VS 2018 O Level Pure Chemistry P1+P2 Answer Scheme

2018 Answer1 Victoria School2018 Chemistry O Levels / 6092Suggested Answer Key PAPER 1Q1Q2Q3Q4Q5Q6Q7Q8Q9Q10Q11Q12Q13Q14Q15Q16Q17Q18Q19Q20BBCDAABCDBCCBBADCDBAQ21Q22Q23Q24Q25Q26Q27Q28Q29Q30Q31Q32Q33Q34Q35Q36Q37Q38Q39Q40BAAADBDDADCDADBBDCCCPAPER 2 (SECTION A)QuestionAnswerMarkA1aiAmmonia, carbon dioxide1aiiSilicon dioxide1aiiiAmmonia1aivLead(II) nitrate1avAmmonia1bFalseTrueTrueFalse[Every 2 correct – 1 mark]2

2018 Answer2 A2ChlorineCopperMolten sodium chloride111A3Row 1: formulae of reagents used: CuO(s) / CuCO3(s) / Cu(OH)2(s)Row 2:formulae of salt: KNO3(s)method used: titration of KOH(aq) with HNO3(aq)Row 3:formulae of reagents used: Pb(NO3)2(aq) and H2SO4(aq) / (NH4)2SO4(aq) / Na2SO4(aq) / K2SO4(aq) or any Group I sulfate method used: ionic precipitation / addition of the two aqueous reagents / solutions, filtration to obtain the residue 1111A4aMetals contain positive ions in a ‘sea’ of delocalised electrons. The electrons can move to conduct electricity. In solutions of ionic compounds, the ions dissociate from the giant ionic lattice structure and can move to conduct electricity. 11bMetals remain unchanged when electricity is passed through.Solutions of ionic compounds break down into their elements/simpler substances when electricity is passed through.1A5aNitrogen: 78Oxygen: 21 1biThe density of helium is the lowest (0.17 g/dm3) as compared to the other gases and it would easily rise to the upper atmosphere and escape into outer space.1biiDensity of argon = 40/24 = 1.67 g/dm3 (to 3 s.f.)1

2018 Answer3 biiiThe percentage of helium in dry air is 0%, which means it cannot be obtained from fractional distillation of liquid air to be used for different purposes. Helium is also a finite resource on earth and should be conserved and used for more important purposes like MRI scanners in hospitals which will benefit patients, rather than on filling party balloons which is for leisure and not important. 11ci Helium nucleus contains2 protons and 2 neutrons. Argon nucleus contains 18 protons and 22 neutrons.1m – helium1m – argon 2 (ii)Both helium and argon are chemically inert. This is because both helium and argon have fully filled outermost electron shells of 2 and 8 electrons respectively. Hence they do not need to lose, gain or share electrons.11A6aThe temperature of 450 °C used in Stage 2 is lower than the temperature of 800 °C used in Stage 1. Hence, less energy is required to maintain a lower temperature in the reactor in Stage 2. The reaction in Stage 1 is endothermic so energy is taken in from the surroundings, compared to the exothermic reaction in Stage 2 which releases energy to the surroundings. Hence, the reaction in Stage 2 requires less energy. 11bCarbon monoxide is a toxic gas which reacts readily and irreversibly with haemoglobin to form carboxyhaemoglobin. This reduces the oxygen carrying capacity of the blood, leading to dizziness/fatigue and eventually death. 1

2018 Answer4 Burning carbon monoxide will produce carbon dioxide which is less toxic and harmful. 1ciThe percentage by volume of nitrogen to hydrogen is 1:3, which is the same as the stoichiometric/reacting ratio of nitrogen to hydrogen in the equation in Stage 2. 11ciiBoth gases have different relative molecular masses (Mr). Nitrogen has an Mr of 28 while hydrogen has an Mr of 2. However, one mole of each gas occupy the same volume of 24 dm3 at room temperature and pressure.1ciiiRecycling the unreacted gases helps allows them to react again, improving the yield of ammonia.Obtaining nitrogen from fractional distillation of liquid air and hydrogen from cracking of crude oil fractions is costly as a lot of energy is used in these processes. Recycling the unreacted gases helps to save cost.Hydrogen is obtained from the cracking of crude oil fractions. Crude oil is a non-renewable resource. Hence, recycling the unreacted hydrogen helps to conserve crude oil. [any 2 of the above]2di 1m – bonding electrons 1m – non-bonding electrons2 diiThe coating on the catalyst reduces the surface area of the catalyst in contact with the reactants, hence decreasing the frequency of effective collisions between the reacting particles and the surface of catalyst, decreasing the rate of reaction. 1A7aBoth have 1 valence electron and tends to lose the electron to form positive ions with a charge of 1+.2

2018 Answer5 Both have low densities.Both have low melting points.Both are good reducing agents. [any 2 of the above]bHydrogen is a gas but the Group I elements are solids/metals at room temperature and pressure.Hydrogen can gain 1 valence electron to form a negative hydride ion with a 1- charge but the Group I elements can only lose 1 valence electron to form a positive ion with a 1+ charge. Hydrogen exists as a diatomic molecule / has a simple molecular structure but Group I elements exist as positive ions in a ‘sea’ of delocalised electrons / have giant metallic structures. Hydrogen cannot conduct electricity in all states but the Group I elements can conduct electricity in solid and molten states. Hydrogen cannot react with water but Group I elements react with water to form alkalis and hydrogen gas. [any 2 of the above]2ciAg2O(s) + H2(g) → 2Ag(l) + H2O(g)1m – balanced equation1m – state symbols2ciiH2 is oxidised. Ag2O is reduced. The oxidation state of hydrogen increases from 0 in H2 to +1 in H2O / H2 gains an oxygen atom to form H2O. The oxidation state of silver decreases from +1 in Ag2O to 0 in Ag / Ag2O loses an oxygen atom to form 2Ag. 111ciiiobservations: For magnesium oxide, there is no visible reaction.For copper(II) oxide, the black solid turns reddish brown. The more reactive a metal is, the more difficult it is to reduce the metal oxide by hydrogen. Magnesium is more reactive than copper and above hydrogen gas in the reactivity series so magnesium oxide cannot be reduced by hydrogen. Copper is below hydrogen gas in the reactivity series and copper(II) oxide is reduced by hydrogen to form copper metal and steam.111

2018 Answer6 PAPER 2 (SECTION B)QuestionAnswerMarkB8aIt will be difficult to distinguish between potassium, rubidium and caesium because the flame colours of these elements are different shades of violet, which are quite similar. The intense yellow-orange colour flame of sodium will also result in difficulty in observing the flame colours of the other Group I elements as it will most likely overpower the other colours. 11biThe mixture contains sodium and rubidium. This is the emission spectra of these elements are present in the emission spectrum of the mixture. The mixture does not contain lithium and potassium because the emission spectra of these elements are not present in the emission spectrum of the mixture. There are some unknown elements present in the mixture as there are some emission spectra present which do not belong to any Group I elements.111biiThe emission spectra of all the Group I elements (caesium and francium) will be needed.1ciNaClNaBrNa2SO4[3 correct – 2 marks, 2 correct – 1 mark, 1 correct – 0 marks]2ciiConcentration of Mg2+ ion = (0.15 / 0.6) x 0.00420 mol/dm3 = 0.00105 mol/dm3 (to 3 s.f.)Concentration of Na+ ion = (1.0 / 0.6) x 0.00420 mol/dm3 11

2018 Answer7 = 0.00700 mol/dm3 (to 3 s.f.)dIon chromatograms are able to identify any ions, even those which contain multiple atoms. However, emission spectrums are only able to identify the types of elements present. In addition, ion chromatograms are able to measure the relative amount or concentration of each ion present in the mixture. 11B9aVegetable oils contains three ester linkages / functional groups.1bCOO CH3 2cThe untreated oils contains acids which will react with the potassium hydroxide catalyst.This reduces the amount of potassium hydroxide catalyst present to catalyse the reaction to make biodiesel, hence the rate of the reaction will decrease.11dAverage volume of KOH used = (21.50 + 21.60)