JAWAPAN KERTAS 1 : PERCUBAAN SPM 2008

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JAWAPAN KERTAS 1 :ujian 1SOALANJAWABANSOALAN JAWABAN

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CB

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DA

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Jawapan kertas 2

(a)(i)80oC110

(ii)Solid and liquid1

(iii)Heat energy is used to break the bond between particles2

(iv)

1

(b)(i)Sum of the number of proton and neutron in the atom1

(ii)D and E1

(iii)Both of the atoms have same proton number but different nucleon number1

(iv)2 . 41

(v)1

(a)(a)(i)Stanum110

(b)atom1

(c) (i)1

(ii) Saiz lain & label1+1

(b)(d)Atoms of other element added to the pure metal to make an alloy are different in size.

The size of tin atoms which are bigger than copper in bronze disrupts the orderly arrangement of atoms in pure metal.

When force is applied to an alloy, the presence of added other atoms prevent layers of atoms from sliding.

1+1+1

(e)To increase the resistance to corrosion of a pure metals

To improve the appearance of a pure metal1+1

(a)(i)Stanum1

(c)(b)atom1

(c) (i)1

(a) Increasing order of proton number1

(b)(i)Sodium / Na1

(ii)Chlorine / Cl1

(iii)Used in advertising light / television tubes1

(c)(i)Ionic1

(ii)2

(iii)Soluble in water // can conduct electricity in molten or aqueous state // high melting and boiling points1

(d)(i)Copper / Cu1

(ii)Formed a coloured compound // Act as catalyst // Formed a complex ion // Have various oxidation number110

(a) Cu2+, H+, SO42-, OH-1

(b)(i)SO42-, OH-1

(ii)OH-1

(iii)4OH- ( 2H2O + O2 + 4e1

(iv)Cu2+1

(v)Cu2+ is chosen to be discharged at Y because the position of Cu2+ in electrochemical series is lower than H+ // Oxidation number of copper decrease from +2 to 01

(c)Brown solid deposited1

(d)Blue colour is faded1

(e)Oxidation and reduction occurs simultaneously19

(a)Uniform scale for both Y and X axes

Correct transferring of points

Smooth curve of the graph3

(b)20 s1

(c)Na2S2O3 + H2SO4 ( Na2SO4 + S + SO2 + SO2 + H2O2

(d)Pressure // size of reactant // concentration of solution // presence of catalyst (Any 2)2

(e)When the temperature increases the kinetic energy of particle increases

Frequency of collision increases

Frequency of effective collision increases

Rate of reaction increases (Any 3)311

aSebatian karbon yang mengandungi unsure karbon dan hydrogen sahaja

2

b i)Alkana1

ii)butana1

iii)C 4H10110

c i)581

ii)0.0725 g ( tunjukkan carakira)4

MARKING SCHEME PAPER 2 SECTION B

NoAnswerMarks

(a)Number of mole in 16 g of oxygen = 16/32 // 0.05 mole

Volume occupied by 16 g of oxygen

= 0.05 mole x 24 dm3 // 12 dm3Number of mole in 22 g of CO2 = 22/44 // 0.05 mole

Volume occupied by 22 g of CO20.05 moles x 24 dm3 // 12 dm3

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1

1

1

4

(b)Able to determine the empirical formula and molecular formula of caffeine correctly

4 Empirical formula = C4H5N2O

5 [C4H5N2O]n = 194

6 [ 97 ]n = 94

7 n = 194/97 // 2

8 Molecular formula = C8H10N4O21

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8

(c)Able to calculate the molar mass and the percentage of nitrogen by mass in each of the three fertilisers and choose the best fertiliser.

Molar mass of ammonium sulphate = 132 g/mol

Percentage of nitrogen in ammonium sulphate = 28/132 x 100% // 21.2%

Molar mass of urea = 60 g/mol

Percentage of nitropgen in urea = 28/60 x 100% // 46.7%

Molar mass of hydrazine = 32 g/mol

Percentage of nitrogen in hydrazine = 28/132 x 100% // 87.5%

Hydrazine has the richest source of nitrogen compares with other fertilisers.

The farmer should choose hydrazine

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Total20

8(a)(i)

(ii)

(iii)

(iv)Blue solution X = Copper (II) sulphate

Colurless solution Y = potassium carbonate // sodium carbonate // ammonium carbonate

Double decomposition method

CuSO4 + K2CO3 ( CuCO3 + K2SO4 //

CuSO4 + Na2CO3 ( CuCO3 + Na2SO4 //

CuSO4 + (NH4 )2CO3 ( CuCO3 + (NH4 )2SO4

Correct rectants and products

Balanced equations

Add sodium hydroxide solution (until excess)

Blue precipitate formed

//

Add ammonia aqueous / ammonium hydroxide solution (until excess)

Blue precipitate soluble in excess1

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12

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2

(b)(i)Copper (II) oxide11

(ii)Materials : [25 100] cm3 of [0.5 2.0] moldm-3 copper (II)sulphate solution(any suitable answer) [25 100] cm3 of [0.5 2.0] moldm-3 sodium carbonate solution

(any suitable answer) Filter paper

Apparatus : Filter funnel, beakers, retort stand and clamp, glass rod and 100cm3 measuring cylinder.

Procedures :

About [25 100] cm3 of [0.5 2.0] moldm-3 copper (II) sulphate solution is measured into a beaker.

About [25 100] cm3 of [0.5 2.0] moldm-3 sodium carbonate solution is measured and mixed with the solution in the beaker.

The mixture is stirred with a glass rod.

The precipitate formed is removed by filtration.

The precipitate is rinsed with distilled water.

The precipitate is dried between the filter paper

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110

(c)2AgNO3 + MgCl2 ( 2AgCl + Mg(NO3)2No of moles AgNO3 = 50 x 1.0 / 1000 = 0.05 mol

2 mol AgNO3 ( 2 mol AgCl from the reaction

0.05 mol AgNO3 ( 0.05 mol AgCl

Mass of AgCl = 0.05 x 143.5

= 7.175 g1

12

9(a)(i)Atom R is located in Group 17, Period 3 1

(ii)Electron arrangement of atom R is 2.8.7. It is located in Group 17 because it has seven valence electron. It is in Period 3 because it has three shells filled with electron 1 11

(b)(i)Atoms P and R form covalent bond. To achieve the stable electron arrangement, atom P needs 4 electrons while atom R needs one electron. Thus, atom P shares 4 pairs of electrons with 4 atoms of R,forming a molecule with the formula PR4 // diagram

11111

(ii)Atom Q and atom R form ionic bond. Atom Q has the electron arrangement 2.8.1. and atom R has the electron arrangement 2.8.7. To achieve a stable (octet )electron arrangement, atom Q donates 1 electron to form a positive ion// equation Q Q+ + eAtom R receives an electron to form ion R-//equation and achieve a stable octet electron arrangement. R + e R- Ion Q+ and ion R- are pulled together by the strong electrostatic forces to form a compound with the formula QR// diagram

111111

(c)The ionic compound/ (b)(ii) dissolves in water while the covalent compound / (b)(i)does not dissolve in water. Water is a polar solvent that can cause the ionic compound to dissociate into ions. Covalent compounds are non-polar and can only dissolve in organic solvents.ORThe melting point of the ionic compound/ (b)(ii) is higher than that of the covalent compound/ (b)(i) . This is because in ionic compounds ions are held by strong electrostatic forces. High energy is needed to overcome these forces. In covalent compounds, molecules are held by weak intermolecular forces. Only a little energy is required to overcome the attractive forces.ORThe ionic compound/(b)(ii) conducts electricity in the molten or aqueous state whereas the covalent compound/(b)(i) does not conduct electricity. This is because in the molten or aqueous state, ionic compounds consist of freely moving ions. Covalent compounds are made up of molecules only1111111111111

Total

10(a)1.1V, 2.7V , 0.4 V3

(b)(i)Type of metal3

(ii)The cell voltage and the negative terminal

(iii)The type and concentration of electrolyte, copper electrode

(c)By using different pairs of metals and determining the negative terminal and cell voltage, an electrochemical series can be constructed3

(d)Suggested answer:

What you do, what you observed / measure

When two different pairs of metals are dipped into an electrolyte and the negative terminal is determined , a potential difference is produced3

(e)Magnesium, Zinc, Lead, Copper3

(f)(i)Blue to colourless/dark blue to light blue3

(ii)discharge of Cu2+- copper is deposited , concentration of Cu2+ in the electrolyte decreases3

(iii)Size of copper electrode increases when the time increases3

(g)More than 1.1 V but less than 2.7V // between 1.1 -2.7V3

(h)2,2

Pb + Cu2+ Cu + Pb2+ 3

3(a)(i)Electrodes

Anode

Cathode

Product

Oxygen gas

Hydrogen gas

Half-equation

4OH- O2+2H2O+4e

2H+ + 2e H2

4

(ii)Concentration of hydrochloric acid increases

Cl- and OH- ions attracted to the anode

OH- ions selectively discharged lower position in the electrochemical series

H+ ions attracted to the anode

H+ ions are discharged one type of ions only

Therefore only left with H+ and Cl- ions1

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1

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ElementCHNO1Mass / g0.480.050.280.162Number of mole0.48/12 //0.040.05/1 //0.050.28/14 //0.020.16/16 //0.013The simplest ratio0.04/0.01 // 40.05/0.01 // 50.02/0.01 // 20.01/0.01 // 1

R

R

R

R

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Q

R

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