giant
molecule. The arrangement of ions in a regular, geometric structure is called a crystal
lattice. So in fact NaCl does not contain one Na and one Cl ion, but rather a lot of these
two ions arranged in a crystal lattice where the ratio of Na to Cl ions is 1:1. The structure
of the crystal lattice is shown below.
Na
Cl
ESABZ
115
6.5
Exercise 6 - 3
More practice
(1.) 0028
(2.) 0029
video solutions
or help at www.everythingscience.co.za
(3.) 002a
Metallic Bonding
The nature of the metallic bond
ESACA
ESACB
The structure of a metallic bond is quite different from covalent and ionic bonds. In a
metallic bond, the valence electrons are delocalised, meaning that an atoms electrons do
not stay around that one nucleus. In a metallic bond, the positive atomic nuclei (sometimes
called the atomic kernels) are surrounded by a sea of delocalised electrons which are
attracted to the nuclei (see gure below).
See video: VPaxw at www.everythingscience.co.za
116
6.5
Properties of metals
ESACC
Activity:
Building models
Using coloured balls (or jellytots) and sticks (or toothpicks) build models of each
type of bonding. Think about how to represent each kind of bonding. For example,
covalent bonding could be represented by simply connecting the balls with sticks
to represent the molecules, while for ionic bonding you may wish to construct part
of the crystal lattice.
117
6.5
Do some research on types of crystal lattices (although the section on ionic bonding only showed the crystal lattice for
sodium chloride, many other types of lattices exist) and try to build some of these.
Share your ndings with your class and
compare notes to see what types of crystal lattices they found. How would you
show metallic bonding?
Exercise 6 - 4
1. Give two examples of everyday objects that contain:
a. covalent bonds
b. ionic bonds
c. metallic bonds
2. Complete the table which compares the different types of bonding:
Covalent
Ionic
Metallic
3. Complete the table below by identifying the type of bond (covalent, ionic
or metallic) in each of the compounds:
Molecular formula
Type of bond
H2 SO4
FeS
NaI
MgCl2
Zn
118
6.6
4. Use your knowledge of the different types of bonding to explain the following statements:
a. A sodium chloride crystal does not conduct electricity.
b. Most jewellery items are made from metals.
c. It is very hard to break a diamond.
d. Pots are made from metals, but their handles are made from plastic.
More practice
(1.) 002b
(2.) 002c
video solutions
(3.) 002d
or help at www.everythingscience.co.za
(4.) 002e
Writing formulae
ESACD
In chapter 2 you learnt about the writing of chemical formulae. Table 6.2 shows some of
the common anions and cations that you should know.
Name of compound ion
formula
formula
Acetate (ethanoate)
CH3 COO
Manganate
MnO2
4
Ammonium
NH+
4
Nitrate
NO
3
Carbonate
CO2
3
Nitrite
NO
2
Chlorate
ClO3
Oxalate
C2 O2
4
Chromate
CrO2
4
Oxide
O2
Cyanide
CN
Permanganate
MnO
4
Dihydrogen phosphate
H2 PO
4
Peroxide
O2
2
Hydrogen carbonate
HCO
3
Phosphate
PO3
4
Hydrogen phosphate
HPO3
4
Phosphide
P3
Hydrogen sulphate
HSO
4
Sulphate
SO2
4
Hydrogen sulphite
HSO
3
Sulphide
S2
Hydroxide
OH
Sulphite
SO2
3
Hypochlorite
ClO
Thiosulphate
S2 O2
3
Table 6.2: Table showing common compound ions and their formulae
Chemistry: Matter and Materials
119
6.6
ESACE
In chapter 4 you learnt about atomic masses. In this chapter we have learnt that atoms
can combine to form compounds. Molecules are formed when atoms combine through
covalent bonding, for example ammonia is a molecule made up of three hydrogen atoms
and one nitrogen atom. The relative molecular mass (M) of ammonia (NH3 ) is:
relative atomic mass of one nitrogen + relative atomic mass of three hydrogens
17, 03
One molecule of NH3 will have a mass of 17, 03 units. When sodium reacts with chlorine
to form sodium chloride, we do not get a molecule of sodium chloride, but rather a sodium
chloride crystal lattice. Remember that in ionic bonding molecules are not formed. We
can also calculate the mass of one unit of such a crystal. We call this a formula unit and
the mass is called the formula mass. The formula mass for sodium chloride is:
M
relative atomic mass of one sodium atom + relative atomic mass of one chlorine atom
23, 0 + 35, 45
58, 45
Exercise 6 - 5
1. Write the chemical formulae for each of the following compounds and
calculate the relative molecular mass or formula mass:
a. hydrogen cyanide
b. carbon dioxide
c. sodium carbonate
d. ammonium hydroxide
e. barium sulphate
f. copper (II) nitrate
2. Complete the following table. The cations at the top combine with the
anions on the left. The rst row is done for you. Also include the names
of the compounds formed and the anions.
120
6.6
3+
Na+
Mg2+
Al
NH+
4
H+
Br
NaBr
MgBr2
AlBr3
(NH4 )Br
HBr
name:
sodium
magnesium
aluminium
ammonium
hydrogen
bromide
bromide
bromide
bromide
bromide
S2
name:
P3
name:
MnO
4
name:
Cr2 O2
7
name:
HPO2
4
name:
More practice
(1.) 002f
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(2.) 002g
Chapter 6 | Summary
See the summary presentation (
A chemical bond is the physical process that causes atoms and molecules to be
attracted to each other and held together in more stable chemical compounds.
Atoms are more reactive, and therefore more likely to bond, when their outer electron orbitals are not full. Atoms are less reactive when these outer orbitals contain
the maximum number of electrons. This explains why the noble gases do not react.
Lewis notation is one way of representing molecular structure. In Lewis notation,
dots and crosses are used to represent the valence electrons around the central atom.
When atoms bond, electrons are either shared or exchanged.
Chemistry: Matter and Materials
121
6.6
Covalent bonding occurs between the atoms of non-metals and involves a sharing of
electrons so that the orbitals of the outermost energy levels in the atoms are lled.
A double or triple bond occurs if there are two or three electron pairs that are shared
between the same two atoms.
The valency is the number of electrons in the outer shell of an atom which are able
to be used to form bonds with other atoms.
Covalent compounds have lower melting and boiling points than ionic compounds.
Covalent compounds are also generally exible, are generally not soluble in water
and do not conduct electricity.
An ionic bond occurs between atoms where there is a large difference in electronegativity. An exchange of electrons takes place and the atoms are held together by the
electrostatic force of attraction between the resulting oppositely-charged ions.
Chapter 6
b. Ne + Ne Ne2
c. Cl + Cl Cl2
6.6
a. calcium
b. iodine
c. hydrogen bromide (HBr)
d. nitrogen dioxide (NO2 )
5. Given the following Lewis structure, where X and Y each represent a different element:
Ca2+
NH+
4
OH
O2
NO
3
PO3
4
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(5.) 002m
(6.) 002n
(7.) 002p
123
Transverse pulses
ESACF
This chapter forms the basis of the discussion into mechanical waves in the following chapters. We begin by discussing pulses. Pulses are disturbances in a medium. If you tap water
in a bucket with your nger, notice that a ripple moves away from the point where you
touched the water. The ripple is a pulse moving away from where you touched the water.
See introductory video: ( Video: VPchw at www.everythingscience.co.za)
ESACG
What is a medium?
A medium is the substance or material
through which a pulse moves. The medium
carries the pulse from one place to another.
The medium does not create the pulse and
the medium is not the pulse. Therefore the
medium does not travel with the pulse as the
pulse moves through it.
In each medium, the particles that make
up the medium are moved temporarily from
their rest position. In order for a pulse to
travel, the different parts of the medium must
be able to interact with each other.
A disturbance in water
DEFINITION: Medium
A medium is the substance or material in which a pulse will move.
124
7.2
ESACH
ESACI
Take a heavy rope. Have two people hold the rope stretched out
horizontally. Flick the rope at one end only once.
What happens to the disturbance that you created in the rope? Does
it stay at the place where it was created or does it move down the length
of the rope?
In the activity, we created a pulse. A pulse is a single disturbance that moves through a
medium. In a transverse pulse the displacement of the medium is perpendicular to the
direction of motion of the pulse. Figure 7.2 shows an example of a transverse pulse. In
the activity, the rope or spring was held horizontally and the pulse moved the rope up and
down. This was an example of a transverse pulse.
DEFINITION: Pulse
A pulse is a single disturbance that moves through a medium.
125
7.2
DEFINITION: Amplitude
The amplitude of a pulse is the maximum disturbance or distance the medium
is displaced from its rest (equilibrium) position.
Quantity: Amplitude (A)
Unit symbol: m
amplitude (A)
position of rest
pulse length
126
7.2
t=0 s
p
A
t=1 s
p
A
t=2 s
p
A
t=3 s
p
Use your ruler to measure the lengths of a and p. Fill your answers
in the table.
Sign
Symbol
Meaning
Sign
Symbol
Meaning
Time
t=0s
t=1s
t=2s
t=3s
127
7.2
Pulse speed
Unit
Speed is dened as the distance travelled per unit time (this will be covered in more detail
in Motion in One Dimension). If the pulse travels a distance D in a time t, then the pulse
speed v is:
v=
D
t
SOLUTION
Step 1 : Analyse the question
We are given:
the distance travelled by the pulse: D = 2 m
the time taken to travel 2 m: t = 4 s
D
t
128
7.2
D
t
2m
=
4s
= 0, 5 m s1
v=
Tip
The pulse speed depends on the properties of the medium
and not on the amplitude or pulse length
Exercise 7 - 1
of the pulse.
5. The diagram shows two pulses in the same medium. Which has the higher
speed? Explain your answer.
B
A
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(5.) 002u
129
7.3
Superposition of pulses
ESACJ
After pulses pass through each other, each pulse continues along its original direction of
travel, and their original amplitudes remain unchanged.
Constructive interference takes place when two pulses meet each other to create a larger
pulse. The amplitude of the resulting pulse is the sum of the amplitudes of the two initial pulses. This could be two crests meeting or two troughs meeting. This is shown in
Figure 7.3.
130
7.3
Destructive interference takes place when two pulses meet and result in a smaller amplitude disturbance. The amplitude of the resulting pulse is the sum of the amplitudes of
the two initial pulses, but the one amplitude will be a negative number. This is shown in
Figure 7.4. In general, amplitudes of individual pulses are summed together to give the
amplitude of the resultant pulse.
Figure 7.4: Superposition of two pulses. The left-hand series of images demonstrates destructive interference, since the pulses cancel each other. The right-hand series of images
demonstrate a partial cancellation of two pulses, as their amplitudes are not the same in
magnitude.
Physics: Waves, Sound and Light
131
7.3
132
7.3
amplitude (m)
2
A
1
0
0
4
5
distance (m)
SOLUTION
Step 1 : After 1 s
After 1 s, pulse A has moved 1 m to the right and pulse B has moved
amplitude (m)
1 m to the left.
2
A
1
0
0
4
5
distance (m)
Step 2 : After 2 s
amplitude (m)
After 1 s more, pulse A has moved 1 m to the right and pulse B has
moved 1 m to the left.
A+B
2
1
0
0
4
5
distance (m)
Step 3 : After 5 s
133
7.3
amplitude (m)
2
B
1
0
0
4
5
distance (m)
Method:
1. Set up the ripple tank
2. Produce a single pulse and observe what happens (you can do this any means,
tapping the water with a nger, dropping a small object into the water, tapping
a ruler or even using a electronic vibrator)
3. Produce two pulses simultaneously and observe what happens
4. Produce two pulses at slightly different times and observe what happens
134
7.3
Results and conclusion: You should observe that when you produce two pulses
simultaneously you see them interfere constructively and when you produce two
pulses at slightly different times you see them interfere destructively.
Exercise 7 - 2
1. For the following pulse, draw the resulting wave forms after 1 s, 2 s, 3 s,
4 s and 5 s. Each pulse is travelling at 1 m s1 . Each block represents 1 m.
Amplitude (m)
The pulses are shown as thick black lines and the undisplaced medium as
dashed lines.
t=0 s
Position (m)
2. For the following pulse, draw the resulting wave forms after 1 s, 2 s, 3 s,
4 s and 5 s. Each pulse is travelling at 1 m s1 . Each block represents 1 m.
The pulses are shown as thick black lines and the undisplaced medium as
Amplitude (m)
dashed lines.
t=0 s
Position (m)
3. For the following pulse, draw the resulting wave forms after 1 s, 2 s, 3 s,
4 s and 5 s. Each pulse is travelling at 1 m s1 . Each block represents 1 m.
The pulses are shown as thick black lines and the undisplaced medium as
Amplitude (m)
dashed lines.
1
0
t=0 s
1
Position (m)
135
7.3
4. For the following pulse, draw the resulting wave forms after 1 s, 2 s, 3 s,
4 s and 5 s. Each pulse is travelling at 1 m s1 . Each block represents 1 m.
Amplitude (m)
The pulses are shown as thick black lines and the undisplaced medium as
dashed lines.
t=0 s
Position (m)
5. For the following pulse, draw the resulting wave forms after 1 s, 2 s, 3 s,
4 s and 5 s. Each pulse is travelling at 1 m s1 . Each block represents 1 m.
Amplitude (m)
The pulses are shown as thick black lines and the undisplaced medium as
dashed lines.
t=0 s
Position (m)
6. For the following pulse, draw the resulting wave forms after 1 s, 2 s, 3 s,
4 s and 5 s. Each pulse is travelling at 1 m s1 . Each block represents 1 m.
The pulses are shown as thick black lines and the undisplaced medium as
Amplitude (m)
dashed lines.
1
0
t=0 s
1
Position (m)
7. What is superposition of waves?
8. What is constructive interference?
9. What is destructive interference?
More practice
136
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7.3
(1.) 002v
(2.) 002w
(3.) 002x
(7.) 0031
(8.) 0032
(4.) 002y
(5.) 002z
(6.) 0030
(9.) 0033
Chapter 7 | Summary
See the summary presentation (
Destructive interference is when two pulses meet and and result in a smaller pulse.
Physical Quantities
Quantity
Unit name
Unit symbol
Amplitude (A)
metre
m s1
Chapter 7
1. A heavy rope is icked upwards, creating a single pulse in the rope. Make
a drawing of the rope and indicate the following in your drawing:
a. The direction of motion of the pulse
Physics: Waves, Sound and Light
137
7.3
b. Amplitude
c. Pulse length
d. Position of rest
2. A pulse has a speed of 2, 5 m s1 . How far will it have travelled in 6 s?
3. A pulse covers a distance of 75 cm in 2, 5 s. What is the speed of the
pulse?
4. How long does it take a pulse to cover a distance of 200 mm if its speed is
4 m s1 ?
More practice
(1.) 0034
138
(2.) 0035
video solutions
(3.) 0036
or help at www.everythingscience.co.za
(4.) 0037
Transverse waves
Introduction
ESACK
Waves occur frequently in nature. The most obvious examples are waves in water, on a
dam, in the ocean, or in a bucket. We are interested in the properties that waves have. All
waves have the same properties.
Waves do not only occur in water, they occur in any kind of medium. Earthquakes release enough energy to create waves that are powerful enough to travel through the rock of
the Earth. When your friend speaks to you sound waves are produced that travel through
the air to your ears. A wave is simply the disturbance of a medium by moving energy
but how is it different from a pulse?.
See introductory video: ( Video: VPgio at
www.everythingscience.co.za)
What is a transverse
wave?
ESACL
DEFINITION: Wave
A wave is a periodic, continuous disturbance that consists of a train of
pulses.
139
8.2
Activity:
Transverse waves
Take a rope or slinky spring. Have two people hold the rope or spring stretched out
horizontally. Flick the one end of the rope up and down continuously to create a
train of pulses.
In the activity, you created waves. The medium through which these waves propagated was
the rope, which is obviously made up of a very large number of particles (atoms). From the
activity, you would have noticed that the wave travelled from one side to the other, but the
particles (the ribbon) moved only up and down.
140
8.3
particle motion
wave motion
Figure 8.2: A transverse wave, showing the direction of motion of the wave perpendicular
to the direction in which the particles move.
When the particles of a medium move at right angles to the direction of propagation of a
wave, the wave is called transverse. For waves, there is no net displacement of the particles
of the medium (they return to their equilibrium position), but there is a net displacement of
the wave. There are thus two different motions: the motion of the particles of the medium
and the motion of the wave.
The following simulation will help you understand more about waves. Select the oscillate option and then observe what happens. See simulation: ( Simulation: VPnue at
www.everythingscience.co.za)
ESACM
Waves have moving crests (or peaks) and troughs. A crest is the highest point the medium
rises to and a trough is the lowest point the medium sinks to.
crests and troughs on a transverse wave are shown in Figure 8.3.
Crests
equilibrium
Troughs
Figure 8.3: Crests and troughs in a transverse wave.
141
8.4
Amplitude
Activity:
ESACN
Amplitude
equilibrium
b
Fill in the table below by measuring the distance between the equilibrium and
each crest and trough in the wave above. Use your ruler to measure the distances.
Crest/Trough
Measurement (cm)
a
b
c
d
e
f
1. What can you say about your results?
2. Are the distances between the equilibrium position and each crest equal?
3. Are the distances between the equilibrium position and each trough equal?
4. Is the distance between the equilibrium position and crest equal to the distance
between equilibrium and trough?
As we have seen in the activity on amplitude, the distance between the crest and the equilibrium position is equal to the distance between the trough and the equilibrium position.
This distance is known as the amplitude of the wave, and is the characteristic height of
the wave, above or below the equilibrium position. Normally the symbol A is used to
represent the amplitude of a wave. The SI unit of amplitude is the metre (m).
142
8.4
FACT
DEFINITION: Amplitude
A tsunami is a series
of sea waves caused
by
an
underwater
earthquake,
slide,
or
landvolcanic
eruption.
When
Amplitude
may
be
30
than
cm
2 x Amplitude
surface
Amplitude
and
can
travel at speeds up
to 700 km hr1 . In
moves
than
the
faster
bottom,
QUESTION
much as 30 m. The
wavelength can be as
If the crest of a wave measures 2 m above the still water mark in the harbour, what
is the amplitude of the wave?
SOLUTION
the
earthquake,
earthquake
crashed
143
8.4
Activity:
Wavelength
c
equilibrium
Fill in the table below by measuring the distance between crests and troughs in
the wave above.
Distance(cm)
a
b
c
d
1. What can you say about your results?
2. Are the distances between crests equal?
3. Are the distances between troughs equal?
4. Is the distance between crests equal to the distance between troughs?
As we have seen in the activity on wavelength, the distance between two adjacent crests
is the same no matter which two adjacent crests you choose. There is a xed distance
between the crests. Similarly, we have seen that there is a xed distance between the
troughs, no matter which two troughs you look at. More importantly, the distance between
two adjacent crests is the same as the distance between two adjacent troughs. This distance
is called the wavelength of the wave.
The symbol for the wavelength is (the Greek letter lambda) and wavelength is measured
in metres (m).
144
8.4
Example 2: Wavelength
QUESTION
The total distance between 4 consecutive crests of a transverse wave is 6 m. What
is the wavelength of the wave?
SOLUTION
6m
equilibrium
=
=
6m
6m
3
2m
145
8.5
Points in phase
Activity:
ESACO
Points in phase
Fill in the table by measuring the distance between the indicated points.
D
H
C
G
B
E
Points
Distance (cm)
A to F
B to G
C to H
D to I
E to J
What do you nd?
In the activity the distance between the indicated points was equal. These points are then
said to be in phase. Two points in phase are separate by a whole (1, 2, 3, . . .) number
multiple of whole wave cycles or wavelengths. The points in phase do not have to be
crests or troughs, but they must be separated by a complete number of wavelengths.
We then have an alternate denition of the wavelength as the distance between any two
adjacent points which are in phase.
146
8.6
Points that are not in phase, those that are not separated by a complete number of wavelengths, are called out of phase. Examples of points like these would be A and C, or D
and E, or B and H in the Activity.
ESACP
Imagine you are sitting next to a pond and you watch the waves going past you. First
one crest arrives, then a trough, and then another crest. Suppose you measure the time
taken between one crest arriving and then the next. This time will be the same for any two
successive crests passing you. We call this time the period, and it is a characteristic of the
wave.
The symbol T is used to represent the period. The period is measured in seconds (s).
DEFINITION: Period
The period is the time taken for two successive crests (or troughs) to pass a
xed point.
Quantity: Period (T )
Unit name: second
Unit symbol: s
Imagine the pond again. Just as a crest passes you, you start your stopwatch and count
each crest going past. After 1 second you stop the clock and stop counting. The number of
Physics: Waves, Sound and Light
147
8.6
crests that you have counted in the 1 second is the frequency of the wave.
DEFINITION: Frequency
The frequency is the number of successive crests (or troughs) passing a given
point in 1 second.
Quantity: Frequency (f )
Unit symbol: Hz
The frequency and the period are related to each other. As the period is the time taken for
1
1 crest to pass, then the number of crests passing the point in 1 second is T . But this is the
frequency. So
f=
1
T
T =
1
.
f
or alternatively,
For example, if the time between two consecutive crests passing a xed point is
the period of the wave is
1
2
1
2
s, then
=
=
=
1
T
1
1
2 s
2 s1
SOLUTION
148
8.7
1
f
1
f
1
10 Hz
0, 1 s
=
=
Step 4 : Write the answer
Speed of a transverse
wave
ESACQ
Unit
The distance between two successive crests is 1 wavelength, . Thus in a time of 1 period,
the wave will travel 1 wavelength in distance. Thus the speed of the wave, v, is:
v=
However, f =
1
T
distance travelled
=
time taken
T
T
1
T
=
=
Physics: Waves, Sound and Light
f
149
8.7
We call this equation the wave equation. To summarise, we have that v = f where
v = speed in m s1
= wavelength in m
f = frequency in Hz
Wave equation
or
v =f
v=
SOLUTION
Step 1 : Determine what is given and what is required
frequency of wave: f = 10 Hz
wavelength of wave: = 0,25 m
We are required to calculate the speed of the wave as it travels along
the string.
All quantities are in SI units.
Step 2 : Determine how to approach the problem
We know that the speed of a wave is:
v =f
and we are given all the necessary quantities.
Step 3 : Substituting in the values
150
8.7
(10 Hz)(0, 25 m)
(10 s1 )(0, 25 m)
2, 5 m s1
SOLUTION
D
v
(from v =
D
)
t
We know that
v =f
Physics: Waves, Sound and Light
151
8.7
D
f
(8.1)
=
=
=
=
d
f
2m
(1 Hz)(0, 2 m)
2m
1 )(0, 2 m)
(1 s
10 s
Exercise 8 - 1
3. Consider the diagram below and answer the questions that follow:
8.7
1
0
1
1
Draw the following:
a. A wave with twice the amplitude of the given wave.
b. A wave with half the amplitude of the given wave.
c. A wave travelling at the same speed with twice the frequency of the
given wave.
d. A wave travelling at the same speed with half the frequency of the
given wave.
e. A wave with twice the wavelength of the given wave.
f. A wave with half the wavelength of the given wave.
g. A wave travelling at the same speed with twice the period of the given
wave.
h. A wave travelling at the same speed with half the period of the given
wave.
6. A transverse wave travelling at the same speed with an amplitude of 5 cm
has a frequency of 15 Hz. The horizontal distance from a crest to the
nearest trough is measured to be 2,5 cm. Find the
a. period of the wave.
b. speed of the wave.
7. A y aps its wings back and forth 200 times each second. Calculate the
period of a wing ap.
8. As the period of a wave increases, the frequency increases/decreases/does
not change.
9. Calculate the frequency of rotation of the second hand on a clock.
10. Microwave ovens produce radiation with a frequency of 2 450 MHz (1 MHz
= 106 Hz) and a wavelength of 0,122 m. What is the wave speed of the
radiation?
153
8.7
C
B
D
E
I
H
P
O
More practice
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(1.) 0039
(2.) 003a
(3.) 003b
(4.) 003c
(5.) 003d
(6.) 003e
(7.) 003f
(8.) 003g
(9.) 003h
(10.) 003i
(11.) 003j
(12.) 003k
Chapter 8 | Summary
See the summary presentation (
In a transverse wave, the particles move perpendicular to the motion of the wave.
The amplitude (A) is the maximum distance from equilibrium position to a crest (or
trough), or the maximum displacement of a particle in a wave from its position of
rest.
154
8.7
The wavelength () is the distance between any two adjacent points on a wave that
are in phase. It is measured in metres.
The period (T ) of a wave is the time it takes a wavelength to pass a xed point. It is
measured in seconds (s).
The frequency (f ) of a wave is how many waves pass a point in a second. It is
measured in hertz (Hz) or s1 .
Frequency: f =
Period: T =
1
T
1
f
Speed: v = f or v =
T.
Physical Quantities
Quantity
Unit name
Unit symbol
Amplitude (A)
metre
Wavelength ()
metre
second
hertz
Hz (s1 )
m s1
Period (T )
Frequency (f )
Wave speed (v)
Chapter 8
155
8.7
b. Write down the letters that indicate any TWO points that are:
i. in phase
ii. out of phase
iii. Represent ONE wavelength.
c. Calculate the amplitude of the wave.
d. Show that the period of the wave is 0,625 s.
e. Calculate the frequency of the waves.
f. Calculate the velocity of the waves.
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156
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(2.) 003n
Longitudinal waves
ESACR
We have already studied transverse pulses and waves. In this chapter we look at another
type of wave called a longitudinal wave. In transverse waves, the motion of the particles
in the medium was perpendicular to the direction of the wave. In longitudinal waves,
the particles in the medium move parallel (in the same direction as) to the motion of the
wave. Examples of transverse waves (discussed in the previous chapter) are water waves.
An example of a longitudinal wave is a sound wave. See introductory video: (
Video:
VPdim at www.everythingscience.co.za)
ESACS
When we studied transverse waves we looked at two different motions: the motion of
the particles of the medium and the motion of the wave itself. We will do the same for
longitudinal waves.
The question is how do we construct such a wave?
A longitudinal wave is seen best in a slinky spring. Do the following investigation to nd
out more about longitudinal waves.
157
9.2
Activity:
ribbon
Tie a ribbon to the middle of the spring. Watch carefully what happens to the
ribbon when the end of the spring is icked. Describe the motion of the ribbon.
Flick the spring back and forth continuously to set up a train of pulses, a longitudinal wave.
From the investigation you will have noticed that the disturbance moves parallel to the
direction in which the spring was pulled. The ribbon in the investigation represents one
particle in the medium. The particles in the medium move in the same direction as the
wave.
As in the case of transverse waves the following properties can be dened for longitudinal
waves: wavelength, amplitude, period, frequency and wave speed.
Compression and
rarefaction
ESACT
However instead of crests and troughs, longitudinal waves have compressions and rarefactions.
158
9.3
DEFINITION: Compression
A compression is a region in a longitudinal wave where the particles are
closest together.
DEFINITION: Rarefaction
A rarefaction is a region in a longitudinal wave where the particles are
furthest apart.
compressions
rarefactions
Figure 9.3: Compressions and rarefactions on a longitudinal wave
Wavelength and
amplitude
ESACU
DEFINITION: Wavelength
The wavelength in a longitudinal wave is the distance between two consecutive points that are in phase.
159
9.4
The wavelength in a longitudinal wave refers to the distance between two consecutive
compressions or between two consecutive rarefactions.
DEFINITION: Amplitude
The amplitude is the maximum displacement from equilibrium. For a longitudinal wave which is a pressure wave this would be the maximum increase
(or decrease) in pressure from the equilibrium pressure that is cause when a
compression (or rarefaction) passes a point.
The amplitude is the distance from the equilibrium position of the medium to a compression or a rarefaction.
ESACV
DEFINITION: Period
The period of a wave is the time taken by the wave to move one wavelength.
DEFINITION: Frequency
The frequency of a wave is the number of wavelengths per second.
160
9.5
The period of a longitudinal wave is the time taken by the wave to move one wavelength.
As for transverse waves, the symbol T is used to represent period and period is measured
in seconds (s).
The frequencyf of a wave is the number of wavelengths per second. Using this denition
and the fact that the period is the time taken for 1 wavelength, we can dene:
f=
1
T
T =
1
f
or alternately,
Speed of a longitudinal
wave
ESACW
The speed of a longitudinal wave is dened in the same was as the speed of transverse
waves:
Unit: m s1
v=
However, f =
1
T
distance travelled
=
time taken
T
161
9.5
T
1
T
We call this equation the wave equation. To summarise, we have that v = f where
v = speed in m s1
= wavelength in m
f = frequency in Hz
SOLUTION
440 Hz
0, 784 m
=
=
162
9.5
SOLUTION
163
9.5
Chapter 9 | Summary
See the summary presentation (
A longitudinal wave is a wave where the particles in the medium move parallel to
1
1
or f =
f
T
The relationship between wave speed (v), frequency (f ) and wavelength () is given
by
v = f
Physical Quantities
Quantity
Unit name
Unit symbol
Amplitude (A)
metre
Wavelength ()
metre
second
hertz
Hz (s1 )
m s1
Period (T )
Frequency (f )
Wave speed (v)
Chapter 9
9.5
a. solid
b. liquid
c. gas
d. vacuum
3. A longitudinal wave has a compression to compression distance of 10 m.
It takes the wave 5 s to pass a point.
a. What is the wavelength of the longitudinal wave?
b. What is the speed of the wave?
4. A ute produces a musical sound travelling at a speed of 320 m s1 . The
frequency of the note is 256 Hz. Calculate:
a. the period of the note
b. the wavelength of the note
More practice
(1.) 003p
(2.) 003q
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(3.) 003r
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(4.) 003s
165
Sound
10
Introduction
ESACX
Have you ever thought about how amazing your sense of hearing is? It is actually pretty
remarkable that we can hear the huge range of sounds and determine direction so quickly.
How does something actually make a sound that you can hear? Anything that generates a
disturbance in the air creates a pulse that travels away from the place where is was created.
If this pulse enters your ear it can cause your ear drum to vibrate which is how you hear.
If the source of the pulse creates a train of pulses then the disturbance is a wave.
introductory video: ( Video: VPdxu at www.everythingscience.co.za)
See
We generally say that sound is a wave. Sound waves are longitudinal, pressure waves, that
means that the waves consists of compressions and rarefactions of the pressure of the air.
Sound waves
ESACY
A tuning fork is an instrument used by musicians to create sound waves of a specic frequency. They are often used to tune musical instruments.
Sound waves coming from a tuning fork are
caused by the vibrations of the tuning fork
which push against the air particles in front of
Tuning fork
it. As the air particles are pushed together a
compression is formed. The particles behind
the compression move further apart causing
a rarefaction. As the particles continue to
push against each other, the sound wave travels through the air. Due to this motion of the
particles, there is a constant variation in the
pressure in the air. Sound waves are therefore
Photo by amonya on Flickr.com
pressure waves. This means that in media
where the particles are closer together, sound
waves will travel faster.
See video: VPdya at www.everythingscience.co.za
166
10.1
Sound waves travel faster through liquids, like water, than through the air because water is
denser than air (the particles are closer together). Sound waves travel faster in solids than
in liquids.
Tip
rarefactions
vibrating
tuning
fork
sound
pressure
This
wave
is
wave.
means
that
compressions
sure
Figure 10.2: Sound waves are pressure waves and need a medium through which to travel.
(compressions)
and
low
pressure
(rarefactions)
are
Activity:
These
Have you ever wondered if you can actually use tin cans or cups to
brates longitudinally
and
nal
the
uctuations.
String
Toothpicks or small sticks
Try This:
1. Tie a toothpick on each end of a length of string.
2. Make a hole in the base of the can or cup. Poke the toothpick
through the end of the can. Pull the string tight so the toothpick
rests on the inside bottom of the can. Put one can at each end
of the string. (You may want to experiment with different cans or
cups and strings or wires to see what works best.)
3. Hold the string tight and talk into one of the cans. The person at
the other end should be able to hear you. Why does the string
have to be tight?
4. Try to make a party line by tying a third string and can or cup
onto the middle of the string. Can everybody talk to everybody
else?
Sound waves need something to travel through. Usually they travel
through air, but they can travel much faster and farther through a string.
The string has to be tight or else the sound wave cannot travel through
it. The cup helps to amplify the sound on the other end.
167
longitudi-
motion
produces
vibrates.
compressions
of
air
pressure
10.2
Speed of sound
ESACZ
The speed of sound depends on the medium the sound is travelling in. Sound travels faster
in solids than in liquids, and faster in liquids than in gases. This is because the density
of solids is higher than that of liquids which means that the particles are closer together.
FACT
The speed of sound
Substance
v (m s1 )
aluminium
6420
brick
3650
copper
4760
glass
5100
gold
3240
lead
2160
water, sea
1531
air, 0 o C
331
air, 20 o C
343
pressure is higher than it would at places high Table 10.1: The speed of sound in different
materials.
above sea level.
168
10.2
Stopwatch
Method: The speed of sound can be measured because light travels much faster
than sound. Light travels at about 300 000 m s1 (you will learn more about the
speed of light in the next chapter) while sound only travels at about 300 m s1 .
This difference means that over a distance of 300 m, the light from an event will
reach your eyes almost instantly but there will be an approximate half a second lag
before you hear the sound produced. Thus if a starters pistol is red from a great
distance, you will see the smoke immediately but there will be a lag before you
hear the sound. If you know the distance and the time then you can calculate the
speed (distance divided by time). You dont need a gun but anything that you can
see producing a loud sound.
Try this:
1. Find a place where you know the precise, straight-line distance between two
points (maybe an athletics track)
2. Someone needs to stand at the one point to produce the sound
3. Another person needs to stand at the other point with the stop watches
4. The person with the stopwatch should start the stopwatch when they see the
other person make the sound and stop the stopwatch when they hear the
sound (do this a few times and write the times down)
Results:
Time (s)
Distance (m)
v (m s1 )
If you took
D
t
Averages
169
10.2
ESADA
When the sound waves collide with an object they are reected. You can think of the
individual particles that are oscillating about their equilibrium position colliding into the
object when the wave passes. They bounce off the object causing the wave to be reected.
In a space with many small objects there are reections at every surface but they are too
small and too mixed up to have an outcome that a human can hear. However, when there
is an open space that has only large surfaces, for example a school hall that is empty, then
the reected sound can actually be heard. The sound wave is reected in such a wave that
the wave looks the same but is moving in the opposite direction.
www.everythingscience.co.za
This means that if you stand in a hall and loudly say hello you will hear yourself say
hello a split second later. This is an echo. This can also happen outdoors in a wide open
space with a large reecting surface nearby, like standing near a mountain cliff in an area
with no trees or bushes.
This is a very useful property of waves.
SONAR
ESADB
10.2
Example 1: SONAR
QUESTION
A ship sends a signal to the bottom of the ocean to determine the depth of the
ocean. The speed of sound in sea water is 1450 m s1 If the signal is received 1,5
seconds later, how deep is the ocean at that point?
SOLUTION
1450 m.s1
0, 75 s one way
Distance
speed time
st
=
=
1450 m s1 0, 75s
1087, 5 m
Echolocation
ESADC
Animals like dolphins and bats make use of sounds waves to nd their way. Just like ships
on the ocean, bats use sonar to navigate. Waves that are sent out are reected off the objects
around the animal. Bats, or dolphins, then use the reected sounds to form a picture of
their surroundings. This is called echolocation.
Physics: Waves, Sound and Light
171
10.3
Characteristics of a
sound wave
ESADD
Since sound is a wave, we can relate the properties of sound to the properties of a wave.
The basic properties of sound are: pitch, loudness and tone.
Sound A
Sound B
Sound C
Figure 10.3: Pitch and loudness of sound. Sound B has a lower pitch (lower frequency)
than Sound A and is softer (smaller amplitude) than Sound C.
Pitch
The frequency of a sound wave is what your ear understands as pitch. A higher frequency
sound has a higher pitch, and a lower frequency sound has a lower pitch. In Figure 10.3
sound A has a higher pitch than sound B. For instance, the chirp of a bird would have a
high pitch, but the roar of a lion would have a low pitch.
The human ear can detect a wide range of frequencies. Frequencies from 20 to 20 000 Hz
are audible to the human ear. Any sound with a frequency below 20 Hz is known as an
infrasound and any sound with a frequency above 20 000 Hz is known as an ultrasound.
Table 10.2 lists the ranges of some common animals compared to humans.
172
10.3
20
20 000
Dogs
50
45 000
Cats
45
85 000
Bats
20
120 000
Dolphins
0,25
200 000
Elephants
Activity:
10 000
Range of wavelengths
Using the information given in Table 10.2, calculate the lower and upper wavelengths that each species can hear. Assume the speed of sound in air is 344 m s1 .
Loudness
The amplitude of a sound wave determines its loudness or volume. A larger amplitude
means a louder sound, and a smaller amplitude means a softer sound. In Figure 10.3 sound
C is louder than sound B. The vibration of a source sets the amplitude of a wave. It transmits
energy into the medium through its vibration. More energetic vibration corresponds to
larger amplitude. The molecules move back and forth more vigorously.
The loudness of a sound is also determined by the sensitivity of the ear. The human ear is
more sensitive to some frequencies than to others. The volume we receive thus depends
on both the amplitude of a sound wave and whether its frequency lies in a region where
the ear is more or less sensitive.
Exercise 10 - 1
Study the following diagram representing a musical note. Redraw the diagram
for a note
1. with a higher pitch
2. that is louder
Physics: Waves, Sound and Light
173
10.3
3. that is softer
More practice
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(1.) 003t
Activity:
The size and shape of instruments inuences the sounds that they are able to produce. Find some instruments that have different physical characteristics and compare their sounds. You could:
Option 1: Vuvuzelas:
Compare the sounds made by blowing through vuvuzelas of different
sizes. You will need to nd a few different vuvuzelas. Take turns blowing
the different ones, one at a time and
record which you think is louder (amplitude), which is of higher pitch (frequency).
174
10.3
Oscilloscope
The microphone can then pick up the
sound and convert it to an electrical
signal which can be displayed on the
oscilloscope.
An oscilloscope
Note: The display of the oscilloscope will show you a transverse wave pattern. This does not
mean that sound waves are transverse waves but just shows that
the pressure being measured is
uctuating because of a pressure
wave.
You will be able to experiment with different amplitudes and frequencies using
the function generator and see what impact the changes have on the waveform
picked up by the microphone.
175
10.3
ESADE
Intensity is one indicator of amplitude. Intensity is the energy transmitted over a unit of
area each second.
The unit of intensity is the decibel (symbol: dB).
Source
Intensity (dB)
Rocket Launch
180
1018
Jet Plane
140
1014
Threshold of Pain
120
1012
Rock Band
110
1011
Factory
80
108
City Trafc
70
107
Normal Conversation
60
106
Library
40
104
Whisper
20
102
Threshold of hearing
176
10.4
Frequency (Hz)
At ear
Bell opening
1m
2m
125
36
62
38
35
250
92
106
82
85
500
103
121
102
101
1 000
106
122
108
100
2 000
101
122
110
101
4 000
97
109
110
102
5 000
93
111
109
100
8 000
87
110
107
98
Table 10.4: Average vuvuzela intensity measurements across frequencies at 4 distinct distances from the bell end of the vuvuzela (dBA) taken from South African Medical Journal
(Cape Town, South Africa) 100 (4): 192
Ultrasound
ESADF
Ultrasound is sound with a frequency that is higher than 20 kHz. Some animals, such as
dogs, dolphins, and bats, have an upper limit that is greater than that of the human ear and
can hear ultrasound.
Application
20
40
50
500
Welding of plastics
15
40
Tumour ablation
250
2000
177
10.4
FACT
Ultrasound generator-
that
with
they
away
claims
frighten
rodents
Ultrasound in medicine can visualise muscle and soft tissue, making them useful for
and
Ultrasound sources may be used to generate local heating in biological tissue, with appli-
cations in physical therapy and cancer treatment. Focused ultrasound sources may be used
quickly
the
learn
that
speakers
are
harmless.
Ultrasonic cleaners, sometimes called supersonic cleaners, are used at frequencies from 2040 kHz for jewellery, lenses and other optical parts, watches, dental instruments, surgical
instruments and industrial parts. These cleaners consist of containers with a uid in which
the object to be cleaned is placed. Ultrasonic waves are then sent into the uid. The main
mechanism for cleaning action in an ultrasonic cleaner is actually the energy released from
the collapse of millions of microscopic bubbles occurring in the liquid of the cleaner.
ESADG
The human ear is divided into three main sections: the outer, middle, and inner ear. Lets
follow the journey of a sound wave from the pinna (outermost part) to the auditory nerve
(innermost part) which transmits a signal to the brain. The pinna is the part of the ear we
typically think of when we refer to the ear. Its main function is to collect and focus a sound
wave. The wave then travels through the ear canal until it meets the eardrum. The pressure
uctuations of the sound wave make the eardrum vibrate. The three very small bones of
the middle ear, the malleus (hammer), the incus (anvil), and the stapes (stirrup), transmit the
178
10.4
179
10.4
Chapter 10 | Summary
See the summary presentation (
The frequency of a sound is an indication of how high or low the pitch of the sound
is.
The human ear can hear frequencies from 20 to 20 000 Hz. Infrasound waves
have frequencies lower than 20 Hz. Ultrasound waves have frequencies higher than
20 000 Hz.
Sound travels faster in a solid than a liquid and faster in a liquid than in a gas.
Sound travels faster at sea level where the air pressure is higher.
The intensity of a sound is the energy transmitted over a certain area. Intensity is a
measure of frequency.
Ultrasound can be used to form pictures of things we cannot see, like unborn babies
or tumours.
Echolocation is used by animals such as dolphins and bats to see their surroundings
by using ultrasound.
Ships use sonar to determine how deep the ocean is or to locate shoals of sh.
Physical Quantities
Quantity
Unit name
Unit symbol
m s1
Wavelength ()
metre
Amplitude (A)
metre
second
hertz
Hz (s1 )
Velocity (v)
Period (T )
Frequency (f )
180
10.4
Chapter 10
1. Choose a word from column B that best describes the concept in column
A.
Column A
Column B
1. pitch of sound
A. amplitude
2. loudness of sound
B. frequency
3. quality of sound
C. speed
D. waveform
c. 20 Hz 20 000 Hz
5. X and Y are different wave motions. In air, X travels much faster than Y
but has a much shorter wavelength. Which types of wave motion could X
and Y be?
181
10.4
1.
microwaves
red light
2.
radio
infra red
3.
red light
sound
4.
sound
ultraviolet
5.
ultraviolet
radio
110 m
cliff 2
cliff 1
Assuming that the velocity of sound is 330 m s1 , what will be the time
interval between the two loudest echoes?
a.
b.
c.
2
3
1
6
5
6
s
s
s
d. 1 s
1
e. 3 s
8. A dolphin emits an ultrasonic wave with frequency of 0,15 MHz. The
speed of the ultrasonic wave in water is 1500 m s1 . What is the wavelength of this wave in water?
a. 0,1 mm
b. 1 cm
c. 10 cm
d. 10 m
e. 100 m
9. The amplitude and frequency of a sound wave are both increased. How
are the loudness and pitch of the sound affected?
182
10.4
loudness
pitch
increased
raised
increased
unchanged
increased
lowered
decreased
raised
decreased
lowered
10. A jet ghter travels slower than the speed of sound. Its speed is said to be:
a. Mach 1
b. supersonic
c. subsonic
d. hypersonic
e. infrasonic
11. A sound wave is different from a light wave in that a sound wave is:
a. produced by a vibrating object and a light wave is not.
b. not capable of travelling through a vacuum.
c. not capable of diffracting and a light wave is.
d. capable of existing with a variety of frequencies and a light wave has
a single frequency.
12. At the same temperature, sound waves have the fastest speed in:
a. rock
b. milk
c. oxygen
d. sand
13. Two sound waves are travelling through a container of nitrogen gas. The
rst wave has a wavelength of 1,5 m, while the second wave has a wavelength of 4,5 m. The velocity of the second wave must be:
a.
b.
1
9
1
3
183
10.4
how long does it take the sound to reach the person standing at the gate?
16. Person 1 speaks to person 2. Explain how the sound is created by person
1 and how it is possible for person 2 to hear the conversation.
17. Sound cannot travel in space. Discuss what other modes of communication astronauts can use when they are outside the space shuttle?
18. An automatic focus camera uses an ultrasonic sound wave to focus on
objects. The camera sends out sound waves which are reected off distant
objects and return to the camera. A sensor detects the time it takes for the
waves to return and then determines the distance an object is from the
camera. If a sound wave (speed = 344 ms1 ) returns to the camera
0,150 s after leaving the camera, how far away is the object?
19. Calculate the frequency (in Hz) and wavelength of the annoying sound
made by a mosquito when it beats its wings at the average rate of 600 wing
beats per second. Assume the speed of the sound waves is 344 ms1 .
20. How does halving the frequency of a wave source affect the speed of the
waves?
21. Humans can detect frequencies as high as 20 000 Hz. Assuming the
speed of sound in air is 344 ms1 , calculate the wavelength of the sound
corresponding to the upper range of audible hearing.
22. An elephant trumpets at 10 Hz. Assuming the speed of sound in air is
344 ms1 , calculate the wavelength of this infrasonic sound wave made
by the elephant.
23. A ship sends a signal out to determine the depth of the ocean. The signal
returns 2,5 seconds later. If sound travels at 1450 m.s1 in sea water, how
deep is the ocean at that point?
24. A person shouts at a cliff and hears an echo from the cliff 1 s later. If the
speed of sound is 344 m s1 , how far away is the cliff?
25. Select a word from Column B that best ts the description in Column A:
184
10.4
Column A
Column B
A. longitudinal waves
B. frequency
C. period
D. amplitude
E. sound waves
F. standing waves
G. transverse waves
H. wavelength
I. music
J. sounds
K. wave speed
More practice
video solutions
or help at www.everythingscience.co.za
(1.) 003u
(2.) 003v
(3.) 003w
(4.) 003x
(5.) 003y
(6.) 003z
(7.) 0040
(8.) 0041
(9.) 0042
(10.) 0043
(11.) 0044
(12.) 0045
(13.) 0046
(14.) 0048
(15.) 0049
(16.) 004c
(17.) 004d
(18.) 004e
(19.) 004f
(20.) 004g
(21.) 004h
(22.) 004i
(23.) 004j
(24.) 004k
(25.) 004m
185
Electromagnetic
radiation
11
What is electromagnetic
radiation?
ESADH
The most common example of electromagnetic (EM) radiation is visible light. Everyone is
very familiar with light in everyday life, you can only see things because light bounces
off them and enters your eyes. This alone makes it worthwhile to learn about it but
there are also very many other applications of EM radiation. It is called electromagnetic because there are electric and magnetic elds making up the radiation. We will
look at this in more detail a little later. See introductory video: ( Video: VPefg at
www.everythingscience.co.za)
In everyday experience, light doesnt seem to have many special properties but it does:
A huge spectrum: The light we can see (visible EM radiation) is only a small part of
all of the EM radiation (electromagnetic spectrum) that exists.
Natures speed limit: Nothing moves faster than the speed of light.
Wave nature: All EM radiation has the ability to behave like a wave which we call
wave-like behaviour.
Particle nature: All EM radiation has the ability to behave like a particle which we
call particle-like behaviour.
186
Wave-like nature of EM
radiation
11.2
ESADI
If you watch a colony of ants walking up the wall, they look like a thin continuous black
line. But as you look closer, you see that the line is made up of thousands of separated
black ants.
Light and all other types of electromagnetic radiation seem like a continuous wave at rst,
but when one performs experiments with light, one can notice that light can have both
wave- and particle-like properties. Just like the individual ants, light can also be made up
of individual bundles of energy, or quanta of light.
Light has both wave-like and particle-like properties (waveparticle duality), but only shows
one or the other, depending on the kind of experiment we perform. A wave-type experiment shows the wave nature, and a particle-type experiment shows particle nature. One
cannot test the wave and the particle nature at the same time. A particle of light is called a
photon.
DEFINITION: Photon
A photon is a quantum (energy packet) of light.
Fields
ESADJ
187
11.2
by c. We will use 3 108 m s1 for all of our calculations in this book. Although an
electromagnetic wave can travel through empty space, it can also travel through a medium
(such as water and air). When an electromagnetic wave travels through a medium, it travels
slower than it would through empty space (vacuum).
x
B
Figure 11.2: A diagram showing the mutually regenerating electric eld (red (solid) line)
and magnetic eld (blue (dashed) line).
v =f
Except that we can replace v with c:
c=f
188
11.2
Example 1: EM radiation I
QUESTION
Calculate the frequency of an electromagnetic wave with a wavelength of
4, 2 107 m.
SOLUTION
3 108 m s1
f 4, 2 107 m
7, 14 1014 Hz
Example 2: EM Radiation II
QUESTION
An electromagnetic wave has a wavelength of 200 nm. What is the frequency of
the radiation?
SOLUTION
189
11.3
3 108 m s1
f 200 109 m
1.5 1015 Hz
Electromagnetic
spectrum
ESADK
EM radiation is classied into types according to the frequency of the wave: these types
include, in order of increasing frequency, radio waves, microwaves, infrared radiation,
visible light, ultraviolet radiation, X-rays and gamma rays.
See video: VPemp at www.everythingscience.co.za
Table 11.3 lists the wavelength and frequency ranges of the divisions of the electromagnetic
spectrum.
190
11.3
Category
gamma rays
<1
> 3 1019
X-rays
1-10
3 1017 -3 1019
ultraviolet light
10-400
7, 5 1014 -3 1017
visible light
400-700
infrared
700-105
microwave
105 108
3 109 -3 1012
radio waves
> 108
< 3 109
Uses
gamma rays
X-rays
ultraviolet light
bees can see into the ultraviolet because owers stand out more clearly at this frequency
visible light
infrared
microwave
radio waves
Exercise 11 - 1
1. Arrange the following types of EM radiation in order of increasing frequency: infrared, X-rays, ultraviolet, visible, gamma.
191
11.3
More practice
(1.) 004q
(2.) 004r
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(3.) 004s
Figure 11.3: The electromagnetic spectrum as a function of frequency. The different types
according to wavelength are shown as well as everyday comparisons.
EM radiation in the visible part of the spectrum is scattered off all of the objects around
us. This EM radiation provides the information to our eyes that allows us to see. The frequencies of radiation the human eye is sensitive to constitute only a very small part of all
possible frequencies of EM radiation. The full set of EM radiation is called the electromagnetic spectrum. To simplify things the EM spectrum divided into sections (such as radio,
microwave, infrared, visible, ultraviolet, X-rays and gamma-rays).
See video: VPfdy at
www.everythingscience.co.za
The EM spectrum is continuous (has no gaps) and innite. Due to technological limitations,
192
11.4
we can only use electromagnetic radiation with wavelengths between 1014 m and 1015 m.
Penetrating ability of
EM radiation
ESADL
193
11.4
The SPF rating does not specify protection against UVA radiation. Some sunscreen lotion
now includes compounds such as titanium dioxide which helps protect against UVA rays.
Other UVA-blocking compounds found in sunscreen include zinc oxide and avobenzone.
What makes a good sunscreen?
UVB protection: Padimate O, Homosalate, Octisalate (octyl salicylate), Octi-
194
11.4
Gamma-rays
Due to their high energies, gamma-rays are able to cause serious damage when absorbed
by living cells.
Gamma-rays are not stopped by the skin and can induce DNA alteration by interfering
with the genetic material of the cell. DNA double-strand breaks are generally accepted to
be the most biologically signicant lesion by which ionising radiation causes cancer and
hereditary disease.
A study done on Russian nuclear workers exposed to external whole-body gamma-radiation
at high doses shows a link between radiation exposure and death from leukaemia, lung,
liver, skeletal and other solid cancers.
Cellphone users are recommended to minimise their exposure to the radiation, by for example:
1. Use hands-free to decrease the radiation to the head.
2. Keep the mobile phone away from the body.
3. Do not use a cellphone in a car without an external antenna.
195
11.5
Exercise 11 - 2
More practice
(1.) 004t
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(2.) 004u
Particle-like nature of
EM radiation
ESADM
196
c
,
11.5
where E is the energy of the photon in joules (J), h is Plancks constant, c is the speed of
light, f is the frequency in hertz (Hz) and is the wavelength in metres (m).
The higher the frequency of EM radiation, the higher the energy.
SOLUTION
hf
6, 63 1034 J s 3 1018 Hz
2 1015 J
197
11.5
QUESTION
What is the energy of an ultraviolet photon with a wavelength of 200 nm?
SOLUTION
We can substitute this into the equation for the energy of a photon,
E = hf , allowing us to deduce:
E=h
c
h
=
=
6, 63 1034 J s
3108 ms1
200109 m
9, 939 1010 J
198
11.5
Exercise 11 - 3
More practice
(1.) 004v
(2.) 004w
video solutions
or help at www.everythingscience.co.za
(3.) 004x
ESADN
People have believed that animals can predict earthquakes and other natural disasters for
centuries. As early as 373 B.C., historians recorded a massive exodus of animals, including
rats, snakes and weasels, from the Greek city of Helice days before a quake struck causing
massive devastation.
This topic is much debated and different behaviours are sometimes seen for different kinds
of animals, for example:
Dogs and cats: are believed by pet owners to howl or bite their owners before natural
disasters, they cite factors like a much stronger sense of smell.
Sharks: researchers in Florida have reported that sharks are observed to move to
deeper water before hurricanes, possibly because of a sensitivity to changes in the air
pressure preceding the hurricane.
Rodents: rodents that live underground will often ee their holes and burrows before
a disaster. Scientists from the California Institute of Technology have noted that there
are many changes preceding earthquakes such as tilting of the Earth. Rodents are
often more sensitive to such small changes and will react to these changes.
199
11.5
Elephants: will allegedly trumpet and ee to higher ground before a tsunami arrived.
This is attributed to their being more sensitive to vibrations on the Earths surface.
Many researchers argue that animals detect certain natural signals, such as the early tremblings of an earthquake, long before humans. This means that the animals have opportunity
to react before we can. However it can be said that they exhibit no special understanding,
they just ee as would any person hearing a shout of re.
Another problem cited with these seemingly clairvoyant animals is that their psychic powers often are based on behaviours that people only recall after the event. Some animal
behaviours happen frequently, but are not remembered unless an earthquake, tsunami, or
mud slide follows. For example, if you see a dog cross a road, you just remember you saw
a dog cross the road. But if an earthquake shook your neighbourhood ve minutes later,
would you say the dog was eeing?
Activity:
What do people believe leads to this behaviour? i.e. do the animals have some
mystic ability or are they more sensitive to anything then we are (such as low
frequency radiation)
http://biology.about.com/od/animalbehavior/a/aa123104a.htm
http://news.nationalgeographic.com/news/2003/11/1111_031111_earthquakeanimals_2.h
Bats sing, mice giggle by Karen Shanor and Jagmeet Kanwal
http://www.sheldrake.org/homepage.html
http://nationalzoo.si.edu/SCBI/AnimalCare/News/earthquake.cfm
http://www.animalvoice.com/animalssixthsense.htm
Present your ndings to your class. Critically analyse all the information you collect
and decide what you believe.
200
11.5
Chapter 11 | Summary
See the summary presentation (
Physical Quantities
Quantity
Unit name
Unit symbol
Energy (E)
joule
Wavelength ()
metre
second
hertz
Hz (s1 )
m s1
Period (T )
Frequency (f )
Speed of light (c)
Chapter 11
201
11.5
More practice
video solutions
or help at www.everythingscience.co.za
(1.) 004y
(3.) 01v4
(4.) 01v5
(5.) 01v6
(6.) 0050
(7.) 0051
202
(2.) 004z
(8.) 0052
(9.) 0053
(10.) 0054
(11.) 0055
(12.) 0056
12
ESADO
ESADP
We have seen that different materials have different properties. But what would we nd
if we were to break down a material into the parts that make it up (i.e. its microscopic
structure)? And how is it that this microscopic structure is able to give matter all its different
properties?
See introductory video: (
The answer lies in the smallest building block of matter: the atom. It is the type of atoms, and the way in
which they are arranged in a material, that affects the
properties of that substance. This is similar to building
materials. We can use bricks, steel, cement, wood,
straw (thatch), mud and many other things to build
structures from. The choice of atoms affects the properties of matter in the same way as the choice of building
material affects the properties of the structure,
It is not often that substances are found in atomic form (just as you seldom nd a building
or structure made from one building material). Normally, atoms are bonded (joined) to
other atoms to form compounds or molecules. It is only in the noble gases (e.g. helium,
neon and argon) that atoms are found individually and are not bonded to other atoms. We
looked at some of the reasons for this in earlier chapters.
203
12.1
Compounds
ESADQ
DEFINITION: Compound
A compound is a group of two or more different atoms that are attracted to
each other by relatively strong forces or bonds. The atoms are combined in
denite proportions.
COMPOUNDS
Covalent molecular structures
water (H2 O)
oxygen (O2 )
sulphur (S8 )
buckyballs (C60 )
Network structures
covalent network
ionic network
metallic network
structures
structures
structures
copper (Cu)
diamond (C)
sodium chloride (NaCl)
graphite (C)
barium sulphate (BaSO4 )
iron (Fe)
silver iodide (AgI)
gold (Au)
silica (SiO2 )
204
12.1
sulphur
buckminsterfullerene
Figure 12.2: Examples of covalent molecular structures
Network structures
Compounds that exist as giant repeating lattice structures are called network structures. Examples include covalent molecules such as diamond, graphite and silica. Ionic substances
are also network structures, for example a sodium chloride crystal is a huge lattice of repeating units made of sodium and chloride ions. All substances formed as a result of ionic
bonding are network structures. Metals exist as large continuous lattice structures and are
also classied as network structures. For example copper, zinc and iron can be seen as a
giant crystals and are therefore considered to be network structures.
See video: VPazc
at www.everythingscience.co.za
covalent network
ionic network
metallic network
Figure 12.3: Examples of network structures
205
12.1
Representing molecules
ESADR
The structure of a molecule can be shown in many different ways. Sometimes it is easiest
to show what a molecule looks like by using different types of diagrams, but at other times,
we may decide to simply represent a molecule using its chemical formula or its written
name.
Note
The empirical and the
For example in gure 12.4 the molecular formula of 2-methyl propane is C4 H10 . This tells
molecular
formulae
us that there are 4 carbon atoms and 10 hydrogen atoms in this molecule, i.e the ratio of
carbon to hydrogen is 4 : 10. But we can simplify this ratio to: 2 : 5. This gives us the
empirical formula of the molecule.
example in carbon
dioxide the molecular
formula is CO2 . This
is also the empirical
206
12.1
The empirical formula is useful when we want to write the formula for network structures.
Since network structures may consist of millions of atoms, it is impossible to say exactly
how many atoms are in each unit. It makes sense then to represent these units using their
empirical formula. So, in the case of a metal such as copper, we would simply write Cu, or
if we were to represent a unit of sodium chloride, we would simply write NaCl. Chemical
formulae (i.e. the molecular or the empirical formula) therefore tell us something about the
types of atoms that are in a compound and the ratio in which these atoms occur in the
compound, but they dont give us any idea of what the compound actually looks like, in
other words its shape. To show the shape of compounds we have to use diagrams. The
simplest type of diagram that can be used to describe a compound is its structural formula.
The structural formula for 2-methyl propane is shown in gure 12.4.
CH3
CH
(a) C4 H10
(b) C2 H5
(c)
CH3 CH3
Figure 12.4: Diagram showing (a) the chemical, (b) the empirical and (c) the structural
formula of 2-methyl propane
In this model, the bonds between atoms are shown as sticks. These sticks are
coloured to show which atoms are bonding.
This is a 3-dimensional molecular model that uses balls to represent atoms and
sticks to represent the bonds between them. The centres of the atoms (the balls)
are connected by straight lines which represent the bonds between them.
Space-lling models
This is also a 3-dimensional molecular model. The atoms are represented by spheres.
Table 12.1 shows examples of the different types of models for all the types of compounds.
207
12.1
Covalent molecular
Covalent network
Ionic network
Metallic network
Name
of
compound
glucose
graphite
silver chloride
zinc
Formula
C6 H12 O6 or CH2 O
AgCl
Zn
Stick model
Ball-andstick model
Space-lling
model
Table 12.1: Different representations for compounds
Activity:
Representing compounds
A list of substances is given below. Make use of atomic model kits, play dough and
toothpicks, or coloured polystyrene balls and skewer sticks to represent each of the
substances in three dimensional structures.
glucose (C6 H12 O6 )
silica (SiO2 )
sodium chloride (NaCl)
sulphur (S8 )
diamond (C)
Note
graphite (C)
buckyballs( C60 )
sucrose (C12 H22 O11 )
CanvasMol (VPdve) is
a website that allows
you to view several
copper (Cu)
compounds. You do
not need to know
these
this
compounds,
is
simply
to
208
General experiment:
12.1
pounds
209
12.1
connecting wire
tape
table salt or sodium sul-
phate
image by Nevit Dilmen
Method: Set up the apparatus as shown above. Observe what happens.
Results: You should observe bubbles forming at the tips of the pencils. Oxygen
gas is formed at the positive side and hydrogen at the negative side.
210
Chapter 12 | Summary
See the summary presentation (
12.1
The smallest unit of matter is the atom. Atoms can combine to form compounds.
A compound is a group of two or more different atoms that are attracted to each other
by relatively strong forces or bonds. The atoms are combined in denite proportions.
In a compound, atoms are held together by chemical bonds. Covalent bonds, ionic
The molecular formula is a concise way of expressing information about the atoms
The empirical formula is a way of expressing the relative number of each type of atom
in a chemical compound. The empirical formula does not show the exact number of
atoms, but rather the simplest ratio of the atoms in the compound.
A ball-and-stick model is a 3-dimensional molecular model that uses balls to represent atoms and sticks to represent the bonds between them.
A space-lling model is also a 3-dimensional molecular model. The atoms are represented by spheres.
Chapter 12
211
12.1
d. empirical formula
e. ball-and-stick model
3. Ammonia, an ingredient in household cleaners, is made up of one part nitrogen (N) and three parts hydrogen (H). Answer the following questions:
a. is ammonia a covalent, ionic or metallic substance?
b. write down the molecular formula for ammonia
c. draw a ball-and-stick diagram
d. draw a space-lling diagram
4. In each of the following, say whether the chemical substance is made
up of covalent, molecular structures, covalent network structures, ionic
network structures or metallic structures:
a. ammonia gas (NH3 )
b. zinc metal (Zn)
c. graphite (C)
d. nitric acid (HNO3 )
e. potassium bromide (KBr)
5. Refer to the diagram below and then answer the questions that follow:
More practice
(1.) 01uq
212
(2.) 01ur
video solutions
(3.) 01us
(4.) 01ut
or help at www.everythingscience.co.za
(5.) 01uu
(6.) 01uv
13
Introduction
ESADS
Matter is all around us. The desks we sit at, the air we breathe and the water we drink are
all examples of matter. But matter doesnt always stay the same. It can change in many
different ways. In this chapter, we are going to take a closer look at physical and chemical
changes that occur in matter.
See introductory video: (
ESADT
A physical change is one where the particles of the substances that are involved in the
change are not broken up in any way. When water is heated for example, the temperature
and energy of the water molecules increases and the liquid water evaporates to form water
vapour. When this happens, some kind of change has taken place, but the molecular
structure of the water has not changed. This is an example of a physical change. All
changes in state are physical changes.
H2 O () H2 O (g)
Conduction (the transfer of energy through a material) is another example of a physical
change. As energy is transferred from one material to another, the energy of each material
is changed, but not its chemical makeup. Dissolving one substance in another is also a
physical change.
213
13.1
There are some important things to remember about physical changes in matter:
1. Arrangement of particles When a physical change occurs, the compounds may rearrange themselves, but the bonds in between the atoms will not break. For example
when liquid water boils, the molecules will move apart but the molecule will stay
intact. In other words water will not break up into hydrogen and oxygen atoms.
Figure 13.1 shows this phase change. Note that the water molecules themselves stay
the same, but their arrangement changed.
FACT
liquid
gas
Figure 13.1: The arrangement of water molecules in the liquid and gas phase
2. Conservation of mass
In a physical change, the total mass, the number of atoms and the number of molecules
will always stay the same. In other words you will always have the same number of
molecules or atoms at the end of the change as you had at the beginning.
3. Energy changes
Energy changes may take place when there is a physical change in matter, but these
energy changes are normally smaller than the energy changes that take place during
a chemical change.
4. Reversibility
Physical changes in matter are usually easier to reverse than chemical changes. Methods such as ltration and distillation can be used to reverse the change. Changing
the temperature is another way to reverse a physical change. For example, a mixture
of salt dissolved in water can be separated by ltration, ice can be changed to liquid
water and back again by changing the temperature.
Activity:
Physical change
Use plastic pellets or marbles to represent water in the solid state. What do you
need to do to the pellets to represent the change from solid to liquid?
Make a mixture of sand and water. Filter this mixture. What do you observe?
Make a mixture of iron lings and sulphur. Can you separate the mixture with a
magnet?
214
13.1
Aim: To demonstrate the synthesis of iron sulphide from iron and sulphur.
Apparatus: 5, 6 g iron lings and 3, 2 g powdered sulphur; porcelain dish; test
tube; Bunsen burner
Method:
1. Measure the quantity of iron and sulphur that you need and mix them in a
porcelain dish.
2. Take some of this mixture and place it in the test tube. The test tube should
be about one third full.
3. Heat the test tube containing the mixture over the Bunsen burner. Increase
the heat if no reaction takes place. Once the reaction begins, you will need
to remove the test tube from the ame. Record your observations.
4. Wait for the product to cool before breaking the test tube with a hammer.
Make sure that the test tube is rolled in paper before you do this, otherwise
the glass will shatter everywhere and you may be hurt.
5. What does the product look like? Does it look anything like the original reactants? Does it have any of the properties of the reactants (e.g. the magnetism
of iron)?
Warning:
When working with a Bunsen burner, work in a well ventilated space and
ensure that there are no ammable substances close by. Always tuck
loose clothing in and ensure that long hair is tied back.
Results:
After you removed the test tube from the ame, the mixture glowed a
bright red colour. The reaction is exothermic and produces heats. The product,
iron sulphide, is a dark colour and does not share any of the properties of the
original reactants. It is an entirely new product.
Conclusions: A synthesis reaction has taken place. The equation for the reaction
215
13.1
ESADU
When a chemical change takes place, new substances are formed in a chemical reaction.
These new products may have very different properties from the substances that were there
at the start of the reaction.
See video: VPbfo at www.everythingscience.co.za
We will consider two examples of chemical change: the decomposition (breaking down)
of hydrogen peroxide and the synthesis (forming) of water.
O
H
H
O
216
General experiment:
13.1
peroxide
Warning:
Hydrogen peroxide can cause chemical burns. Work carefully with it.
Method:
1. Put a small amount (about 5 ml) of hydrogen peroxide in a test tube.
2. Set up the apparatus as shown above.
3. Very carefully add a small amount (about 0, 5 g) of manganese dioxide to the
test tube containing hydrogen peroxide.
Results:
You should observe a gas bubbling up into the second test tube. This
217
13.1
Note
chemical reaction.)
The above experiment can be very vigorous and produce a lot of oxygen very rapidly. For
this reason you should use dilute hydrogen peroxide and only a small amount of manganese
istry.
dioxide.
The oxygen
The synthesis (forming) of water (H2 O) from hydrogen gas (H2 ) and oxygen gas (O2 ) is
another example of chemical change. A simplied diagram of this reaction is shown in
gure 13.3. The chemical bonds between O in O2 and between H in H2 are broken and
new bonds between H and O (to form H2 O) are formed. A chemical change has taken
place.
wards.
H
H H
FACT
H H
H
O
H
O
without
toothpaste
reaction.
Aim:
To observe the synthesis of water.
Apparatus:
Hydrogen gas; balloon; string; candle; long stick: ear plugs; safety glasses
218
13.1
Warning:
This reaction can be highly explosive, for this reason it is best done
outdoors. Always ensure that you wear ear protection or block your ears.
Always have more oxygen than hydrogen in the balloon.
Results:
When you bring the candle close to the balloon you should see a ame and hear a
loud bang.
Conclusions:
When a mixture of hydrogen and oxygen gas is set alight with a candle a chemical
change occurs. Water is made according to the following equation:
2H2 (g) + O2 (g) 2H2 O()
FACT
A mixture of hydrogen and oxygen gas is
1. Arrangement of particles
During a chemical change, the particles themselves are changed in some way. In
the example of hydrogen peroxide that was used earlier, the H2 O2 molecules were
split up into their component atoms. The number of particles will change because
each H2 O2 molecule breaks down into two water molecules (H2 O) and one oxygen
molecule (O2 ).
2. Energy changes
The energy changes that take place during a chemical reaction are much greater than
those that take place during a physical change in matter. During a chemical reaction,
energy is used up in order to break bonds and then energy is released when the new
Chemistry: Chemical change
219
13.1
product is formed.
3. Reversibility
Chemical changes are far more difcult to reverse than physical changes. When
hydrogen peroxide decomposes into water and oxygen, it is almost impossible to get
back to hydrogen peroxide.
4. Mass conservation
Mass is conserved during a chemical change, but the number of molecules may
change. In the example of the decomposition of hydrogen peroxide, for every two
molecules of hydrogen peroxide that decomposes, three molecules are formed (two
water and one oxygen).
Table 13.1 highlights these concepts for the decomposition of hydrogen peroxide.
2H2 O2 2H2 O + O2
Molecules
two molecules
three molecules
Energy changes
broken
formed
Mass is conserved
atoms
atoms
Exercise 13 - 1
For each of the following say whether a chemical or a physical change occurs.
1. Melting candle wax.
2. Mixing sodium chloride (NaCl) and silver nitrate (AgNO3 ) to form silver
chloride (AgCl).
3. Mixing hydrochloric acid (HCl) and magnesium ribbon (Mg) to form magnesium chloride (MgCl2 ).
4. Dissolving salt in water.
5. Tearing a piece of magnesium ribbon.
More practice
220
video solutions
or help at www.everythingscience.co.za
13.2
Conservation of atoms
and mass in reactions
ESADV
In a chemical reaction the total mass of all the substances taking part in the reaction remains
the same. Also, the number of atoms in a reaction remains the same. Mass cannot be
created or destroyed in a chemical reaction.
See video: VPbkj at www.everythingscience.co.za
Table 13.1 illustrates this law for the decomposition of hydrogen peroxide.
Activity:
We will use the reaction of hydrogen and oxygen to form water in this activity.
Materials: Coloured modelling clay rolled into balls or marbles and prestik to
represent atoms. Each colour will represent a different element.
Method:
1. Build your reactants. Use marbles and prestik or modelling clay to represent
the reactants and put these on one side of your table. Make at least ten (H2 )
units and at least ve (O2 ) units.
2. Place the H2 and O2 units on a table. The table represents the test tube
where the reaction is going to take place.
3. Now count the number of atoms (H and O) you have in your test tube. Fill
in the reactants column in the table below. Refer to table 13.1 to help you ll
in the mass row.
4. Let the reaction take place. Each person can now take the H and O unit and
use them to make water units. Break the H and O units apart and build H2 O
units with the parts. These are the products. Place the products on the table.
221
13.2
Products
Number of molecules
Mass
Number of atoms
Discussion: You should have noticed that the number of atoms in the reactants
is the same as the number of atoms in the product. The number of atoms is conserved during the reaction. However, you will also see that the number of molecules
in the reactants and products are not the same. The number of molecules is not conserved during the reaction.
222
13.2
Warning:
Always be careful when handling chemicals (particularly strong acids like
hydrochloric acid) as you can burn yourself badly.
Method:
Reaction 1
1. Solution 1: In one of the beakers dissolve 5 g of silver
nitrate in 100 ml of water.
2. Solution 2: In a second beaker, dissolve 4.5 g of
sodium iodide in 100 ml of water.
3. Determine the mass of each of the reactants.
AgNO3
NaI
NaOH
HCl
223
13.2
H2 O
Reaction 2
Reaction 3
Reactants
Products
Add the masses for the reactants for each reaction. Do the same for the products.
For each reaction compare the mass of the reactants to the mass of the products.
What do you notice? Is the mass conserved?
In the experiment above you should have found that the total mass at the start of the reaction
is the same as the mass at the end of the reaction. Mass does not appear or disappear in
chemical reactions. Mass is conserved, in other words, the total mass you start with is the
total mass you will end with.
Exercise 13 - 2
Complete the following chemical reactions to show that atoms and mass are
224
13.3
conserved. For each reaction give the total molecular mass of the reactants and
the products.
1. Hydrogen gas combines with nitrogen gas to form ammonia.
More practice
(1.) 0058
(2.) 0059
video solutions
or help at www.everythingscience.co.za
(3.) 005a
Law of constant
composition
ESADW
In any given chemical compound, the elements always combine in the same proportion
with each other. This is the law of constant composition.
The law of constant composition says that, in any particular chemical compound, all samples of that compound will be made up of the same elements in the same proportion or
ratio. For example, any water molecule is always made up of two hydrogen atoms and one
oxygen atom in a 2 : 1 ratio. If we look at the relative masses of oxygen and hydrogen in
Chemistry: Chemical change
225
13.3
a water molecule, we see that 94% of the mass of a water molecule is accounted for by
oxygen and the remaining 6% is the mass of hydrogen. This mass proportion will be the
same for any water molecule.
This does not mean that hydrogen and oxygen always combine in a 2 : 1 ratio to form
H2 O. Multiple proportions are possible. For example, hydrogen and oxygen may combine
in different proportions to form H2 O2 rather than H2 O. In H2 O2 , the H : O ratio is 1 : 1
and the mass ratio of hydrogen to oxygen is 1 : 16. This will be the same for any molecule
of hydrogen peroxide.
silver
nitrate
(AgNO3 )
0, 1 M sodium chloride
(NaCl)
0, 1 M sodium hydroxide
(NaOH)
9 large test tubes
3 propettes
Method:
226
13.3
will remain in the nal solution, together with the products formed.
ESADX
In a chemical reaction between gases, the relative volumes of the gases in the reaction are
present in a ratio of small whole numbers if all the gases are at the same temperature and
pressure. This relationship is also known as Gay-Lussacs Law.
For example, in the reaction between hydrogen and oxygen to produce water, two volumes
of H2 react with 1 volume of O2 to produce 2 volumes of H2 O.
2H2 (g) + O2 (g) 2H2 O()
In the reaction to produce ammonia, one volume of nitrogen gas reacts with three volumes
of hydrogen gas to produce two volumes of ammonia gas.
N2 (g) + 3H2 (g) 2NH3 (g)
Chapter 13 | Summary
See the summary presentation (
Matter does not stay the same. It may undergo physical or chemical changes.
A physical change is a change that can be seen or felt, but that does not involve the
break up of the particles in the reaction. During a physical change, the form of matter
Physical changes involve small changes in energy and are easily reversible.
A chemical change occurs when one or more substances change into other mate-
rials. A chemical reaction involves the formation of new substances with different
227
13.3
part in a chemical reaction is conserved and the number of atoms of each element in
the reaction does not change when a new product is formed.
The law of constant composition states that in any particular compound, all samples
of that compound will be made up of the same elements in the same proportion or
ratio.
Gay-Lussacs Law states that in a chemical reaction between gases, the relative volumes of the gases in the reaction are present in a ratio of small whole numbers if all
the gases are at the same temperature and pressure.
Chapter 13
Physical or chemical
13.3
5. For the following equation: CaCO3 (s) CaO + CO2 show that the law
of conservation of mass applies. Draw sub-microscopic diagrams to represent this reaction.
More practice
(1.) 005b
(2.) 005c
video solutions
(3.) 005d
(4.) 005e
or help at www.everythingscience.co.za
(5.) 005f
229
Representing chemical
change
14
Introduction
ESADY
As we have already mentioned, a number of changes can occur when elements are combined with one another. These changes may either be physical or chemical. In this chapter
we will look at chemical changes. One way of representing chemical changes is through
balanced chemical equations. A chemical equation describes a chemical reaction by using symbols for the elements involved. For example, if we look at the reaction between
iron (Fe) and sulphur (S) to form iron sulphide (FeS), we could represent these changes in a
sentence, in a word equation or using chemical symbols:
Sentence: Iron reacts with sulphur to form iron sulphide. Word equation: Iron +
sulphur iron sulphide. Chemical symbols: Fe + S FeS
Another example would be:
Sentence: Ammonia reacts with oxygen to form nitrogen monoxide and water. Word
equation: Ammonia + oxygen nitrogen monoxide + water. Chemical symbols:
4NH3 + 5O2 4NO + 6H2 O
See introductory video: (
Compounds on the left of the arrow are called the reactants and these are needed for the
reaction to take place. The compounds on the right are called the products and these are
what is formed from the reaction.
In order to be able to write a balanced chemical equation, there are a number of important
things that need to be done:
1. Know the chemical symbols for the elements involved in the reaction
2. Be able to write the chemical formulae for different reactants and products
3. Balance chemical equations by understanding the laws that govern chemical change
4. Know the state symbols for the equation
We will look at each of these steps separately in the next sections.
230
Chemical symbols
14.1
ESADZ
It is very important to know the chemical symbols for common elements in the periodic table, so that you are able to write chemical equations and to recognise different compounds.
Activity:
Write down the chemical symbols and names of all the elements that you
know.
Compare your list with another learner and add any symbols and names that
class and then exchange tests with them so that you each have the chance to
answer a test.
ESAEA
A chemical formula is a concise way of giving information about the atoms that make up a
particular chemical compound. A chemical formula shows each element by its symbol and
also shows how many atoms of each element are found in that compound. The number of
atoms (if greater than one) is shown as a subscript.
The following exercise serves as revision. If you do not recall how to write chemical
formulae refer back to chapter 2.
231
14.2
Exercise 14 - 1
1. Write down the chemical formula for each of the following compounds:
a. iron (III) chloride
b. zinc nitrate
c. aluminium sulphate
d. calcium hydroxide
e. magnesium carbonate
h. potassium oxide
i. copper (II) bromide
j. potassium dichromate
d. BaF2
e. Cr(HSO4 )3
c. (NH4 )2 SO4
f. CH4
More practice
(1.) 02u2
video solutions
or help at www.everythingscience.co.za
(2.) 02u3
Balancing chemical
equations
The law of conservation of mass
ESAEB
ESAEC
In order to balance a chemical equation, it is important to understand the law of conservation of mass.
232
14.2
Tip
Iron is a metal. When
we represent it in
For any chemical equation (in a closed system) the mass of the reactants must be equal
to the mass of the products. In order to make sure that this is the case, the number of
atoms of each element in the reactants must be equal to the number of atoms of those
same elements in the products. An example is shown below:
a balanced chemical
equation,
we write
Fe + S FeS
Fe is 55, 8
S is 32, 1
Mass of products is
87, 9
To calculate the mass of the molecules we use the relative atomic masses for iron and
sulphur, as seen in table 14.2. You will notice that the mass of the reactants equals the
mass of the product. A chemical equation that is balanced will always reect the law of
conservation of mass and the law of conservation of atoms.
Activity:
1. You will need: coloured balls (or marbles), prestik, a sheet of paper and coloured
pens.
We will try to balance the following equation:
Al + O2 Al2 O3
Take one ball of one colour. This represents a molecule of Al. Take two balls of
another colour and stick them together. This represents a molecule of O2 . Place
these molecules on your left. Now take two balls of the rst colour and three balls
of the second colour to form Al2 O3 . Place this compound on your right. On a piece
of paper draw coloured circles to represent the balls. Draw a line down the centre
of the paper to represent the molecules on the left and on the right.
Count the number of balls on the left and the number on the right. Do you
233
14.2
chemical equation. Make sure that your atoms are balanced. Use the same colour
jelly tots for the same atoms.
C + H2 O CO2 + CO + H2
Add compounds until the atoms are balanced. Write the equation down and
use a coefcient to indicate how many compounds you used. For example if you
had to use three water molecules then write 3H2 O 3 Use ball and stick drawings to
balance the atoms in the following reaction:
NH3 + O2 NO + H2 O
Use your drawings to write a balanced chemical equation for the reaction.
4 Lead (Pb), lead (IV) oxide (PbO2 ) and sulphuric acid (H2 SO4 ) are used in car
batteries. The following reaction takes place: Pb + PbO2 + H2 SO4 PbSO4 + H2 O
Cut out circles from four different colours of paper to represent each of the
atoms. Build a few of the compounds (Pb PbO2 H2 SO4 ). These are the reactants.
Do not build the products. Rearrange the atoms so that the products are formed.
Add more reactants if needed to balance the atoms (e.g. you will need two H2 SO4
molecules). Use what you have learnt to write a balanced equation for the reaction.
ESAED
When balancing a chemical equation, there are a number of steps that need to be followed.
234
14.2
Step 1: Identify the reactants and the products in the reaction and write their chemical formulae.
Step 2: Write the equation by putting the reactants on the left of the arrow and the products
on the right.
Step 3: Count the number of atoms of each element in the reactants and the number of atoms
of each element in the products.
Step 4: If the equation is not balanced, change the coefcients of the molecules until the
number of atoms of each element on either side of the equation balance.
Step 5: Check that the atoms are in fact balanced.
Step 6: (we will look at this a little later): Add any extra details to the equation e.g. phase
symbols.
SOLUTION
Step 1 : Identify the reactants and products
This has been done in the question.
Step 2 : Write the equation for the reaction
This has been done in the question.
Step 3 : Count the number of atoms of each element in the reactants and
products
Reactants: Mg = 1 atom; H = 1 atom; Cl = 1 atom
Products: Mg = 1 atom; H = 2 atoms; Cl = 2 atoms
Step 4 : Balance the equation
The equation is not balanced since there are two chlorine atoms in the
product and only one in the reactants. If we add a coefcient of two to
the HCl to increase the number of H and Cl atoms in the reactants, the
equation will look like this:
235
14.2
SOLUTION
Step 1 : Count the number of atoms of each element in the reactants and
products
Reactants: C = 1; H = 4; O = 2
Products: C = 1; H = 2; O = 3
Step 2 : Balance the equation
If we add a coefcient of 2 to H2 O, then the number of hydrogen
atoms in the products will be 4, which is the same as for the reactants.
The equation will be:
CH4 + O2 CO2 + 2H2 O
Step 3 : Check that the atoms balance
Reactants: C = 1; H = 4; O = 2
236
14.2
Products: C = 1; H = 4; O = 4
You will see that, although the number of hydrogen atoms now
balances, there are more oxygen atoms in the products. You now need
to repeat the previous step. If we put a coefcient of 2 in front of O2 ,
then we will increase the number of oxygen atoms in the reactants by
2. The new equation is:
CH4 + 2O2 CO2 + 2H2 O
When we check the number of atoms again, we nd that the number of atoms of each element in the reactants is the same as the number
in the products. The equation is now balanced.
SOLUTION
Step 1 : Identify the reactants and products in the reaction.
Reactants: sugar (C6 H12 O6 ) and oxygen (O2 )
Products: carbon dioxide (CO2 ) and water (H2 O)
Step 2 : Write the equation
C6 H12 O6 + O2 CO2 + H2 O
Step 3 : Count the number of atoms of each element in the reactants and in
the products
Reactants: C = 6; H = 12; O = 8
Products: C = 1; H = 2; O = 3
Step 4 : Balance the equation
It is easier to start with carbon as it only appears once on each side. If
we add a 6 in front of CO2 , the equation looks like this:
C6 H12 O6 + O2 6CO2 + H2 O
Reactants: C = 6; H = 12; O = 8
237
14.2
Products: C = 6; H = 12; O = 18
Step 6 : Now we just need to balance the oxygen atoms.
C6 H12 O6 + 6O2 6CO2 + 6H2 O
Reactants: C = 6; H = 12; O = 18
Products: C = 6; H = 12; O = 18
See simulation: (
Exercise 14 - 2
2. Ca + H2 O Ca(OH)2 + H2
3. CuCO3 + H2 SO4 CuSO4 + H2 O + CO2
4. CaCl2 + Na2 CO3 CaCO3 + NaCl
10. The synthesis of ammonia (NH3 ), made famous by the German chemist
Fritz Haber in the early 20th century, is one of the most important reactions in the chemical industry. Balance the following equation used to
produce ammonia: N2 (g) + H2 (g) NH3 (g)
238
More practice
video solutions
(1.) 005g
(2.) 005h
(3.) 005i
(4.) 005j
(7.) 005n
(8.) 005p
(9.) 005q
14.2
or help at www.everythingscience.co.za
(5.) 005k
(6.) 005m
(10.) 005r
ESAEE
The state (phase) of compounds can be expressed in the chemical equation. This is done
by placing the correct label on the right hand side of the formula. The following four labels
can be used:
SOLUTION
Step 1 : Identify the reactants and products
The reactants are zinc (Zn) and hydrochloric acid (HCl). The products
are zinc chloride (ZnCl2 ) and hydrogen (H2 ).
Chemistry: Chemical change
239
14.2
Exercise 14 - 3
Write balanced equations for each of the following reactions, include state symbols:
1. Lead (II) nitrate solution reacts with a potassium iodide solution to form a
precipitate (solid) of lead iodide while potassium nitrate remains in solution.
2. When heated, aluminium metal reacts with solid copper oxide to produce
copper metal and aluminium oxide (Al2 O3 ).
3. When calcium chloride solution is mixed with silver nitrate solution, a
white precipitate (solid) of silver chloride appears. Calcium nitrate (Ca(NO3 )2 )
is also produced in the solution.
4. Solid ammonium carbonate decomposes to form three gaseous products.
More practice
240
video solutions
or help at www.everythingscience.co.za
(1.) 00b7
(2.) 00b8
(3.) 00b9
14.2
(4.) 00ba
Aim: To investigate the relationship between the amount of product and the amount
of reactant.
Apparatus:
ask
measuring cylinder
water bowl
delivery tube
hydrogen
carbonate
(NaHCO3 ) powder
dilute sulphuric acid (H2 SO4 )
Method:
1 Weigh 20 g of NaHCO3 and place it into a ask.
2 Set up the above apparatus.
3 Measure out 5 ml of H2 SO4 and carefully pour this into the funnel (make sure
that the stopcock is closed).
4 Slowly add the H2 SO4 to the NaHCO3 by opening the stopcock.
5 Observe what happens.
6 Record the volume of gas collected in the measuring cylinder.
7 Repeat the above steps but this time use 10 ml of H2 SO4 .
8 Write a balanced equation for this reaction. (Hint: carbon dioxide gas is
formed, as well as water and sodium sulphate.)
Results and discussion: You should observe that more gas is formed when using
more H2 SO4 .
241
14.2
Chapter 14 | Summary
See the summary presentation (
In a chemical equation, reactants are written on the left hand side of the equation
and the products on the right. The arrow is used to show the direction of the reaction.
When representing chemical change, it is important to be able to write the chemical
formula of a compound.
The law of conservation of mass states that the mass of a closed system of substances
will remain constant, regardless of the processes acting inside the system. Matter can
change form, but cannot be created or destroyed.
In any chemical reaction, the law of conservation of mass applies. This means that
the total atomic mass of the reactants must be the same as the total atomic mass of
the products. This also means that the total number of atoms of the reactants must be
the same as the total number of atoms of the product.
If the number of atoms of each element in the reactants is the same as the number of
atoms of each element in the product, then the equation is balanced.
If the number of atoms of each element in the reactants is not the same as the number
of atoms of each element in the product, then the equation is not balanced.
and products until the number of atoms of each element is the same on both sides of
the equation.
The state of the compounds in a chemical reaction can be expressed in the chem-
ical equation by using one of four symbols. The symbols are g (gas), (liquid), s
(solid) and aq (aqueous solutions). These symbols are written in brackets after the
compound.
Chapter 14
1. Propane is a fuel that is commonly used as a heat source for engines and
homes. Balance the following equation for the combustion of propane:
C3 H8 () + O2 (g) CO2 (g) + H2 O()
242
14.2
H2 S + SO2 S + H2 O
video solutions
or help at www.everythingscience.co.za
243
14.2
(1.) 005s
(2.) 005t
(3.) 005u
(4.) 005v
(5.) 005w
(6.) 005x
(7.) 005y
244
Magnetism
15
Introduction
ESAEF
Magnetism is an interaction that allows certain kinds of objects, which are called magnetic objects, to exert forces on each other
without physically touching. A magnetic object is surrounded by a magnetic eld that
gets weaker as one moves further away from
the object. A second object can feel a mag-
FACT
magnes,
probably
from
important source of
Magnetic elds
ESAEG
A magnetic eld is a region in space where a magnet or object made of magnetic material
will experience a non-contact, magnetic force.
A moving charged particle has a magnetic elds associated with it. One example of a
charged particle that occurs in most matter is the electron. Electrons are in constant motion
inside material, orbiting the nucleus in the atom which may also be moving, rotating or
vibrating.
Physics: Electricity and Magnetism
Magnesia
in
245
lodestone.
15.1
So electrons inside any object are moving and have magnetic elds associated with them.
In most materials these elds point in various directions, so the net magnetic eld is zero.
For example, in the plastic ball below, the directions of the magnetic elds of the electrons (shown by the arrows) are pointing in different directions and cancel each other out.
Therefore the plastic ball is not magnetic and has no magnetic eld.
In some materials (e.g. iron), called ferromagnetic materials, there are regions called domains, where the electrons magnetic elds line up with each other. All the atoms in each
domain are grouped together so that the magnetic elds from their electrons point the same
way. The picture shows a piece of an iron needle zoomed in to show the domains with the
electric elds lined up inside them.
iron needle
In permanent magnets, many domains are lined up, resulting in a net magnetic eld. Objects made from ferromagnetic materials can be magnetised, for example by rubbing a
magnet along the object in one direction. This causes the magnetic elds of most, or all,
of the domains to line up in one direction. As a result the object as a whole will have a net
magnetic eld. It is magnetic. Once a ferromagnetic object has been magnetised, it can
stay magnetic without another magnet being nearby (i.e. without being in another magnetic
eld). In the picture below, the needle has been magnetised because the magnetic elds
in all the domains are pointing in the same direction.
246
15.1
A permanent magnet
iron needle
zoomed-in part
of needle
Investigation:
tion
1. Find 2 paper clips. Put the paper clips close together and observe
what happens.
a. What happens to the paper clips?
b. Are the paper clips magnetic?
2. Now take a permanent bar magnet and rub it once along 1 of the
paper clips. Remove the magnet and put the paper clip which was
touched by the magnet close to the other paper clip and observe
what happens. Does the untouched paper clip experience a force
on it? If so, is the force attractive or repulsive?
3. Rub the same paper clip a few more times with the bar magnet, in
the same direction as before. Put the paper clip close to the other
one and observe what happens.
a. Is there any difference to what happened in step 2?
b. If there is a difference, what is the reason for it?
c. Is the paper clip which was rubbed repeatedly by the magnet
now magnetised?
d. What is the difference between the two paper clips at the level
of their atoms and electrons?
4. Now, nd a metal knitting needle, or a metal ruler, or other metal
object. Rub the bar magnet along the knitting needle a few times
in the same direction. Now put the knitting needle close to the
paper clips and observe what happens.
a. Does the knitting needle attract the paper clips?
247
15.1
A ferromagnetic material is a substance that shows spontaneous magnetisation. Spontaneous means self-generated or to happen without external cause. This means that a ferromagnetic material has a magnetic eld without any external factors being required.
Permanent magnets
ESAEH
ESAEI
Because the domains in a permanent magnet all line up in a particular direction, the magnet
has a pair of opposite poles, called north (usually shortened to N) and south (usually
shortened to S). Even if the magnet is cut into tiny pieces, each piece will still have both
a N and a S pole. These magnetic poles always occur in pairs. In nature, we never nd a
north magnetic pole or south magnetic pole on its own.
See video: VPm at www.everythingscience.co.za
S
... after breaking in half ...
S N
In nature, positive and negative electric charges can be found on their own, but you never
nd just a north magnetic pole or south magnetic pole on its own. On the very small scale,
zooming in to the size of atoms, magnetic elds are caused by moving charges (i.e. the
negatively charged electrons).
248
15.1
ESAEJ
Like (identical) poles of magnets repel one another whilst unlike (opposite) poles attract.
This means that two N poles or two S poles will push away from each other while a N pole
and a S pole will be drawn towards each other.
Do you think the following magnets will repel or be attracted to each other?
We are given two magnets with the N pole of one approaching the N pole of the other.
Since both poles are the same, the magnets will repel each other.
We are given two magnets with the N pole of one approaching the S pole of the other.
Since both poles are the different, the magnets will be attracted to each other.
ESAEK
Magnetic elds can be represented using magnetic eld lines starting at the North pole
and ending at the South pole. Although the magnetic eld of a permanent magnet is
everywhere surrounding the magnet (in all three dimensions), we draw only some of the
eld lines to represent the eld (usually only a two-dimensional cross-section is shown in
drawings).
See video: VPfmc at www.everythingscience.co.za
249
15.1
Tip
1. Field
lines
3-dimensional representation
never cross.
2. Arrows drawn
on
the
lines
2-dimensional representation
eld
indicate
the direction of
the eld.
3. A
magnetic
eld
points
the
pole
south
of
In areas where the magnetic eld is strong, the eld lines are closer together. Where the
eld is weaker, the eld lines are drawn further apart. The number of eld lines drawn
crossing a given two-dimensional surface is referred to as the magnetic ux. The magnetic
ux is used as a measure of the strength of the magnetic eld through that surface.
magnet.
N
N
250
15.1
As the investigation shows, one can map the magnetic eld of a magnet by placing it underneath a piece of paper and sprinkling iron lings on top. The iron lings line themselves
up parallel to the magnetic eld.
Take two bar magnets and place them a short distance apart such
that they are repelling each other. Place a sheet of white paper over
the bar magnets and sprinkle some iron lings onto the paper. Give the
paper a shake to evenly distribute the iron lings. In your workbook,
draw both the bar magnets and the pattern formed by the iron lings.
Repeat the procedure for two bar magnets attracting each other and
draw what the pattern looks like for this situation. Make a note of the
shape of the lines formed by the iron lings, as well as their size and
their direction for both arrangements of the bar magnet. What does the
pattern look like when you place both bar magnets side by side?
251
15.1
magnet
magnet
S
magnet
N
Arrangement 3
magnet
Arrangement 2
magnet
magnet
magnet
magnet
Arrangement 1
Arrangement 4
As already stated, opposite poles of a magnet attract each other and bringing them together
causes their magnetic eld lines to converge (come together). Like poles of a magnet repel
each other and bringing them together causes their magnetic eld lines to diverge (bend
out from each other).
252
15.2
Ferromagnetism
ESAEL
Ferromagnetism is a phenomenon shown by materials like iron, nickel or cobalt. These materials can form permanent magnets. They always magnetise so as to be attracted to a magnet, no matter which magnetic pole is brought toward the unmagnetised iron/nickel/cobalt.
ESAEM
The ability of a ferromagnetic material to retain its magnetisation after an external eld is
removed is called its retentivity.
Paramagnetic materials are materials like aluminium or platinum, which become magnetised in an external magnetic eld in a similar way to ferromagnetic materials. However,
they lose their magnetism when the external magnetic eld is removed.
Diamagnetism is shown by materials like copper or bismuth, which become magnetised in
a magnetic eld with a polarity opposite to the external magnetic eld. Unlike iron, they
are slightly repelled by a magnet.
The compass
ESAEN
NW
magnetised needle
NE
E
pivot
SW
SE
253
15.2
FACT
Lodestone, a magnetised form of ironoxide, was found to
orientate itself in a
north-south direction
if left free to rotate
by suspension on a
string or on a oat
in water.
Lodestone
A compass
ESAEO
In the picture below, you can see a representation of the Earths magnetic eld which is
very similar to the magnetic eld of a giant bar magnet like the one on the right of the
picture. The Earth has two magnetic poles, a north and a south pole just like a bar magnet.
In addition to the magnetic poles the Earth also has two geographic poles. The two geographic poles are the points on the Earths surface where the line of the Earths axis of
rotation meets the surface. To visualise this you could take any round fruit (lemon, orange
etc.) and stick a pencil through the middle so it comes out the other side. Turn the pencil,
the pencil is the axis of rotation and the geographic poles are where the pencil enters and
254
15.2
exits the fruit. We call the geographic north pole true north.
The Earths magnetic eld has been measured very precisely and scientists have found that
the magnetic poles do not correspond exactly to the geographic poles.
So the Earth has two north poles and two south poles: geographic poles and magnetic
poles.
Magnetic North pole
11.5o
Geographic North pole
FACT
The direction of the
Earths magnetic eld
ips direction about
once every 200 000
years!
The Earths magnetic eld is thought to be caused by owing liquid metals in the outer core
of the planet which causes electric currents and a magnetic eld. From the picture you can
see that the direction of magnetic north and true north are not identical. The geographic
north pole is about 11,5o away from the direction of the magnetic north pole (which is
where a compass will point). However, the magnetic poles shift slightly all the time.
know that magnetic eld lines always point from north to south, then the compass tells us
that what we call the magnetic north pole is actually the south pole of the bar magnet!
ESAEP
Another interesting thing to note is that if we think of the Earth as a big bar magnet, and we
255
15.2
these particles come close to the Earth, they are deected by the Earths magnetic eld and
cannot shower down to the surface where they can harm living organisms. Astronauts in
space are at risk of being irradiated by the solar wind because they are outside the zones
where the charged particles are trapped.
The region above Earths atmosphere in
which charged particles are affected the
Earths magnetic eld is called the magnetoVisualisation of the magnetosphere
Photograph by seishin17 on
Flickr.com
Photograph by Trodel on Flickr.com
256
15.2
As this only happens close to the North and South poles, we cannot see the aurorae from
South Africa. However, people living in the high Northern latitudes in Canada, Sweden,
and Finland, for example, often see the Northern lights.
See simulation: ( Simulation: VPfkg at www.everythingscience.co.za)
Chapter 15 | Summary
See the summary presentation (
A compass can be used to nd the magnetic north pole and help us nd our direction.
The Earths magnetic eld protects us from being bombarded by high energy charged
particles which are emitted by the Sun.
Chapter 15
257
15.2
More practice
video solutions
or help at www.everythingscience.co.za
(1.) 005z
(3.) 0061
(4.) 0062
(5.) 0063
(6.) 0064
(7.) 0065
258
(2.) 0060
(8.) 0066
(9.) 0067
(10.) 0068
(11.) 0069
(12.) 006a
Electrostatics
16
ESAEQ
Electrostatics is the study of electric charge which is at rest or static (not moving). In this
chapter we will look at some of the basic principles of electrostatics as well as the principle
of conservation of charge.
See introductory video: (
ESAER
All objects surrounding us (including people!) contain large amounts of electric charge.
There are two types of electric charge: positive charge and negative charge. If the same
amounts of negative and positive charge are found in an object, there is no net charge and
the object is electrically neutral. If there is more of one type of charge than the other on
the object then the object is said to be electrically charged. The picture below shows what
the distribution of charges might look like for a neutral, positively charged and negatively
charged object.
There are:
6 positive charges and
6 negative charges
6 + (-6) = 0
++
+ - +
-+
+
There is zero net charge:
The object is neutral
-+ + +- + +
+ +
- +
- -+
-+ +
- +
+ +-
Positive charge is carried by the protons in material and negative charge by electrons. The
overall charge of an object is usually due to changes in the number of electrons. To make
an object:
259
16.2
Positively charged: electrons are removed making the object electron decient.
Negatively charged: electrons are added giving the object an excess of electrons.
So in practise what happens is that the number of positive charges (protons) remains the
same and the number of electrons changes:
There are:
6 positive charges and
6 negative charges
6 + (-6) = 0
++
+ - +
-+
+
There is zero net charge:
The object is neutral
+ +
- - +
+
- -+
-+ +
- +
+ +-
Tribo-electric charging
ESAES
Objects may become charged in many ways, including by contact with or being rubbed by
other objects. This means that they can gain or lose negative charge. For example, charging
Tip
happens when you rub your feet against the carpet. When you then touch something
We say
metallic or another person, you feel a shock as the excess charge that you have collected is
discharged.
When you rub your feet against the carpet, negative charge is transferred to you from the
served.
carpet. The carpet will then become positively charged by the same amount.
See video: VPfns at www.everythingscience.co.za
Another example is to take two neutral objects such as a plastic ruler and a cotton cloth
(handkerchief). To begin, the two objects are neutral (i.e. have the same amounts of positive
and negative charge).
Note: We represent the positive charge with a + and the negative charge with a -. This is
just to illustrate the balance and changes that occur, not the actual location of the positive
and negative charges. The charges are spread throughout the material and the real change
happens by increasing or decreasing electrons on the surface of the materials.
260
16.2
BEFORE rubbing:
-+
-+
+
+
+-
- + - + - +- + - + - + - + - + - +
The ruler has 9 positive charges and
9 negative charges
Now, if the cotton cloth is used to rub the ruler, negative charge is transferred from the
cloth to the ruler. The ruler is now negatively charged (i.e. has an excess of electrons) and
the cloth is positively charged (i.e. is electron decient). If you count up all the positive
and negative charges at the beginning and the end, there are still the same amount, i.e.
total charge has been conserved !
AFTER rubbing:
+
+
+
+
+ -
SOLUTION
Step 1 : Analyse the information provided
There are two materials provided and they will be rubbed together.
This means we are dealing with the interaction between the materials.
The question is related to the charge on the materials which we can
assume were neutral to begin with. This means that we are dealing
electrostatics and the interaction of materials leading to the materials
Physics: Electricity and Magnetism
261
16.2
Note that in this example the numbers are made up to be easy to calculate. In the real
world only a tiny fraction of the charges would move from one object to the other, but the
total charge would still be conserved.
See simulation: (
The process of materials becoming charged when they come into contact with other materials is known as tribo-electric charging. Materials can be arranged in a tribo-electric series
according to the likelihood of them gaining or losing electrons.
If a material has equal numbers of positive and negative charges we describe it as being
neutral (not favouring positive or negative overall charge).
If a neutral material loses electrons it becomes electron decient and has an overall positive
charge. If a neutral material gains electrons it has excess electrons and has an overall
negative charge. For this reason we describe the ordering of materials in the tribo-electric
series as more positive or more negative depending on whether they are more likely to lose
or gain electrons.
Amber
photos/amber_byJohnAlanElson_wikimedia.jpg
262
Aluminium
16.2
Material
Glass
photos/Aluminium_wikimedia.jpg
Human hair
Tribo-electric series
Very positive
Nylon
Photograph on wikimedia
Wool
Fur
Lead
Silk
Aluminium
Paper
Cotton
Steel
Wood
Amber
Hard rubber
Nickel, Copper
Gold, Platinum
Polyester
Polyurethane
Silicon
Polypropylene
Silicon
photos/Silicon_wikimedia.jpg
Teon
Very negative
ESAET
The force exerted by non-moving (static) charges on each other is called the electrostatic
force. The electrostatic force between:
Physics: Electricity and Magnetism
263
16.2
attractive force
repulsive force
repulsive force
The closer together the charges are, the stronger the electrostatic force between them.
FACT
+ F
F +
it to pick up bits of
straw.
QUESTION
Two charged metal spheres hang from strings and are free to move as shown in the
picture below. The right hand sphere is positively charged. The charge on the left
hand sphere is unknown.
SOLUTION
264
16.3
Conservation of charge
ESAEU
In all of the examples weve looked at charge was not created or destroyed but it moved
from one material to another.
265
16.3
ESAEV
Some materials allow electrons to move relatively freely through them (e.g. most metals,
the human body). These materials are called conductors.
See video: VPfnt at www.everythingscience.co.za
Tip
The
effect
of
the
is
the
Other materials do not allow the charge carriers, the electrons, to move through them (e.g.
plastic, glass). The electrons are bound to the atoms in the material. These materials are
called non-conductors or insulators.
identical
If an excess of charge is placed on an insulator, it will stay where it is put and there will
conductors
for
the
sharing of charge.
is placed on a conductor, the like charges will repel each other and spread out over the
outside surface of the object. When two conductors are made to touch, the total charge on
them is shared between the two. If the two conductors are identical, then each conductor
will be left with half of the total net charge.
Arrangement of charge
ESAEW
FACT
This
collection
of
The electrostatic force determines the arrangement of charge on the surface of conductors.
This is possible because charges can move inside a conductive material. When we place
a charge on a spherical conductor the repulsive forces between the individual like charges
cause them to spread uniformly over the surface of the sphere. However, for conductors
with irregular shapes, there is a concentration of charge near the point or points of the
object.
building
has
Notice in Figure 16.1 that we show a concentration of charge with more or + signs,
- - - - - -
---
---
insulator
since
266
16.4
When two identical conducting spheres on insulating stands are allowed to touch they
share the charge evenly between them. If the initial charge on the rst sphere is Q1 and
the initial charge on the second sphere is Q2 , then the nal charge on the two spheres after
they have been brought into contact is:
Q=
Q1 + Q2
2
Quantisation of charge
ESAEX
The basic unit of charge, called the elementary charge, e, is the amount of charge carried
by one electron.
Unit of charge
ESAEY
FACT
19
and
Harvey
Fletcher
Q = nqe
the
measured
charge
on
an
Charge is measured in units called coulombs (C). A coulomb of charge is a very large
charge. In electrostatics we therefore often work with charge in micro-coulombs (1 C =
1 106 C) and nanocoulombs (1 nC = 1 109 C).
QUESTION
force
267
the
to
determine
charge
electron.
on
an
16.4
SOLUTION
=
=
1, 92 1017
1, 6 1019
120 electrons
SOLUTION
268
16.4
4 nC
on each sphere.
into contact and then separated, what charge will each have?
SOLUTION
269
16.4
SOLUTION
Step 1 : Analyse the question
We need to determine what will happen to the charge when the spheres
touch. They are metal spheres so we know they will be conductors.
This means that the charge is able to move so when they touch it is
possible for the charge on each sphere to change. We know that charge
will redistribute evenly across the two spheres because of the forces
between the charges. We need to know the charge on each sphere, we
have been given one.
Step 2 : Identify the principles involved
This problem is similar to the earlier worked example. This time we
have to determine the total charge given a certain number of protons.
We know that charge is quantised and that protons carry the base unit
of charge and are positive so it is +1, 6 1019 C.
Step 3 : Apply the principles
The total charge will therefore be:
Q2
=
=
60 1, 6 1019 C
9, 6 1018 C
As the spheres are identical in material, size and shape the charge will
redistribute across the two spheres so that it is shared evenly. Each
270
16.4
=
=
=
Q1 + Q2
2
9, 6 1018 + 9, 6 1018
2
0C
SOLUTION
271
16.4
Q=
7, 2 1018
1, 6 1019
= 45 electrons
272
+ +
++ +
+
++
- - - - - - ++
--
16.4
charged rod
metal plate
++
+
+ ++
++ ++
glass container
273
16.4
+
+- - - +-+- -+
-
+
+
++
- -- - - -+-
metal plate
Polarisation
ESAEZ
Unlike conductors, the electrons in insulators (non-conductors) are bound to the atoms
of the insulator and cannot move around freely through the material. However, a charged
object can still exert a force on a neutral insulator due to a phenomenon called polarisation.
If a positively charged rod is brought close to a neutral insulator such as polystyrene, it can
attract the bound electrons to move round to the side of the atoms which is closest to the
rod and cause the positive nuclei to move slightly to the opposite side of the atoms. This
process is called polarisation. Although it is a very small (microscopic) effect, if there are
many atoms and the polarised object is light (e.g. a small polystyrene ball), it can add up
to enough force to cause the object to be attracted onto the charged rod. Remember, that
the polystyrene is only polarised, not charged. The polystyrene ball is still neutral since
no charge was added or removed from it. The picture shows a not-to-scale view of the
polarised atoms in the polystyrene ball:
274
16.4
+
positively
charged rod
+
+
+- ++- +- + ++- +- - ++
+- +- + - ++- +- ++- +-
+
+
+
+ + ++
polarised
polystyrene ball
Some materials are made up of molecules which are already polarised. These are molecules
which have a more positive and a more negative side but are still neutral overall. Just as
a polarised polystyrene ball can be attracted to a charged rod, these materials are also
affected if brought close to a charged object.
You can easily test that like charges repel and unlike charges attract
each other by doing a very simple experiment.
Take a glass ball and rub it with a piece of silk, then hang it from its
middle with a piece string so that it is free to move. If you then bring
another glass rod which you have also charged in the same way next to
it, you will see the ball on the string move away from the rod in your
hand i.e. it is repelled. If, however, you take a plastic rod, rub it with
a piece of fur and then bring it close to the ball on the string, you will
see the rod on the string move towards the rod in your hand i.e. it is
attracted.
////////////////////
+
+
275
16.4
////////////////////
+ +
+ +
++++
////////////////////
- - - -
+ +
+ +
This happens because when you rub the glass with silk, tiny amounts
of negative charge are transferred from the glass onto the silk, which
causes the glass to have less negative charge than positive charge, making it positively charged. When you rub the plastic rod with the fur,
you transfer tiny amounts of negative charge onto the rod and so it has
more negative charge than positive charge on it, making it negatively
charged.
Photograph by NASA
Chapter 16 | Summary
276
16.4
Objects that are neutral have equal numbers of positive and negative charge.
Unlike charges are attracted to each other and like charges are repelled from each
other.
Charge is neither created nor destroyed, it can only be transferred.
Q1 + Q2
2
Physical Quantities
Quantity
Unit name
Unit symbol
Charge (Q)
coulomb
coulomb
Chapter 16
while
.
277
16.4
b. If I now move the balls so that they are 1 m apart, what happens to
the strength of the electrostatic force between them?
5. I have 2 charged spheres each hanging from string as shown in the picture
below.
Choose the correct answer from the options below: The spheres will
a. swing towards each other due to the attractive electrostatic force between them.
b. swing away from each other due to the attractive electrostatic force
between them.
c. swing towards each other due to the repulsive electrostatic force between them.
d. swing away from each other due to the repulsive electrostatic force
between them.
6. Describe how objects (insulators) can be charged by contact or rubbing.
7. You are given a perspex ruler and a piece of cloth.
a. How would you charge the perspex ruler?
b. Explain how the ruler becomes charged in terms of charge.
c. How does the charged ruler attract small pieces of paper?
8. (IEB 2005/11 HG) An uncharged hollow metal sphere is placed on an insulating stand. A positively charged rod is brought up to touch the hollow
metal sphere at P as shown in the diagram below. It is then moved away
from the sphere.
+++
Where is the excess charge distributed on the sphere after the rod has
been removed?
a. It is still located at point P where the rod touched the sphere.
b. It is evenly distributed over the outer surface of the hollow sphere.
c. It is evenly distributed over the outer and inner surfaces of the hollow
sphere.
278
16.4
16. Two identical, insulated spheres have different charges. Sphere 1 has a
charge of 96 1018 C. Sphere 2 has 60 excess electrons. If the two
spheres are brought into contact and then separated, what charge will
each have?
17. Two identical, metal spheres have different charges. Sphere 1 has a charge
of 4, 8 1018 C. Sphere 2 has 30 excess protons. If the two spheres
are brought into contact and then separated, what charge will each have?
How many electrons or protons does this correspond to?
More practice
video solutions
or help at www.everythingscience.co.za
(1.) 006b
(2.) 006c
(3.) 006d
(4.) 006e
(5.) 006f
(6.) 006g
(7.) 006h
(8.) 006i
(9.) 006j
(10.) 006k
(11.) 006m
(12.) 006n
(13.) 006p
(14.) 006q
(15.) 006r
(16.) 006s
(17.) 006t
279
Electric circuits
17
ESAFA
When a circuit is connected and complete, charge can move through the circuit. Charge
will not move unless there is a reason, a force to drive it round the circuit. Think of it as
though charge is at rest and something has to push it along. This means that work needs
to be done to make charge move. A force acts on the charges, doing work, to make them
move. The force is provided by the battery in the circuit.
A battery has the potential to drive charge round a closed circuit, the battery has potential
energy that can be converted into electrical energy by doing work on the charge in the
circuit to make it move.
See introductory video: ( Video: VPfqj at www.everythingscience.co.za)
W
q
tential difference are the volt (V) which is dened as one joule per coulomb.
Quantity: Potential difference (V)
Unit name: volt
Unit symbol:
V
ESAFB
Voltmeter
A voltmeter is an instrument for measuring
A voltmeter
V
Photography by windy_ on Flickr.com
280
17.1
EMF
ESAFC
FACT
that is not in a complete circuit you are measuring the emf of the battery. This is the maxi-
(17451827).
Batteries
One lead of the voltmeter is connected to
one end of the battery and the other lead
is connected to the opposite end. The voltmeter may also be used to measure the voltage across a resistor or any other component
of a circuit but must be connected in parallel.
Photography on Flickr.com
281
17.2
Current
ESAFD
Flow of charge
FACT
Benjamin
ESAFE
Franklin
direction
charge
ow
of
when
rough
wool.
from
positive
negative)
which
this,
Due
electrons
objects
which
When we talk about current we talk about how much charge moves past a xed point in
circuit in one second. Think of charges being pushed around the circuit by the battery,
there are charges in the wires but unless there is a battery they wont move.
See video: VPfva at www.everythingscience.co.za
When one charge moves the charges next to
Copper wire
marble
of electrons. By the
instantaneously.
Photography on Flickr.com
tion of electron ow
was discovered, the
convention of posi-
DEFINITION: Current
Current is the rate at which charges moves past a xed point in a circuit.
The units of current are the ampere (A) which is dened as one coulomb per
second.
change it.
282
Unit symbol: A
17.2
We use the symbol I to show current and it is
measured in amperes (A). One ampere is one
Plasma ball
Ammeter
ESAFF
Ammeter
Activity:
Constructing circuits
Construct circuits to measure the emf and the terminal potential difference for a battery. Some common elements (components) which can be found in electrical circuits include light bulbs, batteries, connecting leads, switches, resistors, voltmeters
and ammeters. You have learnt about many of these already. Below is a table with
283
17.2
Symbol
Usage
light bulb
battery
switch
resistor
voltmeter
ammeter
connecting lead
Tip
A battery does not
produce
the
same
amount of current no
matter what is connected to it.
While
The table below summarises the use of each measuring instrument that we discussed and
the way it should be connected to a circuit component.
the
Instrument
depends on what is
in the circuit.
Activity:
Proper Connection
Voltage
In Parallel
Ammeter
of current supplied
Measured Quantity
Voltmeter
amount
Current
In Series
Using meters
If possible, connect meters in circuits to get used to the use of meters to measure
electrical quantities. If the meters have more than one scale, always connect to the
largest scale rst so that the meter will not be damaged by having to measure values
that exceed its limits.
284
17.2
SOLUTION
Step 1 : Analyse the question
We are given an amount of charge and a time and asked to calculate
the current. We know that current is the rate at which charge moves
past a xed point in a circuit so we have all the information we need.
We have quantities in the correct units already.
Step 2 : Apply the principles
We know that:
I
Q
t
45 C
1s
45 C s1
45 A
285
17.2
SOLUTION
Step 1 : Analyse the question
We are given an amount of charge and a time and asked to calculate
the current. We know that current is the rate at which charge moves
past a xed point in a circuit so we have all the information we need.
We have quantities in the correct units already.
Step 2 : Apply the principles
We know that:
I
Q
t
53 C
2s
26, 5 C s1
26, 5 A
SOLUTION
Step 1 : Analyse the question
We are given a number of charged particles that move past a xed point
and the time that it takes. We know that current is the rate at which
charge moves past a xed point in a circuit so we have to determine
the charge. In the last chapter we learnt that the charge carried by an
286
17.3
electron is 1, 6 1019 C.
Step 2 : Apply the principles: determine the charge
We know that each electron carries a charge of 1, 61019 C, therefore
the total charge is:
Q
95 1, 6 1019 C
1, 52 1017 C
Q
t
1, 52 1017 C
1
10 s
1, 52 1017 C
1
1
1
10 s
1, 52 1017 C
10
1
1s
1, 52 1016 C s1
1, 52 1016 A
Resistance
ESAFG
287
17.3
ESAFH
Examples of resistors
DEFINITION: Resistance
Resistance slows down the ow of charge in a circuit. The unit of resistance
is the ohm () which is dened as a volt per ampere of current.
Quantity: Resistance R
Unit: ohm
Unit symbol:
FACT
Fluorescent
bulbs
do
lightnot
use
1 ohm = 1
volt
ampere
The wires connecting the lamp to the cell or battery hardly even get warm while conducting
the same amount of current. This is because of their much lower resistance due to their
larger cross-section (they are thicker).
288
17.3
An important effect of a resistor is that it converts electrical energy into other forms of heat
energy. Light energy is a by-product of the heat that is produced.
FACT
There is a special type
of conductor, called a
superconductor that
has
no
resistance,
Length: if a resistor is increased in length its resistance will increase. Typically if you
increase the length of a resistor by a certain factor you will increase the resistance by
the same factor.
superconductors
only start superconducting at very low
temperatures.
The
highest temperature
superconductor
mercury
is
barium
is
super-
conducting
for
temperatures
L
R ,
A
ESAFI
It is important to understand what effect adding resistors to a circuit has on the total resistance of a circuit and on the current that can ow in the circuit.
289
of
17.4
Series resistors
ESAFJ
The voltage is divided across the resistors. The voltage across the battery in the
circuit is equal to the sum of voltages across the series resistors:
Vbattery = V1 + V2 + . . .
The resistance to the ow of current increases. The total resistance, RS is given
by:
RS = R 1 + R 2 + . . .
We will revisit each of these features of series circuits in more detail below.
When resistors are in series, one after the other, there is a potential difference across each
resistor. The total potential difference across a set of resistors in series is the sum of the
potential differences across each of the resistors in the set.
Vbattery = V1 + V2 + . . .
Look at the circuits below. If we measured the potential difference between the black dots
in all of these circuits it would be the same as we saw earlier. So we now know the total
potential difference is the same across one, two or three resistors. We also know that some
work is required to make charge ow through each resistor.
Let us look at this in a bit more detail. In the picture below you can see what the different
measurements for 3 identical resistors in series could look like. The total voltage across all
three resistors is the sum of the voltages across the individual resistors. Resistors in series
are known as voltage dividers because the total voltage across all the resistors is divided
amongst the individual resistors.
290
17.4
zooming in
V = 5V
V = 5V
V = 5V
V
V = 15V
Consider the diagram below. On the left there is a circuit with a single resistor and a
battery. No matter where we measure the current, it is the same in a series circuit. On
the right, we have added a second resistor in series to the circuit. The total resistance of
the circuit has increased and you can see from the reading on the ammeter that the current
in the circuit has decreased.
R=2
R=2
I=1A
A
V=2V
I=1A
V=2V
I = 0.67 A
(the current is
smaller)
R=1
I = 0.67 A
(the current is
smaller)
Aim: Test what happens to the current and the voltage in series circuits when additional resistors are added.
Apparatus:
A battery
A voltmeter
An ammeter
Wires
Resistors
Physics: Electricity and Magnetism
291
17.4
R3
R1
R1
R2
R1
R2
292
17.4
5
17
SOLUTION
Step 1 : Analyse the question
We are told that the circuit is a series circuit and that we need to calculate the total resistance. The values of the two resistors have been given
in the correct units, .
Step 2 : Apply the relevant principles
The total resistance for resistors in series is the sum of the individual
resistances. We can use
R S = R1 + R 2 + . . .
. We have only two resistors and we now the resistances. In this case
we have that:
R S = R1 + R 2 + . . .
RS = R1 + R 2
= 5 + 17
= 22
Step 3 : Quote the nal result
The total resistance of the resistors in series is 22 .
293
17.4
11
0, 5
7, 5
SOLUTION
Step 1 : Analyse the question
We are told that the circuit is a series circuit and that we need to calculate the total resistance. The values of the three resistors have been
given in the correct units, .
Step 2 : Apply the relevant principles
The total resistance for resistors in series is the sum of the individual
resistances. We can use
RS = R 1 + R 2 + . . .
. We have three resistors and we now the resistances. In this case we
have that:
RS = R 1 + R2 + . . .
RS = R 1 + R2 + R3
= 0, 5 + 7, 5 + 11
= 19
Step 3 : Quote the nal result
294
17.4
750 k
1, 7 M
SOLUTION
Step 1 : Analyse the question
We are told that the circuit is a series circuit and that we need to calculate the total resistance. The values of the two resistors have been given
in the correct units, .
Step 2 : Apply the relevant principles
The total resistance for resistors in series is the sum of the individual
resistances. We can use
R S = R1 + R 2 + . . .
. We have only two resistors and we now the resistances. In this case
295
17.5
= 0, 75 106 + 1, 7 106
= 2, 45 106 = 2, 45 M
Parallel resistors
ESAFK
296
17.5
Look at the following circuit diagrams. The battery is the same in all cases, all that changes
is more resistors are added between the points marked by the black dots. If we were to
measure the potential difference between the two dots in these circuits we would get the
same answer for all three cases.
Lets look at two resistors in parallel more closely. When you construct a circuit you use
wires and you might think that measuring the voltage in different places on the wires will
make a difference. This is not true. The potential difference or voltage measurement will
only be different if you measure a different set of components. All points on the wires that
have no circuit components between them will give you the same measurements.
All three of the measurements shown in the picture below (i.e. AB, CD and EF) will give
you the same voltage. The different measurement points on the left have no components
between them so there is no change in potential energy. Exactly the same applies to the
different points on the right. When you measure the potential difference between the
points on the left and right you will get the same answer.
V=5V
A
E
C
:
ing in
zoom
B
D
B
D
A
C
F
V
V=5V
V
V=5V
297
17.5
Example 7: Voltages I
QUESTION
Consider this circuit diagram:
2V
V1
SOLUTION
Step 1 : Check what you have and the units
We have a circuit with a battery and one resistor. We know the voltage
across the battery. We want to nd that voltage across the resistor.
Vbattery = 2 V
Step 2 : Applicable principles
We know that the voltage across the battery must be equal to the total
voltage across all other circuit components.
Vbattery = Vtotal
There is only one other circuit component, the resistor.
Vtotal = V1
This means that the voltage across the battery is the same as the voltage
across the resistor.
Vbattery = Vtotal = V1
Vbattery = Vtotal = V1
V1 = 2V
298
17.5
Example 8: Voltages II
QUESTION
Consider this circuit:
2V
VA
VB = 1V
What is the voltage across the unknown resistor in the circuit shown?
SOLUTION
Step 1 : Check what you have and the units
We have a circuit with a battery and two resistors. We know the voltage
across the battery and one of the resistors. We want to nd that voltage
across the resistor.
Vbattery = 2V
VB = 1V
Step 2 : Applicable principles
We know that the voltage across the battery must be equal to the total
voltage across all other circuit components that are in series.
Vbattery = Vtotal
The total voltage in the circuit is the sum of the voltages across the
individual resistors
Vtotal = VA + VB
Using the relationship between the voltage across the battery and total
voltage across the resistors
Vbattery = Vtotal
299
17.5
Vbattery = V1 + Vresistor
2V = V1 + 1V
V1 = 1V
7V
VB
VC = 4V
What is the voltage across the unknown resistor in the circuit shown?
SOLUTION
Step 1 : Check what you have and the units
We have a circuit with a battery and three resistors. We know the
voltage across the battery and two of the resistors. We want to nd that
voltage across the unknown resistor.
Vbattery
7V
Vknown
VA + VC
1V + 4V
300
17.5
individual resistors
Vtotal = VB + Vknown
Using the relationship between the voltage across the battery and total
voltage across the resistors
Vbattery
Vtotal
Vbattery
VB + Vknown
7V
VB + 5V
VB
2V
7V
1V
What is the voltage across the parallel resistor combination in the circuit shown?
Hint: the rest of the circuit is the same as the previous problem.
SOLUTION
Step 1 : Quick Answer
The circuit is the same as the previous example and we know that the
voltage difference between two points in a circuit does not depend on
what is between them so the answer is the same as above Vparallel =
2V .
Step 2 : Check what you have and the units - long answer
We have a circuit with a battery and ve resistors (two in series and
301
17.5
Vbattery
Vparallel + Vknown
7V
Vparallel + 5V
Vparallel
2V
In contrast to the series case, when we add resistors in parallel, we create more paths along
which current can ow. By doing this we decrease the total resistance of the circuit!
Take a look at the diagram below. On the left we have the same circuit as in the previous
section with a battery and a resistor. The ammeter shows a current of 1 A. On the right
we have added a second resistor in parallel to the rst resistor. This has increased the
number of paths (branches) the charge can take through the circuit - the total resistance has
decreased. You can see that the current in the circuit has increased. Also notice that the
current in the different branches can be different.
302
17.5
I=2A
R=2
R=1
A
A
R=2
I=1A
V=2V
I=1A
I=3A
The current
is bigger
V=2V
The total resistance of a number of parallel resistors is NOT the sum of the individual
resistances as the overall resistance decreases with more paths for the current. The total
resistance for parallel resistors is given by:
1
1
1
=
+
+ ...
RP
R1
R2
Let us consider the case where we have two resistors in parallel and work out what the nal
resistance would be. This situation is shown in the diagram below:
R2
R1
RP
R1
R2
R2
R1
1
R2
R1
=
+
RP
R1 R2
R 1 R2
303
17.5
Rearrange
R2 + R 1
1
=
RP
R1 R2
1
R1 + R 2
=
RP
R1 R2
R1 R2
RP =
R1 + R 2
R1 R2
product of resistances
=
sum of resistances
R 1 + R2
Aim: Test what happens to the current and the voltage in series circuits when additional resistors are added.
Apparatus:
A battery
A voltmeter
An ammeter
Wires
Resistors
Method:
Connect each circuit shown below
Measure the voltage across each resistor in the circuit.
Measure the current before and after each resistor in the circuit and before
and after the parallel branches.
304
17.5
15
SOLUTION
Step 1 : Analyse the question
We are told that the resistors in the circuit are in parallel circuit and that
we need to calculate the total resistance. The values of the two resistors
have been given in the correct units, .
Step 2 : Apply the relevant principles
The total resistance for resistors in parallel has been shown to be the
305
17.5
R1 R2
R 1 + R2
. We have only two resistors and we now the resistances. In this case
we have that:
R1 R2
R1 + R 2
(15 )(7 )
=
15 + 7
105 2
=
22
= 4, 77
RP =
15
SOLUTION
Step 1 : Analyse the question
We are told that the resistors in the circuit are in parallel circuit and that
we need to calculate the total resistance. The value of the additional
resistor has been given in the correct units, .
306
17.5
RP
R1
R2 R3
R2
R1 R3
R3
R1 R2
1
R2 R 3
R1 R3
R1 R2
=
+
+
RP
R1 R 2 R 3
R1 R2 R3
R1 R2 R3
rearrange
R2 R 3 + R1 R3 + R 2 R 3
1
=
RP
R1 R2 R3
R1 R2 R3
RP =
R2 R 3 + R1 R3 + R 2 R 3
(15 )(7 )(3 )
RP =
(7 )(3 ) + (15 )(3 ) + (7 )(15 )
315 3
RP =
21 2 + 45 2 + 105 2
315 3
RP =
171 2
RP = 1, 84
Step 3 : Quote the nal result
The total resistance of the resistors in parallel is 1, 84 . It makes sense
that this is less than in the previous example because we added an
additional path and reduced the overall resistance in the circuit.
When calculating the resistance for complex resistor congurations, you can start with any
combination of two resistors (in series or parallel) and calculate their total resistance. Then
you can replace them with a single resistor that has the total resistance you calculated.
Now use this new resistor in combination with any other resistor and repeat the process
until there is only one resistor left. In the above example we could just have used the
answer from the rst example in parallel with the new resistor and we would get the same
answer.
Physics: Electricity and Magnetism
307
17.5
15
SOLUTION
Step 1 : Analyse the question
We are told that the resistors in the circuit are in parallel circuit and that
we need to calculate the total resistance. The value of the additional
resistor has been given in the correct units, .
Step 2 : Apply the relevant principles
We can swap the resistors around without changing the circuit:
15
4, 77
308
17.5
Step 3 : Calculate the total resistance for the next pair of resistors
Then we use the formula for two parallel resistors again to get the total
resistance for this new circuit:
R1 R2
R1 + R2
(4, 77 )(3 )
=
4, 77 + 3
14, 31 2
=
11, 77
RP =
= 1, 84
Step 4 : Quote the nal result
The total resistance of the resistors in parallel is 1, 84 . This is the
same result as when we added all three resistors together at once.
Exercise 17 - 1
More practice
(1.) 006u
See simulation: (
(2.) 006v
video solutions
(3.) 006w
or help at www.everythingscience.co.za
(4.) 006x
309
17.5
Chapter 17 | Summary
See the summary presentation (
The potential difference across the terminals of a battery when it is not in a complete
Current is the rate at which charge moves/ows and is measured in amperes (A) which
is equivalent to C s1 .
Conventional current ows from the positive terminal of a battery, through a circuit,
to the negative terminal.
Resistance of circuit elements is related to the material from which they are made as
well as the physical characteristics of length and cross-sectional area.
Current is constant through resistors in series and they are called voltage dividers as
the sum of the voltages is equal to the voltage across the entire set of resistors.
The total resistance of resistors in series is the sum of the individual resistances,
RS = R1 + R2 + . . ..
Voltage is constant across resistors in parallel and they are called current divides
because the sum of the current through each is the same as the total current through
1
RP
1
R1
1
R2
+...
Physical Quantities
Quantity
Unit name
Unit symbol
Potential difference (V )
volt
emf
volt
Voltage (V )
volt
Current (I)
ampere
ohm
Resistance (R)
310
Chapter 17
17.5
Symbol
Unit of measurement
Symbol of unit
e.g. Distance
e.g. D
e.g. kilometre
e.g. km
Resistance
Current
Potential difference
4. [SC 2003/11] The emf of a battery can best be explained as the . . .
a. rate of energy delivered per unit current
b. rate at which charge is delivered
c. rate at which energy is delivered
d. charge per unit of energy delivered by the battery
5. [IEB 2002/11 HG1] Which of the following is the correct denition of the
emf of a battery?
a. It is the product of current and the external resistance of the circuit.
b. It is a measure of the cells ability to conduct an electric current.
c. It is equal to the lost volts in the internal resistance of the circuit.
d. It is the power supplied by the battery per unit current passing through
the battery.
6. [IEB 2005/11 HG] Three identical light bulbs A, B and C are connected in
an electric circuit as shown in the diagram below.
311
17.5
Current in B
(a)
decreases
increases
(b)
decreases
decreases
(c)
increases
increases
(d)
increases
decreases
17.5
More practice
video solutions
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(1.) 006y
(2.) 006z
(3.) 0070
(4.) 0071
(5.) 0072
(6.) 0073
(7.) 0074
(8.) 01uw
(9.) 01ux
(10.) 01uy
(11.) 01uz
(12.) 01v0
(13.) 01v1
(14.) 01v2
(15.) 01v3
313
Reactions in aqueous
solution
18
Introduction
ESAFL
Many reactions in chemistry and all biological reactions (reactions in living systems) take
place in water. We say that these reactions take place in aqueous solution. Water has many
unique properties and is plentiful on Earth. For these reasons reactions in aqueous solutions
occur frequently. In this chapter we will look at some of these reactions in detail. Almost
all the reactions that occur in aqueous solutions involve ions. We will look at three main
types of reactions that occur in aqueous solutions, namely precipitation reactions, acidbase reactions and redox reactions. Before we can learn about the types of reactions, we
need to rst look at ions in aqueous solutions and electrical conductivity. See introductory
video: ( Video: VPbls at www.everythingscience.co.za)
ESAFM
Water is seldom pure. Because of the structure of the water molecule, substances can
dissolve easily in it. This is very important because if water wasnt able to do this, life
would not be possible on Earth. In rivers and the oceans for example, dissolved oxygen
means that organisms (such as sh) are able to respire (breathe). For plants, dissolved
nutrients are available in a form which they can absorb. In the human body, water is able
to carry dissolved substances from one part of the body to another.
Dissociation in water
ESAFN
Water is a polar molecule. If we represent water using Lewis structures we will get the
following:
314
18.2
H O
You will notice that there are two electron pairs that do not take part in bonding. This side
of the water molecule has a higher electron density than the other side where the hydrogen
atoms are bonded. This side of the water molecule is more negative than the side where
the hydrogen atoms are bonded. We say this side is the delta negative () side and the
hydrogen side is the delta positive (+) side. This means that one part of the molecule has
a slightly positive charge (positive pole) and the other part has a slightly negative charge
(negative pole). We say such a molecule is a dipole. It has two poles. Figure 18.2 shows
this.
+
Figure 18.2: Water is a polar molecule
ESAFO
It is the polar nature of water that allows ionic compounds to dissolve in it. In the case
of sodium chloride (NaCl) for example, the positive sodium ions (Na+ ) are attracted to the
negative pole of the water molecule, while the negative chloride ions (Cl ) are attracted
to the positive pole of the water molecule. When sodium chloride is dissolved in water,
the polar water molecules are able to work their way in between the individual ions in the
lattice. The water molecules surround the negative chloride ions and positive sodium ions
and pull them away into the solution. This process is called dissociation. Note that the
positive side of the water molecule will be attracted to the negative chlorine ion and the
negative side of the water molecule to the positive sodium ions. A simplied representation
of this is shown in Figure 18.3. We say that dissolution of a substance has occurred when
a substance dissociates or dissolves. Dissolving is a physical change that takes place. It can
be reversed by removing (evaporating) the water.
315
18.2
DEFINITION: Dissociation
Dissociation is a general process in which ionic compounds separate into
smaller ions, usually in a reversible manner.
DEFINITION: Dissolution
Dissolution or dissolving is the process where ionic crystals break up into
ions in water.
DEFINITION: Hydration
Hydration is the process where ions become surrounded with water molecules.
Cl
Na
18.3
There are exceptions to this and some molecular substances will form ions when they
dissolve. Hydrogen chloride for example can ionise to form hydrogen and chloride ions.
Exercise 18 - 1
1. For each of the following, say whether the substance is ionic or molecular.
a. potassium nitrate (KNO3 )
b. ethanol (C2 H5 OH)
c. sucrose (a type of sugar) (C12 H22 O11 )
d. sodium bromide (NaBr)
2. Write a balanced equation to show how each of the following ionic compounds dissociate in water.
a. sodium sulphate (Na2 SO4 )
b. potassium bromide (KBr)
c. potassium permanganate (KMnO4 )
d. sodium phosphate (Na3 PO4 )
3. Draw a diagram to show how KCl dissolves in water.
More practice
(1.) 0075
(2.) 0076
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(3.) 0077
Electrolytes, ionisation
and conductivity
ESAFP
You have learnt that water is a polar molecule and that it can dissolve ionic substances in
water. When ions are present in water, the water is able to conduct electricity. The solution
is known as an electrolyte.
317
18.3
DEFINITION: Electrolyte
An electrolyte is a substance that contains free ions and behaves as an electrically conductive medium.
Because electrolytes generally consist of ions in solution, they are also known as ionic
solutions. A strong electrolyte is one where many ions are present in the solution and a
weak electrolyte is one where few ions are present. Strong electrolytes are good conductors
of electricity and weak electrolytes are weak conductors of electricity. Non-electrolytes do
not conduct electricity at all. Conductivity in aqueous solutions, is a measure of the ability
of water to conduct an electric current. The more ions there are in the solution, the higher
its conductivity. Also the more ions there are in solution, the stronger the electrolyte.
ESAFQ
trolyte (e.g. potassium nitrate, KNO3 ), a weak electrolyte (e.g. acetic acid, CH3 COOH)
or a non-electrolyte (e.g. sugar, alcohol, oil) will affect the conductivity of water because the concentration of ions in solution will be different in each case. Strong
electrolytes form ions easily, weak electrolytes do not form ions easily and nonelectrolytes do not form ions in solution.
See video: VPbon at www.everythingscience.co.za
Temperature. The warmer the solution, the higher the solubility of the material being
dissolved and therefore the higher the conductivity as well.
318
18.3
Apparatus:
Solid salt (NaCl) crystals
different liquids such as distilled water, tap water, seawater, sugar, oil and
alcohol
solutions of salts e.g. NaCl, KBr, CaCl2 , NH4 Cl
battery
battery
Ammeter
Ammeter
test substance
X
crocodile clip
crocodile clip
X
test substance
Ammeter reading
319
18.4
Remember that for electricity to ow, there needs to be a movement of charged particles
e.g. ions. With the solid NaCl crystals, there was no ow of electricity recorded on the
ammeter. Although the solid is made up of ions, they are held together very tightly within
the crystal lattice and therefore no current will ow. Distilled water, oil and alcohol also
dont conduct a current because they are covalent compounds and therefore do not contain
ions.
The ammeter should have recorded a current when the salt solutions and the acid and base
solutions were connected in the circuit. In solution, salts dissociate into their ions, so that
these are free to move in the solution. Look at the following examples:
Dissociation of potassium bromide:
KBr (s) K+ (aq) + Br (aq)
Dissociation of table salt:
Precipitation reactions
ESAFR
Sometimes, ions in solution may react with each other to form a new substance that is
insoluble. This is called a precipitate. The reaction is called a precipitation reaction.
See video: VPbpr at www.everythingscience.co.za
320
18.4
DEFINITION: Precipitate
A precipitate is the solid that forms in a solution during a chemical reaction.
CuCl2
Na2 CO3
CuCl2
Na2 SO4
Method:
1. Prepare 2 test tubes with approximately 5 ml of dilute copper(II) chloride
solution in each
2. Prepare 1 test tube with 5 ml sodium carbonate solution
3. Prepare 1 test tube with 5 ml sodium sulphate solution
4. Carefully pour the sodium carbonate solution into one of the test tubes containing copper(II) chloride and observe what happens
5. Carefully pour the sodium sulphate solution into the second test tube containing copper(II) chloride and observe what happens
Results:
1. A light blue precipitate forms when sodium carbonate reacts with copper(II)
chloride.
2. No precipitate forms when sodium sulphate reacts with copper(II) chloride.
The solution is light blue.
321
18.4
For reaction 1 you have the following ions in your solution: Cu2+ , Cl , Na+ and CO2 .
3
A precipitate will form if any combination of cations and anions can become a solid. The
following table summarises which combination will form solids (precipitates) in solution.
Salt
Solubility
Nitrates
Sulphates
All are soluble except lead(II) sulphate, barium sulphate and calcium sulphate
Carbonates
Compounds with uorine
Almost all are soluble except those of magnesium, calcium, strontium (II), barium (II) and lead (II)
Chlorates
Tip
Salts of carbonates,
phosphates, oxalates,
chromates and sulphides are generally
If you look under carbonates in the table it states that all carbonates are insoluble except
insoluble.
potassium sodium and ammonium. This means that Na2 CO3 will dissolve in water or
remain in solution, but CuCO3 will form a precipitate. The precipitate that was observed
in the reaction must therefore be CuCO3 . The balanced chemical equation is:
2Na+ (aq) + CO2 (aq) + Cu2+ (aq) + 2Cl (aq) CuCO3 (s) + 2Na+ (aq) + 2Cl (aq)
3
Note that sodium chloride does not precipitate and we write it as ions in the equation. For
reaction 2 we have Cu2+ , Cl , Na+ and SO2 in solution. Most chlorides and sulphates
4
are soluble according to the table. The balanced chemical equation is:
2Na+ (aq)+SO2 (aq)+Cu2+ (aq)+2Cl (aq) 2Na+ (aq)+SO2 (aq)+Cu2+ (aq)+2Cl (aq)
4
4
Both of these reactions are ion exchange reactions.
322
18.4
ESAFS
We often want to know which ions are present in solution. If we know which salts precipitate, we can derive tests to identify ions in solution. Given below are a few such tests.
See video: VPbqd at www.everythingscience.co.za
323
18.4
SO2 (aq) + Ba2+ (aq) + Cl (aq) BaSO4 (s) + Cl (aq) (BaSO4 is a white precipitate)
4
CO2 (aq) + Ba2+ (aq) + Cl (aq) BaCO3 (s) + Cl (aq) (BaCO3 is a white precipitate)
3
If the precipitate is treated with nitric acid, it is possible to distinguish whether the salt is a
sulphate or a carbonate (as in the test for a chloride).
BaSO4 (s) + HNO3 () (no reaction; precipitate is unchanged)
BaCO3 (s) + 2HNO3 () Ba2+ (aq) + 2NO (aq) + H2 O() + CO2 (g) (precipitate disappears)
3
Test for a carbonate
If a sample of the dry salt is treated with a small amount of acid, the production of carbon
dioxide is a positive test for a carbonate.
2HCl + K2 CO3 (aq) CO2 (g) + 2KCl (aq) + H2 O
If the gas is passed through limewater (an aqueous solution of calcium hydroxide) and the
solution becomes milky, the gas is carbon dioxide.
Ca2+ (aq)+2OH (aq)+CO2 (g) CaCO3 (s)+H2 O() (It is the insoluble CaCO3 precipitate
that makes the limewater go milky)
Exercise 18 - 2
1. Silver nitrate (AgNO3 ) reacts with potassium chloride (KCl) and a white
precipitate is formed.
a. Write a balanced equation for the reaction that takes place. Include
the state symbols.
b. What is the name of the insoluble salt that forms?
c. Which of the salts in this reaction are soluble?
2. Barium chloride reacts with sulphuric acid to produce barium sulphate
and hydrochloric acid.
a. Write a balanced equation for the reaction that takes place. Include
the state symbols.
b. Does a precipitate form during the reaction?
c. Describe a test that could be used to test for the presence of barium
sulphate in the products.
324
18.5
3. A test tube contains a clear, colourless salt solution. A few drops of silver
nitrate solution are added to the solution and a pale yellow precipitate
forms. Chlorine water and carbon tetrachloride were added, which resulted in a purple solution. Which one of the following salts was dissolved
in the original solution? Write the balanced equation for the reaction that
took place between the salt and silver nitrate.
a. NaI
b. KCl
c. K2 CO3
d. Na2 SO4
More practice
(1.) 0078
(2.) 0079
video solutions
or help at www.everythingscience.co.za
(3.) 007a
ESAFT
We will look at two types of reactions that occur in aqueous solutions. These are ionexchange reactions and redox reactions. Ion exchange reactions include precipitation reactions, gas forming reactions and acid-base reactions. Redox reactions are electron transfer
reactions. It is important to remember the difference between these two types of reactions.
In ion exchange reactions ions are exchanged, in electron transfer reactions electrons are
transferred. These terms will be explained further in the following sections.
ESAFU
AB(aq) + CD (aq) AD + CB
Either AD or CB may be a solid or a gas. When a solid forms this is known as a precipitation
reaction. If a gas is formed then this may be called a gas forming reaction. Acid-base
reactions are a special class of ion exchange reactions and we will look at them separately.
The formation of a precipitate or a gas helps to make the reaction happen. We say that the
325
18.5
reaction is driven by the formation of a precipitate or a gas. All chemical reactions will only
take place if there is something to make them happen. For some reactions this happens
easily and for others it is harder to make the reaction occur.
FACT
Ion
exchange
actions
in
ion
are
reused
A type of reaction where the positive ions exchange their respective negative
exchange
chromatography.
Ion exchange chromatography is used to
purify water and as
a means of softening
water.
Often when
Redox reactions
ESAFV
Redox reactions involve the exchange of electrons. One ion loses electrons and becomes
more positive, while the other ion gains electrons and becomes more negative. To decide
if a redox reaction has occurred we look at the charge of the atoms, ions or molecules
involved. If one of them has become more positive and the other one has become more
326
18.5
negative then a redox reaction has occurred. For example, sodium metal is oxidised to
form sodium oxide (and sometimes sodium peroxide as well). The balanced equation for
this is:
4Na + O2 2Na2 O
In the above reaction sodium and oxygen are both neutral and so have no charge. In the
products however, the sodium atom has a charge of +1 and the oxygen atom has a charge
of 2. This tells us that the sodium has lost electrons and the oxygen has gained electrons.
Since one species has become more positive and one more negative we can conclude that
a redox reaction has occurred. We could also say that electrons have been transferred from
one species to the other. (In this case the electrons were transferred from the sodium to the
oxygen).
Warning:
Sodium metal is very reactive. Sodium
metal reacts vigorously with water and
should never be placed in water. Be very
careful when handling sodium metal.
327
18.5
328
18.5
In the experiment above, you should have seen how each reaction type differs from the
others. For example, a gas forming reaction leads to bubbles in the solution, a precipitation
reaction leads to a precipitate forming, an acid-base reaction can be seen by adding a
suitable indicator and a redox reaction can be seen by one metal disappearing and a deposit
forming in the solution.
Chapter 18 | Summary
See the summary presentation (
The polar nature of water means that ionic compounds dissociate easily in aqueous
solution into their component ions.
Dissociation is a general process in which ionic compounds separate into smaller
ions, usually in a reversible manner.
Dissolution or dissolving is the process where ionic crystals break up into ions in
water.
Hydration is the process where ions become surrounded with water molecules.
Conductivity is a measure of a solutions ability to conduct an electric current.
An electrolyte is a substance that contains free ions and is therefore able to conduct
an electric current. Electrolytes can be divided into strong and weak electrolytes,
based on the extent to which the substance ionises in solution.
A non-electrolyte cannot conduct an electric current because it does not contain free
ions.
The type of substance, the concentration of ions and the temperature of the solution
Precipitation and acid-base reactions are sometimes known as ion exchange reactions. Ion exchange reactions also include gas forming reactions. Ion exchange reac-
tions are a type of reaction where the positive ions exchange their respective negative
ions due to a driving force.
A precipitate is formed when ions in solution react with each other to form an insol-
uble product. Solubility rules help to identify the precipitate that has been formed.
A number of tests can be used to identify whether certain anions (chlorides, bromides, iodides, carbonates, sulphates) are present in a solution.
An acid-base reaction is one in which an acid reacts with a base to form a salt and
water.
329
18.5
A redox reaction is one in which electrons are transferred from one substance to
another.
Chapter 18
Column B
1. A polar molecule
A. H2 SO4
2. Molecular solution
B. CaCO3
C. NaOH
D. salt water
5. A strong electrolyte
E. calcium
6. A white precipitate
F. carbon dioxide
7. A non-conductor of electricity
G. potassium nitrate
H. sugar water
I. O2
18.5
c. carbon dioxide
d. hexane (C6 H14 )
e. lithium uoride (LiF)
f. magnesium chloride
6. Three test tubes (X, Y and Z) each contain a solution of an unknown
potassium salt. The following observations were made during a practical
investigation to identify the solutions in the test tubes:
A: A white precipitate formed when silver nitrate (AgNO3 ) was added to
test tube Z.
B: A white precipitate formed in test tubes X and Y when barium chloride
(BaCl2 ) was added.
C: The precipitate in test tube X dissolved in hydrochloric acid (HCl) and
a gas was released.
D: The precipitate in test tube Y was insoluble in hydrochloric acid.
a. Use the above information to identify the solutions in each of the test
tubes X, Y and Z.
b. Write a chemical equation for the reaction that took place in test tube
X before hydrochloric acid was added.
(DoE Exemplar Paper 2 2007)
More practice
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(5.) 007f
(6.) 007g
331
Quantitative aspects of
chemical change
19
ESAFW
An equation for a chemical reaction can provide us with a lot of useful information. It tells
us what the reactants and the products are in the reaction, and it also tells us the ratio in
which the reactants combine to form products. Look at the equation below:
Fe + S FeS
In this reaction, every atom of iron (Fe) will react with a single atom of sulphur (S) to form
iron sulphide (FeS). However, what the equation does not tell us, is the quantities or the
amount of each substance that is involved. You may for example be given a small sample
of iron for the reaction. How will you know how many atoms of iron are in this sample?
And how many atoms of sulphur will you need for the reaction to use up all the iron you
have? Is there a way of knowing what mass of iron sulphide will be produced at the end
of the reaction? These are all very important questions, especially when the reaction is an
industrial one, where it is important to know the quantities of reactants that are needed,
and the quantity of product that will be formed. This chapter will look at how to quantify
the changes that take place in chemical reactions.
See introductory video: (
The Mole
ESAFX
Sometimes it is important to know exactly how many particles (e.g. atoms or molecules)
are in a sample of a substance, or what quantity of a substance is needed for a chemical
reaction to take place.
The amount of substance is so important in chemistry that it is given its own name, which
is the mole.
332
19.1
DEFINITION: Mole
The mole (abbreviation mol) is the SI (Standard International) unit for
amount of substance.
The mole is a counting unit just like hours or days. We can easily count one second or one
minute or one hour. If we want bigger units of time, we refer to days, months and years.
Even longer time periods are centuries and millennia. The mole is even bigger than these
numbers. The mole is 602 204 500 000 000 000 000 000 or 6, 022 1023 particles. This is a
very big number! We call this number Avogadros number.
FACT
The original hypothesis that was proposed
by Amadeo Avogadro
was that equal volumes of gases,
at
If we had this number of cold drink cans, then we could cover the surface of the earth to a
depth of over 300 km! If you could count atoms at a rate of 10 million per second, then it
would take you 2 billion years to count the atoms in one mole!
We use Avogadros number and the mole in chemistry to help us quantify what happens in
chemical reaction. The mole is a very special number. If we measure 12, 0 g of carbon we
have one mole or 6, 022 10 carbon atoms. 63, 5 g of copper is one mole of copper or
6, 022 1023 copper atoms. In fact, if we measure the relative atomic mass of any element
23
Exercise 19 - 1
Avogadros number.
d. 24, 3 g of magnesium
e. 24, 0 g of carbon
c. 5 moles of phosphorus
2. Complete the following table:
333
19.1
Element
Hydrogen
1, 01
1, 01
Magnesium
24, 3
24, 3
Carbon
12, 0
24, 0
Chlorine
35, 45
70, 9
Nitrogen
14, 0
42, 0
More practice
(1.) 007h
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(2.) 007i
Molar Mass
ESAFY
You will remember that when the mass, in grams, of an element is equal to its relative
atomic mass, the sample contains one mole of that element. This mass is called the molar
mass of that element.
written as Mm . We
will use M in this
book, but you should
It is worth remembering the following: On the periodic table, the relative atomic mass that
nate notation.
1. The mass (in grams) of a single, average atom of that element relative to the mass of
an atom of carbon.
2. The average atomic mass of all the isotopes of that element. This use is the relative
atomic mass.
3. The mass of one mole of the element. This third use is the molar mass of the element.
334
19.1
Element
Magnesium
24, 3
24, 3
24, 3
Lithium
6, 94
6, 94
6, 94
Oxygen
16, 0
16, 0
16, 0
Nitrogen
14, 0
14, 0
14, 0
Iron
55, 8
55, 8
55, 8
Table 19.1: The relationship between relative atomic mass, molar mass and the mass of
one mole for a number of elements.
SOLUTION
=
=
=
111, 7 g
1
55, 8 g mol
111, 7 g mol
55, 8 g
2 mol
335
19.1
SOLUTION
Exercise 19 - 2
1. Give the molar mass of each of the following elements:
a. hydrogen gas
b. nitrogen gas
c. bromine gas
2. Calculate the number of moles in each of the following samples:
a. 21, 6 g of boron (B)
b. 54, 9 g of manganese (Mn)
c. 100, 3 g of mercury (Hg)
d. 50 g of barium (Ba)
e. 40 g of lead (Pb)
More practice
336
video solutions
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(1.) 007j
19.1
(2.) 007k
ESAFZ
mass (g)
mole (mol)
m
M
The following diagram may help to remember the relationship between these three variables. You need to imagine that the horizontal line is like a division sign and that the
vertical line is like a multiplication sign. So, for example, if you want to calculate M, then
the remaining two letters in the triangle are m and n and m is above n with a division sign
between them. Your calculation will then be M =
m
n
Tip
Remember that when
you use the equation
n =
m
,
M
the mass
SOLUTION
337
19.1
n=
m
M
127 g
1
63, 5g mol
= 2 mol
SOLUTION
n=
m
81 g
=
1 = 3 mol
M
27, 0 g mol
338
19.1
Exercise 19 - 3
1. Calculate the number of moles in each of the following samples:
a. 5, 6 g of calcium
b. 0, 02 g of manganese
c. 40 g of aluminium
2. A lead sinker has a mass of 5 g.
a. Calculate the number of moles of lead the sinker contains.
b. How many lead atoms are in the sinker?
3. Calculate the mass of each of the following samples:
a. 2, 5 mol magnesium
b. 12 mol lithium
c. 4, 5 1025 atoms of silicon
More practice
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(2.) 007n
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(3.) 007p
Compounds
ESAGA
So far, we have only discussed moles, mass and molar mass in relation to elements. But
what happens if we are dealing with a compound? Do the same concepts and rules apply?
The answer is yes. However, you need to remember that all your calculations will apply to
the whole compound. So, when you calculate the molar mass of a covalent compound,
you will need to add the molar mass of each atom in that compound. The number of moles
will also apply to the whole molecule. For example, if you have one mole of nitric acid
1
(HNO3 ) the molar mass is 63, 01 g mol and there are 6, 022 1023 molecules of nitric
acid. For network structures we have to use the formula mass. This is the mass of all the
atoms in one formula unit of the compound. For example, one mole of sodium chloride
1
In a balanced chemical equation, the number that is written in front of the element or
compound, shows the mole ratio in which the reactants combine to form a product. If
there are no numbers in front of the element symbol, this means the number is 1.
Chemistry: Chemical change
339
19.1
SOLUTION
Hydrogen = 1, 01 g mol
1
Sulphur = 32, 1 g mol
Oxygen = 16, 0 g mol
)+(32, 1 gmol
)+(416, 0 gmol
) = 98, 12 gmol
SOLUTION
340
19.1
+ (2 35, 45 g mol
) = 95, 2g mol
1 000 g
95, 2 g mol
= 10, 5 mol
Exercise 19 - 4
341
19.2
a. KOH
b. FeCl3
c. Mg(OH)2
2. How many moles are present in:
a. 10 g of Na2 SO4
b. 34 g of Ca(OH)2
c. 2, 45 1023 molecules of CH4 ?
More practice
(1.) 007q
(2.) 007r
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(3.) 007s
(4.) 007t
Composition
or help at www.everythingscience.co.za
(5.) 007u
ESAGB
Knowing either the empirical or molecular formula of a compound, can help to determine
its composition in more detail. The opposite is also true. Knowing the composition of a
substance can help you to determine its formula. There are four different types of composition problems that you might come across:
1. Problems where you will be given the formula of the substance and asked to calculate the percentage by mass of each element in the substance.
2. Problems where you will be given the percentage composition and asked to calculate the formula.
3. Problems where you will be given the products of a chemical reaction and asked to
calculate the formula of one of the reactants. These are often referred to as combustion analysis problems.
342
19.2
SOLUTION
Step 2 : Use the calculations in the previous step to calculate the molecular
mass of sulphuric acid.
1
Mass = 2, 02 gmol
+32, 1 gmol
+64, 0 gmol
= 98, 12 gmol
atomic mass
100%
molecular mass of H2 SO4
2, 02 g mol
1
1
98, 12 g mol
100% = 2, 0587%
Sulphur
32, 1 g mol
98, 12 g mol
Oxygen
64, 0 g mol
98, 12 g mol
343
19.2
SOLUTION
n=
m
M
Therefore:
n (Carbon) =
n (Hydrogen) =
n (Oxygen) =
52, 2 g
12, 0 g mol
13, 0 g
1, 01 g mol
34, 8 g
16, 0 g mol
= 4, 35mol
= 12, 871mol
= 2, 175mol
19.2
4, 35
=2
2, 175
Hydrogen
12, 871
=6
2, 175
Oxygen
2, 175
=1
2, 175
Therefore the empirical formula of this substance is: C2 H6 O.
SOLUTION
345
19.2
n=
m
M
Lead
n=
Oxygen
n=
207 g
207, 2 g mol
32 g
1
16, 0 g mol
= 1 mol
= 2 mol
SOLUTION
n=
19.2
m
M
nC
nH
nO
39, 9 g
12, 0 g mol
6, 7 g
1, 01 g mol
53, 4 g
16, 0 g mol
= 3, 325 mol
= 6, 6337 mol
= 3, 3375 mol
before heating was 5 g. What is the number of moles of water molecules in the
347
19.2
SOLUTION
133, 35 g mol
2, 2 g
18, 02 g mol
0, 02099 . . . : 0, 12 . . .
Next we convert the ratio to whole numbers by dividing both sides
by the smaller amount:
AlCl3 : H2 O
0, 020997375 : 0, 12208657
0, 021 0, 122
:
0, 021 0, 021
1:6
The mole ratio of aluminium trichloride to water is: 1 : 6
Step 4 : Write the nal answer
And now we know that there are 6 moles of water molecules in the
crystal. The formula is AlCl3 6H2 O.
348
19.2
Exercise 19 - 5
1. Calcium chloride is produced as the product of a chemical reaction.
a. What is the formula of calcium chloride?
b. What is the percentage mass of each of the elements in a molecule of
calcium chloride?
c. If the sample contains 5 g of calcium chloride, what is the mass of
calcium in the sample?
d. How many moles of calcium chloride are in the sample?
2. 13 g of zinc combines with 6, 4 g of sulphur.
a. What is the empirical formula of zinc sulphide?
b. What mass of zinc sulphide will be produced?
c. What is the percentage mass of each of the elements in zinc sulphide?
1
. De-
taining 5, 0 g of magnesium sulphate was heated until all the water had
evaporated. The nal mass was found to be 2, 6 g. How many water
molecules were in the original sample?
More practice
(1.) 007v
(2.) 007w
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(3.) 007x
(4.) 007y
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(5.) 007z
349
19.3
Amount of substance
ESAGC
ESAGD
This applies to any gas that is at standard temperature and pressure. In grade 11 you will
learn more about this and the gas laws.
3
For example, 2 mol of H2 gas will occupy a volume of 44, 8 dm at standard temperature
3
and pressure (S.T.P.). and 67, 2 dm of ammonia gas (NH3 ) contains 3 mol of ammonia.
ESAGE
A typical solution is made by dissolving some solid substance in a liquid. The amount of
substance that is dissolved in a given volume of liquid is known as the concentration of the
liquid. Mathematically, concentration (C) is dened as moles of solute (n) per unit volume
(V) of solution.
See video: VPbur at www.everythingscience.co.za
C=
n
V
n
C
350
19.3
For this equation, the units for volume are dm3 (which is equal to litres). Therefore, the
3
unit of concentration is mol dm .
DEFINITION: Concentration
Concentration is a measure of the amount of solute that is dissolved in a
given volume of liquid. It is measured in mol dm
SOLUTION
3, 5 g
m
=
1 = 0, 0875 mol
M
40, 01 g mol
n
0, 0875 mol
3
= 0, 035 mol dm
=
3
V
2, 5 dm
3
351
19.3
You have a 1 dm container in which to prepare a solution of potassium permanganate (KMnO4 ). What mass of KMnO4 is needed to make a solution with a
3
concentration of 0, 2 mol dm
SOLUTION
1dm
= 0, 2 mol
SOLUTION
n = C V = 0, 01 mol dm
0, 5 dm
19.4
= 0, 005 mol
Exercise 19 - 6
More practice
(1.) 0080
(2.) 0081
Stoichiometric
calculations
video solutions
(3.) 0082
(4.) 0083
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(5.) 0084
ESAGF
353
19.4
Reactants
Balanced equation
mass
molar mass
volume
concentration
Products
of propane (C3 H8 )? (Hint: CO2 and H2 O are the products in this reaction (and in
all combustion reactions))
SOLUTION
354
19.4
SOLUTION
m
5, 6 g
=
1 = 0, 1 mol
M
55, 8 g mol
= 8, 79 g
The mass of iron (II) sulphide that is produced during this reaction is
8, 79 g.
Theoretical yield
ESAGG
When we are given a known mass of a reactant and are asked to work out how much product is formed, we are working out the theoretical yield of the reaction. In the laboratory,
chemists almost never get this amount of product. In each step of a reaction a small amount
Chemistry: Chemical change
355
19.4
of product and reactants is lost either because a reactant did not completely react or some
other unwanted products are formed. This amount of product that you actually got is called
the actual yield. You can calculate the percentage yield with the following equation:
% yield =
actual yield
100
theoretical yield
SOLUTION
m
2 000 g
=
1 = 20, 38320424 mol
M
98, 12 g mol
= 2 324, 477 g
The maximum amount of ammonium sulphate that can be produced is 2, 324 kg.
Step 5 : Calculate the % yield
356
% yield =
19.4
actual yield
2, 2
100 =
100 = 94, 64%
theoretical yield
2, 324
SOLUTION
n=
m
2g
=
1 = 0, 0096 mol
M
208, 2 g mol
= 0, 94 g
(answer to 1)
357
19.4
Exercise 19 - 7
1. Diborane, B2 H6 , was once considered for use as a rocket fuel. The combustion reaction for diborane is:
B2 H6 (g) + 3O2 (g) 2HBO2 (g) + 2H2 O ()
If we react 2, 37 g of diborane, how many grams of water would we expect
to produce?
2. Sodium azide is a commonly used compound in airbags. When triggered,
it has the following reaction:
2NaN3 (s) 2N a (s) + 3N2 (g)
If 23, 4 g of sodium azide is used, how many moles of nitrogen gas would
we expect to produce? What volume would this nitrogen gas occupy at
STP?
3. Photosynthesis is a chemical reaction that is vital to the existence of life on
Earth. During photosynthesis, plants and bacteria convert carbon dioxide
gas, liquid water, and light into glucose (C6 H12 O6 ) and oxygen gas.
a. Write down the equation for the photosynthesis reaction.
b. Balance the equation.
c. If 3mol of carbon dioxide are used up in the photosynthesis reaction,
what mass of glucose will be produced?
More practice
(1.) 0085
358
(2.) 0086
video solutions
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(3.) 0087
19.4
Chapter 19 | Summary
See the summary presentation (
The mole (n) (abbreviation mol) is the SI (Standard International) unit for amount of
substance.
The number of particles in a mole is called Avogadros number and its value is
6, 022 1023 . These particles could be atoms, molecules or other particle units,
per mole or g mol1 . The numerical value of an elements molar mass is the same
as its relative atomic mass. For a covalent compound, the molar mass has the same
numerical value as the molecular mass of that compound. For an ionic substance,
the molar mass has the same numerical value as the formula mass of the substance.
The relationship between moles (n), mass in grams (m) and molar mass (M) is dened
by the following equation:
n=
m
M
In a balanced chemical equation, the number in front of the chemical symbols describes the mole ratio of the reactants and products.
The empirical formula of a compound is an expression of the relative number of
We can use the products of a reaction to determine the formula of one of the reactants.
We can nd the number of moles of waters of crystallisation.
3
n
V
3
). The concen-
Stoichiometry is the calculation of the quantities of reactants and products in chemical reactions. It is also the numerical relationship between reactants and products.
The theoretical yield of a reaction is the maximum amount of product that we expect
359
19.4
360
Chapter 19
19.4
a. 0, 09 mol dm
3
b. 5, 7 104 mol dm
3
c. 0, 57 mol dm
d. 0, 03 mol dm
361
19.4
c. 5, 2 mol of helium
d. 0, 05 mol of copper (II) chloride (CuCl2 )
e. 31, 31 1023 molecules of carbon monoxide (CO)
8. Calculate the percentage that each element contributes to the overall mass
of:
a. Chloro-benzene (C6 H5 Cl)
b. Lithium hydroxide (LiOH)
9. CFCs (chlorouorocarbons) are one of the gases that contribute to the
depletion of the ozone layer. A chemist analysed a CFC and found that it
contained 58, 64% chlorine, 31, 43% uorine and 9, 93% carbon. What is
the empirical formula?
10. 14 g of nitrogen combines with oxygen to form 46 g of a nitrogen oxide.
Use this information to work out the formula of the oxide.
11. Iodine can exist as one of three oxides (I2 O4 ; I2 O5 ; I4 O9 ). A chemist has
produced one of these oxides and wishes to know which one they have.
If he started with 508 g of iodine and formed 652 g of the oxide, which
oxide has he produced?
12. A uorinated hydrocarbon (a hydrocarbon is a chemical compound containing hydrogen and carbon) was analysed and found to contain 8, 57%
H, 51, 05% C and 40, 38% F.
a. What is its empirical formula?
b. What is the molecular formula if the molar mass is 94, 1 g mol
13. Copper sulphate crystals often include water. A chemist is trying to determine the number of moles of water in the copper sulphate crystals.
She weighs out 3 g of copper sulphate and heats this. After heating, she
nds that the mass is 1, 9 g. What is the number of moles of water in the
crystals? (Copper sulphate is represented by CuSO4 xH2 O).
3
14. 300 cm3 of a 0, 1 moldm solution of sulphuric acid is added to 200 cm3
3
of a 0, 5 mol dm solution of sodium hydroxide.
a. Write down a balanced equation for the reaction which takes place
when these two solutions are mixed.
of concentration 0, 5 mol dm
b. Using an accurate balance the learner accurately measures the correct mass of the NaOH pellets. To the pellets he now adds exactly
200 cm3 of pure water. Will his solution have the correct concentration? Explain your answer.
362
19.4
a. What mass of zinc will you need for the reaction, if all the sulphur is
to be used up?
b. Calculate the theoretical yield for this reaction.
c. It is found that 275 g of zinc sulphide was produced. Calculate the %
yield.
17. Calcium chloride reacts with carbonic acid to produce calcium carbonate
and hydrochloric acid according to the following equation:
CaCl2 + H2 CO3 CaCO3 + 2HCl
More practice
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(1.) 0088
(2.) 0089
(3.) 008a
(4.) 008b
(5.) 008c
(6.) 008d
(7.) 008e
(8.) 008f
(9.) 008g
(10.) 008h
(11.) 008i
(12.) 008j
(13.) 008k
(14.) 008m
(15.) 008n
(16.) 008p
(17.) 008q
363
20
Introduction to vectors
and scalars
ESAGH
We come into contact with many physical quantities in the natural world on a daily basis.
For example, things like time, mass, weight, force, and electric charge, are physical quantities with which we are all familiar. We know that time passes and physical objects have
mass. Things have weight due to gravity. We exert forces when we open doors, walk along
the street and kick balls. We experience electric charge directly through static shocks in
winter and through using anything which runs on electricity.
There are many physical quantities in nature, and we can divide them up into two broad
groups called vectors and scalars. See introductory video: ( Video: VPgao at www.everythingscience.co
ESAGI
Scalars are physical quantities which have only a number value or a size (magnitude). A
scalar tells you how much of something there is.
DEFINITION: Scalar
A scalar is a physical quantity that has only a magnitude (size).
For example, a person buys a tub of margarine which is labelled with a mass of 500 g. The
mass of the tub of margarine is a scalar quantity. It only needs one number to describe it,
in this case, 500 g.
Vectors are different because they are physical quantities which have a size and a direction.
A vector tells you how much of something there is and which direction it is in.
364
Physics: Mechanics
20.1
DEFINITION: Vector
A vector is a physical quantity that has both a magnitude and a direction.
For example, a car is travelling east along a freeway at 100 km h1 . What we have here is
a vector called the velocity. The car is moving at 100 km h1 (this is the magnitude) and
we know where it is going east (this is the direction). These two quantities, the speed and
direction of the car, (a magnitude and a direction) together form a vector we call velocity.
Examples of scalar quantities:
mass has only a value, no direction
electric charge has only a value, no direction
Examples of vector quantities:
force has a value and a direction. You push or pull something with some strength
(magnitude) in a particular direction
weight has a value and a direction. Your weight is proportional to your mass (magnitude) and is always in the direction towards the centre of the earth.
Exercise 20 - 1
More practice
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(1.-7.) 008r
Physics: Mechanics
365
20.1
Vector notation
ESAGJ
Vectors are different to scalars and must have their own notation. There are many ways of
writing the symbol for a vector. In this book vectors will be shown by symbols with an
arrow pointing to the right above it. For example, F , W and v represent the vectors of
force, weight and velocity, meaning they have both a magnitude and a direction.
Sometimes just the magnitude of a vector is needed. In this case, the arrow is omitted. For
the case of the force vector:
Graphical
representation of vectors
ESAGK
Vectors are drawn as arrows. An arrow has both a magnitude (how long it is) and a direction
(the direction in which it points). The starting point of a vector is known as the tail and the
end point is known as the head.
See video: VPgdw at www.everythingscience.co.za
magnitude
tail
head
Figure 20.3: Parts of a vector
366
Physics: Mechanics
20.1
Directions
ESAGL
There are many acceptable methods of writing vectors. As long as the vector has a magnitude and a direction, it is most likely acceptable. These different methods come from the
different methods of representing a direction for a vector.
Relative Directions
The simplest way to show direction is with relative directions: to the left, to the right,
forward, backward, up and down.
Compass Directions
Another common method of expressing directions is to use the points of a compass: North,
South, East, and West. If a vector does not point exactly in one of the compass directions,
then we use an angle. For example, we can have a vector pointing 40 North of West. Start
with the vector pointing along the West direction (look at the dashed arrow below), then
rotate the vector towards the north until there is a 40 angle between the vector and the
West direction (the solid arrow below). The direction of this vector can also be described
as: W 40 N (West 40 North); or N 50 W (North 50 West).
40
Bearing
A further method of expressing direction is to use a bearing. A bearing is a direction relative
to a xed point. Given just an angle, the convention is to dene the angle clockwise with
respect to North. So, a vector with a direction of 110 has been rotated clockwise 110
relative to North. A bearing is always written as a three digit number, for example 275 or
080 (for 80 ).
110
Physics: Mechanics
367
20.1
Exercise 20 - 2
d. 075 , 2 cm
1
e. 100 k h
, 0
2. Use two different notations to write down the direction of the vector in
each of the following diagrams:
a.
c.
b.
40
60
More practice
(1.) 008s
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(2.) 008t
Drawing Vectors
ESAGM
In order to draw a vector accurately we must represent its magnitude properly and include
a reference direction in the diagram. A scale allows us to translate the length of the arrow
into the vectors magnitude. For instance if one chooses a scale of 1 cm = 2 N (1 cm
represents 2 N), a force of 20 N towards the East would be represented as an arrow 10 cm
long pointing towards the right. The points of a compass are often used to show direction
or alternatively an arrow pointing in the reference direction.
20 N
Physics: Mechanics
20.1
SOLUTION
N
North will point up the page
Step 3 : Determine the length of the arrow at the specic scale.
If 1 cm = 2 m s1 , then 6 m s1 = 3 cm
Step 4 : Draw the vector as an arrow.
Scale used: 1 cm = 2 m s1
6 m s1
Physics: Mechanics
369
20.1
SOLUTION
N
16 m
Exercise 20 - 3
Draw each of the following vectors to scale. Indicate the scale that you have
used:
1. 12 km south
2. 1,5 m N 45 W
3. 1 m s1 , 20 East of North
More practice
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4. 50 km h1 , 085
5. 5 mm, 225
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(1.-5.) 008u
370
Physics: Mechanics
20.2
Properties of vectors
ESAGN
Vectors are mathematical objects and we will now study some of their mathematical properties.
If two vectors have the same magnitude (size) and the same direction, then we call them
equal to each other. For example, if we have two forces, F1 = 20 N in the upward direction
Just like scalars which can have positive or negative values, vectors can also be positive
or negative. A negative vector is a vector which points in the direction opposite to the
reference positive direction. For example, if in a particular situation, we dene the upward
F1 =
indicates that the direction of F1 is opposite to that of the reference positive direction.
Like scalars, vectors can also be added and subtracted. We will investigate how to do this
next.
Physics: Mechanics
371
20.2
ESAGO
When vectors are added, we need to take into account both their magnitudes and directions.
For example, imagine the following. You and a friend are trying to move a heavy box. You
stand behind it and push forwards with a force F1 and your friend stands in front and pulls
it towards them with a force F2 . The two forces are in the same direction (i.e. forwards)
and so the total force acting on the box is:
F1
F2
FT ot =F1 + F2
It is very easy to understand the concept of vector addition through an activity using the
displacement vector.
Displacement is the vector which describes the change in an objects position. It is a vector
that points from the initial position to the nal position.
Activity:
Adding vectors
372
Physics: Mechanics
20.2
total. Therefore, if we add the displacement vectors for 2 steps and 3 steps, we
should get a total of 5 steps in the forward direction.
2. It does not matter whether you take 3 steps forward and then 2 steps forward,
or two steps followed by another 3 steps forward. Your nal position is the
same! The order of the addition does not matter!
We can represent vector addition graphically, based on the activity above. Draw the vector
for the rst two steps forward, followed by the vector with the next three steps forward.
2 steps
3 steps
=
5 steps
We add the second vector at the end of the rst vector, since this is where we now are after
the rst vector has acted. The vector from the tail of the rst vector (the starting point) to
the head of the second vector (the end point) is then the sum of the vectors.
As you can convince yourself, the order in which you add vectors does not matter. In the
example above, if you decided to rst go 3 steps forward and then another 2 steps forward,
the end result would still be 5 steps forward.
Subtracting vectors
Lets go back to the problem of the heavy box that you and your friend are trying to move.
If you didnt communicate properly rst, you both might think that you should pull in your
own directions! Imagine you stand behind the box and pull it towards you with a force
F1 and your friend stands at the front of the box and pulls it towards them with a force
F2 . In this case the two forces are in opposite directions. If we dene the direction your
friend is pulling in as positive then the force you are exerting must be negative since it is
in the opposite direction. We can write the total force exerted on the box as the sum of the
individual forces:
F1
F2
FT ot
F2 +( F1 )
F2 F1
What you have done here is actually to subtract two vectors! This is the same as adding
two vectors which have opposite directions.
As we did before, we can illustrate vector subtraction nicely using displacement vectors. If
you take 5 steps forward and then subtract 3 steps forward you are left with only two steps
forward:
5 steps
Physics: Mechanics
3 steps
2 steps
373
20.2
What did you physically do to subtract 3 steps? You originally took 5 steps forward but
then you took 3 steps backward to land up back with only 2 steps forward. That backward
displacement is represented by an arrow pointing to the left (backwards) with length 3. The
net result of adding these two vectors is 2 steps forward:
5 steps
Tip
3 steps
2 steps
Subtracting a vector
Thus, subtracting a vector from another is the same as adding a vector in the opposite
direction (i.e. subtracting 3 steps forwards is the same as adding 3 steps backwards).
We can illustrate the concept of the resultant vector by considering our two situations in
using forces to move the heavy box. In the rst case (on the left), you and your friend
are applying forces in the same direction. The resultant force will be the sum of your two
applied forces in that direction. In the second case (on the right), the forces are applied in
opposite directions. The resultant vector will again be the sum of your two applied forces,
however after choosing a positive direction, one force will be positive and the other will be
negative and the sign of the resultant force will just depend on which direction you chose
as positive. For clarity look at the diagrams below.
Forces are applied in opposite directions
Forces are applied in the same direction
(positive direction to the right)
F2
F1
(20 N)
(15 N)
F2
(15 N)
F1
(20 N)
F2 +(F2 )
20 N + 15 N
35 N to the right
F2 F 1
15 N 20 N
FR
5 N
F1 + F2
=
=
FR
=
374
5 N to the left
Physics: Mechanics
20.3
FR
(35 N)
FR
(5 N)
There is a special name for the vector which has the same magnitude as the resultant
vector but the opposite direction: the equilibrant. If you add the resultant vector and the
equilibrant vectors together, the answer is always zero because the equilibrant cancels the
resultant out.
DEFINITION: Equilibrant
The equilibrant is the vector which has the same magnitude but opposite
direction to the resultant vector.
If you refer to the pictures of the heavy box before, the equilibrant forces for the two
situations would look like:
FE
(35 N)
FE
FR
(35 N)
=
=
FR
35 N to the left
Techniques of vector
addition
FR
(5N)
FE
=
=
FE
(5N)
FR
5 N to the right
ESAGP
Now that you have learned about the mathematical properties of vectors, we return to
vector addition in more detail. There are a number of techniques of vector addition. These
techniques fall into two main categories - graphical and algebraic techniques.
Graphical techniques
ESAGQ
Graphical techniques involve drawing accurate scale diagrams to denote individual vectors
and their resultants. We will look at just one graphical method: the head-to-tail method.
Physics: Mechanics
375
20.3
1 step
1 step
2 steps
2 steps
This example says 1 step backward and then another step backward is the same as an arrow
twice as long two steps backward.
1 step
1 step
2 steps
2 steps
It is sometimes possible that you end up back where you started. In this case the net result
of what you have done is that you have gone nowhere (your start and end points are at the
same place). In this case, your resultant displacement is a vector with length zero units.
We use the symbol 0 to denote such a vector:
1 step
1 step
1 step
1 step
1 step
1 step
1 step
1 step
=0
=0
Check the following examples in the same way. Arrows up the page can be seen as steps
left and arrows down the page as steps right.
376
Physics: Mechanics
20.3
+ = =
+ =
+ = =
= 0
+ =
= 0
It is important to realise that the directions are not special forward and backwards or left
and right are treated in the same way. The same is true of any set of parallel directions:
= 0
= 0
In the above examples the separate displacements were parallel to one another. However the same head-to-tail technique of vector addition can be applied to vectors in any
direction.
SOLUTION
50 N
45 N
Physics: Mechanics
377
20.3
50 N
Step 4 : Take the next vector and draw it starting at the arrowhead of the
previous vector.
Since your friend is pushing in the same direction as you, your force
vectors must point in the same direction. Using the scale, this arrow
should be 4,5 cm long.
50 N
45 N
Step 5 : Draw the resultant, measure its length and nd its direction
There are only two vectors in this problem, so the resultant vector must
be drawn from the tail (i.e. starting point) of the rst vector to the head
of the second vector.
50 N + 45 N = 95 N
The resultant vector measures 9,5 cm and points to the right. Therefore
the resultant force must be 95 N in the positive direction (or to the right).
player if two players on his team are pushing him forwards with forces of F1 = 60 N
and F2 = 90 N respectively and two players from the opposing team are pushing
SOLUTION
Physics: Mechanics
20.3
We will start with drawing the vector F1 = 60 N, pointing in the positive direction. Using our scale of 0,5 cm = 10 N, the length of the
arrow must be 3 cm pointing to the right.
F1 = 60 N
Step 3 : Take the next vector and draw it starting at the arrowhead of the
previous vector
F1 = 60 N
F2 = 90 N
Step 4 : Take the next vector and draw it starting at the arrowhead of the
previous vector
F1 = 60 N
F2 = 90 N
F3 = 100 N
Step 5 : Take the next vector and draw it starting at the arrowhead of the
previous vector
F1 = 60 N
F4 = 65 N
F2 = 90 N
F3 = 100 N
Step 6 : Draw the resultant, measure its length and nd its direction
We have now drawn all the force vectors that are being applied to the
player. The resultant vector is the arrow which starts at the tail of the
rst vector and ends at the head of the last drawn vector.
FR = 15 N
F1 = 60 N
F4 = 65 N
Physics: Mechanics
F2 = 90 N
F3 = 100 N
379
20.3
Algebraic techniques
ESAGR
SOLUTION
380
Physics: Mechanics
20.3
10 m
2,5 m
Wall
Start
Step 2 : Decide which method to use to calculate the resultant
We know that the resultant displacement of the ball (xR ) is equal to the
sum of the balls separate displacements (x1 and x2 ):
xR
x1 + x2
Since the motion of the ball is in a straight line (i.e. the ball moves
towards and away from the wall), we can use the method of algebraic
addition just explained.
Step 3 : Choose a positive direction
Lets choose the positive direction to be towards the wall. This means
that the negative direction is away from the wall.
Step 4 : Now dene our vectors algebraically
With right positive:
x1
+10, 0 m
x2
2, 5 m
(+10 m) + (2, 5 m)
(+7, 5) m
Physics: Mechanics
381
20.3
SOLUTION
3 m s1
2 m s1
Wall
Start
Step 2 : Decide which method to use to calculate the resultant
Remember that velocity is a vector. The change in the velocity of the
ball is equal to the difference between the balls initial and nal velocities:
v = vf vi
Since the ball moves along a straight line (i.e. left and right), we
can use the algebraic technique of vector subtraction just discussed.
Step 3 : Choose a positive direction
Choose the positive direction to be towards the wall. This means that
the negative direction is away from the wall.
Step 4 : Now dene our vectors algebraically
vi
vf
+3 m s1
2 m s1
382
Physics: Mechanics
=
=
20.3
(2 m s1 ) (+3 m s1 )
(5) m s1
SOLUTION
5N
2N
Step 2 : Decide which method to use to calculate the resultant
Remember that force is a vector. Since the crate moves along a straight
line (i.e. left and right), we can use the algebraic technique of vector
addition just discussed.
Step 3 : Choose a positive direction
Choose the positive direction to be towards the crate (i.e. in the same
direction that the man is pushing). This means that the negative direction is away from the crate (i.e. against the direction that the man is
pushing).
Step 4 : Now dene our vectors algebraically
Physics: Mechanics
383
20.3
Fman
+5 N
Fcrate
2 N
(5 N) + (2 N)
7N
Fman + Fcrate
Remember that the technique of addition and subtraction just discussed can only be applied to vectors acting along a straight line. When vectors are not in a straight line, i.e. at
an angle to each other then simple geometric and trigonometric techniques can be used to
nd resultant vectors.
Exercise 20 - 4
384
Physics: Mechanics
20.3
5. Mpihlonhle walks to the shop by walking 500 m Northwest and then 400
m N 30 E. Determine her resultant displacement by doing appropriate
calculations.
More practice
(1.) 008v
video solutions
(2.) 008w
(3.) 008x
(4.) 008y
or help at www.everythingscience.co.za
(5.) 008z
Chapter 20 | Summary
See the summary presentation (
point (e.g. 30 from the river bank); using a compass (e.g. N 30 W); or bearing (e.g.
053 ).
The resultant vector is the single vector whose effect is the same as the individual
vectors acting together.
Chapter 20
385
20.3
More practice
(1.) 0092
386
(2.) 0094
video solutions
(3.) 0095
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(4.) 0096
Physics: Mechanics
Motion in one
dimension
21
Introduction
ESAGS
FACT
the
changing.
are:
Reference frame
ESAGT
The rst thing to focus on when studying motion of an object or person is their position.
The word position describes your location (where you are). However, saying that you
are here or there is meaningless, you have to use known points (reference points) to help
specify your position.
Physics: Mechanics
387
acceleration
is
21.2
This still
negative () direction
origin
You can dene different frames of reference for the same problem but the outcome, the
physical results, will be the same. For example, a boy is standing still inside a train as it
pulls out of a station. Both you and the boy dene your location as the point of reference
and the direction train is moving as a where you are standing as the point of reference and
the direction the train is moving in as forward.
388
Physics: Mechanics
21.2
You are standing on the platform watching the train move from left to right. To you it looks
as if the boy is moving from left to right, because relative to where you are standing (the
platform), he is moving. According to the boy, and his frame of reference (the train), he is
not moving.
A frame of reference must have an origin (where you are standing on the platform) and
at least a positive direction. The train was moving from left to right, making to your right
positive and to your left negative. If someone else was looking at the same boy, his frame
of reference will be different. For example, if he was standing on the other side of the
platform, the boy will be moving from right to left.
boy is standing still
From your frame of reference the boy is moving from left to right.
For this chapter, we will only use frames of reference in the x-direction. By doing this we
restrict ourselves to one dimensional motion. We can use the sign of the position value
(positive or negative) to indicate the direction relative to the origin.
For example the blue dot in the gure below can only move along the x-axis.
origin
Physics: Mechanics
389
21.2
Position
ESAGU
DEFINITION: Position
Position is a measurement of a location, with reference to an origin.
Quantity: Position (x)
Unit symbol: m
A position is a measurement of a location within a reference frame. This means that positions can be negative or positive depending on the choice for the reference frames coordinate system.
Depending on which reference point we choose, we can say that the school is 300 m from
Kosmas house (with Kosmas house as the reference point or origin) or 500 m from Kevins
house (with Kevins house as the reference point or origin).
School
Komal
100 m
100 m
Kogis
Kosma
Kholo
100 m
100 m
Shop
Kevin
100 m
100 m
The shop is also 300 m from Kosmas house, but in the opposite direction as the school.
When we choose a reference point, we have a positive direction and a negative direction.
If we choose the direction towards the school as negative, then the direction towards the
shop is positive. A negative direction is always opposite to the direction chosen as positive.
School
Kosmas house
(reference point)
Shop
x (m)
The origin is at Kosmas house and the position of the school is 300 m. Positions towards
the left are dened as negative and positions towards the right are dened as positive.
Note that we could also choose the positive direction to be towards the school. In this case
Kosmas house is still 300 m away from the school, but it is now in the positive direction.
390
Physics: Mechanics
21.2
Kosmas house
(reference point)
School
Shop
x (m)
The origin is at Kosmas house and the position of the school is +300 m. Positions towards
the left are dened as positive and positions towards the right are dened as negative.
Exercise 21 - 1
1. Write down the positions for objects at A, B, D and E. Do not forget the
units.
reference point
B
D
A
-4
-3
-2
-1
E
2
x (m)
2. Write down the positions for objects at F, G, H and J. Do not forget the
units.
Physics: Mechanics
391
21.2
reference point
G
H
F
4
-1
J
-2
-3
x (m)
-4
20 m
20 m
20 m
20 m
a. Draw a frame of reference with house A as the origin and write down
the positions of houses B, C, D and E.
b. You live in house C. What is your position relative to house E?
c. What are the positions of houses A, B and D, if house B is taken as
the reference point?
More practice
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(2.) 009h
video solutions
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(3.) 009i
ESAGV
DEFINITION: Distance
Tip
The symbol is read
out as delta.
Distance is the total length of the path taken in going from the initial posi-
is
Unit symbol: m
in Mathematics and
Science to indicate a
change in a certain
quantity, or a nal
value minus an initial
value.
For example,
In the simple map below you can see the path that winds because of a number of hills from
a school to a nearby shop. The path is shown by a dashed line. The initial point, xi , is the
school and the nal point, xf , is the shop.
x means change in
x while t means
change in t.
392
Physics: Mechanics
21.2
Start
(School)
Finish
(Shop)
DEFINITION: Displacement
Displacement is the change in an objects position. It is a vector that points
from the initial position (xi ) to the nal position (xf ).
Quantity: Displacement (x)
Unit symbol: m
The displacement of an object is dened as its change in position (nal position minus
initial position). Displacement has a magnitude and direction and is therefore a vector. For
example, if the initial position of a car is xi and it moves to a nal position of xf , then the
displacement is:
x = xf xi
To help visualise what the displacement vector looks like think back to the tail-to-head
method. The displacement is the vector you add to the initial position vector to get a vector
to the nal position. However, subtracting an initial quantity from a nal quantity happens
often in Physics, so we use the shortcut to mean nal - initial. Therefore, displacement
can be written:
x = xf xi
The following diagram illustrates the concept of displacement:
xi
Tip
xf
For example, if you roll a ball 5 m along a oor, in a straight line, then its displacement is
5 m, taking the direction of motion as positive, and the initial position as 0 m.
Displacement does not depend on the path travelled, but only on the initial and nal
positions. We use the word distance to describe how far an object travels along a particular
path.
Physics: Mechanics
393
21.2
Tip
We will use D in this
book, but you may
see d used in other
books.
Tip
We use the expres-
If we go back to the simple map repeated below you can see the path as before shown by
a dashed line.
Distance is the length of dashed line. The displacement is different. Displacement is the
straight-line distance from the starting point
to the endpoint from the school to the shop
Start
(School)
in the gure as shown by the solid arrow.
See video: VPgmo at www.everythingscience.co.za
Finish
(Shop)
If we use the same situation as earlier we can explore the concepts in more detail. Consider
our description of the location of the houses, school and the shop.
School
Komal
100 m
100 m
Kogis
Kosma
Kholo
100 m
100 m
Shop
Kevin
100 m
100 m
Komal walks to meet Kevin at his house before walking to school. What is Komals displacement and what distance did he cover if he walks to school via Kevins house?
Komal covers a distance of 400 m to Kevins house and another 500 m from Kevins house
to the school. He covers a total distance of 900 m. His displacement, however, is only
100 m towards the school. This is because displacement only looks at the starting position
(his house) and the end position (the school). It does not depend on the path he travelled.
To calculate his distance and displacement, we need to choose a reference point and a
direction. Lets choose Komals house as the reference point, and towards Kevins house as
the positive direction (which means that towards the school is negative). We would do the
calculations as follows:
=
Displacement(x)
path travelled
400 m + 500 m
Distance (D)
900 m
=
=
xf xi
100 m + 0 m
100 m
Very often in calculations you will get a negative answer. For example, Komals displacement in the example above, is calculated as 100 m. The minus sign in front of the answer
means that his displacement is 100 m in the opposite direction (opposite to the direction
chosen as positive in the beginning of the question). When we start a calculation we
choose a frame of reference and a positive direction. In the rst example above, the reference point is Komals house and the positive direction is towards Kevins house. Therefore
Komals displacement is 100 m towards the school. Notice that distance has no direction,
394
Physics: Mechanics
21.2
Displacement(x)
path travelled
500 m + 500 m
0m+0m
1000 m
0m
xf xi
Displacement
2. always positive
3. is a scalar
3. is a vector
Exercise 21 - 2
Komal
100 m
100 m
Kogis
Kosma
Kholo
100 m
100 m
Shop
Kevin
100 m
100 m
a. Kogis walks to Kosmas house and then to school, what is her distance
and displacement?
b. Kholo walks to Kosmas house and then to school, what is her distance and displacement?
c. Komal walks to the shop and then to school, what is his distance and
displacement?
d. What reference point did you use for each of the above questions?
2. You stand at the front door of your house (displacement, x = 0 m). The
street is 10 m away from the front door. You walk to the street and back
again.
Physics: Mechanics
395
21.3
More practice
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video solutions
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(2.) 009k
ESAGW
Unit
Unit
Before moving on review the difference between distance and displacement. Sometimes
the average speed can be a very big number while the average velocity is zero.
See
video:
at www.everythingscience.co.za
Average velocity is the rate of change of position. It tells us how much an objects position
changes per unit of time. Velocity is a vector. We use the symbol vav for average velocity.
396
Physics: Mechanics
21.3
vav
Velocity can be positive or negative. A positive velocity points in the direction you chose
as positive in your coordinate system. A negative velocity points in the direction opposite
to the positive direction.
Average speed (symbol vav ) is the distance travelled (D) divided by the time taken (t) for
the journey. Distance and time are scalars and therefore speed will also be a scalar. Speed
is calculated as follows:
distance (in m)
time (in s)
D
t
2 km
SOLUTION
Step 1 : Identify what information is given and what is asked for
The question explicitly gives
the distance and time out (2 km in 30 minutes)
the distance and time back (2 km in 30 minutes)
397
21.3
1 000 m
2 km
2 000 m
30 min =
60s
1 800 s (multiply both sides by 30)
vav
=
=
=
D
t
4 000 m
3 600 s
1, 11 m s1
vav
=
=
=
398
x
t
0m
3 600 s
0 m s1
Physics: Mechanics
21.3
ESAGX
Velocity
2. always positive
3. is a scalar
3. is a vector
4. no dependence on direction
Additionally, an object that makes a round trip, i.e. travels away from its starting point and
then returns to the same point has zero velocity but travels at a non-zero speed.
Exercise 21 - 3
1. Bongani has to walk to the shop to buy some milk. After walking 100 m,
he realises that he does not have enough money, and goes back home. If
it took him two minutes to leave and come back, calculate the following:
a. How long was he out of the house (the time interval t in seconds)?
b. How far did he walk (distance (D))?
c. What was his displacement (x)?
d. What was his average velocity (in m s1 )?
e. What was his average speed (in m s1 )?
shop
home
Physics: Mechanics
399
21.3
x
t ),
right.
b. If Bridget measures the velocity of a red Golf to be 16, 67 m s1 , in
which direction was the Golf travelling? Bridget leaves her stopwatch
running, and notices that at t = 5, 0 s, a taxi passes the left pole at the
same time as a bus passes the right pole. At time t = 7, 5 s the taxi
passes the right pole. At time t = 9, 0 s, the bus passes the left pole.
c. How long did it take the taxi and the bus to travel the distance between the poles? (Calculate the time interval (t) for both the taxi
and the bus).
d. What was the average velocity of the taxi and the bus?
e. What was the average speed of the taxi and the bus?
f. What was the average speed of taxi and the bus in km h1 ?
50 m
3s
t=9s
t=5s
t = 7,5 s
t=5s
3m
car
3m
3m
100 m
a. If the car is travelling at 120 kmh1 , what is the cars speed in ms1 .
b. How long will it take the a car to travel 100 m?
c. If the rabbit is running at 10 km h1 , what is its speed in m s1 ?
400
Physics: Mechanics
21.3
d. If the freeway has 3 lanes, and each lane is 3 m wide, how long will
it take for the rabbit to cross all three lanes?
e. If the car is travelling in the furthermost lane from the rabbit, will the
rabbit be able to cross all 3 lanes of the freeway safely?
More practice
(1.) 009m
(2.) 009n
video solutions
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(3.) 009p
Does the colour of your car determine the speed you are travelling at?
Any other relevant question that you would like to investigate.
Discussion
Physics: Mechanics
401
21.4
Acceleration
ESAGY
Acceleration is a measure of how fast the velocity of an object changes in time. If we have
a change in velocity (v) over a time interval (t), then the average acceleration (aav ) is
dened as:
average acceleration (in m s2 ) =
aav =
v
t
We only deal with problems with constant acceleration. This means that the average acceleration and the instantaneous acceleration are the same. To make things simpler we will
402
Physics: Mechanics
21.4
only talk about acceleration and not average or instantaneous. This is represented as a. We
can also have the magnitude of the acceleration. This is:
a=
v
t
Acceleration is a vector. Acceleration does not provide any information about the motion,
but only about how the motion changes. It is not possible to tell how fast an object is
moving or in which direction from the acceleration alone.
Tip
Avoid the use of the
word deceleration to
refer to a negative ac-
acceleration and the velocity are the same, the object is speeding up. If both velocity and
acceleration are positive, the object is speeding up in a positive direction. If both velocity
and acceleration are negative, the object is speeding up in a negative direction. We can
see this in the following diagram:
a
speeding up
a
slowing down
deceleration
a
speeding up
negative acceleration
down or not.
If velocity is positive and acceleration is negative, then the object is slowing down. Similarly, if the velocity is negative and the acceleration is positive the object is slowing down.
This is illustrated in the following worked example.
Example 2: Acceleration
QUESTION
A car accelerates uniformly from and initial velocity of 2 ms1 to a nal velocity
of 10 ms1 in 8 seconds. It then slows down uniformly to a nal velocity of 4 ms1
in 6 seconds. Calculate the acceleration of the car during the rst 8 seconds and
during the last 6 seconds.
SOLUTION
Physics: Mechanics
403
21.4
vf
ti
tf
2 m s1
10 m s1
vi
vf
0s
ti
8s
8s
tf
14 s
10 m s1
4 m s1
v
t
v
t
10
4
=
8 s0 s
14 s 8 s
= 1 m s2
= 1 m s2
During the rst 8 seconds the car had a positive acceleration. This
=
means that its velocity increased. The velocity is positive so the car is
speeding up. During the next 6 seconds the car had a negative acceleration. This means that its velocity decreased. The velocity is positive so
the car is slowing down.
Exercise 21 - 4
1. An athlete is accelerating uniformly from an initial velocity of 0 ms1 to a
nal velocity of 4 ms1 in 2 seconds. Calculate his acceleration. Let the
direction that the athlete is running in be the positive direction.
2. A bus accelerates uniformly from an initial velocity of 15 ms1 to a nal
velocity of 7 ms1 in 4 seconds. Calculate the acceleration of the bus. Let
the direction of motion of the bus be the positive direction.
3. An aeroplane accelerates uniformly from an initial velocity of 100 ms1 to
a velocity of 200 ms1 in 10 seconds. It then accelerates uniformly to a
nal velocity of 240 ms1 in 20 seconds. Let the direction of motion of
the aeroplane be the positive direction.
a. Calculate the acceleration of the aeroplane during the rst 10 seconds
404
Physics: Mechanics
21.5
of the motion.
b. Calculate the acceleration of the aeroplane during the next 20 seconds of its motion.
More practice
(1.) 009q
(2.) 009r
video solutions
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(3.) 009s
Instantaneous velocity
and speed
Sprinters taking off
ESAGZ
Physics: Mechanics
405
Tip
An instant in time
is different from the
time taken or the time
interval.
It is there-
21.6
Description of motion
ESAHA
The purpose of this chapter is to describe motion, and now that we understand the denitions of displacement, distance, velocity, speed and acceleration, we are ready to start
using these ideas to describe how an object or person is moving. We will look at three
ways of describing motion:
1. words
2. diagrams
3. graphs
These methods will be described in this section.
We will consider three types of motion: when the object is not moving (stationary object),
when the object is moving at a constant velocity (uniform motion) and when the object is
moving at a constant acceleration (motion at constant acceleration).
Stationary Object
ESAHB
The simplest motion that we can come across is that of a stationary object. A stationary
object does not move and so its position does not change.
406
Physics: Mechanics
21.6
Vivian stands at a stop sign.
1
0
60
120
time (s)
(a)
60
(b)
120
time (s)
acceleration a (ms2 )
velocity v (m s1 )
position x (m)
60
120
(c)
Figure 21.2: Graphs for a stationary object (a) position vs. time (b) velocity vs. time (c)
acceleration vs. time.
Vivians position is 2 metres in the positive direction from the stop street. If the stop street is
taken as the reference point, her position remains at 2 metres for 120 seconds. The graph
is a horizontal line at 2 m. The velocity and acceleration graphs are also shown. They
are both horizontal lines on the x-axis. Since her position is not changing, her velocity is
0 m s1 and since velocity is not changing, acceleration is 0 m s2 .
DEFINITION: Gradient
(Recall from Mathematics) The gradient, m, of a line can be calculated by
dividing the change in the y-value (dependent variable) by the change in the
x-value (independent variable). m =
Physics: Mechanics
y
x
407
time (s)
21.6
Tip
The gradient of a position vs. time graph
gives the average ve-
Since we know that velocity is the rate of change of position, we can conrm the value for
the velocity vs. time graph, by calculating the gradient of the x vs. t graph.
If we calculate the gradient of the x vs. t graph for a stationary object we get:
x
t
xf xi
=
tf ti
2m2m
=
(initial position = nal position)
120 s 60 s
v=
Tip
The gradient of a velocity vs. time graph
gives the average acceleration, while the
tangent of a velocity
vs. time graph gives
Similarly, we can conrm the value of the acceleration by calculating the gradient of the
velocity vs. time graph.
If we calculate the gradient of the v vs. t graph for a stationary object we get:
v
t
vf vi
=
tf ti
a=
0 m s1 0 m s1
120 s 60 s
= 0 m s2
Tip
Additionally, because the velocity vs. time graph is related to the position vs. time graph,
vs.
time
gives
the
displacement.
we can use the area under the velocity vs. time graph to calculate the displacement of an
object.
The displacement of the object is given by the area under the graph, which is 0 m. This is
obvious, because the object is not moving.
ESAHC
Motion at a constant velocity or uniform motion means that the position of the object is
changing at the same rate.
Assume that Vivian takes 100 s to walk the 100 m to the taxi-stop every morning. If we
assume that Vivians house is the origin and the direction to the taxi is positive, then
408
Physics: Mechanics
21.6
v=
= 1 m s1
Vivians velocity is 1 m s1 . This means that she walked 1 m in the rst second, another
metre in the second second, and another in the third second, and so on. For example, after
t=0s
x=0m
t = 50 s
x = 50 m
t = 100 s
x = 100 m
We can now draw graphs of position vs.time (x vs. t), velocity vs. time (v vs. t) and acceleration vs.time (a vs. t) for Vivian moving at a constant velocity. The graphs are shown
here:
80
60
x
40
20
acceleration a (ms2 )
velocity v (ms1 )
position x (m)
100
0
0
20
40
60
time t (s)
80
100
0
0
20
40
60
time t (s)
80
100
20
40
60
80
100
time t (s)
Graphs for motion at constant velocity (a) position vs. time (b) velocity vs. time (c) acceleration vs. time. The area of the shaded portion in the v vs. t graph corresponds to the
objects displacement.
In the evening Vivian walks 100 m from the bus stop to her house in 100 s. Assume that
Vivians house is the origin. The following graphs can be drawn to describe the motion.
Physics: Mechanics
409
time t (s)
t
80
20
60
x
40
40
60
80
100
velocity v (ms1 )
position x (m)
100
20
0
0
20
40
60
time t (s)
80
100
acceleration a (ms2 )
21.6
0
0
20
40
60
80
100
time t (s)
Graphs for motion with a constant negative velocity. The area of the shaded portion in the
v vs.t graph corresponds to the objects displacement.
We see that the v vs. t graph is a horizontal line. If the velocity vs. time graph is a horizontal
line, it means that the velocity is constant (not changing). Motion at a constant velocity is
known as uniform motion. We can use the x vs. t to calculate the velocity by nding the
gradient of the line.
x
t
xf xi
=
tf t i
0 m 100 m
=
100 s 0 s
v=
= 1 m s1
Vivian has a velocity of 1 m s1 , or 1 m s1 towards her house. You will notice that the
v vs. t graph is a horizontal line corresponding to a velocity of 1 m s1 . The horizontal
line means that the velocity stays the same (remains constant) during the motion. This is
uniform velocity.
We can use the v vs. t to calculate the acceleration by nding the gradient of the line.
v
t
vf vi
=
tf ti
a=
1 m s1 1 m s1
100 s 0 s
= 0 m s2
Vivian has an acceleration of 0 ms2 . You will notice that the graph of a vs. t is a horizontal
line corresponding to an acceleration value of 0 m s2 . There is no acceleration during
the motion because his velocity does not change.
We can use the v vs. t graph to calculate the displacement by nding the area under the
graph.
410
Physics: Mechanics
21.6
Exercise 21 - 5
More practice
(1.) 009t
video solutions
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(2.) 009u
Physics: Mechanics
411
21.6
Aim:
To measure the position and time during motion at constant velocity and determine
the average velocity as the gradient of a Position vs. Time graph.
Apparatus:
A battery operated toy car, stopwatch, meter stick or measuring tape.
Method:
1. Work with a friend. Copy the table below into your workbook.
2. Complete the table by timing the car as it travels each distance.
3. Time the car twice for each distance and take the average value as your accepted time.
4. Use the distance and average time values to plot a graph of Distance vs.
Time onto graph paper. Stick the graph paper into your workbook. (Remember that A vs. B always means y vs. x).
5. Insert all axis labels and units onto your graph.
6. Draw the best straight line through your data points.
7. Find the gradient of the straight line. This is the average velocity.
Results:
Time (s)
Distance (m)
1
Ave.
0
0,5
1,0
1,5
2,0
2,5
3,0
Conclusions:
Answer the following questions in your workbook:
1. Did the car travel with a constant velocity?
412
Physics: Mechanics
21.6
2. How can you tell by looking at the Distance vs. Time graph if the velocity
is constant?
3. How would the Distance vs. Time graph look for a car with a faster velocity?
4. How would the Distance vs. Time graph look for a car with a slower velocity?
ESAHD
The nal situation we will be studying is motion at constant acceleration. We know that
acceleration is the rate of change of velocity. So, if we have a constant acceleration, this
means that the velocity changes at a constant rate.
Lets look at our rst example of Vivian waiting at the taxi stop again. A taxi arrived and
Vivian got in. The taxi stopped at the stop street and then accelerated in the positive
direction as follows: After 1 s the taxi covered a distance of 2, 5 m, after 2 s it covered
10 m, after 3 s it covered 22, 5 m and after 4 s it covered 40 m. The taxi is covering a larger
distance every second. This means that it is accelerating.
STOP
2,5 m
t=1s
10 m
t=2s
22,5 m
t=3s
40 m
t=4s
To calculate the velocity of the taxi you need to calculate the gradient of the line at each
second:
x
t
xf xi
=
tf ti
5m0m
=
1, 5 s 0, 5 s
v1s =
= 5 m s1
Physics: Mechanics
413
21.6
x
t
xf xi
=
tf ti
15 m 5 m
=
2, 5 s 1, 5 s
v2s =
= 10 m s1
x
t
xf xi
=
tf ti
30 m 15 m
=
3, 5 s 2, 5 s
v3s =
= 15 m s1
From these velocities, we can draw the velocity-time graph which forms a straight line.
The acceleration is the gradient of the v vs. t graph and can be calculated as follows:
v
t
vf vi
=
tf t i
a=
15 m s1 5 m s1
3s1s
= 5 m s2
The acceleration does not change during the motion (the gradient stays constant). This is
motion at constant or uniform acceleration.
20
15
10
5
15
10
0
0
414
acceleration a (ms2 )
velocity v (ms1 )
position x (m)
9
8
7
6
5
4
3
2
1
0
0
time t (s)
time t (s)
time t (s)
Graphs for motion with a constant acceleration starting from rest.
Physics: Mechanics
21.6
= 5 m s2 2 s
= 10 m s1
Summary of Graphs
ESAHE
The relation between graphs of position, velocity and acceleration as functions of time is
summarised in the next gure.
x (m)
v (m s1 )
a (ms2 )
Stationary
object
x (m)
t (s)
v (m s1 )
t (s)
t (s)
a (ms2 )
Uniform
motion
x (m)
t (s)
v (m s1 )
t (s)
t (s)
a (ms2 )
Motion
with
con-
stant acceleration
t (s)
t (s)
t (s)
Physics: Mechanics
415
21.6
Tip
The description of the
motion
represented
by a graph should
You will also often be required to draw graphs based on a description of the motion in
words or from a diagram. Remember that these are just different methods of presenting the
same information. If you keep in mind the general shapes of the graphs for the different
types of motion, there should not be any difculty with explaining what is happening.
the
object
moving
is
in
the
or
timer
positive
negative
direction
2. whether
object
rest,
at
the
is
Aim:
To measure the position and time during motion and to use that data
at
moving
constant
velocity
Apparatus:
Trolley, ticker tape apparatus, tape, graph paper, ruler, ramp
or
moving at constant
positive
acceleration
(speeding
or
up)
constant
negative
ac-
celeration
(slowing down)
Method:
1. Work with a friend. Copy the table below into your workbook.
2. Attach a length of tape to the trolley.
3. Run the other end of the tape through the ticker timer.
4. Start the ticker timer going and roll the trolley down the ramp.
5. Repeat steps 1 - 3.
6. On each piece of tape, measure the distance between successive
dots. Note these distances in the table below.
7. Use the frequency of the ticker timer to work out the time intervals
between successive dots. Note these times in the table below,
8. Work out the average values for distance and time.
9. Use the average distance and average time values to plot a graph
of Distance vs. Time onto graph paper. Stick the graph paper
into your workbook. (Remember that A vs. B always means y
vs. x).
10. Insert all axis labels and units onto your graph.
11. Draw the best straight line through your data points.
Results:
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Physics: Mechanics
21.6
Distance (m)
1
Ave.
Time (s)
1
Ave.
Discussion:
Describe the motion of the trolley down the ramp.
Worked examples
ESAHF
The worked examples in this section demonstrate the types of questions that can be asked
about graphs.
Physics: Mechanics
417
21.6
5
4
3
2
1
t (s)
0
0
SOLUTION
Step 1 : Identify what information is given and what is asked for
The question gives a position vs. time graph and the following three
things are required:
1. Draw a v vs. t graph.
2. Draw an a vs. t graph.
3. Describe the motion of the car.
To answer these questions, break the motion up into three sections: 0
2 seconds, 2 4 seconds and 4 6 seconds.
Step 2 : Velocity vs. time graph for 0 2 seconds
For the rst 2 seconds we can see that the position (and hence the
displacement) remains constant - so the object is not moving, thus it has
zero velocity during this time. We can reach this conclusion by another
path too: remember that the gradient of a displacement vs. time graph
is the velocity. For the rst 2 seconds we can see that the displacement
vs. time graph is a horizontal line, i.e.. it has a gradient of zero. Thus
the velocity during this time is zero and the object is stationary.
Step 3 : Velocity vs. time graph for 2 4 seconds
For the next 2 seconds, displacement is increasing with time so the
object is moving. Looking at the gradient of the displacement graph
we can see that it is not constant. In fact, the slope is getting steeper
(the gradient is increasing) as time goes on. Thus, remembering that the
gradient of a displacement vs. time graph is the velocity, the velocity
must be increasing with time during this phase.
Step 4 : Velocity vs. time graph for 4 6 seconds
For the nal 2 seconds we see that displacement is still increasing with
time, but this time the gradient is constant, so we know that the object
418
Physics: Mechanics
21.6
is now travelling at a constant velocity, thus the velocity vs. time graph
will be a horizontal line during this stage. We can now draw the graphs:
So our velocity vs. time graph looks like this one below. Because
we havent been given any values on the vertical axis of the displacement vs. time graph, we cannot gure out what the exact gradients are
and therefore what the values of the velocities are. In this type of question it is just important to show whether velocities are positive or negative, increasing, decreasing or constant.
v (m s1 )
t (s)
0
Once we have the velocity vs. time graph its much easier to get the
acceleration vs. time graph as we know that the gradient of a velocity
vs. time graph is the just the acceleration.
Step 5 : Acceleration vs. time graph for 0 2 seconds
For the rst 2 seconds the velocity vs. time graph is horizontal and has
a value of zero, thus it has a gradient of zero and there is no acceleration during this time. (This makes sense because we know from the
displacement time graph that the object is stationary during this time,
so it cant be accelerating).
Step 6 : Acceleration vs. time graph for 2 4 seconds
For the next 2 seconds the velocity vs. time graph has a positive gradient. This gradient is not changing (i.e. its constant) throughout these 2
seconds so there must be a constant positive acceleration.
Step 7 : Acceleration vs. time graph for 4 6 seconds
For the nal 2 seconds the object is travelling with a constant velocity.
During this time the gradient of the velocity vs. time graph is once again
zero, and thus the object is not accelerating. The acceleration vs. time
graph looks like this:
a (ms2 )
t (s)
Physics: Mechanics
419
21.6
0
1
10
11
12
13
14
15
1
2
SOLUTION
Step 1 : Decide how to tackle the problem
We are asked to calculate the distance and displacement of the car.
All we need to remember here is that we can use the area between the
velocity vs. time graph and the time axis to determine the distance and
displacement.
Step 2 : Determine the area under the velocity vs. time graph
Break the motion up: 0 5 seconds, 5 12 seconds, 12 14 seconds
and 14 15 seconds.
420
Physics: Mechanics
1
bh
2
1
5 s 4 m s1
2
10 m
=
=
=
1
bh
2
1
2 s 4 m s1
2
4m
=
=
=
21.6
For 5 12 seconds: The displacement is equal to the area of
the rectangle:
Area
=
=
=
7 s 4 m s1
28 m2
=
=
=
1
bh
2
1
1 s 2 m s1
2
1m
10 m + 28 m + 4 m + 1 m
43 m
=
=
Physics: Mechanics
10 m + 28 m + 4 m 1 m
421
21.6
0
0
SOLUTION
=
=
=
x
t
4m0m
4s0s
1 m s1
422
Physics: Mechanics
21.6
0
1
1
2
SOLUTION
Step 1 : Calculate the velocity values by using the area under each part of the
graph.
The motion of the car can be divided into three time sections: 0 2
seconds; 2 4 seconds and 4 6 seconds. To be able to draw the
velocity vs. time graph, the velocity for each time section needs to be
calculated. The velocity is equal to the area of the square under the
graph:
For 0 2 seconds:
For 2 4 seconds:
Area
Area
=
=
=
2s 2ms
=
=
4 m s1
2s 0ms
For 4 6 seconds:
Area
2
=
=
0 m s1
2 s 2 m s2
4 m s1
4 ms1 at t = 2s.
0 m s1 from t = 2 s to
t = 4 s.
decreasing. It starts at a
velocity of 4 m s1 and
decreases to 0 m s1 .
Step 2 : Now use the values to draw the velocity vs. time graph.
Physics: Mechanics
423
21.6
v (m s1 )
4
3
2
1
t (s)
0
0
Exercise 21 - 6
1. A car is parked 10 m from home for 10 minutes. Draw a displacementtime, velocity-time and acceleration-time graphs for the motion. Label all
the axes.
2. A bus travels at a constant velocity of 12 m s1 for 6 seconds. Draw
the displacement-time, velocity-time and acceleration-time graph for the
motion. Label all the axes.
3. An athlete runs with a constant acceleration of 1 m s2 for 4 s. Draw
4. The following velocity-time graph describes the motion of a car. Draw the
displacement-time graph and the acceleration-time graph and explain the
motion of the car according to the three graphs.
v (m s1 )
6
0
t (s)
Physics: Mechanics
21.7
v (m s1 )
8
More practice
(1.) 009v
video solutions
(2.) 009w
See simulation: (
(3.) 009x
t (s)
or help at www.everythingscience.co.za
(4.) 009y
(5.) 009z
Equations of motion
ESAHG
In this section we will look at the third way to describe motion. We have looked at describing motion in terms of words and graphs. In this section we examine equations that can be
used to describe motion.
This section is about solving problems relating to uniformly accelerated motion. In other
words, motion at constant acceleration.
The following are the variables that will be used in this section:
initial velocity ( m s1 ) at t = 0 s
vi
vf
displacement (m)
time (s)
acceleration (m s2 )
An alternate convention for some of the variables exists that you will likely encounter so
Physics: Mechanics
425
21.7
FACT
Galileo
Galilei
of
u
v
initial velocity ( m s1 ) at t = 0 s
nal velocity (m s1 ) at time t
displacement (m)
total
distance
rest, is proportional
to the square of the
time.
He also con-
vf
refuting
the
accepted Aristotelian
2
vf
(21.1)
vi + at
(vi + vf )
t
2
1
vi t + at2
2
2
vi + 2ax
(21.2)
(21.3)
(21.4)
laws
of
The questions can vary a lot, but the following method for answering them will always
work. Use this when attempting a question that involves motion with constant acceleration.
You need any three known quantities (vi , vf , x, t or a) to be able to calculate the fourth
one.
Problem solving strategy:
1. Read the question carefully to identify the quantities that are given. Write them down.
2. Identify the equation to use. Write it down!!!
3. Ensure that all the values are in the correct units and ll them in your equation.
4. Calculate the answer and check your units.
SOLUTION
Step 1 : Identify what information is given and what is asked for
426
Physics: Mechanics
21.7
We are given:
10 m s1
vi
725 m
10 s
725 m =
725 m 100 m =
a
1
vi t + at2
2
1
(10 m s1 10 s) + a (10 s)2
2
(50 s2 ) a
12, 5 m s2
Physics: Mechanics
427
21.7
SOLUTION
Step 1 : Identify what information is given and what is asked for
We are given:
0 m s1 (because the object starts from rest.)
vi
64 m
4s
vf
64 m
64 m
1
vi t + at2
2
1
(0 m s1 4 s) + a (4 s)2
2
(8 s2 )a
8 m s2 East
Step 4 : Final velocity: Find a suitable equation to calculate the nal velocity
We can use Equation 21.1 - remember we now also know the acceleration of the object.
vf = vi + at
428
Physics: Mechanics
21.7
vf
vi + at
vf
0 m s1 + (8 m s2 )(4 s)
32 m s1 East
Step 6 : Time at half the distance: Find an equation to calculate the time
We can use Equation 21.3:
1
vi + at2
2
32 m
32 m
t2
2, 83 s
8s
1
(0 m s1 )t + (8 m s2 )(t)2
2
0 + (4 m s2 )t2
Step 7 : Distance at half the time: Find an equation to relate the distance and
time
Half the time is 2 s, thus we have vi , a and t - all in the correct units.
We can use Equation 21.3 to get the distance:
x
=
=
=
1
vi t + at2
2
1
(0 m s1 )(2 s) + (8 m s2 )(2 s)2
2
16 m East
Exercise 21 - 7
1. A car starts off at 10 ms1 and accelerates at 1 ms2 for 10 s. What is its
nal velocity?
Physics: Mechanics
429
21.7
2. A train starts from rest, and accelerates at 1 ms2 for 10 s. How far does
it move?
3. A bus is going 30 ms1 and stops in 5 s. What is its stopping distance for
this speed?
4. A racing car going at 20 ms1 stops in a distance of 20 m. What is its
acceleration?
5. A ball has a uniform acceleration of 4 ms1 . Assume the ball starts from
rest. Determine the velocity and displacement at the end of 10 s.
6. A motorcycle has a uniform acceleration of 4 ms1 . Assume the motorcycle has an initial velocity of 20 ms1 . Determine the velocity and
displacement at the end of 12 s.
7. An aeroplane accelerates uniformly such that it goes from rest to 144
kmhr1 in 8 s. Calculate the acceleration required and the total distance
that it has travelled in this time.
More practice
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(2.) 00a1
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(3.) 00a2
(5.) 00a4
(4.) 00a3
(6.) 00a5
(7.) 00a6
ESAHH
The following does not form part of the syllabus and can be considered additional information.
Derivation of 21.1
According to the denition of acceleration:
430
Physics: Mechanics
a=
21.7
v
t
vf vi
t
vf
vi + at
Derivation of 21.2
We have seen that displacement can be calculated from the area under a velocity vs. time
graph. For uniformly accelerated motion the most complicated velocity vs. time graph we
can have is a straight line. Look at the graph below - it represents an object with a starting
velocity of vi , accelerating to a nal velocity vf over a total time t.
v (m s1 )
vf
vi
t (s)
To calculate the nal displacement we must calculate the area under the graph - this is just
the area of the rectangle added to the area of the triangle. This portion of the graph has
been shaded for clarity.
Area
1
2b
=
=
=
Area
1
2t
(vf vi )
1
2 vf t
1
2 vi t
t vi
=
Physics: Mechanics
vi t
431
21.7
Displacement
Area + Area
1
1
vi t + 2 vf t 2 vi t
(vi +vf )
t
2
Derivation of 21.3
This equation is simply derived by eliminating the nal velocity vf in 21.2. Remembering
from 21.1 that
vf = vi + at
then 21.2 becomes
vi +vi +at
t
2
2vi t+at2
2
vi t + 1 at2
2
Derivation of 21.4
This equation is just derived by eliminating the time variable in the above equation. From
21.1 we know
t=
vf vi
a
432
Physics: Mechanics
=
=
vi vf
a
vf vi
a
vi
vi vf
a
2
vi
a
21.7
+ 1a
2
+ 1a
2
2
vi
a
2
vf
2a
vf vi
a
2
2
vf 2vi vf +vi
a2
vi vf
a
2ax
=
=
2
vi
2a
2
2
2
2vi + vf + vi
2
vf
2
vi + 2ax
This gives us the nal velocity in terms of the initial velocity, acceleration and displacement
and is independent of the time variable.
ESAHI
What we have learnt in this chapter can be directly applied to road safety. We can analyse
the relationship between speed and stopping distance. The following worked example
illustrates this application.
child 50 m in front of him in the road. He hits the brakes to stop the truck. The
truck accelerates at a rate of -1.25 ms2 . His reaction time to hit the brakes is 0,5
seconds. Will the truck hit the child?
SOLUTION
Physics: Mechanics
433
21.7
child is here
0,5 s
C
negative acceleration
constant v
50 m
10
d
t
d
0, 5
5m
vi
vf
1, 25 m s2
0 m s1
0 =
10 m s1
vi + at
10 m s1 + (1, 25 m s2 )t
(1, 25 m s2 )t
8s
434
Physics: Mechanics
21.7
(vi + vf )
t
2
10 + 0
(8)
s
40 m
Chapter 21 | Summary
See the summary presentation (
The displacement of an object is how far it is from the reference point. It is the
shortest distance between the object and the reference point. It has magnitude and
direction because it is a vector.
The distance of an object is the length of the path travelled from the starting point to
the end point. It has magnitude only because it is a scalar.
Speed (v) is the distance covered (D) divided by the time taken (t):
v=
D
t
Average velocity (vav ) is the displacement (x) divided by the time taken (t):
vav =
x
t
435
21.7
Acceleration (a) is the change in velocity (v) over a time interval (t):
a=
v
t
The area under a velocity - time graph (v vs. t) gives the displacement.
The area under an acceleration - time graph (a vs. t) gives the velocity.
vi + at
(vi +vf )
t
2
1
vi t + 2 at2
2
vf
2
vi + 2ax
Physical Quantities
Quantity
Vector
Unit name
Unit symbol
Position (x)
metre
Distance (D)
metre
metre
m s1
m s1
m s1
m s1
m s2
m s2
m s2
Displacement (x)
Speed (vav )
436
Physics: Mechanics
Chapter 21
21.7
Column B
gradient
area
velocity
displacement
acceleration
slope
3. Indicate whether the following statements are TRUE or FALSE. Write only
true or false. If the statement is false, write down the correct statement.
a. A scalar is the displacement of an object over a time interval.
b. The position of an object is where it is located.
c. The sign of the velocity of an object tells us in which direction it is
travelling.
d. The acceleration of an object is the change of its displacement over a
period in time.
4. [SC 2003/11] A body accelerates uniformly from rest for t0 seconds after
which it continues with a constant velocity. Which graph is the correct
representation of the bodys motion?
Physics: Mechanics
437
21.7
t0
t0
(a)
t0
(b)
t0
(c)
(d)
v (m s1 )
P
6
5
4
3
2
1
0
t (s)
0 1 2 3 4
The difference in the distance travelled by the two cars (in m) after 4 s is
...
(a) 12
(b) 6
(c) 2
(d) 0
6. [IEB 2005/11 HG] The graph that follows shows how the speed of an
athlete varies with time as he sprints for 100 m.
speed (m s1 )
10
11
time (s)
1
(b) 100 = (10)(11) + 2 (10)t
1
(c) 100 = 10t + 2 (10)t2
1
(d) 100 = 1 (0)t + 2 (10)t2
2
7. [SC 2002/03 HG1] In which one of the following cases will the distance
covered and the magnitude of the displacement be the same?
(a) A girl climbs a spiral staircase.
(b) An athlete completes one lap in a race.
(c) A raindrop falls in still air.
438
Physics: Mechanics
21.7
x (m)
motorcycle
car
375
300
X 10
15
t (s)
(a) Use the graph to nd the magnitude of the constant velocity of the
car.
(b) Use the information from the graph to show by means of calculation
that the magnitude of the acceleration of the motorcycle, for the rst
10 s of its motion is 7,5 ms2 .
(c) Calculate how long (in seconds) it will take the motorcycle to catch
up with the car (point X on the time axis).
(d) How far behind the motorcycle will the car be after 15 seconds?
9. [IEB 2005/11 HG] Which of the following statements is true of a body that
accelerates uniformly?
(a) Its rate of change of position with time remains constant.
(b) Its position changes by the same amount in equal time intervals.
(c) Its velocity increases by increasing amounts in equal time intervals.
(d) Its rate of change of velocity with time remains constant.
10. [IEB 2003/11 HG1] The velocity-time graph for a car moving along a
straight horizontal road is shown below.
v (m s1 )
20
12
Area A
Area B
0
Physics: Mechanics
t (s)
439
21.7
Area A
t
Area A + Area B
t
Area B
t
Area A Area B
t
(a) Calculate the magnitude of the minimum acceleration which the car
must have to avoid exceeding the speed limit, if the municipal speed
limit is 16.6 m s1 .
(b) Calculate the time from the instant the driver applied the brakes un-
til he reaches the speed trap. Assume that the cars velocity, when
reaching the trap, is 16.6 m s1 .
12. A trafc ofcer is watching his speed trap equipment at the bottom of
a valley. He can see cars as they enter the valley 1 km to his left until
they leave the valley 1 km to his right. Nelson is recording the times of
cars entering and leaving the valley for a school project. Nelson notices a
white Toyota enter the valley at 11:01:30 and leave the valley at 11:02:42.
Afterwards, Nelson hears that the trafc ofcer recorded the Toyota doing
140 kmhr1 .
(a) What was the time interval (t) for the Toyota to travel through the
valley?
(b) What was the average speed of the Toyota?
(c) Convert this speed to kmhr1 .
(d) Discuss whether the Toyota could have been travelling at 140kmhr1
at the bottom of the valley.
(e) Discuss the differences between the instantaneous speed (as measured by the speed trap) and average speed (as measured by Nelson).
13. [IEB 2003/11HG] A velocity-time graph for a ball rolling along a track is
shown below. The graph has been divided up into 3 sections, A, B and C
for easy reference. (Disregard any effects of friction.)
velocity (m s1 )
0,6
10
t1
12
time (s)
-0,2
440
Physics: Mechanics
21.7
(b) How long does it take you to walk a distance equal to the width of
the average car?
(c) What is the speed in m s1 of a car travelling at the speed limit in a
town?
(d) How many metres does a car travelling at the speed limit travel, in
the same time that it takes you to walk a distance equal to the width
of car?
(e) Why is the answer to the previous question important?
(f) If you see a car driving toward you, and it is 28 m away (the same as
the length of 8 cars), is it safe to walk across the road?
(g) How far away must a car be, before you think it might be safe to cross?
How many car-lengths is this distance?
15. A bus on a straight road starts from rest at a bus stop and accelerates at
2 ms2 until it reaches a speed of 20 m s1 . Then the bus travels for 20 s
at a constant speed until the driver sees the next bus stop in the distance.
The driver applies the brakes, stopping the bus in a uniform manner in
5 s.
(a) How long does the bus take to travel from the rst bus stop to the
second bus stop?
(b) What is the average velocity of the bus during the trip?
More practice
Physics: Mechanics
video solutions
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441
21.7
(1.) 00a7
(3.) 00a9
(4.) 00aa
(5.) 00ab
(6.) 00ac
(7.) 00ad
(8.) 00ae
(9.) 00af
(10.) 00ag
(11.) 00ah
(12.) 00ai
(13.) 00aj
442
(2.) 00a8
(14.) 00ak
(15.) 00am
Physics: Mechanics
Mechanical energy
22
Introduction
ESAHJ
All objects have energy. The word energy comes from the Greek word energeia (),
meaning activity or operation. Energy is closely linked to mass and cannot be created or
destroyed. In this chapter we will consider gravitational potential and kinetic energy. See
introductory video: ( Video: VPgjm at www.everythingscience.co.za)
Potential energy
ESAHK
The potential energy of an object is generally dened as the energy an object has because
of its position relative to other objects that it interacts with. There are different kinds of
potential energy such as gravitational potential energy, chemical potential energy, electrical
potential energy, to name a few. In this section we will be looking at gravitational potential
See video: VPfhw at www.everythingscience.co.za
energy.
Physics: Mechanics
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22.2
In the case of Earth, gravitational potential energy is the energy of an object due to its position above the surface of the Earth. The symbol EP is used to refer to gravitational potential
energy. You will often nd that the words potential energy are used where gravitational
potential energy is meant. We can dene gravitational potential energy as:
EP = mgh
where EP = potential energy (measured in joules, J)
Tip
You can treat the gravitational acceleration, g, as a constant and you will learn more about
books.
Lets look at the case of a suitcase, with a mass of 1 kg, which is placed at the top of a
2 m high cupboard. By lifting the suitcase against the force of gravity, we give the suitcase
potential energy. We can calculate its gravitational potential energy using the equation
dened above as:
EP
mgh
If the suitcase falls off the cupboard, it will lose its potential energy. Halfway down to the
oor, the suitcase will have lost half its potential energy and will have only 9, 8 J left.
EP
mgh
(1kg)(9, 8m s2 )(1m) = 9, 8 J
At the bottom of the cupboard the suitcase will have lost all its potential energy and its
potential energy will be equal to zero.
EP
mgh
(1kg)(9, 8m s2 )(0m) = 0 J
This example shows us that objects have maximum potential energy at a maximum height
and will lose their potential energy as they fall.
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Physics: Mechanics
22.2
SOLUTION
ground again.
Step 3 : Use the denition of gravitational potential energy to calculate the
answer
Physics: Mechanics
445
22.2
EP
mgh
(1kg)(9, 8m s2 )(4m)
39, 2 J
SOLUTION
446
Physics: Mechanics
22.2
EP = mgh
First we need to calculate h. The height of the ball above the ground
when the girl shoots for goal is h = (1,7 + 0,5) = 2,2 m.
Now we can use this information in the equation for gravitational
potential energy:
EP
mgh
10, 78 J
Step 4 : Calculate the potential energy of the ball at the height of the net
Again we use the denition of gravitational potential energy to solve
this:
EP
mgh
12, 25 J
EP
mgh
0J
Exercise 22 - 1
1. Describe the relationship between an objects gravitational potential energy and its:
a. mass and
b. height above a reference point.
2. A boy, of mass 30 kg, climbs onto the roof of a garage. The roof is 2, 5 m
from the ground.
Physics: Mechanics
447
22.3
a. How much potential energy did the boy gain by climbing onto the
roof?
b. The boy now jumps down. What is the potential energy of the boy
when he is 1m from the ground?
c. What is the potential energy of the boy when he lands on the ground?
3. A hiker, of mass 70 kg, walks up a mountain, 800 m above sea level, to
spend the night at the top in the rst overnight hut. The second day she
walks to the second overnight hut, 500 m above sea level. The third day
she returns to her starting point, 200 m above sea level.
a. What is the potential energy of the hiker at the rst hut (relative to sea
level)?
b. How much potential energy has the hiker lost during the second day?
c. How much potential energy did the hiker have when she started her
journey (relative to sea level)?
d. How much potential energy did the hiker have at the end of her journey when she reached her original starting position?
More practice
(1.) 00an
(2.) 00ap
video solutions
or help at www.everythingscience.co.za
(3.) 00aq
Kinetic energy
ESAHL
Unit symbol:
Kinetic energy is the energy an object has because of its motion. This means that any
moving object has kinetic energy. Kinetic energy is dened as:
EK =
448
1
mv 2
2
Physics: Mechanics
22.3
Tip
Therefore the kinetic energy EK depends on the mass and velocity of an object. The faster
it moves, and the more massive it is, the more kinetic energy it has. A truck of 2000 kg,
moving at 100 km hr1 will have more kinetic energy than a car of 500 kg, also moving at
100 km hr1 .
kinetic
energy
Consider the 1 kg suitcase on the cupboard that was discussed earlier. When it is on the
top of the cupboard, it will not have any kinetic energy because it is not moving:
EK
=
=
EK
=
=
1
mv 2
2
1
(1 kg)(0 m s1 )2 = 0 J.
2
other books.
When the suitcase falls, its velocity increases (falls faster), until it reaches the ground with a
maximum velocity. As its velocity velocity increases, it will gain kinetic energy. Its kinetic
energy will increase until it is a maximum when the suitcase reaches the ground. If it has
a velocity of 6,26 m s1 when it reaches the ground, its kinetic energy will be:
1
mv 2
2
1
(1 kg)(6, 26 m s1 )2 = 19, 6 J.
2
Physics: Mechanics
see
449
22.3
SOLUTION
Step 1 : Analyse the question to determine what information is provided
The mass of the brick m = 1 kg
The velocity of the brick at the bottom v = 8,85 m s1
These are both in the correct units so we do not have to worry about
unit
conversions.
Step 2 : Analyse the question to determine what is being asked
We are asked to nd the kinetic energy of the brick at the top and the
bottom. From the denition we know that to work out EK , we need to
know the mass and the velocity of the object and we are given both of
these values.
Step 3 : Calculate the kinetic energy at the top
Since the brick is not moving at the top, its kinetic energy is zero.
Step 4 : Substitute and calculate the kinetic energy
EK
=
=
=
450
1
mv 2
2
1
(1 kg)(8, 85 m s1 )2
2
39, 2 J
Physics: Mechanics
22.3
SOLUTION
EK
=
=
=
1
mv 2
2
1
(80 kg)(2, 7 m s1 )2
2
291, 6 J
EK
=
=
=
1
mv 2
2
1
(25 kg)(2, 7 m s1 )2
2
91, 13 J
Note: Even though the sheep and the lamb are running at the same
velocity, due to their different masses, they have different amounts of
kinetic energy. The sheep has more than the lamb because it has a
Physics: Mechanics
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22.3
Checking units
ESAHM
According to the equation for kinetic energy, the unit should be kg m2 s2 . We can prove
that this unit is equal to the joule, the unit for energy.
2
kg m s2 m
N m
(kg) m s1
We can do the same to prove that the unit for potential energy is equal to the joule:
(kg) m s2 (m)
N m
SOLUTION
Step 1 : Analyse the question to determine what information is provided
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Physics: Mechanics
22.3
Mass in kg
0, 150 kg
=
=
=
1
mv 2
2
1
(0, 150 kg)(960 m s1 )2
2
69 120 J
Exercise 22 - 2
Physics: Mechanics
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22.4
More practice
(1.) 00ar
(2.) 00as
video solutions
or help at www.everythingscience.co.za
(3.) 00at
Mechanical energy
ESAHN
Mechanical energy, EM , is simply the sum of gravitational potential energy (EP ) and the
kinetic energy (EK ). Mechanical energy is dened as:
Tip
EM = EP + EK
You may see mechan-
1
EM = mgh + mv 2
2
tation is sometimes
used.
QUESTION
Calculate the total mechanical energy for a ball of mass 0,15 kg which has a kinetic
energy of 20 J and is 2 m above the ground.
SOLUTION
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Physics: Mechanics
22.5
EM
=
=
=
=
EP + E K
1
mgh + mv 2
2
mgh + 20
(0, 15 kg)(9, 8 m s1 )(2 m) + 20 J
2, 94 J + 20 J
22, 94 J
Conservation of
mechanical energy
ESAHO
Tip
In problems involving
So far we have looked at two types of energy: gravitational potential energy and kinetic
energy. The sum of the gravitational potential energy and kinetic energy is called the
mechanical energy. In a closed system, one where there are no external dissipative forces
acting, the mechanical energy will remain constant. In other words, it will not change
of
energy,
The only
important
(become more or less). This is called the Law of Conservation of Mechanical Energy.
quantities
Physics: Mechanics
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22.5
This means that potential energy can become kinetic energy, or vice versa, but energy
cannot disappear. For example, in the absence of air resistance, the mechanical energy
of an object moving through the air in the Earths gravitational eld, remains constant (is
conserved).
See simulation: (
ESAHP
Mechanical energy is conserved (in the absence of friction). Therefore we can say that the
sum of the EP and the EK anywhere during the motion must be equal to the sum of the
EP and the EK anywhere else in the motion.
We can now apply this to the example of the suitcase on the cupboard. Consider the
mechanical energy of the suitcase at the top and at the bottom. We can say:
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Physics: Mechanics
22.5
EM 1
EM 2
EP 1 + EK1
1
mgh + mv 2
2
EP 2 + EK2
1
mgh + mv 2
2
1
0 + (1 kg)(v 2 )
2
1 2
(v )
2
39, 2m2 s2
(1 kg)(9, 8 m s1 )(2 m) + 0 =
19, 6 =
v2
6, 26 m s1
Activity:
Conversion of energy
Materials:
A length of plastic pipe with diameter approximately 20 mm, a marble, some
masking tape and a measuring tape.
To do (1):
First put one end of the pipe on the table top so that it is parallel to the top of
the table and tape it in position with the masking tape.
Lift the other end of the pipe upwards and hold it at a steady height not too high
above the table.
Measure the vertical height from the table top to the top opening of the pipe.
Now put the marble at the top of the pipe and let it go so that it travels through
the pipe and out the other end.
Questions:
What is the velocity (i.e. fast, slow, not moving) of the marble when you rst
put it into the top of the pipe and what does this mean for its gravitational
potential and kinetic energy?
Physics: Mechanics
457
22.5
the other end of the pipe and rolls onto the desk? What does this mean for its
gravitational potential and kinetic energy?
To do (2):
Now lift the top of the pipe as high as it will go.
Measure the vertical height of the top of the pipe above the table top.
Put the marble into the top opening and let it roll through the pipe onto the
table.
Questions:
What is the velocity (i.e. fast, slow, not moving) of the marble when you put it
into the top of the pipe, and what does this mean for its gravitational potential
and kinetic energy?
Compared to the rst attempt, what was different about the height of the top
of the tube? How do you think this affects the gravitational potential energy of
the marble?
Compared to your rst attempt, was the marble moving faster or slower when
it came out of the bottom of the pipe the second time? What does this mean
for the kinetic energy of the marble?
The activity with the marble rolling down the pipe shows very nicely the conversion between gravitational potential energy and kinetic energy. In the rst instance, the pipe was
held relatively low and therefore the gravitational potential energy was also relatively low.
The kinetic energy at this point was zero since the marble wasnt moving yet. When the
marble rolled out of the other end of the pipe, it was moving relatively slowly, and therefore
its kinetic energy was also relatively low. At this point its gravitational potential energy was
zero since it was at zero height above the table top.
In the second instance, the marble started off higher up and therefore its gravitational
potential energy was higher. By the time it got to the bottom of the pipe, its gravitational
potential energy was zero (zero height above the table) but its kinetic energy was high since
it was moving much faster than the rst time. Therefore, the gravitational potential energy
was converted completely to kinetic energy (if we ignore friction with the pipe).
In the case of the pipe being held higher, the gravitational potential energy at the start was
higher, and the kinetic energy (and velocity) of the marble was higher at the end. In other
words, the total mechanical energy was higher and and only depended on the height you
held the pipe above the table top and not on the distance the marble had to travel through
the pipe.
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Physics: Mechanics
22.5
waterfall
5m
5 m high.
If air resistance is ignored, calculate:
1. the potential energy of the tree trunk at the top of the waterfall.
2. the kinetic energy of the tree trunk at the bottom of the waterfall.
3. the magnitude of the velocity of the tree trunk at the bottom of
the waterfall.
SOLUTION
Step 1 : Analyse the question to determine what information is provided
The mass of the tree trunk m = 100 kg
The height of the waterfall h = 5 m.
EP
mgh
4900 J
Physics: Mechanics
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22.5
98
1
mv 2
2
1
(100 kg)(v 2 )
2
v2
9, 899 . . . m s1
9, 90 m s1
Example 8: Pendulum
QUESTION
A 2 kg metal ball is suspended from a rope as a pendulum. If it is released from
point A and swings down to the point B (the bottom of its arc):
1. show that the velocity of the ball is independent of its mass,
2. calculate the velocity of the ball at point B.
A
0.5m
B
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Physics: Mechanics
22.5
SOLUTION
Step 1 : Analyse the question to determine what information is provided
The mass of the metal ball is m = 2 kg
The change in height going from point A to point B is h = 0,5 m
The ball is released from point A so the velocity at point, vA =
0 m s1 .
EM 2
EP 1 + EK1
1
mgh1 + m(v1 )2
2
EP 2 + EK2
1
mgh2 + m(v2 )2
2
1
0 + m(v2 )2
2
1
m(v2 )2
2
mgh1 + 0 =
mgh1
The mass of the ball m appears on both sides of the equation so it can
be eliminated so that the equation becomes:
gh1
2gh1
1
(v2 )2
2
(v2 )2
This proves that the velocity of the ball is independent of its mass. It
does not matter what its mass is, it will always have the same velocity
when it falls through this height.
Step 4 : Calculate the velocity of the ball at point B
We can use the equation above, or do the calculation from rst prin-
Physics: Mechanics
461
22.5
2gh1
(v2 )2
(v2 )2
(2)(9.8 m s1 )(0, 5 m)
v2
v2
9, 8 m s1 m
9, 8 m2 s2
3, 13 m s1
=
=
mgh1 + 0 =
(v2 )2
(v2 )2
v2
v2
EK2 + EP 2
1
mgh2 + m(v2 )2
2
1
0 + m(v2 )2
2
2mgh1
m
2(2 kg)(9, 8 m s2 )(0, 5 m)
2 kg
9, 8 m2 s2
3, 13 m s1
Roller coaster
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Physics: Mechanics
22.5
1. the velocity of the roller coaster when it reaches the top of the
loop
2. the velocity of the roller coaster at the bottom of the loop (i.e.
ground level)
SOLUTION
EM 2
EK1 + EP 1
0 + mgh1
EK2 + EP 2
1
m(v2 )2 + mgh2
2
We can eliminate the mass, m, from the equation by dividing both sides
by m.
gh1
(v2 )2
v2
(v2 )
Physics: Mechanics
1
(v2 )2 + gh2
2
2(gh1 gh2 )
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22.5
EM 3
EK1 + EP 1
1
m1 (0)2 + mgh1
2
mgh1
(v3 )2
EK3 + EP 3
1
m(v3 )2 + mg(0)
2
1
m(v3 )2
2
2gh1
v3
(v3 )
2(9, 8 m s2 )(50 m)
31, 30 m s1
SOLUTION
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Physics: Mechanics
22.5
EM 1
EM 2
EK1 + EP 1
1
m(v1 )2 + mgh1
2
0 + mgh1
(v2 )2
(v2 )2
EK2 + EP 2
1
m(v2 )2 + mgh2
2
1
m(v2 )2 + 0
2
2mgh
m
2gh
(v2 )2
v2
(2)(9, 8 m s2 )(10 m)
14 m s1
Note: the distance that the bottle travelled (i.e. 100 m) does not play
any role in calculating the energies. It is only the height difference that
is important in calculating potential energy.
Step 4 : Calculate the difference between the climbers potential energy at
the top of the slope and her potential energy at the bottom of the slope
At the top of the slope, her potential energy is:
EP 1
mgh1
=
Physics: Mechanics
5880 J
465
22.5
mgh1
0J
Therefore the difference in her potential energy when moving from the
top of the slope to the bottom is:
EP 1 EP 2 = 5880 0 = 5880 J
Exercise 22 - 3
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Physics: Mechanics
22.5
6. Prove that the velocity of an object, in free fall, in a closed system, is independent of its mass.
More practice
(1.) 00au
(2.) 02u1
Physics: Mechanics
video solutions
(3.) 00av
(4.) 00aw
or help at www.everythingscience.co.za
(5.) 00ax
(6.) 00ay
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22.5
Chapter 22 | Summary
See the summary presentation (
The gravitational potential energy of an object is the energy the object has because
of its position in the gravitational eld relative to some reference point.
The kinetic energy of an object is the energy the object has due to its motion.
The mechanical energy of an object is the sum of the potential energy and kinetic
energy of the object.
The Law of Conservation of Mechanical Energy states that the total mechanical energy of an isolated system (i.e. no friction or air resistance) remains constant.
EP = mgh
Kinetic Energy
1
EK = 2 mv 2
Mechanical Energy
EM = EK + E P
Physical Quantities
Quantity
Unit name
Unit symbol
joule
joule
joule
Chapter 22
Physics: Mechanics
22.5
3. A man res a rock out of a slingshot directly upward. The rock has an
initial velocity of 15 m s1 .
b. Draw graphs to show how the potential energy, kinetic energy and
mechanical energy of the rock changes as it moves to its highest point.
4. A metal ball of mass 200 g is tied to a light string to make a pendulum.
The ball is pulled to the side to a height (A), 10 cm above the lowest point
of the swing (B). Air friction and the mass of the string can be ignored.
The ball is let go to swing freely.
a. Calculate the potential energy of the ball at point A.
b. Calculate the kinetic energy of the ball at point B.
c. What is the maximum velocity that the ball will reach during its motion?
5. A truck of mass 1, 2 tons is parked at the top of a hill, 150 m high. The
truck driver lets the truck run freely down the hill to the bottom.
a. What is the maximum velocity that the truck can achieve at the bottom of the hill?
b. Will the truck achieve this velocity? Why/why not?
6. A stone is dropped from a window, 6 m above the ground. The mass of
the stone is 25 g. Use the Principle of Conservation of Energy to determine
the speed with which the stone strikes the ground.
More practice
(1.) 00az
(2.) 00b0
Physics: Mechanics
video solutions
(3.) 00b1
(4.) 00b2
or help at www.everythingscience.co.za
(5.) 00b3
(6.) 00b4
469
The hydrosphere
23
Introduction
ESAHQ
The Earth
suitable for life to exist. Also, the Earths atmosphere has exactly
the right type of gases in the right amounts for life to survive.
Our planet also has water on its surface, which is something
very unique. In fact, Earth is often called the Blue Planet
because most of it is covered in water. This water is made up of
freshwater in rivers and lakes, the saltwater of the oceans and
Photo by NASA on
Flickr.com
Interactions of the
hydrosphere
FACT
The total mass of
the
hydrosphere
is
approximately
1, 4
1018
tonnes!
ESAHR
It is important to realise that the hydrosphere is not an isolated system, but rather interacts
with other global systems, including the atmosphere, lithosphere and biosphere. These
interactions are sometimes known collectively as the water cycle.
See video: VPbzy at www.everythingscience.co.za
Atmosphere When water is heated (e.g. by energy from the sun), it evaporates and
forms water vapour. When water vapour cools again, it condenses to form liquid
water which eventually returns to the surface by precipitation e.g. rain or snow. This
900 A4 textbooks!))
cycle of water moving through the atmosphere and the energy changes that accompany it, is what drives weather patterns on earth.
Lithosphere
In the lithosphere (the ocean and continental crust at the Earths surface), water is
an important weathering agent, which means that it helps to break rock down into
rock fragments and then soil. These fragments may then be transported by water to
another place, where they are deposited. These two processes (weathering and the
470
23.3
transporting of fragments) are collectively called erosion. Erosion helps to shape the
earths surface. For example, you can see this in rivers. In the upper streams, rocks
are eroded and sediments are transported down the river and deposited on the wide
ood plains lower down.
On a bigger scale, river valleys in mountains have been
River valley
important for the plants and animals that live in the photo by AlanVernon on Flickr.com
water.
Biosphere
In the biosphere, land plants absorb water through their roots and then transport this
through their vascular (transport) system to stems and leaves. This water is needed in
photosynthesis, the food production process in plants. Transpiration (evaporation of
water from the leaf surface) then returns water back to the atmosphere.
Exploring the
hydrosphere
ESAHS
The large amount of water on our planet is something quite unique. In fact, about 71%
of the earth is covered by water. Of this, almost 97% is found in the oceans as saltwater,
about 2, 2% occurs as a solid in ice sheets, while the remaining amount (less than 1%) is
available as freshwater. So from a human perspective, despite the vast amount of water
on the planet, only a very small amount is actually available for human consumption (e.g.
drinking water).
In chapter 18 we looked at some of the reactions that occur in aqueous solution and saw
some of the chemistry of water, in this section
we are going to spend some time exploring a
part of the hydrosphere in order to start appreciating what a complex and beautiful part
of the world it is. After completing the following investigation, you should start to see just
how important it is to know about the chemistry of water.
A lake
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23.3
Site 2
Site 3
Temperature
pH
Conductivity
Dissolved oxygen
Clarity
Animals
Plants
Interpreting the data Once you have collected and recorded your
data, think about the following questions:
How does the data you have collected vary at different sites?
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23.4
ESAHT
It is so easy sometimes to take our hydrosphere for granted and we seldom take the time to
really think about the role that this part of the planet plays in keeping us alive. Below are
just some of the important functions of water in the hydrosphere:
Water is a part of living cells Each cell in a living organism is made up of almost 75%
water, and this allows the cell to function normally. In fact, most of the chemical
reactions that occur in life, involve substances that are dissolved in water. Without
water, cells would not be able to carry out their normal functions and life could not
exist.
Water provides a habitat The hydrosphere provides an important place for many
animals and plants to live. Many gases (e.g. CO2 , O2 ), nutrients e.g. nitrate (NO ),
3
nitrite (NO ) and ammonium (NH+ ) ions, as well as other ions (e.g. Mg2+ and Ca2+ )
2
4
are dissolved in water. The presence of these substances is critical for life to exist in
water.
Regulating climate One of waters unique characteristics is its high specic heat. This
means that water takes a long time to heat up and also a long time to cool down. This
is important in helping to regulate temperatures on earth so that they stay within a
range that is acceptable for life to exist. Ocean currents also help to disperse heat.
Human needs Humans use water in a number of ways. Drinking water is obviously
very important, but water is also used domestically (e.g. washing and cleaning) and
in industry. Water can also be used to generate electricity through hydropower.
These are just a few of the functions that water plays on our planet. Many of the functions
of water relate to its chemistry and to the way in which it is able to dissolve substances in
it.
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23.5
Threats to the
hydrosphere
ESAHU
It should be clear by now that the hydrosphere plays an extremely important role in the
survival of life on Earth and that the unique properties of water allow various important
chemical processes to take place which would otherwise not be possible. Unfortunately
for us however, there are a number of factors that threaten our hydrosphere and most of
these threats are because of human activities. We are going to focus on two of these issues:
pollution and overuse and look at ways in which these problems can possibly be overcome.
See video: VPcbf at www.everythingscience.co.za
1. Pollution
Pollution of the hydrosphere is a major problem. When we think of pollution, we
sometimes only think of things like plastic, bottles, oil and so on. But any chemical
that is present in the hydrosphere in an amount that is not what it should be is a pollutant. Animals and plants that live in the Earths water bodies are specially adapted
to surviving within a certain range of conditions. If these conditions are changed
(e.g. through pollution), these organisms may not be able to survive. Pollution then,
can affect entire aquatic ecosystems. The most common forms of pollution in the
hydrosphere are waste products from humans and from industries, nutrient pollution
e.g. fertiliser runoff which causes eutrophication (an excess of nutrients in the water
leading to excessive plant growth) and toxic trace elements such as aluminium, mercury and copper to name a few. Most of these elements come from mines or from
industries.
2. Overuse of water
We mentioned earlier that only a very small percentage of the hydrospheres water
is available as freshwater. However, despite this, humans continue to use more and
more water to the point where water consumption is fast approaching the amount of
water that is available. The situation is a serious one, particularly in countries such
as South Africa which are naturally dry and where water resources are limited. It is
estimated that between 2020 and 2040, water supplies in South Africa will no longer
be able to meet the growing demand for water in this country. This is partly due
to population growth, but also because of the increasing needs of industries as they
expand and develop. For each of us, this should be a very scary thought. Try to
imagine a day without water... difcult isnt it? Water is so much a part of our lives,
that we are hardly aware of the huge part that it plays in our daily lives.
As populations grow, so do the demands that are placed on dwindling water resources.
While many people argue that building dams helps to solve this water-shortage problem,
there is evidence that dams are only a temporary solution and that they often end up doing
far more ecological damage than good. The only sustainable solution is to reduce the
demand for water, so that water supplies are sufcient to meet this. The more important
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23.5
475
23.5
of people you represent: (Remember to take notes dur- Photo by owcomm on Flickr.com
ing your discussions and nominate a spokesperson to
give feedback to the rest of the class on behalf of your
group)
What steps could be taken by your group to conserve water?
Why do you think these steps are not being taken?
A dam
Photo by Redeo on
Flickr.com
If possible talk to people who have lived in the area for a long time
and try to get their opinion on how life changed since the dam was
built. If it is not possible to talk to people in the area, then look for
relevant literature on the area.
Try to nd out if any environmental impact assessments (this is
where people study the environment and see what effect the pro-
posed project has on the environment) were done before the dam
was built. Why do you think this is important?
476
23.6
Look at how the ecology has changed. What was the ecology of
the river before the dam was built? What is the current ecology?
Do you think it has changed in a good way or a bad way? You
could interview people in the community who lived there long
ago before the dam was built.
It is important to realise that our hydrosphere exists in a delicate balance with other systems
and that disturbing this balance can have serious consequences for life on this planet.
Project: School Action Project
There is a lot that can be done within a school to save water. As a class, discuss
what actions could be taken by your class to make people more aware of how
important it is to conserve water. Also consider what ways your school can
save water. If possible, try to put some of these ideas into action and see if they
really do conserve water. During break walk around the school and make a list
of ll the places where water is being wasted.
ESAHV
When you drink a glass of water you are not just drinking water, but many other substances
that are dissolved into the water. Some of these come from the process of making the
water safe for humans to drink, while others come from the environment. Even if you took
water from a mountain stream (which is often considered pure and bottled for people to
consume), the water would still have impurities in it. Water pollution increases the amount
of impurities in the water and sometimes makes the water unsafe for drinking. In this
section we will look at a few of the substances that make water impure and how we can
make pure water. We will also look at the pH of water.
In chapter 18 we saw how compounds can dissolve in water. Most of these compounds
(e.g. Na+ , Cl , Ca2+ , Mg2+ , etc.) are safe for humans to consume in the small amounts
that are naturally present in water. It is only when the amounts of these ions rise above the
safe levels that the water is considered to be polluted.
You may have noticed sometimes that when you pour a glass a water straight from the tap, it
has a sharp smell. This smell is the same smell that you notice around swimming pools and
Chemistry: Chemical systems
477
23.6
is due to chlorine in the water. Chlorine is the most common compound added to water
to make it safe for humans to use. Chlorine helps to remove bacteria and other biological
contaminants in the water. Other methods to purify water include ltration (passing the
water through a very ne mesh) and occulation (a process of adding chemicals to the
water to help remove small particles).
pH of water is also important. Water that is too basic (pH greater than 7) or too acidic
(pH less than 7) may present problems when humans consume the water. If you have ever
noticed after swimming that your eyes are red or your skin is itchy, then the pH of the
swimming pool was probably too basic or too acidic. This shows you just how sensitive
we are to the smallest changes in our environment. The pH of water depends on what ions
are dissolved in the water. Adding chlorine to water often lowers the pH. You will learn
more about pH in grade 11.
sea
river
rain
1. Look at each water sample and note if the water is clear or cloudy.
2. Examine each water sample under a microscope and note what you see.
3. Test the pH of each of the water samples.
4. Pour some of the water from each sample through lter paper.
5. Refer to chapter 18 for the details of common anion tests. Test for chloride,
sulphate, carbonate, bromide and iodide in each of the water samples.
Results:
Write down what you saw when you just looked at the water samples.
Write down what you saw when you looked at the water samples under a microscope. Where there any dissolved particles? Or other things in the water? Was there
a difference in what you saw with just looking and with looking with a microscope?
478
tap
23.6
Write down the pH of each water sample. Look at the lter paper from each sample.
Is there sand or other particles on it? Which anions did you nd in each sample?
Discussion: Write a report on what you observed. Draw some conclusions on the
purity of the water and how you can tell if water is pure or not.
Conclusion: You should have seen that water is not pure, but rather has many
substances dissolved in it.
Chapter 23 | Summary
See the summary presentation (
The hydrosphere includes all the water that is on Earth. Sources of water include
freshwater (e.g. rivers, lakes), saltwater (e.g. oceans), groundwater (e.g. boreholes)
and water vapour. Ice (e.g. glaciers) is also part of the hydrosphere.
The hydrosphere interacts with other global systems, including the atmosphere, litho-
479
23.6
Chapter 23
1. What is the hydrosphere? How does it interact with other global systems?
2. Why is the hydrosphere important?
3. Write a one page essay on the importance of water and what can be done
to ensure that we still have drinkable water in 50 years time.
More practice
(1.) 00b5
480
(2.) 00b6
video solutions
or help at www.everythingscience.co.za
(3.) 024v
481
24
ESAHW
Physical Quantities
Quantity
Unit name
Unit symbol
metre
m s2
metre
m s1
coulomb
mol dm3
Current (I)
ampere
Density (d)
g cm3
Displacement (x)
metre
Distance (D)
metre
Energy (E)
Joule
Frequency (f )
Hertz
Hz
m s2
m s1
m s1
deciBel
dB
m s2
gram
g mol1
mole
mol
Amplitude (A)
Atomic mass unit (amu)
Average acceleration (aav )
Average speed (vav )
Average velocity (vav )
Charge (Q)
Concentration (C)
Intensity (I)
Magnitude of acceleration (a)
Mass (m)
Molar mass (M )
Mole (n)
482
24.1
Physical Quantities
Quantity
Unit name
Unit symbol
Period (T )
second
Position (x)
metre
Volt
m s1
Resistance (R)
Ohm
degrees
decimetre cubed
dm3
Wavelength ()
metre
Wavespeed (v)
m s1
Potential difference (V )
Temperature (T)
Volume (V )
483
Exercise solutions
25
Science skills
ESAHX
Exercise 1-1
1.
ESAHY
(c) 5 107 m
(c) 59, 8 kN
(b) 37, 83
2.
(a) 3, 63 106
(d) 2, 50 107 m
(d) 2, 5 mA
(e) 3, 5 102 g
(e) 7, 5 km
(c) 6, 3 104
(a) 5, 11 10 V
3.
(b) 1, 0 101
(a) 0, 1602 aC
(b) 1, 992 MJ
Exercise 1-2
1.
ESAHZ
(a) 1, 01 104 Pa
(b) 9, 8 102 m s2
(c) 1, 256 10
2.
(a) 1, 23 106 N
(c) 7, 2 Mm
(b) 4, 17 10 kg
3.
(b) 1 000 mg
(d) 11 n
(c) 2, 47 105 A
(d) 8, 80 104 mm
484
4.
(a) 1, 01s
5. 234, 44 m s1
6. 373 K
25.2
Classication of matter
ESAIA
(b) A
2. a
(c) E
ESAIB
3.
(a) compound
8.
(d) B
(b) compound
(e) C
(c) heterogeneous
5.
mixture
(d) solution
6.
(b) BF3
(c) KMnO4
(e) solution
(d) ZnCl2
9.
(a) friction
(b) durable
(b) KBr
(f) compound
(a) FeSO4
(c) CO2
(c) durable
(d) shiny
(h) solution
(a) D
(f) brous
(g) element
4.
7.
ESAIC
(a) sublimation
(b) evaporation
(b) carbon
3. see denition
4.
2.
ESAID
(c) helium
5.
solid
liquid
gas
485
25.4
The atom
ESAIE
(a) atomic
5. A
ber
6. C
(b) electron
7. B
orbital
8.
[Ne]3s 3p
(a)
(c)
(c) true
(d)
(d) false
9.
(e) 5; 6; 5
[He]2s2 2p2
(b) true
3. B
(d) 3; 4; 3
(c)
(b)
(a) false
(b)
4. B
mass num-
2.
ESAIF
11.
(c) 9
[He]2s2 2p6
12. B
(e)
[Ne]3s2 3p6
10.
(a)
[Ne]3s2 3p1
13. 63, 62 u
ESAIG
(a) false
ESAIH
(c) Cl > Br
(b) true
(a) Rh
(b) 17
(d)
5
4 Be
12
6 C
48
22 Ti
19
9 F
(d) Br > Cl
(c) false
(d) false
2.
4.
(a)
(c)
(b) period
(c) halogens
3.
(b) Cl > Br
486
(d)
(a) Br > Cl
(b)
25.6
Chemical bonding
ESAII
ESAIJ
2. B
3. B
(b) 3
7.
Ca
2+
NH+
4
OH
5.
(a) 1
NH4 OH
O2
K2 O
CaO
(NH4 )2 O
KNO3
Ca(NO3 )2
NH4 NO3
PO3
4
4.
Ca(OH)2
NO
3
6.
KOH
K3 PO4
Ca3 (P O4 )2
(NH4 )3 PO4
Transverse pulses
ESAIK
ESAIL
1.
2. 15 m
3. 0, 3 m s1
4. 0, 05 s
487
25.8
Transverse waves
ESAIM
(i) BD
(c) 0, 75 m
(f) 14, 4 m
(b) 1, 33 s
2.
(a) 0, 2 m
(b)
ESAIN
(ii) AB
(d) 0, 625 s
(a) 3
(iii) BD
(e) 1, 60 Hz
Longitudinal waves
ESAIO
3.
2. D
ESAIP
4.
a. 10 m
b. 2 m s
1
256
ESAIQ
a.
b. 1, 25 m
Sound
1.
ESAIR
3. D
8. C
13. C
18. 25, 8 m
4. C
9. A
14. 1 700 m
19. 0, 57 m
5. C
10. C
15. 0, 15 s
waveform 6. B
11. B
16.
7. E
12. A
17. radios
frequency
(b)
amplitude
(c)
2. E
488
600 Hz
20. increase
wavelength
21. 17 mm
25.
22. 34, 4 m
waves
(f) wave
frequency
(g) wave
(c)
sound
waves
transverse
speed
(h)
amplitude
(d)
(b)
24. 172 m
(e)
wavelength
(a)
23.
1 812, 5 mm
25.11
period
speed
EM radiation
ESAIS
ESAIT
1. 2.0 1025 J
E = 3, 7 1019 J
8, 6 1024 J
102 m.
3. 8, 6 1021 J
ESAIU
ESAIV
(a) molecule
(c)
(3)
(d)
(a) covalent
(b) NH3
(4)
489
25.13
ESAIW
ESAIX
(4)
(a) decomposition
(b) synthesis
(3)
(c) decomposition
Representing chemical
change
ESAIY
4CO2 (g) +
Magnetism
End of chapter exercises
490
6H2 O()
(d) synthesis
(4) 2C2 H6 (g) + 7O2 (g)
ESAIZ
ESAJA
ESAJB
25.16
(12)
Electrostatics
ESAJC
of 4
gether
(3) repulsive, attractive
(8) B
(4)
(9) polarisation
(a) repulsive
(5) D
ESAJD
trons
Electric circuits
ESAJE
ESAJF
(2)
(c) B
(4) 1
(7) a
(5) d
(11) 1, 4 V
(12) 18
(a) brighter
(8) 4, 25 V
(13) 7, 5
(6)
(9) 17 V
(14) 15 V
same, C is off
(10) 7 V
(15) 2 V
491
25.18
Reactions in aqueous
solutions
ESAJG
ESAJH
(1)
(e) ionic
(e) C
(a) condensation
(b) ion
(f) ionic
(f) B
(c) hardness
(g) I
(6)
(a) X: carbonate,
(5)
(a) B
(a) molecular
ride
(b) CO2 + Ba2+ +
3
(b) H
(b) ionic
(c) E
(c) molecular
(d) A
(d) molecular
Cl BaCO3 +
Cl
Quantitative aspects of
chemical change
ESAJI
(a) molar
ESAJJ
(b) 0, 09
H: 4, 48
(b) Li:
(c) 0, 04
mass
(d) 5, 06
(b)
Avogadros
Y:
sulphate, Z: chlo-
(7)
(a) 11, 2
(b) 0, 03
(15)
(a) 4
(c) yes
29, 17
O: 66, 67
(b) incorrect
H: 4, 17
(b) 21, 62
number
(14)
(c) 0, 1mols
(a) 196, 14
(2) D
(c) 20, 8
(10) NO2
(3) A
(d) 6, 72
(11) I4 O9
(b) 292, 32
(4) A
(e) 145, 66
(12)
(c) 94, 1
(8)
(5) C
(6)
492
(a) 0, 31
(b) C4 H8 F2
64, 02
(a) C2 H4 F
(13) 5
25.20
Vectors
ESAJK
(3) A
(2) D
ESAJL
(4) A
Motion in one
dimension
ESAJM
(5) D
(c) acceleration
(8)
(f) instantaneous
(a) 30 m s1
(b) 1, 8 s
(d) 30 m
(d) 300 m
(e) determines
(c) 16, 67 m s1
(9) D
(c) velocity
(11)
(f) no
(c) true
(d) false
(a) 1, 75 m s2
(b) 4, 8 s
(a) false
(12)
safe
distance
(10) A
(d) velocity
(b) true
(a) 1, 11 m s1
(c) 8 s
(a) displacement
(b) acceleration
(14)
(b) 7, 5 m s2
ve-
locity
(3)
(b) 11, 5 s
(7) C
(e) velocity
(a) 0, 6 m s1 1, 5 m
0, 4 m s2
(6) A
(2)
(13)
(4) B
(b) vector
(1)
ESAJN
(a) 72 s
(b) 27, 8 m s1
(g) 60 m
(15)
(a) 35 s
(b) 15, 71 m s1
(c) 100 km h1
493
25.22
Mechanical energy
ESAJO
ESAJP
(b) false
(b) 0, 196 J
(b) joules
(c) true
(d) false
(c) 1, 4 m s1
(1)
(e) true
(e) mass
(2)
494
(3)
(a) 11, 48 m
(a) false
(4)
(a) 0, 196 J
(5)
(a) 54, 22 m s1
(b) no
(6) 7, 7 m s1
25.22
495