chemistry trial-exam spm 2012 marking scheme...

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1 BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN KEMENTERIAN PELAJARAN MALAYSIA CHEMISTRY TRIAL-EXAM SPM 2012 MARKING SCHEME PAPER 1 PAPER 2 PAPER 3

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  • 1

    BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH

    DAN SEKOLAH KECEMERLANGAN

    KEMENTERIAN PELAJARAN MALAYSIA

    CHEMISTRY

    TRIAL-EXAM

    SPM 2012

    MARKING SCHEME

    PAPER 1

    PAPER 2

    PAPER 3

  • 2

    SKEMA KERTAS 1

    PENTAKSIRAN DIAGNOSTIK AKADEMIK

    SBP 2012

    CHEMISTRY 4541/1

    1 A 26 C

    2 A 27 D

    3 C 28 C

    4 A 29 D

    5 A 30 B

    6 D 31 B

    7 B 32 C

    8 A 33 A

    9 B 34 C

    10 A 35 C

    11 B 36 D

    12 A 37 A

    13 D 38 B

    14 C 39 B

    15 D 40 D

    16 A 41 D

    17 B 42 D

    18 D 43 A

    19 C 44 B

    20 B 45 C

    21 C 46 B

    22 D 47 C

    23 D 48 B

    24 B 49 D

    25 A 50 C

  • 3

    MARKING SCHEME

    CHEMISTRY PAPER 2 SECTION A (4541/2)

    Question Mark scheme Sub

    Mark

    Total

    Mark

    1(a) A: Detergents

    B: Soap

    r : sodium salt

    1

    1

    2

    1(b)(i) magnesium ion // or calcium ion

    r : Mg2+

    , Ca2+

    1

    2 1(b)(ii) Soaps are biodegradable 1

    1(c) Analgesics

    Psychotherapeutic

    r: wrong spelling

    1

    1

    2

    1(d)(i) Sugar

    Aspartame

    1

    1

    3

    1(d)(ii) To add / restore the colour of food // To enhance its visual

    appeal / appearance// more attractive

    1

    TOTAL 9

    Question Mark scheme Sub

    Mark

    Total

    Mark

    2(a)(i) Magnesium

    r:Mg 1

    2

    2(a)(ii) Has 3 shell occupied with electrons 1

    2 (b)

    1. Proton number of chlorine is higher than magnesium // the

    number of proton of chlorine atom higher than magnesium atom

    // total positive charged in nucleus higher than magnesium

    2. The attractive force between the nucleus and the electrons in

    chlorine atom stronger than magnesium atom // nuclei attraction

    towards electrons stronger.

    1

    1

    2

    2(c) Light bulb // in welding process 1 1

    2 (d) Argon 1 1

    2(e)(i)

    3 Cl2 (g) + 2 Fe (s) 2 FeCl3(s)

    1

    1

    2(e)(ii) Mass iron (III) chloride = 0.05 x 161 // 8.05 g 1 1

  • 4

    2(e)(ii)

    1

    1

    Total 9

    Question Mark scheme Sub

    Mark

    Total

    Mark

    3(a)(i) 6 1

    4

    3(a)(ii) To estimate the age of fossils and artifacts 1

    3(a)(iii) C-12 // C-13

    1

    3(a)(iv) 7 / 6 1

    3(b)(i) A: covalent

    r: covalent bond

    B: ionic

    r: ionic bond

    1

    1

    6

    3(b)(ii) 2.8.8.1 1

    3(b)(iii) High melting point and boiling point // conduct electricity in

    molten or aqueous solution // soluble in water // insoluble in

    organic solvent.

    [Any one]

    1

    3(c)(iv) 2K + Cl2 2KCl

    1. Formula of reactants and products correct

    2. Balance the chemical equation

    1

    1

    TOTAL 10

    13

    C 6

    12

    C 6

    Or

    Hot iron wool

    Wul besi panas

    Heat

    Panaskan

    Chlorine gas

    Gas klorin

  • 5

    Question Mark scheme Sub

    Mark

    Total

    Mark

    4(a)(i) Pipette

    r: wrong spelling 1

    3

    4(a)(ii) Phenolphthalein // methyl orange

    r: wrong spelling 1

    4(a)(iii)

    Phenolphthalein : pink to colourless //

    methyl orange : yellow to orange

    1

    4(b)(i)

    2KOH + H2 SO4 K2 SO4 + 2H2O

    Formula of reactants and products correct

    Balance the chemical equation

    1

    1

    2

    4(b)(ii)

    1. Mol of KOH = (1.0)(25) / 1000

    = 0.025 mol

    2. Molarity of H2 SO4 = (0.0125)(1000) / 10

    = 1.25 mol dm-3

    //

    Molarity of H2SO4 = 0.0125 /0.01 = 1.25 mol dm-3

    Or

    b

    a

    VM

    VM

    bb

    aa ; Molarity of H2SO4 = 10

    )25(1

    2

    1x

    = 1.25 mol dm-3

    r: wrong unit or without unit

    1

    1

    2

    4(c)(i) Hydrogen ion

    r: H+ (symbol ion)

    1

    3 4(c)(ii)

    1. no water

    2. contain of molecule // no hydrogen ion, H+

    1

    1

    TOTAL 10

  • 6

    Question Mark scheme Sub

    Mark

    Total

    Mark

    5(a) Chemical formula that shows the simplest whole number ratio of

    atom of each element in the compound.

    1 1

    5(b)(i) Mass of Mg = 2.4 g

    Mass of O = 1.6 g

    1

    6

    5(b)(ii) Mole of Mg = 2.4 / 24 = 0.1

    Mol of O = 1.6 / 16 = 0.1

    Mg : O = 1 : 1

    1

    1

    5(b)(iii) MgO 1

    5(b)(iv) to allow oxygen enter the crucible 1

    5(b)(v) Repeat the process heating, cooling and weighing until a

    constant mass is obtained.

    1

    5(c)

    Apparatus set-up correct and functional

    Labeled : magnesium ribbon, heat

    1

    1

    2

    5(d)(i) Zinc oxide // aluminium oxide

    r: formula

    1

    2

    5(d)(ii) Silver is less reactive / not reactive metal 1

    TOTAL 11

    Magnesium

    ribbon

    Heat

    crucible

  • 7

    Question Mark scheme Sub

    Mark

    Total

    Mark

    6(a) Hydrogenation 1

    5

    6(b) C4H8 + H2 C4H10 1

    6(c)(i)

    Mole of butene = 2.8 /56 = 0.05 mol

    1

    6(c)(ii) P1. 1 mol of butene burns in oxygen gas produce 4 mol of

    carbon dioxide.

    Therefore 0.05 mol of butene burns in oxygen gas produce 0.2

    mol of carbon dioxide //

    C4H8 : CO2

    1 : 4

    0.05 : 0.2

    P2. volume of CO2 = 0.2 x 24 = 4.8 dm3

    1

    1

    6(d) P1: 2 cm3 of but-2-ene and 2 cm

    3 of butane gas are filled in two

    different test tubes.

    P2 : 2 -3 drops of acidified potassium manganate (VII) solution

    is added to both test tubes.

    P3: But-2-ene decolourises purple acidified KMnO4

    Purple acidified KMnO4 in butane remains unchange.

    Or

    P1: 2 cm3 of but-2-ene and 2 cm

    3 of butane gas are filled in two

    different test tubes.

    P2 : 2-3 drops of bromine water is added to both test tubes.

    P3: But-2-ene decolourises brown bromine water.

    Brown bromine water in butane remains unchange.

    1

    1

    1

    Or

    1

    1

    1

    3

    6(e)(i) A: unvulcanised rubber

    B: vulcanised rubber

    1

    1

    3

    6(e)(ii) Rubber type B/ vulcanised rubber is more elastic than rubber

    type A/ unvulcanised rubber

    1

    TOTAL 11

  • 8

    SECTION B

    Question Mark scheme Mark Σ

    Mark

    7 (a)

    (i) Ascending order : Cu, Y,X, W

    (ii)

    P1 : Positive terminal : Cu

    P2 : Potential difference : 1.3V

    P3: Copper is less electropositive // X is more electropositive

    correct value and unit

    1

    1

    1

    1

    4

    7 ( b) (i)

    Experiment I : Pb 2+

    , Cl-

    Experiment II: H +

    . OH- , Cl

    r: lead(II) ion, chloride ion

    Hydrogen ion, hydroxide ion, chloride ion

    1

    1

    2

    7(b)(ii)

    Experiment 1 Experiment II

    Product at anode:

    Chlorine gas

    Products at cathode:

    Oxygen gas

    Reason:

    P1: Cl-

    is discharged

    P2: the only

    anion presence

    and discharged at anode

    Reason:

    P1:OH- is selected to be

    discharged

    P2:the position of OH- is lower

    than Cl- in electrochemical series

    .Half equation:

    2Cl-

    Cl 2 + 2e

    P1:Correct formula of

    reactant and product :

    P2: Balance equation

    . Half equation:

    4OH- 2 H2O + O2 + 4e

    P1: Correct formula of reactant

    and product

    P2: Balance equation

    1 + 1

    1 + 1

    1 + 1

    1 + 1

    1 + 1

    10

  • 9

    7(c)

    P1: Positive terminal: R

    P2: Negative terminal:Cu

    P3: Suitable metal for R : Silver

    P4: Suitable solution for R : Silver nitate solution

    1

    1

    1

    1

    4

    Total 20

    Question Mark scheme Mark Σ Mark

    8(a) P1. Smaller size has larger total surface area.

    P2. Absorb heat faster.

    P3. Bigger size has smaller total surface area.

    P4. Absorb heat slower

    1

    1

    1

    1

    4

    (b)(i) Copper(II) sulphate 1 1

    (b)(ii) 1. Experiment I

    Rate of reaction = 40/2 = 20 cm3 min

    -1

    2. Experiment II

    Rate of reaction = 60/2 = 30 cm3min

    -1

    1

    1

    2

    (b)(iii) P1. Rate of reaction in Experiment II is higher than

    Experiment I.

    P2. Substance X used in Experiment II is a catalyst.

    P3. Catalyst provided an alternative path with requires a

    lower

    activation energy.

    P4. More particles are able to achieve lower activation

    energy.

    P5. Frequency of effective collisions between zinc atoms and

    hydrogen ions are higher.

    1

    1

    1

    1

    1

    5

  • 10

    (b)(iv) 1. Label of axes and unit

    2. Correct curve and label

    1

    1

    2

    (v) 1. Correct formula of reactants and product

    2. Balanced equation

    Zn + 2H+ Zn

    2+ + H2

    1

    1

    2

    (vi) 1. Rate of reaction using sulphuric acid is higher.

    2. Volume of hydrogen gas released is doubled.

    3. Sulphuric acid is a diprotic acid.

    4. Concentration of hydrogen ions in sulphuric acid is double

    than that in hydrochloric acid.

    1

    1

    1

    1

    4

    Total 20

    Volume of

    gas / cm3

    Time / min

    Exp II

    Exp I

    2

  • 11

    Question Mark scheme Mark Σ

    Mark

    9(a) P1: Magnesium atom undergoes oxidation

    P2: oxidation number increases from 0 to 2+

    P3: Copper (II) ion undergoes reduction

    P4: oxidation number decreases from +2 to 0

    P5: The reaction involving oxidation and reduction

    1

    1

    1

    1

    1

    MAX

    4

    (b) Experiment I

    L can reduce copper(II) oxide// L can react with copper(II) oxide

    L is more reactive than copper.

    Experiment II

    M can reduce copper(II) oxide//M can react with copper(II)

    oxide

    M is more reactive than copper.

    Experiment III

    M cannot reduce L oxide // M cannot react with L oxide.

    M is less reactive than L//L is more reactive than M.

    The arrangement in order of increasing reactivity toward oxygen is Cu,

    M and L.

    1

    1

    1

    1

    1

    1

    Max

    5

    1

    6

    (c)

    Procedure :

    P1. Pour 2 cm3 of potassium bromide solution into a test tube.

    P2. Add 2 cm3 of chlorine water to the test tube and shake the mixture.

    P3. Add 2 cm3 of 1,1,1 trichloroethane to the test tube and shake the

    mixture.

    P4. Record the observation

    P5. Repeat steps 1-4 using another halogens and halide solution.

    .

    1

    1

    1

    1

    1

  • 12

    Result :

    Chlorine water Bromine water Iodine water

    Potassium

    chloride

    X X

    Potassium

    bromide X

    Potassium

    iodide

    Ionic equation:

    1. Cl2 + 2Br- 2Cl

    - + Br2

    2. Cl2 + 2I- 2Cl

    - + I2

    3. Br2 + 2I- 2Br

    - + I2

    1

    1

    1

    1

    1

    10

    20

  • 13

    Question Mark scheme Mark Σ

    Mark

    10(a)

    Precipitation / double decomposition reaction

    Barium nitrate solution/barium chloride solution [Any sulphate solution]

    Example: sodium sulphate, potasium sulphate, sulphuric

    acid

    Reject : Lead(II) sulphate, calcium sulphate

    Ba2+ + SO42 BaSO4

    1

    1

    1

    1

    4

    10(b)(i) Cation : Iron(II) ion / Fe2+

    Anion: Chloride ion / Cl

    1

    1

    2

    10(b)(ii)

    Test for NO3

    P1: Add 2 cm3

    of dilute sulphuric acid into the test tube follow

    by 2 cm3 of iron(II) sulphate solution.

    P2: Add a few drops of concentrated sulphuric acid

    P3: carefully and slowly along the side of slanting test tube

    into the mixture.

    P2: A brown ring is formed.

    1

    1

    1

    1

    4

    10(c)

    Procedure:

    P1. Add zinc nitrate solution to sodium carbonate solution in a beaker.

    P2. Stir the mixture. P3. Filter the white precipitate/solid zinc carbonate formed.

    P4. Add zinc carbonate to sulphuric acid in a beaker until some zinc carbonate solid no longer dissolve.

    P5. Filter the mixture. P6. Transfer the filtrate to a evaporating dish. P7. Heat the filtrate(zinc sulphate solution) until saturated//

    Heat the filtrate to about one-third (1/3) of its initial

    volume

    P8. Allow the saturated solution to cool at room temperature. P9. Filter the crystals formed. P10. Dry the crystals by pressing it between two sheets of

    filter papers.

    1

    1

    1

    1

    1

    1

    1

    1

    1

    1

    10

    Total 20