Pure Chemistry Compiled Nov 2020 P2 to Nov 2012 P2 ANS

1 GCE ‘O’ Level Examinations 2020 Chemistry 6092 Paper 2 Section A [50 marks] 1 (a) (i) calcium hydroxide 1 (ii) hydrogen and iron 1 (iii) argon / graphite 1 (iv) ethanol 1 (v) graphite 1 (vi) graphite and hydrogen 1 (b) true true false false 2 2 (a) The temperature in the car engine is very high and nitrogen and oxygen molecules gain kinetic energy and move faster. The frequency of collisions increases . More n itrogen and oxygen molecules collide with energy greater than activation energy and the frequency of effective collisions increases. For a reaction to take place, the reactant particles must collide with energy that is equal to or greater than the activation energy. At room temperature, the molecules do not have sufficient energy that is equal to or greater than the activation energy, hence, there is no reaction at room temperature. 1 1 1 (b) (i) Equation 1: 2NO(g) + O2 (g) → 2NO2(g) Equation 2: 4NO2(g) + 2H2O (l) + O2 → 4HNO3 (aq) 1 1 (ii) Sulfur dioxide 1 3 (a) (b) In metals, the sea of electrons are delocalised and are free to move about / mobile around the positive ions. In ionic compounds, they are able to conduct electricity only in molten or aqueous states. Both positively charged cations and negatively charged anions are free to move about in molten and aqueous state and act as charged carriers. 1 1

2 (c) (i) Mild steel is an alloy of iron with carbon atoms. Iron and carbon atoms are of different sizes. The regular arrangement of atoms in the pure iron is disrupted with the inclusion of carbon atoms. The layers of atoms cannot slide over each other easily when the force is applied. As a result, mild steel is stronger than pure iron. 1 1 (ii) With reference to the diagram, the size of an iron atom is smaller than the size of a carbon atom. However, this is not accurate as the atomic radius of an iron atom (0.126 nm) is larger than the atomic radius of a carbon atom (0.077 nm). Mild steel is an alloy consisting of iron (99%), carbon (0.25%) and manganese (0.4%) atoms. However, manganese atoms are absent in the diagram. With reference to the diagram, the proportion / percentage of iron atoms (24 atoms) to carbon atoms (4 atoms) is much lesser/lower, which is not an accurate representation of the ratio 99 : 0.25. 2 4 (a) An electrolytic cell is being set up. The anode will be the tin metal and is connected to the positive terminal of the battery. The cathode will be the iron can and is connected to the negative terminal of the battery. Both anode and cathode are pla ced in an electrolyte which is the aqueous tin(II) nitrate solution. 1 1 1 (b) Sn(s) + 2 H+ (aq) → Sn2+ (aq) + H2 (g) 1 mark – correct ionic equation. 1 mark – correct state symbol 2 (c) (i) Poly(ethene) will not corrode and produce toxic substances so that it is safer for consumers to use. 1 (ii) Macromolecule – A very large molecule that are made up of many atoms joined together by covalent bonds. Addition polymerisation – A process where many unsaturated monomers join together to form a single large molecule, without losing any molecules or atoms. 1 1 1

3 5 (a) Carbon dioxide has a simple molecular structure while silicon dioxide has a giant molecular structure. The weak intermolecular forces of attraction between carbon dioxide molecules can be easily overcome by small amount of energy. In silicon dioxide, there are a network of strong covalent bonds between silicon and oxygen atoms in a three -dimensional structure of silicon dioxide. As a lot of energy is required to break strong covalent bonds between the atoms, Hence carbon dioxide has low melting and boiling points while silicon dioxide has a high melting and boiling points. 1 1 1 (b) Acidic oxide – CO2 / NO2 / P2O5 / SO2 / SiO2 Basic oxide – Li2O / Na2O / MgO 1 (c) (i) MgO 1 mark – correct electronic configuration and number of cations (Mg) 1 mark – correct electronic configuration and number of anions (O) O2 1 mark – correct number of bonding electrons 1 mark – correct number of electrons not involved in bonding 2 2 (ii) In MgO, A magnesium atom loses two outer electrons to an oxygen atom to form Mg2+ and O 2-. The magnesium cations and oxide anions are held by strong electrostatic forces of attraction. In O2, Each oxygen atom shares two of its outer electrons with another oxygen atom. A double covalent bond is formed between the oxygen atoms within the molecule. 1 1

4 6 (a) 2M + H2 → 2MH 1 (b) (i) Number of moles of LiH = 8.0g / (7 +1) = 1 mol Number of moles of NaH = 24.0g / (23+1) = 1 mol Number of moles of KH = 40.0g / (39+1) = 1 mol Although the mass of metal hydride produced increases as we go down the group, t he number of moles of metal hydride produce d from 1 g of hydrogen remains the same. Hence Beth is correct and Jean is wrong. 1 1 (ii) No. of moles of H2 = 1/(1 + 1) = 0.5 mol No. of moles of RbH = 2 x 0.5 = 1 mol Mass of RbH = 1 x (85 + 1) = 86.0 g Mass of RbH depends on the molar mass of the substance, hence 86.0 g of rubidium hydride should be produced instead of 56.0 g. Hence Ryan is incorrect. 1 1 (c) Potassium reacts even more violently with hydrogen. The metal glows more brightly and more heat is produced. As we go down Group I, the size of the atom increases. It is easier to lose the valence electron as it is further away from the nucleus. Hence, reactivity increases down Group I. Potassium is more reactive than sodium as potassium has a higher tendency to lose electrons than sodium. 1 1 (d) Oxidation state of oxygen decreases from 0 in O 2 to -2 in H 2O. Thus, oxygen is reduced. Oxidation state of sodium increases from 0 in Na to +1 in NaH. Thus, sodium is oxidised. 1 1

5 Section B [30 marks] 7 (a) From Table 7.1, 30g of sodium chloride can dissolve in 100 cm3. 100 cm3 → 30g 1 dm3 → 30 × 10 = 300g In 1 dm3 of water, mass of sodium chloride that can be dissolved is 300g. Moles of sodium chloride in 1 dm3 of water = 300 / (23+35.5) = 300 / 58.5 = 5.13 mol Moles of sodium chloride = 399 / 58.5 = 6.82 mol (range of answers 5.13 to 6.82 mol) 1 (b) Similarities: 1. Both salts have the same freezing point of -20 ºC when 20g of salts is added respectively. 2. For the first 20g of salts added, the freezing point decreases. Differences: 1. When 40g of CaCl 2 is added to the water, it increases the freezing point to +12ºC from ̶ 45ºC. Whereas when 40g of NaCl is added, it is not fully dissolved. 2. CaCl2 has the lower possible freezing point, ̶ 45ºC, when it is added to water compared to NaCl, ̶ 20ºC. 3. After 20g of salts added, the freezing point with the addition of sodium chloride increases from -20 to -15 C, while the freezing point with the addition of calcium chloride continues to decrease from -20 to -45 C. 3 (c) The enthalpy change of solution, ΔH sol, is negative for both CaCl2 and MgCl2. This means that the dissolution of the salts in water is a exothermic reaction and the heat energy released helps to melt the solid ice. Both CaCl2 and MgCl2 are also hygroscopic, which attracts water vapour from the air and uses it to form a solution on the surface of the ice, which allows the de-icers to dissolve. 1 1 (d) (i) 1 mark – shape of graph, with energy level of products higher than reactants in urea energy level diagram with correctly labelled products. 3

6 1 mark – shape of graph, with energy level of products lower than reactants in calcium chloride energy level diagram with correctly labelled products. 1 mark – correctly labelled enthalpy change (+15.0 kJ/mol) for urea an