2022 SASS A-Z Chem Prelims Ans Pt4

1 A – Z School’s Preliminary Papers 4 Express Chemistry ANSWERS: Part 4 P Paya Lebar Methodist Girls’ School Q Queensway Secondary School R Riverside Secondary School S Singapore Chinese Girls’ School. T Temasek Secondary School PAYA LEBAR METHODIST GIRLS’ SCHOOL Paper 1 1 C 11 C 21 A 31 A 2 B 12 C 22 C 32 C 3 D 13 B 23 A 33 A 4 A 14 A 24 A 34 C 5 C 15 C 25 B 35 D 6 B 16 B 26 D 36 B 7 B 17 D 27 B 37 C 8 C 18 B 28 D 38 D 9 D 19 C 29 C 39 C 10 D 20 D 30 C 40 A Paper 2 A1(a) Cu, Fe (must give both to score 1m)(accept if students write name) [1] (b) Mg [1] (c) H2 (Al accepted as AlCl3 is covalent- students may not know) [1] (d) Cu or Fe [1] (e) Fe [1] A2 (a)(i) Accept any value from 6.00 to 9.00 Marker’s Comments: Units are required (several students did not write the units). However, students were not penalised for not writing units. [1] (a)(ii) There is no trend in the mpt of the Group V elements [1] (b)(i) Compound R is an ionic compound with a giant crystal lattice structure and the strong electrostatic forces of attraction between the Na+ and P3- ions (or oppositely charged ions) require lots of energy to overcome, hence high melting point. [1] [1] SASChemistry 2022edition

2 Compound S is a covalent compound with simple molecular structure. The weak intermolecular forces of attraction between the molecules require little energy to overcome (or easily overcome), hence low melting point. Marker’s comments Question says structure and bonding, hence students are required to specify ionic bonding AND giant crystal lattice structure (however, 1 m was given if students stated either ionic or giant crystal lattice structure) ; covalent compound AND simple molecular structure. For compound R: Students are required to specify the Na+ and P3- ions (or oppositely charged ions)-answer according to the question! The qn says high mpt, so it is not a comparison with compound S, hence “lots of energy to overcome” and NOT more energy is required. For compound S: The qn says low mpt, so it is not a comparison with compound R, hence “little energy overcome” and NOT lesser energy is required. [1] [1] (b)(ii) [1]- correct no of electrons for P [1]-correct no of electrons shared with 3 H [1] [1] (c)(i) Mr of P4O10 = (31 x 4) + (16 x10) = 284 % of P = (31 x 4) /284 x 100% = 43.7% (3sf) Marker’s comments Mr has no units. If students write units, they will be penalised. [1] [1] (ii) Oxidation state of S in Compound S = -3 P4O10 = +5 [1] [1] A3 (a) To prevent the iron from reacting with the oxygen in the air/being oxidised. Marker’s comments Several students thought it was to ensure the reaction was complete. Do remember most hot metals will react with oxygen in the air! [1]

3 It is required to write “oxygen” and not any gas(other gases) in air as this is a specific reaction- oxidation. (b)(i) Mass of iron = 56.04 – 51.56 = 4.48 g Mass of chlorine = 64.56 – 56.04 =8.52 g [1] [1] (b)(ii) No of moles of iron