0620/23 Oct/Nov 2025 Worked Answers (IGCSE Chemistry 0620 Extended)
40 questions · 40 marks · 45 minutes
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Worked answers for 40 questions
- Step 1: At constant temperature, squeezing a gas to a higher pressure pushes the particles closer together, so the volume decreases. Step 2: At constant pressure, heating a gas makes the particles move faster and spread out, so the volume increases. Step 3: So increasing the pressure decreases the volume, while increasing the temperature increases the volume.Method:Treat the two changes separately and recall how each one moves the gas particles.Examiner tips
- Pressure up at constant temperature: volume down. Temperature up at constant pressure: volume up.
- Step 1: Lighter gas molecules move faster at a given temperature, so the gas with the smallest relative molecular mass diffuses most quickly. Step 2: The relative molecular masses are CH₄ = 16, CO = 28, C₂H₆ = 30 and SO₂ = 64. Step 3: Methane has the smallest M_r (16), so it diffuses most quickly.Method:Calculate the relative molecular mass of each gas and pick the smallest.Examiner tips
- Work out each M_r; the smallest one diffuses fastest.
- Step 1: The nucleus of an atom contains protons and neutrons, which are together called nucleons. Step 2: The nucleon number counts all of these nucleons, that is the protons plus the neutrons. Step 3: So the nucleon number is the total number of protons and neutrons in the nucleus.Method:Recall that nucleons are the nuclear particles and add up protons and neutrons.Examiner tips
- Nucleon number = protons + neutrons (everything in the nucleus).
- 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.Method:Calculate the neutrons in each atom and compare with its protons.Examiner tips
- Subtract the proton number from the nucleon number to count neutrons.
- 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.Method:Link the positive charge to having more protons than electrons.Examiner tips
- Positive charge means protons outnumber electrons.
- 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.Method:Use the weighted-average idea and check which abundances give a mean of 63.5.
- 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.Method:Use the normal numbers of bonds for carbon (4) and oxygen (2) to assemble CH₃OH.Examiner tips
- Build CH₃OH as a CH₃ group bonded to an OH group: carbon with 4 bonds, oxygen with 2.
- 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.Method:Decide which property actually explains each use and discard the irrelevant true facts.Examiner tips
- Match the property to the use: hardness for cutting, slipperiness for lubrication.
- 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.Method:Work out the empirical formula of each compound and find the one unchanged from its molecular formula.Examiner tips
- Divide by the highest common factor; a factor of 1 means molecular = empirical.
- 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⁻.Method:Start from the known sulfate charge and balance the total charges to find the chromium charge.Examiner tips
- Remember sulfate is SO₄²⁻, then balance the charges in the formula.
- 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.Method:Choose the option about the maximum dissolved solute, not the solvent or hydrocarbons.Examiner tips
- Saturated solution = holds the most dissolved solute possible at that temperature.
- 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.Method:Convert mass to moles, use the equation ratio, then convert moles to molecules.Examiner tips
- moles = mass ÷ M_r, then molecules = moles × 6.02 × 10²³.
- 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.Method:Apply the discharge rules at each electrode for a concentrated chloride solution.Examiner tips
- Concentrated NaCl(aq): hydrogen at cathode, chlorine at anode.
- 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).Method:Follow the electrons from where they are released (anode) round the external circuit to the cathode.Examiner tips
- Electrons flow anode → wire → cathode; the electrolyte conducts by ions, not electrons.
- 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.Method:Recall the names of a hydrogen-oxygen fuel cell to identify fuel and oxidising agent.Examiner tips
- Hydrogen-oxygen fuel cell: hydrogen is the fuel, oxygen the oxidising agent, water the product.
- 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.Method:Compare the energy of bond breaking with bond making to decide exothermic or endothermic.Examiner tips
- Bond breaking takes energy in; bond making gives energy out; compare the two for the overall change.
- 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.Method:Add up the bond energies broken, add up the bond energies made, then subtract made from broken.Examiner tips
- List bonds broken and bonds made carefully, then subtract: in minus out.
- 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.Method:Separate what changes the collisions (concentration, temperature) from the fixed activation energy.Examiner tips
- Activation energy is fixed; concentration and temperature change the collisions, not the barrier.
- 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.Method:Apply le Chatelier ideas to each change and check the claimed effect.Examiner tips
- Count gas moles each side for pressure; use exo/endo for temperature; catalyst never changes yield.
- 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.Method:Recall the named Contact process conditions and distinguish them from the Haber process.Examiner tips
- Contact process: vanadium(V) oxide catalyst, about 200 kPa pressure.
- 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.Method:Pick the reaction that is characteristic of bases and reject the acid properties.Examiner tips
- Base + ammonium salt → ammonia gas; the carbonate and litmus clues point to acids.
- 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.Method:Assign oxidation numbers before and after and find the element whose number decreases.Examiner tips
- Reduction = decrease in oxidation number; nitrogen falls from +5 to +4.
- 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.Method:Use the position (metal side) and the group number (outer electrons) to describe the element.Examiner tips
- Group number tells you outer electrons; position (left/right) tells you metal/non-metal.
- 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.Method:Use the equal final volumes to rule out amount, then link the faster rate to stronger acid.Examiner tips
- Strong acid = more ionised = faster reaction; equal concentration and equal final volume isolate the strength.
- 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.Method:Identify which oxide is amphoteric and so reacts with the alkali sodium hydroxide.Examiner tips
- Amphoteric oxides like aluminium oxide react with both acids and alkalis.
- 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.Method:Recall the typical physical properties of transition metals.Examiner tips
- Transition elements: high melting point and high density.
- 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.Method:Test each statement against the known trends and the halogen displacement reaction.Examiner tips
- Chlorine displaces bromide → orange bromine; recall Group I/VII trends for the distractors.
- 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.Method:Match each described reaction to a position in the reactivity series and assign the metals.Examiner tips
- No acid reaction → below hydrogen; rapid with cold water → very reactive (Na/K/Ca).
- 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.Method:Pick the equation that converts the harmful gases into harmless products.Examiner tips
- Catalytic converter: CO + NO → harmless CO₂ + N₂.
- 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.Method:Check each statement against how the greenhouse effect works; here all three are true.Examiner tips
- All three describe how greenhouse gases trap and re-radiate the Earth's heat.
- 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.Method:Recognise that a saturated alkane reacts with a halogen by substitution.Examiner tips
- Alkane + halogen in UV light = substitution.
- 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.Method:Trace the carbon: burning to CO₂, then CO₂ to CO, then CO reducing the iron ore.Examiner tips
- Sequence: coke → CO₂, then CO₂ → CO, then CO reduces the iron oxide.
- 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.Method:Identify the oxidation steps (electrons lost) for the anode and the reduction step for the cathode.Examiner tips
- Anode reactions: oxide ions oxidised to oxygen, and that oxygen burns the carbon anode.
- 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.Method:Apply the typical reactions of acids and pick the carbonate reaction.Examiner tips
- Carboxylic acids fizz with carbonates, releasing carbon dioxide.
- 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.Method:Check each monomer for two end groups that react with –COOH, then count those that work.Examiner tips
- –COOH reacts with –OH or –NH₂; a monomer with a spare –COOH end cannot extend the diacid chain.
- 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.Method:Work out each addition product of propene and match it to the structures shown.Examiner tips
- In addition, the two new atoms go onto the two carbons that were double-bonded (adjacent).
- 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.Method:Classify each compound by functional group and group together those in the same series.Examiner tips
- Alkanes (single bonds, CₙH₂ₙ₊₂) form one series; ethane and propane are both alkanes.
- 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.Method:Check each formation/reaction of ethanoic acid for correct chemistry and naming.Examiner tips
- Ethanol → ethanoic acid by oxidation (manganate(VII) or bacteria); name the ester from acid + alcohol carefully.
- 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.Method:Find the row where the flame colour and the precipitate both point to the same cation.Examiner tips
- Copper(II): blue-green flame and a light blue hydroxide precipitate insoluble in excess.
- 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.Method:Sequence the practical steps logically, leaving the locating agent until the colourless spots need revealing.Examiner tips
- Order: baseline → spot → run → remove → mark front → spray locating agent.
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