NSW INDEPENDENT TRIAL EXAMS – 2012
CHEMISTRY PRELIMINARY HSC EXAMINATION
MARKING GUIDELINES
Section I
1
2
3
D B A
4
C
5
D
6
A
7
D
8
B
9
D
10
C
11
C
12
B
13
C
14
A
15
C
16
D
17
D
Section II
Question 21(a)
Criteria
 Correctly calculates the empirical formula
 Assigns the atomic mass for each element and determines the number of moles for
each element
 Assigns the atomic mass for each element OR determines the number of moles for
each element
Answer may include:
In 100 g of the compound: Na = 32.4 g
S = 22.5 g
O = 45.1 g
Na =
32.4
23
Na = 1.4
S=
22.5
32
O=
S = 0.7
18
B
19
A
20
B
Mark
3
2
1
45.1
16
O = 2.8
Mole ratio – divide by the smallest number: 0.7
Na = 2
S=1
O=4
Therefore the empirical formula is Na2SO4.
Question 21(b)
Criteria
Mark
2
 Explains that the empirical formula gives the ratio of atoms in the lattice because an
ionic compound is a repeating lattice
1
 Explains that the empirical formula occurs because an ionic compound is a repeating
lattice
Answer may include: An ionic compound consists of a repeating three-dimensional lattice of ions. The
formula is the ratio of atoms in the lattice, which is the same as the empirical formula.
Question 22(a)
Criteria
 Draws the Lewis dot structure for the chlorine molecule (1 mark)
 Draws the Lewis dot structure for the chloride ion (1 mark)
Answer may include:
Cl Cl
Cl
Chlorine molecule
Mark
2
OR
Chloride ion
Question 22(b) (i)
Criteria
 Writes a correctly balanced equation between a metal and chlorine gas
Answer may include:
Mg + Cl2 → MgCl2
Question 22(b) (ii)
Criteria
 Writes the correctly balanced half-equation for chlorine
Answer:
Cl2 + 2e- → 2Cl-
Mark
1
Mark
1
NSW Independent Trial Exams 2012 – Chemistry Yr 11 Preliminary Examination: Marking Suggested Responses –
Page 1
Question 23(a)
Criteria
 Draws the carbon dioxide molecule (1 mark)
 Draws at least TWO units of the silicon dioxide covalent array (1 mark)
Answer may include:
O=C=O
O
O
Si
O
Carbon dioxide
O
O
Si
Mark
2
O
O
Silicon dioxide
Question 23(b)
Criteria
Mark
3
 Identifies the differences in melting points and solubility (1 mark)
 Explains melting points in terms of structure (1 mark)
 Explains solubilities in terms of structure (1 mark)
Answer may include: Carbon dioxide consists of small non-polar molecules with weak intermolecular
forces. It melts at a low temperature and is able to dissolve in water. Silicon dioxide is a strongly
bound covalent array. The strong covalent bonds give it a high melting point and prevent it dissolving
in water to the same extent as carbon dioxide.
Question 24(a)
Criteria
Mark
2
 Outlines the work of Gay-Lussac (1 mark)
 Outlines the work of Avogadro (1 mark)
Answer may include: Gay-Lussac showed that the volumes of gases in a chemical reaction have simple
whole number ratios to one another (assuming constant temperature and pressure). Avogadro proposed
that equal volumes of gases contain the same number of molecules at the same temperature and
pressure.
Question 24(b)(i)
Criteria
 Correctly calculates volume of oxygen
Answer may include: 50.0 L SO2 = 2 volumes; volume O2 = 1 volume = 25.0 L
Mark
1
Question 24(b)(ii)
Criteria
Mark
1
 Correctly calculates volume of sulfur trioxide
Answer may include: Limiting reagent is SO2. 2 volumes SO2 produces 2 volumes SO3 = 15.0L.
Question 25(a)
Criteria
Mark
1
 Correctly defines first ionisation energy
Answer may include: The energy required to remove an electron from the neutral atom of an element
when in the gaseous state
Question 25(b)
Criteria
Mark
2
 Predicts correct group for X and justifies the decision
1
 Predicts correct group for X OR justifies the decision
Answer may include: X would most likely be found in Group 1. It is more reactive than magnesium
and calcium from Group 2 as it reacts rapidly with water at room temperature. It also has a lower
ionisation energy than both calcium and magnesium. As ionisation energy increases across a period
and decreases down a group, it would be expected that element X would be in group 1.
NSW Independent Trial Exams 2012 – Chemistry Yr 11 Preliminary Examination: Marking Suggested Responses –
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Question 25(c)
Criteria
 Constructs TWO correct half-equations (states not necessary)
Answer may include: Mg(s) → Mg2+ + 2e–
2H+ + 2e– → H2(g)
Mark
2
Question 26(a)
Criteria
Mark
2
 Outlines at least 4 main steps for recycling of aluminium
1
 Outlines at least 2 or 3 main steps for recycling of aluminium
Answer may include:
1. Aluminium cans are collected and stored at local collection centers.
2. The cans are screened to remove unwanted substances, e.g., food, labels and ,magnets are
used to remove steel.
3. The cans are then crushed into large blocks and melted in a furnace.
4. The liquid aluminium is poured into moulds and cooled to produce ingots.
5. Aluminium ingots are then rolled thinly and made into new cans.
Question 26(b)
Criteria
 Correctly calculates mass of aluminium hydroxide
 Calculates the moles of alumina OR applies the 1:2 mole ratio for aluminium
hydroxide
Answer may include:
Moles alumina = 742 ÷ 101.96 = 7.27 mol.
Mole ratio is 1:2, so moles aluminium hydroxide = 7.27 x 2 = 14.54 mol.
Mass aluminium hydroxide = 14.54 x 78.004 = 1134 g
Mark
2
1
Question 27(a)
Criteria
Mark
2
 A practical method is described and includes an observation indicating the solubility
of silver chloride
1
 A correct observation about silver chloride’s solubility is indicated
Answer may include: A drop of a silver nitrate solution was placed on a clear plastic sheet using an eye
dropper. A drop of a solution containing sodium chloride solution was added. The mixture of drops
would take on a white, cloudy appearance if silver chloride is insoluble.
Question 27(b)
Criteria
 A correct net ionic equation is given (solid must be indicated)
Answer may include: Ba2+ + SO42– → BaSO4(s)
Mark
1
Question 27(c)(i)
Criteria
 Identifies sodium carbonate
Answer may include: Sodium carbonate being insoluble
Mark
1
Question 27(c)(ii)
Criteria
 A correct set of solutions is named
Answer may include: Sodium chloride and potassium carbonate solutions
Mark
1
NSW Independent Trial Exams 2012 – Chemistry Yr 11 Preliminary Examination: Marking Suggested Responses –
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Question 28(a)
Criteria
 Correct Lewis electron dot diagrams are drawn and the shapes are described as bent
 Correct Lewis electron dot diagrams drawn OR the shapes are described as bent
Answer may include:
O
H H
Mark
2
1
S
H H
The molecules have similar structures and both have a bent shape.
Question 28(b)
Criteria
Mark
2
 Identifies water as being more polar leading to hydrogen bonds and that these stronger
intermolecular bonds raises the BP of water
1
 Identifies water as being more polar leading to hydrogen bonds OR that stronger
intermolecular bonds raises the BP of water
Answer may include: Water is more highly polar than hydrogen sulfide, leading to strong hydrogen
bonding between its molecules. With much stronger intermolecular forces, it has a much higher
boiling point.
Question 29(a)
Criteria
 Correctly calculates the molar concentration of calcium chloride
 Applies the mole ratio 1:2 correctly
Answer may include:
n = m/M = 10 g/110.98 g mol–1 = 0.0901 mol CaCl2
c = n/V = 0.0901 mol/0.250 L = 0.360 mol L–1
Cl– ions = 0.721 mol L–1
Question 29(b)
Criteria
 Draws a diagram showing the correct orientation of water molecules surrounding a
chloride ion
Answer may include:
Mark
2
1
Mark
1
Question 29(c)
Criteria
Mark
2
 Identifies the non-polar nature of carbon tetrachloride and applies the “like dissolves
like” principle
1
 Identifies the non-polar nature of carbon tetrachloride OR applies the “like dissolves
like” principle
Answer may include: A solution forms only when solute and solvent have similar intermolecular
forces (“like dissolves like”).
Carbon tetrachloride is a non-polar substance with only weak intermolecular forces (dispersion forces
only) and so will not dissolve in water, which is highly polar.
NSW Independent Trial Exams 2012 – Chemistry Yr 11 Preliminary Examination: Marking Suggested Responses –
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Question 30(a) (i)
Criteria
 Identifies water
Answer: Water
Mark
1
Question 30(a) (ii)
Criteria
 Correctly states the effect of an increase in temperature on viscosity
Answer may include: The substances would flow faster.
Mark
1
30(b)
Criteria
Mark
2
 Explanation includes forces (cohesive or hydrogen bonds or polarity) and
a minimum surface area
1
 Explanation includes forces (cohesive or hydrogen bonds or polarity) OR
a minimum surface area
Answer may include: Water molecules form relatively strong cohesive forces (hydrogen bonds)
between them. Unbalanced forces at the surface of the water cause it to form a sphere to minimise its
surface tension.
Question 31(a)
Criteria
Mark
2
 Procedure contains three or more correct steps and names a silver salt
1
 Procedure contains three or more correct steps OR names a silver salt
Answer may include:
1. Prepare a sample of wet silver chloride.
2. Quickly spread the AgCl over a filter paper with a stirring rod.
3. Place the filter paper flat on a white tile and cover half the filter paper with black cardboard.
4. Leave the other half exposed to bright light (eg. direct sunlight).
5. After 10 minutes remove the black cardboard and compare the two halves of the filter paper.
Question 31(b)
Criteria
 Identifies one correct safety precaution related to the investigation
Answer may include: Wearing gloves to avoid skin contact with silver nitrate.
Mark
1
Question 31(c)
Criteria
 Identifies one practical application related to light on a silver salt
Answer may include: photography.
Mark
1
Question 32
Criteria
 Calculates the mass of silver chloride formed and includes correct units
 Calculation is incorrect but working includes a correct equation and a correct use of a
mole formula
 Calculation is incorrect but working includes a correct equation OR a correct use of a
mole formula
Answer:
Mark
3
2
1
NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq)
nNaCl = c x V = 0.10 x 0.10 = 0.01
nAgNO3 = c x V = 0.10 x 0.10 = 0.01
mole ratio is 1:1
NSW Independent Trial Exams 2012 – Chemistry Yr 11 Preliminary Examination: Marking Suggested Responses –
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∴ n AgCl formed = 0.01
∴ m AgCl formed = n x M = 0.01 x (107.9 + 35.45) = 1.43 g
Question 33(a)
Criteria
 Calculates the heat released during the reaction and includes correct units
 Calculates the heat released during the reaction OR demonstrates that the correct
enthalpy formula was used.
Answer:
H = -mCT = 100 x 4.18 x 1.40
= 585.2 J
Mark
2
1
Question 33(b)
Criteria
Mark
1
 One relevant and correct limitation is suggested
Answer may include: The apparatus allows heat to be lost directly to the surroundings through the top
of the foam cup.
End of paper
NSW Independent Trial Exams 2012 – Chemistry Yr 11 Preliminary Examination: Marking Suggested Responses –
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