May/June 2025 Paper 31 Worked Answers (IGCSE Chemistry 0620 Core)

53 questions · 80 marks · 75 minutes

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1. Question 1(a)

   1 marksElement that forms a 1+ ion
   Step 1: Potassium is in Group I, so an atom has one outer electron. Step 2: Losing that single electron gives one extra positive charge, so potassium forms a 1+ ion. Step 3: Magnesium forms a 2+ ion, aluminium a 3+ ion and oxygen a 2- ion, so only potassium forms a 1+ ion.
   Method:

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

   • Group I metals form 1+ ions by losing their single outer electron.

2. Question 1(b)

   1 marksMetal extracted in the blast furnace
   Step 1: The blast furnace is used to extract iron from iron ore by reduction with carbon. Step 2: Iron is below carbon in the reactivity series, so it can be reduced by carbon in the furnace. Step 3: Aluminium, potassium and magnesium are more reactive than carbon and are extracted by electrolysis, so the blast-furnace metal is iron.
   Method:

   Match the extraction method to the metal: carbon reduction for iron.
   Examiner tips

   • Metals below carbon (e.g. iron) are extracted by reduction with carbon in a blast furnace.

3. Question 1(c)

   1 marksSoft metal more reactive than sodium
   Step 1: Reactivity increases down Group I, and potassium is below sodium in Group I. Step 2: Potassium is therefore more reactive than sodium and, like other Group I metals, is soft enough to cut with a knife. Step 3: Iron, magnesium and aluminium are all less reactive than sodium, so the answer is potassium.
   Method:

   Use the trend in Group I reactivity to find the metal more reactive than sodium.
   Examiner tips

   • Group I metals are soft and get more reactive going down the group.

4. Question 1(d)

   1 marksMain gas in 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 78% is therefore nitrogen. Step 3: Oxygen is about 21%, while carbon dioxide and argon are present in small or trace amounts, so the answer is nitrogen.
   Method:

   Recall the composition of clean dry air and pick the gas at about 78%.
   Examiner tips

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

5. Question 1(e)

   1 marksGas identified using damp red litmus paper
   Step 1: Ammonia is an alkaline gas, so it is the gas tested with damp red litmus paper. Step 2: Ammonia turns the damp red litmus paper blue, identifying it. Step 3: Carbon dioxide, oxygen and nitrogen have different tests, so the gas identified with damp red litmus is ammonia.
   Method:

   Match each gas to its standard test and pick the one using damp red litmus.
   Examiner tips

   • Ammonia is the only common alkaline gas: it turns damp red litmus blue.

6. Question 1(f)

   1 marksLiquid element at room temperature
   Step 1: Going down Group VII the elements change from gases to liquids to solids. Step 2: Bromine is the halogen that is a red-brown liquid at room temperature and pressure. Step 3: Fluorine and oxygen are gases and iron is a solid, so the liquid element is bromine.
   Method:

   Recall the states of the elements and pick the one that is liquid.
   Examiner tips

   • Down Group VII: fluorine and chlorine gases, bromine liquid, iodine solid.

7. Question 1(g)

   1 marksMetal used in food containers
   Step 1: Aluminium forms a thin protective oxide layer that stops further corrosion. Step 2: This resistance to corrosion makes aluminium suitable for food containers and drink cans. Step 3: Iron rusts easily, and potassium and magnesium are too reactive, so the answer is aluminium.
   Method:

   Pick the element whose corrosion resistance makes it useful for food containers.
   Examiner tips

   • Aluminium resists corrosion thanks to its protective oxide layer.

8. Question 1(h)

   1 marksIdentifying a transition element
   Step 1: Transition elements are the block of metals between Group II and Group III in the centre of the Periodic Table. Step 2: Iron sits in this central transition block, so iron is a transition element. Step 3: Magnesium (Group II), potassium (Group I) and aluminium (Group III) are not in the transition block, so iron is the transition element.
   Method:

   Locate each element on the Periodic Table and pick the one in the central transition block.
   Examiner tips

   • Iron, copper and zinc are common transition elements in the central block.

9. Question 2(a)(i)

   1 marksNaming a compound from its formula
   Step 1: The formula $\text{CaCO}_3$ contains a calcium ion and a carbonate ion, $\text{CO}_3^{2-}$. Step 2: A compound of calcium with the carbonate ion is named calcium carbonate. Step 3: Calcium oxide is CaO and calcium chloride is $\text{CaCl}_2$, so $\text{CaCO}_3$ is calcium carbonate.
   Method:

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

   • $\text{CO}_3$ = carbonate, O alone = oxide, Cl = chloride.

10. 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 in 1750 cm³ = 19.0 × (1750 ÷ 1000) = 19.0 × 1.75. Step 3: This gives 33.25 g of compounds in 1750 cm³.
    Method:

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

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

11. Question 2(a)(iii)

    1 marksSolubility of common salts
    Step 1: Using the solubility rules, all common potassium compounds are soluble, and so are all hydroxides of Group I metals. Step 2: Potassium hydroxide is therefore soluble in water. Step 3: Lead(II) chloride and silver chloride are insoluble chlorides, and most carbonates (magnesium carbonate) are insoluble, so the soluble compound is potassium hydroxide.
    Method:

    Apply the solubility rules to each compound and pick the soluble one.
    Examiner tips

    • All sodium, potassium and ammonium salts are soluble; most carbonates are not.

12. Question 2(b)

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

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

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

13. Question 2(c)

    2 marksElectronic configuration and charge of a sodium ion
    Step 1: A sodium atom has 11 electrons arranged 2,8,1. Step 2: Losing the single outer electron leaves 10 electrons arranged 2,8. Step 3: Losing one negative electron leaves the ion with a charge of 1+, so the ion is 2,8 with a 1+ charge.
    Method:

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

    • A Group I atom loses one electron to form a 1+ ion with a full outer shell.

14. Question 2(d)

    2 marksArrangement and motion of particles in a liquid
    Step 1: In a liquid the particles are still close together but they are arranged irregularly, with no fixed pattern. Step 2: They have enough energy to move around and slide over each other. Step 3: So the particles in a liquid are close together but irregular and slide over one another.
    Method:

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

    • Liquid = close together + irregular + sliding/flowing over each other.

15. Question 2(e)

    2 marksProperties of ionic compounds
    Step 1: Ionic compounds have strong forces between oppositely charged ions, giving high melting and boiling points. Step 2: When molten or dissolved, the ions are free to move and carry charge, so the compound conducts electricity. Step 3: So two properties are a high melting point and conducting electricity when molten or in aqueous solution.
    Method:

    Recall the typical physical properties of ionic compounds and choose two correct ones.
    Examiner tips

    • Ionic compounds: high melting point + conduct when molten/aqueous.

16. Question 2(f)

    1 marksGas essential for aquatic life
    Step 1: Aquatic animals such as fish respire using dissolved oxygen in the water. Step 2: The gas essential for aquatic life is therefore oxygen. Step 3: Nitrogen is unreactive, carbon dioxide is a product of respiration and hydrogen is not normally dissolved, so the answer is oxygen.
    Method:

    Identify the dissolved gas used by aquatic animals in respiration.
    Examiner tips

    • Dissolved oxygen is essential for aquatic life.

17. Question 3(a)(i)

    1 marksGroup number from outer electrons
    Step 1: The group number of a main-group element equals the number of electrons in its outer shell. Step 2: Sulfur is in Group VI because it has 6 electrons in its outer shell. Step 3: The number of shells gives the period, and neutrons and relative atomic mass do not set the group, so the reason is 6 outer electrons.
    Method:

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

    • Outer-shell electrons = group number for main-group elements.

18. Question 3(a)(ii)

    3 marksSub-atomic particles in isotopes
    Step 1: The lower number (16) is the proton number, so each isotope has 16 protons and, being neutral atoms, 16 electrons. Step 2: Neutrons = mass number − proton number, so $^{33}\text{S}$ has 33 − 16 = 17 neutrons and $^{36}\text{S}$ has 36 − 16 = 20 neutrons. Step 3: So $^{33}\text{S}$ is 16 protons, 17 neutrons, 16 electrons and $^{36}\text{S}$ is 16 protons, 20 neutrons, 16 electrons.
    Method:

    Read the proton number from the bottom figure and subtract it from the mass number for neutrons.
    Examiner tips

    • Bottom number = protons (= electrons); top number = protons + neutrons.

19. Question 3(b)(i)

    1 marksSource of sulfur dioxide in air
    Step 1: Fossil fuels such as coal and oil contain sulfur compounds. Step 2: When these fuels are burned in power stations and engines, the sulfur is oxidised to sulfur dioxide, which enters the air. Step 3: Photosynthesis, evaporation and respiration do not produce sulfur dioxide, so the source is burning fossil fuels.
    Method:

    Recall how sulfur dioxide is produced and pick the matching source.
    Examiner tips

    • Burning fossil fuels that contain sulfur is the main source of sulfur dioxide.

20. Question 3(b)(ii)

    1 marksAdverse effect of sulfur dioxide
    Step 1: Sulfur dioxide dissolves in rain water and is oxidised to form sulfuric acid. Step 2: This makes the rain acidic, producing acid rain that damages plants, lakes and buildings. Step 3: Sulfur dioxide is not a major greenhouse gas and does not form ozone or help breathing, so its adverse effect is acid rain.
    Method:

    Recall the environmental problem linked to sulfur dioxide.
    Examiner tips

    • Sulfur dioxide → acid rain, which damages buildings, plants and lakes.

21. Question 3(b)(iii)

    2 marksBalancing a symbol equation
    Step 1: Count the atoms on the left: sulfur = 1 (from $\text{SO}_2$) + 2 (from 2$\text{H}_2$S) = 3, hydrogen = 4, oxygen = 2. Step 2: To balance sulfur, 3 S are needed; to balance the 4 hydrogen and 2 oxygen, 2 $\text{H}_2\text{O}$ are needed (4 H and 2 O). Step 3: So the balanced equation is $\text{SO}_2 + 2\text{H}_2\text{S} \rightarrow 3\text{S} + 2\text{H}_2\text{O}$.
    Method:

    Count atoms of each element on both sides and choose coefficients that balance them.
    Examiner tips

    • Balance the element appearing in fewest places first, then check H and O.

22. Question 3(b)(iv)

    1 marksType of bonding in sulfur dioxide
    Step 1: Sulfur and oxygen are both non-metals. Step 2: Non-metal atoms bond by sharing pairs of electrons, which is covalent bonding. Step 3: Ionic bonding is between metals and non-metals, and metallic bonding is in metals, so sulfur dioxide has covalent bonding.
    Method:

    Use the types of elements present to decide the bonding.
    Examiner tips

    • Non-metal + non-metal = covalent bonding (shared electrons).

23. Question 3(c)

    2 marksRelative formula mass
    Step 1: There are 2 Al = 2 × 27 = 54, and 3 sulfate groups contain 3 S = 3 × 32 = 96 and 12 O = 12 × 16 = 192. Step 2: Add them: 54 + 96 + 192. Step 3: This gives a relative formula mass of 342.
    Method:

    Multiply each element atom count by its relative atomic mass and add the totals.
    Examiner tips

    • $(\text{SO}_4)_3$ contains 3 S and 12 O atoms in total.

24. Question 4(a)(i)

    1 marksSeparating petroleum
    Step 1: Petroleum is separated by using the different boiling points of its hydrocarbons. Step 2: This separation of a liquid mixture by boiling point is called fractional distillation. Step 3: Cracking breaks molecules up, polymerisation joins them and electrolysis splits ionic compounds, so the separation process is fractional distillation.
    Method:

    Name the boiling-point separation used on crude oil.
    Examiner tips

    • Fractional distillation separates petroleum using boiling point differences.

25. Question 4(a)(ii)

    1 marksUse of fuel oil
    Step 1: Fuel oil is a heavier fraction of petroleum used for large-scale heating. Step 2: It is used as a fuel for ships and in home heating systems. Step 3: Drinks are not made from fuel oil, jet engines use kerosene and roads are surfaced with bitumen, so the use is fuel for ships or home heating.
    Method:

    Recall the typical use of the fuel oil fraction.
    Examiner tips

    • Fuel oil → fuel for ships and home heating.

26. Question 4(a)(iii)

    1 marksMain constituent of natural gas
    Step 1: Natural gas is mostly made up of the smallest alkane. Step 2: The main constituent of natural gas is methane, $\text{CH}_4$. Step 3: Ethane is only a minor component, ethanol is an alcohol and carbon dioxide is not a fuel, so the main constituent is methane.
    Method:

    Recall the main compound found in natural gas.
    Examiner tips

    • Natural gas ≈ methane, the simplest alkane.

27. Question 4(b)(i)

    4 marksCracking process, conditions and products
    Step 1: Breaking long chain hydrocarbons into shorter ones is called cracking. Step 2: Cracking needs a catalyst and a high temperature to break the strong carbon-carbon bonds. Step 3: The products are a smaller, more useful alkane together with an alkene (and sometimes hydrogen), so the correct option names cracking with catalyst, high temperature, alkane + alkene.
    Method:

    Identify the process, then state its conditions and products together.
    Examiner tips

    • Cracking = heat + catalyst to break long chains into a smaller alkane + an alkene.

28. Question 4(b)(ii)

    1 marksReason for cracking
    Step 1: Cracking produces shorter hydrocarbons such as petrol and useful alkenes. Step 2: These shorter molecules are in greater demand and are more useful than the surplus long chain molecules. Step 3: So a reason for cracking is that the products are more useful and in greater demand.
    Method:

    Recall the economic reason for cracking long hydrocarbons.
    Examiner tips

    • We crack because the smaller products are more useful and in greater demand.

29. Question 4(c)

    1 marksDisplayed formula of ethane
    Step 1: Ethane is a saturated alkane with formula $\text{C}_2\text{H}_6$. Step 2: The two carbon atoms are joined by a single C-C bond, and each carbon forms three more bonds to hydrogen, giving six C-H bonds. Step 3: This uses all four bonds on each carbon, so the displayed formula shows a C-C single bond with three H on each carbon.
    Method:

    Give each carbon four single bonds: one to the other carbon and three to hydrogen.
    Examiner tips

    • Each carbon forms 4 bonds; in ethane that is 1 C-C and 3 C-H.

30. Question 4(d)(i)

    2 marksConditions for fermentation
    Step 1: Fermentation uses yeast, which provides the enzymes that convert glucose to ethanol. Step 2: It works best at a warm temperature of about 25-35 °C and in the absence of oxygen (anaerobic). Step 3: A metal catalyst, high temperatures, excess oxygen or freezing are not used, so the correct conditions are yeast and a warm 25-35 °C.
    Method:

    Recall the conditions under which yeast ferments glucose to ethanol.
    Examiner tips

    • Fermentation: yeast + warm (25-35 °C) + anaerobic conditions.

31. Question 4(d)(ii)

    1 marksUse of ethanol
    Step 1: Ethanol mixes with many substances and burns cleanly. Step 2: It is therefore widely used as a solvent (for example in perfumes) and as a fuel. Step 3: Ethanol is not a metal, a balloon gas or an acid, so a correct use is as a solvent or fuel.
    Method:

    Recall a common everyday use of ethanol.
    Examiner tips

    • Ethanol uses: solvent and fuel.

32. Question 4(e)(i)

    1 marksDeducing a molecular formula
    Step 1: The molecular formula is found by counting each type of atom in the displayed formula. Step 2: There are 6 carbon atoms, 10 hydrogen atoms and 3 oxygen atoms. Step 3: Writing these as a molecular formula gives $\text{C}_6\text{H}_{10}\text{O}_3$.
    Method:

    Count every carbon, hydrogen and oxygen atom and write them as a molecular formula.
    Examiner tips

    • Molecular formula = a simple count of each atom shown in the displayed formula.

33. Question 4(e)(ii)

    1 marksMeaning of unsaturated
    Step 1: An unsaturated molecule is one that contains a carbon-carbon double bond. Step 2: This C=C bond can open up to add other atoms, which is the test for an unsaturated compound. Step 3: Single bonds only would be saturated, and oxygen atoms or the H-to-C ratio do not decide saturation, so it is unsaturated because of the C=C double bond.
    Method:

    Link the term unsaturated to the presence of a C=C double bond.
    Examiner tips

    • Unsaturated molecules have at least one C=C double bond.

34. Question 5(a)(i)

    1 marksOrder of reactivity from reaction with acid
    Step 1: The more vigorously a metal reacts with acid (the more bubbles of hydrogen), the more reactive it is. Step 2: D fizzes most, so D is most reactive, followed by A, then B. Step 3: C fizzes least, so it is least reactive, giving the order D, A, B, C.
    Method:

    Use the amount of bubbling to rank the metals from most to least reactive.
    Examiner tips

    • Rank metals by the vigour of their reaction with acid.

35. Question 5(a)(ii)

    2 marksIncreasing the rate of a reaction
    Step 1: Rate increases when particles collide more often or more energetically. Step 2: Increasing the temperature gives particles more energy, and using smaller pieces of metal increases the surface area, so both speed up the reaction (adding a catalyst would too). Step 3: Cooling, using one large lump, or diluting the acid all slow the reaction, so the correct pair is higher temperature and larger surface area.
    Method:

    Recall the factors that increase reaction rate, excluding concentration which is already given.
    Examiner tips

    • Rate factors: temperature, concentration, surface area, catalyst.

36. Question 5(a)(iii)

    1 marksTest for hydrogen gas
    Step 1: Hydrogen is tested using a lighted splint. Step 2: The hydrogen burns rapidly with a characteristic squeaky pop. Step 3: A glowing splint relighting tests for oxygen, limewater tests for carbon dioxide and damp red litmus tests for ammonia, so hydrogen gives a squeaky pop with a lighted splint.
    Method:

    Match the gas to its test and observation: hydrogen gives a squeaky pop.
    Examiner tips

    • The squeaky pop with a lighted splint is the hydrogen test.

37. Question 5(a)(iv)

    1 marksIon present in all acids
    Step 1: Acids produce hydrogen ions when dissolved in water. Step 2: So the ion present in all aqueous acids is the hydrogen ion, $\text{H}^+$. Step 3: Hydroxide ions are found in alkalis, and chloride and oxide ions are not what make a solution acidic, so the ion is $\text{H}^+$.
    Method:

    Recall the ion that defines an acidic solution.
    Examiner tips

    • Acid = source of $\text{H}^+$; alkali = source of $\text{OH}^-$.

38. Question 5(b)(i)

    1 marksMeaning of the term alloy
    Step 1: An alloy is made by mixing a metal with other elements, usually other metals. Step 2: So an alloy is a mixture of a metal with one or more other elements. Step 3: It is not a compound, not a pure metal and not a coating, so the correct meaning is a mixture of a metal with other elements.
    Method:

    Recall the definition of an alloy as a mixture.
    Examiner tips

    • Alloy = mixture of a metal with one or more other elements.

39. Question 5(b)(ii)

    1 marksProperty of stainless steel for cutlery
    Step 1: Cutlery must stay strong and not corrode when washed and used with food. Step 2: Stainless steel is hard and resists rusting, so it stays clean and keeps its edge. Step 3: Rusting, a low melting point or dissolving would all make it unsuitable, so the useful property is being hard and rust-resistant.
    Method:

    Match a property of stainless steel to the needs of cutlery.
    Examiner tips

    • Stainless steel: hard + rust-resistant = good for cutlery.

40. Question 6(a)

    3 marksAcid and carbonate reaction products
    Step 1: An acid reacting with a metal carbonate gives a salt, carbon dioxide and water. Step 2: The salt is named from the metal (magnesium) and the acid (nitric acid → nitrate), giving magnesium nitrate. Step 3: So the products are magnesium nitrate, carbon dioxide and water.
    Method:

    Apply the acid-plus-carbonate pattern and name the salt from the acid used.
    Examiner tips

    • Acid + carbonate always makes salt + CO₂ + water.

41. Question 6(b)

    3 marksPreparing pure dry salt crystals
    Step 1: Excess copper(II) oxide is filtered off so only the copper(II) chloride solution remains. Step 2: The solution is heated and evaporated to the point of crystallisation (until crystals just start to form). Step 3: The crystals are then removed and dried between sheets of filter paper, giving pure, dry copper(II) chloride.
    Method:

    Remove the excess solid, crystallise by evaporation, then dry the crystals.
    Examiner tips

    • Salt prep with excess base: filter → evaporate to crystallisation → dry.

42. Question 6(c)(i)

    2 marksIdentifying the anode in electrolysis
    Step 1: The anode is the positive electrode in electrolysis. Step 2: The right-hand electrode is connected to the positive terminal, so it is the anode. Step 3: The left-hand electrode (negative terminal) is the cathode, and heat position does not decide it, so the anode is the right-hand (positive) electrode.
    Method:

    Identify the positive electrode as the anode.
    Examiner tips

    • Anode is positive; cathode is negative ("an ox, red cat").

43. Question 6(c)(ii)

    1 marksInert electrode material
    Step 1: An inert electrode does not react during electrolysis. Step 2: Besides platinum, graphite (carbon) is the common inert electrode material because it conducts electricity and is unreactive. Step 3: Copper and iron are reactive metals used as non-inert electrodes, and sodium is far too reactive, so the answer is graphite/carbon.
    Method:

    Recall the common inert electrode materials.
    Examiner tips

    • Inert electrodes: platinum and graphite (carbon).

44. Question 6(c)(iii)

    2 marksProducts of electrolysis of molten lithium iodide
    Step 1: In molten lithium iodide the ions are $\text{Li}^+$ and $\text{I}^-$. Step 2: Positive iodide ions are attracted to the negative electrode? No: negative iodide ions move to the positive electrode (anode) and form iodine, while positive lithium ions move to the negative electrode (cathode) and form lithium. Step 3: So iodine forms at the positive electrode and lithium at the negative electrode.
    Method:

    Send positive metal ions to the cathode and negative non-metal ions to the anode.
    Examiner tips

    • Molten binary salt: metal at cathode (−), non-metal at anode (+).

45. Question 6(d)

    3 marksIdentifying simple molecular compounds
    Step 1: Simple molecular compounds have low melting points (weak forces between molecules) and do not conduct electricity because they have no free ions or electrons. Step 2: Compound B (−157 °C, non-conducting) and compound E (−83 °C, non-conducting) both fit this description. Step 3: Compounds A, C and D have high melting points or conduct when molten, so they are not simple molecules; the two simple molecular compounds are therefore B and E.
    Method:

    Select the compounds with both a low melting point and no conductivity when molten.
    Examiner tips

    • Pick the compounds that are BOTH low-melting AND non-conducting when molten.

46. Question 7(a)(i)

    1 marksGreatest temperature change
    Step 1: Work out the size of each temperature change: experiment 1 = 6 °C, experiment 2 = 8 °C, experiment 3 = 4 °C, experiment 4 = 7 °C. Step 2: The largest of these is 8 °C. Step 3: That is experiment 2, so it shows the greatest temperature change.
    Method:

    Calculate each temperature change and choose the largest.
    Examiner tips

    • Greatest temperature change = largest difference (ignore the direction).

47. Question 7(a)(ii)

    2 marksBalancing the zinc and acid equation
    Step 1: $\text{ZnCl}_2$ contains 2 chlorine atoms, so 2 HCl are needed to supply 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 $\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2$.
    Method:

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

    • Balance chlorine first (2 Cl needs 2 HCl), then the hydrogen forms $\text{H}_2$.

48. Question 7(a)(iii)

    2 marksLabelling a reaction pathway diagram
    Step 1: The vertical axis of a reaction pathway diagram is energy. Step 2: The reactants are drawn on the left and the products on the right. Step 3: Because the reaction is exothermic, the products are at a lower energy than the reactants, so the products line is lower on the right.
    Method:

    Label energy on the y-axis and place reactants left, products lower right.
    Examiner tips

    • Exothermic: products lower than reactants on an energy (y) vs progress (x) diagram.

49. Question 7(b)

    2 marksNaming changes of state
    Step 1: Change G goes from liquid to gas, which is evaporation (boiling). Step 2: Change H goes from liquid to solid, which is freezing. Step 3: Condensation is gas to liquid and melting is solid to liquid, so G is evaporation and H is freezing.
    Method:

    Identify the start and end states of each change and name it.
    Examiner tips

    • Name the change by its start and end state: liquid→gas evaporation, liquid→solid freezing.

50. Question 8(a)

    2 marksStages in water treatment
    Step 1: Activated carbon is added to absorb substances that cause unpleasant tastes and odours. Step 2: Chlorination adds chlorine to kill harmful microbes (bacteria) so the water is safe to drink. Step 3: So carbon removes tastes and odours while chlorination kills microbes.
    Method:

    Match each treatment stage to the problem it solves.
    Examiner tips

    • Water treatment: carbon for taste/odour, chlorine to kill microbes.

51. Question 8(b)

    2 marksTesting water purity by boiling point
    Step 1: Pure water has a fixed boiling point of exactly 100 °C at normal pressure. Step 2: Heat the sample and measure the temperature at which it boils. Step 3: If it boils at exactly 100 °C it is pure; if it boils above 100 °C it contains dissolved impurities, so measuring the boiling point tests the purity.
    Method:

    Measure the boiling point and compare it with the value for pure water.
    Examiner tips

    • Purity test: measure boiling point; pure water boils at exactly 100 °C.

52. Question 8(c)

    1 marksWhy distilled water is used
    Step 1: Tap water contains dissolved substances such as salts and chlorine. Step 2: Distilling removes these, so distilled water contains far fewer impurities. Step 3: Fewer impurities means they cannot interfere with reactions or results, which is why distilled water is used rather than tap water.
    Method:

    Explain the preference in terms of the impurity content of the water.
    Examiner tips

    • Distilled water = very few impurities, so it does not interfere with experiments.

53. Question 8(d)

    2 marksHarmful effects of air pollutants
    Step 1: Carbon monoxide is a toxic gas that combines with the blood and reduces oxygen transport, so it is poisonous. Step 2: Methane is a greenhouse gas that traps heat and causes global warming (climate change). Step 3: So carbon monoxide is toxic and methane causes global warming.
    Method:

    Recall the harmful effect linked to each named pollutant.
    Examiner tips

    • Carbon monoxide = poisonous; methane = greenhouse gas (global warming).