BMSS Prelim Ans

Bukit Merah Secondary School Preliminary Examination 2024 Chemistry 6092 Paper 1 1 D 6 C 11 B 16 B 2 C 7 B 12 C 17 C 3 B 8 A 13 A 18 C 4 A 9 C 14 C 19 C 5 A 10 D 15 B 20 A 21 D 26 C 31 C 36 B 22 B 27 D 32 C 37 B 23 B 28 A 33 A 38 C 24 D 29 D 34 D 39 C 25 C 30 D 35 A 40 C Paper 2 Qn Suggested Answers Remarks 1 (a) potassium manganate (VII) (b) methane (c) potassium iodide (d) calcium oxide (e) ethene (f) aluminium oxide [1] each 2 (a) 0.5 (b) Yellow: remains in middle; Blue: Furthest from middle; Red: In between yellow and blue (c) At 80% alcohol, all 3 dyes have the same Rf values of 0.4, which shows up as one spot on the chromatogram. [1] [1]-Y [1]-B&R [1] 3 (a) (i) 14, 14 14, 15 14,16 (ii) Ar = 28.11 (2 dp) (b) Silicon dioxide has giant molecular structure. Large amount of heat energy is needed to overcome the strong covalent bonds between the atoms. Hence it has high melting point (heat- resistant). As there is no free moving electrons / ions available, it is electrically insulator. [1] [1] [1] – structure [1] – energy/bond [1]

4 (a) Cl H N Pb Mass 46.7 1.76 6.14 45.4 Ar 35.5 1 14 207 Mol 1.3155 1.76 0.43857 0.21932 / smallest 6 8 2 1 Cl6H8N2Pb (b) Anion: PbCl62- Cation: NH4+ (c) +4 [1] - mol [1] – final [1] each [1] 5 (a) (b) NaH2PO4 (c) Phenolphthalein The equivalence point for stage B is within the pH range where phenolphthalein changes colour. [1] [1] [1] [1] 6 (a) Lower the ionisation energy, the more reactive the Group 1 metal is. This is because, as atomic size increases, the valence electron is further away from the nucleus (protons), requiring lesser amount of energy to remove. (b) Accept any value more than 0.526 Fluorine (2,7) has more protons as compared to lithium (2,1) hence requires more energy to remove one valence electron. [1] – trend [1]- explanation [1] [1] 7 (a) Energy taken in during bond breaking = 602 + 193 = 795 kJ/mol Energy release during bond forming = 349 + 2(276) = 901 kJ/mol dH = 795 – 901 = -106 kJ/mol. (b) Correct shape (exothermic) Correctly label with correct arrows [1] [1] [1] [1] [1]

8 (a) Experiment A as the initial gradient is steeper (indicating faster reaction). At higher temperature, particles gain energy and move faster resulting in higher rate of effective collisions. (b) Rate = 0.135 / 2 = 0.0675 g/min (c) carbon dioxide is released. (d) mol of CO2 = 0.5 / 44 = 0.0113636 mass of PbCO3 = 0.0113636 x 267 = 3.03 g (e) use of catalyst (must be named) / higher concentration of nitric acid / finer particle size of lead carbonate (f) Must end at 0.25 g [1] [1] [1] [1] [1] [1] [1] [1] [1]

9 (a) Experiment 1: White solid formed Mg + H2O  MgO + H2 Experiment 2: Reddish-brown solid formed (from black solid). (Good to hav