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1 a i Haber process 1ii Temperature: 450°C

Pressure: 200 atmCatalyst: ferum

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iii N2 + 3H2 → 2NH3 2Iv 1 mol of N2 react with 3 mol of H2 to produce 2 mol of NH3. 1

b i Vanadium (V) oxide 2ii oleum 1

2 a i 8 1ii 2.6 1iii Group 18

Period 211

b Ionic bond 1ii P2Q 1iii High melting and boiling point // dissolve in water but not dissolve in organic

solvent // conduct electricity in molten state or aqueous state.1

c i RQ2 1ii 2

d Atom S has achieved stable octet electron arrangement. 1

3 a i Water 1ii 1

b i Purple acidifie d potassium manganate (VII) decolourised 1ii Calcium butanoate // water // carbon dioxide (either one product) 1

c i -OH or hydroxyl group 1ii Esterification 1iii 2

4 a i Speed at which reactant are converted into product in a chemical reaction 1b Zn + 2HCl → ZnCl2 +H2 2c i No of mol of H2 gas (from the graph) = 0.030 / 24 = 0.00125mol

From the equation, 1 mol of H2 is produce by 2 mol of HCl 0.00125 mol of of H2 is produce by 0.0025 mol of HClConcentration of HCl = 0.0025 x 1000 / 50 = 005 mol dm-3

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ii Average rate of reaction = 30 cm3 / 4 min = 7.5 cm3 min-1 1

Negeri sembilan 2013 Trial Page 1

d The concentration of HCl decrease with time //The number of mol of H+ decrease with time.The frequency of collision between H+ and Zn decrease with time.The frequency of effective collision between H+ and Zn decrease with time.

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e ii 1

5 a i Ag+, NO3-, OH-, H+ 1

b The iron spoon should be connected to cathode not anode. 1c i A shiny silver deposit on iron spoon 1

ii Ag → Ag+ + e- 1d i Draw arrow from iron spoon to silver tod 1

ii Iron spoon. Iron is more electropositive than silver . 2iii Ag+

iv Increase. The distance between zinc and silver is further than the distance between ferum and silver in electrochemical series.OrThe zinc is more electropositive than ferum.

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6 a i 1

ii Hard water contain magnesium ions (Mg2+)or calcium ions (Ca2+).Soap will form scum with hard water but detergent will not form scum with hard water.Amount of detergent is more than soap in the cleansing action.

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1b i Analgesics 1

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iii Causes internal bleeding in stomach // ulceration 1 iv Paracetamol 1


7 a i Element C H OMass (g) 40 6.7 53.5No of mol 3.33 6.7 3.33Simplest mol ratio 1 2 1

Empirical formula = CH2O

Determine molecular formula

(CH2O)n = 60

30n = 60

n=2

molecular formula = C2H4O2

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ii 2CH3COOH + 2Na → 2CH3COONa + H2 2b i 2.8.8.1 1

ii Group 1 because contain 1 valence electronPeriod 4 because contain 4 shells filled with electrons

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c i 1. A small piece of Q is cut by using a knife2. The oil on the surface of K is removed by rolling it on a piece of filter paper3. The Q is placed slowly onto the water surface in water through by using a forceps. 4. Observed the reactivity of Q on the water surface.

Observation: Q will move vigorously and randomly on the water surface and produce a “hiss” and “pop” sound. An alkaline solution is produced.

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1ii Q is more reactive than sodium when react with water

The number of proton of Q is more than Na // The size of Q is bigger than NaThe distance between the nucleus and the valence electron in Q is further than the Na.The force of attraction between the nucleus and valence electron in Q is weaker than Na.Valence electron in Q can be release more easily than Na.

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8 a Ethanoic acid is a weak acid which ionised partially in water to produce low concentration of hydrogen ions. Thus it pH value is higher.

Nitric acid is a strong acid which will ionised completely in water to produce high concentration of hydrogen ions. Thus it pH value is lower.

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b Ammonia need water to show it alkali propertyAmmonia will only ionise in water to produce OH- which can turn red litmus paper to blue.Ammonia in propanone is in molecule form.No OH- presents to change the colour of red litmus paper to blue.

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c i ZnCO3 + H2SO4 → ZnSO4 + CO2 + H2O 1


1. 50 cm3 of 0.5 mol dm-3 of H2SO4 is measure by using a measuring cylinder and pour in a beaker. Heat the acid

2. A spatula of ZnCO3 powder is added bit by bit to the hot acid until it cannot dissolve.

3. To unreacted ZnCO3 is filtered and the filtrate is pour in evaporating dish.4. The salt solution is gently heated to evaporate the solution to obtain a saturated

solution5. The hot saturated solution is allowed to cool for crystallisation take place.6. The crystal formed are filtered out, washed with a little cold distilled water and

dried between by pressing between two sheets of filter paper.7. The crystal can be recrystallised to obtain a pure crystal of ZnSO4.

ii ZnCO3 + H2SO4 → ZnSO4 + CO2 + H2O 100 cm3 mass=? 1.0 mol dm-3

No of mol of H2SO4 = MV /1000 = 1 x 100 / 1000 = 0.1 molBased on equation, 1 mol of H2SO4 produce 1 mol of ZnSO4

0.1 mol of H2SO4 produce 0.1 mol of ZnSO4

Mass of ZnSO4 = 0.1 x [64 + 32 + 4 (16)] = 16.1g

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9 a i Sodium hydroxide solution 1

ii 3

iii The reaction between HCl and NaOH to produce NaCl and H2O is exothermic reactionThe temperature of the mixture increaseThe total energy of 1 mol of HCl and 1 mol NaOH is more than the 1 mol NaCl and 1 mol H2O.When 1 mol of HCl react with 1 mol NaOH to produce 1 mol NaCl and 1 mol H2O , 57kJ of heat is release.

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b Suitable example: silver nitrate solution and sodium chloride solution

Apparatus and materials :50 cm3 of 0.5 mol dm-3 of sodium chloride solution, 50 cm3 of 0.5 mol dm-3 of silver nitrate, polystyrene cup, thermometer, measuring cylinder.

Procedure:1. 50 cm3 of 0.5 mol dm-3 of silver nitrate solution is measure and pour in a

polystyrene cup.2. The initial temperature of silver nitrate solution is measured after a few minutes.3. 50 cm3 of 0.5 mol dm-3 of sodium chloride solution is measured and pour in a

polystyrene cup.4. The initial temperature of the sodium chloride solution is measured after a few

minutes.5. The sodium chloride solution is added quickly and carefully to the silver nitrate

solution. The mixture is stirred with a thermometer.

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HCl + NaOH

NaCl + H2O

= -57kJ/mol

6. The highest temperature of the reacting mixture is measured and recorded.

To calculate the heat released:No of mol of Ag+ = 50 x 1 / 1000 = 0.05 molNo of mol of Cl- = 50 x 1 / 1000 = 0.05 mol

Temperature increased after reaction = θ Heat released , H = mcθ = (50+50) x 4.2 x θ = 420θJ

Ag+ + Cl- → AgCl

0.05 mol of Ag+ react with 0.05 mol of Cl- and released 420θJ of heat1 mol of Ag+ react with 1 mol of Cl- and released 8400θJ of heat

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10 a Oxidation: zincBecause zinc gain oxygen

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b i P is chlorineOxidation: Fe2+ → Fe3+ + e-

Reduction: Cl2 + 2e- →Cl-

Fe2+ undergo oxidaiton by releasing 1 electron and Cl2 undergo reduction by gaining 2 electrons.Fe2+ is reducing agent and Cl2 is oxidising agent.

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1ii Q is zinc.

Oxidation : Zn → Zn2 + 2e-

Reduction: Fe3+ + e- → Fe2+

Zinc is reducing agent and Fe3+ is oxidising agent.Zinc undergo oxidaiton by releasing 1 electron and Fe3+ undergo reduction by gaining 1 electron.

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c Oxidising agent: chlorine water (any other suitable answer)Reducing agent: iron (II) sulphate solution (any other suitable answer)

1. A U-tube is clamped to a retort stand

2. Dilute sulphuric acid is poured in the U-tube until it level are 6 cm away from the mouth of the U-tube

3. Using a dropper, 0.5 mol dm-3 of FeSO4 solution is carefully added to one of the arms of the U-tube until the layer of FeSO4 solution reach the height of 3 cm arms.

4. In a similar manner of step 3, chlorine water is added to the other arm of the U-

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tube.5. A carbon electrode is placed in each arm of the U-tube.6. The electrode are connected to a voltmeter7. Based on the deflection of the voltmeter, it can prove that electrical energy can be

produced by redox reaction by transfering of electrons at a distance.

ObservationPositive terminal: The pale yellow chlorine water decolourisedNegative terminal: Pale green FeSO4 solution change to yellow.

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Paper 3

1 a 77.0, 71.0, 66.0, 65.0, 65.0, 65.0, 65.0, 59.0 3b i Time (min) 0 1 2 3 4 5 6 7

Temperature (°C)

77.0 71.0 66.0 65.0 65.0 65.0 65.0 59.03

ii To get full marks, make sure your graphSize of graph more than halfLabel the axis with unitUniform scaleAll points transfer correctly

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iii To get full marks, make sure you got draw a dotted line on the melting point on your graph and label it.

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c i The temperature decrease for the first 2 minutes. At the 3rd until 6th minutes, the temperature become costant. After 6th minute, the temperature start to decrease again.

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ii For the first 2 minutes, the liquid start to cool down.At the 3rd minute until 6th minutes, the temperature become constant because heat loss to surrounding is balance by the heat energy released by the particle to form solid.After 6th minute, the solid X release heat energy and so continue to decrease in temperature.

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d Molecule: glucose, methylbenzene, ammoniaIon: potassium chloride, copper (II) sulphate

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2 a Manipulated variable: type of metalResponding varible: brighteness of the flameConstant variable: mass of metal used

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b The higher the reactivity of metal, the brighter the flame will be produced when react with oxygen.

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c Glow dimly 3d When magnesium powder is heated with oxygen gas, it will burn with a bright 3


flame.


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Negeri sembilan 2013