IGCSE Chemistry 0620/23 October/November 2025 Past Paper with Answers

Question 1

A fixed mass of gas is studied. The pressure on the gas and the temperature of the gas are each increased separately (one at a time). What happens to the volume of the gas in each case?

1. increasing the pressure decreases the volume, and increasing the temperature increases the volume
2. increasing the pressure increases the volume, and increasing the temperature increases the volume
3. increasing the pressure has no effect on the volume, and increasing the temperature decreases the volume
4. increasing the pressure has no effect on the volume, and increasing the temperature increases the volume

Question 2

Four gases are compared at room temperature and pressure: methane ($\text{CH}_4$, $M_r = 16$), ethane ($\text{C}_2\text{H}_6$, $M_r = 30$), carbon monoxide ($\text{CO}$, $M_r = 28$) and sulfur dioxide ($\text{SO}_2$, $M_r = 64$). Which gas diffuses most quickly?

1. methane, CH₄
2. ethane, C₂H₆
3. carbon monoxide, CO
4. sulfur dioxide, SO₂

Question 3

Which statement correctly defines the nucleon number (mass number) of an atom?

1. it is the total number of protons and neutrons in the nucleus of an atom
2. it is the total number of protons in the nucleus of an atom
3. it is the total number of electrons surrounding the nucleus of an atom
4. it is the total number of neutrons in the nucleus of an atom

Question 4 · Answer A

Step 1: Work out the neutrons for each atom: atom 1 has 12 − 6 = 6 neutrons (equal to 6 protons), atom 2 has 14 − 6 = 8 neutrons (more than 6 protons). Step 2: Atom 3 has 14 − 7 = 7 neutrons (equal to 7 protons), and atom 4 has 29 − 14 = 15 neutrons (more than 14 protons). Step 3: Only atoms 2 and 4 have more neutrons than protons.

Question 5 · Answer A

Step 1: A positive ion forms when an atom loses one or more electrons, so it has fewer electrons than it started with. Step 2: The number of protons does not change, so the positive ion now has more protons than electrons. Step 3: This extra positive charge from the protons gives the ion its overall positive charge, so all positive ions have more protons than electrons.

Question 6 · Answer A

Step 1: The relative atomic mass is the weighted average of the isotope masses, so the value 63.5 lies between 63 and 65 but closer to 63. Step 2: Testing 75% copper-63 and 25% copper-65 gives (0.75 × 63) + (0.25 × 65) = 47.25 + 16.25 = 63.5. Step 3: This matches the relative atomic mass, so the abundances are 75% copper-63 and 25% copper-65.

Question 7 · Answer A

Step 1: In methanol the carbon atom forms four single covalent bonds: three to hydrogen atoms and one to the oxygen atom. Step 2: The oxygen atom forms two single bonds, one to the carbon atom and one to a hydrogen atom, giving the O–H group. Step 3: So methanol is a carbon atom bonded to three hydrogen atoms and to an oxygen atom, with that oxygen also bonded to one hydrogen atom.

Question 8 · Answer A

Step 1: Diamond is a giant covalent structure with every carbon strongly bonded to four others, making it very hard, which is why it is used for a drill tip (statement 2 correct). Step 2: Graphite has layers that can slide over each other, so it is soft and slippery and works as a lubricant (statement 4 correct). Step 3: Diamond being a non-conductor and graphite being a conductor are true facts but are not the reasons for these particular uses, so statements 1 and 3 are wrong, leaving 2 and 4.

Question 9 · Answer A

Step 1: The empirical formula divides the atom numbers by their highest common factor. For butan-1-ol C₄H₁₀O the numbers 4, 10 and 1 have a highest common factor of 1, so the empirical formula stays C₄H₁₀O. Step 2: Butane C₄H₁₀ simplifies to C₂H₅, but-1-ene C₄H₈ simplifies to CH₂, and butanoic acid C₄H₈O₂ simplifies to C₂H₄O. Step 3: Only butan-1-ol cannot be simplified, so its empirical formula equals its molecular formula.

Question 10 · Answer A

Step 1: The sulfate ion is a familiar ion with the formula SO₄²⁻, so it carries a 2− charge. Step 2: The formula has two chromium ions to three sulfate ions; the three sulfates carry a total charge of 3 × 2− = 6−. Step 3: To balance this, the two chromium ions must carry a total of 6+, so each chromium ion is 3+, giving Cr³⁺ and SO₄²⁻.

Question 11 · Answer A

Step 1: A saturated solution is one in which no more solute can dissolve at a given temperature. Step 2: This means it holds the maximum amount of dissolved solute possible at that temperature. Step 3: The other options describe the solvent or use the word "saturated" for hydrocarbons, which is a different meaning, so the correct description is the maximum concentration of dissolved solute.

Question 12 · Answer A

Step 1: Moles of CaCO₃ = mass ÷ relative formula mass = 10.0 ÷ 100 = 0.100 mol. Step 2: The equation shows 1 mole of CaCO₃ gives 1 mole of CO₂ gas, so 0.100 mol of CO₂ is produced. Step 3: Number of molecules = moles × Avogadro constant = 0.100 × 6.02 × 10²³ = 6.02 × 10²² molecules.

Question 13 · Answer A

Step 1: At the cathode the choice is between hydrogen and sodium; the less reactive element is produced, so hydrogen is released rather than the reactive sodium. Step 2: At the anode the choice is between chlorine and oxygen; with a concentrated chloride solution the chloride ions are discharged, so chlorine is produced. Step 3: So hydrogen forms at the cathode and chlorine at the anode.

Question 14 · Answer A

Step 1: At the anode X, bromide ions are oxidised and release electrons into the electrode. Step 2: These electrons travel through the external wire and the power supply round to the cathode Y, where lead ions gain them. Step 3: So electrons move from X to Y through the external wire; electrons never travel through the electrolyte (ions carry the charge there).

Question 15 · Answer A

Step 1: In a hydrogen-oxygen fuel cell the fuel that is oxidised is hydrogen. Step 2: The substance that brings about the oxidation (the oxidising agent) is oxygen. Step 3: Hydrogen and oxygen react to form water, so the fuel is hydrogen and the oxidising agent is oxygen.

Question 16 · Answer A

Step 1: Breaking bonds takes in energy (endothermic) and making bonds releases energy (exothermic). Step 2: In an exothermic reaction more energy is given out making bonds than is taken in breaking bonds, so the overall change releases energy. Step 3: This means the energy to break bonds is less than the energy released making bonds, which is the correct statement.

Question 17 · Answer A

Step 1: Energy to break the bonds in the reactants = C=C + 4(C–H) + H–H = 614 + (4 × 410) + 436 = 614 + 1640 + 436 = 2690 kJ/mol. Step 2: Energy released making the bonds in ethane = C–C + 6(C–H) = 350 + (6 × 410) = 350 + 2460 = 2810 kJ/mol. Step 3: Enthalpy change = energy in − energy out = 2690 − 2810 = −120 kJ/mol.

Question 18 · Answer A

Step 1: Increasing the concentration puts more acid particles in the same volume, so collisions happen more often (a higher collision rate). Step 2: The activation energy is fixed for a given reaction and is not changed by concentration or temperature. Step 3: So increasing the concentration raises the collision rate but does not change the activation energy.

Question 19 · Answer A

Step 1: A catalyst speeds up the forward and backward reactions equally, so it does not change the equilibrium yield (statement 1 correct). Step 2: The forward reaction is exothermic, so lowering the temperature shifts the equilibrium forwards and increases the yield (statement 2 correct). Step 3: The forward reaction goes from 2 moles of gas to 1 mole of gas, so decreasing the pressure shifts the equilibrium backwards (towards more moles) and decreases the yield, making statement 3 wrong, leaving 1 and 2.

Question 20 · Answer A

Step 1: The Contact process uses vanadium(V) oxide as the catalyst (iron is the catalyst for the Haber process, not the Contact process). Step 2: The reaction already gives a high yield, so only a modest pressure of about 200 kPa (about 2 atmospheres) is needed. Step 3: So the conditions are a pressure of 200 kPa and a vanadium(V) oxide catalyst.

Question 21 · Answer A

Step 1: A base reacts with an ammonium salt such as ammonium chloride, on warming, to release ammonia gas. Step 2: Reacting with a carbonate to give carbon dioxide, and turning blue litmus red, are properties of acids, not bases. Step 3: Bases do not react with copper to give hydrogen, so the correct property of a base is reacting with ammonium chloride to produce ammonia.

Question 22 · Answer A

Step 1: Reduction is a decrease in oxidation number. In HNO₃ nitrogen has an oxidation number of +5, and in NO₂ it is +4. Step 2: Nitrogen goes from +5 to +4, a decrease, so nitrogen is reduced. Step 3: Copper goes from 0 to +2 (oxidised), while oxygen and hydrogen keep their oxidation numbers, so the element reduced is nitrogen.

Question 23 · Answer A

Step 1: Elements on the left-hand side of the Periodic Table are metals, so a Group II element is a metal. Step 2: The group number gives the number of electrons in the outer shell, so a Group II element has 2 outer-shell electrons. Step 3: So the element is a metal with 2 electrons in its outer shell.

Question 24 · Answer A

Step 1: Both acids have the same concentration and produce the same final volume of hydrogen, so the same total amount of acid reacts in each case. Step 2: Acid R reacts faster (a steeper curve), which means it produces hydrogen ions more readily, so it is more fully ionised. Step 3: A more fully ionised acid is a stronger acid, so acid R is a stronger acid than acid S.

Question 25 · Answer A

Step 1: An oxide that reacts with an alkali such as sodium hydroxide must be acidic or amphoteric. Step 2: Aluminium oxide is amphoteric, so it reacts with both acids and alkalis, including sodium hydroxide, to form a salt and water. Step 3: Calcium oxide, copper(II) oxide and sodium oxide are basic oxides that do not react with sodium hydroxide, so the answer is aluminium oxide.

Question 26 · Answer A

Step 1: Transition elements are typical metals with strong metallic bonding, which gives them high melting points. Step 2: Their atoms are packed closely with relatively high masses, so they have high densities. Step 3: So a transition element has both a high melting point and a high density.

Question 27 · Answer A

Step 1: Chlorine is more reactive than bromine, so it displaces bromine from potassium bromide, forming bromine which gives an orange solution (correct). Step 2: Group VII elements exist as diatomic molecules (like Cl₂), not monatomic, and Group I reactivity increases down the group, so caesium is more reactive than lithium. Step 3: The melting point of Group I metals decreases down the group, so the only correct statement is that potassium bromide reacts with chlorine to give an orange solution.

Question 28 · Answer A

Step 1: P does not react with dilute acid, so it is below hydrogen in the reactivity series; copper fits. Step 2: Q is more reactive than magnesium, so potassium (a very reactive metal) fits, and R reacts rapidly with cold water, which calcium does. Step 3: S reacts with dilute acid but not cold water, which magnesium does, so P = copper, Q = potassium, R = calcium and S = magnesium.

Question 29 · Answer A

Step 1: A catalytic converter removes the harmful gases carbon monoxide (CO) and nitrogen oxides (NO) from the exhaust. Step 2: It converts them into harmless gases: carbon dioxide and nitrogen. Step 3: The equation 2CO + 2NO → 2CO₂ + N₂ shows the harmful gases being changed into harmless carbon dioxide and nitrogen, so it is the correct reaction.

Question 30 · Answer A

Step 1: Greenhouse gas molecules absorb thermal energy and then re-emit it, so statement 1 is correct. Step 2: The warm Earth radiates thermal energy back out, so the Earth is a source of thermal energy and statement 2 is correct. Step 3: When a greenhouse gas molecule emits thermal energy it does so in all directions, including back towards the Earth, so statement 3 is correct, making all three correct.

Question 31 · Answer A

Step 1: Methane is a saturated alkane with only single bonds, so it cannot undergo addition. Step 2: In ultraviolet light, a chlorine atom takes the place of a hydrogen atom on the methane, which is a substitution reaction. Step 3: So the reaction between methane and chlorine in ultraviolet light is substitution.

Question 32 · Answer A

Step 1: First the coke burns completely in the hot air blast, so carbon is oxidised to carbon dioxide (step 1). Step 2: This carbon dioxide then rises and reacts with more hot coke, being reduced to carbon monoxide (step 3). Step 3: The carbon monoxide then reduces the iron(III) oxide to iron (step 2), giving the order 1, 3, 2.

Question 33 · Answer A

Step 1: Oxidation (loss of electrons) happens at the anode. Reaction 3, 2O²⁻ → O₂ + 4e⁻, is the oxidation of oxide ions, so it occurs at the anode. Step 2: The oxygen produced reacts with the hot carbon anode, C + O₂ → CO₂, so reaction 2 also occurs at the anode. Step 3: Reaction 1 is the gain of electrons by aluminium ions (reduction), which happens at the cathode, so the anode reactions are 2 and 3.

Question 34 · Answer A

Step 1: Ethanoic acid is a weak acid, and acids react with carbonates to give a salt, water and carbon dioxide. Step 2: So aqueous ethanoic acid reacts with sodium carbonate to produce carbon dioxide gas (correct). Step 3: Acids react with magnesium to give hydrogen (not oxygen), turn blue litmus red (not red litmus blue), and turn methyl orange red (not yellow), so the only correct statement is the reaction with sodium carbonate.

Question 35 · Answer A

Step 1: A –COOH group reacts with an –NH₂ group or an –OH group. To keep a chain growing with the diacid, a monomer needs two such reactive groups at its ends. Step 2: Monomer 1 (–NH₂ at both ends), monomer 3 (–OH and –NH₂) and monomer 4 (–OH at both ends) each have two groups that react with –COOH, so all three form polymers. Step 3: Monomer 2 has a –COOH at one end, which will not react with the diacid's –COOH, so the chain cannot grow; only 3 monomers form a polymer.

Question 36 · Answer A

Step 1: Adding hydrogen to propene gives propane (structure 3), where the double bond becomes a single bond and two hydrogen atoms are added across it. Step 2: Adding steam to propene gives propanol (structure 4), with an –OH group added across the double bond. Step 3: Adding bromine gives 1,2-dibromopropane, with bromine on adjacent carbons; structures 1 and 2 have the bromine atoms in the wrong positions, so only structures 3 and 4 are genuine products.

Question 37 · Answer A

Step 1: Members of the same homologous series have the same functional group and general formula. Step 2: Ethane (compound 1) and propane (compound 4) are both alkanes with only single bonds and the general formula CₙH₂ₙ₊₂. Step 3: Ethene is an alkene and ethanol is an alcohol, which are different series, so only compounds 1 and 4 are in the same homologous series.

Question 38 · Answer A

Step 1: Ethanol is oxidised to ethanoic acid by acidified potassium manganate(VII), so this statement is correct. Step 2: Bacteria oxidise (not reduce) ethanol to ethanoic acid, so that statement is wrong. Step 3: Magnesium ethanoate has the formula (CH₃COO)₂Mg, not CH₃COOMg, and ethanoic acid with methanol gives methyl ethanoate (not ethyl methanoate), so only the manganate(VII) oxidation statement is correct.

Question 39 · Answer A

Step 1: A blue-green flame is the flame test result for copper(II) ions. Step 2: Copper(II) ions also give a light blue precipitate of copper(II) hydroxide with sodium hydroxide, which is insoluble in excess. Step 3: Both observations point to the same cation, copper(II), so this combination is possible; the other rows pair a flame colour with a precipitate that belongs to a different cation.

Question 40 · Answer A

Step 1: First draw the pencil baseline (5) and place the spot of mixture on it (4), then stand the paper in the solvent (2). Step 2: Remove the paper when the solvent front is near the top (1) and mark the solvent front with a pencil line (6) before it evaporates. Step 3: Because the mixture is colourless, finally spray with a locating agent to reveal the spots (3), giving the order 5, 4, 2, 1, 6, 3.