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CONFIDENTIAL*/SULIT*
[Turn over(Lihat sebelah)
SULIT *
Jawapan Peperiksaan Percubaan STPM 2010
Kimia Kertas 1
1 D 11 B 21 D 31 D 41 D
2 A 12 A 22 D 32 C 42 B
3 A 13 C 23 D 33 B 43 C
4 C 14 C 24 C 34 B 44 B
5 B 15 D 25 B 35 B 45 A
6 A 16 D 26 A 36 B 46 B
7 D 17 D 27 C 37 D 47 A
8 D 18 A 28 C 38 A 48 D
9 C 19 C 29 B 39 B 49 B
10 B 20 B 30 C 40 B 50 B
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CONFIDENTIAL*/SULIT*
Paper 2 Answer Chemistry STPM Trial 2010
Section A
1 (a) (i) 1
(ii) 1 Bent / V-shape
(iii) 1 Right shape of sp3orbitals1 Label sp
3hybrid orbital of S atom and 1s orbital of H atom
(b) (i) 1 O is more electronegative than S /
Bonding pair electrons in the OH bond are drawn closer to the O atom compared to the
bonding pair electrons in the SH bond.
1 Repulsion between bonding pair electronbonding pair electrons in H2O in H2S1 bond angle H-O-H > bond angle H-S-H
(ii) 1 van der Waals forces between H2S molecules
1 Hydrogen bonds between H2O molecules
1 Hydrogen bonds are stronger than van der Waals forces
2 (a) (i) 1 pH is the measure of hydrogen ion concentration // pH = - log [H+]
(ii) pH = - log [H+] ; 3.5 = - log [H
+]
1 [H+] = 3.16 x 10
-4mol dm
-3
(b) (i) HX + NaOHNaX + H2O
111
10.05.270.25 x
Mx
1 M = 0.11 mol dm-3
(ii) 1 Orange juice is a weak acid. // Orange juice dissociates partially.
1 The concentration of the H+released is not the same as the concentration of the acid in the orange
juice.
(iii) 1 cK]H[ a
111.0
)10x16.3(
c
]H[K
242
a
= 9.1 x 1O-7
mol dm-3
(iv) 1 Orange juice is a weak acid (pH = 3.5) and NaOH is a strong base, the pHrange of phenolphthalein will enable it to produce a sharp end point.
1 Colour change is more prominent from colourless to pink
3 (a) (i) 1 Ge4+
>Sn4+
>Pb4+
(ii) 1 Ge2+
(iii) 1 Pb2+
1 The electrode potential for Pb4+
/Pb2+
is positive, this shows that Pb4+
is readily converted to Pb2+
.
(b) (i) 1 SiO2 / silica / silicon dioxide
(ii) 1 Telecommunication / endoscope.
(iii) 1 Optical fibres can transmit more information than copper wires.
1 Optical fibres have 100% efficiency/ no loss of signal
(c) 1 In glass, there is no regular arrangement of particles.
1 In diamond, the atoms are arranged in an orderly/ giant tetrahedral and closely-packed structure.
SH
H
H
S
H
sp3hybrid orbital of S atom [1]
1sorbital of H atom [1]
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CONFIDENTIAL*/SULIT* 3
1 Trigonal bipyramidal
(iv) 1 5 big iodine atoms cause steric hindrance
6 (a) (i) Zn(s) + Cu2+
(aq) Cu(s)+ Zn2+
(aq)From data booklet:
1 Zn2+
+ 2e- Zn - 0.76 V
Cu2+
+ 2e- Cu + 0.34 V
1 E.m.f . cell , E
cell
= + 0.34(- 0.76) = + 1.10 V
1 At equilibrium, Ecell = 0 V
Overall electrochemical equation is a follows.
Zn(s) + Cu2+
(aq) Zn2+
(aq) + Cu(s) E
cell= +1.10 V
1 0 = 1.10][
][log
2
059.02
2
Cu
Zn
1 K = 1.94 x 1037
(ii) Using ba
dc
cellcellBA
DC
nEE
][][
][][log
059.0
4.0
8.0log
2
059.010.11
cellE
09.11 cell
E V
(b) 1 2H+(aq) + SO3
2-(aq) H2O (l) + SO2(g) H
H2(g) + S(s) + 3/2O2 2H+(aq) + SO3
2-(aq) (i) - 623
S(s) + O2(g) SO2(g) (ii) -296
H2(g) + O2(g) H2O (l) (iii) -286
1 (iii) + (ii)(i) //
{[H2(g) + O2(g)+ [S(s) + O2(g)][H2(g) + S(s) + 3/2O2]}
{[H2O (l)] + [SO2(g)][2H+(aq) + SO3
2-(aq)]}
0 {[H2O (l)] + [SO2(g)][2H+(aq) + SO3
2-(aq)]}
1 H = (-286) + (-296)(-623) = + 41 kJ
(c) 1 A buffer solution is capable of maintaining the pH of its solution when a
small amount of acid or alkali is added to it.
1 H2CO3 H++ HCO3
-
1 NaHCO3 Na++ HCO3
-
1 When a small amount of acid is added, the hydrogen ion reacts with the HCO3-
ion //H
++ HCO3
- H2CO3
1 When a small amount of alkali is added, the hydroxide ion reacts with the
acid H2CO3 // OH-+ H2CO3H2O + HCO3
-
7 (a) 1 MCO3(s) MO(s) + CO2(g)* (*Physical state is optional)1 Thermal stability of carbonate increases as descending Group 2
1 Charge density of cation, M2+
decreases as descending Group 2 //Ionic charge of cation remains as 2+ and the cationic size increases as descending Group 2
1 The carbonate anion, CO32-
has relatively large electron cloud
1 Polarising power of cation, M2+
over the electron cloud of CO32-
ion gradually weaker //Electron cloud of carbonate anion is less polarised as descending Group
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CONFIDENTIAL*/SULIT* 4
(b) (i) 1 Electrolysis of molten bauxite with graphite anode and cathode
1 Cryolite (or) Na3AlF6; at about 950oC to 1000
oC
1 Cathode: Al3+
is reduced to Al
Al3+
+ 3e Al
1 Anode: O2
is oxidised to O2
2O2
O2+ 4e
(ii) 1 Light weight and strong //resistance to corrosion
(c) 1 Colourless gas is carbon dioxide.
1 Aluminium ion has high charge density.
1 The aluminium sulphate aqueous solution contains Al(H2O)63+
hydrated ions //
Al3+
(aq) + 6H2O(l) [Al(H2O)6]3+
(aq)
1 Al(H2O)63+
ion hydrolyses to produce H+/ H3O
+ // the salt solution is acidic //
Al(H2O)6]3+
(aq) + H2O(l) [Al(H2O)5 (OH)]2+
(aq) + H3O+(aq)
1 2Al(H2O)63+
(aq) + 3CO32
(aq) 2Al(OH)3(H2O)3(s) + 3H2O (l) + 3CO2(g)
8 (a) 1 The high temperature in the car engines N2to combine with O2to form NO(g).
1 N2(g) + O2(g) 2NO(g)
1 2NO + 2CO N2+ 2CO2// 2NO N2+ O2
(b) 1 HCl is thermally more stable than HI
1 Purple colour = I2(g)1 2HI H2 + I2
(c) (i) Co : N : Cl : H
19.5852.23 :
0.1400.28 :
5.3553.42 :
0.1
95.5
0.40 : 2.0 : 1.20 : 6.0
1 1 : 5 : 3 : 15
1 Empirical formula of Q = CoCl3(NH3)5
1 (CoCl3(NH3)5)n = 250 ; n =1
1 Q = [CoCl(NH3)5]2+
. 2Cl
(ii) 1 Co3+
= [Ar] 3d6, 3dorbitals are partially filled.
1 Under the influence of ligands, the 3dorbitals are split into 2 groups1 dd transition happens
1 Energy difference falls in the visible region
9 (a) 1 CH3COOH, HCOOH, ClCH2COOH
1 RCOOH(aq) + H2O(l) RCOO(aq) + H3O
+(aq)
Acid Conjugate base
1 CH3is electron releasing group and Cl is electron withdrawing group
1 CH3of ethanoic acid causes OH bond of theCOOH group to be less
polarised, thus, less able to ionise. //
CH3of ethanoic acid destabilises the CH3COOconjugate base and
shifting the equilibrium position to the left.
1 Cl of chloroethanoic acid causes OH bond of theCOOH group to be
more polarised, thus more able to ionise. //
Cl of the chloroethanoic acid stabilises the CH3COO conjugate base, thus, shifting the
equilibrium position to the right.
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CONFIDENTIAL*/SULIT* 5
(b) (i) 1 Reagent and condition: Acidified KMnO4solution; room temperature.
1 Cyclohexene decolourises the purple colour of acidified KMnO4solution while no obvious
observation for cyclohexane.
1 OH
+ (O) + H2O
OH
(ii) 1 Reagent and condition: Acidified KMnO4solution; Heat
1 Toluene decolourises the purple colour of acidified KMnO4solution while
no obvious observation for benzene.
1
(c) 1 Free radical substitution
1 Cl2 2Cl
1 CH3CH2CH3+ Cl (CH3)2 CH + HCl
1 (CH3)2 CH + Cl2 (CH3)2 CHCl + Cl
10 (a) (i) 1 phenol1 amide
(ii) 1 Alkaline hydrolysis
1 NaO NH2
1 CH3COONa
(b)
NO2 NO2
50oC-60
oC
CH3[1] + [1] [1] [1]
(c) H2N-*CH-COOH H2N*CHCOOH
CH2 CH2
OH COOH
[1] [1]
1 Has a carbon chiral
1 shown as * in the structure
1 Mirror images are non-superimposable1 rotates plane-polarised light
COOH + H2O
CH3Cl / FeCl3Conc H2SO4 /
Conc HNO3
CH3 + 3(O)