EJC 2023 8873 'A' Level H1 Chemistry Suggested Solutions P1

Eunoia Junior College 9729 H1 Chemistry 2023 Paper 1 Suggested Solution 1 A ✓: The relative charge of a proton, +1, is –1 times the relative charge of an electron, –1. B : A proton has a relative charge of +1, while a neutron has no charge. C : The relative mass of an electron is 31 27 9.11 10 1 1 1.67 10 1833 2000 − −  = times the relative mass of a neutron . D : The relative masses of protons and neutrons are the same, which are about 200 times the relative mass of an electron.  A 2 3s 3p Al 1s2 2s2 2p6   1 Si 1s2 2s2 2p6    2 P 1s2 2s2 2p6     3  A 3 Rate of reaction is affected by the rate constant which in turn, is affected by the activation energy of the reaction. The lower the activation energy, the higher the rate constant, the faster the reaction. The C–Cl bond is more polar than the C–I bond as Cl is more electronegative than I. We would expect chloroethane to react faster with the negatively charged OH – in view of the polarity, which is not the case. Thus, the weaker C–I bond (240 kJ mol –1) compared to C –Cl bond (340 kJ mol –1) must have lowered the activation energy sufficiently to render the reaction of iodoethane faster  D 4 P4 + aHNO3 → 4H3PO4 + …H2O + …NO2 P4 + aHNO3 → 4H3PO4 + …H2O + aNO2 P4 + aHNO3 → 4H3PO4 + (a–16) H2O + aNO2 Balancing the number of H: a = 12 + 2(a–16) a = 20 HNO3 + bHBr → 2 b Br2 + …H2O + …NO HNO3 + bHBr → 2 b Br2 + …H2O + 1NO HNO3 + bHBr → 2 b Br2 + 2H2O + 1NO Balancing the number of H: b + 1 = 4 b = 3  C 5 To calculate the relative molecular mass, which is the average mass of one molecule of the compound, relative to 1 12 the mass of 12C, the molecular formula which shows the actual number of each atom is needed and not the empirical formula which shows the simplest ratio. Since the average mass of one molecule of the compound is required, the relative atomic masses of the elements in the compound, which is averaged over the isotopic distribution, is needed and not the exact relative isotopic masses.  C 6 X is a 3s orbital and Y is a 3p orbital. For the same principal quantum shell, energy of the orbitals: ns < np < nd < nf There are 1 ns orbital and 3 np orbitals for each principal quantum number n  2.  A 7 1 ✓: r 60.1 69 39.9 71 69.8100A  + = 2 ✓: The atomic number of Ga is 31, hence both isotopes have 31 protons. 3 ✓: 69Ga has 69 nucleons (neutrons + protons), while 71Ga has 71.  A 8  A 9 pH = –lg [H+] = 1.4 1.4 3H 10 0.0398 mol dm+ − − == H 0.0398 4 0.159 moln + =  = H2X → 2H+ + X2– 2HX H 1 0.0796 mol2nn +==  B 10 Halogen exists as non-polar diatomic X 2 molecules held together by instantaneous dipole-induced dipole attractions. Volatility, i.e. boiling, involves overcoming the intermolecular forces of attraction and not breaking of the X–X bond. Instantaneous dipole -induced dipole attractions get stronger from F2 to I2 due