YSS 4E Pure Chem Paper 1 Prelim 2024 MS

1 Yishun Secondary School Mark Scheme Preliminary Examination 2024 Secondary 4 Pure Chemistry 6092 Paper 1 Date of exam: 28/08/2024 Setter: Widayah Answers to Paper 1 Qn Answer Explanation 1 A For the experiment, the following apparatus is needed: • 2.35 g of a salt: electronic balance • 25.0 cm3 of water: measuring cylinder • Endothermic reaction: thermometer • Stopwatch is not needed as no time interval is required 2 C Mr of: carbon monoxide, CO = 12 + 16 = 28 ethane, C2H6 = 30 helium, He = 4 sulfur dioxide, SO2 = 64 Gas which has Mr lower than air (average Mr = 28), will go through the porous pot and increase the volume in the porous pot, causing the air to leave the set -up thru the delivery tube, causing bubbles to form. 3 D • Locating agents are needed only used for the chromatography of colourless substances. They are chemicals that react with the colourless substances to form coloured spots. • As the substance at level X is travels more distance from the start line this means that it has higher Rf value and is more soluble than the substance at level Y in the solvent. 4 C At stage R to S, the temperature is decreasing towards the freezing point. Volume of liquid is decreasing as temp reaches S, the liquid turned to solid. 5 A Ne10 20 N7 15 3– O8 16 2– F9 17 ‾ no of p 10 7 8 9 no of e 10 10 10 10 no of n 10 8 8 8 6 C Mr of C2H3Br3 = 12 × 2 + 3 × 1 + 79 × 3 Mr of C2H3Br3 = 12 × 2 + 3 × 1 + 81 × 3 Mr of C2H3Br3 = 12 × 2 + 3 × 1 + 79 × 2 + 81 Mr of C2H3Br3 = 12 × 2 + 3 × 1 + 79 × 1 + 81 × 2 7 D silver chloride is insoluble in water and exist as solid at rtp. Hence, cannot conduct electricity as there is no free -moving ions when in water or at rtp. 8 A Diamond and silicon dioxide have similar tetrahedral structure. 9 A A The molecule contains a total of 10 electrons.

2 B W and X are both non-metals: Covalent bonding happens between non-metals C W is in Group 15 of the Periodic Table : W has 5 valence electrons, hence is from Grup 15 D X is hydrogen: correct 10 C Copper is a metal which conducts electricity due to the delocalised electrons. 11 B A 0.2 mol of magnesium metal, Mg = 0.2 × 6.02 × 1023 = 1.204 × 1023 B 0.3 mol of ammonia gas, NH3 = 4 × 0.3 × 6.02 × 1023 = 7.224 × 1023 C 0.4 mol of hydrogen chloride, HCl = 2 × 0.4 × 6.02 × 1023 = 4.816 × 1023 D 1.2 mol of carbon dioxide gas, CO2 = = 3 × 1.2 × 6.02 × 1023 = 2.1672× 1024 12 A SO2 H2SO4 percentage of sulfur by mass 32 32+2(16) x 100% = 50% 32 2×1+32+4(16) x 100% = 32.65% Na2S PbS2 percentage of sulfur by mass 32 2×23 +32 x 100% = 41.03 % 32 ×2 207+2(32) x 100% = 23.62% 13 B no of mole of CN = 600 24 000 = 0.025 mol Mr = 1.3 0.025 = 52 Mr of CN = 12 + 14 = 26 n = 52 26 = 2, molecular formula is B 14 B At rtp, no of mole of any gases and no of molecules are the same. mass of 1mole of any gasses is different 15 C mol of = 0.025 × 1.00 = 0.025 mol Na2SO4 : BaSO4 1 : 1 0.025 mol : 0.025 mol theoretical mass = 0.025 × (137 + 32 + 4 × 16) = 5.825 % yield = 5.36 5.825 × 100% = 92.0% 16 A the ionic equation between zinc metal and acid . H + ions will react with Zn to form hydrogen gas. 17 A Aluminium ions forms white precipitate, insoluble in excess aqueous ammonia Copper (II) ions forms blue precipitate, soluble in excess forming a dark blue solution zinc ions forms white precipitate, soluble in excess forming a colourless solution 18 C calcium sulfate, lead(II) sulfate, silver chloride are all insoluble salts. Method of preparation is precipitation 19 C initial pH – hydrochloric acid (low pH) end of experiment – excess sodium hydroxide (high pH) 20 D A adding more iron catalyst to the reaction mixture – metals are expensive B decreasing the volumes of nitrogen and hydrogen gases added – the yield will be low and the reaction may not move forward

3 C increasing the pressure exerted on the reaction mixture – increasing the pressure is costly. D pumping unreacted nitrogen and hydrogen gases back for further reaction – recycling unreacted nitrogen and hydrogen will save cost. 21 B Since Y is an oxidising agent, Y will react with reducing agents, KI and not with KMNO4 22 C stage 1: 2NO + O2 → 2NO2 stage 2: NO2 + SO2 → SO3 + NO A NO2 is reduced to NO. , NO2 is an oxidising agent. B NO is oxidised to NO2 C Oxygen is reduced as its oxidation state decreased from 0 in NO2 to -2 in NO2 D SO2 is oxidised to SO3. Hence, SO2 is a reducing agent. 23 B Copper is a reactive electrode. During electrolysis, copper electrode oxidised to form copper(II) ions. This will replenishes Cu2+ ions which are reduced at the cathode Concentration of Cu2+ ions remain unchanged Colour intensity and concentration of copper(II) sulfate solution remains the same 24 C For dilute copper(II) chloride solution, cathode/ negative electrode: copper metal will be discharged anode/ positive electrode: oxygen and water will be discharged 25 B option C has the highest difference in reactivity between the 2 metal electrodes which will contribute to the biggest potential difference 26 D Down a group, charge on the ion – stays the same number of outer shell electrons – same number of protons – increasing total number of inner shell electrons – increasing 27 A element X is more reactive than Y, as X displaced Y from a salt of Y chlorine is more reactive than iodine in group 17 28 A reactions of group 1 metals in water forming an alkali and hydrogen gas. When the alkali reacts with acid, neutralisation happens forming salt and water. 29 C Properties of transition metals: 1 High melting points and high densities. 2 Variable oxidation states 3 Form coloured compounds 4 Good catalysts 30 D Sacrificial protection occurs where the more reactive metal has the tendency to oxidise and form metallic ions. experiment 1 experiment 2 Fe(s) → Fe2+(aq) + 2e− Mg(s) → Mg2+(aq) + 2e− 31 C 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(g) Bond breaking (endothermic): 12 × N-H + 3 × O=O = 12 x 391 + 3 x 498 = 4692 + 1494 = + 6186 kJ

4 Bond forming (exothermic): 2 × N≡N + 12 x O-H = 2 x 945 + 12 x 467 = 1890 + 5604 = - 7494 kJ enthalpy change = - 1308 kJ 32 C A Activation energy of the forward reaction is given by +50 kJ/mol B Enthalpy change of the forward reaction is given by 20 + 50 = +70 kJ/mol. C Activation energy of the backward reaction is given by +50 kJ/mol. D Enthalpy change of the backward reaction is given by − 70 kJ/mol. 33 B same concentration of hydrogen peroxide with different particle size 34 D A At each level in the distillation column, a mixture of hydrocarbons is collected. B The higher up the distillation column, the lower the boiling points of the hydrocarbon collected. C The molecules collected at the bottom of the distillation column are the least flammable. D The molecules reaching the top of the distillation column have the smallest relative molecular masses. 35 B C23H48 → C3H8 + 4C2H4 + 4C3H6 36 C Option A, B and D are the same structural formulae 37 D Serine has the carboxylic acid group which will react with sodium carbonate to form salt, water and carbon dioxide. 38 C repeat unit of nylon 39 C P – respiration (give out energy) → exothermic, Q – photosynthesis (absorb light energy) → endothermic, R – combustion (give out heat) → exothermic 40 C Characteristics of catalytic converters