October/November 2025 Paper 21 Worked Answers (IGCSE Chemistry 0620 Extended)
40 questions · 40 marks · 45 minutes
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Worked answers for 40 questions
- Step 1: Heating the air gives the particles more energy, so their kinetic energy increases and they move faster. Step 2: The faster particles collide with the balloon walls harder and more often, so the pressure increases. Step 3: So both the pressure and the kinetic energy of the particles increase.Method:Link the temperature rise to faster particles, then to harder, more frequent collisions and higher pressure.Examiner tips
- Higher temperature = more kinetic energy = more frequent, harder collisions = higher pressure.
- Step 1: Lighter molecules diffuse faster than heavier ones. Ammonia (Mr 17) is lighter than hydrogen chloride (Mr 36.5). Step 2: So the ammonia diffuses faster and travels further along the tube before the gases meet. Step 3: The meeting point, where the white ring forms, is therefore nearer the slower, heavier hydrogen chloride, because hydrogen chloride has the greater molecular mass.Method:Compare the molecular masses of the two gases and link the slower (heavier) gas to where the ring forms.Examiner tips
- Lower molecular mass = faster diffusion; the product forms nearer the heavier gas.
- Step 1: Neutrons = nucleon number − protons, so W (Na) = 23 − 11 = 12 and Y (O) = 16 − 8 = 8. Step 2: Na⁺ has lost one electron, so X = 11 − 1 = 10 electrons. Step 3: O²⁻ has gained two electrons, so Z = 8 + 2 = 10 electrons, giving W = 12, X = 10, Y = 8, Z = 10.Method:Calculate neutrons from nucleon − protons, then adjust the electron count for the ion charges.Examiner tips
- Positive ions have fewer electrons than protons; negative ions have more.
- 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.Method:Match the bonding pattern of each structure to the property (sliding vs hardness) it gives.Examiner tips
- Layers + weak forces between them = lubricant; rigid 3D strong bonds = very hard cutting tool.
- 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.Method:Calculate the neutron number of every particle and find the matching pair.Examiner tips
- The 2+ charge changes electrons, not neutrons; always use nucleon − protons for neutrons.
- 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.Method:Identify the metal and non-metal, then state which atom gives the electron.Examiner tips
- Metal atoms donate electrons; non-metal atoms gain them to form ions.
- 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.Method:Recall the metallic structure and identify the mobile charged particle (delocalised electrons).Examiner tips
- Metallic conduction = delocalised electrons moving; the positive ions stay in place.
- 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.Method:Find the moles of lithium hydroxide, use the 1 : 1 ratio for the acid, then test each statement.Examiner tips
- Always convert cm³ to dm³ by dividing by 1000 before using moles = c × V.
- 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.Method:Check each formula to see whether its atom ratio can be simplified, then pick those that cannot.Examiner tips
- Try to divide the subscripts by a common factor; if you cannot, it is already empirical.
- 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.Method:Test each statement with mole calculations based on the balanced equation.Examiner tips
- Convert masses to moles, apply the equation ratio, then compare.
- 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.Method:Recall that one mole equals the Avogadro number of atoms for a monatomic gas.Examiner tips
- Remember the Avogadro constant: 6.02 × 10²³ particles per mole.
- 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.Method:Track copper moving from the anode into solution and from solution onto the cathode.Examiner tips
- Active copper electrodes: anode loses mass, cathode gains mass, solution unchanged.
- 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.Method:Decide the products (O₂ and H₂), then assign the oxidation half-equation to the anode and the reduction to the cathode.Examiner tips
- At the anode OH⁻ is oxidised to O₂; at the cathode H⁺ is reduced to H₂.
- 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.Method:Apply the energy rules for endothermic and exothermic reactions to each statement.Examiner tips
- Exothermic: bond-forming energy out > bond-breaking energy in; reactants higher than products.
- 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.Method:Identify each labelled distance on the energy profile and match it to Eₐ and ΔH.Examiner tips
- Eₐ is always measured from the reactants up to the top of the curve; ΔH is reactants − products.
- 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.Method:Read off the foam heights and identify which oxide gives the greatest rate.Examiner tips
- Compare the foam heights: the tallest in the same time is the most effective catalyst.
- 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).Method:Recall the conversion of sulfur dioxide to sulfur trioxide and check the formulae and states.Examiner tips
- Learn the key Contact-process step: 2SO₂ + O₂ ⇌ 2SO₃ (all gases).
- 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.Method:Assign oxidation numbers to manganese and iodine before and after, then state the changes.Examiner tips
- Oxidation numbers are always per atom; the uncombined element is 0.
- 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.Method:Apply the equilibrium rules for pressure (moles of gas) and temperature (exothermic direction).Examiner tips
- Yield rules: high pressure (fewer product moles) and low temperature (exothermic forward).
- 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.Method:Recall the meaning of a weak acid and reject statements that wrongly deny normal acid reactions.Examiner tips
- "Weak" describes how much the acid ionises in water, not how completely it reacts.
- 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.Method:Match the single insoluble salt to X and check all of W, Y and Z are soluble.Examiner tips
- Sodium, potassium and ammonium salts and all nitrates are soluble; calcium sulfate and lead(II) chloride are insoluble.
- 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.Method:Recall which oxides are amphoteric and reject the basic oxides.Examiner tips
- Amphoteric oxides (Al₂O₃ and ZnO) react with both acids and alkalis.
- 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.Method:Use each ion charge to deduce the number of outer electrons and hence the group.Examiner tips
- Negative ion charge = 8 − group number; positive ion charge = group number.
- 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.Method:Recall the three Group I trends and select the one that decreases.Examiner tips
- Group I trends: melting point down, reactivity up, density up.
- 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.Method:Compare the reactivity of the added halogen with the halide and decide if displacement occurs.Examiner tips
- Displacement happens only when the added halogen is more reactive than the halide present.
- 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.Method:Recall the reactivity series order and arrange the metals from least to most reactive.Examiner tips
- Know the order calcium > magnesium > zinc > copper, then reverse if asked least to most.
- 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.Method:Use the displacement to rank reactivity, then check the oxidation and agent roles.Examiner tips
- Displacement shows the displacing metal is more reactive and a better reducing agent.
- 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.Method:Decide for each treatment whether it keeps water and oxygen away from the iron.Examiner tips
- Prevent rust by keeping oxygen and water off the iron with a coating.
- 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.Method:Check each statement against the roles of cryolite and the electrode reactions.Examiner tips
- Cryolite lowers the melting point; carbon anodes burn away as carbon dioxide.
- 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.Method:Identify the products and recall which one has a harmful environmental effect.Examiner tips
- Catalytic converters still emit carbon dioxide, which is a greenhouse gas.
- 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.Method:Separate the natural air gases from the pollutants and pick the matching pair.Examiner tips
- Clean dry air: nitrogen, oxygen, argon, carbon dioxide. Pollutants: SO₂, NO₂, CO.
- 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.Method:Recall the uses of each petroleum fraction and test the statements.Examiner tips
- Match each fraction to its use: refinery gas (bottled gas), kerosene (jet fuel), lubricating (polishes, waxes).
- 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.Method:Identify the functional group in each structure and pick the one with -OH.Examiner tips
- Functional groups: -OH alcohol, C=C alkene, -COOH carboxylic acid.
- 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.Method:Identify ethane as a saturated alkane, then give the substitution products.Examiner tips
- Alkane + halogen → substitution, releasing the hydrogen halide (e.g. HCl).
- 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".Method:Open the double bond of propene and draw the matching single-bonded repeat unit.Examiner tips
- The repeat unit of an addition polymer matches the monomer but with the double bond opened.
- 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.Method:Recall the catalyst specific to the hydrogenation of alkenes.Examiner tips
- Match the catalyst to the process: nickel (hydrogenation), iron (Haber), vanadium(V) oxide (Contact), yeast (fermentation).
- 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.Method:Recall that microbial oxidation of ethanol forms ethanoic acid (vinegar).Examiner tips
- Bacterial oxidation of ethanol gives ethanoic acid, the acid in vinegar.
- 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.Method:Test each statement against the definitions of addition and condensation polymerisation.Examiner tips
- Condensation polymers lose a small molecule (water); addition polymers do not.
- 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.Method:Match each flame colour to its cation using the standard flame-test results.Examiner tips
- Learn the flame colours: Na yellow, K lilac, Cu²⁺ blue-green, Ca orange-red.
- 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.Method:Recognise two miscible liquids with different boiling points and choose fractional distillation.Examiner tips
- Use fractional distillation to separate miscible liquids by their boiling points.
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