AJC 2013_ALevel Chem P1 solns (2020 version)

1 H2 Chemistry 9647 2013 ‘A’ Level P1 Worked Solutions 1 Two H2O molecules are eliminated when three H3PO4 molecules undergo condensation. The molecular formula of triphosphoric acid produced can be deduced by construct ing a balanced chemical equation as shown: 3H3PO4 H5P3O10 + 2H2O FYI: A water molecule is eliminated between two H3PO4 molecules as shown: HO P O O OH H O P OH O OH H B 2 NO, H, D, 14 7 16 8 1 1 2 1 D3O+ H3O+ NH2– OD– no. of protons 3 + 8 = 11 3 + 8 = 11 7 + 2 = 9 8 + 1 = 9 no. of electrons 3 + 8 – 1 = 10 3 + 8 – 1 = 10 7 + 2 + 1 = 10 8 + 1 + 1 = 10 no. of neutrons 3 + 8 = 11 0 + 8 = 8 7 + 0 = 7 8 + 1 = 9 B 3 C+ N Si– P+ no. of electrons on gaining an e– 5 + 1 = 6 7 + 1 = 8 15 + 1 = 16 14 + 1 = 15 electronic config. 1s2 2s2 2p2 1s2 2s2 2p4 1s2 2s2 2p6 3s2 3p4 1s2 2s2 2p6 3s2 3p3 D 4 Largest increase from 4th to 5th I.E (4550 kJ mol–1)  5th electron is removed from the next inner quantum shell  4 electrons in outermost (valence) shell  M is a Group 4 element  M forms a chloride with formula MCl4 C 5 Solid graphite (giant covalent) conducts electricity (due to presence of delocalised e–). A 6 The behaviour of a gas is most ideal at low pressure and high temperature. At low pressure, the gas molecules are far apart. The volume of the gas molecules therefore becomes insignificant compared to the volume of container. Similarly, the intermolecular forces of attraction are negligible. At high temperature, the gas molecules move faster (higher K.E.) and the intermolecular forces of attraction become negligible. C

2 7 Hr = Hf (products) – Hf (reactants) reactants: CH3CO2Na(aq) and H2O(l) products: CH3CO2Na.3H2O(s) A 8 Temperature falls  endothermic reaction, i.e. H is positive Vigorous reaction occurs  reaction is spontaneous, i.e. G is negative Since G = H – TS –TS = G – H TS = –(G – H) S must be positive as –(G – H) is always positive. B 9 Au3+ + 3e– Au n(Au) = 6.0 / 197.0 = 0.03045 mol n(e–) = 3 x 0.03045 = 0.09135 mol Q = neF = 0.09135 x 96500 = 8815 C t = Q/I = 8815 / 0.10 = 8.8 x 104 s D 10 Ionic product of CaCO3 = [Ca2+][CO32–] = 0.10 x 1 x 10–9 = 1 x 10–10 < Ksp(CaCO3)  no ppt Ionic product of FeCO3 = [Fe2+][CO32–] = 0.10 x 1 x 10–9 = 1 x 10–10 > Ksp(FeCO3)  ppt forms Ionic product of MnCO3 = [Mn2+][CO32–] = 0.10 x 1 x 10–9 = 1 x 10–10 > Ksp(MnCO3)  ppt forms  FeCO3 and MnCO3 only are precipitated. C 11 • [H+] = [OH–] at all temperatures for a neutral solution • Position of equilibrium (POE) lies furthest to the right (towards ionisation of water) at 50 C since Kw at this temperature is the largest (since the larger the K w, the greater the extent