Unit 2 - Introduction To Matter
Unit 2 - Introduction To Matter
Unit 2 - Introduction To Matter
Part 1: Matter
states of matter
● solid
○ general properties
■ crystalline – ex salt, diamond
■ amorphous – ex glass, rubber, plastic
● liquid
○ general properties
● gases
○ general properties
○ gas vs vapor
Classifying Matter
● pure substance
○ element
○ compound
■ general properties
● fixed composition
● own unique properties
● can only be broken by chemical means
(resulting in changed composition)
mixtures
● general properties
■ Contains two or more pure substances
■ Physically combined
● retain the properties of the original substances
■ separate by physical means
● homogeneous mixture Matter classification
● heterogeneous mixture
○ Varying composition throughout
B 1) element
___
C 2) compound
___
A 3) mixture of elements
___
Not
E- 4) mixture of compounds
___
Element
compound
not
touching
___
D 5) mixture of compounds & elements
compound
B /___
A 6) only atoms; no molecules
C
/ ___
E- 7) only molecules, no separate atoms
Physical vs Chemical Properties
● properties
● physical property
○ examples – odor, texture, color, density, melting point, mass
● chemical property
○ examples – flammability of gasoline, reactivity of alkali metals, ability
of iron to rust
Practice Problem
Determine whether each property is physical or chemical.
a) The explosiveness of hydrogen gas Chemical
b) The bronze color of copper chemical
c) The shiny appearance of silver physical
d) The ability of dry ice to sublime physical
Physical vs Chemical Changes
● physical change
○ examples – phase changes, grinding, cutting
● chemical change
○ examples – rusting, combustion, digestion, cooking
○ chemical reaction "" "" & "° "
■ Reactants
■ Products g
Ex: 4 Fe(s) + 3 O2(g) ->2 Fe2O3(s)
2 C4H10(l) + 13 O2(g) -> 8 CO2(g) + 10 H2O(g)
How to recognize chemical changes
We use indicators of a chemical reaction
%
1) color change ,
49cg %
2) heat transfer
Or ✗
c¥É%¥a¥se%ne
go
"
3) gas formation ¥9s %
,
%¥k¢
,
4) precipitate formation on
5) Sometimes irreversible
Researcher: Separation of Mixtures
With the partner assigned below, research the following lab techniques:
distillation, filtration, chromatography, and evaporation.
Using the results of your research, complete the Task on the next slide with your
group.
Task: You have been hired as a chemical consultant for BISS Industries. BISS
supplies major fast food restaurants with the basic ingredients for hamburger
patties. During a recent accident the silos housing the ingredients were filled
beyond capacity and the structures failed, allowing the contents of all the silos
to mix. BISS needs your help to determine if mixtures have chemically bonded and
if there is hope for separating the components so that they can continue their
thriving business. Using intelligent reasoning your partnership will work to
devise a plan to separate the given “sample” into its 4 components and identify
if the “sample” is a pure substance or mixture at each stage of separation.
Hint: The 4 components of the mixture are Beans, Iron, Sand, and Salt
Separating substances through physical means
● decanting
● distillation
○ separation of two miscible (soluble) liquids or soluble solute in a
liquid solvent
○ process uses vaporization and condensation
○ volatile liquid – easily vaporizes
● filtration
MIXTURE
+ mass of substance 2
+ mass of substance 3
Mass of mixture
are different
1 k¥ I # ☒ ±
899 KJ
4; ✗
3. go
4,06k¥ ! 73%6%3.6=204×10 -6kWh
2. 04 ✗ 10 -4kWh
Energy and chemical and Physical Change
● physical change vs chemical change
● exothermic reaction - activated complex
-
unstable
hybrid molecule
chemical reaction
○ ∆H = ¢3
exothermic to
-
needed create a chemical
↳
proles activation reaction .
energy
in heat is the AH is t
taking
↳ potential
-
when
○ hot energy AH -0
↳ ↳
potential emerge " sin
○ A -> B + energy is
form
lost
of
in the
heat
( SH / ) Potential
energy
change
bonds are
particrecolitionsg
have to
-
particles Glide
↳ reactants
AH -
SH -
motion
Endothermic
○ gains or absorbs energy Transition State
Energy lost
R: E-
" tant
"
from Ea →
groundings
Reverse reaction
gaining Transfer
energy
○ ∆H = +
↓
At =
-
J
Energy lost AH -
AH +
↳ difference between
○ cold
products and reactants
○ solid ~
-> liquid -> gas
AH +
¥+7
\/
they take in energy be
kinetic
they need more
energy
?⃝
Temperature
● temperature average kinetic
energy
● heat
● Fahrenheit (oF) scale
○ water freezes at 32oF and boils at 212oF
● Celsius (oC) scale
○ water freezes at 0oC and boils at 100oC
● Kelvin (K) scale
○ no negative temperatures
○ absolute zero ( 0k )
Practice Problems
1) Convert the following to Kelvin
a) 52.0oC b) -14.0oC
325
259k
2380 250
Heat Capacity, Specific Heat Capacity, Calculations
● heat (q) J or RJ
= J/oC or q/oC
● specific heat capacity(C) Fe
Tg
Fe
50g
Fe
5-009
↳ tells
÷ = J/goC or q/goC
us how much
\ I /
energy to increase
specific heat
↳ 1cal
/g°c
Capacity .
Jules grams (
heat
q = m • C • ∆T
( mass ) ( Change in
A in C°= A in k° Container
( heat ) ( specific
Temp , rake / ✓ in
heat )
o
Note: ∆T in C is equal to ∆T in K
Practice Problems
1) When 335 J of heat is added to 2.25 g of vegetable oil
at 25.0oC, the temperature increases to 75.0oC. What is
the specific heat of the vegetable oil.
9 = m .
c. AT
335J =
2.259
'
C .
50.0¢
9 335J
2.98519°C
=
= C
G. 2s g) ( 50.0 C)
◦
MIT
-13,5091
J
10%0
0.903
g- = ✗
✗
DT -10.0C
-
, ¢
( =
0.903J / gk
9- =
1219.05J → 1220J
3) A chunk of gold has a heat capacity of 34.5 J/oC and a
mass of 145 g. Calculate the specific heat of gold.
34.5J
° -238 J
=
/ goc
14 Sgoc
3. Select the “Energy Symbols”
and “Link Heaters”
Describe what occurred when the hot iron piece was added to the water
(think of the E’s and the temperature changes).
4) A small marble chip with a mass of 3.25 g is heated and
placed in a foam cup calorimeter containing 28.0 mL
(density of water is 1.00 g/mL) of water at 22.0oC. The
water reaches a maximum temperature of 25.7oC. How many
joules of heat were released by the marble chip? What is
the specific heat capacity of marble chip?(Starting
temperature of the marble chip was 77.0oC)
AT 9- chip =
f) 9-1*0
M C .
AT
,
3.25
51.3°C
=
-51,3°C
9-
-
water
Chip
m c AT 9-s-m.c.AT
9. =
28.0g .
4,18 J/g°C .
3.70
9-
9- Hzos 433.0J = C
m.IT
28-0/9 4.18 J
3.7¢ -433.05J
× c
,
433.048J
,
=
1 ☒% / (3.259×-51.30)
-
433.05J
=
C
-166.72 goc
9- Ito -
433.0J
2.59%4 :C
9- chips -433,0J
5) A 46.2 g sample of copper is heated to 95.4oC and then
placed in a calorimeter containing 75.0 g of water at
19.6oC. The final temperature of both the water and the
copper is 21.8oC. What is the specific heat of copper?
water
water
9- ÷ 75.019 4.18J 2.2¢
✗ ×
copper I ,g,¢ 1
Mass 46,29 75.09 9- =
689.7J
C 0.20%4 4.18519°C
2. zco
Copper
AT -73.6C
9-
9- c=
-689.7J 689.7J
m.AT
( = 0.20J / got
-689.7J
[=
-3400,329°C