Mark schemePaper 3 Chemistry (set 3.7)

SCIENCE (CHEMISTRY) – Paper 3 Chemistry (set 3.7) – mark scheme

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: 65 (Section A: 55 marks; Section B: one question of 10 marks)

Section A – Question 1  [3]

PartAnswer and mark pointsMarks
1(a)
• condenser drawn in the dashed box, connected to the side arm and to the delivery tube leading to the beaker, with a water jacket (a tube inside a larger tube / a coiled tube in a jacket) [1]
• cold water in labelled at the lower end (the end nearer the beaker) and cold water out labelled at the upper end (the end nearer the flask) [1]
heatthermometerseawaterpure watercold water incold water out
Accept any recognisable condenser (straight, coiled or bulbed) with a jacket. For the second mark the inlet must be below the outlet. Reject: a single line or an open tube with no jacket; the water entering at the top; arrows with no labels; a condenser that is not joined to the side arm.
2
1(b)100 (accept 99 to 101)
The thermometer is in the vapour of the pure water, which condenses at its boiling point. Reject: 0; room temperature; a value above 101 (the thermometer is in the vapour of pure water, not in the seawater).
1

Section A – Question 2  [4]

PartAnswer and mark pointsMarks
2(a)7 electrons (crosses) in the outer (third) shell
17+17+−chlorine atomchloride ion
Reject: any other number of electrons in the third shell; electrons added to the first or second shell.
1
2(b)8 electrons (crosses) in the outer (third) shell and the charge 1− (accept −) written outside the bracket
Both are needed for the mark. The ion has 17 protons and 18 electrons. Reject: 1+; 8 electrons with no charge.
1
2(c)B
A calcium atom has 20 protons (nucleus 20+) and loses two electrons, leaving 18 electrons (2,8,8). Reject: A (a calcium atom, 20 electrons); C (an argon atom, nucleus 18+); D (nucleus 19+, a potassium ion).
1
2(d)3517Cl: 18 neutrons and 3717Cl: 20 neutrons (both needed)
35 − 17 = 18; 37 − 17 = 20. Reject: 35 and 37; 17 and 17.
1

Section A – Question 3  [6]

PartAnswer and mark pointsMarks
3(a)
• row 1 (It conducts electricity when solid): stainless steel [1]
• row 2 (It conducts electricity when molten): calcium chloride, stainless steel [1]
• row 3 (It is made of simple molecules): chlorine [1]
• 1 mark for each row. No mark for a row with a missing or an extra tick.
3
3(b)
• in the solid the ions are held in fixed positions in the lattice / cannot move
• when molten the ions are free to move (mobile) and carry the charge
• Reject: “electrons move”; “the molecules move”; “there are no ions in the solid”; “ions are free in the solid”.
2
3(c)any two from: good conductor of electricity; good conductor of heat; malleable; ductile; high melting point / boiling point; high density; shiny (lustrous); strong
Both properties are needed for the mark. Reject: “hard” or “brittle”; low melting point; two statements of the same property.
1

Section A – Question 4  [5]

PartAnswer and mark pointsMarks
4(a)
• acidic (pH 0 to 6), neutral (pH 7) and alkaline (pH 8 to 14) written in the correct three boxes [1]
• green written in the box for pH 7 [1]
acidicneutralalkaline01234567891011121314redorangeyellowbluepurplegreenpH
Accept “acid” and “alkali” for acidic and alkaline. Reject: blue-green or yellow-green for pH 7; the labels in the wrong boxes.
2
4(b)5 or 6 (accept any value from 5 to 6)
Reject: 7; any value below 5 or above 6.
1
4(c)it decreases / goes down / becomes lower
Reject: the pH decreases; the hydroxide ions decrease; “neutralised” with no change in hydrogen ion concentration.
1
4(d)the concentration of hydrogen ions (H+) equals the concentration of hydroxide ions (OH−) / equal numbers of H+ and OH− ions
Reject: “no ions”; “pH 7” alone; “neither an acid nor an alkali”.
1

Section A – Question 5  [5]

PartAnswer and mark pointsMarks
5(a)hold the powder in a small tube / container / on a piece of paper inside the flask, fit the plug, then tip the powder into the acid (and start the stopwatch then)
Accept any sensible method that closes the flask before, or at the moment, the powder meets the acid. Reject: “repeat the experiment”; “use a bigger flask”; “use a gas syringe” (this is not an improvement to the way the powder is added); “use less powder”.
1
5(b)0.90 ÷ 60 = 0.015 g / s (accept 1.5 × 10−2; unit g/s or g s−1)
Both the value and the unit are needed. Reject: 15; 0.015 with no unit or a wrong unit (g, s, g/min).
1
5(c)
• the particles move faster / have more (kinetic) energy
• so they collide more frequently (more collisions per second) (accept: more successful collisions per second)
• 1 mark for each point. Reject: “more particles”; “more collisions” with no reference to time.
2
5(d)a reaction that gives out (transfers) heat energy to the surroundings (accept: the temperature of the surroundings rises)
Reject: “the reaction needs heat”; “heat is taken in”; “gives out gas”.
1

Section A – Question 6  [5]

PartAnswer and mark pointsMarks
6(a)
• carbon monoxide, source: incomplete combustion (burning) of the fuel / burning the fuel in a limited supply of oxygen [1]
• sulfur dioxide, effect: forms acid rain (accept: damages buildings / harms fish and plants in lakes; irritates the lungs) [1]
• nitrogen dioxide (accept: oxides of nitrogen; NOx) [1]
• Reject: “burning the fuel” with no reference to incomplete combustion or limited oxygen; “global warming” or “ozone depletion” for the effect of sulfur dioxide; “nitrogen” or “nitrous oxide” for the pollutant; nitrogen monoxide / nitric oxide (the formula given is NO2).
3
6(b)photosynthesis
Reject: respiration; combustion; “the carbon cycle”.
1
6(c)the carbon dioxide released is (about) the same as the amount taken in by the sugar cane by photosynthesis as it grew, so it is offset / balanced
Reject: “bioethanol does not produce carbon dioxide”; “bioethanol burns more cleanly”; no reference to the plants taking in carbon dioxide.
1

Section A – Question 7  [10]

PartAnswer and mark pointsMarks
7(a)
Wcationcopper(II), Cu2+blue solution; light blue precipitate (Data Sheet: copper(II) hydroxide is the only light blue hydroxide) that dissolves in excess ammonia only
Wanionsulfate, SO42−white precipitate with nitric acid then barium nitrate
Xcationcalcium, Ca2+white precipitate with sodium hydroxide that does not dissolve in excess; no precipitate with aqueous ammonia (aluminium and zinc would give a precipitate with ammonia)
Xanionchloride, Cl−white precipitate with nitric acid then silver nitrate
Yanioncarbonate, CO32−effervescence with dilute acid; the gas turns limewater milky (carbon dioxide)
Zcationaluminium, Al3+white precipitate that dissolves in excess sodium hydroxide, and does not dissolve in excess ammonia (zinc would dissolve in both)
Zanionsulfate, SO42−white precipitate with nitric acid then barium nitrate
1 mark each (7 marks): accept the name or the correct formula of the ion. Accept “copper”, “calcium”, “aluminium” without the charge or oxidation state. Reject the formula of the compound (e.g. CuSO4) for a single ion; sulfite or sulfide for sulfate; zinc for Z; iron(II), iron(III) or ammonium for any cation.
7
7(b)
• correct formulae of all reactants and products [1]; balanced [1]; correct state symbols [1]
• Missing or incorrect state symbols lose the third mark only. Correct formulae but unbalanced: 1 mark. A wrong formula loses the first mark; the balancing mark may still be awarded if the equation as written is balanced. Ionic equations with state symbols are also accepted for 3 marks.
• CuSO4(aq) + 2NaOH(aq) → Cu(OH)2(s) + Na2SO4(aq)  (ionic: Cu2+(aq) + 2OH−(aq) → Cu(OH)2(s))
• or CuSO4(aq) + 2NH3(aq) + 2H2O(l) → Cu(OH)2(s) + (NH4)2SO4(aq)
• or CuSO4(aq) + Ba(NO3)2(aq) → BaSO4(s) + Cu(NO3)2(aq)  (ionic: Ba2+(aq) + SO42−(aq) → BaSO4(s))
• or CaCl2(aq) + 2NaOH(aq) → Ca(OH)2(s) + 2NaCl(aq)  (ionic: Ca2+(aq) + 2OH−(aq) → Ca(OH)2(s))
• or CaCl2(aq) + 2AgNO3(aq) → 2AgCl(s) + Ca(NO3)2(aq)  (ionic: Ag+(aq) + Cl−(aq) → AgCl(s))
• or Na2CO3(aq) + 2HNO3(aq) → 2NaNO3(aq) + H2O(l) + CO2(g)  (ionic: CO32−(aq) + 2H+(aq) → H2O(l) + CO2(g))
• or Al2(SO4)3(aq) + 6NaOH(aq) → 2Al(OH)3(s) + 3Na2SO4(aq)  (ionic: Al3+(aq) + 3OH−(aq) → Al(OH)3(s))
• or Al2(SO4)3(aq) + 3Ba(NO3)2(aq) → 3BaSO4(s) + 2Al(NO3)3(aq)
• or Al2(SO4)3(aq) + 6NH3(aq) + 6H2O(l) → 2Al(OH)3(s) + 3(NH4)2SO4(aq)
3

Section A – Question 8  [7]

PartAnswer and mark pointsMarks
8(a)the sum of the relative atomic masses of all the atoms in a molecule / in the formula (accept: the average mass of a molecule compared with one twelfth of the mass of a carbon-12 atom)
Reject: the mass of one mole; the mass of a molecule in grams; “the total mass of the atoms” with no reference to relative atomic mass.
1
8(b)final volume = 48 cm3 = 0.048 dm3; 0.048 ÷ 24 = 0.00200 mol (2.00 × 10−3)
Reject: 48 ÷ 24 = 2.0 (the cm3 not converted).
1
8(c)0.00200 mol (ecf from (b)) – the equation shows 1 mol of Ca(HCO3)2 gives 1 mol of CO2
Reject: 0.00400 (ratio 1 : 2 used).
1
8(d)
• 50.0 cm3 = 0.0500 dm3 [1]
• 0.00200 ÷ 0.0500 = 0.0400 mol / dm3 [1] (ecf from (c))
• Reject: 0.00200 ÷ 50 = 4.0 × 10−5 (cm3 not converted).
2
8(e)40 + 2 × (1 + 12 + 3 × 16) = 1621
8(f)0.0400 × 162 = 6.48 g / dm3 (accept 6.5; ecf from (d) and (e))1

Section A – Question 9  [10]

PartAnswer and mark pointsMarks
9(a)
• statement 1: incorrect [1]
• statement 2: correct [1]
• statement 3: incorrect [1]
• statement 4: incorrect [1]
• 1 mark for each row. No mark for a row with no tick or with two ticks.
4
9(b)
• statement 1: ethene is unsaturated because its molecules contain a carbon-to-carbon double bond (C=C) [1]
• statement 3: the polymer contains no C=C double bonds: the double bond in each ethene molecule opens to join the molecules together (addition polymerisation) [1]
• statement 4: ethanol is not a hydrocarbon: a hydrocarbon contains only carbon and hydrogen, but ethanol also contains oxygen [1]
• Each correction must give the reason, not only the opposite word. Reject for statement 1: “ethene is unsaturated” with no reference to the double bond.
3
9(c)
• general formula of alkanes: CnH2n+2 [1]
• name of the alkene with the formula C3H6: propene [1]
• molecular formula of methanol: CH3OH (accept CH4O) [1]
• Accept prop-1-ene for propene. Reject: CnH2n or CnH2n−2 for alkanes; propane for the alkene; CH3O or CH4OH for methanol.
3

Section B – Question 10  [10]

PartAnswer and mark pointsMarks
10(a)
• chlorine is a gas (its boiling point is below 28 °C)
• bromine is a liquid and iodine is a solid (bromine melts below and boils above 28 °C; iodine melts above 28 °C). All three correct = 2 marks; two correct = 1 mark; fewer = 0. Reject: vapour for gas; "melted" or "boiling" in place of a state.
2
10(b)a cross at period 5 and 185 °C (accept 175 to 195 °C, i.e. within half a small square) and a straight line (ruled) joining it to the boiling point of bromine. Both needed. Reject: a point at period 4 or 6; a line drawn from the cross to another point; a curve or a freehand line.1
10(c)any value from 250 to 350 °C (the boiling points rise by 93 to 154 °C from one element to the next, so 185 + 93 = 278 to 185 + 154 = 339). Accept a value in this range with the unit omitted. Reject: a value of 185 °C or less; a value above 350 °C; "higher than iodine" with no value.1
10(d)the solution turns orange (accept yellow, yellow-brown, orange-brown, brown). Chlorine displaces bromine from the bromide. Reject: purple; the solution stays colourless; "bromine forms" with no colour or other observation; green gas.1
10(e)(i)fluorine (accept F). Reactivity decreases down Group 17, so the element at the top of the group (period 2) is the most reactive. Reject: chlorine; iodine; astatine; "halogen" or "Group 17" alone.1
10(e)(ii)seven (7) (all Group 17 elements have seven outer shell electrons). Reject: 17; 85; eight; 2.8.7.1
10(f)
• one shared pair of electrons (one dot and one cross) in the overlap between the hydrogen atom and the fluorine atom; hydrogen has no other electrons
• fluorine has three non-bonding pairs (lone pairs), so eight electrons in its outer shell including the shared pair
• Accept without outer circles drawn; accept all electrons as dots or all as crosses if the shared pair is clear. Do not credit: ions drawn (H+ and F−); a double bond; seven electrons around fluorine; extra electrons on hydrogen.
HF
2
10(g)chlorine is an oxidising agent (it oxidises iodide ions to iodine, which is brown) or chlorine is more reactive than iodine / chlorine displaces iodine from the iodide. Accept: chlorine takes electrons from the iodide ions / chlorine is reduced. Reject: chlorine is a reducing agent; chlorine is a gas; "iodine forms" alone.1

Section B – Question 11  [10]

PartAnswer and mark pointsMarks
11(a)
• box A: lead(II) oxide and carbon (accept a mixture of lead(II) oxide and carbon / PbO and C; both needed). Reject: lead; carbon only; limewater
• box B: limewater (accept calcium hydroxide solution / aqueous calcium hydroxide). Reject: water; lime; calcium oxide
2
11(b)a tube (one or two lines) drawn from the bung, leading into the limewater with its open end below the surface of the limewater. Reject: a tube that ends above the liquid; a tube drawn through the glass wall; a tube that does not join the bung.1
11(c)a different mass of carbon was used with calcium oxide (1.0 g instead of 0.5 g). Accept: not the same amount of carbon in all three tests. Reject: the oxides are different; different observations; different masses of oxide (they are all 2.0 g).1
11(d)calcium oxide, because calcium is more reactive than carbon (so carbon cannot remove the oxygen). Both the oxide and the reason are needed. Accept: calcium is higher in the reactivity series than carbon. Reject: calcium is unreactive; zinc oxide; lead(II) oxide; a reason with no comparison with carbon.1
11(e)(i)the limewater turns milky / cloudy / a white precipitate forms. Reject: bubbles only; the limewater stays clear; a colour change other than milky / cloudy / white.1
11(e)(ii)zinc oxide (zinc is the most reactive of the three metals, so its oxide is the hardest to reduce; Table 11.1 shows zinc oxide needs very strong heating). Reject: lead(II) oxide; iron(III) oxide.1
11(f)
• two shared pairs of electrons, one between the sulfur atom and each hydrogen atom (each pair one dot and one cross); hydrogen has no other electrons
• sulfur has two non-bonding pairs (lone pairs), so eight electrons in its outer shell including the shared pairs
• Accept without outer circles drawn; accept all electrons as dots or all as crosses if the two shared pairs are clear. Do not credit: ions drawn; double bonds; a missing lone pair on sulfur; hydrogen atoms joined to each other.
HSH
2
11(g)lead(II) oxide (accept PbO), because it loses oxygen. Both the substance and the reason are needed. Accept: lead(II) oxide, because lead(II) ions gain electrons. Reject: lead; carbon; carbon dioxide; "gains oxygen".1