The problems are on the atmosphere and Earth resources worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Air is about 78 % nitrogen and 21 % oxygen. These worked answers correct the 20 problems of the Chemistry 9-10 worksheet on atmosphere and Earth resources, from reading the air bar to desalination or groundwater: each solution shows the formula, the substitution and the final answer with its unit.
Each answer gives the key idea, the working and the final result in bold with its unit. Check that you can explain the chemistry behind each number, because exam questions often ask you to justify a conclusion as well as calculate it.
Level: Chemistry grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: HS-ESS3-6, GCSE C9-C10
Problem 1 Reading the air bar
1. A is nitrogen, about 78 %.2. B is oxygen, about 21 %.
3. 0.21 × 200 = 42 cm³.
4. Argon (or carbon dioxide).
Problem 2 Greenhouse gas or not
1. Carbon dioxide, methane and water vapour.2. Cattle farming, rice fields or landfill sites.
3. Burning fossil fuels or cutting down forests.
Problem 3 Match the pollutants
Carbon monoxide: c and s. Sulfur dioxide: a and p. Nitrogen oxides: b and r. Particulates: d and q.Problem 4 Potable or pure
1. Water that is safe to drink.2. It contains dissolved minerals and added chlorine, so it is a mixture.
3. By distillation (evaporate and condense).
Problem 5 Water treatment stages
1. E, sterilise with chlorine.2. D (the sand and gravel filter), and also C for the fine solids that settle.
3. Chlorine (or ozone or ultraviolet light).
Problem 6 Photosynthesis and oxygen
1. 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂. Carbon 6 = 6, hydrogen 12 = 12, oxygen 18 = 6 + 12.2. Algae and plants removed carbon dioxide and released oxygen.
3. In sedimentary rocks (limestone), fossil fuels or shells.
Problem 7 Carbon dioxide levels
1. 420 − 280 = 140 ppm.2. 140 ÷ 280 × 100 = 50 %.
3. 420 ÷ 1 000 000 × 100 = 0.042 %.
Problem 8 Explain the greenhouse effect
1. A: sunlight reaching the surface. B: infrared radiation from the surface. C: infrared sent back down by greenhouse gases. D: infrared that escapes to space.2. More gas absorbs more infrared, so C increases, less escapes and the average temperature rises.
3. Without it the average surface temperature would be far below freezing, so liquid water and life could not exist.
Problem 9 Commuting emissions
1. 12 × 2 × 5 = 120 km.2. Car: 120 × 170 = 20 400 g = 20.4 kg. Bus: 120 × 80 = 9600 g = 9.6 kg.
3. 20.4 − 9.6 = 10.8 kg.
Problem 10 Nitrogen oxides
1. N₂ + O₂ → 2NO.2. 2NO + O₂ → 2NO₂.
3. 2CO + 2NO → 2CO₂ + N₂. Carbon 2, oxygen 4, nitrogen 2 on each side.
4. Nitrogen is very unreactive. The very high temperature in the engine gives it enough energy to react with oxygen.
Problem 11 Sulfur dioxide from fuel
1. 0.015 × 40 = 0.60 t.2. S + O₂ → SO₂; M(SO₂) = 64, M(S) = 32, so mass = 0.60 × 2 = 1.2 t.
3. 2SO₂ + O₂ + 2H₂O → 2H₂SO₄. Sulfur 2 = 2, hydrogen 4 = 4, oxygen 4 + 2 + 2 = 8 = 8.
Problem 12 Treating fresh water
1. Particles can shield microbes from the steriliser and use up chlorine, so solids are removed first.2. It is cheap and keeps working in the pipes after leaving the works.
3. Distillation needs a very large amount of energy and is too costly for a whole town when fresh water is available.
Problem 13 Desalination plant
1. 0.40 × 5000 = 2000 m³.2. 2000 × 1000 = 2 000 000 dm³.
3. 2 000 000 ÷ 150 = 13 333, so about 13 000 people for a day. The brine is 5000 − 2000 = 3000 m³.
Problem 14 Waste water stages
1. Screening, sedimentation, aerobic biological treatment of the liquid; the sludge from sedimentation goes to sludge digestion.2. Large objects such as grit and rags.
3. Methane (biogas), which can be burnt for energy.
Problem 15 Trees and a car
1. 1200 × 2.3 = 2760 kg.2. 2760 ÷ 20 = 138 trees.
Problem 16 Carbon dioxide from petrol
1. 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O. Oxygen: 32 + 18 = 50 = 25 × 2.2. M = 8 × 12 + 18 = 114; n = 1000 ÷ 114 = 8.77 mol.
3. CO₂ = 8 × 8.77 = 70.2 mol; mass = 70.2 × 44 = 3088 g, about 3.1 kg.
Problem 17 Oxygen in air experiment
1. Oxygen used = 21 cm³, so 21 %.2. 21 ÷ 24 000 = 0.000875 mol.
3. Copper = 2 × 0.000875 = 0.00175 mol; mass = 0.00175 × 63.5 = 0.11 g.
Problem 18 Bottle life cycle
1. 4.8 ÷ 20 + 0.30 = 0.24 + 0.30 = 0.54 MJ per use.2. Plastic is 1.1 MJ per use, so the glass bottle uses about half the energy (0.54 ÷ 1.1 = 0.49).
3. The comparison ignores transport mass, recycling of plastic and pollution, and the number of reuses is an estimate.
Problem 19 Faulty atmosphere notes
1. All four are wrong.2. (a) Global warming is caused by greenhouse gases absorbing infrared radiation. The ozone hole is a separate problem. (b) Carbon monoxide forms in incomplete combustion. (c) Potable water is safe to drink but contains dissolved substances. (d) Oxygen is second at 21 %; argon is third at about 0.9 %.
Problem 20 Desalination or groundwater
1. Desalination: 3.5 × 1.0 × 10⁶ = 3.5 × 10⁶ kWh. Groundwater: 0.5 × 1.0 × 10⁶ = 5.0 × 10⁵ kWh.2. 3.5 × 10⁶ − 0.5 × 10⁶ = 3.0 × 10⁶ kWh.
3. The groundwater may run out or be contaminated by salt, while sea water is a reliable supply.
