Question Answers Additional comments/Guidelines Mark
02.1
M1 (oxide ions react with water to) form/produce hydroxide ions M2 sodium hydroxide
... [Show More] more soluble than magnesium hydroxide
M1 O2– + H2O 2OH–
Ignore all non-ionic equations
M2 ideas that more sodium hydroxide dissolves
/ dissociates
Allow sodium oxide more soluble / dissociates more than magnesium oxide NOT ‘molecules’ or ‘atoms’
1
1
02.2
P4O10 + 6H2O 4H3PO4
Allow multiples and fractions Allow ionic products
NOT P2O5
1
02.3
M1 V2O5 + SO2 V2O4 + SO3
M2 V2O4 + 1 O2 V2O5
2
Allow 1 mark if both equations correct, but in wrong order
ALLOW multiples
1
1
Question Answers Additional comments/Guidelines Mark
03.1
M1 absorb (some) wavelengths/frequencies/colours/energies of (visible) light
M2 to promote/excite electrons in d-orbitals
M3 remaining/complementary wavelengths/frequencies/colours/energies of (visible) light reflected/transmitted (to give colour seen)
wavelengths/frequencies/colours/energies of (visible) light only needed once in the answer
Allow absorption of a photon of light NOT uv light
Allow d-subshell / d-energy level / d-electrons Reference to ‘d’ can appear anywhere in the answer
NOT emissions/emitting or ‘give out’
1
1
1
03.2
M1 ()E = ℎ𝑐
𝜆
M2 490 × 10−9
8
M3. = (6.63 × 10−34 × 3.00 × 10 ) = 4.06 × 10−19 J
490 × 10−9
Allow in two stages / expressed in words M2 for conversion
Correct answer scores 3 marks
4.06 × 10−𝑛 scores 2 marks (no M2)
9.75 × 10−32 = 1 mark (M2)
1
1
1
03.3
M1 measure absorbance for (a range of) known concentrations
M2 plot graph absorbance v concentration
M3 read value of concentration for the measured absorbance from this graph
Insist on description of taking measurements Allow concentration v absorbance
If no M1, must mention both variables Need to describe HOW they use the graph
1
1
1
03.4 M1 amount of iron in each tablet = 4.66 × 10−3 × 250
1000
(= 0.001165 mol)
M2 mass of iron in each tablet
= 4.66 × 10−3 × 250 × 55.8 = 0.0650 𝑔 = 65 𝑚𝑔
1000
Correct answer = 2 marks Allow M2 for (M1 x 55.8 x 1000)
1
1
Question Answers Additional comments/Guidelines Mark
04.1
DNA Replication
NOT mitosis
NOT DNA synthesis
Ignore terms relating to cell division processes Ignore ‘damages DNA’
Ignore DNA transcription Ignore ’cell replication’
1
04.2
[Pt(NH3)2Cl2] + H2O [Pt(NH3)2Cl(H2O)]+ + Cl
M1 Correct formula and charge of B
M2 Correct balancing and charges in equation
Allow M2 if the only error in complex B is the charge (M1 not awarded) with Cl only
ALLOW complexes without [ ] and/or ( ) around H2O
IGNORE ( ) around Cl
NOT any additional different species (loses M2) (allow uncancelled water on both sides)
1
1
04.3
M1 Pt in a cis-diammine complex bonded to the correct nitrogen atoms
Pt must have the two ammonia ligands shown NOT if drawn as trans
IGNORE any charge on Pt
Ignore any wedges and dashes (3D representations)
M2 both lone pairs shown OR two arrows indicating co-ordinate bonds
Allow M2 if bonds to platinum are from the incorrect nitrogen atoms
1
1
04.4
M1 plot concentration (y-axis) against time (x-axis)
and take tangents / (calculate the) gradients (to calculate rates) M2 Plot rate/gradients against conc
M3 straight line through origin / directly proportional confirms first order
Allow concentration-time graph
NOT time-concentration graph (unless clarified in words or sketch) but mark on
allow first order if rate halves/doubles when conc halves/doubles
Alternatives to M2 and M3:
M2 Plot a graph of log rate vs log conc M3 (Straight) line of gradient = 1
M2 measure (at least) two half-lives (in this case, tangents not required for M1)
M3 constant half-life means first order
M2 compare rates/gradients at different concentrations
M3 first order if rate halves when conc halves
1
1
1
04.5 temperature,
T / K 1 / K1
T rate constant,
k / s1
ln k
Allow 3.14 x 103
1
1
318 0.00314 6.63 x 107 14.2
04.6
Vertical axis with sensible scales (plotted points must take up more than half the grid)
NOT M1 if y-axis in wrong direction
all points plotted correctly (within ±0.5 small square)
Best fit straight line based on the student’s data (ignoring anomalous point if relevant)
Gradient calculated within range: 12876 - 13598
Mark is for their (gradient x 8.31) and
conversion into kJmol-1
Ea in the range: 107 – 114 kJ mol1 NOT a negative activation energy
1
1
1
Gradient = 13 125 1
(-13125 = Ea )
R
Ea = 13 125 x 8.31 = 109 069
= 109 (kJ mol1) 1
Question Answers Additional comments/Guidelines Mark
Amount of hexane = 2 = 0.0233 mol
ecf = M1 x 4154 ecf = M2/12.4
If no other marks awarded, allow one mark for 4154/12.4 = 335
1
86
05.1 q = 4154 x 0.0233 ( = 96.6 - 96.8 kJ) 1
Ccal = 96.6 = 7.79 - 7.81 (kJ K1) 1
12.4
q = CcalT = 7.79 x 12.2 = 95.0 kJ
(amount of octane = 2 = 0.0175 mol)
114
heat change per mole = 95.0 = 5417 kJ mol1
0.0175
Ecf for 05.1 x 12.2
If candidate converted 12.4 into kelvin in 05.1, ignore conversion to kelvin in 05.2
1
05.2 1
Allow 5420 kJ mol1 Using the value given:
6.52 x 12.2 = 79.54(4)
79.54/0.0175 = 4545
05.3
pressure not constant in bomb calorimeter
Allow enthalpy change requires constant pressure
1
100 x 0.2 = 1.64%
12.2
use bigger mass of fuel (so T greater)
Allow 1.6%
1
Allow 2% if working shown
05.4 NOT 2.0%
Allow octane or hexane as the fuel 1
Allow more / greater volume of fuel
Question Answers Additional comments/Guidelines Mark
06.1
H2(g) AND 100kPa
1 mol dm3 AND HCl/HNO3/H+
Pt electrode AND temperature of 298 K (25oC)
Allow 1 bar
NOT 1 atm or 101kPa
0.5 mol dm3 and H2SO4
1
1
1
06.2 This question is marked using levels of response. Refer to the Mark Scheme Instructions for examiners for guidance on how to mark this question Indicative Chemistry content Stage 1 Preparing solution
(1a) Weigh 7.995 / 8.00 g TiOSO4
(1b) Dissolve in / add (allow react with) (0.50 moldm-3) sulfuric acid
(1c) transfer to volumetric flask and make up to the mark
Stage 2 Set up cell
Content can be shown in a labelled diagram
(2a) piece of Ti immersed in (1 mol dm3 acidified)
TiO2+(aq) / the solution
(2b) (connect solutions with) salt bridge or description
(2c) (connect metals through high R) voltmeter
Stage 3 measurements and calculation
(3a) record voltage/potential difference/emf of the cell
(3b) Ecell = ERHS – ELHS
Ecell = Ecopper – Etitanium
(3c) ELHS = ERHS – Ecell OR Ecell should be +1.22 V if Cu on RHS (or −1.22 if Cu electrode on LHS)
Level 3
5-6 marks All stages are covered and the explanation of each stage is correct and virtually complete
Answer communicates the whole explanation, coherently and shows a logical progression through all three stages. ‘Coherence’ requires clear practical details (e.g. weighing into beaker/ by difference/ plus washings, not straight into volumetric flask, saturated solution chosen for salt bridge, salt bridge solution is suitable)
Level 2
3-4 marks All stages are covered but the explanation of each stage may be incomplete or may contain inaccuracies
OR two stages covered and the explanations are generally correct and virtually complete
Answer is coherent and shows some progression through all three stages. Some steps in each stage may be incomplete
Level 1
1-2 marks Two stages are covered but the explanation of each stage may be incomplete or may contain inaccuracies
OR only one stage is covered but the explanation is generally correct and virtually complete
Answer shows some progression between two stages
Level 0
0 marks Insufficient correct Chemistry to warrant a mark
06.3
TiO2+ + 2H+ + 4e() → Ti + H2O
Allow reverse reaction Ignore state symbols Allow multiples or fractions
Allow equilibrium arrow
1
06.4
( (+)0.34 compared with 0.00 shows that)
Ecell for Cu + 2H+ → H2 + Cu2+ / reaction of copper with most acids is negative / 0.34
/ (+)0.34 shows Cu less powerful reducing agent than H2
( (+)0.96 compared with (+)0.34 shows that)
Ecell for reaction of Cu with nitrate/nitric acid is positive
/ (+)0.62 V
2NO + 8H+ + 3Cu → 2NO + 4H O + 3Cu2+
3 2
OR M1 (Eo) H+/H2 (or the hydrogen electrode) less +ve/< than (Eo) Cu2+/Cu (or the copper electrode) so H+ cannot oxidise Cu to Cu2+ / H+ poorer oxidising agent (or reverse argument)
–
M2 (Eo) NO3 /NO (or the nitrate/nitric acid
electrode) more +ve/> than (Eo) Cu2+/Cu (or the
– 2+
copper electrode) so NO3 can oxidise Cu to Cu
(or reverse argument)
Allow multiples or fractions Ignore state symbols
1
1
1
07 B 1
08 A 1
09 B 1
10 D 1
11 D 1
12 B 1
13 B 1
14 A 1
15 B 1
16 C 1
17 B 1
18 A 1
19 D 1
20 A 1
21 C 1
22 A 1
23 C 1
24 D 1
25 B 1
26 A 1
27 B 1
28 D 1
29 A 1
30 D 1
31 C 1
32 D 1
33 D 1
34 C 1
35 D 1
36 D 1 [Show Less]