Pasir Ris 2024 4E Prelim MS

SEC 4 EXPRESS – CHEMISTRY 6092 PRELIM EXAM 2024 - ANSWERS Paper 1 – MCQ 1 D 11 C 21 D 31 D 2 A 12 D 22 A 32 B 3 B 13 C 23 C 33 C 4 D 14 B 24 B 34 B 5 B 15 D 25 B 35 A 6 D 16 A 26 D 36 C 7 C 17 B 27 B 37 A 8 D 18 D 28 D 38 C 9 B 19 B 29 B 39 D 10 A 20 C 30 D 40 D Paper 2 – Section A – Structured Questions No Suggested Answers Marks 1(a) silicon dioxide [1] 1(b) calcium oxide / copper(II) oxide [1] 1(c) zinc oxide [1] (i) 2(a) (ii) B: iron(II) hydroxide / Fe(OH)2 [1] (iii) C: iron(III) hydroxide / Fe(OH)3 [1] (iv) D: ammonia gas / NH3 [1] (v) E: silver chloride / AgCl [1] [4] 2(b) iron(II) chloride / FeCl3 [1] ammonium chloride / NH4Cl [1] iron(II) nitrate / ammonium nitrate also accepted e.c.f. allowed; follow part (a) answer [2] 3(a) (i) covalent bond [1] There is sharing of valence electrons between atoms. [1] wrong choice of terms -0.5 (ii) There are only six valence electrons surrounding aluminium atom. or Aluminium has not achieved stable noble gas configuration /valence shell not fully filled. or aluminium usually loses electrons but shares electrons with hydrogen instead [1] [3]

NOTE: students should use proper terms e.g. loses electrons instead of giving away electrons / gains electrons instead of taking in electrons. 3(b) (i) correct magnesium ion – [1]: 0.5 for charge, 0.5 for electrons correct hydride ion x 2 – [1]: 0.5 for charge, 0.5 for electrons −[0.5] for missing legend students should note that magnesium ion should be drawn between hydride ions since there are two hydride ions. (ii) Magnesium hydride has a giant ionic structure^ while aluminium hydride has a simple covalent structure^. A large amount of energy # is needed to overcome the strong electrostatic forces of attraction between oppositely charged ions*, while only a small amount of energy# is needed to overcome the weak intermolecular forces of attraction between aluminium hydride molecules*. Hence, the melting point of magnesium hydride is so much higher than the melting point of aluminium hydride. Compare type of structures^ – [1] Compare amount of energy# – [1] Compare bonds/forces* – [1] penalised if wrong choice of words e.g. atoms / ions / molecules for the wrong examples. misconception: covalent bonds are weak / weaker than ionic bonds intermolecular forces between aluminium and hydrogen atoms [5] 4(a) Sulfur dioxide [0.5] is released into the atmosphere, it will be oxidised by the atmospheric oxygen and dissolved into rainwater to form acid rain [0.5]. Acid rain will corrode limestone buildings and metal infrastructures / destroy crops and vegetation / kill lakes [1]. (mentions at least 2 effects) students confused between silicon dioxide and sulfur dioxide they tend to ignore environmental issue and gave health issues instead. [2] 4(b) (i) aqueous copper(II) sulfate / chloride / nitrate [1] copper(II) hydroxide is rejected since it is insoluble in water [5]

rejected if (II) was not stated (ii) copper plate: Cu(s) → Cu2+(aq) + 2e– blister copper: Cu2+(aq) + 2e– → Cu(s) many students gave the opposite reactions Minus half for this question if state symbols missing. (iii) Although the copper(II) cations are discharged at the cathode, the copper anode ionises away to form copper( II) ions at the same time. [1] - ecf accepted The concentration of copper(II) ions remains the same. [1] quite a number of students did not compare the processes at BOTH electrodes, but only mentioned one only. They should specify the electrodes where the reactions occur. They also did not mention the concentration of copper(II) ions in the electrolyte. OR The rate of discharge of copper(II) ions at the cathode [1] is equals to the rate of oxidation of copper atoms at the anode [1].

4(c) Mass of copper in blister copper = 0.96 x 2.00 tonnes = 1.92 tonnes [0.5] Mr of chalcopyrite = 64 + 56 + 2(32) = 184 [0.5] Since CuFeS2  Cu, mass of chalcopyrite used = 184 64 × 1.92 tonnes = 5.52 tonnes [1] OR Mass of copper in blister copper = 0.96 x 2.00 tonnes = 1.92 tonnes [0.5] Mr of chalcopyrite = 64 + 56 + 2(32) = 184 [0.5] no. of moles of copper = 1920000 64 = 30000 moles [0.5] no. of moles of chalcopyrite = no. of moles of copper = 30000 moles mass of chalcopyrite used = 30000 × 184 = 5520000 g = 5.52 tonnes [0.5] e.c.f. allowed majority of students did not consider the purity of the copper to be 96%. otherwise, most of them could complete the steps mark was awarded for the Mr (0.5) they should also note that the calculation of mass in mole calculations should be in grams, not tonnes/kg. [2] 4(d) (i) Zinc atoms have a different atomic size compared to the copper atoms. [0.5] This disrupts the orderly arrangement of atoms in copper. [0.5] The layers of atoms [0.5] of different sizes cannot slide over one another easily, [0.5] making the bronze harder and stronger. generally well done except for some students who failed to mention the layers / unable to use the correct terms e.g. ions/particles/molecules (ii) Presence of free moving / mobile electrons to act as charged carriers. [1] generally well done [3] 5(a) (i) There are two times more H+ ions in experiment 1 than experiment 2 in a given volume / more reactant particles in a given volume. [1] penalised (0.5) for inability to mention per unit volume The particles in experiment 1 will collide more frequently than in experiment 2. This increases the frequency of effective collisions in experiment 1 than in experiment 2. [1] penalised for the use of chance/probability instead of frequency/per unit time/ rate of effective collisions (ii) Granulated zinc in experiment 3 is bigger in size than powdered zinc in experiment 2. There is less (total) surface area (to volume ratio) of contact between reactant particles / less surface area exposed for reactant particles to collide in experiment 3 than in experiment 2. [1] The particles in experiment 3 will collide less frequently than in experiment 2. This decreases the frequency of effective collisions. [1] students are reminded to explain according to the question rather than explain the opposite and expect marker to imply [4]

5(b) (i) H2SO4 + Zn → ZnSO4 + H2 [1] Number of moles of H2SO4 = 25/1000 x 0.5 = 0.0125 mol Molar ratio of H2SO4 : H2 = 1 : 1 Therefore, number of moles of H2 = 0.0125 mol [1] Maximum volume of hydrogen gas = 0.0125 x 24 = 0.300 dm3 OR 300 cm3 [1] students are reminded not to include state symbols unless specified in the question. penalised (0.5) if state symbols are written BELOW the formulae. penalised (0.5 per paper) if no 3 sf for final answer [5] (ii) experiment 1 – steeper gradient, twice the volume of gas – [1] experiment 3 – gentler gradient, same volume of gas – [1] generally well done except the volume of gas for expt 1 5(c) Catalyst increases the rate of reaction , [0.5] hence the time taken to collect hydrogen gas will be lower than 20 s. [0.5] generally not well done as students are unable to quote 20s / mentioned time taken ‘faster’ which is conceptually incorrect [1] 6(a) fractional distillation of liquefied air [1] reject if no mention of fractional distillation / liquid / liquefied [1] 6(b) (i) Yield of ammonia decreases at higher temperature/ thermal decomposition of ammonia. [1] (ii) Rate of formation of ammonia will decrease / takes a longer time to obtain the ammonia / reaction will slow down.[1] generally well done [2] volume of hydrogen gas / cm 3 time taken / s experiment 3 experiment 2 experiment 1

6(c) Ammonium ions from the ammonium fertilisers will react with the (excess) calcium oxide to produce ammonia. [1] The nitrogen in the form of ammoni