Use this to mark your own attempt. Accept any correct alternative that answers the question as set. Marks are for the ideas, not the exact words. Unless stated, ecf (error carried forward) applies in calculations. Chermistry practice paper – not an official examination paper.
Total marks: 80 (Section A: 70 marks; Section B: one question of 10 marks)
| Part | Answer and mark points | Marks |
|---|---|---|
| 1(a) | P: ionic; Q: simple molecular; R: giant covalent; S: metallic [2] for all four correct; [1] for two or three correct. Accept “giant ionic lattice”, “giant metallic lattice”. P: only substance that conducts as a liquid but not as a solid. S: conducts as a solid. Q and R: do not conduct; the low melting point of Q and the high melting point of R separate them. | 2 |
| 1(b) | • (Q is made of) molecules, which have no overall charge / are uncharged / there are no ions [1] • there are no free (mobile) ions or delocalised electrons to carry the charge [1] • Reject: “there are no electrons”; “covalent bonds do not conduct” without reference to charge carriers. | 2 |
| 1(c) | • silicon(IV) oxide has a giant covalent structure: many strong covalent bonds must be broken to melt it, needing a large amount of energy [1] • carbon dioxide has a simple molecular structure: only weak forces between molecules need to be overcome, needing little energy [1] • Both a structure and the matching energy / force idea are needed for each mark. Reject: “the covalent bonds in carbon dioxide are weak”. | 2 |
| 1(d)(i) | diagram 1 (it contains atoms of a different size / a different element) | 1 |
| 1(d)(ii) | the different-sized atoms disrupt the regular arrangement (layers) of atoms, so the layers cannot slide over each other easily (either idea, stated clearly, earns the mark). Reject: the atoms are held by stronger bonds; “alloys have more electrons”. | 1 |
| Part | Answer and mark points | Marks |
|---|---|---|
| 2(a) | 64 30Zn: 34 neutrons; 66 30Zn2+: 28 electrons; 80 35Br−: 36 electrons; 84 36Kr: 48 neutrons [2] for all four correct; [1] for two or three correct. 64 − 30 = 34. 30 − 2 = 28 electrons. 35 + 1 = 36 electrons. 84 − 36 = 48. | 2 |
| 2(b) | Br− and Kr (both have 36 electrons) Accept: the full nuclide notation. Both are needed. | 1 |
| 2(c)(i) | a shared pair of electrons between two atoms (nuclei) Accept: a pair of electrons shared by two atoms. Reject: “sharing of electrons” with no pair; transfer of electrons. | 1 |
| 2(c)(ii) | two single shared pairs, one between O and each Cl, each shown with one dot and one cross [1] two lone pairs on O and three lone pairs on each Cl, so that each atom has eight outer shell electrons [1] Circles (overlapping shells) are optional and inner shells need not be drawn. Accept dots and crosses interchanged, as long as the two electrons of each shared pair are shown differently. Reject: a double bond; shared pairs missing. | 2 |
| 2(d) | it is a mixture (of two different dyes / colours) Accept: it contains (at least) two dyes. Reject: “it is a compound”; “it is impure” without reference to two dyes. | 1 |
| Part | Answer and mark points | Marks |
|---|---|---|
| 3(a) | WO3(s) + 3H2(g) → W(s) + 3H2O(g) correct formulae and balancing [1]; correct state symbols [1] Steam is (g). Accept multiples. Any wrong formula: 0 for the first mark. | 2 |
| 3(b) | 184 + (3 × 16) = 184 + 48 = 232 | 1 |
| 3(c) | • amount of WO3 = 69.6 ÷ 232 = 0.300 mol [1] • amount of H2 = 14.4 ÷ 24 = 0.600 mol [1] • 0.300 mol of WO3 needs 3 × 0.300 = 0.900 mol of H2, but only 0.600 mol is supplied, so hydrogen is the limiting reactant [1] • Alternative: 0.600 mol H2 reacts with only 0.200 mol WO3 (less than 0.300 mol). ecf from (b) for the first mark. Final mark needs a stated conclusion. | 3 |
| 3(d) | the hydrogen limits: 3 mol H2 form 1 mol W, so 0.600 ÷ 3 = 0.200 mol W [1]; 0.200 × 184 = 36.8 g [1] ecf: if (c) wrongly concludes that WO3 is limiting, 0.300 × 184 = 55.2 g scores [1] (method) and 55.2 may be carried into (e): 33.12 ÷ 55.2 × 100 = 60.0%. A correct 36.8 g scores [2] without working. | 2 |
| 3(e) | 33.12 ÷ 36.8 × 100 = 90.0 % (ecf from (d); accept 90) | 1 |
| Part | Answer and mark points | Marks |
|---|---|---|
| 4(a) | • nitric acid [1] • it has the lower pH (a higher concentration of H+ ions) even though its solution is much less concentrated; so a greater proportion of its molecules is ionised (it is fully ionised) [1] • The second mark is awarded only if the first is correct. Accept: dissociates for ionises. Reject: an answer that refers to pH or concentration alone without the comparison. | 2 |
| 4(b)(i) | the excess potassium hydroxide is dissolved (in the solution) so cannot be filtered off / removed by filtration Accept: the solution would contain excess alkali / both are soluble. | 1 |
| 4(b)(ii) | • titration: measure a fixed volume of potassium hydroxide (using a pipette) and add the acid from a burette with an indicator until the indicator just changes colour (neutralisation point); note the volume of acid [1] • repeat with the same volumes of alkali and acid but no indicator [1] • evaporate (heat) the solution until it is saturated (crystallisation point), cool to crystallise; filter off the crystals and dry them between filter paper [1] • Each mark is a separate credit point. Accept pH meter / probe in place of an indicator. Accept: wash the crystals with a little cold distilled water before drying. Reject: drying in an oven / strong heating; evaporating to dryness; adding the indicator to the final mixture to be crystallised. | 3 |
| 4(c) | any soluble barium salt (barium nitrate / barium chloride) and a soluble sulfate or sulfuric acid (sodium sulfate / potassium sulfate / dilute sulfuric acid) [1] (both needed) Reject: barium carbonate (insoluble); insoluble sulfates such as calcium sulfate or lead(II) sulfate; barium sulfate itself. | 1 |
| 4(d) | H2SO4: 0.0500 × 20.0 ÷ 1000 = 0.00100 mol; KOH: 0.00100 × 2 = 0.00200 mol [1]; 0.00200 ÷ 0.0250 = 0.0800 mol / dm3 [1] Accept 0.08. If the 1 : 2 ratio is not used, 0.0400 scores [1] (ecf). | 2 |
| Part | Answer and mark points | Marks |
|---|---|---|
| 5(a) | C2H4 (ethene) Balance: C9H20 − C7H16 = C2H4. Accept: C2H4, ethene, CH2=CH2. Reject: C2H6 (an alkane). | 1 |
| 5(b)(i) | addition Reject: substitution; polymerisation; “bromination” alone. | 1 |
| 5(b)(ii) | one Br atom on each carbon atom, each carbon atom has two H atoms, a single C–C bond, all atoms and all bonds shown Reject: 1,1-dibromoethane (both Br on one carbon); a C=C bond; Br written as Br2 between the carbons. | 1 |
| 5(c) | it has a carbon–carbon double bond (C=C) Reject: a double bond between carbon and hydrogen; “it contains a double bond” with no mention of carbon. | 1 |
| 5(d) | 1 mark: carbon–carbon single bonds in the backbone, no C=C remaining, and continuation bonds at both ends (open bonds beyond the first and last carbon atoms) 1 mark: two repeat units, each of two backbone carbon atoms, one with two H atoms and the other with one H atom and one Cl atom, so that the Cl atoms are on alternate carbon atoms of the backbone Reject: any C=C; no continuation bonds; Cl inside the chain; Cl atoms on adjacent carbon atoms (head-to-head); only one repeat unit, even if brackets and the subscript n are used. | 2 |
| 5(e) | C=C with two F atoms on each carbon atom, all atoms and all bonds shown Reject: a structure with no C=C; a monomer with fewer or more than four F atoms. | 1 |
| 5(f) | methylpropane: a central carbon atom bonded to three CH3 groups and one H atom, all atoms and all bonds shown Reject: butane itself (unbranched); a structure that has 4 C but not 10 H; any C=C. | 1 |
| Part | Answer and mark points | Marks |
|---|---|---|
| 6(a) | exothermic and the products are at a lower energy than the reactants (in Fig. 6.1) / energy is given out to the surroundings Both the term and the reason are needed. Accept: ΔH is negative as the reason. Reject: “gives out energy to the reactants”. | 1 |
| 6(b) | 1 mark: arrow for Ea drawn upwards from the level of the reactants to the top of the curve, and labelled 1 mark: arrow for ΔH drawn downwards from the level of the reactants to the level of the products (arrow head pointing down), and labelled Reject: an Ea arrow that starts at the products or at the axis; a ΔH arrow pointing upwards; arrows with no labels. | 2 |
| 6(c) | a second curve (dashed on the answer figure above) that starts at the reactants and ends at the products (same levels as the original curve) and has a lower peak than the original curve; labelled C Reject: a curve that ends at a different energy level; a curve with a higher peak. | 1 |
| 6(d) | the energy given out when bonds form (are made) is greater than the energy taken in to break bonds The comparison must be stated. Reject: “energy is released when bonds break”; “bond breaking is exothermic”; a statement of the energy released with no comparison. | 1 |
| 6(e) | • the rate is greater (increases) because the particles move faster / have more (kinetic) energy • so the frequency of collisions is higher / collisions are more frequent, or a greater proportion of collisions have energy at least equal to the activation energy (more successful collisions per second) • Reject: “more collisions” with no reference to time or frequency; “more particles”; “lower activation energy” (a catalyst does that). The direction (greater / higher / more frequent) must be stated. | 2 |
| 6(f) | mass of CO2 lost = 155.00 − 154.12 = 0.88 g; moles = 0.88 ÷ 44 = 0.0200 mol [1] volume = 0.0200 × 24 000 = 480 cm3 [1] ecf from the first step. Accept 0.48 dm3 as an equivalent value if the unit is given. The loss in mass is the carbon dioxide that escapes. | 2 |
| Part | Answer and mark points | Marks |
|---|---|---|
| 7(a) | aluminium: +3 and iodine: −1 (both needed) Accept: 3+ and 1−; III and −I; Al3+ and I−. Reject: the signs the wrong way round; +3 and −3. | 1 |
| 7(b) | • iodine (I2) • iodine gains electrons (from aluminium) / each iodine atom gains one electron to become an iodide ion • The second mark depends on iodine being identified. Accept: electrons are transferred from aluminium to iodine. Reject: iodine loses electrons; aluminium as the oxidising agent; an explanation in terms of oxidation state only. | 2 |
| 7(c) | 27 + (3 × 127) = 408 Reject: 154 (only one iodine atom counted); 816 (the formula Al2I6 used). | 1 |
| 7(d) | yellow precipitate Accept: pale yellow precipitate. Reject: a white or cream precipitate; “yellow solution” with no precipitate. | 1 |
| 7(e) | aluminium gains oxygen Accept: aluminium atoms lose electrons; the oxidation state of aluminium increases from 0 to +3. Reject: “oxygen is oxidised”; “aluminium loses oxygen”. | 1 |
| 7(f)(i) | 2,8,3 and Group 13 (both needed) Accept: 2.8.3; Group III (the older numbering). Reject: Group 3 (a different group in the 1–18 numbering). | 1 |
| 7(f)(ii) | an aluminium atom has three outer electrons, which it loses (to form Al3+) Reject: gains three electrons; “three shells”. | 1 |
| 7(f)(iii) | from metallic to non-metallic Accept: metals on the left, non-metals on the right; becomes less metallic / more non-metallic. Reject: reactivity trends. | 1 |
| Part | Answer and mark points | Marks |
|---|---|---|
| 8(a) | they have different boiling points (accept: different boiling point ranges) Reject: different densities; different sizes alone; “melting points”. | 1 |
| 8(b) | Mr of methane = 12 + 4 = 16; moles of methane = 3.2 ÷ 16 = 0.200 mol; the ratio CH4 : CO2 is 1 : 1, so 0.200 mol of CO2 [1] volume = 0.200 × 24 = 4.8 dm3 [1] ecf from the first step. Accept 4800 cm3 as an equivalent value if the unit is given. Reject 9.6 dm3 (a 1 : 2 ratio from the coefficient of O2 or H2O). | 2 |
| 8(c) | nitrogen and oxygen (from the air) react at the high temperature in the engine Both gases must be named. Accept: spark / high temperature causes nitrogen to react with oxygen. Reject: the burning of fuel only; nitrogen from the fuel. | 1 |
| 8(d) | it reacts with (erodes / corrodes / dissolves) limestone or marble / metal structures corrode Accept: damages stonework or metal. Reject: kills fish; damages crops (not buildings); causes global warming. | 1 |
| 8(e) | • correct formulae for all four substances: C2H4 + O2 → CO2 + H2O • balanced: C2H4 + 3O2 → 2CO2 + 2H2O • Accept any correct multiple. State symbols are not required. The second mark is awarded only if all four formulae are correct. | 2 |
| 8(f) | three carbon atoms in a chain with the –OH group on the end carbon atom, all atoms and all bonds shown (including the O–H bond) Reject: propan-2-ol (the OH on the middle carbon); the OH group drawn as C–HO; a structure with missing C–H bonds. | 1 |
| 8(g) | magnesium ethanoate and hydrogen (both needed) Accept: Mg(CH3CO2)2 and H2 as formulae. Reject: magnesium acetate; carbon dioxide; water; oxygen. | 1 |
| 8(h) | • carboxylic acid: propanoic acid • alcohol: methanol • Accept: the correct formulae, CH3CH2CO2H and CH3OH. Reject: ethanoic acid or ethanol (carbon atoms miscounted); the answers the wrong way round. | 2 |
| Part | Answer and mark points | Marks |
|---|---|---|
| 9(a) | it contains ions that are free to move / mobile (and carry the charge). Accept: ions can move to the electrodes. Reject: "electrons move through the solution"; "it contains ions" with no mention of movement. | 1 |
| 9(b) | • cathode: hydrogen (accept H2). Reject: hydrogen ions; water • anode: oxygen (accept O2). Reject: sulfur / sulfur dioxide / sulfate; hydroxide | 2 |
| 9(c) | 2H2O → O2 + 4H+ + 4e−. Accept either this or 4OH− → 2H2O + O2 + 4e− as equal alternatives; accept multiples (e.g. 2OH− → H2O + ½O2 + 2e−); state symbols not required. Reject: electrons on the left; unbalanced equations; O (not O2). | 1 |
| 9(d) | hydrogen ions gain electrons / electrons are gained. Accept: the oxidation state of hydrogen decreases from +1 to 0. Reject: "loses oxygen"; "loses electrons". | 1 |
| 9(e) | • [1] magnesium is more reactive than copper, with reason (magnesium displaces copper; copper does not displace magnesium) • [1] copper is more reactive than silver, with reason (copper displaces silver; silver does not displace copper). If the correct order magnesium, copper, silver is given with no reasoning, or with only one comparison supported, award [1] in total | 2 |
| 9(f) | Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s). All species, balanced atoms and charge, and correct state symbols required. Reject: unbalanced charge; nitrate ions included in a wrong equation. | 1 |
| 9(g)(i) | carbon is less reactive than magnesium (magnesium is higher in the reactivity series than carbon), so carbon cannot remove the oxygen / reduce magnesium oxide. Accept: magnesium holds on to oxygen more strongly than carbon does. Reject: "magnesium is too reactive" with no comparison with carbon. | 1 |
| 9(g)(ii) | yes, silver is less reactive than copper (silver is below copper in the reactivity series), and copper(II) oxide is reduced by carbon / silver oxide is easier to reduce. Prediction and reason both needed. Accept ecf from the order given in (e). Reject: "no" (unless consistent with an incorrect order in (e)). | 1 |
| Part | Answer and mark points | Marks |
|---|---|---|
| 10(a) | in the solid, the ions are held in fixed positions in a lattice (cannot move); when molten the ions are free to move (and carry charge). Both the solid and the molten state must be addressed. Reject: "electrons move in the molten liquid"; "the solid has no ions". | 1 |
| 10(b) | • cathode: lead (accept Pb). Reject: lead(II) ions; bromine • anode: bromine (accept Br2). Reject: bromide; bromine ions; hydrogen | 2 |
| 10(c) | 2Br− → Br2 + 2e−. Accept: multiples; state symbols not required. Reject: electrons on the left; Br (not Br2); unbalanced charge. | 1 |
| 10(d) | lead(II) ions gain electrons / electrons are gained. Accept: the oxidation state of lead decreases from +2 to 0. Reject: "loses oxygen"; "loses electrons". | 1 |
| 10(e) | • melting point lower than 63 °C (accept any value below 63 °C, e.g. 39 °C). Reject: a value of 63 °C or higher • rubidium reacts more vigorously / faster than potassium (accept: explodes; bursts into flame; faster / more rapid fizzing; flame). Reject: a less vigorous reaction; "reacts" with no comparison | 2 |
| 10(f) | 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g). Formulae, balancing and state symbols all correct. Reject: LiO2; unbalanced equations; missing state symbols. | 1 |
| 10(g)(i) | high melting point OR high density (either property earns the mark; a comparison or data is not required). Do not credit colour or coloured compounds (a property of compounds, asked for in (ii)). | 1 |
| 10(g)(ii) | they are coloured / form coloured compounds (iron(II) chloride solution is pale green, but potassium chloride solution is colourless). Reject: high melting point or high density (properties of the elements, asked for in (i)); catalyst; variable oxidation state (not shown by the data). | 1 |