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 have) Water droplets on side of test tube CuO + H2  Cu + H2O (b) To prevent hot copper metal from oxidising / forming back copper(II) oxide / reacting with oxygen in air. (c) (i) Experiment 1: copper is too unreactive / low in reactivity series, will not react with steam. Experiment 2: magnesium is more reactive / higher in reactivity series than H2, Mg in MgO will not be reduced by H2. (ii) Silver is less reactive than copper, hence silver(I) oxide is less (thermally) stable than copper(II) oxide. [1] [1] [1] [1] [1] [1] [1] [1] 10 (a) (b) No. of moles of repeating units in PTT = 1000 ÷ [11(12) + 10(1) + 4(16)] = 4.85437 mol = 4.85 mol (to 3 s.f.) No. of moles of repeating units in poly(isoprene) = 1000 ÷ [5(12) + 8(1)] = 14.7059 mol = 14.7 mol (to 3 s.f.) (c) No. of moles of CO2 formed from burning 1.0 kg of poly(isoprene) = 14.7059 x 5 = 73.5 mol (to 3 s.f.) [1] each [1] each [1] for both calculations

No. of mol of CO2 formed from burning 1.0 kg of PTT = 4.85437 x 11 = 53.4 mol (to 3 s.f.) Burning 1.0 kg of poly(isoprene) produces more carbon dioxide than burning 1.0 kg of PTT. Carbon dioxide is a greenhouse gas which contributes to global warming. (d) (i) Adding water in presence of acid catalyst. (ii) recycling [1] [1] [1] 11 (a) Anode: Ag (s)  Ag+ (aq) + e- Cathode: 2H+ (aq) + 2e-  H2 (g) (b) oxidation state of Ag increases from 0 in Ag to +1 in AgCl hence Ag is oxidised. Oxidation state of O decreases from 0 on O2 to -2 in H2O hence O is reduced. Since oxidation and reduction occur simultaneously, it is a redox reaction. (c) Ensure O2 is continuously dissolve in the water for more accurate reading / even distribution of water sample. (d) Silver; silver chloride is formed. (e) D as it has the lowest ammeter reading indicating low concentration of oxygen in blood. With intake of carbon monoxide, haemoglobin will bind to CO to prevent further intake of oxygen gas in the body. (f) Manganese is not used up in the experiment, exhibiting the use of catalyst behavior / multiple oxidation states of Mn (g) Mol of sodium thiosulfate = 0.01 x 0.0112 = 0.000112 mol ratio of I2 = 0.000112 / 2 = 0.000056 mol of Mn(OH)3 = 0.000056 x 2 = 0.000112 mol of O2 = 0.000112 / 4 = 0.000028 Concentration of O2 = 0.000028 / 0.1 = 0.00028 mol/dm3 [1] [1] [1] [1] [1] [1] each [1] – trend [1] – reason [1] [1] [1] [1] 12 (a) (i) methoxybutane, CH3-O-C4H9 (ii) Propanol, [1] each [1] each

(iii) Propyl ethanoate (b) (i) A: Isomerisation because it has same number of carbon and hydrogen atoms as octane. B & D: Cracking because octane is broken down into smaller molecule (lesser carbon and hydrogen atoms). C: Substitution because one of the hydrogen in octane is replaced with bromine. (ii) B reacted with steam under 300 oC, 60 atm and phosphoric acid catalyst. [1] each [1] [1] [1] [1] 13 (a) (i) alkene CnH2n (ii) Any isomer (b) (i) Test: Bubble into bromine solution Isobutylene: Turns brown bromine solution to colourless. X: Brown bromine solution remains brown. Or use any acid reactions (ii) Isobutylene can be added to steam under 300 oC, 60 atm and phosphoric acid catalyst to form isobutanol which is further oxidised by adding acidified potassium manganate (VII) to form X. (c) Isobutylene has simple molecular structure with small amount of heat energy needed to overcome the weak intermolecular forces of attraction between the molecules. After polymerization, many small molecules of isobutylene joined together to form a macromolecule (polymer). Hence more energy is now required to over the stronger intermolecular forces of attraction between the molecules. [1] [1] [1] [1] [1] [1] [1] [1] [1]

Paper 3 Qn Skill Suggested Answers Remarks 1(a)(i) PDO Results table: - Records initial burette readings, final burette readings and volume added with correct headings and units in a titration table. - All burette readings for all accurate titres in titration table are recorded to nearest 0.05 cm3. Titration results: - accuracy for the average titre (of consistent readings) within 0.20 cm3 of Supervisor’s average value score 2 marks. for the average titre (of consistent readings) within 0.30 cm3 of Supervisor’s average value score 1 mark - concordance at least two titre values are within 0.20 cm3 Teacher’s result: 21.40 cm3 [1] [1] [2] [1] (ii) MMO Appropriate average volume in 2 d.p. from the closest titre values (should be identified either in the table by a tick or in a calculation) [1] (b)(i) ACE number of moles of sodium hydroxide = 0.150 x 0.025 = 0.00375 mol [1] (ii) A