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 = 4.48/56 =0.0800 mol No of moles of chlorine = 8.52/35.5 =0.2400 mol Mole ratio of no of moles of iron: no of moles of chlorine atoms = 0.0800: 0.240 = 1: 3 Marker’s comments: Common mistake was to calculate and compare the no of moles of chlorine gas instead of chlorine atoms. From the ratio, the formula of the iron chloride compound is derived, which students need to write the balanced chemical equation in (iii). [1]- for both Fe & Cl, [1]-ratio [1]- for both Fe & Cl [1]-ratio (b)(iii) 2Fe + 3Cl2 → 2FeCl3 Marker’s comments: To write balanced chemical equation, chlorine is Cl2, not Cl as chlorine as a diatomic molecule. [1] A4(a) Volume of hydrogen gas = 45 cm3 No of moles of hydrogen gas = 45/ 24 000 = 0.001875 mol Mole ratio Mg: H2 = 1: 1 No of moles of Mg = 0.001875 mol Mass of Mg = 0.001875 x 24 = 0.045 g [1]-mol of gas, [1]- mol & mass of Mg [1]-mol of gas [1]- mol & mass of Mg (b) (c) (Expt 2) [1]- Ea [1]- ∆H [1]-exo + rxts & pdts Note that the arrows are single direction, not double headed! For ∆H, the enthalpy sign “-“ is required [1]- Ea [1]- ∆H [1]-exo + rxts & pdts [1]- (ci) Ea Energy/ kJ Mg + 2HCl ∆H = “-“ MgCl2 + H2 MgCl2 + H2 Progress of reac;on Enthalpy change will be halved as the no of moles of product produced is halved. Enthalpy change is based on the balanced chemical equa;on (actual no of moles of product formed). Ea remains unchanged. Ea is only affected by presence of catalyst.

4 (d) Note: volume of hydrogen gas should be at 22.5 cm3 The gradient is gentler and the time reaction ends must be longer than the original of 3.5 mins. [1] (e) As the temperature increases, the kinetic energy of the particles increases. (“kinetic” energy must be written to gain 1m) There are more particles with energy greater than or equal to the Ea. (many students missed out this point) Frequency of effective collisions increases and rate of reaction increases. [1] [1] [1] A5 (a) A weak acid ionises partially in water to produce few H+ ions in solution. Marker’s comments: Several students missed out the important key words like “water” and “H+ ions”, besides ionise partially. [1] (b) SO2 + H2O → H2SO3 [1] (c) SO2 is produced from volcanic eruptions or burning of fossil fuels. [1] (d) SO2 dissolves in the rain water to produce acid rain. Acid rain corrodes metal structures and limestone buildings/ destroys aquatic animals and plants. Marker’s comments: Need to talk about harmful effect to gain the 1m. Acid rain is insufficient. “Metal” structures or “limestone” buildings are required to show that acid rain react with these chemicals. [1] (d)(i) Thermal decomposition Note: thermal is required [1] Expt 3

5 (d)(ii) The oxidation state of sulfur increased from +4 in SO2 to +6 in CaSO4, hence SO2 is oxidised. Note: Both oxidation states must be correct to gain full credit. No partial credit given. [2] (d)(iii) E absorbed = + [635 + 297 + (498/2)] = + 1181 kJ E released = - 1434 kJ ∆H = + 1181 – 1434 = - 253 kJ/mol Marker’s comments: Proper statements and units are expected. However, wrong units were not penalised. [1] [1] A6 (a) The negative yellow coloured CrO4 2– ions move (get attracted) to the positive electrode. Marker’s comments: Several students said that CrO42- is “discharged” which is incorrect concept. The ion that is discharged is the ion that undergoes reaction, in this case, since it is the anode, it will be oxidation. If CrO42- is oxidized, it will no longer be in solution and the solution will not be yellow. [1] (b) 4OH- (aq) → 2H2O (l) + O2 (g) + 4e- Marker’s comments: State symbols are required for all half equations!!!! [1] (c) Cu2+(aq) + 2e- → Cu (s) -Cu2+ ions are preferentially discharged instead of H+ ions as they are less stable in solution OR copper is below hydrogen in the reactivity series. -The Cu2+ ions gain electrons and are reduced to form copper metal. Marker’s comments: State symbols are required for all half equations!!!! For explanation, both parts are expected, ie why Cu2+ is preferentially discharged and Cu2+ ions gain electrons and are reduced. However, if either 1 point is given, students are given the 1 m. [1] [1] (d) Positive electrode: bubbles seen Negative electrode: reddish brown deposit/solid [1] [1] (e) Blue ppt soluble in excess aq ammonia to form a dark blue solution. [Accept: green ppt soluble in excess aq ammonia to form a dark green solution; Blue ppt soluble in excess aq ammonia to form a dark green solution.] 1m- ppt 1m-colour of soln, must mention ppt soluble in excess Marker’s comments: Many students did not say “ppt soluble in excess aq ammonia to form….” This is a test of QA, hence this phrase is important. [2]

6 B7(a) H2, O2 …(any diatomic element) [1] (b) Metals have few valence electrons and they tend to lose electrons rather than attract electrons to achieve the stable noble gas electro