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Name : ………………………………………………… NRIC : …………………………
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA
MARKING SCHEME
This answer paper consists of 17 printed pages
CHEMISTRY
PAPER 1 & 2
TRIAL962/1
JABATAN PELAJARAN
NEGERI TERENGGANU
TRIAL EXAMINATION 2009
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OTI 2 2009 STPM
Marking scheme OTI 22009
2
STPM CHEMISTRY 962/1
OTI 2 2009
MARKING SCHEME PAPER 1
QuestionNo.
Answer Explanation
1 A These are the isotopes of iron. Different isotopes have different numbers of
neutron
2 D Of the 4 gasses given hydrogen gas is most likely to exhibit ideal behavior.
Refer to any graph of ( pV/nRT ) against p.
3 D Carbon dioxide cannot be solidified at 1.0 atm if the temperature is greatly
reduced. At 1.0 atm CO2 exists in both solid and gaseous forms.
4 C Actual electronic arrangement is 1s2
2s2
2p6
3s2
3p1. Total number of
electron is 13. Thus the proton number is 13
5 C Na+
is the biggest as it has the smallest number of nuclear charge. The
smaller the nuclear charge, the bigger the ionic size.6 D Only phosphorus can exist as monatomic solid with acidic properties in its
oxides.
7 C Across period 3 from left to right, conductivity increases for the first 3
elements, then starts to decrease as the elements become metalloid, and
decreases further as elements are non-metals.
8 C In the CH3CHO molecule, hydrogen atom is bonded to carbon not oxygen.
So no intermolecular hydrogen bonds.
9 B No energy gap between the two bands, so it belongs to a metal, which is
magnesium.
10 C There are only repulsive forces exist between electron orbitals. So the lone
pair electrons push each other so that the angle between them is maximized.
11 A The graph is reversible reaction. The rate of the forward reaction is equal tothe rate of the backward reaction over time. The formation of ester is a
reversible reaction.
12 C e.g.: if the original sample = 100g
100g → 50g → 25g → 12.5g → 6.25g
t 1/2 t 1/2 t 1/2 t 1/2
Percentage remaining after the 4th
half life = 6.25 x 100 = 6.25%
100
13 B
Since the forward reaction is endothermic, an increase in temperature shifts
the equilibrium to the right according to Le Chatelier’s principle, by
absorbing the excess heat. In the case, more gas particles are produced andthere is an increase in volume. Furthermore, heating will also cause the gases
to expand.
14 A
AgBr (s) Ag + (aq) + Br-(aq)
Ksp = [ Ag+
] [Br-]
= 5 x 10-13
mol2dm
-3
But [ Ag+
] = [Br-]
[ Ag+ ] = ( 5 x 10-13
)1/2
= 7.1 x 10-7
moldm-3
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Question
No.
Answer Explanation
15 B
mB = MB x Po
B
mW MW x Po
W
mB = 156.6 x (101 x 85)
100 18 x 85
= 163.66
% bromobenzene = 163.66 x 100
(1663.66 + 100)
= 62%
16
BThe very negative Eo value of Mg shows that Mg2+ is very difficult to bereduced to Mg. Therefore, normal chemical reduction is not suitable. A
feasible way would be the electrolysis of molten Mg2+
, in this case, MgCl2
(MgCl2 has a lower m.p. than MgO).
17 A
Quantity of electricity used = I x t
= 8 x (100 x 60)
= 48000 C
Amount of electrons passed = 48000C96000 Cmol-1
= 0.5 mol
Amount of O2 evolved = 0.5
4
= 0.125 mol
Volume of O2 evolved = 0.125 x 22.4
= 2.8 dm3
18 A The enthalpy change of formation of CO is the enthalpy change when 1
mol of gaseous CO is formed from its constituent elements, that is C and O2,
at their standard states.
19 A
H = -52 + (-85) = - 137 kJmol-1
20 A
For cis-trans isomers, 2 different groups must be bonded to the same C at the
double bond.
B : 1 and 3 are identical
C : 1 and 4 are structural isomersD : 2 and 4 are structural isomers
2O2-
(l) O2(g) + 4e
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Question
No.
Answer Explanation
21 C Upon reacting with Br2 , but-1- ene gives 1,2- dibromobutane and
but -2-ene gives 2,3- dibromobutane.22
BBeing an alkene, hex-1-ene undergoes electrophilic addition readily with Br2
in CCl4 and decolourised it. Methylbenzene is unable to do so.
CH3(CH2)3CH=CH2 + Br2 CH3(CH3)3CHBr—CH2Br
23 A
24 ARemember: Ethanoyl chloride (acyl chloride) will react with alcohols but
not with carboxylic acids.
25 B CH3CH2COOH All carbonyl compounds, which are aldehyde and ketone
react with 2,4-dinitrophenyihy- drazine to form orange precipitate that has a
high melting point.
26 A Ethanoic acid is a weak acid and it partially ionises in water to produce
hydroxonium ion, H+.
CH3COOH + H20 CH3COO-+ H
If ethanoic acid is dissolved in a base like ammonia that is able to extract the
proton (H+
) from ethanoic acid more effectively, more ethanoic acidmolecules will ionise. Thus H
+ions are produced are the strength of ethanoic
acid will increase. Thus ethanoic is a stronger acid in liquid ammonia than in
water.
27 D The reaction equation is as follows.
Remember: This molecule contains phenyihydrazine carbonyl group.
28 C Hydrogen bonds are formed between ethanol and water molecules.
Thus ethanol is soluble in water.
Whereas ethyl ethanoate (ester) that has a long carbon chain does not formhydrogen bonds with water, thus it is less soluble in water
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Question
No.
Answer Explanation
29 C
30 D chloride
An aromatic amine with an amino group at its side chain will have similar
chemical properties as an aliphatic amine.
31 D
At pH 2, an acidic solution and 2-amino- propanoic acid (alanine)will exist
as positive ions.
32 B
The reaction results in a nucleophilic addition to the double bond and theanionic polymerisation process takes place.
33 ANaH – ion, SiH4 - covalent, H2S – covalent, HCl – covalent.
The sodium atom wiyh the biggest size among the four elements has
the lowest ionization energy and can lose electron readily to form
ionic compound.
34 AAluminium is a amphoteric. It si soluble in aqueous sodium
hydroxide.Al2O3 (p) + 2OH
-(aq) + 3H2O(l) → 2[(OH)4]
-(aq)
35 DXCl4 decomposes at room temperature showing that the X-Cl bond is
very weak. Hence, X must have the biggest size in Group 14.
PbCl4(l) → PbCl2 (s) + Cl2 (g)
36 DThe inert pair effect increases going down the Group 14. Hence, the
+2 state becomes progressively more stable while the + 4 state
becomes less stable.
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Question
No.
Answer Explanation
37 C The Lewis structure of N2O4 is as follows :
It is a symmetrical molecule. Hence is non-polar.
There are 3 bond pairs with no lone pairs surrounding the nitrogen
atom. Hence, its shape is triginal planar and not ‘V’ shape.
As can be seen from the Lewis diagram all the electrons in N2O4 are
paired. N2O
4is formed from the dimerisation of NO
2.2NO2 N2O4
38 C The lone pair electrons in NH3 pushes the three bonding pairs closer to
one another. Hence, its bond angle is slightly less than that of a perfect
tetrahedron, 109.5o
.
39 D KI reacts with H2SO4 to produce iodine, which is a purple fume.
KI + H2SO4 → KHSO4 + HI(g)
2HI(g) + H2SO4 → I2(g) + SO2(g) + 2H2O
KI reacts with silver nitrate to produce silver iodide, which is yellow.
KI(aq) + AgNO3 (aq) → KNO3 (aq) + AgI(s)
AgI is insoluble in either dilute or concentrated ammonia.
40 B Heterogeneous catalyst uses their empty d orbitals to form temporary bonds
with the reacting molecules. This process is known as adsorption.
41 B 1 Elevation of boiling point of solvent
2 Depression of freezing point of solvent
42 B(1,2)
According to Le Chatliers Principle,
When OH-ions are neutralized by the acid, position of eq will shift to the
right, making the solid Ca5(PO4)3OH dissolved.
When the PO43-
ions accepts H+
, the eq will shift to the right, making the
solid Ca5(PO4)3OH dissolved.
43 A( 1 )
Halogen is an donating group which decreases the acid strength.( Ka is
greater)
44 D(1,2,3)
Partition law can only be used under the following condition- the solutions must be dilute
- the solute cannot undergo dissociation or association in one solvent and
not the other
- the temperature must be fixed
45 D The Grignard reagent has the general formula, RMgX. With ketones, a
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OTI 2 2009 STPM
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7
Question
No.
Answer Explanation
tertiary alcohol is produced.
Remember: The C—Mg and C—Li bonds are very polar. When either the
C—Mg or the C—Cl bonds undergoes fission, the carbon acts
as a nucleophile
46 A To form a diazonium salt, nitric(III) acid must react with a primary aromatic
amine solution at a temperature of 0 — 5°C. 2-methylphenylamine, is a
primary aromatic amine.
47 A Going down group 2, the size of the cation increases causing the polarizing
power to decrease. Statement 2 only indicates that the oxides are more stable
than the carbonates. It does not explain about the stability trend of the
carbonates.
48 C Zn(OH)2 and Cu(OH)2 are soluble in excess ammonia.
49 B
50 B Mol ratio of Ni : N : H = 13.98/59 : 13.27/14 : 2.79/1
= 0.236 : 0.947 : 3.79
= 1 : 4 : 16
:. Ratio of H2 : N = 8 : 4 = 2 : 1
The complex does not contain polydentate ligands. Hence, it cannot exhibits
optical isomerism.
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8
STPM CHEMISTRY
OTI 2 2009/TRIAL EXAM
MARKING SCHEME
PAPER 2
SECTION A ( Structured Question )
QUESTION
NO SUGGESTED ANSWERS SUGGESTED
MARKS 1(a)(i) The element X contains isotopes. 1
1(a)(ii) Lets the percentage abundance of 79
X = x%
Percentage abundance of 81
X = (100-x)%
79.99 =
X = 50.5
Percentage abundance of isotopes79
X = 50.5% Percentage abundance of isotopes
81X = 49.5%
1 or 1
1
1
Any 2
1(a)(iii)
Axis labeled
and unit 1
Draw a correct
peak 1
1(b)(i)
At low pressure, the molecules are far apart , so repulsive forces among the molecules of CO2
1 1
1(b)(ii) Positive deviation 1
1(b)(iii)
Forces of repulsion between the H2 gas molecules
which causes the speed of H2 molecule collision with the vessel wall to
increase.
So the pressure of H2 is higher, causing > 1 (positive deviation)
1 1
Total 10
79 80 81 m/e
Abundance(%)
50.5
49.5
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9
QUESTION
NO SUGGESTED ANSWERS SUGGESTED
MARKS 2 (a) (i) Zn(s) + 2Ag
+ (aq) → Zn
2+(aq) + 2Ag(s) 1
2 (a) (ii) Half cell reaction E 0 /V
Zn2+
(aq) + 2e→ Zn(s) -0.762Ag
+(aq) + 2e → 2Ag(s) +0.8
Zn(s) + 2Ag+
(aq) → Zn2+
(aq) + 2Ag(s)
E.m.f. of cell = 0.80 + 0.76 = 1.56 V
11
1
2 (a) (iii) E cell — E ceii° - 0.059 log [Zn2+
]
2 [Ag]2
1
2 (b) (i)
2 (b) (ii) -487 = 178 + 279 + (590 + 1150) + 337 + Lattice energy
Lattice energy = -3021 kJ mol-1
5
For every
single level
energyshown,one
mark will be
given
1+1
Total 10
3 (a) (i) CH3COCl + CH3CH2CH2OH → CH3COO CH2CH2CH3 + HCl
Propyl ethanoate1
1
3 (a) (ii) CH3COCl + C6H5NH2 → CH3CONH C6H5 + HCl
N-phenyl ethanamide
1
1
3 (b) Heat with dilute sulphuric acid.
CH3CONH C6H5 + H2O → CH3COOH + C6H5NH2
1
3 (c) Structural formulae of the products :
HOCH2CHOH and NaOOCCH2COONa
│ CH3
1+1
3 (d) (i) Heat 2-hydroxybenzoic acid with ethanoyl chloride. 1
1
3 (d) (ii) X is aspirin. It is an analgesic. 1
Total 10
Latice energy
CaS(s)
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10
QUESTION
NO SUGGESTED ANSWERS SUGGESTED
MARKS 4 (a) (i) Dinitrogen oxide, N2O 1
4 (a) (ii) NH4NO3(s) → N2O(g) + 2H2O(l) 1
4 (b)
Explanation:
Na2O and MgO, ionic with strong electrostatic forces;Al2O3, ionic
with
Covalent character; SiO2 macromolecular with high melting
point;
Oxides of P and S,simples molecules
1
1
1
1
4 (c) (i) 1
4 (c) (ii) 3d electrons absorb energy in the visible wavelengths except the
green wavelength which it reflects and jump to the higher set of vacant
3d orbitals. Therefore, Cr3+
ions appear green.
1+1
1
Total 10
Mg
Melting point of oxides
P SSiAlNa
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11
QUESTION
NO SUGGESTED ANSWERS SUGGESTED
MARKS 5(a) The hydrogen emission spectrum consist of discrete lines
whereas the spectrum produced by the tungsten filament bulb is a
continuos spectrum.
1 1
5(b) 1s,2s,2px,2py The difference between 1s and 2s is that the size of 2s is bigger than 1s. or The difference between 2s and 2p is that 2s is spherical but that 2p is dumb-bell shape.
1
1 or 1
………… Max 2
………... 5(c) (i) Hund`s rule:
When electrons are placed in a set of orbital with equal energies, the
electrons must occupy them singly with parallel spin before they can
occupy the orbital in pairs.
Pauli exclusion principle: Each orbital can only be filled with two electrons with opposite spins.
Aubau principle:
Electrons occupy orbitals in order of the energy levels of the orbitals.
Orbitals with the lowest energy are always occupied first.
1
1
1
5(c) (ii) Number of electrons in O2-
ion = 8+2=10 Step 1: Apply Pauli exclusion principle and Aufbau principle.Fill 1s
orbital with two electrons.
1s 2s 2p Step 2:Fill 2s orbital with two electrons
1s 2s 2p Step 3: Apply Hund`s rule.
Fill 2p x,and 2p y orbitals wih three electrons. The electrons must be in parallel spins.
1s 2s 2p Step 4: Fill the remaining three electrons in 2p x,2p y and 2pz orbitals .
Each pair of electrons must be in parallel spins.
1s 2s 2p
1
1
1
1 Describe or
show the filling
using a
diagram Any 3
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12
QUESTION
NO SUGGESTED ANSWERS SUGGESTED
MARKS 5(d) (i) Atomic radius is defined as half the distance between the nuclei of two
closest and identical atoms.
1
5(d) (ii) When across a period(from lithium to neon), atomic radius decreases. Because the nuclear charge increases(number of protons) But the screening effect remains almost constant as the number of
shells remain the same. The attraction of the valence electrons by the nucleus increases.
1 1 1
1
Total 15
6 (a) NH3 Valence electron, N : 5e
3H : 3e
------------- 8e 3 bond pair, 1 lone pair
-------------
.. N
Or trigonal pyramidal 1070
NH4+
Valence electron, N : 5e 4H : 4e
Positive charge : -1e (less 1 electron)
------------- 8e 4 bond pair, no lone pair
--------------
Or tetrahedral
109.5)
The bond angle of NH3 is 1070
wheares the bond angle of NH4+
1
Draw or state
the shape 1
1
Draw or state
the shape 1
Any 4
H H H
N
H+
H H H
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13
QUESTION
NO SUGGESTED ANSWERS SUGGESTED
MARKS 6 (b) (i) H2O
..
Overlapping
1
label
1
6 (b) (ii)C2H4
Overlapping 1
label 1
6 (c) Boiling point of H2O is higher than HF because each HF molecule forms 2 intermolecular hydrogen bonds compared to 4 intermolecular hydrogen bonds formed by each H2O
molecule.
1 1
6 (d) (i)
From the graph, the half- life of the reaction is 13.5 minutes.
Label the axis
and unit 1
Plot the graph 1+1
Indicate on thegraph
t1/2
1
H(1 s)
H(1 s)
sp3
orbitals
H(1s)
H(1s)
H(1s)
H(1s)
π
δ
C C
Volume of KMnO4
10 20 30 t/min
40
30
20
10
O
t1/2= 13.5 t1/2= 13.5
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14
QUESTION
NO SUGGESTED ANSWERS SUGGESTED
MARKS 6 (d) (ii) From the graph, half life doesnot depend on the concentration of
hydrogen peroxide. Thus, the reaction is first order. 1
Total 15
7(a) When an equilibrium is disturbed then equilibrium will shift in thedirection that will reduce the effect of the disturbance.
2 m
7(b)(i) Yield of NO decreases.
When pressure is increased, equilibrium will shift to the left because
the backward reaction will reduce the number of gaseous molecules.
1 m
1 m
7(b)(ii) Value of Kc increases.
The forward reaction is exothermic, the Kc value decreases with
increase in temperature.
1 m
1 m
7(c) Let x be the number of mol of NO produced
Total number of moles at equilibrium = 0.40 – x + 0.50 – 1.25x + x +1.5x or
1.5xx1.25x0.5x0.4
x
+++
= 30 %
x =0.29 mol
1 m
1 m
7(d) (ii) (ii)
]
7(e)(i)
]A[
1= kt +
0]A[
1
k =25
0.20.12 −= 0.40 mol
–1dm
3min
–1
1 m
1 m
7(e)(ii)
0]A[
1= 2.0
[A]0 = 0.50 mol dm–3 1m
7(f) A + A → M slow step
M → 2B fast step
1 m
1 m
Total 15
Rat
[A
0
A
t0
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15
QUESTION
NO SUGGESTED ANSWERS SUGGESTED
MARKS 8 (a) Boiling point of tetrachlorides increases down the group
All the tetrachlorides are non-polar and are simple molecules.
Molecular size increases from top to bottom.
Strength of van der Waals forces increases down the group.
1
1
1
1
8 (b) SiC14 undergoes hydrolysis whereas CCI4 does not.
This is because the Si atom has empty d orbitals which can be attacked
by lone pair electrons from water molecules. The product of
hydrolysis is silicon(IV) oxide and hydrochloric acid.
, C atom in tetrachioromethane does not contain empty
d orbitals, hence does not undergo hydrolysis.
1
1
11
8 (c) (i) Semiconductor
Semimetal, electrical conductivity increases when temperature is
raised.
Ceramic
The Si—O covalent bond in the giant covalent network is strong.
In fire extinguishers. Does not support combustion
1
1
1
8 (d) An aluminium factory has to be located near a port so that the raw
material, bauxite ore, can be easily transported to it. This will reduce
cost.
As the electrolysis of molten aluminium oxide involves the use of a
large amount of electricity, the aluminium factory must be located
close to a hydroelectric dam for a supply of cheap electricity.
These two factors can help to reduce the cost of producing aluminium.
One bad effect on the environment caused by the extraction of
aluminium is the release of poisonous hydrogen fluoride and fluorine
gas into the air, which are formed during electrolysis of a mixture of aluminium oxide and cryolite.
1
1
1
1
Total 15
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16
QUESTION
NO SUGGESTED ANSWERS SUGGESTED
MARKS
9(a) CH3-CH=CH-CH3
CH3- CH- CH
2-CH
3
Cl
2m
9(b) A: CH3CH2Cl D : CH2Br- CH2Br
B : CH3CH2OH E : CH2OH – CH2OH
C : CH2=CH2 5m
9(c) Ethanolic KOH 1m
9(d) monomer for polyester 1m
9 (e) - heat with ethanolic silver nitrate
1-iodohexane : yellow precipitate of AgI
1- chlorohexane : white precipitate of AgCl
- heat with ethanolic silver nitrate
Chlorocyclohexane : white precipitate
Chlorobenzene : no precipitate
- add acidified KMnO4
1- chloro-1- butane : decolourisation of KMnO4
1- chlorobutane : KMnO4 solution is not decolourised.
1 + 1
1+ 1
1+ 1
Total 15
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17
QUESTION
NO SUGGESTED ANSWERS SUGGESTED
MARKS 10 (a) Tetrahydroisoquinoline is the stronger base. In tetrahydroquinoline, the
unshared electron pair is delocalised into the aromatic ring, making it
less available to accept a proton, whereas tetrahydroisoquinoline
resembles an alkylamine.
1
1
10 (b) 3,4-Difluorophenylamine is a weaker base than phenylamine because
the two F atoms are electron-withdrawing and they reduce the
availability of the lone pair of electrons on the N atom to accept a
proton.
1
1
10 (c) An ethyl group, CH3CH2 is transferred from triethylaluminium to
titanium(IV) chloride to produce a complex,
Ethene molecules act as Lewis bases and are bonded to titanium which
has vacant d orbitals. These ethene monomers are inserted between the
titanium and the ethyl group to form a polymer.
The process is terminated when a hydrogen atom is added to a titanium
atom, and the poly(ethene) chain is separated from titanium.
1
1
1
1
1
1
10 (d) The poly(ethene) polymer formed by the addition polymerisation
process using the Ziegler-Natta catalyst is a linear polymer and of high
density and high melting point.
1
1+1
10 (e) Proteins and polypeptides.
Nylon has peptide linkages, CONH, which are also found in
proteins and polypeptides.
1
1
Total 15