May/June 2025 Paper 32 Worked Answers (IGCSE Chemistry 0620 Core)

53 questions · 80 marks · 75 minutes

Question papers and mark schemes are copyright Cambridge International. We do not reproduce them: the worked answers here are written by The Practice Book. Have the paper open alongside. Get the official paper from Cambridge

1. Question 1(a)

   1 marksElement forming a 2- ion
   Step 1: Oxygen is in Group VI, so an atom has 6 outer electrons and gains 2 electrons to reach a full outer shell. Step 2: Gaining 2 electrons gives the atom two extra negative charges, so it forms an ion with a 2- charge. Step 3: Bromine forms a 1- ion, aluminium a 3+ ion and potassium a 1+ ion, so only oxygen forms a 2- ion.
   Method:

   Use each element group number to work out the charge of the ion it forms.
   Examiner tips

   • Group number gives the outer electrons; a non-metal gains (8 - group) electrons.

2. Question 1(b)

   1 marksIdentifying an ionic compound
   Step 1: Ionic compounds are formed when a metal combines with a non-metal. Step 2: Magnesium oxide is made from the metal magnesium and the non-metal oxygen, so it is an ionic compound. Step 3: Carbon dioxide and sulfur dioxide are made from non-metals only (covalent), and oxygen is an element, so the ionic compound is magnesium oxide.
   Method:

   Pick the compound made from a metal combined with a non-metal.
   Examiner tips

   • Ionic = metal + non-metal; covalent = non-metal + non-metal.

3. Question 1(c)

   1 marksMetal used as a catalyst
   Step 1: A catalyst speeds up a reaction without being used up. Step 2: Iron is the catalyst used in the Haber process to make ammonia. Step 3: Neon, oxygen and carbon dioxide are not used as catalysts here, so the catalyst is iron.
   Method:

   Recall which metal is used as a catalyst and pick it from the list.
   Examiner tips

   • Iron catalyses the Haber process; transition metals are common catalysts.

4. Question 1(d)

   1 marksGas at about 21% of clean dry air
   Step 1: Clean dry air is about 78% nitrogen and 21% oxygen, with small amounts of other gases. Step 2: The gas making up roughly 21% is therefore oxygen. Step 3: Carbon dioxide is about 0.04%, neon is a trace gas and sulfur dioxide is a pollutant, so the answer is oxygen.
   Method:

   Recall the composition of clean dry air and pick the gas at about 21%.
   Examiner tips

   • Remember the 78% nitrogen / 21% oxygen split of clean dry air.

5. Question 1(e)

   1 marksGas needed for photosynthesis
   Step 1: In photosynthesis, plants take in carbon dioxide and water and use light energy to make glucose and oxygen. Step 2: The gas needed (a reactant) is carbon dioxide. Step 3: Oxygen is a product (released), neon is unreactive and sulfur dioxide is a pollutant, so the gas needed for photosynthesis is carbon dioxide.
   Method:

   Recall the reactants of photosynthesis and pick the gas taken in.
   Examiner tips

   • Photosynthesis: carbon dioxide + water → glucose + oxygen.

6. Question 1(f)

   1 marksMost reactive metal
   Step 1: In the reactivity series, potassium is near the top and reacts very vigorously. Step 2: Potassium reacts violently even with cold water, more vigorously than calcium, magnesium, aluminium or iron. Step 3: So the most reactive metal in the list is potassium.
   Method:

   Order the metals by the reactivity series and pick the highest one.
   Examiner tips

   • Order: potassium > calcium > magnesium > aluminium > iron.

7. Question 1(g)

   1 marksGas tested with acidified potassium manganate(VII)
   Step 1: Acidified aqueous potassium manganate(VII) is purple and is decolourised by a reducing gas. Step 2: Sulfur dioxide is a reducing gas, so it turns acidified potassium manganate(VII) from purple to colourless. Step 3: Carbon dioxide (limewater), oxygen (glowing splint) and neon have different or no tests, so the gas tested with potassium manganate(VII) is sulfur dioxide.
   Method:

   Match each gas to its standard test and pick the one tested with potassium manganate(VII).
   Examiner tips

   • Sulfur dioxide is a reducing gas: it decolourises purple potassium manganate(VII).

8. Question 1(h)

   1 marksIdentifying a noble gas
   Step 1: Noble gases are the unreactive elements in Group VIII (Group 0) of the Periodic Table. Step 2: Neon is in Group VIII, so neon is a noble gas. Step 3: Oxygen is a reactive Group VI element, carbon dioxide is a compound and potassium is a metal, so the noble gas is neon.
   Method:

   Locate each substance and pick the unreactive Group VIII element.
   Examiner tips

   • Noble gases (helium, neon, argon...) are in Group VIII (0) and are unreactive.

9. Question 1(i)

   1 marksElement in Group VII
   Step 1: Group VII contains the halogens, such as chlorine, bromine and iodine. Step 2: Bromine is a halogen, so it is in Group VII. Step 3: Calcium is in Group II, potassium in Group I and aluminium in Group III, so the Group VII element is bromine.
   Method:

   Identify the group of each element and pick the Group VII halogen.
   Examiner tips

   • Group VII = the halogens: fluorine, chlorine, bromine, iodine.

10. Question 2(a)(i)

    1 marksNaming a compound from its formula
    Step 1: The formula $\text{MgSO}_4$ contains a magnesium ion and a sulfate ion, $\text{SO}_4^{2-}$. Step 2: A compound of magnesium with the sulfate ion is named magnesium sulfate. Step 3: Magnesium sulfide would be MgS and magnesium sulfite would be $\text{MgSO}_3$, so $\text{MgSO}_4$ is magnesium sulfate.
    Method:

    Identify the ions in the formula and name the compound from them.
    Examiner tips

    • $\text{SO}_4$ = sulfate, $\text{SO}_3$ = sulfite, S alone = sulfide.

11. Question 2(a)(ii)

    1 marksProportion calculation of dissolved mass
    Step 1: The mass of compounds is proportional to the volume of sea water. Step 2: The mass from 750 cm³ = 12.0 × (750 ÷ 500) = 12.0 × 1.5. Step 3: This gives 18.0 g of compounds from 750 cm³.
    Method:

    Set up the proportion of masses to volumes and solve for the new mass.
    Examiner tips

    • Mass is proportional to volume: multiply by (new volume ÷ original volume).

12. Question 2(a)(iii)

    1 marksCarbonate reacting with acid
    Step 1: Carbonates react with dilute acids to form a salt, water and carbon dioxide. Step 2: Calcium carbonate is a carbonate, so it reacts with acid to produce carbon dioxide. Step 3: Sodium chloride, magnesium sulfate and potassium bromide are not carbonates, so the compound that produces carbon dioxide is calcium carbonate.
    Method:

    Identify the carbonate, which is the compound that fizzes with acid.
    Examiner tips

    • Only carbonates fizz with dilute acid, releasing carbon dioxide.

13. Question 2(b)

    2 marksTest for bromide ions
    Step 1: Halide ions are tested by acidifying with dilute nitric acid and then adding aqueous silver nitrate. Step 2: With bromide ions a cream precipitate of silver bromide forms. Step 3: A white precipitate indicates chloride, a yellow precipitate indicates iodide and limewater tests for carbon dioxide, so the correct test gives a cream precipitate with silver nitrate.
    Method:

    Recall the silver nitrate halide test and match the colour to bromide.
    Examiner tips

    • Chloride → white, bromide → cream, iodide → yellow precipitate with silver nitrate.

14. Question 2(c)

    2 marksElectronic configuration and charge of a calcium ion
    Step 1: A calcium atom has 20 electrons arranged 2,8,8,2. Step 2: Losing the two outer electrons leaves 18 electrons arranged 2,8,8. Step 3: Losing two negative electrons leaves the ion with a charge of 2+, so the ion is 2,8,8 with a 2+ charge.
    Method:

    Take the atom configuration, remove the outer electrons, and assign the matching charge.
    Examiner tips

    • A Group II atom loses two electrons to form a 2+ ion with a full outer shell.

15. Question 2(d)

    2 marksArrangement and motion of particles in a solid
    Step 1: In a solid the particles are packed closely in a regular arrangement (rows or a lattice). Step 2: They have only enough energy to vibrate about fixed positions, so they cannot move from place to place. Step 3: So the particles in a solid are in a regular arrangement and vibrate about fixed positions.
    Method:

    Recall the kinetic particle description of a solid for both arrangement and motion.
    Examiner tips

    • Solid = regular arrangement + vibrating in fixed positions.

16. Question 2(e)

    1 marksSource of nitrate ions in water
    Step 1: Nitrate ions are found in fertilisers (NPK fertilisers contain nitrogen). Step 2: When fertilisers are washed off farmland by rain, the nitrate ions are carried into rivers and the sea. Step 3: Dissolved gases and insoluble rock do not provide nitrate ions, so fertilisers are a correct source.
    Method:

    Recall what nitrate ions are used in and pick the matching source.
    Examiner tips

    • Nitrates (and phosphates) in water usually come from fertilisers.

17. Question 3(a)(i)

    1 marksGroup number and outer electrons
    Step 1: The group number of an element equals the number of electrons in its outer shell. Step 2: Phosphorus is in Group V, so a phosphorus atom has 5 electrons in its outer shell. Step 3: The number of shells gives the period (not the group), and neutrons or relative atomic mass do not decide the group, so the reason is that it has 5 outer electrons.
    Method:

    Link the group number to the number of outer-shell electrons.
    Examiner tips

    • Group number tells you the outer-shell electrons; period number tells you the shells.

18. Question 3(a)(ii)

    3 marksProtons, neutrons and electrons in isotopes
    Step 1: The proton number is 15, so each atom has 15 protons and (being neutral atoms) 15 electrons. Step 2: Neutrons = mass number − proton number, so phosphorus-31 has 31 − 15 = 16 neutrons and phosphorus-32 has 32 − 15 = 17 neutrons. Step 3: So phosphorus-31 is 15 protons, 16 neutrons, 15 electrons and phosphorus-32 is 15 protons, 17 neutrons, 15 electrons.
    Method:

    Use the proton number for protons and electrons, then subtract it from the mass number for neutrons.
    Examiner tips

    • Isotopes have the same protons and electrons but different numbers of neutrons.

19. Question 3(a)(iii)

    2 marksBalancing a combustion equation
    Step 1: The product $\text{P}_4\text{O}_{10}$ contains 4 phosphorus atoms, so 4 P are needed on the left. Step 2: It contains 10 oxygen atoms; each $\text{O}_2$ has 2 oxygen atoms, so 10 ÷ 2 = 5 molecules of $\text{O}_2$ are needed. Step 3: So the balanced equation is $4\text{P} + 5\text{O}_2 \rightarrow \text{P}_4\text{O}_{10}$.
    Method:

    Match the atoms in the product to the reactants by choosing coefficients.
    Examiner tips

    • Count atoms on the product side, then choose coefficients so both sides match.

20. Question 3(a)(iv)

    1 marksAcidic or basic oxide
    Step 1: Non-metal oxides are generally acidic, while metal oxides are generally basic. Step 2: Phosphorus is a non-metal, so phosphorus(V) oxide is an acidic oxide. Step 3: It is therefore acidic because phosphorus is a non-metal.
    Method:

    Classify phosphorus as a metal or non-metal, then apply the oxide rule.
    Examiner tips

    • Metal oxide → basic; non-metal oxide → acidic.

21. Question 3(b)(i)

    1 marksElement in NPK fertilisers
    Step 1: NPK fertilisers contain nitrogen (N), phosphorus (P) and potassium (K). Step 2: The K in NPK stands for potassium. Step 3: Krypton, nickel and platinum are not part of NPK fertilisers, so the other element is potassium.
    Method:

    Recall what the letters N, P and K stand for and pick the matching element.
    Examiner tips

    • NPK = Nitrogen, Phosphorus, Potassium.

22. Question 3(b)(ii)

    1 marksWhy farmers use fertilisers
    Step 1: Crops take up nutrients such as nitrogen, phosphorus and potassium from the soil as they grow. Step 2: Fertilisers replace these elements, allowing crops to grow better and increasing the yield. Step 3: So farmers use fertilisers to increase crop growth by replacing elements removed from the soil.
    Method:

    Link the role of fertilisers to replacing soil nutrients for better crop growth.
    Examiner tips

    • Fertilisers add back nutrients (N, P, K) removed by previous crops.

23. Question 3(c)

    2 marksRelative formula mass of a compound
    Step 1: Count the atoms in $\text{Ca}_3(\text{PO}_4)_2$: 3 Ca, 2 P and 8 O. Step 2: Multiply by the $A_r$ values: Ca = 3 × 40 = 120, P = 2 × 31 = 62, O = 8 × 16 = 128. Step 3: Add them: 120 + 62 + 128 = 310, so the relative formula mass is 310.
    Method:

    Count each type of atom, multiply by its Ar, then sum.
    Examiner tips

    • The bracket subscript multiplies everything inside: (PO₄)₂ = 2 P and 8 O.

24. Question 4(a)(i)

    1 marksMeaning of hydrocarbon
    Step 1: A hydrocarbon is a type of organic compound. Step 2: It contains only the two elements hydrogen and carbon. Step 3: Compounds containing oxygen as well, elements, or mixtures do not fit the definition, so a hydrocarbon is a compound of hydrogen and carbon only.
    Method:

    Recall the precise definition limiting a hydrocarbon to two elements.
    Examiner tips

    • Hydrocarbon = hydrogen + carbon only (no other elements).

25. Question 4(a)(ii)

    4 marksFractional distillation of petroleum
    Step 1: Petroleum is separated by fractional distillation, which relies on the fractions having different boiling points. Step 2: The petroleum is heated and vaporised, and the fractionating column has a temperature gradient (hot at the bottom, cooler at the top). Step 3: As the vapours rise and cool, each fraction condenses when it reaches its own boiling point, so the correct outline uses fractional distillation and different boiling points.
    Method:

    Name the technique, state the property (boiling point), and describe the temperature gradient and condensing.
    Examiner tips

    • Key ideas: fractional distillation + different boiling points + temperature gradient + condense.

26. Question 4(a)(iii)

    2 marksUses of petroleum fractions
    Step 1: Kerosene is a fuel for aircraft, so it is used as jet fuel. Step 2: Bitumen is a thick, tarry fraction used for surfacing and making roads, and gasoline (petrol) is used as fuel for cars. Step 3: So the correct matching is kerosene → jet fuel, bitumen → making roads, gasoline → fuel for cars.
    Method:

    Recall the typical use of each named fraction and match them up.
    Examiner tips

    • Heavy bitumen → roads; gasoline → cars; kerosene → jet fuel.

27. Question 4(b)(i)

    2 marksDisplayed formula of ethanol
    Step 1: Ethanol is $\text{C}_2\text{H}_5\text{OH}$: a CH3 group joined to a CH2 group joined to an OH group. Step 2: Every bond is a single covalent bond drawn as a line, including the C-O bond and the O-H bond. Step 3: So the correct displayed formula shows CH3-CH2-O-H with all bonds drawn, including the O-H bond.
    Method:

    Draw the two carbons with their hydrogens, then add the C-O and O-H bonds of the OH group.
    Examiner tips

    • Alcohols contain an -OH group; show the O-H bond explicitly in a displayed formula.

28. Question 4(b)(ii)

    2 marksConditions for making ethanol from ethene
    Step 1: Ethanol is made by the catalytic addition of steam to ethene. Step 2: The conditions are a high temperature (about 300 °C), a high pressure (about 60 atm) and an acid catalyst (phosphoric acid). Step 3: So two correct conditions are a high temperature of about 300 °C and an acid catalyst.
    Method:

    Recall the industrial conditions for the hydration of ethene.
    Examiner tips

    • Hydration of ethene: ~300 °C, ~60 atm, phosphoric acid catalyst.

29. Question 4(c)(i)

    1 marksDeducing a molecular formula
    Step 1: Tetramethylsilane has one silicon atom bonded to four CH3 groups. Step 2: Four CH3 groups contain 4 carbon atoms and 4 × 3 = 12 hydrogen atoms. Step 3: Adding the one silicon atom gives the molecular formula $\text{SiC}_4\text{H}_{12}$.
    Method:

    Count the atoms in all four CH3 groups, then add the central silicon.
    Examiner tips

    • Count every atom: 1 Si + 4 C + 12 H.

30. Question 4(c)(ii)

    1 marksElements in the same group
    Step 1: Carbon and silicon are both in Group IV of the Periodic Table, each with 4 outer electrons. Step 2: Hydrogen sits separately at the top and is not in Group IV. Step 3: So the two elements in the same group are silicon and carbon.
    Method:

    Find the group of each element and pick the two that share a group.
    Examiner tips

    • Group IV: carbon, silicon, germanium... all have 4 outer electrons.

31. Question 5(a)(i)

    2 marksWord equation for a metal oxide and acid
    Step 1: A metal oxide (a base) reacts with an acid to form a salt and water. Step 2: Zinc oxide with sulfuric acid gives the salt zinc sulfate plus water. Step 3: Sulfuric acid gives sulfate salts (not sulfide or chloride), and a metal oxide with acid gives water (not hydrogen), so the products are zinc sulfate + water.
    Method:

    Apply the base + acid rule and use the correct salt name for sulfuric acid.
    Examiner tips

    • Sulfuric acid → sulfate; oxide + acid → salt + water.

32. Question 5(a)(ii)

    1 marksRemoving an insoluble solid
    Step 1: The excess zinc oxide is an insoluble solid in a solution. Step 2: Filtration separates an insoluble solid from a liquid: the solid stays on the filter paper as residue and the solution passes through as filtrate. Step 3: So the method used to remove the excess solid zinc oxide is filtration.
    Method:

    Match the separation method to removing an insoluble solid from a liquid.
    Examiner tips

    • Insoluble solid + liquid → separate by filtration.

33. Question 5(a)(iii)

    2 marksObtaining crystals from a salt solution
    Step 1: The salt solution is heated to evaporate some of the water until it reaches the point of crystallisation (a saturated solution). Step 2: It is then left to cool so that crystals form, and the crystals are filtered off. Step 3: The crystals are dried (for example between filter papers), so the correct method is evaporate to crystallisation point, then filter and dry the crystals.
    Method:

    Describe controlled evaporation to a saturated solution, then crystallisation, filtering and drying.
    Examiner tips

    • Evaporate to crystallisation point (not to dryness), then leave to crystallise.

34. Question 5(b)(i)

    2 marksLabelling an electrolysis cell
    Step 1: The cathode is the electrode connected to the negative terminal of the power supply. Step 2: The electrolyte is the substance that conducts electricity and is broken down, which here is the molten zinc chloride. Step 3: So the cathode is the negative electrode and the electrolyte is the molten zinc chloride.
    Method:

    Identify the negative electrode as the cathode and the molten liquid as the electrolyte.
    Examiner tips

    • Cathode = negative; anode = positive; electrolyte = the molten/aqueous substance.

35. Question 5(b)(ii)

    1 marksMaterial for inert electrodes
    Step 1: An inert electrode must conduct electricity but not react with the electrolyte or its products. Step 2: Graphite (carbon) conducts electricity and is inert, so it is a suitable electrode material (platinum is also used). Step 3: Zinc and sodium would react, and wood does not conduct, so the suitable material is graphite (carbon).
    Method:

    Pick a conducting material that does not react with the electrolyte.
    Examiner tips

    • Inert electrode = conducts but does not react: graphite or platinum.

36. Question 5(b)(iii)

    2 marksProducts of electrolysis of molten zinc chloride
    Step 1: In molten zinc chloride the ions are $\text{Zn}^{2+}$ and $\text{Cl}^-$. Step 2: Positive metal ions ($\text{Zn}^{2+}$) go to the negative electrode (cathode) and form zinc; negative chloride ions go to the positive electrode (anode) and form chlorine. Step 3: So chlorine forms at the positive electrode and zinc at the negative electrode.
    Method:

    Send positive ions to the negative electrode and negative ions to the positive electrode.
    Examiner tips

    • Molten salt: metal at cathode (negative), non-metal at anode (positive).

37. Question 5(c)

    2 marksMeaning of ionic bond
    Step 1: An ionic bond forms when a metal transfers electrons to a non-metal, making positive and negative ions. Step 2: These oppositely charged ions attract each other by a strong electrostatic force. Step 3: So an ionic bond is the strong electrostatic attraction between oppositely charged ions.
    Method:

    Define the bond as the electrostatic attraction between oppositely charged ions.
    Examiner tips

    • Ionic = electrostatic attraction between + and − ions; covalent = shared electrons.

38. Question 6(a)(i)

    1 marksOrder of reactivity from reactions with acid
    Step 1: The more vigorously a metal reacts with acid (more bubbles of hydrogen), the more reactive it is. Step 2: B bubbles most (most reactive), then C, then D, and A produces no bubbles (least reactive). Step 3: So the order from most to least reactive is B, C, D, A.
    Method:

    Compare how vigorously each metal bubbles with acid to rank them.
    Examiner tips

    • Rank by vigour of bubbling: most bubbles = most reactive; no bubbles = least reactive.

39. Question 6(a)(ii)

    2 marksWays to decrease reaction rate
    Step 1: A reaction slows when particles collide less often or less successfully. Step 2: Decreasing the temperature gives slower, less frequent collisions, and using larger pieces of metal reduces the surface area for collisions (lowering the concentration would also work). Step 3: So two ways to decrease the rate are lowering the temperature and using larger pieces of metal.
    Method:

    Reverse the factors that speed up a reaction to find ways to slow it down.
    Examiner tips

    • To slow a reaction: lower temperature, lower concentration, or use larger lumps.

40. Question 6(b)

    1 marksFlame test colour for potassium
    Step 1: Flame tests give characteristic colours for different metal ions. Step 2: Potassium ions give a lilac (light purple) flame colour. Step 3: Red is lithium, yellow is sodium and blue-green is copper, so potassium gives a lilac flame.
    Method:

    Recall the flame-test colour associated with potassium ions.
    Examiner tips

    • Lithium red, sodium yellow, potassium lilac, copper blue-green.

41. Question 6(c)(i)

    1 marksName for a mixture of metals
    Step 1: A mixture of a metal with one or more other elements (usually metals) is called an alloy. Step 2: Brass is a mixture of two metals, so brass is an alloy. Step 3: A compound has chemically bonded elements, an element is a single substance and an isotope refers to atoms of one element, so the correct name is an alloy.
    Method:

    Recall the term for a metal mixture and apply it to brass.
    Examiner tips

    • Alloy = a mixture of a metal with other elements.

42. Question 6(c)(ii)

    2 marksMetals in brass
    Step 1: Brass is an alloy made from two metals. Step 2: Those two metals are copper and zinc. Step 3: Copper and tin make bronze, iron and carbon make steel and magnesium and aluminium make magnalium, so brass is copper and zinc.
    Method:

    Recall the two metals that make up brass.
    Examiner tips

    • Brass = copper + zinc; bronze = copper + tin.

43. Question 6(c)(iii)

    1 marksAdvantage of an alloy over a pure metal
    Step 1: In an alloy, atoms of different sizes disrupt the regular layers, making it harder for layers to slide. Step 2: This makes magnalium harder and stronger than pure magnesium. Step 3: So the useful property is that magnalium is harder/stronger than pure magnesium.
    Method:

    Recall how alloying changes hardness and strength compared with the pure metal.
    Examiner tips

    • Mixing in different-sized atoms makes an alloy harder/stronger than the pure metal.

44. Question 7(a)

    2 marksIdentifying chemical changes
    Step 1: A chemical change forms a new substance and is usually hard to reverse. Step 2: Burning methane forms new substances (carbon dioxide and water) and rusting of iron forms a new substance (iron oxide), so both are chemical changes. Step 3: Boiling water, dissolving salt and mixing ink and water are physical changes (no new substance formed), so the two chemical changes are burning methane and rusting of iron.
    Method:

    Pick the processes that form a new substance.
    Examiner tips

    • Chemical change = new substance formed (burning, rusting); physical change = no new substance.

45. Question 7(b)

    1 marksSymbol for a reversible reaction
    Step 1: A reversible reaction can go forwards and backwards. Step 2: This is shown by a double half-arrow, $\rightleftharpoons$, with one arrow pointing each way. Step 3: A single arrow shows a one-way reaction, an equals sign and a plus sign are not reaction-direction symbols, so the reversible symbol is $\rightleftharpoons$.
    Method:

    Recall the two-way arrow symbol used for reversible reactions.
    Examiner tips

    • Reversible reaction = $\rightleftharpoons$ (double half-arrow).

46. Question 7(c)(i)

    1 marksGreatest temperature change
    Step 1: The temperature change is the difference between the final and initial temperatures, ignoring sign. Step 2: Experiment 1 changes by 5, experiment 2 by 2, experiment 3 by 4 and experiment 4 by 10 (29 − 19). Step 3: The largest change is 10 °C, so experiment 4 shows the greatest temperature change.
    Method:

    Calculate the size of each temperature change and compare.
    Examiner tips

    • Greatest change = largest |final − initial|, regardless of direction.

47. Question 7(c)(ii)

    1 marksMost endothermic experiment
    Step 1: In an endothermic reaction, energy is taken in, so the temperature falls. Step 2: Experiments 2 and 3 show falls (2 and 3 fall in temperature); experiment 3 falls the most (18 → 14, a fall of 4 °C). Step 3: So the most endothermic experiment is experiment 3.
    Method:

    Find the experiments where the temperature falls and pick the largest fall.
    Examiner tips

    • Endothermic = takes in energy = temperature decreases.

48. Question 7(c)(iii)

    2 marksBalancing a metal-acid equation
    Step 1: $\text{MgCl}_2$ contains 2 chlorine atoms, so 2 HCl are needed on the left to provide them. Step 2: The 2 HCl provide 2 hydrogen atoms, which form one molecule of hydrogen gas, $\text{H}_2$. Step 3: So the balanced equation is $\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$.
    Method:

    Balance the chlorine atoms, then deduce the hydrogen gas product.
    Examiner tips

    • Balance chlorine first (2 HCl), then the hydrogen forms one H₂ molecule.

49. Question 7(c)(iv)

    2 marksReaction pathway diagram for an endothermic reaction
    Step 1: A reaction pathway diagram has energy on the vertical axis and progress of reaction on the horizontal axis. Step 2: In an endothermic reaction, energy is taken in, so the products end up at a higher energy than the reactants. Step 3: So the vertical axis is labelled energy, with the reactants on the lower line and the products on the higher line.
    Method:

    Label energy on the vertical axis and place the products above the reactants for an endothermic reaction.
    Examiner tips

    • Endothermic pathway: products higher than reactants; axis = energy.

50. Question 8(a)

    2 marksStages in water treatment
    Step 1: Activated carbon adsorbs substances that cause tastes and odours, so addition of carbon removes tastes and odours. Step 2: Sedimentation lets insoluble solids settle out, removing solids, and chlorination adds chlorine to kill microbes. Step 3: So the correct matching is carbon → tastes/odours, sedimentation → solids, chlorination → microbes.
    Method:

    Match each water-treatment stage to the purpose it serves.
    Examiner tips

    • Sedimentation settles solids; carbon adsorbs tastes/odours; chlorine kills microbes.

51. Question 8(b)

    2 marksTesting purity using melting point
    Step 1: A pure substance has a sharp, fixed melting point. Step 2: Pure water melts sharply at exactly 0 °C, so if the sample melts sharply at 0 °C it is pure. Step 3: An impurity lowers the melting point and makes it melt over a range, so impure water melts below 0 °C or over a range, which is how melting point tests purity.
    Method:

    Compare the sample melting behaviour with the sharp 0 °C melting point of pure water.
    Examiner tips

    • Pure substance = sharp fixed melting point; impurity = lower and spread-out melting.

52. Question 8(c)(i)

    2 marksHarmful effects of air pollutants
    Step 1: Sulfur dioxide dissolves in rain water to form an acid, causing acid rain. Step 2: Particulates (small solid particles) can be breathed in and cause respiratory problems (and can cause global dimming). Step 3: So sulfur dioxide causes acid rain and particulates cause respiratory problems.
    Method:

    Match each pollutant to its main harmful effect.
    Examiner tips

    • Acidic gas (sulfur dioxide) → acid rain; particulates → respiratory problems.

53. Question 8(c)(ii)

    2 marksFlue gas desulfurisation
    Step 1: Sulfur dioxide is an acidic gas. Step 2: In flue gas desulfurisation, a base such as calcium oxide (or calcium carbonate) reacts with the sulfur dioxide. Step 3: This neutralisation removes the sulfur dioxide from the flue gases, so it is reduced by neutralising the acidic gas with a base such as calcium oxide.
    Method:

    Use the acid-base idea: a base neutralises the acidic sulfur dioxide.
    Examiner tips

    • Acidic gas + base (calcium oxide) → neutralisation removes the gas.