0620/31 Oct/Nov 2025 Worked Answers (IGCSE Chemistry 0620 Core)

58 questions · 80 marks · 75 minutes

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1. Question 1(a)

   1 marksElement giving a red flame colour
   Step 1: Flame tests give characteristic colours for different metal ions. Step 2: Lithium ions give a red flame colour, so the symbol is Li. Step 3: Potassium gives lilac and copper gives blue-green, and sodium is not in the available list, so the answer is Li.
   Method:

   Match each metal ion to its flame colour and pick the one giving red.
   Examiner tips

   • Lithium red, sodium yellow, potassium lilac, copper blue-green.

2. Question 1(b)

   1 marksElement present in diamond
   Step 1: Diamond is one of the giant covalent structures formed by carbon, in which each carbon atom is bonded to four others. Step 2: So the element present in diamond is carbon, whose symbol is C. Step 3: Silicon forms silicon(IV) oxide structures, while aluminium and sulfur are not part of diamond, so the element present in diamond is carbon.
   Method:

   Recall which element makes up diamond.
   Examiner tips

   • Diamond and graphite are both giant covalent forms of carbon.

3. Question 1(c)

   1 marksElement forming a 2- ion
   Step 1: Sulfur is in Group VI, so its atom has 6 outer electrons and gains 2 electrons to reach a full outer shell. Step 2: Gaining 2 electrons gives the sulfide ion a 2- charge, so the symbol is S. Step 3: Chlorine forms a 1- ion, nitrogen a 3- ion and magnesium a 2+ ion, so only S forms a 2- ion.
   Method:

   Use the group number to work out the charge of the ion each element forms.
   Examiner tips

   • Group VI non-metals gain 2 electrons to form 2- ions.

4. Question 1(d)

   1 marksMetal extracted from bauxite
   Step 1: Bauxite is the main ore of aluminium. Step 2: Aluminium is extracted from bauxite by electrolysis, so the symbol is Al. Step 3: Iron comes from haematite and copper from copper ores, so the element extracted from bauxite is Al.
   Method:

   Match each metal to the ore it is extracted from.
   Examiner tips

   • Bauxite -> aluminium; haematite -> iron.

5. Question 1(e)

   1 marksMetal used in electrical wiring
   Step 1: Electrical wiring needs a metal that conducts electricity well and can be drawn into wires. Step 2: Copper is an excellent electrical conductor and is ductile, so the symbol is Cu. Step 3: Sulfur is a non-metal, potassium is too reactive and titanium is used for strength, so the answer is Cu.
   Method:

   Pick the metal that conducts electricity well and can be drawn into wires.
   Examiner tips

   • Copper: good conductor + ductile = ideal for electrical wiring.

6. Question 2(a)

   2 marksProtons and neutrons in an argon atom
   Step 1: The proton number (18) equals the number of protons, so there are 18 protons. Step 2: The number of neutrons = mass number - proton number = 40 - 18 = 22. Step 3: So one atom has 18 protons and 22 neutrons.
   Method:

   Read the proton number for protons and subtract it from the mass number for neutrons.
   Examiner tips

   • Protons = proton number; neutrons = mass number - proton number.

7. Question 2(b)(i)

   1 marksMeaning of monatomic
   Step 1: The prefix mon- means one, so monatomic means one atom. Step 2: A monatomic element exists as single, separate atoms that are not joined to others. Step 3: Diatomic means two atoms joined, so monatomic means existing as single atoms.
   Method:

   Break down the word monatomic and describe the particles present.
   Examiner tips

   • Noble gases such as argon are monatomic — they exist as single atoms.

8. Question 2(b)(ii)

   1 marksWhy argon is unreactive
   Step 1: Atoms react to gain a full (stable) outer shell of electrons. Step 2: Argon already has a full outer shell, so it has no tendency to gain, lose or share electrons. Step 3: Having a full outer shell makes argon stable and therefore unreactive.
   Method:

   Link the full outer shell to having no tendency to react.
   Examiner tips

   • Noble gases are unreactive because their outer shells are full.

9. Question 2(c)

   2 marksEffect of temperature and pressure on gas volume
   Step 1: Heating a gas gives its particles more energy so they move faster and spread out, pushing the plunger out, so the volume increases. Step 2: Increasing the pressure pushes the particles closer together, so the volume decreases. Step 3: So volume increases on heating and decreases when pressure is raised.
   Method:

   Use particle theory to predict how volume responds to temperature and pressure.
   Examiner tips

   • Heat -> bigger volume; squeeze (more pressure) -> smaller volume.

10. Question 3(a)(i)

    1 marksMost concentrated positive ion in river water
    Step 1: Compare the masses of the positive ions only. Step 2: Sodium has the largest mass, 8.3 mg per 1000 cm³, which is more than lithium (4.3), calcium (1.4) or ammonium (0.5). Step 3: So the positive ion with the highest concentration is sodium.
    Method:

    Identify the positive ions, then compare their masses to find the largest.
    Examiner tips

    • Read off only the positive (cation) masses and choose the biggest.

11. Question 3(a)(ii)

    1 marksFormula of the sulfate ion
    The formula for the sulfate ion is $\text{SO}_4^{2-}$.
    Method:

    Recall the formula and charge of the sulfate ion.
    Examiner tips

    • Sulfate $\text{SO}_4^{2-}$, sulfite $\text{SO}_3^{2-}$, sulfide $\text{S}^{2-}$.

12. Question 3(a)(iii)

    2 marksTest for iodide ions
    Step 1: Halide ions are tested by acidifying with dilute nitric acid and adding aqueous silver nitrate. Step 2: With iodide ions a yellow precipitate of silver iodide forms. Step 3: A white precipitate shows chloride, the hydroxide test detects ammonium and limewater tests for carbon dioxide, so iodide gives a yellow precipitate.
    Method:

    Recall the silver nitrate halide test and match the colour to iodide.
    Examiner tips

    • Chloride -> white, bromide -> cream, iodide -> yellow precipitate with silver nitrate.

13. Question 3(a)(iv)

    1 marksMass of nitrate ions in a volume of water
    Step 1: The mass of nitrate is proportional to the volume of water. Step 2: Mass in 200 cm³ = 9.5 × (200 ÷ 1000) = 9.5 × 0.20. Step 3: This gives 1.9 mg of nitrate ions in 200 cm³.
    Method:

    Set up the proportion of mass to volume and solve for the new mass.
    Examiner tips

    • Mass is proportional to volume: multiply by (new volume ÷ original volume).

14. Question 3(b)

    1 marksIons in an NPK fertiliser
    Step 1: NPK stands for nitrogen, phosphorus and potassium. Step 2: With potassium already provided, nitrogen comes from the nitrate ion $\text{NO}_3^{-}$ and phosphorus comes from the phosphate ion $\text{PO}_4^{3-}$. Step 3: So the pair that supplies N and P is nitrate and phosphate.
    Method:

    Match each letter of NPK to the ion that provides that element.
    Examiner tips

    • NPK = nitrogen (nitrate or ammonium) + phosphorus (phosphate) + potassium.

15. Question 3(c)

    2 marksIons causing deoxygenation of water
    Step 1: Nitrate and phosphate ions are plant nutrients found in fertilisers. Step 2: When they wash into rivers they cause rapid growth of algae; the algae later decompose, using up dissolved oxygen (deoxygenation), which harms aquatic life. Step 3: So the ions responsible are nitrate and phosphate.
    Method:

    Recall which fertiliser ions cause eutrophication in water.
    Examiner tips

    • Nitrate + phosphate -> algal growth -> oxygen used up (eutrophication).

16. Question 3(d)

    1 marksSeparation method to obtain pure water
    Step 1: The process boils the water to form vapour and then cools the vapour back to liquid. Step 2: Boiling a liquid and condensing the vapour to collect the pure liquid is called distillation. Step 3: Filtration removes insoluble solids and chromatography separates dissolved substances, so this is distillation.
    Method:

    Recognise the evaporate-then-condense process as distillation.
    Examiner tips

    • Distillation = evaporate then condense to get the pure solvent.

17. Question 4(a)

    1 marksMolecular formula from a displayed formula
    Step 1: Count each type of atom in the displayed formula. Step 2: There are 10 carbon atoms, 16 hydrogen atoms and 2 oxygen atoms. Step 3: Writing these together gives the molecular formula $\text{C}_{10}\text{H}_{16}\text{O}_2$.
    Method:

    Count each element in the displayed formula and combine into a molecular formula.
    Examiner tips

    • Carefully tally C, H and O atoms one by one from the structure.

18. Question 4(b)

    1 marksFeature showing a molecule is unsaturated
    Step 1: Unsaturated means the molecule contains a carbon-carbon double bond. Step 2: The feature that shows unsaturation is therefore the C=C double bond. Step 3: A C=O, O-H or C-H bond does not make a molecule unsaturated, so the answer is the C=C double bond.
    Method:

    Recall that unsaturation means a carbon-carbon double bond.
    Examiner tips

    • Unsaturation in IGCSE organic chemistry means a carbon-carbon double bond.

19. Question 4(c)

    2 marksRelative molecular mass of an organic compound
    Step 1: Carbon: 11 × 12 = 132. Step 2: Hydrogen: 14 × 1 = 14, and oxygen: 3 × 16 = 48. Step 3: Adding gives 132 + 14 + 48 = 194, so the relative molecular mass is 194.
    Method:

    Multiply each atom count by its Ar and add the totals.
    Examiner tips

    • Relative molecular mass = sum of (number of atoms × relative atomic mass).

20. Question 4(d)

    1 marksMeaning of hydrocarbon
    Step 1: The word hydrocarbon combines hydrogen and carbon. Step 2: A hydrocarbon is a compound that contains only the elements hydrogen and carbon. Step 3: If oxygen or other elements are present it is not a hydrocarbon, so the correct meaning is a compound of hydrogen and carbon only.
    Method:

    Define hydrocarbon as a compound of hydrogen and carbon only.
    Examiner tips

    • Hydrocarbons contain only hydrogen and carbon — nothing else.

21. Question 4(e)(i)

    1 marksFraction with the lowest boiling point
    Step 1: Smaller molecules have lower boiling points and are collected near the top of the fractionating column. Step 2: Of the listed fractions, gasoline (petrol) contains the smallest molecules, so it has the lowest boiling point. Step 3: Lubricating oil, diesel oil and fuel oil have larger molecules and higher boiling points, so the answer is gasoline.
    Method:

    Link molecule size to boiling point and pick the lightest fraction.
    Examiner tips

    • Top of column = smallest molecules = lowest boiling point.

22. Question 4(e)(ii)

    1 marksFraction with the highest viscosity
    Step 1: Viscosity (thickness) increases as the hydrocarbon molecules get larger. Step 2: Lubricating oil is the heaviest of the listed fractions, with the largest molecules, so it has the highest viscosity. Step 3: Gasoline, naphtha and kerosene are lighter fractions and flow more easily, so the answer is lubricating oil.
    Method:

    Link molecule size to viscosity and choose the heaviest fraction.
    Examiner tips

    • Heavier fractions (bottom of column) are more viscous.

23. Question 4(e)(iii)

    1 marksFraction used for jet fuel
    Step 1: Each fraction has typical uses based on its molecule size. Step 2: Kerosene (paraffin) is the fraction used as jet fuel. Step 3: Gasoline fuels cars, diesel oil fuels lorries and lubricating oil makes lubricants, so jet fuel is kerosene.
    Method:

    Recall the use of each petroleum fraction and pick the jet-fuel one.
    Examiner tips

    • Kerosene = jet fuel; gasoline = cars; diesel = lorries.

24. Question 4(f)(i)

    1 marksMolecular formula of heptane
    Step 1: Alkanes have the general formula $\text{C}_n\text{H}_{2n+2}$. Step 2: With n = 7, the number of hydrogen atoms is (2 × 7) + 2 = 16. Step 3: So heptane is $\text{C}_7\text{H}_{16}$.
    Method:

    Apply the alkane general formula with seven carbon atoms.
    Examiner tips

    • Alkanes: $\text{C}_n\text{H}_{2n+2}$; for 7 carbons, H = 16.

25. Question 4(f)(ii)

    1 marksPredicting the melting point of octane
    Step 1: Octane has 8 carbon atoms, between heptane (7) and nonane (9). Step 2: Its melting point should lie between heptane (-90 °C) and nonane (-54 °C). Step 3: A value such as -57 °C fits this range, so the best prediction is about -57 °C.
    Method:

    Use the trend in the homologous series to interpolate octane between its neighbours.
    Examiner tips

    • Interpolate: the value lies between the neighbours in the series.

26. Question 4(g)(i)

    1 marksName of the reaction that breaks large molecules
    Step 1: Heating a large hydrocarbon with a catalyst to break it into smaller molecules is called cracking. Step 2: Cracking is endothermic and produces smaller alkanes and alkenes. Step 3: Fractional distillation only separates molecules, polymerisation joins them and combustion burns them, so the reaction is cracking.
    Method:

    Name the thermal-decomposition process that breaks large hydrocarbons.
    Examiner tips

    • Cracking = breaking long chains into shorter chains (plus alkenes).

27. Question 4(g)(ii)

    2 marksDisplayed formula of ethene
    Step 1: Ethene is $\text{C}_2\text{H}_4$, an alkene, so it has a carbon-carbon double bond. Step 2: Each carbon forms four bonds: two go into the C=C double bond region and two are C-H bonds, so each carbon bonds to two hydrogen atoms. Step 3: So the displayed formula shows a C=C double bond with two hydrogen atoms on each carbon.
    Method:

    Draw the C=C double bond first, then add hydrogens so each carbon has four bonds.
    Examiner tips

    • Each carbon makes 4 bonds: in ethene that is one C=C and two C-H.

28. Question 4(h)(i)

    1 marksType of polymerisation forming poly(ethene)
    Step 1: Ethene molecules contain a C=C double bond and add together without losing any atoms. Step 2: Joining many unsaturated monomers into a polymer with no small molecule lost is addition polymerisation. Step 3: Condensation polymerisation loses a small molecule (such as water), so forming poly(ethene) is addition polymerisation.
    Method:

    Identify the polymerisation type from the alkene monomer and lack of by-product.
    Examiner tips

    • Addition polymerisation: many alkene monomers, no small molecule lost.

29. Question 4(h)(ii)

    2 marksEnvironmental problems of plastic disposal
    Step 1: Most plastics are non-biodegradable, so they build up in landfill sites and accumulate in the oceans. Step 2: If they are burned, they can release toxic or acidic gases. Step 3: So two genuine challenges are filling landfill and producing toxic gases on burning.
    Method:

    Recall the main environmental problems linked to disposing of plastics.
    Examiner tips

    • Key disposal issues: non-biodegradable (landfill/oceans) + toxic gases on burning.

30. Question 5(a)

    2 marksColour and state of chlorine
    Step 1: Down Group VII the colour darkens and the state changes from gas to solid. Step 2: Chlorine, near the top of the group, is a pale yellow-green gas at room temperature and pressure. Step 3: Bromine is a red-brown liquid and iodine a grey-black solid, so chlorine is a pale yellow-green gas.
    Method:

    Recall the colour and state trend down Group VII for chlorine.
    Examiner tips

    • Group VII: chlorine yellow-green gas, bromine red-brown liquid, iodine grey-black solid.

31. Question 5(b)

    1 marksUse of chlorine in water treatment
    Step 1: Drinking water must be safe from disease-causing microbes. Step 2: Chlorine is added to kill bacteria and other microbes, making the water safe to drink. Step 3: Colour and taste are dealt with by other steps, so the role of chlorine is to kill microbes.
    Method:

    Recall why chlorine is added during water treatment.
    Examiner tips

    • Chlorination = killing microbes to make water safe to drink.

32. Question 5(c)

    2 marksPhysical state from melting and boiling points
    Step 1: A substance is a gas at temperatures above its boiling point. Step 2: The boiling point of iodine is 184 °C, and 195 °C is higher than this. Step 3: So at 195 °C iodine is a gas, because the temperature is above its boiling point.
    Method:

    Compare the temperature with the melting and boiling points to decide the state.
    Examiner tips

    • Below mp = solid; between mp and bp = liquid; above bp = gas.

33. Question 5(d)

    1 marksPhysical property of potassium
    Step 1: Group I metals are unusual metals because they are soft. Step 2: Potassium is soft enough to be cut with a knife, and like other metals it is shiny when freshly cut and conducts electricity. Step 3: It is not hard, it does conduct electricity and it is a solid, so the correct extra property is that it is soft.
    Method:

    Recall the unusual physical properties of Group I metals.
    Examiner tips

    • Group I metals are soft, shiny when cut, and good conductors.

34. Question 5(e)(i)

    3 marksObservations when potassium reacts with water
    Step 1: Potassium is very reactive and less dense than water, so it floats and moves about on the surface. Step 2: The reaction releases heat, melting the potassium into a ball, and produces hydrogen gas, causing fizzing. Step 3: The heat ignites the hydrogen, giving a lilac flame, so all these observations are seen.
    Method:

    Recall the vigorous observations of a Group I metal reacting with water.
    Examiner tips

    • Group I + water: floats, melts, fizzes, moves, lilac flame for potassium.

35. Question 5(e)(ii)

    1 marksTest for hydrogen gas
    Step 1: Hydrogen is a flammable gas. Step 2: When a lighted splint is held to hydrogen it burns rapidly with a characteristic squeaky pop. Step 3: A glowing splint relighting tests for oxygen and limewater tests for carbon dioxide, so the squeaky pop confirms hydrogen.
    Method:

    Recall the standard squeaky-pop test for hydrogen.
    Examiner tips

    • Lighted splint + squeaky pop = hydrogen.

36. Question 5(f)

    2 marksBalancing the equation for potassium and oxygen
    Step 1: The product is $2\text{K}_2\text{O}$, which contains 4 potassium atoms and 2 oxygen atoms. Step 2: To balance, the left side needs 4 K atoms (so 4K) and 2 O atoms (so one $\text{O}_2$ molecule). Step 3: This gives $4\text{K} + \text{O}_2 \rightarrow 2\text{K}_2\text{O}$, balanced on both sides.
    Method:

    Count the atoms in the product and balance the potassium and oxygen on the left.
    Examiner tips

    • Balance the product first, then adjust the reactants.

37. Question 5(g)

    1 markspH of an alkaline solution
    Step 1: Potassium oxide is a metal oxide that dissolves to form an alkali. Step 2: Alkaline solutions have a pH greater than 7, usually around pH 11 to 14. Step 3: Of the choices, pH 12 is alkaline, while pH 7 is neutral and pH 5 and pH 1 are acidic, so the answer is pH 12.
    Method:

    Recall that alkalis have pH greater than 7 and pick the alkaline value.
    Examiner tips

    • Alkali = pH > 7; a strong alkali is around pH 12 to 14.

38. Question 5(h)

    3 marksIons in potassium chloride
    Step 1: Potassium (2,8,8,1) loses its single outer electron to become 2,8,8 with a 1+ charge. Step 2: Chlorine (2,8,7) gains that electron to become 2,8,8 with a 1- charge. Step 3: So both ions have the configuration 2,8,8, with potassium 1+ and chloride 1-.
    Method:

    Transfer one electron from potassium to chlorine, then write each ion configuration and charge.
    Examiner tips

    • Group I loses 1 electron (1+); Group VII gains 1 electron (1-); both reach 2,8,8.

39. Question 6(a)(i)

    2 marksReasons carbon is used in the blast furnace
    Step 1: Carbon (coke) burns in the hot air blast, releasing heat to keep the furnace very hot. Step 2: Burning carbon also produces carbon dioxide, which then reacts with more carbon to make carbon monoxide, the reducing agent. Step 3: So carbon is used both as a fuel to heat the furnace and to produce the gas that leads to reduction of the iron ore.
    Method:

    Recall the two roles of coke in the blast furnace: fuel and gas production.
    Examiner tips

    • Coke = fuel (heat) + source of carbon dioxide/monoxide for reduction.

40. Question 6(a)(ii)

    2 marksHow slag is produced
    Step 1: In the furnace, calcium carbonate (limestone) thermally decomposes to calcium oxide and carbon dioxide. Step 2: The calcium oxide is a base and reacts with the acidic silicon(IV) oxide (sand) impurity. Step 3: This reaction forms molten calcium silicate, which is the slag, so slag comes from calcium oxide reacting with silicon(IV) oxide.
    Method:

    Trace limestone through decomposition and reaction with the sandy impurity.
    Examiner tips

    • Slag = calcium oxide + silicon(IV) oxide -> calcium silicate.

41. Question 6(a)(iii)

    1 marksOxidation in terms of oxygen
    Step 1: Oxidation in terms of oxygen means gaining oxygen. Step 2: In the equation, CO (one oxygen) becomes $\text{CO}_2$ (two oxygens), so the carbon monoxide has gained oxygen. Step 3: Gaining oxygen means it is oxidised, while the iron(III) oxide loses oxygen and is reduced.
    Method:

    Compare the oxygen in CO and CO2 to show oxygen is gained.
    Examiner tips

    • OIL RIG / "oxygen gained = oxidised": CO -> CO2.

42. Question 6(b)(i)

    2 marksMetals in stainless steel
    Step 1: Stainless steel is iron mixed with other metals to stop it rusting. Step 2: The two metals most commonly added are chromium and nickel. Step 3: Copper/tin make bronze and zinc is used in brass or galvanising, so the answer is chromium and nickel.
    Method:

    Recall the two alloying metals in stainless steel.
    Examiner tips

    • Chromium gives the rust resistance; nickel adds strength in stainless steel.

43. Question 6(b)(ii)

    1 marksUse of stainless steel
    Step 1: Stainless steel is strong and resists rusting, so it is used where contact with water or food occurs. Step 2: Cutlery is a common use because it stays shiny and does not rust. Step 3: Wiring uses copper, windows use glass and balloons use rubber, so a correct use is making cutlery.
    Method:

    Match the rust-resistant property to a suitable everyday use.
    Examiner tips

    • Rust-resistant + strong -> cutlery, sinks, surgical instruments.

44. Question 6(c)(i)

    2 marksOrder of reactivity from reactions with acid
    Step 1: The more vigorously a metal reacts with the acid, the more reactive it is. Step 2: Titanium does not react (least reactive), zinc reacts slowly, magnesium reacts quickly and calcium reacts most vigorously (most reactive). Step 3: So from least to most reactive the order is titanium, zinc, magnesium, calcium.
    Method:

    Compare how vigorously each metal reacts to order them by reactivity.
    Examiner tips

    • Rank by vigour: no bubbles < slow bubbles < quick bubbles < very rapid bubbles.

45. Question 6(c)(ii)

    2 marksProducts of a metal and an acid
    Step 1: A reactive metal reacting with an acid produces a salt and hydrogen. Step 2: The salt of ethanoic acid is an ethanoate, so calcium gives calcium ethanoate, and hydrogen gas is released. Step 3: So the products are calcium ethanoate and hydrogen.
    Method:

    Apply the metal-plus-acid pattern and name the ethanoate salt.
    Examiner tips

    • Metal + acid -> salt + hydrogen; name the salt from the acid.

46. Question 7(a)

    2 marksMeaning of electrolysis
    Step 1: Electrolysis needs an ionic compound that is molten or dissolved in water so the ions can move. Step 2: An electric current is passed through this electrolyte. Step 3: The current breaks the compound down into simpler substances at the electrodes, so electrolysis is the decomposition of an ionic compound by electricity.
    Method:

    Define electrolysis as decomposition of an electrolyte by an electric current.
    Examiner tips

    • Key words: decomposition + ionic compound + molten/aqueous + electric current.

47. Question 7(b)

    1 marksIdentifying the cathode
    Step 1: In electrolysis the cathode is the negative electrode. Step 2: It is connected to the negative terminal of the power supply, and positive ions (cations) move towards it. Step 3: The positive electrode is the anode, so the cathode is the one joined to the negative terminal.
    Method:

    Recall that the cathode is the negative electrode in electrolysis.
    Examiner tips

    • Cathode is negative; anode is positive (remember: cat = negative).

48. Question 7(c)

    1 marksProduct at the anode in dilute sulfuric acid
    Step 1: During the electrolysis of dilute sulfuric acid, hydrogen forms at the cathode and oxygen forms at the anode. Step 2: At the anode, hydroxide ions from the water are discharged to release oxygen. Step 3: So the product at the anode is oxygen, not hydrogen, sulfur or chlorine.
    Method:

    Recall the products of electrolysing dilute sulfuric acid at each electrode.
    Examiner tips

    • Anode (positive) of dilute sulfuric acid gives oxygen.

49. Question 7(d)(i)

    1 marksMeaning of inert
    Step 1: An inert electrode is one that takes no part in the chemical reaction. Step 2: It conducts the current but does not react with the electrolyte or the products. Step 3: So inert means unreactive, not that it dissolves or acts as a catalyst.
    Method:

    Define inert as unreactive in the context of electrodes.
    Examiner tips

    • Inert electrode = conducts current but does not react.

50. Question 7(d)(ii)

    1 marksUse of graphite
    Step 1: In graphite the layers of carbon atoms can slide over each other. Step 2: This makes graphite slippery, so it is used as a lubricant (and in pencil leads). Step 3: Graphite conducts electricity (so it is not an insulator) and is soft (so it is not used for cutting), so a correct use is as a lubricant.
    Method:

    Recall a use of graphite that comes from its sliding layers.
    Examiner tips

    • Graphite: sliding layers -> lubricant; also used in pencils and electrodes.

51. Question 7(e)(i)

    1 marksChemical versus physical change
    Step 1: In a chemical change, new substances with different properties are made. Step 2: Here copper(II) sulfate, water and carbon dioxide are produced, which are new substances. Step 3: So this is a chemical change because new substances are formed.
    Method:

    Decide whether new substances are made to classify the change.
    Examiner tips

    • Chemical change = new substances; physical change = same substance, different form.

52. Question 7(e)(ii)

    1 marksRemoving excess insoluble solid
    Step 1: The excess copper(II) carbonate is an insoluble solid in a solution. Step 2: Filtration separates an insoluble solid from a liquid: the solid stays on the filter paper and the solution passes through. Step 3: So filtration removes the excess copper(II) carbonate.
    Method:

    Choose the method that separates an insoluble solid from a liquid.
    Examiner tips

    • Filtration removes insoluble solids from a liquid.

53. Question 7(e)(iii)

    1 marksReactant to make copper(II) sulfate
    Step 1: Copper(II) sulfate is made by reacting a copper compound with sulfuric acid. Step 2: Copper(II) oxide is a base that reacts with sulfuric acid to give copper(II) sulfate and water. Step 3: Sodium chloride, calcium carbonate and magnesium oxide would not give copper(II) sulfate, so the answer is copper(II) oxide.
    Method:

    Pick a copper-containing base that reacts with sulfuric acid.
    Examiner tips

    • Copper(II) oxide or copper(II) hydroxide + sulfuric acid -> copper(II) sulfate.

54. Question 7(f)

    2 marksForms of copper(II) sulfate
    Step 1: Aqueous copper(II) sulfate is the compound dissolved in water. Step 2: Hydrated copper(II) sulfate is the solid that is chemically combined with water (water of crystallisation). Step 3: Anhydrous copper(II) sulfate is the solid containing no water, so the correct matching is aqueous = dissolved, hydrated = combined with water, anhydrous = no water.
    Method:

    Match each term to the correct state of copper(II) sulfate and its water content.
    Examiner tips

    • Anhydrous = no water; hydrated = water of crystallisation; aqueous = dissolved.

55. Question 8(a)

    1 marksReading the time for a reaction to finish
    Step 1: A reaction has finished when no more gas is produced, so the curve becomes a horizontal (constant) line. Step 2: The line first becomes horizontal at about 119 s, where the volume stays at 250 cm³. Step 3: So the reaction finishes at 119 s.
    Method:

    Find where the curve first levels off and read the time on the axis.
    Examiner tips

    • Reaction finishes when the volume stops increasing (curve goes flat).

56. Question 8(b)(i)

    1 marksEffect of larger pieces on reaction time
    Step 1: Larger pieces of the same mass have a smaller total surface area. Step 2: A smaller surface area means fewer collisions per second between acid particles and the magnesium, so the reaction is slower. Step 3: A slower reaction takes longer to finish, so the time increases.
    Method:

    Link surface area to collision frequency and then to reaction time.
    Examiner tips

    • Smaller surface area slows the reaction, so it takes more time.

57. Question 8(b)(ii)

    1 marksEffect of piece size on total volume of gas
    Step 1: The total volume of hydrogen depends on the amount (mass) of magnesium that reacts, not on the size of the pieces. Step 2: The same total mass of magnesium is used, so the same number of moles of magnesium reacts. Step 3: Therefore the total volume of hydrogen produced is unchanged.
    Method:

    Recognise that total product depends on amount of reactant, not on rate.
    Examiner tips

    • Total gas volume depends on amount of reactant, not surface area.

58. Question 8(c)

    2 marksEnergy change from a reaction pathway diagram
    Step 1: If the products are at a lower energy level than the reactants, energy has been released to the surroundings. Step 2: Releasing energy to the surroundings is an exothermic change. Step 3: So this is exothermic, because the products are lower in energy than the reactants.
    Method:

    Compare the energy levels of reactants and products to classify the change.
    Examiner tips

    • Exothermic: products below reactants on the energy profile.