YSS 4E Pure Chem Paper 2 Prelim 2024 MS

Page 1 of 6 Yishun Secondary School Mark Scheme 2024 4E Chemistry Preliminary Examination Date of exam: P2 – 23 Aug 2024 Setter: Paper 2 – Mr Rizal Answers to Paper 2 Qn Key Answers Remarks 1(a) graphite [1] 1(b) calcium chloride / sodium chloride / copper(II) carbonate (reject chemical formula) [1] 1(c) copper(II) carbonate [1] 1(d) copper, zinc [1] 1(e) helium [1] 1(f) zinc [1] 2(a) Experiment 1 Anode: 4OH ̶ (aq) → 2H2O (l) + O2 (g) + 4e ̶ Cathode: Ag+ (aq) + e ̶ → Ag (s) Experiment 2 Anode: Ag (s) → Ag+ (aq) + e ̶ Cathode: Ag+ (aq) + e ̶ → Ag (s) both cathode reactions [1] [1] [1] 2(b) When aqueous silver nitrate reacts with aqueous sodium chloride it will form a white precipitate of silver chloride [1] 2(c) In Experiment 1, no more silver was deposited on the object as all the silver ions in the electrolyte have been discharged. In Experiment 2, this will not be a problem as the silver ions discharged at the cathode are continuously being replaced by the silver anode which ionised / dissociate into silver ions. [1] [1] 2(d) Iron being higher in the reactivity series / more reactive than silver, will be able to displace silver from aqueous silver nitrate. Silver metal is deposited on the iron. But gold is lower than silver in the reactivity series, thus it cannot displace silver from aqueous silver nitrate. [1] [1] 3(ai) salt reagent 1 reagent 2 potassium sulfate sulfuric acid potassium hydroxide lead(II) nitrate nitric acid lead(II) oxide / lead(II) carbonate all correct [2] 2 – 3 correct [1] 0 – 1 correct [0] [2]

Page 2 of 6 Qn Key Answers Remarks 3(aii) K2SO4 + Pb(NO3)2 → PbSO4 + 2KNO3 no. of mole of K2SO4 = 15 1000 x 2 = 0.03 mol K2SO4 : PbSO4 1 : 1 0.03 mol : 0.03 mol mass of PbSO4 = 0.03 x (207+32+4(16)) = 9.09 g [1] [1] [1] 3(bi) willemite: SiO44- castorite: Si4O104- chrysotile: Si2O52- 1 – 2 correct [1] all correct [2] [2] 3(bii) Na Al Si O % by mass 8.7 10.3 32.3 48.7 molar mass 23 27 28 16 no. of moles 8.7 23 = 0.378 10.3 27 = 0.381 32.3 28 = 1.154 48.7 16 = 3.04 smallest ratio 1 1 3 8 Empirical Formula = NaAlSi3O8 (shown) [1] [1] 4(a) calcium [1] 4(b) number of electron shells number of valence electrons 7 4 [1] 4(c) Similarity: They have the same number of protons which is 114. Difference: 289Fl has 175 neutrons but 288Fl has 174 neutrons (must state both) [1] [1] 4(di) correct diagram (ratio of cations : electrons must be 1 : 1 ) correct label [1] [1] 4(dii) good conductor of electricity / good conductor of heat / malleable / ductile / high melting and boiling point [1] 4(diii) It reacts with acids to form salt and water [1] 5(a) Rf value of phaeophytin = 5 / 8 = 0.625 [1] (sea of delocalised) electrons

Page 3 of 6 Qn Key Answers Remarks 5(b) Extract from frozen spinach contains four pigments which are chlorophyll, xantophyll, phaeophytin and an unknown pigment. [1] 5(c) Extract from market is fresher as it does not contain phaeophytin as presence of phaeophytin shows that chlorophyll has decomposed. (reverse argument accepted) [1] [1] 6(a) Dissolve fertiliser in water, filter and use the filtrate for the following test. Add aqueous sodium hydroxide into both solutions. if white precipitate formed, the fertiliser is calcium dihydrogenphosphate if no white precipitate formed, the fertiliser is ammonium phosphate (must give both results) [1] [1] [1] 6(bi) A decreasing temperature leads to a higher yield of ammonia [1] 6(bii) Advantage: A higher pressure of 300 atm leads to a higher yield of ammonia Disadvantage: Maintaining a higher pressure of 300 atm incurs a larger cost / more expensive [1] [1] 6(c) Plant growth depends on soil acidity or pH / plants have optimum pH (for growth) Add Ca(OH)2 / slaked lime to reduce the acidity in soil. [1] [1] 7(ai) Y, iron, X [1] 7(aii) green solution turns colourless grey deposit/solid formed X is more reactive than iron as X displaced iron from iron(II) sulfate solution to form iron [1] [1] [1] 7(bi) greater volume of water on soil surface which will react with iron in the steel post [1] 7(bii) (Galvanising coats the steel with a more reactive metal zinc) Zinc will corrode in place of the iron in steel, It is preferred because even when the layer is scratched, the iron in steel will still not rust. [1] [1] 8(a) Addition polymers are formed from monomers that are unsaturated / have C=C monomers while condensation polymers are formed from monomers that have different functional groups. Addition polymers are formed without losing any molecules while condensation polymers are formed with the removal of small molecules such as water. [1] [1] 8(bi) [2]

Page 4 of 6 Qn Key Answers Remarks 8(bii) Mr of a repeated unit of nylon-6,6 (C12H22N2O2) = 12(12)+22(1)+2(14)+2(16)= 226 For Mr = 12 000, the number of repeated units (n) = 12 000 ÷ 226 = 53.1 For Mr = 20 000, the number of repeated units (n) = 20 000 ÷ 226 = 88.5 Range of average number of repeating units, n, is 54 ≤ n ≤ 88 / 54 to 88 [1] [1] 8(ci) [1] 8(cii) similarity: Both contain the amide linkage. [1] 9(a) 1,2,3 [1] 9(b) Going down Group 2, the atomic radii increase as the number of electron shell increases from one member to the next. [1] [1] 9(c) When a beryllium atom forms a beryllium ion, the radius decreases to less than half that of the atom. [1] 9(d) Barium would react vigorously with cold water. (reject explosively) As the reactivity increases down the Group. [1] [1] 9(e) Beryllium hydroxide is amphoteric in nature. It can react with alkaline solution. [1] [1] 9(f) Effervescence will be observed. Calcium reacts with water moderately/quickly/readily. Ca(s) + H2O(l) → Ca(OH)2(aq) + H2(g) correct balanced chemical equation correct state symbols (awarded if balanced chem equation is written) [1] [1] [1] [1] 10(ai) enegy require for bond breaking (use new numbers) = 6(413)+347+276+891 = +3992kJ energy released for bond forming = 5(413)+2(347)+891+363 = - 4013kJ enthalpy change = 3992+(-4013) = -21 kJ/mol ecf (incorrect energy for bond breaking and forming but subtract correctly) [1] [1]

Page 5 of 6 Qn Key Answers Remarks 10(aii) correct axes and correct labelling the reactants & products (must be chemical formulae of balanced equation) correct enthalpy change and shape (refer to level of reactants and products) correct activation energy ecf from wrong H in (b)(i) ecf based on shape of curve that students draw [1] [1] [1] 10(aiii) HBr dissociates in water to produce hydrogen/H+ ions. [1] 10(bi) There is a difference of a CH2 unit from one member to the next consecutive member in a homologous series / same functional group [1] 10(bii) CnH2n [1] 10(biii) Substitution Presence of light is needed in this reaction. [1] [1] 11(a) acidified potassium manganate (VII) (reject oxygen) [1] 11(bi) name of series alcohols carboxylic acid name of third member propanol propanoic acid structural formula of third member [1] [1] 11(bii) [1] 11(ci) C2H4 + H2O →C2H5OH Any two high temperature / high pressure / catalyst of phosphoric acid [1] [1] -21kJ/mol

Page 6 of 6 Qn Key Answers Remarks 11(cii) no. of moles of glucose = 36 x 106 / 180 = 2 x 105 mol glucose : ethanol 1 : 2 0.2 x 106 mol : 0.4 x 106 mol mass of ethanol = 0.4 x 106 x 46 = 18.4 tonnes [1] [1] 11(ciii) ethene obtained from crude oil / petroleum / fossil fuels which is a finite resource / non-renewable / will run out glucose obtained from plants so continuous supply /