IGCSE Chemistry 0620/63 October/November 2025 Past Paper with Answers

Question 1

A fuel is burned and the gases produced are drawn through cooled apparatus by a suction pump to show that water is a product. To collect the hot gases from directly above the burning fuel and channel them into a narrow connecting tube, a wide-mouthed cone-shaped item of apparatus, labelled W, is held over the flame. What is this item of apparatus?

1. a funnel, whose wide mouth collects the gases and whose narrow stem channels them onward
2. a condenser, a long water-jacketed tube used to cool and collect a distillate
3. a measuring cylinder, a tall straight-sided tube marked with a volume scale
4. a crucible, a small open dish used to heat solids strongly over a flame

Question 2

When the fuel is burned, the steam (water vapour) produced passes into a U-tube that is surrounded by ice. Why is the U-tube surrounded by ice?

1. to cool the steam so that the water vapour condenses to liquid water that can be collected and tested
2. to speed up the burning of the fuel so that more water vapour is produced each second
3. to keep the suction pump cold so that it can pull the gases through more strongly
4. to freeze the water vapour into ice so that its mass can be weighed directly

Question 3

A solid in the cooled U-tube changes colour when liquid water collects on it, which confirms that water is present. Which solid and colour change correctly show the presence of water?

1. anhydrous copper(II) sulfate, changing from white to blue (or anhydrous cobalt(II) chloride, changing from blue to pink)
2. anhydrous copper(II) sulfate, changing from blue to white (or anhydrous cobalt(II) chloride, changing from pink to blue)
3. hydrated copper(II) sulfate, changing from blue to white as it gives up its water of crystallisation
4. limewater, changing from colourless to milky as the water passes through it

Question 4 · Answer A

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.

Question 5 · Answer A

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.

Question 6 · Answer A

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.

Question 7 · Answer A

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.

Question 8 · Answer A

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.

Question 9 · Answer A

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.

Question 10 · Answer A

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.

Question 11 · Answer A

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.

Question 12 · Answer A

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.

Question 13 · Answer A

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.

Question 14 · Answer A

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.

Question 15 · Answer A

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.

Question 16 · Answer A

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.

Question 17 · Answer A

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.

Question 18 · Answer A

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.

Question 19 · Answer A

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.

Question 20 · Answer A

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.

Question 21 · Answer A

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.

Question 22 · Answer A

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.