IGCSE Chemistry 0620 Paper 4: Air, Water, Fertilisers and Environmental Chemistry

Level: Year 10-11 (IGCSE) | Subject: Chemistry 0620 | Time: 60 minutes | Total: 100 marks | Parts: 30

Paper at a glance

Paper	Time	Marks	Calculator
Paper 4 (Extended, structured)	1 h 15 min	75	Not allowed

Questions

Question 1 (16 marks) – Composition of air and air pollutants

(a) Name the two most abundant gases in clean dry air and state the percentage by volume of each. (2 marks)

(b) State one main source of carbon monoxide, of sulfur dioxide and of nitrogen oxides, and give one effect of each on human health. (3 marks)

(c) A city burns 2.0 x 10^5 kg of diesel containing 1.5% sulfur by mass each day. Assuming all the sulfur is converted to sulfur dioxide, calculate the mass of sulfur dioxide produced per day in kg (Ar: S = 32.0, O = 16.0). (4 marks)

(d) A car catalytic converter converts carbon monoxide and nitrogen monoxide: 2CO + 2NO gives 2CO2 + N2. If 56.0 g of carbon monoxide reacts completely, calculate the number of moles of nitrogen monoxide needed and the mass of carbon dioxide formed (Mr: CO = 28.0, CO2 = 44.0). (4 marks)

(e) Explain why nitrogen oxide concentrations are usually highest on sunny afternoons and suggest one way of reducing nitrogen oxide emissions from vehicles. (3 marks)

Question 2 (17 marks) – Water quality testing and sewage treatment

(a) State three main stages in the treatment of drinking water and give the purpose of each. (3 marks)

(b) A 25.0 cm3 water sample is titrated with 0.0100 mol/dm3 EDTA solution and 12.5 cm3 is used to reach the end-point. EDTA reacts with calcium ions in a 1 to 1 mole ratio. Calculate the concentration of calcium ions in the sample in mol/dm3. (4 marks)

(c) Compare the behaviour of hard water and soft water when soap is added, and state one practical disadvantage of hard water. (3 marks)

(d) The dissolved oxygen in a river is 8.0 mg/dm3 before discharge and falls to 3.2 mg/dm3 after five days. The discharge flow is 2000 dm3 per day. Calculate the biochemical oxygen demand per day in g. (4 marks)

(e) Explain how run-off from farmland containing a high concentration of nitrates leads to eutrophication in a river. (3 marks)

Question 3 (16 marks) – Fertilisers, soil and agricultural chemistry

(a) Explain what the letters N, P and K on a fertiliser label stand for and give one role of each element in plant growth. (3 marks)

(b) Ammonium nitrate (NH4NO3) is a common nitrogen fertiliser. Calculate the percentage by mass of nitrogen in it and the mass of nitrogen in 100 kg of the fertiliser (Ar: N = 14.0, H = 1.0, O = 16.0). (4 marks)

(c) Ammonium sulfate ((NH4)2SO4) contains 21.2% nitrogen. Ammonium nitrate costs 6.00 dollars per kg and ammonium sulfate costs 4.00 dollars per kg. Calculate the cost per kg of nitrogen for each fertiliser and state which is better value (Ar: N = 14.0, H = 1.0, S = 32.0, O = 16.0). (4 marks)

(d) State two effects on the environment of applying too much nitrogen fertiliser. (3 marks)

(e) Explain why farmers add lime (calcium carbonate or calcium oxide) to acidic soil. (2 marks)

Question 4 (17 marks) – Limestone and its industrial uses

(a) Write the balanced equation for the thermal decomposition of limestone (calcium carbonate) and state one industrial use of a product. (3 marks)

(b) 1000 kg of limestone, assumed to be pure calcium carbonate, is decomposed completely by strong heating. Calculate the mass of calcium oxide and of carbon dioxide formed, in kg (Mr: CaCO3 = 100.0, CaO = 56.0, CO2 = 44.0). (4 marks)

(c) 2.20 g of carbon dioxide is bubbled through excess limewater. Calculate the mass of calcium carbonate precipitate formed (Mr: CO2 = 44.0, CaCO3 = 100.0). (4 marks)

(d) Write the equation for the reaction of quicklime with water and describe one safety risk of this process on a building site. (3 marks)

(e) Explain why injecting powdered limestone into the flue of a coal-fired power station reduces sulfur dioxide emissions, and write the reaction involved. (3 marks)

Question 5 (17 marks) – Greenhouse gases, energy and carbon emissions

(a) Explain how the greenhouse effect arises and name two greenhouse gases. (3 marks)

(b) A household uses 4500 kWh of electricity a year. The grid emits 0.70 kg of carbon dioxide per kWh. Calculate the household annual carbon dioxide emissions in tonnes, and the emissions if 20% of the electricity came from solar panels. (4 marks)

(c) A solar panel system generates 3200 kWh a year and replaces grid electricity entirely. Using 0.70 kg of carbon dioxide per kWh, calculate the total carbon dioxide avoided over 25 years in tonnes and state one assumption in the calculation. (4 marks)

(d) Coal-fired electricity emits 0.90 kg of carbon dioxide per kWh while gas-fired electricity emits 0.35 kg. Calculate the percentage reduction in emissions when gas replaces coal and give one reason for it. (3 marks)

(e) State two practical ways for an individual to reduce their carbon footprint and give one limitation of each. (3 marks)

Question 6 (17 marks) – Synoptic: acid rain and data analysis

(a) A lake holds 1.0 x 10^6 dm3 of water containing sulfuric acid at a concentration of 0.0010 mol/dm3. Calculate the mass of calcium carbonate needed to neutralise the acid completely, in kg (Mr: CaCO3 = 100.0). (4 marks)

(b) Unpolluted rain has a pH of about 5.6 while rain in one district has a pH of 4.0. Calculate the ratio of their hydrogen ion concentrations and explain what a difference of 1.6 pH units means. (4 marks)

(c) Name the two main gases that cause acid rain, state their sources and give one effect of acid rain on buildings. (3 marks)

(d) The annual mean sulfur dioxide concentration in a city fell from 80 μg/m3 in 1990 to 16 μg/m3 in 2020. Calculate the percentage decrease and suggest two possible causes. (3 marks)

(e) A government plans to subsidise the replacement of cars with electric vehicles in order to cut roadside nitrogen oxides. From a data analysis point of view, state two factors that must be considered when judging whether the policy worked. (3 marks)

Answers and marking notes

Question 1 – Composition of air and air pollutants

(a) Nitrogen makes up about 78% (one mark); oxygen makes up about 21% (one mark). [2 marks]

(b) Carbon monoxide comes from the incomplete combustion of fossil fuels and binds to haemoglobin, reducing the oxygen-carrying capacity of the blood (one mark); sulfur dioxide comes from burning sulfur-containing fuels and irritates the airways, triggering asthma (one mark); nitrogen oxides come from engines and other high-temperature combustion and irritate the lungs and form photochemical smog (one mark). [3 marks]

(c) Mass of sulfur = 2.0 x 10^5 x 1.5% = 3000 kg (one mark); n(S) = 3 000 000 / 32.0 = 93 750 mol (one mark); S + O2 gives SO2, so n(SO2) = 93 750 mol (one mark); m(SO2) = 93 750 x 64.0 = 6.0 x 10^6 g = 6000 kg (one mark). [4 marks]

(d) n(CO) = 56.0 / 28.0 = 2.00 mol (one mark); from the equation n(NO) = 2.00 mol (one mark); n(CO2) = 2.00 mol (one mark); m(CO2) = 2.00 x 44.0 = 88.0 g (one mark). [4 marks]

(e) The high temperature inside the engine makes nitrogen and oxygen from the air combine to form nitrogen monoxide (one mark); strong sunlight on a clear day oxidises nitrogen monoxide to nitrogen dioxide and drives photochemical reactions, so pollutants build up during the afternoon (one mark); a suitable method is to fit a catalytic converter, improve engine combustion efficiency or switch to electric vehicles (one mark). [3 marks]

Question 2 – Water quality testing and sewage treatment

(a) A coagulant such as alum is added so that fine suspended particles clump together and settle (one mark); filtration removes the remaining suspended solids (one mark); chlorine or sodium chlorate(I) is added to disinfect the water and kill pathogens (one mark). [3 marks]

(b) n(EDTA) = 0.0100 x 12.5 / 1000 = 1.25 x 10^-4 mol (one mark); the mole ratio is 1 to 1, so n(Ca2+) = 1.25 x 10^-4 mol (one mark); the volume in dm3 is 0.0250 dm3 (one mark); concentration = 1.25 x 10^-4 / 0.0250 = 5.00 x 10^-3 mol/dm3 (one mark). [4 marks]

(c) Hard water contains more calcium and magnesium ions (one mark); these ions form an insoluble scum with soap so little lather forms, whereas soft water lathers readily (one mark); a disadvantage is that more soap is needed and scale forms in heaters and pipes, lowering heating efficiency (one mark). [3 marks]

(d) The fall in dissolved oxygen is 8.0 – 3.2 = 4.8 mg/dm3 (one mark); the biochemical oxygen demand is therefore 4.8 mg/dm3 (one mark); oxygen used per day = 4.8 x 2000 = 9600 mg (one mark); that is 9.6 g (one mark). [4 marks]

(e) Nitrates are nutrients for algae and aquatic plants, so algae grow rapidly (one mark); the algae cover the surface and block light, so submerged plants die (one mark); bacteria decompose the dead algae and plants, using up the dissolved oxygen so that fish and other organisms suffocate (one mark). [3 marks]

Question 3 – Fertilisers, soil and agricultural chemistry

(a) N stands for nitrogen, which is used to make proteins and chlorophyll and promotes leaf growth (one mark); P stands for phosphorus, which promotes root development and flowering (one mark); K stands for potassium, which regulates water balance and improves disease resistance (one mark). [3 marks]

(b) Mr = 14.0 x 2 + 1.0 x 4 + 16.0 x 3 = 80.0 (one mark); each formula unit contains 28.0 of nitrogen (one mark); percentage = 28.0 / 80.0 x 100% = 35.0% (one mark); 100 kg x 35.0% = 35.0 kg of nitrogen (one mark). [4 marks]

(c) Ammonium nitrate costs 6.00 / 0.350 = 17.1 dollars per kg of nitrogen (one mark); for ammonium sulfate Mr = 28.0 + 8.0 + 32.0 + 64.0 = 132.0, so the nitrogen content is 28.0 / 132.0 = 21.2% (one mark); ammonium sulfate costs 4.00 / 0.212 = 18.9 dollars per kg of nitrogen (one mark); ammonium nitrate is cheaper per kg of nitrogen and is better value (one mark). [4 marks]

(d) Unused nitrates are washed into rivers and lakes by rain and cause eutrophication (one mark); nitrates seep into groundwater and contaminate drinking water supplies (one mark); applying and manufacturing fertiliser also releases greenhouse gases such as dinitrogen oxide (one mark). [3 marks]

(e) Lime is alkaline and neutralises the acidity of the soil, raising its pH (one mark); at the higher pH nutrients are more readily absorbed by plants and aluminium ions are less toxic to roots (one mark). [2 marks]

Question 4 – Limestone and its industrial uses

(a) CaCO3 gives CaO + CO2 (one mark); the condition is strong heating (one mark); calcium oxide is used to make cement, to neutralise acidic soil or to remove sulfur dioxide from flue gases (one mark). [3 marks]

(b) n(CaCO3) = 1.0 x 10^6 / 100.0 = 1.00 x 10^4 mol (one mark); m(CaO) = 1.00 x 10^4 x 56.0 = 5.60 x 10^5 g = 560 kg (one mark); m(CO2) = 1.00 x 10^4 x 44.0 = 4.40 x 10^5 g = 440 kg (one mark); checking, 560 + 440 = 1000 kg, so mass is conserved (one mark). [4 marks]

(c) n(CO2) = 2.20 / 44.0 = 0.0500 mol (one mark); Ca(OH)2 + CO2 gives CaCO3 + H2O (one mark); n(CaCO3) = 0.0500 mol (one mark); m = 0.0500 x 100.0 = 5.00 g (one mark). [4 marks]

(d) CaO + H2O gives Ca(OH)2 (one mark); the reaction is vigorous and releases a large amount of heat (one mark); splashes of hot alkali can cause serious burns to skin and eyes, so workers must wear goggles and gloves (one mark). [3 marks]

(e) The limestone decomposes on heating to calcium oxide (one mark); the calcium oxide reacts with sulfur dioxide to form calcium sulfite, CaO + SO2 gives CaSO3 (one mark); the solid product is collected by the dust precipitator, so sulfur dioxide is not released into the atmosphere (one mark). [3 marks]

Question 5 – Greenhouse gases, energy and carbon emissions

(a) Solar radiation enters the atmosphere as short-wave radiation and is mostly absorbed by the ground (one mark); the warmed ground re-emits long-wave infrared radiation, which greenhouse gases absorb and re-emit in all directions, some of it back towards the ground (one mark); common greenhouse gases include carbon dioxide, methane and water vapour (one mark). [3 marks]

(b) Annual emissions = 4500 x 0.70 = 3150 kg (one mark); that is 3.15 tonnes (one mark); emissions from 20% of the electricity = 3150 x 0.20 = 630 kg (one mark); the remaining emissions are 3150 – 630 = 2520 kg, that is 2.52 tonnes (one mark). [4 marks]

(c) Avoided each year = 3200 x 0.70 = 2240 kg (one mark); over 25 years = 2240 x 25 = 56 000 kg (one mark); that is 56.0 tonnes (one mark); the calculation assumes the grid emission factor stays unchanged for 25 years and that the output of the panels does not decline (one mark). [4 marks]

(d) The difference is 0.90 – 0.35 = 0.55 kg per kWh (one mark); the percentage reduction is 0.55 / 0.90 x 100% = 61% (one mark); gas has a higher hydrogen-to-carbon ratio, so less carbon dioxide is produced per unit of energy, and gas plants usually have higher thermal efficiency (one mark). [3 marks]

(e) Cutting household electricity use, for example switching off appliances on standby and using efficient lamps, though household electricity is only part of total emissions (one mark); using public transport or walking, though public transport is limited in some areas (one mark); eating less meat, though dietary habits are shaped by culture and price and change slowly (one mark). [3 marks]

Question 6 – Synoptic: acid rain and data analysis

(a) n(H2SO4) = 1.0 x 10^6 x 0.0010 = 1.0 x 10^3 mol (one mark); CaCO3 + H2SO4 gives CaSO4 + CO2 + H2O (one mark); n(CaCO3) = 1.0 x 10^3 mol (one mark); m = 1.0 x 10^3 x 100.0 = 1.0 x 10^5 g = 100 kg (one mark). [4 marks]

(b) pH is the negative logarithm of the hydrogen ion concentration, so the ratio is 10^(5.6 – 4.0) = 10^1.6 (one mark); 10^1.6 is about 40 (one mark); a fall of one pH unit means the hydrogen ion concentration is ten times greater (one mark); therefore 1.6 units corresponds to about a forty-fold increase in acidity (one mark). [4 marks]

(c) Sulfur dioxide comes from burning fossil fuels that contain sulfur (one mark); nitrogen oxides come from vehicle engines and other high-temperature combustion (one mark); these gases are oxidised and dissolve in rain to form acids that react with limestone and marble, eroding the stonework and carvings of buildings (one mark). [3 marks]

(d) The decrease is 80 – 16 = 64 μg/m3 (one mark); the percentage decrease is 64 / 80 x 100% = 80% (one mark); possible causes include switching to low-sulfur fuel and gas-fired generation, fitting flue gas desulfurisation to power stations, moving polluting industry out of the city and tightening vehicle emission standards (one mark). [3 marks]

(e) Data from a comparison period or a comparison area are needed, and the effects of weather, traffic volume and economic activity must be excluded before the fall in concentration can be attributed to the policy (one mark); the electricity used to charge the vehicles may still cause emissions at power stations, so total emissions rather than roadside concentrations alone must be counted (one mark); the cost of the subsidy must be compared with the emission reduction achieved, and the replacement rate must be checked to see whether it can lower concentrations appreciably (one mark). [3 marks]

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