2024 Y5 DHS H2 Chemistry Timed Practice 1 Suggested Answers

1 © DHS 2024 DUNMAN HIGH SCHOOL 2024 YEAR 5 H2 CHEMISTRY TIME PRACTICE 1 Mole Concept, Redox & Atomic Structure Section A – Multiple Choice Questions MCQ No. 1 2 3 4 5 Answer C C B C B 1 Answer: C 423Co (SO ) 20.3= = 0.0500 mol406 n A✓ 3+ 423Co (SO )Co 2nn = and 2 4234 Co (SO )SO = 3nn − 3+ 2 4234ions Co (SO )Co SO= + = 5 = 5 0.0500 = 0.250 moln n n n −  23 ionsno. of ions 1.50 10nL=  =  B✓ 3+ 423Co (SO )Co = 2 = 2 0.0500 = 0.100 molnn  C 2 58.9mass % of Co = = 29.0 %0 46 100 %   D✓ 4 3 16.0mass % of O = = 47.3 %0 46 100 %   2 Answer: C Let the oxidation state of X in the product be n. Reduction: Fe3+ + e– → Fe2+ Oxidation: X3+ → Xn+ + (n – 3)e– No of moles of Fe3+ = 24 1000 x 0.0300 = 0.000720 mol = number of moles of electrons No of moles of X3+ = 20 1000 x 0.0120 = 0.000240 mol 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑒− 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑋3+ = 𝑛 − 3 1 = 0.000720 0.000240 n – 3 = 3 n = 6 Option A: n + (-2) = +1, n = 3 Option B: n + (-2) = +2, n = 4 Option C: n + 2(-2) = +2, n = 6 Option D: 2n + 2(-2) = +2, n = 3

2 © DHS 2024 3 Answer: B M shows the largest increase from the 2nd to 3rd IE, so it has 2 valence electrons and it is from group 2. N shows the largest increase from the 7 th to 8th IE, so it has 7 valence electrons and it is from group 17. M2+ and N– ions form MN2. 4 Answer: C angle of deflection  charge mass for 1H+, 𝑞 𝑚 = +1 1 for 2D–, 𝑞 𝑚 = −1 2 Hence, angle of deflection for 2D– = –2o 5 Answer: B The conversion results in the loss of a proton. So, proton/atomic number decrease by 1. Since the proton is converted into a neutron, mass/nucleon number remains unchanged. 111 111 53 52 Te→I Section B – Structured Questions 6 (a) (i) Reduction: O2 + 4H+ + 4e− → 2H2O Oxidation: Cu → Cu2+ + 2e− Balanced overall equation: 2Cu + O2 + 4H+ → 2Cu2+ + 2H2O (ii) amount of copper foil = 5 63.5 = 0.07874 mol amount of O2 used in the reaction = 0.0787 2 1  = 0.03937 mol Volume of O2 used in the reaction at r.t.p. = 0.0394  24 = 0.945 dm3 (3 s.f.) (b) (i) mass of water = 0.163  50 = 8.15 g amount of [(CH3COO)2Cu]2.Cu(OH)2 = 50 – 8.15 460.5 = 0.09088 mol = 0.0909 mol (3 s.f.)

3 © DHS 2024 (iii) amount of water present = 8.15 18.0 = 0.4528 mol 0. 0.4528 = 09088 = 4.98 = (nearest whole number)5 x 7 2 2 2C H (g) + + O (g) CO (g) + H O(l) 42 →  xy yyxx The contraction of 40 cm3 is due to removal of acidic CO2 by NaOH. CxHy ≡ xCO2 1 10= 40x x = 4 The first contraction of is due to the difference in volume of gaseous reactants and gaseous products. Note that water is a liquid at r.t.p. ( ) ( )2 2 2C H initial O CO formed excess O+ + = 30 –V V V V xy ( )22initial O excess O– +10 – 40 = 30VV 2 3 O used = 60 cmV 2C H + O 4    xy yx 1 10= 60 + 4 yx y = 8 The molecular formula of the hydrocarbon is C4H8. 8 (a) The oxidation number of chromium increases from +3 in Cr(OH)4 ‒ to +6 in CrO4 2‒. The oxidation number of oxygen decreases from –1 in H2O2 to –2 in H2O. H2O2 is an oxidising agent. (b) (i) volume of Na2S2O3 used = 24.85 + 24.95 2 = 24.90 cm3

4 © DHS 2024 (ii) Cr2O7 2‒ ≡ 3I2 ≡ 6S2O32– n(S2O3 2–) used = 24.90 24.8 1000 248.2  = 2.49 10–3 mol n(K2Cr2O7) in 20.0 cm3 aliquot = –31 2.49 10 6  = 4.15  10–4 mol n(K2Cr2O7) in 250.0 cm3 = –4250.0 4.15 10 20.0  = 5.18 10–3 mol mass of K2Cr2O7 = 5.18  10–3  294.2 = 1.52 g % purity of K2Cr2O7 = 1.52 100 % 1.65  = 92.4 % 9 (a) (i) K+: 1s2 2s2 2p63s2 3p6 (ii) Number of electron shells increase from Na+ to K+, which leads to an increase in screening effect. Hence, outermost electrons are increasingly further away from the nucleus, despite the increase in nuclear charge Ionic radius of K+ is thus larger than that of Na+. (iii) 3s 3px 3py 3pz

5 © DHS 2024 (b) log (2nd IE) O P Q R S T U V W elements Fig. 9.1 (b) (i) There is a large drop in 2nd IE from Q to R. Electron removed from Q+ comes from an inner shell which is nearer to the nucleus. Hence, it experiences stronger attraction from the nucleus and requires more energy to remove. Q has 1 valence electron, i.e. Q is in group 1. (ii) S+: ns2 T+: ns2 np1 Smaller amount of energy is required to remove the p electron in T which is of higher energy than the s electron in S. (iii) (see above)