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Our Chemistry Team has worked hard to bring you the worked solutions to the 2015 Chemistry HSC Paper's multiple choice section.
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Do you struggle with past paper multiple-choice sections? It’s one thing to be able to see what the right answer is when NESA releases the past papers, it’s another to understand how they got to that answer. Fear not! Matrix has got your back! In this post, our Chemistry Team give you the worked solutions for the 2015 Chemistry HSC Paper multiple-choice questions!
Read on and learn why you got it right or wrong, so you can get it right every time!
| Question | Answer | Explanation |
| 1 | B | The UV protective ozone layer is located in the stratosphere. |
| 2 | D | A burette is used for this purpose in a titration. The known volume of the other solution has been measured using a pipette and transferred to a conical flask. |
| 3 | B | Copper ions give a blue-green flame test. |
| 4 | C | Fe2+ reacts to form Fe3+. This requires the loss of one electron: Fe2+(aq) → Fe3+(aq) + e–, which is an oxidation process. |
| 5 | A | Non-metal oxides tend to form acidic solutions (pH less than 7) while metal oxides tend to form basic solutions (pH greater than 7). |
| 6 | B | CH2F2 is a HFC that does not contain chlorine or bromine atoms, therefore it will not release ozone-destroying free radicals upon photodissociation in the stratosphere. CH4 also doesn’t contain chlorine, but since it is not a haloalkane it will not have the appropriate properties to act as an effective replacement for CFCs in its applications (aerosols, refrigeration etc.). |
| 7 | B | Copper is more active than silver (higher on the Table of Standard Reduction Potentials), therefore copper atoms from the electrode are oxidised and will lose electrons. |
| 8 | C | Octane is a non-polar hydrocarbon, therefore it only forms dispersion forces with other substances. Ethanol can form dispersion forces, dipole-dipole interactions and hydrogen bonds, depending on what the other substance is. |
| 9 | C | The structure given is an ester. The side with the doubly bonded oxygen comes from the alkanoic acid (butanoic acid), the other carbon chain comes from the alkanol (propan-1-ol). During esterification a water molecule is condensed out to form the ester functionality. |
| 10 | A | Incomplete combustion produces incompletely oxidised carbon products (carbon soot C and carbon monoxide CO). While (D) also produces the correct products, it is not a balanced equation. |
| 11 | D | Condensation involves the elimination of a small molecule such as water. Combining the monomers shown produces structure (D) as the condensation polymer. |
| 12 | A | The equation must be balanced so the sum of the top numbers (mass numbers) are the same on both sides, and the sum of the bottom numbers (atomic numbers) are the same on both sides. For this to occur X must be a neutron (10n) and Y must be an electron (0-1e). |
| 13 | B | The highest pH solution will be the least acidic. Acetic acid is weaker than hydrochloric acid, so it will have a higher pH. 0.10 M is the lower concentration. |
| 14 | D | The indicator colour change range should match the titration equivalence point (at the steepest part of the titration curve). This is in the 8.3-10.0 region. |
| 15 | A | X is near farmland and a village.
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| 16 | C | To increase the yield of Z, the equilibrium needs to shift to the right. This can be predicted using Le Chatelier’s principle:
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| 17 | D | C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(g) n(C6H12O6) = m / MM = 10.3 / (12.01 x 6 + 1.008 x 12 + 16 x 6) = 0.05717 mol n(CO2) = 0.05717 x 2 = 0.114345 mol V(CO2) = n x MV = 0.114345 x 24.79 = 2.83462 L = 2.83 L (3 s.f.) |
| 18 | A | Reliability can be improved by repeating the same procedure, removing outliers and averaging the concordant results. |
| 19 | C | Pb2+(aq) + 2Cl–(aq) → PbCl2(s) n(PbCl2) = m / MM = 0.13 / (207.2 + 35.45 x 2) = 4.6746 x 10-4 mol n(Pb2+) = 4.6746 x 10-4 mol c(Pb2+) = n / V = 4.6746 x 10-4 / 0.05 = 9.349 x 10-3 mol L-1 = 9.3 x 10-3 mol L-1 (2 s.f.) Note that for this gravimetric analysis procedure to work, an excess of chloride ions must be added to precipitate all of the lead ions present in the sample. You can check this from the information given in the question: n(Cl-) = c x V = 0.2 x 0.025 = 5 x 10-3 mol, which can precipitate a maximum of 2.5 x 10-3 mol lead ions. |
| 20 | D | Heptan-1-ol contains 7 carbon atoms, so ΔHc = 4638 kJ mol-1. n(C7H15OH) = m / MM = 1 / (12.01 x 7 + 1.008 x 16 + 16) = 8.606 x 10-3 mol q = ΔH x n = 39.9146 kJ q = mcΔT m = q / cΔT = 39914.6 / (4.18 x 103 x (45 – 20)) = 0.38196 kg = 0.38 kg (2 s.f.) |
Written by Michelle Wong
Michelle is the Senior Chemistry Teacher at Matrix Education, and has been tutoring Chemistry and Mathematics for over 15 years.
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