Latihan Kimia F5 ( DLP )

Chemistry Form 5 Chapter 5 Consumer and Industrial Chemistry © Penerbitan Pelangi Sdn. Bhd. 96 (b) Diagram 5.2 shows a registered Malaysia’s wastewater and sanitation company. It is a government-owned company which develops and maintains a modern and efficient wastewater system in Peninsular Malaysia. Rajah 5.2 menunjukkan syarikat air sisa dan sanitasi Malaysia yang berdaftar. Syarikat milik kerajaan ini membangunkan dan mengekalkan sistem air sisa secara moden dan berkesan di Semenanjung Malaysia. Diagram 5.2 / Rajah 5.2 (i) What is wastewater? Apakah sisa air? [1 mark / 1 markah] (ii) State three main source of wastewater. Nyatakan tiga sumber utama sisa air. [3 marks / 3 markah] (iii) Suggest three benefits of efficient wastewater treatment. Cadangkan tiga kelebihan rawatan air sisa secara berkesan. [3 marks / 3 markah] (iv) State two impacts of improper wastewater treatment on human health. Nyatakan dua kesan rawatan air sisa yang tidak terurus terhadap kesihatan manusia. [2 marks / 2 markah] HOTS Challenge Diagram below shows a bottle of mineral supplement given to Darren who had a bone fracture. Explain how this supplement can help him to resolve the health problem faster and better. Rajah di bawah menunjukkan sebotol mineral tambahan yang diberikan kepada Darren yang mengalami patah tulang. Terangkan bagaimana makanan tambahan ini dapat membantu menyelesaikan masalah kesihatan Darren dengan lebih cepat dan baik. PAK-21 Quiz 5

97 © Penerbitan Pelangi Sdn. Bhd. SPM Model Paper Paper 1 / Kertas 1 [1 hour 15 minutes / 1 jam 15 minit] [40 marks / 40 markah] Answer all questions in this section. Jawab semua soalan dalam bahagian ini. 1. Which of the following is a natural polymer? Antara berikut, yang manakah polimer semula jadi? A Nylon / Nilon B Polystyrene / Polistirena C Starch / Kanji 2. Which of the following are the correct safety precautions when handling bromine in an experiment? Antara berikut, yang manakah langkah keselamatan yang betul apabila mengendalikan bromin di dalam eksperimen? I Wear goggles Pakai kaca mata keselamatan II Wear a mask Pakai topeng muka III Conduct the experiment on a workbench Menjalankan eksperimen di atas meja kerja IV Conduct the experiment on a table Menjalankan eksperimen di atas meja A I and II C II and IV I dan II II dan IV B I and III D III and IV I dan III IIII dan IV 3. Which of the following statement is true about molten silver iodide? Antara berikut, pernyataan yang manakah benar tentang leburan argentum iodida? A Molten silver iodide consists of silver atoms and iodine atoms. Leburan argentum iodida terdiri daripada atom argentum dan atom iodin. B Molten silver iodide consists of silver ions and iodide ions. Leburan argentum iodida terdiri daripada ion argentum dan ion iodida. C Molten silver iodide consists of silver atoms and iodine ions. Leburan argentum iodida terdiri daripada atom argentum dan ion iodin. D Molten silver iodide consists of silver iodide molecules that move freely. Leburan argentum iodida terdiri daripada molekul argentum iodida yang bergerak bebas. 4. Who discovered protons? Siapakah yang menemui proton? A J.J. Thomson B James Chadwick C Ernest Rutherford 5. The nucleon number of atom X is 23 and it has 12 neutrons in its nucleus. What is the electron arrangement of ion X? Nombor nukleon atom X ialah 23 dan mempunyai 12 neutron dalam nukleusnya. Apakah susunan elektron ion X? A 2.8 B 2.8.1 C 2.8.2 D 2.8.8.5 6. Which of the following is correctly matched the isotopes and its use? Antara berikut, yang manakah pasangan isotop dan kegunaannya yang betul? A Iodine-131: To treat thyroid cancer Iodin-131: Untuk merawat kanser tiroid B Gamma rays: To detect leaks in underground pipes Sinaran gamma: Untuk mengesan kebocoran paip di bawah tanah C Carbon-13: To estimate the age of artefacts Karbon-13: Untuk menganggar usia artifak D Sodium-24: To generate nuclear energy Natrium-24: Untuk menghasilkan tenaga nuklear 7. Which of the following device can be used to detect isotopes? Antara berikut, alat yang manakah boleh digunakan untuk mengesan isotop? A Geiger counter Kaunter Geiger B X-ray machine Mesin sinar-X C Nuclear reactor Reaktor nuklear

Chemistry Form 5 SPM Model Paper © Penerbitan Pelangi Sdn. Bhd. 98 11. A compound has the following percentage of composition, 49.48% carbon, 5.15% hydrogen, 28.87% nitrogen and the remaining is oxygen. Given that the relative molecular mass of the compound is 194. What is the molecular formula of the compound? [Relative atomic mass: H = 1, C = 12, N = 14, O = 16] Suatu sebatian mempunyai peratus komposisi berikut, 49.48% karbon, 5.15% hidrogen, 28.87% nitrogen dan selebihnya ialah oksigen. Diberi jisim molekul relatif sebatian tersebut ialah 194. Apakah formula molekul sebatian tersebut? [Jisim atom relatif: H = 1, C = 12, N = 14, O = 16] A C4 H5 N2 B C8 H10N4 C C4 H5 N2 O D C8 H10N4 O2 12. Atom M and atom N are two elements from the same group. Given that the proton number of atom M is 8 and atom N is located below atom M in the Periodic Table of Elements. What is the possible electron arrangement of ion N? Atom M dan atom N merupakan dua unsur dari kumpulan yang sama. Diberi nombor proton atom M ialah 8 dan atom N terletak di bawah atom M dalam Jadual Berkala Unsur. Apakah susunan elektron yang mungkin bagi ion N? A 2.6 C 2.8.6 B 2.8 D 2.8.8 13. Molecule P is usually used as a disinfectant. It also eliminates bacteria, molds and algae that commonly grow in water supply reservoirs. Which of the following is the correct drawing of electron arrangement for atom P? Molekul P biasanya digunakan sebagai pembasmi kuman. Molekul P juga menghilangkan bakteria, kulat dan alga yang biasanya tumbuh di takungan air. Antara berikut, yang manakah lukisan susunan elektron yang betul untuk atom P? A P B P C P – D P – 8. Which of the following chemical name and molar mass are correctly matched? [Relative atomic mass: K = 39, O = 16, H = 1, Mg = 24, Cl = 35.5, Ag = 108, N = 14, Pb = 207, I = 127] Antara berikut, yang manakah pasangan nama kimia dan jisim molar yang dipadankan dengan betul? [Jisim atom relatif: K = 39, O = 16, H = 1, Mg = 24, Cl = 35.5, Ag = 108, N = 14, Pb = 207, I = 127] Chemical name Nama kimia Molar mass (g mol–1) Jisim molar (g mol–1) A Potassium oxide Kalium oksida 55 B Magnesium chloride Magnesium klorida 59 C Silver nitrate Argentum nitrat 232 D Lead(II) iodide Plumbum(II) iodida 461 9. The relative molecular mass of substance X2 Y2 Z3 is 160. If the relative atomic mass of atom X is 24 and atom Y is 32, calculate the relative atomic mass of atom Z. Jisim molekul relatif suatu bahan X2 Y2 Z3 ialah 160. Jika jisim atom relatif atom X ialah 24 dan atom Y ialah 32, hitung jisim atom relatif atom Z. A 16 C 39 B 32 D 64 10. A pack of instant coffee contains 125 mg of caffeine, C8 H10N4 O2 . Calculate the number of caffeine molecules in the coffee. [Relative atomic mass: H=1, C=12, N=14, O=16; Avogadro constant = 6.02 × 1023 mol–1] Satu paket kopi segera mengandungi 125 mg kafein, C8 H10N4 O2 . Hitung bilangan molekul kafein di dalam kopi itu. [Jisim atom relatif: H = 1, C = 12, N = 14, O = 16; Pemalar Avogadro = 6.02 × 1023 mol–1] A 3.88 × 1020 B 3.88 × 1021 C 3.88 × 1022 D 3.88 × 1023

Chemistry Form 5 SPM Model Paper 99 © Penerbitan Pelangi Sdn. Bhd. 16. Amethyst is a purple germstone that consists of silicone dioxide and manganese. Which of the following statement best explains the colour formed in amethyst? Batu kecubung ialah batu permata berwarna ungu yang terdiri daripada silikon dioksida dan mangan. Antara berikut, pernyataan yang manakah paling tepat menerangkan tentang pembentukan warna pada batu kecubung? A Germstone normally contains colour. Batu permata biasanya mengandungi warna. B Presence of impurities causes the formation of colour. Kehadiran bendasing menghasilkan warna. C Manganese is a coloured metal. Mangan ialah logam berwarna. D Manganese is a transition metal that can form a coloured compound. Mangan ialah logam peralihan yang boleh membentuk sebatian berwarna. 17. Which of the following statement is true about the hydroxonium ion, H3 O+? Antara berikut, yang manakah benar tentang ion hidroksonium ion, H3 O+? A Hydroxonium ion is formed by a hydrogen bond between hydrogen gas and oxide ion. Ion hidroksonium terbentuk melalui ikatan hidrogen antara gas hidrogen dengan ion oksida. 14. Diagram 1 shows an apparatus set-up to investigate the reaction between chlorine gas and iron wool. Rajah 1 menunjukkan susunan radas untuk mengkaji tindak balas antara gas klorin dengan wul besi. Heat Panaskan Potassium manganate(VII) crystals Hablur kalium manganat(VII) Concentrated hydrochloric acid Asid hidroklorik pekat Iron wool Wul besi Combustion tube Tiub pembakaran Conical flask Kelalang kon Soda lime Soda kapur Retort stand Kaki retort Diagram 1 / Rajah 1 Which of the following statements are incorrect about the reaction? Antara berikut, pernyataan yang manakah tidak benar tentang tindak balas tersebut? I Hot iron wool burns very brightly. Wul besi panas terbakar dengan sangat terang. II A brown solid is formed. Pepejal perang terbentuk. III Soda lime is used to absorb moisture in the conical flask. Soda kapur digunakan untuk menyerap lembapan di dalam kelalang kon. IV The experiment is conducted in a fume chamber because chlorine gas is explosive. Eksperimen dijalankan di dalam kebuk wasap kerana gas klorin boleh meletup. A I and II I dan II B I and III I dan III C II and IV II dan IV D III and IV IIII dan IV 15. Chlorine gas is channelled into a pail of water. Then, a rusty nail is immersed into the pail and left for a day. The rusty nail turns shiny grey the next day. Explain this situation. Gas klorin disalurkan ke dalam sebuah baldi berisi air. Kemudian, paku yang berkarat dicelupkan ke dalam baldi tersebut dan dibiarkan selama sehari. Paku berkarat bertukar menjadi kelabu berkilat pada keesokan harinya. Terangkan keadaan ini. A Chlorine water can act as a cleaning agent. Air klorin boleh bertindak sebagai sebagai bahan pencuci. B Chlorine water is acidic. Air klorin bersifat asid. C Chlorine water has a bleaching effect. Air klorin mempunyai kesan pelunturan. D Chlorine water is a strong acid that dissolves iron(III) oxide on the nail. Air klorin ialah asid kuat yang melarutkan ferum(III) oksida pada paku.

Chemistry Form 5 SPM Model Paper © Penerbitan Pelangi Sdn. Bhd. 100 B Hydroxonium ion is formed by a metallic bond. Ion hidroksonium terbentuk melalui ikatan logam. C Hydroxonium ion is the formation of a covalent compound. Ion hidroksonium merupakan pembentukan sebatian kovalen. D Hydroxonium ion is formed by a dative bond between water molecule with hydrogen ion through sharing of lone pair of electrons. Ion hidroksonium terbentuk daripada ikatan datif antara molekul air dengan ion hidrogen melalui perkongsian sepasang elektron tunggal. 18. Tungsten filament is used in a light bulb because it conducts electricity very well. What causes the conductivity of electricity in tungsten? Filamen tungsten digunakan di dalam mentol kerana tungsten mengkonduksikan elektrik dengan baik. Apakah yang menyebabkan kekonduksian elektrik dalam tungsten? A Free-moving ions in tungsten Ion-ion yang bergerak bebas dalam tungsten B Free-moving electrons in tungsten Elektron-elektron yang bergerak bebas dalam tungsten C Protons and electrons in tungsten Proton dan elektron dalam tungsten D Vibrating atoms in tungsten Atom-atom yang bergetar dalam tungsten 19. Idell is requested to prepare a standard solution of 1 dm3 of 1 mol dm–3 copper(II) sulphate solution. He is given a bottle of copper(II) sulphate crystals, 1 dm3 of distilled water, a volumetric flask and other apparatus. Calculate the mass of copper(II) sulphate crystals needed to prepare the standard solution. [Relative atomic mass: O = 16, S = 32, Cu = 64] Idell diminta untuk menyediakan 1 dm3 larutan piawai kuprum(II) sulfat 1 mol dm–3. Dia dibekalkan dengan sebotol hablur kuprum(II) sulfat, 1 dm3 air suling, kelalang piawai dan radas yang lain. Hitung jisim hablur kuprum(II) sulfat yang diperlukan untuk menyediakan larutan piawai tersebut. [Jisim atom relatif: O = 16, S = 32, Cu = 64] A 20 g C 80 g B 40 g D 160 g 20. 500 cm3 of 0.2 mol dm–3 hydrochloric acid is added into excess magnesium powder to produce magnesium chloride and hydrogen gas. What is the maximum volume of hydrogen gas liberated at room condition? [Molar volume of gas = 24 dm3 mol–1 at room condition] 500 cm3 asid hidroklorik 0.2 mol dm–3 ditambahkan ke dalam serbuk magnesium yang berlebihan untuk membentuk magnesium klorida dan gas hidrogen. Berapakah isi padu maksimum gas hidrogen yang terhasil pada keadaan bilik? [Isi padu molar gas = 24 dm3 mol–1 pada keadaan bilik] A 2.4 dm3 C 1.2 dm3 B 24 dm3 D 12 dm3 21. Given that the concentration of sulphuric acid is 0.05 mol dm–3. Calculate its pH value. Diberi bahawa asid sulfurik mempunyai kepekatan 0.05 mol dm–3. Hitung nilai pHnya. A 0.1 B 1.0 C 0.13 D 1.3 22. Oxalic acid is found in spinach. Which of the following statement is true about oxalic acid? Asid oksalik boleh dijumpai di dalam sayur bayam. Antara berikut, pernyataan yang manakah benar mengenai asid oksalik? A pH value of oxalic acid is more than 8. Nilai pH asid oksalik lebih daripada 8. B Oxalic acid cannot react with magnesium and calcium. Asid oksalik tidak dapat bertindak balas dengan magnesium dan kalsium. C Oxalic acid turns moist red litmus paper to blue. Asid oksalik menukarkan kertas litmus merah lembap kepada biru. D Oxalic acid ionises partially in water to produce hydrogen ions. Asid oksalik mengion separa di dalam air membentuk ion hidrogen. 23. Which of the following is not the observation when pH indicator is added into potassium hydroxide solution? Antara berikut, yang manakah bukan pemerhatian apabila penunjuk pH dimasukkan ke dalam larutan kalium hidroksida?

Chemistry Form 5 SPM Model Paper 101 © Penerbitan Pelangi Sdn. Bhd. A Red litmus paper – blue Kertas litmus merah – biru B Methyl orange – orange Metil jingga – jingga C Universal indicator – purple Penunjuk universal – ungu D Phenolphthalein – pink Fenolftalein – merah jambu 24. Which of the following chemical substance and its colour is not correctly matched? Antara berikut, yang manakah bahan kimia dan warnanya tidak dipadankan dengan betul? A Silver nitrate – white precipitate Argentum nitrat – mendakan putih B Lead(II) iodide – yellow precipitate Plumbum(II) iodida – mendakan kuning C Barium sulphate – white precipitate Barium sulfat – mendakan putih D Lead(II) chromate – yellow precipitate Plumbum(II) kromat – mendakan kuning A Hydrogen gas is formed at the end of the experiment. Gas hidrogen terhasil pada akhir eksperimen. B Manganese(IV) oxide dissolves completely at the end of the experiment. Mangan(IV) oksida larut sepenuhnya pada akhir eksperimen. C Hydrogen peroxide turns purple at the end of experiment. Hidrogen peroksida menjadi warna ungu pada akhir eksperimen. D Time taken to obtain the maximum volume of oxygen gas is reduced. Masa yang diambil untuk mendapatkan isi padu maksimum gas oksigen berkurang. 27. Composite material Q is used in the camera lens as shown in Diagram 3. The properties of material Q are shown in Table 2. Bahan komposit Q digunakan pada kanta kamera seperti yang ditunjukkan dalam Rajah 3. Sifat-sifat bahan Q ditunjukkan dalam Jadual 2. Diagram 3 / Rajah 3 • Transparent Lut sinar • Absorbs UV rays Menyerap sinaran UV • The absorption of UV rays depends on light intensity Penyerapan sinar UV bergantung pada keamatan cahaya Table 2 / Jadual 2 25. Table 1 shows the steps in an anion verification. Jadual 1 menunjukkan langkah-langkah dalam pengesahan suatu anion. • Add 1 cm3 dilute HNO3 solution into a test tube with 1 cm3 solution Y. Tambahkan 1 cm3 larutan HNO3 cair ke dalam tabung uji berisi 1 cm3 larutan Y. • Add 1 cm3 dilute sulphric acid followed by 1 cm3 of iron(II) sulphate solution Tambahkan 1 cm3 asid sulfurik cair diikuti dengan 1 cm3 larutan ferum (II) sulfat. • Tilt the test tube slightly, add few drops of concentrated sulphuric acid slowly on the side of the test tube. Condongkan tabung uji sedikit, tambahkan beberapa titis asid sulfurik pekat secara perlahan-lahan pada sisi tabung uji. Table 1 / Jadual 1 What is the anion present in solution Y? Apakah anion yang hadir di dalam larutan Y? A SO4 2– C NO3 – B CO3 2– D Cl– 26. Diagram 2 shows the apparatus set-up of an experiment to determine the rate of reaction. Rajah 2 menunjukkan susunan radas bagi satu eksperimen untuk menentukan kadar tindak balas. Hydrogen peroxide solution Larutan hidrogen peroksida Manganese(IV) oxide Mangan(IV) oksida Water Air Burette Buret Diagram 2 / Rajah 2 Which of the following is true about this experiment? Antara berikut, yang manakah benar tentang eksperimen ini?

Chemistry Form 5 SPM Model Paper © Penerbitan Pelangi Sdn. Bhd. 102 What is Q? Apakah Q? A Fiber optic Gentian optik B Photochromic glass Kaca fotokromik C Fiber glass Gentian kaca 28. Which of the following is true about the balanced chemical equation below? Antara berikut, yang manakah benar mengenai persamaan kimia seimbang di bawah? CuO(s) + H2 (g) → Cu(s) + H2 O(l) CuO(p) + H2 (g) → Cu(p) + H2 O(ce) A Copper(II) oxide undergoes oxidation by releasing oxygen gas. Kuprum(II) oksida mengalami pengoksidaan dengan membebaskan gas oksigen. B Hydrogen gas undergoes reduction by gaining oxygen gas to form water. Gas hidrogen mengalami penurunan dengan menerima gas oksigen untuk membentuk air. C Black solid turns brown is observed. Pepejal hitam bertukar menjadi perang diperhatikan. D Copper(II) oxide is a reducing agent. Kuprum(II) oksida ialah agen penurunan. 29. Which of the following substance cannot convert iron(II) ion, Fe2+ to iron(III) ion, Fe3+? Antara berikut, bahan yang manakah tidak dapat menukarkan ion ferum(II), Fe2+ menjadi ion ferum(III), Fe3+? A Bromine water Air bromin B Potassium iodide Kalium iodida C Chlorine water Air klorin D Acidified potassium manganate(VII) solution Larutan kalium manganate(VII) berasid 30. Which of the following are not true about standard electrode potential (Eo )? Antara berikut, yang manakah tidak benar tentang keupayaan elektrod piawai (Eo )? I The more positive the Eo value, the weaker the oxidising power of an oxidising agent. Semakin positif nilai Eo , semakin lemah kuasa pengoksidaan suatu agen pengoksidaan. II The more positive the Eo value, the higher the ability of atom to receive electrons. Semakin positif nilai Eo , semakin tinggi keupayaan atom untuk menerima elektron. III The more negative the Eo value, the weaker the reducing power of a reducing agent. Semakin negatif nilai Eo , semakin lemah kuasa penurunan suatu agen penurunan. IV The more negative the Eo value, the higher the ability of atom to release electrons. Semakin negatif nilai Eo , semakin tinggi keupayaan atom untuk menderma elektron. A I and II I dan II B I and III I dan III C II and IV II dan IV D III and IV IIII dan IV 31. Diagram 4 shows the apparatus set-up of a voltaic cell. The needle in the voltmeter deflects. Rajah 4 menunjukkan susunan radas suatu sel kimia. Jarum di dalam voltmeter terpesong. Copper electrode Elektrod kuprum Copper(II) nitrate solution Larutan kuprum(II) sulfat Iron electrode Elektrod ferum Iron(II) nitrate solution Larutan ferum(II) nitrat V Diagram 4 / Rajah 4 Which of the following statements are true about Diagram 4? Antara berikut, pernyataan yang manakah benar mengenai Rajah 4? I Iron acts as the anode and turns thinner. Ferum bertindak sebagai anod dan menipis. II Blue copper(II) nitrate solution turns dark blue. Larutan biru kuprum(II) nitrat menjadi biru tua. III Electrons flow from iron to copper through the external circuit. Elektron mengalir dari ferum ke kuprum melalui litar luar. IV Green iron(II) nitrate solution turns light green. Larutan hijau ferum(II) nitrat menjadi hijau muda. A I and II I dan II B I and III I dan III C II and IV II dan IV D III and IV IIII dan IV

Chemistry Form 5 SPM Model Paper 103 © Penerbitan Pelangi Sdn. Bhd. 32. Diagram 5 shows the structural formulae of two hydrocarbons. Rajah 5 menunjukkan formula struktur bagi dua hidrokarbon. H H & & C"C!C!H & & & H H H H H H & & & H!C!C!C!H & & & H H H Diagram 5 / Rajah 5 What is the similarity between these two hydrocarbons? Apakah persamaan antara kedua-dua hidrokarbon ini? A Both react with bromine water. Kedua-duanya bertindak balas dengan air bromin. B Both are soluble in water. Kedua-duanya larut di dalam air. C Both cannot conduct electricity in any state. Kedua-duanya tidak mengkonduksikan elektrik dalam semua keadaan. D Both cannot oxidise to alcohol. Kedua-duanya tidak boleh teroksida kepada alkohol. 33. Compound P has the following properties. Sebatian P memiliki sifat-sifat berikut: • Reacts with aluminium strips to form colourless bubbles Bertindak balas dengan kepingan aluminium untuk membentuk gelembung tak berwarna • Turns a moist blue litmus paper to red Menukarkan kertas litmus biru lembap menjadi merah • Reacts with egg shells and form colourless bubbles Bertindak balas dengan kulit telur dan membentuk gelembung yang tak berwarna What is the homologous series of compound P? Apakah siri homolog bagi sebatian P? A Alkene Alkena B Carboxylic acid Asid karboksilik C Alcohol Alkohol 34. Diagram 6 shows the thermometer readings when excess magnesium powder is added into a polystyrene cup that contains 200 cm3 of 1.0 mol dm–3 copper(II) sulphate solution. Rajah 6 menunjukkan bacaan termometer apabila serbuk magnesium berlebihan ditambahkan ke dalam cawan polistirena yang mengandungi 100 cm3 larutan kuprum(II) sulfat 1.0 mol dm–3. Plastic cup Cawan plastik Thermometer Termometer Magnesium powder Serbuk magnesium Weighing bottle Botol penimbang Copper(III) sulphate solution Larutan kuprum(II) sulfat Lid Penutup Stir Kacau Plastic cup Cawan plastik Diagram 6 / Rajah 6 Given that the heat of displacement for this reaction is –226.8 kJ mol–1. Calculate the temperature change of this experiment. [Specific heat capacity of water = 4.2 J g–1 °C–1; Density of water = 1.0 g cm–3] Diberi bahawa haba penyesaran untuk tindak balas ini ialah –226.8 kJ mol–1. Hitung perubahan suhu eksperimen ini. [Muatan haba tentu air = 4.2 J g–1 °C–1; Ketumpatan air = 1.0 g cm–3] A 17 °C C 34 °C B 27 °C D 54 °C 35. Diagram 7 shows the apparatus set-up to investigate the heat of combustion for propanol. Rajah 7 menunjukkan susunan radas untuk mengkaji haba pembakaran bagi propanol. Spirit lamp Lampu spirit Windshield Penghadang angin Wooden block Bongkah kayu Propanol Propanol Water Air Thermometer Termometer Copper can Tin kuprum Tripod stand Tungku kaki tiga Diagram 7 / Rajah 7 Which of the following are used to increase the accuracy of heat released by propanol? Antara berikut, yang manakah digunakan untuk meningkatkan kejituan haba yang dibebaskan oleh propanol? I Stir the water and observe the maximum level of mercury achieved. Kacau air dan perhatikan tahap maksimum merkuri yang dicapai.

Chemistry Form 5 SPM Model Paper © Penerbitan Pelangi Sdn. Bhd. 104 II Add a windshield around the apparatus set-up. Tambahkan pengadang angin di sekeliling susunan radas. III Add more volume of propanol. Tambahkan isi padu propanol. IV Add more volume of water. Tambahkan isi padu air. A I and II I dan II B I and III I dan III C II and IV II dan IV D III and IV IIII dan IV 36. Given that the heat of combustion of ethanol is –1368 kJ mol–1. What is the fuel value of ethanol? [Relative atomic mass: H = 1, C = 12, O = 16] Diberi bahawa haba pembakaran etanol ialah –1368 kJ mol–1. Berapakah nilai bahan api etanol? [Jisim atom relatif: H = 1, C = 12, O = 1] A 29.74 kJ g–1 B 30.40 kJ g–1 C 45.60 kJ g–1 D 42.75 kJ g–1 37. Diagram 8 shows the structural formula for a part of polyvinyl chloride. Which of the following statement is true about polyvinyl chloride? Rajah 8 menunjukkan formula struktur bagi bahagian polivinil klorida. Antara berikut, pernyataan yang manakah benar mengenai polivinil klorida? H H H H H & & & & & !C!C!C!C!C! & & & & & Cl H Cl H Cl Diagram 8 / Rajah 8 38. Which of the following is the correct uses of thermoset? Antara berikut, yang manakah benar mengenai kegunaan termoset? I Plastic chair Kerusi plastik II Disc brake piston Omboh cakera brek III Compact disc (CD) and digital video disc (DVD) Cakera padat (CD) dan cakera video digital (DVD) IV Microwavable container Bekas gelombang mikro A I and II I dan II B I and III I dan III C II and IV II dan IV D III and IV IIII dan IV I Is an example of natural polymer Merupakan contoh polimer semula jadi II Environmentally friendly product Produk mesra alam III Produced through addition polymerisation Dihasilkan melalui pempolimeran penambahan IV Commonly applied in building, construction and piping Biasanya diaplikasikan dalam bangunan, pembinaan dan perpaipan A I and II I dan II B I and III I dan III C II and IV II dan IV D III and IV IIII dan IV 39. Which of the following is true about saturated fats and unsaturated fats? Antara berikut, yang manakah benar mengenai lemak tepu dan lemak tak tepu? Saturated fat Lemak tepu Unsaturated fat Lemak tak tepu A Usually found in fresh fish Biasanya dijumpai dalam ikan segar Usually found in burgers Biasanya dijumpai dalam burger B Can be found in cheesy pizzas Boleh didapati dalam pizza berkeju Can be found in a glass of fresh avocado juice Boleh didapati di dalam segelas jus avocado C Very helpful in protecting the heart Amat berguna untuk melindungi jantung Easier to cause high blood pressure Mudah menyebabkan tekanan darah tinggi D As a seasoning in salad Sebagai perasa dalam salad Use to deep-fry chicken Digunakan untuk menggoreng ayam

Chemistry Form 5 SPM Model Paper 105 © Penerbitan Pelangi Sdn. Bhd. 40. Diagram 9 shows the lady is applying cosmetic P on her lips. Rajah 9 menunjukkan seorang wanita memakai kosmetik P pada bibirnya. Cosmetic P Kosmetik P Diagram 9 / Rajah 9 C Cosmetic P must be removed thoroughly before going to bed. Kosmetik P mesti dibersihkan dengan sepenuhnya sebelum tidur. D Cosmetic P can be used by anyone without causing any side effects or allergies. Kosmetik P boleh digunakan oleh sesiapa sahaja tanpa menyebabkan sebarang kesan sampingan atau alahan. Which of the following is true about cosmetic P? Antara berikut, yang manakah benar mengenai kosmetik P? A Cosmetic P can be applied on the lips overnight. Kosmetik P boleh dipakai pada bibir semalaman. B Cosmetic P can be used even over 2 years. Kosmetik P boleh digunakan lagi walaupun sudah melebihi 2 tahun. Paper 2 / Kertas 2 [2 hours 30 minutes / 2 jam 30 minit] Section A / Bahagian A [60 marks / 60 markah] Answer all questions in this section. Jawab semua soalan dalam bahagian ini. 1. (a) Diagram 1.1 shows the arrangement of atoms in substance J and K. Rajah 1.1 menunjukkan susunan atom dalam bahan J dan K. Substance J J Substance K K Diagram 1.1 / Rajah 1.1 (i) Identify which of the substance in Diagram 1.1 is pure metal and pure alloy. Kenal pasti bahan yang manakah dalam Rajah 1.1 ialah logam tulen dan aloi tulen. Pure metal / Logam tulen: Alloy / Aloi: [2 marks / 2 markah]

Chemistry Form 5 SPM Model Paper © Penerbitan Pelangi Sdn. Bhd. 106 (ii) State the property of substance J based on the atomic arrangement in Diagram 1.1. Nyatakan sifat bahan J berdasarkan susunan atom dalam Rajah 1.1. [1 mark / 1 markah] (b) Diagram 1.2 shows the National Monument, a bronze sculpture which was built in remembrance of the brave soldiers who died fighting for the independence of our country. Rajah 1.2 menunjukkan Tugu Negara, arca gangsa yang dibina untuk memperingati askar-askar berani yang terkorban demi memerdekakan negara kita. Diagram 1.2 / Rajah 1.2 State the reason why bronze is used instead of pure copper. Nyatakan sebab mengapa gangsa digunakan dan bukannya kuprum tulen. [2 marks / 2 markah] 2. Diagram 2 shows the apparatus set-up used in the experiment to determine the heat of displacement of copper by magnesium. The initial temperature of copper(II) sulphate is 28.0 °C and the highest temperature achieved is 45.0 °C. Rajah 2 menunjukkan susunan radas yang digunakan dalam eksperimen untuk menentukan haba penyesaran kuprum oleh magnesium. Suhu awal kuprum(II) sulfat ialah 28.0 °C dan suhu tertinggi yang dicapai ialah 45.0 °C. 50.0 cm3 0.2 mol dm–3 copper(II) sulphate solution 50.0 cm3 kuprum(II) klorida 0.5 mol dm–3 Excess zink powder Serbuk zink berlebihan Thermometer Termometer Polystryrene cup Cawan polistirena Diagram 2 / Rajah 2

Chemistry Form 5 SPM Model Paper 107 © Penerbitan Pelangi Sdn. Bhd. (a) State the type of reaction that occurred based on the temperature change in the experiment. Nyatakan jenis tindak balas yang berlaku berdasarkan perubahan suhu dalam eksperimen tersebut. [1 mark / 1 markah] (b) Write a balanced chemical equation for the reaction. Tuliskan persamaan kimia seimbang bagi tindak balas tersebut. [1 mark / 1 markah] (c) (i) Calculate the number of moles of copper(II) sulphate solution. Hitung bilangan mol larutan kuprum(II) sulfat. [1 mark / 1 markah] (ii) Calculate the heat released in this experiment. Hitung haba yang dibebaskan dalam eksperimen ini. [1 mark / 1 markah] (iii) Calculate the heat of displacement of copper by magnesium in this experiment. Hitung haba penyesaran kuprum oleh magnesium dalam eksperimen ini. [1 mark / 1 markah]

Chemistry Form 5 SPM Model Paper © Penerbitan Pelangi Sdn. Bhd. 108 (d) Predict the value of heat of displacement if magnesium powder is used in this experiment to replace zinc powder. Give a reason. Ramalkan nilai haba penyesaran sekiranya serbuk magnesium digunakan dalam eksperimen ini untuk menggantikan serbuk zink. Berikan alasan. [2 marks / 2 markah] 3. Diagram 3 shows the information of atoms represented by unknown M, N, Q, R, X and Y in the Periodic Table of Elements. Rajah 3 menunjukkan maklumat atom yang ditunjukkan oleh M, N, Q, R, X dan Y yang tidak diketahui dalam Jadual Berkala Unsur. R X Y M Q N Diagram 3 / Rajah 3 Based on Diagram 3, Berdasarkan Rajah 3, (a) Suggest a name for element M. Cadangkan nama bagi unsur M. [1 mark / 1 markah] (b) (i) Identify the transition metal. Kenal pasti logam peralihan. [1 mark / 1 markah] (ii) State one special feature for answer in 3(b)(i). Nyatakan satu ciri khas bagi jawapan di 3(b)(i). [1 mark / 1 markah]

Chemistry Form 5 SPM Model Paper 109 © Penerbitan Pelangi Sdn. Bhd. (c) Write the electron arrangement for atom Y. Tuliskan susunan elektron bagi atom Y. [1 mark / 1 markah] (d) Write the formula of ion formed from atom Q. Tuliskan formula ion yang terbentuk daripada atom Q. [1 mark / 1 markah] (e) When a small piece of element Y is burnt in gas Q, a reaction occurred to produce an ionic compound. Write a chemical equation for the reaction. Apabila sekeping kecil unsur Y dibakar dalam gas Q, tindak balas berlaku untuk menghasilkan sebatian ion. Tuliskan persamaan kimia bagi tindak balas tersebut. [2 marsk / 2 markah] 4. Table 1 shows the number of protons and neutrons for atoms P and Q. Jadual 1 menunjukkan bilangan proton dan neutron bagi atom P dan Q. Atom Atom Number of protons Bilangan proton Number of neutrons Bilangan neutron P 12 12 Q 11 12 Table 1 / Jadual 1 (a) (i) Write the electron arrangement of atom Q. Tuliskan susunan elektron atom Q. [1 mark / 1 markah] (ii) State the position of atom Q in the Periodic Table of Elements. Nyatakan kedudukan atom Q dalam Jadual Berkala Unsur. [1 mark / 1 markah]

Chemistry Form 5 SPM Model Paper © Penerbitan Pelangi Sdn. Bhd. 110 (iii) Give a reason for your answer in 4(a)(i). Berikan alasan untuk jawapan anda di 4(a)(i). [2 marsk / 2 markah] (b) (i) What is the nucleon number of atom P? Apakah nombor nukleon bagi atom P? [1 mark / 1 markah] (ii) Write the electron arrangement of ion P. Tuliskan susunan elektron ion P. [1 mark / 1 markah] (iii) Draw the electron arrangement of ion P. Lukiskan susunan elektron ion P. [2 marks / 2 markah] 5. Diagram 4 shows the mass spectrum of element X. Rajah 4 menunjukkan jisim spektrum untuk unsur X. Intensity (% abundance) Keamatan (% kelimpahan) Isotopic mass Jisim isotop 82.8 8.1 9.1 24 25 26 Diagram 4 / Rajah 4

Chemistry Form 5 SPM Model Paper 111 © Penerbitan Pelangi Sdn. Bhd. (a) How many naturally occurring isotopes does the element contain? Berapakah isotop semula jadi yang terdapat dalam unsur ini? [1 mark / 1 markah] (b) Define natural abundance. Takrifkan kelimpahan semula jadi. [1 mark / 1 markah] (c) State the natural abundance for the isotopes of element X. Nyatakan kelimpahan semula jadi bagi isotop unsur X. [2 marks / 2 markah] (d) Write the symbols for the isotopes of element X. Tuliskan simbol bagi semua isotop unsur X. [2 marks / 2 markah] (e) Calculate the relative atomic mass of element X. Hitungkan jisim atom relatif unsur X. [2 marks / 2 markah] 6. Diagram 5 shows two sets of experiments carried out to study the effect of catalyst on the rate of reaction between zinc and sulphuric acid. Rajah 5 menunjukkan dua set eksperimen yang dijalankan untuk mengkaji kesan mangkin terhadap kadar tindak balas antara zink dengan asid sulfurik. Set I Set II Water Air Burette Buret Sulphuric acid solution Larutan asid sulfurik Excess zinc granule Ketulan zink berlebihan Water Air Burette Buret Sulphuric acid + catalyst Asid sulfurik + mangkin Excess zinc granule Ketulan zink berlebihan Diagram 5 / Rajah 5

Chemistry Form 5 SPM Model Paper © Penerbitan Pelangi Sdn. Bhd. 112 Table 2 shows the results obtained from the experiment. Jadual 2 menunjukkan hasil yang diperoleh daripada eksperimen tersebut. Set Set Total volume of gas collected in 3 minutes (cm3 ) Isipadu gas terkumpul dalam 3 minit (cm3 ) Temperature (°C) Suhu (°C) I 40.00 30.0 II 50.00 30.0 Table 2 / Jadual 2 (a) Sketch a graph of volume of gas (cm3 ) against time (s) for both sets on the same axis. Lakarkan graf isi padu gas (cm3 ) melawan masa (s) untuk kedua-dua set pada paksi yang sama. [2 marks / 2 markah] (b) Calculate the average rate of reaction for the first 3 minutes for Set I and Set II, in cm3 min–1 (in two decimal places). Hitung kadar tindak balas purata bagi 3 minit pertama untuk Set I dan Set II, dalam cm3 min–1 (dalam dua titik perpuluhan). (i) Set I / Set I (ii) Set II / Set II [2 marks / 2 markah]

Chemistry Form 5 SPM Model Paper 113 © Penerbitan Pelangi Sdn. Bhd. (iii) Compare the average rate of reaction for the first 3 minutes for Set I and Set II. Bandingkan kadar tindak balas purata selama 3 minit pertama untuk Set I dan Set II. [1 mark / 1 markah] (iv) Based on the Collision Theory, explain how catalyst affects the rate of reaction. Berdasarkan Teori Perlanggaran, terangkan bagaimana mangkin mempengaruhi kadar tindak balas. [3 marks / 3 markah] 7. Diagram 6 shows the U-tube consisting iron(II) sulphate solution and bromine water. Rajah 6 menunjukkan tiub-U berisi larutan ferum(II) sulfat dan air bromin. G FeSO4 (aq) FeSO4 (ak) Carbon electrodes Elektrod karbon H2 SO4 (aq) H2 SO4 (ak) Bromine water(aq) Air bromin(ak) Diagram 6 / Rajah 6 (a) (i) Identify the anode and the cathode. Kenal pasti anod dan katod. Anode: Anod: Cathode: Katod: [2 marks / 2 markah]

Chemistry Form 5 SPM Model Paper © Penerbitan Pelangi Sdn. Bhd. 114 (ii) Write a half equation to support your answer in 7(a)(i). Tuliskan persamaan setengah untuk menyokong jawapan anda di 7(a)(i). Anode / Anod: Cathode / Katod: [2 marks / 2 markah] (b) Identify the oxidising agent and the reducing agent. Kenal pasti agen pengoksidaan dan agen penurunan. Oxidising agent / Agen pengoksidaan: Reducing agent / Agen penurunan: [2 marks / 2 markah] (c) Write the overall ionic equation. Tuliskan persamaan ion keseluruhan. [2 marks / 2 markah] 8. Diagram 7 shows the apparatus set-up for two types of cells, Cell A and Cell B. Rajah 7 menunjukkan susunan radas bagi dua jenis sel, Sel A dan Sel B. Zinc electrode Elektrod zink Zinc sulphate solution Larutan zink sulfat Copper electrodes Elektrod kuprum Porous pot Pasu berliang Copper(II) sulphate solution Larutan kuprum(II) sulfat V Cell A Sel A Cell B Sel B Diagram 7 / Rajah 7

Chemistry Form 5 SPM Model Paper 115 © Penerbitan Pelangi Sdn. Bhd. (a) (i) Name Cell A and Cell B. Namakan Sel A dan Sel B. [2 marks / 2 markah] (ii) State the energy conversion in Cell B. Nyatakan penukaran tenaga dalam Sel B. [1 mark / 1 markah] (b) (i) Compare the observations on copper(II) sulphate solution in Cell A and Cell B after 30 minutes. Bandingkan pemerhatian terhadap larutan kuprum(II) sulfat dalam Sel A dengan Sel B selepas 30 minit. [2 marks / 2 markah] (ii) Explain your answer in 8(b)(i). Jelaskan jawapan anda di 8(b)(i). [2 marks / 2 markah] (c) (i) Suggest a method to increase the potential difference of Cell A. Cadangkan kaedah untuk meningkatkan perbezaan keupayaan Sel A. [1 mark / 1 markah] (ii) Explain your answer. Terangkan jawapan anda. [1 mark / 1 markah]

Chemistry Form 5 SPM Model Paper © Penerbitan Pelangi Sdn. Bhd. 116 Section B / Bahagian B [20 marks / 20 markah] Answer any one question from this section. Jawab mana-mana satu soalan dalam bahagian ini. 9. In an investigaton, Bryan and Derrick carried out three experiments as shown in Table 3. Dalam suatu penyiasatan, Bryan dan Derrick menjalankan tiga eksperimen seperti yang ditunjukkan dalam Jadual 3. Experiment Eksperimen Condition Keadaan 1 100 cm3 of 0.1 mol dm–3 nitric acid acid + iron powder at room condition 100 cm3 0.1 mol dm–3 asid nitrik + serbuk besi pada keadaan bilik 2 100 cm3 of 0.1 mol dm–3 nitric acid + iron powder at 50 o C 100 cm3 0.1 mol dm–3 asid nitrik + serbuk besi pada 50 o C 3 100 cm3 of 0.1 mol dm–3 nitric acid + iron powder in 50 o C + copper(II) sulphate solution 100 cm3 0.1 mol dm–3 asid nitrik + serbuk besi pada 50 o C + larutan kuprum(II) sulfat Table 3 / Jadual 3 (a) Write an ionic equation to represent all the three experiments in Table 3. Tuliskan persamaan ion untuk mewakili ketiga-tiga eksperimen dalam Jadual 3. [2 marks / 2 markah] (b) Explain each factor that you have identified in Experiment 1, Experiment 2 and Experiment 3. Terangkan setiap faktor yang telah anda kenal pasti dalam Eksperimen 1, Eksperimen 2 dan Eksperimen 3. [3 marks / 3 markah] (c) Based on the Collision Theory, compare and explain: Berdasarkan Teori Perlanggaran, bandingkan dan terangkan: (i) Experiment 1 and Experiment 2 Eksperimen 1 dan Eksperimen 2. [5 marks / 5 markah] (ii) Experiment 2 and Experiment 3 Eksperimen 2 dan Eksperimen 3. [6 marks / 6 markah] (d) Then, sketch a graph to show the comparison between experiments in 9(c)(i) and 9(c)(ii) on the same axes. Kemudian, lakarkan graf untuk menunjukkan perbandingan antara eksperimen di 9(c)(i) dengan 9(c)(ii) pada paksi yang sama. [4 marks / 4 markah]

Chemistry Form 5 SPM Model Paper 117 © Penerbitan Pelangi Sdn. Bhd. 10. (a) Diagram 8.1 shows the flow chart involving the formation of ethene gas from a cup of pineapple juice. Rajah 8.1 menunjukkan carta alir yang melibatkan pembentukan gas etena daripada secawan jus nanas. Ethanol Etanol Ethene Etena I II Diagram 8.1 / Rajah 8.1 (i) Name process I and process II. Then, write a balanced chemical equation for the conversion of process I and process II. Namakan proses I dan proses II. Kemudian, tulis persamaan kimia yang seimbang bagi penukaran proses I dan proses II. [6 marks / 6 markah] (ii) Briefly describe the formation of ethene gas from ethanol. Explain how you verify the formation of ethene gas using a suitable reagent solution. Huraikan secara ringkas pembentukan gas etena daripada etanol. Terangkan bagaimana anda mengesahkan pembentukan gas etena menggunakan larutan reagen yang sesuai. [8 marks / 8 markah] (b) Diagram 8.2 shows the coagulated latex in a cup left for more than 4 hours. Rajah 8.2 menunjukkan getah beku di dalam cawan yang dibiarkan selama lebih daripada 4 jam. Diagram 8.2 / Rajah 8.2 (i) Explain this situation. Terangkan keadaan ini. [4 marks / 4 markah] (ii) Suggest a solution that can slow down the formation of the white solid as shown in Diagram 8.2. Give your reason. Cadangkan larutan yang dapat melambatkan pembentukan pepejal putih yang ditunjukkan dalam Rajah 8.2. Berikan alasan anda. [2 marks / 2 markah]

Chemistry Form 5 SPM Model Paper © Penerbitan Pelangi Sdn. Bhd. 118 Section C / Bahagian C [20 marks / 20 markah] Answer all questions in this section. Jawab semua soalan dalam bahagian ini. 11. Diagram 9 shows the conversation between Janelle and Allen. Rajah 9 menunjukkan perbualan antara Janelle dan Allen. I added 1 cm3 of barium chloride solution but cannot differentiate between potassium carbonate solution and potassium sulphate solution. Why? Saya menambahkan 1 cm3 larutan barium klorida tetapi tidak dapat membezakan antara larutan kalium karbonat dengan larutan kalium sulfat. Mengapa? Janelle, are you able to differentiate between potassium carbonate solution and potassium sulphate solution using the barium chloride solution given by Madam Chin? Janelle, adakah anda dapat membezakan larutan kalium karbonat dengan larutan kalium sulfat menggunakan larutan barium klorida yang diberikan oleh Puan Chin? Diagram 9 / Rajah 9 (a) Give an explanation to their problem. Suggest a change that should be done to the procedure in order for the test to be successful. In your explanation, include ionic equations. Berikan penjelasan tentang masalah mereka. Cadangkan perubahan yang harus dilakukan terhadap prosedur supaya ujian tersebut berjaya. Dalam penjelasan anda, sertakan persamaan ion. [10 marks / 10 markah] (b) Suzannah and Danish are given different solutions. They are assigned to describe chemical tests that can be carried out in the laboratory to differentiate the solutions. Suzannah dan Danish diberikan larutan yang berbeza. Mereka ditugaskan untuk menerangkan ujian kimia yang boleh dilakukan di makmal untuk membezakan larutan-larutan tersebut. (i) Sodium chloride solution and sodium iodide solution. Larutan natrium klorida dan larutan natrium iodida. (ii) Ammonium sulphate solution and sodium sulphate solution. Larutan ammonium sulfat dan larutan natrium sulfat. [10 marks / 10 markah]

A1 © Penerbitan Pelangi Sdn. Bhd. 1 CHAPTER Redox Equilibrium Keseimbangan Redoks Oxidation and Reduction 1.1 Pengoksidaan dan Penurunan 1. Definition Definisi Oxidation Pengoksidaan Reduction Penurunan (a) Transfer of oxygen Pemindahan oksigen Gain Penerimaan Loss Kehilangan (b) Transfer of hydrogen Pemindahan hidrogen Loss Kehilangan Gain Penerimaan (c) Transfer of electron Pemindahan elektron Release Kehilangan Receive Penerimaan (d) Change in oxidation number Perubahan nombor pengoksidaan Increase Bertambah Decrease Berkurang Answers 2. (a) NH4 + NH4 = +1 N + 4(+1) = +1 N = –3 (b) CO3 2– CO3 = –2 C + 3(–2) = –2 C = +4 (c) K2Cr2O7 K2Cr2O7 = 0 2(+1) + 2Cr + 7(–2) = 0 Cr = +6 (d) Na2S2O3 Na2S2O3 = 0 2(+1) + 2S + 3(–2) = 0 S = +2 Standard Electrode Potential 1.2 Keupayaan Elektrod Piawai 1. half cells / setengah sel 2. (a) Eº cell / Eº sel = (–0.45) – (–0.76) = 0.31 V (b) Eº cell / Eº sel = (+0.80) – (+0.34) = 0.46 V (c) Eº cell / Eº sel = (–0.76) – (–2.37) = 1.61 V (d) Eº cell / Eº sel = (+0.34) – (–0.13) = 0.47 V Voltaic Cell 1.3 Sel Kimia 1. chemical, electrical kimia, elektrik 2. (a) (i) Zinc / Zink (ii) Zinc, copper / Zink, kuprum (b) (i) Copper / Kuprum (ii) Copper, zinc Kuprum, zink (c) (i) Zn → Zn2+ + 2e– (ii) Cu2+ + 2e– → Cu (d) (i) Zinc / Zink (ii) Copper(II) ion / Cu2+ Ion kuprum(II) / Cu2+ (e) (i) Zinc / Zink (ii) Copper / Kuprum (iii) Blue, lighter blue biru, biru muda (f) (i) Zinc, zinc ion / Zink, ion zink (ii) Copper, copper Kuprum, kuprum (iii) decreases / berkurang (g) zinc, copper / zink, kuprum (h) (i) Salt bridge / Titian garam (ii) – Dilute hydrochloric acid / Asid hidroklorik cair – Dilute sodium chloride solution / Larutan natrium klorida cair – Dilute sodium sulphate solution / Larutan natrium sulfat cair (iii) Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Zn(p) + Cu2+(ak) → Zn2+(ak) + Cu(p) Electrolytic Cell 1.4 Sel Elektrolisis 1. aqueous solution, molten state, electric current larutan akueus, keadaan leburan, arus elektrik 2. free-moving ions, chemical, passes ion bergerak bebas, kimia 3. molecules / molekul-molekul 4. electrical, chemical / elektrik, kimia 5. Electrolyte Elektrolit Nonelectrolyte Bukan elektrolit – Potassium chloride solution Larutan kalium klorida – Molten silver chloride Leburan argentum klorida – Nitric acid Asid nitrik – Sucrose solution Larutan sukrosa – Lead(II) bromide Plumbum(II) bromida – Molten acetamide Leburan asetamida Extraction of Metal from Its Ore Pengesktrakan Logam daripada 1.5 Bijihnya 1. (a) sodium, magnesium, aluminium natrium, magnesium, aluminium (b) iron, tin, zinc ferum, stanum, zink 2. – Cheap / Murah – Easily available / Mudah diperoleh – Solid in room condition / Pepejal dalam keadaan bilik – An effective reducing agent / Agen penurunan yang berkesan 3. (i) Consuming extremely high amount of electricity during extraction causing pollution. Penggunaan jumlah elektrik yang sangat tinggi semasa pengekstrakan menyebabkan pencemaran. (ii) Removal of native vegetation in mining area resulting in soil erosion. Pembuangan tumbuhan asal di kawasan perlombongan mengakibatkan hakisan tanah. (iii) Greenhouse gas emission causing global warming. Pembebasan gas rumah hijau menyebabkan pemanasan global. Rusting 1.6 Pengaratan 1. oxidation, loss, positively-charged pengoksidaan, kehilangan, bercas positif 2. iron, steel / besi, keluli 3. oxygen, water / oksigen, air

Chemistry Form 5 Answers © Penerbitan Pelangi Sdn. Bhd. A2 4. slow down, zinc / magnesium / iron / aluminium melambatkan, zink / magnesium / ferum / aluminium 5. speed up, tin/ lead/ copper/ silver mempercepatkan, stanum / plumbum / kuprum / argentum SPM Practice 1 Paper 1 1. C 2. D 3. A 4. C 5. B 6. D 7. C 8. C 9. D 10. C 11. D 12. D 13. D 14. D 15. D 16. C 17. B 18. B 19. A 20. C 21. C 22. C 23. D 24. C 25. D 26. D 27. C 28. A Paper 2 Section A / Bahagian A 1. (a – Concentration of ions 1.0 mol dm–3 Kepekatan ion 1.0 mol dm–3 – Temperature 25 °C or 298 K Suhu 25 °C atau 298 K – Pressure of 1 atm or 101 kPa Tekanan pada 1 atm atau 101 kPa – Platinum is used as inert electrode Platinum digunakan sebagai elektrod lengai [any three answers / manamana tiga jawapan] (b) (i) Cl2  Ag+  Cu2+  Zn2+ (ii) Cl– , Ag , Cu , Zn (c) Eº cell = Eº cathode – Eº anode Eº sel = Eº katod – Eº anod = (+0.34) – (–0.76) = 1.10 V 2. (a) Anode / Anod: 2O2– → O2 + 4e– Cathode / Katod: Al3+ + 3e– → Al (b) (i) Oxide ion undergoes oxidation reaction by releasing electrons to form oxygen gas. Ion oksida menjalani tindak balas pengoksidaan dengan membebaskan elektron membentuk gas oksigen. (ii) Aluminium ion undergoes reduction reaction by receiving electrons to form aluminium. Ion aluminium menjalani tindak balas penurunan dengan menerima elektron membentuk aluminium. (c) (i) Aluminium ion / Ion aluminium (ii) Oxide ion / Ion oksida 3. (a) Oxygen / Oksigen (b) (i) 4OH– → 2H2O + O2 + 4e– (ii) Hydroxide ion / Ion hidroksida (iii) – Bring near glowing splinter into test tube with gas X / oxygen gas. Dekatkan kayu uji berbara ke dalam tabung uji berisi gas X / gas oksigen. – Splinter relights indicates the presence of gas X / oxygen gas. Kayu uji menyala menunjukkan kehadiran gas X / gas oksigen. (c) (i) Copper(II) ion / Ion kuprum(II) (ii) Brown solid is formed. / Pepejal perang terbentuk. (iii) Copper is deposited on the cathode. Kuprum terenap pada katod. (d) 4OH– + 2Cu2+ → 2H2O + O2 + 2Cu 4. (a) (i) Anode: Carbon immersed in iron(II) sulphate solution. Anod: Karbon yang dicelup di dalam larutan ferum(II) sulfat. Cathode: Carbon immersed in acidified potassium dichromate(VI) solution. Katod: Karbon yang dicelup di dalam larutan dikromat(VI) berasid. (ii) Anode / Anod: Fe2+ → Fe3+ + e– Cathode / Katod: Cr2O7 2– + 14H+ + 6e– → 2Cr3+ + 7H2O (b) Oxidising agent / Agen pengoksidaan: Acidified dichromate(VI) ion / Ion dikromat(VI) berasid Reducing agent / Agen penurunan: Iron(II) ion / Ion ferum(II) (c) Electron flows from carbon immersed in iron(II) sulphate solution to carbon immersed in acidified potassium dichromate(VI) solution through external circuit. Elektron mengalir dari karbon yang dicelup di dalam larutan ferum(II) sulfat ke karbon yang dicelup di dalam larutan kalium dikromat(VI) berasid melalui litar luar. (d) 6Fe2+ + Cr2O7 2– + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O Section B / Bahagian B 5. (a) Presence of water and oxygen gas. Kehadiran air dan gas oksigen. (b) – The middle region of water droplet that covers the iron block is the anode. Kawasan tengah titisan air yang menutupi blok besi ialah anod. – In this area, the concentration of oxygen gas is lower. Di kawasan ini, kepekatan gas oksigen lebih rendah. – Iron undergoes oxidation reaction Ferum menjalani tindak balas pengoksidaan – by releasing electrons to form iron(II) ions. dengan membebaskan elektron membentuk ion ferum(II). – The edge region of water droplet that covers the iron block is the cathode. Kawasan pinggir air yang menutupi blok besi ialah katod. – In this area, the concentration of oxygen gas is higher. Di kawasan ini, kepekatan gas oksigen lebih tinggi. – Oxygen gas and water undergo reduction reaction Gas oksigen dan air menjalani tindak balas penurunan – by receiving electrons to form hydroxide ions. dengan menerima elektron membentuk ion hidroksida. (c) – Half-equation in anode / Persamaan setengah di anod: Fe → Fe2+ + 2e– – Iron serves as a reducing agent. Ferum bertindak sebagai agen penurunan. – Half-equation / Persamaan setengah di katod: O2 + 2H2O + 4e– → 4OH– – Oxygen gas serves as an oxidising agent. Gas oksigen bertindak sebagai agen pengoksidaan. (d) – Coil magnesium on the iron block as a sacrificial metal. Lilitkan magnesium pada blok besi sebagai logam korban. – Magnesium is more electropositive than iron, thus it will be oxidised to protect iron. Magnesium lebih elektropositif daripada besi, maka magnesium akan teroksida untuk melindungi besi. – Apply paint on the surface of iron. Sapukan cat pada permukaan besi.

Chemistry Form 5 Answers A3 © Penerbitan Pelangi Sdn. Bhd. – Paint prevents the penetration of water and oxygen gas into the iron block. Cat menghalang penembusan air dan gas oksigen pada blok besi. Section C / Bahagian C 6. (a) Electrolyte: Silver nitrate solution Elektrolit: Larutan argentum nitrat Anode: Pure silver rod Anod: Rod argentum tulen Procedure / Prosedur: 1. The iron key is cleaned using sandpaper. Kunci besi dibersihkan dengan kertas pasir. 2. 150 cm3 1.0 mol dm–3 silver nitrate solution is measured and poured into a beaker. 150 cm3 larutan argentum nitrat 1.0 mol dm–3 disukat dan dituang ke dalam sebuah bikar. 3. Pure silver is connected to the positive terminal of a battery and cleaned iron key is connected to the negative terminal of the battery using a connecting wire. Argentum tulen disambung ke terminal positif bateri dan kunci yang bersih ke terminal negatif bateri dengan menggunakan wayar penyambung. 4. Both electrodes are immersed into silver nitrate solution. Kedua-dua elektrod direndam ke dalam larutan argentum nitrat sulfat. 5. The switch is turned on and the electricity is allowed to flow for 30 minutes. Suis dihidupkan dan elektrik dibiarkan mengalir selama 30 minit. 6. All changes that occur at the anode, cathode and electrolyte are recorded. Semua perubahan yang berlaku pada anod, katod dan elektrolit dicatatkan. Observations / Pemerhatian: – Pure silver anode becomes thinner. Anod argentum tulen semakin nipis. – A layer of shiny grey metal is coated on the antique key. Lapisan logam kelabu berkilat menyaluti kunci antik. 3. Each iron nail is placed into a test tube respectively. Then, three drops of phenolphthalein indicator and three drops of potassium hexacyanoferrate(III) solution are added into the hot agar solution and stirred well. Setiap paku besi diletakkan ke dalam tabung uji masingmasing. Kemudian, tiga titis petunjuk fenolftalein dan tiga titis larutan kalium heksasianoferat(III) ditambahkan ke dalam larutan agar panas dan dikacau hingga sebati. 4. The hot agar solution is poured into each test tube until the iron nails are covered with hot agar solution. Larutan agar panas itu dituangkan ke dalam setiap tabung uji sehingga semua paku besi ditutupi dengan larutan agar panas. 5. The test tubes are left on the rack for a day. All observations are recorded. Semua tabung uji tersebut dibiarkan selama sehari. Semua pemerhatian dicatatkan. (b) Materials / Bahan: Iron nails, zinc strip, magnesium strip, copper strip, sandpaper, hot agar solution, phenolphthalein indicator and 0.1 mol dm–3 potassium hexacyanoferrate(III) solution Paku besi, jalur zink jalur magnesium, jalur kuprum, kertas pasir, larutan agar panas, petunjuk fenolftalein, dan larutan kalium heksasianoferat(III) 0.1 mol dm–3 Apparatus / Radas: Test tubes, test tube rack, and dropper Tabung uji, rak tabung uji dan penitis Procedure / Prosedur: 1. All the metals and iron nail are cleaned with sandpaper. Semua logam dan paku besi dibersihkan menggunakan kertas pasir. 2. Thin strip of magnesium is coiled on iron nail B, zinc on iron nail C, and copper on iron nail D. Jalur magnesium yang nipis dililit pada paku besi B, jalur zink pada paku besi C dan jalur kuprum pada paku besi D. Result / Keputusan: Test tube Tabung uji Observation Pemerhatian Inference Inferens A Fe • Dark blue spots are observed on the agar gel. Bintik-bintik biru tua kelihatan di atas agar gel • Rusting occurs Pengaratan berlaku B Fe + Mg • Pink colouration is found on agar gel Warna merah jambu terbentuk di atas agar gel • Rusting does not occur Pengaratan tidak berlaku C Fe + Zn • Pink colouration is found on agar gel Warna merah jambu terbentuk di atas agar gel • Rusting does not occur Pengaratan tidak berlaku D Fe + Cu • The whole agar gel turns dark blue Keseluruhan agar gel menjadi biru tua • Rusting occurs the fastest Pengaratan berlaku paling cepat HOTS Challenge – Conditions to cause rusting: Presence of oxygen gas and water. Keadaan yang menyebabkan pengaratan: Kehadiran gas oksigen dan air. – Polish the rusted grill with sandpaper to remove all the brown solid on the grill. Gosokkan jeriji yang berkarat dengan kertas pasir untuk menyingkirkan semua pepejal perang pada jeriji tersebut.

Chemistry Form 5 Answers © Penerbitan Pelangi Sdn. Bhd. A4 (d) C3H7OH(l) + 3O2(g) → 3CO(g) + 4H2O(g) // C3H7OH(ce) + 3O2(g) → 3CO(g) + 4H2O(g) // C3H7OH(l) + 3 2O2(g) → 3C(s) + 4H2O(g) C3H7OH(ce) + 3 2 O2(g) → 3C(p) + 4H2O(g) (e) C2H4(g) + H2O(g) → C2H5OH(l) C2H4(g) + H2O(g) → C2H5OH(ce) (f) C2H5OH(l) + 2[O] → CH3COOH(l) + H2O(l) C2H5OH(ce) + 2[O] → CH3COOH(ce) + H2O(ce) (g) C2H5OH(l) + C3H7COOH(l) → C3H7COOC2H5(l) + H2O(l) C2H5OH(ce) + C3H7COOH(ce) → C3H7COOC2H5(ce) + H2O(ce) 2. (a) Ethanol / Etanol (b) 1,2-dibromopropane / 1,2-dibromopropana (c) Ethene / Etena 3. (a) Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g) Mg(p) + 2H+(ak) → Mg2+(ak) + H2(g) Colorless bubbles are observed. / Gelembung tak berwarna kelihatan. (b) CaCO3(s) + 2H+(aq) → Ca2+(aq) + CO2(g) + H2O(l) CaCO3(p) + 2H+(ak) → Ca2+(ak) + CO2(g) + H2O(ce) Colorless bubbles are observed. / Gelembung tak berwarna kelihatan. (c) H+(aq) + OH– (aq) → H2O(l) H+(ak) + OH– (ak) → H2O(ce) Blue cobalt(II) chloride paper turns pink. Kertas kobalt(II) klorida yang berwarna biru bertukar menjadi merah jambu. 4. (a) Esterification / Pengesteran (b) Concentrated sulphuric acid / Asid sulfurik pekat (c) Methyl ethanoate / Metil etanoat (d) – sweet / wangi – low / rendah – Insoluble / Tidak larut Isomers and Naming Based on IUPAC Nomenclature Isomer dan Penamaan mengikut 2.4 IUPAC 1. chemical formula, structural formula formula kimia, formula struktur – Paint the cleaned iron grill. Catkan jeriji besi yang telah dibersihkan. – Paint can prevent the contact of water and oxygen gas with the iron grill. Cat dapat mengelakkan sentuhan air dan gas oksigen dengan jeriji besi. 2CHAPTER Carbon Compound Sebatian Karbon Types of Carbon Compound 2.1 Jenis-jenis Sebatian Karbon 1. carbon / karbon 2. (a) carbon, living / karbon, hidup (b) saturated, unsaturated / tepu, tak tepu (c) natural gas, petroleum / gas asli, petroleum 3. Organic compounds Sebatian organik Characteristics Ciri-ciri Inorganic compounds Sebatian tak organik Living things Benda hidup Source Sumber Non-living things Benda bukan hidup Lower Lebih rendah Melting point and boiling point Takat lebur dan takat didih Higher Lebih tinggi Homologous Series 2.2 Siri Homolog 1. General formula Formula am CnH2n+2 CnH2n CnH2n–2 Functional group Kumpulan berfungsi Carbon-carbon single bond Ikatan karbon tunggal Carbon-carbon double bond Ikatan karbon ganda dua Carbon-carbon triple bond Ikatan karbon ganda tiga 2. (a) Pentane / Pentana (b) Propanol / Propanol (c) Hexanoic acid / Asid heksanoik (d) Butyne / Butuna 3. Octane Oktana Odourless Tidak berbau Heptene Heptena Flammable Mudah terbakar Pentyne Pentuna Soury smell Berbau masam Ethanol Etanol Fragrant smell Berbau wangi Methanoic acid Asid metanoik Soluble in water Larut dalam air Ethyl butanoate Etil butanoate Very volatile Mudah meruap Chemical Properties and Interconversion of Compounds between Homologous Series Sifat Kimia dan Saling Pertukaran 2.3 Sebatian antara Siri Homolog 1. (a) C4H10(l) + 13 2 O2(g) → 4CO2(g) + 5H2O(g) C4H10(ce) + 13 2 O2(g) → 4CO2(g) + 5H2O(g) (b) C3H6(l) + 3O2(g) → 3CO(g) + 3H2O(g) // C3H6(ce) + 3O2(g) → 3CO(g) + 3H2O(g) // C3H6(l) + 3 2O2(g) → 3C(s) + 3H2O(g) C3H6(ce) + 3 2 O2(g) → 3C(p) + 3H2O(g) (c) C5H8(l) + 7O2(g) → 5CO2(g) + 4H2O(g) C5H8(ce) + 7O2(g) → 5CO2(g) + 4H2O(g)

Chemistry Form 5 Answers A5 © Penerbitan Pelangi Sdn. Bhd. 2. Butyne / Butuna H H & & H!C#C!C!C!H & & H H 1–butyne 1–butuna 1,3–butadiene 1,3–butadiena H!C & H C !H & H C!C & & ' ' H H 3. (a) (i) Ethyl butanoate / Etil butanoat (ii) Butanoic acid / Asid butanoik (iii) Ethanol / Etanol (b) (i) Methyl ethanoate / Metil etanoat (ii) Ethanoic acid / Asid etanoik (iii) Methanol / Metanol SPM Practice 2 Paper 1 1. A 2. D 3. B 4. B 5. D 6. A 7. C 8. D 9. B 10. B 11. D 12. C 13. D 14. B 15. A 16. C 17. A 18. B 19. C 20. D 21. A 22. C 23. B 24. C 25. C 26. C 27. C 28. B Paper 2 Section A / Bahagian A 1. (a) X: CnH2n+2 Y: CnH2n Z: CnH2n–2 (b) X: Propane / Propana Y: Propene / Propena Z: Propyne / Propuna (c) (i) C3H4(g) + 4O2(g) → 3CO2(g) + 2H2O(g) C3H4(g) + 4O2(g) → 3CO2(g) + 2H2O(g ) (ii) – Number of moles of propyne / Bilangan mol propuna = 4 [3(12) + 4(1)] = 0.1 mol – Mole ratio / Nisbah mol = 1 : 3 = 0.1 : 0.3 – Volume of gas / Isi padu gas = 0.3 × 24 = 7.2 dm3 (iii) Carbon dioxide / Karbon dioksida 2. (a) P: Ethene / Etena Q: Ethanol / Etanol R: Ethanoic acid / Asid etanoik (b) I: Hydration / Penghidratan II: Oxidation / Pengoksidaan (c) (i) Brown bromine water turns colourless. Warna perang air bromin menjadi tak berwarna. (ii) Compound P has carboncarbon double bond. Sebatian P mempunyai ikatan karbon ganda dua. (d) (i) Orange to green / Jingga kepada hijau (ii) C2H5OH(l) + 2[O] → CH3COOH(aq) + H2O(l) C2H5OH(ce) + 2[O] → CH3COOH(ak) + H2O(ce) 3. (a) (i) Carbon, hydrogen and oxygen Karbon, hydrogen dan oksigen (ii) Molecular formula / Formula molekul: C9H8O4 Empirical formula / Formula empirik: C9H8O4 (iii) Molar mass / Jisim molar = 9(12) + 8(1) + 4(16) = 180 g mol–1 (b) (i) Colorless bubbles are formed. Gelembung gas tak berwarna terbentuk. (ii) – Aspirin dissolved in warm water releases hydrogen ion. Aspirin yang larut di dalam air suam membebaskan ion hidrogen. – Hydrogen ion reacts with calcium carbonate from the eggshells to produce carbon dioxide gas. Ion hidrogen bertindak balas dengan kalsium karbonat daripada kulit telur untuk menghasilkan gas karbon dioksida. Section B / Bahagian B 4. (a) (i) – Structural formula of butene / Formula struktur butena: H H & & H!C"C!C!C!H & & H H – C4H8(l) + 6O2(g) → 4CO2(g) + 4H2O(g) C4H8(ce) + 6O2(g) → 4CO2(g) + 4H2O(g) (ii) Catalyst X / Mangkin X: Nikel powder / Serbuk nikel Temperature Y / Suhu Y: 180 o C Compound Z / Sebatian Z: Butane / Butana (iii) H H H H & & & & H!C!C!C!C!H & & & & H H H H H & H!C!H & H & H & & & H!C!C!C!H & & & H H H Butane / Butana 2-methylpropane / 2-metilpropana (b) Butyne Butuna Butene Butena Butanol Butanol Butanoic acid Asid butanoik Homologous series Siri homolog Alkyne Alkuna Alkene Alkena Alcohol Alkohol Carboxylic acid Asid karbosilik General formula Formula am CnH2n–2 CnH2n CnH2n+1OH CnH2n+1COOH Functional group Kumpulan berfungsi Carboncarbon triple bond Ikatan karbon ganda tiga Carboncarbon double bond Ikatan karbon ganda dua Hydroxyl group Kumpulan hidroksil Carboxyl group Kumpulan karboksil Section C / Bahagian C 5. (a) (i) Element Unsur C H O Composition by mass Komposisi dengan jisim 52.3 13.3 34.4 Number of moles Bilangan mol 52.3 12 = 4.36 13.3 1 = 13.3 34.4 16 = 2.15 Mole ratio Nisbah mol 4.36 2.15 = 2.03 13.3 2.15 = 6.19 2.15 2.15 = 1 Simplest ratio Nisbah teringkas 2 6 1

Chemistry Form 5 Answers © Penerbitan Pelangi Sdn. Bhd. A6 3CHAPTER Thermochemistry Termokimia Heat Change in Reactions 3.1 Perubahan Haba dalam Tindak Balas 1. (a) energy changes / perubahan tenaga (b) heat changes / perubahan haba (c) (i) Exothermic / eksotermik (ii) Endothermic / endotermik (d) released, absorbed dibebaskan, diserap Heat of Reaction 3.2 Haba Tindak Balas 1. (a) precipitation, double decomposition pemendakan, penguraian ganda dua (b) 1 mole of precipitate is formed from the ions in the solution. 1 mol mendakan terbentuk daripada ion-ion di dalam larutan. 2. (a) AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq) AgNO3(ak) + KCl(ak) → AgCl(p) + KNO3(ak) (b) Ag+(aq) + Cl– (aq) → AgCl(s) Ag+(ak) + Cl– (ak) → AgCl(p) (c) – Number of moles of AgNO3 / Bilangan mol AgNO3 = (0.5)(50) 1 000 = 0.025 mol – Number of moles of KCl / Bilangan mol KCl = (0.5)(50) 1 000 = 0.025 mol (d) Average initial temperature / Purata suhu awal = (28.0 + 29.0) 2 = 28.5 o C Temperature change Perubahan suhu = (37.5 – 28.5) = 9.0 o C (e) Heat change, Q Perubahan haba, Q = mcθ = (50 + 50)(4.2)(9.0) = 3780 J = 3.78 kJ (f) Heat of precipitation, ΔH Haba pemendakan, ΔH = 3.78 0.025 = –151.2 kJ mol–1 – Empirical formula / Formula empirik = C2H6O – n[C2H6O]= n= 1 – Molecular formula / Formula molekul = C2H6O (ii) – General formula / Formula am: CnH2n+1OH – Homologous series: Alcohol Siri homolog: Alkohol (iii) – Struktural formula / Formula struktur: H H & & H!C!C!O!H & & H H – IUPAC nomenclature: Ethanol Penamaan IUPAC: Etanol (b) (i) 1. Glass wool is soaked in ethanol / compound M and inserted into the end of a combustion tube. Wul kaca direndam di dalam etanol / sebatian M dan dimasukkan ke dalam hujung tiub pembakaran. 2. Porcelain chips are arranged in the middle of the combustion tube and the combustion tube is sealed with a rubber stopper connected to a delivery tube. Serpihan porselin disusun di tengah tiub pembakaran dan tiub pembakaran ditutup dengan penyumbat getah yang disambungkan kepada tiub penghantar. 3. The porcelain chips are heated strongly until hot red, while the glass wool is heated gently. Serpihan porselin dipanaskan dengan kuat sehingga merah panas manakala wul kaca dipanaskan secara perlahan. Ethene / compound N is collected using water displacement technique in an inverted test tube placed inside the water. Etena / sebatian N dikumpulkan menggunakan teknik sesaran air di dalam tabung uji yang diterbalikkan di dalam air. 46 [2(12) + 6(1) + 16] (ii) C2H5OH(l) → C2H4(g) + H2O(g) C2H5OH(ce) → C2H4(g) + H2O(g) Compound N: Ethene Sebatian N: Etena (iii) Test 1 / Ujian 1 – 2 cm3 of bromine water is added into a test tube containing ethene / compound N and shaken. 2 cm3 air bromin ditambahkan ke dalam tabung uji yang berisi etena / sebatian N dan digoncang. – Brown bromine water turns colourless. Air bromin perang bertukar menjadi tak berwarna. Test 2 / Ujian 2 – 2 cm3 of acidified potassium maganate(VII) solution is added into a test tube containing ethene / compound N and shaken. 2 cm3 larutan kalium manganate(VII) berasid ditambahkan ke dalam tabung uji yang berisi etena / sebatian N dan digoncang. – Purple acidified potassium manganate(VII) solution turns colourless. Larutan ungu kalium manganate(VII) berasid bertukar menjadi tak berwarna. (iv) – Soluble in water Larut dalam air – Low melting point and boiling point Takat lebur dan takat didih yang rendah HOTS Challenge – Diesel / Diesel – Percentage of carbon in diesel, C12H24 Peratusan karbon dalam diesel, C12H24 = 12(12) [12(12) + 24(1)] × 100% = 85.71% – Percentage of carbon in RON 95 petrol, C8H20 Peratusan karbon dalam petrol RON 95, C8H20 = 8(12) [8(12) + 20(1)] × 100% = 82.76%

Chemistry Form 5 Answers A7 © Penerbitan Pelangi Sdn. Bhd. (g) Energy / Tenaga Ag+(aq) + Cl– (aq) Ag+(ak) + Cl– (ak) AgCl(s) AgCl(p) ΔH = –151.2 kJ mol–1 (h) – Heat of precipitation / Haba pemendakan – silver chloride / argentum klorida – number of moles / Bilangan mol 3. (a) more, less lebih, kurang (b) 1 mole of metal is displaced from its metal ion solution by a more electropositive metal 1 mol logam disesarkan daripada larutan ion logamnya oleh logam yang lebih elektropositif 4. (a) Zn(s) + Pb(NO3)2(aq) → Zn(NO3)2(aq) + Pb(s) Zn(p) + Pb(NO3)2(ak) → Zn(NO3)2(ak) + Pb(p) (b) Zn(s) + Pb2+(aq) → Zn2+(aq) + Pb(s) Zn(p) + Pb2+(ak) → Zn2+(ak) + Pb(p) (c) Number of moles of Pb(NO3)2 / Bilangan mol Pb(NO3)2 = (1.0)(200) 1 000 = 0.2 mol (d) Heat change, Q / Perubahan haba, Q = mcθ = (200)(4.2)(54.7 – 28.0) = 22428 J = 22.428 kJ (e) Heat of displacement, ΔH Haba penyesaran, ΔH = 22.428 0.2 = –112.14 kJ mol–1 (f) Energy / Tenaga Zn2+(aq) + Pb(s) Zn2+(ak) + Pb(p) ΔH = –112.14 kJ mol–1 Zn(s) + Pb2+(aq) Zn(p) + Pb2+(ak) 5. (a) alkali, base, salt, water alkali, bes, garam, air (b) when 1 mole of water is formed through the neutralization reaction between acid solution and alkali solution apabila 1 mol air terbentuk melalui tindak balas peneutralan antara larutan asid dan larutan alkali 6. (a) HCl(aq) + KOH(aq) → KCl(aq) + H2O(l) HCl(ak) + KOH(ak) → KCl(ak) + H2O(ce) (b) H+(aq) + OH– (aq) → H2O(l) H+(ak) + OH– (ak) → H2O(ce) (c) Number of moles of HCl / Bilangan mol HCl = (1.0)(100) 1 000 = 0.1 mol Number of moles of KOH / Bilangan mol KOH = (1.0)(100) 1 000 = 0.1 mol (d) Heat change, Q / Perubahan haba, Q = mcθ = (100 + 100)(4.2)(6.5) = 5460 J = 5.46 kJ (e) Heat of neutralisation, ΔH Haba peneutralan, ΔH = 5.46 0.1 = –54.6 kJ mol–1 (f) Energy / Tenaga KCl(aq) + H2 O(l) HCl(ak) + KOH(ce) ΔH = –54.6 kJ mol–1 HCl(aq) + KOH(aq) HCl(ak) + KOH(ak) (g) Some heat is loss to the environment. Sebahagian haba terbebas ke persekitaran. 7. (a) carbon dioxide gas, water gas karbon dioksida, air (b) carbon monoxide gas, carbon, water. gas karbon monoksida, karbon, air (c) Alcohol, alkane, alkene Alkohol, alkana, alkena (d) 1 mole of fuel is burnt in excess oxygen gas to form carbon dioxide gas and water 1 mol bahan api terbakar dalam gas oksigen yang berlebihan untuk membentuk gas karbon dioksida dan air 8. (a) CH3OH(l) + 3 2O2(g) → CO2(g) + 2H2O(g) CH3OH(ce) + 3 2O2(g) → CO2(g) + 2H2O(g) (b) Mass of methanol / Jisim metanol = 54.87 – 53.75 = 1.12 g Number of moles of methanol / Bilangan mol metanol = 1.12 [12 + 3(1) + 16 + 1] = 0.035 mol Heat change, Q / Perubahan suhu, Q = mcθ = 500 × 4.2 × (40 – 29) = 23100 J = 23.1 kJ Heat of combustion, ΔH / Haba pembakaran, ΔH = 23.1 0.035 = –660 kJ mol–1 (c) Energy / Tenaga CO2 (g) + 2H2 O(g) CO2 (g) + 2H2 O(g) ΔH = –660 kJ mol–1 CH3 OH(l) + O2 (g) CH3 OH(ce) + O2 (g) 3 —2 —3 2 Application of Endothermic and Exothermic Reactions in Daily Life Aplikasi Tindak Balas Endotermik dan 3.3 Eksotermik dalam Kehidupan Harian 1. (a) Fuel / Bahan api (b) fuel value / nilai bahan api (c) the amount of heat energy released when 1 g of fuel is completely burnt in excess oxygen gas jumlah tenaga haba yang dibebaskan ketika 1 g bahan api terbakar dengan lengkap dalam gas oksigen yang berlebihan

Chemistry Form 5 Answers © Penerbitan Pelangi Sdn. Bhd. A8 2. (a) Molar mass / Jisim molar C4H10 = 58 g mol–1 Fuel value / Nilai bahan api = 2 880 58 = 49.66 kJ g–1 (b) Molar mass / Jisim molar CH4 = 16 g mol–1 Fuel value / Nilai bahan api = 52 × 16 = –832 kJ mol–1 (c) Molar mass / Jisim molar C2H5OH = 46 g mol–1 Fuel value / Nilai bahan api = 1 370 46 = 29.78 kJ g–1 (d) Molar mass / Jisim molar C3H8 = 44 g mol–1 Fuel value / Nilai bahan api = 51 × 44 = –2244 kJ mol–1 SPM Practice 3 Paper 1 1. A 2. C 3. C 4. C 5. B 6. B 7. D 8. C 9. C 10. A 11. A 12. C 13. D 14. A 15. B 16. A Paper 2 Section A / Bahagian A 1. (a) Pb2+(aq) + SO4 2–(aq) → PbSO4(s) Pb2+(ak) + SO4 2–(ak) → PbSO4(p) (b) Heat is released. / Haba dibebaskan. (c) Temperature of the reaction increases. / Suhu tindak balas meningkat. (d) – Exothermic / Eksotermik – Heat is released. / Haba dibebaskan. (e) Heat change, Q / Perubahan haba, Q = 50 × 0.5 = 25 kJ (f) Energy / Tenaga PbSO4 (s) PbSO4 (p) ΔH = –50 kJ mol–1 Pb2+(aq) + SO4 2–(aq) Pb2+(ak) + SO4 2–(ak) 2. (a) Mg(NO3)2(aq) + K2CO3(aq) → MgCO3(s) + 2KNO3(aq) Mg(NO3)2(ak) + K2CO3(ak) → MgCO3(p) + 2KNO3(ak) (b) Mg2+(aq) + CO3 2–(aq) → MgCO3(s) Mg2+(ak) + CO3 2–(ak) → MgCO3(p) (c) Average of initial temperature of Mg(NO3)2 and K2CO3 solutions Purata suhu awal larutan Mg(NO3)2 dan K2CO3 = (28.0 + 29.0) 2 = 28.5 o C Temperature change, θ / Perubahan suhu, θ = 28.5 – 16.5 = 12 o C Heat change, Q / Perubahan haba, Q = mcθ = (50 + 50)(4.2)(12) = 5040 J (d) Number of moles of Mg(NO3)2 / Bilangan mol Mg(NO3)2, n = (2.0)(50) 1 000 n = 0.1 mol Number of moles of K2CO3 / Bilangan mol K2CO3, n = (2.0)(50) 1 000 n = 0.1 mol Number of moles of MgCO3 formed / Bilangan mol MgCO3 yang terbentuk = Number of moles of Mg2+, CO3 2– / Bilangan mol Mg2+, CO3 2– = 0.1 mol Heat of precipitation of MgCO3 / Haba pemendakan MgCO3 = +50.4 kJ mol–1 (e) Mg(NO3)2(aq) + K2CO3(aq) → MgCO3(s) + 2KNO3(aq) ΔH = +50.4 kJ mol–1 // Mg(NO3)2(ak) + K2CO3(ak) → MgCO3(p) + 2KNO3(ak) ΔH = +50.4 kJ mol–1 // Mg2+(aq) + CO3 2–(aq) → MgCO3(s) ΔH = +50.4 kJ mol–1 Mg2+(ak) + CO3 2–(ak) → MgCO3(p) ΔH = +50.4 kJ mol–1 (f) Energy / Tenaga MgCO3 (s) MgCO3 (p) ΔH = +50.4 kJ mol–1 Mg2+(aq) + CO3 2–(aq) Mg2+(ak) + CO3 2–(ak) 3. (a) Heat change when 1 mole of water is formed from the neutralization between potassium hydroxide solution and hydrochloric acid. Perubahan haba apabila 1 mol air terbentuk daripada peneutralan antara larutan kalium hidroksida dengan asid hidroklorik. (b) (i) Final thermometer reading is higher than the initial thermometer reading. Bacaan termometer akhir adalah lebih tinggi daripada bacaan termometer awal. (ii) • Heat of neutralisation for Experiment II is higher / doubled the heat of neutralisation for Experiment I. Haba peneutralan bagi Eksperimen II lebih tinggi / dua kali ganda haba peneutralan bagi Eksperimen I. • Sulphuric acid is a strong diprotic acid that ionises completely in water to produce a higher / double the concentration of hydrogen ions. Asid sulfurik ialah asid diprotik kuat yang mengion sepenuhnya di dalam air untuk menghasilkan kepekatan ion hidrogen yang lebih tinggi / berganda. • More heat is produced when a higher number of moles of water is formed. Lebih banyak haba dihasilkan apabila bilangan mol air yang terbentuk lebih tinggi. (c) • Nitric acid / Asid nitrik • Nitric acid is a strong monoprotic acid that ionises completely in water to form a higher concentration of

Chemistry Form 5 Answers A9 © Penerbitan Pelangi Sdn. Bhd. hydrogen ions just like hydrochloric acid. Asid nitrik ialah asid monoprotik kuat yang mengion sepenuhnya di dalam air untuk membentuk kepekatan ion hidrogen yang lebih tinggi seperti asid hidroklorik. • The same number of moles of water formed as hydrochloric acid. Bilangan mol air yang sama terbentuk seperti asid hidroklorik. Section B / Bahagian B 4. (a) (i) • Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s) Fe(p) + Cu2+(ak) → Fe2+(ak) + Cu(p) • Blue copper(II) sulphate solution turns green. Larutan kuprum(II) sulfat yang berwarna biru bertukar menjadi hijau. • Concentration of Cu2+ ions decreases. Kepekatan ion Cu2+ menurun. • Concentration of Fe2+ ions increases. Kepekatan ion Fe2+ meningkat. • Brown solid – copper Pepejal perang – kuprum (ii) • Number of moles of copper formed / Bilangan mol kuprum yang terbentuk = 3.2 6.4 = 0.05 mol • Mole ratio between Cu and CuSO4 Bilangan mol antara Cu dengan CuSO4 = 1 : 1 = 0.05 : 0.05 • Heat change, Q / Perubahan haba, Q = mcθ = 100 × 4.2 × (57.5 – 28.5) = 12180 J = 12.18 kJ • Heat of displacement, ΔH / Haba penyesaran, ΔH = 12.18 0.05 = –243.6 kJ mol–1 (iii) Concentration / Kepekatan = 0.05 × 1 000 100 = 0.5 mol dm–3 (b) (i) • T1 is higher than 29.0 o C. T1 lebih tinggi daripada 29.0 o C. • Zinc is more electropositive than iron. Zink lebih elektropositif daripada besi. • More heat energy is produced at the end of the reaction. Lebih banyak tenaga haba dihasilkan pada akhir tindak balas. • T2 remains as 29.0 o C. T2 masih kekal pada 29.0 o C. • Silver is less electropositive than copper. Argentum kurang elektropositif daripada kuprum. • Silver cannot displace copper from copper(II) sulphate solution. Argentum tidak dapat menyesarkan kuprum daripada larutan kuprum(II) sulfat. (ii) • Magnesium / Magnesium • Blue copper(II) sulphate solution turns lighter blue / colourless. Larutan biru kuprum(II) sulfat bertukar kepada biru muda / tak berwarna. • Final thermometer reading is higher than the initial thermometer reading. Bacaan termometer akhir lebih tinggi daripada bacaan termometer awal. • Brown solid is formed. Pepejal perang terbentuk. Section C / Bahagian C 5. (a) • Correct scales, labels and unit in both axes. Skala, label dan unit yang betul pada kedua-dua paksi. • Transferring data correctly onto graph. Pemindahan data yang betul ke atas graf. • Shape of the graph. Bentuk graf. • Showing dotted line on the graph to determine the heat of combustion for x. Menunjukkan garis putus-putus pada graf untuk menentukan haba pembakaran bagi x. • x = –3350 kJ mol–1 1 2 3 4 5 6 4000 3000 2000 1000 0 Number of carbon atoms per alcohol molecule Bilangan atom karbon per molekul alkohol Heat of combustion (kJ mol–1) Haba pembakaran (kJ mol–1) x (b) Materials and apparatus / Bahan dan radas: Pentane, water, copper can, thermometers, spirit lamp, windshield, wooden block, tripod stand, electric weighing balance and 50 cm3 measuring cylinder Pentana, air, tin kuprum, termometer, lampu pelita, penghadang angin, blok kayu, tungku kaki tiga, penimbang elektrik dan silinder penyukat 50 cm3 Procedures / Prosedur: 1. Measure and pour 500 cm3 of water into a copper can and leave it for 5 minutes. Sukat dan tuangkan 500 cm3 air ke dalam tin kuprum dan biarkan selama 5 minit. 2. Place the copper can filled with water on a tripod stand. Then, insert a thermometer into the copper can to measure the initial thermometer reading and record as T1. Letakkan tin kuprum berisi air di atas tungku kaki tiga. Kemudian, masukkan termometer ke dalam tin kuprum untuk menyukat bacaan awal termometer dan catatkan sebagai T1. 3. Fill up a spirit lamp with pentane and measure its initial mass as m1. Isikan lampu pelita dengan pentana dan timbang jisim awalnya sebagai m1. 4. Place the spirit lamp containing pentane on a wooden block under the copper can. Put a windshield around the copper can and the spirit lamp. Letakkan lampu pelita berisi pentana di atas blok kayu di

Chemistry Form 5 Answers © Penerbitan Pelangi Sdn. Bhd. A10 bawah tin kuprum. Letakkan penghadang angin di sekeliling tin kuprum dan lampu pelita. 5. Light up the wick in the spirit lamp to heat water in the copper can and stir the water in the copper can throughout the experiment until the final thermometer reading achieves 70 o C. Nyalakan sumbu di dalam lampu pelita untuk memanaskan air di dalam tin kuprum dan kacau air di dalam tin kuprum sepanjang eksperimen sehingga bacaan termometer akhir mencapai 70 o C. 6. Put out the flame in the spirit lamp and measure the final mass of spirit lamp with pentane again as m2. Matikan nyalaan lampu spirit dan timbang jisim akhir lampu spirit bersama pentana sebagai m2. Calculation for heat of combustion / Penghitungan haba pembakaran: – Number of moles of pentane Bilangan mol petana = m1 – m2 72 – Heat change, Q / Perubahan suhu, Q = mcθ = 500 × c × (70 – T1) – Heat of combustion / Haba pembakaran = –[500 × c × (70 – T1)] 1000 (m1 – m2) 72 kJ mol–1 (c) (i) Number of moles of pentane Bilangan mol petana = 1.8 72 = 0.025 mol Heat change, Q / Perubahan suhu, Q = mcθ = 500 × 4.2 × (70 – 28) = 88200 J = 88.2 kJ – Heat of combustion / Haba pembakaran = 88.2 0.025 = –3528 kJ mol–1 (ii) Energy / Tenaga 5CO2 (g) + 6H2 O(g) 5CO2 (g) + 6H2 O(g) ΔH = –3528 kJ mol–1 C5 H12(l) + 8O2 (g) C5 H12(ce) + 8O2 (g) HOTS Challenge – Mak Minah’s selection – Palm oil Pilihan Mak Minah – Minyak kelapa sawit – Reason 1 – Palm oil has a higher fuel value that is able to produce more heat when 1 g of palm oil is burnt. Alasan 1 – Minyak kepala sawit memiliki nilai bahan api yang lebih tinggi yang boleh menghasilkan lebih banyak tenaga apabila 1 g minyak kelapa sawit terbakar. – Reason 2 – The cost price of palm oil is cheaper than olive oil. Alasan 2 – Harga minyak kelapa sawit lebih murah berbanding minyak zaitun. 4CHAPTER Polymer Chemistry Kimia Polimer Polymer 4.1 Polimer 1. natural, synthetic semula jadi, sintetik 2. plants, animals tumbuh-tumbuhan, haiwan 3. man-made, petroleum buatan manusia, petroleum 4. additional, condensation penambahan, kondensasi 5. long-chained molecules, monomers molekul berantai panjang, monomer 6. (a) Glucose / Glukosa (b) Natural / Semula jadi (c) Protein / Protein (d) Isoprene / Isoprena (e) Natural / Semula jadi (f) Polyvinyl chloride / Polivinil klorida (g) Styrene / Stirena (h) Synthetic / Sintetik 7. (a) Polypropylene Polipropilena (b) Food packaging Pembungkus makanan (c) Epoxy glue Gam epoksi (d) Waterproof jacket Jaket kalis air Natural Rubber 4.2 Getah Asli 1 2-methybut-1,3-diene 2-metilbut-1,3-diena 2. CH3 & H!C"C!C"CH2 & & H H 3. Bakteria, negative charges, protein membrane Bakteria, cas-cas negatif, membran protein 4. (a) methanoic acid / ethanoic acid / formic acid asid metanoik / asid etanoik / asid formik (b) ammonia solution larutan ammonia 5. heating the natural rubber with sulphur, crosslinks pemanasan getah asli dengan sulfur, taut silang 6. Vulcanised rubber Getah tervulkan Properties Sifat-sifat Unvulcanised rubber Getah tak tervulkan Harder Lebuh keras Hardness Kekerasan Softer Lebih lembut More elastic Lebih kenyal Elasticity Kekenyalan Less elastic Kurang kenyal More resistant to heat Lebih tahan haba Resistance to heat Daya tahan haba Less resistant to heat Kurang tahan haba More resistant to oxidation Lebih tahan terhadap pengoksidaan Resistance to oxidation Daya tahan pengoksidaan Less resistant to oxidation Kurang tahan terhadap pengoksidaan

Chemistry Form 5 Answers A11 © Penerbitan Pelangi Sdn. Bhd. Synthetic Rubber 4.3 Getah Sintetik 1. a type of artificial elastomer which is synthesised from petroleum byproducts sejenis elastomer tiruan yang disintesis daripada hasil sampingan petroleum 2. thermal stability, resistance to oils kestabilan terma, ketahanan terhadap minyak 3. styrene, buta-1,3-diene stirena, buta-1,3-diena 4. (a) Elastic / Kenyal (b) Less elastic / Kurang kenyal (c) Less / Kurang (d) More / Lebih (e) Easily / Mudah (f) Not easily / Tidak mudah SPM Practice 4 Paper 1 1. C 2. A 3. A 4. B 5. B 6. C 7. C 8. C 9. A 10. D 11. D 12. B 13. D 14. B 15. B 16. C 17. D 18. D Paper 2 Section A / Bahagian A 1. (a) (i) Both undergo additional polymerisation. Kedua-duanya menjalani pempolimeran penambahan. (ii) – Monomer A: 2-methybuta-1,3-diene Monomer A: 2-metilbuta1,3-diena – Monomer B: chloroethene / vinyl chloride Monomer B: kloroetena / vinil klorida (iii) – Natural polymer: natural rubber Polimer semula jadi: getah asli – Synthetic polymer: polychloroethane / polyvinyl chloride Polimer sintetik: polikloroetana / polivinil klorida (b) (i) – As a water pipe / Sebagai paip air – As an electrical wire case / Sebagai kotak wayar elektrik (ii) – Polychloroethane / Polyvinyl chloride (PVC) is the most environmental damaging polymer. Polikloroetana / polivinil klorida (PVC) ialah polimer yang paling merosakkan alam sekitar. – Disposal of PVC releases toxic chemicals and chlorine to the environment. Pembuangan PVC membebaskan bahan kimia beracun dan klorin kepada alam sekitar. 2. (a) Petroleum / Petroleum (b) (i) – Plastic A does not have a crosslink between polymer chains. Plastik A tidak mempunyai taut silang di antara rantai polimer. – Plastic B has crosslinks between polymer chains. Plastik B mempunyai taut silang di antara rantai polimer. (ii) – When thermoplastic is heated, polymer chains slide over one another causing the plastics to soften and melt. Apabila termoplastik dipanaskan, rantai polimer menggelongsor antara satu sama lain dan menyebabkan plastik menjadi lembut dan cair. – When thermoset is heated, the crosslinks prevent the polymer chains from sliding over each other. This will prevent the plastic from melting. Apabila termoplastik dipanaskan, taut silang menghalang rantai polimer daripada saling menggelongsor antara satu sama lain. Hal ini dapat mengelakkan plastik daripada menjadi cair. (c) (i) Characteristic Ciri-ciri Thermoplastic Termoplastik Thermoset Termoset Hardness Kekerasan Soft Lembut Hard Keras Ability to be moulded Keupayaan untuk dibentuk Can be moulded repeatedly Boleh dibentuk berulang kali Can only be moulded once Hanya boleh dibentuk sekali sahaja Effect on heat Kesan terhadap haba Melt when heated and harden again when cooled Cair apabila dipanaskan dan mengeras semula apabila disejukkan Do not melt when heated Tidak cair apabila dipanaskan (ii) – Plastic A: Thermoplastic Plastik A: Termoplastik – Plastic B: Thermoset Plastik B: Termoset Section B / Bahagian B 3. (a) (i) Polyethene / Polietena – Type: thermoplastic Jenis: termoplastik – Monomer: ethene Monomer: etena – Use: plastic bag Kegunaan: beg plastik SBR – Type: synthetic rubber Jenis: getah sintetik – Monomer: styrene and butadiene Monomer: stirena dan butadiena – Use: synthetic tyres Kegunaan: tayar sintetik (ii) Polyethene / Polietena – Recycle the polyethene by heating, melting and remoulding it into a new product. Kitar semula polietena dengan memanaskan, mencairkan dan membentuknya menjadi produk baru. – Reuse the polyethene. Guna semula polietena. SBR – Donate to Fisheries Department as building blocks for artificial reefs. Sumbangkan kepada Jabatan Perikanan sebagai binaan tukun tiruan. – Sell back to tyre manufacturers for recycling.

Chemistry Form 5 Answers © Penerbitan Pelangi Sdn. Bhd. A12 Menjual semula kepada pengeluar tayar untuk dikitar semula. (b) (i) – Plastics substitute woods and papers to prevent a massive deforestation. Plastik menggantikan kayu dan kertas untuk mengelakkan penebangan hutan secara berleluasa. – Plastics substitute metals to conserve natural resources from the core of the Earth. Plastik menggantikan logam untuk memulihara sumber semula jadi dari teras Bumi. – Plastics substitute some parts of the vehicles to make them lighter and improve fuel efficiency. Plastik menggantikan beberapa bahagian pada kenderaan supaya lebih ringan dan memperbaiki kecekapan bahan api. (ii) – Clogged drains causing flash flood. Longkang yang tersumbat menyebabkan banjir kilat. – Marine animals like turtles mistake the plastics for food causing them to die of food poisoning. Haiwan laut seperti penyu menganggap plastik sebagai makanan dan menyebabkannya mati akibat keracunan makanan. – Clogged soil prevents the free flow of water in soil, thus depleting the soil fertility. Tanah yang tersumbat mencegah pengaliran air yang lancar di dalam tanah dan mengakibatkan kesuburan tanah berkurang. – Production of toxic gases during open burning. Penghasilan gas toksik semasa pembakaran terbuka. (iii) Recycle / Kitar semula – Return the aluminium cans to manufacturers to reprocess into new aluminium cans. Kembalikan tin aluminium kepada pengilang untuk diproses semula menjadi tin aluminium baru. Reuse / Guna semula – Reuse the plastic bags for shopping. Gunakan semula beg plastik untuk membelibelah. Reduce / Kurangkan • Reduce the usage of seedling plastic bags and substitute with biodegradable seedling bags. Kurangkan penggunaan beg plastik semaian dan gantikan dengan beg semaian terbiodegradasi. Section C / Bahagian C 4. (a) Experiment I / Eksperimen I – Ethanoic acid contains hydrogen ions. Asid etanoik mengandungi ion hidrogen. – Hydrogen ions neutralise negative charges on the protein membrane surface of rubber particles. Ion hidrogen meneutralkan cas negatif pada permukaan membran protein zarah getah. – Neutral rubber particles are formed. Zarah getah yang neutral terbentuk. – When neutral particles collide among each other, protein membranes break. Apabila zarah neutral berlanggaran antara satu sama lain, membran protein pecah. – Rubber polymers are released, entangled and coagulated to form a white solid. Polimer getah terbebas, bersimpul dan bergumpal untuk membentuk pepejal putih. Experiment II / Eksperimen II – Ammonia solution contains hydroxide ions. Larutan ammonia mengandungi ion hidroksida. – Hydroxide ions neutralise hydrogen ions produced by bacteria. Ion hidroksida meneutralkan ion hidrogen yang dihasilkan oleh bakteria. – Negative charges remain on the protein membrane surface. Cas negatif kekal pada permukaan membran protein. – Repulsive force prevents the collision among rubber particles. Daya tolakan menghalang perlanggaran antara zarah getah. – Protein membranes do not break, rubber polymers are not released, no coagulation occurs and white liquid remains white. Membran protein tidak pecah, molekul getah tidak terbebas, penggumpalan tidak berlaku dan cecair putih kekal putih. (b) Material / Bahan: – 5 cm rubber strip, disulphur dichloride in methylbenzene Jalur getah 5 cm, disulfur diklorida dalam metilbenzena Apparatus / Radas: – Bulldog clips, beakers, retort stand and clamp, 50 g weight and meter ruler Klip bulldog, bikar, kaki retort dan pengapit, pemberat 50 g dan pembaris meter Procedure / Prosedur: 1. Dip a 5 cm of rubber strip into disulphur dichloride in methylbenzene for 5 minutes to produce a vulcanised rubber. Celupkan jalur getah 5 cm ke dalam disulfur diklorida dalam metilbenzena selama 5 minit untuk menghasilkan getah tervulkan. 2. Hang the vulcanised rubber strip using a retort stand and a clamp. Gantung jalur getah tervulkan dengan menggunakan kaki retort dan pengapit. 3. Measure and record the initial length of the rubber strip. Ukur dan catatkan panjang awal jalur getah itu. 4. Add 50 g weight to the rubber strip and leave it for 3 hours. Tambahkan pemberat 50 g pada jalur getah dan biarkan selama 3 jam. 5. Remove the 50 g weight. Then, measure and record the final length of the rubber strip. Alihkan pemberat 50 g kemudian ukur dan catatkan panjang akhir jalur getah. 6. Repeat step 1-5 with unvulcanised rubber. Ulangi langkah 1-5 menggunakan getah tak tervulkan.

Chemistry Form 5 Answers A13 © Penerbitan Pelangi Sdn. Bhd. Tabulation of data / Penjadualan data: Type of rubber strip Jenis jalur getah Initial length of rubber strip (cm) Panjang awal jalur getah (cm) Final length of rubber strip (cm) Panjang akhir jalur getah (cm) Length of rubber strip after removal of 50 g weight Panjang jalur getah selepas pemberat 50 g dialihkan Unvulcanised rubber Getah tak tervulkan 5.0 10.2 5.2 Vulcanised rubber Getah tervulkan 5.0 10.0 5.0 HOTS Challenge – Dip the latex gloves into disulphur dichloride in methylbenzene. Celupkan sarung tangan getah ke dalam disulfur diklorida dalam metilbenzena. – Vulcanisation of latex gloves will occur. Pemvulkanan sarung tangan lateks akan berlaku. – Texture of the latex gloves will become stronger and more elastic. Tekstur sarung tangan getah akan menjadi lebih kuat dan lebih kenyal. 5CHAPTER Consumer and Industrial Chemistry Kimia Konsumer dan Industri Oils and Fats 5.1 Minyak dan Lemak 1. carbon, hydrogen, oxygen karbon, hidrogen, oksigen 2. carbon-carbon double bond ikatan karbon ganda dua 3. unsaturated, carbon-carbon double bond tak tepu, ikatan karbon ganda dua 4. solid, liquid pepejal, cecair 5. Hydrogenation, unsaturated, saturated penghidrogenan, tak tepu, tepu 6. (a) Energy supply to the body Bekalan tenaga kepada badan (b) Supports cell growth Menyokong pertumbuhan sel (c) Protects internal organs Melindungi organ dalaman (d) Keeps the body warm Memanaskan badan (e) Absorbs of nutrients and vitamins A,D,E, K Menyerap nutrien dan vitamin A, D, E, K (f) Produces hormones Menghasilkan hormon 7. (a) Heart diseases / Penyakit jantung (b) Stroke / Strok (c) Weight gain / Obesity Pertambahan berat badan / Kegemukan (d) Hypertension / Tekanan darah tinggi 8. renewable, biodegradable, non-toxic diperbaharui, boleh terbiodegradasi, tidak beracun Cleaning Agents 5.2 Bahan Pencuci 1. (a) long-chain fatty acid, ROOK+ asid lemak berantai panjang, ROO– K+ (b) alkyl / alkil (c) animal fats, vegetable oils lemak haiwan, minyak sayuran (d) saponification / saponifikasi 2. (a) vegetable / olive, sodium hydroxide / potassium hydroxide sayur / zaitun, natrium hidroksida / kalium hidroksida (b) heat / panaskan (c) room temperature / suhu bilik (d) sodium chloride / natrium klorida (e) soap / sabun 3. (a) petroleum fractions / pecahan petroleum (b) (i) Sodium alkyl sulphate detergent Detergen natrium alkil sulfat (ii) Sodium alkylbenzene sulphonate detergent Detergen natrium alkilbenzena sulfonat 4. (a) (i) Hydrophobic / Hidrofobik (ii) Hydrophilic / Hidrofilik (b) (i) Hydrophobic / Hidrofobik (ii) Hydrophilic / Hidrofilik 5. Soap Sabun Comparison Perbandingan Detergent Detergen Plants and animals Tumbuhan dan haiwan Source Sumber Petroleum fraction Pecahan petroleum Not effective Tidak berkesan Effectiveness in hard water Keberkesanan dalam air liat Effective Berkesan Biodegradable Terbiodegradasi Biodegradability Kebolehan terbiodegradasi Non-biodegradable Tidak terbiodegradasi Food Additives 5.3 Bahan Tambah Makanan 1. synthetic, spoilage, appearance / sintetik, kerosakan, rupa 2. Preservatives Pengawet Prevent the oxidation that causes rancidity of oils and fats and fruit browning Menghalang pengoksidaan yang menyebabkan minyak dan lemak berbau tengik dan buah bertukar menjadi warna perang Antioxidants Pengantioksida Prevent an emulsion from separating out Mengelakkan pemisahan emulsi Flavours Perisa Stabilise an emulsion Menstabilkan emulsi Stabilisers Penstabil Improve or restore the taste loss due to processing Memperbaiki atau memulihkan rasa yang hilang disebabkan oleh pemprosesan Dyes Pewarna Improve or restore the colour loss due to processing Memperbaiki atau memulihkan warna yang hilang disebabkan oleh pemprosesan Thickeners Pemekat Slow down the bacterial growth Melambatkan pertumbuhan bakteria Emulsifiers Pengemulsi Thicken the food texture Memekatkan tekstur makanan

Chemistry Form 5 Answers © Penerbitan Pelangi Sdn. Bhd. A14 Medicines and Cosmetics 5.4 Ubat-ubatan dan Bahan Kosmetik 1. (a) pain / kesakitan (b) bacterial growth / pertumbuhan bakteria (c) thinking, behaviours / pemikiran, tingkah laku (d) allergic effects / kesan alergi (e) inflammation / keradangan 2. Treat skin problems or wounds Merawat masalah kulit atau luka Treat bloated stomach Merawat perut kembung Reduce high blood pressure Mengurangkan tekanan darah tinggi Treat inflammation Merawat keradangan Build up stamina and reduce fatigue Membina stamina dan mengurangkan keletihan 3. Example Contoh Type of cosmetic Jenis kosmetik Function Fungsi Make-up cosmetic Kosmetik rias Products to treats body or face Produk untuk merawat tubuh atau muka Fragrance Pewangi Products to remove body odour. Produk untuk mewangikan badan. Treatment cosmetic Kosmetik perawatan Products to beautify the face. Produk hiasan untuk wajah. Application of Nanotechnology in Industry 5.5 Aplikasi Nanoteknologi dalam Industri 1. 1-100 / 1-100 2. very small scale skala yang sangat kecil 3. carbon, two-dimensional honeycomb lattice karbon, kekisi sarang lebah dua dimensi 4. (a) ✓ (b) ✓ (c) ✓ Application of Green Technology in Industrial Waste Manangement 5.6 Aplikasi Teknologi Hijau dalam Pengurusan Sisa Industri 1. Green Technology is defined as the development and application of products, equipment and systems to preserve the environment and nature as well as to minimise the negative effects of human activities. Teknologi Hijau didefinisikan sebagai pembangunan dan aplikasi produk, peralatan serta sistem untuk memelihara alam sekitar dan alam semula jadi serta meminimumkan kesan negatif akibat aktiviti manusia. 2. human waste, soap sisa manusia, sabun 3. Reduce, Reuse and Recycle, Restore Kurangkan, Guna semula, Kitar semula dan Pulihkan 4. Type of wastewater Jenis air sisa Example Contoh Domestic wastewater Air sisa domestik Industrial wastewater Air sisa daripada perindustrian Stormwater runoff Air larian ribut SPM Practice 5 Paper 1 1. B 2. A 3. B 4. C 5. C 6. D 7. A 8. C 9. C 10. B 11. D 12. C 13. C 14. B 15. C 16. C 17. A 18. B 19. A 20. B 21. A Paper 2 Section A / Bahagian A 1. (a) (i) – Soft water: tap water Air lembut: air paip – Hard water: well water Air liat: air perigi (ii) A – Soap is an effective cleansing agent in soft water. Sabun ialah agen pembersih yang berkesan dalam air lembut. B – Soap is not an effective cleansing agent in hard water. Sabun bukan agen pembersih yang berkesan dalam air liat. (b) – Hydrophobic part of soap dissolves in the oil while hydrophilic part of soap dissolves in water. Bahagian hidrofobik sabun larut di dalam minyak manakala bahagian hidrofilik sabun larut di dalam air.

Chemistry Form 5 Answers A15 © Penerbitan Pelangi Sdn. Bhd. – Agitation breaks the oil into smaller oil droplets. Kocakan memecahkan minyak menjadi titisan minyak yang lebih kecil. – Oil droplets are suspended and rinsed off easily. Titisan minyak diapungkan dan dibilas dengan mudah. 2. (a) Sunburn / Selaran matahari (b) (i) Aloe vera / Lidah buaya (ii) Leave / Daun (iii) – As a moisturiser / Sebagai pelembap – Promotes hair growth / Menggalakkan pertumbuhan rambut (c) – Make use of natural products / Memanfaatkan produk semula jadi Has less side effects / Mempunyai kurang kesan sampingan 3. (a) (i) Substance of the same element that exists in more than one form of arrangements. Bahan bagi unsur yang sama wujud dalam keadaan lebih daripada satu bentuk susunan. (ii) Diamond and graphite / Berlian dan grafit (b) (i) – To produce semiconductor Untuk menghasilkan semikonduktor – To produce grapheneinfused carbon fibre helmet Untuk menghasilkan helmet gentian karbon grafen (ii) – Best electrical conductor / Konduktor elektrik terbaik – Thin but strong / Nipis tetapi kuat [any one answer / manamana satu jawapan] Section B / Bahagian B 4. (a) (i) – When the soap molecules dissolve in water, Apabila molekul sabun larut dalam air – the surface tension of water reduces / the ability of water to wet the handkerchief increases. ketegangan permukaan air berkurang / keupayaan air membasahi sapu tangan meningkat. (b) (i) – P: Preservative / Pengawet Q: Preservative / Pengawet R: Preservative / Pengawet – P is sugar. / P ialah gula. – that draws water out of the microorganisms’ cells to retard the growth of microorganisms. yang menyingkirkan air daripada sel mikroorganisma untuk merencatkan pertumbuhan mikroorganisma. – Q is vinegar. / Q ialah cuka. – which is acidic and inhibits the growth of microorganisms. yang bersifat asid dan menghalang pertumbuhan mikroorganisma. – R is sodium nitrite / nitrate R ialah natrium nitrik / nitrat – that slows down the growth of microorganisms. yang melambatkan pertumbuhan mikroorganisma. (ii) May cause cancer / asthma / allergies / hyperactive Boleh menyebabkan kanser / asma / alahan / hiperaktif [any one answer / manamana satu jawapan] Section C / Bahagian C 5. (a) (i) Plastic pollution / Pencemaran plastik (ii) – Formation of breeding sites for Aedes mosquitoes Pembentukan tapak pembiakan nyamuk Aedes – Open burning of the plastic bottles produces toxic gases Pembakaran botol plastik secara terbuka menghasilkan gas-gas beracun (iii) – Reduce / Kurangkan – Reduce the purchase of materials / foods / drinks in plastic bottles / containers / packagings. Consumers are encouraged to bring – The hydrophilic part of soap dissolves in water / The hydrophilic part of soap is attracted to water molecules. Bahagian hidrofilik sabun larut di dalam air / Bahagian hidrofilik sabun tertarik kepada molekul air. – The hydrophobic part of soap dissolves in grease. Bahagian hidrofobik sabun larut di dalam gris. – Mechanical agitation helps to pull the grease free and Kocakan mekanikal membantu menarik keluar gris dan – break the grease into smaller droplets. memecahkan minyak menjadi titisan yang lebih kecil. – The grease droplets do not coagulate and redeposit on the surface of the handkerchief due to the repulsion between negative charges on the surface. Titisan minyak tidak bergumpal dan terkumpul pada permukaan sapu tangan kerana tolakan antara cas negatif pada permukaannya. – These droplets are suspended in water forming an emulsion. Titisan ini terapung dalam air membentuk emulsi. – Rinse away these droplets and the handkerchief is clean. Bilaskan titisan ini dan sapu tangan menjadi bersih. (ii) – Soap is biodegradable. / Soap can be decomposed naturally by bacteria or microorganisms. Sabun terbiodegradasi. / Sabun boleh terurai secara semula jadi oleh bakteria atau mikroorganisma. – Soap does not cause pollution. / Soap is environmental friendly. Sabun tidak menyebabkan pencemaran. / Sabun bersifat mesra alam.

Chemistry Form 5 Answers © Penerbitan Pelangi Sdn. Bhd. A16 their own bottles / containers when they are going to the restaurants / shops. Kurangkan pembelian bahan / makanan / minuman di dalam botol plastik / bekas / pembungkus. Pengguna digalakkan membawa botol / bekas sendiri semasa berkunjung ke restoran / kedai. – Reuse / Guna semula Consumers can use plastic containers to keep snack / plastic bottles for sauce / oil / plastic bag as dustbin liners. Pengguna boleh menggunakan bekas plastik untuk menyimpan makanan ringan / botol plastik untuk kuah / minyak / beg plastik sampah. – Recycle / Kitar semula Consumers are encouraged to return / resell used plastic bottles / containers to manufacturers to reprocess into new products. Pengguna digalakkan untuk mengembalikan / menjual kembali botol / bekas plastik terpakai kepada pengeluar untuk diproses semula menjadi produk baru. – Restore / Pulihkan Plastic manufacturers restore the used plastic bottles / containers by converting them into fuels / industrial products instead of piling them up on landfills. Pengilang plastik menukarkan botol / bekas plastik terpakai kepada bahan bakar / produk industri daripada mengakibatkan timbunan di tempat pembuangan sampah. (b) (i) Wastewater is water that has been used including human wastes, food wastes, soaps, oils and chemical wastes. Air sisa ialah air yang telah digunakan termasuklah sisa manusia, sisa makanan, sabun, minyak dan bahan kimia. (ii) – Domestic wastewater Air sisa domestik – Industrial wastewater Air sisa daripada perindustrian – Stormwater runoff Air larian ribut (iii) – Produces alternative energy through methane gas collection Menghasilkan tenaga alternatif melalui pengumpulan gas metana – Produces clean and reusable water Menghasilkan air yang bersih dan boleh digunakan semula – Produces natural fertilisers Menghasilkan baja semula jadi (iv) – Increases in waterborne diseases like cholera or typhoid fever. Peningkatan penyakit bawaan air seperti kolera atau demam kepialu. – Water source contamination by E. coli bacteria that causes diarrhea / cramping stomach pain / vomiting. Pencemaran sumber air oleh bakteria E. coli yang menyebabkan cirit-birit / sakit perut / muntah. HOTS Challenge – Mineral supplement contains nanosize calcium ions and magnesium ions. Mineral tambahan mengandungi ion kalsium dan ion magnesium bersaiz nano. – This allows a more efficient calcium ions and magnesium ions absorption into the bloodstream due to their nano-size. Membenarkan penyerapan ion kalsium dan ion magnesium dengan lebih berkesan ke dalam saluran darah kerana bersaiz nano. – Enrichment of magnesium helps building up stronger bones for a better absorption of calcium ions. Kaya dengan magnesium membantu membina tulang yang lebih kuat untuk penyerapan ion kalsium yang lebih baik. SPM Model Paper Paper 1 1. C 2. A 3. B 4. C 5. A 6. A 7. A 8. D 9. A 10. A 11. D 12. D 13. B 14. D 15. D 16. D 17. D 18. B 19. D 20. C 21. B 22. D 23. B 24. A 25. C 26. D 27. B 28. C 29. B 30. B 31. B 32. C 33. B 34. B 35. A 36. A 37. D 38. C 39. B 40. C Paper 2 Section A / Bahagian A 1. (a) (i) Pure metal / Logam tulen: J Alloy / Aloi: K (ii) Substance J is ductile. Bahan J bersifat mulur. (b) – The presence of zinc atoms in different sizes disrupt the orderly arrangement of copper atoms. Kehadiran atom zink yang berbeza saiz mengganggu susunan atom kuprum yang teratur. – Minimise the layers of copper atoms from sliding. Meminimumkan lapisan atom kuprum daripada menggelongsor. 2. (a) Exothermic reaction / Tindak balas eksotermik (b) Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s) Zn(p) + CuSO4(ak) → ZnSO4(ak) + Cu(p) (c) (i) Number of moles / Bilangan mol = 0.2 × 50 1 000 = 0.01 mol (ii) Heat released / Haba terbebas = 50 × 4.2 × (45.0 – 28.0) = 3570 J = 3.57 kJ (iii) Heat of displacement of copper Haba penyesaran kuprum = 3.57 0.01 = 3.57 kJ mol–1 (d) – Heat of displacement is higher than –357 kJ mol–1. Haba sesaran lebih tinggi daripada –357 kJ mol–1. – Magnesium is more electropositive than zinc. Magnesium lebih elektropositif daripada zink. 3. (a) Aluminium / Aluminium (b) (i) R

Chemistry Form 5 Answers A17 © Penerbitan Pelangi Sdn. Bhd. (ii) – Has more than one oxidation number Mempunyai lebih daripada satu nombor pengoksidaan – Can form a coloured ion / compound Boleh membentuk ion / sebatian berwarna – Can form a complex ion / compound Boleh membentuk ion / sebatian kompleks – Can act as a catalyst Boleh bertindak sebagai mangkin [any one answer / mana-mana satu jawapan] (c) 2.8.1 (d) Q– (e) 2Y(s) + Q2(g) → 2YQ(s) 2Y(p) + Q2(g) → 2YQ(p) 4. (a) (i) 2.8.1 (ii) Group 1, Period 3 Kumpulan 1, Kala 3 (iii) – Group 1 because it has one valence electron. Kumpulan 1 kerana mempunyai satu elektron valens. – Period 3 because it has three electron-filled shells. Kala 3 kerana mempunyai tiga petala berisi elektron. (b) (i) 24 (ii) 2.8 (iii) 5. (a) 3 (b) Natural abundance is the percentage of isotopes presents in a natural sample of an element. Kelimpahan semula jadi ialah peratusan isotop yang hadir dalam suatu sampel semula jadi unsur. (c) 24 12X – 82.8%, 54 12X – 8.1%, 26 12X – 9.1% (d) 24 12X, 25 12X, 26 12X (e) The relative atomic mass of element X Jisim atom relatif unsur X = [(24 × 82.8) + (25 × 8.1) + (26 × 9.1)] 1 000 = 24.263 6. (a) Volume of gas (cm3 ) Isi padu gas (cm3 ) Time (minute) Masa (minit) 3 Set II / Set II Set I / Set I 50.00 40.00 (b) (i) Set I / Set I = 40 3 = 13.33 cm3 min–1 (ii) Set II / Set II = 50 3 = 16.67 cm3 min–1 (iii) Rate of reaction for Set II is higher than Set I. Kadar tindak balas bagi Set II lebih tinggi daripada Set I. (iv) – Catalyst provides an alternative pathway with lower activation energy. Mangkin menyediakan lintasan alternatif dengan tenaga pengaktifan yang lebih rendah. – Frequency of collision between zinc atoms and hydrogen ions is higher. Frekuensi perlanggaran antara atom zink dengan ion hidrogen lebih tinggi. – Frequency of effective collision between zinc atoms and hydrogen ions is higher. Frekuensi perlanggaran berkesan antara atom zink dengan ion hidrogen lebih tinggi. 7. (a) (i) Anode: Carbon that is immersed in iron(II) sulphate solution. Anod: Karbon yang dicelup di dalam larutan ferum(II) sulfat. Cathode: Carbon that is immersed in bromine water. Katod: Karbon yang dicelup di dalam air bromin. (ii) Anode / Anod: Fe2+ → Fe3+ + e– Cathode / Katod: Br2 + 2e– → 2Br– (b) Oxidising agent / Agen pengoksidaan: Bromine water / Air bromin Reducing agent / Agen penurunan: Iron(II) ion / Ion ferum(II) (c) 2Fe2+ + Br2 → 2Fe3+ + 2Br– 8. (a) (i) Cell A / Sel A: Voltaic cell / Sel kimia Cell B / Sel B: Electrolytic cell / Sel elektrolisis (ii) Electrical energy to chemical energy Tenaga elektrik kepada tenaga kimia (b) (i) Cell A – Blue copper(II) sulphate solution turns light blue / colorless. Sel A – Larutan biru kuprum(II) sulfat bertukar menjadi biru muda / tak berwarna. Cell B – Blue copper(II) sulphate solution remains blue. Sel B – Larutan biru kuprum(II) sulfat kekal biru. (ii) Cell A – Concentration of Cu2+ ion decreases. Sel A – Kepekatan ion Cu2+ berkurang. Cell B – Concentration of Cu2+ ion remains the same. Sel B – Kepekatan ion Cu2+ kekal sama. (c) (i) Replace zinc with magnesium / aluminium. Gantikan zink menjadi magnesium / aluminium. (ii) Magnesium / aluminium is more electropositive than zinc. Magnesium / aluminium lebih elektropositif daripada zink. Section B / Bahagian B 9. (a) 2H+(aq) + Fe(s) → Fe2+(aq) + H2(g) 2H+(ak) + Fe(p) → Fe2+(ak) + H2(g) (b) Experiment 1 – When powdered iron is used, total surface area exposed to hydrogen ions is larger. Eksperimen 1 – Apabila serbuk ferum digunakan, jumlah luas permukaan yang terdedah kepada ion hidrogen lebih besar. Experiment 2 – When HNO3 is heated to 50 °C, the kinetic energy of hydrogen ion is higher. Eksperimen 2 – Apabila HNO3 dipanaskan sehingga 50 °C, tenaga kinetik ion hidrogen lebih tinggi. P 2+

Chemistry Form 5 Answers © Penerbitan Pelangi Sdn. Bhd. A18 Experiment 3 – When CuSO4 is added as a catalyst, an alternative pathway with lower activation energy is provided. Eksperimen 3 – Apabila CuSO4 ditambahkan sebagai mangkin, lintasan alternatif dengan tenaga pengaktifan yang lebih rendah disediakan. (c) (i) In Experiment 2 / Dalam Eksperimen 2, – Rate of reaction is higher. Kadar tindak balas lebih tinggi. – Temperature of nitric acid, HNO3 is higher. Suhu asid nitrik, HNO3 lebih tinggi. – Kinetic energy of hydrogen ion is higher. Tenaga kinetik ion hidrogen lebih tinggi. – Frequency of collision between iron atoms and hydrogen ions is higher. Frekuensi perlanggaran antara atom ferum dengan ion hidrogen lebih tinggi. – Frequency of effective collision between iron atoms and hydrogen ions is higher. Frekuensi perlanggaran berkesan antara atom ferum dengan ion hidrogen lebih tinggi. (ii) In Experiment 3 / Dalam Eksperimen 3, – Rate of reaction is higher. Kadar tindak balas lebih tinggi. – Copper(II) sulphate solution acts as a catalyst. Larutan kuprum(II) sulfat bertindak sebagai mangkin. – Catalyst provides an alternative pathway with lower activation energy. Mangkin menyediakan lintasan alternatif yang mempunyai tenaga pengaktifan yang lebih rendah. – More iron atoms and hydrogen ions achieve the activation energy. Lebih banyak atom ferum dan ion hidrogen mencapai tenaga pengaktifan. – Frequency of collision between iron atoms and hydrogen ions is higher. Frekuensi perlanggaran antara atom ferum dengan ion hidrogen lebih tinggi. – Frequency of effective collision between iron atoms and hydrogen ions is higher. Frekuensi perlanggaran berkesan antara atom ferum dengan ion hidrogen lebih tinggi. (d) Volume of H2 (cm3 ) Isi padu H2 (cm3 ) Time taken (s) Masa yang diambil (s) 2 1 Volume of H2 (cm3 ) Isi padu H2 (cm3 ) Time taken (s) Masa yang diambil (s) 3 2 10. (a) (i) – Process I is fermentation. Proses I ialah penapaian. – Process II is dehydration. Proses II ialah pendehidratan. – Chemical equation for process I: Persamaan kimia untuk proses I: C6H12O6(l) yeast C2H5OH(l) + 2H2O(l) C6H12O6(ce) yis C2H5OH(ce) + 2H2O(ce) – Chemical equation for process II: Persamaan kimia untuk proses II: C2H5OH(l) heated Al2O3 CH2=CH2(g) + H2O(g) C2H5OH(ce) Al2O3 dipanaskan CH2=CH2(g) + H2O(g) (ii) 1. Soak a glass wool in ethanol and insert it into the end of a combustion tube. Rendamkan wul kaca dalam etanol dan masukkan ke dalam hujung tiub pembakaran. 2. Clamp the combustion tube horizontally using a retort stand and clamp. Pasangkan tiub pembakaran secara mendatar dengan menggunakan kaki retort dan pengapit. 3. Put aluminium oxide in the centre of the combustion tube and close the combustion tube with a rubber stopper connected to a delivery tube. Letakkan aluminium oksida di dalam bahagian tengah tiub pembakaran dan tutup tiub pembakaran dengan penyumbat getah yang disambungkan ke salur penghantar. 4. Insert the other end of the delivery tube into an inverted test tube. Masukkan hujung salur penghantar yang lain ke dalam tabung uji yang diterbalikkan. 5. Heat the aluminium oxide strongly until brick red and gently heat the glass wool with ethanol. Panaskan aluminium oksida dengan kuat sehingga merah bata dan panaskan wul kaca dengan etanol secara perlahan. 6. Collect the colourless bubbles via inverted water displacement. Kumpulkan gelembunggelembung tak berwarna melalui penyesaran air berbalik. 7. Add a few drops of bromine water into the test tube containing the colorless gas. Tambahkan beberapa titis air bromin ke dalam tabung uji yang mengandungi gas tak berwarna. 8. Brown bromine water that turns colourless indicates the presence of ethene gas.

Chemistry Form 5 Answers A19 © Penerbitan Pelangi Sdn. Bhd. Air bromin berwarna perang yang bertukar menjadi tak berwarna menunjukkan kehadiran gas etena. (b) (i) – Bacteria / microorganisms present in the air. Bakteria / mikroorganisma hadir dalam udara. – Bacteria / microorganisms in the air produce hydrogen ions that neutralise the negative charges around the protein membranes of rubber molecules. Bakteria / mikroorganisma dalam udara menghasilkan ion hidrogen yang meneutralkan cas negatif di sekitar membran protein molekul getah. – Neutral rubber molecules collide among one another and break the protein membranes. Molekul getah neutral berlanggar antara satu sama lain dan memecahkan membran protein. – Rubber polymers are released and entangled causing the formation of the white solid. Polimer getah terbebas dan bersimpul menyebabkan pembentukan pepejal putih. (ii) – Ammonia solution Larutan ammonia – Hydroxide ions from ammonia solution neutralise the hydrogen ions produced by the bacteria / microorganisms. Ion hidroksida daripada larutan ammonia meneutralkan ion hidrogen yang dihasilkan oleh bakteria / mikroorganisma. Section C / Bahagian C 11. (a) – Barium ion reacts with carbonate ion to form white precipitate, barium carbonate. Ion barium bertindak balas dengan ion karbonat untuk membentuk mendakan putih, barium karbonat. – Ba2+(aq) + CO3 2–(aq) → BaCO3(s) Ba2+(ak) + CO3 2–(ak) → BaCO3(p) – Barium ion reacts with sulphate ion to form white precipitate, barium sulphate. Ion barium bertindak balas dengan ion sulfat untuk membentuk mendakan putih, barium sulfat. – Ba2+(aq) + SO4 2–(aq) → BaSO4(s) Ba2+(ak) + SO4 2–(ak) → BaSO4(p) Changes that should be done, Perubahan yang harus dilakukan, 1. Measure and pour 1 cm3 of potassium carbonate solution and potassium sulphate solution into two different test tubes. Sukat dan tuangkan 1 cm3 larutan kalium karbonat dan larutan kalium sulfat ke dalam dua tabung uji yang berbeza. 2. Add 1 cm3 of nitric acid into both test tubes. Tambahkan 1 cm3 asid nitrik ke dalam kedua-dua tabung uji. 3. Add 1 cm3 of barium chloride solution into both test tubes. Tambahkan 1 cm3 larutan barium klorida ke dalam keduadua tabung uji. 4. White precipitate is formed in the test tube containing potassium sulphate solution. Mendakan putih terbentuk di dalam tabung uji yang berisi larutan kalium sulfat. 5. Colourless solution is formed in the test tube containing potassium carbonate solution. Larutan tak berwarna terbentuk di dalam tabung uji yang berisi larutan kalium karbonat. 6. Barium carbonate formed is soluble in nitric acid. Barium karbonat yang terbentuk larut dalam asid nitrik. (b) (i) 1. Measure and pour 1 cm3 of both solutions into two test tubes separately. Sukat dan tuangkan 1 cm3 kedua-dua larutan tersebut dalam dua tabung uji yang berasingan. 2. Add 1 cm3 of nitric acid into both test tubes. Tambahkan 1 cm3 asid nitrik ke dalam kedua-dua tabung uji. 3. Add 1 cm3 silver nitrate solution into both test tubes. Tambahkan 1 cm3 larutan argentum nitrat ke dalam kedua-dua tabung uji. 4. White precipitate is formed in the test tube containing sodium chloride solution. Mendakan putih terbentuk di dalam tabung uji yang berisi larutan natirum klorida. 5. Yellow precipitate is formed in the test tube containing sodium iodide solution. Mendakan kuning terbentuk di dalam tabung uji yang berisi larutan natrium iodida. (ii) 1. Measure and pour 1 cm3 of both solutions into two test tubes separately. Sukat dan tuangkan 1 cm3 kedua-dua larutan tersebut dalam dua tabung uji yang berasingan. 2. Add 1 cm3 of sodium hydroxide solution into both test tubes. Then, heat the mixture solution. Tambahkan 1 cm3 larutan natrium hidroksida ke dalam kedua-dua tabung uji. Kemudian, panaskan larutan campuran tersebut. 3. Test tube with pungent smell gas produced contains ammonium sulphate solution. Tabung uji dengan gas berbau sengit yang terhasil mengandungi larutan ammonium sulfat. 4. A moist red litmus paper is inserted into the test tube containing the pungent smell gas. Moist red litmus turns blue. Kertas litmus merah lembap dimasukkan ke dalam tabung uji yang berisi gas berbau sengit itu. Kertas litmus merah lembap bertukar menjadi biru. 5. Test tube with no observable change contains sodium sulphate solution. Tabung uji tanpa perubahan yang kelihatan mengandungi larutan natrium sulfat.

© Penerbitan Pelangi Sdn. Bhd. A20Transition metals Period 1 HHydrogen 1 1234567 1 HHydrogen 1 3 Li Lithium 7 4 Be Beryllium 9 11 Na Sodium 23 12 Mg Magnesium 24 19 KPotassium 39 20 Ca Calsium 40 37 Rb Rubidium 85.5 38 Sr Strontium 88 55 Cs Cesium 133 56 Ba Barium 137 87 Fr Francium 88 Ra Radium 105 Db Dubnium 106 Sg Seaborgium 107 Bh Bohrium 108 Hs Hassium 109 Mt Meitnerium 110 Ds Darmstadtium 111 Rg Roentgenium 112 Cn Copernicium 21 Sc Scandium 45 39 YYtrium 89 22 Ti Titanium 48 40 Zr Zirconium 91 72 Hf Hafnium 178.5 23 VVanadium 51 41 Nb Niobium 93 73 Ta Tantalum 181 24 Cr Cromium 52 42 Mo Molybdenum 96 74 WTungsten 184 25 Mn Manganese 55 43 Tc Technetium 75 Re Rhenium 186 26 FeIron 56 44 Ru Ruthenium 101 76 Os Osmium 190 27 Co Cobalt 59 45 Rh Rodium 103 77 Ir Iridium 192 28 Ni Nickel 59 46 Pd Palladium 106 78 Pt Platinum 195 29 Cu Copper 64 47 AgSilver 108 79 AuGold 197 30 ZnZinc 65 48 Cd Cadmium 112 80 Hg Mercury 201 31 Ga Gallium 70 49 InIndium 115 81 Tl Thallium 204 113 Nh Nohonium 114 Fl Flerovium 115 Mc Moscovium 116 Lv Livermorium 117 Ts Tennessine 118 Og Oganesson 32 Ge Germanium 73 50 SnTin 119 82 PbLead 207 33 As Arsenic 75 51 Sb Antimony 122 83 Bi Bismuth 209 34 Se Selenium 79 52 Te Tellurium 128 84 Po Polonium 210 35 Br Bromine 80 53 IIodine 127 85 At Astatine 210 36 Kr Krypton 84 54 XeXenon 131 86 RnRadon 222 5 BBoron 11 13 Al Aluminium 27 6 CCarbon 12 14 Si Silicon 28 7 NNitrogen 14 15 PPhosphorus 31 8 OOxygen 16 16 SSulphur 32 9 FFluorine 19 17 Cl Chlorine 35.5 2 He Helium4 10 NeNeon 20 18 Ar Argon 40 58 Ce Cerium 140 90 Th Thorium 232 57 La Lanthanum 139 89 Ac Actinium 59 Pr Praseodymium 141 91 Pa Protactinium 231 60 Nd Neodymium 144 92 UUranium 238 61PmPromethium 145 93Np Neptunium 62 SmSamarium 150 94Pu Plutonium 63 Eu Europium 152 95AmAmericium 64 Gd Gadolinium 157 96CmCurium 65 TbTerbium 159 97Bk Berkelium 66 Dy Dysprosium 162.5 98 Cf Californium 67 Ho Holmium 165 99 Es Einsteinium 68 Er Erbium 167 100FmFermium 69 TmThulium 169 101 Md Mendelevium 70 Yb Ytterbium 173 102No Nobelium 71 Lu Lutetium 175 103 Lr Lawrencium Proton number 1 Alkali metals 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Metal Semi-metal Non-metal Symbol of element Name of element Relative atomic mass 104 Rf Rutherfordium 57 – 71 Lanthanoids 89 – 103 Actinoids Key: Group Alkaline earth metals Halogens Inert gases Lanthanides Actinides Periodic Table of Elements