Section I part A

The views expressed in this Answers and Comments document are those of the author who had nothing to do with setting or marking the paper. The answers are not in any way 'official'. They are simply the responses of a competent chemist familiar with the syllabus and the textbooks in common use. The answers are probably more detailed than would be required to gain full marks from HSC examiners but are a worthy goal for students to aspire to.

You need a copy of the exam paper to make sense of the answers given here. If you do not already have one you can get a copy from the Board of Studies web site www.boardofstudies.nsw.edu.au . Click on 'HSC exams' in the left hand column, select '2012 HSC exam papers' then find 'Chemistry' in the alphabetical list. Click on 'Examination Paper' to open it or right click on it to save it.

Notes from the Marking Centre, Sample Answers and Marking Guidelines are also available there. You might like to look at these. Answers are not given for all questions, some of the samples are not complete answers and some do not seem to answer the precise question asked, but overall they are of considerable help in giving you an idea of what the examiners were expecting – often a much lower level of sophistication than perhaps the question implied.

Some general comments

The 2012 exam paper continues the practice of putting heavy emphasis upon recall of factual knowledge (often with some interpretation of the recalled material being required) as opposed to problem-solving. This year recall of experimental procedures only occurs in the options. The table below gives a breakdown of marks versus type of exam question over several years.

Type of question		Marks
Type of question		2005	2006	2010	2012
1	numerical calculation^{a b}	19 to 23	22	18 to 21	19 to 22
2	non-numerical problem-solving^{b c}	11 to 14	14 to 16	17 to 21	12 to 16
3	straight recall of factual material (describe, outline, summarise ...)	32 to 34	30 to 36	34 to 48	42 to 47
4	recall of factual material with some interpretation (evaluate, assess, analyse, discuss, compare ...)	28 to 31	29 to 35	8 to 20	18
5	descriptions of experiments (including risk assessments)	5	6	5 to 9	0 to 5
6	number of extended response questions^d	5	6	3	2
7	marks for extended response questions (included in categories 3 and 4 above)	30	22	17	13

^a Includes drawing graph if required
^b There are ranges because different options distribute marks somewhat differently
^c Includes writing an equation or drawing a structure not given in a textbook
^d Parts (a), (b), (c) and (d) in options questions count as separate questions

Averaged over these four years, about 35% of the marks have been for problem-solving (qualitative and quantitative) and 65% for recall of information (including description of experiments) with some interpretation of it.

Extended response questions

Extended response questions are the ones that require the most careful attention to make sure that you give the information asked for. With no guidance about how much emphasis to place on each aspect of the question, it is easy to give too much information about one aspect and neglect another equally important one. In tackling extended-response questions (questions that are just one or two sentences but worth 5 to 7 marks) you should start from the rule of thumb that the question requires one significant piece of information per mark, so a six mark question will require six pieces of relevant information; that probably means at least six sentences, often more. And it is going to require about 1.8 minutes per mark to answer as fully as the examiners want (11 minutes for a 6 mark question). Analyse the question carefully before you start writing to make sure you understand exactly what the question is asking for. If it contains two verbs such as saying 'describe' something and 'discuss' its importance or role in ..., then you need to split the marks between the two parts, say 3 for 'describe' and 3 for 'discuss'. If you feel that it is difficult to give 3 significant facts for the 'discuss' part, perhaps you could do a 4/2 split but a 5/1or 6/0 split would not be answering the question as asked. Even if you could easily give 6 significant facts for the 'describe' part, you must give some for the 'discuss' part even though that might be a struggle. Note that verbs such as evaluate and assess require you to make some judgment, so even if you cannot give a deep and insightful one, at least give some judgment even if it seems trivial, such as it's a good thing, it's really important or it's of no great significance, because the examiners are looking for some sort of judgment and they do not have to agree with your judgment to give you the mark (but they cannot give you the mark if no judgment is given).

Questions from the 2012 exam paper that fit into this extended-response category are Question 33 and part (e) of each option. Although Questions 26 and 29 each carry 5 marks, they are not 'extended-response' questions, because they request very specific bits of information.

If you finish a question with specific instructions ahead of time, then you have extra time to give to questions that are causing you difficulties. However if you seem to have finished an extended-response question in much less time than was allocated to it, and have used only a small proportion of the space allocated to the question in the exam book, then look back over it very carefully; you may have omitted an aspect of the question. If not, then it is very likely that you have not given as much detail as the examiner is really expecting.

With extended-response questions be very careful that you answer the question asked and not answer a related question that you know a good answer for. Try to answer the question exactly as asked. If you cannot give enough detail about that question to fill the time allocation, then perhaps you could provide related but not strictly asked-for information, but do that only as a last resort. One reason for suggesting this is that sometimes it is not possible to spend the allocated time on the question asked; it was not a good question.

Answers and comments

Section I part A (Multiple Choice questions)

With one mark per question, these multiple-choice questions should require 1.8 minutes per question. However the last few questions are more difficult (or at least more time-consuming) than the first few, so you need to proceed as quickly as you can through the early ones and not relax if you seem to be ahead of time, because you will need that saved time for the last few questions. Plan on spending 36 minutes in total on these multiple-choice questions rather than the same amount of time (1.8 minutes) for each question.

Question & Answer		Comment (and relevant page in CCHSC)
1.	B	Something you need to know (p.266-7)
2.	D	You need to consider the question in two parts: which of the four given compounds can be converted to a polymer, and which of those compounds can be made from C₂H₄, ethylene. Cellulose is already a polymer so A is not the answer. Ethanol cannot be converted directly to a polymer so B is out. Both glucose and styrene can be converted to polymers but only styrene can be made from ethylene (effectively by reacting it with benzene), so D is the answer. This question relies heavily on your knowledge of the polymers of ethylene. (p. 11, 14-6)
3.	A	Something you have to know. Generally a catalyst decreases the activation energy so D was meant to catch out careless readers. Catalysts have no effect on the position of equilibrium and they have no effect on the heat of reaction. (CCPC p.291-2, 296, CCHSC p. 202)
4.	C	The key piece of glassware for preparing a standard solution is a volumetric flask. C is the only answer that includes this so C is the correct answer. To confirm that C is correct, note that you have to weigh out an accurate amount of solid so a beaker could be used, and in washing the solid from the beaker to the volumetric flask a filter funnel is a useful precaution against spilling any of the solution. (p.157-60)
5.	B	All the equations are balanced so there is no clue in finding an unbalanced one. You need to know that fermentation takes place in the absence of air (no oxygen), so C and D are out, and that fermentation of glucose produces ethanol: hence B is the answer. (p. 33-5)
6.	C	There are many free neutrons in a nuclear reactor in which uranium-235 is being split into smaller nuclei in a chain reaction involving neutrons so this is the most convenient machine to use to make cobalt-60. Particle accelerators and cyclotrons (which are a type of particle accelerator) accelerate charged particles and so do not involve free neutrons. A scintillator is an instrument that detects alpha, beta and gamma rays so is of no use to make an isotope. (p. 81-2, 84-5)
7.	D	You need both methyl orange and bromothymol blue to be yellow (and phenolphthalein to be colourless) for the mixed solution to look yellow. Reading from the chart this is between 4.5 and 6. For pH 3 to 4.5 the solution will have a tinge of red in it. (The gaps between the heavy black lines in the chart are regions in which the colour is changing from one colour to the other, that is a mixed colour.) For 6 to 7.5 the solution will have a touch of blue in it. Choosing Answer A would reveal a complete lack of understanding of what such an indicator chart means. (p. 108-10, 139)
8.	D	A buffer is a solution that contains significant amounts of a moderately weak acid and of its moderately weak conjugate base. The dihydrogen phosphate, hydrogen phosphate pair is such a mixture. Nitric acid is a strong acid so C cannot be a buffer (HNO₃ will always be completely dissociated no matter how much nitrate is present). A and B are conjugates but A involves a strong acid H₃O⁺ and B a strong base OH^– so neither can act as a buffer. (p. 166-8)
9.	B	You need to know that a coordinate covalent bond is one formed by the sharing of a pair of electrons that originally came from just one of the partners – in this case the lone pair of electrons on the ammonia molecule is used to form a bond with a hydrogen ion. (p. 242-3)
10.	B	You need to know your solubility rules: BaCl₂ is soluble, BaSO₄ is insoluble so there will be no precipitate with chloride but there will be a precipitate with sulfate: hence answer B without having to worry about the flame colour. However if you know the common flame colours then the green flame confirms your choice. (CCPC p. 205, CCHSC p. 206-8 and inside back cover)
11.	B	The highest pH (lowest hydrogen ion concentration) will come from the weakest acid. HCl is a strong acid so is eliminated. Both acetic and citric acids are weak acids with acetic being the weakest so the answer is B. While I think it is unreasonable to expect HSC students to know that acetic acid is weaker than citric, there is a specific syllabus reference to the relative strengths of citric and acetic acids in Section 9.3.3 so I guess you have to memorise such details. It is a symptom of the current syllabus that students are required to memorise more and more facts and be able to recall them in HSC exams: I preferred the older (and more educationally sound) approach of testing students' ability to reason, to be able to solve problems, both numerical and non-numerical ones. (p.140-1)
12.		Actually none of the given answers is correct! The correct name for this compound is 2-chloro-3-fluorobutane. However proceeding logically you would have eliminated A because the F and Cl atoms are on different C atoms (2 ... 2 says they are on the same C atom). B and C are wrong because they do not list the substituents alphabetically, so that leaves D. And if Conquering Chemistry was your text and you had not checked the on-line errors page, you would have concluded that D was correct from Rule 5 on page 250. Unfortunately that rule is not correct. It should be If Rules 1 to 4 produce two names then the correct name is the one that gives the smaller number to the first-named of the alphabetically listed substituents. Hence 2-chloro-3-fluorobutane. A reader drew my attention to this error in the middle of 2012 (surprising no one had pointed it out to me earlier as the book has been available since 2005) and I posted a correction on the Conq Chem website in June 2012. Students who were aware of that correction or who had been taught the correct rule would have been in a dilemma over this question: no answer was correct. I imagine examiners took D as correct and probably accepted no answer as correct also (since there is no opportunity to lodge a 'none of the above' response on the answer sheet. Answers A, B and C are definitely wrong. The Board of Studies has recently drawn teachers attention to this matter in an Official Notice dated 1 March 2013 and has referred to a Royal Australian Chemical Institute (RACI) document on naming organic compounds.
13.	D	The oxidising agent is the substance which brings about oxidation, that is, which takes electrons from another reagent. In this case Fe²⁺ is losing electrons to form Fe³⁺ and dichromate Cr₂O₇^2– is bringing this about so dichromate is the oxidising agent. Surprisingly there is no mention of oxidising or reducing agents in the syllabus document for the core modules: it appears in the Industrial Chemistry option. I think it was remiss of me not to include the terms on p 42. (CCPC p. 117-8 CCHSC p. 42, 339)
14.	A	You either subtract the standard electrode potentials (1.36 – 0.77) or you add the half cell emfs (1.36 + (–0.77)). Answer D comes from adding the standard electrode potentials, B comes from using the wrong standard electrode potential for the iron species and adding the electrode potentials instead of subtracting them. It is not obvious how answer C arises. (p. 66-70)
15.	C	You just have to know which oxides are neutral (CO, N₂O, NO, H₂O), amphoteric (oxides of Al, Zn Pb, Sn, Cr), basic (metal oxides except the amphoteric ones) and acidic (those of non-metals that are not neutral), (p.115-6)
16.	A	First you need to identify the curves. The one starting from the origin is the product, methanol. Of the other two, the solid line is CO and the dotted line H₂ (because twice as much hydrogen disappears as does CO). At T₁ the concentration of H₂ instantly increases, meaning that H₂ is added then; at T₂ methanol instantly drops which means that methanol is removed. Hence answer A.
17.	C	Note that heat of combustion is per mole of propan-1-ol whereas the equation is in terms of two moles. 2021 kJ heat is released when 4 moles of water are formed. Moles of water formed when 1530 kJ heat released = (1530/2021) X 4 = 3.02 mol Molar mass of water is 18.0 so mass of water formed = 3.02 X 18.0 = 54.4 g. B assumes that only one mole of water is formed per mole of propanol. Mass of water = (1530/2021) X 1 X 18.0 = 13.6 g. D ignores heat data completely: says (wrongly) 1 mol propan-1-ol gives 8 mol water so mass of water formed = 8 X 18.0 = 144 g A uses 1/4 moles of water instead of 4. (CCPC p. 277-80, CCHSC p. 36)
18.	C	The pH change is easy to determine: dilute the acetic acid solution by a factor of two and the concentrations of all species will decrease so pH increases. The next bit is not quite so easy. Acetic acid is a weak acid so it is only partially ionised – that is, only some of the molecules of acetic acid have split into hydrogen ions and acetate ions. When we dilute the solution the ionisation equilibrium CH₃COOH + H₂O H₃O⁺ + CH₃COO^– moves to the right – we have decreased the concentrations of products, H⁺ and CH₃COO^– and increased the 'concentration' of water. Hence the equilibrium moves to the right (Le Chatelier's principle) and so the extent of ionisation is greater; that is a greater fraction of the acetic acid molecules has split into ions. So the answer is C. (p. 140-3)
19.	A	Moles of acetic acid to be neutralised = 0.10 X 0.50 = 0.0500 mol The equation is 2CH₃COOH + Na₂CO₃ ® H₂O + CO₂(g) + 2CH₃COO^– + 2Na⁺ so moles of Na₂CO₃ needed = ½ X 0.0500 = 0.0250 mol Mass of Na₂CO₃ needed = 0.0250 X (2 X 23.0 + 12.0 + 3 X 16.0) = 2.65 g Answer B assumes they react in a 1:1 ratio; answer C says moles of Na₂CO₃is twice the moles of acetic acid (instead of ½) and answer D says that the number of moles of acetic acid is 0.500 (confuses moles and molarity). (p. 157-62)
20.	A	Molar mass of lead chloride, PbCl₂ = 207.2 + 2 X 35.5 = 278.2 Mass of lead in the precipitate = (207.2/278.2) X 0.595 = 0.443 g Concentration of lead in the solution = 0.443 / 0.050 = 8.86 g/L B is obtained by taking the formula of lead chloride as PbCl. C calculates the concentration of lead chloride in the solution (0.595 / 0.050).It is not obvious how D was obtained. (CCPC p. 142-6, 217-9, CCHSC p. 220-1)