trial 2014 chem p2 ms

SULIT 4541/2 (PP)

SMK SULTAN ABDUL JALIL, KLUANG, JOHOR______________________________________________

PEPERIKSAAN PERCUBAAN SPM 2014 4541/2TINGKATAN LIMA

ChemistryKertas 2Peraturan PemarkahanSeptember 2014

UNTUK KEGUNAAN PEMERIKSA SAHAJA

___________________________________________________________________________Peraturan pemarkahan ini mengandungi 11 halaman bercetak

Lihat halaman sebelah4541/2 (PP) © 2014 Hak Cipta Panitia Kimia SMKSAJ SULIT

AMARAN

Peraturan pemarkahan ini SULIT dan Hak Cipta Panitia Kimia SMKSAJ. Kegunaannya khusus untuk pemeriksa yang berkenaan sahaja. Sebarang maklumat dalam peraturan pemarkahan ini tidak boleh dimaklumkan kepada sesiapa. Peraturan pemarkahan ini tidak boleh dikeluarkan dalam apa-apa bentuk media.

SULIT 4541/2 (PP)

Question 1

No. of Q Explanation Marks Total Mark

1 (a) (i) Melting point is the temperature at which solid change into liquid at standard temperature and pressure.

1

(a) (ii) 83 oC 1

(a) (iii) 136 oC 1

(a) (iv) (a)

Solid + liquid 1

(a) (iv) (b)

Gas 1

(a) (v) 1. Heat energy loss to the surrounding

2. balanced by heat energy liberated by particles to attract

each other.

1

1

(a) (vi)1

(b) (i) Able to state the meaning of diffusionSample answer:A process when particles of a substance /gas P/Q move between the particles of another substance/gas Q/P

1

(b) (ii) Able to state the colour correctlyAnswer:Brown // Light brown

1

Total 10

Question 2

No. of Q Explanation Marks

Total Mark

2 (a) (i) 2.8.4 1

(a) (ii) 14 1

(b) (i) Decrease // become smaller 1


SULIT 4541/2 (PP)

No. of Q Explanation Marks

Total Mark

(b) (ii) Proton number / positive charge increase // force attraction increase

1

(c) Achieve octet electron arrangement // has 8 valence electronDo not accept or share electron

11

(d) (i) Al // Aluminium 1

(d) (ii) Al2O3 1

(e)

Pt 1: Label nucleus and correct number of shellPt 2: Octet electron arrangement and correct charges

11

Total 10

Question 3


3 (a) Is a representation of a chemical substance using letters for atoms and subscripts for each type of atoms present 1 1

(b) (i) [ Able to name suitable acid and metal and its equation ]For exampleHydrochloric acid and zinc metal

1 1

(ii) Zn + 2HCl à ZnCl2 + H2 1 1

(c) Hydrogen gas must be flowed/through/into the combustion tube for a few minutes before heating/ The flow of hydrogen gas must be continuous throughout the experiment/ [ Accept any one answer]

1 1

(d) (i) Number of mole of copper = 1.62 64 = 0.025mole

1 1


SULIT 4541/2 (PP)


(ii) Number of mole of copper = 0.40 16 = 0.025mole

1 1

(iii) Number of mole of copper: Number of mole of oxygen 0.025 : 0.025The simplest ratio 1 : 1The empirical formula of copper(II) oxide is CuO

11

2

(e) Iron(II) oxide / Tin(II) oxide / Lead(II) 1 1

(f) Burning of metal in excess oxygen 1 1

Total 10

Question 4


4(a) C4H10 1 1

(b)

1+1 2

(c) - Butene is an unsaturated hydrocarbon/ has double bond- Butane is a saturated hydrocarbon/ has single bond

11 2

C4H8 + H2 à C4H10- correct reactants and products

- balanced equation1

1

(e) (i)


SULIT 4541/2 (PP)


- Functional diagram - Labelled diagram

11 2

(ii) Hexene produced more sootMore carbon in one molecule of hexene compare to one molecule of hexane

11 2

Total 10

Question 5


5 (a) (i) Acid (substance) which ionises partially// has lower degreeof dissociation in water to produce low concentration of H + ion

1

(a) (ii) The concentration of hydrogen ions in hydrochloric acid is higher 1The higher the concentration of hydrogen ions, the lower the pH value

1

(b)

(i) Neutralisation 1

(b)

(ii) Sodium chloride 1

(b)

(iii)

Pink to colourless 1

(b)

(iv) HCI + NaOH à NaCI + H 2O

No of mole of NaOH = 0.1 (25) 1000 = 0.0025 mol 1


SULIT 4541/2 (PP)


1 mol of NaOH react completely with 1 mol of HCl0.0025 mol of NaOH react completely with 0.0025 mol HCI

Molarity of HCI = 0.0025 x 1000 12.5

= 0.2 mol drri3

1

1

Total 10

Question 6

No. of Q Explanation MarksTotal Mark

6 (a) (i) Ethanol 1

(ii) The heat released when one mole of ethanol completely burnt in Oxygen (under standard condition) is 1260 kJ

1

(b)

(i) No of moles of alcohol = 0.23 /46 = 0.005 mol1 mol of alcohol burnt released 1260 kJThus 0.005 mol of alcohol burnt released 6.3 kJ

1

11

(ii) ᶿ = H mc = 6300 200 x 4.2 = 7.5 o C

1

1

(c) Heat is lost to the surrounding // Heat is absorbed by the apparatus or containers // Incomplete combustion of alcohol

1

(d)

Label energy and diagram has 2 different energy lavelsBalanced chemical equation


SULIT 4541/2 (PP)

No. of Q Explanation MarksTotal Mark

Total 10

Question 7


7 (a) (i) Electrode P : Iodide ion / I- ion and hydroxide ion / OH- ionElectrode Q : Hydrogen ion / H+ ion and Potassium ion / K+ ion

11 2

(ii) Electrode P : Oxygen molecule gas // Oxygen gasElectrode Q : Hydrogen molecule gas // hydrogen gasR : formula

11 2

(iii) Electrode P : iodine molecule // iodineElectrode Q : hydrogen molecule // hydrogen gas

Half equation :Electrode P : 2 I- I2 + 2eElectrode Q: 2 H+ + 2 e H2

Confirmatory test at P :Starch solution is added into the solution around electrode PBlue colouration / precipitate is formed

11

11

11 6

(b) Cell X Cell YType of cell Electrolytic cell Voltaic cellThe energy change

Electrical energy to chemical energy

Chemical energy to electrical energy

The terminal of cell

Anode : Copper electrode

Cathode : Copper electrode

Anode : Aluminium electrode

Cathode : Copper electrode

Ions present in the electrolyte

Cu2+ , H+

SO42- , OH-

Observation Anode : Become thinner

Anode : become thinner

1

1

1

1

1

1


SULIT 4541/2 (PP)


Cathode : Brown deposit // brown solid is deposited // become thicker

Cathode : Brown deposited // brown solid deposited // become thicker

Half equation for both electrodes

Anode :Cu Cu2+ + 2e

Cathode:Cu2+ + 2e Cu

Anode :Al Al3+ + 3e

Cathode :Cu2+ + 2e Cu

Name of the process occurred at both electrodes

Anode : Oxidation processCathode : Reduction process

1

1

1

1

10

Total 20

Question 8


8 (a) The temperature in a refrigerator is lowerBacterial activity is lower, less toxin is produced by bacteriaIn a kitchen cabinat, the temperature is higher , bacteria activity is higherTherefore the rate of food spoilage is faster in a kitchen cabinet than in a refrigerator.

11

1

1 4

(b) (i) Volume of gas released = 50 cm3

Time taken = 55 sAverage rate of reaction = 50/55 = 0.91 cm3s-1

11 2

(b) (ii) Experiment I and Experiment IIThe temperature for the reaction II is higher than experiment IIncrease the kinetic energy of the reacting particleIncrease the rate of collision between calcium carbonate

11

1


SULIT 4541/2 (PP)


particles and hydrochloric acid particlesIncrease the rate of effective collision.Hence, experiment II has a higher rate of reaction than experiment I

Experiment II and Experiment IIIThe size of calcium carbonate in reaction III is smaller than reaction IISmaller size of reactant has bigger total surface area exposedIncrease the rate of collision between calcium carbonate particles and hydrochloric acid particlesIncrease the rate of effective collision.Hence, experiment III has a higher rate of reaction than experiment II

1

1

11

11

1

10

(b) (iii) No. moles of HCl = 0.5 x 30 1000 = 0.015 mol

2 mol of HCl produce 1 mol of CO2

Therefore 0.015 mol of HCl produce 0.0075 mol CO2

Volume = 0.0075 x 24 = 0.18 dm3

1

11

1 4

Total 20

Question 9


9 (a) (i) NaOH 1 1

(ii) Materials : Solid sodium hydroxide and distilled waterApparatus : 50cm3 beaker, 250cm3 volumetric flask, electronic balance, glass rod, filter funnel.

Calculation : Determine the mass of sodium hydroxide, NaOH:

No. of moles of NaOH = MV = 1 x 250 = 0.25 mol 1000 1000

11

1


SULIT 4541/2 (PP)


Mass of NaOH needed = No. of moles x Molar mass, of NaOH = 0.25 x [23 + 16 + 1 ] = 10 g

Steps :1. Using an electronic balance, 10 g of sodium hydroxide is exactly

weighed and placed into a beaker,2. Distilled water is added to the beaker to dissolve all the solid sodium

hydroxide,3. Then the solution is poured into a 250cm3 volumetric flask. The beaker

is rinsed with distilled water and the solution is poured into the volumetric flask.

4. The solution in the volumetric flask is topped up with distilled water until its calibration mark.

11

1

1

1

19

(ii) To identify Mg 2+ ion-Magnesium nitrate solution is poured into a test tube-NaOH solution is added until excess-white precipitate formed and insoluble in excess and-test with NH3 solution-same step and same result will obtained when test with NH3solution

NO3- ion

-Magnesium nitrate solution is poured into a test tube-dilute sulfuric acid is added-followed by iron(II) sulphate solution-shake the test tube-concentrated sulphuric acid is added slowly/through the wall of the test tube

-brown ring formed

1111

1111

11 10

Total 20

Question 10


10 (a) Part X – hydrophobic/hydrocarbon 1


SULIT 4541/2 (PP)


Part Y – hydrophilic/ionicPart X – dissolves in greasePart Y – dissolves in water

111 4

(b) 1.The cloth in experiment II is clean whereas the cloth in Experiment I is still dirty.2.In hard water,soap react with magnesium ion / calcium ion3.to form scum4.Detergent are more effective in hard water5.Detergent does not form scum6.Detergent are better cleansing agen then soap to remove oily stain.

11111

1 6

(c)(i) Pure metal atoms has similar size and shapeEasily to slide

11 2

(c)(ii) Draw for pure copperDraw for its alloy and labels for copper and zinc

11 2

(c)(iii) Increase the strength and hardness of metalPrevent the corrosion of metalImprove the appearance

111 3

(c)(iv) DuraluminIts stronger / harderCan withstand compression

111 3

Total 20

END OF MARKING SCHEMEPERATURAN PEMARKAHAN TAMAT


trial 2014 chem p2 ms

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