Ahs Preliminary Examination 2008 2

AHS PRELIMINARY EXAMINATION 2008

ANSWER SCHEME SEC 4 CHEMISTRY (PAPER 1); 5072 / 1 1

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3

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10

C

C

A

C

D

D

B

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D

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C

C

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21

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C

C

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B

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31

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40

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A

SEC 4 CHEMISTRY (PAPER 2); 5072 / 2 SECTION A: A1.

(a)

A2.

(a)

Methane; (b) chlorofluorocarbons; (c) lead(II) nitrate; (d) phosphorus(V) oxide.

39

40

Number of protons

Number of electrons

Number of neutrons

19

K

19

19

20

19

K

19

19

21

(b)

A very vigorous/violent reaction takes place with the occurrence of an effervescence. Potassium floats on the surface of water and the gas evolved burns with a blue flame.

(c)

(i) (ii) (iii) (iv)

2K = 2OH- , # mol of K = 0.195/39 = 0.00500 # mol of OH- = 0.00500 # mol of H+ = 0.100 x 100 = 0.0100 (3 s.f.) 1000 H+ + OH- → H2O pH = 1 or 2 since an excess of 0.005 mol of H + ions were left behind after the neutralization has taken place completely.

A3.

A4.

(a)

Correct structure with outer shell electrons only and bonding .

(b)

H2O2 is reduced. The oxidation state of oxygen is reduced from -1 in H 2O2 to -2 in H2O. I- is oxidized to become I2 because there is an increase in the oxidation state of iodine from -1 to 0. Thus, it is a redox reaction.

(a)

Ethanoic acid is a weak acid/partially ionises. There is a low concentration of ions with fewer number of ions available to conduct electricity. CH3COOH (aq) + NaOH (aq) → CH3COONa (aq) + H2 O (I) All state symbols must be correct. The salt CH3COONa formed is soluble/fully ionises. Thus, there is a higher concentration/numerous number of mobile ions that results in an increase in conductivity. In addition, NaOH is a strong alkali /fully ionises, so there are numerous number of mobile ions present to conduct electricity. 40 cm3

(b) (c)

(d) A5.

(a) (b)

Zn (s) + I2 (aq) → Zn2+ (aq) + 2I- (aq) (i) Zinc because the final mass of zinc > 0. (ii) Final mass of zinc level higher, gradient less steep

G

(c)

Addition of aqueous ammonia to the sample. White ppt, soluble in excess aq. ammonia. (Both must be correct.)

2

A6.

(a) (b)

To allow the ions to be transported between the two aqueous electrolytes. B

(c)

(i) Cu2+ (aq) + 2e → Cu (s) (ii) Zn(s) → Zn2+ (aq) + 2e Blue solution gradually lightens and becomes colourless / decolourised. A reddish-brown deposit / pink deposit is observed.

(d) A7.

A8.

(a) (b)

(c)

C2H4 + H2 → C2H6 Energy absorbed to break the bonds in C2H4 & H2 = (600 + 4 x 435) + 436 = 2776kJ Energy released when bonds in ethane are formed = 348 + (6 x 435) = 2958kJ Enthalpy change, ΔH = + 2776 - 2958 = - 182 (kJ) Correct energy profile diagram with Ea & ΔH correctly labelled.

(d)

(i)

A reaction in which a large number of similar small molecules with C=C bonds joined up together to form a gigantic molecule or Many small unsaturated molecules chemically combined to form a gigantic molecule.

(ii)

Correct structure of ethene molecules & polyethene shown,

(a) (b) (c)

(e)

A gas, carbon dioxide, is produced and its escape into the air causes the loss in mass. To absorb the acid spray to prevent any loss of mass which leads to the inaccuracy in measurement. (i) Experiment II; There was a greater loss in mass of 8.3g from the Experiment II than that from the Experiment I. (ii) Experiment II; Since Experiment II had the faster speed of reaction and higher concentration of the acid led to faster reaction as a result of more collisions happened. Powdered calcium carbonate will react faster than lumps of calcium carbonate. Powdered form has larger surface area for contact and will lead to more frequent collisions between reacting particles and hence a faster reaction takes place. 3

A9.

(a)

Carbon monoxide might be formed due to the incomplete combustion of hydrocarbon. CuO is used to convert carbon monoxide to carbon dioxide through a redox reaction. (b) To remove/absorb all water moisture formed during combustion. (c) CO2 is acidic and sodium hydroxide is basic/alkaline. (d) Oxygen

SECTION B: B10. (a) (b) (c) (d)

B11. (a) (b)

Eddium, Alexium, Davium, Bobium, Charium Da2O3 2Da+ 3O 2key: x – electron of oxygen; o – electron of Da. (i) Forms coloured compound. (ii) Iron (iii) Fe + 2HCl → FeCl2 + H2 (iv) Add dilute sodium hydroxide to a sample of the green solution. When this mixture is filtered, the residue is the green precipitate [iron (II) hydroxide] and the filtrate is a colourless solution (NaCl). (v) Iron(II) hydroxide is oxidized by oxygen in air to form iron(III) hydroxide. H+ (i)

(ii)

(iii)

(c)

(i) (ii)

# moles of Mg = 0.24 = 0.0100 (mol) 24 # moles of HCl = 5.0 x 2.0 = 0.0100 (mol) 1000 But Mg Ξ 2HCl 1 mol 2mol Hence, only 0.00500 mole of Mg reacts with 0.0100 mole of HCl. Mg is in excess. Mg + 2HCl → MgCl2 + H2 1mol 1 mol 0.00500mol 0.00500 mol Mass of 0.00500 mole of MgCl2 = 0.00500 x (24 + 35.5 x2) = 0.475 (g) [3 significant figures] # mole of CH3COOH used = 5.0 x 2.0 = 0.0100 (mol) 1000 Hence, the same number of moles of CH3COOH was used to produce the same volume of H2 gas. Ethanoic acid (CH3 COOH) is a weak acid and it ionises partially to give a low concentration of H+ ions. Hence, the reaction is slower than that using HCl(aq). Na2CO3 + 2CH3COOH → 2CH3COONa + H2O + CO2 The carbonate dissolved and an effervescence occurred rapidly.

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EITHER B12. (a) (b) (c)

(d)

Halogens. Their electronic structures have seven valence/outer electrons.

WCl2 has a gigantic ionic crystal lattice structure with positive ion, W2+, and negative ion, Cl- , held together strongly by electrostatic forces of attraction. Thus it requires a lot more heat energy to break down this giant structure.that results in having a very high boiling point. YCl2 has a simple molecular structure with molecules held by weak intermolecular forces which requires little heat energy to separate these molecules, thus it has a relatively low melting point.

OR B12. (a)

(i)

(ii) (iii) (b)

(i) (iii)

(iv) (v) (vi)

In solid sodium chloride, the ions are held together tightly by strong electrostatic forces and hence the ions are immobile. In aqueous sodium chloride, all the ions become mobile that thus leads to its ability to conduct electricity. Electrode X : 2Cl- → Cl2 + 2e Electrode Y : 2H+ + 2e → H2 Red litmus solution turns blue. Hydroxide ions left in the electrolyte cause this change in colour of the litmus solution. Butanal, CH3CH2CH2CHO H │ H─C─C=O │ │ H O─H Reaction with steam in the presence of phosphoric(V) acid as a catalyst at a high temperature of 300oC and high pressure of 70atm. Using acidified potassium dichromate(VI) or potassium permanganate(VII). Propyl ethanoate.

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