PAPER 2 (ANSWER) - CHEMISTRY TRIAL SPM (2018)

SKEMA JAWAPAN PEP. PERCUBAAN KIMIA KERTAS 2 (2018) Question 1 (a) (i)

(ii) (b)

(iii) (i)

(ii)

Subtotal Total 1 2 1

1

1

1

1

Silica/ silicon dioxide Silika / silikon dioksida Borosilicate glass // lead (crystal) glass Kaca borosilikat // kaca (kristal) plumbum Type: Flavouring agent Jenis: perisa makanan

1

1

1

1

1

2

Function: to improve the taste / aroma of food Fungsi: untuk meningkatkan rasa / aroma makanan To reduce the effect of insect bite / headache / nausea / windy condition / stomach discomfort / toothache Mengurangkan kesan gigitan serangga / sakit kepala / loya / keadaan berangin / ketidakselesaan perut / sakit gigi // Relieving joint / muscle pain Melegakan kesakitan sendi / otot Total

1

(i)

(ii)

(c)

1. Long-chained molecules Molekul berantai panjang 2. made up from a large number of small repeating identical unit of monomer. terbentuk daripada gabungan banyak unit monomer yang sama Vinyl chloride / chloroethene Vinil klorida / kloroetena

4541/2

1

1

9

1

SKEMA JAWAPAN PEP. PERCUBAAN KIMIA KERTAS 2 (2018) Question 2 (a)

(b)

(c)

(i) (ii)

(d)

A : Sodium -24 Natrium-24 B : Iodine-131 Iodin-131 Atoms of the same element with the same proton number but different nucleon number.

Atom unsur-unsur yang sama yang mempunyai nombor proton yang sama tetapi nombor nukleon yang berbeza // Atoms of the same element with the same number of protons but different number of neutrons.

Atom unsur-unsur yang sama yang mempunyai bilangan proton yang sama tetapi bilangan neutron yang berbeza Diffusion Resapan 1. The odour of durians’ particles are tiny and discrete Zarah-zarah bau durian halus dan diskrit 2. moving randomly from high concentration region to low concentration region bergerak rawak dari kepekatan tinggi ke kepekatan rendah 3. diffuse into the empty spaces between air particles meresap ke ruang-ruang kosong antara zarahzarah udara 1. The number of protons / positive charge increases when going across the period Bilangan proton / cas positif meningkat apabila merentasi kala // 2. The attraction force between nucleus and valence electron becomes stronger Daya tarikan antara nukleus dan elektron valens semakin kuat 3. Electrons are pulled towards the nucleus Elektron ditarik ke arah nukleus Total

4541/2 Subtotal Total 1 2 1 1

1

1

1

1

Max. 2

1

1

1

3

1

1 9

2

SKEMA JAWAPAN PEP. PERCUBAAN KIMIA KERTAS 2 (2018) Question 3 (a) (b)

(c)

(i) (ii)

(iii)

Alkali metal Logam alkali 1. Y 2. Atomic size of Y is the largest.

Saiz atom Y paling besar.

3. The force of attraction between nucleus and valence electron atom Y is the weakest.

Daya tarikan antara nukleus dan elektron valens atom Y paling lemah.

4. It is easier for atom Y to release its valence electron Lebih mudah untuk atom Y melepaskan elektron valensnya Note : atom Y must be mentioned at least once in P2 / P3 / P4 Blue litmus paper is bleached Warna kertas litmus biru dilunturkan.

1. Correct formula of reactants and products Formula bahan tindak balas dan hasil tindak balas yang betul 2. Balanced chemical equation Persamaan kimia yang seimbang Answer: Cl2 + 2NaOH → NaCl + NaOCl + H2O 1. Number of mole of NaOH Bil. mol NaOH = MV /1000 = (0.5)(10) / 1000 = 0.005 mol

4541/2 Subtotal Total 1 1 1 1

Max. 3

1

1

1

1

1

2

1

1

2. Mole ratio 2 mole of NaOH reacts with 1 mole of Cl2 0.005 mole of NaOH reacts with 0.0025 mole of Cl2 2 mol NaOH bertindak balas dengan 1 mol Cl2 0.005 mol NaOH bertindak balas dengan 0.0025 mol Cl2

1

3. Volume of Cl2 Isipadu Cl2 = 0.0025 x 24 = 0.06 dm3 / 60 cm3

1

Total

3

10

3

SKEMA JAWAPAN PEP. PERCUBAAN KIMIA KERTAS 2 (2018) Question 3 (a) (b)

(c)

(i) (ii)

(iii)

Alkali metal Logam alkali 1. Y 2. Atomic size of Y is the largest.

Saiz atom Y paling besar. a : size of atoms increases when going down the group saiz atom meningkat apabila menuruni kumpulan 3. The force of attraction between nucleus and valence electron atom Y is the weakest.

Daya tarikan antara nukleus dan elektron valens atom Y paling lemah.

4. It is easier for atom Y to release its valence electron Lebih mudah untuk atom Y melepaskan elektron valensnya Note : atom Y must be mentioned at least once in P2 / P3 / P4 Blue litmus paper is bleached Warna kertas litmus biru dilunturkan.

1. Correct formula of reactants and products Formula bahan tindak balas dan hasil tindak balas yang betul 2. Balanced chemical equation Persamaan kimia yang seimbang Answer: Cl2 + 2NaOH → NaCl + NaOCl + H2O 1. Number of mole of NaOH Bil. mol NaOH = MV /1000 = (0.5)(10) / 1000 = 0.005 mol

4541/2 Subtotal Total 1 1 1 1

Max. 3

1

1

1

1

1

2

1

1

2. Mole ratio 1 mole of NaOH reacts with 1 mole of Cl2 0.005 mole of NaOH reacts with 0.005 mole of Cl2 1 mol NaOH bertindak balas dengan 1 mol Cl2 0.005 mol NaOH bertindak balas dengan 0.005 mol Cl2

1

3. Volume of Cl2 Isipadu Cl2 = 0.005 x 24 = 0.12 dm3

1

Total

3

10

4

SKEMA JAWAPAN PEP. PERCUBAAN KIMIA KERTAS 2 (2018) Question 4 (a) (i) (ii)

(b)

(iii) (i) (ii) (iii) (iv)

(c)

(i)

2+

+

-

-

Cu , H , OH , NO3 [Both cells must be correct to get 1 mark] Cell / Sel A : Electrical energy to chemical energy Tenaga elektrik kepada tenaga kimia Cell / Sel B : Chemical energy to electrical energy Tenaga kimia kepada tenaga elektrik 2+ Mg → Mg + 2e2I- → I2 + 2eX2 + 2e- → 2XOxidizing agent Agen pengoksidaan Bromine water / Chlorine water Air bromin / Air klorin 1. Zinc is more electropositive than iron Zink lebih elektropositif daripada besi // Zinc is above iron in the electrochemical series Zink terletak di atas ferum dalam siri elektrokimia // Zinc has a higher tendency to release / donate electron compared to iron Zink lebih cenderung melepaskan / mendermakan elektron berbanding ferum

4541/2 Subtotal Total 1 1 1 1

1 1 1 1

1 1 1 1

1

1

1

3

2. Zinc is corroded / sacrificed / ionized / oxidized / loss electron Zink terkakis / dikorbankan / terion / dioksidakan / kehilangan electron

1

3. Zinc prevents the iron/ steel form contacting with oxygen from the air and water.

Zink menghalang besi / keluli daripada bersentuhan dengan oksigen dari udara dan air.

Total

1

10

5

SKEMA JAWAPAN PEP. PERCUBAAN KIMIA KERTAS 2 (2018) Question 5 (a) (i) (ii) (iii)

(iv)

(b)

(i)

(ii)

Zinc nitrate / zink nitrat / Zn(NO3)2 Residue / Baki E : Zinc oxide / Zink oksida / ZnO Acid / Asid G : Nitric acid / Asid nitrik / HNO3 [Any one of the answer] Colourless gas bubbles released Gelembung udara tidak bewarna terbebas // Gas turns lime water cloudy Gas yang terbebas menyebabkan air kapur menjadi keruh // The residue is yellow when hot and white when cold Baki kuning ketika panas sejuk bewarna putih.

1. Pour dilute sulphuric acid follow by iron(II) sulphate solution.

Tuangkan asid sulfurik cair dan larutan ferum(II) sulfat 2. Drop/slowly/carefully add concentrated sulphuric acid in slanted Titiskan perlahan-lahan asid sulfurik pekat secara condong 3. Brown ring is formed Cincin perang terbentuk 1. Correct formula of reactants and products Formula bahan tindak balas dan hasil tindak balas yang betul 2. Balanced chemical equation Persamaan kimia yang seimbang Answer: 2NH3 + H2SO4 → (NH4)2SO4 1. 28 × 100 132 2. 21.2 %

4541/2 Subtotal Total 1 1 1 2 1 1 1

1

3

1

1 1

2

1

1

2

1 Total

11

6

SKEMA JAWAPAN PEP. PERCUBAAN KIMIA KERTAS 2 (2018) Question 6 (a) (i)

(b)

(c)

(i)

(ii)

(iii)

As catalyst Sebagai mangkin // Increases the rate of reaction Meningkatkan kadar tindak balas // Speeds of the reaction Mempercepatkan tindak balas Brown gas produced Gas perang dihasilkan // Damp blue limus paper turns red Kertas litmus biru lembap bertukar menjadi merah 1. Correct formula of reactants and products Formula bahan tindak balas dan hasil tindak balas yang betul 2. Balanced chemical equation Persamaan kimia yang seimbang Answer: 5O2 + 4NH3 → 4NO + 6H2O 1. Y-axis drawn and labelled ‘Energy Paksi-Y dilukis dan dilabel ‘Tenaga’ 2. Two different levels of energy drawn Dua aras tenaga berbeza dilukis

1. Platinum increases the rate of reaction Platinum meningkatkan kadar tindak balas 2. By lowering the activation energy Dengan merendahkan tenaga pengaktifan 3. More particles are able to achieve the (lower) activation energy Lebih banyak zarah dapat mencapai tenaga pengaktifan (yang lebih rendah) 4. Frequency of effective collision between oxygen gas / molecule and ammonia gas / molecule increases

4541/2 Subtotal Total 1 1

1

1

1

2

1

1

2

1

1

4

1 1

1

7

SKEMA JAWAPAN PEP. PERCUBAAN KIMIA KERTAS 2 (2018)

(ii)

Frekuensi perlanggaran berkesan antara gas / molekul oksigen dan gas / molekul ammonia meningkat→ (NH4)2SO4 Catalyst is specific to a certain reaction Mangkin spesifik terhadap tindak balas tertentu a : Only platinum can be used as the catalyst in the reaction Hanya platinum boleh digunakan sebagai mangkin dalam tindak balas itu Total

4541/2

1

1

11

8

SKEMA JAWAPAN PEP. PERCUBAAN KIMIA KERTAS 2 (2018) Question 7 (a)

[Any two structural formulae with 2 correct corresponding names]

4541/2 Subtotal Total 2+2 4

1.

2. But – 1 – ene But – 1 – ena

3.

4. But – 2 – ene But – 2 – ena

5.

6. 2 – methylprop – 1 – ene 2 – metilprop – 1 – ena (b) (i)

1. X= propanol 2. Y= propan-1,2-diol propan-1,2-diol 3. Z= propane propana

1 1

3

1

9

SKEMA JAWAPAN PEP. PERCUBAAN KIMIA KERTAS 2 (2018) (ii)

(c)

1. I : C3H6 + H2O → C3H7 OH 2. II: C3H6 + H2O + [O] → C3H6 (OH)2 3. III: C3H6 + H2 → C3H8 Hexane Hexene Heksana Heksena Test / Ujian 1 1. Saturated 2. Unsaturated hydrocarbon hydrocarbon Hidrokarbon tepu Hidrokarbon tak tepu 3. Hydrocarbon consists 4. Compound consists of single bond between double bond between carbon atoms carbon atoms.

Hidrokarbon yang Hidrokarbon yang mengandungi ikatan mengandungi ikatan tunggal antara atom ganda dua antara atom karbon karbon 5. Does not undergo 6. Undergoes addition addition reaction reaction Tidak mengalami Mengalami tindak tindak balas balas penambahan penambahan Test 2/ Ujian 2 7. Lower percentage of 8. Higher percentage of carbon atoms by mass carbon by mass Kurang jelaga kerana Lebih jelaga kerana peritus karbon peritus karbon mengikut jisim lebih mengikut jisim lebih rendah tinggi 9.

72 × 100 86 = 83.7%

10. 72 × 100 84 = 85.7%

4541/2 1 1 1

3

1+1

10

1+1

1+1

1+1

1+1 Total

20

10

SKEMA JAWAPAN PEP. PERCUBAAN KIMIA KERTAS 2 (2018) Question 8 (a)

1. Correct formula of reactants and products Formula bahan tindak balas dan hasil tindak balas yang betul 2. Balanced chemical equation Persamaan kimia yang seimbang

4541/2 Subtotal Total 1 6

1

Answer: HNO3 + NaOH → NaNO3 + H2O 3. Number of mole of HNO3 Bil. mol NaOH = MV /1000 = (0.1)(22) / 1000 = 0.0022 mol

1

4. Mole ratio 1 mole of HNO3 reacts with 1 mole of NaOH 0.0022 mole of HNO3 reacts with 0.0022 mole of NaOH 1 mol HNO3 bertindak balas dengan 1 mol NaOH 0.0022 mol HNO3 bertindak balas dengan 0.0022 mol NaOH

1

5. Calculation of molarity of NaOH Pengiraan kemolaran NaOH = 0.0022 Ă— 1000 25 6. Answer of molarity of NaOH + unit Jawapan kemolaran NaOH + unit = 0.088 mol dm-3 (b)

(c)

1. Vinegar Cuka 2. Weak acid, food substances Asid lemah, bahan makanan 3. Neutralise the jelly fish sting Meneutralkan sengatan obor-obor 4. Safe and natural remedies Penawar yang selamat dan semulajadi 1. Volumetric flask used is 250 cm3 Kelalang volumetrik yang digunakan ialah 250 cm3 2. Volume of NaOH needed, M1V1 = M2V2 Isipadu NaOH yang diperlukan, M1V1 = M2V2 = 25 cm3 3. Pipet 25cm3 of NaOH, 1.0 mol dm-3 and add into a beaker Pipetkan 25 cm3 larutan NaOH 1.0 mol dm-3 dan masukkan ke dalam bikar

1

1

1

4

1 1 1 1

10

1

1

11

SKEMA JAWAPAN PEP. PERCUBAAN KIMIA KERTAS 2 (2018) 4. Add some water Tambahkan air suling 5. Stir using a glass rod Kacau menggunakan rod kaca 6. Pour the solution into volumetric flask Tuang larutan ke dalam kelalang volumetrik 7. Rinse beaker, glass rod and filter funnel Bilas bikar, rod kaca dan corong turas 8. Add water into the volumetric flask Tambahkan air ke dalam kelalang volumetrik 9. When near the graduation mark, add water drop by drop Apabila hampir dengan tanda senggatan, tambahkan air suling setitis demi setitis 10. Stopper the volumetric flask and invert the flask to obtain a homogenous solution Tutup kelalang volumetrik dan goncang (terbalikkan) supaya mendapat larutan yang homogen Total

4541/2 1

1 1 1 1

1

1

20

12

SKEMA JAWAPAN PEP. PERCUBAAN KIMIA KERTAS 2 (2018) Question 9 (a)

(b)

(c)

1. CO2 // any covalent compound CO2 // mana-mana sebatian kovalen 2. Particles are attracted by weak intermolecular force.

Zarah-zarah tertarik oleh daya intermolekul yang lemah.

3. Less heat energy is needed to overcome the force of attraction Sedikit tenaga haba yang diperlukan untuk mengatasi daya tarikan 1. Electron arrangement of atom X = 2.1 / [any suitable electron arrangement] Susunan elektron atom X = 2.1 [mana-mana susunan elektron yang sesuai] 2. Electron arrangement of atom Y = 2.7 / [any suitable electron arrangement] Susunan elektron atom Y = 2.7 / [mana-mana susunan elektron yang sesuai] 3. Ionic bond Ikatan ion 4. Atom of X donates 1 electron to form X+ ion Atom X menderma 1 elektron untuk membentuk ion X+.

5. Atom of Y receives 1 electron to form Y- ion Atom Y menerima 1 elektron untuk membentuk ion Y-.

6. To achieve stable octet electron arrangement Untuk mencapai susunan electron octet yang stabil 7. One X+ ion and one Y- ion are attracted together by strong electrostatic force.

Satu ion X+ dan satu ion Y- ditarik bersama oleh daya tarikan elektrostatik.

1. Materials and apparatus: Bahan dan radas: Compound XY, carbon electrodes, dry cells, connecting wire with crocodile clip, crucible, light bulb/ ammeter/ galvanometer Sebatian XY, elektrod karbon, sel kering, wayar penyambung dengan klip buaya, mangkuk pijar, mentol/ ammeter/ galvanometer Procedure: Prosedur: 2. Place one spatula of solid XY into a crucible.

Letakkan satu spatula pepejal XY ke dalam sebuah mangkuk pijar.

3. Dip two carbon electrodes into the crucible.

Celupkan dua elektrod karbon ke dalam mangkuk pijar.

4. Connect the carbon electrodes with connecting wire and bulb to the dry cells.

4541/2 Subtotal Total 1 3 1

1

1

7

1

1 1 1 1 1

1

Max. 10

1

1

1

13

SKEMA JAWAPAN PEP. PERCUBAAN KIMIA KERTAS 2 (2018) Sambungkan elektrod karbon dengan wayar penyambung dan mentol kepada sel kering.

5. Turn on the switch.

Tutupkan suis.

6. Observe whether the light bulb glow and record observation.

Perhatikan sama ada mentol menyala dan mencatat pemerhatian.

7. Heat the solid XY in the crucible until it melts.

Panaskan pepejal XY dalam mangkuk pijar sehingga lebur.

8. Observe whether the light bulb glow and record observation.

Perhatikan sama ada mentol menyala dan mencatat pemerhatian.

9. Observation: Pemerhatian: State of compound XY Observation Keadaan sebatian XY Pemerhatian Solid Bulb does not light up.

Pepejal Mentol tidak menyala Molten Bulb lights up.

Leburan Mentol menyala.

10. Functional diagram Rajah berfungsi Pipe clay triangle, arrow for heat Segi tiga tahan liat, anak panah untuk panaskan 11. Label Label Carbon electrodes, molten XY, heat Elektrod karbon, leburan XY, panaskan

4541/2

1 1

1

1

1

1

1

OR 1. Materials and apparatus: Bahan dan radas: Compound XY, water, carbon electrodes, dry cells, connecting wire with crocodile clip, beaker, light bulb/ ammeter/ galvanometer

1

14

SKEMA JAWAPAN PEP. PERCUBAAN KIMIA KERTAS 2 (2018)

4541/2

Sebatian XY, air, elektrod karbon, sel kering, wayar penyambung dengan klip buaya, bikar, mentol/ ammeter/ galvanometer Procedure: Prosedur: 2. Put one spatula of solid XY into a beaker.

Letakkan satu spatula pepejal XY ke dalam bikar 3. Add water Tambah air 4. Stir the solution Kacau larutan itu 5. Dip two carbon electrodes into the beaker Celupkan dua elektrod karbon ke dalam bikar 6. Connect the carbon electrodes with connecting wire and bulb to the dry cells.

Sambungkan elektrod karbon dengan wayar penyambung dan mentol kepada sel kering.

7. Turn on the switch.

Tutupkan suis.

8. Observe whether the light bulb glow and record observation.

Perhatikan sama ada mentol menyala dan mencatat pemerhatian.

9. Repeat steps 2 – 8 using solid XY to replace solution XY Ulang langkah 2 – 8 menggunakan pepejal XY untuk menggantikan larutan XY 10. Observation: Pemerhatian: State of compound XY Keadaan sebatian XY Solution Larutan Solid Pepejal

1 1 1 1 1

1 1

1

1 Observation Pemerhatian Bulb lights up.

Mentol menyala.

Bulb does not light up.

Mentol tidak menyala

11. Functional diagram Rajah berfungsi Dashes in solution Tanda sempang dalam larutan 12. Label Label Carbon electrodes, solution XY Elektrod karbon, larutan XY

1

1

15

SKEMA JAWAPAN PEP. PERCUBAAN KIMIA KERTAS 2 (2018)

Total

Question 10 (a)

Rubric Reaction I 1. Exothermic reaction Tindak balas eksotermik 2. Energy level of reactants is higher than products Aras tenaga bahan tindak balas lebih tinggi daripada hasil tindakbalas 4. Heat released during bond formation is higher than heat is absorbed during breaking of bond Haba yang dibebaskan semasa pembentukan ikatan lebih tinggi daripada haba diserap semasa pemutusan ikatan 6. Heat is released from the surrounding Haba dibebaskan daripada persekitaran // Temperature increases Suhu meningkat

Reaction II Endothermic reaction Tindak balas endotermik 3. Energy level of reactants is lower than products Aras tenaga bahan tindak balas lebih tinggi daripada hasil tindakbalas 5. Heat released during bond formation is lower than heat is absorbed during breaking of bond Haba yang dibebaskan semasa pembentukan ikatan lebih tinggi daripada haba diserap semasa pemutusan ikatan

7. Heat is absorbed from the surrounding Haba diserap daripada persekitaran // Temperature decreases Suhu menurun

4541/2

20

Subtotal Total Max. 1 5

1+1

1+1

1+1

16

SKEMA JAWAPAN PEP. PERCUBAAN KIMIA KERTAS 2 (2018) (b)

(c)

4541/2

1. Number of mole of copper(II) sulphate Bilangan mol kuprum(II) sulfat = 1 × 50 1000 = 0.05 mol

1

2. Heat released Haba dibebaskan = 0.05 x 21000 J // 1050 J

1

3. Temperature change Perubahan suhu = 1050 50 × 4.2

1

4. Answer + unit Jawapan + unit = 5 °C

1

1. Correct formula of reactants and products Formula bahan tindak balas dan hasil tindak balas yang betul 2. Balanced chemical equation Persamaan kimia yang seimbang Possible answers : Pb (NO3)2 + 2NaCl → PbCl2 + NaNO3 Pb (NO3)2 + Na2SO4 → Pb SO4 + NaNO3 Pb (NO3)2 + 2KCl → PbCl2+ KNO3 Ag NO3 + NaCl → AgCl + NaNO3 Ag NO3 + KCl → AgCl + KNO3

1

Procedure / Prosedur : 3. Measure 25 cm3 of 1.0 mol dm-3 [silver nitrate solution] and pour into polystyrene cup Sukat 25 cm3 argentum nitrat 1.0 mol dm-3 dan tuangkan ke dalam cawan polisterina 4. Record the initial temperature of [silver nitrate solution] Rekodkan suhu awal [larutan argentum nitrat] 5. Measure 25 cm3 of 1.0 mol dm-3 [sodium chloride solution] pour into another polystyrene cup.

Sukat 25 cm3 [larutan natrium klorida] 1.0 mol dm-3 dan tuangkan ke dalam cawan polisterina yang lain 6. Record the initial temperature of solution Rekodkan suhu awal [larutan natrium klorida] 7. Pour the [sodium chloride solution] quickly and carefully to the cup containing [silver nitrate solution]

1

4

Max. 11

1

1 1

1

1

17

SKEMA JAWAPAN PEP. PERCUBAAN KIMIA KERTAS 2 (2018) Tuangkan [larutan natrium klorida] dengan cepat dan cermat ke dalam cawan polisterena yang mengandungi [larutan argentum nitrat] 8. Stir the solution with thermometer Kacau campuran dengan termometer 9. Record the highest temperature of the mixture.

Rekodkan suhu tertinggi yang tercapai 10. Data Initial temperature of [silver nitrate solution] Suhu awal [larutan argentum nitrat] Initial temperature of [sodium chloride solution] Suhu awal [larutan natrium klorida] Average initial temperature of mixture Purata suhu awal campuran Highest temperature of mixture Suhu campuran tertinggi Temperature change, θ Perubahan suhu

4541/2

1 1 1

T1 0C

T20C

T30C T40C (T4 - T3) °C

11. Heat given out / released Haba yang dibebaskan *depends on volume of solutions (refer P3 & P5) = (25 + 25) × 4.2 × (T4 - T3) J / X J

1

12. Number of moles of precipitate Bilangan mol mendakan *depends on molarity & volume of solutions (refer P3 & P5) = MV = Y mol 1000 13. Heat of precipitation Haba pemendakan = X = - Z kJ mol-1 Y

1

1 Total

20

18