Practice paper, not an official KPM paperKertas latihan, bukan kertas rasmi KPM

Peperiksaan Percubaan SPM (Model)SPM Trial Examination (Model)

KimiaChemistry

Tingkatan 5, 2 jam / 2 hours, 100 marksmarkah

Nama / NameNama

Kelas / ClassKelas

Tarikh / DateTarikh

Markah / marksmarkah: / 100

InstructionsArahan

Kertas soalan ini mengandungi empat bahagian: Bahagian A, Bahagian B, Bahagian C dan Bahagian D.This question paper consists of four sections: Section A, Section B, Section C and Section D.
Bahagian A: pilih satu jawapan bagi setiap soalan. Bahagian B: jawab semua soalan. Bahagian C: jawab satu soalan sahaja. Bahagian D: soalan mesti dijawab.Section A: choose one answer for each question. Section B: answer all questions. Section C: answer one question only. Section D: this question must be answered.
Kerja mengira anda mesti ditunjukkan.Your working must be shown.
Rajah yang mengiringi soalan tidak dilukis mengikut skala kecuali dinyatakan.The diagrams are not drawn to scale unless stated.
Jawapan boleh ditulis dalam Bahasa Melayu atau Bahasa Inggeris.Answers may be written in Bahasa Melayu or English.
Jadual Berkala Unsur dan kalkulator saintifik dibenarkan.The Periodic Table and a scientific calculator are allowed.

Answer schemes are in English (DLP).Skema jawapan dalam Bahasa Inggeris (DLP).

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.

Bahagian ASection A

20 marksmarkah

Jawab semua soalan. Setiap soalan diikuti oleh empat pilihan jawapan, A, B, C dan D. Pilih satu jawapan sahaja.Answer all questions. Each question is followed by four options, A, B, C and D. Choose one answer only.

Apakah nombor pengoksidaan fosforus dalam asid fosforik, H₃PO₄?What is the oxidation number of phosphorus in phosphoric acid, H₃PO₄?

+3
+5
+7
-3

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H = +1 and O = -2, so 3(+1) + x + 4(-2) = 0 and x = +5.

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

TextbookBuku teks p.ms. 8

Antara tindak balas berikut, yang manakah ialah tindak balas redoks?Which of the following reactions is a redox reaction?

NaOH + HNO₃ → NaNO₃ + H₂O
Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃
CaCO₃ → CaO + CO₂
Zn + 2HCl → ZnCl₂ + H₂

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Only in Zn + 2HCl does an oxidation number change: Zn goes from 0 to +2 and H from +1 to 0. A is neutralisation, B is precipitation and C is thermal decomposition; none of them has a change in oxidation number.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 1, standard 1.1

TextbookBuku teks p.ms. 12

Kuprum tak tulen dijadikan anod dan kuprum tulen dijadikan katod dalam larutan kuprum(II) sulfat. Apakah pemerhatian pada anod dan larutan?Impure copper is made the anode and pure copper is made the cathode in copper(II) sulphate solution. What are the observations at the anode and in the solution?

Anod menjadi nipis; warna biru larutan tidak berubahThe anode becomes thinner; the blue colour of the solution does not change
Anod menjadi tebal; warna biru larutan tidak berubahThe anode becomes thicker; the blue colour of the solution does not change
Anod menjadi nipis; warna biru larutan semakin pudarThe anode becomes thinner; the blue colour of the solution fades
Gelembung gas dibebaskan pada anod; warna biru larutan semakin pudarBubbles of gas are released at the anode; the blue colour of the solution fades

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At the anode Cu → Cu²⁺ + 2e⁻, so the impure anode dissolves and becomes thinner. At the cathode Cu²⁺ is discharged at the same rate, so the concentration of Cu²⁺ and the blue colour stay the same.

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

TextbookBuku teks p.ms. 44

Mengapakah aluminium diekstrak daripada aluminium oksida melalui elektrolisis dan bukan melalui pemanasan dengan karbon?Why is aluminium extracted from aluminium oxide by electrolysis and not by heating with carbon?

Aluminium oksida larut dalam air.Aluminium oxide dissolves in water.
Karbon lebih mahal daripada elektrik.Carbon is more expensive than electricity.
Aluminium lebih elektropositif daripada karbon, maka karbon tidak dapat menurunkan aluminium oksida.Aluminium is more electropositive than carbon, so carbon cannot reduce aluminium oxide.
Aluminium mempunyai takat lebur yang lebih rendah daripada karbon.Aluminium has a lower melting point than carbon.

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Carbon can only reduce the oxides of metals below it in the reactivity series. Aluminium is above carbon, so a stronger reducing method, electrolysis, is needed.

For the teacherUntuk guru: UnderstandingMemahami, mediumsederhana difficultyaras mediumsederhana, ChapterBab 1, standard 1.5

TextbookBuku teks p.ms. 48

Nilai E° bagi Mn²⁺/Mn, Fe²⁺/Fe, Cu²⁺/Cu dan Ag⁺/Ag masing-masing ialah -1.18 V, -0.44 V, +0.34 V dan +0.80 V. Logam yang manakah ialah agen penurunan yang paling kuat?The E° values of Mn²⁺/Mn, Fe²⁺/Fe, Cu²⁺/Cu and Ag⁺/Ag are -1.18 V, -0.44 V, +0.34 V and +0.80 V respectively. Which metal is the strongest reducing agent?

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The more negative the E°, the more easily the metal releases electrons, so the stronger it is as a reducing agent. Mn has the most negative E°.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 1, standard 1.2

TextbookBuku teks p.ms. 25

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

-OH
C=C
-COOH
-COO-

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

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

TextbookBuku teks p.ms. 72

Etena dialirkan ke dalam larutan kalium manganat(VII) berasid. Apakah hasil tindak balas itu?Ethene is bubbled into acidified potassium manganate(VII) solution. What is the product of the reaction?

EtanolEthanol
Etana-1,2-diolEthane-1,2-diol
Asid etanoikEthanoic acid
EtanaEthane

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Ethene is oxidised by acidified potassium manganate(VII) (the purple colour is decolourised) to give ethane-1,2-diol, which has two -OH groups. Ethanol comes from hydration with steam, and ethane from hydrogenation.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 2, standard 2.3

TextbookBuku teks p.ms. 86

Sebatian X mengandungi 54.5% karbon, 9.1% hidrogen dan 36.4% oksigen mengikut jisim, dan jisim molekul relatifnya ialah 88. X bertindak balas dengan natrium karbonat untuk membebaskan gas. Apakah X? [Jisim atom relatif: H = 1, C = 12, O = 16]Compound X contains 54.5% carbon, 9.1% hydrogen and 36.4% oxygen by mass, and its relative molecular mass is 88. X reacts with sodium carbonate to release a gas. What is X? [Relative atomic mass: H = 1, C = 12, O = 16]

Asid propanoikPropanoic acid
Etil etanoatEthyl ethanoate
Asid pentanoikPentanoic acid
Asid butanoikButanoic acid

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Mole ratio C : H : O = 4.54 : 9.1 : 2.28 = 2 : 4 : 1, so the empirical formula is C₂H₄O (mass 44). Mr = 88 gives C₄H₈O₂. A compound that releases gas with a carbonate is an acid, so X is butanoic acid. Ethyl ethanoate also has the formula C₄H₈O₂ but does not react with carbonates.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 2, standard 2.3

TextbookBuku teks p.ms. 98

Haba pembakaran ialah perubahan haba apabilaThe heat of combustion is the heat change when

1 mol air terbentuk daripada ion hidrogen dan ion hidroksida1 mole of water is formed from hydrogen ions and hydroxide ions
1 mol mendakan terbentuk daripada ion dalam larutan akueus1 mole of precipitate is formed from ions in aqueous solution
1 mol logam disesarkan daripada larutan garamnya1 mole of metal is displaced from its salt solution
1 mol bahan terbakar lengkap dalam oksigen berlebihan1 mole of a substance burns completely in excess oxygen

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Heat of combustion: 1 mol of a substance burns completely in excess oxygen. A is the heat of neutralisation, B the heat of precipitation and C the heat of displacement.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 3, standard 3.2

TextbookBuku teks p.ms. 135

Persamaan bagi tindak balas hidrogen dengan klorin ialah H₂ + Cl₂ → 2HCl. Tenaga ikatan: H–H = 436, Cl–Cl = 243, H–Cl = 432 kJ mol⁻¹. Hitung perubahan haba bagi tindak balas ini.The equation for the reaction of hydrogen with chlorine is H₂ + Cl₂ → 2HCl. Bond energies: H–H = 436, Cl–Cl = 243, H–Cl = 432 kJ mol⁻¹. Calculate the heat change for this reaction.

-185 kJ
+185 kJ
-92.5 kJ
+1 543 kJ

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Bonds broken: 436 + 243 = 679 kJ. Bonds formed: 2 × 432 = 864 kJ. ΔH = 679 - 864 = -185 kJ (exothermic). -92.5 kJ is the value per mole of HCl.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 3, standard 3.1

TextbookBuku teks p.ms. 119

25 cm³ larutan barium klorida 1.0 mol dm⁻³ dicampurkan dengan 25 cm³ larutan natrium sulfat 1.0 mol dm⁻³. Suhu campuran naik sebanyak 3.5 °C. Apakah haba pemendakan barium sulfat? [Muatan haba tentu larutan = 4.2 J g⁻¹ °C⁻¹; ketumpatan larutan = 1 g cm⁻³]25 cm³ of 1.0 mol dm⁻³ barium chloride solution is mixed with 25 cm³ of 1.0 mol dm⁻³ sodium sulphate solution. The temperature of the mixture rises by 3.5 °C. What is the heat of precipitation of barium sulphate? [Specific heat capacity of the solution = 4.2 J g⁻¹ °C⁻¹; density of the solution = 1 g cm⁻³]

-14.7 kJ mol⁻¹
-0.735 kJ mol⁻¹
-29.4 kJ mol⁻¹
-58.8 kJ mol⁻¹

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Q = 50 × 4.2 × 3.5 = 735 J = 0.735 kJ. n(BaSO₄) = 1.0 × 25/1000 = 0.025 mol. ΔH = -0.735 ÷ 0.025 = -29.4 kJ mol⁻¹. Using 0.05 mol gives -14.7 and forgetting to divide by moles gives -0.735.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 3, standard 3.2

TextbookBuku teks p.ms. 122

Jadual 1 menunjukkan kenaikan suhu apabila 50 cm³ larutan natrium hidroksida 2.0 mol dm⁻³ dicampurkan dengan 50 cm³ setiap asid 1.0 mol dm⁻³. Pernyataan yang manakah betul?

Asid	HCl	HNO₃	CH₃COOH	H₂SO₄
Kenaikan suhu (°C)	6.5	6.5	5.2	13.0

Jadual 1Table 1 shows the temperature rise when 50 cm³ of 2.0 mol dm⁻³ sodium hydroxide solution is mixed with 50 cm³ of each 1.0 mol dm⁻³ acid. Which statement is correct?

Acid	HCl	HNO₃	CH₃COOH	H₂SO₄
Temperature rise (°C)	6.5	6.5	5.2	13.0

Table 1

Haba peneutralan asid sulfurik dua kali ganda haba peneutralan asid hidroklorik kerana kenaikan suhunya lebih tinggi.The heat of neutralisation of sulphuric acid is twice that of hydrochloric acid because its temperature rise is higher.
Asid sulfurik memberi kenaikan suhu yang paling tinggi kerana ia menghasilkan bilangan mol air dua kali ganda berbanding asid hidroklorik.Sulphuric acid gives the highest temperature rise because it produces twice as many moles of water as hydrochloric acid.
Asid etanoik memberi kenaikan suhu yang paling rendah kerana ia asid yang lemah, maka kurang air terbentuk.Ethanoic acid gives the lowest temperature rise because it is a weak acid, so less water is formed.
Asid nitrik dan asid etanoik memberi kenaikan suhu yang sama kerana kedua-duanya asid monoprotik.Nitric acid and ethanoic acid give the same temperature rise because both are monoprotic acids.

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NaOH is in excess, so the acid is the limiting reactant: 0.050 mol H⁺ from HCl but 0.100 mol H⁺ from H₂SO₄, giving twice the moles of water and twice the heat in the same volume. The heat of neutralisation per mole of water is the same (A is wrong). For ethanoic acid the same moles of water form, but some heat is used to ionise the weak acid completely (C is wrong). D is wrong because 6.5 °C and 5.2 °C are different.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 3, standard 3.2

TextbookBuku teks p.ms. 131

Polimer yang manakah digunakan untuk membuat kotak pembungkus makanan yang menebat haba?Which polymer is used to make food packaging boxes that insulate against heat?

Polivinil klorida (PVC)Polyvinyl chloride (PVC)
PolistirenaPolystyrene
NilonNylon
PolipropenaPolypropene

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Polystyrene is light and a good heat insulator, so it is used for food packaging. PVC is used for water pipes and insulation of cables, nylon for ropes and clothes, and polypropene for toys and textiles.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 4, standard 4.1

TextbookBuku teks p.ms. 148

Persamaan n CH₂=CHCl → (-CH₂-CHCl-)ₙ mewakiliThe equation n CH₂=CHCl → (-CH₂-CHCl-)ₙ represents

pempolimeran penambahanaddition polymerisation
pempolimeran kondensasicondensation polymerisation
pengesteranesterification
penghidrogenanhydrogenation

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The monomer has a C=C double bond that opens so that the monomers join in a chain, and only the polymer is formed: addition polymerisation.

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

TextbookBuku teks p.ms. 146

Baris yang manakah betul bagi polimer, jenis pempolimeran dan molekul kecil yang terbebas?Which row is correct for the polymer, its type of polymerisation and the small molecule released?

Polipropena; kondensasi; airPolypropene; condensation; water
PVC; penambahan; hidrogen kloridaPVC; addition; hydrogen chloride
Terilena; penambahan; airTerylene; addition; water
Nilon; kondensasi; hidrogen kloridaNylon; condensation; hydrogen chloride

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Each option gives the polymer, its type of polymerisation and the small molecule released. Nylon, made from 1,6-hexanediamine and decanedioyl dichloride, is a condensation polymer and releases hydrogen chloride. Polypropene and PVC are addition polymers with no by-product. Terylene is a condensation polymer, and releases water.

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

TextbookBuku teks p.ms. 147

Lesitin ditambahkan kepada marjerin. Apakah fungsinya?Lecithin is added to margarine. What is its function?

Menghalang pertumbuhan bakteriaTo prevent the growth of bacteria
Memperlahankan pengoksidaan lemakTo slow down the oxidation of fats
Membolehkan minyak dan air bercampurTo allow oil and water to mix
Menambah warna kuningTo add a yellow colour

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Lecithin is an emulsifier: it lets two immiscible liquids such as oil and water mix. A describes a preservative, B an antioxidant and D a dye.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 5, standard 5.3

TextbookBuku teks p.ms. 181

Pasangan ubat dan kesan sampingnya yang manakah betul?Which pair of medicine and its side effect is correct?

Aspirin; boleh menyebabkan ulser perutAspirin; may cause stomach ulcers
Antihistamin; kerosakan hati apabila diambil berlebihanAntihistamine; liver damage when taken in excess
Parasetamol; muka bulat (moon face)Paracetamol; moon face
Penisilin; mengantukPenicillin; drowsiness

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Aspirin is acidic and can cause stomach ulcers. Paracetamol overdose damages the liver (not moon face, which is caused by long-term corticosteroids), antihistamines cause drowsiness, and penicillin may cause a fatal allergy.

For the teacherUntuk guru: UnderstandingMemahami, mediumsederhana difficultyaras mediumsederhana, ChapterBab 5, standard 5.4

TextbookBuku teks p.ms. 183

Pernyataan yang manakah menerangkan satu aplikasi nanoteknologi?Which statement describes an application of nanotechnology?

Menambah ammonia kepada lateks supaya tidak menggumpalAdding ammonia to latex so that it does not coagulate
Mencampurkan etanol dengan petrol sebagai bahan apiMixing ethanol with petrol as a fuel
Menggunakan klorin untuk membunuh bakteria dalam kolam renangUsing chlorine to kill bacteria in a swimming pool
Menambah nanozarah titanium dioksida dalam losyen pelindung matahari untuk menyerap sinar UV tanpa meninggalkan kesan putihAdding titanium dioxide nanoparticles to sunscreen lotion to absorb UV light without leaving a white mark

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Nanoparticles are 1 to 100 nm in size and coat the skin evenly and are transparent, which is a cosmetics application. The others use ordinary chemicals (ammonia, ethanol, chlorine) and do not involve nanoscale particles.

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

TextbookBuku teks p.ms. 188

Antara berikut, yang manakah ialah amalan Teknologi Hijau dalam pengurusan sisa pepejal?Which of the following is a Green Technology practice in solid waste management?

Membakar sampah secara terbukaBurning rubbish in the open
Mengumpul gas metana dari tapak pelupusan untuk menjana elektrikCollecting methane gas from landfills to generate electricity
Membuang sisa kimia ke dalam sungai selepas dicairkanDischarging chemical waste into rivers after diluting it
Menanam semua sisa dalam satu tapak tanpa dipisahkanBurying all waste in one site without sorting

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Methane recovery is a Green Technology practice: it stops a greenhouse gas escaping and the methane is used as an energy source. The other options pollute air, water or land.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 5, standard 5.6

TextbookBuku teks p.ms. 193

Bahan kimia manakah yang sering ditambah secara haram ke dalam krim pemutih kulit dan boleh merosakkan buah pinggang dan sistem saraf?Which chemical is sometimes added illegally to skin-whitening creams and can damage the kidneys and nervous system?

HidrokuinonHydroquinone
Betametason valeratBetamethasone valerate
MerkuriMercury
Tretinoin

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Mercury damages the kidneys and nervous system. Hydroquinone causes hypersensitive skin, and tretinoin causes redness and peeling.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 5, standard 5.4

TextbookBuku teks p.ms. 184

Bahagian BSection B

40 marksmarkah

Jawab semua soalan dalam bahagian ini.Answer all questions in this section.

Rajah 1 menunjukkan susunan radas untuk menyadur sebuah sudu besi dengan argentum. Elektrod X dan Y ditimbang sebelum dan selepas elektrolisis selama 30 minit. Jadual 2 menunjukkan keputusannya. [Jisim atom relatif: Ag = 108]

Elektrod	Jisim sebelum (g)	Jisim selepas (g)
X	31.20	30.66
Y	24.86	25.40

Jadual 2Diagram 1 shows the apparatus used to electroplate an iron spoon with silver. Electrodes X and Y are weighed before and after electrolysis for 30 minutes. Table 2 shows the results. [Relative atomic mass: Ag = 108]

Electrode	Mass before (g)	Mass after (g)
X	31.20	30.66
Y	24.86	25.40

Table 2

A beaker of 1.0 mol per dm3 silver nitrate solution holds electrode X on the left, connected to the positive terminal of a cell, and electrode Y, a spoon, on the right, connected to the negative terminal — Rajah 1Diagram 1

[1 markmarkah] (a)Elektrod X ialah __________.Electrode X is the __________.

anodanodekatodcathode

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anode

X is connected to the positive terminal of the cell, so it is the anode. The spoon Y is the cathode.

[1 markmarkah] (b)Tulis setengah persamaan pada elektrod Y.Write the half equation at electrode Y.

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Ag⁺ + e⁻ → Ag1

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[1 markmarkah] (c)Tulis setengah persamaan pada elektrod X.Write the half equation at electrode X.

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Ag → Ag⁺ + e⁻1

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[2 marksmarkah] (d)Nyatakan perubahan kepekatan larutan argentum nitrat semasa elektrolisis dan berikan satu sebab.State the change in the concentration of the silver nitrate solution during electrolysis and give one reason.

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The concentration does not change [1]1
The rate at which Ag⁺ ions are discharged at Y equals the rate at which Ag atoms ionise at X [1]1

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[2 marksmarkah] (e)Daripada Jadual 2, hitung (i) perubahan jisim elektrod Y dan (ii) bilangan mol argentum yang dienapkan.From Table 2, calculate (i) the change in mass of electrode Y and (ii) the number of moles of silver deposited.

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(i) 25.40 - 24.86 = 0.54 g [1]1
(ii) 0.54 ÷ 108 = 0.0050 mol [1]1

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[1 markmarkah] (f)Elektrod X digantikan dengan elektrod karbon. Namakan gas yang terbebas pada elektrod karbon.Electrode X is replaced with a carbon electrode. Name the gas released at the carbon electrode.

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Oxygen1

OH⁻ ions are discharged at the anode: 4OH⁻ → 2H₂O + O₂ + 4e⁻, because nitrate ions are not discharged.

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[1 markmarkah] (g)Nyatakan satu tujuan menyadur sudu besi dengan argentum.State one purpose of electroplating an iron spoon with silver.

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To make it look more attractive / to prevent rusting / to make it more resistant to corrosion1

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

TextbookBuku teks p.ms. 44

Jadual 3 menunjukkan maklumat bagi empat polimer sintetik, P, Q, R dan S.

Polimer	Bilangan monomer berbeza	Molekul kecil yang terbebas	Kesan pemanasan
P	1 (mengandungi C=C)	Tiada	Melembut
Q	2	Air	Melembut
R	2	Air	Tidak melembut; menjadi hitam
S	1 (etena)	Tiada	Melembut

Jadual 3Table 3 shows information about four synthetic polymers, P, Q, R and S.

Polymer	Number of different monomers	Small molecule released	Effect of heating
P	1 (contains C=C)	None	Softens
Q	2	Water	Softens
R	2	Water	Does not soften; turns black
S	1 (ethene)	None	Softens

Table 3

[1 markmarkah] (a)Antara P, Q, R dan S, polimer manakah dihasilkan melalui pempolimeran penambahan?Which of P, Q, R and S are made by addition polymerisation?

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P and S1

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[2 marksmarkah] (b)Terangkan mengapa Q terhasil melalui pempolimeran kondensasi.Explain why Q is formed by condensation polymerisation.

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It is made from two different monomers, each with two functional groups [1]1
A small molecule, water, is released each time two units join [1]1

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[2 marksmarkah] (c)Kelaskan R sebagai termoplastik atau termoset dan terangkan jawapan anda berdasarkan struktur.Classify R as a thermoplastic or a thermoset and explain your answer in terms of structure.

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Thermoset [1]1
The chains are held by strong cross-links (covalent bonds) so they cannot slide past one another and the plastic does not soften [1]1

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[1 markmarkah] (d)Polimer manakah boleh dikitar semula dengan melebur dan membentuk semula?Which polymers can be recycled by melting and remoulding?

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P, Q and S (thermoplastics)1

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[1 markmarkah] (e)Tulis persamaan untuk pempolimeran S menggunakan n molekul monomer.Write the equation for the polymerisation of S using n molecules of the monomer.

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n CH₂=CH₂ → (-CH₂-CH₂-)ₙ1

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[1 markmarkah] (f)Nyatakan satu kegunaan R yang memerlukan sifatnya apabila dipanaskan.State one use of R that depends on its behaviour when heated.

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Handles of cooking pots / electrical sockets and plugs / kettle handles, because it does not soften or melt1

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

TextbookBuku teks p.ms. 147

Jadual 4 menunjukkan takat didih empat anggota pertama siri homolog asid karboksilik.

Asid	Formula molekul	Takat didih (°C)
Asid metanoik	HCOOH	101
Asid etanoik	CH₃COOH	118
Asid propanoik	C₂H₅COOH	141
Asid butanoik	C₃H₇COOH	164

Jadual 4Table 4 shows the boiling points of the first four members of the homologous series of carboxylic acids.

Acid	Molecular formula	Boiling point (°C)
Methanoic acid	HCOOH	101
Ethanoic acid	CH₃COOH	118
Propanoic acid	C₂H₅COOH	141
Butanoic acid	C₃H₇COOH	164

Table 4

[1 markmarkah] (a)Tulis formula am bagi siri homolog ini.Write the general formula of this homologous series.

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CₙH₂ₙ₊₁COOH (or CₙH₂ₙO₂)1

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[2 marksmarkah] (b)Nyatakan corak perubahan takat didih dalam Jadual 4 dan terangkan sebabnya.State the trend in boiling point in Table 4 and explain the reason.

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The boiling point increases as the number of carbon atoms increases [1]1
Molecules get larger, so the forces between molecules are stronger and more heat energy is needed to overcome them [1]1

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[1 markmarkah] (c)Ramalkan takat didih asid pentanoik. Beri nilai dalam julat 180 °C hingga 190 °C.Predict the boiling point of pentanoic acid. Give a value in the range 180 °C to 190 °C.

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About 185 °C (accept 180 to 190 °C; each step adds about 23 °C)1

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[3 marksmarkah] (d)Asid propanoik bertindak balas dengan serbuk magnesium. Namakan garam yang terbentuk dan tulis persamaan kimia yang seimbang.Propanoic acid reacts with magnesium powder. Name the salt formed and write the balanced chemical equation.

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Magnesium propanoate [1]1
Correct formulae of reactants and products [1]1
2C₂H₅COOH + Mg → (C₂H₅COO)₂Mg + H₂ (balanced) [1]1

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For the teacherUntuk guru: AnalysingMenganalisis, lowrendah difficultyaras lowrendah, ChapterBab 2, standard 2.2.3

TextbookBuku teks p.ms. 75

Tin makanan yang memanaskan sendiri mengandungi kalsium oksida, CaO, dalam satu petak dan air dalam petak lain. Apabila pengguna menekan butang, kalsium oksida bercampur dengan air: CaO + H₂O → Ca(OH)₂, ΔH = -65 kJ mol⁻¹. Dalam satu ujian, 5.6 g kalsium oksida digunakan untuk memanaskan 150 g air. [Jisim atom relatif: O = 16, Ca = 40; muatan haba tentu air = 4.2 J g⁻¹ °C⁻¹]A self-heating food can contains calcium oxide, CaO, in one compartment and water in another. When the user presses a button, calcium oxide mixes with water: CaO + H₂O → Ca(OH)₂, ΔH = -65 kJ mol⁻¹. In a test, 5.6 g of calcium oxide is used to heat 150 g of water. [Relative atomic mass: O = 16, Ca = 40; specific heat capacity of water = 4.2 J g⁻¹ °C⁻¹]

[1 markmarkah] (a)Nyatakan sama ada tindak balas ini eksotermik atau endotermik dan berikan satu sebab.State whether this reaction is exothermic or endothermic and give one reason.

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Exothermic, because ΔH is negative / heat is released1

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[1 markmarkah] (b)Hitung bilangan mol kalsium oksida yang digunakan.Calculate the number of moles of calcium oxide used.

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5.6 ÷ 56 = 0.10 mol1

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[1 markmarkah] (c)Hitung haba yang dibebaskan.Calculate the heat released.

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0.10 × 65 = 6.5 kJ (6 500 J)1

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[2 marksmarkah] (d)Hitung kenaikan suhu maksimum air, dengan menganggap semua haba diserap oleh air.Calculate the maximum temperature rise of the water, assuming all the heat is absorbed by the water.

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Q = mcθ, so θ = 6 500 ÷ (150 × 4.2) [1]1
θ = 10.3 °C [1]1

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[1 markmarkah] (e)Kenaikan suhu sebenar lebih rendah daripada nilai dalam (d). Berikan satu sebab.The actual temperature rise is lower than the value in (d). Give one reason.

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Some heat is lost to the surroundings / absorbed by the can and the food1

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 3, standard 3.3

TextbookBuku teks p.ms. 137

Bahagian CSection C

20 marksmarkah

Jawab satu soalan sahaja.Answer one question only.

26Choose 1Pilih 1

Sebuah syarikat gas mahu melindungi paip keluli (besi) bawah tanah daripada berkarat dengan melekatkan blok logam pada paip itu. Jadual 5 menunjukkan nilai E° bagi beberapa logam.

Setengah sel	Mg²⁺/Mg	Zn²⁺/Zn	Fe²⁺/Fe	Sn²⁺/Sn	Cu²⁺/Cu	Ni²⁺/Ni
E° (V)	-2.38	-0.76	-0.44	-0.14	+0.34	-0.25

Jadual 5A gas company wants to protect an underground steel (iron) pipe from rusting by attaching blocks of metal to the pipe. Table 5 shows the E° values of some metals.

Half-cell	Mg²⁺/Mg	Zn²⁺/Zn	Fe²⁺/Fe	Sn²⁺/Sn	Cu²⁺/Cu	Ni²⁺/Ni
E° (V)	-2.38	-0.76	-0.44	-0.14	+0.34	-0.25

Table 5

[6 marksmarkah] (a)Bandingkan sel kimia (sel voltan) dengan sel elektrolisis berdasarkan (i) perubahan tenaga, (ii) kutub negatif dan positif elektrod, dan (iii) tindak balas pada anod dan katod. Beri satu contoh penggunaan bagi setiap jenis sel.Compare a chemical (voltaic) cell with an electrolytic cell in terms of (i) the energy change, (ii) the negative and positive terminals, and (iii) the reactions at the anode and cathode. Give one use of each type of cell.

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Any 6 pointsMana-mana 6 isi

Voltaic: chemical energy changes to electrical energy [1]1
Electrolytic: electrical energy changes to chemical energy [1]1
Voltaic: the anode (where oxidation occurs) is the negative terminal and the cathode is the positive terminal [1]1
Electrolytic: the anode is connected to the positive terminal of the cell and the cathode to the negative terminal [1]1
Both: oxidation occurs at the anode and reduction occurs at the cathode [1]1
Voltaic cell is used in a torch / battery / fuel cell [1]1
Electrolytic cell is used in electroplating / extraction / purification of metals [1]1

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[6 marksmarkah] (b)Sebuah sel kimia dibina menggunakan zink dalam larutan zink sulfat dan nikel dalam larutan nikel(II) sulfat, dihubungkan oleh titian garam. (i) Nyatakan terminal negatif. (ii) Tulis setengah persamaan di setiap terminal. (iii) Hitung voltan sel. (iv) Tulis tatatanda sel. (v) Nyatakan arah pengaliran elektron dalam wayar luar.A chemical cell is made using zinc in zinc sulphate solution and nickel in nickel(II) sulphate solution, joined by a salt bridge. (i) State the negative terminal. (ii) Write the half equation at each terminal. (iii) Calculate the cell voltage. (iv) Write the cell notation. (v) State the direction of electron flow in the external wire.

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(i) Zinc (more negative E°) [1]1
(ii) Negative: Zn → Zn²⁺ + 2e⁻ [1]1
(ii) Positive: Ni²⁺ + 2e⁻ → Ni [1]1
(iii) (-0.25) - (-0.76) = +0.51 V [1]1
(iv) Zn | Zn²⁺ || Ni²⁺ | Ni [1]1
(v) From zinc to nickel [1]1

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[8 marksmarkah] (c)(i) Antara logam dalam Jadual 5, cadangkan dua logam yang sesuai untuk melindungi paip besi dan berikan sebab. (ii) Dengan merujuk kepada satu logam yang anda pilih, terangkan bagaimana paip dilindungi, termasuk setengah persamaan. (iii) Pada pendapat anda, antara dua logam yang anda pilih, yang manakah lebih sesuai? Wajarkan jawapan anda. (iv) Mengapakah timah dan kuprum tidak sesuai?(i) From the metals in Table 5, suggest two metals that are suitable to protect the iron pipe and give the reason. (ii) With reference to one metal you chose, explain how the pipe is protected, including a half equation. (iii) In your opinion, which of the two metals you chose is more suitable? Justify your answer. (iv) Why are tin and copper not suitable?

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(i) Magnesium and zinc [1]1
(i) They are more electropositive than iron (E° more negative than -0.44 V) [1]1
(ii) Mg → Mg²⁺ + 2e⁻ (or Zn → Zn²⁺ + 2e⁻) [1]1
(ii) The metal is oxidised in preference to iron; electrons flow to the iron, so iron is not oxidised [1]1
(ii) Sacrificial protection: oxygen is reduced at the iron surface, O₂ + 2H₂O + 4e⁻ → 4OH⁻ [1]1
(iii) Zinc: it is less reactive, corrodes slowly and the blocks last longer, so replacement is less frequent, OR magnesium: larger difference in E° gives stronger protection, but it corrodes faster and must be replaced often [1]1
(iii) A clear choice with a reason related to the cost or the life of the blocks [1]1
(iv) Tin and copper are less electropositive than iron, so iron becomes the negative terminal and rusts faster [1]1

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

For the teacherUntuk guru: EvaluatingMenilai, lowrendah difficultyaras lowrendah, ChapterBab 1, standard 1.6

TextbookBuku teks p.ms. 55

27Choose 1Pilih 1

Sebuah gerai menjual pisang goreng dan menggoreng makanan menggunakan minyak yang sama berulang kali. Pemilik gerai ingin memilih minyak masak yang paling sesuai. Jadual 6 menunjukkan komposisi asid lemak bagi tiga jenis minyak.

Minyak	Asid lemak tepu (%)	Asid lemak tak tepu (%)
Minyak kelapa	91	9
Minyak sawit	49	51
Minyak jagung	14	86

Jadual 6A stall sells fried bananas and fries food using the same oil again and again. The owner wants to choose the most suitable cooking oil. Table 6 shows the fatty acid composition of three oils.

Oil	Saturated fatty acids (%)	Unsaturated fatty acids (%)
Coconut oil	91	9
Palm oil	49	51
Corn oil	14	86

Table 6

[6 marksmarkah] (a)Bandingkan minyak dengan lemak berdasarkan (i) sumber, (ii) keadaan fizik pada suhu bilik dan (iii) peratus asid lemak tak tepu. Berikan satu contoh bagi setiap jenis.Compare oils and fats in terms of (i) source, (ii) physical state at room temperature and (iii) the percentage of unsaturated fatty acids. Give one example of each.

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Oils come from plants; fats come from animals [1]1
Oils are liquid at room temperature; fats are solid [1]1
Oils have a high percentage of unsaturated fatty acids; fats have a high percentage of saturated fatty acids [1]1
Oils have lower melting points than fats [1]1
Example of oil: palm oil / corn oil / olive oil [1]1
Example of fat: butter / lard / ghee [1]1

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

[6 marksmarkah] (b)Asid oleik, C₁₇H₃₃COOH, ialah asid lemak tak tepu. (i) Tulis persamaan kimia bagi penghidrogenan asid oleik. (ii) Nyatakan mangkin dan suhu yang digunakan. (iii) Hitung jisim hidrogen yang diperlukan untuk menepukan 28.2 g asid oleik. [Jisim atom relatif: H = 1, C = 12, O = 16] (iv) Terangkan mengapa penghidrogenan digunakan untuk membuat marjerin.Oleic acid, C₁₇H₃₃COOH, is an unsaturated fatty acid. (i) Write the chemical equation for the hydrogenation of oleic acid. (ii) State the catalyst and the temperature used. (iii) Calculate the mass of hydrogen needed to saturate 28.2 g of oleic acid. [Relative atomic mass: H = 1, C = 12, O = 16] (iv) Explain why hydrogenation is used to make margarine.

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(i) C₁₇H₃₃COOH + H₂ → C₁₇H₃₅COOH [1]1
(ii) Nickel as the catalyst [1]1
(ii) About 180 °C [1]1
(iii) M of oleic acid = 282; n = 28.2 ÷ 282 = 0.10 mol [1]1
(iii) 1 mol acid needs 1 mol H₂; mass = 0.10 × 2 = 0.20 g [1]1
(iv) Hydrogenation changes liquid oil to a saturated, solid fat with a higher melting point, so margarine is solid and spreadable [1]1

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

[8 marksmarkah] (c)(i) Antara tiga minyak dalam Jadual 6, yang manakah paling mudah menjadi tengik? Beri sebab. (ii) Pada pendapat anda, minyak yang manakah paling sesuai untuk gerai itu? Wajarkan jawapan anda dengan dua sebab. (iii) Cadangkan dua cara untuk melambatkan minyak menjadi tengik. (iv) Mengapakah minyak yang sama tidak sesuai digunakan berulang kali?(i) Of the three oils in Table 6, which turns rancid most easily? Give a reason. (ii) In your opinion, which oil is most suitable for the stall? Justify your answer with two reasons. (iii) Suggest two ways to slow down the oil turning rancid. (iv) Why is it not suitable to use the same oil again and again?

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(i) Corn oil [1]1
(i) It has the highest percentage of unsaturated fatty acids, and the C=C bonds are easily oxidised by oxygen [1]1
(ii) Palm oil (or corn oil, if the owner changes the oil often) [1]1
(ii) Palm oil has a balanced composition, so it is more resistant to oxidation than corn oil at high frying temperature [1]1
(ii) It is not solid at room temperature and has less saturated fat than coconut oil, so it is healthier and easier to use than coconut oil [1]1
(iii) Add an antioxidant such as vitamin E or ascorbic acid [1]1
(iii) Store in a closed, dark and cool container [1]1
(iv) Repeated heating oxidises the oil and forms harmful substances (that may cause cancer), and the oil becomes rancid with a bad smell [1]1

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

For the teacherUntuk guru: EvaluatingMenilai, mediumsederhana difficultyaras mediumsederhana, ChapterBab 5, standard 5.1

TextbookBuku teks p.ms. 167

Bahagian DSection D

20 marksmarkah

Soalan ini mesti dijawab.This question must be answered.

Seorang murid mengkaji empat sampel plastik, P, Q, R dan S, yang sama saiz. Setiap jalur plastik diletakkan dalam ketuhar pada suhu 160 °C selama 10 minit, seperti dalam Rajah 2. Jalur dikeluarkan dan disejukkan, kemudian murid cuba membengkokkannya kepada bentuk baharu. Jadual 7 menunjukkan keputusan.

Sampel	Selepas 10 minit pada 160 °C	Boleh dibentuk semula selepas disejukkan?
P	Melembut	Ya
Q	Melembut	Ya
R	Kekal keras; permukaan menjadi hitam	Tidak
S	Melembut	Ya

Jadual 7
Sampel P ialah polistirena yang mempunyai jisim molekul relatif 104 000. [Jisim atom relatif: H = 1, C = 12]A pupil investigates four plastic samples, P, Q, R and S, of the same size. Each plastic strip is placed in an oven at 160 °C for 10 minutes, as in Diagram 2. The strip is taken out and cooled, then the pupil tries to bend it into a new shape. Table 7 shows the results.

Sample	After 10 minutes at 160 °C	Can be reshaped after cooling?
P	Softens	Yes
Q	Softens	Yes
R	Stays hard; the surface turns black	No
S	Softens	Yes

Table 7
Sample P is polystyrene, which has a relative molecular mass of 104 000. [Relative atomic mass: H = 1, C = 12]

An oven at 160 degrees Celsius containing four plastic strips labelled P, Q, R and S on a shelf, heated for 10 minutes — Rajah 2Diagram 2

[3 marksmarkah] (a)(i) Nyatakan pemboleh ubah dimanipulasikan dan pemboleh ubah bergerak balas dalam eksperimen ini. (ii) Nyatakan satu pemboleh ubah yang dimalarkan.(i) State the manipulated variable and the responding variable in this experiment. (ii) State one variable that is kept constant.

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(i) Manipulated: type of plastic [1]1
(i) Responding: whether the plastic softens / can be reshaped after heating [1]1
(ii) Temperature of the oven / heating time / size of the plastic strip [1]1

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

[3 marksmarkah] (b)Kelaskan R sebagai termoplastik atau termoset. Terangkan mengapa R berbeza daripada P dalam eksperimen ini berdasarkan struktur polimer.Classify R as a thermoplastic or a thermoset. Explain why R behaves differently from P in this experiment, in terms of the structure of the polymer.

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R is a thermoset [1]1
In R the polymer chains are joined by strong cross-links (covalent bonds), so the chains cannot move/slide, and R chars instead of softening [1]1
In P the polymer chains are separate and held by weak forces, so they slide past each other when heated and P softens [1]1

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

[5 marksmarkah] (c)(i) Tulis persamaan bagi pempolimeran stirena, C₆H₅CH=CH₂, untuk membentuk polistirena. (ii) Hitung bilangan unit monomer, n, dalam satu molekul polistirena P. (iii) Namakan jenis pempolimeran ini.(i) Write the equation for the polymerisation of styrene, C₆H₅CH=CH₂, to form polystyrene. (ii) Calculate the number of monomer units, n, in one molecule of polystyrene P. (iii) Name this type of polymerisation.

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(i) Correct formulae of monomer and repeat unit [1]1
(i) n C₆H₅CH=CH₂ → (-CH₂-CH(C₆H₅)-)ₙ [1]1
(ii) Mr of styrene, C₈H₈ = 8(12) + 8(1) = 104 [1]1
(ii) n = 104 000 ÷ 104 = 1 000 [1]1
(iii) Addition polymerisation [1]1

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

[5 marksmarkah] (d)Jadual 8 menunjukkan sifat tiga plastik.

Plastik	Jenis	Suhu melembut (°C)	Ketumpatan (g cm⁻³)	Boleh dikitar semula?
Polipropena	Termoplastik	160	0.90	Ya
Polistirena	Termoplastik	100	1.05	Ya
Melamina	Termoset	Tidak melembut	1.50	Tidak

Jadual 8 Sebuah kilang ingin membuat pemegang cerek elektrik. (i) Plastik yang manakah paling sesuai? (ii) Beri dua sebab. (iii) Berikan satu kelemahan menggunakan plastik itu dan cadangkan satu cara untuk mengurangkan kesan kelemahan itu.Table 8 shows the properties of three plastics.

Plastic	Type	Softening temperature (°C)	Density (g cm⁻³)	Can be recycled?
Polypropene	Thermoplastic	160	0.90	Yes
Polystyrene	Thermoplastic	100	1.05	Yes
Melamine	Thermoset	Does not soften	1.50	No

Table 8 A factory wants to make the handle of an electric kettle. (i) Which plastic is the most suitable? (ii) Give two reasons. (iii) State one disadvantage of using this plastic and suggest one way to reduce the effect of the disadvantage.

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(i) Melamine [1]1
(ii) It does not soften or melt when it is hot (kettle water at 100 °C and heating element near the handle) [1]1
(ii) Polypropene softens at 160 °C and polystyrene at 100 °C, which is too low; melamine is also hard and a good electrical insulator [1]1
(iii) It cannot be recycled, so waste builds up in landfills [1]1
(iii) Design the kettle to last long and repair it instead of throwing it away / reuse the product as long as possible [1]1

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

[4 marksmarkah] (e)Kantin sekolah menggunakan kotak polistirena sekali guna. Cadangkan satu bahan alternatif dan wajarkan cadangan anda. Nyatakan satu kesan alam sekitar yang timbul apabila sisa polistirena dibakar.A school canteen uses single-use polystyrene boxes. Suggest one alternative material and justify your suggestion. State one environmental effect that arises when polystyrene waste is burnt.

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Alternative: biodegradable plastic / paper box / reusable stainless steel container / banana leaf [1]1
Justification: it can be broken down by bacteria (biodegradable) or reused many times, so less waste is dumped [1]1
Polystyrene is not biodegradable, so it stays in landfills for a very long time [1]1
Burning produces carbon dioxide / carbon monoxide / soot / toxic gases that pollute the air and cause global warming [1]1

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

For the teacherUntuk guru: EvaluatingMenilai, hightinggi difficultyaras hightinggi, ChapterBab 4, standard 4.1

TextbookBuku teks p.ms. 147

Test specification (JSU)Jadual Spesifikasi Ujian (JSU)

ChapterBab	RememberingMengingat	UnderstandingMemahami	ApplyingMengaplikasi	AnalysingMenganalisis	EvaluatingMenilai	CreatingMencipta	Items	marksmarkah
1. Redox equilibrium	1 (1)	2 (2)	3 (11)	–	1 (20)	–	7	34
2. Carbon compound	1 (1)	–	1 (1)	2 (8)	–	–	4	10
3. Thermochemistry	1 (1)	–	3 (8)	1 (1)	–	–	5	10
4. Polymer	1 (1)	2 (2)	–	1 (8)	1 (20)	–	5	31
5. Consumer and industrial chemistry	2 (2)	3 (3)	–	1 (10)	1 (20)	–	7	35
TotalJumlah	6 (6)	7 (7)	7 (20)	5 (27)	3 (60)	0 (0)	28	120

Each cell shows items (marks).Setiap petak menunjukkan item (markah).

KimiaChemistry

Bahagian ASection A

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Bahagian BSection B

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Bahagian CSection C

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Bahagian DSection D

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