4.10 Using resources — revision question pack

11 specification points · notes, questions, answers and worked methods

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.

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4.10.1.1 · Using the Earth's resources and sustainable development

Explanation

Natural resources provide warmth, shelter, food and transport; farmed timber and cotton can supplement wild supplies, while synthetic fibres and rubber can replace natural materials.
Classify a resource as finite if it is used much faster than it is replaced, and as renewable if natural processes replace it on a human timescale.
Sustainable development meets present needs without reducing the ability of future generations to meet theirs, so comparisons should include material use, energy, waste and environmental effects.
A common error is to call every natural resource renewable: crude oil and metal ores are natural but finite, whereas sustainably managed timber can be renewable.
Whether a supply is renewable also depends on the rate of use and management, not simply whether the material originated in nature.

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.

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.

Tier 1 · Easy

A factory uses crude oil to make polymer fibres and timber from a replanted forest to make boards. Classify each raw material as finite or renewable.
[2 marks]
Total for this question: 2
Cotton fibres from a farm and polyester fibres made from crude oil are both used for clothing. Identify which fibre is an agricultural product and which is a synthetic product.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A company replaces crude-oil-based packaging with timber packaging from a replanted forest. Explain why this change does not automatically make the packaging sustainable.
[3 marks]
Total for this question: 3
A managed woodland produces $900$ tonnes of new timber each year, but $1200$ tonnes are harvested each year. Explain why the timber is renewable but this rate of use is not sustainable.
[3 marks]
Total for this question: 3
Wild rubber trees supply natural rubber, while a factory uses crude oil to make synthetic rubber. Describe how farmed rubber trees and synthetic rubber can supplement the wild supply, and identify the finite raw material.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A building company compares insulation for a $40$ -year project. Straw panels use a renewable crop, last $20$ years, require $12$ tonnes per installation and use $1.1$ GJ of processing energy per tonne. Polymer panels use a finite raw material, last $40$ years, require $8$ tonnes and use $4.6$ GJ per tonne. Evaluate which choice is more sustainable. Use calculations.
[5 marks]
Total for this question: 5
A region uses $3.0\times10^9$ kg of finite natural gas each year. A biogas plant using food waste could supply the same energy as $4.5\times10^6$ kg of natural gas each year. Calculate the percentage of the present fuel demand replaced and evaluate the claim that this change makes the region's fuel supply sustainable.
[5 marks]
Total for this question: 5
A metal ore reserve contains $9.0\times10^{10}$ kg of usable ore. Mining removes $1.8\times10^8$ kg each year, while geological processes form $3.0\times10^3$ kg each year. Calculate how long the reserve would last at the stated mining rate, determine the approximate number of orders of magnitude by which removal exceeds formation, and explain why the present use is not sustainable.
[5 marks]
Total for this question: 5
A managed forest contains $18\,400$ tonnes of usable timber. It grows by $1150$ tonnes each year and a company plans to harvest $1420$ tonnes each year for $12$ years. The forest must contain at least $16\,000$ tonnes after $12$ years. Calculate the final timber stock, decide whether the plan meets the target, and determine the greatest constant annual harvest that would meet it.
[6 marks]
Total for this question: 6
A clothing company needs $2400$ tonnes of fibre each year. It plans to obtain $65\%$ from a replanted cotton crop and the rest as polyester made from crude oil. A drought limits the cotton harvest to $1320$ tonnes, so polyester must supply the shortfall. Calculate the actual percentage supplied by polyester and the mass of crude oil used if each tonne of polyester requires $1.8$ tonnes of crude oil. Evaluate the company's claim that its plan relies mainly on a renewable resource.
[5 marks]
Total for this question: 5

4.10.1.2 · Potable water

Explanation

Potable water is safe to drink because it has sufficiently low levels of dissolved salts and microbes; it is not chemically pure because dissolved substances remain.
In the UK, potable water is usually made by selecting a suitable fresh-water source, passing the water through filter beds and sterilising it with chlorine, ozone or ultraviolet light.
In water analysis, measure pH and dissolved solids and use distillation for purification; where fresh water is scarce, distillation or reverse osmosis can desalinate salty water but requires substantial energy.
A common error is to say filtration sterilises water: filter beds remove suspended solids, whereas sterilisation kills harmful microorganisms.
A potable-water process must be selected for the source: fresh water generally needs less energy to treat than seawater.

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.

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.

Tier 1 · Easy

Explain why potable water can be safe to drink without being pure water in the chemical sense.
[2 marks]
Total for this question: 2
Name one sterilising agent other than chlorine that can be used in potable-water treatment and state its purpose.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A water company passes river water through filter beds and then declares it potable. Evaluate this treatment and state one suitable additional process.
[3 marks]
Total for this question: 3
Describe how to determine the mass of dissolved solids in a $50.0$ cm³ water sample using an evaporating basin and a balance.
[4 marks]
Total for this question: 4
Describe how distillation can separate pure water from a sample containing dissolved salt and how the water is collected.
[4 marks]
Total for this question: 4

Tier 3 · Hard

An island must desalinate $1.5\times10^6$ litres of sea water each day. Distillation would use $2.8$ MJ per litre and reverse osmosis would use $0.018$ MJ per litre. Calculate the daily energy saved by reverse osmosis and evaluate its use if membranes need regular replacement.
[5 marks]
Total for this question: 5
Source P is fresh water containing suspended clay and harmful microorganisms but few dissolved salts. Source Q is sea water containing many dissolved salts but no suspended clay or harmful microorganisms. Describe a suitable treatment for each source and explain why producing potable water from Q requires more energy.
[5 marks]
Total for this question: 5
Evaporating $25.0$ cm³ of water sample A leaves $0.090$ g of dissolved solids. Evaporating $100$ cm³ of sample B leaves $0.210$ g. Calculate the concentration of dissolved solids in each sample in g dm^-3, identify the sample with the lower concentration, and state one suitable further treatment and its purpose if that sample contains harmful microorganisms.
[5 marks]
Total for this question: 5
A reverse-osmosis plant treats $800\,000$ litres of sea water each day. The feed contains $34.0$ g of dissolved salts per litre and the membrane reduces this concentration by $98.4\%$ . Potable water from this plant must contain no more than $0.60$ g of dissolved salts per litre. Calculate the salt concentration after treatment and the daily energy use if the process requires $0.021$ MJ per litre. Evaluate whether these data alone show that the product is potable.
[5 marks]
Total for this question: 5
Quality-control results for reservoir water are shown after three stages. Raw water contains suspended solids, $4.8$ g dm^-3 of dissolved salts and $260$ microorganism colonies per sample. After filter beds it contains no suspended solids, $4.8$ g dm^-3 of dissolved salts and $245$ colonies. After ultraviolet treatment it still contains $4.8$ g dm^-3 of dissolved salts but no colonies. The dissolved-salt limit is $2.5$ g dm^-3. Evaluate each stage and specify the additional treatment needed before the water meets all three stated requirements.
[5 marks]
Total for this question: 5

4.10.1.3 · Waste water treatment

Explanation

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.

Tier 1 · Easy

State the two treatment stages used before sewage is allowed to settle and name the two products of sedimentation.
[2 marks]
Total for this question: 2
Give one reason why industrial waste water may need an additional treatment that ordinary sewage does not need.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A student sends sewage sludge for aerobic treatment and liquid effluent for anaerobic digestion. Correct both parts of this plan.
[2 marks]
Total for this question: 2
Air is bubbled through liquid effluent during sewage treatment. Explain why air is supplied and what happens to the organic matter.
[3 marks]
Total for this question: 3
Compare the substances that must be removed from agricultural waste water and industrial waste water before either is released into the environment.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A treatment works processes $9.5\times10^3$ m³ of sewage in one day. Sedimentation separates $7.0\%$ of this volume as sludge. Calculate the sludge volume and explain why making potable water from treated waste water is generally harder than making it from fresh ground water.
[5 marks]
Total for this question: 5
Compare the treatment needed to obtain potable water from three sources: fresh ground water with few contaminants, waste water containing organic matter and microbes, and sea water containing a high concentration of dissolved salts.
[5 marks]
Total for this question: 5
After screening, grit removal and sedimentation, one day's liquid effluent contains $250$ kg of organic matter. Aerobic biological treatment removes $88.0\%$ of this organic matter. Calculate the mass of organic matter remaining and explain the role of the air and microorganisms in this treatment.
[5 marks]
Total for this question: 5
An industrial site has $50\,000$ dm³ of waste water containing $2.4$ g dm^-3 of organic matter and $0.060$ g dm^-3 of a harmful chemical. Biological treatment removes $92\%$ of the organic matter but none of the chemical. Calculate the mass of each contaminant remaining. The chemical concentration must be reduced to $0.0020$ g dm^-3 before release; calculate the minimum percentage of the chemical that an additional treatment must remove.
[6 marks]
Total for this question: 6
Waste water entering a factory treatment works contains $500$ kg of organic matter and $40$ kg of a harmful chemical. Biological stage B removes $96\%$ of the organic matter but only $10\%$ of the chemical. Chemical stage C removes $95\%$ of the harmful chemical but only $10\%$ of the organic matter. Release limits are $30$ kg of organic matter and $4.0$ kg of the chemical. Use the data to show why neither stage alone is sufficient and evaluate a treatment train that uses both specialised stages.
[5 marks]
Total for this question: 5

4.10.1.4 · Alternative methods of extracting metals (HT only)

Explanation

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.

Tier 1 · Easy

Describe how phytomining produces material containing copper compounds.
[3 marks]
Total for this question: 3
State the role of bacteria in bioleaching and name the copper-containing product formed.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Compare phytomining with bioleaching by naming the organism used and the copper-containing product obtained from each. Explain why neither method produces pure copper directly.
[5 marks]
Total for this question: 5
Copper sulfate leachate is electrolysed using inert electrodes. Predict the product at the negative electrode and explain how it forms.
[3 marks]
Total for this question: 3
A phytomining crop produces $1.80$ tonnes of dry plant material. Burning the crop leaves ash equal to $4.0\%$ of the dry mass, and copper compounds make up $12\%$ of the ash. Calculate the mass of copper compounds in the ash.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A conventional process yields $760$ kg of copper using $5.4$ GJ and moves $320$ tonnes of rock in $2$ days. A bioleaching process yields $690$ kg using $1.7$ GJ, moves $38$ tonnes of residue and takes $60$ days. Calculate the energy used per kilogram of copper for each process and evaluate bioleaching.
[5 marks]
Total for this question: 5
A leachate contains enough copper compounds to produce $1200$ kg of copper. Electrolysis recovers $94\%$ of the copper and uses $2.6$ GJ. Displacement with scrap iron recovers $81\%$ , uses $0.40$ GJ and produces less-pure copper. Calculate the copper recovered by each route and evaluate the two routes.
[5 marks]
Total for this question: 5
A site can use either one phytomining plot or one bioleaching vessel for $200$ days. The plot produces one crop containing enough compounds for $62$ kg of copper and uses $410$ MJ. The vessel produces enough leachate for $14$ kg of copper every $35$ days and uses $126$ MJ per completed cycle. Calculate the maximum copper output and energy use for each method, then evaluate the two choices.
[6 marks]
Total for this question: 6
A bioleaching vessel produces leachate containing copper compounds equivalent to $3.20$ kg of copper per m³. Electrolysis recovers $87.5\%$ of this copper. Determine the minimum whole number of cubic metres of leachate needed to obtain $420$ kg of copper and calculate the electrolysis energy used if each cubic metre requires $9.0$ MJ. Explain why the bacteria have not produced the copper metal directly.
[5 marks]
Total for this question: 5
A phytomining site has $60$ hectares available. Crop A absorbs copper compounds equivalent to $8.0$ kg of copper per hectare per crop and can produce three crops each year. Crop B absorbs the equivalent of $15$ kg per hectare per crop but produces only one crop each year. After the plants are harvested and burned, the recovery process obtains $75\%$ of the copper represented in the ash. Calculate the annual recovered copper for each crop and evaluate which crop should be chosen.
[6 marks]
Total for this question: 6

4.10.2.1 · Life cycle assessment

Explanation

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.

Tier 1 · Easy

State the four main stages considered in a life cycle assessment.
[4 marks]
Total for this question: 4
Identify the life-cycle stage represented by each event: extracting iron ore for a food can, and sending the used can to landfill.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An industry-funded life cycle assessment compares two bottles but omits transport and disposal and reports only total energy use. Give two reasons why its conclusion may be unreliable.
[2 marks]
Total for this question: 2
An LCA records manufacturing energy as $240$ kWh but assigns river damage a score of $6$ . Explain why the energy value is more objective and why different analysts might assign different river-damage scores.
[3 marks]
Total for this question: 3
Two hundred paper bags each have mass $55$ g and $80\%$ of their mass is recycled after use. Two hundred plastic bags each have mass $8.0$ g and $10\%$ is recycled. Calculate the mass sent for disposal for each type and explain why this disposal result alone is not a complete life cycle assessment.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Two protective packages perform the same job. Package A uses $1.8$ MJ in manufacture, $0.40$ MJ in transport and produces $60$ g of end-of-life waste; its pollutant-effect score is $2$ . Package B uses $0.90$ MJ in manufacture, $1.1$ MJ in transport and produces $25$ g of waste; its pollutant-effect score is $7$ . Evaluate the packages and explain one limitation of the comparison.
[5 marks]
Total for this question: 5
Two roof tiles have the same lifetime. Tile R has mass $18$ kg and travels $40$ km from factory to site. Tile S has mass $6.0$ kg and travels $900$ km. Use mass $\times$ distance as a transport-impact index to compare the tiles, then explain why this result alone cannot decide which tile has the lower life-cycle impact.
[5 marks]
Total for this question: 5
A reusable shopping bag uses $3.60$ MJ when it is manufactured and $0.025$ MJ for cleaning after each use. A disposable bag uses $0.180$ MJ for manufacture and disposal each time it is used. Determine the minimum whole number of uses after which the reusable bag has the lower stated energy impact, and explain one limitation of this comparison.
[5 marks]
Total for this question: 5
An LCA gives Product X energy impacts of $18$ MJ for raw materials, $12$ MJ for manufacture, $4.0$ MJ during use and $6.0$ MJ for disposal. Product Y uses $11$ MJ for raw materials, $9.0$ MJ for manufacture and $5.0$ MJ during use, but its disposal impact is omitted. Determine Y's break-even disposal energy and the condition needed for its total to be lower than X's. Test the conclusion if Y's disposal energy is first estimated as $8.0$ MJ and later measured as $17$ MJ, and explain what the change shows.
[5 marks]
Total for this question: 5
An LCA compares two wall panels that cover the same area for the same lifetime. Panel P uses $75$ dm³ of water, produces $4.8$ kg of solid waste and uses $520$ MJ of energy. Panel Q uses $120$ dm³ of water, produces $3.1$ kg of solid waste and uses $410$ MJ of energy. Explain why the data do not give a single overall conclusion without a value judgement. State one weighting choice that would favour P and one that would favour Q.
[6 marks]
Total for this question: 6

4.10.2.2 · Ways of reducing the use of resources

Explanation

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.

Tier 1 · Easy

Give one example each of reducing, reusing and recycling a material resource.
[3 marks]
Total for this question: 3
State two limited resources whose use can be reduced when scrap steel replaces some iron made from ore.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A glass bottle can either be washed and refilled or crushed, melted and made into a new bottle. Identify which option is reuse and which is recycling, and explain why reuse will usually require less processing.
[3 marks]
Total for this question: 3
A recycling plant receives a mixture of clear glass, coloured glass and metal lids. Explain why separation is needed before new colourless bottles are made.
[3 marks]
Total for this question: 3
A drinks company reduces the mass of each glass bottle from $320$ g to $280$ g. Calculate the mass of glass saved when $50\,000$ bottles are made, and give two environmental benefits of reducing material use at manufacture.
[4 marks]
Total for this question: 4

Tier 3 · Hard

For $6000$ deliveries, a company can use $0.12$ kg single-use trays or reusable crates. Each $9.0$ kg crate lasts $120$ deliveries. Single-use trays require $3.0$ MJ kg^-1 to make. Crates require $5.5$ MJ kg^-1 to make and $0.10$ MJ of washing energy per delivery. Evaluate the switch to crates using raw-material mass and energy.
[5 marks]
Total for this question: 5
A company makes $24000$ aluminium cans, each of mass $15$ g. The aluminium is $65\%$ recycled. Production uses $12$ MJ kg^-1 for recycled aluminium and $180$ MJ kg^-1 for new aluminium. Calculate the total production energy and the energy saved compared with using only new aluminium.
[6 marks]
Total for this question: 6
A demolished building contains $72\,000$ bricks. A contractor can clean and reuse $65\%$ of them. The remaining bricks can be crushed to replace $2.1$ kg of newly quarried aggregate per brick. Calculate the number of bricks reused and the mass of aggregate replaced, then evaluate these choices compared with sending every brick to landfill.
[5 marks]
Total for this question: 5
A manufacturer needs $500$ kg of a metal product. Making the metal from new ore uses $40$ MJ kg^-1, while using sorted recycled metal uses $8.0$ MJ kg^-1. The production-energy limit is $12\,000$ MJ. Determine the minimum mass and percentage of recycled metal needed to meet the limit, and explain one reason why the real environmental impact is not zero even at that recycled fraction.
[5 marks]
Total for this question: 5
A town discards $5000$ glass bottles, each with mass $0.40$ kg. It collects $90\%$ of them, sorting retains $80\%$ of the collected glass, and remelting turns $95\%$ of the sorted glass into usable glass. New bottles each have mass $0.36$ kg. Calculate the number of new bottles that can be made entirely from the recovered glass and the mass of new glass needed to make $5000$ new bottles. Evaluate the claim that recycling creates a completely closed material loop.
[6 marks]
Total for this question: 6

4.10.3.1 · Corrosion and its prevention (chemistry only)

Explanation

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.

Tier 1 · Easy

State the two substances that must both be present for iron to rust.
[2 marks]
Total for this question: 2
Give one barrier method used to protect iron from corrosion and explain how the barrier works.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A scratch through paint exposes steel and rust begins, but a scratch through a zinc coating can remain protected. Explain both observations.
[4 marks]
Total for this question: 4
Choose from greasing and electroplating to protect a moving steel bicycle chain and a decorative steel door handle, using each method once. Explain each choice.
[4 marks]
Total for this question: 4
Iron continues to corrode after its surface starts to rust, but aluminium often stops corroding after a thin surface layer forms. Explain this difference.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Iron harbour bolts can be coated with metal X or copper. The reactivity order is magnesium, X, iron, copper. Evaluate the protection given by each coating if it is scratched through to the iron.
[5 marks]
Total for this question: 5
A $12.0$ kg magnesium block is attached to a buried steel pipe. The block loses $0.75$ kg each year and must be replaced when only $20\%$ of its original mass remains. Calculate its maximum service time and explain how it protects the pipe.
[5 marks]
Total for this question: 5
Three uncoated steel plates lose $0.81$ g, $0.80$ g and $0.79$ g during a corrosion test. Three identical coated plates lose $0.31$ g, $0.30$ g and $0.29$ g. Calculate the mean mass loss for each group and the percentage reduction produced by the coating. Evaluate the claim that the coating reduces corrosion by at least $55\%$ .
[5 marks]
Total for this question: 5
A council must protect $60$ steel railings for $20$ years. Painting each railing uses $0.80$ L of paint, costs £7.00 per litre and must be repeated every $4$ years, including at the start. A zinc coating remains effective for the full period, uses $1.5$ kg of zinc per railing and costs £13.00 per kilogram. Calculate the material used and cost for each plan over $20$ years, then evaluate the plans if scratches are likely.
[6 marks]
Total for this question: 6
A painted steel buoy would lose $2.5$ kg of steel per year if unpainted. Its intact paint barrier reduces this loss by $80\%$ . An attached magnesium block prevents the remaining steel corrosion while it is active; the block starts at $8.0$ kg, loses $1.2$ kg per year and must be replaced when $2.0$ kg remains. Determine the replacement interval, the number of blocks needed for continuous protection over $15$ years, and the steel mass the paint alone would have allowed to corrode. Explain how the two protections work together.
[6 marks]
Total for this question: 6

4.10.3.2 · Alloys as useful materials (chemistry only)

Explanation

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.

Tier 1 · Easy

Name the two metals in bronze and the two metals in brass.
[2 marks]
Total for this question: 2
A sculptor is making a statue and a plumber is making a water fitting. Choose bronze or brass for each use.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Identify the most suitable steel — high-carbon, low-carbon or stainless — for each use: a cutting blade, a pressed car-body panel and a kitchen sink. Justify each choice.
[6 marks]
Total for this question: 6
Pure aluminium has density $2.70$ g cm^-3 and strength $90$ MPa. An aluminium alloy has density $2.80$ g cm^-3 and strength $310$ MPa. Explain why the alloy is more suitable for a lightweight aircraft panel.
[3 marks]
Total for this question: 3
An alloy contains $84\%$ copper and $16\%$ tin. Identify the alloy and explain why changing the percentages can change its properties without making it a different type of substance.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A $2.40$ kg brass fitting contains $68.0\%$ copper by mass and the rest is zinc. Calculate the mass of zinc. Explain why a manufacturer might use this brass rather than pure copper when data show brass is harder but slightly less electrically conductive.
[4 marks]
Total for this question: 4
A marine fitting must have strength of at least $320$ MPa and corrosion loss below $0.50$ mm per year. Alloy A has density $8.5$ g cm^-3, strength $350$ MPa and corrosion loss $0.30$ mm per year. Alloy B has density $2.8$ g cm^-3, strength $310$ MPa and corrosion loss $1.4$ mm per year. Choose the suitable alloy and calculate the mass of a $250$ cm³ fitting made from it.
[5 marks]
Total for this question: 5
A jeweller melts together $15.0$ g of $18$ carat gold and $5.0$ g of $24$ carat gold. Use $18$ carat $=75\%$ gold and $24$ carat $=100\%$ gold. Calculate the mass and percentage of gold in the new mixture, and explain why the product is still an alloy rather than pure gold.
[5 marks]
Total for this question: 5
A foundry has $960$ kg of a copper-zinc alloy containing $72\%$ copper. It adds only zinc so that copper becomes $60\%$ of the final mixture. Calculate the mass of copper already present, the required final mass and the mass of zinc that must be added. Identify the resulting alloy and explain why its composition can be changed in this way.
[5 marks]
Total for this question: 5
Two bronze compositions are tested for a gear. Bronze P contains $8.0\%$ tin, has hardness $110$ and elongation before breaking of $18\%$ . Bronze Q contains $20\%$ tin, has hardness $180$ and elongation of $4.0\%$ . The gear requires hardness of at least $160$ and elongation of at least $3.0\%$ . Choose the suitable bronze and calculate the masses of copper and tin needed for a $750$ kg batch. Explain how the data illustrate a feature of alloys.
[5 marks]
Total for this question: 5

4.10.3.3 · Ceramics, polymers and composites (chemistry only)

Explanation

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.

Tier 1 · Easy

Glass fibres are surrounded and held together by a polymer resin in a composite. Identify the reinforcement and the matrix.
[2 marks]
Total for this question: 2
State the two compounds, other than sand, that are heated to make soda-lime glass.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A carbon-fibre polymer contains carbon fibres held in a polymer resin. Identify the reinforcement and the matrix, then explain why the composite is useful where high strength and low mass are required.
[4 marks]
Total for this question: 4
Explain how low-density poly(ethene) and high-density poly(ethene) can both be produced from ethene but have different properties.
[3 marks]
Total for this question: 3
Describe how a clay brick is made from raw clay, and explain why the clay must be shaped before the final heating stage.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A transparent oven window must work at $720$ °C. Soda-lime glass softens at $650$ °C. Borosilicate glass softens at $900$ °C, is transparent and has density $2.25$ g cm^-3. An opaque clay ceramic works to $1100$ °C. Identify the most suitable material and calculate the mass of an $80$ cm³ window.
[5 marks]
Total for this question: 5
A platform panel has volume $0.080$ m³ and must have strength above $300$ MPa. A glass-fibre composite has density $1900$ kg m^-3 and strength $340$ MPa. Steel has density $7800$ kg m^-3 and strength $500$ MPa. Calculate the mass of each panel and choose the better material where low mass is the priority.
[5 marks]
Total for this question: 5
A polymer-resin panel has mass $2.4$ kg and breaks under a load of $1.8$ kN. A fibre-reinforced panel has mass $3.0$ kg and breaks under $7.2$ kN. Calculate the breaking load per kilogram for each panel, compare the values, and explain the roles of the fibres and resin in the composite.
[5 marks]
Total for this question: 5
Samples P and Q are both made only from ethene. P has density $0.92$ g cm^-3 and is flexible; Q has density $0.96$ g cm^-3 and is more rigid. Identify the form of poly(ethene) represented by each sample, choose the better sample for a squeezable bottle and for a rigid pipe, and explain how one monomer can produce these different materials.
[5 marks]
Total for this question: 5
A clip made from a thermosoftening polymer becomes soft and slowly bends while holding a hot pipe. Explain this failure in terms of polymer-chain movement. Choose and justify either a thermosoftening or a thermosetting polymer for a replacement clip that must keep its shape at the same temperature.
[5 marks]
Total for this question: 5

4.10.4.1 · The Haber process (chemistry only)

Explanation

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.

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.

Tier 1 · Easy

State the source of nitrogen, the usual source of hydrogen and the catalyst used in the Haber process.
[3 marks]
Total for this question: 3
Give the approximate temperature and pressure used inside an industrial Haber reactor.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Explain how ammonia is separated from the gases leaving a Haber reactor and why the remaining gases are returned to the reactor.
[4 marks]
Total for this question: 4
Write a balanced equation for ammonia formation from its elements. Show that the reaction is reversible.
[3 marks]
Total for this question: 3
Explain how the temperature, pressure and iron catalyst used in the Haber reactor each help ammonia to be produced at a useful rate.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A plant sends $800$ kg of a nitrogen-hydrogen mixture through a Haber reactor. After one pass, $184$ kg of ammonia is separated. Calculate the percentage of the feed converted to ammonia in this pass and explain what happens to the remaining gases.
[4 marks]
Total for this question: 4
Higher Tier: An engineer proposes operating the Haber process at $350$ °C and $300$ atmospheres without an iron catalyst instead of $450$ °C and $200$ atmospheres with iron. Evaluate the proposal in terms of equilibrium yield, rate and cost.
[6 marks]
Total for this question: 6
A proposed ammonia plant passes nitrogen and hydrogen through an empty reactor at $80$ °C and $1$ atmosphere, then releases every gas leaving the reactor. Describe the changes needed to make the process resemble the industrial Haber process and give the purpose of each change.
[6 marks]
Total for this question: 6
A Haber plant starts with $1000$ kg of a correctly mixed nitrogen-hydrogen feed. On each pass, $25.0\%$ of the reactant mixture entering the reactor is converted into ammonia. The ammonia is removed and all unreacted gases are recycled, with no fresh feed added. Calculate the ammonia removed on each of the first three passes, the total ammonia collected and the mass of reactants remaining after the third pass. Explain the purpose of cooling between passes.
[6 marks]
Total for this question: 6
Two Haber lines each receive $600$ kg of correctly mixed nitrogen and hydrogen. Line A uses an iron catalyst and converts $30\%$ of the feed to ammonia in one pass, but cooling collects only $95\%$ of that ammonia. Line B has no catalyst and converts $12\%$ in the same time, while its cooling stage collects all the ammonia formed. Calculate the mass collected by each line and evaluate the two stages that limit production.
[6 marks]
Total for this question: 6

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

Explanation

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.

Tier 1 · Easy

State what the letters N, P and K represent on an NPK fertiliser label and explain why phosphate rock is not used directly as a fertiliser.
[4 marks]
Total for this question: 4
State one mined compound that supplies potassium for NPK fertiliser and name the substance used to manufacture ammonium salts.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An NPK fertiliser contains ammonium nitrate, a treated phosphate salt and potassium chloride in controlled proportions. State which ingredient supplies N, P and K, and explain why the product is a formulation.
[4 marks]
Total for this question: 4
A manufacturer requires triple superphosphate rather than single superphosphate. Choose the acid used to make each product from phosphate rock and explain why either treated product is more useful as a fertiliser than untreated phosphate rock.
[3 marks]
Total for this question: 3
Describe how to prepare pure, dry ammonium sulfate crystals from aqueous ammonia and dilute sulfuric acid in a laboratory.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A laboratory batch makes $35$ g of ammonium salt every $45$ minutes. An integrated industrial process operates continuously at $2.4$ kg per minute. Calculate the mass each route makes in $3.0$ hours and compare the routes for large-scale fertiliser production.
[5 marks]
Total for this question: 5
A factory makes $1.60$ tonnes of an NPK fertiliser containing $14\%$ nitrogen, $6.0\%$ phosphorus and $10\%$ potassium by mass. Calculate the mass of each nutrient and explain why industrial NPK production is an integrated formulation process rather than a laboratory batch preparation.
[6 marks]
Total for this question: 6
A farm needs at least $34$ kg of nitrogen and $22$ kg of potassium. An NPK fertiliser contains $12\%$ nitrogen and $8.0\%$ potassium by mass and is sold in $25$ kg bags. Determine the minimum whole number of bags needed, calculate the mass of each nutrient supplied by this number of bags, and explain why the fertiliser is a formulation.
[5 marks]
Total for this question: 5
A factory blends $500$ kg of an ammonium material containing $30\%$ nitrogen, $375$ kg of a treated phosphate material containing $20\%$ phosphorus, $250$ kg of a potassium salt containing $40\%$ potassium and $125$ kg of filler. Calculate the percentage of N, P and K in the final product, check whether the label $12$ - $6$ - $8$ is correct, and explain why the product is a formulation.
[6 marks]
Total for this question: 6
Fertiliser F is labelled $18$ - $4$ - $4$ and fertiliser G is labelled $10$ - $10$ - $10$ , where the numbers are the mass percentages of N, P and K. Both are sold in $50$ kg bags. A field needs at least $90$ kg N, $40$ kg P and $40$ kg K. Determine the minimum total number of bags when both formulations may be used, give one combination that achieves this minimum, and calculate the nutrient masses it supplies.
[6 marks]
Total for this question: 6

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

4.10.1.1 · Using the Earth's resources and sustainable development

Tier 1 · Easy

Mark scheme for 4.10.1.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Crude oil is finite. Timber from the replanted forest is renewable.	Crude oil forms far more slowly than it is extracted, so its supply is finite. Trees can be replaced by replanting and growth on a human timescale, so this timber supply is renewable.	2
Total Question 1			2
02.1	Cotton is the agricultural product. Polyester is the synthetic product.	Cotton is obtained by growing a crop, so it is supplied by agriculture. Polyester is manufactured by chemical processing, so it is synthetic.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.10.1.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Any three of the following. Timber is renewable because trees can be regrown. Using it is sustainable only if trees are replaced and regrow at least as fast as they are harvested. Processing energy and transport still carry an environmental cost. Waste and other environmental effects must also be compared.	Do not equate natural or renewable with sustainable. Check how the forest is managed and compare the full material, energy, waste and environmental consequences.	3
Total Question 1			3
02.1	Trees can be replanted and regrown on a human timescale, so timber is renewable. Harvesting exceeds regrowth by $300$ tonnes each year, so the woodland stock decreases. Continuing this rate would leave less timber for future generations, so it is not sustainable.	Renewability describes whether a resource can be replaced. Sustainability also depends on the rate of use: $1200-900=300$ tonnes more timber is removed than replaced each year.	3
Total Question 2			3
03.1	Farmed rubber trees are an agricultural source that increases the supply of natural rubber. Synthetic rubber can replace natural rubber for some uses. The crude oil used by the factory is the finite raw material.	Separate the two ways of increasing supply. Agriculture grows more of the natural product, whereas chemical manufacture provides a substitute. The usual feedstock for the synthetic product is crude oil, which is finite.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.10.1.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Over $40$ years straw uses $24$ tonnes and $26.4$ GJ; polymer uses $8$ tonnes and $36.8$ GJ. Straw uses renewable feedstock and less processing energy, but needs replacement and more material, so the data do not give one unqualified best choice.	Straw needs two installations, so its mass is $2\times12=24$ tonnes and its energy is $24\times1.1=26.4$ GJ. Polymer needs one installation, so its energy is $8\times4.6=36.8$ GJ. Straw favours renewable supply and lower stated energy; polymer favours durability and lower material mass. A supported judgement must weigh both advantages and note that other impacts, such as transport and disposal, are not given.	5
Total Question 1			5
02.1	The biogas replaces $0.15\%$ of the present fuel demand. Using food waste can reduce demand for a finite resource and can provide a renewable fuel if the feedstock is replaced. The claim is not supported because $99.85\%$ of the stated demand would still be met by finite natural gas; other production and transport impacts are also not given.	The percentage replaced is $(4.5\times10^6/3.0\times10^9)\times100=0.15\%$ . Biogas has a renewable-resource advantage and uses a waste feedstock, but the three-order-of-magnitude difference shows that the proposal replaces only a very small fraction of present finite-fuel use.	5
Total Question 2			5
03.1	The reserve would last $500$ years at the stated mining rate. Removal is $6.0\times10^4$ times formation, so it is approximately five orders of magnitude greater. The reserve is depleted far faster than it is replaced, reducing the resource available to future generations.	Calculate $(9.0\times10^{10})/(1.8\times10^8)=5.0\times10^2=500$ years. The rate ratio is $(1.8\times10^8)/(3.0\times10^3)=6.0\times10^4$ , which is closer to $10^5$ than $10^4$ . Such a large imbalance means the resource is effectively finite on a human timescale and the stated rate cannot continue indefinitely.	5
Total Question 3			5
04.1	The planned harvest exceeds annual growth by $270$ tonnes. Over $12$ years the stock decreases by $3240$ tonnes. The final stock would be $15\,160$ tonnes. The plan misses the $16\,000$ -tonne target by $840$ tonnes. The greatest constant annual harvest that meets the target is $1350$ tonnes. The timber can be renewable, but the proposed rate is not sustainable against the stated future-stock target.	The annual deficit is $1420-1150=270$ tonnes, so the planned final stock is $18\,400-(12\times270)=15\,160$ tonnes. To preserve $16\,000$ tonnes, at most $18\,400+(12\times1150)-16\,000=16\,200$ tonnes may be harvested over the period. Dividing by $12$ gives $1350$ tonnes per year. This reverse calculation tests a future resource constraint rather than merely labelling timber renewable.	6
Total Question 4			6
05.1	Polyester must supply $1080$ tonnes of fibre. Polyester supplies $45.0\%$ of the actual fibre demand. Producing this polyester uses $1944$ tonnes of crude oil. Cotton can be renewable because it can be replanted, but its replacement rate and yield must keep pace with use. Most fibre still comes from cotton, but the drought makes reliance on finite crude oil greater than the planned $35\%$ , so the claim needs qualification.	The polyester shortfall is $2400-1320=1080$ tonnes. Its share is $(1080/2400)\times100=45.0\%$ , and its crude-oil demand is $1080\times1.8=1944$ tonnes. The company still obtains $55\%$ from cotton, but the calculation shows that renewable supply can depend on management and growing conditions while polyester continues to consume a finite natural resource.	5
Total Question 5			5

4.10.1.2 · Potable water

Tier 1 · Easy

Mark scheme for 4.10.1.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	It contains sufficiently low levels of harmful microbes and dissolved salts, but it still contains some dissolved substances.	Use the two different criteria. Potable describes safety for drinking; pure describes a single substance. A safe sample may therefore remain a mixture of water and small amounts of dissolved material.	2
Total Question 1			2
02.1	Ozone (or ultraviolet light). It kills harmful microorganisms or reduces them to a safe level.	Accepted sterilising treatments include chlorine, ozone and ultraviolet light. Sterilisation addresses microbes; it does not remove suspended solids or dissolved salts.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.10.1.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Filtration removes suspended solids but does not reliably remove harmful microorganisms. The water must also be sterilised, for example using chlorine, ozone or ultraviolet light.	Match each treatment to its purpose. Filter beds address insoluble particles; a separate sterilisation step reduces microbes to a safe level.	3
Total Question 1			3
02.1	Measure the mass of the clean, dry evaporating basin. Add $50.0$ cm³ of the sample and heat it so the water evaporates. Allow the basin and residue to cool, then reweigh them. Subtract the original basin mass from the final mass to obtain the mass of dissolved solids.	Only the water is removed during careful evaporation, leaving the dissolved material as residue. Cooling before the final weighing avoids a balance error caused by hot apparatus.	4
Total Question 2			4
03.1	Heat the sample so that water boils and forms vapour. The dissolved salt remains in the heated container. Pass the water vapour through a condenser, where it cools and condenses. Collect the liquid distillate as pure water.	Use the difference in volatility. Water vaporises, but the dissolved salt does not. Cooling the separated vapour changes it back into liquid water, leaving the salt behind.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.10.1.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$4.17\times10^6$ MJ saved per day; reverse osmosis greatly reduces energy demand, but membrane manufacture, replacement and disposal must also be considered.	The energy difference is $2.8-0.018=2.782$ MJ per litre. The daily saving is $(2.782)(1.5\times10^6)=4.173\times10^6$ MJ, or $4.17\times10^6$ MJ to three significant figures. This strongly favours reverse osmosis on energy, but a complete evaluation also includes membrane cost, material use, waste and the treatment of concentrated brine.	5
Total Question 1			5
02.1	Pass P through filter beds to remove the suspended clay, then sterilise it with chlorine, ozone or ultraviolet light to reduce harmful microorganisms. Desalinate Q by distillation or reverse osmosis to remove dissolved salts. Desalination needs substantial energy for boiling and condensing water or for forcing it through a membrane at high pressure.	Choose each process for the contaminant present. Filtration removes insoluble clay and sterilisation treats microbes in P. Dissolved salts pass through ordinary filter beds, so Q needs an energy-intensive desalination process.	5
Total Question 2			5
03.1	Sample A contains $3.60$ g dm^-3 of dissolved solids. Sample B contains $2.10$ g dm^-3, so B has the lower concentration. Sample B must be sterilised, for example using chlorine, ozone or ultraviolet light, to kill harmful microorganisms or reduce them to a safe level.	Convert the tested volumes to dm³. For A, $0.090/0.0250=3.60$ g dm^-3. For B, $0.210/0.100=2.10$ g dm^-3. Dissolved-solids data do not establish that water is free of harmful microorganisms, so sterilisation is still required.	5
Total Question 3			5
04.1	$1.6\%$ of the dissolved salts remain after treatment. The treated-water salt concentration is $0.544$ g per litre. This is below the stated $0.60$ g per litre limit. The process uses $16\,800$ MJ each day. The salt and energy data do not show that harmful microorganisms are at safe levels, so sterilisation or microbiological evidence may still be needed.	The remaining fraction is $100.0-98.4=1.6\%$ . Multiplying gives $34.0\times0.016=0.544$ g per litre, which meets the stated dissolved-salt limit. Daily energy is $800\,000\times0.021=16\,800$ MJ. Potability also requires sufficiently low levels of harmful microbes, which these measurements do not establish.	5
Total Question 4			5
05.1	The filter beds remove the suspended solids. Filtration does not remove the dissolved salts and does not reliably sterilise the water. Ultraviolet treatment sterilises the water because the colony count falls to zero. The dissolved-salt concentration remains $2.3$ g dm^-3 above the limit. Distillation or reverse osmosis is also needed to reduce the dissolved-salt concentration.	Compare each result with the contaminant that the stage is designed to remove. The filter eliminates suspended material but leaves essentially the same microbial and dissolved-salt measurements. Ultraviolet light removes the microbiological failure, but $4.8-2.5=2.3$ g dm^-3 of excess dissolved salts remains. Desalination is therefore required in addition to filtration and sterilisation.	5
Total Question 5			5

4.10.1.3 · Waste water treatment

Tier 1 · Easy

Mark scheme for 4.10.1.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Screening and grit removal; sedimentation produces sewage sludge and effluent.	Large objects are removed by screens and small dense particles are removed as grit. Settling then separates the denser sludge from the liquid effluent.	2
Total Question 1			2
02.1	Industrial waste water may contain harmful chemicals, so these chemicals must be removed before the water is released.	Sewage treatment targets organic matter and harmful microbes. An industrial process can introduce harmful chemicals, creating an extra contaminant that needs its own removal step.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.10.1.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The sludge should undergo anaerobic digestion. The liquid effluent should undergo aerobic biological treatment.	Keep the two sedimentation outputs distinct: microorganisms digest sludge without oxygen, while air is supplied to microorganisms treating the liquid effluent.	2
Total Question 1			2
02.1	The air supplies oxygen for aerobic microorganisms. The microorganisms break down the organic matter in the effluent.	The liquid route is aerobic biological treatment. Supplying oxygen allows aerobic microorganisms to respire and break down remaining organic material.	3
Total Question 2			3
03.1	Organic matter must be removed from both types of waste water. Agricultural waste water may contain harmful microbes that must be removed. Industrial waste water may contain harmful chemicals that must be removed.	Give the shared contaminant before the difference. Both streams can contain organic matter, but agricultural waste water is associated with harmful microbes whereas industrial processes may add harmful chemicals.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.10.1.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$665$ m³; waste water contains more organic matter and harmful microbes, and may contain harmful chemicals, so it needs more treatment.	The sludge volume is $0.070\times9.5\times10^3=665$ m³. Compared with fresh ground water, waste water usually has a larger and more varied contaminant load. Organic matter and microbes must be removed, and industrial inputs may require removal of harmful chemicals before final potable-water treatment.	5
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02.1	Fresh ground water is generally easiest because it has few contaminants, although it may still need filtration and sterilisation. Waste water needs removal of organic matter and harmful microbes by sewage treatment before further potable-water treatment; harmful chemicals may also need removal. Sea water needs desalination by distillation or reverse osmosis, which requires large amounts of energy to remove dissolved salts.	Compare contaminant type rather than treating all water alike. Ground water usually needs the least treatment, waste water needs biological and possibly chemical treatment, and sea water needs energy-intensive salt removal.	5
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03.1	$30.0$ kg of organic matter remains. Air supplies oxygen for aerobic microorganisms. The microorganisms break down organic matter in the liquid effluent.	The fraction remaining is $100.0-88.0=12.0\%$ . Calculate $0.120\times250=30.0$ kg. This is the liquid-effluent route, so oxygen is supplied to aerobic microorganisms rather than the sludge being treated anaerobically.	5
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04.1	The initial organic-matter mass is $120$ kg. $9.6$ kg of organic matter remains after biological treatment. The harmful chemical has an initial and remaining mass of $3.0$ kg. The permitted chemical mass is $0.100$ kg. The additional treatment must remove at least $2.9$ kg of the chemical. This is a minimum removal of $96.7\%$ to three significant figures.	The organic load is $50\,000\times2.4=120\,000$ g, and $8\%$ remains, giving $9.6$ kg. The chemical load is $50\,000\times0.060=3000$ g or $3.0$ kg because the biological stage does not remove it. The permitted mass is $50\,000\times0.0020=100$ g or $0.100$ kg. Therefore at least $2.9$ kg, or $(2.9/3.0)\times100=96.7\%$ , needs separate chemical treatment.	6
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05.1	Stage B leaves $20$ kg of organic matter, so it meets the organic-matter limit. Stage B leaves $36$ kg of harmful chemical, so it fails the chemical limit. Stage C alone leaves $450$ kg of organic matter, so it fails the organic-matter limit. Stage C alone leaves $2.0$ kg of harmful chemical, so it meets the chemical limit. Applying B and then C leaves $18$ kg of organic matter and $1.8$ kg of harmful chemical, so the combined treatment meets both limits.	Stage B leaves $0.04\times500=20$ kg of organic matter and $0.90\times40=36$ kg of chemical. Stage C alone leaves $0.90\times500=450$ kg of organic matter and $0.05\times40=2.0$ kg of chemical. In sequence, C removes $10\%$ of the remaining organic matter and $95\%$ of the remaining chemical, leaving $0.90\times20=18$ kg and $0.05\times36=1.8$ kg respectively.	5
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4.10.1.4 · Alternative methods of extracting metals (HT only)

Tier 1 · Easy

Mark scheme for 4.10.1.4 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Plants absorb copper compounds, are harvested and are burned to leave ash containing copper compounds.	Grow suitable plants on low-grade copper ore so their tissues take up copper compounds. Harvest the plants and burn the biomass. The smaller mass of ash contains a higher concentration of copper compounds for further processing.	3
Total Question 1			3
02.1	Bacteria convert copper compounds in low-grade ore into soluble compounds. A leachate solution containing copper compounds is formed.	Bioleaching uses bacteria rather than plants. The copper is transferred into a leachate as compounds; copper metal is not produced at this stage.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.10.1.4 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Phytomining uses plants, which are harvested and burned to give ash rich in copper compounds. Bioleaching uses bacteria to produce a leachate containing copper compounds. Both products need further processing, such as electrolysis or displacement, to obtain copper metal.	Follow the material produced by each biological route. Plants concentrate compounds into burnable biomass and ash; bacteria transfer compounds into solution. Neither step reduces copper ions all the way to metal.	5
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02.1	Copper metal forms at the negative electrode. Positive copper ions move towards the negative electrode and gain electrons.	The negative electrode attracts the positively charged copper ions in solution. Reduction occurs when these ions gain electrons, depositing copper atoms on the electrode.	3
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03.1	The dry plant mass is $1800$ kg and the ash mass is $72$ kg. The ash contains $8.64$ kg of copper compounds.	Convert tonnes to kilograms, then apply the two percentages in sequence. The ash mass is $0.040\times1800=72$ kg. The mass of copper compounds is $0.12\times72=8.64$ kg.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.10.1.4 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Conventional: $7.1$ MJ kg^-1; bioleaching: $2.5$ MJ kg^-1. Bioleaching uses less energy and moves less material but is much slower and gives less copper in the stated run.	Convert GJ to MJ. Conventional energy per kilogram is $5400/760=7.11$ MJ kg^-1. Bioleaching energy per kilogram is $1700/690=2.46$ MJ kg^-1. Bioleaching therefore reduces stated energy per kilogram and material movement, but its $60$ -day duration is far longer than $2$ days and its output is lower. The preferred process depends on whether environmental impact, rate or output has greatest importance.	5
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02.1	Electrolysis recovers $1128$ kg of copper. Displacement recovers $972$ kg of copper. Electrolysis gives $156$ kg more and purer copper but uses more energy; displacement uses much less energy and scrap iron but gives less, less-pure copper.	Electrolysis gives $0.94\times1200=1128$ kg. Displacement gives $0.81\times1200=972$ kg, a difference of $156$ kg. The preferred route depends on whether energy use, recovery and purity are most important.	5
Total Question 2			5
03.1	Phytomining produces $62$ kg of copper-equivalent material and uses $410$ MJ. Bioleaching completes five cycles, producing $70$ kg of copper-equivalent material and using $630$ MJ. Bioleaching gives $8$ kg more copper in the stated time, while phytomining uses $220$ MJ less energy. The preferred method depends on whether output or energy use has greater importance.	Only complete cycles count: $200/35=5$ complete cycles with $25$ days left. Bioleaching output is $5\times14=70$ kg and its energy use is $5\times126=630$ MJ. Compare these with the plot's $62$ kg and $410$ MJ. Both methods can exploit low-grade ores without moving large amounts of rock, but the data favour different choices for output and energy.	6
Total Question 3			6
04.1	Each cubic metre of leachate yields $2.80$ kg of copper. $150$ m³ of leachate is required. Electrolysing this volume uses $1350$ MJ. The bacteria produce a leachate containing copper compounds rather than copper metal. Copper ions must gain electrons during electrolysis to form copper atoms.	The recovered copper per cubic metre is $3.20\times0.875=2.80$ kg. The required volume is $420/2.80=150$ m³, and its energy use is $150\times9.0=1350$ MJ. Bioleaching transfers copper compounds into solution; reduction at the negative electrode is the separate stage that forms copper metal.	5
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05.1	Crop A contains copper compounds equivalent to $1440$ kg of copper per year. Crop A gives $1080$ kg of recovered copper per year. Crop B contains copper compounds equivalent to $900$ kg of copper per year. Crop B gives $675$ kg of recovered copper per year. Crop A gives $405$ kg more recovered copper each year, so it is preferable if annual output is the priority. Crop B needs fewer growing, harvesting and burning cycles, so energy, cost and land-management data are needed for a complete choice.	For Crop A, annual copper-equivalent uptake is $60\times8.0\times3=1440$ kg and recovery is $0.75\times1440=1080$ kg. For Crop B, uptake is $60\times15\times1=900$ kg and recovery is $0.75\times900=675$ kg. Crop A wins on annual copper output by $405$ kg, but the three annual crop cycles may require more operations than Crop B's single cycle.	6
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4.10.2.1 · Life cycle assessment

Tier 1 · Easy

Mark scheme for 4.10.2.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Raw-material extraction and processing; manufacture and packaging; use during the product's lifetime; disposal at the end of its useful life.	Follow the product from obtaining its raw materials, through making and using it, to its end of life. Transport and distribution are included within the relevant stages.	4
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02.1	Extracting the ore is the raw-material extraction and processing stage. Sending the can to landfill is the end-of-life disposal stage.	Place each event on the product timeline. Obtaining ore happens before manufacture; landfill follows use and belongs to disposal.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.10.2.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Any two of the following. Omitting disposal leaves the life-cycle boundary incomplete, and excluding transport from every stage understates each product's total impact — either could change which bottle appears preferable. The funder's interests may influence which stages or impacts are selected. Reporting only total energy ignores other impacts, whose weighting requires value judgements.	Check both coverage and objectivity. A fair LCA includes equivalent stages for both products, and its choices about omitted stages and environmental weighting should be transparent.	2
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02.1	Energy use can be measured and quantified directly in kWh. A river-damage score requires choices about the importance of different pollutant effects. Those value judgements can differ between analysts, so the score is not purely objective.	Distinguish a measured quantity from an assigned environmental rating. The first has a physical unit; the second depends on how effects are selected and weighted.	3
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03.1	$2.20$ kg of paper is sent for disposal. $1.44$ kg of plastic is sent for disposal. A complete comparison must also consider raw-material extraction, manufacture and use, together with transport at each stage and other environmental impacts.	The total paper mass is $200\times55=11000$ g; $20\%$ is not recycled, giving $2200$ g or $2.20$ kg. The total plastic mass is $200\times8.0=1600$ g; $90\%$ is not recycled, giving $1440$ g or $1.44$ kg. Disposal is only one life-cycle stage.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.10.2.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	A uses $2.2$ MJ and B uses $2.0$ MJ. B has slightly lower stated energy and less waste, while A has the lower pollutant score. The preferred package depends on how impacts are weighted, and the pollutant score may involve subjective value judgements.	Add the stated energy figures: A uses $1.8+0.40=2.2$ MJ and B uses $0.90+1.1=2.0$ MJ. B saves $0.2$ MJ and creates $35$ g less waste, but A has a much lower stated pollutant score. No single product wins every category. The score's construction, omitted water or raw-material data, and subjective weighting limit a definite conclusion.	5
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02.1	Tile R has a transport-impact index of $720$ kg km and Tile S has an index of $5400$ kg km. Tile R has the lower stated transport impact even though it is heavier, because its journey is much shorter. A complete LCA must also compare raw-material extraction, manufacture, use, disposal and impacts such as energy, water, waste and pollution.	Calculate $18\times40=720$ kg km for R and $6.0\times900=5400$ kg km for S. Transport can occur at each of the four life-cycle stages, so a lower transport index cannot establish the overall result without equivalent data for raw materials, manufacture, use and disposal.	5
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03.1	The reusable bag first has the lower stated energy impact at $24$ uses. At $23$ uses it uses $4.175$ MJ compared with $4.140$ MJ for disposable bags; at $24$ uses it uses $4.200$ MJ compared with $4.320$ MJ. The comparison omits impacts such as raw-material use, water use, transport, waste or pollution.	For $n$ uses, the reusable total is $3.60+0.025n$ MJ and the disposable total is $0.180n$ MJ. Checking consecutive whole values gives reusable slightly higher at $n=23$ but $0.120$ MJ lower at $n=24$ . A single energy total cannot represent every LCA impact.	5
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04.1	Product X has a total stated energy impact of $40$ MJ. Product Y has $25$ MJ before disposal, so break-even is $15$ MJ and its disposal energy must be below this for its total to be lower than X's. With an $8.0$ MJ disposal estimate, Y totals $33$ MJ and appears preferable on energy. With a $17$ MJ disposal measurement, Y totals $42$ MJ and X has the lower energy impact. Omitting or selectively estimating one life-cycle stage can reverse the conclusion, so both products need consistent system boundaries.	Add X's four stages to obtain $18+12+4.0+6.0=40$ MJ. Y's known subtotal is $11+9.0+5.0=25$ MJ, leaving a strict allowance below $15$ MJ if Y is to remain below $40$ MJ. The two disposal values give totals of $33$ MJ and $42$ MJ, demonstrating why a selective LCA can be misleading.	5
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05.1	Panel P uses less water than Panel Q. Panel Q produces less solid waste than Panel P. Panel Q uses less energy than Panel P. The three impacts have different units, so combining them into one score requires judgements about their relative importance. Giving sufficiently high weight to water use would favour Panel P. Giving greater weight to energy use or solid waste would favour Panel Q.	Compare each impact category separately. P is better on water use, whereas Q is better on solid waste and energy use, so neither panel is lower in every category. Water, waste and energy cannot be added directly because their units differ. An overall choice therefore depends on the weights assigned to the categories: prioritising water can favour P, while prioritising energy or waste can favour Q.	6
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4.10.2.2 · Ways of reducing the use of resources

Tier 1 · Easy

Mark scheme for 4.10.2.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	For example: use less packaging; refill a glass jar; melt and recast scrap metal.	Reducing avoids material use, reusing keeps the same product in service, and recycling reprocesses its material. Any clear, correctly classified example earns the corresponding mark.	3
Total Question 1			3
02.1	Iron ore. Fossil fuels used to supply process energy, for example coal, oil or natural gas.	Recycled steel supplies some of the required metal without extracting as much new iron ore. Recycling also usually reduces the process energy obtained from limited fuels.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.10.2.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Washing and refilling is reuse; crushing and remelting is recycling. Reuse keeps the same product in service, whereas recycling needs collection and energy to process the material into a new product.	Classify by whether the original object remains intact. Reuse avoids remanufacture; recycling recovers the material but still requires sorting and energy-intensive processing.	3
Total Question 1			3
02.1	Metal lids must be removed because they are a different material and would contaminate the glass. Coloured glass must be separated because it would change the colour of the new bottles. The amount of separation depends on the properties required of the final product, here colourless glass.	Recycling does not automatically accept every mixed material. The feed must be separated enough for the remelted glass to have the required composition and appearance.	3
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03.1	$2000$ kg of glass is saved. Less limited raw material needs to be obtained by quarrying or mining. Less energy is needed to manufacture or transport the bottles, and less waste is produced at the end of use.	Each bottle uses $320-280=40$ g less glass. The total saving is $50\,000\times40=2\,000\,000$ g, which is $2000$ kg. Reduction avoids some extraction and processing before reuse or recycling is even considered.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.10.2.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Single-use trays require $720$ kg and $2160$ MJ. Fifty crates require $450$ kg and $3075$ MJ including washing. Crates reduce material and discarded items, but use more stated energy over these $6000$ deliveries.	Single-use mass is $6000\times0.12=720$ kg and energy is $720\times3.0=2160$ MJ. The company needs $6000/120=50$ crates, with mass $50\times9.0=450$ kg. Crate manufacture uses $450\times5.5=2475$ MJ and washing uses $6000\times0.10=600$ MJ, totalling $3075$ MJ. Crates save $270$ kg of material but use $915$ MJ more energy on the stated boundary; longer life or recycling data could change the judgement.	5
Total Question 1			5
02.1	The cans contain $360$ kg of aluminium: $234$ kg recycled and $126$ kg new. The total production energy is $25488$ MJ. Using only new aluminium would require $64800$ MJ, so the saving is $39312$ MJ.	The total mass is $(24000\times15)/1000=360$ kg. Recycled mass is $0.65\times360=234$ kg, leaving $126$ kg new. Energy is $(234\times12)+(126\times180)=2808+22680=25488$ MJ. All-new production would use $360\times180=64800$ MJ, so $64800-25488=39312$ MJ is saved.	6
Total Question 2			6
03.1	$46\,800$ bricks are reused. The remaining $25\,200$ bricks replace $52\,920$ kg of newly quarried aggregate. Reuse avoids remaking products and crushing recycles the remaining material, so both reduce landfill and demand for limited raw materials, although cleaning, transport and crushing still use energy.	Calculate $0.65\times72\,000=46\,800$ reusable bricks. The remainder is $72\,000-46\,800=25\,200$ , which replaces $25\,200\times2.1=52\,920$ kg of aggregate. The environmental gain must be weighed against the energy and transport needed to recover the materials.	5
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04.1	Using only new metal would require $20\,000$ MJ. Replacing one kilogram of new metal with recycled metal saves $32$ MJ. A total saving of $8000$ MJ is required, so at least $250$ kg must be recycled metal. The minimum recycled fraction is $50\%$ by mass. Collection, transport, separation and reprocessing of the scrap still use energy and can cause environmental impacts.	All-new production uses $500\times40=20\,000$ MJ, so the required saving is $20\,000-12\,000=8000$ MJ. Each kilogram switched from new to recycled saves $40-8.0=32$ MJ. Therefore $8000/32=250$ kg must be recycled, which is $(250/500)\times100=50\%$ .	5
Total Question 4			5
05.1	The discarded bottles contain $2000$ kg of glass. $1800$ kg is collected and $1440$ kg remains after sorting. Remelting produces $1368$ kg of usable recycled glass. This is enough to make $3800$ new bottles. Making $5000$ new bottles requires $1800$ kg, so $432$ kg of new glass is needed. The loop is not completely closed because uncollected bottles, sorting rejects and remelting losses reduce the recovered mass, although recycling still reduces raw-material use and waste.	Initial glass mass is $5000\times0.40=2000$ kg. Successive recovery stages give $2000\times0.90=1800$ kg, then $1800\times0.80=1440$ kg, then $1440\times0.95=1368$ kg. At $0.36$ kg per new bottle this makes $1368/0.36=3800$ bottles. A full run of $5000$ needs $1800$ kg, leaving $1800-1368=432$ kg to come from new material.	6
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4.10.3.1 · Corrosion and its prevention (chemistry only)

Tier 1 · Easy

Mark scheme for 4.10.3.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Oxygen and water.	Rusting is a corrosion reaction of iron that requires oxygen from air and water. Removing either condition prevents rust formation.	2
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02.1	For example, painting, greasing or electroplating. The coating prevents oxygen and water from reaching the iron.	Rusting needs both oxygen and water. An intact coating separates the iron from these substances, so the rusting reaction cannot occur.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.10.3.1 Tier 2 · Standard
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01.1	Paint is only a barrier, so a scratch lets oxygen and water reach the steel. Zinc is more reactive than iron and corrodes sacrificially, protecting the exposed iron.	Distinguish barrier protection from sacrificial protection. A broken barrier no longer excludes air and water; a more-reactive zinc coating can still supply protection at a scratch by oxidising instead of the iron.	4
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02.1	Use grease or oil on the chain because it forms a flexible barrier that remains effective while the links move. Electroplate the door handle because the metal coating forms a durable barrier and gives a suitable appearance.	Both choices keep air and water away, but the use affects the coating selected. A moving chain needs a coating that does not crack as the links move; a handle can use a hard decorative coating.	4
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03.1	Aluminium forms an oxide coating that adheres to and protects the metal beneath. The coating acts as a barrier to substances in the environment. Rust does not provide the same protective barrier, so oxygen and water can continue to reach iron.	Compare the effect of the corrosion products. Aluminium oxide forms a protective surface coating, whereas rust does not seal the iron away from air and water.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.10.3.1 Tier 3 · Hard
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01.1	Both intact coatings are barriers. At a scratch, X gives sacrificial protection because X is more reactive than iron; copper does not, so the exposed iron corrodes preferentially. X is the better protective coating.	An unbroken layer of either metal separates iron from air and water. When scratched, electrical contact remains between the two metals. X is above iron in the reactivity order, so X oxidises instead of the iron. Copper is below iron, so iron is the more reactive metal and corrodes at the scratch. Therefore X remains protective after local damage, unlike copper.	5
Total Question 1			5
02.1	The block can lose $9.6$ kg before replacement, so its maximum service time is $12.8$ years. Magnesium is more reactive than iron, so it corrodes sacrificially instead of the steel pipe.	The mass remaining at replacement is $0.20\times12.0=2.4$ kg, so the usable loss is $12.0-2.4=9.6$ kg. The time is $9.6/0.75=12.8$ years. Electrical contact lets the more reactive magnesium provide sacrificial protection to the iron in the steel.	5
Total Question 2			5
03.1	The mean loss is $0.80$ g for uncoated steel and $0.30$ g for coated steel. The coating reduces mean mass loss by $62.5\%$ . The results support the claim because $62.5\%$ is $7.5$ percentage points above $55\%$ , and the repeats are closely grouped.	The means are $(0.81+0.80+0.79)/3=0.80$ g and $(0.31+0.30+0.29)/3=0.30$ g. The reduction is $((0.80-0.30)/0.80)\times100=62.5\%$ . Identical plates and conditions make the comparison fair, and the small spread supports the reliability of the means.	5
Total Question 3			5
04.1	Painting is needed five times, at years $0$ , $4$ , $8$ , $12$ and $16$ . The painting plan uses $240$ L of paint. The painting plan costs £ $1680$ . The zinc plan uses $90$ kg of zinc and costs £ $1170$ . Zinc costs £ $510$ less over the stated period. Paint is only a barrier at a scratch, whereas zinc is more reactive than iron and can protect exposed steel sacrificially, so zinc is favoured by the stated cost and damage risk.	There are five applications before the end of year $20$ . Paint use is $60\times0.80\times5=240$ L and cost is $240\times7.00=£1680$ . Zinc use is $60\times1.5=90$ kg and cost is $90\times13.00=£1170$ , a difference of £ $510$ . The chemical distinction strengthens the choice: a paint scratch exposes iron to oxygen and water, while zinc can continue to oxidise instead of the iron.	6
Total Question 4			6
05.1	Each magnesium block can lose $6.0$ kg before replacement. Each block protects the buoy for $5.0$ years. Three blocks are needed for continuous protection over $15$ years. Paint alone would reduce the corrosion rate to $0.50$ kg per year. Paint alone would allow $7.5$ kg of steel to corrode over $15$ years. The paint limits contact with oxygen and water, while the more reactive magnesium corrodes sacrificially to protect steel exposed at defects in the barrier.	The usable magnesium mass is $8.0-2.0=6.0$ kg, so a block lasts $6.0/1.2=5.0$ years and three consecutive blocks cover $15$ years. Paint leaves $20\%$ of the unpainted corrosion rate, or $0.20\times2.5=0.50$ kg per year; over $15$ years that would be $7.5$ kg without sacrificial protection. The methods are complementary because one is a barrier and the other protects exposed iron chemically.	6
Total Question 5			6

4.10.3.2 · Alloys as useful materials (chemistry only)

Tier 1 · Easy

Mark scheme for 4.10.3.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Bronze is copper and tin; brass is copper and zinc.	Recall the specified copper alloys. Both contain copper; tin makes bronze and zinc makes brass.	2
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02.1	Use bronze for the statue. Use brass for the fitting.	Bronze is the specified copper alloy used for statues, while brass is used for fittings such as taps and door fittings.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.10.3.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Cutting blade: high-carbon steel because it is strong. Car-body panel: low-carbon steel because it is softer and easily shaped. Kitchen sink: stainless steel because it is hard and resistant to corrosion.	Match required properties to the specified steels. A blade needs high strength, a pressed panel needs a softer steel that can be shaped, and a wet sink needs hardness and resistance to corrosion. These point to high-carbon, low-carbon and stainless steel respectively.	6
Total Question 1			6
02.1	The alloy is much stronger, at $310$ MPa compared with $90$ MPa. Its density is only slightly higher and remains low, so a panel can be both strong and lightweight.	Compare both relevant properties rather than selecting the lowest density alone. The small density increase is outweighed by the large strength increase for a load-bearing aircraft part.	3
Total Question 2			3
03.1	The alloy is bronze. Bronze is a mixture of copper and tin rather than a compound with a fixed composition. The proportions in an alloy can vary, and changing them changes the arrangement of atoms and therefore the properties.	Recognise copper mixed with tin as bronze. Alloys are mixtures, so their compositions are not fixed like those of compounds; different proportions can produce different useful properties.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.10.3.2 Tier 3 · Hard
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01.1	$0.768$ kg zinc; brass is preferred when hardness and resistance to deformation matter more than maximum conductivity.	The zinc percentage is $100.0-68.0=32.0\%$ . Its mass is $0.320\times2.40=0.768$ kg. The alloy's greater hardness can make the fitting more durable, while its lower conductivity is unimportant if the fitting is not required to carry electrical current.	4
Total Question 1			4
02.1	Alloy A is suitable because it meets both the minimum strength and maximum corrosion-loss requirements; alloy B meets neither requirement. The mass of the Alloy A fitting is $2125$ g (or $2.125$ kg).	Check both thresholds before considering density. Alloy A has $350>320$ MPa and $0.30<0.50$ mm per year. Its mass is $m=\rho V=(8.5)(250)=2125$ g. Alloy B is lighter but its strength is too low and its corrosion loss is too high.	5
Total Question 2			5
03.1	The mixture contains $16.25$ g of gold. Gold makes up $81.25\%$ of the $20.0$ g mixture. The remaining $18.75\%$ consists of other metals, so the product is an alloy and not pure gold.	The $18$ carat portion contains $0.75\times15.0=11.25$ g of gold, and the pure portion adds $5.0$ g. Total gold is $16.25$ g in a total mass of $20.0$ g, so $(16.25/20.0)\times100=81.25\%$ . Because other metals remain, the product is a mixture of metals.	5
Total Question 3			5
04.1	The original alloy contains $691.2$ kg of copper. A mixture in which this is $60\%$ copper has a final mass of $1152$ kg. $192$ kg of zinc must be added. The copper-zinc alloy is brass. Brass is a mixture rather than a compound with a fixed formula, so its metal proportions can vary and the change can alter its properties.	Copper mass stays constant at $0.72\times960=691.2$ kg. If that mass is $60\%$ of the final alloy, the total must be $691.2/0.60=1152$ kg. The added zinc is therefore $1152-960=192$ kg. Copper mixed with zinc is brass, whose variable composition is characteristic of an alloy mixture.	5
Total Question 4			5
05.1	Bronze Q meets both requirements; P fails the hardness requirement. A $750$ kg batch of Q needs $150$ kg of tin. The batch needs $600$ kg of copper. Bronze is an alloy mixture of copper and tin whose proportions can vary. The different tin percentages produce different properties, so composition must be matched to the intended use.	Check both thresholds: Q has hardness $180\ge160$ and elongation $4.0\%\ge3.0\%$ , while P is too soft. Q contains $0.20\times750=150$ kg tin and the remaining $0.80\times750=600$ kg copper. The contrasting data show that changing an alloy's composition changes its useful properties.	5
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4.10.3.3 · Ceramics, polymers and composites (chemistry only)

Tier 1 · Easy

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01.1	The glass fibres are the reinforcement and the polymer resin is the matrix or binder.	The material embedded as fibres or fragments is the reinforcement. The continuous material that surrounds and binds it is the matrix, so the glass and resin have those roles respectively.	2
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02.1	Sodium carbonate. Limestone.	Soda-lime glass is made by heating sand with sodium carbonate and limestone.	2
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Tier 2 · Standard

Mark scheme for 4.10.3.3 Tier 2 · Standard
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01.1	The carbon fibres are the reinforcement and the polymer resin is the matrix. The fibres provide high strength while the polymer binds them together in a low-density material.	A composite combines roles rather than forming one uniform substance. The reinforcement carries load; the surrounding matrix holds it in position and transfers forces while keeping the structure relatively light.	4
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02.1	Both polymers are made by joining ethene monomers. Different reaction conditions produce different arrangements of polymer chains. The different structures give the two forms different physical properties, including density.	The monomer identity does not change, so both products are poly(ethene). Changing the polymerisation conditions changes the structure formed and therefore the material properties.	3
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03.1	Wet clay is shaped into the required brick form. The shaped clay is heated in a furnace. The wet clay can be shaped before heating; after furnace heating the ceramic is hard and cannot be reshaped in the same way.	Put the manufacturing stages in order. Water makes the clay workable, so shaping comes first. Furnace heating then produces the rigid ceramic object.	3
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Tier 3 · Hard

Mark scheme for 4.10.3.3 Tier 3 · Hard
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01.1	Borosilicate glass; $180$ g. It remains solid at $720$ °C and is transparent, unlike the other unsuitable options.	Soda-lime glass is unsuitable because $720>650$ , so it would soften. The clay ceramic tolerates the temperature but is opaque. Borosilicate glass satisfies both the temperature and transparency requirements. Its mass is $m=\rho V=(2.25)(80)=180$ g.	5
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02.1	The composite panel has mass $152$ kg and the steel panel has mass $624$ kg. Both exceed the required strength, but the glass-fibre composite is better where low mass is the priority because it is $472$ kg lighter.	Use $m=\rho V$ . Composite mass is $1900\times0.080=152$ kg and steel mass is $7800\times0.080=624$ kg. Both strengths exceed $300$ MPa, so the lower-density composite meets the stated priority, although steel is stronger.	5
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03.1	The resin panel supports $0.75$ kN kg^-1 and the composite supports $2.4$ kN kg^-1. The composite's breaking load per kilogram is $3.2$ times as large. The fibres are the reinforcement that increases strength, while the resin is the matrix that surrounds and binds the fibres.	Divide breaking load by mass: $1.8/2.4=0.75$ kN kg^-1 and $7.2/3.0=2.4$ kN kg^-1. The ratio is $2.4/0.75=3.2$ . Link the improved property to the reinforcing fibres and identify the surrounding resin as the matrix or binder.	5
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04.1	P is low-density poly(ethene), or LDPE. Q is high-density poly(ethene), or HDPE. P is more suitable for the squeezable bottle because it is flexible. Q is more suitable for the rigid pipe because it is more rigid. Different polymerisation conditions produce different arrangements of chains from the same ethene monomer, giving different densities and properties.	Use both density and mechanical behaviour to identify the samples. The lower-density flexible material is LDPE and the denser, more rigid material is HDPE. Both remain poly(ethene), but changing reaction conditions changes how the chains are arranged and therefore changes the material's properties and suitable uses.	5
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05.1	The original clip contains separate polymer chains without cross-links between them. Heating allows the chains to move relative to one another. The load makes the chains slide past each other, so the clip changes shape. Choose a thermosetting polymer for the replacement clip. Strong covalent cross-links restrict movement between its chains, so it does not soften through chain sliding at the working temperature.	Link the observed deformation to structure. Thermosoftening chains are not joined into a cross-linked network, so warming lets them move past one another under the applied load. A thermosetting replacement has covalent cross-links between chains; these connections prevent the sliding that caused the original clip to bend.	5
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4.10.4.1 · The Haber process (chemistry only)

Tier 1 · Easy

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01.1	Air; natural gas; iron.	Nitrogen is separated from air. Hydrogen is commonly manufactured from natural gas. The purified reactant gases are passed over an iron catalyst.	3
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02.1	About $450$ °C. About $200$ atmospheres.	The commercially used conditions are a high temperature of about $450$ °C and a high pressure of about $200$ atmospheres.	2
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Tier 2 · Standard

Mark scheme for 4.10.4.1 Tier 2 · Standard
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01.1	The mixture is cooled so ammonia liquefies and is removed; unreacted nitrogen and hydrogen remain gaseous and are recycled to increase overall conversion and reduce waste.	The reactor does not convert all reactants in one pass because the reaction is reversible. Cooling condenses ammonia at a temperature where nitrogen and hydrogen remain gases. Removing liquid ammonia separates the product, and recycling the remaining reactants gives them further opportunities to react.	4
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02.1	$\mathrm{N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)}$ .	Nitrogen and hydrogen are diatomic. Using coefficients $1:3:2$ gives two nitrogen atoms and six hydrogen atoms on each side, and the reversible symbol shows that ammonia can decompose back into the reactants.	3
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03.1	The high temperature gives particles more kinetic energy, so successful collisions occur more often. The high pressure places the reacting gases closer together, increasing collision frequency. Iron acts as a catalyst and provides an alternative reaction pathway with a lower activation energy. The catalyst increases reaction rate without being used up.	Apply collision theory. Temperature changes particle energy, pressure changes the concentration of reacting gases, and the iron catalyst lowers the activation energy.	4
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Tier 3 · Hard

Mark scheme for 4.10.4.1 Tier 3 · Hard
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01.1	$23.0\%$ ; unreacted nitrogen and hydrogen are recycled through the reactor to reduce waste and increase the overall conversion.	The percentage converted in this pass is $(184/800)\times100=23.0\%$ . Because the reaction is reversible, not all nitrogen and hydrogen react in one pass. Cooling removes liquid ammonia, while the unreacted gases remain gaseous and are returned to the reactor so less reactant is wasted and more is converted overall.	4
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02.1	The lower temperature would increase the equilibrium yield of ammonia because the forward reaction is exothermic, but it would decrease the reaction rate. The higher pressure would increase ammonia yield and rate because the equilibrium shifts to the side with fewer gas molecules, but compression and stronger equipment increase energy and plant costs. Removing the catalyst would make equilibrium take longer to reach without changing its position, so keeping the iron catalyst is preferable.	Assess each change separately. Lower temperature favours the exothermic forward reaction but slows collisions. Higher pressure favours two gas molecules of ammonia over four reactant gas molecules and raises collision frequency, but costs more. Iron speeds both directions equally and does not alter equilibrium yield, so omitting it has no yield advantage.	6
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03.1	Raise the temperature to about $450$ °C to increase the reaction rate. Raise the pressure to about $200$ atmospheres so gas particles collide more frequently. Pass the purified gases over an iron catalyst to increase the reaction rate. Cool the exit mixture so ammonia liquefies and can be removed, then recycle unreacted nitrogen and hydrogen instead of releasing them.	Compare each part of the proposal with the industrial process. The reactor needs the specified high temperature, high pressure and iron catalyst for a useful rate. After reaction, cooling separates liquid ammonia from gaseous reactants, and recycling reduces waste by giving the unreacted gases further chances to react.	6
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04.1	The first pass produces $250$ kg of ammonia. The second pass produces $187.5$ kg of ammonia. The third pass produces $140.625$ kg of ammonia. The total ammonia collected is $578.125$ kg. $421.875$ kg of unreacted nitrogen and hydrogen remains after three passes. Cooling liquefies ammonia so it can be separated while unreacted nitrogen and hydrogen remain gaseous for recycling.	After the first pass, $750$ kg remains. The second pass converts $0.25\times750=187.5$ kg, leaving $562.5$ kg. The third converts $0.25\times562.5=140.625$ kg, leaving $421.875$ kg. Adding the three product masses gives $578.125$ kg. Cooling separates the product and allows the unreacted gases to make further passes.	6
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05.1	Line A forms $180$ kg of ammonia in the reactor. Line A collects $171$ kg of ammonia after cooling. Line B forms and collects $72$ kg of ammonia. Line A collects $99$ kg more ammonia in the stated time. Iron increases reaction rate by providing a lower-activation-energy pathway and is not used up. Line A's separation stage leaves $9.0$ kg of formed ammonia uncollected, while Line B's main limitation is the slower uncatalysed reaction, so A should keep the catalyst and improve cooling or recovery.	Line A forms $0.30\times600=180$ kg and collects $0.95\times180=171$ kg, leaving $9.0$ kg uncollected. Line B forms $0.12\times600=72$ kg and collects all of it. The $171-72=99$ kg output advantage shows the importance of reaction rate, while A's incomplete collection separately identifies a product-separation loss.	6
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4.10.4.2 · Production and uses of NPK fertilisers (chemistry only)

Tier 1 · Easy

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01.1	Nitrogen, phosphorus and potassium; phosphate rock is insoluble, so plants cannot absorb its phosphorus compounds from solution.	Decode each element symbol. Plants take up mineral ions in solution, so the phosphorus in insoluble phosphate rock is not readily available for absorption and the rock must first be chemically treated.	4
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02.1	Potassium chloride (or potassium sulfate) supplies potassium. Ammonia is used to manufacture ammonium salts.	The specified mined potassium sources are potassium chloride and potassium sulfate. Haber-process ammonia is a starting material for ammonium fertiliser salts.	2
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Tier 2 · Standard

Mark scheme for 4.10.4.2 Tier 2 · Standard
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01.1	Ammonium nitrate supplies nitrogen, the phosphate salt supplies phosphorus, and potassium chloride supplies potassium. It is a formulation because the components are mixed in measured proportions to give the required nutrient percentages.	Map each nutrient symbol to the element present in its salt. A controlled mixture designed to deliver specified properties or percentages is a formulation, not one pure compound.	4
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02.1	Use phosphoric acid to make triple superphosphate. Use sulfuric acid to make single superphosphate. Treatment produces soluble salts, making phosphorus compounds available for plant uptake.	Untreated phosphate rock is insoluble. The named acid treatments form useful soluble fertiliser products: sulfuric acid gives single superphosphate and phosphoric acid gives triple superphosphate.	3
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03.1	Use a titration with an indicator to determine the volumes of ammonia solution and sulfuric acid needed for neutralisation. Repeat using the same volumes but without indicator, so the product is not contaminated by indicator. Gently evaporate some water from the neutral solution and leave it to cool so crystals form. Filter the crystals and dry them.	A preliminary titration finds the neutralising volumes because both reactants and the ammonium sulfate product are soluble. Repeating without indicator gives a pure salt solution, which is concentrated and cooled before the crystals are separated and dried.	4
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Tier 3 · Hard

Mark scheme for 4.10.4.2 Tier 3 · Hard
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01.1	Laboratory: $140$ g; industrial: $432$ kg. The continuous industrial route is far faster and avoids repeated small batches, so it is more suitable for large-scale production.	Three hours is $180$ minutes. The laboratory completes $180/45=4$ batches and makes $4\times35=140$ g. Industry makes $2.4\times180=432$ kg. The industrial process has much greater scale and continuous throughput and can integrate ammonia, acid and salt manufacture; the laboratory method is useful for controlled small preparation but not bulk supply.	5
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02.1	The fertiliser contains $224$ kg of nitrogen, $96$ kg of phosphorus and $160$ kg of potassium. It is a formulation because different salts are combined in controlled proportions to give the required nutrient percentages. Industrial production links several large-scale processes and raw materials, often continuously, whereas laboratory preparation makes a small batch using separate controlled steps such as reaction and crystallisation.	Convert $1.60$ tonnes to $1600$ kg. The nutrient masses are $0.14\times1600=224$ kg, $0.060\times1600=96$ kg and $0.10\times1600=160$ kg. These nutrients come from different manufactured or mined salts that must be integrated and blended to a specified composition; this differs from preparing one small salt batch in a laboratory.	6
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03.1	Each bag supplies $3.0$ kg of nitrogen and $2.0$ kg of potassium. The minimum is $12$ bags, supplying $36$ kg of nitrogen and $24$ kg of potassium. The fertiliser is a formulation because different salts are mixed in controlled proportions to give specified nutrient percentages.	One bag supplies $0.12\times25=3.0$ kg nitrogen and $0.080\times25=2.0$ kg potassium. Nitrogen requires more than $34/3.0=11.33$ bags and potassium requires $22/2.0=11$ bags, so $12$ whole bags satisfy both limits. Check: $12\times3.0=36$ kg nitrogen and $12\times2.0=24$ kg potassium.	5
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04.1	The total fertiliser mass is $1250$ kg. It contains $150$ kg of nitrogen, which is $12\%$ of the product. It contains $75$ kg of phosphorus, which is $6.0\%$ of the product. It contains $100$ kg of potassium, which is $8.0\%$ of the product. The $12$ - $6$ - $8$ label is correct. The fertiliser is a formulation because different nutrient salts and filler are blended in controlled proportions to give specified properties and percentages.	Total mass is $500+375+250+125=1250$ kg. The nutrient masses are $0.30\times500=150$ kg nitrogen, $0.20\times375=75$ kg phosphorus and $0.40\times250=100$ kg potassium. Dividing each by $1250$ and multiplying by $100$ gives $12\%$ , $6.0\%$ and $8.0\%$ , matching the label.	6
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05.1	Each bag of F supplies $9$ kg N, $2$ kg P and $2$ kg K. Each bag of G supplies $5$ kg of each nutrient. Twelve bags cannot meet all three limits: enough F for nitrogen leaves too little G for phosphorus and potassium. A minimum combination is seven bags of F and six bags of G, for $13$ bags in total. This combination supplies $93$ kg of nitrogen. It supplies $44$ kg of phosphorus and $44$ kg of potassium, so all three requirements are met.	One F bag supplies $(0.18,0.04,0.04)\times50=(9,2,2)$ kg of N, P and K; one G bag supplies $(5,5,5)$ kg. With $12$ total bags, nitrogen requires at least eight F bags, but then at most four G bags give only $36$ kg each of P and K. With $13$ bags, seven F and six G supply nitrogen $7\times9+6\times5=93$ kg and phosphorus and potassium $7\times2+6\times5=44$ kg each.	6
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