0620/62 May/June 2025 Worked Answers (IGCSE Chemistry 0620 Extended)

22 questions · 40 marks · 60 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 marksNaming apparatus for collecting a gas
   Step 1: A piece of apparatus that collects a gas and measures its volume is a gas syringe, whose plunger moves out as gas enters. Step 2: A measuring cylinder measures liquids, a conical flask holds reaction mixtures and a funnel is for filtering. Step 3: So the apparatus used to collect and measure the hydrogen is a gas syringe.
   Method:

   Match the function "collect and measure a gas" to the correct named apparatus.
   Examiner tips

   • A gas syringe collects a gas and measures its volume.

2. Question 1(b)

   1 marksCleaning magnesium ribbon before use
   Step 1: The dull coating on magnesium is a layer of magnesium oxide that must be removed so fresh metal can react. Step 2: Rubbing the ribbon with sand or emery (abrasive) paper scrapes off this oxide layer. Step 3: So the ribbon should be cleaned by rubbing it with sand paper or emery paper.
   Method:

   Recall that an oxide coating is removed mechanically with abrasive paper.
   Examiner tips

   • Remove an oxide layer by rubbing with abrasive paper.

3. Question 1(c)

   1 marksPurpose of damp mineral wool in a steam reaction
   Step 1: The water must stay at the heated end so that warming it produces steam to pass over the magnesium. Step 2: Mineral wool soaks up and holds the water in place, stopping liquid water running along the tube to the hot metal. Step 3: So the purpose of the mineral wool is to hold and absorb the water.
   Method:

   Link the wool to keeping the water at the heated end to make steam.
   Examiner tips

   • Mineral wool holds the water in place so it can be boiled to steam.

4. Question 1(d)

   2 marksWhere the apparatus must be heated
   Step 1: Heating the damp mineral wool boils the water there to produce steam. Step 2: The steam then passes over the magnesium, which must also be heated so it is hot enough to react with the steam. Step 3: So the two places to heat are the damp mineral wool (to make steam) and the magnesium ribbon.
   Method:

   Identify the two stages that need heat: making the steam and reacting the metal.
   Examiner tips

   • Heat the water to make steam and heat the metal so it reacts.

5. Question 1(e)

   1 marksIdentity of a condensed colourless liquid
   Step 1: The steam supplied to the reaction is simply water in the gas state. Step 2: Steam that does not react cools further along the tube and condenses back to liquid water. Step 3: So the colourless liquid that collects is water.
   Method:

   Recognise that condensed steam is water.
   Examiner tips

   • Condensed steam is just liquid water.

6. Question 1(f)

   1 marksWhy a collected gas is impure
   Step 1: Before heating, the tube and syringe are full of ordinary air. Step 2: As hydrogen is produced it pushes this air ahead of it, so the first gas collected is a mixture of air and hydrogen. Step 3: So the collected gas is impure because it contains air that was in the apparatus at the start.
   Method:

   Account for the gas already present in the apparatus before the reaction.
   Examiner tips

   • The apparatus is full of air at the start, and that air mixes with the hydrogen.

7. Question 2(a)

   4 marksCalculating temperature changes for a results table
   Step 1: Each temperature change is the highest temperature minus the initial temperature. Step 2: This gives 27.0 − 22.5 = 4.5, 29.0 − 22.5 = 6.5, 30.5 − 22.5 = 8.0, 32.5 − 22.0 = 10.5 and 35.5 − 22.0 = 13.5 °C, each kept to one decimal place. Step 3: So the temperature changes are 4.5, 6.5, 8.0, 10.5 and 13.5 °C.
   Method:

   Subtract each initial temperature from its highest temperature, keeping one decimal place.
   Examiner tips

   • Temperature change = highest − initial; match the decimal places of the readings.

8. Question 2(b)

   4 marksPlotting results and drawing a line of best fit
   Step 1: A good plot uses a linear scale (equal steps) chosen so the points fill more than half the available grid. Step 2: Every result is plotted accurately, then one smooth line of best fit is drawn through the trend. Step 3: So the correct approach is a linear y-axis scale filling over half the grid, all points plotted accurately and one line of best fit.
   Method:

   Recall the rules for a good line graph: linear scale, points spread out, single best-fit line.
   Examiner tips

   • Choose a scale so the points use more than half the grid; draw one best-fit line.

9. Question 2(c)

   3 marksExtrapolating a best-fit line beyond the data
   Step 1: A volume of 45 cm³ is just beyond the plotted results, so the line must be continued (extrapolated) smoothly past the last point. Step 2: From 45 on the volume axis, a vertical line is drawn up to the extended line and then across to the temperature-change axis. Step 3: So the value is found by extending the best-fit line to 45 cm³ and reading across to the temperature-change axis.
   Method:

   Extrapolate the best-fit line past the data and read the temperature change at 45 cm³.
   Examiner tips

   • Extrapolation = extend the line beyond the data, then read off the value.

10. Question 2(d)

    1 marksCalculating the energy given out when a solid dissolves
    Step 1: Substitute the values into energy = temperature change × 4.2 × volume of water, giving 13.5 × 4.2 × 15. Step 2: 13.5 × 4.2 = 56.7, and 56.7 × 15 = 850.5. Step 3: So the energy given out is 850.5 J.
    Method:

    Substitute 13.5, 4.2 and 15 into the given equation and evaluate.
    Examiner tips

    • Multiply temperature change × 4.2 × volume of water and keep the unit J.

11. Question 2(e)

    2 marksPredicting the effect of doubling the mass of solid
    Step 1: Doubling the mass of lithium chloride from 2.0 g to 4.0 g means twice as much solid dissolves in the same volume of water. Step 2: Twice as much solid releases about twice as much energy to the same mass of water, so the temperature change roughly doubles. Step 3: So the temperature change is estimated at about 9 °C (about double 4.5 °C).
    Method:

    Relate the amount of solid dissolved to the energy released and the temperature change.
    Examiner tips

    • Energy released is proportional to the amount of solid dissolved.

12. Question 2(f)

    2 marksWhy a polystyrene cup gives more accurate energy results
    Step 1: Dissolving lithium chloride is exothermic, so heat can escape to the surroundings during the experiment. Step 2: Polystyrene is a better insulator than glass, so it reduces this heat loss. Step 3: With less heat lost, the measured temperature change is closer to the true value, so the results are more accurate.
    Method:

    Link the insulating property of polystyrene to reduced heat loss and a more accurate result.
    Examiner tips

    • Better insulation means less heat loss and a more accurate temperature change.

13. Question 2(g)(i)

    1 marksWhy a burette is more accurate than a measuring cylinder
    Step 1: A measuring cylinder reads volume only to the nearest whole division, which is fairly coarse. Step 2: A burette has a finer scale and is read to a smaller division, so it measures volume more precisely. Step 3: So using a burette is an improvement because it measures the volume of water more accurately.
    Method:

    Compare the precision of a burette and a measuring cylinder for measuring volume.
    Examiner tips

    • A burette is more accurate than a measuring cylinder for measuring volume.

14. Question 2(g)(ii)

    1 marksWhy a water-bath is not an improvement here
    Step 1: The whole point of the investigation is to measure the temperature change caused by the dissolving. Step 2: A water-bath holds the contents at a fixed temperature, so it would stop the temperature from rising. Step 3: So a water-bath is not an improvement because it would keep the temperature constant and prevent the temperature change being observed.
    Method:

    Recognise that controlling temperature defeats the purpose of measuring a temperature change.
    Examiner tips

    • You cannot measure a temperature change if a water-bath holds the temperature constant.

15. Question 3(a)

    2 marksEffect of sodium hydroxide on an iron(II) salt
    Step 1: Iron(II) ions react with sodium hydroxide to form a green precipitate of iron(II) hydroxide. Step 2: Iron(II) hydroxide is insoluble in excess sodium hydroxide, so the precipitate remains when more is added. Step 3: So a green precipitate forms dropwise and stays insoluble in excess.
    Method:

    Match the green-precipitate-insoluble-in-excess result to an iron(II) ion.
    Examiner tips

    • Iron(II) with sodium hydroxide: green precipitate, insoluble in excess.

16. Question 3(b)

    1 marksTesting for sulfate ions with barium nitrate
    Step 1: Acidified barium nitrate gives a white precipitate of barium sulfate if sulfate ions are present. Step 2: Here no precipitate forms, so there are no sulfate ions in the solution. Step 3: So the observation is no change, showing that sulfate ions are absent.
    Method:

    Apply the barium nitrate test for sulfate and interpret the absence of a precipitate.
    Examiner tips

    • Barium nitrate + sulfate = white ppt; no ppt = no sulfate.

17. Question 3(c)

    1 marksSilver nitrate test giving a yellow precipitate
    Step 1: Acidified silver nitrate gives a precipitate whose colour identifies the halide: white for chloride, cream for bromide, yellow for iodide. Step 2: This solution contains iodide ions, which give a yellow precipitate. Step 3: So the observation is a yellow precipitate.
    Method:

    Match the silver nitrate precipitate colour to the halide ion present.
    Examiner tips

    • Silver nitrate: chloride white, bromide cream, iodide yellow.

18. Question 3(d)

    1 marksDisplacement of iodine by chlorine
    Step 1: Chlorine is more reactive than iodine, so it displaces iodide ions from solution. Step 2: The displaced iodine colours the solution yellow to brown. Step 3: So adding aqueous chlorine to the iodide solution turns it yellow to brown.
    Method:

    Apply the halogen reactivity order to predict displacement and the colour change.
    Examiner tips

    • A more reactive halogen displaces a less reactive one from its salt.

19. Question 3(e)

    1 marksConcluding there is no ammonium ion
    Step 1: If ammonium ions were present, warming with sodium hydroxide would give off ammonia, which turns damp red litmus paper blue. Step 2: Here the litmus paper stays red, so no ammonia is produced. Step 3: So the conclusion is that solid K does not contain ammonium ions.
    Method:

    Interpret the negative ammonia test as showing no ammonium ion is present.
    Examiner tips

    • Litmus staying red means no ammonia, so no ammonium ion.

20. Question 3(f)

    1 marksIdentifying the gas from the nitrate test
    Step 1: Warming a nitrate with sodium hydroxide and aluminium foil reduces the nitrate to ammonia. Step 2: Ammonia is the only common gas here that turns damp red litmus paper blue. Step 3: So the gas given off is ammonia.
    Method:

    Match the damp-red-litmus-blue result to ammonia.
    Examiner tips

    • Ammonia is the gas that turns damp red litmus paper blue.

21. Question 3(g)

    2 marksIdentifying a solid from flame and nitrate tests
    Step 1: A lilac flame shows a potassium ion. Step 2: Giving off ammonia with sodium hydroxide and aluminium foil shows a nitrate ion. Step 3: Combining a potassium ion with a nitrate ion gives potassium nitrate, $\text{KNO}_3$.
    Method:

    Deduce the cation from the flame test and the anion from the nitrate test, then name the salt.
    Examiner tips

    • Identify the metal from the flame test and the anion from the nitrate test, then combine them.

22. Question 4

    6 marksPlanning a determination of solubility
    Step 1: To find a solubility you fix the volume of water (100 cm³) and the temperature (50 °C, maintained), because solubility depends on both. Step 2: Magnesium sulfate is added in portions with stirring until no more dissolves, so the water is saturated at 50 °C. Step 3: The known mass that has dissolved (or filtering off and allowing for the undissolved solid) gives the solubility in g per 100 cm³, so this is the correct plan.
    Method:

    Design a fair plan that fixes the water volume and temperature, saturates the solution, and measures the mass dissolved.
    Examiner tips

    • Solubility needs a known volume of water, a fixed temperature, saturation, and the mass dissolved.