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  • 8/13/2019 [Done Edu.joshuatly.com] Kedah STPM Trial 2010 Chemistry [w Ans] [33E48B52]

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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, 3d