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the value of the extracted metal can exceed the costs of mining, processing and waste disposal."},{"text":"It contains a low enough concentration of a metal compound.","correct":false,"explanation":"A low concentration makes extraction less economical because more rock must be mined and processed for each unit of metal. The metal compound’s concentration must be high enough."},{"text":"It contains a high enough concentration of a non-metal compound.","correct":false,"explanation":"The concentration of non-metal compounds does not determine whether metal extraction is worthwhile. The rock must contain a high enough concentration of a metal compound."},{"text":"It contains a high enough concentration of waste rock.","correct":false,"explanation":"A high concentration of waste rock increases processing and disposal costs rather than making extraction worthwhile. The valuable metal compound must be sufficiently concentrated."}]},{"id":"49c8e770-9667-448a-bcd4-b53d6e29517a","cardType":"cloze","points":null,"front":"A metal [ ] can be extracted by heating its oxide with carbon.","back":"A metal less reactive than carbon can be extracted by heating its oxide with carbon.","mcqOptions":[{"text":"less reactive than carbon","correct":true,"explanation":"Carbon can remove oxygen from the oxides of metals below it in the reactivity series, because these metals are less reactive than carbon."},{"text":"more reactive than carbon","correct":false,"explanation":"Metals more reactive than carbon form oxides that carbon cannot reduce. They generally require electrolysis rather than heating their oxides with carbon."},{"text":"equally reactive to carbon","correct":false,"explanation":"A metal with the same reactivity as carbon would not have a favourable difference in oxygen affinity. Carbon extraction applies to metals less reactive than carbon."},{"text":"more reactive than aluminium","correct":false,"explanation":"Metals more reactive than aluminium are also above carbon in the reactivity series, so carbon cannot remove oxygen from their oxides. The metal must be less reactive than carbon."}]},{"id":"d718bb00-f9ad-4e1c-8c2b-378edd2d2c80","cardType":"basic","points":1,"front":"Why does carbon act as a reducing agent in metal extraction?","back":"Carbon removes oxygen from the metal oxide.","mcqOptions":[{"text":"Carbon removes oxygen from the metal oxide.","correct":true,"explanation":"Reduction involves removing oxygen from a substance. Carbon removes oxygen from the metal oxide, leaving the metal behind, so carbon acts as the reducing agent."},{"text":"Carbon adds oxygen to the metal oxide.","correct":false,"explanation":"Adding oxygen would oxidise the metal rather than reduce its oxide. Carbon is the reducing agent because it removes oxygen from the metal oxide."},{"text":"Carbon removes metal from the metal oxide.","correct":false,"explanation":"The metal is the desired product and is not removed by the reducing agent. Carbon removes oxygen from the metal oxide, producing the metal."},{"text":"Carbon removes electrons from the metal oxide.","correct":false,"explanation":"Supplying electrons describes reduction in electronic terms, but carbon’s role in this extraction is identified by its removal of oxygen from the metal oxide."}]},{"id":"366d3185-ba3c-42ea-ada3-f4ad37d449ee","cardType":"basic","points":1,"front":"What happens to carbon when it removes oxygen from a metal oxide?","back":"Carbon is oxidised.","mcqOptions":[{"text":"Carbon is oxidised.","correct":true,"explanation":"Carbon gains oxygen when it removes oxygen from the metal oxide. Gaining oxygen is oxidation, so the carbon is oxidised during the reaction."},{"text":"Carbon is reduced.","correct":false,"explanation":"The metal oxide is reduced because it loses oxygen. Carbon instead gains that oxygen, and gaining oxygen means that carbon is oxidised."},{"text":"Carbon is neutralised.","correct":false,"explanation":"Neutralisation is a reaction between an acid and a base, not the transfer of oxygen here. Carbon gains oxygen and is therefore oxidised."},{"text":"Carbon is decomposed.","correct":false,"explanation":"Carbon does not break down into simpler substances during this reaction. It combines with oxygen, so it undergoes oxidation rather than decomposition."}]},{"id":"25e95456-ad93-4bcb-b36a-e33b8bf90ce2","cardType":"cloze","points":null,"front":"Metals [ ], such as aluminium, must be extracted by electrolysis.","back":"Metals more reactive than carbon, such as aluminium, must be extracted by electrolysis.","mcqOptions":[{"text":"more reactive than carbon","correct":true,"explanation":"Carbon cannot remove oxygen from the oxides of metals above it in the reactivity series. These more reactive metals must therefore be extracted using electrolysis."},{"text":"less reactive than carbon","correct":false,"explanation":"Carbon can reduce the oxides of metals below it in the reactivity series, so electrolysis is generally unnecessary. Electrolysis is required for metals more reactive than carbon."},{"text":"equally reactive to carbon","correct":false,"explanation":"A metal at the same reactivity as carbon would not match aluminium’s position above carbon. Aluminium requires electrolysis because it is more reactive than carbon."},{"text":"more reactive than aluminium","correct":false,"explanation":"Aluminium is not more reactive than itself, and this comparison does not state the extraction threshold. Metals more reactive than carbon must be extracted by electrolysis."}]},{"id":"019e9a9a-74ca-488f-8049-74b04b981a74","cardType":"basic","points":1,"front":"Why must an ore be molten or dissolved for electrolysis?","back":"Its ions must be free to move.","mcqOptions":[{"text":"Its ions must be free to move.","correct":true,"explanation":"Electrolysis requires charged ions to travel through the electrolyte towards oppositely charged electrodes. Melting or dissolving the ionic substance frees its ions to move."},{"text":"Its ions must be fixed in place.","correct":false,"explanation":"Fixed ions cannot carry charge through the electrolyte, so a solid ionic ore cannot be electrolysed. Melting or dissolving it allows the ions to move freely."},{"text":"Its electrons must be free to move.","correct":false,"explanation":"Electrons flow through the external circuit, while ions carry charge through the electrolyte. The ore is molten or dissolved so that its ions can move."},{"text":"Its molecules must be free to move.","correct":false,"explanation":"Electrolysis depends on mobile charged particles, not freely moving neutral molecules. Melting or dissolving the ore makes its ions free to move between the electrodes."}]},{"id":"56f3e34d-a0f6-4a2a-8a19-144567a0d5fc","cardType":"basic","points":1,"front":"What happens to positive metal ions at the cathode during electrolysis?","back":"They gain electrons to form metal atoms.","mcqOptions":[{"text":"They gain electrons to form metal atoms.","correct":true,"explanation":"Positive metal ions are attracted to the negative cathode, where they gain electrons. This reduction converts the ions into neutral metal atoms."},{"text":"They lose electrons to form metal atoms.","correct":false,"explanation":"Losing electrons would make particles more positively charged rather than forming neutral atoms. Positive metal ions must gain electrons at the cathode."},{"text":"They gain protons to form metal atoms.","correct":false,"explanation":"The number of protons in a nucleus does not change during electrolysis. Positive metal ions form atoms by gaining electrons at the cathode."},{"text":"They share electrons to form metal atoms.","correct":false,"explanation":"Metal ions are discharged at the cathode by electron transfer, not electron sharing. They gain electrons to become neutral metal atoms."}]},{"id":"46b8c7af-c4dc-45ce-8afc-b3da1eed7c50","cardType":"basic","points":2,"front":"Why are biological extraction methods useful for low-grade ores?","back":"Traditional mining produces too much waste rock and is uneconomical.","mcqOptions":[{"text":"Traditional mining produces too much waste rock and is uneconomical.","correct":true,"explanation":"Low-grade ores contain little valuable metal, so conventional mining creates large amounts of waste rock for a small yield. Biological methods can therefore be more economical."},{"text":"Traditional mining produces too little waste rock and is uneconomical.","correct":false,"explanation":"Low-grade ores produce large, not small, quantities of waste rock relative to the metal obtained. This excessive waste makes traditional mining uneconomical."},{"text":"Traditional mining produces too much valuable metal and is uneconomical.","correct":false,"explanation":"Low-grade ores yield relatively little valuable metal, not too much. Biological methods are useful because conventional processing generates excessive waste rock for that small yield."},{"text":"Traditional mining produces too much concentrated ore and is uneconomical.","correct":false,"explanation":"Concentrated ore is useful rather than wasteful, and low-grade ores contain relatively little of it. Traditional mining becomes uneconomical because it produces too much waste rock."}]},{"id":"cbc036cd-bbf1-4743-8427-bca4d231616a","cardType":"cloze","points":null,"front":"In phytoextraction, plants absorb metal ions through their [ ] and concentrate them in their [ ].","back":"In phytoextraction, plants absorb metal ions through their roots and concentrate them in their leaves and shoots.","mcqOptions":[{"text":"In phytoextraction, plants absorb metal ions through their roots and concentrate them in their leaves and shoots.","correct":true,"explanation":"Phytoextraction uses plant roots to take up dissolved metal ions, which are transported and concentrated mainly in the leaves and shoots for later harvesting."},{"text":"In phytoextraction, plants absorb metal ions through their stems and concentrate them in their flowers.","correct":false,"explanation":"Plants do not absorb the metal ions through stems or store them mainly in flowers. Uptake occurs through roots, with concentration chiefly in leaves and shoots."},{"text":"In phytoextraction, plants absorb metal ions through their leaves and shoots and concentrate them in their roots.","correct":false,"explanation":"Leaves and shoots are the main storage sites, not the absorption structures, and roots are not where the ions are chiefly concentrated. The correct sequence is roots, then leaves and shoots."},{"text":"In phytoextraction, plants absorb metal ions through their seeds and concentrate them in their fruits.","correct":false,"explanation":"Seeds are not the main route for absorbing metal ions, and fruits are not the principal concentration site. Phytoextraction involves uptake by roots and storage in leaves and shoots."}]},{"id":"e8fd4c91-ecae-4cf7-b84c-28a2add2ad15","cardType":"basic","points":1,"front":"What is the metal-rich solution produced in bioleaching called?","back":"A leachate.","mcqOptions":[{"text":"A leachate.","correct":true,"explanation":"The metal-rich solution formed during bioleaching is called a leachate. It contains dissolved metal compounds released from the low-grade ore."},{"text":"A concentrate.","correct":false,"explanation":"A concentrate is material enriched in a desired substance, not the solution produced by bioleaching. The metal-containing solution itself is called a leachate."},{"text":"A residue.","correct":false,"explanation":"The residue is the solid material left behind after treatment, whereas bioleaching produces a metal-rich liquid called a leachate."},{"text":"An electrolyte.","correct":false,"explanation":"An electrolyte conducts electricity in electrolysis, but it is not the specific name for the metal-rich solution made by bioleaching. That solution is a leachate."}]},{"id":"d94daa75-a71c-4be6-8bd3-d10b41fc0805","cardType":"basic","points":1,"front":"How can metal be extracted from a bioleaching leachate?","back":"Use scrap iron displacement or electrolysis.","mcqOptions":[{"text":"Use scrap iron displacement or electrolysis.","correct":true,"explanation":"Scrap iron can displace a less reactive metal from its ions, or electrolysis can reduce the metal ions at an electrode. Both methods recover metal from the leachate."},{"text":"Use copper displacement or filtration.","correct":false,"explanation":"Copper cannot generally displace the desired metal from its ions in this process, and filtration only separates insoluble solids. Scrap iron displacement or electrolysis is required."},{"text":"Use neutralisation or evaporation.","correct":false,"explanation":"Neutralisation changes acidity but does not extract the metal, while evaporation removes solvent without necessarily reducing metal ions to the metal. The valid methods are scrap iron displacement or electrolysis."},{"text":"Use oxidation or crystallisation.","correct":false,"explanation":"Adding oxygen does not reduce dissolved metal ions, and crystallisation produces compounds rather than extracting the metal itself. Scrap iron displacement or electrolysis gives the metal."}]},{"id":"f28de36d-8ea8-47d5-8006-6305fe568c7f","cardType":"cloze","points":null,"front":"Dynamic equilibrium occurs in a [ ] when forward and reverse reaction rates are [ ].","back":"Dynamic equilibrium occurs in a closed system when forward and reverse reaction rates are equal.","mcqOptions":[{"text":"Dynamic equilibrium occurs in a closed system when forward and reverse reaction rates are equal.","correct":true,"explanation":"A dynamic equilibrium requires a closed system so substances cannot escape, and the forward and reverse reactions must continue at equal rates."},{"text":"Dynamic equilibrium occurs in an open system when forward and reverse reaction rates are equal.","correct":false,"explanation":"An open system allows reactants or products to escape, preventing equilibrium from being maintained, even if the reaction rates briefly become equal."},{"text":"Dynamic equilibrium occurs in a closed system when forward and reverse reaction rates are zero.","correct":false,"explanation":"The reactions at equilibrium continue in both directions rather than stopping. A closed system is needed, and the forward and reverse rates are equal, not zero."},{"text":"Dynamic equilibrium occurs in a closed system when forward and reverse reaction rates are unequal.","correct":false,"explanation":"Unequal rates mean the amounts of reactants and products are still changing. Dynamic equilibrium requires a closed system and equal forward and reverse reaction rates."}]},{"id":"a78328ad-cbdf-4d8a-b734-3c82b911dbad","cardType":"basic","points":2,"front":"Why does lowering temperature increase ammonia yield in the Haber process?","back":"The forward reaction is exothermic, so equilibrium shifts right.","mcqOptions":[{"text":"The forward reaction is exothermic, so equilibrium shifts right.","correct":true,"explanation":"The forward Haber reaction is exothermic, so lowering temperature favours the forward direction and shifts equilibrium right, increasing the ammonia yield."},{"text":"The forward reaction is endothermic, so equilibrium shifts left.","correct":false,"explanation":"The forward reaction is exothermic, not endothermic. Lowering temperature therefore favours the forward reaction and shifts equilibrium right rather than left."},{"text":"The forward reaction is exothermic, so equilibrium shifts left.","correct":false,"explanation":"Lowering temperature does not shift an exothermic equilibrium left. It favours the forward exothermic reaction, shifting equilibrium right and increasing ammonia yield."},{"text":"The forward reaction is faster, so equilibrium shifts right.","correct":false,"explanation":"Temperature does not increase ammonia yield by making the forward reaction faster. Lower temperature slows reactions but shifts the exothermic equilibrium right, raising the equilibrium yield."}]},{"id":"92efb889-b567-48b0-b762-d754b2a29770","cardType":"basic","points":2,"front":"Why is $450\\text{ }^\\circ\\text{C}$ used in the Haber process?","back":"It gives a fast rate and a reasonable ammonia yield.","mcqOptions":[{"text":"It gives a fast rate and a reasonable ammonia yield.","correct":true,"explanation":"At 450 °C, the reaction rate is sufficiently fast for industrial production while the ammonia yield remains reasonable. Lower temperatures improve yield but make the reaction too slow."},{"text":"It gives a fast rate and a high ammonia yield.","correct":false,"explanation":"A high temperature gives a fast rate but lowers the equilibrium yield because the forward reaction is exothermic. The industrial compromise is a reasonable, not high, yield."},{"text":"It gives a slow rate and a reasonable ammonia yield.","correct":false,"explanation":"450 °C does not give a slow rate; it is selected to make the reaction fast enough for industry while retaining a reasonable ammonia yield."},{"text":"It gives a slow rate and a low ammonia yield.","correct":false,"explanation":"The temperature is not chosen because it gives a slow rate and low yield. It balances a sufficiently fast reaction with a reasonable equilibrium yield."}]},{"id":"c35b5038-6020-47fa-b4cd-6d2504ac2f87","cardType":"cloze","points":null,"front":"In the Haber process, higher pressure shifts equilibrium to the side with [ ].","back":"In the Haber process, higher pressure shifts equilibrium to the side with fewer gas molecules.","mcqOptions":[{"text":"fewer gas molecules","correct":true,"explanation":"The Haber reaction produces fewer gas molecules on the product side, so increasing pressure shifts equilibrium towards ammonia, the side with fewer gas molecules."},{"text":"more gas molecules","correct":false,"explanation":"Increasing pressure favours the side with fewer gas molecules, not more. In the Haber process, ammonia formation reduces the total number of gas molecules."},{"text":"the same number of gas molecules","correct":false,"explanation":"The reactant and product sides do not contain the same number of gas molecules in the Haber process. The product side has fewer gas molecules and is pressure-favoured."},{"text":"fewer liquid molecules","correct":false,"explanation":"Pressure equilibrium rules compare gas molecules, not liquid molecules. The Haber equilibrium shifts towards the side containing fewer gas molecules."}]},{"id":"b221f9dc-7632-4e56-9968-22cb429271bc","cardType":"basic","points":2,"front":"Why is extremely high pressure not used in the Haber process?","back":"It is expensive and increases explosion risk.","mcqOptions":[{"text":"It is expensive and increases explosion risk.","correct":true,"explanation":"Very high pressure would require costly, strong equipment and increases the danger of an explosion. Industry therefore uses a compromise pressure rather than the maximum possible pressure."},{"text":"It is cheap and increases explosion risk.","correct":false,"explanation":"Very high pressure is not cheap: it requires strong vessels and substantial energy. It also increases explosion risk, so this reverses the main economic problem."},{"text":"It is expensive and reduces explosion risk.","correct":false,"explanation":"High pressure does not remove explosion risk. Although it can improve yield, extremely high pressure is avoided because equipment is expensive and the danger increases."},{"text":"It is cheap and reduces explosion risk.","correct":false,"explanation":"Extremely high pressure requires expensive equipment and does not make the process safer. The increased explosion risk is one reason industry avoids using it."}]},{"id":"adb01060-f04b-4ff9-8606-2d95e1708a26","cardType":"basic","points":1,"front":"What effect does an iron catalyst have on Haber equilibrium yield?","back":"It has no effect on percentage yield.","mcqOptions":[{"text":"It has no effect on percentage yield.","correct":true,"explanation":"A catalyst speeds up both the forward and reverse reactions equally, so equilibrium is reached faster without changing its position or the percentage yield."},{"text":"It causes a higher equilibrium percentage yield.","correct":false,"explanation":"Catalysts increase reaction rates, but they do not shift equilibrium towards ammonia. The iron catalyst therefore has no effect on percentage yield."},{"text":"It causes a lower equilibrium percentage yield.","correct":false,"explanation":"Confusing faster reaction with reduced yield leads to this error. Iron accelerates both reaction directions equally, so it has no effect on percentage yield."},{"text":"It takes longer to reach percentage yield.","correct":false,"explanation":"Iron allows equilibrium to be reached more quickly, not more slowly. Because it accelerates both directions equally, it has no effect on percentage yield."}]},{"id":"cc96cba5-6d19-4f85-9bb9-6097892121ce","cardType":"basic","points":2,"front":"Why are alloys harder than pure metals?","back":"Different-sized atoms disrupt layers, stopping them sliding easily.","mcqOptions":[{"text":"Different-sized atoms disrupt layers, stopping them sliding easily.","correct":true,"explanation":"Different-sized atoms distort the regular arrangement of metal atoms. This prevents layers from sliding over one another easily, making the alloy harder than the pure metal."},{"text":"Different-sized atoms form layers, allowing them to slide easily.","correct":false,"explanation":"Alloys do not contain perfectly aligned layers because their atoms have different sizes. These atoms disrupt the layers and stop them sliding easily."},{"text":"Different-sized atoms speed up layers, making them slide more easily.","correct":false,"explanation":"The greater hardness is not caused by atoms sliding more quickly. Different-sized atoms disrupt the layers, making it more difficult for them to slide."},{"text":"Same-sized atoms maintain layers, stopping them sliding easily.","correct":false,"explanation":"Layers sliding easily is characteristic of softer pure metals, not harder alloys. Different-sized atoms disrupt the layers and prevent this movement."}]},{"id":"4837f2ba-a61f-445f-803a-0d67e64a26b6","cardType":"cloze","points":null,"front":"Iron only rusts when both [ ] and [ ] are present.","back":"Iron only rusts when both oxygen and water are present.","mcqOptions":[{"text":"Iron only rusts when both oxygen and water are present.","correct":true,"explanation":"Rusting is an oxidation process that requires oxygen and water together. Removing either substance prevents iron from forming hydrated iron(III) oxide."},{"text":"Iron only rusts when both oxygen and carbon dioxide are present.","correct":false,"explanation":"Carbon dioxide can affect acidity but is not an essential condition for rusting. Iron requires both oxygen and water to rust."},{"text":"Iron only rusts when both nitrogen and water are present.","correct":false,"explanation":"Nitrogen is abundant in air but does not cause rusting under ordinary conditions. The two essential substances are oxygen and water."},{"text":"Iron only rusts when both hydrogen and carbon dioxide are present.","correct":false,"explanation":"Hydrogen and carbon dioxide are not the required pair for rusting. Iron rusts only when both oxygen and water are available."}]},{"id":"9f2c1d10-7d53-4066-8df2-a2627633ec8b","cardType":"basic","points":2,"front":"How does sacrificial protection stop iron corroding?","back":"A more reactive metal oxidises first, protecting the iron.","mcqOptions":[{"text":"A more reactive metal oxidises first, protecting the iron.","correct":true,"explanation":"The attached metal is more reactive than iron, so it loses electrons and is oxidised preferentially. This prevents the iron from being oxidised."},{"text":"A less reactive metal oxidises first, protecting the iron.","correct":false,"explanation":"A less reactive metal will not be preferentially oxidised instead of iron. Sacrificial protection requires a more reactive metal to oxidise first."},{"text":"A more reactive metal reduces first, protecting the iron.","correct":false,"explanation":"Sacrificial protection relies on oxidation, not reduction, of the attached metal. A more reactive metal loses electrons first and protects the iron from oxidation."},{"text":"The exposed iron oxidises first, protecting the reactive metal.","correct":false,"explanation":"Iron must not oxidise first because its oxidation is the corrosion being prevented. The more reactive attached metal oxidises instead of the iron."}]},{"id":"d080f4f1-956c-490e-a814-b6d01e8dbab1","cardType":"basic","points":2,"front":"Why does galvanising still protect iron if scratched?","back":"Remaining zinc sacrifices itself to protect exposed iron.","mcqOptions":[{"text":"Remaining zinc sacrifices itself to protect exposed iron.","correct":true,"explanation":"Zinc is more reactive than iron, so remaining zinc is oxidised preferentially even when the coating is damaged. This protects the exposed iron sacrificially."},{"text":"Remaining zinc forms a barrier to protect exposed iron.","correct":false,"explanation":"A scratch means the coating no longer forms a complete physical barrier. Protection continues because zinc sacrifices itself by oxidising instead of the exposed iron."},{"text":"Exposed iron sacrifices itself to protect remaining zinc.","correct":false,"explanation":"Iron oxidising is corrosion, so this would not protect it. The more reactive zinc oxidises sacrificially and supplies electrons to the exposed iron."},{"text":"Remaining zinc seals the scratch to protect exposed iron.","correct":false,"explanation":"Zinc does not protect a scratch by physically repairing the coating. Remaining zinc is more reactive and sacrifices itself to prevent oxidation of exposed iron."}]},{"id":"ab46d1ae-f453-407d-ac8d-1c8cf8f6b480","cardType":"basic","points":1,"front":"What formula gives strength-to-density ratio?","back":"$$\\text{Ratio}=\\frac{\\text{Tensile Strength}}{\\text{Density}}$","mcqOptions":[{"text":"$$\\text{Ratio}=\\frac{\\text{Tensile Strength}}{\\text{Density}}$","correct":true,"explanation":"Strength-to-density ratio compares a material’s tensile strength with its mass per unit volume. It is calculated by dividing tensile strength by density."},{"text":"$$\\text{Ratio}=\\frac{\\text{Density}}{\\text{Tensile Strength}}$","correct":false,"explanation":"This reverses the required division and calculates density per unit tensile strength. Strength-to-density ratio is tensile strength divided by density."},{"text":"$$\\text{Ratio}=\\text{Tensile Strength}\\times\\text{Density}$","correct":false,"explanation":"Multiplying the quantities does not compare strength relative to mass per unit volume. The required ratio is tensile strength divided by density."},{"text":"$$\\text{Ratio}=\\text{Tensile Strength}+\\text{Density}$","correct":false,"explanation":"Adding the two quantities is invalid because they measure different properties and have different units. Strength-to-density ratio divides tensile strength by density."}]},{"id":"6f46d8ef-24c5-4f3d-a3f1-e47c2425c77e","cardType":"basic","points":1,"front":"Which example material has the highest strength-to-density ratio for a bicycle frame?","back":"Carbon-fibre composite.","mcqOptions":[{"text":"Carbon-fibre composite.","correct":true,"explanation":"Carbon-fibre composite combines very high tensile strength with low density, giving it the highest strength-to-density ratio among the example bicycle-frame materials."},{"text":"High-carbon steel.","correct":false,"explanation":"Steel has high tensile strength but also has a high density, reducing its strength-to-density ratio. Carbon-fibre composite provides the highest ratio."},{"text":"Aluminium alloy.","correct":false,"explanation":"Aluminium alloy has a relatively low density, but its tensile strength is insufficient to exceed the ratio of carbon-fibre composite in this comparison."},{"text":"Titanium alloy.","correct":false,"explanation":"Titanium alloy has a strong strength-to-density performance, but it does not exceed the example value for carbon-fibre composite, which has the highest ratio."}]},{"id":"7258b72a-4a13-4595-84f6-a46985eb7c2d","cardType":"basic","points":1,"front":"What does an LCA assess across a product’s lifespan?","back":"Its environmental impact.","mcqOptions":[{"text":"Its environmental impact.","correct":true,"explanation":"A life cycle assessment evaluates environmental effects from obtaining raw materials through manufacture and use to disposal, recycling or reuse."},{"text":"Its total financial cost.","correct":false,"explanation":"Financial cost may influence product decisions, but it is not the focus of an LCA. An LCA assesses environmental impact throughout the product’s lifespan."},{"text":"Its commercial profitability.","correct":false,"explanation":"Commercial profitability is a business measure rather than a life cycle assessment measure. An LCA considers the product’s environmental impact from raw materials to disposal."},{"text":"Its manufacturing speed.","correct":false,"explanation":"Manufacturing speed covers only one stage and does not measure environmental effects. An LCA assesses environmental impact across the product’s entire lifespan."}]},{"id":"fa6bba9b-e2fb-4b49-8341-e7d1d8a57147","cardType":"basic","points":1,"front":"Why can recycling reduce environmental impact?","back":"It uses less energy than extracting virgin materials.","mcqOptions":[{"text":"It uses less energy than extracting virgin materials.","correct":true,"explanation":"Reprocessing existing materials generally requires less energy than extracting and processing virgin resources, reducing fossil fuel use and associated greenhouse gas emissions."},{"text":"It makes all waste biodegradable after processing.","correct":false,"explanation":"Recycling does not make waste naturally decomposable; biodegradability depends on a material’s chemical structure. Its environmental benefit comes from using less energy than extracting virgin materials."},{"text":"It produces no greenhouse gases during processing.","correct":false,"explanation":"Recycling still requires energy for collection, sorting and reprocessing, so its emissions are not eliminated. It reduces impact because this usually uses less energy than virgin material extraction."},{"text":"It removes all need for extracting virgin materials.","correct":false,"explanation":"Some virgin resources remain necessary because materials can be contaminated or degrade after repeated recycling. Recycling reduces impact mainly by using less energy than extracting virgin materials."}]}]}],"examCountdown":{"examDate":"2027-05-17","paperLabel":"Paper 1"},"heading":"Improving processes and products","title":"Improving processes and products Flashcards","topbar":{"backHref":"/gcse/chemistry/ocr/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"f888cc15-6319-535c-8206-b87fc465d259","walled":"$L39"}],"$L3a","$L3b"]