Pat Chm Skima123

8/10/2019 Pat Chm Skima123

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PEPERIKSAAN AKHIR TAHUN 2014

SKIMA JAWAPAN KIMIA TINGKATAN 4

KERTAS 1

1C 6A 11A 16A 21D 26D 31C 36D 41D 4A62B 7A 12B 17D 22D 27D 32B 37A 42B 47A

3C 8A 13A 18A 23B 28A 33A 38C 43A 48C

4D 9B 14B 19C 24A 29C 34B 39B 44D 49B

5B 10D 15D 20D 25A 30B 35D 40C 45D 50A

KERTAS 2

ANSWER MARKS

1(a)(i) Total number of proton and neutron in an atom.

(ii) 18

(iii)

P 2.8.2

(iv) 20

(b) Refer graph

1

1

2

1

5

Total: 10

2 (a)(i) No. of moles = 6.0/24 = 0.25 molNo. of molecules = 0.25 x 6.02 x 1023= 1.505 x 1023

(ii) Mass = mol x RMM = 0.25 x 44 = 11 g

(b)(i) Chemical formula that shows the simplest ratio for each element in acompound.

(ii) Magnesium: 2.4 g Oxygen: 1.6 g(iii) 1:1

(iv) MgO(v) Copper is a less reactive metal than hydrogen in the reactivity series.(vi) To allow oxygen to enter the crucible for complete combustion.

2

1

1

21

111

Total: 10

3 (a)(i) Atoms of an element which have the same number of proton but different

number of neutron.(ii) To estimate the age of fossils and artifacts

(b) Proton

(c) 4

1

1

1

1


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2

(d) 612 P [A: proton number, Z: nucleon number]

(e)(i) 2Mg + O2 2MgO(ii) 2+ 2-

Mg O

1

23

Total: 10

4 (a) [Shows the direction on the diagram: Zinc rod Copper rod]

(b) Zinc

(c) Chemical energy Electrical energy

(d) Copper ion, Cu2+

(e)(i) Copper ions from X solution discharged to become copper atoms anddeposited at copper rod.

(ii) Cu2+

+ 2e Cu

(f)(i) The reading of the voltmeter is greater.

(ii) The distance between Mg and Cu in the reactivity series is greater than thatbetween Zn and Cu.

(g) Zn + Cu2+

Zn2+

+ Cu

1

1

1

1

1

1

1

2

1

Total: 10

5(a) P: 2.8.2 Q: 2.4

(b)(i) Ionic compound

(ii) Atom P donates 2 electrons, atom R receives 2 electrons

(iii) No, because there are no freely moving ions in solid state.

(c)(i)

R Q R

(ii) Molecules in a covalent compound are bonded with weak molecular forcesof attraction. Thus less energy required to break the bond.

1

1

2

2

2

2

Total: 10

6(a)(i) S

(ii) The higher the pH value, the higher the concentration of hydroxide ion,OH-which indicates the strongest alkaline solution.

(b)(i) Neutralisation(ii) Phenolphtalein // methyl orange

(iii) Pink to colourless // yellow to orange(iv) H+ + OH- H2O

1

2

11

11


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(c)(i) M1V1 = M2V2(0.1)(0.5) = M2(20.0)

M2 = 0.5 / 20

= 0.025 moldm-3

(ii) No.ofmol= 0.1 moldm-3/ 0.5 dm-3No.of ions = 0.2mol x6.02x1023before dilution = 0.2 mol = 1.204x1023

No.ofmol=0.025 moldm-3/20.0 dm-3No.of ions = 1.25x10-3x6.02x1023after dilution = 1.25 x 10

-3mol = 7.525x10

20

1

2

Total: 10

7(a)(i) 2.5(ii) Group 15 because the number of valence electron is 5. In Period 2 because

the number of shell-filled with electrons is 2.

(b)(i) - big and bright flame is produce

- explosion occur(ii) 2Rb + 2H2O 2RbOH + H2(iii) During handling the Group 1 metals, avoid direct contact with skin.

(c)(i)

Ionic compound Covalent compound

MgONaCl

C2H5OHC6H14

(ii)

Physicalproperties

Ionic compound (MgO) Covalent compound (C6H14)

Melting point andboiling point Ions are attracted by astrong electrostatic forceof attraction. More energy

required to overcome theforces thus the melting and

boiling points are high.

Molecules are bonded withweak molecular forces ofattraction. Less energy

required to break the bondthus the melting and boiling

points are low.

Electricalconductivity

In molten or aqueousstates, ions are freelymoving thus can conduct

electricity.

There are no freely movingions in all states thus cannotconduct electricity.

14

2

21

2

8

Total: 20


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8(a)(i) Q: 2.7 R: 2.4

(ii) Number of neutron = 10, Number of electron = 9

(b) Both Q and R are non-metal elements.

The reaction forms a covalent compound through the sharing of electron.Atom R with electron arrangement 2.4 has 4 valence electrons, thus need 4

more electrons to achieve octet electron arrangement.Atom Q with electron arrangement 2.7 has 7 valence electrons, thus need onemore electron to achieve octet electron arrangement.

One atom R contributes 4 electrons, while 4 atoms Q contribute one electroneach in the sharing. The compound formed is RQ4

(c) Group 1When going down the group, the number of shell increases.

The distance between nucleus and valence electron increases.Thus the strength of nucleus to attract electrons decreases down the group.

The reactivity increases as elements have a higher tendency to lose electrondue to weaker attractive forces from the nucleus.

Group 17When going down the group, the number of shell increases.

The distance between nucleus and valence electron increases.Thus the strength of nucleus to attract electrons decreases down the group.The reactivity decreases as elements have a lower tendency to gain electrondue to weaker attractive forces from the nucleus.

2

2

8

8

Total: 20

9(a)(i)

Glass cover

Gas jarBromine gas

(ii)Bromine gas

Gas jar

(iii) Procedures:

-

A few drops of liquid bromine were placed into a gas jar. The gas jarimmediately covered and set aside for a few minutes.

- Another gas jar containing air was inverted on top of the gas jar whichcontained bromine vapour, Br2

- The cover between the two jars was removed. The apparatus was setaside for a few minutes and observation was recorded.

-

Wear glove when handle bromine gas.

- Activity carried out in a fume chamber.

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(iv) Bromine gas contains discrete particles which possess high kinetic energy.

The particles are moving freely as the spaces between particles are furtherapart.

(b) Materials and apparatus- boiling tube, beaker, thermometer, tripod stand, retort stand and clamp,

Bunsen burner, wire gauze, stopwatch, conical flask.

Diagram:

I. Heating of naphthalene II. Cooling of naphthalene

Procedure

I. Heating of naphthalene1. A boiling tube was filled with naphthalene to a depth 3 cm and a

thermometer was put into it.

2. The boiling tube was suspended in a beaker half-filled with water using aretort stand and a clamp. The level of naphthalene in the boiling tube

below the level of water in the beaker.3.

The water was heated and the naphthalene was stirred slowly with the

thermometer until the temperature of the naphthalene reached 90o

C.

II. Cooling of naphthalene

1. The boiling tube in ( I) was removed from the water bath. The outersurface of the boiling tube was dried and immediately it was put into aconical flask. The naphthalene was stirred continuously.

2. The temperature of the naphthalene was recorded at half-minute intervalsuntil the temperature dropped to about 60oC.

Result

Time/s 0 30 60 90 120 150 180 210 240

Temperature/oC 90 86 82 78 78 78 75 72 60

2

1

2

5

2

Total:20


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10 (a). The stages in Contact Process:

IProduction of sulphur dioxide

Suphur is heated in dry air to produce sulphur dioxide

S + O2 SO2

IIProduction of sulphur trioxide

Sulphur dioxide is further oxidized by oxygen over vanadium(V) oxide, V2O5catalyst to produce sulphur trioxide at a temperature of 450 to 500oC.

V2O5

2SO2 + O2 2SO3450 - 500oC

1 atm

IIIProduction of sulphuric acid

Sulphur trioxide is dissolved in concentrated sulphuric acid to form oleum,H2S2O7

SO3 + H2SO4 H2S2O7

Oleum is then diluted with appropriate amounts of water to form sulphuric acid ina large scale.

H2S2O7 + H2O 2H2SO4

(b) Experiment to show the hardness of the alloy compared to its pure metal.

Apparatus set up:

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10


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KERTAS AMALI

ANSWER MARKS

1.

Titration 1 2 3

Initial burettereading (cm

3)

50.0 50.0 24.4

Final burette

reading (cm3)

24.2 24.3 49.0

Volume of P2

used (cm3)

25.8 25.7 24.6

(a) ( 25.8 + 25.7 + 24.6) / 3 = 25.4 cm3

(b) (i) NaOH + HCl NaCl + H2O

(ii) MHCl= 1.0 moldm-1

VHCl= 25.4 cm3

nHCl= (1.0) (25.4) / 1000= 0.0254 mol

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2

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4

Procedure:

1.

A steel ball is stuck to a brass block using cellophane tape.

2.

A weight of 1 kg is hung at a height of 50 cm from the top of the brass

block.

3. The weight is released so that it falls on the metal ball.

4.

The diameter of depression formed on top of the brass block is measured

using a metre ruler and the measurement is recorded in Table 9.5

5. Steps 1 to 4 are repeated twice on different spots on the brass block to

obtain an average of the diameter

6. The experiment is repeated using a pure copper block.

Tabulation of data:

Type ofBlock

Diameter of depression Averagediameter (mm)1 2 3

Brass

CopperTotal: 20


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From equation, 1 molHCl required 1 molNaOH

So, 0.0254 molHCl requires 0.0254 molNaOH

MNaOH = 0.0254 mol / 0.025 dm3

= 1.0 moldm-3

(c)

Pink to colourless

(d) When HCl is added slowly into NaOH solution mixed with

phenolphthalein, the pink solution turns colourless shows that the mixturehas neutralized.

(e) 12.7 cm3 (half of the volume of HCl)

1

2

2

Total: 20

2

Pair of electrodes Potential

difference (V)

Negative terminal

of the cellMagnesium andcopper

1.4 Magnesium

Metal X andcopper

1.0 Metal X

Iron and copper 0.5 Iron

Metal Y andcopper

0.1 Metal Y

Aluminium andcopper

0.2 Aluminium

Arrangement of metals:

Copper, Magnesium, metal X, Iron, Aluminium, metal Y

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1

Total : 10

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