370 minutes
362 marks
Page 1 of 40
M1.          moles NaOH used          = vol / 1000 × conc (1) = 21.7 (if uses 25 here only scores                                         first of first 4 marks)/ 1000 × 0.112                                         = 0.00243 (1) (consider 0.0024 as an arithmetic                                         error loses 1 mark)                                         (range 0.00242 to 0.00244) moles HCl in 25 cm3       = 0.00243 (1) (or 1 mol HCl reacts with 1 mol NaOH)
moles of HCl in 250 cm3 = 0.0243 (1)
moles ZCl 4                        = 0.0243 / 4 = 0.006075 (1) (or 0.006076 or 0.006 mark
                                        is for / 4) M
r                                     = mass / no. Moles (1) (method mark also 1.304 / 0.006075)
                                        = 214.7 (1) (or 0.006 gives 217) (allow 214 to 215) A
r                                      = 214.7 - 142 = 72.7 (1) (217 gives 75, 142 is 35.5 × 4)
Therefore element is       Germanium (1) (allow conseq correct from A r )
                                        (75 gives As) If not / 4 C.E. from there on but can score 2 independent marks for (mass / moles / method and identity of element) (for candidates who use m
1 v
1 = m
2 v
allow 1st 3 marks
   
2
1
          = 1.39 (1) Allow 1.390 to 1.395 ignore units even if incorrect answer = 1.4 loses last mark
3
          =  (1)
OR M, mol/dm3, mol.l–1
allow 2.928 to 2.93 Note unit mark tied to current answer but allow unit mark if answer = 2.9 or 3
2
1 = m
2 v
         moles NH 3 in 30.8 cm3 = 0.0310 × 2 = 0.0620  (1)
Mark is for ×2 CE if × 2 not used
Allow 2.010 to 2.015 No units OK, wrong units lose last mark
(ii)     moles (NH 4 )
4 = 0.310 (1)
Allow consequential wrong moles in part (i) if clear H 2 SO
4 =(NH
2 SO
if moles of (NH 4 )
2 SO
6
2 [16]
(e)     (i)      moles H 2 SO
4 =  × 1.24 = 0.0310 (1)
         moles of NH 3 in 1 dm3 = 0.620 ×  = 2.01 (1) (mol dm–3)
Page 3 of 40
= 2.15 (mol dm–3) (1)
min 2 d.p. 2.14 to 2.15 Ignore wrong units A.E. lose one mark
3
Conseq
Conseq
(iv)    Relative atomic mass of M: 138 - 60 = 78 (1)
Conseq If 78 = M
r then M = selenium
Concentration =  (1)
(ii)      = 2.85 × 10–3 (mol) (1)
(iii)      = 138 (1)
(ii)     Moles ethanol = n = 1.36/46 (=0.0296 mol) (1)
if V = p/nRT lose M3 and M4
         = 8.996 × 10–4 (m3) (1)
= 899 (900) cm3 (1)        range = 895 – 905
If final answer = 0.899 award (2 + M1); if = 0.899 dm3 or if = 912 award (3 + M1) Note: If 1.36 or 46 or 46/1.36 used as number of moles (n) then M2 and M4 not available Note: If pressure = 100 then, unless answer = 0.899 dm3, deduct M3 and mark consequentially
5
2 + 6H
Number of molecules of NH 3 = 0.0155 × 6.02 × 1023 (1)
[mark conseq] = 9.31 × 1021 (1)
[range 9.2 × 1021 to 9.4 × 1021]
Conseq (min 2 sig fig) 4
(c)     Moles NaCl = 800/58.5             (= 13.68) (1)
Moles of NaHCO 3 = 13.68 (1)
Moles of Na2CO3 = 13.68/2     = 6.84 (1)
Mass of Na 2 CO
3 = 6.84 × 106 = 725 g (1)   [range = 724 – 727]
[1450 g (range 1448 – 1454) is worth 3 marks] Accept valid calculation method, e.g. reacting masses or calculations via the mass of sodium present. Also, candidates may deduce a direct 2:1 ratio for NaCl:Na
2 CO
Page 5 of 40
M5.          (a)     (i)      75.0 × 10–3 × 0.500 = 0.0375 (mol) (1)
accept 0.037 or 0.038
(ii)     21.6 × 10–3 × 0.500 = 0.0108 (mol) (1)
accept 0.011 If both (i) and (ii) answers wrong, allow ONE process mark for both correct processes
(iii)     0.0375 - 0.0108 = 0.0267 (mol) (1) Not conseq – must use figures shown
(iv)    Moles of MgCO 3 = 0.0267/2          = 0.01335 (mol) (1)
allow 0.0134 - 0.0133
3
mark conseq (check for inversion)
= 90% (1) mark conseq
                               range = 89.5 - 90.5% If % expression inverted, lose M4 and M5
8
Na = 36.5/23          S = 25.5/32(.1)              O = 38.0/16 (1)
     = 1.587              = 0.794                         = 2.375
     = 2:1:3 (1) If no % of oxygen Max 1 (allow M2 only) If % for Na and S transposed, or atomic numbers used, M1 only available
(ii)     Na 2 SO
3 + 2HCl → 2NaCl + H
2
1
(iii)     2.43 × 10–2
(iv)    3.01/2.43 × 10–2
(mark conseq on (a)(iii)) 1
         124 (Do not allow 124 without evidence of appropriate calculation in (a) (iii))
1
2 O) = 250 –106 = 144        (mark conseq on M1)
x = 8                                            (mark conseq on M2) (Penalise sf errors once only)
3
(ii)     Moles A r = 325/39.9 = 8.15
(accept M r = 40)
Range = 4.02 × 106 Pa to 4.04 × 106 Pa
(If equation incorrectly rearranged, M3 & M4 = 0 If n =325, lose M2) (Allow M1 if gas law in (ii) if not given in (i))
2 [12]
Page 7 of 40
M7.          (a)     (i)      100 ×10–3 × 0.500    =     5.00 × 10–2 (mol)
accept 5 ×10–2 / 0.05
1
(ii)     27.3 ×10–3× 0.600    =     1.64 × 10–2 / 1.638 × 10–2 (mol) only
1
(iii)     1.64 ×10–2 (mol)
Mark conseq on (ii) 1
(iv)    5.00 × 10–2   -   1.64 × 10–2   =   3.36 × 10–2 (mol)
Mark conseq on (i) & (iii) 1
(v)     3.36 × 10–2 × ½ = 1.68 × 10–2 (mol)
If 2.78 × 10–2 used 1.39 × 10–2
Mark conseq on (iv) 1
1.68 × 10–2 × 132(.1) or 1.39 × 10–2 × 132(.1)
Mark for M r
(b)     pV = nRT 1
1
1
= 408.5 – 410.5 (K) Mark conseq on moles Note Sig. fig. penalty - apply once if single sf given, unless calc works exactly
1 [11]
Page 8 of 40
M8.          (a)     moles HNO 3 = 175 × 10–3 × 1.5 = (0.2625 mol);
1
1
2 = 331.2 x 0.131=43.5 g;
(accept 43.2 - 43.8) (M1 & M2 are process marks. If error in M1, or in M2, do not mark M4 consequentially, i.e. do not award M4) (if atomic numbers used in M3, do not award M4)
1
1
1
[mark is for use of 4/5] 1
          = 2.89 × 10–3 OR 1.78×10–3;
1
OR  0.0821 (g); (if atomic numbers used, M3 = M4 = 0)
1 [11]
Page 9 of 40
1
2
1
(mark answer first – check back if wrong) (transcription error lose M3, mark M4 conseq on error) (if ‘untraceable’ figures used M3=M4=0) (if wrong temp conversion – lose M3 – conseq M4) (if n = RT/pV CE, lose M3 and M4)
(b)     (i)      simplest/lowest ratio of atoms of each / element/s in a compound / substance / species / entity / molecule
1
1               1               2               KNO2      (1)
(M3 tied to M2), (M3 can be transferred from equation if ratio correct but EF not given) (if calc inverted, lose M2 and M3), (if
used At N 1 / wrong No for Ar then CE, lose M2 and M3) (if % of O missing, award M2 only)
3
[10]
                 (1)                                                (1)
M10.          (penalty for sig fig error =1 mark per question)
(a)     neutron:          relative mass = 1          relative charge = 0 (not ‘neutral’)
1
5.56 × 10–4 → 0 relative charge = –1 1
(b)     17O/O17                           mass number               (Do not accept 17.0)
1
          oxygen symbol ‘O’ (if ‘oxygen’ + — ‘mass number = 17’(1)) (if ‘oxygen’+ — ‘mass number = 17’(0))
(if at N 0 given but ≠ 8, treat as ‘con’ for M2)
(if lp on Be, diagram = 0) (ignore bond angles) (not dot and cross diagrams)
1
(c)    
2
            QoL Linear (1)                        bent / V-shaped / angular (1)
(mark name and shape independently) (accept (distorted) tetrahedral) (if balls instead of symbols, lose M1 – can award M2) (penalise missing ‘Cl’ once only) (not ‘non-linear’)
2
Page 11 of 40
(d)     Mr (Mg(NO 3 )
2 = 58(.3) (if At N 0 used, lose M1 and M2)
1
          moles Mg(OH) 2 = 0.0172 (conseq on wrong M2) (answer to 3+ s.f.)
1
          moles HCl = 2 × 0.0172 = 0.0344 or 0.0343 (mol) (process mark) 1
(if candidate used 0.017 or 0.0171 lose M2) (just answer with no working, if in range = (4). if, say, 34 then =(2)) (if not 2:1 ratio, lose M3 and M4) (if work on HCl, CE = 0/4)
1 [12]
[Not ‘atoms’] 1
[Not molecules]
[Accept answers via 4 separate mole calculations] 1
= 0.316 – 0.317 mol [answer to 3 + sf] [Mark conseq on errors in M1/M2]   (1)
1
Page 12 of 40
(iii)     Moles of nitroglycerine = 4 × 1.09 × 10–2     (= 0.0438 mol)
[Mark conseq on their moles of O 2 ]
1
1
Moles of nitroglycerine = 227 × 0.0438 = 9.90 – 9.93(g) [answer to 3+ sf] [If string OK but final answer wrong then allow M6 but AE for M7] [Mark conseq on error in M
r ]     [Penalise wrong units]
1
1
[ignore s.f.] 1
          units = Pa or kPa or MPa               (as appropriate)
[If error in conversion from Pa, treat as a contradiction of the units mark] [If transfer error, mark conseq but penalise M2] [If data from outside of above used, penalise M2 and M3] [If pV expression incorrectly rearranged, penalise M2 and M3] [if T = 1373 K used, penalise M2]
1 [11]
p =  =
2
1
[Not ionic bonding] 1
(iii)     (Here) the ionic bonding in NaCl is stronger/requires more energy to break than the metallic bonding in Na
QoL  Accept ‘bonding/forces of attraction in NaCl is stronger than in Na’
[If IMF/molecules/van der Waals’/dipole–dipole mentioned in parts(i) or (ii), then CE = 0 for parts (i) and/or(ii) and CE = 0 for part(iii)]
1
(c)     Comparison: Sodium conducts and sodium chloride does NOT conduct
Allow ‘only Na conducts’ Accept ‘Na conducts, NaCl only conducts when molten’ [Do not accept sodium conducts better than sodium chloride etc.]
1
1
Allow e– move/carry current/are charge carriers/transfer charge.
[Not ‘electrons carry electricity’] [Not ‘NaCl has no free charged particles’]
Ions can’t move in solid salt 1
Page 14 of 40
(d)     Layers can slide over each other – idea that ions/atoms/particles move [Not molecules] [Not layers separate]
1
1
0.9(39)               0.9(38)               2.8(2) Hence:     1                        1                      3 Accept backwards calculation, i.e. from formula to % composition, and also accept route via M
r to 23; 35.5; 48, and then to 1:1:3
[If % values incorrectly copied, allow M1 only]
[If any wrong A r
values/atomic numbers used = CE = 0]
1
3 + 3H
2 O
1 [12]
M13.          (a)     (i)      Prevents release of toxic CO More energy efficient (releases more energy on combustion)
1
         Balanced equation 1
(iii)     Detect CO gas or C (soot or particles) in exhaust gases 1
(b)     CH 3 CH
2 CH
                                   
(iii)     Can be made into polymers (or alcohols etc) 1
(d)     (i)      % atom economy = mass CH 2 Cl
2 /total mass
(ii)     Because expensive chlorine is not incorperated into desired product Raise money by selling HCl
1 [14]
         PV = nRT 1
= 7.43 × 10–2 m3
1
(b)     (i)      any two from:
exhaust gases hot so would boil the solution away solution would splash reaction might be too slow would need continuous supply of solution and/or replacement of products
2
environmental disadvantage         generation of electricity                                                        likely to lead                                                        to release of CO
2
5
1
2 + O
2
1
(e)     Proportion of O 2 increased leading to higher T (or more
complete combustion) 1
   
M15.          (a)     Single bonds only /no double or multiple bonds; 1
          Contains carbon and hydrogen only; C and H only not C and H molecules
1
(b)     (1) Fractions or hydrocarbons or compounds have different boiling points/ separation depends on bp;
Ignore mp and vdw 1
          (2) bp depends on size/ M r / chain length;
If refer to bond breaking/cracking/ blast furnace/oxygen/air 2 max 1
(3) Temp gradient in tower or column / cooler at top of column or vice versa;
QWC 1
(4) Higher bp / larger or heavier molecules at bottom (of column) or vice versa;
Not increasing size of fraction Not gases at top
1
Page 17 of 40
(c)     Large molecules or compounds or long chain hydrocarbons (broken) into smaller molecules or compounds or smaller chain hydrocarbons;
QWC 1
          C 14
Only 1
Smaller chain molecules are in more demand or have higher value or vice versa;
Insufficient to say more useful/have more uses 1
(d)     C 8 H
18 + 8½ O
Allow multiples 1
          Rh/ Pd/Pt/lr or in words; Penalise contradiction of name and symbol
1
2 / 2CO + O
          Greenhouse gas/ absorbs infrared radiation; 1
(e)     car less powerful/ car stops/ reduced performance/ won’t run smoothly/ can’t accelerate;
Not incomplete combustion or bad effect on engine Not doesn’t go as far.
1
Page 18 of 40
(f)      (compounds with) same molecular formula / same no and type of atoms; Not atoms/elements with same molecular formula. If same chemical formula, can allow M2
1
          And different structure/ structural formula; M2 consequential on M1 Allow displayed formula for M2
1
1 [20]
1
         = 35.37(%) allow 35.0 – 35.4%; Allow 35% Allow 2 marks if correct %
1
(ii)     Sell the HCl or sell the other product or sell the acid (formed in the reaction);
Need a financial gain 1
One mark for M r = 189.9
         allow 0.86 – 0.87;; Ignore units
2
1
1
         Accept 90.6 to 92% 1
[8]
          Same functional group;
          Trend in physical properties eg inc bp as M r increases;
          Contains an additional CH 2 group;
Any two points. 2 max
(b)     (i)
1
6 Cl;
Allow any order of elements. Do not allow EF consequential on their wrong displayed formula.
1
12 C
of atoms; 1
Different structural formula/ different structure/ different displayed formula;
Not atoms or elements with same MF CE=O. Allow different C skeleton. If same chemical formula can allow M2 only. M2 insufficient to say atoms arranged differently. M2 consequential on M1.
1
1
Allow 67.98 or 68.0 or 68%. 1
(d)     (i)      Bp increases with increasing (molecular) size/ increasing M r /
increasing no of electrons/increasing chain length; Atoms CE =0.
1
Increased VDW forces (between molecules) (when larger molecule)/ bigger IMFs;
QWC Not dipole-dipole or hydrogen bonds. If VDW between atoms in M2 CE = 0.
1
[11]
1
132
0.0238 Allow 0.024 Allow 0.0237 Penalise less than 2 sig fig once in (a)
1
0.0474-0.0476 Allow (a) (i) × 2
(iii)     1.21 Allow consequential from (a) (ii) ie allow (a) (ii) × 1000/39.30 Ignore units even if wrong
1
           = 67.98%;
Page 21 of 40
Allow mass or Mr of desired product times one hundred divided by total mass or Mr of reactants/products If 34/212.1 seen correctly award M1
1
1
1
If rearranged incorrectly lose M1 and M3 1
M2 for mark for converting P and T into correct units in any expression
1
= 0.59(4) Allow 0.593 M3 consequential on transcription error only not on incorrect P and T
1
1
Alternative method gives180 for water part = 2 marks 1
x = 10 X = 10 = 3 marks 10.02 = 2 marks
1 [13]
n =
M19.          (a)     (i)      shared pair of electrons
Can have one electron from each atom contributes to the bond Not both electrons from one atom
1
(b)
1
   OR  
2 , watch for Cl in centre- it must be C
Ignore wrong bond angles Representations of lone pairs allowed are the two examples shown with or without the electrons in the lobe. Also they can show the lone pair for either structure by two crosses/dots or a line with two crosses/dots on it e.g.
 or Or a structure with 3 bp and 2 lp
1
1
Page 23 of 40
(d)     (i)      Coordinate/dative (covalent) If wrong CE = 0/3 but if ‘covalent’ or left top line blank, mark on.
1
CE if lone pair is from B 1
Donated from F–/fluoride or donated to the BF 3
Must have the – sign on the F ie F–
Ignore Fl– M3 dependent on M2
1
(ii)     109° to 109.5° 1
(e)     For 1 mark allow 238 as numerator and 438 as denominator or correct strings
1
= 54.3% 2 marks if correct answer to 3 sig figs. 54% or greater than 3 sig figs = 1 mark
1 [11]
1
Only
(ii)     2.49 × 10–3
Allow answer to (a)(i) ÷ 2 Allow answers to 2 or more significant figures
1
(iii)     2.49 × 10–2
Allow (a)(ii) × 10 Allow answers to 2 or more significant figures
1
(iv)    138.2 3.44 divided by the candidate.s answer to (a)(iii) 138.2 or 138.1 (i.e. to 1 d.p.)
1
Page 24 of 40
(v)     (138 – 60) ÷ 2 = 39.1 Allow 39 – 39.1 Allow ((a)(iv) – 60) ÷ 2
1
K/potassium Allow consequential on candidate’s answer to (a)(iv) and (a)(v) if a group 1 metal Ignore + sign
1
(b)     PV = n RT or rearranged If incorrectly rearranged CE = 0
1
Correct M2 also scores M1 1
402(.3) K (or 129 °C) allow 402-403K or 129-130 °C do not penalise °K M3 must include units for mark
1
Penalise incorrect gas 1
If 84 used, max 1 1
6.27 = 0.074(4)
84.3 CE if not 84 or 84.3 Allow answers to 2 or more significant figures M2 = 0.074-0.075
1
T =
4 = 120(.4)
1
M2     Expected mass MgSO 4 = 0.074(4) × 120(.4) = 8.96 g
Allow 8.8 – 9.0 or candidate’s answer to (d)(i) × 120(.4) 1
Allow 8.3 – 8.6 M3 dependent on M2
Alternative method
M2     0.074(4) × 95/100 = 0.0707
M3     0.0707 × 120(.4) = 8.51 g Allow (d)(i) × 95/100 Allow 8.3 – 8.6 M3 dependent on M2
1 [15]
2n+2
(b)    
2
Page 26 of 40
(c)     C 9 H
Only 1
To break the (C-C and/or C-H) bonds M2=0 if break C=C
1
To make products which are in greater demand / higher value / make alkenes
Not more useful products Allow specific answers relating to question
1
Allow other balanced equations which give C and CO/CO 2
1
Apply list principle Ignore causes cancer / toxic
1
1
Page 27 of 40
M22.          (a)     (i)      0.0212
Need 3 sig figs
Allow correct answer to 3 sig figs eg 2.12 x 10-2
1
(ii)     0.0106 Mark is for (a)(i) divided by 2 leading to correct answer 2 sig figs
1
(iii)     M r = 100.1
1.06 g Allow 100.1 as ‘string’ Need 3 sig figs or more Consequential on (a)(ii) x 100(.1)
2
(iv)    Neutralisation or acid / base reaction Allow acid / alkali reaction Apply list principle
1
(b)     (i)      T = 304(K) and P = 100 000 (Pa) Only T and P correctly converted
1
1
1
(ii)     0.0276 – 0.0278(mol) Allow answer to (b)(i) divided by 5 leading to a correct answer Allow 0.028
1
2
1
Ca(NO 3 )
2 H
2 O
Mark is for dividing by the correct Mr values M2 and M3 dependent on correct M1
0.0256           0.102 M2 can be awarded here instead
1         :        3.98
x = 4
Credit alternative method which gives x = 4 1
[12]
M1 and M2 are process marks 1
Mol HNO 3 = 0.0393 × 8 / 3 = 0.105 mol
Allow mark for M1 × 8/3 or M1 × 2.67 1
Vol HNO 3 = 0.105 / 2 = 0.0524 (dm3)
Accept range 0.0520 to 0.0530 No consequential marking for M3 Answer to 3 sig figs required
1
1
n = pV/RT             (= 101000 × 638 × 10−6)
                             (   8.31 × 298                ) Can score M2 with incorrect conversion of p and V If T incorrect lose M1 and M3
1
Page 29 of 40
0.026(0) (mol) If answer correct then award 3 marks Allow answers to 2 sig figs or more 26.02 = 1 If transcription error lose M3 only
1
2 (g) + (1)O
(ii)     Decomposition not complete / side reactions / by-products / some (NO 2 )
escapes / not all reacts / impure Pb(NO 3 )
2
(iii)    Hard to separate O 2 from NO
2 / hard to separate the 2 gases
Allow mixture of gases Not ‘all products are gases’
1 [9]
1
3
1
(=2.02       = 1.35)
1
If Ca 3 N
1
(d)              
3 N
Accept multiples 1
(ii)     Not enough oxygen or air (available for complete combustion) / lack of oxygen or air / too much octane
Ignore poor ventilation, low temp, poor mixing, incomplete combustion
1
2
(ii)     Pt / Pd / Rh / Ir or names Apply list principle
1
Big(ger) surface area / increased reaction rate / removes more of the gases / ensures complete reaction
Allow (ceramic) withstands high temperatures 1
(c)     (i)     Acid rain Allow consequence of acid rain Ignore greenhouse gas / global warming / ozone
1
Allow chemical names 1
Neutralises the gas or words to that effect/it is basic/ SO 2 is acidic
Allow ‘reacts with it’ or ‘it is alkaline’ Ignore ‘absorb’
1 [8]
Do not allow ‘petrol’ 1
(a)      (i)     C 8 H
18 + 8 O
2  → 8CO + 9H
1
1
(ii)     Chain (isomerism) Allow branched chain / chain branched / side chain (isomerism) Ignore position (isomerism) Do not allow straight chain / geometric / branched / function
1
Only 1
(ii)     Thermal cracking If not thermal cracking, CE = 0/2 If blank mark on
1
High temperature Allow ‘high heat’ for ‘high temperature’
(400°C < T < 900°C) or (650 K < T < 1200 K) Not ‘heat’ alone If no T, units must be 650 – 900
and
High pressure (> 10 atm, > 1 MPa, >1000 kPa) 1
(iii)    To produce substances which are (more) in demand / produce products with a high value / products worth more
Ignore ‘to make more useful substances’ 1
(d)     (i)     Corrosive or diagram to show this hazard symbol Ignore irritant, acidic, toxic, harmful
1
=76.75(%) or 76.8(%) Allow answers > 3 sig figs
1
1
 
M27.          (a)     P = 100 000 (Pa) and V = 5.00 x 10–3 (m3)
M1 is for correctly converting P and V in any expression or list Allow 100 (kPa) and 5 (dm3) for M1.
1
M2 is correct rearrangement of PV = nRT 1
= 0.202 moles (of gas produced) This would score M1 and M2.
M3 is for their answer divided by 5 1
Mass of B 2 O
3 = 0.0404 x 69.6
M4 is for their answer to M3 x 69.6 1
= 2.81 (g) M5 is for their answer to 3 sig figures. 2.81 (g) gets 5 marks.
1
3
Accept multiples. 1
3 bonds 1
Pairs repel equally/ by the same amount Do not allow any lone pairs if a diagram is shown.
1
1
3 moles x 3
3
= 2.20 to 2.22 mol dm–3
Allow 2.2 Allow 2 significant figures or more
1
Allow alternative balanced equations to form acid salts. Allow H
3 BO
1
Mark is for both M r values correctly as numerator and
denominator. 1
1
1
2:4 ratio 1
2 Cl
4
Allow 4 marks for correct answer with working shown. Do not allow (BCl
2 )
2
M28. (a)    P = 100 000 Pa and T = 298 K
Wrong conversion of V or incorrect conversion of P / T lose M1 + M3
1
  
If not rearranged correctly then cannot score M2 and M3 1
n(total) = 174(.044) 1
n (NO) = 69.6 Allow student’s M3 × 4 / 10 but must be to 3 significant figures
1
1
176.5 Allow 176 − 177 But if answer = 0.176 − 0.18 (from 3 / 17) then allow 1 mark
1
(ii)     176.47 × 46 = 8117.62 M1 is for the answer to (b)(i) × 46. But lose this mark if 46 ÷ 2 at any stage However if 92 ÷ 2 allow M1
1
Page 35 of 40
  
M3 is for the answer to M2 ÷ 1000 to min 2 significant figures (kg)
OR
  
1
(d)     NO 2 contributes to acid rain / is an acid gas / forms HNO
3 / NO
photochemical smog Ignore references to water, breathing problems and ozone layer. Not greenhouse gas
1
(e)     Ensure the ammonia is used up / ensure complete reaction or combustion
OR
(f)     Neutralisation Allow acid vs alkali or acid base reaction
1 [14]
4 (aq) + 8NO(g) + 4H
− + 3e−   2H 2 O + NO
Ignore state symbols. Credit multiples of this equation only. Ignore absence of charge on the electron.
1
4 2− + 8e− + 8H+
Ignore state symbols. Credit multiples of this equation only. Ignore absence of charge on the electron.
1
(b)     M1 add scrap / recycled / waste iron (or steel) to the aqueous solution If M1 refers to iron / steel, but does not make it clear in the text that it is “scrap” / “waste” / “recycled”, penalise M1 but mark on.
M2 the iron is a more reactive metal OR Fe is a better reducing agent Credit zinc or magnesium as an alternative to iron for M2, M3 and M4 only, penalising M1
M3 Cu2+ / copper ions are reduced / gain electrons
OR Cu2+ + 2e−   Cu
OR copper / Cu is displaced by Fe Ignore absence of charge on the electron.
M4 Fe + Cu 2+   Fe 2+ + Cu ONLY
For M4, ignore state symbols 4
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M30. (a)  
If x = 7 with working then award 3 marks. Allow alternative methods. If M1 incorrect due to AE, M3 must be an integer.
1
Percentage of H 2 O = 44%
  If there is an AE in M1 then can score M2 and M3 If M
r incorrect can only score M1
1
mol ZnCl 2 = 0.06(0) OR 0.12 / 2
1
If M2 incorrect then CE and cannot score M2, M3 and M4.
mass ZnCl 2 = 0.06 × 136.4
Allow 65.4 + (2 × 35.5) for 136.4 1
= 8.18(4) (g) OR 8.2 (g) Must be to 2 significant figures or more. Ignore units.
1
OR moles Zn = 0.0784
Mass Zn reacting = 0.0784 × 65.4 = (5.13 g) M2 is for their M1 × 65.4
1
(= 0.0784)
M3 is M2 × 100 / 5.68 provided M2 is < 5.68 1
= 90.2% OR 90.3% Allow alternative methods. M1 = Moles ZnCl
2 = 10.7 (= 0.0784)
136.4 M2 = Theoretical moles Zn = 5.68 (= 0.0869)
65.4 M3 = M1 × 100 / M2 = (0.0784 × 100 / 0.0869) M4 = 90.2% OR 90.3%
1
1
between oppositely charged ions / + and − ions / F− and Zn2+ ions
If IMF, molecules, metallic bonding implied CE = 0/3 1
[14]