Practice paper, not an official KPM paper

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 marks

Nama / Name

Kelas / Class

Tarikh / Date

Markah / Marks: / 100

Instructions

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).

Practice shows answers as you go. Exam times you and hides answers until you submit.

Bahagian ASection A

20 marks

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.

Seorang murid menguji hipotesis, "Semakin pekat cuka, semakin cepat kulit telur larut." Apakah pemboleh ubah bergerak balas dalam eksperimen ini?A pupil tests the hypothesis, "The more concentrated the vinegar, the faster an eggshell dissolves." What is the responding variable in this experiment?

Kepekatan cukaConcentration of the vinegar
Masa untuk kulit telur larutTime taken for the eggshell to dissolve
Jisim kulit telurMass of the eggshell
Isi padu cukaVolume of the vinegar

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The concentration is changed on purpose (manipulated variable). The time taken depends on it, so it is measured: the responding variable. Mass of eggshell and volume of vinegar are fixed variables.

For the teacher: Applying, low difficulty, Chapter 1, standard 1.2.1

Textbook p. 9

Ion Y³⁺ mempunyai 10 elektron dan nombor nukleon 27. Berapakah bilangan neutron dalam satu atom Y?An ion Y³⁺ has 10 electrons and nucleon number 27. How many neutrons are in one atom of Y?

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Y³⁺ has lost 3 electrons, so the atom had 10 + 3 = 13 electrons = 13 protons. Neutrons = nucleon number − proton number = 27 − 13 = 14.

For the teacher: Applying, medium difficulty, Chapter 2, standard 2.3.2

Textbook p. 33

Klorin semula jadi mengandungi 75% ³⁵Cl dan 25% ³⁷Cl. Berapakah jisim atom relatif klorin?Natural chlorine contains 75% ³⁵Cl and 25% ³⁷Cl. What is the relative atomic mass of chlorine?

35.5
36.0
36.5
37.0

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(75 × 35 + 25 × 37) ÷ 100 = 35.5. The simple average 36.0 ignores the abundances.

For the teacher: Applying, medium difficulty, Chapter 2, standard 2.4.1

Textbook p. 37

Berapakah jisim formula relatif ammonium sulfat, (NH₄)₂SO₄? [Jisim atom relatif: H = 1, N = 14, O = 16, S = 32]What is the relative formula mass of ammonium sulphate, (NH₄)₂SO₄? [Relative atomic mass: H = 1, N = 14, O = 16, S = 32]

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2 × (14 + 4 × 1) + 32 + 4 × 16 = 36 + 32 + 64 = 132.

For the teacher: Applying, low difficulty, Chapter 3, standard 3.1.2

Textbook p. 47

Berapakah bilangan atom oksigen dalam 0.10 mol asid sulfurik, H₂SO₄? [Pemalar Avogadro = 6.02 × 10²³ mol⁻¹]How many oxygen atoms are in 0.10 mol of sulphuric acid, H₂SO₄? [Avogadro constant = 6.02 × 10²³ mol⁻¹]

1.2 × 10²³
2.4 × 10²³
4.8 × 10²³
9.6 × 10²³

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Each molecule has 4 oxygen atoms: 0.10 × 4 × 6.02 × 10²³ = 2.4 × 10²³.

For the teacher: Applying, medium difficulty, Chapter 3, standard 3.2.2

Textbook p. 50

Berapakah jisim 2.4 dm³ gas ammonia, NH₃, pada keadaan bilik? [Jisim atom relatif: H = 1, N = 14; isi padu molar = 24 dm³ mol⁻¹]What is the mass of 2.4 dm³ of ammonia gas, NH₃, at room conditions? [Relative atomic mass: H = 1, N = 14; molar volume = 24 dm³ mol⁻¹]

0.10 g
0.17 g
1.7 g
17 g

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Moles = 2.4 ÷ 24 = 0.10 mol. Mass = 0.10 × 17 = 1.7 g. (0.10 g is the number of moles, not the mass.)

For the teacher: Applying, medium difficulty, Chapter 3, standard 3.2.4

Textbook p. 53

Aluminium terbakar dalam oksigen: 4Al + 3O₂ → 2Al₂O₃. Berapakah jisim aluminium oksida yang terbentuk apabila 5.4 g aluminium terbakar sepenuhnya? [Jisim atom relatif: O = 16, Al = 27]Aluminium burns in oxygen: 4Al + 3O₂ → 2Al₂O₃. What mass of aluminium oxide is formed when 5.4 g of aluminium burns completely? [Relative atomic mass: O = 16, Al = 27]

10.2 g
5.1 g
20.4 g
40.8 g

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n(Al) = 5.4 ÷ 27 = 0.20 mol. 4 mol Al gives 2 mol Al₂O₃, so n(Al₂O₃) = 0.10 mol. Mass = 0.10 × 102 = 10.2 g. Forgetting the ratio gives 20.4 g.

For the teacher: Applying, high difficulty, Chapter 3, standard 3.4.3

Textbook p. 73

Susunan elektron atom unsur Z ialah 2.8.8.1. Di manakah kedudukan Z dalam Jadual Berkala Unsur?The electron arrangement of an atom of element Z is 2.8.8.1. Where is Z in the Periodic Table of Elements?

Kumpulan 4, Kala 1Group 4, Period 1
Kumpulan 17, Kala 4Group 17, Period 4
Kumpulan 1, Kala 3Group 1, Period 3
Kumpulan 1, Kala 4Group 1, Period 4

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One valence electron gives Group 1. Four shells occupied gives Period 4.

For the teacher: Understanding, low difficulty, Chapter 4, standard 4.2.2

Textbook p. 83

Gas klorin dialirkan ke dalam larutan natrium hidroksida sejuk. Apakah hasilnya?Chlorine gas is passed into cold sodium hydroxide solution. What are the products?

Natrium klorida dan hidrogenSodium chloride and hydrogen
Natrium klorida, natrium klorat(I) dan airSodium chloride, sodium chlorate(I) and water
Natrium klorat(I) dan hidrogenSodium chlorate(I) and hydrogen
Natrium oksida dan hidrogen kloridaSodium oxide and hydrogen chloride

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Cl₂ + 2NaOH → NaCl + NaOCl + H₂O. No hydrogen gas is released in this reaction.

For the teacher: Understanding, medium difficulty, Chapter 4, standard 4.5.3

Textbook p. 95

Sebatian yang manakah ialah sebatian kovalen?Which compound is a covalent compound?

Magnesium oksida, MgOMagnesium oxide, MgO
Kalium bromida, KBrPotassium bromide, KBr
Tetraklorometana, CCl₄Tetrachloromethane, CCl₄
Natrium oksida, Na₂OSodium oxide, Na₂O

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CCl₄ is formed from two non-metals (carbon and chlorine) that share electrons. The others are formed from a metal and a non-metal, so they are ionic.

For the teacher: Remembering, low difficulty, Chapter 5, standard 5.7.1

Textbook p. 125

Natrium klorida melebur pada 801 °C manakala tetraklorometana melebur pada −23 °C. Mengapakah?Sodium chloride melts at 801 °C but tetrachloromethane melts at −23 °C. Why?

Daya tarikan elektrostatik yang kuat antara ion memerlukan banyak tenaga untuk diatasi, tetapi daya antara molekul lemah.The strong electrostatic forces between ions need a lot of energy to overcome, but the forces between molecules are weak.
Ikatan kovalen dalam natrium klorida lebih kuat daripada ikatan kovalen dalam tetraklorometana.The covalent bonds in sodium chloride are stronger than those in tetrachloromethane.
Tetraklorometana mengandungi ion yang bebas bergerak.Tetrachloromethane contains ions that are free to move.
Atom natrium klorida lebih besar daripada atom tetraklorometana.Atoms of sodium chloride are larger than atoms of tetrachloromethane.

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Melting an ionic compound means separating oppositely charged ions held by strong electrostatic forces. A covalent compound only needs the weak intermolecular forces overcome; its covalent bonds stay intact.

For the teacher: Understanding, medium difficulty, Chapter 5, standard 5.7.1

Textbook p. 126

Atom J mempunyai susunan elektron 2.8.3 dan atom L mempunyai susunan elektron 2.6. Apakah formula sebatian yang terbentuk apabila J bertindak balas dengan L?Atom J has the electron arrangement 2.8.3 and atom L has the electron arrangement 2.6. What is the formula of the compound formed when J reacts with L?

JL
J₂L
J₃L₂
J₂L₃

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J loses 3 electrons to form J³⁺; L gains 2 electrons to form L²⁻. Charges balance with 2 J³⁺ (+6) and 3 L²⁻ (−6): J₂L₃.

For the teacher: Applying, medium difficulty, Chapter 5, standard 5.2.1

Textbook p. 113

Berapakah nilai pH larutan barium hidroksida, Ba(OH)₂, 0.005 mol dm⁻³?What is the pH of 0.005 mol dm⁻³ barium hydroxide solution, Ba(OH)₂?

2.0
11.7
12.0
12.3

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Ba(OH)₂ is a strong diprotic alkali: [OH⁻] = 2 × 0.005 = 0.010 mol dm⁻³, pOH = 2, pH = 14 − 2 = 12.0. 11.7 comes from forgetting the factor of 2.

For the teacher: Applying, medium difficulty, Chapter 6, standard 6.2.2

Textbook p. 145

25.0 cm³ asid sulfurik 0.20 mol dm⁻³ dineutralkan oleh larutan natrium hidroksida 0.50 mol dm⁻³. Berapakah isi padu natrium hidroksida yang diperlukan?25.0 cm³ of 0.20 mol dm⁻³ sulphuric acid is neutralised by 0.50 mol dm⁻³ sodium hydroxide solution. What volume of sodium hydroxide solution is needed?

10.0 cm³
20.0 cm³
30.0 cm³
40.0 cm³

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H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. n(H₂SO₄) = 0.20 × 25.0 ÷ 1000 = 0.0050 mol; n(NaOH) = 0.010 mol; V = 0.010 ÷ 0.50 = 0.0200 dm³ = 20.0 cm³.

For the teacher: Applying, high difficulty, Chapter 6, standard 6.7.3

Textbook p. 172

Tindakan yang manakah akan mengurangkan kadar tindak balas antara zink dengan asid sulfurik?Which action will decrease the rate of the reaction between zinc and sulphuric acid?

Menggunakan serbuk zink menggantikan ketulan zinkUsing zinc powder instead of zinc granules
Memanaskan asid ituWarming the acid
Menambah larutan kuprum(II) sulfatAdding copper(II) sulphate solution
Menambah air suling kepada asid ituAdding distilled water to the acid

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Water dilutes the acid, so there are fewer H⁺ ions per unit volume and fewer effective collisions. Powder (larger surface area), warming (higher temperature) and copper(II) sulphate (a catalyst) all increase the rate.

For the teacher: Understanding, low difficulty, Chapter 7, standard 7.2.1

Textbook p. 232

Jadual berikut menunjukkan empat eksperimen yang menggunakan 5.0 g magnesium dan 50 cm³ asid hidroklorik. Pasangan eksperimen yang manakah menunjukkan kesan saiz bahan tindak balas ke atas kadar tindak balas?

Eksperimen	Bentuk magnesium	Kepekatan asid (mol dm⁻³)	Suhu (°C)	Masa (s)
I	Pita	0.5	30	120
II	Pita	0.5	40	60
III	Serbuk	0.5	40	20
IV	Serbuk	1.0	40	10

The table shows four experiments that use 5.0 g of magnesium and 50 cm³ of hydrochloric acid. Which pair of experiments shows the effect of the size of the reactant on the rate of reaction?

Experiment	Form of magnesium	Concentration of acid (mol dm⁻³)	Temperature (°C)	Time (s)
I	Ribbon	0.5	30	120
II	Ribbon	0.5	40	60
III	Powder	0.5	40	20
IV	Powder	1.0	40	10

II dan IIIII and III
I dan III and II
III dan IVIII and IV
I dan IVI and IV

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To test one factor only the other factors must be the same. II and III differ only in the form of magnesium (ribbon or powder). I and II differ in temperature; III and IV differ in concentration; I and IV differ in three factors.

For the teacher: Analysing, high difficulty, Chapter 7, standard 7.2.1

Textbook p. 230

Apakah komposisi keluli tahan karat?What is the composition of stainless steel?

Kuprum dan stanumCopper and tin
Kuprum dan zinkCopper and zinc
Ferum, kromium, nikel dan karbonIron, chromium, nickel and carbon
Aluminium, kuprum, magnesium dan manganAluminium, copper, magnesium and manganese

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Stainless steel is iron with carbon, chromium and nickel. Copper and tin make bronze, copper and zinc make brass, and the last option is duralumin.

For the teacher: Remembering, low difficulty, Chapter 8, standard 8.1.2

Textbook p. 257

Pasangan jenis kaca dan kegunaannya yang manakah betul?Which pair of glass type and its use is correct?

Kaca soda kapur: pinggan kalis ketuharSoda-lime glass: oven-proof dishes
Kaca terlakur: kanta teleskop dan gentian optikFused glass: telescope lenses and optical fibres
Kaca hablur plumbum: kelalang makmalLead crystal glass: laboratory flasks
Kaca borosilikat: kaca tingkapBorosilicate glass: window panes

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Fused glass (almost pure silica) is transparent to light and heat-resistant, so it is used for lenses and optical fibres. Oven dishes and lab flasks are borosilicate glass; windows are soda-lime glass; lead crystal is for decorative items and lenses.

For the teacher: Remembering, low difficulty, Chapter 8, standard 8.2.1

Textbook p. 261

Mengapakah seramik digunakan untuk melapik relau?Why is ceramic used to line a furnace?

Ia mengalirkan haba dengan baik supaya relau cepat panas.It conducts heat well so the furnace heats up quickly.
Ia lembut dan mudah dibentuk.It is soft and easy to shape.
Ia kenyal dan tidak pecah apabila dihentak.It is elastic and does not break when struck.
Ia mempunyai takat lebur yang sangat tinggi dan pengalir haba yang lemah.It has a very high melting point and is a poor conductor of heat.

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Ceramics withstand very high temperatures and are heat insulators. They are hard and brittle, not soft or elastic.

For the teacher: Understanding, low difficulty, Chapter 8, standard 8.3.1

Textbook p. 263

Jadual menunjukkan sifat empat bahan yang dipertimbangkan untuk membuat sebuah kuali. Bahan yang manakah paling sesuai?

Bahan	Takat lebur (°C)	Pengaliran haba	Kerapuhan	Ketoksikan
P	1450	Baik	Tidak rapuh	Tidak toksik
Q	330	Baik	Tidak rapuh	Toksik
R	1700	Lemah	Rapuh	Tidak toksik
S	1100	Baik	Rapuh	Tidak toksik

The table shows the properties of four materials considered for making a frying pan. Which material is the most suitable?

Material	Melting point (°C)	Heat conduction	Brittleness	Toxicity
P	1450	Good	Not brittle	Non-toxic
Q	330	Good	Not brittle	Toxic
R	1700	Poor	Brittle	Non-toxic
S	1100	Good	Brittle	Non-toxic

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A frying pan must have a high melting point, conduct heat well, not break when knocked and be safe for food. Only P meets all of these. Q melts too easily and is toxic, R conducts heat poorly and is brittle, and S is brittle.

For the teacher: Evaluating, high difficulty, Chapter 8, standard 8.4.3

Textbook p. 269

Bahagian BSection B

40 marks

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

Sebuah bot porselin ditimbang ketika kosong, bersama ferum oksida, dan bersama ferum yang tinggal selepas ferum oksida dipanaskan dalam aliran gas hidrogen sehingga jisim malar. Keputusan ditunjukkan dalam jadual. [Jisim atom relatif: O = 16, Fe = 56]

Penimbangan	Jisim (g)
Bot kosong	25.40
Bot + ferum oksida	30.04
Bot + ferum selepas pemanasan	28.76

A porcelain boat is weighed empty, with iron oxide, and with the iron left after the oxide is heated in a stream of hydrogen gas until the mass is constant. The results are in the table. [Relative atomic mass: O = 16, Fe = 56]

Weighing	Mass (g)
Empty boat	25.40
Boat + iron oxide	30.04
Boat + iron after heating	28.76

[1 mark] (a)Mengapakah pemanasan, penyejukan dan penimbangan diulang sehingga jisim malar diperoleh?Why are heating, cooling and weighing repeated until a constant mass is obtained?

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To make sure the reaction is complete / all the iron oxide has been reduced to iron1

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[2 marks] (b)Hitung jisim ferum dan jisim oksigen dalam sampel itu.Calculate the mass of iron and the mass of oxygen in the sample.

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Mass of iron = 28.76 − 25.40 = 3.36 g [1 mark]1
Mass of oxygen = 30.04 − 28.76 = 1.28 g [1 mark]1

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[3 marks] (c)Tentukan formula empirik ferum oksida itu. Tunjukkan kerja mengira anda.Determine the empirical formula of the iron oxide. Show your working.

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Fe₃O₄

Moles: Fe = 3.36 ÷ 56 = 0.060 mol; O = 1.28 ÷ 16 = 0.080 mol [1 mark]1
Simplest mole ratio Fe : O = 0.060 : 0.080 = 3 : 4 [1 mark]1
Empirical formula = Fe₃O₄ [1 mark]1

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[1 mark] (d)Aliran hidrogen diteruskan semasa ferum disejukkan. Mengapa?The stream of hydrogen is continued while the iron cools. Why?

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To prevent the hot iron from reacting with oxygen in the air again (re-oxidation), which would increase the mass1

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[1 mark] (e)Mengapakah kaedah ini tidak boleh digunakan untuk menentukan formula empirik magnesium oksida?Why can this method not be used to find the empirical formula of magnesium oxide?

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Magnesium is more reactive than hydrogen, so hydrogen cannot remove the oxygen from magnesium oxide1

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For the teacher: Applying, medium difficulty, Chapter 3, standard 3.3.4

Textbook p. 65

Jadual menunjukkan sifat tiga bahan W, X dan Y (W, X dan Y bukan simbol sebenar).

Bahan	Takat lebur (°C)	Takat didih (°C)	Mengalirkan elektrik ketika pepejal	Mengalirkan elektrik ketika leburan
W	801	1413	Tidak	Ya
X	−23	77	Tidak	Tidak
Y	1085	2562	Ya	Ya

The table shows the properties of three substances W, X and Y (W, X and Y are not the actual symbols).

Substance	Melting point (°C)	Boiling point (°C)	Conducts electricity as a solid	Conducts electricity when molten
W	801	1413	No	Yes
X	−23	77	No	No
Y	1085	2562	Yes	Yes

[3 marks] (a)Kenal pasti bahan yang merupakan (i) sebatian ion, (ii) sebatian kovalen, (iii) logam.Identify the substance that is (i) an ionic compound, (ii) a covalent compound, (iii) a metal.

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(i) W [1 mark]1
(ii) X [1 mark]1
(iii) Y [1 mark]1

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[2 marks] (b)Terangkan mengapa W tidak mengalirkan elektrik ketika pepejal tetapi mengalirkan elektrik ketika leburan.Explain why W does not conduct electricity as a solid but conducts electricity when molten.

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In the solid, the ions are held in fixed positions and cannot move [1 mark]1
When molten, the ions are free to move and carry the charge [1 mark]1

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[1 mark] (c)Terangkan mengapa takat lebur X rendah.Explain why the melting point of X is low.

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X has weak intermolecular forces (between its simple molecules), which need little energy to overcome1

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[2 marks] (d)Unsur J (susunan elektron 2.8.8.1) bertindak balas dengan unsur L (susunan elektron 2.8.7) untuk membentuk W. Huraikan pembentukan ikatan dalam W.Element J (electron arrangement 2.8.8.1) reacts with element L (electron arrangement 2.8.7) to form W. Describe the formation of the bond in W.

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J donates its one valence electron to form the J⁺ ion; L receives one electron to form the L⁻ ion (both achieve the stable octet) [1 mark]1
The oppositely charged ions attract each other by strong electrostatic forces, forming an ionic bond [1 mark]1

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For the teacher: Understanding, low difficulty, Chapter 5, standard 5.7.1

Textbook p. 127

Barium sulfat, BaSO₄, disediakan dengan menambah 50.0 cm³ larutan barium klorida, BaCl₂, 0.20 mol dm⁻³ kepada larutan natrium sulfat, Na₂SO₄, yang berlebihan. Satu mendakan putih terbentuk. [Jisim atom relatif: O = 16, S = 32, Ba = 137]Barium sulphate, BaSO₄, is prepared by adding 50.0 cm³ of 0.20 mol dm⁻³ barium chloride solution, BaCl₂, to excess sodium sulphate solution, Na₂SO₄. A white precipitate forms. [Relative atomic mass: O = 16, S = 32, Ba = 137]

[1 mark] (a)Namakan jenis tindak balas ini.Name the type of this reaction.

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Double decomposition (precipitation)1

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[2 marks] (b)Tulis persamaan ion bagi pembentukan mendakan itu, dengan simbol keadaan.Write the ionic equation for the formation of the precipitate, with state symbols.

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Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

Ba²⁺ + SO₄²⁻ → BaSO₄ [1 mark]1
State symbols: (aq), (aq) and (s) [1 mark]1

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[2 marks] (c)Huraikan bagaimana mendakan itu dipisahkan untuk mendapat serbuk barium sulfat yang kering dan tulen.Describe how the precipitate is separated to obtain dry, pure barium sulphate powder.

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Filter the mixture and rinse the residue with distilled water [1 mark]1
Dry the residue between sheets of filter paper (or in an oven) [1 mark]1

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[3 marks] (d)Hitung jisim maksimum barium sulfat yang boleh terbentuk.Calculate the maximum mass of barium sulphate that can be formed.

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2.33 g

n(BaCl₂) = 0.20 × 50.0 ÷ 1000 = 0.010 mol [1 mark]1
1 mol BaCl₂ gives 1 mol BaSO₄, so n(BaSO₄) = 0.010 mol [1 mark]1
Mass = 0.010 × 233 = 2.33 g [1 mark]1

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For the teacher: Applying, medium difficulty, Chapter 6, standard 6.9.4

Textbook p. 188

Serpihan marmar (kalsium karbonat) yang berlebihan ditambah kepada 25.0 cm³ asid hidroklorik 0.20 mol dm⁻³. Gas karbon dioksida yang terbebas dikumpulkan dalam picagari gas. Rajah 1 menunjukkan graf isi padu gas melawan masa. [Isi padu molar = 24 dm³ mol⁻¹ pada keadaan bilik]Excess marble chips (calcium carbonate) are added to 25.0 cm³ of 0.20 mol dm⁻³ hydrochloric acid. The carbon dioxide gas released is collected in a gas syringe. Diagram 1 shows the graph of the volume of gas against time. [Molar volume = 24 dm³ mol⁻¹ at room conditions]

Line graph of volume of carbon dioxide (cm3) against time (s): 0 at 0 s, 24 at 20, 40 at 40, 50 at 60, 56 at 80, 59 at 100, 60 at 120 and 60 at 140. The curve levels off at 60 cm3 from 120 s. — Rajah 1Diagram 1

[2 marks] (a)Hitung (i) kadar tindak balas purata dalam 20 s pertama dan (ii) kadar tindak balas purata keseluruhan.Calculate (i) the average rate of reaction in the first 20 s and (ii) the overall average rate of reaction.

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(i) 24 ÷ 20 = 1.2 cm³ s⁻¹ [1 mark]1
(ii) 60 ÷ 120 = 0.50 cm³ s⁻¹ [1 mark]1

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[2 marks] (b)Terangkan mengapa kadar tindak balas berkurang dengan masa.Explain why the rate of reaction decreases with time.

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The acid is used up, so its concentration decreases (fewer H⁺ ions per unit volume) [1 mark]1
The frequency of effective collisions between H⁺ ions and calcium carbonate decreases [1 mark]1

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[2 marks] (c)Tunjukkan dengan pengiraan bahawa isi padu maksimum karbon dioksida ialah 60 cm³. Persamaan: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂Show by calculation that the maximum volume of carbon dioxide is 60 cm³. Equation: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

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n(HCl) = 0.20 × 25.0 ÷ 1000 = 0.0050 mol; n(CO₂) = 0.0050 ÷ 2 = 0.0025 mol [1 mark]1
Volume = 0.0025 × 24 000 = 60 cm³ (or 0.0025 × 24 = 0.060 dm³) [1 mark]1

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[2 marks] (d)Eksperimen diulang dengan serbuk marmar yang berlebihan, dengan semua keadaan lain sama. Nyatakan dan terangkan kesan ke atas (i) kadar tindak balas dan (ii) isi padu maksimum gas.The experiment is repeated with excess marble powder, with all other conditions the same. State and explain the effect on (i) the rate of reaction and (ii) the maximum volume of gas.

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(i) Higher rate: the powder has a larger total surface area, so the frequency of collisions increases [1 mark]1
(ii) The maximum volume stays 60 cm³: the amount of acid (limiting reactant) is unchanged [1 mark]1

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For the teacher: Analysing, medium difficulty, Chapter 7, standard 7.1.5

Textbook p. 227

Cecair asetamida panas disejukkan perlahan-lahan dan suhunya dicatat setiap 2 minit. Keputusannya ditunjukkan dalam jadual.

Masa (min)	0	2	4	6	8	10	12	14
Suhu (°C)	96	90	84	82	82	82	76	70

Hot liquid acetamide is cooled slowly and its temperature is recorded every 2 minutes. The results are in the table.

Time (min)	0	2	4	6	8	10	12	14
Temperature (°C)	96	90	84	82	82	82	76	70

[1 mark] (a)Apakah takat beku asetamida?What is the freezing point of acetamide?

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82 °C1

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[1 mark] (b)Nyatakan keadaan jirim asetamida pada minit ke-4.State the physical state of acetamide at the 4th minute.

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Liquid (the temperature is above the freezing point of 82 °C)1

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[2 marks] (c)Terangkan mengapa suhu tidak berubah dari minit ke-6 hingga minit ke-10.Explain why the temperature does not change from the 6th to the 10th minute.

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The liquid is freezing: the particles attract one another and release heat energy as they form the solid [1 mark]1
This heat released balances the heat lost to the surroundings, so the temperature stays constant [1 mark]1

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[2 marks] (d)Huraikan susunan dan pergerakan zarah asetamida pada minit ke-12.Describe the arrangement and movement of the particles of acetamide at the 12th minute.

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Particles are packed closely together in an orderly arrangement (the substance is a solid) [1 mark]1
They vibrate about fixed positions (they cannot move from place to place) [1 mark]1

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[2 marks] (e)Oksigen mempunyai isotop ¹⁶O dan ¹⁸O. Nyatakan satu persamaan dan satu perbezaan antara atom ¹⁶O dengan atom ¹⁸O.Oxygen has the isotopes ¹⁶O and ¹⁸O. State one similarity and one difference between an atom of ¹⁶O and an atom of ¹⁸O.

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Similarity: same number of protons (8) and electrons / same chemical properties [1 mark]1
Difference: different number of neutrons (8 and 10) / different nucleon number / different mass [1 mark]1

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For the teacher: Understanding, low difficulty, Chapter 2, standard 2.1.2

Textbook p. 25

Bahagian CSection C

20 marks

Jawab mana-mana satu soalan. Tulis jawapan anda di ruang yang disediakan.Answer any one question. Write your answers in the spaces provided.

26Choose 1

Seorang murid ingin mengkaji sama ada suhu air mempengaruhi masa yang diambil oleh sebiji tablet berbuih (vitamin C) untuk larut sepenuhnya. Dia menggunakan 100 cm³ air pada setiap suhu dan satu tablet yang sama saiz setiap kali. Keputusannya ditunjukkan dalam Jadual 1.

Suhu air (°C)	20	30	40	50	60
Masa untuk larut (s)	110	78	54	38	26

A pupil wants to find out whether the temperature of water affects the time taken for an effervescent (vitamin C) tablet to dissolve completely. He uses 100 cm³ of water at each temperature and a tablet of the same size each time. His results are shown in Table 1.

Temperature of water (°C)	20	30	40	50	60
Time to dissolve (s)	110	78	54	38	26

[3 marks] (a)Nyatakan (i) pemboleh ubah dimanipulasikan, (ii) pemboleh ubah bergerak balas dan (iii) satu pemboleh ubah dimalarkan.State (i) the manipulated variable, (ii) the responding variable and (iii) one fixed variable.

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(i) Temperature of the water [1 mark]1
(ii) Time taken for the tablet to dissolve [1 mark]1
(iii) Volume of water / size (mass) of the tablet / stirring (any one) [1 mark]1

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[1 mark] (b)Nyatakan tujuan eksperimen ini.State the aim of this experiment.

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To investigate the effect of the temperature of water on the time taken for an effervescent tablet to dissolve1

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[1 mark] (c)Nyatakan satu hipotesis bagi eksperimen ini.State a hypothesis for this experiment.

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The higher the temperature of the water, the shorter the time taken for the tablet to dissolve1

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[2 marks] (d)Namakan dua alat yang diperlukan untuk mengukur pemboleh ubah dimanipulasikan dan pemboleh ubah bergerak balas.Name two pieces of apparatus needed to measure the manipulated variable and the responding variable.

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Thermometer [1 mark]1
Stopwatch [1 mark]1

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[4 marks] (e)Huraikan prosedur eksperimen ini secara ringkas.Describe briefly the procedure of this experiment.

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Measure 100 cm³ of water with a measuring cylinder, pour into a beaker and adjust it to 20 °C [1 mark]1
Drop one tablet into the water and start the stopwatch at the same time [1 mark]1
Stop the stopwatch when the tablet has dissolved completely and record the time [1 mark]1
Repeat with water at 30, 40, 50 and 60 °C, using a new tablet of the same size each time [1 mark]1

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[2 marks] (f)Bina satu jadual untuk merekod keputusan eksperimen ini dengan tajuk dan unit yang betul.Construct a table to record the results of this experiment with correct headings and units.

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Two columns/rows with headings: Temperature of water (°C) and Time taken (s) [1 mark]1
Five temperatures 20, 30, 40, 50, 60 listed with space for times [1 mark]1

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[2 marks] (g)Berdasarkan Jadual 1, nyatakan satu inferens dan satu kesimpulan.Based on Table 1, state one inference and one conclusion.

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Inference: a higher temperature makes the tablet dissolve faster (the reaction is faster at higher temperature) [1 mark]1
Conclusion: the higher the temperature of the water, the shorter the time taken; the hypothesis is accepted [1 mark]1

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[1 mark] (h)Ramalkan masa yang diambil oleh tablet itu untuk larut pada 45 °C.Predict the time taken for the tablet to dissolve at 45 °C.

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A value between 38 s and 54 s (e.g. 46 s)1

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[2 marks] (i)Nyatakan dua langkah berjaga-jaga yang perlu diambil semasa menggunakan air panas dalam eksperimen ini.State two safety precautions to be taken when using hot water in this experiment.

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Any 2 points

Wear safety goggles / heat-resistant gloves1
Hold the hot beaker with tongs / do not touch hot water or the hot plate with bare hands1
Use a stable stand; keep water away from electrical sockets1

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[2 marks] (j)Cadangkan dua cara untuk menjadikan keputusan eksperimen ini lebih boleh dipercayai.Suggest two ways to make the results of this experiment more reliable.

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Any 2 points

Repeat the experiment at each temperature at least three times and use the average time1
Use a magnetic or fixed stirring rate / use a water bath to keep the temperature constant1
Use more temperature values1

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For the teacher: Applying, low difficulty, Chapter 1, standard 1.2.2

Textbook p. 10

27Choose 1

Jadual 2 menunjukkan data bagi beberapa unsur Kumpulan 1 dan Kumpulan 17. Jadual 3 menunjukkan jejari atom unsur Kala 3.

Unsur Kumpulan 1	Takat lebur (°C)
Litium	186
Natrium	98
Kalium	64

Unsur Kumpulan 17	Takat lebur (°C)	Takat didih (°C)
Klorin	−101	−34
Bromin	−7	59
Iodin	114	184

Unsur Kala 3	Na	Mg	Al	Si	P	S	Cl
Jejari atom (nm)	0.186	0.160	0.143	0.118	0.110	0.104	0.100

Table 2 shows data for some Group 1 and Group 17 elements. Table 3 shows the atomic radius of the Period 3 elements.

Group 1 element	Melting point (°C)
Lithium	186
Sodium	98
Potassium	64

Group 17 element	Melting point (°C)	Boiling point (°C)
Chlorine	−101	−34
Bromine	−7	59
Iodine	114	184

Period 3 element	Na	Mg	Al	Si	P	S	Cl
Atomic radius (nm)	0.186	0.160	0.143	0.118	0.110	0.104	0.100

[2 marks] (a)Nyatakan corak perubahan takat lebur unsur Kumpulan 1 menuruni kumpulan, dan ramalkan takat lebur rubidium berbanding kalium.State the trend in the melting point of Group 1 elements down the group, and predict the melting point of rubidium compared with potassium.

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The melting point decreases down the group [1 mark]1
Rubidium has a lower melting point than 64 °C [1 mark]1

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[3 marks] (b)Kalium bertindak balas dengan air. Nyatakan dua pemerhatian dan tulis persamaan kimia yang seimbang bagi tindak balas itu.Potassium reacts with water. State two observations and write the balanced chemical equation for the reaction.

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Any two observations: moves very quickly on the water surface; burns with a lilac flame; hissing/effervescence; dissolves/disappears2
2K + 2H₂O → 2KOH + H₂ [1 mark]1

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[2 marks] (c)Terangkan mengapa kalium lebih reaktif daripada natrium.Explain why potassium is more reactive than sodium.

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Potassium atom is larger: its valence electron is further from the nucleus, so the attraction to the nucleus is weaker [1 mark]1
The electron is donated more easily to form K⁺ [1 mark]1

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[2 marks] (d)Nyatakan keadaan jirim bromin pada suhu bilik (25 °C). Ramalkan sama ada takat didih astatin lebih tinggi atau lebih rendah daripada 184 °C.State the physical state of bromine at room temperature (25 °C). Predict whether the boiling point of astatine is higher or lower than 184 °C.

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Liquid (−7 °C < 25 °C < 59 °C) [1 mark]1
Higher than 184 °C (boiling points increase down the group) [1 mark]1

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[3 marks] (e)Terangkan mengapa kereaktifan unsur Kumpulan 17 berkurang menuruni kumpulan.Explain why the reactivity of Group 17 elements decreases down the group.

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The atomic size increases down the group [1 mark]1
The attraction of the nucleus on the incoming electron becomes weaker [1 mark]1
It becomes harder to receive one electron to form the halide ion [1 mark]1

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[2 marks] (f)Nyatakan corak jejari atom merentasi Kala 3 dan terangkan jawapan anda.State the trend in atomic radius across Period 3 and explain your answer.

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The atomic radius decreases from sodium to chlorine [1 mark]1
Proton number increases, so the nuclear attraction on the valence electrons increases and pulls the shells closer (same number of shells) [1 mark]1

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[3 marks] (g)Natrium oksida, aluminium oksida dan sulfur dioksida ialah oksida unsur Kala 3. Klasifikasikan setiap oksida sebagai oksida asid, oksida bes atau oksida amfoterik.Sodium oxide, aluminium oxide and sulphur dioxide are oxides of Period 3 elements. Classify each oxide as an acidic, basic or amphoteric oxide.

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Sodium oxide: basic oxide [1 mark]1
Aluminium oxide: amphoteric oxide [1 mark]1
Sulphur dioxide: acidic oxide [1 mark]1

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[3 marks] (h)Argon, unsur Kumpulan 18, digunakan untuk mengisi mentol lampu berfilamen. Terangkan mengapa argon sesuai digunakan.Argon, a Group 18 element, is used to fill filament light bulbs. Explain why argon is suitable.

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Argon atom has a stable octet electron arrangement [1 mark]1
So it is chemically unreactive / inert and does not react with the hot tungsten filament [1 mark]1
This prevents the filament from burning (oxidising) in air, so the bulb lasts longer [1 mark]1

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For the teacher: Analysing, medium difficulty, Chapter 4, standard 4.4.3

Textbook p. 91

Bahagian DSection D

20 marks

Jawab soalan ini. Tulis jawapan anda di ruang yang disediakan.Answer this question. Write your answers in the spaces provided.

Sebuah kilang di Kuching mengeluarkan peralatan dapur dan bot nelayan kecil. Jurutera kilang itu perlu memilih bahan yang betul untuk setiap produk. Produk 1 ialah pinggan kalis ketuhar. Produk 2 ialah mangkuk hiasan yang berkilau. Produk 3 ialah pelapik relau pembakaran. Produk 4 ialah kipas bot daripada gangsa (kuprum 90%, stanum 10%). Produk 5 ialah badan bot daripada gentian kaca. Jadual 4 menunjukkan empat bahan yang dipertimbangkan untuk badan bot.

Bahan	Ketumpatan (g cm⁻³)	Kekuatan	Tindak balas dengan air laut	Kos
Keluli lembut	7.9	Kuat	Berkarat dengan cepat	Rendah
Gentian kaca	1.9	Kuat dan lentur	Tiada	Sederhana
Aluminium tulen	2.7	Lembut	Tiada	Sederhana
Seramik	3.9	Sangat keras tetapi rapuh	Tiada	Tinggi

A factory in Kuching makes kitchenware and small fishing boats. The factory engineer must choose the right material for each product. Product 1 is an oven-proof dish. Product 2 is a sparkling decorative bowl. Product 3 is the lining of a kiln. Product 4 is a boat propeller made of bronze (90% copper, 10% tin). Product 5 is a boat hull made of fibre glass. Table 4 shows four materials considered for the hull.

Material	Density (g cm⁻³)	Strength	Reaction with sea water	Cost
Mild steel	7.9	Strong	Rusts quickly	Low
Fibre glass	1.9	Strong and flexible	None	Medium
Pure aluminium	2.7	Soft	None	Medium
Ceramic	3.9	Very hard but brittle	None	High

[2 marks] (a)Produk 1 dibuat daripada kaca borosilikat. Nyatakan satu komponen kaca ini selain silika dan satu sifatnya yang menjadikannya sesuai.Product 1 is made of borosilicate glass. State one component of this glass other than silica and one property that makes it suitable.

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Boron oxide [1 mark]1
Low coefficient of expansion: does not crack when heated or cooled suddenly / heat-resistant [1 mark]1

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[3 marks] (b)Produk 2 dibuat daripada kaca hablur plumbum. Nyatakan dua sifat yang menjadikannya sesuai untuk mangkuk hiasan, dan satu sebab ia tidak sesuai untuk botol susu bayi.Product 2 is made of lead crystal glass. State two properties that make it suitable for a decorative bowl, and one reason it is not suitable for a baby's milk bottle.

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High refractive index: sparkles / looks brilliant [1 mark]1
Heavy and easy to cut or shape into designs / has a clear appearance [1 mark]1
It contains lead(II) oxide; lead is toxic and may dissolve into the milk [1 mark]1

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[3 marks] (c)Produk 3 dibuat daripada seramik. Nyatakan dua sifat seramik yang menjadikannya sesuai dan satu kelemahannya.Product 3 is made of ceramic. State two properties of ceramic that make it suitable and one weakness.

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Very high melting point / withstands high temperatures [1 mark]1
Poor conductor of heat (heat insulator) / hard / chemically inert [1 mark]1
Weakness: brittle (breaks or cracks easily when a force is applied) [1 mark]1

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[3 marks] (d)Terangkan mengapa gangsa dalam Produk 4 lebih keras daripada kuprum tulen, berdasarkan susunan atom.Explain, using the arrangement of atoms, why the bronze in Product 4 is harder than pure copper.

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In pure copper the atoms are of the same size in an orderly arrangement, so the layers slide over each other easily [1 mark]1
In bronze, tin atoms are of a different (larger) size and disrupt the orderly arrangement [1 mark]1
This prevents the layers of atoms from sliding easily when a force is applied, so bronze is harder [1 mark]1

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[4 marks] (e)Gentian kaca ialah bahan komposit. (i) Namakan matriks dan bahan penguat dalam gentian kaca. (ii) Terangkan mengapa gentian kaca lebih sesuai untuk badan bot berbanding kaca sahaja.Fibre glass is a composite material. (i) Name the matrix and the strengthening substance in fibre glass. (ii) Explain why fibre glass is more suitable for a boat hull than glass alone.

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(i) Matrix: plastic (polyester resin) [1 mark]1
Strengthening substance: glass fibres [1 mark]1
(ii) Glass alone is brittle and breaks easily; the plastic matrix holds the fibres and makes the hull flexible and light [1 mark]1
The glass fibres give strength, so the composite is stronger than plastic alone [1 mark]1

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[5 marks] (f)Berdasarkan Jadual 4, pilih bahan yang paling sesuai untuk badan bot. Berikan dua sebab bagi pilihan anda dan satu sebab masing-masing mengapa dua bahan lain ditolak.Based on Table 4, choose the most suitable material for the boat hull. Give two reasons for your choice and one reason each why two other materials are rejected.

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Choice: fibre glass [1 mark]1
Lowest density (1.9 g cm⁻³): the hull is light, so the boat moves easily and can carry more load [1 mark]1
Strong and flexible / does not react with sea water, so it will not corrode [1 mark]1
Mild steel is rejected: it rusts quickly in sea water and is very heavy [1 mark]1
Ceramic is rejected because it is brittle and costly, or pure aluminium because it is soft [1 mark]1

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For the teacher: Evaluating, high difficulty, Chapter 8, standard 8.4.3

Textbook p. 269

Test specification (JSU)

Chapter	Remembering	Understanding	Applying	Analysing	Evaluating	Creating	Items	marks
1. Introduction to Chemistry	–	–	2 (21)	–	–	–	2	21
2. Matter and the Atomic Structure	–	1 (8)	2 (2)	–	–	–	3	10
3. The Mole Concept, Chemical Formula and Equation	–	–	5 (12)	–	–	–	5	12
4. The Periodic Table of Elements	–	2 (2)	–	1 (20)	–	–	3	22
5. Chemical Bond	1 (1)	2 (9)	1 (1)	–	–	–	4	11
6. Acid, Base and Salt	–	–	3 (10)	–	–	–	3	10
7. Rate of Reaction	–	1 (1)	–	2 (9)	–	–	3	10
8. Manufactured Substances in Industry	2 (2)	1 (1)	–	–	2 (21)	–	5	24
Total	3 (3)	7 (21)	13 (46)	3 (29)	2 (21)	0 (0)	28	120

Each cell shows items (marks).

KimiaChemistry

Bahagian ASection A

Question 1

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Bahagian BSection B

Question 21

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Bahagian CSection C

Question 26

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Bahagian DSection D

Question 28