ujian pengesanan pertengahan tahun 2009_mark scheme chemistry 1 and 2

7/31/2019 Ujian Pengesanan Pertengahan Tahun 2009_Mark Scheme Chemistry 1 and 2

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SULIT

4541/2

Chemistry

Paper 1

Paper 2

May 2009

Marking Scheme

JABATAN PELAJARAN MELAKA

__________________________________________________________________________________________

UJIAN PENGESANAN PERTENGAHAN TAHUN

SIJIL PELAJARAN MALAYSIA 2009

MARKING SCHEME

CHEMISTRY

Paper 1 and Paper 2

Kertas soalan ini mengandungi 10 halaman bercetak

4541/2 [Lihat sebelah


2/10

SKEMA PERMARKAHAN

CHEMISTRY 1

4541/1

No

soalanJawapan

No

soalanJawapan

No

soalanJawapan

No

soalanJawapan

No

soalanJawapan

1 D 11 D 21 D 31 B 41 B

2 C 12 A 22 C 32 B 42 A

3 A 13 B 23 A 33 C 43 A

4 D 14 C 24 C 34 C 44 B

5 C 15 B 25 B 35 B 45 D

6 D 16 D 26 A 36 D 46 A

7 A 17 C 27 A 37 A 47 A

8 B 18 D 28 C 38 A 48 D

9 B 19 D 29 B 39 C 49 D

10 B 20 B 30 C 40 C 50 A

2


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PAPER 2 SECTION A

No Marking Scheme Mark 1(a)

(b)

(i)

(ii)

(iii)

(i)

(ii)

14

B

Correct number of shells

Correct number of electrons at each shellsCorrect number of proton in nucleus

Correct number of neutron in nucleus

Examples:

P: liquid

S: solid

Has weak force of attraction between moleculesNeed low energy to overcome the force of attraction

1

1

1

11

1

1

1

11

1

1

4

2

2

10

2 (a) 1. Formula that shows the simplest ratio2. the number of atoms for each element in the compound. 1

(b) Anhydrous calcium chloride//concentrated sulphuric acid 1

(c) To remove all the air in the tube 1

(d) 1. Repeat the process of heating, cooling and weighing2. until a constant mass is obtained

1

1 2

(e) 1. [Determine the mass of Pb and O correctly]

2. [Calculate the no. of mole of Pb and O correctly]

3. [Determine the simplest ratio of mole of Pb and O correctly]4. [State the empirical formula correctly]

Examples:

Pb OMass/g 4.14 g 4.78 4.14 //

0.64 g

No.of mole 4.14/207 // 0.02 0.64/16 // 0.04

Simplest ratio 0.02/0.02 = 1 0.04/0.02 = 2

Empirical formula PbO2

1

1

11 4

(f) Magnesium is more reactive than hydrogen // Hydrogen is less reactive thanmagnesium // Hydrogen cannot reduce the magnesium oxide // Hydrogen is

below magnesiumin reactivity series.

1

3


4/10

10

3(a)

(b)

(c)

(d) (i)

(ii)

(iii)

1. Correct formula of reactant

2. Correct formula of product

3. Balance of equation

Examples:4Na + O2 2 Na2O

Colourless turn to pink

Burn very rapidly//White fume are produced//produced white solid

when cooled.

Have 3 shells occupied/fill with electrons

Decrease

Number of shell filled with electron still same/threeNumber of proton and electron increase

The nucleus attraction will increase // nuclei attraction on the valence

electron increases

1

1

1

1

1

1

1

11

1

3

1

1

1

1

3

10

4 (a) 1. Substances that conduct electricity

2. in molten state or aqueous solution

3. undergoes chemical changes

1

(b) Mark (+) at electrode P and (-) at electrode Q correctly 1

(c) (i) Brown gas 1

(ii) Bromine 1

(d) (i) Na+ , Cl- , H+ , OH- 1

(ii)

Electrode R Electrode S

Cl- , OH- Na+ , H+

1 mark 1 mark 1+1 2

(iii) Chloride 1

(iv) Concentration of chloride ions is higher than hydroxide ions 1

(v) 2Cl- Cl2 + 2e 1

10

4


5/10

5(a)

(b)

(c)

(d)

(i)

(ii)

1. Functional apparatus and labeling of water

2. [Correct apparatus to collect gas] e.g measuring cylinder

1. Correct axis2. Correct scale

3. Transferring of point and smooth curve

1. Able to draw the correct tangent at 40 s

2. Able to state the rate of reaction with correct unit

Range : 0.85 k 0.88

1. [Able to draw the graf which steeper than the first graph

and has the same maximum volume of gas collected]

2. Labelled the graph.

Mass of zinc decrease.

1

1

11

1

1

1

1

1

1

2

3

2

2

1

10

6 (a) (i) Alcohol 1

(ii) OH H H H H OH

l l l l l lH- C C C H or H C C C H

l l l l l l

H H H H H H

1

(b) 1. Correct formula of reactants and products

2. Balance the equationExample:

C3H7OH + 9/2O2 3CO2 + 4H2O

1

1 2

(c) Propanoic acid 1

(d) (i) Dehydration 1(ii) 1. [Functional diagram]

2. [Labeled Al2O3, glass wool + propan-1-ol and heat]

1

1 2

(e) 1. Hexene burns with more sooty flame

2. Percentage of carbon by mass in hexane is higher// [show

calculation]

1

1 2

10

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PAPER 2 SECTION B

7

(a)

(b)(i)

(ii)

(iii)

(iv)

Change of quantity of reactant/ productTime taken

1. Formula of reactants2. Formula of products

3. Balance chemical equation

Mg + 2HCI MgCI2 + H2

1mol Mg = 1 mol H2

0.2g X 24 = 0.2 dm3

24

1. Both axes labeled with correct unit

2. [Experiment III is more steeper than Experiment II ,Experiment I is the less steeper]

3. Same maximum volume of gas collected

Experiment

I

II

III

0.2 / 0.00450

0.2 / 0.01020

0.2 / 0.01315

1.Concentration of acid in experiment II is higher than

experiment I.2. Number of particles per unit volume is higher in Exp. II.

3. Collision between particles is higher in Exp. II.

4. Frequency of collision between particles is higher inExp. II.

5. Frequency of effective collision also higher in Exp. II.

6. Experiment III using catalyst but not in Experiment II

7. Catalyst lower the activation energy.

8. Frequency of effective collision in Exp. III is higher thanin Exp. II.

1

11

1

1

1

1

1

1

1+1+1

1

1

1

1

1

1

1

1

1

5

3

3

8

20

6


7/10

8 (a) (i) 1. [Example of strong acid]H2SO4 // Sulphuric acid // HCl // Hydrochloric acid // HNO3 //

Nitric acid

2. Strong acid ionised completely in water

3. formed H+ with higher concentration4. [Example of weak acid acid]

CH3COOH // ethanoic acid5. Weak acid ionised partially in water6. formed H+ with lower concentration

11

1

11

11

Max 4

(b) (i) 1. Correct formula of reactants

2. Correct formula of products

3. Balance

H2SO4 + 2 KOH K2SO4 + 2 H2O

4. The no. of mol of H2SO4 = 22 x 0.1/1000 // 0.0022

5. Mol of KOH = 2 x 0.0022 // 0.0044

6. Molarity of KOH = 0.0044x 1000 /25 //

= 0.176 mol dm -3

OR

0.1 x 22 1

4. =

Mb x 25 2

5. Mb = 2 x 0.1 x 2225

6. 0.176 mol dm-3

1

1

1

1

1

1 6

7


8/10

8 (c) [chemical substances - 1 mark]

[ apparatus - 1 mark]

[procedure - 7 marks]

[chemical equation - 1 mark]

Sample answer :1. [chemical substances]

magnesium powder /magnesium oxide(Hydroxide, carbonate)

and sulphuric acid

2. [ apparatus]

Beaker, filter funnel, filter paper, glass rod

[procedure]3. [20 100] cm3 sulphuric acid [0.5 2.0 ] mol dm-3 is poured

into a beaker

4. magnesium powder is added to sulphuric acid untilundissolved /excess / no more effervescence

5. Stir6. the mixture is filtered

7. the filtrate is evaporated until saturate (one-third of it original

volume)8. the saturated solution is then allowed to cool to room

temperature.

9. crystals is dried between filter paper

10. [chemical equation]

Mg + H2SO4 MgSO4 + H2 //MgO + H2SO4 MgSO4 + H2O//

MgCO3 + H2SO4 MgSO4 + CO2 + H2O

1

1

1

1

1

11

1

1

1 10

20

8


9/10

9 (a) Substance/compound that contains carbon and hydrogen only 1

(b) 1. n(CH2) = 56

12n + 2n = 562. n = 4

3. Molecular formula C4H8

1

1

1 3

(c) 1. [Correct structural formula]2. [ Correct name]

Examples:

But-1-ene But-2-ene 2-Methylpropene

1+1+11+1+1 6

(d) [chemical substances and quantity - 1 mark][ apparatus - 1 mark]

[procedure - 5 marks]

[Observation 1 mark]

[chemical equation - 2 mark]

1. [2-5] cm3 ethanol and [2-5] cm3 butanoic acid

2. boiling tube

3. Add ethanol to butanoic acid

4. Shake the mixture

5. Add a little drop of concentrated sulphuric acid

6. Heat it

7. Poured into a beaker of water8. The product form has a fruity smell

9. Correct formula of reactants

10. Correct formula of products

C3H7COOH + C2H5OH C3H7COOC2H5 + H2O

11

1

11

1

1

11+1 10

9


10/10

10

(a)

(b)

(c)

(d)

Haber process

i) Formula of reactant

ii) Formula of productiii) Balance of equation

1. N2 + 3H2 2 NH3

2. 1 volume of nitrogen gas to 3 volume of hydrogen gas

3. Pressure at 200 atm.

4. Temperature at 4500 C5. Catalyst : Iron

Apparatus : Retort stand with clamp, [1 kg] weight, steel ballbearing metre rule.

Material : Copper block, bronze block, cellophane tape,

thread.

1. Stick a stell ball bearing onto the copper block usingcellophane tape.

2. Hang 1 kg weight at a height of 50 cm above the ballbearing.

3. Drop the weight so that it hits the ball bearing.

4. Measure the diameter of the dent made on the copper

block.5. Repeat step1 to 4 twice on the other parts of the copper

block.

6. Repeat step 1 to 5 using the bronze block to replace the

copper block.7. [Diagram ] - Functional diagram

8. [Diagram ] labelling of copper/bronze block, steel

Ball bearing, metre rule, weight

1. Fibre glass

2. High tensile strength // low density// easily moulded

1

1

11

1

1

11

1

1

1

11

1

1

11

1

1

1

1

7

10

2

20

END OF MARKING SCHEME

10

ujian pengesanan pertengahan tahun 2009_mark scheme chemistry 1 and 2

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