2022 EFSS Chem Prelims P2 Ans

1 EDGEFIELD SECONDARY SCHOOL 2022 PRELIMINARY EXAM Secondary 4 Express CHEMISTRY 6092 August 2022 Answer scheme Paper 2 (80 marks) Qn. Answers Marks A1 a) A 1 5 b) D 1 c) B 1 d) F 1 e) I 1 A2 a) water and salts have different boiling points (1) water evaporates AND salts / residues / impurities / solids left in flask (1) water condenses / turns to liquid in the condenser (1) 3 10 b) i) Mg2+ and Cl– ii) 1.26/95 = 0.01325 mol (1) 0.01325 x 2 = 0.0265 mol/ dm3 (1) iii) white ppt/ solid/ deposit 2 c) i) Ba2+(aq) + SO42–(aq) à BaSO4(s) (1) ii) conc of SO42– = 1.24/96 = 0.0129 mol/dm3 (1) mol of SO42– = 0.129 x 50/1000 = 6.46 x 10–4 mol (1) mol of BaSO4 = 6.46 x 10–4 mol mass of BaSO4 = 6.46 x 10–4 x 233 = 0.150 (1) mass of SO42– in 50 cm3 seawater = 1.24 x 50/1000 = 0.062 g (1) mol of SO42– = 0.062/96 = 6.46 x 10–4 mol (1) mol of BaSO4 = 6.46 x 10–4 mol mass of BaSO4 = 6.46 x 10–4 x 233 = 0.150 (1) - Many students failed to notice that the concentration is given in g/dm3 so attempted to find moles of SO42– by multiplying 1.24 x 50/1000. 3

2 A3 a) 2.8.4 1 8 b) Si + 2Cl2 → SiCl4 - Many students attempted to give the state symbols but got it wrong (not penalised) - Do note that state symbols are not required unless stated in qns. Wrong state symbols may be penalised. 2 c) does not conduct electricity (1) low melting point / low boiling point (1) - Low density is not accepted since density depends on the actual state of the substance i.e. solid, liquid or gas. e.g. iodine has SMS but density is not low. 2 d) minus one mark for any mistake: missing legend/ non-bonding electrons 2 A4 a) Dividing % by mass by atomic mass N = 12.0/14 H = 3.4/1 O = 41.0/16 V = 43.6 /51 Dividing correctly by smallest to give correct ratio: N = 0.857 H = 3.4 O = 2.56 V = 0.855 Empirical formula: H 4NO3V (order of elements not considered in marking) 1 1 1 8 b) coloured Accept: can act as catalysts 1 c) NH4+ (1) VO3– (1) 2 d) (Y is an) oxidising agent (1) the oxidation number of iodine increases / iodide loses electrons / Y gains electrons (1) Reject: Y is an OA since it reacts with KI which is an RA (need to elaborate) Y is an OA since it oxidises KI to I2 (how do you know?) 2

3 e) i) Ammonia (1) ii) 2NH4VO3 → V2O5 + H2O + 2NH3 (1) 2 A5 a) C2H4O (1) Reject C12H24O6 1 9 b) 1 c) 2Na + Cl2 → 2NaCl - Many students attempted to give the state symbols but got it wrong (not penalised) - Do note that state symbols are not required unless stated in qns. Wrong state symbols may be penalised. d) sodium chloride has giant ionic lattice structure (1) a lot of energy required to overcome the strong electrostatic forces of attraction between oppositely charged ions. (1) chlorine has simple molecular structure (1) a small amount of energy is required to overcome the weak intermolecular forces of attraction between chlorine molecules.(1) 4 e) At 600 °C it is solid so ions cannot move / at 600 °C ions are in fixed position in a solid (1) NOTE: reference needed to solid as well as lack of movement of ions At 1000 °C it is molten/ liquid so ions can move / at 1000 °C it is molten/ liquid so ions are mobile / At 1000 °C it is molten/ liquid because the ions are free (1) NOTE: reference needed to temperature, liquid/ solid as well as movement of ions 2

4 A6 a) - reactants labelled on left and products labelled on the right AND product level below reactant level (1) - enthalpy change labelled and shown by downward arrow value (1) - activation energy shown as upward arrow from left hand energy level to energy ‘hump’ above the highest energy levels of both products and reactants (1) 3 9 b) (i) CO binds irreversibly to haemoglobin in blood; blocks oxygen uptake in blood. Reject: death, breathing difficulties - Many students missed out on the keyword: irreversibly. (ii) H2O (iii) Pd oxidation states +2 to 0; (1) C oxidation states +2 to +4; (1) (iv) palladium has been reduced since OS decreased (1) C has been oxidized since OS increased. (1) e.c.f. from (iii) 6 c) - Extraction of iron 2 B7 a) sodium > barium > magnesium > nickel > copper (Nickel spelt wrongly - zero marks) 1 10 b) - the more reactive the metal, the more negative is the reduction potential OR - the less reactive the metal, the more positive is the reduction potential 1 CH3OH + CO CH3COOH

5 c i) - +0.80 V (1) - silver is less reactive than copper; hence reduction potential will be more positive than copper (1) 2 c ii) Ag+(aq) + e− ⇌ Ag(s) (no state symbols - minus 1 mark) 2 d) metal chromium tin aqueous solution of nickel(II) ions 🗸🗸 dilute nitric acid 🗸🗸 🗸🗸 Each row - correct answer(s) - [1] 2 e i) The more positive/ less negative the electrode potential, the lower is the ion in the electrochemical series. 1 ii) positive: chlorine gas/ Cl2 2Cl–(l) → Cl2(g) + 2e– negative: iron/ Fe Fe 3+(l) + 3e– → Fe(l) Each row - correct answer(s) - [1] 2 B8 a i) concentration of ethanoate = 0.45 mol / dm3 (1) mass = 0.45 × 59 × 200/1000 = 5.31 g (1) 2 10 a ii) concentration of ethanoate ions at 300 s = 0.17 mol / dm3 (1) [ 0.16 mol / dm3 accepted] average rate = 0.17/300 = 5.67 × 10–4 (mol/dm3/ s) or 5.33 x 10–4 (mol/dm3/ s) [1] 1 a iii) rate of reaction decreases with time / reaction slows down (1) gradient becomes less steep with time (1) 2 a iv) concentration of reactants decreases (1) no. of particles per unit volume decreases (1) frequency of collision decreases + frequency of effective collision decreases (1) 3 b) ethanol has a simple molecular structure (1) Only stating covalent bond not accepted. Need to mention simple covalent bond. no free mobile electrons (1) 2 B9 EITHE R a) fractional distillation (1) cracking (1) 1 10

6 b) Electrolysis of dilute aqueous sodium chloride solution will produce oxygen gas at the anode and not chlorine gas. 1 c) 1 d) hydrogen chloride [hydrochloric acid not accepted] 1 e) Accept: chlorine atoms at bottom. correct repeating unit (1) 3 repeat units with continuation bonds at ends (1) No ending lines from the carbon atoms on both ends - minus [1] 2 d i) max mass = 2250 tonnes 1 d ii) % yield = 2175/2250 x 100 (1) = 96.7% (1) 2 B9 OR a) alcohol and carboxylic acid [both need to be correct to get 1 mark] 1 10 b) 2HOCH2COOH + Na2CO3 → 2HOCH2COONa + CO2 + H2O correct products (1) correct coefficients (1) 2 c) - oxygen has been removed from oxalic acid - hydrogen has been added to oxalic acid ALLOW oxidation number of carbon decreases 1 d i) - condensation polymer Any of the following: - because water has been removed - monomer does not have a carbon-carbon double bond - has ester linkage which is formed by condensation) [both need to be correct to get 1 mark] 1 d ii) - Terylene 1 e i) - Less landfill/ pollution - Fewer air pollution/ less poisonous fumes - Small mammals/ birds not trapped/ harmed 1

7 e ii) - plastic bags - any other valid answers Plastics not accepted 1 e iii) correct repeating unit (1) 3 repeat units with continuation bonds at ends (1) No ending lines from the carbon atoms on both ends - minus [1] 2

EXAM: PRELIM EXAM YEAR: 2022 Q K/U HI/SP Q K/U HI/SP Q K/U HI/SP 1 Kinetic Particle Theory and Experimental Techniques 0 2 Separation and Purfication Techniques 2 3 3 3 Elements, Compounds and Mixtures 0 4 Atomic Structure 3a 1 1 5 Ionic bonding 2bi, 5bde, 4 2 6 6 Covalent and metallic bonding 3c, 5d 4 2 8b 2 8 7 Writing equations 2c, 3b, 4e, 6aii 5 5 8 Acids and bases 1d 1 1 9 Salts 0 10 Mole Concept and Chemical Calculations 2bii, 2cii, 4a, 5a 8 8ai, 9ei(f), 9or(b) 7 15 11 Periodic Table 4b, 5c 2 2 12 Metals 6c 1 7a,b,d 4 5 13 Qualitative analysis 1a,b, 2