SJI 2023 Y4 Prelim Chemistry Paper 1 & 2 Answers

1 St. Joseph’s Institution 2023 Year 4 OP Preliminary Exam Answer Scheme Paper 1 1 2 3 4 5 6 7 8 9 10 D D A D B A D B A C 11 12 13 14 15 16 17 18 19 20 B D C C B C D B B D 21 22 23 24 25 26 27 28 29 30 C C D A D C C C D C 31 32 33 34 35 36 37 38 39 40 C B D D B D C A A C Paper 2 Section A Qn Suggested answers Mark A1a sodium iodide 1 A1b silver nitrate and sodium iodide 1 A1c aluminium sulfate and zinc nitrate 1 A1d ammonium chloride, silver nitrate and zinc nitrate 1 TOTAL 4 A2a There is more oxygen (from air) mixed with petrol, hence there is less incomplete combustion of petrol, thus producing lower amounts of carbon monoxide. 1 A2bi Lightning activity 1 A2bii Lean burn engine has a lower operating temperature. Less nitrogen gas and oxygen gas will react to form nitrogen oxides, as this reaction only occurs at high temperatures. 1 1 TOTAL 4 A3a There is no trend in the melting point, as it decreases from 2306 °C in boron to 30 °C in gallium then increases to 304 °C in thallium. 1 A3b Aluminium has a higher electrical conductivity than sodium as it has more mobile electrons than sodium. 1 A3ci [1m] – bonding electrons [1m] – non-bonding electrons 2 A3cii Boron has a giant molecular/covalent structure, while boron trichloride has a simple molecular/covalent structure. A lot of energy is needed to overcome the extensive network of strong covalent bonds between the atoms in boron. Little energy is needed to overcome the weak intermolecular forces of attraction between the molecules in boron trichloride. 1 1 1 TOTAL 7 Cl Cl Cl

2 Qn Suggested answers Mark A4a 2H+ + 2e- → H2 Reduction occurs at the silver electrode. The hydrogen ions gain electrons to form hydrogen gas. 1 1 A4b Solution / electrolyte turns from colourless to blue. 1 A4ci The difference in reactivity between lead and silver is greater than the difference in reactivity between copper and silver, thus resulting in a higher voltmeter reading. 1 A4cii The lead anode oxidises / dissolves to form lead(II) ions, which react with the sulfate ions in the electrolyte to form insoluble lead(II) sulfate. The concentration of mobile ions in the solution decreases, thus the electrical conductivity decreases over time. 1 1 TOTAL 6 A5ai sodium, magnesium, X, lead, silver 1 A5aii aluminium / zinc / iron / tin 1 A5aiii 2Al + 3CuO → Al2O3 + 3Cu / Zn + CuO → ZnO + Cu / Fe + CuO → FeO + Cu / Sn + CuO → SnO + Cu 1 A5aiv Aluminium / zinc / iron / tin is more reactive than copper and displaces copper from copper(II) oxide. 1 A5bi Difference (any one of the below): • Iron has a high density but potassium has a low density. • Iron is hard but potassium is soft. • Iron has high melting and boiling points but potassium has low melting and boiling points. 1 A5bii Galvanising with zinc forms a protective barrier and prevents oxygen and moisture / water vapour from coming into contact with iron and causing iron to rust. Zinc is more reactive than iron and hence loses electrons / oxidises more easily than iron and hence corrodes in place of iron. 1 1 A5ci calcium carbonate 1 A5cii The number of moles of each carbonate / percentage by mass of carbon in each carbonate / relative formula mass of each carbonate / relative atomic mass of the metal in each carbonate is different. [Note: the lower the Ar of the metal in the carbonate, the more moles of metal carbonate present in 2.00 g, and the more moles of CO2(g) evolved at the end of heating (i.e. greater mass loss).] 1 TOTAL 9

3 Qn Suggested answers Mark A6ai H2S(g) + CuSO4(aq) → CuS(s) + H2SO4(aq) [1m] – correct equation [1m] – state symbols 2 A6aii • Place aqueous hydrogen sulfide into the burette • Pipette 25.0 cm3 of aqueous potassium hydroxide into the conical flask • Add a few drops of methyl orange indicator • Perform titration to obtain the average volume of aqueous hydrogen sulfide required to react with 25.0 cm3 of aqueous potassium hydroxide. • Repeat the titration without the indicator. • Heat the salt solution till saturation. • Allow the filtrate to cool and crystals to form. • Filter the mixture to obtain the crystals. • Wash the crystals with a little cold distilled water • Dry the crystals between sheets of filter papers. [1m] – choice of suitable indicator [1m] – titration to find average volume and repeating the titration without the indicator [1m] – crystallisation of potassium sulfide [1m] – washing and drying of potassium sulfide 4 A6bi In titration 1, ethanoic acid is a weak acid that partially dissociates in water but in titration 2, hydrochloric acid is a strong acid which fully dissociates in water (to form H+ ions). Thus, (initial) pH in titration 2 is lower than in titration 1. [For the same concentration of acid (0.1 mol/dm3), HCl will have a higher concentration of H+ ions than ethanoic acid.] 1 A6bii (aqueous) sodium hydroxide / potassium hydroxide / any other strong Group I metal hydroxide 1 A6biii Initial pH: between 7 and 12; and final pH: 1 or lower Steep drop in pH after 15 cm3 of dilute H2SO4 added 1 1 TOTAL 10 A7a naphtha fuel in cars lubricating oils 1 1 1 A7b Cracking produces shorter-chain / smaller hydrocarbons which are in greater demand in the oil industry. Cracking forms hydrogen gas, which is used in the Haber Process / the large-scale production of ammonia in industry. 1 1 A7ci 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0 10 20 30 volume of solution (in burette) added / cm3 Titration I Titration II Titration III 1 2 3

4 Qn Suggested answers Mark A7cii 1 A7ciii C2H6 + 2Br2 → C2H4Br2 + 2HBr 1 A7civ 1 A7cv The polymer is non-biodegradable and may cause marine animals to chew on the polymer and choke / get trapped, suffocate and die. 1 TOTAL 10 Paper 2 Section B Qn Suggested answers Mark B8ai The reaction rate is slower at B than at A, as the gradient is less steep at B than at A. 1 B8aii Yes. [must state but no mark] All the nitric acid is used up in the reaction, as its concentration became 0 mol/dm3. Nitric acid is thus the limiting reagent, and Mg is added in excess. 1 B8bi A catalyst provides an alternative pathway with a lower activation energy. More particles would thus have energies equal to or greater than the activation energy. This increases the frequency of effective collisions between reacting particles, and thus the rate of reaction increases. 1 1 B8bii Propanone / CH3COCH3 The oxidation state of iodine decreased from 0 in I2 to -1 in HI (or -1 in CH3COCH2I). Iodine has been reduced by propanone so propanone is the reducing agent. 1 1 B8ci Comparing Experiments A and C, where the concentration of I2 and H+ was kept constant, when the concentration of propanone triples, the rate of reaction triples. As the rate of reaction is directly proportional to the concentration of propanone, the reaction is first order with respect to propanone. 1 1 B8cii Rate of reaction of Experiment D = 3.2 x 10-6 x 2 = 6.4 x 10-6 mol/(dm3 s) 1 B8di Rate of reaction of Experiment E = 3.2 x 10-6 mol/(dm3 s) Experiments A and E have the same rates of reaction, as the gradients of their concentration-time graphs are identical (since both graphs were found to be parallel). 1 1 B8dii zero order 1 TOTAL 12

5 Qn Suggested answers Mark B9a Monomers for PTT: Monomer for poly(isoprene): 1 1 1 B9bi No. of moles of repeating units in PTT = 1000 ÷ [11(12) + 10(1) + 4(16)] = 4.85437 mol = 4.85 mol (to 3 s.f.) No. of moles of repeating units in poly(isoprene) = 1000 ÷ [5(12) + 8(1)] = 14.7059 mol = 14.7 mol (to 3 s.f.) 1 1 B9bii No. of moles of CO2 formed from burning 1