TA S M A N I A N Chemistry - tqa.tas.gov.au · TA S M A N I A N Chemistry CE R T IF ICA T ... undergone ‘excitation’ or ‘an electron being ... To gain full marks, the answer

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T A S M A N I A N Chemistry

C E R T I F I C A T E Subject Code: CHM315109 O F E D U C A T I O N 2011 Assessment Report

2011 Assessment Report

Part 1 – Criterion 5 Question 1 (a) Candidates identified that it was necessary to find the oxidation states, although many

incorrectly attributed incorrect values to H and O in H2O2. It was required to show a change in an oxidiser and reducer. Many candidates thought that demonstrating a change in oxidation state for one set of conjugates was sufficient to answer the question. They also failed to define that a redox reaction is one in which a transfer of electrons is demonstrated.

(b) Often candidates incorrectly concluded that nitrogen was the reducer rather than N2H4. Question 2 Candidates needed to include relevant ½ equations or Eo values. Many candidates believed that adding H2SO4 was supposed to cause a redox reaction not just acidify the permanganate ions. Question 3 Very few candidates did this question well, many were presumably put off by the inclusion of the spectator ion, SO4

2-. Credit was given for the two ½ equations Pb → Pb2+ + 2e- and PbO2 + 4H+ +2e- → Pb2+ + 2H2O and credit was given if the overall equation was derived from the two ½ equations. Question 4 (a) Candidates either achieved full marks or none. The markers tried to award part marks

for candidates' attempts but this was often not possible if the train of thought was wrong. The term ‘reference cell’ was confusing particularly as it was not specified as to which cell that was. A majority of candidates expended too much effort on this question and failed to complete the paper successfully or to give adequate time to final questions. Many candidates were unsure of what format was required for the correct list of reduction potentials. A significant number gave oxidation half equations and others gave a mixture of reduction and oxidation half equations

(b) Many candidates did not recognise that both were oxidisers. Many spoke about whether the Eo would give a spontaneous reaction.

(c) In this part the markers recognised and gave credit for 'errors carried forward' from part (a). Explanations involving statements such as ‘above or below on the table’ are not deemed appropriate.

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Subject Code: CHM315109


(d) Some credit was given if candidates said that metal ions could be oxidisers but needed to recognise that the metal could not.

Question 5 (a) Good in general. However, a half mark was deducted for not recognising the correct

states for the reactants and products (a molten electrolyte should be shown as a liquid and not aqueous). A number of candidates wrote the symbol for magnesium as Mn. Many candidates were unable to assign the electrical symbols for positive and negative and consequently incorrectly labelled cathodes and anodes.

(b) Some missed giving the required half equations or identifying the change in products. Question 6 (a) Generally very well done. (b) Candidates tried to respond to the first sentence in the situation rather than answer the

question. In answering, candidates needed to ensure that an observation was given; (being colourless, it's not possible to see the increase in [Mg2+

(aq)]). Many mistook this situation as electrolysis with water being involved after other ions had been used up.

Question 7 A straight forward question in the first two parts. Candidates found part (c) of the question ambiguous and 1 mark was given for any reasonable response. Question 8 For those candidates with time, this question was answered well. (a) Well answered but some annotation of the diagram was required. Half equations were

expected as well as a statement of the conditions required. Many candidates incorrectly marked the cathode and anode sites.

(b) Well answered by almost all candidates. Grease is not a noble coating – only a water and oxygen barrier. Appropriate half equation were expected and there was a need to discuss both situations. For those who answered this question, many candidates gave more information than the marks allocated deserved.

Part 2 – Criterion 6 In terms of degree of difficulty, this section was very well balanced as was indicated by the distribution of marks clustered around 25 marks (out of forty). This section was also good and

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unambiguous to mark, mainly because the mark distribution fitted well with the expected answers. Questions 11, 14 and 15 offered opportunities for candidates to demonstrate their level of understanding. It was noticed that many candidates showed signs of running out of time towards the end of this section, thus leaving out the last few questions; this may be an indication that they spent too much time on other sections or deciphering the questions. Question 9 (a) Most answered briefly and correctly that ethanol's use as a fuel is because the

combustion reaction is highly exothermic. Some went into comparison of whether products or reactants had higher bond energies – this was unnecessary for one mark.

(b) ∆H = –1368 kJ mol-1. The minus sign was missed in many cases and cost one mark. Lack of units cost half mark. The rest of the calculation was 2 marks, with one mark for correctly finding the mass (and moles) of ethanol (mass = 785 g and n = 17.1 moles) and one mark for completing the rest; E = 23.3 MJ. The most common mistake was that the mass of ethanol was 0.785 g. There were also several kinds of incorrect calculations for E, including the use of E = mc∆T.

(c) This question was answered well. However, mistakes included: o Rewriting and balancing the combustion equation (sometimes incorrectly) even

though it was supplied o Forgetting the C-C bond in C2H5OH o Using C-O bond instead of C=O in CO2 o Several candidates, mistakenly, calculated a positive ∆H, or an answer very far

from -1368 kJ, and made no connection with having discussed the exothermic properties of alcohol combustion in parts (a) and (b) of the question!

(d) Generally very well answered. Question 10 Surprisingly, very few answers were without some error or other. The most common errors were: o Using m =10.6 or even 210.6 (200 + 10.6) in E = mc∆T o Forgetting that when m(water) is in grams and c = 4.18, then E is in joules and not

kilojoules. o Using ∆T = (2.50 + 273) o Having correctly found E, using that as the ∆H value instead of finding kJ per mole of

Na2CO3. o Leaving out the negative sign of ∆H.

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Question 11 (a) Very well answered by majority of the candidates. Some said that the reaction would

go to completion since it was an open system (no equilibrium). But without also showing that HNO3 was the excess reactant, that earned only half a mark.

(b) An excellent question which allowed the good candidates to shine! One mark each was awarded for the following – several candidates made errors in the last

step. 1. The correct shape of the rate of decrease of mass curve. 2. Starting the curve at 63.92g and decreasing – curve starting at 60g was accepted too

with a half mark penalty. 3. Recognising that loss in mass was due to escaping NO2 (and not dissolving Cu which

was a common error) 4. Using n(Cu) to determine the maximum mass of NO2 evolved. Question 12 This question was also answered well by the majority of candidates. It was essential to mention Ea and the fact that at higher temperatures a greater fraction of particles had the threshold energy Ea or higher. [If explained well, full marks were awarded even without the distribution diagram.] Explanations based on merely higher velocity/kinetic Energy at higher temp, got no marks. Question 13 Many creative answers involving increasing surface area of wick by fraying/splaying/chopping, increasing ambient pressure, increasing oxygen proportion, using pure oxygen, incorporating catalyst in wax/wick, etc. One mark, each, was awarded for each method, and one mark for any suitable way in which that method could be applied to the candle/wick. Its worth pointing out that increasing concentration of oxygen in air was not counted as being different from increasing pressure around the burning candle. Question 14 a) ‘to use or not to use [H2O(l)] in the Kc concentration expression stumped a few

candidates. The main misconception was that the ratio of reactants and products must be same at equilibrium - and no marks were given if this was the explanation for Expt 2 and 3 being at equilibrium (which they are).

b) Correctly answered by most, though some did say, incorrectly, that increased yield of ester was due to the product consisting of just ester rather than ester AND water.

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c) This 6 mark question was a saviour since it has a standard well-rehearsed answer and was very well answered by most. One mark was awarded for each dot point and the sixth mark was for any other little nugget of information, usually regarding catalysis being a surface phenomenon/providing alternate reaction pathway, etc

To get a full mark for the first dot point, they had to mention that the rate of both

forward and backward reactions increases. If only the forward reaction was mentioned, only half a mark was earned.

Question 15 This was an 'A' discerning question being quite difficult and thought provoking. It elicited verbose and often very elegant answers. o Calculations which agreed with ‘the candidate’ mentioned in the question got a zero. o Recognising that pH = 9 is alkaline and therefore impossible earned one mark. o If candidates mentioned that water with acid can have pH only less than or equal to 7

this earned two marks. o If the presence of H+ and OH- in water was mentioned as well then this gained 3 marks o If Kw and the effect of disturbing the equilibrium [H+] was discussed, then full marks. Part 3 – Criterion 7 Question 16 (a) Some candidates stated that the ‘lowest energy sub-shells were not filled’ but this was

incorrect as the ground state would have had a partially filled 3p orbital. The atom has undergone ‘excitation’ or ‘an electron being promoted’ were acceptable explanations for 1 mark.

(b) Either Group VI or 16 were accepted but saying Group 6 was incorrect. (c) Many gave the ground state for sulfur having failed to recognise that the question

required the ground state of the sulfide ion. Question 17 Candidates' answers to these questions were often marred by excessive personification in descriptions e.g. ‘Na loves to lose electrons, neon is happy, fluorine is attractive......!’ (a) To gain full marks, the answer required reference to the fact the two elements have

different valence shells. (b) A range of acceptable answers were presented with most referring to the fact that E2

involves the removal of an electron from a Mg+ ion whereas E1 involves the removal of an electron from a neutral Mg atom. Charge ratios were discussed and usually led to good answers too.

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(c) Acceptable answers needed to indicate that low I.E. values are indicative of a reactive metal because metals react by electron LOSS whereas I.E. does not indicate the extent to which non-metals gain electrons.

Question 18 Many candidates failed to base their answer on ‘Explain in terms of electronic configuration’ and then also omitted any discussion as to why fluorine was the one halogen used to synthesise noble gas compounds. Although it was not expected, some candidates attempted to draw a Lewis 'electron-dot' diagram for XeF4 and encountered difficulties. No deductions were made for these attempts. Question 19 (a) Most candidates discussed the effect of the nuclear charge increasing with increasing

atomic number but then lost marks by not emphasising that for elements in the same period, the electrons are being added to the same quantum level or shell.

(b)(i) A majority of candidates deduced that X was magnesium and Y was sulfur although sodium and chlorine were popular but incorrect choices. The justification for these selections were sometimes very brief or vague.

(b)(ii) This question caused confusion and consequently a number of possible answers were accepted.

Many candidates correctly showed the reactions of the oxides with water to give alkaline or acidic solutions. Some gave reactions between MgO and SO2 or SO3. Question 20 There were insufficient marks allocated to this question. IUPAC nomenclature would give 1-ethyl-3-methylbenzene but many candidates didn't list the substituents alphabetically. Being non-IUPAC, 3-ethyltoluene was not acceptable. When determining the molecular formula, many did not take into account that benzene had two side groups in calculating the number of hydrogens. A very significant majority incorrectly answered C9H14 rather than C9H12. In the last part of this question, candidates confused solubility with reactivity. A disturbing number assumed that dissolving meant the breaking of covalent bonds. Many candidates did not justify their choice of solvent but a ‘like dissolves like’ answer was deemed acceptable for 1 mark.

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Question 21 (a) This question was not well done with the vast majority of candidates ending up with

but-1-ene rather than but-2-ene.

Many stated that reaction 1 indicated that the hydrocarbon was an alkene and failed to consider that it could be an alkyne. Reaction 2 was interpreted correctly by most candidates but it was reaction 3 that was not considered carefully enough. This ruled out the alkyne possibility because of the 1:1 mole reacting quantities. There being one addition product only meant that the alkene must have been but-2-ene giving a symmetric addition situation. (Markownikoff's Rule is not covered by the course)

(b) Using the 'error carried forward' rule, most candidates achieved full marks for this part.

Some gave 1-chlorobutane and some gave 2-chlorobutane although names were not asked for.

(c) A significant number of candidates had little idea of showing the polymerisation

reaction and often ended up with a poly(ethene) structure i.e. with no side groups. Question 22 (a)(i) This was very well done with most candidates selecting 1-chlorobutane but 1-chloro-2-

methylpropane was acceptable too. (ii) NaCl(aq) as a product was often omitted and for full marks subscripts were expected.

(b) This question showed that many candidates have a poor understanding of the influence of intermolecular bonding (especially hydrogen bonding) on the properties of organic molecules. Many had 1-chlorobutane with the highest B.P.

Question 23 (a) This was very well done with most candidates selecting ethanoic acid and methyl

methanoate. A range of other acceptable answers were presented too with unsaturated diols being a popular choice.

(b) This question was interpreted by many as hydrolysis of the ester where NaOH was

serving a catalytic role. A more acceptable answer was one where the parent carboxylic acid subsequently reacted with the NaOH to form sodium methanoate. Those who selected ethanoic acid as the unknown compound and had this reacting with sodium hydroxide gained some marks despite there not being two organic products.

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Part 4 – Criterion 8 As in previous years, there are a number of key points that need to be emphasised, especially for this 'calculation' section where many candidates lost marks due to poor answering technique or errors in expressing their answers. - 'Significant figure' misuse was frequently encountered and usually penalised by a ½

mark. As a general rule, expressing final answers to 3 significant figures is usually sufficient. Many gave answers to 1 or sometimes 9 sig figs! In general, significant figures need more attention, as it would appear that some candidates believe that 0.009 is correct to 3 significant figures rather than 0.00924

- Candidates must take more care with setting out calculations. All calculations should be labelled (e.g. n(HCl) = ... rather than n = ...) and calculation steps should follow sequentially.

- Candidates should avoid rounding off numerical answers prematurely; e.g. in 250.0 mL of 0.400 mol L-1 X(aq) the n(X(aq)) = 0.100 mol not 0.1 mol - For most chemical equations, the reactant and product states should be indicated as

subscripts. This was frequently overlooked. - Candidates should be advised not to cross out any answer part until they are sure it has

been replaced by a preferred answer. In some cases we saw correct answers that had been crossed out but not replaced by anything else.

- Candidates need to be reminded to consider whether their answer is within the bounds of sensibility. Amongst some of the remarkable answers submitted were:

(i) ‘from 12.66 g of Pb3O4 the mass of O2 obtained was 2.23 x 1022 g.’ (ii) ‘from 3.50 g of KI, the mass of PbI2 precipitate was 1.34 tonnes.’ (iii) ‘the time of electrolysis was 1.57 x 1019 seconds.’ Question 24 (a)(i) This was very well done with most candidates getting the volume of chlorine gas to be

7.12 L. The most common error was to confuse STP with SLC and 1 mark was lost for this mistake. A small but disturbing number of candidates failed to convert temperatures to kelvin. (ii) For full marks candidates need to mention the significant effects of the intermolecular forces between chlorine molecules at this low temperature.

(b) This should have been an easy question but few scored full marks. Failure to mention that the moles of gas products was half the initial reactant moles of gas was common and many thought that the temperature had been altered despite the information given. Some candidates resorted to answers based on there being an equilibrium system present.

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Question 25 This question was answered very well by the majority of candidates. The most common errors were to obtain ∆H = +73 kJ or to omit the kJ units. Question 26 This question was an easy 4 marks and most gained close to full marks. Some assumed it was Au3+ and then showed that the mass produced was consistent with this. A surprising number of candidates gave the half equation using Ag3+. Question 27 (a) This was an easy gas law question (PV = nRT) but errors were frequently encountered

with the following being common: - not converting 240 mL to litres - using the mass of oxygen as 12.66 g - not converting 27oC to 300 K - using M(O2) = 16.00 g mol-1 - using n(Pb3O4) in PV = nRT (b) This was a fairly difficult question and many candidates had some attempt at it and then

gave up and declared the empirical formula as PbO. Without clear workings, this scored no marks.

By using the n(O2) from part (a) and determining the initial n(Pb3O4), candidates were often able to achieve correct answers. Part marks were awarded for valid attempts.

Question 28 (a) This question was seldom answered correctly as many candidates failed to realise that

the final pH had to be determined by knowing the concentration of excess KOH(aq). An alarming number of candidates didn't know the formula for nitric acid, providing an assortment of weird and wonderful answers including: NOH, NH4, H2NO3, NH3, H5NO3, H2SO4 2-. When and where the excess n(OH-

(aq)) had been determined, many candidates. When and where the excess n(OH-

(aq)) had been determined, many candidates then forgot to convert this to a concentration by dividing by the final volume of 0.0600 L. The need to then use Kw was often overlooked and a pH of 1.50 was thus a common error. Some candidates used pOH correctly and this made the calculation easier.

(b) About 1% of candidates answered this question correctly. Almost all candidates assumed that the extent of dissociation of the weak acid would not affect its initial concentration significantly. This would have normally been acceptable but in this case

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the initial concentration was so small that the acid was about 15% dissociated. Half marks were awarded to candidates who used the approximation method despite it being invalid in this instance.

Question 29 (a) Many candidates thought that copper would react with dilute H2SO4 as well as the zinc.

Some wrote equations with a metal reactant of ‘CuZn’! Sulfur dioxide was commonly given as one of the products.

(b) Candidates who used Mr(H2) = 2.0 rather than 2.02 lost half a mark for incorrect significant figure usage but candidates who used Mr(H2) = 1.01 lost 2 marks. Some candidates performed calculations based on the sum of the Ar of Cu and Zn and ended up with 50% of each metal in the alloy. No marks were awarded in these cases.

Question 30 This question was very well done although a few candidates gave extraordinary answers such as [KI(aq)] = 79.3 mol L-1 rather than 0.0793 mol L-1 . Question 31 Most candidates made a very good showing with this question. The more common errors included: - formula for lead(II) iodide being PbI - not balancing the equation correctly - giving the number of moles of the reactants to only 1 significant figure - not realising that there was a slight excess of KI - the question asked for the amount of PbI2 so mass calculations were not required but

usually given. Unexpected answers were sometimes presented with the largest mass of precipitate being 1.34 tonnes!

TASMANIAN QUALIFICATIONS AUTHORITY

CHM315109 Chemistry

ASSESSMENT PANEL REPORT

Award Distribution

Student Distribution (SA or better)

EA HA CA SA Total

This year 16% (86) 29% (tss) 2t% (LL4) 33% (L78) s33

Last year 20% (106) 27% (L4L) 22% (tL3) 3L% (L64) 524

Last year (allexaminedsubjects)

LL% 20% 39% 30%

Previous 5 years 24% 24% 3L% 2L%

Previous 5 years(all examinedsubjects)

LL% L9% 40% 30%

Male Female Year LL Year L2

This year s8% (307) 42% (226) 0% (0) L00% (s33)

Last year s2% (270) 48% (2s4) 0% $) L00% (s23)

Previous 5 years 5L% 49% 0% 1.00%

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TA S M A N I A N Chemistry - tqa.tas.gov.au · TA S M A N I A N Chemistry CE R T IF ICA T ... undergone ‘excitation’ or ‘an electron being ... To gain full marks, the answer