KimiaChemistry

Print questions onlyCetak soalan sahaja Print with answersCetak dengan skema Print answer key onlyCetak kunci jawapan sahaja Practice shows answers as you go. Exam times you and hides answers until you submit.Latihan menunjukkan jawapan semasa menjawab. Peperiksaan menetapkan masa dan menyembunyikan jawapan sehingga dihantar.

1

QuestionSoalan 1

Apakah nombor pengoksidaan nitrogen dalam ion nitrat(III), NO₂⁻?What is the oxidation number of nitrogen in the nitrate(III) ion, NO₂⁻?

• A+4
• B+5
• C+3
• D-3

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C

The sum of the oxidation numbers equals the charge on the ion: x + 2(-2) = -1, so x = +3. +5 is nitrogen in the nitrate(V) ion, and +4 comes from ignoring the charge.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 1, standard 1.1

TextbookBuku teks p.ms. 8

2

QuestionSoalan 2

Larutan argentum nitrat, 0.5 mol dm⁻³, dielektrolisiskan menggunakan elektrod karbon. Apakah hasil di katod dan di anod, serta setengah persamaan di anod?Silver nitrate solution, 0.5 mol dm⁻³, is electrolysed using carbon electrodes. What are the products at the cathode and at the anode, and the half equation at the anode?

• AKatod: argentum; anod: oksigen, 4OH⁻ → 2H₂O + O₂ + 4e⁻Cathode: silver; anode: oxygen, 4OH⁻ → 2H₂O + O₂ + 4e⁻
• BKatod: hidrogen; anod: oksigen, 4OH⁻ → 2H₂O + O₂ + 4e⁻Cathode: hydrogen; anode: oxygen, 4OH⁻ → 2H₂O + O₂ + 4e⁻
• CKatod: argentum; anod: oksigen, O₂ + 2H₂O + 4e⁻ → 4OH⁻Cathode: silver; anode: oxygen, O₂ + 2H₂O + 4e⁻ → 4OH⁻
• DKatod: argentum; anod: nitrogen dioksidaCathode: silver; anode: nitrogen dioxide

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A

Ag⁺ (E° = +0.80 V) is discharged ahead of H⁺ at the cathode. At the anode OH⁻ is discharged in preference to NO₃⁻ and oxygen is released by oxidation, so electrons appear on the right of the half equation. Option C is a reduction, so it is wrong for an anode.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 1, standard 1.4

TextbookBuku teks p.ms. 38

3

QuestionSoalan 3

Larutan kuprum(II) sulfat dielektrolisiskan menggunakan elektrod kuprum. Selepas beberapa ketika, jisim anod berkurang sebanyak 0.64 g. Apakah perubahan pada katod dan pada larutan?Copper(II) sulphate solution is electrolysed using copper electrodes. After some time, the mass of the anode has decreased by 0.64 g. What are the changes at the cathode and in the solution?

• AKatod bertambah 0.32 g; keamatan warna biru berkurangCathode gains 0.32 g; the intensity of the blue colour decreases
• BKatod tidak berubah; keamatan warna biru bertambahCathode does not change; the intensity of the blue colour increases
• CKatod bertambah 0.64 g; keamatan warna biru berkurangCathode gains 0.64 g; the intensity of the blue colour decreases
• DKatod bertambah 0.64 g; keamatan warna biru tidak berubahCathode gains 0.64 g; the intensity of the blue colour does not change

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D

Each copper atom that leaves the anode as Cu²⁺ is matched by one Cu²⁺ that is deposited at the cathode, so the cathode gains 0.64 g and the concentration of Cu²⁺, hence the blue colour, stays the same.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 1, standard 1.4

TextbookBuku teks p.ms. 44

4

QuestionSoalan 4

Apakah kumpulan berfungsi bagi asid karboksilik?What is the functional group of carboxylic acids?

• A–OH
• B–COOH
• C–COO–
• D–C=C–

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B

Carboxylic acids contain the carboxyl group, -COOH. -OH is the group of alcohols, -COO- of esters and C=C of alkenes.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 2, standard 2.2

TextbookBuku teks p.ms. 73

5

QuestionSoalan 5

Gas etena dialirkan ke dalam larutan kalium manganat(VII) berasid. Apakah pemerhatian yang dibuat?Ethene gas is passed into acidified potassium manganate(VII) solution. What is observed?

• AWarna ungu menjadi tanpa warnaThe purple colour turns colourless
• BWarna jingga menjadi hijauThe orange colour turns green
• CWarna perang menjadi tanpa warnaThe brown colour turns colourless
• DTiada perubahan yang dapat diperhatikanNo change is observed

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A

Ethene is an unsaturated hydrocarbon and is oxidised by the manganate(VII) ion, so the purple solution is decolourised. Orange to green is the change of dichromate(VI) and brown to colourless is the change of bromine water.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 2, standard 2.3

TextbookBuku teks p.ms. 86

6

QuestionSoalan 6

Apakah nama IUPAC bagi sebatian dengan formula struktur CH₃CH=CHCH(CH₃)CH₃?What is the IUPAC name of the compound with the structural formula CH₃CH=CHCH(CH₃)CH₃?

• A2-metilpent-3-ena2-methylpent-3-ene
• B4-metilpent-3-ena4-methylpent-3-ene
• C4-metilpent-2-ena4-methylpent-2-ene
• D2-metilpent-2-ena2-methylpent-2-ene

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C

The longest chain containing C=C has five carbons (pentene). Number from the end that gives the double bond the lower number: C=C starts at carbon 2, so the methyl branch is on carbon 4: 4-methylpent-2-ene. Numbering from the other end would put the double bond on carbon 3.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 2, standard 2.4

TextbookBuku teks p.ms. 104

7

QuestionSoalan 7

2.3 g asid metanoik, HCOOH, dipanaskan dengan etanol berlebihan dan beberapa titis asid sulfurik pekat. Apakah jisim maksimum ester yang boleh terbentuk? [Jisim atom relatif: H = 1, C = 12, O = 16]2.3 g of methanoic acid, HCOOH, is heated with excess ethanol and a few drops of concentrated sulphuric acid. What is the maximum mass of ester that can be formed? [Relative atomic mass: H = 1, C = 12, O = 16]

• A1.85 g
• B2.3 g
• C7.4 g
• D3.7 g

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D

n(HCOOH) = 2.3 ÷ 46 = 0.05 mol. 1 mol of acid gives 1 mol of ethyl methanoate, HCOOC₂H₅ (M = 74). Mass = 0.05 × 74 = 3.7 g.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 2, standard 2.3.2

TextbookBuku teks p.ms. 96

8

QuestionSoalan 8

Jadual 1 menunjukkan hasil ujian ke atas dua sebatian, W dan X. Kedua-duanya mempunyai formula molekul C₃H₆O₂. Pernyataan yang manakah betul?

Ujian	W	X
Ditambah natrium karbonat	Gas yang mengeruhkan air kapur terbebas	Tiada gas terbebas
Bau	Berbau cuka	Berbau manis

Jadual 1Table 1 shows the results of tests on two compounds, W and X. Both have the molecular formula C₃H₆O₂. Which statement is correct?

Test	W	X
Sodium carbonate added	A gas that turns limewater cloudy is released	No gas is released
Smell	Smells like vinegar	Sweet smell

Table 1

• AW dan X ialah isomer kedudukanW and X are position isomers
• BW ialah ester dan X ialah asid karboksilikW is an ester and X is a carboxylic acid
• CW ialah asid propanoik dan X ialah ester; W dan X ialah isomerW is propanoic acid and X is an ester; W and X are isomers
• DW ialah propan-1-ol dan X ialah asid etanoikW is propan-1-ol and X is ethanoic acid

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C

W releases carbon dioxide with a carbonate and has an acid smell, so it is a carboxylic acid; with three carbons it is propanoic acid. X gives no gas and has a sweet smell, so it is an ester. They share the formula C₃H₆O₂ but differ in structure, so they are isomers.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 2, standard 2.4

TextbookBuku teks p.ms. 101

9

QuestionSoalan 9

Antara pernyataan berikut, yang manakah menerangkan tenaga pengaktifan?Which of the following statements describes the activation energy?

• AJumlah haba yang dibebaskan apabila 1 mol bahan api terbakar lengkapThe heat released when 1 mole of a fuel burns completely
• BTenaga minimum yang mesti dimiliki oleh zarah yang berlanggar supaya tindak balas berlakuThe minimum energy that colliding particles must have for a reaction to occur
• CPerbezaan antara kandungan tenaga hasil dengan kandungan tenaga bahan tindak balasThe difference between the energy content of the products and of the reactants
• DJumlah tenaga yang diserap untuk memecahkan semua ikatan dalam bahan tindak balasThe total energy absorbed to break all the bonds in the reactants

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B

Activation energy is the minimum energy colliding particles must have to react; on an energy profile it is the height from the reactants to the peak. A is the heat of combustion, C is ΔH and D is the energy to break bonds.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 3, standard 3.1.2

TextbookBuku teks p.ms. 119

10

QuestionSoalan 10

25 cm³ asid nitrik 2.0 mol dm⁻³ dicampurkan dengan 25 cm³ larutan kalium hidroksida 2.0 mol dm⁻³. Suhu campuran naik sebanyak 13.0 °C. Apakah haba peneutralan? [Muatan haba tentu larutan = 4.2 J g⁻¹ °C⁻¹; ketumpatan larutan = 1 g cm⁻³]25 cm³ of 2.0 mol dm⁻³ nitric acid is mixed with 25 cm³ of 2.0 mol dm⁻³ potassium hydroxide solution. The temperature of the mixture rises by 13.0 °C. What is the heat of neutralisation? [Specific heat capacity of the solution = 4.2 J g⁻¹ °C⁻¹; density of the solution = 1 g cm⁻³]

• A-54.6 kJ mol⁻¹
• B-27.3 kJ mol⁻¹
• C-109 kJ mol⁻¹
• D-5.46 kJ mol⁻¹

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A

Q = 50 × 4.2 × 13.0 = 2 730 J = 2.73 kJ. n(H₂O) = 2.0 × 25/1000 = 0.050 mol. ΔH = -2.73 ÷ 0.050 = -54.6 kJ mol⁻¹. Using 0.10 mol (adding the two volumes) gives -27.3, and using 0.025 mol gives -109.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 3, standard 3.2.1

TextbookBuku teks p.ms. 129

11

QuestionSoalan 11

Dalam suatu eksperimen untuk menentukan haba pembakaran etanol, nilai yang diperoleh ialah -1 120 kJ mol⁻¹, lebih rendah daripada nilai teori -1 376 kJ mol⁻¹. Pengubahsuaian yang manakah memberikan nilai yang lebih hampir dengan nilai teori?In an experiment to determine the heat of combustion of ethanol, the value obtained is -1 120 kJ mol⁻¹, lower than the theoretical value of -1 376 kJ mol⁻¹. Which modification gives a value closer to the theoretical value?

• AMenjauhkan pelita daripada bekas kuprum supaya bekas tidak dihitamkan oleh jelagaMoving the lamp further away from the copper can so that the can is not blackened by soot
• BMenggantikan bekas kuprum dengan bikar kaca yang tebalReplacing the copper can with a thick glass beaker
• CMembuka semua sisi supaya udara bebas masuk ke dalam nyalaanOpening all sides so that air flows freely around the flame
• DMengelilingi pelita dan bekas kuprum dengan pengadang anginSurrounding the lamp and the copper can with a draught shield

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D

A draught shield reduces the heat lost to the surroundings, so more of the heat is absorbed by the water. A, B and C all increase the heat loss or reduce the heat absorbed.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 3, standard 3.2.4

TextbookBuku teks p.ms. 135

12

QuestionSoalan 12

Persamaan bagi pembakaran metana ialah CH₄ + 2O₂ → CO₂ + 2H₂O. Tenaga ikatan: C–H = 413, O=O = 498, C=O = 805, O–H = 464 kJ mol⁻¹. Hitung perubahan haba bagi tindak balas ini.The equation for the combustion of methane is CH₄ + 2O₂ → CO₂ + 2H₂O. Bond energies: C–H = 413, O=O = 498, C=O = 805, O–H = 464 kJ mol⁻¹. Calculate the heat change for this reaction.

• A+818 kJ
• B-818 kJ
• C+110 kJ
• D-3 466 kJ

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B

Bonds broken: 4(413) + 2(498) = 2 648 kJ. Bonds formed: 2(805) + 4(464) = 3 466 kJ. ΔH = 2 648 - 3 466 = -818 kJ. +110 comes from counting only two O–H bonds, and -3 466 is the energy released alone.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 3, standard 3.1.2

TextbookBuku teks p.ms. 119

13

QuestionSoalan 13

Rajah 1 menunjukkan profil tenaga bagi suatu tindak balas. Apakah tenaga pengaktifan bagi tindak balas songsang?Diagram 1 shows the energy profile of a reaction. What is the activation energy of the reverse reaction?

• A190 kJ
• B80 kJ
• C270 kJ
• D350 kJ

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C

The reverse reaction starts from the products (40 kJ) and must climb to the peak (310 kJ): 310 - 40 = 270 kJ. 190 kJ is the activation energy of the forward reaction (310 - 120) and 80 kJ is the size of ΔH.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 3, standard 3.1.2

TextbookBuku teks p.ms. 118

14

QuestionSoalan 14

Polimer yang manakah dihasilkan daripada propena dan digunakan untuk membuat mainan dan radas makmal?Which polymer is made from propene and used to make toys and laboratory apparatus?

• APolitenaPolyethene
• BPolivinil klorida (PVC)Polyvinyl chloride (PVC)
• CNilonNylon
• DPolipropenaPolypropene

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D

Propene, CH₂=CH–CH₃, polymerises by addition to give polypropene, which is durable and used for toys, textiles and laboratory apparatus. Polyethene comes from ethene, PVC from chloroethene and nylon is a condensation polymer.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 4, standard 4.1.3

TextbookBuku teks p.ms. 147

15

QuestionSoalan 15

5.2 g stirena, C₆H₅CH=CH₂, dipolimerkan sepenuhnya melalui pempolimeran penambahan. Apakah jisim polistirena yang terbentuk?5.2 g of styrene, C₆H₅CH=CH₂, is polymerised completely by addition polymerisation. What is the mass of polystyrene formed?

• A5.2 g
• B2.6 g
• C10.4 g
• D5.0 g

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A

In addition polymerisation the monomers simply join together and no small molecule is given off, so the mass of the polymer equals the mass of the monomer: 5.2 g.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 4, standard 4.1.2

TextbookBuku teks p.ms. 146

16

QuestionSoalan 16

Getah sintetik yang manakah tahan suhu yang sangat tinggi dan lengai, serta digunakan untuk membuat implan perubatan?Which synthetic rubber withstands very high temperatures, is inert and is used to make medical implants?

• ANeoprenaNeoprene
• BGetah silikonSilicone rubber
• CGetah stirena-butadiena (SBR)Styrene-butadiene rubber (SBR)
• DGetah nitrilNitrile rubber

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B

Silicone rubber resists high temperatures and is chemically inert, so it is used for medical implants and cooking utensils. Neoprene is used for hoses and belts, SBR for tyres, and nitrile rubber for gloves and sealants.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 4, standard 4.3.1

TextbookBuku teks p.ms. 160

17

QuestionSoalan 17

Jadual 2 menunjukkan keputusan ujian ke atas empat sampel plastik, P, Q, R dan S, setiap satu berjisim 10.0 g, yang ditanam dalam tanah selama enam bulan. Sebuah kedai ingin menggunakan beg plastik yang boleh membawa barang seberat 5 kg dan tidak lama mencemarkan alam sekitar. Plastik yang manakah paling sesuai?

Plastik	P	Q	R	S
Jisim selepas 6 bulan (g)	9.9	6.2	9.8	3.1
Beban maksimum sebelum koyak (kg)	8.0	6.0	2.0	1.5

Jadual 2Table 2 shows the results of tests on four plastic samples, P, Q, R and S, each of mass 10.0 g, buried in soil for six months. A shop wants plastic bags that can carry 5 kg of goods and that do not pollute the environment for long. Which plastic is the most suitable?

Plastic	P	Q	R	S
Mass after 6 months (g)	9.9	6.2	9.8	3.1
Maximum load before tearing (kg)	8.0	6.0	2.0	1.5

Table 2

• AP, kerana ia paling kuatP, because it is the strongest
• BS, kerana ia mereput paling banyakS, because it decomposes the most
• CQ, kerana ia boleh membawa 5 kg dan mereput dengan ketaraQ, because it can carry 5 kg and decomposes noticeably
• DR, kerana ia hampir tidak berubah dalam tanahR, because it hardly changes in the soil

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C

The bag must carry 5 kg, so P (8.0 kg) and Q (6.0 kg) qualify; R and S would tear. Of those two, Q loses far more mass (3.8 g) than P (0.1 g), so it is the one that does not persist as pollution.

For the teacherUntuk guru: EvaluatingMenilai, hightinggi difficultyaras hightinggi, ChapterBab 4, standard 4.1.3

TextbookBuku teks p.ms. 148

18

QuestionSoalan 18

Zarah nano argentum ditambah ke dalam plastik bekas makanan. Apakah kelebihan penggunaannya?Silver nanoparticles are added to the plastic of a food container. What is the advantage?

• AMembunuh mikroorganisma supaya makanan tahan lebih lamaThey kill microorganisms so that the food stays fresh for longer
• BMenjadikan makanan lebih manisThey make the food sweeter
• CMenjadikan plastik mudah mereputThey make the plastic decompose easily
• DMenambah warna merah pada makananThey add a red colour to the food

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A

Nanoparticle additives give anti-microbial packaging: the silver kills bacteria and fungi so the food lasts longer.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 5, standard 5.5

TextbookBuku teks p.ms. 187

19

QuestionSoalan 19

Detergen biologi mengandungi enzim lipase. Apakah jenis kotoran yang disingkirkan oleh lipase?A biological detergent contains the enzyme lipase. Which type of stain does lipase remove?

• AKotoran darahBlood stains
• BKotoran minyak masak dan lemakCooking oil and fat stains
• CKotoran karatRust stains
• DKotoran dakwatInk stains

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B

Lipase breaks down fats and oils. Protease removes protein stains such as blood.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 5, standard 5.2

TextbookBuku teks p.ms. 177

20

QuestionSoalan 20

6.12 g natrium stearat, C₁₇H₃₅COONa, dilarutkan dalam air liat yang mengandungi ion Ca²⁺ berlebihan. Apakah jisim maksimum kekat sabun, (C₁₇H₃₅COO)₂Ca, yang terbentuk? [Jisim atom relatif: H = 1, C = 12, O = 16, Na = 23, Ca = 40]6.12 g of sodium stearate, C₁₇H₃₅COONa, is dissolved in hard water containing excess Ca²⁺ ions. What is the maximum mass of scum, (C₁₇H₃₅COO)₂Ca, that is formed? [Relative atomic mass: H = 1, C = 12, O = 16, Na = 23, Ca = 40]

• A6.12 g
• B3.03 g
• C12.12 g
• D6.06 g

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D

M of sodium stearate = 306, so n = 6.12 ÷ 306 = 0.020 mol. 2 mol of stearate gives 1 mol of scum, so n(scum) = 0.010 mol. M of (C₁₇H₃₅COO)₂Ca = 606, so mass = 0.010 × 606 = 6.06 g. 12.12 g forgets the 2 : 1 ratio.

For the teacherUntuk guru: ApplyingMengaplikasi, hightinggi difficultyaras hightinggi, ChapterBab 5, standard 5.2

TextbookBuku teks p.ms. 175

21

QuestionSoalan 21

Tong kitar semula sebuah kantin sekolah mengandungi tiga jenis plastik. Jadual 3 menunjukkan kod kitar semula, monomer dan kegunaan plastik itu. [Jisim atom relatif: H = 1, C = 12]

Kod	Polimer	Monomer	Kegunaan
1	PET (Terilena)	1,2-etanadiol dan asid tereftalik	Botol minuman
2	Politena	Etena	Botol susu
5	Polipropena	Propena	Bekas makanan

Jadual 3The recycling bin in a school canteen holds three types of plastic. Table 3 shows the recycling code, monomers and use of the plastics. [Relative atomic mass: H = 1, C = 12]

Code	Polymer	Monomer	Use
1	PET (Terylene)	1,2-ethanediol and terephthalic acid	Drinks bottle
2	Polyethene	Ethene	Milk bottle
5	Polypropene	Propene	Food container

Table 3

[2 marksmarkah] (a)Polimer yang manakah dihasilkan melalui pempolimeran kondensasi? Namakan molekul kecil yang terbebas.Which polymer is formed by condensation polymerisation? Name the small molecule released.

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• PET / Terylene [1]1
• Water [1]1

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[1 markmarkah] (b)Namakan jenis pempolimeran bagi polipropena.Name the type of polymerisation of polypropene.

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• Addition polymerisation1

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[1 markmarkah] (c)Tulis unit ulangan bagi polipropena.Write the repeat unit of polypropene.

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• -CH₂-CH(CH₃)-1

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[2 marksmarkah] (d)Jisim molekul relatif bagi satu molekul polipropena ialah 42 000. Hitung bilangan unit ulangan dalam molekul itu.The relative molecular mass of one molecule of polypropene is 42 000. Calculate the number of repeat units in the molecule.

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• M of C₃H₆ = 3(12) + 6(1) = 42 [1]1
• 42 000 ÷ 42 = 1 000 units [1]1

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[2 marksmarkah] (e)Botol susu politena dikitar semula dengan cara dileburkan dan dibentuk semula. Terangkan mengapa ia boleh dilakukan.Polyethene milk bottles are recycled by melting and remoulding them. Explain why this is possible.

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• Polyethene is a thermoplastic: its long chains are not joined by cross-links [1]1
• The weak forces between the chains break when heated, so it softens and can be reshaped (and hardens again on cooling) [1]1

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[2 marksmarkah] (f)Nyatakan satu kelebihan dan satu kelemahan menggunakan botol plastik berbanding botol kaca untuk minuman.State one advantage and one disadvantage of using plastic bottles instead of glass bottles for drinks.

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• Advantage: lighter / does not break easily / cheaper [1]1
• Disadvantage: non-biodegradable, so it pollutes the environment / burns to release toxic gases [1]1

Tick each point you got.Tandakan setiap isi yang betul.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 4, standard 4.1

TextbookBuku teks p.ms. 146

22

QuestionSoalan 22

Serbuk kuprum berlebihan ditambahkan ke dalam 50 cm³ larutan argentum nitrat 0.20 mol dm⁻³ dalam cawan polistirena. Suhu naik daripada 29.0 °C kepada 32.5 °C. Rajah 2 menunjukkan gambar rajah aras tenaga bagi tindak balas ini. [Muatan haba tentu larutan = 4.2 J g⁻¹ °C⁻¹; ketumpatan larutan = 1 g cm⁻³]Excess copper powder is added to 50 cm³ of 0.20 mol dm⁻³ silver nitrate solution in a polystyrene cup. The temperature rises from 29.0 °C to 32.5 °C. Diagram 2 shows the energy level diagram for this reaction. [Specific heat capacity of the solution = 4.2 J g⁻¹ °C⁻¹; density of the solution = 1 g cm⁻³]

[1 markmarkah] (a)Tulis persamaan ion bagi tindak balas penyesaran ini.Write the ionic equation for this displacement reaction.

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• Cu + 2Ag⁺ → Cu²⁺ + 2Ag1

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[1 markmarkah] (b)Nyatakan satu pemerhatian dalam tindak balas ini.State one observation in this reaction.

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• A grey / silvery solid is deposited, or the colourless solution turns blue1

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[1 markmarkah] (c)Hitung bilangan mol argentum yang disesarkan.Calculate the number of moles of silver displaced.

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• 0.20 × 50 / 1000 = 0.010 mol1

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[1 markmarkah] (d)Hitung haba yang dibebaskan dalam eksperimen ini.Calculate the heat released in this experiment.

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• 50 × 4.2 × 3.5 = 735 J1

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[2 marksmarkah] (e)Hitung haba penyesaran argentum oleh kuprum. Tulis nilainya dengan tanda dan unit.Calculate the heat of displacement of silver by copper. Write the value with its sign and unit.

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• 0.735 ÷ 0.010 = 73.5 kJ mol⁻¹ [1]1
• ΔH = -73.5 kJ mol⁻¹ (negative because the temperature rises / exothermic) [1]1

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[1 markmarkah] (f)Dalam Rajah 2, Cu(s) + 2Ag⁺(aq) berada pada aras __________.In Diagram 2, Cu(s) + 2Ag⁺(aq) is at level __________.

AMBN

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M

The reaction is exothermic, so the reactants have more energy and sit on the upper level M; the products are on the lower level N.

[2 marksmarkah] (g)Eksperimen diulang menggunakan serbuk zink berlebihan dan bukannya serbuk kuprum dalam keadaan yang sama. Ramalkan kenaikan suhu berbanding 3.5 °C dan berikan satu sebab.The experiment is repeated using excess zinc powder instead of copper powder under the same conditions. Predict the temperature rise compared with 3.5 °C and give one reason.

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• The temperature rise is higher than 3.5 °C [1]1
• Zinc is more electropositive than copper (E° more negative), so the displacement of silver releases more heat [1]1

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 3, standard 3.2.2

TextbookBuku teks p.ms. 125

23

QuestionSoalan 23

Jadual 4 menunjukkan bilangan ikatan ganda dua C=C bagi satu molekul dua trigliserida, P dan Q, dan keadaan fizikalnya pada suhu bilik. [Isi padu molar gas pada suhu bilik = 24 dm³ mol⁻¹]

Trigliserida	Bilangan ikatan C=C bagi satu molekul	Keadaan pada suhu bilik
P	0	Pepejal
Q	2	Cecair

Jadual 4Table 4 shows the number of C=C double bonds in one molecule of two triglycerides, P and Q, and their physical state at room temperature. [Molar volume of a gas at room temperature = 24 dm³ mol⁻¹]

Triglyceride	Number of C=C bonds per molecule	State at room temperature
P	0	Solid
Q	2	Liquid

Table 4

[1 markmarkah] (a)Antara P dan Q, yang manakah ialah lemak? Berikan satu sebab.Which of P and Q is a fat? Give one reason.

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• P, because it is a solid at room temperature / has only saturated fatty acids1

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[1 markmarkah] (b)Terangkan mengapa Q ialah cecair pada suhu bilik.Explain why Q is a liquid at room temperature.

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• Q contains unsaturated fatty acids; the C=C double bonds lower its melting point1

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[2 marksmarkah] (c)Q ditukarkan kepada marjerin melalui tindak balas dengan gas hidrogen. Namakan proses itu dan mangkin yang digunakan.Q is changed into margarine by reacting it with hydrogen gas. Name the process and the catalyst used.

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• Hydrogenation [1]1
• Nickel [1]1

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[2 marksmarkah] (d)Hitung isi padu gas hidrogen pada suhu bilik yang diperlukan untuk menukarkan 0.25 mol Q sepenuhnya kepada hasil tepu.Calculate the volume of hydrogen gas at room temperature needed to convert 0.25 mol of Q completely into a saturated product.

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• Moles of H₂ = 0.25 × 2 = 0.50 mol [1]1
• Volume = 0.50 × 24 = 12 dm³ [1]1

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[1 markmarkah] (e)Nyatakan satu kesan buruk pengambilan lemak tepu yang berlebihan terhadap kesihatan.State one harmful effect on health of eating too much saturated fat.

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• Raises the risk of heart problems / obesity / hardening of the arteries1

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[1 markmarkah] (f)P dipanaskan dengan larutan natrium hidroksida pekat. Namakan tindak balas itu.P is heated with concentrated sodium hydroxide solution. Name the reaction.

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• Saponification1

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For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 5, standard 5.1

TextbookBuku teks p.ms. 167

24

QuestionSoalan 24

Rajah 3 menunjukkan tiub-U yang mengandungi larutan besi(II) sulfat di lengan kiri dan larutan kalium manganat(VII) berasid di lengan kanan. Asid sulfurik cair memisahkan kedua-dua larutan itu. Elektrod karbon X dan Y disambungkan kepada galvanometer. Jarum galvanometer terpesong.Diagram 3 shows a U-tube containing iron(II) sulphate solution in the left limb and acidified potassium manganate(VII) solution in the right limb. Dilute sulphuric acid separates the two solutions. Carbon electrodes X and Y are connected to a galvanometer. The needle of the galvanometer is deflected.

[1 markmarkah] (a)Antara X dan Y, yang manakah terminal negatif? Berikan satu sebab.Which of X and Y is the negative terminal? Give a reason.

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• X, because the Fe²⁺ ions are oxidised there and release electrons to the electrode1

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[1 markmarkah] (b)Tulis setengah persamaan pengoksidaan.Write the half equation for the oxidation.

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• Fe²⁺ → Fe³⁺ + e⁻1

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[1 markmarkah] (c)Tulis setengah persamaan penurunan.Write the half equation for the reduction.

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• MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O1

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[1 markmarkah] (d)Nyatakan perubahan warna larutan kalium manganat(VII) berasid.State the colour change of the acidified potassium manganate(VII) solution.

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• Purple to colourless1

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[1 markmarkah] (e)Nyatakan arah pengaliran elektron dalam wayar luar.State the direction of flow of electrons in the external wire.

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• From X to Y1

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[1 markmarkah] (f)Namakan agen pengoksidaan dalam tindak balas ini.Name the oxidising agent in this reaction.

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• Potassium manganate(VII) / manganate(VII) ion1

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[1 markmarkah] (g)Nyatakan perubahan nombor pengoksidaan mangan.State the change in the oxidation number of manganese.

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• +7 to +21

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 1, standard 1.1.2

TextbookBuku teks p.ms. 12

25

QuestionSoalan 25

Sebatian X mempunyai formula struktur CH₃CH(OH)CH₂CH₃.Compound X has the structural formula CH₃CH(OH)CH₂CH₃.

[1 markmarkah] (a)Namakan kumpulan berfungsi dalam X.Name the functional group in X.

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• Hydroxyl group (-OH)1

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[1 markmarkah] (b)Namakan X mengikut nama IUPAC.Name X using the IUPAC name.

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• Butan-2-ol1

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[1 markmarkah] (c)Namakan isomer kedudukan bagi X.Name the position isomer of X.

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• Butan-1-ol1

The straight-chain alcohol with the -OH group on a different carbon is butan-1-ol.

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[2 marksmarkah] (d)Tulis persamaan kimia seimbang bagi pembakaran lengkap X dalam oksigen berlebihan.Write the balanced chemical equation for the complete combustion of X in excess oxygen.

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• Correct formulae of reactants and products: C₄H₉OH (C₄H₁₀O), O₂, CO₂, H₂O [1]1
• C₄H₉OH + 6O₂ → 4CO₂ + 5H₂O (balanced) [1]1

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[1 markmarkah] (e)Nyatakan satu kegunaan alkohol dalam kehidupan harian.State one use of alcohols in daily life.

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• Solvent (in paints, perfumes) / fuel / antiseptic or disinfectant / making alcoholic drinks (ethanol) / making esters1

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For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 2, standard 2.4

TextbookBuku teks p.ms. 104

26Choose 1Pilih 1

QuestionSoalan 26

Seorang murid menentukan haba pembakaran tiga alkohol dengan membakar setiap alkohol dalam pelita untuk memanaskan 200 g air dalam bekas kuprum. Dalam setiap eksperimen, 0.050 mol alkohol dibakar dan suhu awal air ialah 28.0 °C. Jadual 5 menunjukkan keputusan. [Jisim atom relatif: H = 1, C = 12, O = 16; muatan haba tentu air = 4.2 J g⁻¹ °C⁻¹]

Alkohol	Jisim alkohol yang dibakar (g)	Suhu tertinggi air (°C)
Metanol, CH₃OH	1.60	48.0
Etanol, C₂H₅OH	2.30	56.0
Propan-1-ol, C₃H₇OH	3.00	63.0

Jadual 5A pupil determines the heat of combustion of three alcohols by burning each alcohol in a lamp to heat 200 g of water in a copper can. In each experiment 0.050 mol of the alcohol is burnt and the initial temperature of the water is 28.0 °C. Table 5 shows the results. [Relative atomic mass: H = 1, C = 12, O = 16; specific heat capacity of water = 4.2 J g⁻¹ °C⁻¹]

Alcohol	Mass of alcohol burnt (g)	Highest temperature of water (°C)
Methanol, CH₃OH	1.60	48.0
Ethanol, C₂H₅OH	2.30	56.0
Propan-1-ol, C₃H₇OH	3.00	63.0

Table 5

[1 markmarkah] (a)Apakah yang dimaksudkan dengan haba pembakaran?What is meant by heat of combustion?

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• The heat released when 1 mole of a substance burns completely in excess oxygen1

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[4 marksmarkah] (b)Hitung haba pembakaran etanol berdasarkan keputusan eksperimen. Tunjukkan kerja mengira anda.Calculate the heat of combustion of ethanol from the experimental results. Show your working.

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• Temperature rise = 56.0 - 28.0 = 28.0 °C [1]1
• Heat absorbed by the water = 200 × 4.2 × 28.0 = 23 520 J = 23.52 kJ [1]1
• Heat released per mole = 23.52 ÷ 0.050 = 470.4 kJ [1]1
• ΔH = -470.4 kJ mol⁻¹ (sign and unit) [1]1

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[3 marksmarkah] (c)Hitung haba pembakaran propan-1-ol berdasarkan keputusan eksperimen.Calculate the heat of combustion of propan-1-ol from the experimental results.

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• Temperature rise = 63.0 - 28.0 = 35.0 °C; heat absorbed = 200 × 4.2 × 35.0 = 29 400 J = 29.4 kJ [1]1
• 29.4 ÷ 0.050 = 588 kJ [1]1
• ΔH = -588 kJ mol⁻¹ [1]1

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[2 marksmarkah] (d)Berdasarkan keputusan, nyatakan corak perubahan haba pembakaran apabila bilangan atom karbon dalam alkohol bertambah dan terangkan corak itu.From the results, state the pattern in the heat of combustion as the number of carbon atoms in the alcohol increases and explain the pattern.

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• The heat of combustion increases (methanol -336, ethanol -470.4, propan-1-ol -588 kJ mol⁻¹) [1]1
• More carbon and hydrogen atoms per molecule, so more CO₂ and H₂O molecules are formed and more heat is released when the bonds form [1]1

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[2 marksmarkah] (e)Nilai teori haba pembakaran etanol ialah -1 376 kJ mol⁻¹. Berikan dua sebab nilai eksperimen jauh lebih rendah.The theoretical heat of combustion of ethanol is -1 376 kJ mol⁻¹. Give two reasons why the experimental value is much lower.

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• Some heat is lost to the surroundings / absorbed by the copper can and the thermometer [1]1
• Incomplete combustion of the alcohol (soot forms) / some alcohol evaporates / not all the heat reaches the water [1]1

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[3 marksmarkah] (f)Lukis gambar rajah aras tenaga bagi pembakaran etanol dengan menggunakan nilai yang anda hitung dalam (b). Labelkan paksi, bahan tindak balas, hasil dan ΔH.Draw the energy level diagram for the combustion of ethanol using the value you calculated in (b). Label the axis, the reactants, the products and ΔH.

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• Vertical axis labelled energy; reactants C₂H₅OH + 3O₂ on the upper level and products 2CO₂ + 3H₂O on the lower level [1]1
• Downward arrow between the levels [1]1
• ΔH = -470.4 kJ mol⁻¹ labelled [1]1

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[4 marksmarkah] (g)Sebuah syarikat bas menggantikan sebahagian diesel dengan biodiesel yang dihasilkan daripada minyak masak terpakai yang dikutip daripada restoran. Syarikat itu menyatakan bas-bas itu membebaskan karbon dioksida bersih yang lebih sedikit, tetapi biodiesel lebih mahal bagi setiap liter dan bekalan minyak terpakai terhad. Nilai penggunaan biodiesel itu. Berikan dua kelebihan dan satu kelemahan, kemudian buat satu kesimpulan.A bus company replaces part of its diesel with biodiesel made from used cooking oil collected from restaurants. The company says the buses release less net carbon dioxide, but the biodiesel costs more per litre and the supply of used oil is limited. Evaluate the use of this biodiesel. Give two advantages and one disadvantage, then make one conclusion.

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• Advantage: renewable / made from waste, so less fossil fuel is used [1]1
• Advantage: less net carbon dioxide / less sulphur dioxide, so less global warming and air pollution [1]1
• Disadvantage: more expensive / limited supply of used oil [1]1
• Conclusion: worthwhile as a partial replacement (blend), because it lowers emissions even if it cannot meet all the demand [1]1

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[1 markmarkah] (h)Nyatakan pemboleh ubah dimanipulasikan dalam eksperimen ini.State the manipulated variable in this experiment.

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• The type of alcohol / the number of carbon atoms in the alcohol1

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For the teacherUntuk guru: EvaluatingMenilai, mediumsederhana difficultyaras mediumsederhana, ChapterBab 3, standard 3.2.4

TextbookBuku teks p.ms. 135

27Choose 1Pilih 1

QuestionSoalan 27

Rajah 4 menunjukkan siri tindak balas bagi etena, C₂H₄. P, Q, R dan S ialah sebatian karbon.Diagram 4 shows a series of reactions of ethene, C₂H₄. P, Q, R and S are carbon compounds.

[3 marksmarkah] (a)Namakan sebatian P dan sebatian Q, serta namakan Proses I.Name compound P and compound Q, and name Process I.

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• P: ethanol [1]1
• Q: ethanoic acid [1]1
• Process I: hydration [1]1

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[2 marksmarkah] (b)Namakan Proses IV dan tulis persamaan kimia bagi proses itu.Name Process IV and write the chemical equation for the process.

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• Dehydration [1]1
• C₂H₅OH → C₂H₄ + H₂O [1]1

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[3 marksmarkah] (c)Namakan ester R dan tulis persamaan kimia bagi pembentukannya.Name ester R and write the chemical equation for its formation.

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• Ethyl propanoate [1]1
• Correct formulae of reactants and products [1]1
• C₂H₅COOH + C₂H₅OH → C₂H₅COOC₂H₅ + H₂O (balanced) [1]1

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[4 marksmarkah] (d)11.5 g P bertindak balas dengan asid propanoik berlebihan dalam Proses III. Peratus hasil ester R ialah 60%. Hitung jisim ester R yang diperoleh. [Jisim atom relatif: H = 1, C = 12, O = 16]11.5 g of P reacts with excess propanoic acid in Process III. The percentage yield of ester R is 60%. Calculate the mass of ester R obtained. [Relative atomic mass: H = 1, C = 12, O = 16]

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• n(P) = 11.5 ÷ 46 = 0.25 mol [1]1
• 1 mol P gives 1 mol R, so n(R) = 0.25 mol; M of C₅H₁₀O₂ = 102 [1]1
• Theoretical mass = 0.25 × 102 = 25.5 g [1]1
• Actual mass = 60/100 × 25.5 = 15.3 g [1]1

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[2 marksmarkah] (e)Huraikan satu ujian kimia untuk membezakan etena daripada S.Describe one chemical test to distinguish ethene from S.

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• Shake each gas with bromine water (or acidified potassium manganate(VII) solution) [1]1
• Ethene decolourises the brown (purple) colour; S (ethane) causes no change [1]1

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[6 marksmarkah] (f)Seorang murid menyediakan etena di makmal dengan memanaskan wap etanol di atas serpihan porselin dan mengumpul gas itu di atas air. (i) Nyatakan fungsi serpihan porselin. (ii) Mengapakah etena boleh dikumpul di atas air? (iii) Nyatakan satu langkah berjaga-jaga apabila pemanasan dihentikan. (iv) Huraikan cara mengesahkan bahawa gas itu ialah alkena. (v) Gas itu terbakar dengan nyalaan kuning yang berjelaga. Terangkan sebabnya. (vi) Industri menghasilkan etena melalui peretakan pecahan petroleum dan bukan melalui pendehidratan etanol. Wajarkan kaedah industri itu.A pupil prepares ethene in the laboratory by heating ethanol vapour over porcelain chips and collecting the gas over water. (i) State the function of the porcelain chips. (ii) Why can ethene be collected over water? (iii) State one precaution to take when the heating is stopped. (iv) Describe how to confirm that the gas is an alkene. (v) The gas burns with a yellow, sooty flame. Explain why. (vi) Industry makes ethene by cracking petroleum fractions instead of dehydrating ethanol. Justify the industrial method.

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One mark for each of (i) to (vi). Total 6 marks.

• (i) Catalyst (it also holds the ethanol vapour)1
• (ii) Ethene is insoluble in water1
• (iii) Remove the delivery tube from the water before stopping the heating (to prevent suck-back)1
• (iv) Bubble the gas into bromine water: the brown colour is decolourised1
• (v) Ethene has a high percentage of carbon, so combustion in air is incomplete and soot (carbon) forms1
• (vi) Cracking uses abundant, cheap petroleum fractions and gives large quantities of ethene at lower cost than ethanol1

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For the teacherUntuk guru: EvaluatingMenilai, lowrendah difficultyaras lowrendah, ChapterBab 2, standard 2.3.2

TextbookBuku teks p.ms. 96

28

QuestionSoalan 28

Rajah 5 menunjukkan dua sel elektrolisis. Eksperimen I menggunakan elektrod karbon, P dan Q. Eksperimen II menggunakan elektrod kuprum, R dan S. Kedua-dua sel mengandungi larutan kuprum(II) sulfat 0.5 mol dm⁻³. Elektrod P dan R disambungkan kepada terminal negatif bateri. Selepas arus dialirkan selama 30 minit, kedua-dua elektrod dalam Eksperimen II ditimbang semula. Jadual 6 menunjukkan keputusannya. [Jisim atom relatif: Cu = 64; pemalar Avogadro = 6.02 × 10²³ mol⁻¹]

Elektrod	Jisim sebelum (g)	Jisim selepas (g)
R	8.20	8.52
S	12.46	12.14

Jadual 6Diagram 5 shows two electrolytic cells. Experiment I uses carbon electrodes, P and Q. Experiment II uses copper electrodes, R and S. Both cells contain 0.5 mol dm⁻³ copper(II) sulphate solution. Electrodes P and R are connected to the negative terminal of the battery. After a current flows for 30 minutes, the two electrodes in Experiment II are weighed again. Table 6 shows the results. [Relative atomic mass: Cu = 64; Avogadro constant = 6.02 × 10²³ mol⁻¹]

Electrode	Mass before (g)	Mass after (g)
R	8.20	8.52
S	12.46	12.14

Table 6

[2 marksmarkah] (a)Namakan semua ion yang hadir dalam larutan kuprum(II) sulfat.Name all the ions present in the copper(II) sulphate solution.

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• Copper(II) ion, Cu²⁺, and hydrogen ion, H⁺ [1]1
• Sulphate ion, SO₄²⁻, and hydroxide ion, OH⁻ [1]1

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[3 marksmarkah] (b)Dalam Eksperimen I, namakan ion yang dinyahcas di katod dan di anod. Terangkan jawapan anda.In Experiment I, name the ion discharged at the cathode and at the anode. Explain your answers.

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• Cathode: Cu²⁺ ion [1]1
• Cu²⁺ is less electropositive than H⁺ (E° = +0.34 V compared with 0.00 V), so it is discharged more easily [1]1
• Anode: OH⁻ ion; it is lower than SO₄²⁻ in the electrochemical series, so SO₄²⁻ is not discharged [1]1

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[2 marksmarkah] (c)Tulis setengah persamaan di katod dan di anod dalam Eksperimen I.Write the half equations at the cathode and at the anode in Experiment I.

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• Cathode: Cu²⁺ + 2e⁻ → Cu [1]1
• Anode: 4OH⁻ → 2H₂O + O₂ + 4e⁻ [1]1

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[1 markmarkah] (d)Nyatakan satu pemerhatian di katod dalam Eksperimen I.State one observation at the cathode in Experiment I.

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• A brown (pink) solid is deposited1

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[3 marksmarkah] (e)Dalam Eksperimen II, tulis setengah persamaan di anod dan di katod. Terangkan mengapa keamatan warna biru larutan tidak berubah.In Experiment II, write the half equations at the anode and at the cathode. Explain why the intensity of the blue colour of the solution does not change.

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• Anode (S): Cu → Cu²⁺ + 2e⁻ [1]1
• Cathode (R): Cu²⁺ + 2e⁻ → Cu [1]1
• The rate at which Cu²⁺ ions form at the anode equals the rate at which they are discharged at the cathode, so the concentration of Cu²⁺ stays constant [1]1

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[3 marksmarkah] (f)Berdasarkan Jadual 6, hitung perubahan jisim bagi elektrod R dan elektrod S, dan terangkan hubungan antara kedua-dua nilai itu.From Table 6, calculate the change in mass of electrode R and of electrode S, and explain the relationship between the two values.

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• R: 8.52 - 8.20 = +0.32 g (gains mass) [1]1
• S: 12.14 - 12.46 = -0.32 g (loses mass) [1]1
• Equal: the number of copper atoms that dissolve at S equals the number of Cu²⁺ ions deposited at R, because the same number of electrons flows through both electrodes [1]1

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[3 marksmarkah] (g)Hitung (i) bilangan mol kuprum yang dilarutkan dari anod, (ii) bilangan mol elektron yang mengalir melalui litar, dan (iii) bilangan elektron itu.Calculate (i) the number of moles of copper dissolved from the anode, (ii) the number of moles of electrons that flowed through the circuit, and (iii) the number of electrons.

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• (i) 0.32 ÷ 64 = 0.0050 mol1
• (ii) Cu → Cu²⁺ + 2e⁻, so 0.0050 × 2 = 0.010 mol of electrons1
• (iii) 0.010 × 6.02 × 10²³ = 6.02 × 10²¹ electrons1

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[3 marksmarkah] (h)Seorang murid ingin menyadur paku besi dengan kuprum menggunakan set radas Eksperimen II. (i) Elektrod yang manakah perlu digantikan dengan paku besi dan kepada terminal bateri yang mana ia disambungkan? (ii) Cadangkan satu perubahan untuk mendapatkan lapisan kuprum yang lebih tebal. (iii) Cadangkan satu langkah untuk memastikan lapisan kuprum melekat dengan baik.A pupil wants to electroplate an iron nail with copper using the set-up of Experiment II. (i) Which electrode should be replaced by the iron nail, and to which terminal of the battery is it connected? (ii) Suggest one change to obtain a thicker layer of copper. (iii) Suggest one step to make sure the copper layer sticks well.

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• (i) Replace R (the cathode) with the nail; connect it to the negative terminal [1]1
• (ii) Electrolyse for a longer time / use a larger current [1]1
• (iii) Clean the nail with sandpaper or dilute acid before plating to remove rust and grease [1]1

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For the teacherUntuk guru: EvaluatingMenilai, hightinggi difficultyaras hightinggi, ChapterBab 1, standard 1.4

TextbookBuku teks p.ms. 44

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