0620/33 May/June 2025 Worked Answers (IGCSE Chemistry 0620 Core)

54 questions · 80 marks · 75 minutes

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1. Question 1(a)

   1 marksIon charge from group number
   Step 1: Nitrogen is in Group V, so an atom has 5 outer electrons and needs to gain 3 electrons to reach a full outer shell. Step 2: Gaining 3 electrons gives the atom three extra negative charges, so it forms an ion with a 3- charge. Step 3: Oxygen forms a 2- ion, chlorine a 1- ion and calcium a 2+ ion, so only nitrogen forms a 3- 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 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: Copper sits in this central transition block, so copper is a transition element. Step 3: Calcium (Group II), potassium (Group I) and nitrogen (Group V) are not in the transition block, so copper is the transition element.
   Method:

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

   • Copper, iron and zinc are common transition elements in the central block.

3. Question 1(c)

   1 marksGas identified using limewater
   Step 1: Limewater is the standard test for carbon dioxide. Step 2: When carbon dioxide is bubbled through limewater it turns the limewater milky (cloudy white), so carbon dioxide is the gas identified using limewater. Step 3: Oxygen, chlorine and nitrogen have different tests, so the answer is carbon dioxide.
   Method:

   Match each gas to its standard test and pick the one tested with limewater.
   Examiner tips

   • Carbon dioxide + limewater = milky/cloudy white precipitate.

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 neon are only 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 marksMetal used in electrical wiring
   Step 1: Electrical wiring needs a metal that conducts electricity well and can be drawn into wires. Step 2: Copper is an excellent electrical conductor and is ductile, so it is widely used in electrical wiring. Step 3: Calcium and potassium are too reactive, and carbon is a non-metal, so copper is the correct choice.
   Method:

   Pick the metal that conducts electricity well and can be drawn into wires.
   Examiner tips

   • Copper: good conductor + ductile = ideal for electrical wiring.

6. Question 1(f)

   1 marksLeast reactive metal
   Step 1: In the reactivity series, gold is one of the least reactive metals and is found uncombined (native) in nature. Step 2: Potassium and calcium are very reactive (they react with water), and copper is more reactive than gold. Step 3: So the least reactive metal in the list is gold.
   Method:

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

   • Gold and platinum are at the bottom (least reactive) of the reactivity series.

7. Question 1(g)

   1 marksDiatomic solid halogen
   Step 1: A diatomic molecule contains two atoms joined together, and the question also needs a substance that is solid at room temperature. Step 2: Iodine is a Group VII element that exists as $\text{I}_2$ molecules and is a grey solid at room temperature and pressure. Step 3: Chlorine and nitrogen are diatomic but are gases, and neon exists as single atoms, so the answer is iodine.
   Method:

   Find the substance that is both diatomic and solid at room temperature.
   Examiner tips

   • Down Group VII the state changes: chlorine gas, bromine liquid, iodine solid.

8. Question 1(h)

   1 marksRemoving impurities in the blast furnace
   Step 1: In the blast furnace, limestone decomposes to calcium oxide, which is a basic oxide. Step 2: Calcium oxide reacts with the acidic sandy impurities (silicon dioxide) to form molten slag (calcium silicate), which is removed. Step 3: Carbon is the reducing agent and carbon dioxide is a waste gas, so the compound that forms slag is calcium oxide.
   Method:

   Identify the basic compound that neutralises acidic impurities to form slag.
   Examiner tips

   • Calcium oxide (from limestone) + silicon dioxide impurity → slag.

9. Question 1(i)

   1 marksElement with a given relative atomic mass
   Step 1: The relative atomic mass of an element can be read from the Periodic Table. Step 2: Carbon has a relative atomic mass of 12. Step 3: Oxygen is 16, nitrogen is 14 and calcium is 40, so the element with a relative atomic mass of 12 is carbon.
   Method:

   Recall the relative atomic mass of each element and pick the one equal to 12.
   Examiner tips

   • Read relative atomic masses straight from the Periodic Table.

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 in 800 cm³ = 38.0 × (800 ÷ 2000) = 38.0 × 0.40. Step 3: This gives 15.2 g of compounds in 800 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 marksSolubility of common salts
    Step 1: Using the solubility rules, all common sodium, potassium and ammonium compounds are soluble, and so are all hydroxides of Group I metals. Step 2: Sodium hydroxide is therefore soluble in water. Step 3: Lead(II) sulfate, most carbonates (magnesium carbonate) and silver chloride are insoluble, so the soluble compound is sodium 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.

13. Question 2(b)

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

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

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

14. Question 2(c)

    2 marksElectronic configuration and charge of a potassium ion
    Step 1: A potassium atom has 19 electrons arranged 2,8,8,1. Step 2: Losing the single outer electron leaves 18 electrons arranged 2,8,8. Step 3: Losing one negative electron leaves the ion with a charge of 1+, so the ion is 2,8,8 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.

15. Question 2(d)

    2 marksSeparation and motion of particles in a solid
    Step 1: In a solid the particles are packed very close together, usually touching, in a regular arrangement. 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 close together and vibrate about fixed positions.
    Method:

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

    • Solid = close packed + vibrating in fixed positions.

16. Question 2(e)

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

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

    • Phosphates (and nitrates) in water usually come from fertilisers.

17. Question 2(f)

    2 marksDot-and-cross diagram of water
    Step 1: Water is covalent, so the atoms share electrons. Each hydrogen shares one pair of electrons with the oxygen, forming a single O-H bond. Step 2: Oxygen has 6 outer electrons; two are used in the two bonds, leaving 4 non-bonding electrons (two lone pairs). Step 3: So each hydrogen has one shared pair and the oxygen carries two non-bonding pairs.
    Method:

    Form a single shared pair for each O-H bond, then place oxygen remaining electrons as lone pairs.
    Examiner tips

    • Each single covalent bond is one shared pair; count remaining electrons as lone pairs.

18. Question 3(a)(i)

    1 marksProperty used in fractional distillation
    Step 1: Fractional distillation separates liquids that have different boiling points. Step 2: The hydrocarbons in petroleum have different boiling points because they have different sized molecules, so they condense at different heights in the column. Step 3: So the property used is their different boiling points.
    Method:

    Recall what physical property fractional distillation depends on.
    Examiner tips

    • Fractional distillation = separation by differences in boiling point.

19. Question 3(a)(ii)

    2 marksHow fractional distillation separates petroleum
    Step 1: The petroleum is heated so it vaporises and the vapours rise up the fractionating column. Step 2: The column has a temperature gradient: hot at the bottom and cooler towards the top. Step 3: As the vapours rise and cool, each fraction condenses when it reaches its own boiling point, so different fractions are collected at different heights.
    Method:

    Describe the vaporising, the temperature gradient and the condensing of each fraction.
    Examiner tips

    • Key ideas: vaporise + temperature gradient + condense at boiling points.

20. Question 3(a)(iii)

    1 marksBreaking long hydrocarbons into shorter ones
    Step 1: Cracking uses heat and a catalyst to break large hydrocarbon molecules into smaller ones. Step 2: This produces smaller alkanes and useful alkenes. Step 3: Fractional distillation only separates molecules, polymerisation joins small molecules together and neutralisation is an acid-base reaction, so the process is cracking.
    Method:

    Name the thermal-decomposition process used to break large hydrocarbons.
    Examiner tips

    • Cracking = breaking long chains into short chains (and alkenes).

21. Question 3(a)(iv)

    2 marksUses of petroleum fractions
    Step 1: Bitumen is a thick tarry fraction used for surfacing and making roads. Step 2: Lubricating oil provides waxes and polishes, and naphtha is used as a chemical feedstock for making other chemicals. Step 3: So the correct matching is bitumen → roads, lubricating oil → waxes and polishes, naphtha → chemical feedstock.
    Method:

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

    • Heavy bitumen → roads; light naphtha → chemical feedstock.

22. Question 3(b)(i)

    2 marksDisplayed formula of ethanoic acid
    Step 1: Ethanoic acid is $\text{CH}_3\text{COOH}$, so it has a CH3 group joined to a carboxylic acid group. Step 2: The carboxylic acid carbon has one double bond to oxygen (C=O) and a single bond to an O-H group. Step 3: So the displayed formula shows CH3 joined to a carbon with a C=O and an O-H, with all bonds drawn, including the O-H bond.
    Method:

    Draw the CH3 group, then the carboxylic acid carbon with its C=O and O-H bonds.
    Examiner tips

    • Carboxylic acid group -COOH = C double-bonded to O plus C-O-H.

23. Question 3(b)(ii)

    1 marksSalt from a carboxylic acid and a carbonate
    Step 1: A carboxylic acid reacting with a carbonate makes a salt, water and carbon dioxide. Step 2: The salt is named from the metal (sodium) and the acid (ethanoic acid), giving sodium ethanoate. Step 3: So the salt produced is sodium ethanoate.
    Method:

    Combine the metal name with the -oate ending of the acid.
    Examiner tips

    • Acid name → salt name: ethanoic acid → ethanoate.

24. Question 3(c)

    2 marksComplete combustion of ethane
    Step 1: Complete combustion of a hydrocarbon in excess oxygen gives carbon dioxide and water. Step 2: The carbon in ethane is oxidised to carbon dioxide and the hydrogen is oxidised to water. Step 3: So the two products are carbon dioxide and water.
    Method:

    Recall the products of complete combustion of a hydrocarbon in excess oxygen.
    Examiner tips

    • Excess oxygen = complete combustion = $\text{CO}_2$ + $\text{H}_2\text{O}$.

25. Question 3(d)

    2 marksWays to manufacture ethanol
    Step 1: One way to make ethanol is fermentation, in which yeast converts glucose into ethanol and carbon dioxide. Step 2: The other way is the catalytic addition of steam to ethene (hydration of ethene). Step 3: So the two industrial methods are fermentation and the catalytic addition of steam to ethene.
    Method:

    Recall the two standard industrial routes to ethanol.
    Examiner tips

    • Ethanol: fermentation (renewable, slow) or hydration of ethene (fast, pure).

26. Question 3(e)(i)

    1 marksDeducing a molecular formula from a displayed formula
    Step 1: The molecular formula is found by counting every atom of each element in the displayed formula. Step 2: Counting gives 7 carbon atoms, 12 hydrogen atoms and 4 oxygen atoms. Step 3: So the molecular formula is $\text{C}_7\text{H}_{12}\text{O}_4$.
    Method:

    Count the carbon, hydrogen and oxygen atoms shown to build the molecular formula.
    Examiner tips

    • Carefully count each atom type from the displayed formula.

27. Question 3(e)(ii)

    1 marksMeaning of unsaturated
    Step 1: Unsaturated molecules contain at least one carbon-carbon double bond (C=C). Step 2: A molecule with only single C-C bonds is saturated, so the presence of a C=C is what makes a molecule unsaturated. Step 3: An O-H group or being a hydrocarbon does not decide saturation, so the key feature is a carbon-carbon double bond.
    Method:

    Recall the definition of unsaturated and identify the C=C bond.
    Examiner tips

    • Saturated = all single bonds; unsaturated = at least one C=C.

28. Question 4(a)(i)

    1 marksGroup number and outer electrons
    Step 1: The group number of a main-group element equals the number of electrons in its outer shell. Step 2: Nitrogen has the electronic configuration 2,5, so it has 5 outer electrons. Step 3: Having 5 outer electrons places nitrogen in Group V.
    Method:

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

    • Outer electrons = group number; number of shells = period.

29. Question 4(a)(ii)

    3 marksSubatomic particles in isotopes
    Step 1: Both isotopes have proton number 7, so each atom has 7 protons and, being neutral, 7 electrons. Step 2: Neutrons = nucleon number − proton number, so nitrogen-14 has 14 − 7 = 7 neutrons and nitrogen-15 has 15 − 7 = 8 neutrons. Step 3: So nitrogen-14 is 7 protons, 7 neutrons, 7 electrons and nitrogen-15 is 7 protons, 8 neutrons, 7 electrons.
    Method:

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

    • Isotopes share the proton number but differ in neutron number.

30. Question 4(b)(i)

    1 marksSource of oxides of nitrogen
    Step 1: Oxides of nitrogen form when nitrogen and oxygen from the air react together at high temperatures. Step 2: Car engines reach high enough temperatures for this to happen, so they are a major source of oxides of nitrogen. Step 3: Photosynthesis, evaporation and burning hydrogen do not produce oxides of nitrogen, so car engines are a correct source.
    Method:

    Identify where nitrogen and oxygen get hot enough to react in everyday life.
    Examiner tips

    • Oxides of nitrogen come mainly from high-temperature engines (vehicles).

31. Question 4(b)(ii)

    1 marksBalancing a symbol equation
    Step 1: Count the hydrogen atoms on the left: $4\text{NH}_3$ gives 4 × 3 = 12 hydrogen atoms. Step 2: Each water molecule, $\text{H}_2\text{O}$, contains 2 hydrogen atoms, so 12 ÷ 2 = 6 water molecules are needed. Step 3: So $n = 6$, giving $6\text{H}_2\text{O}$ (and this also balances the oxygen atoms).
    Method:

    Count hydrogen atoms on the left and divide by 2 to find the number of water molecules.
    Examiner tips

    • Match hydrogen atoms: 12 H on the left needs 6 $\text{H}_2\text{O}$.

32. Question 4(b)(iii)

    1 marksBonding in nitrogen dioxide
    Step 1: Nitrogen dioxide is made of two non-metals, nitrogen and oxygen. Step 2: Non-metal atoms bond by sharing electrons, which is covalent bonding. Step 3: Ionic bonding needs a metal and a non-metal, and metallic bonding needs metal atoms, so the bonding in $\text{NO}_2$ is covalent.
    Method:

    Decide the bonding from the types of element present.
    Examiner tips

    • Non-metal + non-metal = covalent bonding.

33. Question 4(b)(iv)

    1 marksClassifying nitrogen dioxide as an oxide
    Step 1: Oxides of non-metals are acidic, while oxides of metals are basic. Step 2: Nitrogen is a non-metal, so its oxide nitrogen dioxide is an acidic oxide. Step 3: So nitrogen dioxide is acidic because nitrogen is a non-metal.
    Method:

    Use the metal/non-metal rule to classify the oxide and justify it.
    Examiner tips

    • Non-metal oxide = acidic; metal oxide = basic.

34. Question 4(c)

    2 marksRelative formula mass calculation
    Step 1: Add the masses: aluminium = 1 × 27 = 27, nitrogen = 3 × 14 = 42, oxygen = 9 × 16 = 144. Step 2: Total relative formula mass = 27 + 42 + 144. Step 3: This gives 213.
    Method:

    Multiply each element relative atomic mass by its number of atoms and add them up.
    Examiner tips

    • Multiply each atom by its Ar and count all atoms, including the 9 oxygens.

35. Question 5(a)

    3 marksReaction of a carbonate with an acid
    Step 1: An acid reacting with a metal carbonate always gives a salt, carbon dioxide and water. Step 2: Nitric acid gives a nitrate salt, so copper(II) carbonate gives copper(II) nitrate, plus carbon dioxide and water. Step 3: So the three products are copper(II) nitrate, carbon dioxide and water.
    Method:

    Apply the general acid + carbonate reaction and name the nitrate salt.
    Examiner tips

    • Carbonate + acid always releases carbon dioxide as well as salt and water.

36. Question 5(b)

    3 marksPreparing pure dry crystals of a soluble salt
    Step 1: First remove the leftover (excess) magnesium oxide by filtering, leaving pure magnesium chloride solution. Step 2: Heat the solution to evaporate water to the point of crystallisation (until crystals just start to form), then allow it to cool so crystals grow. Step 3: Finally dry the crystals between sheets of filter paper, giving pure, dry crystals.
    Method:

    Order the three steps: remove excess solid, crystallise by evaporation, then dry.
    Examiner tips

    • Soluble salt prep: react with excess base, filter, evaporate to crystallisation, dry.

37. Question 5(c)(i)

    2 marksIdentifying the anode and electrolyte
    Step 1: The anode is always the electrode connected to the positive terminal of the power supply. Step 2: The electrolyte is the substance that conducts electricity and is broken down, here the molten cobalt(II) bromide. Step 3: So the anode is the positive electrode and the electrolyte is the molten cobalt(II) bromide.
    Method:

    Identify the positive electrode as the anode and the molten compound as the electrolyte.
    Examiner tips

    • Remember: anode positive, cathode negative; electrolyte is the molten/aqueous compound.

38. Question 5(c)(ii)

    2 marksProducts of electrolysis of a molten salt
    Step 1: Molten cobalt(II) bromide contains $\text{Co}^{2+}$ ions and $\text{Br}^-$ ions, with no water present. Step 2: The positive bromide ions lose electrons at the anode to form bromine, while the cobalt ions gain electrons at the cathode to form cobalt metal. Step 3: So bromine forms at the positive electrode and cobalt forms at the negative electrode.
    Method:

    Identify the ions in the molten salt and send the metal to the cathode, non-metal to the anode.
    Examiner tips

    • Molten compound: metal at cathode (-), non-metal at anode (+).

39. Question 5(d)

    3 marksIdentifying simple molecular compounds from properties
    Step 1: Simple molecular compounds have low melting points because only weak forces between molecules need to be overcome. Step 2: They also do not conduct electricity because they have no ions or free electrons. Step 3: Compounds C (−182 °C) and E (−101 °C) both have low melting points and do not conduct when molten, so C and E are the simple molecular compounds.
    Method:

    Use the two key properties of simple molecules (low mp, non-conducting) to pick the compounds.
    Examiner tips

    • Low melting point + no conduction = simple molecular structure.

40. Question 6(a)

    1 marksIdentifying a chemical change
    Step 1: A chemical change forms one or more new substances and is usually difficult to reverse. Step 2: Decomposing vegetation breaks down into new substances, so it is a chemical change. Step 3: Boiling, dissolving and mixing are physical changes because no new substances are made, so the chemical change is decomposing vegetation.
    Method:

    Decide which process makes a new substance rather than just changing form.
    Examiner tips

    • Chemical change = new substance formed (hard to reverse).

41. Question 6(b)(i)

    1 marksGreatest temperature change
    Step 1: The temperature change is the difference between final and initial temperatures, ignoring direction. Step 2: Experiment 1 changes by 6, Experiment 2 by 8, Experiment 3 by 6 and Experiment 4 by 5. Step 3: The largest change is 8 °C, in Experiment 2.
    Method:

    Calculate the size of each temperature change and select the largest.
    Examiner tips

    • Temperature change = |final − initial|; compare the sizes.

42. Question 6(b)(ii)

    1 marksIdentifying the most endothermic reaction
    Step 1: Endothermic reactions take in heat, so the temperature falls. The most endothermic is the one with the largest temperature drop. Step 2: Experiment 1 falls by 6 °C and Experiment 4 falls by 5 °C, while Experiments 2 and 3 rise (exothermic). Step 3: The largest fall is 6 °C in Experiment 1, so it is the most endothermic.
    Method:

    Pick the experiment with the largest temperature decrease.
    Examiner tips

    • A temperature fall means heat is absorbed (endothermic).

43. Question 6(b)(iii)

    2 marksLabelling an exothermic reaction pathway
    Step 1: On a reaction pathway diagram the vertical axis is labelled energy. Step 2: For an exothermic reaction the products are at a lower energy than the reactants, because energy is released. Step 3: So the reactants are on the left at a higher level and the products on the right at a lower level, with the energy axis labelled.
    Method:

    Label the energy axis, then place reactants high and products low for an exothermic change.
    Examiner tips

    • Energy on the y-axis; exothermic means products end lower than reactants.

44. Question 7(a)

    2 marksStages in water treatment
    Step 1: Water treatment first removes solid particles by sedimentation and filtration, so stage 1 is sedimentation/filtration. Step 2: After the carbon stage (which removes tastes and odours), the final stage kills microbes by chlorination. Step 3: So stage 1 is sedimentation/filtration and stage 3 is chlorination.
    Method:

    Recall the sequence of water-treatment stages and fill the gaps in order.
    Examiner tips

    • Order: filtration/sedimentation → carbon (taste) → chlorination (kills microbes).

45. Question 7(b)

    2 marksChemical test for water
    Step 1: Water is detected with anhydrous copper(II) sulfate (or anhydrous cobalt(II) chloride). Step 2: When water is added, white anhydrous copper(II) sulfate turns blue. Step 3: Limewater tests for carbon dioxide and litmus tests for chlorine, so the correct test is white-to-blue with anhydrous copper(II) sulfate.
    Method:

    Recall the standard chemical test for water and its colour change.
    Examiner tips

    • Water tests: copper(II) sulfate white→blue, or cobalt(II) chloride blue→pink.

46. Question 7(c)

    2 marksHarmful effects of air pollutants
    Step 1: Carbon dioxide is a greenhouse gas, so it causes global warming and climate change. Step 2: Oxides of nitrogen dissolve in rain to form acids, causing acid rain (and they can cause respiratory problems). Step 3: So carbon dioxide causes global warming and oxides of nitrogen cause acid rain.
    Method:

    Match each pollutant to its main environmental effect.
    Examiner tips

    • Greenhouse gas (CO2) → warming; acidic gas (oxides of nitrogen) → acid rain.

47. Question 8(a)(i)

    1 marksOrder of reactivity from reactions with acid
    Step 1: The more vigorously a metal reacts with acid, the more reactive it is. Step 2: A reacts strongly at room temperature (most reactive), D reacts strongly only on warming, C reacts weakly on warming, and B does not react (least reactive). Step 3: So the order from most to least reactive is A, D, C, B.
    Method:

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

    • Rank by vigour of reaction: room-temperature reaction beats only-on-warming; no reaction is least.

48. Question 8(a)(ii)

    2 marksWays to increase reaction rate
    Step 1: A reaction speeds up when particles collide more often or more successfully. Step 2: Increasing the acid concentration gives more frequent collisions, and using smaller pieces of metal increases the surface area for collisions (adding a catalyst would also work). Step 3: So two ways are increasing concentration and increasing surface area.
    Method:

    Recall the factors that raise reaction rate, other than temperature.
    Examiner tips

    • Rate increases with concentration, surface area, temperature and a catalyst.

49. Question 8(b)(i)

    1 marksFlame test colour for lithium
    Step 1: Flame tests give characteristic colours for different metal ions. Step 2: Lithium ions give a red flame colour. Step 3: Lilac is potassium, yellow is sodium and blue-green is copper, so lithium gives a red flame.
    Method:

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

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

50. Question 8(b)(ii)

    1 marksIon present in all alkalis
    Step 1: An alkali is a soluble base that produces ions in water. Step 2: All aqueous alkalis contain hydroxide ions, $\text{OH}^-$, which make the solution alkaline. Step 3: Hydrogen ions make acids, while oxide and chloride ions are not what makes a solution alkaline, so the answer is the hydroxide ion.
    Method:

    Recall the ion that defines an alkaline solution.
    Examiner tips

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

51. Question 8(b)(iii)

    1 marksIndicator colour in an alkali
    Step 1: Thymolphthalein is an indicator that is colourless in acids and neutral solutions. Step 2: In an alkali it turns blue. Step 3: So thymolphthalein is blue in an aqueous alkali.
    Method:

    Recall the colour of thymolphthalein in alkaline solution.
    Examiner tips

    • Thymolphthalein: colourless (acid) → blue (alkali).

52. Question 8(c)(i)

    1 marksParticle structure of an alloy
    Step 1: An alloy contains atoms of different elements, so it has atoms of different sizes. Step 2: The different-sized atoms disrupt the regular layers found in a pure metal. Step 3: So the arrangement that best represents an alloy shows atoms of two different sizes mixed together, not identical atoms in a regular pattern.
    Method:

    Pick the arrangement showing a mixture of different-sized atoms.
    Examiner tips

    • Alloys contain different-sized atoms that break up the regular metal layers.

53. Question 8(c)(ii)

    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 pewter harder and stronger than pure tin. Step 3: So the useful property is that pewter is harder/stronger than pure tin.
    Method:

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

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

54. Question 8(d)

    2 marksPhysical properties of metals
    Step 1: Typical physical properties of metals include being malleable, ductile, good conductors of electricity and having high melting points. Step 2: Two correct properties are that metals are malleable and good conductors of electricity. Step 3: Being brittle, non-conducting, low-melting or gaseous are not typical of metals, so the correct pair is malleable and good electrical conductor.
    Method:

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

    • Metal properties: malleable, ductile, conduct heat and electricity, high melting point.