SASS Chem P1 Revision Ans

Page 1 ST ANDREW’S SECONDARY SCHOOL NAME CLASS CHEMISTRY 6092/01 Paper 1 REVISION PACKAGE + MULTIPLE CHOICE QUESTIONS Solutions 1 B Since forward rxn represents exothermic, the reverse / backward rxn hence represent endothermic. Ea is labelled for the energy difference between the reactants and the apex / peak of the graph. 2 D NbClx + x Na ® Nb + x NaCl No. of moles of Nb produced = mass ÷ molar mass = 18.58 g ÷ 92.9 g mol–1 = 0.2 mol Mass of Cl = 54.08 g – 18.58 g = 35.50 g No. of moles of Cl lost = No. of moles of NaCl formed = 35.50 g ÷ 35.5 g mol–1 = 1.0 mol Nb : NaCl 1 : x 0.2 : 1.0 x = 5 3 C NaOH + HCl ® NaCl + H2O No. of moles of NaOH added = vol × conc = 0.0125 dm3 × 0.05 mol dm–3 = 0.000625 mol No. of moles of HCl added = vol × conc = 0.025 dm3 × 0.1 mol dm–3 = 0.0025 mol ∴HCl is the ER. No. of moles of HCl remain unreacted = 0.0025 mol – 0.000625 mol = 0.001875 mol Combined volume of solution = 0.0125 dm3 + 0.025 dm3 = 0.0375 dm3 Conc of remaining HCl = mol ÷ vol = 0.001875 mol ÷ 0.0375 dm3 = 0.05 mol dm–3 4 C Na2O reacts with water to form NaOH (strong alkali) NaCl dissolves in water to form its solution (neutral) SO2 reacts with water to form H2SO3 (sulfurous acid, strong acid) SiCl4 reacts with water to form HCl (strong acid): SiCl4 + 2 H2O ® SiO2 + 4 HCl [Hydrolysis = Breaking apart with water] Even if you are unaware of the hydrolysis rxn, the other 3 pairs still does not give a pH of 7 each. 5 C Total no. of moles of gases in gaseous mixture = 0.12 dm3 ÷ 24 dm3 mol–1 = 0.005 mol No. of moles of O2 = 0.096 g ÷ 32 g mol–1 = 0.003 mol No. of moles of N2 = 0.056 g ÷ 28 g mol–1 = 0.002 mol 6 B No. of moles of 13C = 0.13 g ÷ 13 g mol–1 = 0.01 mol Each mole of 13C contains L atoms, of which each 13C atom contains 7 neutrons. 7 C Oxide of P = P2O5 (OS of P = +5) Oxide of S = SO2 (OS of S = +4) P4S3 + 8 O2 ® 2 P2O5 + 3 SO2 8 C No. of moles of H2O formed = 3.6 g ÷ 18 g mol–1 = 0.2 mol No. of moles of HCl required = No. of moles of NH3 produced = vol × conc = 0.1 dm3 × 1.0 mol dm–3 = 0.1 mol ⇒ x = 1, y = 2

Page 2 9 D S (Group VI, forms a molecule S8) : simple molecular structure Si (Group IV, forms a lattice similar to C) : giant covalent structure Al (Group III, metal) : giant metallic structure 10 A Atomisation: Forms discreet atoms from molecules (e.g. X2: X–X ® 2 X) through bond breaking = Endothermic rxn Combustion = Burning with O2 = Exothermic rxn Hydration = Addition of water, e.g. water of crystallisation = Exothermic rxn Neutralisation = Acid + Alkali = Exothermic rxn 11 D No. of moles of KMnO4 titrated = vol × conc = 0.0274 dm3 × 0.02 mol dm–3 = 0.000548 mol MnO4– : Fe2+ 1 : 5 0.000548 mol : 0.00274 mol Conc of FeSO4 reacted = mol ÷ vol = 0.00274 mol ÷ 0.025 dm3 = 0.1096 mol dm–3 12 B Q contains NH4+ that accounts for the NH3 evolved. At least 4 H in Q. Molecular