4.10 Using resources — notes on one page

Every AQA GCSE Chemistry 8462 specification point in 4.10, read straight through: 11 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

Checked against AQA 8462 section 4.10. Review basis: the qualification registry sourced from the AQA GCSE Chemistry (8462) specification; registry verification recorded 17 July 2026.

How this checking works

4.10.1.1 · Using the Earth's resources and sustainable development

Definition. Sustainable development is development that meets the needs of current generations without compromising the ability of future generations to meet their own needs. A finite resource is one that is not being replaced, or is being replaced more slowly than it is used; a renewable resource is replaced at least as fast as it is used.

Humans use the Earth's resources for warmth, shelter, food and transport. Natural resources, supplemented by agriculture, provide food, timber, clothing and fuels, while finite resources from the Earth, the oceans and the atmosphere are processed into energy and materials.
Chemistry's role is twofold: improving agricultural and industrial processes to make new products, and making those processes more sustainable.
The distinction the exam tests is almost always the same one — crude oil and metal ores are finite, while timber, wool, cotton and biofuels come from renewable sources, and solar energy is renewable while oil is not.
A good comparison answer names the resource, states finite or renewable, and links that to a consequence such as running out, cost or environmental damage.
Recycling is the other recurring theme: recycling a metal uses less energy than extracting it from its ore, conserves the limited ore, and reduces landfill.

Worked example

A mineral reserve contains $4.8\times10^8$ kg of usable ore. It is extracted at $6.0\times10^7$ kg per year. Calculate how many years the reserve would last if the rate stayed constant, and explain why the ore is finite.

1.Divide the reserve by the annual extraction: $(4.8\times10^8)/(6.0\times10^7)=8.0$ years. Ore deposits take geological timescales to form, so this rate of use cannot be naturally replaced on a human timescale.

Answer: $8.0$ years. The ore is used much faster than geological processes replace it.

Mark-scheme wording to use

“oil is non-renewable/finite or paper comes from a renewable source” — AQA 8462, June 2024 Paper 2F, Q06.7
“less energy; conserves copper ore; less waste/landfill; specified environmental impact. Allow converse extraction statements; ignore cost” — AQA 8462, recycling two-marker, June 2022 Paper 2F, Q08.4
“judgement strongly linked to sufficient correct reasons” — AQA 8462, Level 3 descriptor, June 2025 Paper 2F, Q10.1

What the examiner wants. A stated judgement at the end of any evaluate question. Level 3 requires the judgement to be linked to the reasons given — a balanced list with no conclusion is capped at Level 2.

Common mistakes

Don't fall into the trap of calling every natural resource renewable: crude oil and metal ores are natural but finite, whereas sustainably managed timber can be renewable.
Don't fall into the trap of defining sustainability only as recycling and omitting the needs of future generations.

Exam tip. For a sustainability comparison, consider raw materials, energy, waste and environmental effects over time.

4.10.1.2 · Potable water

Definition. Potable water is water that is safe to drink: it has sufficiently low levels of dissolved salts and microbes. Potable water is not pure water in the chemical sense, because it still contains dissolved substances.

That distinction is the point examiners test most often — pure water contains only H₂O, while potable water is safe to drink but still contains dissolved solids. Which treatment is used depends on the water available.
In the UK, rain supplies fresh water with low levels of dissolved substances, which collects in the ground, in lakes and in rivers, and most potable water is produced in three steps: choose an appropriate source of fresh water, pass it through filter beds to remove insoluble solids, then sterilise it.
The sterilising agents named by the specification are chlorine, ozone and ultraviolet light, and their purpose is to kill harmful microbes.
Filtration cannot remove dissolved ions, because they are in solution and pass straight through the filter.
Where fresh water is scarce, salty water must be desalinated by distillation or reverse osmosis, both of which require large amounts of energy.

Two routes to potable water: fresh water is filtered and sterilised, while salty water must be desalinated at much greater energy cost.

Worked example

Describe how fresh water containing mud particles and harmful bacteria is treated to make potable water. Give the purpose of each treatment step.

1.First select a suitable fresh-water source. Filtration removes insoluble particles such as mud. Sterilisation is a separate step and reduces harmful microorganisms to safe levels; an accepted agent is chlorine, ozone or ultraviolet light.

Answer: Pass it through filter beds to remove suspended solids, then sterilise it using chlorine, ozone or ultraviolet light to kill harmful microorganisms.

Mark-scheme wording to use

“ions pass through/not removed. 1 mark: ions are in solution / smaller than filter pores” — AQA 8462, June 2022 Paper 2H, Q05.3
“harmful microbes present. 1 mark: water sterilised/microbes destroyed” — AQA 8462, June 2022 Paper 2H, Q05.4
“filter beds remove solids; chlorine/ozone/UV sterilises and destroys microbes” — AQA 8462, six-mark indicative content, June 2024 Paper 2F, Q10.3

What the examiner wants. The purpose of each step, not just its name. 'Filtration then sterilisation' scores far less than 'filter beds remove insoluble solids, then chlorine kills harmful microbes'.

Common mistakes

Don't fall into the trap of saying filtration sterilises water: filter beds remove suspended solids, whereas sterilisation kills harmful microorganisms.
Don't fall into the trap of calling potable water chemically pure even though safe concentrations of dissolved substances remain.

Exam tip. Keep the treatment stages distinct: filtration removes solids, sterilisation reduces microbes, and desalination removes dissolved salts.

4.10.1.3 · Waste water treatment

Sewage and agricultural waste water need organic matter and harmful microbes removed; industrial waste water may also contain harmful chemicals requiring extra treatment. Sewage treatment begins with screening and grit removal, followed by sedimentation that separates sewage sludge from liquid effluent.
The sludge is treated by anaerobic digestion, while the effluent receives aerobic biological treatment before it is released.
Fresh ground water is generally easiest to make potable; waste water needs biological treatment and salt water needs energy-intensive desalination.
Do not reverse the anaerobic sludge and aerobic effluent stages.
The amount and type of contamination determine why industrial waste water may need additional chemical treatment.

Sewage treatment separates liquid effluent from sludge before different biological treatments.

Worked example

Explain how the sludge and effluent from a sedimentation tank are treated biologically.

1.Treat the two streams separately. Anaerobic microorganisms break down organic matter in the sludge without oxygen. Air or oxygen is supplied to the effluent so aerobic microorganisms can break down its remaining organic material.

Answer: Sludge undergoes anaerobic digestion by microorganisms without oxygen; effluent undergoes aerobic biological treatment by microorganisms supplied with oxygen.

Common mistakes

Don't fall into the trap of reversing the biological stages by giving aerobic treatment for sludge and anaerobic digestion for effluent.
Don't fall into the trap of saying screening removes dissolved salts or microorganisms instead of large objects and grit.

Exam tip. Write sewage treatment in order from screening to sedimentation, then separate the sludge and effluent routes.

4.10.1.4 · Alternative methods of extracting metals (HT only)

Higher tier: Phytomining uses plants to absorb metal compounds from low-grade ore; the plants are harvested and burned, leaving ash rich in metal compounds.
Bioleaching uses bacteria to produce a leachate solution containing metal compounds, avoiding the movement and disposal of large amounts of rock.
Copper can be recovered from these solutions by electrolysis or by displacement with a metal more reactive than copper, such as scrap iron.
A common error is to claim biological extraction gives pure metal directly: both methods first produce metal compounds that need further processing.
Higher tier: these methods can exploit low-grade ores with less rock movement, but extraction is slow and further processing still uses energy.

Worked example

Bioleaching produces a solution containing copper compounds. Explain how scrap iron can be used to obtain copper from this solution.

1.Place scrap iron in the copper-compound solution. Because iron is above copper in the reactivity series, iron atoms form ions while copper ions gain electrons and become copper metal. The deposited copper can then be separated from the mixture.

Answer: Iron is more reactive than copper, so iron displaces copper from the solution and solid copper is formed.

Common mistakes

Don't fall into the trap of claiming biological extraction gives pure metal directly: both methods first produce metal compounds that need further processing.
Don't fall into the trap of confusing phytomining with bioleaching by assigning bacteria to the plant-uptake process.

Exam tip. Higher tier: name the organism or plant stage, the metal-compound product and the final recovery method.

4.10.2.1 · Life cycle assessment

A life cycle assessment considers extracting and processing raw materials, manufacture and packaging, use during the product's lifetime, and end-of-life disposal, including transport at every stage.
Compare products using available data for energy, water, resources and waste, and keep the functional use the same before making numerical comparisons.
For comparisons such as paper and plastic shopping bags, use the same function: divide a reusable product's lifetime impacts by its uses and add per-use impacts such as washing.
A common error is to treat an LCA as fully objective: pollutant effects can require value judgements, and selective LCAs can omit stages to support a preferred conclusion.
Transport can contribute at every stage, and a fair comparison must use consistent system boundaries for both products.

Worked example

A reusable food container takes $14$ MJ to manufacture, is used $100$ times and takes $0.030$ MJ to wash after each use. A disposable container takes $0.24$ MJ to manufacture and is used once. Calculate the lifetime energy per use for each container.

1.The reusable container uses $14+(100\times0.030)=17$ MJ in its lifetime. Dividing by $100$ gives $0.17$ MJ per use. The disposable container is used once, so its value remains $0.24$ MJ per use.

Answer: Reusable: $0.17$ MJ per use; disposable: $0.24$ MJ per use.

Common mistakes

Don't fall into the trap of treating an LCA as fully objective: pollutant effects can require value judgements, and selective LCAs can omit stages to support a preferred conclusion.
Don't fall into the trap of comparing products that deliver different numbers of uses without converting impacts to the same functional unit.

Exam tip. Structure an LCA answer by stage, then identify where missing data or value judgements limit the conclusion.

4.10.2.2 · Ways of reducing the use of resources

Reducing use, reusing products and recycling materials can lower demand for limited raw materials, energy use, waste and environmental damage.
Reuse keeps a product in service without remaking it; recycling processes the material into a new product and usually requires collection, separation and energy.
Metals can be melted and recast, glass can be crushed and remelted, and scrap steel can replace some newly extracted iron in steel production.
A common error is to assume recycling has no impact: it often saves raw materials and energy but still needs transport, sorting and processing.
The best option depends on the material and product because reduction usually avoids more processing than reuse or recycling.

Worked example

Producing a metal from ore uses $15.2$ MJ kg^-1, while producing it from sorted scrap uses $1.8$ MJ kg^-1. Calculate the energy saved when $840$ kg is produced from scrap.

1.The saving per kilogram is $15.2-1.8=13.4$ MJ. For $840$ kg, the saving is $13.4\times840=11256$ MJ, which is $1.13\times10^4$ MJ to three significant figures.

Answer: $1.13\times10^4$ MJ saved.

Common mistakes

Don't fall into the trap of assuming recycling has no impact: it often saves raw materials and energy but still needs transport, sorting and processing.
Don't fall into the trap of calling reuse and recycling the same process even though recycling requires material reprocessing.

Exam tip. For an ‘evaluate’ question, give a saved resource or impact and also a collection, sorting, transport or processing cost.

4.10.3.1 · Corrosion and its prevention (chemistry only)

Corrosion is the destruction of a material by chemical reactions with substances in its environment; rusting is the corrosion of iron and needs both oxygen and water.
To test rusting conditions, change the presence of air or water while keeping identical iron samples, temperature and time the same.
Barrier methods such as grease, paint and electroplating keep air and water away; aluminium protects itself with an adherent oxide layer.
A common error is to say any metal coating protects a scratch: sacrificial protection works only when the coating metal is more reactive than iron, as zinc is in galvanising.
An experiment establishing rusting conditions needs comparison samples that isolate oxygen and water while all other variables remain controlled.

Worked example

Plan an experiment using three identical iron nails to show that both air and water are needed for rusting.

1.Put one nail in contact with both air and water. Keep a second in dry air using a drying agent so water is absent. Put a third in boiled water and add an oil layer so oxygen cannot re-enter. Use identical nails at the same temperature for the same time. Only the nail with both air and water should rust.

Answer: Compare a nail in air and water with a nail in dry air and a nail in boiled water covered by oil; keep other conditions the same and record rusting.

Common mistakes

Don't fall into the trap of saying any metal coating protects a scratch: sacrificial protection works only when the coating metal is more reactive than iron, as zinc is in galvanising.
Don't fall into the trap of saying galvanising is only a barrier and omitting that zinc can protect exposed iron sacrificially.

Exam tip. For sacrificial protection, compare metal reactivity and state that the more reactive coating oxidises instead of iron.

4.10.3.2 · Alloys as useful materials (chemistry only)

Bronze is copper and tin and is used for statues; brass is copper and zinc and is used for fittings. Jewellery gold is alloyed with silver, copper or zinc: $24$ carat is pure gold and $18$ carat is $75\%$ gold.
High-carbon steel is strong but brittle and suits cutting tools; low-carbon steel is softer and shapeable for car bodies; chromium-nickel stainless steel resists corrosion and suits sinks or cutlery.
Choose an alloy by linking composition to measured properties and the intended use; low-density aluminium alloys are useful for aircraft parts.
A common error is to describe an alloy as a compound: its proportions can vary, and different compositions can give different properties.
Different steels are not interchangeable: carbon content and alloying elements change strength, brittleness, shapeability and corrosion resistance.

Worked example

A gold ring has mass $12.0\,\text{g}$ and is $18$ carat. Calculate the mass of gold in the ring.

1.Use the specification relationship: $18$ carat gold is $75\%$ gold.
2.Convert to a decimal and multiply: $0.75\times12.0\,\text{g}$ .
3.Evaluate the gold mass: $9.0\,\text{g}$ .

Answer: $9.0\,\text{g}$ of gold

Common mistakes

Don't fall into the trap of describing an alloy as a compound: its proportions can vary, and different compositions can give different properties.
Don't fall into the trap of linking an alloy to a use without stating the property that makes the composition suitable.

Exam tip. Use a composition → property → use chain, and convert carat values to percentages when a numerical comparison is required.

4.10.3.3 · Ceramics, polymers and composites (chemistry only)

Soda-lime glass is made by heating sand, sodium carbonate and limestone; borosilicate glass contains sand and boron trioxide and melts at a higher temperature. Clay ceramics are shaped while wet and heated in a furnace; different reaction conditions allow the same ethene monomer to form both LDPE and HDPE.
Thermosoftening polymers melt on heating because chains can move past one another, whereas cross-links between chains prevent thermosetting polymers from melting.
A composite has a matrix or binder surrounding a reinforcement; examples include glass-fibre polymer, carbon-fibre polymer and concrete.
Link the components' combined properties to the use.
Material comparisons may require quantitative data, so values and units should support rather than replace the structure-property explanation.

Worked example

Explain, in terms of structure, why a thermosoftening polymer melts when heated but a thermosetting polymer does not.

1.Compare movement between chains. Heating weakens the attractions between separate thermosoftening chains enough for them to slide, so the solid softens and melts. Covalent cross-links join thermosetting chains into a network and prevent that sliding, so heating does not melt the polymer.

Answer: Thermosoftening polymer chains can move past one another when heated; thermosetting polymers have strong cross-links between chains that stop this movement, so they do not melt.

Common mistakes

Don't fall into the trap of saying thermosetting polymers melt because their chains slide past one another despite the cross-links.
Don't fall into the trap of calling the reinforcement the matrix in a composite and reversing the roles of the two components.

Exam tip. When selecting a material, identify its structure or components, link these to a property, then link the property to the use.

4.10.4.1 · The Haber process (chemistry only)

The Haber process makes ammonia from nitrogen obtained from air and hydrogen commonly obtained from natural gas: $\mathrm{N_2+3H_2\rightleftharpoons2NH_3}$ . Purified gases pass over an iron catalyst at about $450$ °C and $200$ atmospheres; cooling liquefies ammonia, and unreacted nitrogen and hydrogen are recycled.
Higher tier: the equation gives a $1:3:2$ mole ratio and the same gas-volume ratio at equal temperature and pressure, so one amount of nitrogen reacts with three of hydrogen to form two of ammonia.
Higher tier: lower temperature and higher pressure favour ammonia equilibrium yield, but industry compromises for a fast enough rate and acceptable energy and equipment costs; a catalyst changes rate, not equilibrium position.
Removing liquid ammonia and recycling unreacted gases allow continuous production despite incomplete conversion in one pass.
Higher tier: AQA restricts "explain the trade-off between rate of production and position of equilibrium", interpreting condition-versus-rate graphs and applying dynamic equilibrium here, to Higher tier.

Worked example

Higher tier: determine the volumes of hydrogen and ammonia associated with $40\,\text{dm}^3$ of nitrogen at the same temperature and pressure.

1.Read the balanced equation $\mathrm{N_2+3H_2\rightleftharpoons2NH_3}$ .
2.Use the gas-volume ratio $1:3:2$ .
3.Calculate hydrogen: $3\times40=120\,\text{dm}^3$ ; ammonia: $2\times40=80\,\text{dm}^3$ .

Answer: $120\,\text{dm}^3$ of hydrogen reacts and $80\,\text{dm}^3$ of ammonia forms.

Common mistakes

Don't fall into the trap of claiming the iron catalyst increases the equilibrium yield of ammonia rather than increasing the rates of both directions.
Don't fall into the trap of saying the highest possible pressure and lowest possible temperature are used without considering rate, energy and equipment costs.

Exam tip. Higher tier: explain each compromise separately—temperature affects rate and equilibrium, pressure affects equilibrium and cost, while the catalyst affects rate.

4.10.4.2 · Production and uses of NPK fertilisers (chemistry only)

NPK fertilisers are formulations containing salts that supply nitrogen, phosphorus and potassium in suitable percentages to improve agricultural productivity.
Ammonia is used to make ammonium salts and nitric acid; potassium chloride, potassium sulfate and phosphate rock are obtained by mining.
Phosphate rock is insoluble and is treated with acids to make useful salts: nitric acid gives calcium nitrate, sulfuric acid can give calcium sulfate or single superphosphate, and phosphoric acid gives calcium phosphate or triple superphosphate.
A common error is to describe industrial fertiliser production as a scaled-up school crystallisation only: industry uses integrated, often continuous processes with different raw materials and controls.
The required NPK percentages make the fertiliser a formulation rather than a single pure compound.

Worked example

Name one useful salt formed when phosphate rock is treated with each acid: nitric acid, sulfuric acid and phosphoric acid.

1.Match the acid anion to the named calcium salt from phosphate-rock treatment. Nitrate comes from nitric acid, sulfate from sulfuric acid and phosphate from phosphoric acid; the superphosphate names are accepted industrial product names.

Answer: Nitric acid: calcium nitrate; sulfuric acid: calcium sulfate or single superphosphate; phosphoric acid: calcium phosphate or triple superphosphate.

Common mistakes

Don't fall into the trap of describing industrial fertiliser production as a scaled-up school crystallisation only: industry uses integrated, often continuous processes with different raw materials and controls.
Don't fall into the trap of stating that phosphate rock is applied directly as the phosphorus component even though it is insoluble.

Exam tip. Match each acid treatment to the named fertiliser salt and distinguish integrated industrial production from a batch laboratory preparation.