2013 SPM Chemistry Paper 1 Answers

SPM 2013 Chemistry Paper 1 Answers (My Version)

1 D	2 A	3 D	4 C	5 B	6 C	7 B	8 C	9 A	10 B
11 D	12 C	13 C	14 C	15 D	16 D	17 C	18 B	19 A	20 B
21 D	22 A	23 B	24 C	25 C	26 C	27 D	28 D	29 C	30 B
31 A	32 B	33 A	34 D	35 D	36 A	37 A	38 D	39 B	40 A
41 D	42 C	43 A	44 B	45 C	46 D	47 A	48 B	49 A	50 C

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 2^nd and the 3^rd 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)

Comments or email me if you have different answers. Thank you.

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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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