NSW INDEPENDENT TRIAL EXAMS – 2010
CHEMISTRY PRELIMINARY HSC EXAMINATION
MARKING GUIDELINES
Section I
1
D
2
B
3
D
4
C
5
C
6
A
7
C
8
A
9
C
10
B
11
C
12
A
13
B
14
D
15
B
16
D
17
C
18
B
19
C
20
A
Section II
Question 21(a)
Criteria
Mark
2
 Compares the polarity of ethanol and alkanes (petrol) (1 mark)
 Identifies the polar nature and hydrogen bonding of water and applies the “like
dissolves like” principle (1 mark)
Answer may include:
 Ethanol has a polar OH group making its molecules polar (and capable of H bonding).
 Alkanes are non-polar hydrocarbons.
 Water is strongly polar and dissolves other polar substances, such as ethanol. Non-polar substances
are insoluble in water.
Question 21(b)(i)
Criteria
Mark
2
 Removes the lower aqueous layer and retains the upper (petrol) layer (1 mark)
 Identifies a separating funnel as the required apparatus (1 mark)
Answer may include:
 The mixture is shaken in a stoppered separating funnel. Most of the ethanol dissolves into the water,
in which it is more soluble.
 The lower aqueous layer is run off and discarded.
The upper layer is the petrol component and is retained. (No distillation – too dangerous)
Question 21(b)(ii)
Criteria
Mark
2
 Eliminates any fire hazard (1 mark)
 Employs personal protection equipment (1 mark)
Answer may include:
 The experiment must be performed in a room with no naked flames or sources of electric sparks.
 Safety glasses must be worn and the procedures carried out in a fume hood, or well-ventilated area.
Question 21(c)(i)
Criteria
 Methane (1 mark) Formula only (1/2 mark)
Mark
1
Question 21(c)(ii)
Criteria
Mark
1
 Clearly describes the origin of natural gas (1 mark)
Answer may include:
Natural gas (like crude oil) formed from the decomposition by bacteria of marine organisms such as
algae and diatoms. Hence, its original origin is from photosynthesis.
NSW Independent Trial Exams 2010 – Chemistry Yr 11 Preliminary Examination: Marking Suggested Responses –
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Question 22(a)
Criteria
 Constructs Lewis diagrams for Al3+ and O2– (1 mark)
 States configurations for Al3+ and O2– (1 mark)
Answer may include:
..
Al3+
2,8
Mark
2
2–
:O:
..
2,8
Question 22(b)
Criteria
Mark
3
 Describes metallic bonding (1 mark)
 Describes ionic bonding (1 mark)
 Identifies bonding in aluminium and alumina (1 mark)
Answer may include:
 Aluminium metal has metallic bonding, consisting of a regular arrangement of aluminium ions
within a “sea” of delocalised electrons.
 Alumina has ionic bonding in which a lattice of aluminium ions and oxide ions is bound together by
forces of electrical attraction (ionic bonds).
Question 22(c)
Criteria
Mark
2
 Identifies TWO metallic properties of aluminium (1 mark)
 Identifies TWO properties of alumina (1 mark)
Answer may include:
 Aluminium is an excellent conductor and can easily be shaped (flexible, ductile, malleable).
 Alumina is a non-conducting solid which is brittle.
Question 23(a)
Criteria
Mark
2
 Calculates the proportion of Al in alumina from the formula (1 mark)
 Correctly determines the percentage yield as one-half of the above (1 mark)
Answer may include:
 The proportion of aluminium in alumina (Al2O3) is calculated as:
100 x (2  27.0)/(2  27.0 + 3  16.0) = 100  54/102 = 52.9%.
 As bauxite yields 50% alumina, the yield of aluminium from bauxite = ½  52.9 = 26.5%.
Question 23(b)
Criteria
Mark
2
 Relates the high energy requirement for extraction to alumina’s bonding (1 mark)
 States that recycling involves melting only, with low energy input (1 mark)
Answer may include:
 The electrolytic extraction of aluminium requires a large amount of energy to overcome the strong
ionic bonds in alumina.
 Recycling aluminium only requires melting and requires little energy to overcome the relatively
weak metallic bonds.
Question 24
Criteria
 Identifies ONE element with the stated property (1 mark each)
Answer may include:
Property
Element
Is the most reactive nonmetal
F
Consists of monatomic molecules
He, Ne, Ar, Kr (any one)
Is a semi-metal
(B), Si, Ge, (As)
Has the highest ionisation energy
He
Mark
4
NSW Independent Trial Exams 2010 – Chemistry Yr 11 Preliminary Examination: Marking Suggested Responses –
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Question 25(a)
Criteria
 Writes the balanced ionic equation (1 mark)
Answer may include: Li(s) + H+(aq) → Li+(aq) + ½H2(g)
Question 25(b)
Criteria
 Identifies and uses the 2:1 mole ratio (1 mark)
 Uses the Avogadro No. to find the number of molecules (1 mark)
Answer may include:
 As the mole ratio is 2:1 the number of moles of H2(g) n = 0.01/2 = 0.005 mol.
 The number of gas molecules = n  NA = 0.005  6.02  1023 = 3.01  1021 molecules.
Mark
1
Mark
2
Question 25(c)
Criteria
Mark
2
 Identifies and uses the 1:1 mole ratio for magnesium (1 mark)
 Shows that double the volume of gas is formed (1 mark)
Answer may include:
 The line is drawn showing double the volume for lithium.
 Magnesium forms 2+ ions and displaces hydrogen in a 1:1 mole ratio; double that for lithium.
Question 26 (a)
Criteria
Mark
2
 Describes trends in 1st ionisation across a period and down a group (1 mark)
 Relates low ionisation energy to high reactivity for metals (1 mark)
Answer may include: Reactive metals are those which lose their valence electrons most easily, meaning
that these metals have low ionisation energy. In the Periodic Table, ionisation energy increases across
each period and decreases down each group. Correspondingly metal activity decreases across each
period and increases down each group. Within the first 20 elements this means that potassium is the
most reactive metal, and beryllium (or aluminium) the least reactive.
Question 26 (b)
Criteria
 Writes a correct equation for this reaction. (1 mark)
 Correctly calculates the concentration of CaCl2 (1 mark)
Answer:
Ca + 2HCl → CaCl2 + H2
Mark
2
n Ca = 4.2 / 40.08 = 0.1047904
 n CaCl2 = 0.1047904
 conc CaCl2 = 0.1047904 / 0.0164 = 6.39 mol L-1
Question 27(a)
Criteria
Mark
Identifies magnesium hydroxide as the precipitate and writes an equation (1 mark)
1
2+
–
Answer may include: Magnesium hydroxide is precipitated. Mg + 2OH ®-←→ Mg(OH)2(s)
Question 27(b)
Criteria
 Identifies a suitable reagent in solution (1 mark)
 Constructs an ionic equation for the precipitation reaction (1 mark)
Answer may include:
Either: Potassium sulfate forms a precipitate with barium ion, but not zinc ion:
Ba2+(aq) + SO42–(aq) → BaSO4(s)
Or: Sodium hydroxide forms a precipitate with zinc ion, but not barium ion:
Zn2+(aq) + 2OH–(aq) → Zn(OH)2(s)
Mark
2
NSW Independent Trial Exams 2010 – Chemistry Yr 11 Preliminary Examination: Marking Suggested Responses –
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Question 27(c)
Criteria
Mark
2
 Shows the 1:2 effect of dilution. (1 mark)
 Determines the concentration of chloride ions (1 mark)
Answer may include:
 Silver ions are completely formed into a solid and removed from the solution. Dilution x 2 occurs.
The final concentration of chloride ion is 0.05 x 2 = 0.10 mol L–1.
Question 28(a)
Criteria
 Constructs the electron dot structural formula (1 mark)
 Draws the shape and polarity of the water molecule (1 mark)
..
Answer may include:
–
:O:H
..
H
Mark
2
O
H+
H+
Question 28(b)
Criteria
Mark
2
 Relates surface tension of substance to the strength of its intermolecular forces (1 mk)
 Identifies that polarity and hydrogen bonding create strong intermolecular forces in
water (1 mark)
Answer may include:
 The surface of a substance is determined by the strength of its intermolecular forces. The stronger
the forces, the higher the surface tension.
 Water is highly polar and forms extensive hydrogen bonds between molecules, giving a much
higher surface tension than non-polar molecules such as hydrocarbons.
Question 28(c)
Criteria
Mark
2
 Identifies the minerals of the Earth’s crust as containing many ionic compounds (1mk)
 Explains the solvent action of water in dissolving these minerals and contributing to
rock weathering (1 mark)
Answer may include: The Earth’s crust contains minerals with macromolecular and ionic structures.
Water acts as a solvent to dissolve ions out of the latter, especially sodium, potassium, calcium,
chloride and sulfate ions. This contributes to the breakdown of rocks in the process of weathering.
Question 29(a)
Criteria
 Describes the term ‘allotrope’ correctly (1 mark)
Answer may include: An allotrope is a different structural form of the same element.
Mark
1
Question 29(b)
Criteria
Mark
3
 Describes the network structures of diamond and graphite (1 mark)
 Relates each structure to melting point (1 mark)
 Relates each structure to ONE dissimilar property (1 mark)
Answer may include: Diamond and graphite have covalent network structures, each atom covalently
bonded to other carbon atoms. Melting involves breaking these covalent bonds between atoms, hence
diamond and graphite have similar melting points.
In diamond, each carbon atom is bonded to 4 other carbon atoms in a 3-dimensional network structure,
making it the hardest substance known. Since no electrons are delocalised or free to move throughout
the structure, diamond is a non-conductor.
In graphite, each carbon atom is bonded to 3 other carbon atoms in a 2-dimensional layered structure.
The fourth valence electron delocalised, being free to move along the layer, making graphite a good
conductor along the layers, which can slide over each other making graphite very soft.
NSW Independent Trial Exams 2010 – Chemistry Yr 11 Preliminary Examination: Marking Suggested Responses –
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Question 29(c)
Criteria
Mark
2
 Describes ONE use of graphite (1 mark)
 Relates the use to graphite’s properties (1 mark)
Answer may include: Use may be one of pencils and artist’s crayons, lubricant, electrodes, shafts of
sports equipment, turbine blades. Depending on the use chosen, relate to conductivity, softness and
layered structure, strength along covalent networked layers, colour and lustre.
Question 30
Criteria
Mark
3
 Explains how photosynthesis traps solar energy in plants (1 marks)
 Describes the process of formation of fossil fuels from once-living materials (1 mark)
 Explains that energy is stored in the molecules of fossil fuels (1 mark)
Answer may include: Fossil fuels have been produced from plant and animal material that has been
buried under the pressure of other sediments for millions of years.
The original plant and animal material was predominantly made of high energy carbohydrates
produced by the endothermic photosynthesis process that occurs in green plants. Photosynthesis
involves plants absorbing sunlight energy to allow the conversion of CO2 and H2O into carbohydrates,
including glucose, and O2.
Energy + 6CO2(g) + 6H2O(l) → C6H12O6 + 6O2(g)
The absorbed light energy is thus transformed into chemical potential energy, stored in the high energy
bonds in the carbohydrate molecules, which are further changed to hydrocarbons and other fuels. This
energy can be released as useful heat during the combustion of fossil fuels.
E.g.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) + energy
The statement is thus a good description of fossil fuels, especially considering the way in which fossil
fuels are used.
Question 31(a)
Criteria
 Identifies the series as ethenes (1 mark)
Question 31(b)
Criteria
 As molecular mass increases, dispersion forces increase and so density increases.
(1 mark)
Answer must include an increase in density and an increase in dispersion forces
Mark
1
Mark
1
Question 31(c)
Criteria
Mark
1
 Identifies TWO important conditions for safe storage (½ mark each)
Answer may include: These hydrocarbons are combustible gases and should be stored in cool
conditions, a well-ventilated area, and isolated from any source of ignition such as flames or electric
sparks.
NSW Independent Trial Exams 2010 – Chemistry Yr 11 Preliminary Examination: Marking Suggested Responses –
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