Copy of KCPSS 2024 4E Chem Prelim P1&2 MS

KUO CHUAN PRESBYTERIAN SECONDARY SCHOOL SECONDARY FOUR EXPRESS CHEMISTRY PRELIMINARY EXAMINATION 2024 Answer Scheme Paper 1 – Multiple Choice Questions (40 marks) Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 A B B D B B C A D D Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 D B A A C A B D C C Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 B D D B C A C A C C Q31 Q32 Q33 Q34 Q35 Q36 Q37 Q38 Q39 Q40 A A C B C C B C C B

KUO CHUAN PRESBYTERIAN SECONDARY SCHOOL SECONARY FOUR EXPRESS CHEMISTRY 6092 PRELIMINARY EXAMINATION 2024 MARK SCHEME Section A (70 marks) 1 a Ar [1] b Ca, Ar [1] c C, At [1] d Na [1] e Cu, Zn [1] f C, Pt [1] Note: Penalise if incomplete or extra answers are given. Penalise one mark overall if names were given instead of symbols. Marker’s comment: Majority of the students missed out on giving complete answers. Common mistakes for (e) make up of brass is carbon and iron and (f) electrodes is copper and zinc. 2 Mixture separation technique ammonium chloride + sodium chloride Sublimation [1] water + lead(II) sulfate Filtration [1] methanol + glucose solution (fractional/simple) distillation [1] 1m each [3] Marker’s comment: Common mistake is to give preparation methods instead of separation techniques. For lead(II) sulfate, evaporation to dryness was rejected as question asked for the most appropriate technique.

3 a Thermometer, Conical flask, Measuring cylinder All correct 2m, 2 correct 1m. Penalise for spelling once throughout the paper. [2] b No of mol of Mg = 2 / 24 = 0.083333 No of mol of HNO3 = 100 / 1000 x 1.5 = 0.150 Mole ratio of Mg : HNO3 is 1 : 2 For 0.15 mol of HNO 3 used, ½ x 0.15 = 0.075 mol of Mg is needed. As there’s 0.0833 mol of Mg, it is in excess. OR For 0.0833 mol of Mg used, 2 x 0.0833 = 0.167 mol of HNO 3 is needed. Since there is only 0.15 mol of HNO3, it is the limiting reactant and Mg is in excess. [1] [1] [1] Marker’s comment: (a) students incorrectly identify gas jar, missing out that the apparatus has markings. (b) poorer response was the inability to do the link/explain clearly why Mg is in excess. Some calculations were given in fractions and was unclear which value was greater/smaller. 4 a Ar of Cu = (69.15/100 x 63) + (30.85/100 x 65) = 63.617 = 64 (nearest whole number) [1] [1] bi AgNO3: White precipitate forms. NH3: Light blue / Blue precipitate forms, soluble in excess to give a dark blue solution. Penalize once overall if short-form (ppt) was given. [1] [1] [1] bii Precipitation [1] biii Element Cu F K [3]

% in 100g 21.5 38.7 39.8 Ar 64 19 39 No. of mol 0.33594 2.0368 1.0205 Mol ratio 0.33594 / 0.33594 = 1 2.0368 / 0.33594 = 6.06 1.0205 / 0.33594 = 3.04 Simplest ratio 1 6 3 Empirical formula: CuF6K3 1m – Indication of the conversion from % to mass , 1m – workings, 1m – empirical formula (accept any combination of the formula) Markers’ comment: (a) poorer response did not follow instructions to give answers to nearest whole number. (b)(i) majority was not able to give complete answers for the observation. Misconception of ppt turning into solution. (b)(ii) most common mistake was “metal displacement”. Students did not understand the reaction. (b)(iii) common mistake was not converting % to mass and not showing calculations. 5 a Set-up A: Anticlockwise Set-up B: Clockwise [1] bi Zinc electrode in set–up A increased in size, whereas the zinc electrode in set–up B decreased in size. In set-up A, Zn2+ ions are discharged / gained 2e- / reduced to formed Zn metal which causes the electrode to increase in size. In set-up B, Zinc is more reactive than copper and loses electrons to form Zn2+ ions causing the electrode to decrease in size. Set-up A: Zn2+ (aq) + 2e → Zn (s) Set-up B: Zn (s) → Zn2+ (aq) + 2e [1] [1] [1] [1] [1] bii The copper electrode in A will decrease in size whereas the copper electrode in B will increase in size. [1] [1]

The blue aqueous CuSO4 colour will intensify in A whereas the blue aqueous CuSO4 colour will fade in B. c Zn (s) / ZnO (s) / ZnCO3 (s) CuO(s) / CuCO3 (s) Penalize once for missing state symbols. [1] [1] Markers’ comment: (b)(i) students did not read the question carefully that both setup changes in size. Half-equations were missing state- symbols. Explanations were missing the key concept (more reactive metal has a higher tendency to lose electrons) of simple cell. (b)(ii) Students gave only one differences or incomplete comparisons. (c) common mistake was giving salts as answers and students thinking that copper can be used, forgetting that it is an unreactive metal. 6 a Ammonia particles are far apart and disorderly. They move about rapidly in all direction. [1] b As the number of bonds between nitrogen atoms increases from single to triple bond, the bond energy increases from 160 kJ/mol to 941 kJ/mol. This is due to a stronger attraction between the nitrogen atoms due to more electrons shared between them, require more energy to break the bonds. Note: Vice versa accepted. [1] [1] ci Total energy absorbed = 941 + 3(436) = 2249 kJ Total energy released = 2 x 3(391) = 2346 kJ Overall enthalpy change = 2249 – 2346 = –97 kJ [1] [1] [1] cii +97kJ [1] Markers’ comment: (a) incomplete answers. Commonly missing out the idea of “disorderly” or “rapid movement”

(b) misconception that bond were “overcome” when it is broken or that intermolecular forces of attraction were incorrectly discussed showing poor understanding of question. (c)(i) incorrect use of data. N-N data used instead of N≡N Poor statements given and incorrect calculation of the number of bonds. (ii) missing signs and incorrect units given. 2 moles of ammonia decomposes hence it isn’t kJ/mol. 7 a reaction without catalyst reaction with catalyst 1m – same reactant and product height, labelled 1m – correct Ea / Ec 1m – same H (endothermic) [3] bi Empirical formula: CH2 [1] [1] bii 2CH2 + 3O2 → 2CO2 + 2H2O [1] biii No. of mol of poly(propene) = 5000 / (12+2) = 357.14 Mole ratio of CO2 : Poly(propene) = 2 : 2 ∴ No. of mol of CO2 = 357.14 Vol of CO2 = 357.14 x 24 [1] (C3H6)n (C3H6)n C7 – C10 hydrocarbons C7 – C10 hydrocarbons

= 8571.42 = 8570 dm3 (to 3 s.f.) [1] c Advantage: Poly(propene) is durable / does not rust unlike iron. Disadvantage: Poly(propene) is non-biodegradable and would contribute to waste, pollution problems. [1] [1] 8 ai Butanedioic acid, HOOC(CH2)2COOH OR HOOCCH2CH2COOH [1] aii HOOC(CH2)nCOOH OR (CH2)n(COOH)2 [1] b Disagree with the claim. It is unable to undergo condensation polymerization on its own as it only has carboxyl functional group. OR does not contain hydroxyl or amine group. [1] ci The term weak acid means the acid undergoes only partial dissociation in water to form H+ ions. Circle the 2 acidic hydrogen of carboxy functional group. [1] [1] cii Physical property: pH < 7, turns moist blue litmus paper red, turns green Universal Indicator orange/yellow. (any one) Chemical property: Reacts with metal to produce salt and hydrogen, reacts with metal carbonate to produce salt, water and carbon dioxide, undergoes redox reaction with potassium manganate (VII). (any one) [1] [1] ciii Tartaric acid contain 2 carboxyl groups (per molecule) whereas butanoic acid contains only 1 carboxyl group (per molecule). Tartaric acid contains 2 types of functional groups (per molecule), hydroxyl and carboxyl whereas butanoic acid contains only 1 type of function group (per molecule), carboxyl. Tartaric acid contains a hydroxyl functional group (per molecule), whereas butanoic acid does not. [2]

Tartaric acid contains 4 functional groups (per molecule), whereas butanoic acid contains only 1 functional group.