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

Peperiksaan Akhir Tahun Tingkatan 4 (model)Form 4 End-of-Year Examination (model)

KimiaChemistry

Tingkatan 4, 2 jam 30 minit / 2 hours 30 minutes, 100 marksmarkah

Nama / NameNama

Kelas / ClassKelas

Tarikh / DateTarikh

Markah / marksmarkah: / 100

InstructionsArahan

Kertas soalan ini mengandungi empat bahagian: Bahagian A, B, C dan D.This question paper consists of four sections: Section A, B, C and D.
Bahagian A: pilih satu jawapan bagi setiap soalan.Section A: choose one answer for each question.
Bahagian B: jawab semua soalan.Section B: answer all questions.
Bahagian C: jawab mana-mana satu soalan. Bahagian D: jawab soalan yang diberi.Section C: answer any one question. Section D: answer the question given.
Tunjukkan langkah-langkah penting dalam kerja mengira anda. Tulis persamaan kimia yang seimbang.Show your working. Write balanced chemical equations.
Rajah tidak dilukis mengikut skala kecuali dinyatakan. Jadual Berkala Unsur dan kalkulator saintifik dibenarkan.Diagrams are not drawn to scale unless stated. The Periodic Table of Elements 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. [Jisim atom relatif: H = 1, C = 12, N = 14, O = 16, K = 39, Cl = 35.5; pemalar Avogadro = 6.02 × 10²³ mol⁻¹; isi padu molar = 24 dm³ mol⁻¹ pada suhu bilik]Answer all questions. Each question is followed by four options, A, B, C and D. Choose one answer only. [Relative atomic mass: H = 1, C = 12, N = 14, O = 16, K = 39, Cl = 35.5; Avogadro constant = 6.02 × 10²³ mol⁻¹; molar volume = 24 dm³ mol⁻¹ at room conditions]

Seorang murid memerhati bahawa paku besi berkarat lebih cepat dalam larutan garam yang lebih pekat. Pernyataan yang manakah ialah hipotesis yang sesuai?A student observes that iron nails rust faster in more concentrated salt solution. Which statement is a suitable hypothesis?

Paku besi berkarat dalam air garam.Iron nails rust in salty water.
Semakin tinggi kepekatan larutan garam, semakin cepat paku besi berkarat.The higher the concentration of the salt solution, the faster the iron nails rust.
Adakah kepekatan larutan garam mempengaruhi pengaratan?Does the concentration of the salt solution affect rusting?
Kepekatan larutan garam dan masa pengaratan.The concentration of the salt solution and the time for rusting.

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A hypothesis is a statement that links the manipulated variable (concentration of salt solution) to the responding variable (speed of rusting) and gives the direction. A is only an observation, C is a question (the problem statement) and D only lists variables.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 1, standard 1.2.1

TextbookBuku teks p.ms. 9

Pernyataan yang manakah menerangkan susunan zarah dalam cecair dengan betul?Which statement describes the arrangement of particles in a liquid correctly?

Zarah berjauhan dan bergerak bebas ke semua arahParticles are far apart and move freely in all directions
Zarah tersusun rapat dengan teratur dan hanya bergetar pada kedudukan tetapParticles are closely packed in an orderly pattern and only vibrate about fixed positions
Zarah rapat antara satu sama lain, tidak tersusun teratur dan boleh bergerak melintasi satu sama lainParticles are close together, not in an orderly pattern, and can move past one another
Zarah berjauhan dan hanya bergetarParticles are far apart and only vibrate

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In a liquid the particles are close but disordered and can slide past each other. A describes a gas, B a solid.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 2, standard 2.1.2

TextbookBuku teks p.ms. 25

Atom unsur X mempunyai 13 proton dan 14 neutron. Pilihan yang manakah memberikan susunan elektron dan nombor nukleon atom X dengan betul?An atom of element X has 13 protons and 14 neutrons. Which option gives the electron arrangement and the nucleon number of atom X correctly?

Susunan elektron 2.8.3; nombor nukleon 27Electron arrangement 2.8.3; nucleon number 27
Susunan elektron 2.8.3; nombor nukleon 13Electron arrangement 2.8.3; nucleon number 13
Susunan elektron 2.8.2.1; nombor nukleon 27Electron arrangement 2.8.2.1; nucleon number 27
Susunan elektron 2.8.3; nombor nukleon 14Electron arrangement 2.8.3; nucleon number 14

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An atom has equal protons and electrons, so 13 electrons: 2.8.3 (the third shell takes the remaining 3). Nucleon number = 13 + 14 = 27.

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

TextbookBuku teks p.ms. 33

Kuprum mempunyai dua isotop, Cu-63 dan Cu-65. Jisim atom relatif kuprum ialah 63.5. Dalam satu sampel yang mengandungi 400 atom kuprum, berapakah bilangan atom Cu-65?Copper has two isotopes, Cu-63 and Cu-65. The relative atomic mass of copper is 63.5. In a sample containing 400 copper atoms, how many atoms are Cu-65?

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Let the fraction of Cu-65 be w: 63(1 - w) + 65w = 63.5, so 63 + 2w = 63.5 and w = 0.25. Number of Cu-65 atoms = 0.25 × 400 = 100.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 2, standard 2.4.2

TextbookBuku teks p.ms. 38

Sampel yang manakah mengandungi bilangan molekul yang paling banyak? [Jisim atom relatif: H = 1, C = 12, N = 14, O = 16]Which sample contains the greatest number of molecules? [Relative atomic mass: H = 1, C = 12, N = 14, O = 16]

14 g nitrogen, N₂
16 g oksigen, O₂16 g oxygen, O₂
22 g karbon dioksida, CO₂22 g carbon dioxide, CO₂
2 g hidrogen, H₂2 g hydrogen, H₂

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Moles: N₂ 14 ÷ 28 = 0.5; O₂ 16 ÷ 32 = 0.5; CO₂ 22 ÷ 44 = 0.5; H₂ 2 ÷ 2 = 1.0. Equal moles contain equal numbers of molecules, so hydrogen has the most.

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

TextbookBuku teks p.ms. 50

Propana terbakar lengkap dalam oksigen berlebihan: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. Berapakah isi padu gas karbon dioksida yang terbentuk pada suhu bilik apabila 2.2 g propana terbakar lengkap? [Jisim atom relatif: H = 1, C = 12; isi padu molar = 24 dm³ mol⁻¹]Propane burns completely in excess oxygen: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. What is the volume of carbon dioxide gas formed at room conditions when 2.2 g of propane burns completely? [Relative atomic mass: H = 1, C = 12; molar volume = 24 dm³ mol⁻¹]

1.2 dm³
3.6 dm³
6.0 dm³
10.8 dm³

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Molar mass of C₃H₈ = 3(12) + 8(1) = 44 g mol⁻¹, so n = 2.2 ÷ 44 = 0.050 mol. From the equation 1 mol C₃H₈ gives 3 mol CO₂, so n(CO₂) = 0.150 mol and V = 0.150 × 24 = 3.6 dm³. 1.2 dm³ forgets the 1 : 3 mole ratio.

For the teacherUntuk guru: ApplyingMengaplikasi, hightinggi difficultyaras hightinggi, ChapterBab 3, standard 3.4.3

TextbookBuku teks p.ms. 73

Ciri yang manakah ditunjukkan oleh unsur peralihan?Which characteristic is shown by transition elements?

Berketumpatan rendah dan lembutLow density and soft
Ion-ionnya sentiasa tidak berwarnaTheir ions are always colourless
Takat lebur lebih rendah daripada unsur Kumpulan 1Melting points lower than those of Group 1 elements
Membentuk sebatian berwarnaForm coloured compounds

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Transition elements are hard, dense, have high melting points and form coloured compounds and ions. Group 1 elements are the soft, low-density ones with low melting points.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 4, standard 4.7.2

TextbookBuku teks p.ms. 102

Gas klorin dialirkan ke dalam larutan kalium bromida yang tidak berwarna. Apakah pemerhatian yang betul?Chlorine gas is passed into colourless potassium bromide solution. What is the correct observation?

Larutan tidak berwarna bertukar menjadi perangThe colourless solution turns brown
Larutan tidak berwarna bertukar menjadi unguThe colourless solution turns purple
Mendakan putih terbentukA white precipitate forms
Gas hijau kekuningan terbebasA yellowish-green gas is released

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Chlorine is more reactive than bromine, so it displaces bromine from the bromide: Cl₂ + 2KBr → 2KCl + Br₂. Bromine in water is brown. Purple is the colour of iodine in an organic solvent.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 4, standard 4.4.3

TextbookBuku teks p.ms. 91

Oksida unsur Kala 3 yang manakah bertindak balas dengan asid dan juga dengan alkali?The oxide of which Period 3 element reacts with both acids and alkalis?

NatriumSodium
Magnesium
Aluminium
SulfurSulphur

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Aluminium oxide is amphoteric. Sodium and magnesium oxides are basic and sulphur oxides are acidic.

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 4, standard 4.6.2

TextbookBuku teks p.ms. 99

Jadual 1 menunjukkan sifat empat bahan, P, Q, R dan S. Bahan yang manakah ialah sebatian ion?

Jadual 1 / Table 1	Takat lebur / Melting point (°C)	Kekonduksian pepejal / Solid conducts	Kekonduksian leburan / Molten conducts	Keterlarutan dalam air / Solubility in water
P	1083	Ya / Yes	Ya / Yes	Tidak larut / Insoluble
Q	801	Tidak / No	Ya / Yes	Larut / Soluble
R	−95	Tidak / No	Tidak / No	Tidak larut / Insoluble
S	114	Tidak / No	Tidak / No	Tidak larut / Insoluble

Table 1 shows the properties of four substances, P, Q, R and S. Which substance is an ionic compound?

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An ionic compound has a high melting point, does not conduct as a solid (ions fixed in the lattice), conducts when molten (free-moving ions) and usually dissolves in water. P is a metal (conducts as a solid). R and S do not conduct in any state, so they are covalent.

For the teacherUntuk guru: AnalysingMenganalisis, mediumsederhana difficultyaras mediumsederhana, ChapterBab 5, standard 5.7.1

TextbookBuku teks p.ms. 126

Atom kalium (2.8.8.1) bertindak balas dengan atom sulfur (2.8.6). Pilihan yang manakah menerangkan sebatian yang terbentuk dengan betul?A potassium atom (2.8.8.1) reacts with a sulphur atom (2.8.6). Which option describes the compound formed correctly?

KS; satu atom kalium memindahkan satu elektron kepada satu atom sulfurKS; one potassium atom transfers one electron to one sulphur atom
K₂S; dua atom kalium berkongsi dua elektron dengan satu atom sulfurK₂S; two potassium atoms share two electrons with one sulphur atom
KS₂; satu atom kalium memindahkan satu elektron kepada dua atom sulfurKS₂; one potassium atom transfers one electron to two sulphur atoms
K₂S; dua atom kalium masing-masing memindahkan satu elektron kepada satu atom sulfurK₂S; two potassium atoms each transfer one electron to one sulphur atom

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Each K atom has 1 valence electron to give away (K⁺) and S needs 2 electrons to reach 2.8.8 (S²⁻). Two K atoms are needed for one S atom, so the formula is K₂S, an ionic compound formed by electron transfer.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 5, standard 5.2.1

TextbookBuku teks p.ms. 113

Garam yang manakah tidak larut dalam air?Which salt is insoluble in water?

Barium sulfatBarium sulphate
Natrium karbonatSodium carbonate
Kalium nitratPotassium nitrate
Ammonium kloridaAmmonium chloride

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All nitrates, all potassium, sodium and ammonium salts are soluble. Barium sulphate is one of the insoluble sulphates (with lead(II) and calcium sulphates).

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 6, standard 6.7.3

TextbookBuku teks p.ms. 172

Seorang murid ingin menyediakan 500 cm³ asid hidroklorik 0.20 mol dm⁻³ daripada larutan stok 2.0 mol dm⁻³. Berapakah isi padu larutan stok yang diperlukan?A student wants to prepare 500 cm³ of 0.20 mol dm⁻³ hydrochloric acid from a 2.0 mol dm⁻³ stock solution. What volume of the stock solution is needed?

10 cm³
25 cm³
50 cm³
100 cm³

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M₁V₁ = M₂V₂: 2.0 × V₁ = 0.20 × 500, so V₁ = 50 cm³.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 6, standard 6.3.2

TextbookBuku teks p.ms. 150

Berapakah isi padu larutan natrium hidroksida 0.20 mol dm⁻³ yang diperlukan untuk meneutralkan 20.0 cm³ asid sulfurik 0.15 mol dm⁻³? [H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O]What volume of 0.20 mol dm⁻³ sodium hydroxide solution is needed to neutralise 20.0 cm³ of 0.15 mol dm⁻³ sulphuric acid? [H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O]

15.0 cm³
30.0 cm³
45.0 cm³
60.0 cm³

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n(H₂SO₄) = 0.15 × 20.0 ÷ 1000 = 0.0030 mol. From the equation n(NaOH) = 2 × 0.0030 = 0.0060 mol. V = 0.0060 ÷ 0.20 = 0.0300 dm³ = 30.0 cm³. 15.0 cm³ forgets the 1 : 2 mole ratio.

For the teacherUntuk guru: ApplyingMengaplikasi, hightinggi difficultyaras hightinggi, ChapterBab 6, standard 6.9.4

TextbookBuku teks p.ms. 188

Isi padu gas yang terbebas daripada suatu tindak balas diukur dalam cm³ setiap saat. Apakah unit kadar tindak balas itu?The volume of gas released in a reaction is measured in cm³ every second. What is the unit of the rate of reaction?

mol dm⁻³
cm³
g mol⁻¹
cm³ s⁻¹

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Rate = change in volume ÷ time = cm³ ÷ s = cm³ s⁻¹. mol dm⁻³ is the unit of concentration and g mol⁻¹ is molar mass.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 7, standard 7.1.2

TextbookBuku teks p.ms. 221

Zink bertindak balas dengan asid sulfurik. Isi padu asid dan jisim zink adalah sama dalam semua eksperimen. Set keadaan yang manakah memberikan kadar tindak balas yang paling rendah?Zinc reacts with sulphuric acid. The volume of acid and the mass of zinc are the same in all experiments. Which set of conditions gives the lowest rate of reaction?

Asid 0.5 mol dm⁻³, ketulan zink, 30 °C0.5 mol dm⁻³ acid, zinc granules, 30 °C
Asid 1.0 mol dm⁻³, ketulan zink, 30 °C1.0 mol dm⁻³ acid, zinc granules, 30 °C
Asid 0.5 mol dm⁻³, serbuk zink, 30 °C0.5 mol dm⁻³ acid, zinc powder, 30 °C
Asid 1.0 mol dm⁻³, ketulan zink, 40 °C1.0 mol dm⁻³ acid, zinc granules, 40 °C

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The lowest rate comes from the lowest concentration, the larger particles (granules) and the lower temperature. B has a higher concentration, C has a larger total surface area and D is hotter and more concentrated.

For the teacherUntuk guru: AnalysingMenganalisis, mediumsederhana difficultyaras mediumsederhana, ChapterBab 7, standard 7.2.1

TextbookBuku teks p.ms. 232

Jadual 2 menunjukkan masa yang diambil untuk tanda X di bawah kelalang kon hilang apabila natrium tiosulfat bertindak balas dengan asid hidroklorik pada tiga suhu. Berapa kalikah kadar tindak balas dalam eksperimen 3 berbanding eksperimen 1?

Jadual 2 / Table 2	1	2	3
Suhu / Temperature (°C)	30	40	50
Masa / Time (s)	48	24	12

Table 2 shows the time taken for the X mark under a conical flask to disappear when sodium thiosulphate reacts with hydrochloric acid at three temperatures. How many times greater is the rate of reaction in experiment 3 than in experiment 1?

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Rate is proportional to 1 ÷ time. Rate in experiment 3 ÷ rate in experiment 1 = 48 ÷ 12 = 4 times.

For the teacherUntuk guru: AnalysingMenganalisis, mediumsederhana difficultyaras mediumsederhana, ChapterBab 7, standard 7.2.1

TextbookBuku teks p.ms. 233

Pemangkin ditambah kepada suatu tindak balas. Pernyataan yang manakah menerangkan peningkatan kadar tindak balas dengan betul?A catalyst is added to a reaction. Which statement explains the increase in rate of reaction correctly?

Pemangkin meningkatkan tenaga kinetik zarah bahan tindak balasThe catalyst increases the kinetic energy of the reactant particles
Pemangkin meningkatkan bilangan zarah bahan tindak balas dalam setiap unit isi paduThe catalyst increases the number of reactant particles per unit volume
Pemangkin meningkatkan jumlah hasil yang terbentukThe catalyst increases the total amount of product formed
Pemangkin merendahkan tenaga pengaktifan, maka lebih banyak zarah yang berlanggar mempunyai tenaga yang mencukupi untuk perlanggaran berkesanThe catalyst lowers the activation energy, so more colliding particles have enough energy for an effective collision

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A catalyst gives an alternative pathway with lower activation energy. More colliding particles reach this energy, so the frequency of effective collisions rises. It does not change the temperature, the concentration or the amount of product.

For the teacherUntuk guru: UnderstandingMemahami, hightinggi difficultyaras hightinggi, ChapterBab 7, standard 7.4.3

TextbookBuku teks p.ms. 248

Aloi yang manakah dipadankan dengan betul dengan komposisi dan kegunaannya?Which alloy is matched correctly with its composition and use?

Gangsa: kuprum dan stanum; patung dan pingatBronze: copper and tin; statues and medals
Loyang: kuprum dan stanum; alat muzikBrass: copper and tin; musical instruments
Keluli: besi dan zink; jambatanSteel: iron and zinc; bridges
Duralumin: besi dan karbon; badan kapal terbangDuralumin: iron and carbon; aeroplane bodies

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Bronze is copper + tin (statues, medals). Brass is copper + zinc. Steel is iron + carbon. Duralumin is aluminium with small amounts of copper, magnesium and manganese.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 8, standard 8.1.3

TextbookBuku teks p.ms. 258

Suatu bahan keras, rapuh, tahan haba dan bahan kimia serta tidak mengalirkan elektrik. Apakah bahan itu dan kegunaannya yang sesuai?A material is hard, brittle, resists heat and chemicals and does not conduct electricity. What is the material and a suitable use?

Keluli tahan karat: sudu garpuStainless steel: cutlery
Seramik: lapisan relauCeramic: lining of a furnace
Gentian kaca: badan botFibre glass: boat hull
Gangsa: patungBronze: statues

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Ceramics are hard, brittle, chemically inert, heat resistant and electrical insulators, so they line furnaces. Steel and bronze are metals that conduct electricity and are not brittle. Fibre glass is a composite and is not brittle like ceramic.

For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 8, standard 8.3.1

TextbookBuku teks p.ms. 263

Bahagian BSection B

40 marksmarkah

Jawab semua soalan. Tulis jawapan anda di ruang yang disediakan.Answer all questions. Write your answers in the spaces provided.

Jadual 3 menunjukkan pH bagi empat larutan berkepekatan 0.010 mol dm⁻³. P ialah asid nitrik, Q ialah asid etanoik, R ialah larutan kalium hidroksida dan S ialah larutan ammonia.

Jadual 3 / Table 3	P	Q	R	S
pH	2.0	3.4	12.0	10.6

Table 3 shows the pH of four solutions of concentration 0.010 mol dm⁻³. P is nitric acid, Q is ethanoic acid, R is potassium hydroxide solution and S is ammonia solution.

[2 marksmarkah] (a)Mengapakah P ialah asid kuat manakala Q ialah asid lemah?Why is P a strong acid while Q is a weak acid?

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P ionises completely in water to produce a high concentration of hydrogen ions (pH 2.0) [1]1
Q ionises only partially in water, so the concentration of hydrogen ions is lower (pH 3.4) [1]1

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[1 markmarkah] (b)Tulis persamaan pengionan bagi asid nitrik dalam air.Write the equation for the ionisation of nitric acid in water.

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HNO₃ → H⁺ + NO₃⁻ (accept HNO₃ + H₂O → H₃O⁺ + NO₃⁻)1

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[2 marksmarkah] (c)Larutan manakah ialah alkali lemah? Beri satu sebab.Which solution is a weak alkali? Give one reason.

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S (ammonia solution) [1]1
It ionises partially in water, so its pH (10.6) is lower than that of the strong alkali R at the same concentration [1]1

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[2 marksmarkah] (d)Nyatakan jenis tindak balas antara P dengan R dan tulis persamaan ion bagi tindak balas itu.State the type of reaction between P and R and write the ionic equation for the reaction.

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Neutralisation [1]1
H⁺ + OH⁻ → H₂O [1]1

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[1 markmarkah] (e)Tulis persamaan kimia seimbang bagi tindak balas antara Q dengan R dan namakan garam yang terbentuk.Write the balanced chemical equation for the reaction between Q and R and name the salt formed.

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CH₃COOH + KOH → CH₃COOK + H₂O; potassium ethanoate [1 for the correct equation with the salt named]1

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

For the teacherUntuk guru: UnderstandingMemahami, lowrendah difficultyaras lowrendah, ChapterBab 6, standard 6.2.2

TextbookBuku teks p.ms. 145

Dua eksperimen, I dan II, dijalankan untuk mengkaji penguraian 50 cm³ larutan hidrogen peroksida 0.10 mol dm⁻³: 2H₂O₂ → 2H₂O + O₂. Dalam eksperimen II, 0.5 g serbuk mangan(IV) oksida ditambah. Isi padu oksigen yang terkumpul ditunjukkan dalam Rajah 1. [Isi padu molar = 24 dm³ mol⁻¹]Two experiments, I and II, are carried out to study the decomposition of 50 cm³ of 0.10 mol dm⁻³ hydrogen peroxide solution: 2H₂O₂ → 2H₂O + O₂. In experiment II, 0.5 g of manganese(IV) oxide powder is added. The volume of oxygen collected is shown in Diagram 1. [Molar volume = 24 dm³ mol⁻¹]

Volume of oxygen against time. Curve II (with catalyst): 30 cm3 at 10 s, 48 at 20 s, 57 at 30 s, 60 cm3 from 40 s. Curve I: 12 cm3 at 30 s, 22 at 60 s, 31 at 90 s, 38 at 120 s, 44 at 150 s, 49 at 180 s, 53 at 210 s, 56 at 240 s, 58 at 270 s, 60 cm3 at 300 s. — Rajah 1Diagram 1

[2 marksmarkah] (a)Hitung isi padu maksimum oksigen yang boleh terbentuk.Calculate the maximum volume of oxygen that can be formed.

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n(H₂O₂) = 0.10 × 50 ÷ 1000 = 0.0050 mol, so n(O₂) = 0.0025 mol [1]1
V = 0.0025 × 24 000 = 60 cm³ [1]1

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[1 markmarkah] (b)Hitung kadar tindak balas purata bagi eksperimen II dalam 20 s pertama.Calculate the average rate of reaction of experiment II in the first 20 s.

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48 ÷ 20 = 2.4 cm³ s⁻¹1

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[1 markmarkah] (c)Mengapakah isi padu maksimum oksigen sama dalam kedua-dua eksperimen?Why is the maximum volume of oxygen the same in both experiments?

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The same amount (same volume and concentration) of hydrogen peroxide is used, and a catalyst does not change the amount of product1

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[3 marksmarkah] (d)Terangkan, menggunakan teori perlanggaran, mengapa kadar tindak balas dalam eksperimen II lebih tinggi.Explain, using collision theory, why the rate of reaction in experiment II is higher.

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Manganese(IV) oxide is a catalyst that provides an alternative pathway with lower activation energy [1]1
More colliding particles have energy equal to or greater than the activation energy [1]1
Frequency of effective collisions increases, so the rate increases [1]1

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[1 markmarkah] (e)Nyatakan satu cara lain untuk menambahkan kadar tindak balas tanpa mengubah isi padu maksimum oksigen.State one other way to increase the rate of reaction without changing the maximum volume of oxygen.

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Any one: increase the temperature / use more concentrated H₂O₂ with the same amount of solute (e.g. 25 cm³ of 0.20 mol dm⁻³)1

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For the teacherUntuk guru: AnalysingMenganalisis, mediumsederhana difficultyaras mediumsederhana, ChapterBab 7, standard 7.4.3

TextbookBuku teks p.ms. 248

Unsur D dan E mempunyai nombor proton 7 dan 1 masing-masing. D bertindak balas dengan E untuk membentuk sebatian DE₃. Unsur G mempunyai nombor proton 11. Huruf-huruf itu bukan simbol sebenar unsur.Elements D and E have proton numbers 7 and 1 respectively. D reacts with E to form compound DE₃. Element G has proton number 11. The letters are not the actual symbols of the elements.

[2 marksmarkah] (a)Tulis susunan elektron bagi atom D dan atom E.Write the electron arrangement of atom D and atom E.

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D: 2.5 [1]1
E: 1 [1]1

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[2 marksmarkah] (b)Nyatakan jenis ikatan dalam DE₃ dan bilangan pasangan elektron yang dikongsi dalam satu molekul DE₃.State the type of bond in DE₃ and the number of electron pairs shared in one molecule of DE₃.

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Covalent bond [1]1
3 pairs of electrons [1]1

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[1 markmarkah] (c)DE₃ tidak mengalirkan elektrik dalam sebarang keadaan. Terangkan mengapa.DE₃ does not conduct electricity in any state. Explain why.

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It consists of neutral molecules, so there are no free-moving ions or electrons to carry charge1

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[2 marksmarkah] (d)Logam G mengalirkan elektrik dalam keadaan pepejal. Terangkan mengapa dengan merujuk kepada ikatan logam.Metal G conducts electricity in the solid state. Explain why, referring to the metallic bond.

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Metal atoms are held by a metallic bond, with valence electrons delocalised in a 'sea' of electrons around positive metal ions [1]1
The delocalised electrons move freely throughout the metal and carry charge [1]1

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[1 markmarkah] (e)Bandingkan takat lebur sebatian ion dengan takat lebur DE₃ dan beri satu sebab.Compare the melting point of an ionic compound with that of DE₃ and give one reason.

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Ionic compound has a higher melting point because strong electrostatic forces between oppositely charged ions need a lot of energy to overcome, whereas DE₃ has only weak forces between molecules [1]1

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 5, standard 5.7.1

TextbookBuku teks p.ms. 127

Seorang murid mencairkan asid sulfurik pekat di dalam makmal untuk menyediakan larutan yang akan digunakan dalam satu eksperimen.A student dilutes concentrated sulphuric acid in the laboratory to prepare a solution for an experiment.

[1 markmarkah] (a)Nyatakan bahaya asid sulfurik pekat.State the hazard of concentrated sulphuric acid.

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Corrosive1

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[2 marksmarkah] (b)Mengapakah asid pekat mesti dituang perlahan-lahan ke dalam air dan bukan sebaliknya?Why must the concentrated acid be poured slowly into water and not the other way round?

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Dilution releases a large amount of heat [1]1
Adding water to the acid can make the mixture boil and spurt out dangerously / pouring acid into water slowly lets the heat be absorbed by the large amount of water [1]1

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[1 markmarkah] (c)Seorang murid menumpahkan sedikit asid itu pada kulitnya. Apakah tindakan segera yang harus diambil?The student spills a little of the acid on his skin. What immediate action should be taken?

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Wash the skin with plenty of running water (and tell the teacher)1

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[1 markmarkah] (d)Namakan alat yang paling sesuai untuk menyukat 25.0 cm³ larutan dengan tepat.Name the most suitable apparatus to measure 25.0 cm³ of a solution accurately.

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Pipette1

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[2 marksmarkah] (e)Murid itu mengkaji sama ada kepekatan asid sulfurik mempengaruhi kadar tindak balasnya dengan pita magnesium. Nyatakan pemboleh ubah dimanipulasikan dan pemboleh ubah bergerak balas.The student investigates whether the concentration of sulphuric acid affects the rate of its reaction with magnesium ribbon. State the manipulated variable and the responding variable.

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Manipulated variable: concentration of sulphuric acid [1]1
Responding variable: rate of reaction (volume of hydrogen gas released per unit time / time for the magnesium to dissolve) [1]1

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[1 markmarkah] (f)Murid itu menggunakan pita magnesium yang sama panjangnya. Mengapakah ini penting?The student uses magnesium ribbon of the same length each time. Why is this important?

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It keeps the surface area (fixed variable) the same so only the concentration affects the rate / a fair test1

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For the teacherUntuk guru: ApplyingMengaplikasi, lowrendah difficultyaras lowrendah, ChapterBab 1, standard 1.3.1

TextbookBuku teks p.ms. 13

Jadual 4 menunjukkan ketumpatan unsur Kumpulan 18 pada suhu bilik. [Isi padu molar = 24 dm³ mol⁻¹; jisim atom relatif: He = 4, Ne = 20, Ar = 40, Kr = 84, Xe = 131, Rn = 222]

Unsur / Element	Helium	Neon	Argon	Kripton / Krypton	Xenon
Ketumpatan / Density (g dm⁻³)	0.17	0.84	1.66	3.48	5.46

Table 4 shows the density of the Group 18 elements at room conditions. [Molar volume = 24 dm³ mol⁻¹; relative atomic mass: He = 4, Ne = 20, Ar = 40, Kr = 84, Xe = 131, Rn = 222]

[1 markmarkah] (a)Nyatakan corak perubahan ketumpatan menuruni kumpulan.State the trend in density down the group.

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Density increases down the group1

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[2 marksmarkah] (b)Terangkan corak itu.Explain this trend.

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One mole of each gas occupies the same volume (24 dm³) at room conditions [1]1
The mass of one mole (relative atomic mass) increases down the group, so the mass per unit volume increases [1]1

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[2 marksmarkah] (c)Ramalkan ketumpatan radon pada suhu bilik.Predict the density of radon at room conditions.

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Density = relative atomic mass ÷ molar volume = 222 ÷ 24 [1]1
= 9.25 g dm⁻³ [1]1

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[1 markmarkah] (d)Mengapakah unsur Kumpulan 18 wujud sebagai atom tunggal?Why do Group 18 elements exist as single atoms?

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Their atoms have a stable duplet / octet electron arrangement, so they do not donate, accept or share electrons1

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[2 marksmarkah] (e)Helium digunakan untuk mengisi belon udara. Mengapakah helium lebih selamat daripada hidrogen walaupun hidrogen lebih ringan?Helium is used to fill airships and balloons. Why is helium safer than hydrogen even though hydrogen is lighter?

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Helium is unreactive / non-flammable [1]1
Hydrogen is flammable and can burn or explode [1]1

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For the teacherUntuk guru: AnalysingMenganalisis, hightinggi difficultyaras hightinggi, ChapterBab 4, standard 4.2.2

TextbookBuku teks p.ms. 83

Bahagian CSection C

20 marksmarkah

Jawab satu soalan sahaja daripada soalan 26 dan soalan 27. [Jisim atom relatif: H = 1, C = 12, O = 16, Na = 23; pemalar Avogadro = 6.02 × 10²³ mol⁻¹; isi padu molar = 24 dm³ mol⁻¹ pada suhu bilik]Answer one question only from question 26 and question 27. [Relative atomic mass: H = 1, C = 12, O = 16, Na = 23; Avogadro constant = 6.02 × 10²³ mol⁻¹; molar volume = 24 dm³ mol⁻¹ at room conditions]

27Choose 1Pilih 1

Sebuah majlis bandaraya ingin memilih bahan untuk membina kemudahan di sebuah taman tepi laut. Udara di situ mengandungi garam yang menyebabkan banyak logam berkarat. Jadual 5 menunjukkan lima bahan yang dipertimbangkan.

Jadual 5 / Table 5	Bahan / Material	Ketumpatan / Density (g cm⁻³)	Kekuatan / Strength	Udara masin / Salty air	Lut sinar / Transparent	Kos / Cost
P	Keluli lembut / Mild steel	7.8	Kuat / Strong	Berkarat / Rusts	Tidak / No	Rendah / Low
Q	Keluli tahan karat / Stainless steel	7.9	Sangat kuat / Very strong	Tidak berkarat / No rust	Tidak / No	Tinggi / High
R	Kaca soda kapur / Soda-lime glass	2.5	Keras tetapi rapuh / Hard but brittle	Tidak terjejas / Not affected	Ya / Yes	Sederhana / Medium
S	Gentian kaca berplastik / Glass fibre reinforced plastic	1.8	Kuat dan ringan, agak lentur / Strong and light, slightly flexible	Tidak terjejas / Not affected	Tidak / No	Sederhana / Medium
T	Seramik / Ceramic	2.4	Keras tetapi rapuh / Hard but brittle	Tidak terjejas / Not affected	Tidak / No	Rendah / Low

A city council wants to choose materials to build facilities in a seaside park. The air there contains salt that makes many metals rust. Table 5 shows five materials that are considered.

[3 marksmarkah] (a)Bahan Q ialah aloi besi, kromium dan nikel. Terangkan mengapa aloi lebih keras daripada logam tulen.Material Q is an alloy of iron, chromium and nickel. Explain why an alloy is harder than a pure metal.

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Atoms of other elements have different sizes from the iron atoms [1]1
They disrupt the orderly arrangement of the metal atoms [1]1
This reduces the ability of layers of atoms to slide over each other when a force is applied [1]1

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[3 marksmarkah] (b)Pagar tepi laut mesti sangat kuat, tidak berkarat dan kos bukan faktor penghad. Pilih bahan yang paling sesuai dan beri dua sebab.The seafront railing must be very strong and must not rust, and cost is not a limit. Choose the most suitable material and give two reasons.

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Q (stainless steel) [1]1
Very strong [1]1
Does not rust in salty air [1]1

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[2 marksmarkah] (c)Penutup lampu mesti membenarkan cahaya melaluinya dan tidak berkarat. Pilih bahan yang sesuai dan namakan satu bahan mentah utama untuk membuatnya.The lamp cover must let light pass through and must not rust. Choose a suitable material and name one main raw material used to make it.

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R (soda-lime glass) [1]1
Silica / sodium carbonate / calcium carbonate (limestone) [1]1

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[2 marksmarkah] (d)Namakan bahan matriks dan bahan penguat dalam bahan S.Name the matrix and the strengthening substance in material S.

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Matrix: plastic (polyester resin) [1]1
Strengthening substance: glass fibres [1]1

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[3 marksmarkah] (e)Nyatakan dua sifat seramik T yang menjadikannya sesuai untuk lantai berjubin dan satu kelemahannya.State two properties of ceramic T that make it suitable for floor tiles and one of its weaknesses.

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Hard / hard-wearing [1]1
Resistant to chemicals / heat / does not corrode [1]1
Brittle, so it breaks when hit or dropped [1]1

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[3 marksmarkah] (f)Bumbung perhentian bas mesti ringan, kuat dan tidak berkarat. Pilih S dan terangkan mengapa R tidak sesuai.The bus shelter roof must be light, strong and not rust. Choose S and explain why R is not suitable.

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S has a low density (1.8 g cm⁻³), so it is light, and it is strong and does not rust [1]1
R is brittle / breaks easily when struck [1]1
R is denser (heavier) than S and is not needed for a roof that need not be transparent [1]1

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[2 marksmarkah] (g)Keluli lembut P murah. Cadangkan satu cara melindunginya daripada berkarat di taman itu dan nyatakan satu kekurangan cara itu.Mild steel P is cheap. Suggest one way to protect it from rusting in the park and state one disadvantage of that method.

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Any one: paint / coat with plastic / galvanise (coat with zinc) / grease [1]1
Needs repainting or recoating regularly (the coating wears off or scratches) / extra cost and work [1]1

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[2 marksmarkah] (h)Mengapakah kaca soda kapur tidak sesuai untuk membuat peralatan makmal yang dipanaskan?Why is soda-lime glass not suitable for making laboratory apparatus that is heated?

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It expands and contracts a lot when heated or cooled [1]1
So it cracks easily; borosilicate glass is used instead [1]1

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For the teacherUntuk guru: EvaluatingMenilai, hightinggi difficultyaras hightinggi, ChapterBab 8, standard 8.1.1

TextbookBuku teks p.ms. 254

Bahagian DSection D

20 marksmarkah

Jawab soalan ini. Tulis jawapan anda di ruang yang disediakan. [Jisim atom relatif: C = 12]Answer this question. Write your answers in the spaces provided. [Relative atomic mass: C = 12]

Seorang murid mengkaji perubahan keadaan naftalena. Tabung didih yang mengandungi naftalena pepejal direndam dalam air panas dan suhu dicatat setiap minit semasa pemanasan seperti dalam Jadual 6. Naftalena cecair kemudian disejukkan dan lengkung penyejukan ditunjukkan dalam Rajah 2.

Jadual 6 / Table 6: Masa / Time (min)	0	1	2	3	4	5	6	7	8	9	10
Suhu / Temperature (°C)	60	66	72	77	80	80	80	80	84	89	94

A student studies the changes in state of naphthalene. A boiling tube containing solid naphthalene is placed in hot water and the temperature is recorded every minute during heating as in Table 6. The liquid naphthalene is then cooled and the cooling curve is shown in Diagram 2.

Cooling curve of naphthalene: temperature falls from 92 °C at 0 minutes to 80 °C at 3 minutes, stays at 80 °C until 8 minutes, then falls to 77 °C at 9, 73 at 10, 70 at 11 and 67 °C at 12 minutes — Rajah 2Diagram 2

[1 markmarkah] (a)Nyatakan takat lebur naftalena.State the melting point of naphthalene.

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80 °C1

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[2 marksmarkah] (b)Nyatakan keadaan fizikal naftalena pada minit ke-2 dan minit ke-5 semasa pemanasan.State the physical state of naphthalene at the 2nd minute and at the 5th minute during heating.

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2nd minute: solid [1]1
5th minute: solid and liquid (melting) [1]1

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[2 marksmarkah] (c)Terangkan mengapa suhu tidak berubah dari minit ke-4 hingga minit ke-7 walaupun pemanasan diteruskan.Explain why the temperature does not change from the 4th to the 7th minute although heating continues.

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Heat energy is absorbed by the particles and used to overcome the forces of attraction between them [1]1
It is not used to increase the kinetic energy of the particles, so the temperature stays constant [1]1

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[2 marksmarkah] (d)Huraikan susunan dan pergerakan zarah naftalena pada minit ke-9.Describe the arrangement and movement of the naphthalene particles at the 9th minute.

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Particles are close together but not in an orderly pattern [1]1
They move randomly / slide past one another [1]1

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[2 marksmarkah] (e)Berdasarkan Rajah 2, nyatakan takat beku naftalena dan terangkan mengapa nilainya sama dengan takat lebur.Based on Diagram 2, state the freezing point of naphthalene and explain why it equals the melting point.

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80 °C [1]1
A pure substance changes state at a fixed temperature, so the freezing point equals the melting point [1]1

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[2 marksmarkah] (f)Mengapakah tabung didih dipanaskan dalam kukus air dan bukan terus dengan nyalaan?Why is the boiling tube heated in a water bath and not directly with a flame?

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Naphthalene is flammable / its vapour can catch fire [1]1
A water bath heats it slowly and evenly, and 80 °C is below the boiling point of water [1]1

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[1 markmarkah] (g)Mengapakah naftalena dikacau secara berterusan semasa pemanasan?Why is the naphthalene stirred continuously during heating?

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To make sure it is heated evenly so the thermometer shows the true temperature1

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[2 marksmarkah] (h)Karbon mempunyai nombor proton 6 dan hidrogen mempunyai nombor proton 1. Tulis susunan elektron bagi atom karbon dan atom hidrogen dalam naftalena, C₁₀H₈.Carbon has proton number 6 and hydrogen has proton number 1. Write the electron arrangement of a carbon atom and a hydrogen atom in naphthalene, C₁₀H₈.

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Carbon: 2.4 [1]1
Hydrogen: 1 [1]1

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[3 marksmarkah] (i)Karbon semula jadi mengandungi 98.9% karbon-12 dan 1.1% karbon-13. Apakah maksud isotop, dan hitung jisim atom relatif karbon.Natural carbon contains 98.9% carbon-12 and 1.1% carbon-13. What is meant by isotopes, and calculate the relative atomic mass of carbon.

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Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons [1]1
(98.9 × 12 + 1.1 × 13) ÷ 100 [1]1
= 12.01 (12.011) [1]1

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[3 marksmarkah] (j)Naftalena yang tercemar dengan sedikit garam digunakan. Ramalkan kesannya terhadap takat lebur dan nyatakan satu kegunaan karbon-14.A sample of naphthalene contaminated with a little salt is used. Predict the effect on the melting point and state one use of carbon-14.

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The melting point is lower / melts over a range of temperature, and not sharply at 80 °C [1]1
An impurity weakens the regular arrangement/forces between the naphthalene particles [1]1
Carbon-14: dating of archaeological or fossil specimens [1]1

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For the teacherUntuk guru: AnalysingMenganalisis, mediumsederhana difficultyaras mediumsederhana, ChapterBab 2, standard 2.1.3

TextbookBuku teks p.ms. 28

Test specification (JSU)Jadual Spesifikasi Ujian (JSU)

ChapterBab	RememberingMengingat	UnderstandingMemahami	ApplyingMengaplikasi	AnalysingMenganalisis	EvaluatingMenilai	CreatingMencipta	Items	marksmarkah
1. Introduction to Chemistry	–	1 (1)	1 (8)	–	–	–	2	9
2. Matter and the Atomic Structure	1 (1)	–	2 (2)	1 (20)	–	–	4	23
3. The Mole Concept, Chemical Formula and Equation	–	–	3 (22)	–	–	–	3	22
4. The Periodic Table of Elements	1 (1)	1 (1)	1 (1)	1 (8)	–	–	4	11
5. Chemical Bond	–	–	2 (9)	1 (1)	–	–	3	10
6. Acid, Base and Salt	1 (1)	1 (8)	2 (2)	–	–	–	4	11
7. Rate of Reaction	1 (1)	1 (1)	–	3 (10)	–	–	5	12
8. Manufactured Substances in Industry	1 (1)	–	–	1 (1)	1 (20)	–	3	22
TotalJumlah	5 (5)	4 (11)	11 (44)	7 (40)	1 (20)	0 (0)	28	120

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

KimiaChemistry

Bahagian ASection A

QuestionSoalan 1

QuestionSoalan 2

QuestionSoalan 3

QuestionSoalan 4

QuestionSoalan 5

QuestionSoalan 6

QuestionSoalan 7

QuestionSoalan 8

QuestionSoalan 9

QuestionSoalan 10

QuestionSoalan 11

QuestionSoalan 12

QuestionSoalan 13

QuestionSoalan 14

QuestionSoalan 15

QuestionSoalan 16

QuestionSoalan 17

QuestionSoalan 18

QuestionSoalan 19

QuestionSoalan 20

Bahagian BSection B

QuestionSoalan 21

QuestionSoalan 22

QuestionSoalan 23

QuestionSoalan 24

QuestionSoalan 25

Bahagian CSection C

QuestionSoalan 26

QuestionSoalan 27

Bahagian DSection D

QuestionSoalan 28