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: