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Question 1 -- Bonding, Structure and the Odd One Out
Total: 12 marks
A research group in Zurich tabulates data for four elements in order to test how well the simple bonding models used at school actually predict physical behaviour.
element
melting point in °C
electrical conductivity of the solid
behaviour when hammered
sodium
98
very high
flattens easily
magnesium
650
very high
flattens
aluminium
660
very high
flattens
sulfur
115
negligible
shatters to powder
(a)[4]
(i) Sodium and sulfur have melting points that differ by only 17 °C, yet their bonding is completely different. Explain, in terms of what is being overcome on melting, why these two similar values do not indicate similar bond strengths. [3]
(ii) State the one piece of data in the table that most clearly separates the metals from the non-metal, and justify your choice. [1]
Model Answer -- 1(a)
(i) In sodium, melting means partially overcoming the metallic bonding, that is the attraction between the lattice of Na⁺ ions and the sea of delocalised electrons; each ion contributes only one electron and the ion is large, so this attraction is comparatively weak for a metal [1]
(i) In sulfur, melting overcomes only the weak intermolecular forces between separate S₈ molecules; the strong covalent bonds inside each ring are not broken at all [1]
(i) The two melting points are therefore measuring completely different things, so a similar temperature does not imply similar bond strength; comparing the boiling points, 883 °C for sodium against 445 °C for sulfur, shows the true difference more fairly [1]
(ii) The electrical conductivity of the solid, because only a metallic lattice contains delocalised electrons that are free to move while the substance is still solid [1]
⚠ If you missed marks here: The third mark is the discriminating one and is nearly always lost: you must say explicitly that the two melting points refer to different kinds of attraction, so the comparison is not a valid one. Writing that "covalent bonds break when sulfur melts" contradicts the second mark and cancels it.
(b)[4]
(i) Explain, using the bonding model, why the melting point rises steeply from sodium to magnesium but hardly at all from magnesium to aluminium. [3]
(ii) Predict, with a reason, whether aluminium or sodium would be the better conductor of heat. [1]
Model Answer -- 1(b)
(i) Going from sodium to magnesium the ionic charge doubles from 1⁺ to 2⁺ and each atom contributes two electrons instead of one, while the ion also becomes smaller, so the electrostatic attraction is far stronger and much more energy is needed to disrupt the lattice [1]
(i) From magnesium to aluminium the charge rises again to 3⁺ and three electrons are contributed, so a further large rise might be expected [1]
(i) In practice the melting point rises by only 10 °C, which shows that the simple charge argument is not the whole story; melting only loosens the lattice rather than separating the ions completely, and other factors such as the packing of the ions in the liquid also matter [1]
(ii) Aluminium, because each atom releases three electrons into the delocalised sea instead of one, so there are more mobile electrons per unit volume to carry energy through the metal [1]
⚠ If you missed marks here: This part rewards candidates who notice that the data do not fit the expected pattern and say so. An answer that simply asserts "aluminium has a higher charge so it melts higher" ignores the fact that the rise is almost nothing and cannot gain the third mark.
(c)[4]
A fifth sample is found to conduct electricity when solid, to be hammered flat without cracking, and to melt at 232 °C. When a piece is dropped into dilute hydrochloric acid a slow stream of bubbles is seen, and when it is left in copper(II) sulfate solution a pink-brown deposit forms.
(i) Deduce, giving all your reasoning, whether the sample is a metal, a non-metal or an alloy. [2]
(ii) Deduce the range of positions this sample could occupy in the reactivity series, and state the evidence for each limit. [2]
Model Answer -- 1(c)
(i) It must be metallic in structure, because it conducts as a solid and is malleable, which together require a lattice of positive ions with delocalised electrons [1]
(i) The data do not allow a pure metal to be distinguished from an alloy; a sharp melting point would suggest a pure element, whereas a melting range would indicate a mixture, so further information is needed [1]
(ii) It must lie above hydrogen, because it releases hydrogen from dilute hydrochloric acid [1]
(ii) It must lie above copper, because it displaces copper from copper(II) sulfate solution; since it reacts only slowly with acid it is likely to be low in the series, somewhere in the region of iron or tin, but the data fix only the two limits [1]
⚠ If you missed marks here: The second mark rewards intellectual honesty: the evidence given genuinely cannot separate a pure metal from an alloy, and saying so is worth more than guessing. In (ii) note that displacing copper and releasing hydrogen give the same lower limit, so you need both statements for the two marks.
Question 2 -- Building a Reactivity Series from Scratch
Total: 12 marks
A student in Lagos is given three unlabelled metals, J, K and L, together with magnesium and copper as reference metals. All five are tested under identical conditions.
metal
cold water
steam
dilute hydrochloric acid
added to copper(II) sulfate
J
no change
slow reaction, red heat needed
a few bubbles per second
brown coating forms
K
no change
no change
no change
no change
L
very slow bubbles after 10 min
vigorous
very rapid, tube gets hot
brown coating forms rapidly
magnesium
very slow bubbles after 10 min
vigorous
very rapid, tube gets hot
brown coating forms rapidly
copper
no change
no change
no change
no change
(a)[5]
(i) Place J, K and L in order of reactivity, most reactive first, and state where magnesium and copper fit into your order. [2]
(ii) Explain why the results for L and for magnesium do not prove that L is magnesium. [2]
(iii) Suggest one further test that could distinguish L from magnesium. [1]
Model Answer -- 2(a)
(i) L is the most reactive, then J, then K [1]
(i) Magnesium behaves identically to L and so occupies the same position in the order, while copper behaves identically to K and so occupies the lowest position [1]
(ii) Identical behaviour in these four tests shows only that L and magnesium occupy the same region of the reactivity series; the tests are not sensitive enough to separate metals that are close together [1]
(ii) Two different metals can react at very similar rates, so the results are consistent with L being magnesium but do not establish it; identification needs a property unique to the element [1]
(iii) Any one of: measure the density; measure the melting point; carry out a flame test; or find the mass of hydrogen released per gram of metal, which depends on the relative atomic mass [1]
⚠ If you missed marks here: Part (ii) is testing the difference between "consistent with" and "proves", which is a scientific reasoning skill rather than a recall point. In (iii) the suggested test must give a number or observation characteristic of one particular element; repeating a reactivity test more carefully will not do.
(b)[4]
(i) J forms ions with a charge of 2⁺. Write the ionic equation and the two half-equations for the reaction of J with copper(II) sulfate solution. [2]
(ii) Explain, in terms of electron loss, why L reacts faster than J with dilute hydrochloric acid. [2]
Model Answer -- 2(b)
(i) J(s) + Cu²⁺(aq) → J²⁺(aq) + Cu(s), with the sulfate ion omitted as a spectator [1]
(i) J → J²⁺ + 2e⁻ (oxidation) and Cu²⁺ + 2e⁻ → Cu (reduction) [1]
(ii) The position of a metal in the reactivity series measures how readily its atoms lose electrons to form positive ions [1]
(ii) L loses electrons more readily than J, so its atoms are converted to ions faster when they meet H⁺ ions from the acid, and hydrogen is therefore released at a greater rate [1]
⚠ If you missed marks here: Both half-equations are needed for the single mark, and each must show the electrons on the correct side. In (ii) the explanation must be in terms of electrons; saying "L is higher in the series" restates the observation rather than explaining it.
(c)[3]
(i) Predict the method that would be used to extract each of J, K and L from its ore, and justify each prediction. [2]
(ii) When the student repeats the acid test on J using a strip that has been left in a drawer for a month, the bubbles take almost a minute to appear. Diagnose the cause and state what should be done. [1]
Model Answer -- 2(c)
(i) J lies below carbon in the series, since it reacts only slowly with acid and needs red heat with steam, so its oxide could be reduced by heating with carbon or carbon monoxide [1]
(i) L behaves like magnesium and so lies above carbon, meaning electrolysis of the molten compound would be required; K is so unreactive that it may well occur uncombined and simply need physical separation [1]
(ii) The strip has tarnished, so a layer of oxide covers the surface and must be attacked by the acid before the metal beneath is exposed; the strip should be cleaned with sandpaper or emery paper before the test [1]
⚠ If you missed marks here: The boundary that decides the extraction method is carbon, so you must place each metal relative to carbon rather than relative to hydrogen. In (ii) a diagnosis of "the acid was old" does not fit the evidence, because only one variable, the age of the metal strip, has changed.
Question 3 -- A Blast Furnace Mass Balance
Total: 12 marks
A works at Port Talbot charges its blast furnace with 2500 tonnes of ore each day. The ore contains 72.0% hematite, Fe₂O₃, by mass, the remainder being silicon dioxide and other rock. Only 94.0% of the iron present in the hematite is recovered as molten iron. (Aᵣ: Fe = 56, O = 16, C = 12)
(a)[4]
(i) Calculate the mass of hematite in the daily charge. [1]
(ii) Calculate the relative formula mass of Fe₂O₃ and the percentage of iron in it. [1]
(iii) Calculate the theoretical mass of iron in the daily charge. [1]
(iv) Calculate the mass of iron actually obtained each day. [1]
Model Answer -- 3(a)
(i) mass of hematite = 72.0/100 × 2500 = 1800 tonnes [1]
(ii) Mᵣ = (2 × 56) + (3 × 16) = 160, and the iron accounts for 112/160 × 100 = 70.0% [1]
(iii) theoretical mass of iron = 70.0/100 × 1800 = 1260 tonnes [1]
(iv) actual mass = 94.0/100 × 1260 = 1184 tonnes, which is about 1180 tonnes to three significant figures [1]
⚠ If you missed marks here: Work in tonnes throughout; converting to grams adds nothing but risks a factor-of-a-million error. Apply the three percentages one after another in the right order — multiplying 2500 by 0.72, 0.70 and 0.94 in a single line is fine, but only if all three factors appear.
(b)[4]
(i) Write the balanced equation for the reduction of hematite by carbon monoxide. [1]
(ii) Calculate the amount, in tonne-moles, of Fe₂O₃ in the daily charge, and hence the mass of carbon monoxide needed to reduce it completely. [2]
(iii) Calculate the mass of carbon dioxide released by this reduction alone. [1]
Model Answer -- 3(b)
(i) Fe₂O₃ + 3CO → 2Fe + 3CO₂ [1]
(ii) amount of Fe₂O₃ = 1800 / 160 = 11.25 tonne-moles, so the carbon monoxide required is 3 × 11.25 = 33.75 tonne-moles [1]
(ii) mass of CO = 33.75 × 28 = 945 tonnes [1]
(iii) mass of CO₂ = 33.75 × 44 = 1485 tonnes, about 1490 tonnes [1]
⚠ If you missed marks here: The coefficient 3 must be applied to both gases; forgetting it gives answers exactly one third of the correct values. Note that the mass of carbon dioxide is larger than the mass of carbon monoxide even though the number of moles is the same, because each molecule has gained an oxygen atom.
(c)[4]
(i) Suggest two reasons why only 94.0% of the iron is recovered. [2]
(ii) The remaining 28% of the ore is mainly silicon dioxide. Explain what happens to it in the furnace and write the equation. [2]
Model Answer -- 3(c)
(i) Some iron is not fully reduced before the charge reaches the bottom, because the gas and the solid are in contact for a limited time [1]
(i) Some iron is carried away dissolved or trapped in the slag, and a little is lost as dust in the waste gases or left in the furnace lining [1]
(ii) Limestone in the charge decomposes to calcium oxide, a basic oxide, which then reacts with the acidic silicon dioxide to form molten calcium silicate; this slag is less dense than iron so it floats and is tapped off separately [1]
(ii) CaO + SiO₂ → CaSiO₃ [1]
⚠ If you missed marks here: "Human error" and "the reaction is reversible" are not credited here; the losses in an industrial furnace are physical and kinetic. In (ii) remember that the limestone must decompose first, so an equation showing CaCO₃ reacting directly with SiO₂ will not gain the second mark.
Question 4 -- Designing an Alloy
Total: 12 marks
A metallurgist in Pune is asked to choose a material for the turbine blades of a small water pump. The blades must be hard, must not corrode in river water, and must be as light as is practicable.
material
composition
relative hardness
density in g/cm³
behaviour in river water
pure copper
100% Cu
35
8.96
very slight green tarnish
brass
65% Cu, 35% Zn
90
8.47
slight tarnish
mild steel
99.75% Fe, 0.25% C
130
7.85
rusts within days
stainless steel
72% Fe, 18% Cr, 8% Ni, 2% C
190
7.80
no visible change
duralumin
94% Al, 4% Cu, 2% Mg
120
2.80
no visible change
(a)[4]
(i) Using only the data, recommend one material for the blades and justify your recommendation against all three requirements. [3]
(ii) State one piece of information not given in the table that the metallurgist would still need before making a final decision. [1]
Model Answer -- 4(a)
(i) Duralumin is the best compromise: its hardness of 120 is well above that of pure copper and brass and close to that of mild steel [1]
(i) It shows no visible change in river water, so it meets the corrosion requirement as well as stainless steel does, because of the protective aluminium oxide layer [1]
(i) Its density of 2.80 g/cm³ is less than a third of that of any other material listed, so the blades are far lighter; stainless steel is harder but nearly three times as dense (accept a reasoned choice of stainless steel if hardness is argued to matter most) [1]
(ii) Any one of: the cost per kilogram; the strength or resistance to fracture rather than hardness alone; how easily the material can be cast or machined into blade shapes; or how it behaves at the working temperature [1]
⚠ If you missed marks here: An evaluation question needs every stated requirement to be addressed with a figure from the table; naming a material and giving one reason cannot reach three marks. Note that hardness and strength are not the same property, which is why (ii) can be answered in several ways.
(b)[4]
(i) Explain, in terms of structure, why duralumin with only 6% of added elements is more than three times as hard as pure aluminium. [2]
(ii) Mild steel contains only 0.25% carbon, yet its hardness is nearly four times that of pure iron. Explain why such a small proportion has so large an effect. [2]
Model Answer -- 4(b)
(i) Copper and magnesium ions are a different size from aluminium ions, so when they take up positions in the lattice the close-packed layers are no longer flat and regular [1]
(i) The distorted layers catch on one another and can no longer slide when a force is applied, so a much larger force is needed to deform the metal and it is far harder [1]
(ii) Carbon atoms are very much smaller than iron atoms and fit into the gaps between them rather than replacing them, so even a tiny proportion by mass corresponds to a large number of atoms [1]
(ii) Each interstitial atom pins the layers where it sits, and only a scattering of such obstacles is needed to prevent the whole plane from slipping, which is why the effect is out of proportion to the mass added [1]
⚠ If you missed marks here: Part (ii) asks specifically why a very small percentage matters so much, so an answer that only repeats the general "different sizes stop the layers sliding" argument earns at most one mark. The key ideas are the small mass but large number of carbon atoms, and the fact that a few obstacles are enough to block a whole plane.
(c)[4]
A single blade has a volume of 45.0 cm³.
(i) Calculate the mass of one blade if it is made from duralumin, and the mass if it is made from stainless steel. [2]
(ii) The pump has eight blades. Calculate the total mass saved by using duralumin. [1]
(iii) Calculate the mass of copper in one duralumin blade. [1]
Model Answer -- 4(c)
(i) duralumin blade: mass = 45.0 × 2.80 = 126 g [1]
(i) stainless steel blade: mass = 45.0 × 7.80 = 351 g [1]
(ii) saving per blade = 351 − 126 = 225 g, so for eight blades the saving is 8 × 225 = 1800 g, that is 1.80 kg [1]
(iii) mass of copper = 4/100 × 126 = 5.04 g [1]
⚠ If you missed marks here: In (ii) the saving must be multiplied by eight; finding the difference for a single blade and stopping there is the commonest slip. In (iii) apply the percentage to the mass of the duralumin blade, not to the volume and not to the stainless steel figure.
Question 5 -- Investigating Rusting
Total: 10 marks
A student in Cardiff is asked to design an investigation to find out whether salt speeds up the rusting of iron, and to prove at the same time that both water and oxygen are needed.
(a)[4]
(i) Explain which pair of tubes tests whether oxygen is needed, and which pair tests whether water is needed. [2]
(ii) Explain which pair of tubes tests the effect of salt, and state why the other two tubes cannot be used for that comparison. [2]
Model Answer -- 5(a)
(i) Tubes 1 and 3 test oxygen, because both contain liquid water but only tube 3 has air; boiling drives the dissolved air out of tube 1 and the oil seal keeps more from dissolving [1]
(i) Tubes 2 and 3 test water, because both are open to air but the drying agent removes water vapour from tube 2 [1]
(ii) Tubes 3 and 4 test the effect of salt, because they differ only in whether the water contains dissolved sodium chloride [1]
(ii) Tubes 1 and 2 cannot be used for that comparison because each is already missing one of the substances needed for rusting, so no rusting occurs in either and there is nothing to compare [1]
⚠ If you missed marks here: Each comparison must differ in exactly one variable, so name the pair and say what the single difference is. A common error is to pair tube 1 with tube 2, which differ in two ways at once and therefore prove nothing on their own.
(b)[3]
After a week the student records: tube 1 no rust, tube 2 a light dusting of rust, tube 3 moderate rust, tube 4 heavy rust.
(i) State which result is anomalous and explain why. [2]
(ii) Suggest one practical cause of this anomalous result. [1]
Model Answer -- 5(b)
(i) The result for tube 2 is anomalous, because that tube was supposed to contain no water at all [1]
(i) Rusting requires both water and oxygen, so if the drying had been complete no rust whatever should have appeared in tube 2 [1]
(ii) Any one of: the calcium chloride had already absorbed moisture before use and was no longer effective; the bung did not seal so damp air leaked in; or the nail was not completely dry when it was placed in the tube [1]
⚠ If you missed marks here: An anomalous result is one that contradicts the expected pattern, not simply the smallest or the largest reading; tube 4 having the most rust is exactly what the theory predicts. The suggested cause in (ii) must be a specific practical failure, not "an error was made".
(c)[3]
(i) Explain why the tube containing salt solution rusts fastest. [1]
(ii) Suggest two improvements that would make the investigation give quantitative rather than merely descriptive results. [2]
Model Answer -- 5(c)
(i) Dissolved salt makes the water a much better conductor, so the electrons released by the iron can travel more easily to where oxygen is being reduced, and the whole process speeds up [1]
(ii) Weigh each cleaned nail before the experiment and reweigh it after the rust has been removed, so that the loss in mass can be compared [1]
(ii) Repeat each tube several times and average the results, use a fixed range of salt concentrations, and keep all the tubes at the same temperature for exactly the same length of time [1]
⚠ If you missed marks here: Quantitative means producing numbers, so "look more carefully" or "use a better camera" do not qualify. Weighing is the standard route to a numerical measure of how much iron has corroded.
Question 6 -- Protecting a Steel Pipeline
Total: 12 marks
A water authority in Gujarat must protect a buried steel pipeline 4.0 km long. Three options are considered: painting with bitumen, galvanising, and bolting magnesium anodes to the pipe at intervals and connecting them with copper wire.
(a)[4]
(i) Write the half-equation for the corrosion of iron in wet soil, and the half-equation for the magnesium that protects it. [2]
(ii) Explain, using these two half-equations, why the magnesium anodes prevent the pipeline from rusting. [2]
Model Answer -- 6(a)
(i) Fe → Fe²⁺ + 2e⁻ [1]
(i) Mg → Mg²⁺ + 2e⁻ [1]
(ii) Magnesium is much higher in the reactivity series, so it loses electrons far more readily than iron and is oxidised in preference to it [1]
(ii) The electrons released travel along the copper wire into the pipeline, so the iron atoms are held in a condition where they cannot lose electrons; the first half-equation is therefore prevented from occurring and no Fe²⁺ is formed [1]
⚠ If you missed marks here: The examiner is looking for the phrase "in preference to" or an equivalent, plus the direction of electron flow from magnesium to iron. Answers that describe the magnesium as a barrier miss the point entirely, since the anodes touch only a small part of the pipe.
(b)[4]
(i) Compare the three options, giving one advantage and one disadvantage of each. [3]
(ii) Recommend one option for a pipeline that will be buried for fifty years and cannot easily be dug up, and justify your choice. [1]
Model Answer -- 6(b)
(i) Painting is cheap and easy to apply, but it is a barrier only, so any scratch made while the pipe is being laid allows rusting to begin and spread beneath the coating [1]
(i) Galvanising gives both a barrier and sacrificial protection, so a scratch does not matter, but the zinc layer is thin and is eventually consumed, and the pipe would have to be dipped during manufacture [1]
(i) Magnesium anodes give strong sacrificial protection over long distances and can be sized to last for decades, but they are consumed and must be replaced, and installing and wiring them is expensive [1]
(ii) Galvanising combined with anodes, or anodes alone, is the best choice, because protection must continue after inevitable damage to any coating and the pipe cannot be inspected or repainted once buried (a reasoned choice of either is credited) [1]
⚠ If you missed marks here: Each of the three marks requires both an advantage and a disadvantage for that option; a one-sided comment scores nothing. In (ii) the justification must refer to the specific situation described, namely fifty years buried and inaccessible.
(c)[4]
Each magnesium anode has a mass of 12.0 kg and is expected to lose 0.240 kg per year. (Aᵣ: Mg = 24)
(i) Calculate how many years one anode would last if 90.0% of its mass could be used. [2]
(ii) Calculate the amount, in moles, of magnesium lost by one anode in a single year. [1]
(iii) State, with a reason, what would happen to the pipeline if copper anodes were used instead of magnesium. [1]
Model Answer -- 6(c)
(i) usable mass = 90.0/100 × 12.0 = 10.8 kg [1]
(i) lifetime = 10.8 / 0.240 = 45.0 years [1]
(ii) 0.240 kg = 240 g, so amount = 240 / 24 = 10.0 mol [1]
(iii) The pipeline would rust faster than if it had been left bare, because copper is below iron in the reactivity series, so the iron would give up its electrons to the copper and would be oxidised preferentially [1]
⚠ If you missed marks here: In (ii) convert kilograms to grams before dividing by the relative atomic mass, or the answer will be a thousand times too small. In (iii) note that the answer is not simply "no protection" — the corrosion is actively made worse.
Question 7 -- The True Cost of Aluminium
Total: 10 marks
A smelter produces 540 tonnes of aluminium each day by the electrolysis of aluminium oxide dissolved in molten cryolite. All the oxygen released at the anodes reacts with the hot carbon of the electrodes. (Aᵣ: Al = 27, O = 16, C = 12)
(a)[3]
(i) Write the two ionic half-equations and the overall equation for the decomposition of aluminium oxide. [2]
(ii) Explain why the cell must be run at about 950 °C rather than at the melting point of pure aluminium oxide. [1]
(ii) Aluminium oxide alone melts above 2000 °C, and heating a cell of that size to such a temperature would consume enormous amounts of energy; dissolving the oxide in molten cryolite gives a conducting liquid at about 950 °C and so makes the process economic [1]
⚠ If you missed marks here: To obtain the overall equation you must multiply the cathode half-equation by four and the anode half-equation by three so that twelve electrons cancel; simply adding the two as written leaves electrons in the answer.
(b)[4]
(i) Using the overall equation and C + O₂ → CO₂, show that 3 mol of carbon are consumed for every 4 mol of aluminium produced. [1]
(ii) Calculate the mass of carbon consumed for every tonne of aluminium produced. [2]
(iii) Calculate the mass of carbon burned away at the anodes in one day. [1]
Model Answer -- 7(b)
(i) The overall equation gives 3 mol of O₂ for every 4 mol of Al, and each mole of O₂ consumes 1 mol of carbon, so 3 mol of carbon are used for every 4 mol of aluminium [1]
(ii) 4 mol of Al has a mass of 4 × 27 = 108 and 3 mol of C has a mass of 3 × 12 = 36 [1]
(ii) mass of carbon per tonne of aluminium = 36/108 × 1 = 0.333 tonnes [1]
⚠ If you missed marks here: The chain here has three links — oxide to oxygen, oxygen to carbon, and moles to mass — and dropping any one of them gives a plausible but wrong figure. Keep the ratio 36 : 108 rather than converting to a decimal too early, since rounding 0.333 at an intermediate stage introduces error.
(c)[3]
(i) Calculate the mass of carbon dioxide released each day by the burning of the anodes. [1]
(ii) Recycling aluminium uses about 5% of the energy of extraction and produces no anode carbon dioxide at all. Evaluate, using your figures, the case for recycling rather than extracting. [2]
Model Answer -- 7(c)
(i) 180 tonnes of carbon corresponds to 180/12 = 15.0 tonne-moles, and each gives one CO₂, so the mass is 15.0 × 44 = 660 tonnes per day [1]
(ii) Recycling avoids all 660 tonnes of anode carbon dioxide each day and uses only about a twentieth of the electricity, so both the fuel bill and the greenhouse gas emissions fall sharply; the metal recovered is chemically identical to newly extracted metal [1]
(ii) Against this, scrap must be collected, sorted and transported, alloys of different composition must be separated, and there is simply not enough scrap in circulation to meet total demand, so some extraction will still be needed [1]
⚠ If you missed marks here: The word "evaluate" requires arguments on both sides, so a purely enthusiastic answer about recycling can gain only one of the two marks. Quote your own calculated figure for the carbon dioxide, since the question says to use it.
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