XMS SEC 3 WA3 2022 MS

1 SEC 3E CHEMISTRY 6092 TERM 3 WA3 MARK SCHEME Question 1 2 3 4 5 Answer A B C D D SECTION B & C [30 marks] 6 (a) A strong acid ionises completely (in water) whereas a weak acid ionises partially (in water). 2 (b) (i) empirical formula is CH2O element C H O mass 0.96 g 0.16 g 1.28 g Ar 12 1 16 moles 0.08 0.16 0.08 simplest ratio 1 2 1 1 1 (ii) (CH2O)n = 60 30n = 60 n = 2 molecular formula is C2H4O2 1 5 7 (a) silver nitrate / silver sulfate (any soluble silver salt) nitric acid sulfuric acid / sodium sulfate / potassium sulfate (any soluble sulfate) sodium hydroxide / sodium carbonate sulfuric acid / sodium sulfate / potassium sulfate (any soluble sulfate) 5 (b) (i) potassium nitrate and sodium chloride (both correct to score) 1 (ii) silver chloride, barium sulfate, lead sulfate (two correct to score) 1 (c) Mass of SO42- ions in BaSO4 = 96 / 233 × 1000 g = 412 g or 0.412 kg (1) Mass of SO42- ions in PbSO4 = 96 / 303 × 1000 g = 317 g or 0.317 kg (1) OR Moles of BaSO4= 1000 / 233 = 4.29 Moles of PbSO4 = 1000 / 303 = 3.30 (1) Since 1 mole of BaSO4 and 1 mole of PbSO4 each contains 1 mole of SO42, there are 4.29 moles of SO42- ions in 1000 g of BaSO4, compared to 3.30 moles of SO42- ions in 1000 g of PbSO4. (1) 2 (d) Any one of the following: Ag+ + Cl- → AgCl Ba2+ + SO42- → BaSO4 Pb2+ + SO42- → PbSO4 2H+ + CO32- → CO2 + H2O H+ + OH- → H2O 1 10 8 (a) From the data provided, the soil pH is observed to be between 4 – 4.5. (1) Only hydrangea blue will grow optimally at this pH. (1) Determines correct soil pH resulting in identification of suitable plant (award 2) Identifies correct plant OR Determines correct soil pH (award 1) 2 (b) Moles of CaCO3 needed = 10500 / 2 = 5250 (1) Mass of CaCO3 needed = 5250 × 100 = 525000 g (1) No units -1 2 4 9 (a) because the charges on the metal ions / the formulae of the hydroxides are different 1 (b) (i) RbOH + HCl → RbCl + H2O 1 (ii) Sr(OH)2 + 2HCl → SrCl2 + 2H2O 1 (c) (i) MCO3 → MO + CO2 1 (ii) MCO3 + 2HCl → MCl2 + H2O + CO2 1 (d) moles of CO2 = 2.57 / 44 = 0.0584 (1) moles of MCO3 = 0.0584 (1) (can be implied) Mr of MCO3 = 4.86 / 0.0584 = 83.2 (1) Ar of M = 83.2 – 12 – 48 = 23.2

2 M is magnesium (1), (as M is a Group II metal) 4 (e) (i) M(OH)2 + H2SO4 → MSO4 + 2H2O 1 (ii) MCO3 + 2HNO3 → M(NO3)2 + H2O + CO2 1 11