F325 Past paper Answers

27

1.(a)G:CO (1)HCOOH/H2SO4 CO + H2O (1)2

(b)H:C (1)C12H22O11 12C + 11H2O (1)2

(c)I:C4H8O2 (1)2C2H6O2 C4H8O2 + 2H2O (1)

Structure:

O

O

(1)accept any sensible structure of C4H8O23

[7]

2.(a)empirical formulaN : O = 63.64/14 : 36.36/16 (1)= 4.56 : 2.27 = 2 : 1. Empirical formula = N2O (1)molecular formulaMr of gas = 1.833 24 = 44 (1)(calc 43.992)with these two pieces of evidence, assume that molecular formula = N2O3

(b)any chemical that reacts to produce gas:e.g. carbonate and CO2 (1)accept: metal more reactive than Pb and H2

balanced equation to match chemical added (1)2

(c)Mr(Lidocaine) = 236 (1)moles Novocaine = 100 103/236 = 4.24 104 (1)concentration of Novocaine = 4.24 104 (1000/5)= 0.0847/0.0848 mol dm3 (1)3

(d)mass C = 12 3.74/44.0 = 12 0.085 = 1.02 g (1)

mass H = 2/18 0.918 = 0.102 g (1)

mass O = 1.394 (1.020 + 0.102) = 0.272 g

ratio C : H : O = 1.02/12 : 0.102/1 : 0.272/16 (1)

= 0.0850 : 0.102 : 0.0170 / 5 : 6 : 1 / C5H6O (1)

C5H10O has relative mass of 82Mr is 164 = 2 82

molecular formula = C10H12O2 (1)5

[13]

3.(i)4PCl5 + 10MgO P4O10 + 10MgCl2 (1)1

(ii)100g P4O10 =

284

100

/ 0.35(2) mol (1)

moles PCl5 needed = 4 0.352 = 1.408/1.4 mol (1)

mass PCl5 = 1.4(08) 208.5 = 293.568 / 294 g/ 291.9 g (1)

(1) for use of 284 for P4O10 and 208.5 for PCl5

73.4/72.975/72.3 g scores 3 marks (no use of 4 factor)18.35 g from dividing by 4 scores 3 marks4

[5]

4.Sulphuric acid molecules form hydrogen bonds (1)

Diagram showing hydrogen bonds between molecules:

O

S

O

O

O

H

H

O

S

O

O

O

H

H

(1)

or H bond from HO to OH (as in water).hydrogen bonds break (on boiling) (1)3

[3]

5.Correct equation for a metal (1)Correct equation for a carbonate (1)Correct equation for a base (1)3

[3]

6.(i)SO42 H2S: S from +6 to 2 (1)I I2: I from 1 to 0 (1)2

(ii)10H+ + SO42 + 8I 4I2 + H2S + 4H2O (1)1

[3]

7.(i) a solution that minimises/resists/opposes pH changes (1)1

(ii)The buffer must contain both CH3COOH and CH3COONa /CH3COO /weak acid and conjugate base(*) (1)

Solution A is a mixture of CH3COOH(*) and CH3COONa(*) // has an excess of acid /is acidic (1)

Solution B, contains only CH3COONa/ only CH3COO/only the salt/ is neutral (1)

CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l) /acid/alkali has been neutralised/CH3COOH(aq) and NaOH react together (1)4

[5]

8.Moles NH3 required =

10

10

12

10

6

.

7

10

6

.

7

17

10

3

.

1

=

=

mol calc: 7.6470588 1010 mol

Volume CH4 = n(NH3) 10.5

3

12

10

dm

10

1

.

1

24

16

)

2

8

(

10

6

.

7

=

+

OR 0.228 1012 dm3 O2 and 0.912 1012 dm3 N2calc: 1.147058824 1012 dm3OR 0.229411764 1012 dm3 O2 and 0.917647059 1012 dm3 N2

Volume H2O = n(NH3) 11.25

10

7.6

10

18

16

10

= 8.6 1011cm3 calc: 8.60294117 1011 cm3/ 8.60294117 108 dm3

[4]

9.(i)6HNO3 + S H2SO4 + 6NO2 + 2H2O/4HNO3 + S H2SO4 + 4NO2 + H2/2HNO3 + S H2SO4 + NO2 +

2

1

N21

(ii)from (+)5 to (+)4 1

[2]

10.M(AgCl) = 143.5 g mol1 Mass of Cl in AgCl =

143.5

0.610

35.5

/ 0.151 g Mass of Fe in compound = 0.270 0.151 = 0.119 g Ratio Fe : Cl =

5

.

35

151

.

0

:

8

.

55

119

.

0

/ 2.13 103 : 4.25 103

Ratio = 1 : 2; Formula = FeCl2

OR:

M(AgCl) = 143.5 g mol1 n(AgCl) =

143.5

0.610

= 4.25 103 mol n(Cl) = 4.25 103 molMass of Cl = 4.25 103 35.5 = 0.151 g Mass of Fe in compound = 0.270 0.151 = 0.119 g

Ratio Fe : Cl =

3

10

25

.

4

:

55.8

0.119

Ratio = 1 : 2; Formula = FeCl2

[6]

11.(i)I2(aq) + H2S(g) 2HI(aq) + S(s)species and balance (1)state symbols: accept (s) for I2; (aq) for H2S (1)2

(ii)amount I2 reacted = 1.89 mol / HI formed = 3.44 mol (1)theoretical amount HI produced = 3.78 mol/484 g (1)% yield =

484

100

440

or

78

.

3

100

44

.

3

= 91.0 % (1)3

(iii)[HI] =

750

1000

44

.

3

= 4.58/4.59 mol dm3 (1)pH = log 4.59 = 0.66 (1)2

[7]

12.(a)(i)(+)1 (1)1

(ii)

or

N

O

O

N

N

N

Look for atoms bonded together.AND other lone pairs.1

(b)(i)C13H18O2 (1)

(ii)any chemical that reacts to produce gas:e.g. carbonate and CO2 (1)

accept: metal more reactive than Pb and H2

balanced equation to match chemical added (1)3

[5]

13.Amount Cl2 produced = 1.6 1012/71 or 2.25 1010 mol (1)

Amount NaCl required = 2 2.25 1010 or 4.5 1010 mol (1)

ecf moles 2 Cl2

Volume brine = 4.5 1010/4 = 1.125 1010 dm3 (1)

ecf moles Cl2/4

i.e. 1.12 1.13 1010 dm33

[3]

14.(i)mass sucrose = 0.47 43 g or 20.21 g (1)Mr of sucrose = 342 (1)moles sucrose = 0.47 43/342 or 0.059 mol (1)(calc: 0.0590935672)no of sucrose molecules = .059 6.02 1023 = 3.6 1022 (1)4

(ii)C12H22O11(s) + 12 O2(g) 12 CO2(g) + 11 H2O(l) (1)Ignore state symbols

Energy = .059 5640 = 332.76 kJ (1)= 332.76/4.18 = 79.6 Calories (1)(i.e. mol sucrose from (a) 5640/4.18)If 0.059 is missed, 5640/4.18 = 1349 Calories would score 1 mark3

7]

15.Empirical formulaN : O = 63.64/14 : 36.36/16 (1)= 4.56 : 2.27 = 2 : 1. Empirical formula = N2O (1)

Molecular formulaMr of gas = 1.833 24 = 44 (1) (calc 43.992)With these two pieces of evidence, assume that molecular formula = N2O3

[3]

16.moles HCl in 23.2 cm3 = 0.200 23.2/1000 = 4.64 103 (1)moles B in 25 cm3 = moles HCl = 4.64 103 (1)moles B in 250 cm3 = 4.64 103 10 = 4.64 102 (1)4.64 102 mol B has a mass of 4.32 gmolar mass of B = 4.32/4.64 102 = 93 g mol1 (1)93 16 = 77 (1)Therefore B is phenylamine / C6H5NH2 (1)6

There may be other valid structures that are amines. These can be credited provided that everything adds up to 93.Answer could be a primary, secondary or tertiary amines.

[6]

17.(a)(i)Oxidation state of nitrogen goes from +5 to +4 (1);Oxidation state of oxygen goes from 2 to 0 (1);

Correct linking of changes of oxidation state withreduction and with oxidation (1)3

If oxidation state of barium given is incorrect max 1 for the oxidation numbers.

Allow ecf from wrong oxidation states for the correct linking markBoth oxidation and reduction needed

(ii)Correct use of molar ratios (1);Correct cycle (1);(+)1000 (kJ mol1) (1)3

Award full marks for (+) 1000 (kJ mol1)Only allow ecf for final lattice energy answer from a correct cycleAllow 1000 (1), +467 (2), +901 (2), +1558 (2)

(b)(i)Moles of Ba(NO3)2 = 0.005 or 0.00502 (1);Moles of gas made = 0.0125 / 0.0126 (1);Volume of gas = 300 cm3 to 302 cm3 (1)3

Allow ecf within questionIgnore significant figures

(ii)Decomposition temperature may be too high / too muchgas will be produced / to fill a gas syringe need a smalleramount of solid / gas syringe too small (1)1

Allow NO2 is toxic / barium compounds are toxicAnswer is consequential on answer to (i)

[10]

18.[H+] increases (1)H2O ionises more /for H2O H+ + OH, equilibrium moves to the right (1)2

exo/endo is noise

[2]

19.(i)O3: Exp 2 has 4 times [H2] as Exp 1and rate increases by 4 (1),

so order = 1 with respect to O3 (1)

C2H4: Exp 3 has 2 [C2H4] and 2 [O3] as Exp 2;and rate has increased by 4 (1),

so order = 1 with respect to C2H4 (1)

rate = k [O3] [C2H4] (1)5

(ii)use of k = rate / [O3] [C2H4] = 1.0 1012 / (0.5 107 1.0 108)to obtain a calculated value (1)k = 2 103 (1)units: dm3 mol1 s1 (1)3

(iii)rate = 1.0 1012 /4 = 2.5 1013 (mol dm3 s1) (1)1

(iv)rate increases and k increases (1)1

[10]

20.1O2(g) O3(g)/O2(g) + O2(g) O3(g) (1)

NO is a catalyst (1) as it is (used up in step 1 and) regenerated in step 2/not used up in the overall reaction(1)allow 1 mark for O/NO2 with explanation of regeneration.

3]

21.(i)H+/proton donor (1)1

(ii)partially dissociates/ionises (1)1

[2]

22.C6H5OH(aq) + OH(aq) C6H5O(aq) + H2O(l)acid 1base 2base 1acid 2 (1)

[1]

23.(i)Ka = [C6H5O(aq)] [H+(aq)] / [C6H5OH(aq)] (1)1

(ii)Mr C6H5OH = 94 (1)[C6H5OH(aq)] 4.7/94 = 0.050 mol dm3 (1)1.3 1010 [H+(aq)]2 / 0.050 mol dm3 (1) (= sign is acceptable)[H+] = {(1.3 1010) (0.050) } = 2.55 106 mol dm3 (1)pH = log[H+] = log 2.55 106 = 5.59 (1)

3 marks: [H+]; pH expression ; calc of pH from [H+]5

[6]

24.[H+(aq)] = 1.99 109 mol dm3 (1)[C6H5O(aq)] = Ka [C6H5OH(aq)] / [H+(aq)](1)[C6H5O(aq)] = 0.13 mol dm3 (1)

Calculation should use half the original concentration of phenol to find the concentration of sodium phenoxide in the buffer. This should then be doubled back up again.

Do not penalise an approach that uses the original concentration of phenol in the expression above.

[3]

25.(a)rate of forward reaction = rate of reverse reaction (1)concentrations of reactants and products are constant but they areconstantly interchanging (1)2

(b)(i)Kc = [CH3OH] / [CO] [H2]2 (1)1

(ii)use of Kc = [CH3OH] / [CO] [H2]2 and moles toobtain a calculated value (1)

convert moles to concentration by +2: [CO] = 3.10 103 mol dm3;[H2] = 2.60 105 mol dm3; [CH3OH] = 2.40 102 mol dm3 (1)

Kc = [2.60 105] / [3.10 103] [2.40 102]2 = 14.6 / 14.56 (1)

If moles not converted to concentration, calculated Kc value = 3.64(scores 1st and 3rd marks)units: dm6 mol2 (1)4

(c)(i)fewer moles of gas on right hand side (1)1

(ii)None (1)1

(d)(i)moved to left hand side/reactants increase/less products (1)1

(ii)H negative because high temperature favours theendothermic direction (1)1

(e)(i)CH3OH + 1 O2 CO2 + 2H2O(1)1

(ii)adds oxygen/oxygenated (1)1

[13]

26.(a)(i)Ca+ is smaller than Ca/ proton : electron ratio in Ca+ > Ca (1)greater attraction from nucleus (1)2

(ii)oxide ion, O and electron are both negative (1)hence energy is required to overcome repulsion (1)2

(b)completes Born-Haber cycle showing 1st IE 2nd IE 1st EA 2ndEAand LE(1)(1)(1) (lose 1 mark for each error/omission)LE = (1)3451 kJ mol1 (1)5

(c)differences in size of lattice enthalpies linked to ionic sizes/attractionusing more/less exothermic rather than bigger or smaller. (1)

Mg2+ is smaller/Mg2+ has greater charge density(1)hence has stronger attraction for O2 (1)3

[12]

27.(i)525 kJ mol1 (1)1

(ii)193.6 J K1 mol1 (1)1

(iii)uses G = H TS (1)To be feasible, G = 0 or G < 0 (1)minimum T = H / S (1)Converts S from J to kJ/1000 or converts H from kJ to J (1)2712 K/ 2438 C / 2439 C (1) (units essential)5

[7]

28.(i)oxidation: Fe Fe2+ + 2e (1)reduction: V3+ + e V2+ (1)2

(ii)Ecell = 0.18 V (1)1

[3]

29.(i)system III 2 and reversed + system IV (1)2H2 + O2 2H2O/H2 + O2 H2O (1)2

(ii)advantages:only H2O formed/ non-pollutinggreater efficiency (1)

disadvantages:H2 difficult to store (1)H2 difficult to manufactured initially /limited life cycle of H2 adsorber/absorber (1)4

[6]

30.To be handed in for marking

[20]

31.Oxidation because oxidation state of Hg changes from 0 to +2 so oxidation (1)Reduction because oxidation number of O changes from 1 to 2 (1)

Or

Correct identification of all the oxidation numbers (1)Correct identification of oxidation and reduction (1)2

Allow ecf for the identification of oxidation and reduction from wrong oxidation numbers

2]

32.Does not have an incomplete set of d electrons / doesnot have a partially filled d orbital / does not have apartially filled d sub-shell / ora (1)1

Allow use of 3d

[1]

33.(i)(1s22s22p6)3s23p63d6 (1)1

(ii)Octahedral shape with some indication of three dimensions (1);

Bond angle 90 (1)2

Allow use of wedges and dotted lines to indicate three dimensions

Allow three dimensions if at least two bond angles of 90o are shown that clearly demonstrate 3D

If two different bond angles do not award bond angle mark

(iii)Green / olive green / dark-green / green-blue ppt (1)

Allow solid instead of precipitateAllow solid or precipitate to be awarded from the state symbol in Fe(OH)2(s)

Fe2+(aq) + 2OH(aq) Fe(OH)2(s) (1)2

[5]

34.(i)MnO2 + 4H+ + 2Fe2+ Mn2+ + 2H2O + 2Fe3+ (1)1

Ignore state symbols

(ii)Moles of Fe2+ that reacted with MnO2 = 0.02 0.0123 = 0.0077 (1)

Allow ecf within question

Mass of MnO2 = 0.00385 86.9 = 0.335 (1)% purity = 66.4% (1)

Allow 66.4 66.5

Alternatively

Moles of MnO2 in 0.504 = 0.00580So moles of Fe2+ that should react with this is 0.0116 (1)Moles of Fe2+ that reacted with MnO2 = 0.02 0.0123 = 0.0077 (1)% purity = 66.4% (1)3

[4]

35.To be handed in for marking

[12]

36.(a)(i)+31

(ii)Cis and trans forms drawn in 3-D (only award thesemarks if C has been chosen)2

(iii)Type of isomerism is cis-trans/geometric1

(b)(i)(concentrated) hydrochloric acid/sodium chloride/Other suitable named ionic chloride but not just chloride or Cl1

(ii)Ligand substitution / ligand exchange1

[6]

37.(a)Emf/voltage/potential difference (of electrochemical cell)1comprising a (Cu/Cu2+) half cell combined with a standardhydrogen electrode11 atm, 1 mol.dm3, 298K (all 3 needed but can transfer mark if stated in (b))1

(b)Salt bridge and voltmeter1Platinum electrode dipping into 1 mol dm3 H+1Hydrogen gas feed1(Accept a suitable alternative standard electrode)

[6]

38.(i)Decolorised / add starch which is decolorisedAllow blue/black white or brown white1Do not allow colourless

(ii)moles S2O32 = 23.20 0.100/1000 = 0.00232 moles1Cu2+ S2O32 / moles Cu2+ = 0.00232 moles1But 25 cm3 of original = 10 0.00232 = 0.0232 moles1Concentration of original = 1000 0.0232 / 251

(iii)Because concentration of Cu2+ is less than 1 mol dm3 / less than standard1equilibrium moves to left (reducing +ve value of E)1

[7]

39.(a)(i)Stainless steel + corrosion resistance or alloys for tools+ hardness or other named alloy/use/property1Allow chrome plating with attractive or barrier to corrosion

(ii)Chromium 1s22s22p63s23p63d54s1 (allow.4s13d5)1

(b)(i)Cr2O72 + 14H+ + 6Fe2+ 2Cr3+ + 6Fe3+ + 7H2O1Cr2O72 / Cr3+ has more positive electrode potential1Therefore Cr2O72 is the stronger oxidising agent whichoxidises Fe2+ to Fe3+ (ora)1

(ii)Emf = (+) 0.56 V1

[6]

40.To be handed in for marking

[9]

41.(i)1O2(g) O3(g)/O2(g) + O2(g) O3(g) (1)

NO is a catalyst (1) as it is (used up in step 1 and)regenerated in step 2/not used up in the overall reaction(1)allow 1 mark for O/NO2 with explanation of regeneration.3

(ii)Rate = k[NO] [O3] (1)Species in rate equation match those reactants in the slowstep / rate determining step (1)2

[5]

42.(a)Kc =

]

PCl

[

]

][Cl

PCl

[

5

2

3

(1)1

(b)(i)PCl5 > 0.3 mol dm3 ; PCl3 and Cl2 < 0.3 mol dm3 (1)1

(ii)At start, system is out of equilibrium with too much PCl3and Cl2 and not enough PCL5 /

3

.

0

3

.

0

3

.

0

= 0.3 is greater than Kc = 0.245 mol dm3 (1)1

(c)(i)Kc does not change as temperature is the same (1)1

(ii)Fewer moles on left hand side (1)system moves to the left to compensate for increase inpressure by producing less molecules (1)2

(d)(i)Kc decreases (as more reactants than products)(1)1

(ii)Forward reaction is exothermic/reverse reaction is endothermic (1)equilibrium left to oppose increase in energy/because Kc decreases (1)2

[9]

43.(a)(i)Ionic product (1)1

(ii)Kw = [H+(aq)] [OH(aq)] (1) state symbols not needed1

(b)moles of HCl =

1000

35

.

21

10

5

3

-

= 1.067 104 mol (1)

moles of Ca(OH)2 =

2

10

067

.

1

4

= 5.34 105 mol (1)

concentration of Ca(OH)2 = 40 5.34 105 = 2.136 103 mol dm3 (1)2 marks for 4.27 103/ 8.54 103 mol dm3(no factor of 4)3

(c)[OH] = 2 2.7 103 = 5.4 103 mol dm3 (1)

[H+(aq)] =

3

14

w

10

4

.

5

10

0

.

1

aq)]

(

[OH

-

-

-

=

K

= 1.85 1012 mol dm3 (1)

pH = log (1.85 1012) = 11.73/11.7 (1)3

ecf is possible for pH mark providing that the [H+]value has been derived from Kw/[OH]If pOH method is used, pOH = 2.27. would get 1st mark,pH = 14 2.27 = 11.73 gets 2nd mark.Commonest mistake will be to not double OH-- and to use 2.7 103This gives ecf answer of 11.43/11.4, worth 2 marks.pH = 11.13 from dividing by 2: worth 2 marks

(d)8 (1)1

[9]

44.(a)Ca3(PO4)2 + 2H2SO4 Ca(H2PO4)2 + 2CaSO4 (1)1

(b)H2PO4(aq) H+(aq) + HPO42(aq) /

H2PO4(aq) 2H+(aq) + PO43(aq) (1)

(or equivalent with H2O forming H3O+)1

(c)(i)HPO42 (1)1

(ii)H3PO4 (1)1

(iii)H2PO4 produced Ca(H2PO4)2 or on LHS of an attemptedequilibrium equation (1)2 equations/equilibria to shown action of buffer (1)(1)from:H2PO4 + H+ H3PO4 /H2PO4 H+ + HPO42/H2PO4 + OH H2O + HPO42 /H+ + OH H2O3

[7]

45.(i)1s22s22p63s23p63d5 (1)1

(ii)Has an incomplete set of d electrons / partiallyfilled d sub-shell / partially filled d orbital (1)1

Allow partially filled d shell

[2]

46.(i)Has a (lone) pair of electrons that can be donated / lone pair that can forma dative bond / pair of electrons that can form a coordinate bond (1)1

(ii)3D diagram of octahedral structure (1);Bond angle 90 (1)2

Nameoctahedral must be present to score two marksAllow use of wedges and dotted lines to indicate three dimensionsAllow three dimensions if at least two bond angles of 90 are shown that clearly demonstrate 3DIf two different bond angles do not award bond angle mark

[3]

47.(i)Brown / red-brown / foxy-red / rusty / orange ppt (1)1

Allow solid instead of precipitateAllow state symbol (s) for precipitate

(ii)Fe3+(aq) + 3OH(aq) Fe(OH)3(s)Correct equation (1)State symbols for the correct formulae even ifspectator ions are present (1)2

Allow equations using the hydrated iron(III) ion

3]

48.(a)(i)Ionisation energy refers to removing electrons thatare attracted to the nucleus / energy needed toovercome the force of attraction between outerelectrons and nucleus (1)1

(ii)Electron affinity involves an electron (beinggained) experiencing attraction to the nucleus (1)1

(b)(i)Correct state symbols (1);

Allow 1 error or omission in state symbols.Providing formula has correct state symbols once in cycle this is sufficient

Correct formula (1);

Correct cycle with labelling or energy values (1)3

(ii)= +178 + 249 + 798 + (141) + 1150 + 590 + (3459) (1)= 635 kJ mol1 (1)2

Final answer must have correct units+635 kJ mol1 scores 0

(iii)Ionic radius of iron(II) less (than that of calcium ion) /charge density of Fe2+ greater (than that of Ca2+) /. ora (1)1

]

49.Fe2O3 + 3Cl2 + 10OH 2FeO42 + 5H2O + 6Cl (2)2

Allow one mark if electrons shownAllow one mark if correct reactants and products but not balanced

[2]

50.Correct Mr for Fe2O3, 159.6, and of Na2FeO4,165.8 (1)Moles of Fe2O3 = 0.00627 (1);Mass of Na2FeO4 = 2.08 (1);Percentage = 21.6 or 21.7 (%) (1)4

Allow full marks for correct answer with some workingAnswer must have 3 sig figsAllow ecf from wrong moles or wrong mass

[4]

51.Oxidation state of iron changes from +6 to +3 so is reduction (1)

Oxidation state of oxygen changes from 2 to 0 so is oxidation (1)

To get the two marks for oxidation states marks any other oxidation state quoted must be correct.Maximum one mark if any other oxidation number given is wrong

OR

Oxidation state of iron changes from +6 to +3 andoxidation state of oxygen changes from 2 to 0 (1)

Iron is reduced and oxygen is oxidised (1)2

[2]

52.(i)(Oxidised to) iodine so a brown (solution) formed /Fe3+ formed which is yellow or orange / Fe2+ formed which is green (1)1

Allow red/brown or orange

(ii)Nitrogen / N2 (1)1

Allow any correctly named oxide of nitrogen / correct formulae / HNO3 etc.

[2]

53.(a)VO2+1

(b)(i)B and D1

(ii)

V

Salt bridge

VO

2+

/H

+

/V

3+

V

2+

/V

3+

Platinum / carbon

Allow ecf from (b) (i)Solutions can be reversed.4

(iii)298 K / 25 C temperatureall solutions 1 mol dm3Both needed for 1 mark. Ignore any reference to pressure1

[7]

54.20 cm3 of 0.100 mol dm3 VO2+ = 0.002 moles10.002 moles VO2+ = 0.0004 moles MnO410.0004 moles MnO4 are in 16.0 cm31

[3]

55.(a)Ligand able to donate two lone pairs1to form dative covalent / co-ordinate bonds1

(b)

O

C

Cr

3

C

O

O

O

O

O

O

O

O

O

O

O

C

C

C

C

3-D diagram with three ethanedioate ligands used1correct bonding between ligands and Cr3+1correct charge on ion (3)1(Accept O O as minimum for ethanedioate ion)

[5]

56.To be handed in for marking

[8]

57.(a)(i)Curve downwards starting at t = 0 2with slope gradually levelling off with no increase (dont worry about hitting the x axis)

(ii)Tangent shown at start 1

(iii)Half-life is constant1OR: draw tangents and then plot a 2nd graph of tangentor rate against concentration, which is a straight linethrough the origin.

(iv)Straight line through origin 1

[CHBr] mol dm

242

3

0

0

initial rate /

mol dm s

31

(b)4 times [KI], rate increases by 4 ,so order = 1 with respect to KI independent marks2

(c)(i)rate/r = k[C2H4Br2] [KI] or ecf from (b)1

(ii)

0.18

0.50

027

.

0

/

[KI]

]

Br

H

[C

rate

2

4

2

=

k

= 0.3(0) units: dm3 mol1 s1

units dependent on rate equation in (i).

Mark independently.3

[11]

58.To be handed in for marking

[12]

59.proton donor partially dissociates

[2]

60.Ka =

/

[HA])

a

(

]

[H

/

[HCOOH]

]

[H

/

[HCOOH]

]

[H

]

[HCOO

2

=

+

+

+

K

1.58 104 =

025

.

0

]

[H

2

+

/ [H+] = {(1.58 104) (0.025) } = 1.99 103 mol dm3 pH = log[H+] = log 1.99 103 = 2.70 5.4034 (no square root) with working would score 1 mark.

[3]

61.(i)A solution that minimises pH changes/resists pHchanges/opposes pH changes (not pH is kept constant/pH maintained/pH cancelled out.1

(ii)HCOONa/HCOO/ NaOH HCOO is the conjugate base/HCOONa is the salt of the weak acid or HCOOH/HCOONa supplies HCOO 2

(iii)Two points from:Ka /pKa /acid strength/amount of dissociation temperature (but not temperature & pressure)ratio/amounts/concentrations of weak acid andconjugate base/salt (or reverse ratio)(not . concentration of base as it could implyNaOH)2 max

[5]

62.Mass of HNO3 =

g

910

/

100

65

1400

Moles of HNO3 =

4

.

14

63

910

=

pH = log[H+] = log 14.4 = 1.16/1.2 calc 1.15836pH from ignoring 65% pH = 1.35: with working, 2 marks.

[3]

63. CO2 + H2O

Complete correct balanced equation for 2nd mark:2HNO3 + CaCO3 Ca(NO3)2 + CO2 + H2O /2H+ + CaCO3 Ca2+ + CO2 + H2O /2H+ + CO23 CO2 + H2O

[2]

64.Two species differing by H+ . AW

one pair: HNO3 and NO3 other pair: HCOOH and HCOOH2+

[3]

65.moles of NaOH =

1000

23.75

0.1263

/ 3.00 103 mol moles of acid = 3.00 103 mol moles of acid in flask = 10 3.00 103 = 3.00 102 mol molar mass of compound =

86

10

00

.

3

58

.

2

n

mass

2

=

=

4

Molecular formula = C4H6O2

A 4 carbon carboxylic acid(e.g. butanoic acid) shown (bod)

Any 2 possible isomers from:CH2 = C(CH3)COOHCH2 = CHCH2COOHcis CH3CH = CHCOOHtrans CH3CH = CHCOOH4

Accept structural formulae that are unambiguous.

8]

66.Add (aqueous) sodium hydroxide which will give a brown/rusty ppt (1)1

Allow solid for precipitate or (s) in equationAllow Use aqueous thiocyanate ions which gives a (blood) red colouration

[1]

67.(i)Cr2O72 + 14H+ + 6Fe2+ 2Cr3+ + 7H2O + 6Fe3+Correct reactants and products (1);Correct balancing (electrons cancelled out) (1)2

(ii)Moles of dichromate(VI) = 3.53 104 (1);Moles of iron(II) = 2.12 103 (1):Moles of impure iron(II) sulphate = 2.36 103 (1);Percentage purity = 89.8 / 89.8 90.0 (1)4

Allow alternative working out via mass instead of moles e.g. mass of iron in hydrated FeSO4 from percentage composition compared to mass of iron from moles of iron(II).Allow ecf throughout unless percentage is above 100%

[6]

68.(a)(i)(Blue to) yellow (solution) / (blue to) green (solution) (1)1

(ii)Lone pair on chloride ion (1);Donated to copper(II) ion (1)2

Allow dative bond / coordinate bond (1)Allow marks via a diagram that must show lone pairs and the dative bond

(b)(Light) blue precipitate / blue solid (1);With excess (dark) blue solution (1)2

Not just goes blue

[5]

69.Any three fromAmmonia molecule 1 lone pair (and 3 bond pairs) (1);Ammonia ligand 4 bond pairs / lone pair is now a bond pair /ligand does not have a lone pair (1);Lone pairs repel more than bond pairs (1):In complex equal repulsion between electron pairs (1)3

Not bonds repel / atoms repel

[3]

70.(i)61

(ii)Species with (lone) pair of electrons1Capable of being donated / forms a dative covalentbond / co-ordinate bond to a metal ion.1(allow suitable diagram)

[3]

71.(i)[Co(H2O)6]2+ is octahedral[CoCl4]2- is tetrahedral (both needed for 1 mark)1

(ii)pink to blue1

(iii)Ligand substitution / exchange/displacement1

[3]

72.(a)(i)1 mark for correct 3-D diagram of cis isomer11 mark for correct 3-D diagram of trans isomer1

(Allow planar diagrams if two appropriate 90 angles are shown)

Allow any suitable 3-D diagrams. Possibilities to include:

Co

Co

Co

(ii)Geometric / cis trans1

(b)1 mark for using cis isomer11 mark for correct 3-D diagrams which are mirror images of each other.1

(If all diagrams are drawn as non-3d do not penalise in (b))

Allow any suitable 3-D diagrams such as:

Co

Cl

e

or

e

Cl

Co

Cl

Cl

CH

2

H N

NH

2

2

NH

2

H N

2

CH

2

CH

2

CH

2

n

n

[5]

73.(i)Brown solution/brown precipitate/black solidAdd starch to get blue / black colour1

(ii)Titration / volumetric analysis1using sodium thiosulphate(with starch indicator)1(allow from equation)

I2 + 2S2O32 2I + S4O621

1 mol Cr2O72 = 6 mols S2O321

[5]

74.(a)A = Platinum(electrode)B = H+(aq) / HCl(aq) / other suitable acidC = Voltmeter / galvanometerD = Cl2(g)State symbols needed for B and DAll correct = 2, 3 correct = 12

(b)(i)Arrow marked on or close to wire via voltmeter pointing fromhydrogen half cell to chlorine half cell1Electrons flow to half cell with more +ve standard electrodepotential1

(ii)Pressure = 1 Atm / 100 kPaTemp = 298 K / 25CConcentration = 1 mol dm3All 3 correct = 2 marks 2 correct = 1 mark2

(c)The standard electrode potential for ClO3 / Cl2 is more positivethan that of Cl2 / Cl1ClO3 has a greater tendency to gain electrons than Cl2 / ClO3 is abetter oxidising agent than Cl21Alternative:Because E is positive, the reaction will go from left to righttherefore ClO3 is reduced so it must be a better oxidisingagent than chlorine.

[8]

75.(a)partial dissociation: HCOOH H+ + HCOO (1)1

(b)(i)pH = log (1.55 103) = 2.81/2.8 (1)[H+] deals with negative indices over a very wide range/pH makes numbers manageable/removes very small numbers (1)2

(ii)Ka =

HCOOH(aq)]

[

aq)]

(

HCOO

][

aq)

(

H

[

-

+

(1) (state symbols not needed)1

(iii)Ka =

015

.

0

)

10

55

.

1

(

]

[HCOOH(aq)

aq)]

(

H

[

2

3

2

-

+

=

(1)= 1.60 104 (mol dm3)(1)pKa = log Ka = log (1.60 104) = 3.80 (1)3

(iv)Percentage dissociating =

015

.

0

100

)

10

55

.

1

(

3

-

= 10.3 % /10% (1)1

(working not required)

[8]

76.(i)HCOOH + NaOH HCOONa + H2O (1)1

state symbols not needed

(ii)n(HCOOH) = 0.0150 25.00/1000 = 3.75 104 (1)volume of NaOH(aq) that reacts is 30 cm3 (1)so [NaOH] = 3.75 104 1000/30 = 0.0125 mol dm3 (1)2

(iii)Kw = [H+(aq)][OH(aq)] (1)pH = log(1 1014/0.0125) = 12.10/12.1 (1)(calc 12.09691001)3

(iv)metacresol purple (1)pH range coincides with pH change during sharp rise ORpH 6-10 /coincides with equivalence point/end point (1)2

[8]

77.(a)Kc =

]

I

][

H

[

[HI]

2

2

2

(1)1

(b)(i)H2I2HI0.300.2000.140.040.32(1)(1)2

(ii)Kc =

04

.

0

14

.

0

32

.

0

2

= 18.28571429 (1)= 18 (to 2 sig figs) (1)no units (1)(or ecf based on answers to (i) and/or (a))3

(c)Kc is constant (1)

Composition of mixture is the same (1)2

[8]

78.To be handed in for marking

[14]

79.(a)Atomisation of Na = (+)218 / 2 (+) 109 (1);Ionisation of Na = (+)990 / 2 (+)495 (1);Any other two correct enthalpy changes (1);Last two correct enthalpy change (1)4

(b)791 + 141 247 990 218 416 (1);2521 (1)2

Allow ecf from part (a) e.g. 2026 if only 1 mole of Na Na+2412 if only 1 mole ofNa (s) Na (g)1917 if only 1 mole of Na throughoutAllow full marks for 2521 with no working out

(c)Calcium chloride (1)

If wrong salt chosen maximum of 2 marks (the comparison of the ions)

And Br has larger ionic radius than Cl / Br has lower chargedensity than Cl / ora (1);

Not Br has larger radius

K+ has a lower charge than Ca2+ / K+ has lower chargedensity than Ca2+ / K+ has a larger ionic radius than Ca2+ / ora (1);

Not K has lower chargeNot K+ has larger atomic radius

Strongest attraction between ions (when smallest radius andhighest charge) / strongest attraction between ions (with thehighest charge density) / ora (1)4

Penalise use of atoms rather than ions just once in this question

[10]

80.(a)Zn2+ is 1s22s22p63s23p63d10 andCu2+ is 1s22s22p63s23p63d9 (1);

Allow Zn2+ [Ar]3d10 and Cu2+ [Ar]3d9

Copper has at least one ion with an incomplete filled d-orbital(zinc does not) / copper(II) ion has an incomplete set ofd electrons (zinc ion does not) / copper(II) ion has anincomplete d sub-shell (zinc ion does not) / ora (1)2

(b)Cu2+ compounds are coloured but Zn2+ compounds are not (1);Cu2+ compounds may be catalytic but Zn2+ compounds are not (1)2

Allow Cu2+ forms complexes but Zn2+ does notAllow correct chemistry of Cu2+ compared to Zn2+ e.g. Cu2+ and NaOH gives blue ppt but Zn2+ gives white ppt (that redissolves in excess)

4]

81.Moles of hydrogen = 3.17 103 / moles of zinc = 3.17 10-3 (1);

Not 3 103

Mass of zinc = 0.207 g / moles of zinc 65.4 (1);

Not 0.2

Percentage of copper = 83.2 (1)3

Allow ecfFinal answer must be to 3 or 4 sig figsPenalise significant figures just onceAllow values between 82.983.2

[3]

82.(i)Cu Cu2+ + 2e- / Cu 2e- Cu2+ (1)1

(ii)2Cu + O2 + 4H+ 2Cu2+ + 2H2O (1)1

Allow any correct multipleAllow ecf from (a)(i)

2]

83.Mr of [Cu(CH3COO)2]2.Cu(OH)2 = 460.5 (1)

Allow ecf from wrong Mr

Molar ratio [Cu(CH3COO)2]2.Cu(OH)2 : H2O is 0.182 : 0.906 (1)x = 5 (1)3

Not full marks for 5 with no working out

[3]

84.To be handed in for marking

[10]

85.(a)Emf / voltage / potential difference1Half cell combined with standard hydrogen electrode1Standard conditions 298K, 1 mol dm3, 1 atm1(all 3 required for 1 mark)

(b)(i)Diagram shows:Voltmeter + salt bridge + complete circuit1Solution labelled Cu2+ and electrode labelled Ag1

V

Salt bridge

Ag

+

(aq)

Cu

2+

(aq)

Cu(s)

Ag(s)

(ii)Direction from Cu(s) to Ag(s) (must be in / close to wire)1

(iii)0.80 0.34 = 0.46 V1

(iv)Cu + 2Ag+ Cu2+ + 2Ag1

(c)Standard Electrode Potential for chlorine is more positive thanFe3+ therefore it is a better oxidising agent than Fe3+ (do notaccept E is larger or smaller)1Standard Electrode Potential for iodine is less positive thanFe3+ therefore it is a poorer oxidising agent than Fe3+1(Accept release of electrons/equilibrium arguments)

[10]

86.(a)1s22s22p63s23p63d8 (Do not accept [Ar]3d8)1

(b)(i)Ring around O1Ring around N1(Accept ring around O of C=O as an alternative to O)

(ii)Lone pair (of electrons) / non-bonding pair1

[4]

87.(a)(i)Number of dative bonds / co-ordinate bonds formed with thetransition metal (Do not accept number of ligands but allownumber of lone pairs bonded to.)1

(ii)Square planar1

(b)(i)Ligand substitution1

(ii)x = 21y = 01

(c)(i)cis isomer drawn1trans isomer drawn1(ignore any charges)

Pt

NH

3

NH

3

Cl

Cl

Pt

NH

3

Cl

NH

3

Cl

(ii)cis / trans or geometric1

(iii)Binds with DNA (not binds with cell)1Prevents replication/prevents cell dividing/prevents tumourgrowth (do not allow kills cell)1

[10]

88.(a)Moles V2+ = 25.0 0.100 / 1000 = 0.0025 mols1Moles MnO4 = 30.0 0.0500 / 1000 = 0.00150 mols11 mole of MnO4- changes its Oxidation State by 5 to changethe Oxidation State of 1.67 moles of V2+1Oxidation State of V2+ changes by 5 / 1.67 = 31

(b)3MnO4 + 5V2+ + 3H2O 3Mn2+ + 5VO3 + 6H+(1 mark for correct species, 1 mark for balanced)2

[6]

89.(a)(change in) concentration/mass/volume with time1

(b)(i)O2:Exp 2 has 4 [O2] as Exp. 1: rate increases by 4 (1),so order = 1 with respect to O2 (1)

NO:Exp 3 has 3 [NO] as Exp. 3: rate has increases by 9 (1),so order = 2 with respect to NO (1)4

(ii)rate = k[O2] [NO]2 (1)1

(iii)k =

2

2

2

0010

.

0

0010

.

0

10

.

7

NO]

][

[O

rate

=

= 7.10 109 (1)

units: dm6 mol2 s1 (1)2

[8]

90.(i)The slowest step (1)1

(ii)2NO2 NO + NO3 (1)NO3 + CO NO2 + CO2 (1)2(or similar stage involving intermediates)

[3]

91.4NO2 + O2 + 2H2O 4HNO3 (1)

N from +4 to +5O from 0 to 2 (1) Could be below equation2

[2]

92.(a)strength of acid/extent of dissociation/ionisation (1)1

(b)(i)H2SO3(aq) + CH3COOH(aq) HSO3(aq) + CH3COOH2+(aq)acid 1base 2 (1)base 1acid 2 (1)2

1 mark for labels on each side of equation

(ii)CH3COOH is the stronger acid/Ka CH3COOH is greater/CH3COOH is more acidic ORA (1)

C6H5OH(aq) + CH3COOH(aq) C6H5OH2+(aq) + CH3COO(aq) (1)2

(c)For HCl, pH = log[H+] (1) (or with values).Could be awarded below= log 0.045 = 1.35 (1) (accept 1.3)

For CH3COOH, [H+] = (Ka [CH3COOH]) /(1.70 105 0.045) (1)[H+] = 8.75 104 mol dm3 (1)pH = log 8.75 104 = 3.058/3.06 (1) (accept 3.1)5

[10]

93.HCl and CH3COOH have same number of moles/release same number of moles H+ /1 mole of each acid produce mol of H2 (1)

[H+] in CH3COOH < [H+] in HCl/CH3COOH is a weaker acid than HCl (ora) (1)

Mg + 2HCl MgCl2 + H2 (1)Mg + 2CH3COOH (CH3COO)2Mg + H2 (1)

orMg + 2H+ Mg2+ + H2 (1)(1)4

[4]

94.amount of NaOH in titration = 0.175 x 22.05/1000or 3.86 103 (1) (calc: 3.85875 x 103)

amount of A in 25.0 cm3 = 0.5 mol NaOHor 1.93 103 (1) (calc: 1.929375 103)

amount of A in 250 cm3 = 10 1.93 103 or 1.93 102 (1)

1.93 102 mol A has a mass of 2.82 g

molar mass of A = 2.82/1.93 102 = 146 g mol1 (1)(or Mr of A is 146)

Therefore A is adipic acid / HOOC(CH2)4COOH (1)5

[5]

95.(a)(i)Electron affinity -696 (1 mark);Atomisation of Cl2 +244 (1 mark);From top to bottom2nd IE +1150,1st IE +590,atomisation of Ca +178formation -796 (1 mark)3

Allow 244, 1150, 590 and 176 i.e. without plus sign

(ii)-796 - 178 - 590 - 1150 - 244 + 696 (1);

But

-2262 (with no working) (2)2

Allow ecf from the wrong figures on the Born-Haber cycle1 error max one mark2 errors 0 mark

(iii)Magnesium fluoride more exothermic than calcium chloride / ora

Answer must refer to the correct particle.

because

Ionic radius of Mg2+ is less than that of Ca2+ / charge densityof magnesium ion is greater than that of calcium ion / ora (1);Ionic radius of F- is less than that of Cl / charge densityof fluoride ion is greater than that of chloride ion / ora (1);

Not Mg or magnesium has a smaller radius or fluorine has a smaller radius

Stronger (electrostatic) attraction between cation and anionin MgF2 than in CaCl2 / stronger ionic bonds in MgF2 (1)3

Allow magnesium or fluorine has a smaller ionic radius

(b)Any two fromFor second ionisation energy the electron lost is closer to the nucleus / AW (1);For second ionisation energy the electron is lost from a particle thatis already positive (1);For second ionisation energy there is one more proton than electron (1)So outer electron more firmly attracted to the nucleus (1)2

Allow ora

[10]

96.(a)1s22s22p63s23p63d5 (1);Has an incomplete set of 3d electrons (1)2

Allow 3d orbitals are not completely occupied / incomplete 3d sub-shellAllow has half-filled d orbitals

(b)Any two fromVariable oxidation state / variable valency (1);Act as catalysts (1);Form complexes / form complex ions (1);Form coloured compounds (1)2

Not high melting point / good thermal and electrical conductors / high density etc

(c)Iron (II) ions give a green ppt (1);Iron (III) ions give an orange-rust ppt (1)2

Precipitate must be used onceAllow solid instead of ppt

(d)4Fe2+ + O2 + 4H+ 4Fe3+ + 2H2OCorrect reactants and products (1);Correct balancing (1)2

[8]

97.(i)Copper may react with potassium manganate(VII) /iron(III) ions formed in titration may be reduced back toiron(II) ions by the copper (1)1

(ii)MnO4 gains electrons and is reduced / Mn oxidationstate changes from +7 to +2 so it is reduced (1);Fe2+ loses electrons and is oxidised / Fe oxidation statechanges from +2 to +3 so it is oxidised (1)2

(iii)Moles of MnO4 = 4.50 104 (1);Moles of Fe2+ = 5 moles MnO4 / 2.25 103 (1);Mass of Fe = moles of Fe2+ 55.8 / 0.1256 (1);Percentage = 18.6 % (1)4

Allow answers that use 56 for Ar of Fe this gives 18.7Allow ecf

[7]

98.(a)(Pale blue solution) to a (light) blue ppt (1);with excess dark blue solution (1)2

(b)Octahedral shape with clear indication of 3D either byconstruction lines or wedges etc (1);90 (1)2

Ignore mistakes with the ligands question focuses on octahedral and the bond angle

[4]

99.Water molecule 2 lone pairs (and 2 bond pairs) (1);Water ligand 1 lone pair and 3 bond pairs / lone pair is nowa bond pair / water has one less lone pair when it is a ligand (1);Lone pairs repel more than bond pairs (1)3

Not atoms repel

[3]

100.(i)Central ion surrounded by molecules/ions/ligands1

(ii)Molecule/ion with a lone pair of electrons1Able to form a dative covalent or co-ordinate bond /which can be donated1

[3]

101.To be handed in for marking

[8]

102.(i)Cr2O72 + 14H+ + 6I 2Cr3+ + 3I2 + 7H2OAll species correct (ignore electrons for this mark)1Equation balanced (penalise if electrons not cancelled out)1

(ii)Brown colour disappears1S2O32 reacts with I2 (to form colourless I)1Green colour remains due to Cr3+ (must say what gives green colour)1

[5]

103.To be handed in for marking

[15]

104.(a)(i)O3: 1and C2H4 (1)1

(ii)2 (1)1

(iii)rate = k[O3] [C2H4] (1)1

(b)(i)measure gradient/tangent (1)at t = 0/start of reaction (1)2

(ii)k =

]

H

C

][

[O

rate

4

2

2

(1)k =

8

7

12

10

0

.

1

10

0.5

10

1.0

-

-

-

= 2 103 (1) dm3 mol1 s1 (1)3

(iii)2 mol CH2O forms for every 0.5 mol O2 /stoichiometry of CH2O : O2 is not 1:1 (1)1

(iv)rate increases (1)k increases (1)2

[11]

105.(i)Kw = [H+(aq)] [OH(aq)] (1)1

state symbols not needed

(ii)[H+(aq)] = 10pH = 1013.54 = 2.88/2.9 1014 mol dm3 (1)[NaOH] / [OH(aq)] =

14

14

w

10

88

.

2

10

0

.

1

aq)]

(

[H

-

-

+

=

K

= 0.347 / 0.35 mol dm3 (1)2

[3]

106.(i)proton donor (1)1

(ii)partially dissociates (1)1

[2]

107.C6H5OH(aq) + OH(aq) C6H5O(aq) + H2O(l)acid 1base 2 (1)base 1acid 2 (1)

1 mark for each acid-base pair

[2]

108.(i)Ka =

OH]

H

C

[

]

H

][

O

H

C

[

5

6

5

6

+

-

(1)1

(ii)concentration = 38/94 (1) = 0.40 mol dm3 (1)

(first mark for Mr of phenol incorrect answer here will give ecf for remainder of question)

1.3 1010

0.40

]

aq)

(

H

[

2

+

(1)

(= sign is acceptable)

[H+] = {(1.3 1010) (0.40)} = 7.2 106 mol dm3 (1)

pH = log[H+] = log 7.2 106 = 5.14 (1)5

3 marks: [H+] (1); pH expression (1); calc of pH from [H+] (1)

Common errors:Without square root, answer = 10.28 (1)(1)(0)Use of 38 as molar concentration does not score 1st 2 marks.This gives an answer of 4.15 for 3 marks (1)(1)(1)

[6]

109.

O

-

Na

+

O

-

Na

+

CH

2

(CH

2

)

4

CH

3

/ NaOH /Na (1)weak acid/base pair mixture formed (1)2

On structure, 1 mark for O Na on either or both phenol groups.

[2]

110.To be handed in for marking

[12]

111.(a)Emf of a cell / voltage / potential difference / cell potential1Comprising half cell combined with standard hydrogen electrode1Conc = 1 mol.dm3; Pressure (of H2) = 1 atm; Temp = 298K1(all of above = 1 mark)

(b)+0.16 V (unit required)1

[4]

112.(a)(i)2MnO4 + 10Cl + 16H+ 2Mn2+ + 5Cl2 + 8H2O1

correct species on both sides of equation equation balanced1(ignore electrons for first mark, penalise for balance)

(ii)Chlorine 1 01Manganese +7 +21Link to (i) and allow ecf

(iii)Chloride ion oxidised (not chlorine)1Manganate(VII) ion reduced (not manganese)1

(b)0.16 V too small/rate too slow/insufficient activationenergy/not standard conditions1

[7]

113.(a)(i)Zinc1

(ii)Coins + resist corrosion (not rusting) / hard wearingOr statues + resist corrosion/ attractive patinaOr electrical connections + good conductorOr musical instruments + attractive / sonorousOr plumbing fixtures + hard / corrosion resistant1

(b)(i)Sodium carbonate/sodium hydroxide/other suitablenamed alkali (accept correct formulae)1Do not accept alkali on its own

(ii)Starch1

(iii)Just before the end point/when solution turns pale straw1

(c)(i)0.002 mol1

(ii)One (1)1

(iii)0.002 mol1

(iv)0.002 mols Cu2+ contains 0.002 63.5 g of Cu = 0.127 g1250 cm3 of solution contains 10 0.127 g = 1.27 g1% Cu = 1.27/1.65 100 = 77.0%1(Allow 76.9-77.0; allow ecf)

[11]

114.(a)Number of coordinate / dative covalent bonds attachedto metal ion / number of lone pairs accepted (not number of ligands)1

(b)(i)[Co(H2O)6]2+ is octahedral; [CoCl4]2 is tetrahedralDrawings must be 3 dimensional

Acceptable shapes for [Co(H2O)6]2+ include:

Co

Co

Co

Co

Acceptable shapes for [CoCl4]2 include

Co

Co

2

(ii)Pink blue2

(iii)Add water.(Allow other suitable suggestions, e.g. add lead nitrate toprecipitate Cl as PbCl2)1

[6]

115.

(i)pH = log[H+] / log(0.015) (1) = 1.82 / 1.8 (1) (Not 2)2

(ii)[H+] = 0.0075 mol dm3pH = log(0.0075) = 2.12 / 2.1 (1)1

[3]

116.To be handed in for marking

[9]

117.(i)constant half-life (1)1

(ii)rate = k[N2O5] (1)1

Common error will be to use 2 from equation.

(iii)curve downwards getting less steep (1)curve goes through 1200,0.30; 2400,0.15; 3600,0.075 (1)2

(iv)tangent shown on graph at t = 1200 s (1)1

(v)3.7(2) 104 (1) mol dm3 s1 (1)ecf possible from (ii) using [N2O5]x(2nd order answer: 2.2(3) 104)2

[7]

118.(i)slow step (1)1

(ii)(CH3)2C=CH2 + H2O (CH3)3COH (1)1

(iii)H+ is a catalyst (1)

H+ used in first step and formed in second step/regenerated/ not used up (1)2

(iv)rate = k [(CH3)2C=CH2] [H+] (1)common error will be use of H2O instead of H+1

[5]

119.To be handed in for marking

[6]