IGCSE Chemistry 0620/21 October/November 2025 Past Paper with Answers

Question 1

A sealed balloon contains air. The temperature of the balloon is increased. How do the pressure in the balloon and the kinetic energy of the air particles change?

1. the pressure increases and the kinetic energy of the particles increases
2. the pressure decreases and the kinetic energy of the particles decreases
3. the pressure increases and the kinetic energy of the particles decreases
4. the pressure decreases and the kinetic energy of the particles increases

Question 2

In a long glass tube, cotton wool soaked in aqueous ammonia is placed at one end and cotton wool soaked in aqueous hydrogen chloride at the other end. The two gases diffuse along the tube and meet, forming a white ring of solid ammonium chloride that is nearer to the hydrogen chloride end. Which statement explains why the ammonium chloride forms nearer the hydrogen chloride?

1. hydrogen chloride has a greater molecular mass than ammonia, so it diffuses more slowly
2. aqueous ammonia is a base and aqueous hydrogen chloride is an acid
3. ammonia molecules diffuse more slowly than hydrogen chloride molecules
4. only ammonia molecules diffuse in all directions at the same time

Question 3

Consider four particles. A sodium atom Na has nucleon number 23 and 11 protons; W is its number of neutrons. A sodium ion Na⁺ has nucleon number 23 and 11 protons; X is its number of electrons. An oxygen atom O has nucleon number 16 and 8 protons; Y is its number of neutrons. An oxide ion O²⁻ has nucleon number 16 and 8 protons; Z is its number of electrons. What are the values of W, X, Y and Z?

1. W = 12, X = 10, Y = 8 and Z = 10
2. W = 11, X = 10, Y = 10 and Z = 8
3. W = 11, X = 11, Y = 8 and Z = 10
4. W = 12, X = 11, Y = 10 and Z = 8

Question 4 · Answer A

Step 1: In P the weak attractions between layers let the layers slide over each other, which makes it a good lubricant (like graphite). Step 2: In Q the rigid three-dimensional network of strong bonds makes it extremely hard, so it is used as a cutting tool (like diamond). Step 3: So only P is used as a lubricant and only Q is used as a cutting tool.

Question 5 · Answer A

Step 1: Neutrons = nucleon number − proton number, and the charge on an ion does not change the neutron count. Step 2: W = 40 − 20 = 20, X²⁺ = 41 − 20 = 21, Y = 37 − 18 = 19, Z = 37 − 17 = 20. Step 3: W and Z both have 20 neutrons, so they are the pair with the same number of neutrons.

Question 6 · Answer A

Step 1: Lithium is a Group I metal with one outer electron, and bromine is a Group VII non-metal needing one more electron. Step 2: The lithium atom donates (gives away) its single outer electron to the bromine atom. Step 3: This forms a Li⁺ ion and a Br⁻ ion held by ionic bonding, so a lithium atom donates one electron to a bromine atom.

Question 7 · Answer A

Step 1: A metal is a lattice of positive metal ions surrounded by a sea of delocalised (free) electrons. Step 2: Electricity is conducted when charged particles can move; in magnesium the delocalised electrons are free to move through the lattice. Step 3: So magnesium conducts because it has delocalised electrons that can move through the lattice.

Question 8 · Answer A

Step 1: Moles of lithium hydroxide = concentration × volume in dm³ = 0.050 × (20.0 ÷ 1000) = 0.0010 mol. Step 2: The equation is 1 : 1, so the ethanoic acid that reacted is also 0.0010 mol, which makes that statement correct. Step 3: The other statements are wrong: the moles use volume ÷ 1000, the acid concentration is 0.0010 ÷ 0.0125 = 0.080 mol/dm³, and 20.0 cm³ is 0.020 dm³, not 0.20 dm³.

Question 9 · Answer A

Step 1: The empirical formula is the simplest whole-number ratio of atoms. Step 2: CH₄ (1 : 4) and C₃H₈ (3 : 8) cannot be simplified, so their molecular formulae are also their empirical formulae. Step 3: C₂H₆ simplifies to CH₃ and C₃H₆ simplifies to CH₂, so only 1 and 3 are already empirical formulae.

Question 10 · Answer A

Step 1: The ratio is 3 CuO : 2 NH₃, so 240 g CuO reacts with 34 g NH₃; 80.0 g CuO reacts with only about 11.3 g NH₃, so statement 1 is wrong. Step 2: 4.0 g CuO = 0.05 mol, and 3 CuO → 3 Cu, so 0.05 mol Cu = 0.05 × 64 = 3.2 g, so statement 2 is correct. Step 3: 3 H₂O = 54 g and 1 N₂ = 28 g, so the water mass is greater, making statement 3 correct, giving 2 and 3 only.

Question 11 · Answer A

Step 1: One mole of any substance contains the Avogadro number of particles. Step 2: The Avogadro constant is 6.02 × 10²³ per mole. Step 3: Argon is monatomic, so 1.00 mol of argon contains 6.02 × 10²³ atoms.

Question 12 · Answer A

Step 1: With copper electrodes, copper dissolves from the positive electrode (anode) as Cu²⁺ ions and copper is deposited at the negative electrode (cathode). Step 2: Copper ions enter the solution at the same rate as they leave it, so the concentration of copper ions stays the same and the blue colour does not fade. Step 3: No oxygen is released (the anode dissolves instead), and the cathode gains mass, so the correct statement is that the copper ion concentration stays the same.

Question 13 · Answer A

Step 1: In dilute sodium chloride, oxygen is produced at the anode and hydrogen at the cathode. Step 2: At the anode, hydroxide ions lose electrons (oxidation): 4OH⁻ → 2H₂O + O₂ + 4e⁻. Step 3: At the cathode, hydrogen ions gain electrons (reduction): 2H⁺ + 2e⁻ → H₂, so the correct pair is the anode oxygen half-equation with the cathode hydrogen half-equation.

Question 14 · Answer A

Step 1: In an endothermic reaction energy is taken in, so the products store more energy than the reactants, making statement 1 correct. Step 2: The magnesium reaction is exothermic, so the reactants (magnesium and carbon dioxide) are higher in energy than the products (magnesium oxide and carbon), making statement 2 correct. Step 3: In an exothermic reaction more energy is released forming bonds than is taken in breaking them, so statement 3 is the wrong way round, leaving 1 and 2 only.

Question 15 · Answer A

Step 1: The activation energy is the minimum energy the reactants need, measured from the reactants level up to the top of the peak, which is measurement 1. Step 2: The enthalpy change is the energy difference between the reactants and the products, which is measurement 2. Step 3: So the activation energy is measurement 1 and the enthalpy change is measurement 2.

Question 16 · Answer A

Step 1: A bigger foam height in the same time means faster oxygen production, so it shows a greater rate. Step 2: Manganese(IV) oxide gives the tallest foam (5.4 cm), far higher than the others, so it speeds up the reaction the most. Step 3: So the conclusion supported by the data is that manganese(IV) oxide is the best catalyst of the four metal oxides tested.

Question 17 · Answer A

Step 1: In the Contact process, sulfur dioxide is converted to sulfur trioxide by reaction with oxygen. Step 2: The balanced equation is 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), with all species as gases. Step 3: The other equations have the wrong state (aqueous SO₃) or invented formulae such as SO₄, so the correct equation is 2SO₂(g) + O₂(g) → 2SO₃(g).

Question 18 · Answer A

Step 1: In MnO₄⁻ the oxygen is −2 (total −8) and the overall charge is −1, so manganese is +7; in Mn²⁺ it is +2, so manganese is reduced from +7 to +2. Step 2: Iodide ions I⁻ have oxidation number −1, and in I₂ the element has oxidation number 0, so iodine is oxidised from −1 to 0. Step 3: So manganese goes from +7 to +2 and iodine goes from −1 to 0.

Question 19 · Answer A

Step 1: There are 4 moles of gas on the left and 2 on the right, so increasing the pressure shifts the equilibrium towards the side with fewer moles (the ammonia), increasing the yield. Step 2: The forward reaction is exothermic, so lowering the temperature shifts the equilibrium towards the products, increasing the yield. Step 3: So increasing the pressure and decreasing the temperature increase the yield of ammonia.

Question 20 · Answer A

Step 1: Ethanoic acid is a weak acid, which means only some of its molecules split up to form ions in water. Step 2: So it only partially dissociates when added to water. Step 3: As an acid it still reacts fully like other acids, producing carbon dioxide with sodium carbonate and hydrogen with magnesium, and it is neutralised by sodium hydroxide, so the distinctive property is its partial dissociation.

Question 21 · Answer A

Step 1: The results need W, Y and Z to be soluble and only X to be insoluble. Step 2: Sodium carbonate, potassium sulfate and ammonium nitrate are all soluble, while calcium sulfate is insoluble. Step 3: Placing the insoluble calcium sulfate as X and the soluble salts as W, Y and Z matches the results, so W is sodium carbonate, X is calcium sulfate, Y is potassium sulfate and Z is ammonium nitrate.

Question 22 · Answer A

Step 1: An amphoteric oxide reacts with both acids and bases. Step 2: Aluminium oxide (Al₂O₃) and zinc oxide (ZnO) are the common amphoteric oxides. Step 3: Calcium oxide is basic and copper(II) oxide is basic, so the amphoteric oxides are 1 and 4.

Question 23 · Answer A

Step 1: X gains one electron to form a 1− ion, so it needs one more electron to fill its outer shell, placing it in Group VII. Step 2: Y loses two electrons to form a 2+ ion, so it has two outer electrons, placing it in Group II. Step 3: So X is in Group VII and Y is in Group II.

Question 24 · Answer A

Step 1: Going down Group I the melting point decreases, so property 1 decreases. Step 2: The density generally increases down the group, and the reactivity increases down the group, so properties 2 and 3 do not decrease. Step 3: So only the melting point (1) decreases down the group.

Question 25 · Answer A

Step 1: Reactivity decreases down Group VII: chlorine > bromine > iodine. A halogen displaces a halide only if it is more reactive. Step 2: Chlorine displaces bromide (mixture 1) and chlorine displaces iodide (mixture 3), because chlorine is more reactive. Step 3: Iodine cannot displace bromide (mixture 2) or chloride (mixture 4) because iodine is less reactive, so displacement occurs in 1 and 3.

Question 26 · Answer A

Step 1: In the reactivity series, calcium is the most reactive of these, then magnesium, then zinc, and copper is the least reactive. Step 2: Listing from least to most reactive reverses this order. Step 3: So the order from least to most reactive is copper, zinc, magnesium, calcium.

Question 27 · Answer A

Step 1: Aluminium displaces iron from iron(III) oxide, so aluminium is more reactive and more readily forms ions than iron. Step 2: Aluminium loses electrons (is oxidised), so it is the reducing agent, and its oxidation number rises from 0 to +3. Step 3: So the correct statement is that aluminium has a greater tendency to form ions than iron.

Question 28 · Answer A

Step 1: Rusting needs both oxygen and water; a barrier that keeps them off the iron prevents rust. Step 2: Greasing and painting both coat the iron and keep out air and water, so they prevent rusting. Step 3: Washing with distilled water still leaves the iron exposed to water and air, so it does not prevent rusting, giving greasing yes, painting yes, washing no.

Question 29 · Answer A

Step 1: At the anode oxygen is produced, and at the high temperature it reacts with the hot graphite (carbon) anode to form carbon dioxide, so the anodes burn away. Step 2: Cryolite is added to lower the melting point (not raise it), saving energy. Step 3: Oxygen forms at the anode (not the cathode) and aluminium forms at the cathode, so the correct statement is that the carbon anode reacts with oxygen to form carbon dioxide.

Question 30 · Answer A

Step 1: The reaction converts the harmful gases into nitrogen and carbon dioxide. Step 2: Carbon dioxide is a greenhouse gas that contributes to climate change, which is a disadvantage. Step 3: Nitrogen is unreactive, harmless and not acidic, and carbon dioxide (not carbon monoxide) is used in photosynthesis, so the disadvantage is that carbon dioxide is a greenhouse gas.

Question 31 · Answer A

Step 1: Clean, dry air contains nitrogen, oxygen, argon and a small amount of carbon dioxide. Step 2: Nitrogen dioxide and sulfur dioxide are pollutants that are only present in polluted air. Step 3: So argon (a clean-air gas) paired with nitrogen dioxide (a pollutant) is the correct row; carbon dioxide is in clean air too, and sulfur dioxide is a pollutant not in clean air.

Question 32 · Answer A

Step 1: The lubricating fraction provides lubricating oils and waxes, which are used to make polishes. Step 2: Bottled gas comes from the refinery-gas fraction (not naphtha), and jet aircraft are fuelled by kerosene (not diesel). Step 3: Waxes come from the lubricating/bitumen end (not kerosene), so the correct statement is that polishes are made from the lubricating fraction.

Question 33 · Answer A

Step 1: Alcohols are the homologous series with the hydroxyl (-OH) functional group bonded to a carbon. Step 2: The structure with an -OH group on the carbon chain is therefore the alcohol. Step 3: The hydrocarbon (only C and H) is an alkane, the one with a C=C is an alkene, and the one with the carboxyl group is a carboxylic acid, so only the -OH compound is an alcohol.

Question 34 · Answer A

Step 1: Ethane is a saturated alkane, so it reacts with chlorine by substitution, where a chlorine atom replaces a hydrogen atom. Step 2: The replaced hydrogen joins a chlorine atom to form hydrogen chloride, HCl. Step 3: So the reaction is substitution and the products are CH₃CH₂Cl and HCl.

Question 35 · Answer A

Step 1: In addition polymerisation the carbon-carbon double bond of each propene opens up so the monomers can join together. Step 2: The monomer is propene, CH₂=CHCH₃, so the repeat unit is -[CH₂-CH(CH₃)]- with only single bonds and a CH₃ branch. Step 3: So the correct equation uses propene as the monomer and shows single bonds in the repeat unit, which is "n propene molecules (each CH₂=CHCH₃, with a C=C double bond) join to give the repeat unit -[CH₂-CH(CH₃)]- containing only single bonds".

Question 36 · Answer A

Step 1: Adding hydrogen across the double bond of an alkene (hydrogenation) needs a metal catalyst. Step 2: The catalyst used for the hydrogenation of ethene to ethane is nickel. Step 3: Iron is used in the Haber process, vanadium(V) oxide in the Contact process, and yeast in fermentation, so the correct catalyst here is nickel.

Question 37 · Answer A

Step 1: When ethanol is left exposed to air, bacteria oxidise it. Step 2: This oxidation converts the ethanol into ethanoic acid. Step 3: A dilute solution of ethanoic acid is vinegar, so the bacterial oxidation of ethanol produces vinegar.

Question 38 · Answer A

Step 1: Proteins form from amino acids by condensation polymerisation, which releases a small molecule (water) each time a link forms. Step 2: So water is an additional product when a protein forms, making that statement correct. Step 3: Addition polymers can contain other atoms (such as chlorine in PVC), polyesters use a diol and a dicarboxylic acid (not a diamine), and the double bonds open during addition so the main chain has only single bonds, so the other statements are wrong.

Question 39 · Answer A

Step 1: A yellow flame is given by sodium ions, so X contains sodium. Step 2: A lilac flame is given by potassium ions, so Y contains potassium. Step 3: A blue-green flame is given by copper(II) ions, so Z contains copper, giving X sodium, Y potassium, Z copper.

Question 40 · Answer A

Step 1: Ethanol and water are miscible liquids with different boiling points (ethanol 78 °C, water 100 °C). Step 2: Liquids with different boiling points are separated by fractional distillation using a fractionating column. Step 3: The ethanol vapour comes off first and is condensed, so fractional distillation separates the ethanol from the water.