Spm Chemistry Trial 2009 Imes2009answer

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Chemistry

Soalan

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PAPER 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

B B C C D D C B C B A C C B A C B A C B A B A D C

26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50

D C D B A A B A D A C D C B D C B B D D D B A C C

Ulang

kaji

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SOALAN ULANGKAJI SPM 2009 PAPER 2 SECTION A

(d)

(e) (f)

2.(a) (b) (c)(i) (c)(ii) (c)(iii) (d)(i) (ii) (iii)

(e) 3.(a) (b) (c) (d) (e) (f) (g)(i) (ii) (h)

Answer Zn + 2HCl ZnCl2 Yellow powder changes to grey solid. To avoid the hot lead metal from being oxidised to form lead (II) oxide again. Element Mass/g Number of moles Ratio of moles

Pb 35.65-25.30=10.35 10.35÷207=0.05 1 Empirical Formula: PbO

O 36.45-35.65=0.8 0.8÷16=0.05 1

To ensure that all the lead (II) oxide has been reduced to lead completely Allow the hydrogen gas to flow through the apparatus for an interval of time before heating.Continue the flow of hydrogen gas after heating until the hot deposited is cooled. Ethanol Oxidation Ethene gas The purple solution of potassium manganate (VII) is discoloured. Porcelain chips C2H4 + H2O C2H5OH Esterification C2H5OH + CH3COOH CH3COOC2H5 + H2O 1. Ethanoic acid (compound C) reacts with metal carbonates to form salts,carbon dioxide and water. 2. Ethanoic acid reacts with bases to form salts and water. Compound D have sweet odour. Zinc carbonate ZnO + CO2 ZnCO3 Zinc nitrate Nitrogen dioxide, Oxygen 2ZnO + 4NO2 +O2 2Zn(NO3) ZnO + 2HCl ZnCl2 + H2O A white precipitate is produced The white precipitate dissolves. The solution obtained in (f) contains zinc ions (Zn2+)

Mark 1m 1m 1m 1m

1m 1m 1m Total: 10m 1m 1m 1m 1m 1m 1m 1m 1m 1m 1m Total:10m 1m 1m 1m 1m 1m 1m 1m 1m 1m 1m Total:10m

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Question 1.(a) (b) (c)

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4.(a) (b)

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(c)

(d) (e) (f)

(g) 5. (a) (b) (c) (d)(i) (ii) (e)(i) (ii) (f)(i) (ii) 6. (a) (b)(i)

Cu Cathode:Cu2+ + 2e Anode : 2Cl Cl2 + 2e 1. Pinkish brown solid ,copper is deposited at the copper electrode. 2. Green copper (II) chloride solution fades and decolourises. Neutral. Cu2+ ions and Cl- ions are discharged at the electrodes.The resulting solution contains H+ and OH- ions which are water molecules. Thus,it is neutral. The colourless solution of potassium iodide turns brown. 2Cl- + I2 Cl2 + 2IOH- ions, being more concentrated than Cl- ions, are selectively discharged at the anode to produce oxygen gas and water molecule. Use a glowing wooden splinter. Oxygen gas relights the glowing wooden splinter. Dilute sulphuric acid 0 to +2 Zinc foil because zinc is more electropositive than copper Copper foil becomes thicker. Zinc foil becomes thinner. Zinc foil. Copper (II) ion Zn Zn2+ + 2e Cu Cu2+ + 2e ZnSO4 + H2 Zn + H2SO4 Number of moles of H2SO4 that reacts = 50 x 0.2 1000 = 0.01 mol From the equation in (a), 1mol of sulphuric acid that reacts will release 24dm3 of Hydrogen gas at room condition. -Therefore If 0.01mol of sulphuric acid reacts the volume of hydrogen released = 0.01 x 24 = 2.4dm3

1m 1m 1m 1m 1m 1m

1m 1m 1m

1m Total:10m 1m 1m 1m 1m 1m 1m 1m 1m 1m 1m Total: 10m 1m 1m

1m

1m

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SOALAN ULANGKAJI SPM 2009

(c) (d)

Number of mol H2SO4 that reacts =50x0.1 1000 =0.005 mol From the equation in (a), 1 mol of sulphuric acid that reacts will released 24 dm3 of hydrogen gas at room condition. Therefore, If 0.005 mol of sulphuric acid reacts the volume of hydrogen gas released. =0.005 mol x 24 dm3 =0.12 dm3. Experiment 1 Because the concentration of sulphuric acid used is higher. Volume of hydrogen gas/cm3

240

1m

1m 1m 1m

Experimen 1

Experimen 2

0 Time/minutes (e)

Using a hotter dilute sulphuric acid.

2m 1m Total:10m

ANSWER SCHEME: SECTION B

Question 7(a)

Answer 1.Size of solid particles 2. Concentration of the solution 3. Temperature 4. Pressure 5. Catalyst

Mark 1m 1m 1m 1m 1m Total: 5m

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(b)(ii)

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7(b)

(i) 1. They speeds up reactions. 2. They are chemically unchanged at the end of reaction. 3. Only a small amount of catalyst is required. 4. Most catalysts are very specific and only catalyse a particular reaction.

(ii)- When reactions take place,bonds in the reactants must be broken before new bonds can be formed in the products. - This bond breaking requires a minimum energy which is called the activation energy. - In a catalysed reaction, the activation energy is lowered and less energy is required to break the bonds. - As a result,more effective collision occur and bond breaking takes place more easily. - Hence ,catalysed reactions are faster. (iii) Process Contact process Haber process Cracking of petroleum 8.(a)

8(b)

Catalyst Vanadium (VI) oxide Iron Platinum

- Atom U is smaller in size than atom T - Atom U has more protons in the nucleus - Attractive forces between the nucleus and the electrons in the shells are stronger. - Therefore,the electron filled shells are attracted closer to the nucleus. - T is more reactive . - The electron arrangement of T is 2.8.1 and S is 2.1 - The valence electron of atom T is further from its nucleus. - The attractive forces between the nucleus of atom T towards the nucleus of atom T towards the valence electron is weaker. - Therefore the tendency of atom T to donate the valence electron is higher.

1m 1m 1m 1m Total : 4m

1m 1m 1m 1m 1m Total: 5m

6m

1m 1m 1m 1m Total: 4 m 1m 1m 1m

1m 1m Total: 5m

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8(c) 1m Procedure:

1. A spatula of iron filings is put into the combustion tube. 2. Gas U is passed over the hot iron filings. 3. The iron filings are heated very strongly.

1m 1m 1m

- The iron filings glow brightly and a brown solid is formed.

1m

Chemical equation: 2Fe + 3 U2

8(d)

2FeU3

- S has the electron arrangement of 2.1. - S is located in Group 1 - because S has 1 valence electron. - S is located in Period 2. - because S has 2 electron filled shells.

1m Total: 6m 1m 1m 1m 1m 1m Total: 5m

ANSWER SCHEME: SECTION B Question 9(a)

Answer ∙ Copper (II) sulphate is prepared using Method A ∙ Copper (II) sulphate is a soluble salt which requires crystallisation to obtain the salt. ∙ Lead (II) sulphate is prepared using Method B as lead (II) sulphate is an insoluble salt. ∙ Using method A for preparation of lead (II) sulphate would result in incomplete reaction as salt form ed will be deposited on metal oxide,reducing the surface area of the reactants and eventually will stop the reaction.

Mark 1m 1m 1m 1m 1m

1m Total=6m

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Observation:

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9(b)

∙Preparation of copper (II) sulphate using method A: - Add one spatula of copper (II) oxide to 25 cm3 of 1 mol dm-3 of hot dilute sulphuric acid. - When it has reacted,add further amounts until it can no longer dissolve. - Filter the mixture. - Evaporate the filtrate to one third of its volume or until the solution becomes saturated. - Allow the saturated solution to cool so that the salt crystallises out. - Filter to obtain the crystals. - Wash with a bit of distilled water and dry between the sheets of filter paper.

1m 1m 1m 1m 1m 1m 1m 1m 1m

∙ Preparation of lead (II) sulphate using method B: - Mix lead (II) nitrate solution with sodium sulphate solution. - Stir the mixture well. - Filter the mixture. - The residue is barium sulphate salt. - Wash the salt with distilled water. - Then dry the salt between sheets of filter papers.

10(a)

- R is a covalent compound. - The compound has a low melting point. - and it is insoluble in water.

10(b)

- R has a low melting point. - Attractive forces between the molecules are weak. - less heat energy is needed to overcome the weak forces of attraction.

10(c)

- (i) Q has a high melting point -(ii) It can conduct electricity in aqueous solution or molten state. -(iii) It is soluble in water.

1m 1m 1m 1m 1m Total: 14m 1m 1m 1m Total: 3m 1m 1m 1m Total: 3m 1m 1m 1m Total: 3m

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- Apparatus: Spatula,Bunsen burner,tripod stand,batteries,carbon electrodes,electrode holder,bulb,connecting wires with crocodile clips,switch,pipe-clay triangle ,crucible - Material: Powder R,Powder Q

2m - Procedure: 1. One spatula of Q powder is placed in a crucible. 2. Two carbon electrodes are dipped into Q and the circuit is completed by connecting to the batteries and switch. 3. Powder Q is heated until its melts and the switch is turned on again. 4. Steps 1-3 are repeated by replacing powder Q with powder R. - Observation: Compound Physical States Q Solid Molten R Solid Molten 10(e)

Observation The bulb does not light up. The bulb lights up. The bulb does not light up. The bulb does not light up.

- R consists of neutral molecules. - No ions to carry electrical charges.

1m 1m 1m 1m

2m Total: 9m

1m 1m Total: 2m

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1m

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SOALAN ULANGKAJI SPM 2009 PAPER 3 ANSWER SCHEME Question 1(a)(i) 1(a)(ii) 1 (b)

Answer The temperature [1] of naphthalene powder increases[1] when it is heated [1] When molten naphthalene is cooled down,the temperature falls. To ensure an even temperature during the cooling of naphthalene in order to avoid supercooling

Mark 3m 3m 3m

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1(c) (i)

1(c)(ii)

1 (d)(i) 1 (d) (ii) 1(e)(i) 1(e)(ii) 1(e)(iii) 1(f)(i)

A water bath is used to heat naphthalene. Naphthalene is continuosly stirred during heating of cooling. Time taken to heat/cool Temperature of naphthalene Naphthalene

3m 3m 1m 1m 1m 3m

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SOALAN ULANGKAJI SPM 2009

1(g) 2 (a)

2 (b)

2 (c)

3m

The melting point =The freezing point=80oC -Chlorine oxidises Fe2+ ions to Fe3+ ions through the transfer of electronsat a distance from the reducing agent to the oxidising agent. Manipulated: Chlorine and Fe2+ ions Responding : Products of reaction deflection of Galvanometer needle. Constant : Volume of oxidizing and reducing agents Substances: Chlorine water,dilute sulphuric acid,iron (II) sulphate solution,sodium hydroxide solution.

3m 3m

3m

3m

Apparatys: U-tube,carbonelectrodes,electric wires with crocodile clips,galvanometer,stoppers,dropper 2(d)

Procedure of the experiment

:

1. The apparatus is set up as shown in the diagram and left aside for 30 minutes. 2. Observations before an after the experiment are recorded. 3. At the end of the experiment, a little solution is removed from the negative terminal with a dropper and tested with sodium hydroxide solution. 2(e)

Tabulation of data; Experiment Direction of electron flow

Change to FeSO4 Changes to Cl2 Reaction of iron salt with NaOH

3m Observation Deflection of galvanometer shows electron flow from the negative to positive terminal. Light green to yellow in colour. Greenish yellow to colourless Forms brown precipitate,insoluble in excess alkali.

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1(f)(ii)