SPM 2013 Chemistry
Paper 1 Answers (My Version)
1 D
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2 A
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3 D
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4 C
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5 B
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6 C
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7 B
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8 C
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9 A
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10 B
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11 D
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12 C
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13 C
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14 C
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15 D
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16 D
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17 C
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18 B
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19 A
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20 B
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21 D
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22 A
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23 B
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24 C
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25 C
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26 C
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27 D
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28 D
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29 C
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30 B
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31 A
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32 B
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33 A
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34 D
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35 D
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36 A
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37 A
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38 D
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39 B
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40 A
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41 D
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42 C
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43 A
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44 B
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45 C
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46 D
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47 A
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48 B
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49 A
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50 C
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My Comments
Q10 B: Pg 87 of Form 5 Chemistry KBSM Textbook shows that B is certainly correct. However, some felt that D (i.e. fats and oils are chemically different) is also an acceptable answer because - although fats and oils belong to the same homologous series of esters - most fats (the saturated esters) can’t be hydrogenated while oils (the unsaturated esters) can. Let me put it this way: If methanol
can’t be dehydrated into alkene, does that make methanol chemically different
from the rest of the alcohols which can be dehydrated? The answer to this question will decide whether D is also an acceptable answer in addition to B. :)
Q13 C Please see: pg 28 of Form 4 Chemistry KBSM Textbook – beside ‘Titbits’
Q16 D Please see: pg
15 of Form 5 Chemistry KBSM Textbook on ‘Characteristics of Catalyst’: the 2nd
and the 3rd characteristics given therein
Q28 D (E, Z, X, Y): Reason:
Y ] – the most electropositive
{ X ] ...0.7V
1.4V... { Z ])
{ E
) ... 1.2V – the least electropositive
Q35 D: (Generally
boiling point should be related to intermolecular forces – but within
the same Group of the periodic table such as Group 17 for halogens, going down the Group increases the atomic
size and also increases the melting and the boiling points and the converse is also true. Since bromine and chlorine in Q35 are in the same Group 17, candidates who choose A may not be
wrong either! :)
Q41 D: (Mg : N = 72 : 28 = 24 x 3 : 14 x 2.
Therefore, Mg3N2. (Answer:
D)
Q47 A: (HCl
= (20/1000) x 0.1 mol dm-3 = 0.002 mol
From stoichiometric coefficients
of the chemical eqn:
Therefore, Ca(OH)2 =
0.001 mol
So, (50/1000) x M = 0.001
M = 0.001
x (1000/50) = 0.020 mol dm-3 (Answer A)
Q48 B (From
Zn (excess) + 2 AgNO3 à Zn(NO3)2 + 2 Ag,
Excess Zn
will displace x mole of Ag from x moles of AgNO3.
Since, 20.0
cm3 of 0.1 mol dm-3 of AgNO3 = (20/1000) x 0.1
= 0.002 mol,
This means:
0.002 mol of Ag+ in AgNO3 solution will be displaced by
excess zinc to produce 0.002 mol of Ag and
Release heat
energy, E = mcӨ = 20 x 4.2 x 10 J = 840 J (per 0.002 mol)
Hence, the
Heat of Displacement = - 840 J/0.002 mol =- 420 000 J mol-1
= - 420 kJ
mol-1 (Answer B)
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My answers for SPM 2013 Modern Maths Paper 1 (1449/1) are here.
My answers for SPM 2013 Physics Paper 1 are here.
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1) Just email tutortan1@gmail.com; or WhatsApp: 018-3722 482. Act early to avoid disappointment! Currently in May 2016, existing students are about to finish their Summer exams. Their slots are up for grab starting now! All slots are normally taken up by end-June. Act now to avoid disappointment!
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