Copy of SCSS Prelim 2024 Chem P1 & P2 MS

1 Swiss Cottage Secondary School Preliminary Examination 2024 Secondary Four (O Level) Chemistry 6092 (Mark Scheme) – Paper 1 & 2 Paper 1 (40 marks, [1] per qn) 1 B 21 B 2 B 22 B 3 C 23 B 4 B 24 D 5 D 25 A 6 D 26 D 7 A 27 B 8 A 28 A 9 C 29 A 10 B 30 B 11 C 31 C 12 B 32 A 13 B 33 B 14 C 34 C 15 C 35 B 16 B 36 B 17 D 37 D 18 D 38 C 19 D 39 A 20 D 40 B

2 Paper 2 Section A: 70 marks Qn No. Answer marks markers’ comment A1 (a) zinc [1] (b) copper / hydrogen [1] (c) hydrogen / carbon [1] (d) argon [1] (e) fluorine [1] (f) silicon [1] Total [6] A2 (ai) The atomic radii increase down the group due to the increase in the number of electron shells [1] to hold the electrons, thus resulting in a bigger atom. [1] (aii) The atomic radii increase while electronegativity decreases as it is harder for the nucleus to attract electrons to itself. [1] [1] (aiii) 2.25 (accept value between 2.0 to 2.3). It has a larger atomic radii compared to the elements above it in the group . Since the nucleus is further away from the valence shell, it is harder for the nucleus to attract electrons. [1] [1] (bi) Melting point and boiling point decreases down the group [1] (bii) As the atomic radii increases down the group, the negatively-charged valence electrons are further away from the positively-charged nucleus. The electrostatic forces of attraction between valence electrons and the nucleus becomes weaker. [1] Less energy is required to overcome the metallic bonding/forces of attraction [1] between the valence electrons and the nucleus. [2] Total [6] A3 (a) It is because the acid contains oxygen and can dissociate/ionise in aqueous solution to form H+ ions. [1] (b) name of acid oxidation state of chlorine hypochlorous acid +1 chlorous acid +3 chloric acid +5 perchloric acid +7 [2] Every 2 correct answer [1m]

3 (c) As the oxidation state of chlorine increases, the strength of acid increases. [1] When oxidation state of chlorine increases from +1 to +7, the reaction between the acid and Mg become more vigorous. [1] [2] (d) Hypochlorous acid / HClO has the lowest electrical conductivity. It has the least vigorous reaction with magnesium and is therefore the weakest acid. This means that it must have the lowest concentration of mobile ions [1] to act as charge carriers. [1] Total [6] A4 (ai) Air is made up of a mixture of many gases, like oxygen and nitrogen. Hence • oxygen can react with hydrogen to form water if air is used during Haber process. • nitrogen reacts with oxygen to form oxides of nitrogen • air contains oxygen which oxidises iron catalyst to iron(II) oxide/iron(III) oxide • iron react with oxygen and water and rusting occurs. [choose any one point] [1] (aii) No. of mole of H2SO4 = 10.5 1000 X 0.150 = 0.001575 mol 𝑁𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑁𝐻3 𝑁𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐻2𝑆𝑂4 = 2 1 No. of mole of NH3 = 2 1 X 0.001575 = 0.00315 mol Concentration of NH3 = 0.00315 0.002 = 0.1575 = 0.158 mol/dm3 (to 3.s.f) [1] [2] [1]

4 (b) Mr of NH4NO3 = 80 % N in NH4NO3 = 2(14) / 80 x 100% = 35.0% (to 3 s.f) Mr of CO(NH2)2 = 60 %N in CO(NH2)2 = 2(14) / 60 = 46.667 = 46.7% (3 s f) [1] for both correct %N Hence, urea would give more nitrogen per kg. [1] [2] (c) 1. Add excess calcium carbonate to nitric acid and stir. Filter the mixture to obtain a solution of calcium nitrate as the filtrate. 2. Add the calcium nitrate solution to aqueous sodium sulfate/ sulfuric acid/ any solution with sulfates. 3. Filter the mixture to obtain calcium sulfate as the residue. Wash the residue with distilled water and leave the residue to dry on filter paper. [1]: steps 1-2 [1]: step 3 [2] Total [7] A5 (a) A: Iron / Fe [1] B: Iron(III) oxide / Fe2O3 [1] C: Iron(III) sulfate / Fe2(SO4)3 [1] D: Ammonia / NH3 [1] [4] (b) Zn2+ and NH4+ [1] (c) Zn2+ + 2OH- → Zn(OH)2 {Note: Please follow instructions! State symbols are not required. Please do not include. If correct state symbols are given, no extra credit will be given. However, for incomplete or wrong state symbols, marks will be deducted.} [1] (d) Cl- / chloride and I- / iodide are not present in the filtrate. [1] Total [7]

5 A6 (a) ZnCO3 → ZnO + CO2 No. of mole of zinc carbonate = 2.00 125 = 0.016 mol 𝑁𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶𝑂2 𝑁𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑍𝑛𝐶𝑂3 = 1 1 No. of mole of CO2 = 1 1 X 0.016 = 0.016 mol Mass of CO2 = 0.016 x 44 = 0.704g (to s.f.) Actual mass of CO2 produced = 2.00 – 1.35 = 0.65g % yield of CO2 = 0.65𝑔 0.704𝑔 X 100% = 92.330 = 92.3 % ( 3 s.f) [1] [3] (b) It is because • different carbonates have different Mr values, hence the number of moles of carbonate present in the fixed mass of carbonate is also different. OR • different metals in the various carbonates have different Ar values, hence the number of moles of carbonate present in the fixed mass of carbonate is also different. OR • % of carbon in each compound is different, hence the number of moles of carbonate present in the fixed mass of carbonate is also different. [1] (c) Observations: No visible change occurs when hydrogen was passed over heated aluminium oxide. Black solid turned pink / red-brown when hydrogen was passed over heated copper(II) oxide. [1] {Note: water droplets/ colourless liquid is not acceptable in this case. See remarks on the right.} [2] [1] [1]

6 Explanations (method 1- displacement) Hydrogen is less reactive than aluminium, so hydrogen is not able to displace aluminium from aluminium oxide. Hydrogen is more reactive than copper, so hydrogen is able to displace copper from copper(II) oxide. [1] {“to displace copper from copper(II) oxide” here means “to form the copper metal from copper(II) oxide”} Explanations (method 2- reduction) Hydrogen is less reactive than aluminium, so hydrogen is not able to reduce aluminium oxide to aluminium. Hydrogen is more reactive than copper, so hydrogen is able to reduce copper(II) oxide to copper [1] {“to reduce copper(II) oxide to copper” means “to convert copper(II) oxide [before] to copper [after]”} Total [6] A7 (a) Experiment 1: A1 Experiment 2: B1 [1] (b) Copper(II) ions (Cu2+) gain electrons more readily than hydrogen ions (H+), hence copper(II) ions are reduced to form copper solid. OR Copper(II) ions (Cu2+) are preferentially discharged over hydrogen ions to form copper solid [1] Cu2+ + 2e- → Cu [1] [2] (c) 4OH – → 2H2O + O2 + 4e- {Note: Please follow instructions! State symbols are not required. Please do not include. If correct state symbols are given, no extra credit will be given. However, for incomplete or wrong state symbols, marks will be deducted.} [1] (d) Colour of Universal Indicator in expt 1: red [1] pH of electrolyte in expt 1: pH 1 OR 2 [1] OH- ions and Cu2+ ions are selectively discharged (while H+ and SO42- are not). This results in a higher concentration of H+ ions than OH- ions thus resulting in an acidic solution [1] [3] Total [7]

7 A8 (a) B most reactive C D copper A least reactive [1]: first 2 [1]: next 3 [2] (b) Observation 1: Colour of solution will change from blue to colourless./fades to light blue [1] Observation 2: Red-brown solid/pink solid will be deposited. [1] [2] (c) Voltage reading = 0.00 V / 0 V Ethanol exists as molecules and does not hav