2022 SASS A-Z Chem Prelims Ans Pt5

1 A – Z School’s Preliminary Papers 4 Express Chemistry ANSWERS: Part 5 FINAL U YUying Secondary School V Victoria School W West Spring Secondary School X Xinmin Secondary School Y Yuhua Secondary School Z Zhenghua Secondary School YUYING SECONDARY SCHOOL Paper 1 1 B 11 D 21 A 31 D 2 A 12 D 22 B 32 A 3 A 13 C 23 C 33 C 4 C 14 B 24 C 34 D 5 A 15 D 25 D 35 D 6 A 16 B 26 C 36 C 7 C 17 B 27 A 37 C 8 D 18 C 28 C 38 C 9 A 19 D 29 C 39 B 10 C 20 A 30 D 40 A Paper 2 A1a) B b) C c) C d) A e) E A2) before – solid structure of unshaded magnesium atoms, regularly arranged. [1] After – alternating solid structure of unshaded magnesium atoms and shaded oxygen atoms in the 1:1 ratio, regularly arranged. [1] A3a) X marked a the intersection of the two curves. [1] b) Increase the pressure to 50 atm and above to obtain liquid carbon dioxide. [1] c) 1 m double bond, 1 m all other electrons correct. d) Acidic oxide [1]. It reacts with bases to form salts / dissolves in water to form acid. [1] SASChemistry 2022edition The finallap towards ❷❺OCT⓿❽NOVP2 P1

2 A4a) If only water is used, yellow dye is insoluble [1]. If only ethanol is used, red dye is insoluble [1] b) At 80% ethanol, all the dyes have the same solubility, so there will be only one spot formed on the chromatogram. c) blue: 0.5, red: 0.68, yellow: 0.16 2 marks for all 3 correct, 1 mark for 2 correct, ½ mark for 1 correct. A5a) All the 3 valence electrons of boron are used in the forming of boron nitride.[1] It has no delocalized electrons to move about to conduct electricity. [1] b) There are weak van der Waals forces [½] between the hexagonal rings [½] of boron nitride. These forces are easily overcome [½], hence boron nitride is soft and slippery [½]. It is suitable to be used for lipsticks. c) Very high melting point A6a) True, False, True (1m each) b) Add excess water to the mixture. [½] Stir thoroughly to dissolve all the copper(II) halide. [½] Filter the mixture to remove the copper as residue. [½] Wash the copper with distilled water and dry between pieces of filter paper. [½] c) Mole of cu = 96/64 = 1.5 mols [½] mole of CuCl2 = 1.5 mols [½] mass of CuCl2 = 1.5 x 135 [½] = 202.5 g [½] A7a) Q, copper, P, S, R [1] b) 1. The blue [½] solution decolourises. [½] 2. Pinkish brown deposits of copper formed [1] c) The voltage difference between zinc and copper is more than the voltage difference between P and copper [½]. P is a more reactive metal than copper, but less reactive than zinc [½]. Hence, P is extracted from it’s oxide by reduction with carbon. [1] d) metals have a ‘sea of delocalized electrons’ which are free to move about to conduct electricity. A8a) test: add aqueous silver nitrate. [1] With NaCl: a white precipitate is seen. [½] With Na2SO4 : no precipitate seen. [½] Or: can use aqueous barium nitrate and answers given accordingly. b) Add excess dilute nitric acid to dissolve both carbonates. [1] Add aq. NaOH (or aq. ammonia) [1] With Na2CO3 : no precipitate is seen. [½] With ZnCO3 : a white precipitate is seen which dissolves in excess aq. NaOH (or aq ammonia). [½] A9ai) A hydrocarbon is a compound containing hydrogen and carbon atoms only. ii) C2H2n+2 bi) C8H16 ii) C20H24 à C12H24 + C8H16 + H2 iii) temperature of 600oC [½], Al2O3 or SiO2 catalyst [½] iv) It helps to meet the demand for smaller, more useful fractions, produce H2 etc. ci) They have the same functional group / similar chemical properties etc ii) [1] dibromopropane [1] Section B B10ai) The larger the atomic radius, the more reactive the metal. [1] ii) With a larger atomic radius, the attraction of the nucleus to the valence electrons [½] decreases [1]. This makes it easier to lose the valence electrons [½], and hence the metal gets more reactive.

3 b) The last electron shell is removed/lost [½] when the valence electrons are given away [½], hence the atomic radius of the magnesium atom becomes smaller when it forms magnesium ion. ci) Mg2+ ii) The statement is false. [1] Mg2+ has the highest polarizing power, followed by Ca2+ and Sr2+. [1]. However, Mg(OH)2 has the lowest decomposition temperature, followed by Ca(OH)2 and Sr(OH)2. [1] Hence, the relationship is not directly proportional, but inversely proportional. (accept this line as well, if evidence given is sufficient). di) dilute hydrochloric acid ii) 2 HCl + Sr(OH)2 à SrCl2 + 2 H2O [½] correct formulae, [½] balancing B11ai) As it produces the least energy in KJ per 100g/ least energy in KJ per cent Or: As it produces the pollutant sulfur dioxide as a product, while the rest do not. ii) It provides the most energy in KJ per 100g Or: It produces only harmless water vapor as a product. bi) Exothermic [1]. The energy released in the forming of the bonds of 2H2O [½] is more [½] than the energy taken in in the breaking of the bonds of 2H2 and O2. [1] ii) A suitable catalysed curve showing lower activation energy with labelling. [½] mark only awarded if not labelled. ci) From right to left. ii) H2 à 2H+ + 2e- [½] correct formulae, [½] balancing iii) It has charged particles (ions) which can move about to conduct electricity to complete the circuit. iv) Reduction has occurred. [1]. As electrons are accepted [1] Or: As the oxidation state of oxygen decreases from 0 in O2 to -2 in OH-. B12ai) Bi2O3 ii) Add excess [½] bismuth metal to hot dilute nitric acid. [½] Filter the mixture to remove the excess bismuth metal. [½] Heat the filtrate to make it saturated. [½] Leave the filtrate to cool. [½] Crystals of bismuth nitrate are formed. Dry the crystals between pieces of filter paper. [½] iii) Bi(NO3)3 + 3 NaOH à 3 NaNO3 + Bi(OH)3 [1] correct formulae, [1] balancing iv) To make sure all the bismuth nitrate is fully reacted/used up. v) The bismuth hydroxide decomposes to form bismuth oxide. bi) Bismuth shares its valence electrons with oxygen, there are no delocalized electrons available to move about to conduct electricity. (or: Bi2O3 is a simple covalent molecule/substance, it does not have any mobile ions or electrons to move about to conduct electricity.) iii) Amphoteric oxide [½] As it reacts with both citric acid and potassium hydroxide. [½] B13ai) addition polymer – polyethene etc [1] condensation polymer – terylene [1] ii) 1. In addition polymerization, only one single product is formed, while in condensation polymerization, more than one product is formed. [1] 2. Addition polymers are made up from one type of monomer, while condensation polymers are made up of two types of monomers. [1] bi) [1 mark each] ii) Mr of one repeat unit = 182 + 30 + 56 = 268 [1] Hence, Mr of 20000 repeat units = 20000 x 268 = 5360000 [1] (no units)

4 ci) ii) packaging for food VICTORIA SCHOOL Paper 1 Paper 2 1 Victoria School 2022 Secondary 4 Chemistry Prelim Answer Scheme + Markers’ Report Paper 1 1 2 3 4 5 6 7 8 9 10 B A B D C D A A C B 11 12 13 14 15 16 17 18 19 20 A C B D D D D C B B 21 22 23 24 25 26 27 28 29 30 B A C C D C A D D B 31 32 33 34 35 36 37 38 39 40 D C C B B D D D A B Paper 2 Section A Qn Suggested answers Mark A1a SO2 1 A1b C 1 A1c CaCl2 1 A1d Cu(NO3)2 / ZnSO4 1 A1e Cu(NO3)2 1 A2a Isotope 1 A2bi 2D2 + O2 → 2D2O 1 A2bii They have the same number of valence electrons. 1 A2biii Number of moles of water = 7.2/18 = 0.400 mol Number of moles of sodium = 6.9/23 = 0.300 mol ; Since 0.300 mol of sodium will react with 0.300 mole of water, water is in excess and sodium is the limiting reagent. ; Number of moles of hydrogen = 0.300 ÷ 2 = 0.150 mol Volume of hydrogen gas = 0.15 x 24 = 3.60 dm3 ; 1 1 1 A3a Methane has a simple molecular structure while lithium has giant metallic structure. ; Less energy is needed to overcome weak intermolecular forces of attraction in methane as it has a lower boiling point while a larger amount of energy is needed to overcome strong metallic bonds between positive ions and e