ASOE Chemistry 2021 Reduced-FS-1
ASOE Chemistry 2021 Reduced-FS-1
ASOE Chemistry 2021 Reduced-FS-1
CHEMISTRY
TO BE COMPLETED BY THE STUDENT. USE CAPITAL LETTERS.
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2021 Australian Science Olympiad Examination - Chemistry
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2021 AUSTRALIAN SCIENCE OLYMPIAD EXAM
CHEMISTRY
Time Allowed
INSTRUCTIONS
MARKS
SECTION A 15 multiple choice questions 30 marks
SECTION B 3 short answer questions 30 marks each
Integrity of Competition
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Avogadro constant (N) = 6.022 × 1023 mol–1 Velocity of light (c) = 2.998 × 108 m s–1
1 faraday = 96 485 coulombs Density of water at 25 °C = 0.9971 g cm–3
1 coulomb = 1 A s–1 Acceleration due to gravity = 9.81 m s–2
Universal gas constant (R) 1 newton (N) = 1 kg m s–2
8.314 J K–1 mol–1
8.206 × 10–2 L atm K–1 mol–1
Planck’s constant (h) = 6.626 × 10–34 J s 1 pascal (Pa) = 1 N m–2
Molar volume of ideal gas pH = −log10[H+]
• at 0 °C and 100 kPa = 22.71 L pH + pOH = 14.00 at 25°C
• at 25 °C and 100 kPa = 24.79 L Ka = {[H+] [A–]} / [HA]
• at 0 °C and 101.3 kPa = 22.41 L pH = pKa + log10{[A–] / [HA]}
• at 25 °C and 101.3 kPa = 24.47 L PV = nRT
E = hν
Surface area of sphere A = 4πr2 c = νλ
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SECTION A: MULTIPLE CHOICE
USE THE ANSWER SHEET PROVIDED
B. S2−, Cl−, K+
3. A component of diesel fuel is the hydrocarbon C12H24, with density 0.790 g mL−1. What
volume of CO2 (measured at 25 °C and 100 kPa) is produced from the complete
combustion of 2.00 L of C12H24 in excess oxygen?
A. 2.79 L
B. 3.53 L
C. 1400 L
D. 2790 L
E. 3530 L
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4. Which of the following compounds has a trigonal pyramidal geometry?
A. NCl3
B. CH2Cl2
C. COCl2
D. CH4
E. BCl3
5. When the given amounts of each reagent are mixed together, which of the following will
release the largest mass of CO2?
6. Biological tissue samples are often stained with dyes, which are coloured organic salts.
• Basic dyes consist of a coloured cation and a colourless anion.
• Acidic dyes consist of a coloured anion and a colourless cation.
Which of the following dyes are acidic? Select all that apply.
A. C21H22N3Cl
B. C25H33N2O2Cl
C. C16H17N2ClS
D. C19H17N2NaO5S
E. C33H43N3Na2O8S2
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7. What is the total number of valence electrons in the PO23− ion?
A. 15
B. 17
C. 20
D. 30
E. 34
A. SF4
B. SO2Cl2
C. SOCl2
D. SF6
E. S2F10
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10. First ionisation energy is defined as the energy required to remove one mole of electrons
from one mole of gaseous ions. Which of the following lists elements in order of increasing
first ionisation energy?
A. C, F, N, Li
B. C, N, Li, F
C. Li, C, N, F
D. Li, N, F, C
E. F, N, C, Li
F. F, Li, N, C
A. 1.4 × 1022
B. 9.6 × 1022
C. 1.4 × 1025
D. 9.6 × 1025
E. 9.6 × 1028
12. Acrylonitrile (C3H3N) can be synthesised industrially according to the following chemical
equation:
When 100 kg of C3H6, 50 kg of NH3 and 125 kg of O2 are mixed, which of these reactants
is present in excess? Select all that apply.
A. C3H6
B. NH3
C. O2
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13. Which of the following is both an empirical formula and a molecular formula?
A. C3F6
B. C3F8
C. C4F6
D. C4F8
E. C4F10
A. W
B. Se
C. Mo
D. Rh
E. U
15. Of the following elements, which has the highest third ionisation energy?
A. Ar
B. Si
C. Mg
D. Al
E. Cl
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Question 16
A B C D
The mass of each compound, expressed as a percentage of the original mass of A, is recorded
in the table below.
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(d) Identify the formula of Compound C.
(f) Calculate the concentration of the oxalic acid solution (in mol L−1).
Oxalic acid reacts with sodium hydroxide to produce sodium oxalate and water, according to
the following chemical equation:
H2C2O4(aq) + 2NaOH(aq) → Na2C2O4(aq) + 2H2O(l)
20.00 mL of the oxalic acid solution above requires 18.57 mL of a sodium hydroxide solution
for complete reaction.
(g) Calculate the concentration of the sodium hydroxide solution (in mol L−1).
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The ammonium ion content of a salt can be determined using the following procedure. A
1.988 g sample of an ammonium salt is placed in a flask and heated with 50.00 mL of
0.5493 mol L−1 potassium hydroxide solution (a known excess). The ammonium and
hydroxide ions react to produce water and ammonia, which is expelled from the flask by
evaporation:
The potassium hydroxide remaining in the flask after all of the ammonia is expelled is
determined with 0.1032 mol L−1 hydrochloric acid, 23.89 mL of which is required for
complete reaction.
(h) Calculate the amount (in mol or mmol) of hydrochloric acid added.
(i) Calculate the amount (in mol or mmol) of potassium hydroxide added in the original
50.00 mL sample.
(j) Calculate the amount (in mol or mmol) of ammonium ions in the 1.988 g ammonium salt
sample.
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(k) Calculate the percentage by mass of ammonium ions in the 1.988 g ammonium salt
sample.
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In the absence of volumetric glassware, it is possible to use only mass measurements to
determine the composition of solutions.
20.58 g of KHP (KC8H5O4) is dissolved in water, giving a solution with a mass of 118.48 g.
(m) Calculate the mass of water (in g) that must have been added to the KHP to make the
118.48 g solution.
A solution of sodium hydroxide is also prepared. 4.471 g of this sodium hydroxide solution
reacts completely with 5.979 g of the KHP solution above. Sodium hydroxide reacts in a 1:1
mole ratio with KHP.
In a similar reaction, 4.359 g of the sodium hydroxide solution reacts completely with a
5.925 g sample of vinegar (containing acetic acid, CH3COOH). Sodium hydroxide reacts in a
1:1 mole ratio with acetic acid.
(n) Calculate the mass (in g) of the KHP solution required to react completely with 4.359 g
of the sodium hydroxide solution
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(o) Calculate the amount of pure KHP (in mol or mmol) required to react completely with
4.359 g of the sodium hydroxide solution.
(p) Calculate the percentage by mass of acetic acid present in the vinegar sample.
(q) If the density of the sodium hydroxide solution is 1.045 g mL−1, calculate the
concentration of the sodium hydroxide solution (in mol L−1).
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Question 17
The following question will explore the role symmetry has to play in chemistry, and how it
affects the physical and chemical properties of molecules.
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Recall that the electron geometry of a molecule is determined by the number of domains (i.e.
lone pairs or atoms) around the central atom, as given in the table below.
Number of Electron
Top View Side View
domains Geometry
2 Linear
Trigonal
3
Planar
4 Tetrahedral
Trigonal
5
Bipyramidal
6 Octahedral
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For example, the central oxygen in water (drawn below) is bonded to two hydrogen atoms
and contains two electron lone pairs. As such, there are four domains around the central
oxygen and its electron geometry is tetrahedral.
To determine the molecular geometry of a molecule, we only look at the geometry of the
atoms surrounding a particular molecule (not including lone pairs). As an example, the
possible configurations for 4 domains are shown in the table below.
4 0 Tetrahedral
Trigonal
4 1
Pyramidal
4 2 Bent
For example, since water has two lone pairs and four domains around the central oxygen, it
has a bent molecular geometry.
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(e) Identify the number of domains and lone pairs bonded to the central atom in each of the
following molecular geometries.
(i)
Top View Side View
Domains:
Lone pairs:
(ii)
Top View Side View
Domains:
Lone pairs:
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(iii)
Top View Side View
Domains:
Lone pairs:
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Now that we have revised molecular geometries, we can consider the symmetries that
molecules may have. The first type of symmetry we will consider is reflectional symmetry.
To be precise, a molecule has a mirror plane of symmetry if reflecting the molecule through
that plane leaves the molecule exactly the same. For example, water has two planes of
symmetry, shown below.
The dotted rectangle indicates a plane of symmetry in the plane of the page.
However, the following is not a plane of symmetry of water, since it does not leave the
configuration of the atoms the same:
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(f) How many planes of symmetry does a CH2O molecule have?
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The second type of symmetry we consider is rotational symmetry. A molecule has an n-fold
360∘
rotational symmetry about an axis (where n ≥2), if rotating around that axis by an angle 𝑛𝑛
leaves the molecule the same. For example, water has a 2-fold (180°) rotational symmetry
about the vertical axis while ammonia (NH3) has a 3-fold (120°) rotational symmetry about
the vertical axis:
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The following XeF4 molecule has five axes of symmetry.
(k) How many axes of symmetry of each degree, n, does XeF4 have?
(l) Which of the following molecules has four 3-fold axes of symmetry and three 2-fold axes
of symmetry?
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We can now discuss the first application of symmetry to the properties of molecules - dipole
moments. A molecule is non-polar if it contains two or more different rotational axes of
symmetry, or if it contains a mirror plane of symmetry perpendicular to a rotation axis.
Otherwise, the molecule is polar.
First consider ethene, whose molecular geometry is given below. Note that the hydrogen
atoms on each side of the double bond cannot rotate relative to each other.
Now consider the following molecules, which have been formed by replacing some of the
hydrogen atoms in ethene with chlorine atoms.
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Next, consider allene, whose molecular geometry is given below. Note that the hydrogen atoms
on each side of the double bond cannot rotate relative to each other.
Now consider the following molecules, which have been formed by replacing some of the
hydrogen atoms in allene with chlorine atoms. Note that in the diagrams below, the wedged
line refers to the atoms coming out of the page in the side view, while the dashed line refers to
the atoms going into the page.
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(m) Consider a water molecule. Which of the following factors would turn water into a more
polar species?
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At the start of the question, we considered the electron and molecular geometry of molecules
consisting of a central atom connected to up to six electron domains. In the next part of the
question, we use the symmetry elements discussed above to determine the molecular geometry
of XeI82−, an ion with eight electron domains around the central xenon atom. Its Lewis structure
is provided below:
We consider four possible molecular geometries for XeI82−: cubic, hexagonal bipyramidal,
square antiprismatic, and bicapped trigonal prismatic, each pictured below.
Molecular
Top View Side View Polyhedron
Geometry
Cubic
Hexagonal
Bipyramidal
Square
Antiprismatic
Bicapped
Trigonal
Prismatic
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Replacing one iodine atom in XeI82− with chlorine gives a number of different molecules. How
many different molecules would this result in for each of the possible geometries?
Cubic
Hexagonal Bipyramidal
Square Antiprismatic
4 marks
• When one of the iodine atoms is replaced with chlorine, giving XeClI72−, only one
molecule is produced.
• When two of the iodine atoms are replaced with chlorine atoms, giving XeCl2I62−, none
of the molecules are non-polar.
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Question 18
When representing organic compounds, skeletal formula notation is often used for simplicity.
In this notation, bonds are still represented by lines, but the symbol for carbon atoms is not
used. Hence, the end of a line segment or the meeting point of line segments indicate carbon
atoms.
Hydrogen atoms connected to carbon atoms are implied rather than explicitly shown. Any
other elements are shown.
(b) How many hydrogen atoms are present in the following organic compounds?
(i)
(ii)
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Nucleotides are the building blocks of nucleic acid macromolecules such as DNA and other
forms of genetic material. Nucleotides have three main components; a five-membered sugar,
a phosphate group, and a nucleobase. There are different types of nucleobases which produce
different nucleotides.
Adenine and thymine are examples of nucleobases, and their structure is shown below. (The
‘R’ group is shorthand notation for the rest of nucleotide chain that the base is attached to,
which has no relevance to this question.)
One example of an intermolecular interaction between these two nucleobases has been shown
with a dashed line:
When identifying hydrogen bond interactions, a group that can be either a hydrogen bond
donor or acceptor, will first and foremost, be a hydrogen bond donor.
(c) Identify the type of intermolecular force interactions that can occur at the following
positions:
Adenine N-1
(i) Can it be a hydrogen bond donor?
□ yes
□ no
(ii) Can it form any of the following interactions? Select NA if you answered "yes" to
the hydrogen bond donor question above.
□ NA
□ hydrogen bond acceptor
□ no significant interaction
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Adenine C-2
(i) Can it be a hydrogen bond donor?
□ yes
□ no
(ii) Can it form any of the following interactions? Select NA if you answered "yes" to
the hydrogen bond donor question above.
□ NA
□ hydrogen bond acceptor
□ no significant interaction
Adenine N-10
(i) Can it be a hydrogen bond donor?
□ yes
□ no
(ii) Can it form any of the following interactions? Select NA if you answered
"yes" to the hydrogen bond donor question above.
□ NA
□ hydrogen bond acceptor
□ no significant interaction
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The intermolecular interaction depicted in the previous question (and below) is known as
base pairing, and is responsible for holding nucleotide strands together in DNA.
Base pairing is only possible if there are feasible intermolecular interactions between the
nucleobases. This depends on their type of intermolecular interactions possible, and whether
the groups are close enough to interact. In the diagram, there is a base pairing interaction
between adenine N-10 and thymine O-8.
(d) Identify another feasible base pairing interaction between adenine and thymine.
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Cytosine is another nucleobase, two different forms of which have been shown below,
denoted structure 1 and structure 2.
Recall that when identifying hydrogen bond interactions, a group that can be either a
hydrogen bond donor or acceptor, will first and foremost, be a hydrogen bond donor.
(e) Identify the type of intermolecular force interactions that can occur at the following
positions:
Structure 1 N-3
(i) Can it be a hydrogen bond donor?
□ yes
□ no
(ii) Can it form any of the following interactions? Select NA if you answered "yes" to
the hydrogen bond donor question above.
□ NA
□ hydrogen bond acceptor
□ no significant interaction
Structure 2 N-3
(i) Can it be a hydrogen bond donor?
□ yes
□ no
(ii) Can it form any of the following interactions? Select NA if you answered "yes" to
the hydrogen bond donor question above.
□ NA
□ hydrogen bond acceptor
□ no significant interaction
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Structure 1 O-7
(i) Can it be a hydrogen bond donor?
□ yes
□ no
(ii) Can it form any of the following interactions? Select NA if you answered
"yes" to the hydrogen bond donor question above.
□ NA
□ hydrogen bond acceptor
□ no significant interaction
Structure 1 N-8
(i) Can it be a hydrogen bond donor?
□ yes
□ no
(ii) Can it form any of the following interactions? Select NA if you answered
"yes" to the hydrogen bond donor question above.
□ NA
□ hydrogen bond acceptor
□ no significant interaction
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The two forms of cytosine shown in the previous question differ in their protonation state.
Protonation refers to the addition of a proton (H+) to a molecule.
A proton acceptor, A, can exist in one of two different protonation states:
• A
• AH+
Each protonated species has an associated acid dissociation constant (pKa), which describes
how the pH of its environment will affect its protonation state.
• When pH is equal to pKa, the two forms will be present in equal concentrations.
• When pH is less than pKa, the predominant species will be AH+.
• When pH is greater than pKa, the predominant species will be A.
Consider the generic organic compound, and its associated pKa data.
Group pKa
X 6.5
Y 11.0
Z 3.0
(f) Predict the protonation state that each group will be in at pH 10.
(i) X:
□ X
□ XH+
(ii) Y:
□ Y
□ YH+
(iii) Z:
□ Y
□ YH+
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The table below shows pKa data for N-1 and N-10 positions of the nucleobase, adenine.
Group pKa
N-1 4.15
N-10 9.80
(i) (ii)
(iii) (iv)
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Adenine regularly base pairs with thymine, by interacting with both:
1. ideal base pairing interactions
2. appropriate protonation states
One example of a correct combination has been provided below:
The table below shows pKa data for N-1 and N-10 positions of adenine.
Group pKa
N-1 4.15
N-10 9.80
(h) Choose the form or forms of adenine that can interact with thymine and have both:
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The table below shows pKa data for N-1 and N-10 positions of adenine.
Group pKa
N-1 4.15
N-10 9.80
(i) Select the option that best describes the pH range where both adenine and thymine can
form feasible base pairing interactions.
(ii) The pH must be greater than 4.15 and less than 9.8.
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The table below shows pKa data for N-3 and N-8 positions of the nucleobase, cytosine.
Group pKa
N-1 4.6
N-10 12.2
(i) (ii)
(iii) (iv)
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The table below shows pKa data for N-1, N-7 and N-10 positions of the nucleobase, guanine.
Group pKa
N-1 3.3
N-7 9.4
N-10 12.6
(i) (ii)
(iii) (iv)
(v) (vi)
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Cytosine regularly base pairs with guanine.
The template below has been provided to indicate the orientation of the nucleobases when
they undergo base pairing (guanine on the left, cytosine on the right). The actual structures of
these nucleobases are shown in the question options.
The table below shows pKa data for N-3 and N-8 positions of cytosine.
Group pKa
N-3 4.6
N-8 12.2
(l) Choose the form or forms of cytosine that can interact with guanine and have both:
1. ideal base pairing interactions
2. appropriate protonation states
(i) (ii)
(iii) (iv)
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The table below shows pKa data for N-1, N-7 and N-10 positions of the nucleobase, guanine.
Group pKa
N-1 3.3
N-7 9.4
N-10 12.6
(m) Choose the form or forms of cytosine that can interact with guanine and have both:
1. ideal base pairing interactions
2. appropriate protonation states
(i) (ii)
(iii) (iv)
(v) (vi)
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Complete the following sentence with the appropriate choice.
(n) For cytosine and guanine to form feasible base pairing interactions, the pH must be
greater than:
END OF EXAM
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