October/November 2025 Paper 63 Worked Answers (IGCSE Chemistry 0620 Extended)

22 questions · 40 marks · 60 minutes

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1. Question 1(a)

   1 marksIdentifying an item of apparatus
   Step 1: A wide-mouthed cone-shaped item that narrows to a stem to channel gas (or liquid) into a tube is a funnel. Step 2: Held over the flame, the funnel collects the combustion gases and directs them into the connecting tube to the rest of the apparatus. Step 3: A condenser, measuring cylinder or crucible could not collect and channel the gases this way, so item W is a funnel.
   Method:

   Match the role (collect gas above the flame and channel it into a tube) to the named apparatus.
   Examiner tips

   • A wide mouth narrowing to a stem to channel gas/liquid into a tube = a funnel.

2. Question 1(b)

   1 marksCooling to condense water vapour
   Step 1: The gases from the burning fuel contain water as steam (water vapour). Step 2: Surrounding the U-tube with ice cools the steam below its boiling point, so the water vapour condenses to liquid water that collects in the U-tube. Step 3: The ice does not change the burning, cool the pump or make weighable ice, so its purpose is to cool the steam so the water condenses.
   Method:

   Link the low temperature from the ice to condensing the water vapour to liquid.
   Examiner tips

   • Surrounding a tube with ice cools a vapour so it condenses to a liquid.

3. Question 1(c)

   2 marksChemical test for water
   Step 1: The chemical test for water uses anhydrous copper(II) sulfate or anhydrous cobalt(II) chloride. Step 2: Adding water turns anhydrous (white) copper(II) sulfate blue, and turns anhydrous (blue) cobalt(II) chloride pink. Step 3: The reversed colour changes, the hydrated salt losing water, and limewater (the test for carbon dioxide) are all wrong, so solid X is anhydrous copper(II) sulfate (white to blue) or cobalt(II) chloride (blue to pink).
   Method:

   Recall the chemical test for water and the matching colour change of the anhydrous salt.
   Examiner tips

   • Anhydrous copper(II) sulfate white -> blue; anhydrous cobalt(II) chloride blue -> pink with water.

4. Question 1(d)

   1 marksIdentifying an item of apparatus
   Step 1: A wide, strong test-tube that holds a small volume of solution with tubes through a bung is a boiling tube. Step 2: It holds the acidified potassium manganate(VII) so the gases can be bubbled through the solution. Step 3: A conical flask, a burette and an evaporating basin do not match this narrow tube-with-bung set-up, so item Y is a boiling tube.
   Method:

   Match the tube-shaped vessel that holds solution for bubbling gas to the named apparatus.
   Examiner tips

   • A wide, strong test-tube = a boiling tube.

5. Question 1(e)

   1 marksFinding an error in gas-bubbling apparatus
   Step 1: For a gas to bubble through a liquid, the inlet (delivery) tube must dip below the surface of the liquid. Step 2: Here the inlet tube ends above the surface of the manganate(VII), so the gas simply passes over the top of the liquid and out to the pump without reacting. Step 3: The outlet width, the volume of solution and the tightness of the bung do not stop the gas reaching the liquid, so the error is the inlet tube not reaching below the liquid surface.
   Method:

   Work out what must be true for gas to bubble through a liquid, then find what is wrong in the set-up.
   Examiner tips

   • Inlet tube below the surface = gas bubbles through; inlet above the surface = gas passes over the top.

6. Question 1(f)

   1 marksSulfur dioxide decolourising potassium manganate(VII)
   Step 1: Acidified potassium manganate(VII) is purple and is decolourised (turned colourless) only by a reducing agent. Step 2: Burning the fuel produces sulfur dioxide, which is a reducing gas, so it decolourises the manganate(VII). Step 3: Carbon dioxide and oxygen do not decolourise manganate(VII) and dilution would not remove the colour fully, so the change is caused by sulfur dioxide.
   Method:

   Recall what decolourises manganate(VII), then name the reducing gas made when the fuel burns.
   Examiner tips

   • Manganate(VII) decolourised = a reducing gas (sulfur dioxide) is present.

7. Question 2(a)

   4 marksReading thermometers and calculating temperature changes
   Step 1: Temperature increase = (temperature after one minute) − (initial temperature). Step 2: Experiment 1: 29.0 − 21.0 = 8.0; Experiment 2: 36.5 − 21.5 = 15.0; Experiment 3: 33.5 − 21.5 = 12.0. Step 3: Experiment 4: 38.5 − 22.0 = 16.5; Experiment 5: 46.0 − 22.0 = 24.0, giving 8.0, 15.0, 12.0, 16.5 and 24.0 °C.
   Method:

   Subtract the initial temperature from the after-one-minute temperature for each experiment.
   Examiner tips

   • Temperature increase = final − initial, to one decimal place.

8. Question 2(b)

   4 marksPlotting results and drawing a best-fit line
   Step 1: A good plot uses a linear y-axis scale chosen so the points spread over more than half the available height. Step 2: Each (volume, temperature-increase) point is plotted accurately and one smooth best-fit curve is drawn to show the overall trend. Step 3: A non-linear scale, dot-to-dot joining, or swapping the axes are all incorrect, so the correct approach is a linear scale, accurately plotted points and a smooth best-fit curve.
   Method:

   Recall the rules for a good scale, accurate plotting and a best-fit line, then pick the matching plan.
   Examiner tips

   • Best-fit graphs: linear scale, points over half the axis, one smooth line through the trend.

9. Question 2(c)

   2 marksWhy temperature increase changes with volume of acid
   Step 1: The magnesium is the limiting reactant (the acid is in excess) and the same 5 cm length is used each time, so the same amount of heat energy is released in every experiment. Step 2: That fixed amount of heat warms the acid; a larger volume (greater mass) of acid needs more energy per degree, so the temperature rise is smaller. Step 3: The rate, concentration and cooling explanations are wrong, so the temperature increase falls because the same heat is shared through a larger volume of acid.
   Method:

   Recognise the heat released is fixed by the magnesium, then see how volume of acid affects the temperature rise.
   Examiner tips

   • Fixed heat shared through a larger mass of liquid gives a smaller temperature rise.

10. Question 2(d)

    3 marksExtrapolating a best-fit curve
    Step 1: The temperature increase decreases as the volume of acid increases, so beyond 30.0 cm³ the value must be a little below the 8.0 °C found at 30.0 cm³. Step 2: Extending the smooth best-fit curve to 33.0 cm³ gives a temperature increase of about 7 °C. Step 3: Values of 12, 16 or 24 °C belong to smaller volumes of acid, so the extrapolated increase at 33.0 cm³ is about 7 °C.
    Method:

    Continue the falling best-fit curve past 30.0 cm³ and read off the value at 33.0 cm³.
    Examiner tips

    • Extrapolate by continuing the curve smoothly; the falling trend gives a value just below 8 °C.

11. Question 2(e)

    1 marksEffect of using less magnesium on the results
    Step 1: Using 2.5 cm of magnesium instead of 5 cm means only half as much magnesium reacts, so only about half as much heat energy is released. Step 2: Less heat gives a smaller temperature increase at every volume of acid, so the new line lies below the original one. Step 3: Some heat is still released, so the line stays above zero, giving a line that is below the original at all volumes but does not reach zero.
    Method:

    Link the smaller amount of magnesium to less heat, then to a lower temperature increase everywhere.
    Examiner tips

    • Less limiting reactant = less heat = a lower line at every point (but still above zero).

12. Question 2(f)(i)

    1 marksWhy a burette is used instead of a measuring cylinder
    Step 1: The volume of acid affects the temperature increase, so it must be measured accurately. Step 2: A burette has fine, closely spaced graduations and so measures volume more accurately than a measuring cylinder. Step 3: The other statements are untrue, so a burette is used because it measures the volume more accurately.
    Method:

    Compare how precisely each piece of apparatus measures a volume.
    Examiner tips

    • Burette = finer scale = more accurate volume than a measuring cylinder.

13. Question 2(f)(ii)

    1 marksWhy the mixture is stirred
    Step 1: The heat is released where the magnesium reacts, so without mixing the temperature would not be the same throughout the tube. Step 2: Stirring distributes the heat energy evenly so the thermometer measures the true temperature of the whole mixture. Step 3: Stirring does not change the heat made, protect the glass or cool the mixture, so it is done to mix the contents and spread the heat evenly.
    Method:

    Think about why an unstirred mixture would not have one even temperature.
    Examiner tips

    • Stir to even out temperature so the thermometer reads the true value.

14. Question 2(g)

    2 marksReducing heat loss to improve accuracy
    Step 1: In this exothermic reaction some heat is lost to the surroundings, which makes the measured temperature increase too low. Step 2: Wrapping insulation (lagging) around the boiling tube reduces this heat loss, so the temperature increase recorded is closer to the true value. Step 3: A thinner tube, an uncovered metal can or removing the thermometer would each increase heat loss or stop the reading, so insulating the boiling tube improves accuracy.
    Method:

    Identify what makes the reading inaccurate (heat loss), then choose the change that reduces it.
    Examiner tips

    • Insulating (lagging) the container reduces heat loss in temperature-change experiments.

15. Question 3(a)

    2 marksAqueous ammonia test on iron(III) ions
    Step 1: Iron(III) ions give a red-brown precipitate of iron(III) hydroxide with aqueous ammonia. Step 2: This precipitate is insoluble in excess ammonia, so adding more ammonia gives no further change and the red-brown precipitate remains. Step 3: The white (soluble), green (iron(II)) and light-blue to dark-blue (copper(II)) results belong to other cations, so iron(III) gives a red-brown precipitate that remains in excess.
    Method:

    Match the iron(III) ion to its hydroxide colour and its behaviour in excess ammonia.
    Examiner tips

    • Iron(III) hydroxide is red-brown and insoluble in excess ammonia.

16. Question 3(b)(i)

    1 marksSodium hydroxide test on iron(III) ions
    Step 1: Iron(III) ions react with aqueous sodium hydroxide to form iron(III) hydroxide. Step 2: Iron(III) hydroxide is a red-brown precipitate that is insoluble in excess sodium hydroxide. Step 3: A white (soluble), green (iron(II)) or no precipitate would mean other ions, so iron(III) gives a red-brown precipitate.
    Method:

    Recall the colour of iron(III) hydroxide formed with sodium hydroxide.
    Examiner tips

    • Iron(III) hydroxide is red-brown with both sodium hydroxide and ammonia.

17. Question 3(b)(ii)

    1 marksTest for nitrate ions using aluminium and alkali
    Step 1: Warming a nitrate with sodium hydroxide and aluminium foil reduces the nitrate to ammonia gas. Step 2: Ammonia is alkaline, so it turns damp red litmus paper blue. Step 3: Ammonia does not bleach litmus, leave it red or relight a splint, so the damp red litmus paper turns blue.
    Method:

    Recall the reduction of nitrate to ammonia and the effect of ammonia on damp red litmus.
    Examiner tips

    • The aluminium + alkali + warm test for nitrate gives ammonia, turning red litmus blue.

18. Question 3(b)(iii)

    1 marksIdentifying the gas produced
    Step 1: The gas turns damp red litmus paper blue, so it is alkaline. Step 2: The only common alkaline gas is ammonia, formed when the nitrate is reduced by the aluminium and sodium hydroxide. Step 3: Hydrogen, oxygen and carbon dioxide do not turn damp red litmus blue, so the gas is ammonia.
    Method:

    Use the alkaline-gas test result to name the gas as ammonia.
    Examiner tips

    • Damp red litmus turning blue = ammonia.

19. Question 3(c)

    1 marksBarium nitrate test on a sulfate-free solution
    Step 1: Acidified barium nitrate gives a white precipitate only when sulfate ions are present. Step 2: Iron(III) nitrate contains nitrate ions, not sulfate ions, so there is nothing for the barium nitrate to react with. Step 3: With no sulfate present no precipitate forms, so no change is seen.
    Method:

    Recall what the barium nitrate test detects, then decide if that ion is present.
    Examiner tips

    • Acidified barium nitrate gives a white precipitate only with sulfate; otherwise no change.

20. Question 3(d)

    1 marksIdentifying the gas from a carbonate
    Step 1: A gas that turns limewater milky is carbon dioxide. Step 2: The effervescence when a solid reacts with dilute acid to give that gas shows a carbonate releasing carbon dioxide. Step 3: Hydrogen, ammonia and oxygen do not turn limewater milky, so the gas is carbon dioxide.
    Method:

    Match the limewater result to the gas carbon dioxide.
    Examiner tips

    • Limewater turning milky = carbon dioxide.

21. Question 3(e)

    2 marksIdentifying a compound from flame and gas tests
    Step 1: A light green flame colour is given by barium ions, so the metal ion is barium. Step 2: Fizzing with dilute acid to give a gas that turns limewater milky shows the carbonate ion (releasing carbon dioxide). Step 3: Combining barium with carbonate gives barium carbonate; copper carbonate gives a blue-green flame, barium sulfate does not fizz with acid, and calcium gives an orange-red flame, so solid B is barium carbonate.
    Method:

    Identify the metal ion from the flame test and the anion from the acid test, then combine them.
    Examiner tips

    • Combine the flame colour (barium) with the carbonate test result to name the compound.

22. Question 4

    6 marksPlanning to find the percentage of a metal in an alloy
    Step 1: Weigh the alloy sample, then add excess dilute nitric acid and warm so that the copper, magnesium and manganese all dissolve while the aluminium stays as a solid (aluminium does not react with nitric acid). Step 2: Filter off the unreacted aluminium, wash it with distilled water and dry it, then find the mass of the aluminium residue. Step 3: Percentage by mass of aluminium = (mass of aluminium / mass of alloy) x 100; the other plans dissolve the wrong metals or never weigh the aluminium, so this is the correct plan.
    Method:

    Use the reactivity differences to leave aluminium as a weighable residue, then express its mass as a percentage of the alloy.
    Examiner tips

    • Use nitric acid to remove the soluble metals; the aluminium left is weighed as the residue.