CAPE Chemistry June 2004 U2 P2 Ql - Answer
l(a)
Concentration of B2.
l(b)
Concentration of B 2•
l(c)(i)
One.
l(c)(ii)
Zero.
l(d)
Any value in the range 0.0041-0.0044.
l(e)(i)
l(e)(ii)
k = Rate/[A2]. For experiment 1, k = 0.00141/0.1 = 0.0141s·1.
l(f)
Slow step: A2 -> 2A.
Faststeps: B2 ->2B.
2A. + 2B·-> 2AB
CAPE Chemistry June 2004 U2 P2 Q2 - Answer
2(a)
Ligand exchange occurs because one system is more stable. CO replaces 02
in complex therefore haemoglobin loses its oxygen carrying capacity and
body becomes starved of oxygen.
2(b)(i)
-Original pale blue solution.
-pale blue ppt. formed.
-ppt. dissolves.
-dark blue solution formed.
2(b)(ii)
CAPE Chemistry June 2004 U2 P2 Q3 - Answer
3(a)(i)
NO/N02•
3(a)(ii)
HN0 2 or HN03 .
3(a)(iii)
NH3•
3(a)(iv)
2NO + H20 + ½02 --4 2HN0 2•
3(b)(i)
Concentrations 1 to 4 respectively are: 1.33, 1.46, 1.53 and 1.66.
3(b)(ii)
Mean= (1.33 + 1.46 + 1.53 + 1.66)/4 = 1.50 mg/L.
CAPE Chemistry June 2004 U2 P2 Q4 - Answer
4(a)
Ksp= [Ag+] [Cl·].
4(b)
[Ag+]= 1/1000 x 10· 3= 1.0 x 10·6 mol dm·3. [Cl·]= 1.0 x 10·3 mol dm·3.
Ion concentration [Ag•] [Cl·]= 1.0 x 10·6 x 1.0 x 10·3= 1.0 x 10·9 mol2 dm·6 .
Ion product> Ksp· Precipitate of silver chloride forms.
4(c)(i)
No. of moles H· calculated from results of titration.
No of moles OH·= No. of moles H·.
[OH·] can be calculated.
[Ca2•] = ½[OH·] and Ksp= [Ca2•] [OH·]2.
4(c)(ii)
Calculation of no. of moles H·: (11.5 x 0.1)/1000= 1.15 x 10· 3.
Calculation of [OH·]= (1.15 x 10·3 x 103)/25.0= 0.046 mol dm·3 .
Calculation of [Ca 2·] = 0.046/2= 0.023 mol dm·3•
Ksp= 0.023 x (0.046)2= 4.86 x 10·5 mol3 dm·9.
4(d)
-White ppt. of Ca(OHh forms.
-Common ion effect.
-Addition of cone. CaCl 2 increases the cone. of ca2•.
-Ion product exceeds solubility product.
-Solubility of Ca(OHh decreases.
CAPE Chemistry June 2004 U2 P2 QS - Answer
5(a)
Macroscopic properties are constant under the stated conditions of temp.,
press. and initial cone.
Microscopic processes continue but are in balance.
5(b)
5( c)(i)
Brown colour intensifies.
5( c)(ii)
Fon,vard reaction is endothermic. Increasing temperature shifts equilibrium
position to right to remove the added heat.
5(c)(iii)
Colour fades/lightens.
5(c)(iv)
Equilibrium position shifts to left. Increased pressure favours smaller volume.
Smaller no. of moles of gas on left = smaller no. of moles.
5(d)(i)
No effect on K"
5(d)(ii)
Pressure has no effect on KJan equilibrium constant is constant at a
particular temp/temp has not changed. Although equilibrium position shifts
to right, [N02] and [N204 ] have both increased.
5(e)
= (0.1- 0.035) = 0.065 mol.
Equilibrium amt N 02
= 2 x 0.065 = 0.13 mol.
K, = (0.13)2/(0.035) = 0.48 mol dm·3.
CAPE Chemistry June 2004 U2 P2 Q6 - Answer
6(a)
Pattern in variation of oxidation numbers 1 2 3 4 5 6 7 or 1 2 3 4 5 4 3.
Correct names. Explanation: the metal elements form ionic bonds with 0 2 by
losing the no of electrons equivalent to the group no. and the non-metallic
elements share the no. of electrons in the outermost shell.
6( b)(i)
6( b)(ii)
6( c)(i)
Metal chlorides are expected to be neutral.
6( c)(ii)
AICl3 is acidic due to high charge density of Al 3* 10n, an H* 10n from
hydrated ion complex is easily lost.
6(d)(i)
Across the period from Na to Ar, the oxides change from basic to amphoteric
to acidic. This occurs as the nature of bonds change from ionic in giant ionic
lattice structures to ionic/covalent to covalent discrete molecules.
CAPE Chemistry June 2004 U2 P2 Q7 - Answer
7(a)(i)
-Bonding- group IV element sp3 hybridized.
-Overlap of hybrid and atomic orbital of chlorine forms the sigma covalent
bond.
-Oxidation state of +4 is too polarizing to be ionic.
7(a)(ii)
Explanation - related to strength of bonds and stability of +2 state. Covalent
bonds of lead are weaker. +2 state of lead more stable.
7 (b)(i)
7(b)(ii)
-Bonds in SiCl4 are polar.
-Silicon is electron deficient.
-Silicon acts as a site for attack by lone pair on oxygen atom of water
molecule.
7(b)(iii)
Polymerization yields hydrated silica.
7 (c)(i)
-Silica has a giant structure.
-Strong covalent bonds.
-Silicon oxide has semi conductor properties.
-Electron mobility increases with temperature.
7(c)(ii)
Ceramic would be of lower heat and corrosion resistance. Bonds are more
susceptible to distortion.
CAPE Chemistry June 2004 U2 P2 Q8 - Answer
8(a)
(i): Carbon anode. (ii): Carbon cathode. (iii): Cryolite (Na3AIF6)/Al20 3.
(iv): Molten Al tapped off here.
8(b)
Lower the melting point of the Al203 (2015 °c -+ 600 ° C). Energy saving
strategy.
8(c)
It gets oxidized to CO or CO2•
8(d)(i)
Al.
8( d)(ii)
8(e)(i)
8(e)(ii)
8(f)
Pots and pans -Al as a thermal conductor. Any 1 of the 3 remaining -Al as a
thermal insulator.
Explanation: Structure of Al metal 1,vith 3 mobile electrons makes it a good
thermal conductor, increased kinetic energy of electrons. As a thermal
insulator, Al reflects heat inward thereby keeping baby warm or reflects heat
from surface keeping firemen cool or reflecting car headlights.
CAPE Chemistry June 2004 U2 P2 Q9 - Answer
9(a)(i)
-Vaporization of crude.
-Temperature gradient in column.
-Bubble caps offer surface for condensation.
-Fractions are tapped off.
-More volatile fractions ascend column.
9(a)(ii)
-Major source of fuels.
-Feedstock for petrochemical industry.
-Industrial solvent.
-Generally furnished starting materials that chemical industries and
production of structural and engineering materials.
9(b)
-Concerns due to exposure to chemicals increase in incidence of work related
cancers.
-Environmental degradation from chemical pollution e.g. air quality, water
quality.
9( c)(i)
Chemical change initiated by absorption of UV radiation.
9( c)(ii)
Natural biodegradable of plant or animal material, gasoline stations or
vehicular emissions.
9(c)(iii)
During peak traffic period high activity due to% of airborne hydrocarbons.
CAPE Chemistry June 2005 U2 P2 Ql - Answer
l(a)(i)a)
If small quantities of lactic acid are produced, then H· ions are removed by
reaction with Hco3- and pH does not change appreciably.
l(a)(i)b)
If large quantities of lactic acid are produced then the blood loses its buffer
capacity and the pH of the blood falls.
l(a)(ii)
H· ions from H2C0 3 are used to neutralize the excess alkalinity and CO2
produced in the tissues dissolves in the plasma to produce H2C03 .
l(b)
K0 value used to determine the concentration of salt and weak acid needed.
Known mass of salt and weak acid obtained. Solution of known
concentration prepared mixed. Salt dissolved in solution of known acid
concentration. pH meter is used to confirm pH of buffer.
CAPE Chemistry June 2005 U2 P2 Q2 - Answer
2(a)
-Put equal volumes of water into 4 test tubes.
-Weigh equal quantities of each white solid.
-Shake/stir.
-The one that dissolves completely is MgS04•
-To identify the others, filter and weigh solids (or allow test tubes to settle
and measure height of insoluble component).
-From least to most Ca -> Ba.
2(b)(i)
Dim light/dark.
2(b)(ii)
Highly exothermic possibly burnt.
2(b)(iii)
CAPE Chemistry June 2005 U2 P2 Q3 - Answer
3(a)(i)
Carbohydrate content.
3(a)(ii)
Flavour and colour of product.
3{a)(iii)
-Distil equal volumes of the two beverages.
-Measure volume of alcohol yielded as distillate at the boiling point of
ethanol.
3(b)(i)
Stimulates the central nervous system.
3(b)(ii)
-Abuse impacts negatively on health and therefore on the health care
system.
-Alcohol abuse can lead to depression and a decrease in productivity.
-Increase in antisocial behaviour becomes evident in society.
-Breakdown in family structure.
CAPE Chemistry June 2005 U2 P2 Q4 - Answer
4(a)
-According to Bronsted-Lowry theory an acid is a proton donor.
-Acidity due to COOH group.
-Molecule dissociates to yield H· ions due to polarization of the carboxyl
group.
-Classified as a weak acid due to incomplete dissociation.
4(b)(i)
HA.=H•+A·.
pH= -log [H•].
[H·]= antilog-3.26= 5.50 x 10·4 mol dm·3.
4(b)(ii)
[HA]= 2.0 x 10·4 _ 5.50 x 10·4= 1.95 x 10·2 mol dm·3.
4(b)(iii)
Ka= ((5.50 X 10·4)2/1.95 X 1 0 2· )= ((3.03 X 10·7)/(1.95 X 10·2 ))
K a= 1.55 x 1 0 5· mol dm·3.
4( c)(i)
Ka increases. pH decreases. More molecules undergo cleavage of the COOH
bond.
4(c)(ii)
Weak base unsuitable. Solution buffered as titration proceeds. No suitable
indicator.
CAPE Chemistry June 2005 U2 P2 QS - Answer
5(a)(i)
Muscular stimulation due to electrical discharge/muscle tissue contained a
conducting fluid.
5(a)(ii)
Named metals e.g. Zn, Cu, Ag.
Electrolyte.
Lamp or ammeter.
5(a)(iii)
Voltmeter
.
.
M
..
,-
·�
,,
Solution containing M'
(1 mol d m 3· ) 2 5 •c.
-
N
Salt bndge
SOiuti on containing N"
(1 mol dm·•) 25 ' C .
5(b)(i)
Ozone is a more powerful oxidizing agent than Cl2 . EG is more positive.
5(b)(ii)
E,en = 2.08-1.56 = + 0.52 V. Net positive value so the reaction is feasible.
5(c)(i)
Cathode: Pb02(s) +4H·(aq) + 2e·-> Pb2+ + 2H20. -(1)
Anode: Pb(s) -> Pb2+ + 2e·. -(2)
5(c)(ii)
From (1) EG = 1.47 V . From (2) EG = + 0.13 V . E,eu = 1.47 +0.13 = 1.60 V .
5(d)
The useful byproduct is water.
CAPE Chemistry June 2005 U2 P2 Q6 - Answer
6(a)
In cold alkali a mixture of halide (Cl·) and halate(I) C l o · are formed as
Cl 21g) + 2NaOH1aq) ....... NaC11 aq) + NaCIO (aq) + H20( )· By raising the temperature to
I
70 °c the rate of disproportionation of Clo· increases as the halate(V) ions
are formed as 3CI0·1aq) ....... 2Cl·1 aq) + CI03•1aq)·
6(b)
Cl 2, Br2 and 1 2 react with s20/· to give S40&2 · as all EG values are> +0.09 V
giving a net positive EG value e.g.
Cl2(aq) + 2e· ;= 2Cl·1aq)
2S 203 2 · s= S40&2 · + 2e·
EG = +1.36
EG = -0.09
Net EG = +l.25 V .
However, only Cl 2 and Br 2 can react with S 20 / · to produce so32 · because
these are stronger oxidizing agents more positive EG value than 1 2. A net
negative EG is obtained for the 1 2 ....... so/· system so the reaction is not
feasible as
l 2(aq) + 2e· s= 21· (aq)
s 20 3 2· +GOH·.= 2so/· + 3H20 + 4e·
E = -0.04 V.
6(c)
AgCI-white precipitate.
AgBr-cream precipitate.
Agl -yellow precipitate.
EG = 0.54 V
EG = -0.58
CAPE Chemistry June 2005 U2 P2 Q7 - Answer
7(a)(i)
Water molecules form a complex with the cu2• ions. The coordination of the
cu2• with the water ligands causes a split in the energies of the d orbitals.
Electrons in the lower energy orbitals absorb energy in the visible region of
the EMR. The blue colour is the complement of that absorbed.
7(a)(ii)
The c1· ions displace the H20 molecules in the blue [Cu(H20)6]2• complex. The
complex formed [CuCl4] 2 · is yellow. The green colour is due to a mixture of
the blue and yellow complexes.
moles=
ratio=
7(b)(ii)
C
Ni
7(b)(i)
CN
CN
36.1/58.7
0.615
�
�
CN
N
29.5/12
2.46
4
34.4/14
2.46
4
2-
CN
7 (c)(i)
Haemoglobin unit has a central Fe2• ion that coordinates with 0 2 molecules
for transport to tissues. The CO molecules form a complex with a higher
Ks,ab and 02 is displaced.
7(c)(ii)
Using
s=
s=
Cr20/· + 14H•+ Ge· 2Cr3• + 7H 20
Mn04·+ 8H•+ 5e· Mn 2• + 4H 20
E�= +l.33
E�= +l.52
Cr is readily reduced from +6 to +3 as indicated by the positive E� value.
Mn is also readily reduced from +7 to +2.
CAPE Chemistry June 2005 U2 P2 Q8 - Answer
8(a)(i)
-50 2 obtained from roasting of sulfide ores or burning of sulfur.
-0 2 from air.
- 2502(g) + 02(g) s= 250 3(g)·
-Gaseous reactants combine on surface ofV2 05 catalyst.
-50 3 formed absorbed in concentrated H2504.
-Oleum H25207 diluted to yield H 2504•
8(a)(ii)
50 2 is an acidic oxide. Rain becomes acidic due to absorption of acidic gas by
water. 50 2(g) + H 20 11l -+ H 250 3(aq)·
8( b)(i)
CaC031sl + 5021g) -+ Ca503(sl + C021g)· CaC03 neutralizes 502 resulting in
production of a much weaker acid gas.
8( b)(ii)
Volume of 502 present
At R T
. .P . moles 502
=1 m3
=1 X 10 3 dm3
= (1 x 103/24) =41.7 moles
Massof502
=41.7X64
= 2.67X10 3 g
= 2.67 kg.
Mass CaC03 required
=0.100 x (2.67/0.0064)
= 4.17 kg.
CAPE Chemistry June 2005 U2 P2 Q9 - Answer
9(a)(i)
-Ozone in the stratosphere offers protection from harmful UV radiation.
-Exposure can lead to development of skin cancer and cataracts.
-Decrease in photosynthetic activity can also occur, decreasing agricultural
production.
9(a)(ii)
Volatile, non flammable and chemically inert.
9(b)(i)
NH 3 is a pungent gas and an irritant.
9(b)(ii)
CCl2 F2 hf
---> • CCIF2(gl + Cl• (gl (1)
CJ' acts as a catalyst and can react with more ozone.
9( c)(i)
9(c)(ii)
-Principle of dynamic equilibrium.
-The concentration of ozone shows no net change.
-The rate of formation is equal to the rate of destruction.
CAPE Chemistry June 2006 U2 P2 Ql - Answer
l(a)(i)
Effect on rate: Rate doubles. Reason: Reaction is 1st order w. r.t. [H 2(g)].
l(a)(ii)
Effect on rate: Rate x ¼. Reason: Reaction is 2nd order w. r.t. [NO(g)].
l(a)(iii)
Effect on rate: Rate x 27.
l(b)(i)
Vo.lw.
Of
•,t,>
0o1l•cte4
l(b)(ii)
Pressure measurement.
l(b)(iii)
the apparatus must be sealed and the volume of gas evolved must not
exceed the maximum volume which could be held in the syringe used for
collecting and measuring the volume of gas.
CAPE Chemistry June 2006 U2 P2 Q2 - Answer
2(a)(i)
Cl2(aq) + NaOH (aq) -4 NaCl (aq) + NaCIO (aq) + H20 (ll·
2(a)(ii)
+7.
2(b)(i)
Reddish brown colour.
2(b)(ii)
White ppt. soluble in NH 3(aq )·
2(b)(iii)
Violet vapour.
2(c)
2Br· (aq) + Cl 2(aq) -4 2Cl ·(aq) + Br2(,l·
2(d)
Use of fume cupboard.
CAPE Chemistry June 2006 U2 P2 Q3 - Answer
3(a)
Source: air. Process: fractional distillation.
3(b )(i)
High pressure.
3(b)(ii)
3(c)
Use of a catalyst. Decrease the time b y lowering the energy required for the
reaction.
3(d)
Low temperature and high pressure.
3(e)
Red litmus.
CAPE Chemistry June 2006 U2 P2 Q4 - Answer
4(a)(i)
If a system in equilibrium is subjected to a change processes will occur which
will tend to counteract the change imposed.
4(a)(ii)
Microscopic (molecular scale) processes continue but these are in balance,
and, macroscopic properties are constant.
4(a)(iii)
KP =
(pS03)2
(pS02)2 (P02)
4(a)(iv)
The equilibrium position lies well to the right. There is a much greater
concentration of products relative to reagents.
4(b)(i)
Equilibrium position shifts to the left. Endothermic reaction favoured
according to Le Chatelier's Principle (to nullify the effect of change). Volume
of reactants increases and volume of product decreases.
4(b)((ii)
Equilibrium position shifts to the right. Reduction in volume (reduction in
number of moles of gas) favoured. Reactant concentration decreases and
product concentration increases.
4( b)(iii)
No overall effect on position of equilibrium. The effect on both forward and
reverse reactions are the same.
4(c)
Initial partial pressure of S02(gl = mole fraction x total pressure = ½ x 5 =
3.33 atm.
Initial partial pressure of 02(gl = ½ x 5 = 1.67 atm.
For pS03 to be 3 atm. => of S02(gl had reacted and 1.5 atm. 02(gl had
reacted using mole ratios from equation.
Partial pressures of gases at equilibrium:
pS02 = 3.33-3 = 0.33 atm. p0 2 = 1.67 -1.5 = 0.17 atm.
KP = (3)2/((0.33)2(0.17)) = 9/0.0185 = 486.49 atm.
CAPE Chemistry June 2006 U2 P2 QS - Answer
5(a)
A buffer solution resists pH change on addition of small amounts of acid or
alkali.
5(b)(i)
To maintain a pH of 7.
5( b)(ii)
co 32 • reacts with excess H· ions and HC03• reacts with excess OH· ions.
5( c)(i)
Amino acids contain both acidic and basic groups. -NH2 group can react with
excess H· ions. -COOH group can react with excess OH· ions.
5(c)(ii)
Food processing.
5(d)(i)
M, of CH3COONa= 82. 6.56/82= 0.08 mol dm·3.
5(d)(ii)
K.= [H·J [CH3c o o J· so 1.8 x 10·5= [H·J x 0.08 and [H·] = 1.8 x 10·5 x 0.02
[CH3COOH]
0.02
0.08
so [H·] = 4.5 x 10·5 mol dm·3. Assumption is that the concentration of
CH3Coo· ions is equal to the concentration of CH3 COONa.
5(e)(i)
Initial concentration of CH3COONa= 0.08 mol dm·3 .
Concentration aher adding= 0.08 + 0.005 NaOH= 0.085 mol dm·3•
Initial concentration of CH3COOH= 0.02 mol dm·3•
Concentration aher adding NaOH= 0.02 - 0.005= 0.025 mol dm·3.
5(e)(ii)
pH of the new solution does not differ significantly from that of the original
solution. The ethanoic acid present in the buffer reacts with the OH· ions,
thus decreasing its concentration.
CAPE Chemistry June 2006 U2 P2 Q6 - Answer
6(a)(i)
Bonding changes from covalent o ionic. Acid/base character changes from
neutral to amphoteric. Character of element changes from non-metal to
metal. Stability of covalent bonds decreases with increase in atomic radius.
Basic character increases down the group due to the increase in the ionic
nature of the oxide.
6(a)(ii)
CCl4 is unreactive with water. SiCl4 to PbCl4 are hydrolysed. For SiCl4 to
PbCl4, d orbitals are available for coordination with water molecules.
6(b)(i)
J: Pb02 . K: PbS04 -white precipitate. L: 0 2 - relights a glowing splint. J -is an
oxidizing agent. M: Mn04·(aq) -purple colour. 0: PbCl2.
6( b)(ii)
P is tin(II) chloride/SnCl2 or tin ions. P is a reducing agent since it takes
orange Cr 2072• to green Cr 3• removes oxygen from nitrobenzene and
replaces it with hydrogen.
6(b)(iii)
The yellow/brown solution of Fe3• goes to green. Fe3+ -> Fe2•.
CAPE Chemistry June 2006 U2 P2 Q7 - Answer
7(a)(i)
A: giant ionic lattice. B: simple molecular. C: giant covalent/macromolecular.
D: giant metallic. E: simple molecular.
7(a)(ii)
A: Al203. 8: S. C: Si or Si02• D: Al or Mg. E: PCl5 or PCl 3.
7(b)(i)a)
Vigorous reaction with cold water. H2 gas evolved. Alkaline solution formed.
7(b)(i)b)
Reacts with steam. Reacts slowly at room temperature. Basic oxide formed.
H2 gas evolved. Burns with bright flame.
7(b)(i)c)
Dissolves in water. Acid solution formed. Greenish yellow/yellow solution
formed.
7(b)(ii)
CAPE Chemistry June 2006 U2 P2 Q8 - Answer
8(a)
PVC: electrical insulation. PET: roofing tiles. -C-C and C-CI bonds are strong.
8( b)(i)
Breakdown of polymer into monomeric units.
8(b)(ii)
HCI, NaOH. The ester linkages are hydrolysed to release the monomers (acid
and alcohol.)
8( c)(i)
A n organic solvent e.g. dichloromethane.
8( c)(ii)
To increase the surface area for action of solvent.
8( c)(iii)
Protective eye wear and gas masks.
8(c)(iv)
Industrial rain coats.
8(d)
Reduction in plastic sent to landfills. Heat produced from process can be
recovered. Ash generated still requires disposal. CO2 combustion product
contributes to greenhouse effect. Chlorinated products of incineration
provide health hazard.
CAPE Chemistry June 2006 U2 P2 Q9 - Answer
9(a)(i)
Different boiling points of components. Fraction richer in lower boiling point
components rise up column and collected at the top. Fraction richer in
higher boiling point components condenses and runs off at different levels in
the column.
9(a)(ii)
Larger alkanes are broken down to smaller alkanes and alkenes suing high
temperature and/or catalyst.
soo·c
9(b)
C 10H 22
9(c)
Fractional distillation is necessary to provide useful components for fuel and
the petrochemical industry.
9(d)
Oil spillage kill birds, fishes, corals and ruins beaches.
9(e)(i)
Cumulative poison. Causes depression.
9(e)(ii)
Incomplete combustion of fuel for CO. High temperature promotes a
reaction between N 2 and 02. N 2(gl + 202(gl -> 2N0 2(g) ·
9(e)(iii)
Catalytic converter. Electric cars.
Al203
CsH1s + C2 H4,
CAPE Chemistry June 2007 U2 P2 Ql - Answer
l(a)(i)
Br2(aq) + Cl2(aq) s= 2BrCl(aq)·
l(a)(ii)
K, =
l(b)(i)
K eq = (0.0546)2/((0.0389)(0.0111)) =6.90 so equilibrium is atained.
l(b)(ii)
Accurate analysis of the concentration of the reactants and products and
constant environmental conditons i.e. pressure and temperature.
l(b)(iii)
The experiment should b e conducted in fume cupboard.
l(c)
If the reaction mixture is exposed to a temperature greater than room
temperature.
[BrCl]2
[Br2] [Cl2]
CAPE Chemistry June 2007 U2 P2 Q2 - Answer
2(a)(i)
HCI.
2(a)(ii)
2(a)(iii)
(White) choking fumes of gas.
2(a)(iv)
2(a)(v)
Smell of rotten egg/foul smell.
2(b)
Increasing polarizability down the group.
2(c)
Oxidation state change: Cl2 -+ NaCl + NaCIO
0 -+ -1 and +l.
CAPE Chemistry June 2007 U2 P2 Q3 - Answer
3(a)(i)
Organic phosphate pesticides.
3(a)(ii)
Green algal bloom.
3(a)(iii)
Eutrophication.
3(b)(i)
Concentrations are 0. 70, 0.59 and 0.62 for samples 1, 2 and 3 respectively.
3( b)(ii)
Average [P043·] = (0.70 +0.59 + 0.62)/3 = 1.91/3 = 0.64 mg dm· 3 to 2 s.f.
3( b)(iii)
[P043 ·] is within acceptable limits.
CAPE Chemistry June 2007 U2 P2 Q4 - Answer
4(a)
Buffer solution - regulates p H by responding due to small additions of acid
or alkali.
4(b)(i)
H· ions react with anions from the salt. Concentration of H· decreased.
4(b)(ii)
OH· ions react with weak acid molecules. OH· ions neutralized.
4(c)(i)a)
Equilibrium shifts to produce H2C03. H2C03 dissociates to increase H·
concentration in blood.
4( c)(i)b)
Deep rapid breathing clean lungs of C02(g)· Equilibrium shifts to the left to
release CO2 from the blood. H· ions reabsorbed as equilibrium shifts to the
left.
4( d)(i)
K.= [H·] [CH3 Coo·]
[CH 3COOH]
4(d)(ii)
[H•]= Ka [CH 3COOH]
[CH3Coo·]
= (1.75 x 10·5 x 0.025)/0.010= 4.4 x 10· 5 mol dm·3.
pH= -log [H•]
pH= -log (4.4 x 10·5)
pH= 4.4.
CAPE Chemistry June 2007 U2 P2 QS - Answer
5(a)(i)
Using expts 4 and 5, doubling of [ RX]
=> doubling of rate. 1:1
proportionality=> 1st order.
5(a)(ii)
Using expts 1 and 2, doubling of [NaOH] => doubling of rate. 1:1
proportionality=> 1st order.
5(b)
Overall order= m + n = 2. Rate equation= k [R-X] [NaOH].
5( c)
Using expt 1
units=> dm3 mo1·1 s·1.
5(d)
The mechanism is SN 2.
R
""I
H O - _,,,c
H.,...
R
:::::-
X
'H
-
1�
HO···· C•••• X
H/ ',,H
- HO-C
H
./
.,..
','H
+ X
5(e)(i)
Higher temperature causes increase in: Average kinetic energy of particles,
number of collisions with Eact (activation energy), and, rate.
5(e)(ii)
Higher concentration results in: Number of collisions per unit time increases,
probability of favourable collision increases, and, rate increases.
5(e)(iii)
Increased surface area results in: Probability of favourable collisions
increase, and, rate increases.
CAPE Chemistry June 2007 U2 P2 Q6 - Answer
6(a)
Going from top of table to bottom: Ionic, ionic, intermediate, covalent,
covalent, covalent.
6( b)(i)
Basic.
6( b)(ii)
Amphoteric.
6( c)(i)
6( c)(ii)
6( c)(iii)
6(d)
NaOH: pH 13/14.
HCI: pH 1/2.
6(e)
SiCl4 has a simple covalent molecule with weak intermolecular forces.
Si02 has a giant covalent structure. A large number of covalent bonds must
be broken to break down the giant lattice structure.
CAPE Chemistry June 2007 U2 P2 Q7 - Answer
7(a)
As you go down the group, electrical conductivity increases with increasing
metallic character.
7(b)(I)
Covalent.
7 (b)(ii)
Covalent.
7(b)(iii)
Intermediate.
7(b)(iv)
Intermediate.
7(c)
CO2 and Si02 - acidic (covalent structure). GeOi, Sn0 2, Pb02 - Amphoteric
(Intermediate ionic/covalent structures.)
7(d)
Increase in Ee values from Ge - Pb indicates increase in stability of the +2
oxidation state down the group.
EG/V
7(e)
4
Sn • + 2e· - Sn2•
+0.15 V
2
4+
Pb + 2e· - Pb •
+l.80 V
Cr2072. + 14H+ + Ge· - 2Cr3+ + 7H 20 +1.33 V
With Sn2• Eecell= +l.33 · (+0.15)= +l.18 V.
With Pb2+ Eecell= +l.33 - {+1.8) = -0.47 V .
-ve value, Ee,.11 for Pb2• means it will not reduce Cr 2012·.
+ve value, E Gceu for Sn2• supports the reduction of Cr2012·.
7(f)
Si has available 3d orbitals 1,vhile C does not and so CCl4 does not react with
water.
CAPE Chemistry June 2007 U2 P2 Q8 - Answer
8(a)(i)
Fractional distillation.
8(a)(ii)
Fractions are separated according to boiling point. Higher boiling point
fraction at the bottom of the fractionating column.
8( b)(i)
N0 2 is formed from nitrogen and oxygen in the air because of high
temperatures. N2(gl + 202(gl -> 2N02(g)· Sulfur comes from the fuel itself.
S!sl + 02!gl -> S02(g)·
8(b)(ii)
Both N0 2 and S0 2 will form acid rain. Acid rain destroys buildings and
vegetation.
8( c)(i)
Lead is a neurotoxin. CO will compete with 02 for haemoglobin.
8( c)(ii)
Lead comes from the antiknock agent added to gasoline. This forms PbO
when burned. CO is formed from the incomplete combustion of petrol.
8(c)(iii)
Unleaded gasoline is now available on the market.
8(d)
Port/deep harbour facilities.
Good infrastructure e.g. roads.
Isolation from residential sites.
Good po�ver supply.
CAPE Chemistry June 2007 U2 P2 Q9 - Answer
9(a)(i)
Formation of ozone:
02(g) + hv--> O(g) + O(g) and
Along with the formation of ozone, degradation also takes place to keep
levels constant naturally.
9(a)(ii)
The ozone layer protects the earth from harmful uv rays. Prevents cataracts
and skin cancers.
9(b)
Causes breathing problems.
Takes part in the formation of photochemical smog.
Destroys materials such as rubber.
Destroys vegetation.
9( c)(i)
Low flammability, relatively unreactive and low boiling point.
9( c)(ii)
Refrigerants and foaming agents.
9(d)
They have long residence times.
The regeneration of
as it destroys ozone makes it potent.
er
CAPE Chemistry June 2008 U2 P2 - Resit Ql - Answer
l(a)(i)
l(a)(ii)
Nitric acid and concentrated sulfuric acid.
l(a)(iii)
Sn and HCI.
l(a)(iv)
l(a)(v)
Br2(aq)·
l(b)(i)
l(b)(ii)
Cl
and
Cl
l(b)(iii)
N02
'°
Cl
l(c)(i)
Benzene has no substituents so the chlorine can go to any position,
l(c)(ii)
The methyl group is 2,4 directing so chlorine goes to the ortho and para
positions.
l(c)(iii)
The nitro group is 3,5 directing so chlorine goes to the meta position.
l(d)
Nitrobenzene, benzene, methylbenzene.
The nitro group is deactivating and makes the benzene ring less reactive.
The methyl group is activating and makes the benzene ring more reactive.
CAPE Chemistry June 2008 U2 P2 - Resit Q2 - Answer
2(a)
NaHC03 : Can be obtained in high purity, stable in air and non-hygroscopic.
NaOH: hygroscopic.
2(b)(i)
H2S04 + 2NaHC03 -> Na2 S04 + 2C02 + H20
2(b)(ii)
From the balanced equation, mole ratio of NaHC03 : H 2S04 is 2 : 1.
Moles of sulfuric acid used: (6 x 23)/1000 = 0.138 mol.
Moles of NaHC03 needed: 2 x 0.138 mol = 0.276 mol.
Mass of NaHC03 used: 0.276 x 84 g = 23.2 g.
2(c)
Standardize solution of NaOH,
Transfer aliquot of vinegar to conical flask using pipette.
Add two or three drops of appropriate indicator to vinegar.
Titrate vinegar with NaOH as many times as needed for accurate results.
CAPE Chemistry June 2008 U2 P2 - Resit Q3 - Answer
3(a)(i)
Acid rain formation by burning of fossil fuels, washing with detergents in
rivers and releasee of sewage in waterways.
3(a)(ii)
Unsafe drinking water and unsafe water for recreation.
3( b)(i)
Need increased crop production. Increasing population.
3(b )(ii)
Fertilizers are leached into the ground water. Run-off water from rainfall or
irrigation
enters
drains,
streams, rivers
etc.
Eventually
the
polluted/contaminated water reaches the ocean, thus polluting the water.
3(c)
Pb2•: Kl (aq) yellow ppt (Pbl 2)-+ (soluble in excess HN03).
N0 3·: Zn/NaOH(aq) - boil -+ gas evolved turns damp red litmus blue (NH3
evolved).
CAPE Chemistry June 2008 U2 P2 - Resit Q4 - Answer
4(a)
Ethanol and Na2C0 3:
Ethanol and NaOH:
No reaction.
No reaction.
Phenol and Na2C03:
Phenol and Na:
No reaction.
Sodium phenoxide and hydrogen gas.
Ethanoic acid and Na:
Sodium ethanoate and hydrogen gas.
Ethanoic acid and N aOH: Sodium ethanoate and �vater.
4(b)
The given resonance forms of the phenoxide ion sho�v that there is
delocalization of the negative charge on the oxygen into the benzene ring.
This delocalization increases the ease with �vhich a proton is lost from
phenols.
The given structure of the alkoxide ion sho�vs that there is localization of the
negative charge on the oxygen atom, and this negative charge is enhanced
by the positive inductive effect of the alkyl group to which it is bonded. This
increases the tendency of H· ions being bonded to alkoxide ions.
Therefore phenols show more acidic character than alcohols.
4(c)(i)
Members of a homologous series possess the same functional group and
successive members within a series differ by a fixed increment e.g. -CH2.
4( d)(i)
Structural isomerism refers to compounds possessing the same molecular
formula but different bonding arrangement or structure.
4(d)(ii)
H
H
H
I
I
I
H-c-c-C-O-H
I
I
I
H
H
H
H
I
H
I
H
I
H-c-c-c-H
I
I
I
O
H
H
I
H
CAPE Chemistry June 2008 U2 P2 - Resit QS - Answer
5(a)(i)
2: Ionization chamber, 3: acceleration chamber, 4: (large) magnet, 5: ion
collector (ion detector).
5(a)(ii)
To provide high energy electrons to ionize sample.
5(b)
RAM= ((79 X 50.5) +(81 X 49.5))/100 = 79.99 =80.0.
5( c)(i)
"'
C
"'O
C
::,
..0
·-
"'
"'
"'
a::
I
157
5(c)(ii)
160- 79Br-81Br.
158
I
159
160
m/e
I
161
162
CAPE Chemistry June 2008 U2 P2 - Resit Q6 - Answer
6(a)
Potato and barley.
6(b)
Sucrose is converted to glucose and fructose, then glucose is converted to
ethanol and carbon dioxide using yeast.
yeast or
zymase>
0
II
6(c)
CH 3CH20H +[OJ� CH 3C -H +H 20.
6(d)
Depression of the central nervous system and loss of inhibitions.
6(e)(i)
Job creation, wealth creation: exports, enhanced technology, and fuel
production.
6(e)(ii)
Education programmes and restricted use.
CAPE Chemistry June 2009 U2 P2 Ql - Answer
l(a)(i)
l(a)(ii)
55 ·c to 60 ·c.
l(a)(iii)
Sn/HCI.
l(b)(i)
I+
0
H-N-H + 01·1-
0
l(b)(ii)a)
Less basic.
l(b)(ii)b)
The lone pair of nitrogen forms an extended delocalized system with the
delocalized electrons of the benzene ring. This makes the electrons on
nitrogen less available for proton acceptance. Also, the ammonium ion forms
more hydrogen bonds with �vater and is therefore more stable than the
protonated phenylamine.
l(c)(i)
l(c)(ii)
Reagents: NaN02 and HCI.
Condition: temperature of< 5 °c.
l(c)(iii)
Solution of sodium hydroxide.
l(c)(iv)
Yellow.
CAPE Chemistry June 2009 U2 P2 Q2 - Answer
2(a)(i)
The stage at which the reaction is complete or the two solutions have
reacted exactly.
2(a)(ii)
The point at which there is a change in the colour of the indicator.
2(b)
A known excess of Na2C03 is reacted with BaCl2.
The remaining solution of Na 2C03 is determined by titration.
2(c)(i)
2(c)(ii)
Mole ratio of sodium carbonate to hydrochloric acid is 1: 2.
Moles sodium carbonate remaining= 0.004/2= 0.002.
2(c)(iii)
Moles of barium chloride reacted= (0.005-0.002)= 0.003.
2(c)(iv)
Concentration barium ions= 0.003 x 40= 0.120 mol dm·3 •
2(d)(i)
25 cm 3 oxalic acid is placed in a conical flask and 25 cm 3 of dilute sulfuric
acid is added.
Mixture in flask heated to approximately 80 °c.
The mixture in the flask is titrated immediately with KMn04 a q)·
(
2(d)(ii)
KMn04(aq)·
CAPE Chemistry June 2009 U2 P2 Q3 - Answer
3(a)(i)
The conversion of atmospheric nitrogen to nitrogen compounds.
3(a)(ii)
Electrical discharge in the atmosphere (lightning)- atmospheric.
Action of bacteria in plants (legumes)- biological.
3(b)
N2!gl + 02!gl--+ 2N O!g) ·
3(c)
Deforestation and power generation (fossil fuels).
3(d)
Policies are costly and could result in unemployment.
3(e)(i)
Pink.
3(e)(ii)
Concentrated aqueous sodium hydroxide.
3(e)(iii)
Colour: white. Appearance: gelatinous precipitate.
3(e)(iv)
Filtration.
CAPE Chemistry June 2009 U2 P2 Q4 - Answer
4(a)(i)a)
Compounds that differ only in the arrangement of atoms in space.
4(a)(i)b)
A carbon atom which is bonded to four different groups.
4(a)(ii)
H
8
Optical isomerism.
O
�
'-N-* -C�
a/
I "o-B
H-C-H
I
H
4(a)(iii)
H
H
I• c._
I
... ·· \ ····cooH
CH3
NH2
mirror
4(a)(iv)a)
H
0
H +
;
I
'
H -N-C-C
'o-H
I
n/
n,
H/
1
;o
I
'o-
H-C-H
I
I
H
H +
H
I
0
H/
I
'o-
;
H�-c-c
H-C-H
I
n
H
Condensation polymerization.
4(b)(ii)
+
H
H
H/
I
'o-n
H-C-H
I
Amide.
�
O H
H
H
I
;0
I II I
'-r.-c-c-N-C-C
I
'o-n
H/
I
H
0
;
I
'N-C-C
H
4(b)(iii)
H
N -C-C
H-C-H
4(b)(i)
4(a)(iv)c)
4(a)(iv)b)
H-c-n
H-C-B
H
H
I
I
CAPE Chemistry June 2009 U2 P2 QS - Answer
5(a)(i)
Molecules undergo changes in modes of vibration, and absorption of JR
radiation occurs when vibrations cause a net change in dipole moment of
molecules.
5(a)(ii)
Stretching and bending.
5(a)(iii)
Grind sample with an excess of potassium bromide to a fine texture and
press this mixture into the form of a pellet.
5(b)(i)
A: C H
- , B: 0-H and C: C=O.
5(b)(ii)
I
I
H
o
/
H-C-C
5( c)(i)
�0-H
Advantages: Cheap and uses an environmentally friendly solvent.
Disadvantage: Yield will be lower.
5(c)(ii)
Efficiency is increased due to increased surface are with increased length of
the extractor.
CAPE Chemistry June 2009 U2 P2 Q6 - Answer
6(a)(i)
Chlorine, sodium hydroxide and hydrogen.
6(a)(ii)
At the anode chlorine ions are converted to chlorine gas by oxidation:
At the cathode �vater is decomposed to give hydrogen by reduction.
6(b)
Weed killers, antiseptics and chlorinated organic compounds.
6(c)
Function: Sulfur dioxide is a reducing agent. It prevents oxidation of food
thereby retarding food spoilage.
Disadvantage: Sulfur dioxide distorts the taste of food.
CAPE Chemistry June 2010 U2 P2 Ql - Answer
l(a)(i)
Hot acidified Mn04••
l(a)(ii)
l(b)(i)
CH,
CH,
l(b)(ii)
CH,
CH,
HO
OH
CH,
Br
CH,
Br
l(c)(i)
A macromolecule formed from joining many monomers (small molecules).
l(c)(ii)
Addition.
l(c)(iii)
Poly(ethene).
l(c)(iv)
l(c)(v)
l(d)
No, because they do not have >C=C<.
H -.......,
H ,..........
l(e)
I
?
N-C-C
I
R
......... 0- H
Similarity: They both contain the amide linkage
O H
11 I
- C-N-
Difference: The amide linkages are separated by one C atom in proteins, but
in nylon 6,6 they are separated by several C atoms.
CAPE Chemistry June 2010 U2 P2 Q2 - Answer
2(a)(i)
Radiation consisting of oscillating electric and magnetic fields of energy
which can be transmitted through space.
2(a)(ii)
The distance between successive peaks of waves of radiation.
2 (a)(iii)
The number of waves passing a given point per second.
2(b)(i)
2(b)(ii)
9.5 x 107 Hz.
2(b)(iii)
Infrared radiation.
2(b)(iv)
Radio waves.
2(c)(i)
Sample of Xis accurately weighed on an analytical balance.
Sample of X weighed is dissolved in a suitable solvent and a standard
solution made, using a volumetric flask.
Sample is filtered to remove any solid particles.
Absorbance of sample solution is measured at a selected wavelength in the
U V -visible region, after the spectrophotometer had been blanked.
Ensure that sample absorbance does not exceed 1.0. In such a case, dilute
the sample and re-do the measurement.
2(c)(ii)
Ability to produce a colour when reacting with analyte.
Transparent in the U V -visible region of the electromagnetic spectrum.
2(d)(i)
2(d)(ii)
Emole = E molecule XL= 9.9 X 10·19 J X 6.02 X 10
= 600 kJ mo1·1.
23
mol·1 = 6 x 10 5 J mo1-1
CAPE Chemistry June 2010 U2 P2 Q3 - Answer
3(a)(i)
3(a)(ii)
CO2 comes from respiration and N02 from lightning storms.
3(a)(iii)
3(b)(i)
Electric power generation that relies on burning fossil fuels like coal which
can be contaminated with sulfur.
Burning of gasoline in vehicles in heavily populated areas - there is often
incomplete combustion of gasoline generation NOx.
3(b)(ii)
Destroys aquatic life and leaches soil nutrients.
3( c)(i)
Ensure container is clean.
3(c)(ii)
Pb2•: Kl
3(c)(iii)
Pb2·: yellow ppt.
N0 3·: brown ring at junction of cone. acid and rest of solution.
CAPE Chemistry June 2010 U2 P2 Q4 - Answer
4(a)
They have the same molecular formula, CgH80, but different structural
formulae- atoms are linked differently.
4(b)
Alkene and 1° alcohol.
4( c)(i)
HHHHHHH H
l
�
6-6-6
I III I
H
H H
H-6
HO
4(c)(ii)
CH
3
=C -6-H
Cis isomer:
I
H
H
"---C=CI
/
CH
z-CH
2-CH
2
3 - H-CH
f
H
"
CH
3
Trans isomer:
H
CH
20H
CH
20 H
"c=I
/
CH� -CHz-CHz-CH
2
1
CH
3
H
�
There is no asymmetric or chiral carbon therefore no optical isomerism.
4(d)(i)
Cracking is the splitting od larger alkane molecules to smaller ones together
with the formation of an alkene. This requires high temperature or moderate
temperature and a catalyst.
4(d)(ii)b)
Bubble each gas separately into brown bromine in an inert solvent. The
alkane C4H10 will have no effect and the alkene C4H8 will immediately
decolorize it.
CAPE Chemistry June 2010 U2 P2 QS - Answer
5(a)
Distinguish between molecules of similar Mr
Prediction of possible identity of simple organic molecule based on
fragmentation pattern.
Determination of the number of C atoms in the molecule based on M and
M+l peaks.
5( b)(i)
M = 78and M+l =79.
5(b)(ii)
5(b)(iii)
The name of the compound is 2-chloropropane.
5(c)
A,(RAM) =((75.77 /100) x 34.97) +((24.23/100) x 36.96) =26.5 +8.95 =35.45
5(d)
Separation of dyes and pesticide analysis.
CAPE Chemistry June 2010 U2 P2 Q6 - Answer
6(a)
Fractional distillation of liquid air.
6( b)(i)
Hydrolysis.
6(b)(ii)
6(c)
High pressure favours low volume and low temperature favours forward
exothermic reaction.
6(d)
The Haber process uses the compromise pressure of 250 atm and moderate
temperature of 550 °c with Fe catalyst to increase rate.
6(e)
N0 3·: Source - fertilizers. Polluting
bloom/lowers 02 for marine life.
effect -
Eutrophication/algal
P043 ·: Source- Detergent. Polluting effect- Eutrophication/algal bloom.
Pb2·: Source - Car batteries. Polluting effect - Carcinogenic, cumulative
poison.
CAPE Chemistry June 2011 U2 P2 Ql - Answer
l(a)(i)
No responses required.
l(a)(ii)
Test: Add a few drops of acidified potassium permanganate.
Observation: Colour change from purple to colourless with A.
l(a)(iii)
Observation: Silver mirror with 8.
l(a)(iv)
Test: Add a few drops of 2,4-DNP.
Observation: Yellow/orange precipitate �vith B.
l(b)(i)
+
(l)(b)(ii)
Unimolecular nucleophilic substitution.
l(c)(i)
Step 1: sulfuric acid (concentrated).
Step 2: sulfuric acid (concentrated) and water.
Step 3: acidified potassium permanganate.
l(c)(ii)
Oxidation.
Br
CAPE Chemistry June 2011 U2 P2 Q2 - Answer
2(a)
Accuracy refers to how close a measurement is to the true or accepted value
of the quantity being measured. Precision refers to measurements of the
same quantity. Both accuracy and precision refer to how closely t\vo
measurements agree with each other.
2(b)
Pipette, burette, volumetric flask.
2(c)(i)
Student 1: 0.065, Student 2: 1.1.5, Student 3: 3.46, Student 4: 0.082.
2(c)(ii)
Student 1: precise, not accurate. Student 2: accurate, not precise.
Student 3: not accurate, not precise. Student 4: accurate and precise.
2(d)
Weigh a beaker on an analytical balance.
Condition the 10 cm3 pipette.
Fill the pipette to the mark with distilled water room temperature.
Transfer water from pipette to beaker.
Weigh beaker and water.
Record \Veight obtained in table of results.
Repeat steps above until consistent mass values are obtained.
CAPE Chemistry June 2011 U2 P2 Q3 - Answer
3(a)
Heating crude oil, which is a mixture of hydrocarbons, separates components
according to volatility. The components are collected by the use of a
fractionating tower with the more volatile collected first.
3(b)(i)
Breaking up of large molecules into smaller ones.
3( b)(ii)
Rearrangement of atoms in molecules to form new structures.
3(c)
3(d)
In the production of pharmaceuticals, insulators and plastics.
3(e)(i)
A and B are yeast and starch.
3(e)(ii)
Effervescence would be observed in the conical flask and a white precipitate
would be observed in the beaker.
3(e)(iii)
The yeast would be killed.
CAPE Chemistry June 2011 U2 P2 Q4 - Answer
4(a)(i)
Compounds with he same molecular formula but different structural
formula.
4(a)(ii)
Chain isomers, e.g.
H
I
H- C-H
H H
I
I
I
H
I
H
H
I
I
H-C-C-C-C-H
I 7
H
H
I
H -C-c-c- H
I
I I
H H
I
H
I
H
H
Positional isomers, e.g.
H HH
H HH
H-C·C·C·O·H
H·C·C·C ·H
H HH
H OH
I
I
I
I
I
I
I
I
I
I
I
I
I
H
10 cm 3 35cm3
mole ratio
1
3.5
20 cm3
2
:. X = 2
x=y/4=3.5
y/4=3.5-2
y/4=1.Sand y=6
The molecular formula of the hydrocarbon is C2H 6•
CAPE Chemistry June 2011 U2 P2 Q4 - Answer
4(c)
In order of increasing acid strength: alcohols< phenols< carboxylic acids.
Alcohols are weakest since the alkyl group releases electron density,
resulting in a localization of the negative charge on the oxygen, making it less
likely to release a proton.
Phenols are stronger acids than alcohols but still weakly acidic. This is due to
the electron withdrawing effect of the phenyl ring which causes the
phenoxide ion to be stabilized by resonance.
Carboxylate ions are resonance stabilized:
R
oe
I
.,.C-::::­
0
These ions are less likely to hold on to a proton since the highly
electronegative oxygen pulls electron density towards itself and away from
the C H
- bond.
CAPE Chemistry June 2011 U2 P2 QS - Answer
S(a)(i)
S(a)(ii)
Electronic absorption between molecular energy levels.
--�------- a•
•
---�
- ----- ..
S(b)(i)
This is a group of atoms in a molecule responsible for the absorption of
electromagnetic radiation.
S(b)(ii)
Conjugated system of double bonds or benzene structure.
S(b)(iii)
Using A= Eel
1.2= 288X 1 X C
c = 1.2/(288 x 1) = 4.2 x 10·3 mol dm·3.
S(c)
The absorbance of standard solutions is obtained.
A calibration curve of absorbance vs. concentration is plotted.
Using the calibration curve, the concentration of the unknown solution can
be found.
5(d)
Determining the amount of glucose in blood.
Determining the amount of urea in blood.
CAPE Chemistry June 2011 U2 P2 Q6 - Answer
6(a)
Nitrifying bacteria, proteins, soil nitrates and nitrogen oxides, respectively.
6(b)
Deforestation and burning of fossil fuels.
6(c)
Formation of ozone:
Oxygen is dissociated into atoms by UV radiation.
hµ
•
02!gl -+ O!gl + 0 (gl
Ozone is formed by the reaction of oxygen atoms and molecular oxygen.
Breakdown of ozone:
Ozone decomposes to molecular oxygen and oxygen atoms by lower energy
UV radiation.
Ozone molecules and oxygen atoms produce two molecules of oxygen.
CAPE Chemistry June 2012 U2 P2 Ql - Answer
l(a)(i)
H H H H
I
I
I
I
H-C-C-C-C-OH
I
I
I
I
H H H H
l(a)(ii)
H
I
I
H
I
I
H
H
I
I
H
I
I
l(a)(iv)
H-C-C-C-C-H
H
I
I
I
I
H
H
2-methyl propan-1-ol
OH H
I
I
CH3H
H-C-C-C-OH
butan-1-ol
l(a)(iii)
H
H
butan-2-ol
H
CH 3H
H
OH H
I I I
H-C-C-C-H
I I I
2-methyl propan-2-ol
l(b)
Structural isomerism.
l(c)
Stereoisomerism (optical).
l(d)(i)
butan-1-ol
purple to colourless.
l(d)(ii)
2-methyl propan-1-ol
purple to colourless.
l(d)(iii)
butan-2-ol
purple to colourless.
l(d)(iv)
2-methyl propan-2-ol
no colour change.
l(e)
yellow to green.
CAPE Chemistry June 2012 U2 P2 Q2 - Answer
2(a)
Forensic testing.
2(b)
Obtain a beaker and cover its bottom with a solvent (mobile phase).
Apply the mixture to be separated as a spot at a short distance from one end
of the TLC plate.
Dip the end of the plate below the spot into the solvent (make sure the
solvent level does not cover the spot).
Allow the solvent to climb the plate until it nears the top of the TLC plate.
2(c)
Stationary phase: this is a solid and is the material which holds the solute on
the plate.
Mobile phase: this is a liquid and it carries the solute along the stationary
phase on the plate.
2(d)
Silica and alumina.
2(e)(i)
For A, Rt = (2.4/6.8) = 0.35 and for B, Rt = (5.1/6.8) = 0. 75.
2(e)(ii)
A is more attracted to the stationary phase and B is more attracted to the
mobile phase.
2 (e)(iii)
Nature of stationary and mobile phases.
CAPE Chemistry June 2012 U2 P2 Q3 - Answer
3(a)
Reuse newspaper as absorbing material.
Recycle paper by making papier mache.
Reduce use of paper by using email for communications.
3(b)
Waste streams contaminate and degrade land, and radioactive gases - free
radicals-released into air - carcinogenic.
3(c)
Carbon dioxide and hydrogen sulphide.
3( d)(i)
Sucrose is converted to glucose and then to ethanol by enzymes produced by
yeast.
3(d)(ii)
Compound 1: ethanoic acid. Compound 2: ethanol.
Type of reaction: oxidation.
CAPE Chemistry June 2012 U2 P2 Q4 - Answer
4(a)(i)
4(a)(ii)
CH 3 H
I
I
I
I
CH3 H
4(a)(iv)
H
I
I
I
I
I
CH3
\
C=O + H 20 + CO2
I
H 3C-C - C-H
OH
I
Br Br
OH Br
CH3
I
H 3C-C-C-H
H 3C-C - C-H
4(a)(iii)
I
CH 3
OH
4(b)
It does not.
4(c)
Both groups attached to each carbon of the double bond are the same, and
there are no common groups attached to the carbon atoms of the double
bond.
4(d)
Br
H 3C
\
H
I
C=C +
\
I
H
H C
H 3C
H -Br �
3
4(e)
Electrophilic addition.
H
I
\0
C - C-H �
I
H C
3
I
H
Br
I
H
I
H 3C-C - C-H
I
I
CH 3 H
CAPE Chemistry June 2012 U2 P2 QS - Answer
5(a)
Figure 2 is a sintered glass crucible used for filtration by suction and drying of
precipitate in oven. Figure 3 is a suction funnel and is used for filtration by
suction.
5(b)(i)
Partition coefficient (r)
5(b)(ii)
1. Temperature. 2. Solute does not react with solvents.
5(b)(iii)
Organic compounds are generally more soluble in non polar solvents that in
polar solvents like water. Solvents are immiscible. An organic compound is
partitioned between two solvents by shaking and the organic compound is
recovered by distilling of the solvent after separation of two layers.
5(b)(iv)
Concentration of Yin water= (1.6/100= 0.016.
Concentration of Yin ether= (8.4/200)= 0.042.
concentration of ester in toluene
concentration of ester in water
Partition coefficient = (0.042/0.016) = 2.625 with respect to K, ether and
water.
CAPE Chemistry June 2012 U2 P2 Q6 - Answer
6(a)
Expansion possibilities, political situation, readily accessible energy needs
and good transportation.
6(b)
One safety concern is that it is a high pressure process and there is the
possibility of explosions. The safety measure for this concern is that workers
should be fully aware of evacuation procedures. Another safety concern is
that spill of product is likely in transportation. The safety measure for this
concern is that clean up measures should be developed.
6(c)
2CO + 02-> 2C02 and 2CO + 2NO-> 2C02 + N2.
6(d)(i)
6(d)(ii)
Production of smog; adverse effects on respiratory system.
CAPE Chemistry June 2013 U2 P2 Ql - Answer
l(a)(i)
and
H H H
I
I
I
I
I
I
H-C-C-C-H
11
R'-C-0 Na+
OHOHOH
l(a)(ii)
and
H H H
I
I
I
I
I
I
0
II
H-C-C-C-H
Cl-13 - 0- C- R
OHOHOH
l(a)(iii)a)
Saponification.
l(a)(iii)b)
Transesterification.
l(a)(iv)a)
Soap making.
l(a)(iv)b)
As fuel (biodiesel).
l(b)(i)a)
Orange precipitate produced.
l(b)(i)b)
KMn04 decolourised.
l(b)(i)c)
Z is an aldehyde.
l(b)(i)d)
Z is an aromatic aldehyde.
l(b)(ii)
II
�
.0
l(c)
"
c ,H
�
C= O +
/�
.G
I
0
J
+
Cr. -> - C-0 + H
I
CN
I
-C-oH
I
CN
CAPE Chemistry June 2013 U2 P2 Q2 - Answer
2(a)
Bending and stretching vibrations.
2(b)
Laboratory identification of organic compounds and identification of
functional groups.
2(c)(i)a)
Acetone.
2(c)(i)b)
The >C=O at a wavenumber of 1715 cm·1.
2(c)(ii)a)
2-propanol.
2(c)(ii)b)
-OH at a wavenumber of 3350 cm·1.
2(c)(iii)a)
Butanoic acid.
2(c)(iii)b)
>C =O at a wavenumber 1710 cm·1 and -OH at a wavenumber of 3400 cm·1.
2(d)(i)
Grind the solid sample with potassium bromide and form this mixture into a
pellet under pressure. Insert the sample into machine and record readings.
2(d)(ii)
NaCl is transparent to infrared radiation.
CAPE Chemistry June 2013 U2 P2 Q3 - Answer
3(a)(i)
Leaching.
3(a)(ii)
Fungicides and herbicides.
3(b)(i)
Reagents: FeS04(aq); concentrated H2S04•
Observations: A brown ring forms at the interface of the two liquids.
3(b)(ii)
Reagents: Ammonium molybdate.
Observations: Yellow precipitate.
3( c)(i)
Diffusing from atmosphere and aeration from rapid movement.
3( c)(ii)
Excessive nutrients (N0 3• and Pol·) lead to algal growth followed by
bacterial decomposition on death (removal of oxygen).
3(c)(iii)
It causes corrosion in boiler pipes due to acidity due to the presence of CO2.
3(d)
Filtration and chlorination.
CAPE Chemistry June 2013 U2 P2 Q4 - Answer
4(a)(i)
NaOH
4(a)(ii)
Br
Br
HO
Br
4(a)(iii)
11
Cl - C - CH 2 - CH 3
4(b)(i)
Step I: concentrated sulfuric acid and concentrated nitric acid.
Step II: Br2 and FeBr 3.
Step 11I: Sn and concentrated HCI.
4(b)(ii)
4(b)(iii)
Electrophilic substitution.
4(b)(iv)
N0 2 is a meta director and therefore directs the incoming substituent to the
meta position.
4( c)(i)
Nitrobenzene, benzene, methylbenzene.
CAPE Chemistry June 2013 U2 P2 Q4 - Answer
4(c)(ii)
Cl
Cl
Cl
CAPE Chemistry June 2013 U2 P2 QS - Answer
5(a)
5(b)
The ratio of the solute concentrations immersed in solvents at a given
temperature.
Using the formula k= C.,,.,.Jc.,her,
:. (0.854/10)/(0.159/10)= k= 5.36.
Let the mass of acid in water= x g.
(x/10)/((1-x)/20)= 5.36 => 7.36 x =5.36
X= 5.36/7 .36= 0 .728 g.
5(c)(i)
Steam distillation.
5(c)(ii)
Solvent extraction.
5(c)(iii)
Fractional distillation.
5(c)(iv)
Simple distillation.
5(d)
Liquid of composition x when heated, produces a vapour of composition a1.
Vapour on condensing produces a liquid of composition x1 . Liquid has greater
concentration of A than B. Repeated vaporization and condensation
produces liquid A which is the distillate.
CAPE Chemistry June 2013 U2 P2 Q6 - Answer
6(a)
The ore is crushed and treated with sodium hydroxide. The filtrate is
seeded to produce Al(OHh. The Al(OHh is heated to produce aluminium
oxide.
NaAl(OH)4(aq) -> NaOH (aq) +Al(OHh!s)·
6( b)(i)
Red mud.
6( b)(ii)
Harms flora and fauna, and disfigures the environment.
6( c)(i)
Anode reaction: 202--> 02(gl + 4e -.
Cathode reaction: Al3• +3e·-> Al (i) ·
6(c)(ii)a)
The process requires a large quantity of energy.
6(c)(ii)b)
Energy demands and pollution effects are reduced.
CAPE Chemistry June 2014 U2 P2 Ql - Answer
l(a)(i)
To break the bonds in Br 2 molecules and produce bromine radicals.
l(a)(ii)
l(b)(i)
K b = [RNH/] [OH·]
[RNH 2]
l(b)(ii)
Ethylamine is the stronger base.
l(b)(iii)
The ethyl group increases the availability of the lone pair of electrons.
In phenylamine, the nitrogen atom's lone pair of electrons is delocalized into
the benzene ring, decreasing their availability.
l(c)(i)
ltis larger than 9.38.
l(c)(ii)
The availability of nitrogen's lone pair of electrons is further reduced due to
the presence of-CON- system.
l(d)(i)
Vigorous reaction and the evolution of dense white fumes.
l(d)(ii)
Pale yellow precipitate formed.
l(d)(iii)
White precipitate produced.
l(d)(iv)
Br2 is decolourized.
CAPE Chemistry June 2014 U2 P2 Q2 - Answer
Visible
2(a)(i)
radiation
Inc reasing frequency
2(a)(ii)
Infrared: Band X -rays: A.
2(b)
C= VA
=> 3.0 x 108= 4.5 x 101 5 'A.
'A.= (3.0 x 108)/(4.5 x 1015)= 6.7 x 10·8 m.
2(c)
S0 2 in wines/juice.
2(d)(i)
Step 1: To produce S from mixture.
Step 4: To ensure the removal of all traces of water.
2(d)(ii)
Suction flask and suction funnel.
2(d)(iii)
Oven.
2(e)
Mass of washing soda= 6.44 g.
Mass of anhydrous sodium sulfate= (37.09-34.25) g= 2.84 g.
Mass of water= (6.44-2.84) g = 3.60 g.
(2.84/142) moles of Na 2S04 combined with (3.60/18) mol of water.
=> 1 mole
:. x= 10 moles H20.
(3.60/18) X (142/2.84) = 10
CAPE Chemistry June 2014 U2 P2 Q3 - Answer
3(a)(i)
3(a)(ii)
Temp: 400 °c- 500 °c. Pres: 25 atm - 200 atm.
3(a)(iii)
Fractional distillation of air.
3(a)(iv)
Closeness to labour force. Closeness to water source.
3( b)(i)
The process is exothermic. Ant in temp will shift eq'm to left resulting in a
decrease in yield of ammonia.
3(b)(ii)
Forward rxn favours an increase in pressure due to lowering of the total
number of molecules. Eq'm shifts to right. This increases yield of ammonia.
3( c)(i)
A: steam reforming. B: dissolving of CO2 in water. C: condensation.
3(c)(ii)
Fe catalyst (finely divided).
3(c)(iii)
Liquid.
CAPE Chemistry June 2014 U2 P2 Q4 - Answer
4(a)
Primary: one ethyl group attached to the carbon atom bonded to the
halogen.
Secondary: two alkyl groups attached to the carbon atom bonded to the
halogen.
Tertiary: three alkyl groups attached to the carbon atom bonded to the
halogen.
4(b)(i)
2-bromobutane.
4(b)(ii)
The presence of a chiral centre results in optical isomerism.
4(b)(iii)
4( c)(i)
4(c)(ii)
2-bromo-2-methylpentane.
CH,i
H.iC
I .
c+
' \C-Br
r:.
H3 'J
CH
l�
�.::,
C
CH
I a
C+�Oll
i�
I-l3C CI-13
3
fl;1C CJ-13
l-l3 C
\
C-()11
H3C'#'J
HC
3
4( d)(i)
Off-white precipitate produced.
4(d)(ii)
NaBr(aq) + AgN03(aq) -> NaN03(aq) +AgBr(s)
+
Br
CAPE Chemistry June 2014 U2 P2 QS - Answer
5(a)(i)
Electron beam knocks off e- from molecule producing positive ions.
5(a)(ii)
Magnetic field separates ions according to their respective m/z ratios for
detection.
5(a)(iii)
The recorder presents ions as a series of peaks according to their respective
ion currents produced.
5(b)
the ratio of the relative abundance of the M/(M+l) (molecular ion) peaks
indicates the number of carbon atoms in the compound.
5( c)(i)
88.
5(c)(ii)
43.
5(c)(iii)
H H
H O
I II
I I
H-C-C-0-C-C-H
I
I I
H
H H
5(d)
H
H
O
I
,f'
H
I
H-c-c-c
I
IH HI "-o-c-H
I
CAPE Chemistry June 2014 U2 P2 Q6 - Answer
6(a)(i)
Uv radiation breaks down ozone into O radicals and oxygen molecules.
The ozone and oxygen atoms react to give 02 molecules.
6(a)(ii)
CFC's.
6(a)(iii)
Cataracts and skin cancer.
6( b)(i)
Yeast contains the enzyme zymase which catalyses the breakdo�vn of glucose
into ethanol and carbon dioxide. C6H1206 -> 2C2H50H + 2C02•
6( b)(ii)
Fractional distillation.
6(c)
Yes because the sale of beveragesgives revenue and job opportunities.
CAPE Chemistry June 2015 U2 P2 Ql - Answer
l(a)(i)
Thermal cracking.
l(a)(ii)
Temperatures in excess of 600 °c.
l(a)(iii)
8: CH4
C: CH3CH =CH2
l(b)(i)
The 2s electrons are unpaired and one of these is promoted to the empty 2p
orbital. The 2s and three 2p orbitals are hybridized to give four equivalent
sp3 hybrid orbitals.
l(b)(ii)
Aqueous bromine with propane in sunlight.
-
H
H
H
I
I
2ar·
H
H
- C - C - !'Y:' "."\.,. I
I
H
C
H
H
H
H
H
H
H
�)1
-
H
- C - c·
H
! - ! - �r:
H -
H-
H
'
H
H - C - C - C - B,
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
I
I
H
. .,
H
I
H
" - .,
H
"
H
H
H
"I
I
I
c
-c-e-H
- H-c-c.-c-c-c-c-H
H-c-e-d. ·
I
I
I
I
I
I
l(b)(iii)
Free radical substitution.
l(c)(i)
Red brown colour of bromine is decolorized.
l(c)(ii)
Purple colour of permanganate is decolorized.
CAPE Chemistry June 2015 U2 P2 Q2 - Answer
2(a)(i)
Reading burette at eye level increases accuracy.
2(a)(ii)
Adding titrant dropwise close to the end point increases accuracy.
2(b)
Large molar mass and high degree of purity.
2(c)(i)
J4
II
·t
ffl
H
'
+
�
32
.. .
1-
-
�
•
I+
'
it.
30
r
'
10
IS
20
25
Volume orucicJ (cm')
JO
35
40
""45
CAPE Chemistry June 2015 U2 P2 Q2 - Answer
2(c)(ii)
24.5
2(c)(iii)
25 cm3 NaOH contain (2.0 x 25)/1000 = 0.05 mol
Mole ratio of HCI to NaOH is 1: 1 so moles HCI = 0.05
24.5 cm 3 of HCI contain 0.05 mol HCI and
1000 cm3 contain (0.05 x 1000)/24.5 = 2.04 moles HCI
[HCl(aq)] = 2.04 mol dm·3.
2(d)
Pipette 25 cm3 sodium hydroxide solution into a polystyrene cup.
Allow solution to stand for a fe�v minutes and measure its temperature.
HCI added from burette in 5 cm3 portions from burette. The mixture is
stirred and the temperature recorded. This is repeated until 45 cm3 of acid
have been added.
CAPE Chemistry June 2015 U2 P2 Q3 - Answer
3(a)
Increased pressure of reagents, decreased temperature and the use of a
catalyst.
3(b)(i)
If one o r more factors which affect an equilibrium system, the position of the
equilibrium shifts in the direction which opposes the change.
3(b )(ii)
Low temperature and high pressure.
3(b )(iii)
S0 3 is dissolved in concentrated H 2S04 to produce oleum. Oleum is diluted
with water to yield concentrated H2S04•
3(b)(iv)
3(b)(v)
H 2S04 can cause blindness if it gets into eyes, H 2S04 can cause burns to the
skin and flesh, and the reaction between sulfur trioxide and water is highly
exothermic.
CAPE Chemistry June 2015 U2 P2 Q4 - Answer
4(a)
The process of linking one type of monomer with double or triple bonds to
yield the polymer which is the only product of the reaction (addition
polymerization). The process of linking two types of monomers to give both
the polymer and a small molecule (condensation polymerization).
4(b)
Polysaccharides e.g. starch, and proteins.
4(c)
-
' - C'
' ' • OH +
.. ....
H 11
(\ HO - C
be1)%el'le- •I,&+ •dtcor\tQ.'l'�ll('
0-W,.
-1- /'\ H2.o
4(d)
0 H
I
The link between the two monomers is -CI -N-.
I
The remaining parts of the repeating unit are
and
H
H,
I #o
"N-C-C
/ I \
H
H
H'
I
,,0
, N-C-C
'
I
H
CH OH
3
,,0
, N-C-C
'
I
CH 3
H
So the monomers are
H
I
and
CAPE Chemistry June 2015 U2 P2 QS - Answer
5(a)
Chromatography involves the separation of components of a mixture
between two phases i.e. the partitioning of the components of the mixture
between a stationary phase and a mobile phase. Partitioning results because
the mixture's components experience different absorption forces with the
stationary phase and have different solubility with the mobile phase.
5(b)(i)
By the use of a reagent called a visualizing agent or a locating agent. This
reacts with the component and forms a coloured compound.
5(b)(ii)
Using Rf = distance travelled by solute/distance travelled b y solvent,
for Q , Rf = 2.5/13.5 = 0.2 and for R, Rf = 12.5/13.5 = 0.9.
5(b)(iii)
Q is more polar so it is more strongly adsorbed onto the stationary phase
than R which is not as polar as Q.
5(c)
TLC can separate small amounts of compounds and is used to separate
amino acids. Column chromatography is used where large amounts of
substances need to be separated and collected. Fractions from column
chromatography can be collected for analysis.
CAPE Chemistry June 2015 U2 P2 Q6 - Answer
6(a)
Hydrogen
Chlorine
0 t
--
--
Concentrated
sodium chloride
solution
/
Titanium anode
"'
Steel cathode
Diaphragm
l
Sodium hydroxide solution
contaminated with NaCl.
Anode reaction: 2c1·(aq) - 2e·-> Cl21gl
Cathode reaction: 2H·(aq) + 2e·-> H2(g)·
6(b)
The use of the asbestos diaphragm is of concern since asbestos is considered
to be a carcinogen and dry asbestos fibres can be inhaled, causing
respiratory problems.
6( c)(i)
CFC's accelerate the depletion of the ozone layer in the atmosphere.
Destruction of the ozone layer allows harmful ultraviolet radiation to reach
the surface of the earth. CFC's are broken down in the stratosphere to
produce chlorine free radicals which react with ozone causing its conversion
to oxygen.
6( c)(ii)
Carbon dioxide is a product of combustion of fossil fuels and it is a
greenhouse gas. Increased concentrations of carbon dioxide contributes to
the greenhouse effect i.e. global warming. This has a negative effect on the
atmosphere through melting of glaciers and ice caps increase in sea levels.
CAPE Chemistry June 2016 U2 P2 Ql - Answer
l(a)
Type of isomerism in which compound have the same molecular formula but
different structural formulae.
l(b)(i)
Volume of carbon dioxide produced = 30 cm3
Volume of oxygen consumed
= 50 cm3
One mole of hydrocarbon therefore produces 3 moles of CO 2.
Therefore x = 3, in x + (y/4).
Using volume of oxygen consumed, x + (y/4) = 5 which gives y = 8.
Formula of Xis C3 H8 .
l(b)(ii)
Displayed formula:
l(c)(i)
H
H - C -
H
H H H
l I I
H-C-C-C-H
I I I
H H H
H
H
C - C -
H
H
H
C - H
H
H
0
H - C - C - C - H
H
0
H
2-methyl propan-1-ol
H3C
H
,
H
H - C - H
H
l(c)(ii)
H
2-methyl propan-2-ol
CH3 H3C,
/H
H
CH3
/
/c=c,
c i sb
- u t2
- -ene
/c=c,
H
trans-b u t 2
- ene
-
CAPE Chemistry June 2016 U2 P2 Ql - Answer
l(d)
First pair of compounds:
Test:
To both compounds, add alkaline iodine solution.
Observation:
Fine yellow crystals produced with the first
compound. No change seen with the other.
Second pair of compounds:
Test:
To both compounds add a few drops of acidified
potassium permanganate.
Observation:
Potassium permanganate goes from purple to
colourless with the first compound. No change
observed with the other.
CAPE Chemistry June 2016 U2 P2 Q2 - Answer
2(a)
Compounds have natural modes of vibration. Absorption of JR occurs when
the frequency of the radiation equals the natural frequency of the
compound.
2(b)
The compounds must undergo a change 1n dipole on absorption of JR
radiation.
2(c)(i)
3350-3500 cm·1 = -NH 2 and 1680-1800 cm·1 = >C =O.
2(c)(ii)
Name: aminoethanoic acid.
2(c)(ii)
Displayed formula:
H,
7
f>
H
HI
\
N-C-C
'
0H
-
2(d)
Finely ground mixture of the compound and NaCl i s pressed into a tablet.
2(e)
Cannot be used for non-polar compounds.
CAPE Chemistry June 2016 U2 P2 Q3 - Answer
3(b)
Fertilizers leached
industry.
into ground water and mercury from the chlor-alkali
3( c)(i)
Radioactive waste causes cancer and lead causes poisoning.
3(c)(ii)
Washing with phosphate detergents and fertilizers can cause eutrophication
and limit oxygen content.
3( d)(i)
Pb2• test: add dilute hydrochloric acid. Pb2• observation: white precipitate.
N0 3• test: add FeS04 follo�ved by concentrated H2S04 slowly.
N0 3• observation: brown ring.
3(d)(ii)
Pb2·(aql + Cl·(aql -> PbCl 2(s)·
CAPE Chemistry June 2016 U2 P2 Q4 - Answer
4(a)
Reaction I: Electrophilic substitution. Reaction JV: Diazotization.
4(b)
Reaction II: Reagents - concentrated H2S04 and HN03 . Conditions temperature of 55 ·c to 60 ·c.
Reaction IV: Reagents-NaN02( aq) and HCl(aq)· Conditions-temperature of
-5 ·c.
4(c)
4(d)
Q-N=N c1CH3
CH3
CH3
'-::::
'-::::
'-::::
/7
Br+
4(e)(i)
Br
HO
Br
0
Br
H Br -AIBr3•
4(e)(ii)
4(f)
OH
+
o - Na+ 4(e)(iii)
'-:::
Br
� 19
H-C-C
� 'c1
Br
'-::::
0, C ,..CH3
II
CAPE Chemistry June 2016 U2 P2 QS - Answer
S(a)(i)
S(a)(ii)
Mixing occurs with zero heat change and zero volume change.
S(b)(i)
A constant boiling mixture.
S(b)(ii)
The composition of azeotropes changes with pressure.
S(b)(iii)
8.pt
B.pt
"\...,.
M
z
Composition
X
1001:
Liquid of composition X boils at temperature T1 . Vapour has higher
composition of A. As distillation continues, vapour produces azeotrope M.
The temperature of the residue rises producing pure B.
S(c)
Mass of compound in organic solvent= x g.
Mass of compound in water= (5-x) g.
Concentration of compound in water
Concentration of compound in ether
= 1/0.200 = (x/25)/((5-x)/100)= 2.8 g.
CAPE Chemistry June 2016 U2 P2 Q6 - Answer
6(a)(i)
Cryolite and aluminium fluoride are added to alumina to lower the melting
point. The cell is lined with carbon which is the cathode. The carbon anode
dips into the molten electrolyte. Anode reaction: 202· 1,, -> 02(gl + 4e·.
Cathode reaction:Al 3•1,, + 3e·-> Al (s)·
6(a)(ii)
Oxygen produced reacts with the carbon anode to produce carbon dioxide.
6(a)(iii)
6(b)
Availability of energy - source of electrical power, transportation/use of
railroad services and appropriate disposal of toxic waste.
6( c)(i)
Processing materials which would otherwise become waste.
6( c)(ii)
Aluminium is shredded and hot air blown on shreds to remove coatings. The
shreds are melted to molten aluminium which is made into blocks.
6(c)(iii)
Mirror reflectors, in alloys- for aircrafts/cars, electrical cables.
CAPE Chemistry June 2016 U2 P2 Q6 - Answer
6(a)(i)
Cryolite and aluminium fluoride are added to alumina to lower the melting
point. The cell is lined with carbon which is the cathode. The carbon anode
dips into the molten electrolyte. Anode reaction: 202· 1,, -> 02(gl + 4e·.
Cathode reaction:Al 3•1,, + 3e·-> Al (s)·
6(a)(ii)
Oxygen produced reacts with the carbon anode to produce carbon dioxide.
6(a)(iii)
6(b)
Availability of energy - source of electrical power, transportation/use of
railroad services and appropriate disposal of toxic waste.
6( c)(i)
Processing materials which would otherwise become waste.
6( c)(ii)
Aluminium is shredded and hot air blown on shreds to remove coatings. The
shreds are melted to molten aluminium which is made into blocks.
6(c)(iii)
Mirror reflectors, in alloys- for aircrafts/cars, electrical cables.
CAPE Chemistry June 2017 U2 P2 Ql - Answer
l(a)
1. All members contain the same functional group.
2. All member can be represented by a general formula.
3. Physical properties of members vary with increasing number of carbon
atoms.
l(b)(i)
C
H
0
Mass/g
0.40
0.06
0.54
Moles
0.03
0.06
0.03
Simplest ratio
1
2
1
So empirical formula is CH20.
l(b)(ii)
C2H402,
l(c)(i)
Cn H 2n•1COOH.
l(c)(ii)
II
H, ,.,.c, ,..H
,.,.c
0
H I
Cl
l(d)
The pK0 value of A will be larger since the chlorine atom of the substituted
acid withdraws electron density from the 0-H bond making H· more easily
lost.
l(e)(i)
Add alkaline iodine solution to each compound and warm. The first
compound shows no visible change while the second produces yellow
crystals.
l(e)(ii)
Add a few drops of bromine to each. The first compound shows no change
while the second causes the red colour of bromine to disappear.
CAPE Chemistry June 2017 U2 P2 Q2 - Answer
2(a)(i)
The degree to which a measurement is close to the true value.
2(a)(ii)
The degree to which measurements are close to each other.
2(b)
Steam distillation.
2(c)(i)
C, G, H.
2(c)(ii)
B, E.
2(d)(i)
2
2
C a •(aq) + (204 . (aq)-> CaC 204(s)
2(d)(ii)
Molar mass of calcium oxalate= 128 g mo1·1
Moles of calcium oxalate precipitated= (0.619/128) mol= 0.0048 mol
Therefore moles of calcium= 0.0048 mol
Mass of calcium= (0.0048x40)g= 0.193 g
Percentage calcium in sample= (0.193/0.496)x100%= 39.00%
CAPE Chemistry June 2017 U2 P2 Q3 - Answer
3(a)(i)
3(a)(ii)
It absorbs ultraviolet radiation, preventing damaging radiation from reaching
Earth's surface.
hf
02(g)----+ O(g) + O"(g)
O"(g) + 02(g) + M(g) ----+ 03(g) + M'(g);
ozone
3(b)(i)
1. Respiratory illness can result.
2. Reduced crop yield and forest growth.
3(b)(ii)
N02·(g)--+ NO·(g) + O(g)
3(c)
No. 1
hf
Test: Add aqueous sodium hydroxide and zinc metal and warm.
Inference: Nitrate ion present.
No. 2
Inference: lead(11) ion present.
No. 3
Inference: cyanide ion present.
CAPE Chemistry June 2017 U2 P2 Q4 - Answer
4(a)
Primary alcohol:
H
H
H
H
H H
H
H
I I I I
H-C-C-C-C-OH
I I I I
An alcohol in which the -OH group is bonded to a carbon atom which is
bonded to only one other carbon atom.
Secondary alcohol:
H
H
OH H
I
I I I
H-C-C- C-C-H
I
I I I
H H H H
An alcohol in which the -OH group is bonded to a carbon atom which is
bonded to tvvo other carbon atoms.
Tertiary alcohol:
H CH3H
I I I
H-C-C-C-H
I I I
H
OH H
An alcohol in which the -OH group is bonded to a carbon atom which is
bonded to three other carbon atoms.
4(b)(i)
Type of reaction: dehydration.
Structural formula of compound D:
(b)(ii)
CH=CH2
0
Eis formed by the oxidation of a secondary alcohol to a ketone.
Fis formed by the oxidation of a primary alcohol to an aldehyde.
CAPE Chemistry June 2017 U2 P2 Q4 - Answer
4( c)(i)
E:
4(c)(ii)
A:
4(d)
A will rotate plane polarized light; B will not.
0
F:
B:
OH
CAPE Chemistry June 2017 U2 P2 QS - Answer
5(a)(i)
Stable in air at room temperature, available in a state of high purity and has
a large relative formula mass.
5(a)(ii)
It is not considered a primary standard because it absorbs water from the
atmosphere and dissolves in it.
5(b)
From A to B there is a consistent temperature rise as HC03·(aq) and H*(aq) ions
react. When all the HC03·(aq) ions have reacted with H*(aq) ions, the highest
temperature is reached, at B. As excess H*(aq) ions are added, the
temperature decreases from B to C.
5(c)(i)
Moles of carbonate used = 25 x 0.25 x 10·3= 0.00625.
5(c)(ii)
Moles of acid reacted= 0.5 x 25 x 0.25 x 10·3= 0.003125.
5(c)(iii)
Concentration of acid= 3.125/20 =0.16 mol dm·3 .
5(d)
Potentiometric titration.
5(e)
E.m.f. measurements.
CAPE Chemistry June 2017 U2 P2 Q6 - Answer
6(a)(i)
The decomposition of large hydrocarbon molecules into molecules with
smaller numbers of carbon atoms.
6(a)(ii)
The conversion of straight chained or unbranched alkanes into branched
chained and aromatic hydrocarbons.
6(b)
N2( g) + H2(g) ;= 2NH3(g) l'iH = -ve. High pressure favours reduction in the
number of molecules and therefore the forward reaction which produces
ammonia. Low temperature favours the forward reaction since the forward
reaction is exothermic.
6(c)
Actual pressures used are between 200 and 1000 atmospheres. Actual
compromise temperature of 500 °C is used. At low temperatures the rate of
achieving equilibrium (rate of production of ammonia) is slow. An iron
catalyst is used to increase the rate.
6(d)
Ammonia base fertilizers are leached into ground water. This leads to
overgrowth of algae leading to eutrophication.
CAPE Chemistry June 2018 U2 P2 Ql - Answer
l(a)
B: electrophilic addition.
C: nucleophilic substitution.
l(b)(i)
r r ir r
H- 1-1 1-, HH-r-H
-cH I
H
l(b)(ii)
H
H
H H CH3 H 1-1
I
I
I
I
I
I
I
I
I
I
1-1-c-c-c-c-c-1-1
H H OHH M
l(c)
H
CH3
H CH3
I
/
I I
+
H3C-c-c
-,> H 3C C
- C
- C
- H2-C H3
I I
I
H Br
\ "c2H s
H
�
H-
Br
Br·
.....,.,
l(d)
l(e)
Name of test: oxidation. Reagent: acidified potassium permanganate.
Observations:
yellow/brown
colour of bromine fades or
disappears, white precipitate is
produced and an antiseptic
smell is noted.
With NaOHw1:
Structure
product:
of
Br
HO
Structure
product:
expected
0
Br
a.
of
"'
expected
O Na•
CAPE Chemistry June 2018 U2 P2 Q2 - Answer
2(a)
Suction flask-increases rate of filtration.
Suction funnel - allo�vs filtration under vacuum.
Silica crucible - heating or burning solid substances.
Sintered glass crucible - used for filtration, drying and weighing of
substances withouttransfer.
Oven/furnaces-used for drying substances to reproducible weight.
2(b)(i)
Moles of silver chloride= (0.8402/143.5) mol= 0.0059 mol.
2(b)(ii)
Moles of KCI= 0.0059 mol.
2(b)(iii)
Mass of KCI = (0.0059 x 7 4.55) g= 0.4398 g.
% KCI = (0.4398/0.450) x 100= 97 .33 %.
2(c)
Determination of the amount of phosphorous in fertilizer.
2(d)
I, II and Ill are the Monochromator, beam splitter and reference cell,
respectively.
2(e)
Using A= Ect, c= (O. 7/(8400 x 1)) mol dm·3= 8.33 x 10·5 mol dm·3.
CAPE Chemistry June 2018 U2 P2 Q3 - Answer
3(a)(i)
I, II, Ill, IV and V are Cl2(g), H2(g), titanium anode, steel/nickel cathode and
NaOH/NaCI respectively.
3(a)(ii)
At Ill: 2Cl·(aq)-> Cl2(gl + 2e·. At IV: 2H·(aq) + 2e·-> H2(g) ·
3(a)(iii)
Hydrogen and chlorine would spontaneously ignite. H 2(gl + Cl2(gl-> 2HCl (g)·
3(b)
Bleaching agents, herbicides and pesticides.
3(c)
Risk of pollution from asbestos, NaOH, chlorine.
CAPE Chemistry June 2018 U2 P2 Q4 - Answer
4(a)(i)
H3 c
but-1-ene
"'c
I
cis-2-butene
CH3
trans-2-butene
H H H H
I I I I
H-C-C-C-C-H
I I I I
H OH H
4(b)
,..,c� ,..,H
H
C
I
CH3
H
4(a)(iii)
I
,..,C� ,..,H
H
4(a)(ii)
CH3
H
H
H
Optical isomerism because butan-2-ol contains a chiral carbon.
0
0
II
II
-H20
HO -C-(CH2)4-C-OH + H2N-(CH2)s-NH2--..
0
II
0
II
N-C-(CH2)4-C-NH(CH2)6- NH
n
Formation of the amide bond between 1,6-diaminohexane and hexane-1,6dioic acid results in the loss of a water molecule.
CAPE Chemistry June 2018 U2 P2 QS - Answer
5(a)(i)
5(a)(ii)
The M and M+2 peaks suggest the presence of either chlorine or bromine.
Since the peaks at 108 (M) and 110 (M+2) have almost the same intensity,
this says that bromine is present.
5(a)(iii)
The relative isotopic abundances of 79Br and 81Br are 50-50.
5(b)(i)
The vapour pressure of a component of a mixture is equal to the vapour
pressure of the pure component at that temperature multiplied by its mole
fraction in the mixture.
5(b )(ii)
Positive deviation from Raoult's law occurs when the vapour pressure of a
mixture is higher than expected for an ideal mixture. This higher vapour
pressure is as a result of there being weaker intermolecular forces between
ethanol and cyclohexane �vhen compared to the pure substances. These
weaker intermolecular forces in the mixture are as a result of the differences
in polarity between ethanol and cyclohexane molecules.
5(b)(ii)
°
Boiling temperature
°
78 (
81 (
Vapour
°
Liquid
64.8 (
100% EtOH
0.430 EtOH
100%C6H12
0%C6H 12
0.570 EtOH
0% EtOH
CAPE Chemistry June 2018 U2 P2 Q6 - Answer
6(a)
Sucrose is converted to glucose and fructose by the action of enzymes
present in yeast i.e. C 12H 22 011 + H20 -> 2C6H1206 . The fructose and glucose
sugars then react via the action of another enzyme in yeast to produce
ethanol and carbon dioxide i.e. C6H1206 -> 2C2H50H + 2C02•
6( b)
Fractional distillation.
6(c)
The ethanol present in wine can be oxidized to ethanal and ethanoic acid.
6(d)
Two advantages: Glass can be re-used and recycled which leads to a
reduction in solid waste, and recycling glass means that some of the
manufacturing costs are reduced.
Two disadvantages: Glass manufacture is very expensive and glass is non­
biodegradable.
6(e)
Cirrhosis of the liver and heart disease.