October/November 2025 Paper 33 Worked Answers (IGCSE Chemistry 0620 Core)

60 questions · 80 marks · 75 minutes

Question papers and mark schemes are copyright Cambridge International. We do not reproduce them: the worked answers here are written by The Practice Book. Have the paper open alongside. Get the official paper from Cambridge

1. Question 1(a)

   1 marksMetal used in food containers
   Step 1: A metal used for food containers must resist corrosion so it does not contaminate the food. Step 2: Aluminium forms a thin, unreactive oxide layer on its surface that protects it from further corrosion. Step 3: Iron rusts, while sodium and calcium are reactive metals, so aluminium is the corrosion-resistant metal used in food containers.
   Method:

   Pick the metal whose oxide layer makes it resistant to corrosion.
   Examiner tips

   • Aluminium resists corrosion because of its protective oxide layer.

2. Question 1(b)

   1 marksUnreactive metal
   Step 1: A metal that does not react with water, dilute acids or oxygen must be very low in the reactivity series. Step 2: Gold is one of the least reactive metals; it is found uncombined (native) and does not corrode. Step 3: Calcium, iron and sodium all react with at least one of water, acid or oxygen, so the unreactive metal is gold.
   Method:

   Choose the metal lowest in the reactivity series.
   Examiner tips

   • Gold and platinum are the least reactive metals and resist corrosion.

3. Question 1(c)

   1 marksGas reacting with hydrogen in a fuel cell
   Step 1: A hydrogen-oxygen fuel cell uses hydrogen and oxygen as the two reactants. Step 2: Hydrogen reacts with oxygen to form water, releasing energy as an electric current. Step 3: Nitrogen, carbon and helium are not the reactant, so the element that reacts with hydrogen is oxygen.
   Method:

   Recall the two reactants in a hydrogen-oxygen fuel cell.
   Examiner tips

   • Hydrogen + oxygen -> water is the fuel-cell reaction.

4. Question 1(d)

   1 marksElement bonded to hydrogen in methane
   Step 1: Methane has the formula $\text{CH}_4$, so it contains carbon and hydrogen only. Step 2: In methane one carbon atom is bonded to four hydrogen atoms. Step 3: Oxygen, nitrogen and sulfur are not present in methane, so the element bonded to hydrogen is carbon.
   Method:

   Read the formula of methane and identify the element joined to hydrogen.
   Examiner tips

   • Methane $\text{CH}_4$ = one carbon bonded to four hydrogens.

5. Question 2(a)

   2 marksNeutrons and nucleons in an atom
   Step 1: The number of nucleons is the nucleon (mass) number, which is 131. Step 2: The number of neutrons = nucleon number - proton number = 131 - 54 = 77. Step 3: So the atom has 77 neutrons and 131 nucleons.
   Method:

   Use the proton and nucleon numbers to work out neutrons and nucleons.
   Examiner tips

   • Nucleons = protons + neutrons = mass number; neutrons = mass number - protons.

6. Question 2(b)

   2 marksEffect of temperature and pressure on gas volume
   Step 1: When a gas is cooled, the particles move more slowly and take up less space, so decreasing the temperature decreases the volume. Step 2: When the pressure on a gas is reduced, the particles push the plunger out, so decreasing the pressure increases the volume. Step 3: So lowering the temperature decreases the volume while lowering the pressure increases the volume.
   Method:

   Apply the effect of temperature and pressure on the volume of a gas separately.
   Examiner tips

   • Lower temperature -> smaller volume; lower pressure -> larger volume.

7. Question 2(c)

   2 marksArrangement and motion of particles in a solid
   Step 1: In a solid the particles are packed closely together in a regular arrangement (rows or a lattice). Step 2: They have only enough energy to vibrate about fixed positions, so they cannot move from place to place. Step 3: So solid xenon has a regular arrangement of particles that vibrate about fixed positions.
   Method:

   Recall the kinetic particle description of a solid for arrangement and motion.
   Examiner tips

   • Solid = regular arrangement, particles vibrate about fixed positions.

8. Question 3(a)(i)

   1 marksPositive ion with highest concentration
   Step 1: The concentration of each positive ion is given by its mass in the same 600 cm³ sample. Step 2: Calcium has the largest mass at 25.0 mg, compared with 6.2 mg, 2.7 mg and 0.1 mg for the others. Step 3: So the positive ion with the highest concentration is calcium.
   Method:

   Compare the masses of the positive ions and pick the largest.
   Examiner tips

   • In the same volume, the largest mass means the highest concentration.

9. Question 3(a)(ii)

   1 marksName of the phosphate ion
   The ion $\text{PO}_4^{3-}$, made of phosphorus and oxygen with a 3- charge, is named phosphate (e.g. in calcium phosphate and many fertilisers).
   Method:

   Recall the standard name for the $\text{PO}_4^{3-}$ ion.
   Examiner tips

   • $\text{PO}_4^{3-}$ = phosphate, like $\text{SO}_4^{2-}$ = sulfate.

10. Question 3(a)(iii)

    2 marksTest for nitrate ions
    Step 1: To test for nitrate ions, add aqueous sodium hydroxide and a small piece of aluminium (or Devarda's alloy) and warm. Step 2: The nitrate ions are reduced to ammonia gas, which turns damp red litmus paper blue. Step 3: The silver nitrate test is for halides, limewater tests for carbon dioxide and the acid test detects carbonates, so the correct nitrate test produces ammonia.
    Method:

    Recall the sodium hydroxide and aluminium test that reduces nitrate to ammonia.
    Examiner tips

    • Nitrate test: NaOH + aluminium, warm -> ammonia (red litmus turns blue).

11. Question 3(a)(iv)

    1 marksMass of ions in a smaller volume
    Step 1: The mass of ions is proportional to the volume of water. Step 2: Mass in 200 cm³ = 2.7 × (200 ÷ 600) = 2.7 × (1/3). Step 3: This gives 0.9 mg of sodium ions in 200 cm³.
    Method:

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

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

12. Question 3(a)(v)

    1 marksSource of nitrate ions in river water
    Step 1: Nitrate ions are a major component of fertilisers used on farmland. Step 2: When rain washes fertiliser off the land, the nitrate ions run into rivers, raising their concentration. Step 3: Dissolved gases and river-bed minerals do not add nitrate, so the source is fertilisers.
    Method:

    Recall what nitrate-rich substances are spread on farmland.
    Examiner tips

    • Fertiliser run-off is the usual source of nitrate in rivers.

13. Question 3(b)

    1 marksRemoving insoluble substances from water
    Step 1: Insoluble substances are solids that do not dissolve in the water. Step 2: Filtration traps these insoluble solids while the water passes through; sedimentation lets them settle out. Step 3: Chlorination kills microbes and distillation removes dissolved substances, so insoluble solids are removed by filtration (or sedimentation).
    Method:

    Pick the separation method that removes undissolved solids.
    Examiner tips

    • Filtration/sedimentation removes insoluble solids from water.

14. Question 3(c)(i)

    1 marksSource of harmful microbes in water
    Step 1: Harmful microbes such as bacteria come from waste containing them. Step 2: Sewage (human and animal waste) entering a river is a common source of these microbes. Step 3: Dissolved gases and sand do not contain harmful microbes, so the source is sewage.
    Method:

    Identify the waste source that introduces microbes into water.
    Examiner tips

    • Sewage is the usual source of harmful microbes in water.

15. Question 3(c)(ii)

    1 marksTreating water to kill microbes
    Step 1: Microbes in drinking water are killed by adding chlorine to the water. Step 2: This process, chlorination, makes the water safe to drink. Step 3: Filtration and sedimentation only remove solids, and adding fertilisers would pollute the water, so chlorination is the correct treatment.
    Method:

    Recall the water-treatment step that kills microbes.
    Examiner tips

    • Chlorination = adding chlorine to kill microbes in drinking water.

16. Question 3(d)

    1 marksRemoving bad taste and odour from water
    Step 1: Bad tastes and odours come from dissolved substances in the water. Step 2: Carbon (activated carbon) adsorbs these substances, removing the taste and odour. Step 3: Chlorine kills microbes, sand filters solids and sodium chloride is a salt, so carbon is the substance that removes taste and odour.
    Method:

    Recall the substance used to remove taste and odour in water treatment.
    Examiner tips

    • Carbon (activated carbon) removes taste and odour from water.

17. Question 4(a)

    1 marksMolecular formula from a displayed formula
    Step 1: A molecular formula lists the total number of atoms of each element in the molecule. Step 2: Compound B has 10 carbon, 12 hydrogen and 1 oxygen atom. Step 3: Writing these together gives the molecular formula $\text{C}_{10}\text{H}_{12}\text{O}$.
    Method:

    Count the atoms of each element and write the molecular formula.
    Examiner tips

    • Count each element separately and write C, then H, then O.

18. Question 4(b)

    1 marksFeature showing a molecule is unsaturated
    Step 1: An unsaturated molecule contains at least one carbon-carbon double bond. Step 2: The C=C double bond is the feature that shows unsaturation, because it can take part in addition reactions. Step 3: Single bonds (C-H, C-C, O-H) are found in saturated molecules too, so only the C=C double bond shows unsaturation.
    Method:

    Recall the bond type that defines an unsaturated molecule.
    Examiner tips

    • C=C double bond = unsaturated; only single bonds = saturated.

19. Question 4(c)

    2 marksRelative molecular mass
    Step 1: Multiply each element's relative atomic mass by the number of atoms: carbon 9 × 12 = 108, hydrogen 10 × 1 = 10, oxygen 3 × 16 = 48. Step 2: Add these together: 108 + 10 + 48. Step 3: This gives a relative molecular mass of 166.
    Method:

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

    • Mr = sum of (number of atoms × relative atomic mass) for every element.

20. Question 4(d)(i)

    1 marksFraction with the highest boiling point
    Step 1: Fractions with larger molecules have higher boiling points and condense lower in the fractionating column. Step 2: Lubricating oil is collected near the bottom, so it has the largest molecules and the highest boiling point. Step 3: Gasoline, naphtha and kerosene are collected higher up and have lower boiling points, so the highest boiling point fraction is lubricating oil.
    Method:

    Pick the heaviest fraction, collected lowest in the column.
    Examiner tips

    • Bottom of column = biggest molecules = highest boiling point.

21. Question 4(d)(ii)

    1 marksFraction with the shortest chain length
    Step 1: Shorter hydrocarbon chains have smaller molecules, lower boiling points and are collected near the top of the column. Step 2: Gasoline/petrol is collected at the top, so it has the shortest chains. Step 3: Fuel oil, lubricating oil and diesel oil are heavier fractions with longer chains, so the shortest chain length is in gasoline/petrol.
    Method:

    Pick the lightest fraction, collected at the top of the column.
    Examiner tips

    • Top of column = smallest molecules = shortest chains.

22. Question 4(d)(iii)

    1 marksFraction used to make waxes and polishes
    Step 1: Each petroleum fraction has typical uses based on its properties. Step 2: Lubricating oil is the fraction used for lubricants and for making waxes and polishes. Step 3: Gasoline is a fuel for cars, naphtha is a chemical feedstock and kerosene is jet fuel, so waxes and polishes come from lubricating oil.
    Method:

    Match the use waxes and polishes to the correct heavy fraction.
    Examiner tips

    • Lubricating oil -> lubricants, waxes and polishes.

23. Question 4(e)(i)

    1 marksMeaning of homologous series
    Step 1: A homologous series is a family of organic compounds. Step 2: Its members share the same functional group and so have similar chemical properties, with each member differing by $\text{CH}_2$. Step 3: They do not all share one molecular formula, so the correct meaning is a family of compounds with the same functional group and similar chemical properties.
    Method:

    Recall the definition of a homologous series.
    Examiner tips

    • Homologous series: same functional group + similar chemical properties + general formula.

24. Question 4(e)(ii)

    1 marksFormula of an alcohol from a pattern
    Step 1: Each member of this series increases by one carbon and two hydrogens: $\text{C}_n\text{H}_{2n+1}\text{OH}$. Step 2: For octanol, n = 8, so the hydrogen count is 2(8) + 1 = 17. Step 3: So octanol is $\text{C}_8\text{H}_{17}\text{OH}$, fitting between heptanol ($\text{C}_7\text{H}_{15}\text{OH}$) and nonanol ($\text{C}_9\text{H}_{19}\text{OH}$).
    Method:

    Apply the general formula of the series for n = 8 carbons.
    Examiner tips

    • Alcohols follow $\text{C}_n\text{H}_{2n+1}\text{OH}$.

25. Question 4(e)(iii)

    1 marksPredicting a melting point from a trend
    Step 1: Melting points increase steadily up the homologous series. Step 2: Nonanol lies between octanol (-16 °C) and decanol (7 °C), so its melting point should be between these two values. Step 3: A value such as -4 °C lies between -16 °C and 7 °C, so it is a reasonable prediction; values below -16 °C or above 7 °C do not fit the trend.
    Method:

    Pick a value between the two neighbouring melting points.
    Examiner tips

    • Interpolate between the neighbouring members for a missing value.

26. Question 4(f)(i)

    1 marksProduct of complete combustion
    Step 1: Complete combustion of a fuel containing carbon, hydrogen and oxygen happens in excess oxygen. Step 2: The carbon is fully oxidised to carbon dioxide and the hydrogen forms water. Step 3: Carbon monoxide and soot form only in incomplete combustion, so the other product here is carbon dioxide.
    Method:

    Recall the products of complete combustion of a carbon-containing fuel.
    Examiner tips

    • Complete combustion: carbon -> $\text{CO}_2$, hydrogen -> $\text{H}_2\text{O}$.

27. Question 4(f)(ii)

    1 marksTest for water with anhydrous copper(II) sulfate
    Step 1: Anhydrous copper(II) sulfate is white and is used to test for water. Step 2: When water is added it forms hydrated copper(II) sulfate, which is blue. Step 3: So the colour change observed is from white to blue.
    Method:

    Recall the colour change of the anhydrous copper(II) sulfate test for water.
    Examiner tips

    • White anhydrous copper(II) sulfate + water -> blue (test for water).

28. Question 4(g)

    3 marksReactant and conditions for fermentation
    Step 1: Fermentation makes ethanol from a sugar. The reactant is aqueous glucose. Step 2: Yeast is added to provide enzymes, and the mixture is kept warm at about 25-35 °C in the absence of oxygen. Step 3: Ethene + steam is the catalytic (hydration) route, not fermentation, and the other options use wrong reactants or conditions, so the correct set is aqueous glucose with yeast at 25-35 °C.
    Method:

    Recall the reactant and conditions used in fermentation to make ethanol.
    Examiner tips

    • Reactant = glucose; conditions = yeast, warm 25-35 °C, no air.

29. Question 4(h)

    2 marksDisplayed formula of ethanoic acid
    Step 1: Ethanoic acid is $\text{CH}_3\text{COOH}$: a methyl group joined to a carboxylic acid group. Step 2: The carboxylic acid carbon has one double bond to oxygen (C=O) and a single bond to an O-H group. Step 3: So the displayed formula shows CH3 joined to a carbon bearing a C=O and an O-H, with all bonds drawn.
    Method:

    Draw the CH3 group, then the acid carbon with its C=O and O-H bonds.
    Examiner tips

    • Carboxylic acid group -COOH = C double-bonded to O plus C-O-H.

30. Question 5(a)

    1 marksGroup number and outer electrons
    Step 1: The group number of a main-group element equals the number of electrons in its outer shell. Step 2: Silicon is in Group IV because it has 4 electrons in its outer electron shell. Step 3: The number of shells gives the period, not the group, so silicon is in Group IV because of its 4 outer electrons.
    Method:

    Link the group number to the number of outer-shell electrons.
    Examiner tips

    • Outer-shell electrons = group number for main-group elements.

31. Question 5(b)

    2 marksPhysical state from melting and boiling points
    Step 1: A substance is solid below its melting point, liquid between its melting and boiling points, and gas above its boiling point. Step 2: At 175 °C the temperature is below tin's melting point of 232 °C. Step 3: So tin is a solid at 175 °C because it has not yet reached its melting point.
    Method:

    Compare 175 °C with the melting and boiling points to deduce the state.
    Examiner tips

    • Compare the temperature with the melting and boiling points to find the state.

32. Question 5(c)

    2 marksColour and state of iodine
    Step 1: Down Group VII the elements get darker and change state: chlorine is a gas, bromine a liquid and iodine a solid. Step 2: Iodine is a grey-black solid at room temperature and pressure. Step 3: The gaseous and liquid options describe chlorine and bromine, so iodine is a grey-black solid.
    Method:

    Recall the colour and state of iodine among the halogens.
    Examiner tips

    • Down Group VII: chlorine gas, bromine liquid, iodine solid (grey-black).

33. Question 5(d)(i)

    2 marksObservations of calcium reacting with water
    Step 1: Calcium reacts with cold water to form calcium hydroxide and hydrogen gas. Step 2: Bubbles of hydrogen gas are seen (effervescence) and the calcium gradually gets smaller as it reacts; the mixture also warms up and a white solid forms. Step 3: So a correct pair of observations is that bubbles are given off and the calcium gets smaller.
    Method:

    Recall the reaction of calcium with water and its visible signs.
    Examiner tips

    • Calcium + water: effervescence (hydrogen), metal gets smaller, mixture warms, white solid.

34. Question 5(d)(ii)

    2 marksWord equation for calcium and water
    Step 1: A reactive metal reacting with water forms a metal hydroxide and hydrogen gas. Step 2: Calcium reacts with water to form calcium hydroxide and hydrogen. Step 3: Calcium oxide forms with oxygen not water, and oxygen is not a product, so the equation is calcium + water -> calcium hydroxide + hydrogen.
    Method:

    Apply the metal + water reaction pattern to calcium.
    Examiner tips

    • Reactive metal + water -> metal hydroxide + hydrogen.

35. Question 5(e)(i)

    1 markspH of a dilute acid
    Step 1: Acids have a pH below 7; the stronger or more concentrated the acid, the lower the pH. Step 2: Dilute nitric acid is a strong acid, so it has a low pH such as pH 1. Step 3: pH 7 is neutral and pH 9 and pH 13 are alkaline, so the pH of dilute nitric acid is pH 1.
    Method:

    Recall that strong acids have a low pH around 1.
    Examiner tips

    • Acids: pH below 7; strong acids near pH 1.

36. Question 5(e)(ii)

    1 marksCompleting a neutralisation equation
    Step 1: Neutralisation is the reaction of hydrogen ions with hydroxide ions to form water. Step 2: Combining $\text{H}^+$ with $\text{OH}^-$ gives $\text{H}_2\text{O}$, which balances for hydrogen and oxygen. Step 3: The other ions do not balance the equation, so the missing ion is $\text{OH}^-$.
    Method:

    Recall the ionic equation for neutralisation forming water.
    Examiner tips

    • Neutralisation: $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$.

37. Question 5(e)(iii)

    2 marksMethyl orange colour change on neutralisation
    Step 1: Methyl orange is red in acidic solution and yellow in neutral or alkaline solution. Step 2: As lithium carbonate neutralises the nitric acid, the solution stops being acidic, so the methyl orange changes from red to yellow. Step 3: The reverse change and the other colours are wrong for methyl orange, so the change is from red to yellow.
    Method:

    Recall the colours of methyl orange in acid and after neutralisation.
    Examiner tips

    • Methyl orange: red (acid) -> yellow (neutral/alkali).

38. Question 5(f)

    2 marksFlame test for lithium ions
    Step 1: Metal ions can be identified by the colour they give in a flame test. Step 2: Lithium ions give a red flame colour. Step 3: A lilac flame indicates potassium, silver nitrate tests for halides and limewater tests for carbon dioxide, so lithium ions give a red flame.
    Method:

    Recall the flame-test colour for lithium ions.
    Examiner tips

    • Flame tests: lithium red, sodium yellow, potassium lilac.

39. Question 5(g)

    3 marksDot-and-cross diagram of lithium chloride
    Step 1: Lithium chloride is ionic, so electrons are transferred. Lithium (2,1) loses its single outer electron to become $\text{Li}^+$ with no outer electrons. Step 2: Chlorine (2,8,7) gains that electron to reach eight outer electrons, becoming $\text{Cl}^-$. Step 3: So the diagram shows $\text{Li}^+$ with an empty outer shell and $\text{Cl}^-$ with eight outer electrons (seven dots plus one cross), with 1+ and 1- charges.
    Method:

    Transfer lithium's outer electron to chlorine and assign the matching charges.
    Examiner tips

    • Ionic bonding: metal loses electrons (+), non-metal gains them (-).

40. Question 6(a)(i)

    1 marksType of reaction when a carbonate is heated
    Step 1: In the reaction one compound breaks down into two simpler substances when heated. Step 2: Breaking a compound down using heat is called thermal decomposition. Step 3: Neutralisation, combustion and precipitation describe different reactions, so heating calcium carbonate is thermal decomposition.
    Method:

    Identify the reaction type where heat breaks a compound apart.
    Examiner tips

    • Heating a carbonate to break it down = thermal decomposition.

41. Question 6(a)(ii)

    1 marksClassifying calcium oxide
    Step 1: Metal oxides are generally basic, while non-metal oxides are generally acidic. Step 2: Calcium is a metal, so calcium oxide is a metal oxide and is therefore basic. Step 3: It is not a non-metal oxide and does react with acids, so calcium oxide is a basic oxide.
    Method:

    Classify the oxide using whether the element is a metal or non-metal.
    Examiner tips

    • Metal oxides are basic; non-metal oxides are acidic.

42. Question 6(a)(iii)

    1 marksGas removed in flue gas desulfurisation
    Step 1: The name desulfurisation tells us a sulfur-containing gas is being removed. Step 2: Calcium oxide (a base) reacts with acidic sulfur dioxide to remove it from the flue gases. Step 3: Nitrogen, oxygen and carbon monoxide are not the target, so the gas removed is sulfur dioxide.
    Method:

    Use the word desulfurisation to identify the sulfur-containing gas.
    Examiner tips

    • Flue gas desulfurisation removes acidic sulfur dioxide using a base.

43. Question 6(b)(i)

    1 marksReduction as loss of oxygen
    Step 1: Reduction is the loss of oxygen from a substance. Step 2: In the equation, zinc oxide (ZnO) loses its oxygen to become zinc (Zn), while the carbon gains the oxygen to form $\text{CO}_2$. Step 3: Because the zinc oxide loses oxygen, it is reduced.
    Method:

    Apply the loss-of-oxygen definition of reduction to the equation.
    Examiner tips

    • Loss of oxygen = reduction; gain of oxygen = oxidation.

44. Question 6(b)(ii)

    1 marksOther metal in brass
    Step 1: Brass is an alloy made by mixing two metals. Step 2: Brass is a mixture of copper and zinc, so the other metal alongside zinc is copper. Step 3: Iron, tin and lead are not the main metals in brass, so the answer is copper.
    Method:

    Recall the two metals that make up brass.
    Examiner tips

    • Brass = copper + zinc; bronze = copper + tin.

45. Question 6(b)(iii)

    2 marksAdvantages of an alloy over a pure metal
    Step 1: In an alloy, atoms of different sizes disrupt the regular layers, making it harder and stronger than the pure metal. Step 2: Brass is therefore harder, stronger and more resistant to corrosion than pure zinc, which is useful for hard-wearing water taps. Step 3: The other options describe properties brass does not have, so brass is chosen because it is harder, stronger and corrosion-resistant.
    Method:

    Recall how alloying improves hardness, strength and corrosion resistance.
    Examiner tips

    • Alloy advantages: harder, stronger, more resistant to corrosion.

46. Question 6(c)(i)

    1 marksAdverse effect of sulfur dioxide
    Step 1: Sulfur dioxide is an acidic gas that dissolves in rain water. Step 2: This forms acidic rain, which damages buildings, plants and aquatic life. Step 3: Sulfur dioxide does not deplete ozone, add oxygen or make rain alkaline, so its adverse effect is acid rain.
    Method:

    Recall the environmental effect of acidic sulfur dioxide.
    Examiner tips

    • Sulfur dioxide -> acid rain.

47. Question 6(c)(ii)

    1 marksRemoving carbon dioxide from the atmosphere
    Step 1: Plants take in carbon dioxide from the air during photosynthesis. Step 2: Planting more trees increases the amount of carbon dioxide removed from the atmosphere. Step 3: Burning fossil fuels, cutting down forests and using more cars all add carbon dioxide, so the helpful strategy is planting trees.
    Method:

    Choose the strategy that takes carbon dioxide out of the atmosphere.
    Examiner tips

    • Planting trees removes carbon dioxide by photosynthesis.

48. Question 6(d)

    2 marksOrdering metals by reactivity from observations
    Step 1: The faster a metal produces bubbles and heat with acid, the more reactive it is. Step 2: W shows no reaction (least reactive), Fe reacts very slowly, Mg reacts quickly, and Ba reacts most vigorously (most reactive). Step 3: So from least to most reactive the order is W, Fe, Mg, Ba.
    Method:

    Rank the metals by how vigorously each reacts with the acid.
    Examiner tips

    • Rank by vigour of reaction with acid; no reaction = least reactive.

49. Question 7(a)

    1 marksIdentifying the anode in electroplating
    Step 1: In electrolysis the anode is always the electrode connected to the positive terminal of the power supply. Step 2: The nickel object being plated is the cathode (negative electrode), so the anode is the positive electrode (the copper). Step 3: The negative electrode is the cathode and the electrolyte is not an electrode, so the anode is the electrode connected to the positive terminal.
    Method:

    Recall that the anode is the positive electrode in electrolysis.
    Examiner tips

    • Anode = positive electrode; cathode = negative electrode.

50. Question 7(b)

    1 marksChoosing an electrolyte for copper plating
    Step 1: To plate copper onto an object, the electrolyte must contain copper ions. Step 2: A soluble copper salt such as copper(II) sulfate provides $\text{Cu}^{2+}$ ions, which are deposited onto the cathode as copper. Step 3: Sodium chloride, sodium hydroxide and plain sulfuric acid contain no copper ions, so the correct electrolyte is aqueous copper(II) sulfate.
    Method:

    Choose an electrolyte that supplies the ions of the plating metal.
    Examiner tips

    • Electroplating with copper needs an electrolyte containing copper ions.

51. Question 7(c)

    1 marksGas produced at the anode from molten lead(II) bromide
    Step 1: During electrolysis of molten lead(II) bromide, the bromide ions are discharged at the anode. Step 2: This forms bromine, which is seen as a red-brown vapour at the anode. Step 3: Chlorine, hydrogen and sodium are not produced here, so the gas at the anode is red-brown bromine.
    Method:

    Identify the non-metal discharged at the anode and recall its colour.
    Examiner tips

    • Molten lead(II) bromide: bromine (red-brown) at the anode, lead at the cathode.

52. Question 7(d)

    1 marksElectrolysis as a chemical change
    Step 1: A chemical change makes new substances, while a physical change does not. Step 2: Electrolysis of molten lead(II) bromide produces lead and bromine, which are new substances different from the starting compound. Step 3: Because new substances are formed, it is a chemical change.
    Method:

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

    • Making new substances = chemical change.

53. Question 7(e)(i)

    1 marksNaming an acid from its salt
    Step 1: The salt formed is named after the acid: an ethanoate salt comes from ethanoic acid. Step 2: Since the product is cobalt(II) ethanoate, the dilute acid used must be ethanoic acid. Step 3: Hydrochloric, sulfuric and nitric acids give chloride, sulfate and nitrate salts, so the acid here is ethanoic acid.
    Method:

    Work back from the salt name to identify the parent acid.
    Examiner tips

    • Salt name -> acid name: ethanoate comes from ethanoic acid.

54. Question 7(e)(ii)

    1 marksTest for hydrogen gas
    Step 1: Hydrogen is tested using a lighted (burning) splint held near the gas. Step 2: The hydrogen burns rapidly, giving a characteristic squeaky pop. Step 3: Relighting a glowing splint tests for oxygen, limewater tests for carbon dioxide and damp red litmus tests for ammonia, so the hydrogen test gives a squeaky pop.
    Method:

    Recall the lighted-splint test and result for hydrogen.
    Examiner tips

    • Hydrogen + lighted splint = squeaky pop.

55. Question 7(f)

    2 marksForms of hydrated, aqueous and anhydrous salt
    Step 1: Hydrated means the solid contains water of crystallisation, while anhydrous means the solid contains no water. Step 2: Aqueous means the substance is dissolved in water to form a solution. Step 3: So hydrated = solid containing water, anhydrous = solid with no water, and aqueous = dissolved in water.
    Method:

    Match each term to whether the salt is a solid with/without water or dissolved.
    Examiner tips

    • Hydrated (with water) / anhydrous (no water) are solids; aqueous = in solution.

56. Question 8(a)

    1 marksMeaning of the state symbol (l)
    In a chemical equation the state symbol (l) means liquid (for example, $\text{H}_2\text{O(l)}$ is liquid water), while (s) is solid, (g) is gas and (aq) is aqueous.
    Method:

    Recall the meaning of the (l) state symbol.
    Examiner tips

    • State symbols: (s) solid, (l) liquid, (g) gas, (aq) aqueous.

57. Question 8(b)

    1 marksReading a volume from a rate graph
    Step 1: To read the volume at 24 s, follow the time axis up to the curve and across to the volume axis. Step 2: At 24 s the curve is in its steep early region, just above the 50 cm³ gridline, giving about 55 cm³. Step 3: 110 cm³ is the final plateau value and the other readings do not match 24 s, so the volume at 24 s is 55 cm³.
    Method:

    Use the graph to read the volume at the given time.
    Examiner tips

    • Read up from the time, then across to the volume axis.

58. Question 8(c)(i)

    1 marksEffect of concentration on reaction time
    Step 1: A lower concentration means fewer acid particles in the same volume, so there are fewer collisions per second. Step 2: Fewer collisions per second means a slower rate of reaction. Step 3: A slower reaction takes more time to finish, so the time taken increases.
    Method:

    Link lower concentration to collision frequency and reaction time.
    Examiner tips

    • Lower concentration -> fewer collisions per second -> slower -> longer time.

59. Question 8(c)(ii)

    1 marksEffect of concentration on total product
    Step 1: The total volume of carbon dioxide depends on the amount of the limiting reactant, the sodium hydrogencarbonate. Step 2: The acid is still in excess and the amount of sodium hydrogencarbonate is unchanged, so the same number of moles of carbon dioxide is made. Step 3: Therefore the total volume of carbon dioxide does not change; only the rate changes.
    Method:

    Use the amount of limiting reactant to judge the total product.
    Examiner tips

    • Total product depends on the limiting reactant, not on concentration.

60. Question 8(d)

    2 marksIdentifying an endothermic reaction from a pathway diagram
    Step 1: In a reaction pathway diagram, the relative heights show whether energy is taken in or given out. Step 2: Here the products are drawn higher in energy than the reactants, which means energy has been taken in from the surroundings. Step 3: A reaction that takes in energy is endothermic, so the diagram with products above reactants shows an endothermic reaction.
    Method:

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

    • Endothermic: products above reactants (energy absorbed). Exothermic: products below.