SKEMA JAWAPAN KIMIA KERTAS 2

• adp atom
• - + Na
• 1

  1

  11 OSO

  3 Na

  → CH

  3

  (CH

  2

  )

  11 OSO

  3

  1 (c) Cleaning agent Z does not form scum //

  Cleaning agent Z reacts with Mg

  2+

  / Ca

  2+ , forming soluble salt.

  1 (d) (i) Preservative

  2

  1

  1 (ii) Stabiliser // Thickener

  1

  1 (iii) To preserve the food // To prevent the growth of microorganism // to make the tomato sauce last longer// To prevent food spoilage

  1

  1 (iv) -should be used:

  Food additive improve the taste/ texture/ appearance of food//preserve food OR

  It may cause health problem

  1

  1

  1

  1

  2 TOTAL

  9

  )

  (CH

  9 R ^{24 12} NO ANSWER SUB TOTAL MARK MARK

  1 (ii) Group 1

  SKEMA JAWAPAN KIMIA KERTAS 2

  BAHAGIAN A NO ANSWER SUB MARK TOTAL MARK

  1 (a) (i) The number of proton in an atom

  1

  1 (ii)

  1. Proton

  2. Neutron

  3. Electron [ any two]

  1

  1 1 max 2 (iii)

  1

  1 (b) (i)

  2.8.1

  1

  Period 3

  3

  1

  1

  2 (iii) Group 1 because atom S has one valence electron

  Period 3 because atom S has three shells occupied with electron

  1

  1

  2 TOTAL

  9 NO ANSWER SUB MARK TOTAL MARK

  2 (a) (i) Cleaning agent Y

  1

  1 (ii) Saponification // alkaline hydrolysis

  1

  1 (b)

  CH

• shouldn’t be used:

  3 (a) (i) Electrolytic cell

  1

  1 (ii) Electrical energy to chemical energy

  1

  1 (b) To allow the movement of ions

  1

  1

  2+

  (c)

  1

  1 Zn → Zn + 2e (d) (i) The blue solution turns to pale blue in cell A

  1 in cell B the blue solution remains unchanged.

  1

  2

  2+

  (ii)

  1. The concentration of Cu ion decreases in Cell A

  1

  2+

  2. The concentration of Cu ion in Cell B remain unchanged

  1

  2 (e) (i) Replace the Zn electrode with Mg electrode

  1

  1 (ii) The distance between Mg and Cu in electrochemical series is

  1

  1 further.

  TOTAL

  10 NO ANSWER

  SUB TOTAL MARK MARK 4 (a) The increase/change in volume of hydrogen gas per unit time

  1

  1 (b) Zn + 2HNO

  3 3 ) 2 + H

  2

  → Zn(NO Correct formula of reactants and products

  1 Balanced equations

  1

  2 (c) Experiment I:

  22

  3 -1

  Average Rate of reaction= =0.183 cm s

  1

  1

  2

  60  .

  

1

  20

  (d)

  



  Number of moles of HNO

  3 = = 0.002 mol

  1

  

1000

  2 mol of HNO

  3 produce 1 mol of H 2 //

  1 0.002 mol of HNO produce 0.001 mol of H

  3

  2

  1

  3

  3 Maximum volume of H 2 = 0.001 x 24 000 // 24 cm

  (e) (i) Temperature

  1

  1 (ii) At normal body temperature,

  The rate of metabolism is high// insufficient oxygen supply to

  1 organs At low temperature surgery, The rate of metabolisme reduced/low// organs do not require

  1

  2 much oxygen supply

  TOTAL

  10 NO ANSWER SUB TOTAL MARK MARK

  5 (a) (i)

  1. Mass of NaOH used

  1

  2

  2. Volume of NaOH solution

  1 r: volume of water (ii) Place the volumetric flask on a flat plane and put a piece of white paper behind the volumetric flask while the eye is straight to the graduation mark.

  1

  1 (iii)

  1. No of mol = MV/1000 = (0.1) (1000) / 1000 // 0.1

  1

  2

  2. Mass = No of mol X RMM = 0.1 X 40 g //

  1 = 4 g

  (b) (i) Ammonia ionised partially in water to produce low

  1

  1 concentration of hydroxide ion (ii) Tetrachlorometane // chloroform

  1

  1 (iii) Exist as molecule //

  1

  1 No free moving ion (iv) MaVa = 1

  MbVb 2

  1

  2

• 3

  Ma = ½ X (0.1) (15) moldm //

  1

  25

• 3

  = 0.03 moldm (v) sebagai baja

  1

  1 TOTAL

  11 NO ANSWER SUB TOTAL MARK MARK

  6 (a) Exothermic reaction

  1

  1 Tindak balas eksotermik (b) To reduce heat loses to the surrounding // as the insulator

  1

  1 Untuk mengurangkan kehilangan haba ke persekitaran //

  sebagai penebat haba

  (c) Zn + CuSO

4 Cu + ZnSO

  4

  1. Correct chemical formula of reactants and products

  1

  2. Balanced equation

  1

  2 (d) (i) H = 50 x 4.2 x 8 // 1680 J

  1

  1 (ii) No. of mol = 0.2 x 50 /1000 // 0.01 mol

  1

  1

• 1

  (iii) Heat of displacement, H = 1680 /0.01 = - 168 kJmol

  1

  1 (e)

  Energy/Tenaga

• 2

  Zn + Cu

• 1

  H = - 168 kJmol

• 2

  Cu + Zn

  1. Two different energy levels with correct sign of H

  1

  1

  2

  2. Correct chemical / ionic equation

• 1 (f) Heat of displacement is more than 168 kJmol .

  1 Because magnesium metal is more electropositive than zinc

  1

  2 metal in electrochemical series.

• 1 Nilai haba penyesaran lebih besar daripada 168 kJmol .

  Kerana logam magnesium lebih elektropositif daripada logam zink dalam siri elektrokimia.

  TOTAL 11

  11

• H

  1

  2

  [ CH

  2

  ]n = 56 [ 14 ]n = 56 n = 4 Molecular formula of P = C

  4 H

  8

  1

  1

  1

  1

  7.14 14.29 = 2

  5 (ii) Name of compound P: butene

  Homologous series: alkene

  1

  1

  2 (c) (i) Compound R :

  Name: Pentene Fuctional group: Double bond between carbon atoms Compound S : Name: Pentanoic acid Fuctional group: Carboxyl group

  1

  1

  1

  1

  7.14 Empirical formula of P = CH

  1 Mol ratio 7.14 = 1

  BAHAGIAN B Question

  2 O

  No Mark Scheme Sub

  Mark Total Mark 7 (a) (i) Alcohol : Methanol

  Carboxylic asid: Propanoic acid C

  2 H

  5 COOH + CH

  

3 OH → C

  2 H

  5 COOCH

  3

  1

  12 14.29 = 14.29

  1

  2

  4 (ii) Mol ester = 1.32 = 0.15

  88 Mass alcohol = 0.15 x 32 = 4.8 g

  1

  1

  2 (b) Empirical formula

  Unsur Carbon Hydrogen Mass

  85.71

  14.29 No. of mol 85.71 = 7.14

  4

  (ii) H H H H H

  I I I I I H

• – C = C – C – C – C – H

  I I I H H H

  1 H H H H H

  I I I I I H

• – C – C = C – C – C – H

  I I I I H H H H

  1 H H H

  I I I H

• – C = C – C – C – H

  I I H H H

• – C – H

  I

  1 H H H

  I I H

• – C – C = C – C – H

  I I H H H

• – C – H

  I

  1 H H H H H

  I I I I H

• – C = C – C – C – H

  I H H

• – C – H

  I

  1 H

  Max Any three

  3

  3 Total

  20

NO ANSWER

  SUB TOTAL MARK MARKS 8 (a) (i) Q O ₂

  1 PO ²

  1

  2 (ii)

  4Q + O ^{2  2Q 2 O}

  1. Correct formula of reactant and product

  1

  2. Balanced chemical equation

  1

  2 (iii) Number of mole of Q used = mass of Q .

Molar mass of Q

= 9.2 23 = 0.4 mol

  1 4 mol of Q produce 2 mol of Q ^{2 O} 0.4 mol of Q produce 0.2 mol Q ^{2 O}

  1 Molar mass of Q ^{2 O = 23x2 + 16} _-1 = 62 g mol Maximum mass of Q

^{2 O = number of mole of Q}

^{2 O x molar mass} of Q ^{2 O} = 0.2 x 62

  = 12.4 g

  1

  3 (b) S has archived stable octet electron arrangement of 2.8.8. S telah mencapai susunan elektron oktet.2.8.8.

  1 S does not receive, release or share electrons with other atoms. S tidak menerima, melepaskan ataupun berkongsi elektron

  1 dengan atom yang lain R has electron arrangement of 2.8.7

  1 R mempunyai susunan elektron 2.8.7 One atom R share one pair of electron with another atom R to achieve stable octet electron arrangement.

  Satu atom R akan berkongsi satu pasang elektron dengan satu atom R yang lain untuk mencapai susunan elektron oktet yang stabil

  1

  4 (c) (i) - Q and R

  Q dan R

  1

• Q with electron arrangement of 2.8.1 Q dengan susunan elektron 2.8.1

  1

• R with electron arrangement of 2.8.7 R dengan susunan elektron 2.8.7

  1

• Atom of Q will release one valence electron to achieve stable electron arrangement Atom Q akan melepaskan satu elektron valens untuk mencapai

  1 susunan elektron stabil _{+ +}

• Positive ion Q is formed // Q  Q + e _{+ +} Ion positif Q terbentuk // Q  Q + e

  1

• Atom of R will receive one electron from atom Q to achieve stable electron arrangement

  Atom R akan menerima satu elektron daripada atom Q untuk mencapai susunan elektron stabil

  1

• – –
• Negative ion R is formed // R + e  R
• – –

  Ion negative R terbentuk // R + e  R

  1 - An ionic compound with formula QR is formed. Satu sebatian ionik dengan formula QR terbentuk

  1

  1. Correct nucleus and charge for each ion

  1 Max 8

  2. Correct number of electrons in each shell

  1 ⁺

–

(ii) Molten or aqueous state, ion Q and R are able to move freely. ⁺ – Manakala dalam keadaan leburan dan akueus, ion Q and R dapat bergerak bebas.

  1 ⁺ – In solid state, ion Q and R are held in fixed position by strong electrostatic forces. //there are no mobile ion ⁺

–

Dalam keadaan pepejal, ion Q and R adalah diikat pada kedudukan yang tetap oleh daya elektrostatik yang kuat //tiada ion yang bergerak bebas

  1

  2 TOTAL

  20 BAHAGIAN C

  Question Answer:

SUB TOTAL

  Soalan Jawapan Marks MARK 9(a)(i) Sample answer:

  1 Soluble sulphate salts : potassium sulphate // zinc sulphate // any soluble sulphate salt Garam sulfat terlarutkan: kalium sulfat // zink sulfat // mana-mana garam sulfat terlarutkan Insoluble sulphate salts : Lead(II) sulphate //Barium sulphate //

  1

  2 Calcium sulphate Garam sulfat tak terlarutkan: Plumbum(II) sulfat // Barium sulfat // Kalsium sulfat

  1 + 1

  2 (ii) Name of reactants to prepare potassium sulphate: Potassium hydroxide and sulphuric acid

  Nama bahan tindak balas untuk menyediakan kalium sulfat : Kalium hidroksida dan asid sulfurik (iii)

  Salt crystals 1 + 1

• salt solution is poured into an evaporating dish
• the solution is heated gently until saturated

  1

• cool - filter and dry between two filter paper.

  1

  1

• larutan garam dituangkan ke dalam mangkuk penyejat

  1

• -larutan dipanaskan perlahan-lahan sehingga tepu
• - sejukkan

  6

• - turas dan keringkan antara dua helai kertas turas

  3+ b(i) cation test : Fe ion ₃

• pour 2 cm of solution in the test tube
₃

  1

• add 2 cm of sodium hydroxide solution into the test tube

  1

• brown precipitate is formed shows the presence iron (III) ion

  1

  3 3+

  Ujian kation : ion : Fe

  3

• Tuang 2 cm larutan ke tabung uji

  3

• Tambah 2 cm larutan natrium hidroksida ke dalam

  tabung uji itu Mendakan perang terbentuk menunjukkan kehadiran ion - ferum (III).

• ion
• pour 2 cm
• add 2 cm
• add 2 cm

• -drop slowly concentrated sulphuric acid into a slanted test tube

• brown ring is formed shows the presence of nitrate ion Ujian Anion: Ion NO
3<
• Tuang 2 cm

  1

  1 Max 3 TOTAL

  1

  1

  1

  klorida

  3 larutan argentum nitrat

  3 asid nitrik cair ke dalam tabung uji itu

  3 larutan ke tabung uji

  Ujian Anion : Cl

  3 silver nitrate solution - white precipitate shows the presence of chloride ion.

  3 nitric acid into the test tube

  3 of solution in the test tube

  1 Max 4 (iii) anion test : Cl

  1

  1

  1

• Tambah 2 cm

  nitrat

  tabung uji yang dicondongkan

  3 larutan ferum(II) sulfat

  3 asid hidroklorik cair ke dalam tabung uji itu

  3 larutan ke tabung uji

  3 of iron (II) sulphate solution

  3 of dilute sulphuric acid into the test tube

  3 of solution in the test tube

  (ii) anion test : NO ³

• Tambah 2 cm

• - Cincin perang terbentuk menunjukkan kehadiran ion

• ion
• pour 2 cm
• add 2 cm
• add 2 cm
• ion
• Tuang 2 cm
• Tambah 2 cm
• Tambah 2 cm

• - Mendakan putih terbentuk menunjukkan kehadiran ion

• - Titiskan perlahan-lahan asid sulfurik pekat ke dalam

  20

NO ANSWER

  SUB TOTAL MARK MARK 10 (a) Sample answer: HCl + NaOH  NaCl + H O ₂

• + 1 -1 +1 -2 +1 +1 -1 +1 -2

  1 Correct formulae of reactants and products

  1

  2 Balanced equation 1 3 [state the changes in the oxidation 1 numbers of each element]

  4

  4 Oxidation number of all elements in the compound remain 1 … unchanged (b) (i) Sample answer: X: copper

  1 Y: magnesium / aluminium

  1 Z: iron / tin / lead 1 (ii) Carbon is less reactive than Y // Carbon cannot displace Y from 1 oxide of metal Y.

  Carbon is more reactive than X and Z // Carbon is able to displace

  1 the metal X or metal Z from respective oxide of metal. The further the distance between metal and carbon in reactivity

  1 Any 2 series, the brighter the flame or glow produced. X, Z, carbon, Y

  1

  6 (c) More electropositive metal: Mg/ Al/ Zn

  1 Less electropositive metal: Sn/ Pb/ Cu

  1 Procedure:

  1

  1. [more electropositive metal] strip, [less electropositive metal] strip and iron nails are cleaned with sandpaper.

  1

  2. One iron nail is coiled with [more electropositive metal] strip whereas another iron nail is coiled with [less electropositive metal] strip .

  1

  3. Pour the mixture of hot jelly solution and a few drops of potassium hexacyanoferrate(III) solution into two separate test tubes until half full.

  1

  4. Both iron nails that are coiled with different metal strips

  1 are placed into separate test tubes.

  5. Put both test tubes in a test tube rack and leave for a few days/ [1

• – 3 days]

  Results: Test tube Observation Inference ₂₊

Iron nail coiled with Blue colour is Fe present //

1+1 [less electropositive produced iron rusts metal] strip ₂₊

Iron nail coiled with No blue colour can Fe absent //

  1+1 Max:

[more electropositive be seen iron does not

10 metal] strip rust

  TOTAL

  20