Download 2007 Nov Paper 1 - A Level Tuition

2007 Nov Paper 1
1.
5.
[07N P1 Q01 Relative Masses]
92.2 % 28Si, 4.7 % 29Si and 3.1 % 30Si
B
C
1st
(92.2×28)+(4.7×29)+(3.1×30)
92.2+4.7+3.1
∆
© Step-by-Step
[07N P1 Q02 Mole]
C
2nd
3rd
5th
4h
6th
moles of Fe3+(aq)
5
850
900
2100
1200
6300
There is a 'big' jump in I.E. between the 5th and
6th I.E. values, showing that the 6th electron is
removed from the next inner shell. Hence,
element X has 5 valence electrons (Group V
element) and so, forms two chlorides, XCl3 and
XCl5. (ans)
[Ar calculated from options A, C and D are
28.099, 28.068 and 28.071 respectively.]
moles of Fe
1
XCl3
I.E. 950 , 1800, 2700, 4800, 6000, 12300
= 28.109 (ans)
2.
07N-1
[07N P1 Q05 Ionisation Energies]
From the relative abundance given in option B,
Ar of Si =
2007 Nov Paper 1
© Step-by-Step
6.
[07N P1 Q06 Bonding]
The reaction can be represented as follows:
CH3
Fe + 2 Fe3+ → 3 Fe2+
For the resultant solution to contain equal
numbers of moles of Fe3+(aq) and Fe2+(aq), an
excess of Fe3+(aq) is used; i.e. 3 mol in excess.
Hence, the reaction involves 1 mol of Fe and 5
mol of Fe3+(aq), and the resultant mixture
contains 3 mol of Fe3+(aq) and 3 mol of
Fe2+(aq). (ans)
B
O C
CH3
B has the largest dipole (due to presence of a
very electronegative O atom).
Cl
C has a smaller dipole than B
O C
because C=O and C–Cl bonds
Cl
have similar dipoles (since O
and Cl are of similar electronegativity). (ans)
© Step-by-Step
© Step-by-Step
3.
[07N P1 Q03 Periodicity]
C
7.
PCl5
Given: 0.3 mol of XCln gives 0.6 mol of HCl
or 1 mol of XCln gives 2 mol of HCl.
With excess cold water at room temperature,
PCl5 undergoes hydrolysis to give POCl3 and
2HCl.
PCl5 + H2O → POCl3 + 2HCl
Hence the chloride is PCl5. (ans)
Four electrons from each oxygen are
involved in forming hydrogen bonds.
In ice, each water molecule is hydrogen
bonded to two other water molecules.
..
H
H
[07N P1 Q04 Atomic Structure]
D
B
δ+
© Step-by-Step
4.
[07N P1 Q07 Solids]
Both ions have 20 neutrons in their nuclei.
no. of p
no. of n
no. of e–
36 216 S
16
36 – 16 = 20
16 + 2 = 18
37
17 Cl
17
37 – 17 = 20
17 + 1 = 18
∴ Both 36S2– and 37Cl – have 20 neutrons in
their nuclei. (ans)
© Step-by-Step
δ–
δ+
O : ′′′′′′′′′′′′′ H
..
O δ–
H
..
′′′′
′′′′ δ+
′′′′ H
hydrogen
δ–
bonding
O :
H
..
Hence, 2 lone pairs (or 4 electrons) from each
oxygen atom are involved in forming hydrogen
bonds. The bond angle around each oxygen
atom is 109.5° (four bond pairs, no lone pair).
The hydrogen bonds are longer (and weaker)
than the O–H covalent bonds. The open
structure of ice causes ice to be less dense than
water (i.e. ice floats on water). (ans)
© Step-by-Step
A-Level Solutions – Chemistry
07N-2
8.
2007 Nov Paper 1
11. [07N P1 Q11 Ionic Equilibria]
[07N P1 Q08 Electrolysis]
C
The number of moles of Y deposited is half
the number of moles of X deposited.
[H+ ] =
With YCl2, the reactions involved are:
Y 2+(l) + 2e– → Y(s)
2 Cl –(l) → Cl2(g) + 2e–
Hence for the same current and the same time,
mol of X deposited : mol of Y deposited
=2:1
or mol of Y deposited = 12 × mol of X deposited.
(ans)
© Step-by-Step
[07N P1 Q09 Redox]
B
3
mol of Tl + = 0.20 ×
25
1000
mol of MnO4– = 0.10 ×
= 5.0 × 10–3
25
1000
pH < 7
H2O(l)
H+(aq) + OH–(aq)
Kw = [H+ ][OH– ]
With XCl, the reactions involved are:
2 X +(l) + 2e– → 2 X(s)
2 Cl –(l) → Cl2(g) + 2e–
9.
C
= 2.5 × 10–3
= 1.44 × 10−14
= 1.20 × 10–7 mol dm–3
pH = –log10 [H+ ]
= –log10 (1.20 × 10–7) = 6.92
Hence, pH < 7. (ans)
© Step-by-Step
12. [07N P1 Q12 Titration]
14
12
10
pH 8
6 A
4
2
0
2.5 × 10
= 2 mol of Tl +
Since 2 Tl + → 2 Tl 3+ + 4 e–,
∴ 1 mol of MnO4– gains 4 mol of e–,
oxidation number of Mn in MnO4– = +7
∴O.N. of Mn in the reduced form = +7 – 4
= +3 (ans)
© Step-by-Step
D
B
C
0 2 4 6 8 10 12 14 16 18
volume of 0.100 mol dm–3 NaOH added / cm3
−3
∴1 mol of MnO4– reacts with 5.0 × 10 −3
since [H+ ] = [OH– ]
Kw
For a weak acid HA,
HA(aq)
H +(aq) + A –(aq)
[H + ][ A − ]
[HA]
Hence, pH = pKa when Ka = [H+ ];
i.e. when [A– ] = [HA].
This occurs at the point of half-neutralisation;
i.e. where vol of NaOH added = 5 cm3. (ans)
Ka =
© Step-by-Step
10. [07N P1 Q10 Redox]
13. [07N P1 Q13 Buffers]
B
The solution turns brown without
effervescence.
B
H2O2 + 2H+ + 2e– → 2H2O
2 I– → I2 + 2e–
+1.77 V
–0.54 V
H2O2 + 2H+ + 2 I– → I2 + 2H2O
+1.23 V
o
Ecell = +1.77 – 0.54 = +1.23 V
Since Ecell o > 0, reaction is feasible.
Hence, KI(aq) is oxidised to I2 (shown by
solution turning brown) while H2O2 is reduced
to water. No effervescence is seen as O2(g) is
not formed. (ans)
HCO3–
The main buffer in blood plasma is the
hydrogen carbonate and carbonic acid buffer
system, which is linked to the gaseous
exchange of CO2 in the lungs.
Hence, contaminating H+(aq) ions are removed
by the HCO3– ions present in the human body.
HCO3– + H+ → H2CO3
(ans)
© Step-by-Step
© Step-by-Step
A-Level Solutions – Chemistry
2007 Nov Paper 1
14. [07N P1 Q14 Kinetics]
D
17. [07N P1 Q17 Periodicity]
It produces its own catalyst.
pH of resulting
solution
6.5
solubility in water
The graph suggests an autocatalysed reaction,
which is slow at first and then becomes more
rapid as the catalyst (Mn2+) is produced in the
reaction.
very slow: no catalyst
[MnO4–]
fast: catalyst present
slow: reactants
dissolves
B
MgCl2(s) + aq → [Mg(H2O)6]2+ + 2Cl –
[Mg(H2O)5(OH)]+ + H+
[Mg(H2O)6]2+
MgCl2 (an ionic chloride) dissolves in water
with slight hydrolysis (due to the larger
polarising power of Mg2+ ions) to give a
slightly acidic solution (pH 6.5). (ans)
© Step-by-Step
used up
0
(ans)
time
0
07N-3
18. [07N P1 Q18 Bonding]
© Step-by-Step
D
15. [07N P1 Q15 Periodicity]
C
the copper ion in CuO
In CuO, Cu2+ has configuration [Ar] 3d9 and
so, has a single unpaired electron. (ans)
The melting points of the elements
decrease steadily.
© Step-by-Step
Across the period, the melting points of the
elements do not decrease steadily but vary as
shown.
19. [07N P1 Q19 Isomerism]
B
CH3CH(OH)CO2H
melting point
Si
Compound X is optically active and so, has a
chiral carbon (marked by *).
*
CH CH(OH)CO
H
Mg
Al
Na
S
3
P
1 mol of X liberates 1 mol of H2 when reacted
with Na suggests that X contains two –OH
groups.
Ar
Cl
11 12 13 14 15 16 17 18
proton no.
2
(ans)
CH3CHCO2H + 2Na → CH3CHCO2– Na+ + H2
© Step-by-Step
O– Na+
OH
(ans)
© Step-by-Step
16. [07N P1 Q16 Group VII]
inside the cold
tube
black deposit
A
•
with aqueous
silver nitrate
white precipitate
Iodides react with c. H2SO4 to give HI, which
is oxidised by conc. H2SO4 to I2 (seen as black
deposit in the cold tube).
–
–
I + H2SO4 → HI + HSO4
8HI + H2SO4 → 4I2 + H2S + 4H2O
•
C
It is optically active.
Bupropion has a chiral carbon (marked by *)
and so, is optically active.
CH3
*
CH3C NH CH(CH3) CO
CH3
2° amine
Chlorides react with c. H2SO4 to give HCl,
which forms a white precipitate of AgCl when
bubbled into aqueous silver nitrate.
Cl – + H2SO4 → HCl + HSO4–
Ag+(aq) + Cl –(aq) → AgCl(s) (ans)
© Step-by-Step
A-Level Solutions – Chemistry
20. [07N P1 Q20 Isomerism]
ketone
Cl
Bupropion has a secondary amine (not amide)
and a ketone group. It has no reaction with aq.
NaOH (since aryl chloride does not undergo
hydrolysis). (ans)
© Step-by-Step
07N-4
2007 Nov Paper 1
21. [07N P1 Q21 Alkenes] [93N P4 Q22]
B
25. [07N P1 Q25 Halogen Derivatives]
CH3COCH2CH3
CH3CH=CHCH3 + ½O2
A
catalyst
CH3CH2C–CH3
but-2-ene
O (ans)
© Step-by-Step
22. [07N P1 Q22 Alkenes]
C
bromoethane
Bromoethane, CH3CH2Br, reacts with
ammonia to give ethylamine, CH3CH2NH2.
δ–
δ+
..
NH3 + CH3CH2–Br
heat
CH3CH2NH2 + HBr
– a nucleophilic substitution reaction. (ans)
CH3CHCH2I
© Step-by-Step
Cl
26. [07N P1 Q26 Alcohols]
Propene undergoes electrophilic addition.
δ+ δ–
CH3CH=CH2 + I–Cl
C
slow
+
CH3CHCH2I + Cl
fast
CH3CHCH2I
Cl
(ans)
© Step-by-Step
Alcohols with molecular formula C5H12O are:
CH3CH2CH2CH2CH2OH
pentan-1-ol
Butane, CH3CH2CH2CH3, which does not
contain any halogen atoms will not destroy
ozone and so, can be used safely as a
replacement for CFCs. (ans)
© Step-by-Step
OH CH3
3-methylbutan-2-ol
CH3
3-methylbutan-1-ol
CH3CH2CHCH2CH3
OH
pentan-3-ol
CH3
CH3C–CH2OH
CH3
CH3
2,2-dimethylpropan-1-ol
24. [07N P1 Q24 Phenols]
C
C6H5O– Na+
Solid Z dissolves readily in water to give a
weakly alkaline solution suggests that it is a
salt of a strong base and a weak acid. Hence, Z
is C6H5O– Na+, which is formed from NaOH
(strong base) and phenol (weak acid). (ans)
© Step-by-Step
OH
pentan-2-ol
CH3CH CHCH3
CH3CHCH2CH2OH
CH3CH2CH2CH3
CH3CH CH2CH2CH3
CH3CH2CHCH2OH
CH3
2-methylbutan-1-ol
23. [07N P1 Q23 Halogen Derivatives]
D
2
–
CH3CH2C–CH3
OH
2-methylbutan-2-ol
Only the two alcohols with CH3CH(OH)–
group give yellow precipitate of CHI3 with
alkaline aqueous iodine. (ans)
© Step-by-Step
27. [07N P1 Q27 Carbonyl Compounds]
C
CH3COCI3
CH3COCI3
NaOD
NaOD
D2 O
D2 O
O
CH3C
+ CDI3
O– Na+
yellow
ppt
– a triiodomethane (iodoform) reaction. (ans)
© Step-by-Step
A-Level Solutions – Chemistry
2007 Nov Paper 1
28. [07N P1 Q28 Amino Acids]
C
31. [07N P1 Q31 Solids] [89J P3 Q31]
It reacts with ethanoyl chloride to form an
ester.
CO2H
N-methyl-D-aspartic acid has
CH2
a chiral carbon (marked by *)
and so, is optically active.
CH3NHCHCO2H
*
The –CO2H group reacts with ethanol (to form
an ester) and it reacts with PCl5 to form an acyl
chloride.
CO2H
07N-5
CO2C2H5
+ 2C2H5OH
CH2
CH2
c. H2SO4
heat
CH3NHCHCO2H
CH3NHCHCO2C2H5
graphite
sodium
2
3
In graphite, each C atom is bonded covalently
to three other carbon atoms, using three of the
valence electrons. The fourth valence electron
is available for π bonding between adjacent
carbon atoms, resulting in delocalisation
throughout the layer.
Sodium has a giant metallic structure, which
consists of a lattice of sodium ions surrounded
by a sea of delocalised electrons moving
randomly throughout the lattice. (ans)
© Step-by-Step
CO2H
CH2
COCl
+ PCl5
CH3NHCHCO2H
32. [07N P1 Q32 Energetics]
CH2
CH3NHCHCOCl
N-methyl-D-aspartic acid has no alcohol group
and so, cannot react with ethanoyl chloride to
form an ester. (ans)
© Step-by-Step
29. [07N P1 Q29 Amines]
B
o
∆Hr
o
∆Hr = ∑∆Hf (products) – ∑∆Hf (reactants)
= ∆Hf o for Ca(OH)2(s) – 2∆Hf o for H2O(l)
Further information needed is ∆Hf o for H2O(l).
(ans)
Order of base strength:
C2H5NH2 > NH3 > C6H5NH2
© Step-by-Step
∴ strength of conjugate acid:
C2H5NH3+ Cl – < NH4+ Cl – < C6H5NH3+ Cl –
Hence C6H5NH3+ Cl – is most acidic in aqueous
solution. (ans)
© Step-by-Step
30. [07N P1 Q30 Amino Acids]
D H2NCHCO2–
CH2
CO2–
At pH 10 (alkaline medium), the acidic –CO2H
groups react with the OH– ions present to give
the corresponding carboxylate anion.
H2NCHCO2– + 2H2O
CH2
CH2
CO2H
CO2–
(ans)
© Step-by-Step
A-Level Solutions – Chemistry
Ca(s) + 2H2O(l) → Ca(OH)2(s) + H2(g)
Since ∆Hf o is zero for both Ca(s) and H2(g),
∴∆Hf o for Ca(OH)2(s) = ∆Hr + 2∆Hf o for H2O(l)
C6H5NH3+ Cl –
H2NCHCO2H + 2OH–
∆Hf o for H2O(l)
1
33. [07N P1 Q33 Kinetics]
2
3
x
y
z
1
2
–1
–2
+3
+6
Given: rate = k [P] [Q] x
⇒ k = rate x
[P][Q]
units of rate
units of k =
(units of concentration)1+ x
-3 -1
= mol dm -3 s1+ x
(mol dm )
= mol–x dm3x s–1
Given: units of k = moly dmz s–1
By inspection, y = –x and z = 3x
Hence, options 2 and 3. (ans)
© Step-by-Step
07N-6
2007 Nov Paper 1
34. [07N P1 Q34 Group II] [80N P3 Q35]
1
37. [07N P1 Q37 Amides]
The ionic radius of the M 2+ ion increases
from calcium to barium.
1
Ionic radius increases down the group (from Ca
to Ba) as each succeeding element has one
more shell of electrons. Hence, the outer
electrons are progressively further away from
the nucleus. Hydration energy of M 2+ ion
decreases as size of M 2+ increases.
M(g) → M 2+(g) + 2e–
2
•
chlorine
bromine
Br Br
Br
•
–
O2 + 4H + 4e → 2H2O
[Fe(CN)6]3– + e– → [Fe(CN)6]4–
Cr3+ + e– → Cr2+
Fe(OH)3 + e– → Fe(OH)2 + OH–
On heating with NaOH(aq), both phenol and
amide groups react – phenol undergoes
neutralisation while amide undergoes
hydrolysis. Hence, two moles of NaOH are
used up per mole of capsaicin.
Product:
CH3O
A solution of potassium
hexacyanoferrate(III).
Na+ –O
E o value
+1.23 V
+0.36 V
–0.41 V
+1.07 V
When left to stand in the atmosphere, only
solutions with E o value less positive than
+1.23 V will be oxidised. Hence, Cr2+(aq) will
be oxidised to Cr3+(aq), while Fe(OH)2 which
is formed when FeSO4 reacts with NaOH, will
be oxidised to Fe(OH)3. However, [Fe(CN)6]3–
remains unchanged (chemically stable) because
it is already in its highest oxidation state. (ans)
CH2NHCO(CH2)4CH–CHCH(CH3)2
HO
36. [07N P1 Q36 Transition Elements]
+
With Br2 in organic solvent, both phenol and
alkene groups react – phenol undergoes
electrophilic substitution while alkene
undergoes electrophilic addition. Hence, three
atoms of bromine is incorporated into the
molecule.
Product:
CH3O
© Step-by-Step
1
alkene
phenol
35. [07N P1 Q35 Group VII]
Fe3+ + e– → Fe2+
E o = +0.77 V
–
–
E o = +1.36 V
Cl2 + 2e → 2Cl
–
–
Br2 + 2e → 2Br
E o = +1.07 V
–
–
E o = +0.54 V
I2 + 2e → 2I
To oxidise Fe2+(aq) to Fe3+(aq), the oxidising
agent must have a E o value more positive than
+0.77V. Hence, only chlorine and bromine
will oxidise Fe2+(aq) to Fe3+(aq). (ans)
amide
HO
© Step-by-Step
1
2
CH2 NHCO (CH2)4 CH=CH CH(CH3)2
CH3O
∆H = 1st I.E. + 2nd I.E.
∆H decreases from Ca to Ba since ionisation
energy decreases down a group. (ans)
Addition of bromine in an organic solvent
causes three atoms of bromine to be
incorporated into the molecule.
On heating with NaOH(aq), two moles of
NaOH are used up per mole of capsaicin.
•
CH2NH2
+ Na+ –O2C(CH2)4CH=CHCH(CH3)2
On heating under reflux with conc. KMnO4/H+,
one of the possible reaction is oxidation with
cleavage of C=C bond. (CH3)2CHCHO
formed initially is further oxidised to give
(CH3)2CHCO2H.
CH3O
CH2NHCO(CH2)4CH=CHCH(CH3)2
HO
CH3O
CH2NHCO(CH2)4CO2H
+ (CH3)2CHCO2H
HO
(ans)
© Step-by-Step
© Step-by-Step
A-Level Solutions – Chemistry
2007 Nov Paper 1
38. [07N P1 Q38 Phenols]
1
2
•
•
•
40. [07N P1 Q40 Carboxylic acids]
Its aqueous solution is acidic.
It can exist in optically active forms.
Its aqueous solution phenol
is acidic due to the
presence of an acidic O N
2
phenol group.
OH CH3
CHCH2CH3
*
•
An ester is formed when the phenol group
reacts with ethanoyl chloride (and not with
ethanol). (ans)
2
3
CO2CH3
OCOCH3
CN
OCOCH3
In 2, salicylic acid is obtained by acid
hydrolysis of both ester groups.
Dinoseb has a chiral
NO2
carbon (marked by *)
and so, can exist in optically active forms.
CO2CH3
OCOCH3 H O/H+
2
heat
•
CN
OCOCH3 H O/H+
2
39. [07N P1 Q39 Carbonyl Compounds]
heat
•
2,4-dinitrophenylhydrazine reagent
alkaline aqueous iodine
•
H NO2
C
O + H2 N
COCl
NO2
N
Cl
CH3
propanone
CO2H
+ NH4+
OH
+ CH3CO2H
In 1, the aryl chloride group is not hydrolysed
because the aryl–Cl bond is strengthened by
the overlapping of the p-orbital of Cl with the π
orbitals of the benzene ring. Hence, the
hydrolysis reaction is
With 2,4-DNPH, propanone (a carbonyl
compound) reacts to give an orange precipitate.
CH3
CO2H
+ CH3OH
OH
+ CH3CO2H
In 3, salicylic acid is obtained by acid
hydrolysis of both the nitrile and ester groups.
© Step-by-Step
1
2
07N-7
H2O/H
+
CO2H
Cl
+ HCl
heat
2,4-DNPH
and salicylic acid is not formed. (ans)
N
C
CH3
•
© Step-by-Step
H NO2
CH3
NO2 + H2O
N
orange precipitate
With alkaline aqueous I2, propanone (which
has CH3C=O group) reacts to give a yellow
precipitate of CHI3.
O
CH3C
+ 3I2 + NaOH
CH3
propanone
warm
O
CHI3 + CH3
yellow
ppt
C
+ 3HI
O– Na+
[07N P1 MCQ Key]
Q.
Key
Q.
Key
Q.
Key
Q.
Key
1
2
3
4
5
B
C
C
D
C
11
12
13
14
15
C
B
B
D
C
21
22
23
24
25
B
C
D
C
A
31
32
33
34
35
C
D
C
D
B
6
7
8
9
10
B
B
C
B
B
16
17
18
19
20
A
B
D
B
C
26
27
28
29
30
C
C
C
B
D
36
37
38
39
40
D
B
B
B
C
© Step-by-Step
•
Only aldehydes react with Fehling's reagent.
Hence, there is no reaction with both
propanone (a ketone) and pentyl ethanoate (an
ester). (ans)
© Step-by-Step
A-Level Solutions – Chemistry
07N-8
1.
2007 Nov Paper 2
2007 Nov Paper 2
2.
[07N P2 Q01 Mole / Gases]
(a) alkanes. (ans)
[07N P2 Q02 Group VII]
(a)
(i) Cl2 + 2NaOH → NaCl + NaClO + H2O (ans)
(b) 2 C14H30 + 43 O2 → 28 CO2 + 30 H2O (ans)
[or C14H30 +
43
2
O2 → 14 CO2 + 15 H2O]
(c)
(i) mass of C14H30 burnt = (10.8 × 10700) kg
= 10.8 × 10700 tonnes
1000
= 115.56 tonnes
= 116 tonnes (ans)
(ii) Mr of C14H30 = (14 × 12.0) + (30 × 1.0) = 198.0
Mr of CO2 = 12.0 + (2 × 16.0) = 44.0
From the equation in (b),
(2 × 198.0) tonnes of C14H30 gives (28 × 44.0)
tonnes of CO2.
(28 × 44.0)
∴mass of CO2 produced =
× 116
(2 ×198.0)
= 361 tonnes
= 361000 kg (ans)
(d) n =
(ii) 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O
(ans)
(b)
(i)
reactant
products
species
oxidation
number
Cl2
NaCl
0
–1
species
NaClO
oxidation
number
+1
(ans)
(ii) In this reaction, Cl2 undergoes
disproportionation; i.e. Cl2 is simultaneously
oxidised to NaClO and reduced to NaCl. (ans)
(c)
(i) H2O2 → O2 + 2H+ + 2e– (ans)
(ii) H2O2 + 2ClO3– + 2H+ → O2 + 2ClO2 + 2H2O
(ans)
(iii) H2O2 functions as a reducing agent. (ans)
© Step-by-Step
(6 ×100 ×103 ) × (670 ×10−6 )
pV
=
8.31× (20 + 273)
RT
= 0.165 mol (ans)
[NB. Pressure must be converted into Pa (or N m–2)
and volume converted into m3.]
0.165 × 8.31× (5 + 273)
(e) p = nRT =
V
670 ×10−6
= 569000 Pa
= 569 kPa
= 5.69 bar (ans)
(f) No, the tyres should not be deflated before
flight.
maximum safe pressure difference between the
air inside the tyre and the atmospheric pressure
acting on it = 8 – 1.01 = 6.99 bar
pressure difference across the wall of the tyre
in flight = 5.69 – 0.28 = 5.41 bar
Hence, it is not necessary to reduce the air
pressure inside the tyre since difference
between the internal and external pressure
(5.41 bar) is well within the pressure range
(6.99 bar) that the tyre can withstand before it
will burst. (ans)
© Step-by-Step
A-Level Solutions – Chemistry
2007 Nov Paper 2
3.
(d)
(i) stage I: hydrolysis (nucleophilic substitution)
(ans)
stage II: elimination reaction (ans)
[07N P2 Q03 Transition Elements]
(a)
(i) Co-ordination number of Co = 6 (ans)
(ii) No significant colour change is observed
because Co2+(aq) is not oxidised by ClO2.
E o /V
+
–
–
ClO2 + 4H + 5e → Cl + 2H2O
+1.50
Co2+(aq) → Co3+(aq) + e–
–1.82
2+
+
–
07N-9
(ii) Mechanism: nucleophilic substitution with
OH– as the nucleophile.
slow
C
Cl
3+
Cl
OH
C
..
OH–
Cl
ClO2 + 4H + 5Co → Cl + 5Co + 2H2O
Θ
O
O
δ+
δ–
Cl
o
Ecell = +1.50 – 1.82 = –0.32 V
Since Ecell o < 0, reaction is not feasible. (ans)
fast
O
(iii) Cation P is [Co(NH3)6]2+. (ans)
C
Cl
(iv) [Co(NH3)6]3+ is formed. (ans)
OH + Cl –
(ans)
o
ClO2 + 4H+ + 5e– → Cl – + 2H2O
[Co(NH3)6]2+ → [Co(NH3)6]3+ + e–
2+
+
–
E /V
+1.50
–0.11
3+
ClO2 + 4H + 5[Co(NH3)6] → Cl + 5[Co(NH3)6] + 2H2O
o
Ecell = +1.50 – 0.11 = +1.39 V
Since Ecell o > 0, reaction is feasible.
Hence, [Co(NH3)6]2+ is oxidised by ClO2 to
[Co(NH3)6]3+. (ans)
© Step-by-Step
4.
[07N P2 Q04 Halogen Derivatives]
(a) 2CHCl3(l) + O2(g) → 2COCl2(g) + 2HCl(g)
∆H = ∑∆Hf (products) – ∑∆Hf (reactants)
–356 = 2 ∆Hf (COCl2) + 2(–92) – 2(–134) – 0
–356 = 2 ∆Hf (COCl2) + 84
∴ ∆Hf (COCl2) = −356 − 84
2
= –220 kJ mol–1 (ans)
(b) This oxidation is initiated by ultraviolet light
and involves free-radicals. (ans)
(c)
(i)
• ••
•
Cl •
•× •
•• • ×
O C
• • • × ×• •
•
• •Cl••
(ans)
(ii) Cl–C–Cl bond angle is 120°. (ans)
A-Level Solutions – Chemistry
(e) The organic compound formed is
O
Cl
C
OCH2CH3
(ans)
[The other Cl atom in phosgene can also be
replaced by –OCH2CH3 group.]
(f)
(i) Reagent: ammonia, NH3 (ans)
(ii) Urea can act as a base because the lone pair of
electrons on each of the N atoms can accept H+
(from the strong acid) by forming dative bond
with it. (ans)
(g)
(i) 4 CO(NH2)2 + 6 NO2 → 4 CO2 + 8 H2O + 7 N2
(ans)
(ii) NO2 in the atmosphere can result in acid rain
(by dissolving in water vapour in the air to
form nitric acid). (ans)
© Step-by-Step
07N-10
5.
2007 Nov Paper 2
(b)
(i)
[07N P2 Q05 Phenols]
(a)
reagent
OCH3
structural formula of the organic
product
CH2CHCH2Br
CH3C
O
CH2CH=CH2
(ii)
CO2H
O
+
OCH3
Product:
(ans)
OCH3
[Oxidation with cleavage of C=C bond.]
OH
OCH3
O2N
(c)
(i)
CH=CHCH2OH
CH2CH=CH2
(ans)
compound C
–
(ii) stage I
reagent(s)
conditions
+
O Na
OCH3
stage II
reagent(s)
conditions
CH2CH=CH2
compound A
Product:
Tollen's
reagent
CH3C
OH
compound C
Na
OH (ans)
[Reaction of phenol and alkene with aq. Br2.]
O
Product:
CH2CHCH2Br
Br
Product:
dilute
HNO3
OCH3
Br
and
compound C
CH3COCl
OH
OH
Br
compound A, B, C or D /
O
LiAl H4 (ans)
in dry ether (ans)
CH3CO2H (ans)
heat under reflux with a small
amount of conc. H2SO4 (ans)
[or CH3COCl at room temperature.]
CH=CHC
© Step-by-Step
OH
(ans)
A-Level Solutions – Chemistry
2007 Nov Paper 3
1.
[07N P3 Q01 Equilibria / Entropy / RX]
(a) Kp =
pCH CH CH CHO
atm–2 (ans)
( pCH CH=CH )( pCO )( pH )
3
2
3
2
2
2
(i) CH3CH=CH2 + CO + H2
t=0/mol
t=0/atm
1
CH3CH2CH2CHO
1
1
0
120
3
120
3
120
3
0
= 40
= 40
= 40
eqm/atm 40 – 39.6 40 – 39.6 40 – 39.6
= 0.4
= 0.4
0
39.6
= 0.4
∴ partial pressure of CH3CH=CH2 = 0.4 atm (ans)
partial pressure of CO = 0.4 atm (ans)
partial pressure of H2 = 0.4 atm (ans)
pCH CH CH CHO
( pCH CH=CH )( pCO )( pH )
(39.6)
=
= 619 atm–2 (ans)
(0.4)(0.4)(0.4)
(ii) Kp =
3
3
2
(ii)
• To convert butan-1-ol into 1-bromobutane,
react with HBr, which is obtained by heating
NaBr with conc. H2SO4. (ans)
[or react with PBr3, PBr5 or SOBr2 at room
temperature.]
•
2
(iii) CH3CH=CH2 + CO + H2
CH3CH2CH2CHO
A higher pressure would favour the production
of butanal because the forward reaction is
accompanied by a reduction in pressure (3 mol
of gas converted to 1 mol of gas). By Le
Chatelier's principle, an increases in pressure
would shift the position of equilibrium to the
right (the side with fewer gas molecules) so as
to remove some of the excess pressure. (ans)
reaction III – LiAl H4 in dry ether. (ans)
[reaction II is oxidation, III is reduction]
2
2
07N-11
(c)
(i)
• reaction II – heat under reflux with acidified
K2Cr2O7(aq). (ans)
•
(b)
2007 Nov Paper 3
To convert butan-1-ol into 2-bromobutane,
heat with conc. H2SO4 at 180 °C (to give
but-1-ene), followed by reaction with HBr.
(ans)
(d)
(i) With 2-bromobutane, the rate of reaction is
slower than that with 2-iodobutane. This is
because the C–Br bond is shorter and hence,
stronger than C–I bond (since Br atom is
smaller than I atom). (ans)
(ii)
• CH3CH2CHCH3
NaCN
in alcohol
CH3CH2CHCH3
Br
CN
LiAl H4
CH3CH2CHCH3
(iv) ∆S for reaction I is negative because there is a
decrease in the number of gaseous particles
and hence, the system becomes more orderly.
(ans)
(v) ∆G = ∆H – T ∆S
Since ∆S is negative, the term (– T ∆S ) is
positive. Hence for ∆G to be negative, ∆H is
negative (at least as negative as T ∆S). (ans)
(vi) CH3CH=CH2 + CO + H2 → CH3CH2CH2CHO
Bonds broken (∆H )
1 C–C
1 C=C
6 C–H
1 C≡O
1 H–H
+350
+610
6(+410)
+1077
+436
+4933
Bonds formed
3 C–C
8 C–H
1 C=O
∴ ∆H = +4933 + (–5070)
= –137 kJ mol–1 (ans)
(∆H )
3(–350)
8(–410)
–740
–5070
CH2NH2
•
CH3CH2CHCH3
Br
NaOH
in alcohol
CH3CH=CHCH3
+ CH3CH2CH=CH2
CH3CH=CHCH3 + CH3CH2CH=CH2
cold KMnO4(aq)
CH3CH–CHCH3 + CH3CH2CH–CH2
OH OH
OH OH
(ans)
(e) Free-radical substitution occurs and any one of
the H atoms in butane
H H H H
may be replaced by Br
1 2
4
3
H C C C C H
atom. Replacing any
H H H H
one of the six H atoms
on C-1 and C-4 with Br gives 1-bromobutane,
whereas replacing any one of the four H atoms
on C-2 and C-3 gives 2-bromobutane. Hence
ratio of 1-bromobutane : 2-bromobutane = 6 : 4
or 3 : 2. (ans)
© Step-by-Step
A-Level Solutions – Chemistry
07N-12
(a)
If a white ppt is seen which dissolves in excess
NaOH(aq) to give a colourless solution, it
suggests that the presence of Al 3+ and so, the
sample is Al2O3.
[07N P3 Q02 Periodicity / Acid Derivatives]
first ionisation energy
2.
2007 Nov Paper 3
Ar
P
Cl
Mg
Si
Na
Al 3+(aq) + 3OH–(aq) → Al (OH)3(s)
Al (OH)3(s) + OH–(aq) → Al (OH)4–(aq) (ans)
S
Al
11 12 13 14 15 16 17 18
proton number
First ionisation energy (I.E.) increases from Na
to Ar due to the increase in nuclear charge and
decrease in atomic size (while shielding effect
remains almost the same since the electrons all
go into the same shell).
(c) NaCl is an ionic chloride and so, dissolves in
water without further reaction to give a neutral
solution (pH 7).
NaCl(s) + aq → Na+(aq) + Cl –(aq)
•
The discontinuities in the increasing trend are:
•
•
The first I.E. of Al is lower than that of Mg.
Mg [Ne] 3s2
Al [Ne] 3s2 3p1
This is because less energy is required to
remove a 3p electron in Al than a 3s electron in
Mg since the 3p electron is further away from
the nucleus and it also experiences slightly
better shielding (from the 3s electrons).
The first I.E. of S is lower than that of P.
P [Ne] 3s2 3px1 3py1 3pz1
S [Ne] 3s2 3px2 3py1 3pz1
This is because less energy is required to
remove an electron from the paired 3p
electrons in S due to inter-electron repulsion
arising from two electrons occupying the same
orbital. (ans)
(b) Add dilute aqueous acid, e.g. dilute HCl, to the
sample of white powder.
If there is no reaction, the sample is SiO2.
If the white powder dissolves to give a
colourless solution, it is either MgO or Al2O3.
MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l)
Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(l)
•
To the resultant solution, add dilute NaOH(aq)
dropwise until in excess.
If a white precipitate (ppt) is seen which does
not dissolves in excess NaOH(aq), it suggests
the presence Mg2+ and so, the sample is MgO.
Mg2+(aq) + 2OH–(aq) → Mg(OH)2(s)
PCl5 is an acidic chloride (simple covalent) and
so, undergoes complete hydrolysis in water to
give a strongly acidic solution (pH = 1).
PCl5 + 4H2O → H3PO4 + 5HCl (ans)
(d)
(i) One precaution is to perform the experiment in
a fume cupboard since pungent gases (SO2 and
HCl ) are produced. (ans)
[SO2 is poisonous/choking; HCl is acidic.]
(ii) SOCl2 + H2O → SO2 + 2HCl
(ans)
(iii) With AgNO3(aq), a white precipitate of
AgCl(s) is seen.
Ag+(aq) + Cl –(aq) → AgCl(s)
(from HCl )
•
With K2Cr2O7(aq), SO2 (a reducing agent)
would reduce the orange K2Cr2O7(aq) to green
Cr3+(aq) and itself being oxidised to SO42–(aq).
Cr2O72– + 2H+ + 3SO2 → 2Cr3+ + 3SO42– + H2O
(ans)
(e)
(i)
O
O
C
A
C
O
OH
OH
C
B
C
Cl
Cl
(ans)
O
(ii) I: acid-base reaction (ans)
II: nucleophilic substitution (ans)
III: hydrolysis reaction (ans)
(iii) III: heat under reflux with dilute aqueous
H2SO4. (ans)
[or heating under reflux with aqueous NaOH.]
© Step-by-Step
A-Level Solutions – Chemistry
2007 Nov Paper 3
3.
[07N P3 Q03 Energetics / Grp II / Carbonyl]
(a)
(i)
Ca2+(g) + 2F(g) + 2e–
07N-13
(iii) Since ∆Hsol is positive (endothermic), the
dissolution of CaF2 is, therefore, favoured by
high temperature. Hence, CaF2 is more soluble
in hot water than in cold. (ans)
E
(c)
(i) The very low melting point (60 °C) suggests
that Be(NO3)2 is a covalent compound with a
simple molecular structure (due to the high
charge density of Be2+ ion).
2nd I.E.(Ca)
Ca+(g) + 2F(g) + e–
2(–328)
1st I.E.(Ca)
Ca(g) + 2F(g)
Ca(g) + F2(g) B.E.(F–F)
0
Ca2+(g) + 2F –(g)
•
Ca(s) + F2(g) ∆Hat(Ca) = +178
∆Hf
= –1220
L.E.
CaF2(s)
(ans)
By Hess' Law,
L.E. = ∆Hf – [∆Hat (Ca) + B.E.(F–F) + 1st I.E.
+ 2nd I.E. + 2(–328)]
= –1220 – [178 + 158 + 590 + 1150 + 2(–328)]
= –2640 kJ mol–1 (ans)
(ii) L.E. ∝
•
•
q+ q−
(r+ + r− )
(ii) Mg(NO3)2(s) → MgO(s) + 2NO2(g) +
•
(ans)
Ca2+(aq) + 2F–(aq)
–543
2(–333)
Ca(s) + F2(g)
By Hess' Law,
∆Hsol = –543 + 2(–333) – (–1220)
= +11 kJ mol–1 (ans)
A-Level Solutions – Chemistry
Ca ion, being larger in size than Mg ion,
has a smaller charge density and hence, weaker
polarising power. The NO3– ion is, thus, less
polarised by the larger Ca2+ ion. Ca(NO3)2 is,
therefore, more stable than Mg(NO3)2 and so,
decomposes at a higher temperature. (ans)
O
O
,
and CH3C
CH3C
H
Mr 30
CH3
H
Mr 44
Mr 58
which are produced in the ratio 1:2:1. (ans)
[ 2 (CH3CO2)2Ca + 2 (HCO2)2Ca
→ CH3COCH3 + HCHO
+ 2 CH3CHO + 4 CaCO3 ]
(ii) The two chemical tests are:
1. Warm each compound separately with alkaline
aqueous iodine.
Only CH3COCH3 and CH3CHO give yellow
precipitate of CHI3. Hence the compound that
does not give any such precipitate is HCHO.
2×2.3×10–4 mol dm–3
Ksp = [Ca2+] [F – ]2
= (2.3 × 10–4)(2 × 2.3 × 10–4)2
= 4.87 × 10–11 mol3 dm–9 (ans)
–1220
2+
O
Ca2+(aq) + 2F –(aq)
2.3×10–4
∆Hsol
2+
H–C
The lattice energy of CaF2 is numerically
smaller than that of CaO because the
electrostatic forces of attraction between Ca2+
and the singly charged F – ion is weaker than
that between the doubly charged Ca2+ and O2–
ion. (ans)
(ii) CaF2(s) + aq
O2(g)
(ans)
The lattice energy of CaF2 is numerically
greater than that of CaCl2 due to the smaller
ionic size of F – ion (compared to Cl – ion)
while the charge remains the same. (ans)
CaF2(s)
1
2
(d)
(i) The three carbonyl compounds are:
(b)
(i) Ksp = [Ca2+] [F – ]2 mol3 dm–9
•
The very high melting point (561 °C) suggests
that Ca(NO3)2 is an ionic compound with a
giant lattice. (ans)
2.
Having identified HCHO, heat the remaining
two compounds separately with acidified
KMnO4(aq).
CH3CHO is oxidised to CH3CO2H, and the
purple KMnO4(aq) solution is decolourised.
CH3COCH3 is not oxidised and so, no such
colour change is seen. (ans)
[With KMnO4/H+, HCHO is oxidised to give
CO2(g), which forms a white ppt with limewater.]
© Step-by-Step
07N-14
2007 Nov Paper 3
4. [07N P3 Q04 Ionic Eqm/Kinetics/Aldehyde]
(a)
(i) HCN(aq)
H+(aq) + CN–(aq)
Ka =
[H + ][CN - ]
[HCN]
•
Let the rate equation be
rate = k [(CH3)2CO] [NaCN] [HCN]n
Using values from Expt 2 and 4,
1.25 = k (0.050)(0.008)(0.040)n
(I)
0.94 = k (0.050)(0.006)(0.050)n
(II)
k (0.050)(0.008)(0.040)n
(I) ÷ (II), 1.25 =
0.94 k (0.050)(0.006)(0.050) n
1.33 = 1.33 × (0.8) n
(0.8) n = 1
∴ n=0
∴ reaction is zero order w.r.t. HCN. (ans)
•
rate equation is rate = k [(CH3)2CO] [NaCN]
(ans)
(ans)
(ii) Since the degree of dissociation, α , is very
small, [HCN]eqm ≈ [HCN]0
∴Ka =
[H + ]2
since [CN–] = [H+]
[HCN]
[H+] =
K a × [HCN]
4.9 ×10−10 × 0.100
=
= 7.00 × 10–6 mol dm–3
(ans)
−6
∴% HCN molecules ionised = 7.00 ×10 × 100
0.100
= 0.00700 % (ans)
(iii) HCN
H+ + CN–
NaCN → Na+ + CN–
(I)
common ion
The addition of NaCN increases the [CN– ].
The presence of this common ion shifts the
position of equilibrium (I) to the left and so,
less HCN molecules dissociate. Hence the
percentage of HCN molecules ionised is
decreased. (ans)
(iii) Mechanism: nucleophilic addition with CN– as
nucleophile.
The rate equation shows that only one
molecule of (CH3)2CO and one molecule of
NaCN are involved in the slow step.
HCN being a weak acid is a poor source of
CN– ion. Hence, a small amount of NaCN is
added to accelerate the reaction.
NaCN → Na+ + CN–
O–
CH3
..
CN–
δ-
C O
slow
CH3
δ+
C
CH3
CH3
CN
+ HCN
(b)
(i) The order of reaction with respect to a given
reactant is the power to which the
concentration of that reactant is raised to in the
experimentally determined rate equation, e.g.
In the reaction, aA + bB → products
rate equation is rate = k [A]m [B]n
where m = order of reaction with respect to A
n = order of reaction with respect to B
m + n = overall order of reaction. (ans)
(ii) Comparing Expt 1 and 2, [HCN] and [NaCN]
constant, when [(CH3)2CO] increases from 0.4
to 0.5 mol dm–3 (or factor of 00..45 or 54 ), initial
rate also increases by a factor of
1.25
1.00
or
5
4
.
st
∴ reaction is 1 order w.r.t. (CH3)2CO. (ans)
•
Comparing Expt 1 and 3, [(CH3)2CO] and
[HCN] constant, when [NaCN] decreases from
0.008 to 0.006 mol dm–3 (or factor of 00..006
008 or
3
4
OH
CH3
C
CH3 + CN–
(ans)
CN
(iv) Rate of reaction with HCN:
CH3CH2
C
H
O
>
CH3
C
O
CH3
Propanal reacts at a faster rate because the
carbonyl C atom has a larger δ+ charge (it has
one less electron-donating alkyl group) and
there is less steric hindrance than in propanone
(which has two bulky CH3 groups). (ans)
(v) HCN is the reagent for the nucleophilic
addition reaction; it acts as a Bronsted acid.
NaCN acts as a catalyst in the reaction; it
provides the CN– nucleophile initially and is
regenerated in the reaction. (ans)
), initial rate also decreases by a factor of
0.75
1.00
or
3
4
. ∴ reaction is 1st order w.r.t. NaCN.
A-Level Solutions – Chemistry
2007 Nov Paper 3
OH
(c) CH3CH2
C
CH3CH2 *C H
O + HCN
H
CN
5. [07N P3 Q05 Proteins]
(a)
(i) Requirements needed for two molecules to
form a hydrogen bond between them are:
Optical isomerism occurs since the product
obtained has a chiral carbon (marked by *).
Isomers D and E are optical isomers.
CN
CN
C
C
H
OH
1.
2.
H
HO
δ+
OH
•
CH3CH2
hot
C H
Al2O3
CH3CH=CHCN
H
CH3
CN
cis–isomer
•
CN
H
C=C
CH3
H
trans–isomer
Each of the two isomers D and E produce the
same mixture of cyanoalkenes because the
same reaction (dehydration) takes place in both
cases. Both isomers can each lose an H and
the OH from adjacent carbon atoms in two
different ways, thereby giving the cis and trans
compound. (ans)
[or since the reagents used are non-chiral, the
reactions involved are identical.]
© Step-by-Step
A-Level Solutions – Chemistry
δδ+
δ-
H ′′′′′′′′′′′′′ :
Hδ +
Hδ +
N
H
Dehydration occurs, and the product obtained
shows cis-trans isomerism. Isomers F and G
are cis-trans isomers.
C=C
N
H
H
hydrogen bonding
CN
H
an electron-deficient hydrogen atom
bonded to a very electronegative atom
(such as N, O or F) in one molecule, and
a lone pair of electrons on a very
electronegative atom (such as N, O or F)
in the other molecule.
e.g. hydrogen bonds between NH3 molecules
..
CH2CH3
CH2CH3
07N-15
δ+
(ans)
(ii) The intermolecular forces in carboxylic acids
(RCO2H) comprises both van der Waals' forces
(between the hydrocarbon, R, chains) and
hydrogen bonding (between –CO2H groups),
while that between water molecules are mainly
hydrogen bonding.
•
The first four members of
δ -O
the series of carboxylic
H
δ
δ - ′′′ ′′ +
acids are fully
Hδ +
′′
δ + O′
soluble in water
R C δ- δ+
because they
O H ′′′′′′′′′′
form hydrogen
Oδ - δ +
H
bonds with water molecules.
•
As the carbon chain length
increases, the acids become increasingly
insoluble because the molecule becomes less
polar (or more hydrocarbon-like) and
intermolecular van der Waals' forces become
predominant. Hence, energy released when
weak van der Waals' forces are formed
between the acid and water molecules is
insufficient to compensate for the energy
required to break the strong intermolecular
hydrogen bonding between water molecules.
Hence, the dissolution of long chain carboxylic
acids is endothermic and so, does not occur
readily. (ans)
Hδ +
07N-16
2007 Nov Paper 3
(b)
(i) The primary structure of a protein shows the
order (or sequence) of amino acid residues in a
protein (or polypeptide chain). (ans)
(ii) A section of the protein chain of HSA using
serine, cysteine and phenylalanine:
O
(v) Two other types of side-chain interaction:
1. hydrogen bonding between polar groups,
e.g. between serine and lysine (marked by
2.
H C H
H
H
O
C
C
N
C
C
H C H
H
H
O
N
C
C N
H
H
O
S
S
S
H C CH3
CH3
2
CH2
CH2
CH2OH
(ans)
H
CH2
(iii) A polypeptide chain is held in the shape of an
α - helix by hydrogen bonds between the N–H
group of each amino acid unit and the fourth
C=O group following it along
the chain. There are 3.6
amino acids per turn and
the pattern repeats itself
every five turns. The
hydrogen bonds in α - helix
are linear and so, are
α - helix
maximally stable. (ans)
(iv) The thiol group in cysteine is oxidised to
disulphide in the presence of oxidising agents
such as oxygen or hydrogen peroxide.
–CH2–SH + HS–CH2– +
1
2
).
van der Waals' attraction between non-polar
groups, e.g. between phenylalanine and valine
(marked by 2 ).
H
H C H
1
O2
→ –CH2–S–S–CH2– + H2O
(ans)
[or –CH2–SH + HS–CH2– + [O]
→ –CH2–S–S–CH2– + H2O
or –CH2–SH + HS–CH2– → –CH2–S–S–CH2– + 2[H]
1
(CH2)4NH2
S
S
(ans)
[or ionic bonding between charged groups;
e.g. between glutamic acid (RCH2CH2CO2–)
and lysine (RCH2CH2CH2CH2NH3+).]
(vi) The long carbon chain on stearic acid is
hydrophobic. Hence two amino acids which
would interact with the long chain on stearic
acid are leucine and valine. (ans)
[or phenylalanine, all of which interact with
stearic acid by hydrophobic (van der Waals' )
interactions.]
(vii) Three amino acids likely to interact strongly
with water molecules are glutamic acid, lysine
and threonine. (ans)
[or serine, all of which interact strongly with
water molecules by forming hydrogen bonds
with water molecules.]
© Step-by-Step
or –CH2–SH + HS–CH2– → –CH2–S–S–CH2– + H2 ]
[NB. the –CH2– chain was required.]
•
Since there are 36 cysteine residues per
molecule of HSA,
∴Within each HSA molecule, there could be
36
2
or 18 disulphide bridges. (ans)
A-Level Solutions – Chemistry