spm trial 2010 che ans (pahang)

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  • 8/8/2019 SPM Trial 2010 Che Ans (Pahang)

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    ANSWERS FOR SPM TRIAL 2010 SULIT

    PAPER 1

    Question

    Number

    Answer Question

    Number

    Answer

    1 B 26 A

    2 A 27 D

    3 C 28 A

    4 C 29 C

    5 D 30 C

    6 A 31 D

    7 D 32 D

    8 A 33 D

    9 B 34 C

    10 D 35 A

    11 B 36 C12 B 37 B

    13 B 38 D

    14 A 39 A

    15 C 40 D

    16 A 41 D

    17 B 42 B

    18 D 43 B

    19 A 44 D

    20 C 45 A

    21 A 46 A

    22 C 47 C

    23 C 48 A

    24 B 49 B

    25 C 50 C

    SULIT

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    SULIT 4541/2(PP)

    4541/2(PP) @ PKPSM Cawangan Pahang 2010 SULIT

    4541/2

    CHEMISTRY

    Kertas 2

    Peraturan Pemarkahan

    Sept 2010

    PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUASEKOLAH MENENGAH MALAYSIA (PKPSM)

    CAWANGAN PAHANG

    PEPERIKSAAN PERCUBAAN TAHUN 2010

    TINGKATAN 5

    PERATURAN PEMARKAHAN

    _________________________________________________________________________

    Kertas ini mengandungi 11 halaman bercetak

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    SULIT 4541/2(PP)

    4541/2(PP) @ PKPSM Cawangan Pahang 2010 SULIT

    No. Mark Scheme SubMark

    TotalMark

    1 (a)(i)

    (ii)

    X : electronY : nucleusElectron // X

    11 2

    (b)(i)

    (ii)

    (iii)

    (iv)

    (v)

    The number of protons in the nucleus of an atom18

    2.8.7

    Number of shells , 3Number of electrons

    7

    1

    1

    1

    11

    1

    6

    (c ) (i)(ii)

    A and BBecause A and B has different number ofneutrons but same number of protons // hasdifferent nucleon number but same proton

    number

    1

    1

    2

    TOTAL 9

    No. Mark Scheme SubMark

    TotalMark

    2 (a) (i)

    (ii)

    Carbon dioxide

    Limewater turns cloudy/milkychalky//whiteprecipitate formed

    1

    12

    (b) Zinc oxide 1 1

    (c) ZnCO3 ZnO + CO21. correct formula for reactant2. correct formula for products

    11 2

    (d) (i)4

    125 mol // 0.032 mol 1

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    SULIT 4541/2(PP)

    4541/2(PP) @ PKPSM Cawangan Pahang 2010 SULIT

    (ii)

    (iii)

    1. Correct ratio1 mole of ZnCO3 : 1 mole of CO2 //

    0.032 mole : 0.032 mole2. Correct volume of gas with unit

    0.032 x 24 dm3 // 0.768 dm3 // 768 cm3.

    0.032 x 6.02 x 10 23 molecules // 1.926 x 10 22

    molecules.

    1

    1

    1

    4

    TOTAL 9

    No. Mark Scheme SubMark

    TotalMark

    3 (a) P 1 1(b) R 1 1(c) (i)

    (ii)

    The electronegativity is increase from Q to V

    Across the period from Q to V , the atomic size

    decreases,The attraction force between nucleus and valenselectrons increases

    1

    1

    1

    3

    (d) R is more reactive than Q 1 1(e) Because the element has octet electron

    arrangement/Because the outermost shell has occupied 8electrons.

    1 1

    (f) (i)

    (ii)

    TU4 1

    3

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    SULIT 4541/2(PP)

    4541/2(PP) @ PKPSM Cawangan Pahang 2010 SULIT

    Corect diagram, label T and U, correct no. of electron

    Correct valens electron each of the atom

    11

    TOTAL 10 M

    No. Mark Scheme SubMark

    TotalMark

    4 (a) 1. Strong acid is the acid that is completelydissociated in water2. produce higher concentration of hydrogenions

    1

    1 2

    (b)(i)

    (b)(ii)

    (b)(iii)

    HCl + NaOH NaCl + H2O- Correct formula of reactants- Correct formula of products

    No. of mol of HCl = (20.00 x 0.1)/1000= 0.002 mol

    1. From the eq., 1 mol HCl : 1 mol NaOH2. Molarity of NaOH = 0.002 x 1000 /

    11

    11

    2

    2

    3

    (c) Colourless to pink 1 1

    TOTAL 10

    No. Mark Scheme SubMark

    TotalMark

    5 (a)(i)

    (a)(ii)

    C4H9OH

    1.

    2.

    1

    1

    1

    1

    2

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    SULIT 4541/2(PP)

    4541/2(PP) @ PKPSM Cawangan Pahang 2010 SULIT

    (a)(iii) Ethyl butanoate 1 1

    (b)(i)

    (b)(ii)

    Carbon dioxide

    C4H10 + 13/2O2 4CO2 + 5H2O //2C4H10 + 13O2 8CO2 + 10H2O- Correct formula of reactants and products- Balanced equation

    1

    11

    1

    2

    (c)

    1. Functional diagram2. Labeled diagram

    11 2

    (d) 1. Butene is unsaturated hydrocarbon //has double bond

    2. Butane is saturated hydrocarbon //Has single bond

    1

    1 2

    TOTAL 11

    Glass woolsoaked withbutanol

    Porcelainchips

    Butene

    Water

    Heat

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    SULIT 4541/2(PP)

    4541/2(PP) @ PKPSM Cawangan Pahang 2010 SULIT

    No. Mark Scheme SubMark

    TotalMark

    6 (a) To allow the movement of ions 1 1

    (b)(i)

    (b)(ii)

    Brown colour of bromine water turns colourless

    Green colour of iron(II) sulphate solution turnsbrown

    1

    1 2

    (c)(i)

    (c)(ii)(c)(iii)

    Br2 + 2e 2Br

    0 to -1Reduction reaction

    1

    11 3

    (d)(i)

    (d)(ii)

    Fe2+ Fe3+ + e

    Potassium iodide solution, KI //Reactive metals, e.g Zinc, Zn

    1

    1 2

    (e) Br2 + 2Fe2+ 2Br - + 2Fe3+

    - Correct formula of reactants and products 1 1

    (f) Y X 1 1

    (g) Bromine water 1 1

    [TOTAL 11

    SECTION C

    No.

    Mark Scheme SubMar

    k

    Total

    Mark

    7 (a)(i)

    Solution A: covalentSolution B: ionic

    11

    2

    (ii) - In solution A, A exist as molecule // no free movingions

    - Molecule does not carry the charge and the bulb doesnot light up

    - In solution B, ions can move freelyIons carry the charge and the bulb is lighted up

    1

    111

    4

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    SULIT 4541/2(PP)

    4541/2(PP) @ PKPSM Cawangan Pahang 2010 SULIT

    (b)(i)

    Cell J Cell KGas bubbles are released Anode becomes thinner

    Gas oxygen Copper (II) ion

    1 +11 +1

    4

    (b)(ii) Cell J : 4OH-

    O2 + 2H2O + 4eCell K : Mg Mg2+ + 2e 11 2

    (c)

    - correct functional diagram- label

    11

    2

    (d)(i)

    (ii)

    (iii)

    (iv)

    - potassium iodide solution : colourless solution turns brown- acidified potassium manganate (VII) solution : purplesolution decolourised

    potassium iodide solution : 2I

    -

    I2 + 2eacidified potassium manganate (VII) solution :MnO4

    - + 8H+ + 5e Mn2+ + 4H2O

    To allow the flow of ions from both electrolytes // to separatethe reducing agent from the oxidising agent.

    Potassium nitrate solution // sodium chloride chloride solution// potassium sulphate solution [any suitable answer]

    1

    1

    1

    1

    1

    1

    6

    TOTAL 20

    Dilute sulphuric acid

    Acidified potassium

    manganate (VII) solution

    Potassium iodide

    solution

    Carbon electrodes

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    SULIT 4541/2(PP)

    4541/2(PP) @ PKPSM Cawangan Pahang 2010 SULIT

    No. Mark Scheme SubMark

    TotalMark

    8 (a) [Able to classify the type of food additives ingiven situation correctly]

    Foodadditive

    Substance

    Colouring TartazineFlavouring Octyl

    ethanoateAnti oxidant Citric acidSweetener//preservative

    aspartame

    11

    1

    1

    4

    (b) Antibiotic

    1.To make sure all the bacteria are killed

    Otherwise the patient may becomes ill again

    Bacteria become more resistant to theantibiotic.

    As a result, the antibiotic is no longereffective.// Patients need stronger antibiotic tofight the same infection

    1

    1

    1

    1

    1

    5

    (c) Part X hydrophobic/hydrocarbonPart Y hydrophilic/ionicParx X dissolves in greasePart Y dissolves in water

    4 4

    (c) 1.The cloth in experiment II is cleanwhereas the cloth in Experiment I is still dirty.

    2.In hard water,soap react with magnesium ion3.to form scum4.Detergent are more effective in hard water

    5.Detergent does not form scum6.Detergent are better cleansing agen then soapto remove oily stain.

    11111

    1

    1

    7

    TOTAL 20 M

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    SULIT 4541/2(PP)

    4541/2(PP) @ PKPSM Cawangan Pahang 2010 SULIT

    No. Mark Scheme SubMark

    TotalMark

    9 (a) Rate of reaction is the speed at which reactantsare converted into products in a chemicalreaction.

    1 1

    (b)(i)

    (ii)

    (iii)

    30 = 0.25 cm3

    s-1

    2x 60

    45 = 0.375 / 0.38 cm3s-12x 60

    15 = 0.125 / 0.13 cm3s-12x 60

    1

    1

    1

    3

    (c)(i)

    (ii)

    CaCO3 + H2SO4 CaSO4 + CO2 + H2O

    Reactant and product correct

    Balance equation

    The number of moles CaCO3 = 0.2/40 + 12 +(16x3)

    = 0.002 mol

    1 mole of CaCO3 releases 1 mole of CO2. (ratioof CaCO3 to CO2)

    0.002 mole of CaCO3 releases 0.002 mole ofCO2The maximum volume of CO2 = 0.002 x 22.4

    = 0.0448 dm3= 44.8 cm3

    1

    1

    1

    1

    1

    1

    2

    4

    (d)(i)

    - Experiment II has a higher rate of reactioncompared to experiment I

    - The concentration of sulphuric acid,H2SO4 in experiment II is higher t