0620/53 Oct/Nov 2025 Worked Answers (IGCSE Chemistry 0620 Extended)

19 questions · 40 marks · 75 minutes

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1. Question 1(a)

   5 marksRecording temperature-change results in a table
   Step 1: Volumes from a burette are recorded to one decimal place and all temperatures to a consistent number of decimal places, and each temperature increase is found by subtracting the initial temperature from the temperature after one minute. Step 2: The same 5 cm coil releases the same amount of heat each time, so a smaller volume of acid is warmed more: the temperature increase rises from experiment 1 (30.0 cm³) to experiment 5 (10.0 cm³). Step 3: Because experiment 5 uses one third of the volume of experiment 1, its temperature increase is about three times as large.
   Method:

   Recall the recording rules, then reason that the same heat into less acid gives a larger temperature rise.
   Examiner tips

   • Record all volumes to one decimal place and temperatures to a consistent precision.

2. Question 1(b)

   4 marksPlotting a results graph and drawing a best-fit line
   Step 1: A linear y-axis scale chosen so the points fill more than half the grid makes the plot easy to read and reduces plotting error. Step 2: Each of the five points is then plotted as accurately as possible. Step 3: A single straight line of best fit is drawn so that the points are evenly balanced about it, rather than joining them dot-to-dot.
   Method:

   Apply the rules for choosing a scale, plotting points and drawing a line of best fit.
   Examiner tips

   • Use a linear scale, plot every point, and draw one best-fit line.

3. Question 1(c)

   2 marksExplaining why temperature increase changes with acid volume
   Step 1: The magnesium coil is the same length in every experiment and the acid is in excess, so the same number of moles of magnesium reacts and the same amount of heat is released each time. Step 2: That fixed amount of heat is shared among the acid present, so a larger volume of acid is warmed by a smaller temperature increase. Step 3: Therefore the temperature increase falls as the volume of acid rises.
   Method:

   Identify the limiting reactant, conclude the heat released is constant, then link volume to temperature rise.
   Examiner tips

   • When the limiting reactant is fixed, the heat released is fixed; more liquid means a smaller rise.

4. Question 1(d)

   3 marksExtrapolating a best-fit line beyond the plotted points
   Step 1: 33.0 cm³ is beyond the largest plotted volume (30.0 cm³), so the best-fit line is extended smoothly in the same direction (extrapolated). Step 2: The temperature increase is read off where the extended line reaches 33.0 cm³. Step 3: The reading is reported with its unit, °C.
   Method:

   Recognise the value is beyond the data, extrapolate the line, read off and add the unit.
   Examiner tips

   • Extrapolate the line for values beyond the data, and always give the unit.

5. Question 1(e)

   1 marksPredicting the effect of using less magnesium
   Step 1: A 2.5 cm coil contains half as much magnesium, so half as much heat is released in each experiment. Step 2: With less heat released, the temperature increase is smaller at every volume of acid, so the new line lies below the original. Step 3: The reaction is still exothermic, so the temperature still rises and the line stays above zero.
   Method:

   Link less magnesium to less heat, then to a lower line that stays above zero.
   Examiner tips

   • Halving the limiting reactant lowers the whole line but keeps it positive.

6. Question 1(f)(i)

   1 marksChoosing a burette over a measuring cylinder
   Step 1: The volumes of acid (30.0, 25.0, 20.0, 15.0, 10.0 cm³) need to be measured precisely so the comparison between experiments is fair. Step 2: A burette can be read to one decimal place, so it measures these volumes more accurately than a measuring cylinder. Step 3: Therefore a burette is used because it is more accurate.
   Method:

   Recall that a burette gives a more accurate volume reading than a measuring cylinder.
   Examiner tips

   • Choose a burette over a measuring cylinder when an accurate volume is needed.

7. Question 1(f)(ii)

   1 marksReason for stirring the reaction mixture
   Step 1: The reaction releases heat at the magnesium, so without mixing the temperature would not be the same throughout the mixture. Step 2: Stirring distributes the heat evenly so that the thermometer measures the true temperature of the whole mixture. Step 3: Therefore the mixture is stirred to spread the heat evenly and give an accurate reading.
   Method:

   Link stirring to even heat distribution and an accurate temperature measurement.
   Examiner tips

   • Stir to distribute heat evenly for an accurate temperature reading.

8. Question 1(g)

   2 marksImproving accuracy by reducing heat loss
   Step 1: In an open boiling tube some of the heat released escapes to the surroundings, so the measured temperature increase is smaller than the true value. Step 2: Insulating the boiling tube (for example with a lid and lagging) reduces this heat loss. Step 3: With less heat lost, the measured temperature increase is closer to the true value, so the results are more accurate.
   Method:

   Identify heat loss as the source of error, then choose insulation to reduce it.
   Examiner tips

   • Insulate to reduce heat loss when measuring a temperature change.

9. Question 2(a)

   2 marksReaction of iron(III) ions with sodium thiosulfate
   Step 1: Iron(III) ions react with thiosulfate to form a coloured complex, so the solution first turns red. Step 2: This complex is unstable, so on standing the red colour fades to a paler colour. Step 3: Therefore the solution turns red and then the colour fades.
   Method:

   Recall the transient red colour of the iron(III)-thiosulfate reaction.
   Examiner tips

   • Iron(III) plus thiosulfate gives a red colour that does not last.

10. Question 2(b)(i)

    2 marksReducing iron(III) to iron(II) with zinc and acid
    Step 1: Zinc reacts with the dilute sulfuric acid to produce hydrogen, which is seen as effervescence. Step 2: Zinc also reduces the iron(III) ions to iron(II) ions, so the solution changes colour to pale green. Step 3: Therefore effervescence is seen and the solution becomes (pale) green.
    Method:

    Separate the gas from the acid-zinc reaction and the colour change from the reduction of iron(III).
    Examiner tips

    • Zinc and acid reduce iron(III) (yellow/brown) to iron(II) (green) with effervescence.

11. Question 2(b)(ii)

    1 marksSodium hydroxide test for iron(II) ions
    Step 1: Iron(II) ions react with aqueous sodium hydroxide to form iron(II) hydroxide. Step 2: Iron(II) hydroxide is insoluble and green. Step 3: Therefore a green precipitate forms.
    Method:

    Match the iron(II) ion to its green hydroxide precipitate.
    Examiner tips

    • Iron(II): green precipitate with sodium hydroxide.

12. Question 2(c)

    2 marksAqueous ammonia test for iron(III) ions
    Step 1: Iron(III) ions react with aqueous ammonia to form iron(III) hydroxide, a brown (red-brown) precipitate. Step 2: Iron(III) hydroxide is insoluble in excess ammonia, so adding more makes no further change. Step 3: Therefore a brown precipitate forms and remains in excess.
    Method:

    Match the iron(III) ion to its brown hydroxide that is insoluble in excess.
    Examiner tips

    • Iron(III): brown precipitate with ammonia, insoluble in excess.

13. Question 2(d)

    1 marksBarium nitrate test with no sulfate present
    Step 1: Acidified barium nitrate gives a white precipitate only when sulfate ions are present. Step 2: This solution contains no sulfate ions, so no precipitate can form. Step 3: Therefore no change is seen.
    Method:

    Apply the sulfate test and note the absence of sulfate gives no change.
    Examiner tips

    • Acidified barium nitrate stays clear when no sulfate is present.

14. Question 2(e)

    2 marksTest for nitrate ions using sodium hydroxide and aluminium
    Step 1: In warm alkaline solution the aluminium reduces nitrate ions to ammonia gas, which is seen as effervescence. Step 2: Ammonia is alkaline, so it turns damp red litmus paper blue. Step 3: Therefore effervescence is seen and the gas turns damp red litmus paper blue, confirming nitrate ions.
    Method:

    Recall the nitrate test and match ammonia to the damp red litmus result.
    Examiner tips

    • Nitrate test: sodium hydroxide + aluminium, warm, gives ammonia (litmus blue).

15. Question 2(f)(i)

    1 marksOxidation of iodide by iron(III) ions
    Step 1: Iron(III) ions oxidise iodide ions to iodine. Step 2: Iodine in solution is brown. Step 3: Therefore the solution turns brown.
    Method:

    Recognise iron(III) as an oxidising agent that releases brown iodine from iodide.
    Examiner tips

    • Iron(III) + iodide gives brown iodine in solution.

16. Question 2(f)(ii)

    1 marksStarch test for iodine
    Step 1: Starch is the specific indicator for iodine. Step 2: When starch meets iodine it forms a blue-black colour. Step 3: Therefore a blue-black colour forms, confirming iodine is present.
    Method:

    Apply the starch test for iodine.
    Examiner tips

    • Starch + iodine = blue-black.

17. Question 2(f)(iii)

    1 marksRemoving iodine with sodium thiosulfate
    Step 1: Sodium thiosulfate reduces iodine back to colourless iodide ions. Step 2: With the iodine removed, the starch no longer has iodine to colour, so the blue-black colour disappears. Step 3: Therefore the colour disappears, leaving a colourless or pale solution.
    Method:

    Link thiosulfate removing iodine to the loss of the blue-black colour.
    Examiner tips

    • Thiosulfate reduces iodine, removing the blue-black colour.

18. Question 2(g)

    2 marksIdentifying the cation and anion in a salt
    Step 1: A brown precipitate with sodium hydroxide that is insoluble in excess shows iron(III) ions (iron(II) would give a green precipitate). Step 2: Warming with sodium hydroxide and aluminium foil to give a gas that turns damp red litmus paper blue (ammonia) shows nitrate ions. Step 3: Combining the iron(III) cation with the nitrate anion gives iron(III) nitrate.
    Method:

    Read each test result, deduce the ion it indicates, then name both ions.
    Examiner tips

    • Identify the cation and the anion separately, then combine them.

19. Question 3

    6 marksPlanning a quantitative investigation by selective reaction
    Step 1: Find the mass of the duralumin sample, then add excess dilute nitric acid and warm; the copper, magnesium and manganese all react and dissolve, but the aluminium does not react and is left as a solid. Step 2: Filter off the aluminium residue, wash it with distilled water and dry it, then find its mass. Step 3: The percentage by mass of aluminium = (mass of the aluminium residue / mass of the original sample) × 100, which gives the percentage of aluminium in the alloy.
    Method:

    Use selective reactivity to isolate aluminium as a residue, then find its mass fraction of the sample.
    Examiner tips

    • Choose the acid that leaves only the metal of interest as a residue, then weigh it.