KimiaChemistry

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

1

QuestionSoalan 1

Pernyataan P hingga T menunjukkan langkah-langkah dalam kaedah saintifik.
P: Membuat hipotesis
Q: Mengumpul data
R: Mengenal pasti masalah
S: Mengenal pasti pemboleh ubah
T: Membuat kesimpulan
Antara yang berikut, yang manakah ialah susunan yang betul?Statements P to T show steps in the scientific method.
P: Making a hypothesis
Q: Collecting data
R: Identifying the problem
S: Identifying the variables
T: Making a conclusion
Which of the following is the correct order?

• AR, S, P, Q, T
• BR, P, S, Q, T
• CP, R, S, Q, T
• DR, P, S, T, Q

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B

The scientific method runs: identify the problem (R), make a hypothesis (P), identify the variables (S), plan and carry out the experiment to collect data (Q), then analyse the data and make a conclusion (T).

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 1, standard 1.2.2

TextbookBuku teks p.ms. 10

2

QuestionSoalan 2

Seorang teknisian makmal ingin mencairkan asid sulfurik pekat. Prosedur yang manakah betul dan selamat?A laboratory technician wants to dilute concentrated sulphuric acid. Which procedure is correct and safe?

• ATuang air dengan cepat ke dalam asid pekatPour water quickly into the concentrated acid
• BTuang asid pekat dengan cepat ke dalam air panasPour the concentrated acid quickly into hot water
• CTuang air perlahan-lahan ke dalam asid pekat sambil mengacauPour water slowly into the concentrated acid while stirring
• DTuang asid pekat perlahan-lahan ke dalam air sambil mengacauPour the concentrated acid slowly into water while stirring

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D

Diluting concentrated sulphuric acid releases a large amount of heat. The acid is always added slowly into the water with stirring so the heat is absorbed by the large volume of water and the mixture does not spit. Water added to concentrated acid can boil and splash.

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

TextbookBuku teks p.ms. 13

3

QuestionSoalan 3

Bahan yang manakah terdiri daripada molekul?Which substance is made up of molecules?

• AKarbon dioksidaCarbon dioxide
• BNatrium kloridaSodium chloride
• CArgon
• DKuprumCopper

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A

Carbon dioxide is made up of molecules (two non-metals joined by covalent bonds). Sodium chloride is made up of ions, argon is a monatomic gas made up of atoms and copper is a metal made up of atoms.

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

TextbookBuku teks p.ms. 25

4

QuestionSoalan 4

Unsur M mempunyai dua isotop, M-10 dengan kelimpahan semula jadi 20% dan M-11 dengan kelimpahan 80%. Berapakah jisim atom relatif unsur M?Element M has two isotopes, M-10 with natural abundance 20% and M-11 with abundance 80%. What is the relative atomic mass of element M?

• A10.2
• B10.5
• C10.8
• D11.0

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C

Relative atomic mass = (20 × 10 + 80 × 11) ÷ 100 = 1080 ÷ 100 = 10.8. It lies closer to 11 because M-11 is more abundant.

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

TextbookBuku teks p.ms. 38

5

QuestionSoalan 5

Berapakah jisim 0.25 mol kalsium karbonat, CaCO₃?What is the mass of 0.25 mol of calcium carbonate, CaCO₃?

• A25 g
• B100 g
• C12.5 g
• D4.0 g

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A

Molar mass of CaCO₃ = 40 + 12 + 3(16) = 100 g mol⁻¹. Mass = 0.25 × 100 = 25 g. 100 g is the mass of 1 mol.

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

TextbookBuku teks p.ms. 50

6

QuestionSoalan 6

Berapakah bilangan atom oksigen dalam 0.050 mol asid sulfurik, H₂SO₄?What is the number of oxygen atoms in 0.050 mol of sulphuric acid, H₂SO₄?

• A3.0 × 10²²
• B6.0 × 10²²
• C1.2 × 10²³
• D3.0 × 10²³

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C

Molecules = 0.050 × 6.02 × 10²³ = 3.0 × 10²². Each H₂SO₄ molecule has 4 oxygen atoms, so oxygen atoms = 4 × 3.0 × 10²² = 1.2 × 10²³. 3.0 × 10²² counts molecules, not oxygen atoms.

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

TextbookBuku teks p.ms. 51

7

QuestionSoalan 7

Sebanyak 4.8 g magnesium ditambah kepada 0.25 mol asid hidroklorik. Persamaan: Mg + 2HCl → MgCl₂ + H₂. Berapakah isi padu gas hidrogen yang terbebas pada suhu bilik?4.8 g of magnesium is added to 0.25 mol of hydrochloric acid. Equation: Mg + 2HCl → MgCl₂ + H₂. What volume of hydrogen gas is released at room conditions?

• A4.8 dm³
• B2.4 dm³
• C6.0 dm³
• D3.0 dm³

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D

n(Mg) = 4.8 ÷ 24 = 0.20 mol, which needs 0.40 mol HCl, but only 0.25 mol HCl is present, so HCl is the limiting reactant. n(H₂) = 0.25 ÷ 2 = 0.125 mol, volume = 0.125 × 24 = 3.0 dm³. 4.8 dm³ uses the moles of magnesium.

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

TextbookBuku teks p.ms. 73

8

QuestionSoalan 8

Unsur T mempunyai nombor proton 20. Pernyataan yang manakah tentang unsur T adalah betul?Element T has proton number 20. Which statement about element T is correct?

• AMembentuk ion T²⁻ dan oksidanya berasidIt forms T²⁻ ions and its oxide is acidic
• BMembentuk ion T²⁺ dan oksidanya bersifat besIt forms T²⁺ ions and its oxide is basic
• CTerletak dalam Kumpulan 4 dan Kala 2It is in Group 4 and Period 2
• DIalah unsur peralihan yang membentuk sebatian berwarnaIt is a transition element that forms coloured compounds

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B

T has electron arrangement 2.8.8.2: 2 valence electrons and 4 occupied shells, so it is in Group 2, Period 4. It donates 2 electrons to form T²⁺, and a metal oxide is basic. It is a main group metal, not a transition element.

For the teacherUntuk guru: UnderstandingMemahami, mediumsederhana difficultyaras mediumsederhana, ChapterBab 4, standard 4.2.2

TextbookBuku teks p.ms. 83

9

QuestionSoalan 9

Pasangan yang manakah menunjukkan mangkin dan proses yang betul?Which pair shows the correct catalyst and process?

• ABesi, proses SentuhIron, Contact process
• BVanadium(V) oksida, proses HaberVanadium(V) oxide, Haber process
• CBesi, proses HaberIron, Haber process
• DNikel, proses OstwaldNickel, Ostwald process

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C

Iron is the catalyst in the Haber process (ammonia). Vanadium(V) oxide is used in the Contact process (sulphuric acid), and the Ostwald process (nitric acid) uses platinum.

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

TextbookBuku teks p.ms. 102

10

QuestionSoalan 10

Atom unsur X mempunyai nombor proton 9. Apabila X bertindak balas dengan natrium, bagaimanakah atom X mencapai susunan elektron oktet yang stabil?An atom of element X has proton number 9. When X reacts with sodium, how does the atom of X achieve a stable octet electron arrangement?

• AMenerima satu elektron daripada atom natrium untuk membentuk ion X⁻It receives one electron from a sodium atom to form an X⁻ ion
• BMenderma tujuh elektron kepada atom natrium untuk membentuk ion X⁷⁺It donates seven electrons to a sodium atom to form an X⁷⁺ ion
• CMenderma satu elektron kepada atom natrium untuk membentuk ion X⁺It donates one electron to a sodium atom to form an X⁺ ion
• DMenerima dua elektron daripada atom natrium untuk membentuk ion X²⁻It receives two electrons from a sodium atom to form an X²⁻ ion

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A

X has electron arrangement 2.7, so it needs one more electron to reach the octet 2.8. Sodium is a metal that donates one electron, so X receives one electron and forms X⁻.

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

TextbookBuku teks p.ms. 113

11

QuestionSoalan 11

Kuprum pepejal mengalirkan arus elektrik tetapi natrium klorida pepejal tidak. Apakah sebabnya?Solid copper conducts electricity but solid sodium chloride does not. What is the reason?

• AKuprum mempunyai takat lebur yang lebih rendah daripada natrium kloridaCopper has a lower melting point than sodium chloride
• BNatrium klorida mengandungi molekul neutral yang tidak mempunyai casSodium chloride contains neutral molecules with no charge
• CAtom kuprum bergerak bebas dalam pepejal tetapi ion natrium klorida tidakCopper atoms move freely in the solid but sodium chloride ions do not
• DKuprum mempunyai elektron tak berpusat yang bergerak bebas, tetapi ion dalam natrium klorida pepejal berada pada kedudukan tetapCopper has delocalised electrons that move freely, but the ions in solid sodium chloride are held in fixed positions

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D

In a metal the valence electrons are delocalised and carry charge through the solid. In solid sodium chloride the ions are held in fixed positions by strong electrostatic forces, so no charge can flow until it melts or dissolves. NaCl is an ionic compound, not made of molecules.

For the teacherUntuk guru: UnderstandingMemahami, mediumsederhana difficultyaras mediumsederhana, ChapterBab 5, standard 5.6.1

TextbookBuku teks p.ms. 123

12

QuestionSoalan 12

Spesies yang manakah mengandungi ikatan datif?Which species contains a dative bond?

• AMetana, CH₄Methane, CH₄
• BIon ammonium, NH₄⁺Ammonium ion, NH₄⁺
• CAmmonia, NH₃
• DKarbon dioksida, CO₂Carbon dioxide, CO₂

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B

In NH₄⁺ the nitrogen atom of NH₃ donates its lone pair to an H⁺ ion, forming a dative bond. CH₄, NH₃ and CO₂ contain only ordinary covalent bonds where each atom contributes one electron to each pair.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 5, standard 5.5.1

TextbookBuku teks p.ms. 121

13

QuestionSoalan 13

Asid etanoik, CH₃COOH, dititratkan dengan larutan natrium hidroksida. Metil jingga berubah warna pada julat pH 3.1 hingga 4.4 dan fenolftalein pada julat pH 8.3 hingga 10.0. Penunjuk yang manakah sesuai dan mengapa?Ethanoic acid, CH₃COOH, is titrated with sodium hydroxide solution. Methyl orange changes colour over the pH range 3.1 to 4.4 and phenolphthalein over the pH range 8.3 to 10.0. Which indicator is suitable and why?

• AFenolftalein, kerana pH pada takat akhir melebihi 7Phenolphthalein, because the pH at the end point is above 7
• BFenolftalein, kerana asid etanoik ialah asid kuatPhenolphthalein, because ethanoic acid is a strong acid
• CMetil jingga, kerana pH pada takat akhir tepat 7Methyl orange, because the pH at the end point is exactly 7
• DMetil jingga, kerana garam yang terbentuk bersifat asidMethyl orange, because the salt formed is acidic

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A

Ethanoic acid is a weak acid and sodium hydroxide a strong alkali, so the salt formed, sodium ethanoate, hydrolyses to give an alkaline solution (pH above 7) at the end point. Phenolphthalein changes colour in this range, while methyl orange would change too early.

For the teacherUntuk guru: EvaluatingMenilai, mediumsederhana difficultyaras mediumsederhana, ChapterBab 6, standard 6.7.2

TextbookBuku teks p.ms. 171

14

QuestionSoalan 14

Sebanyak 50 cm³ larutan natrium hidroksida 1.0 mol dm⁻³ dicairkan kepada kepekatan 0.25 mol dm⁻³. Berapakah isi padu air yang perlu ditambah?50 cm³ of 1.0 mol dm⁻³ sodium hydroxide solution is diluted to a concentration of 0.25 mol dm⁻³. What volume of water must be added?

• A50 cm³
• B100 cm³
• C200 cm³
• D150 cm³

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D

M₁V₁ = M₂V₂: 1.0 × 50 = 0.25 × V₂, so V₂ = 200 cm³ is the final volume. Water added = 200 - 50 = 150 cm³. 200 cm³ is the final volume, not the volume of water added.

For the teacherUntuk guru: ApplyingMengaplikasi, mediumsederhana difficultyaras mediumsederhana, ChapterBab 6, standard 6.6.3

TextbookBuku teks p.ms. 166

15

QuestionSoalan 15

Berapakah jisim natrium hidroksida, NaOH, yang diperlukan untuk meneutralkan sepenuhnya 25.0 cm³ asid hidroklorik 0.50 mol dm⁻³? [Jisim atom relatif: H = 1, O = 16, Na = 23]What mass of sodium hydroxide, NaOH, is needed to neutralise 25.0 cm³ of 0.50 mol dm⁻³ hydrochloric acid completely? [Relative atomic mass: H = 1, O = 16, Na = 23]

• A0.25 g
• B0.40 g
• C0.50 g
• D1.00 g

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C

n(HCl) = 0.50 × 25.0 ÷ 1000 = 0.0125 mol. HCl : NaOH = 1 : 1, so n(NaOH) = 0.0125 mol. Mass = 0.0125 × 40 = 0.50 g.

For the teacherUntuk guru: ApplyingMengaplikasi, hightinggi difficultyaras hightinggi, ChapterBab 6, standard 6.7.3

TextbookBuku teks p.ms. 172

16

QuestionSoalan 16

Berapakah nilai pH bagi asid sulfurik, H₂SO₄, 0.0050 mol dm⁻³? Anggap asid itu mengion lengkap.What is the pH value of 0.0050 mol dm⁻³ sulphuric acid, H₂SO₄? Assume the acid ionises completely.

• A2.3
• B2.0
• C2.7
• D3.0

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B

H₂SO₄ is diprotic: H₂SO₄ → 2H⁺ + SO₄²⁻, so [H⁺] = 2 × 0.0050 = 0.010 mol dm⁻³ and pH = -log 0.010 = 2.0. pH 2.3 treats the acid as monoprotic.

For the teacherUntuk guru: ApplyingMengaplikasi, hightinggi difficultyaras hightinggi, ChapterBab 6, standard 6.2.2

TextbookBuku teks p.ms. 145

17

QuestionSoalan 17

Perubahan manakah meningkatkan kadar tindak balas dengan merendahkan tenaga pengaktifan?Which change increases the rate of reaction by lowering the activation energy?

• AMenaikkan suhuRaising the temperature
• BMeningkatkan kepekatan bahan tindak balasIncreasing the concentration of the reactant
• CMengurangkan saiz bahan tindak balas pepejalDecreasing the size of the solid reactant
• DMenambah mangkinAdding a catalyst

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D

A catalyst provides an alternative pathway with a lower activation energy. Raising the temperature, increasing the concentration and decreasing the particle size increase the frequency of effective collisions but do not change the activation energy.

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

TextbookBuku teks p.ms. 248

18

QuestionSoalan 18

Rajah 1 menunjukkan graf isi padu gas hidrogen melawan masa bagi dua eksperimen. Dalam Eksperimen I, 50 cm³ asid hidroklorik 0.20 mol dm⁻³ bertindak balas dengan pita magnesium berlebihan pada 30 °C. Apakah yang digunakan dalam Eksperimen II? [Pita magnesium juga berlebihan dalam Eksperimen II]Diagram 1 shows graphs of the volume of hydrogen gas against time for two experiments. In Experiment I, 50 cm³ of 0.20 mol dm⁻³ hydrochloric acid reacts with excess magnesium ribbon at 30 °C. What was used in Experiment II? [Magnesium ribbon is also in excess in Experiment II]

• A25 cm³ asid hidroklorik 0.40 mol dm⁻³ pada 30 °C25 cm³ of 0.40 mol dm⁻³ hydrochloric acid at 30 °C
• B100 cm³ asid hidroklorik 0.10 mol dm⁻³ pada 30 °C100 cm³ of 0.10 mol dm⁻³ hydrochloric acid at 30 °C
• C50 cm³ asid hidroklorik 0.40 mol dm⁻³ pada 30 °C50 cm³ of 0.40 mol dm⁻³ hydrochloric acid at 30 °C
• D50 cm³ asid hidroklorik 0.20 mol dm⁻³ pada 20 °C50 cm³ of 0.20 mol dm⁻³ hydrochloric acid at 20 °C

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A

Both curves end at 120 cm³ (n(HCl) = 0.010 mol gives 0.0050 mol H₂ = 120 cm³), so the same number of moles of acid reacted in Experiment II, but it reacted faster. 25 cm³ of 0.40 mol dm⁻³ is 0.010 mol with a higher concentration. B has the same moles but a lower concentration (slower), C doubles the moles (240 cm³) and D is at a lower temperature (slower).

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

TextbookBuku teks p.ms. 230

19

QuestionSoalan 19

Apakah yang ditambah kepada kaca semasa pembuatannya untuk menghasilkan kaca berwarna?What is added to glass during its manufacture to produce coloured glass?

• ABoron oksidaBoron oxide
• BOksida logam peralihanOxides of transition metals
• CPlumbum(II) oksida berlebihanExcess lead(II) oxide
• DNatrium karbonat berlebihanExcess sodium carbonate

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B

Oxides of transition metals give glass its colour, for example cobalt(II) oxide gives blue glass and chromium(III) oxide gives green glass. Boron oxide makes borosilicate glass heat resistant and lead(II) oxide makes lead crystal glass.

For the teacherUntuk guru: RememberingMengingat, lowrendah difficultyaras lowrendah, ChapterBab 8, standard 8.2.1

TextbookBuku teks p.ms. 261

20

QuestionSoalan 20

Seramik tidak sesuai digunakan untuk membuat tali kabel jambatan penggantung. Apakah sebabnya?Ceramics are not suitable for making the cables of a suspension bridge. What is the reason?

• ASeramik mengalirkan haba dengan baikCeramics conduct heat well
• BSeramik bertindak balas dengan air hujanCeramics react with rain water
• CSeramik rapuh dan mudah patah apabila diregangkan atau dihentamCeramics are brittle and break easily when stretched or struck
• DSeramik mempunyai takat lebur yang rendahCeramics have a low melting point

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C

Ceramics are hard but brittle, so they fracture easily under tension or impact. They are heat insulators, chemically inert and have very high melting points.

For the teacherUntuk guru: UnderstandingMemahami, mediumsederhana difficultyaras mediumsederhana, ChapterBab 8, standard 8.3.1

TextbookBuku teks p.ms. 263

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QuestionSoalan 26

Seorang murid ingin mengkaji sama ada jisim natrium hidrogenkarbonat mempengaruhi isi padu gas karbon dioksida yang terbebas apabila natrium hidrogenkarbonat bertindak balas dengan 50 cm³ asid hidroklorik 0.50 mol dm⁻³ berlebihan. Persamaan: NaHCO₃ + HCl → NaCl + H₂O + CO₂. Murid itu menggunakan 0.10 g, 0.20 g, 0.30 g, 0.40 g dan 0.50 g natrium hidrogenkarbonat. Gas dikumpulkan dalam silinder penyukat 250 cm³ yang diterbalikkan dalam air. Keputusan eksperimen ditunjukkan dalam Jadual 3.

Jisim NaHCO₃ / Mass of NaHCO₃ (g)	0.10	0.20	0.30	0.40	0.50
Isi padu CO₂ / Volume of CO₂ (cm³)	29	57	86	114	143

Jadual 3A student wants to investigate whether the mass of sodium hydrogencarbonate affects the volume of carbon dioxide gas released when sodium hydrogencarbonate reacts with excess 50 cm³ of 0.50 mol dm⁻³ hydrochloric acid. Equation: NaHCO₃ + HCl → NaCl + H₂O + CO₂. The student uses 0.10 g, 0.20 g, 0.30 g, 0.40 g and 0.50 g of sodium hydrogencarbonate. The gas is collected in a 250 cm³ measuring cylinder inverted in water. The results of the experiment are shown in Table 3.

[2 marksmarkah] (a)Nyatakan pernyataan masalah bagi eksperimen ini dalam bentuk soalan.State the problem statement for this experiment in the form of a question.

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• Includes both variables: the mass of sodium hydrogencarbonate and the volume of carbon dioxide gas released [1]1
• In the form of a question, e.g. Does the mass of sodium hydrogencarbonate affect the volume of carbon dioxide gas released? [1]1

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

[3 marksmarkah] (b)Nyatakan pemboleh ubah dimanipulasikan, pemboleh ubah bergerak balas dan satu pemboleh ubah dimalarkan.State the manipulated variable, the responding variable and one fixed variable.

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• Manipulated variable: mass of sodium hydrogencarbonate [1]1
• Responding variable: volume of carbon dioxide gas released [1]1
• Fixed variable (any one): volume of hydrochloric acid / concentration of hydrochloric acid / temperature [1]1

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

[2 marksmarkah] (c)Tulis satu hipotesis bagi eksperimen ini.Write a hypothesis for this experiment.

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• Links the two variables [1]1
• States the direction: the greater the mass of sodium hydrogencarbonate, the greater the volume of carbon dioxide gas released [1]1

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

[2 marksmarkah] (d)Senaraikan bahan dan radas yang diperlukan, selain silinder penyukat 250 cm³.List the materials and apparatus needed, other than the 250 cm³ measuring cylinder.

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• Materials: sodium hydrogencarbonate, 0.50 mol dm⁻³ hydrochloric acid [1]1
• Apparatus (any three): conical flask, delivery tube with stopper, trough/basin of water, electronic balance, 50 cm³ measuring cylinder, spatula [1]1

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

[4 marksmarkah] (e)Huraikan prosedur eksperimen ini.Describe the procedure of this experiment.

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• Measure 50 cm³ of 0.50 mol dm⁻³ hydrochloric acid and pour it into a conical flask [1]1
• Weigh 0.10 g of sodium hydrogencarbonate, add it to the flask and close the stopper quickly with the delivery tube connected to the inverted measuring cylinder [1]1
• Record the volume of gas collected when the reaction has stopped [1]1
• Repeat with 0.20 g, 0.30 g, 0.40 g and 0.50 g of sodium hydrogencarbonate using a fresh 50 cm³ of acid each time [1]1

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

[2 marksmarkah] (f)Berdasarkan Jadual 3, nyatakan hubungan antara jisim natrium hidrogenkarbonat dengan isi padu gas karbon dioksida dan nyatakan sama ada hipotesis diterima.Based on Table 3, state the relationship between the mass of sodium hydrogencarbonate and the volume of carbon dioxide gas and state whether the hypothesis is accepted.

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• The volume of carbon dioxide gas increases as the mass of sodium hydrogencarbonate increases (directly proportional) [1]1
• The hypothesis is accepted [1]1

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

[2 marksmarkah] (g)Hitung isi padu karbon dioksida yang dijangka daripada 0.50 g natrium hidrogenkarbonat dan bandingkan dengan keputusan dalam Jadual 3. [Jisim molar NaHCO₃ = 84 g mol⁻¹]Calculate the expected volume of carbon dioxide from 0.50 g of sodium hydrogencarbonate and compare it with the result in Table 3. [Molar mass of NaHCO₃ = 84 g mol⁻¹]

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• n(NaHCO₃) = 0.50 ÷ 84 = 0.00595 mol, so n(CO₂) = 0.00595 mol; volume = 0.00595 × 24 000 = 143 cm³ [1]1
• The same as the experimental value of 143 cm³, so the acid was in excess [1]1

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

[2 marksmarkah] (h)Nyatakan dua langkah berjaga-jaga keselamatan semasa mengendalikan asid hidroklorik 0.50 mol dm⁻³.State two safety precautions when handling 0.50 mol dm⁻³ hydrochloric acid.

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• Wear safety goggles / gloves / laboratory coat (personal protective equipment) [1]1
• Handle with care to avoid spills; wash hands after the experiment; do not taste or inhale directly [1]1

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[1 markmarkah] (i)Sedikit asid hidroklorik tertumpah pada tangan murid itu. Nyatakan tindakan segera yang perlu diambil.A little hydrochloric acid spills on the student's hand. State the immediate action to be taken.

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• Wash the affected area with plenty of running water (then inform the teacher)1

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

For the teacherUntuk guru: CreatingMencipta, mediumsederhana difficultyaras mediumsederhana, ChapterBab 1, standard 1.2.2

TextbookBuku teks p.ms. 10

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QuestionSoalan 27

Dalam dua eksperimen, 10.0 cm³ asid sulfurik 0.20 mol dm⁻³ bertindak balas dengan zink berlebihan pada suhu bilik. Dalam Eksperimen II, beberapa titik larutan kuprum(II) sulfat ditambah sebagai mangkin. Isi padu gas hidrogen yang terkumpul ditunjukkan dalam Jadual 4.

Masa / Time (s)	0	30	60	90	120	150	180	210	240
Eksperimen I / Experiment I (cm³)	0	12	22	30	37	42	46	48	48
Eksperimen II / Experiment II (cm³)	0	20	33	41	46	48	48	48	48

Jadual 4
Rajah 2 menunjukkan gambar rajah profil tenaga bagi tindak balas ini. Lengkung P ialah tindak balas tanpa mangkin dan lengkung Q ialah tindak balas dengan mangkin.In two experiments, 10.0 cm³ of 0.20 mol dm⁻³ sulphuric acid reacts with excess zinc at room temperature. In Experiment II, a few drops of copper(II) sulphate solution are added as a catalyst. The volume of hydrogen gas collected is shown in Table 4. Diagram 2 shows the energy profile diagram of this reaction. Curve P is the reaction without a catalyst and curve Q is the reaction with a catalyst.

[2 marksmarkah] (a)Tulis persamaan kimia seimbang bagi tindak balas antara zink dengan asid sulfurik.Write the balanced chemical equation for the reaction between zinc and sulphuric acid.

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• Correct formulae [1]1
• Zn + H₂SO₄ → ZnSO₄ + H₂ (balanced) [1]1

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[3 marksmarkah] (b)Hitung isi padu maksimum gas hidrogen yang boleh dihasilkan pada suhu bilik. [Isi padu molar = 24 dm³ mol⁻¹]Calculate the maximum volume of hydrogen gas that can be produced at room conditions. [Molar volume = 24 dm³ mol⁻¹]

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• n(H₂SO₄) = 0.20 × 10.0 ÷ 1000 = 0.0020 mol [1]1
• n(H₂) = 0.0020 mol (ratio 1 : 1) [1]1
• Volume = 0.0020 × 24 000 = 48 cm³ [1]1

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[2 marksmarkah] (c)Hitung kadar tindak balas purata dalam 60 saat pertama bagi Eksperimen I dan Eksperimen II.Calculate the average rate of reaction in the first 60 seconds for Experiment I and Experiment II.

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• Experiment I: 22 ÷ 60 = 0.37 cm³ s⁻¹ [1]1
• Experiment II: 33 ÷ 60 = 0.55 cm³ s⁻¹ [1]1

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[2 marksmarkah] (d)Eksperimen manakah menggunakan mangkin? Terangkan jawapan anda berdasarkan Jadual 4.Which experiment used the catalyst? Explain your answer based on Table 4.

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• Experiment II [1]1
• The reaction is faster (higher volume at each time and reaches 48 cm³ sooner) but the final volume is the same, because a catalyst does not change the amount of product [1]1

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[2 marksmarkah] (e)Mengapakah kadar tindak balas dalam Eksperimen I berkurang apabila masa bertambah?Why does the rate of reaction in Experiment I decrease as time passes?

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• The concentration of sulphuric acid decreases as it is used up [1]1
• The frequency of effective collisions between H⁺ ions and zinc atoms decreases [1]1

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[3 marksmarkah] (f)Dengan menggunakan teori perlanggaran, terangkan bagaimana mangkin meningkatkan kadar tindak balas.Using collision theory, explain how the catalyst increases the rate of reaction.

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• The catalyst provides an alternative pathway with a lower activation energy [1]1
• More colliding particles have energy equal to or greater than the activation energy [1]1
• The frequency of effective collisions increases, so the rate of reaction increases [1]1

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[2 marksmarkah] (g)Berdasarkan Rajah 2, nyatakan tenaga pengaktifan bagi lengkung P dan lengkung Q.Based on Diagram 2, state the activation energy for curve P and curve Q.

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• Curve P: 290 - 200 = 90 kJ mol⁻¹ [1]1
• Curve Q: 250 - 200 = 50 kJ mol⁻¹ [1]1

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[2 marksmarkah] (h)Hitung perubahan haba tindak balas, ΔH, dan nyatakan sama ada tindak balas itu eksotermik atau endotermik.Calculate the heat change of the reaction, ΔH, and state whether the reaction is exothermic or endothermic.

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• ΔH = 40 - 200 = -160 kJ mol⁻¹ [1]1
• Exothermic, because the products have less energy than the reactants / ΔH is negative [1]1

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[1 markmarkah] (i)Adakah nilai ΔH bagi lengkung Q sama dengan, lebih besar atau lebih kecil daripada nilai ΔH bagi lengkung P?Is the value of ΔH for curve Q the same as, larger than or smaller than the value of ΔH for curve P?

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• The same, because a catalyst only lowers the activation energy and does not change the energy of reactants and products1

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[1 markmarkah] (j)Cadangkan satu cara lain, selain menambah mangkin, untuk menjadikan tindak balas dalam Eksperimen I lebih cepat dengan isi padu gas maksimum yang sama.Suggest one other way, besides adding a catalyst, to make the reaction in Experiment I faster with the same maximum volume of gas.

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• Any one: use zinc powder instead of granules / raise the temperature of the acid1

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

TextbookBuku teks p.ms. 248

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