ANSWERS SCHEME
1 PRE-TEST Answer the following questions. 1. Write the formula for the following substances: a) Magnesium bromide: .…MgBr2… b) Aluminium carbonate : …Al2(CO3)3……………. 2. Write the balance chemical equation for the following reaction: a) Sodium react with water ………2Na + 2H2O → 2NaOH + H2………………………………………….. b) Hydrochloric acid react with zinc …………2HCl + Zn → ZnCl2 + H2 …………………………………….. 3. State the type of particle for: a) Nitrogen dioxide : ……………NO2……………………… b) Magnesium ribbon : ………Mg………………………………. c) Sodium oxide : ……Na2O……………………………. 4. Calculate the number of particles in 0.25 mol of iron. number of particles = 0.25 mol x 6.02x 1023 = 1.505 x 1023 5. Calculate the mass of 0.1 mol of glucose, C6H12O6. [ RAM : C, 12; H,1; O,16] mass of glucose = 0.1 mol x ( 12 x 6 + 1 x 12 + 16 x 6) g mol-1 = 18 g 6. Calculate the number of molecules present in 3600 cm3 of carbon dioxide gas at room temperature and pressure. [1 mol of gas = 24 dm3 at room temperature and pressure] Volume = 3600/1000 = 3.6 dm3 Number of moles = 3.6 dm3 / 24 dm3mol-1 = 0.15 mol Number of molecules = 0.15 mol x 6.02 x 1023 mol-1 = 9.03 x 1022
2 7. Calculate the mass of 18.8 dm3 of propane gases, C3H8 at standard temperature and pressure. [ 1 mol of gas = 22.4 dm3 at STP, RAM : C, 12; H, 1] Number of moles = 18.8 dm3 / 22.4 dm3mol-1 = 0.839 mol Mass = 0.839 mol x (12 x 3 + 1 x 8) gmol-1 = 36.916 g 8. 4.0 g solid sodium hydroxide is dissolved in the water and made up 250 cm3 of solution. Calculate the concentration of sodium hydroxide solution in gdm-3 and moldm-3 ? [ RAM : Na, 23; H,1; O,16] Volume = 250/1000 = 0.25 dm3 Concentration of sodium hydroxide solution = 4.0 g 0.25 dm3 = 16 g dm-3 9. The burning of methane is represented by the equation: C2H4 + 3O2 → 2CO2 + 2H2O If 2.6 g of ethene burn in excess oxygen, calculate the mass of carbon dioxide formed. Number of moles = 2.6 g (12 x 2 + 1 x 4) gmol-1 = 0.0928 mol 1 mol of C2H4 : 2 mol of CO2 0.0928 mol of C2H4 : 0.1856 mol of CO2 Mass = 0.1856 mol x (12 + 16 x 2) =8.166 g 10. Lead is extracted according to the following equation. C + 2PbO → CO2 + 2Pb Determine the mass of lead extracted from 0.5 mole of lead (II) oxide. [RAM Pb = 207, O=16] 1 mol of : 2 mol of Pb 0.5 mol of C : 1 mol of Pb Mass = 1 mol x 207 gmol-1 = 207 g
3 2.0 MOLE Concept Konsep Mol 1.1 The relationship between mole and number of particle Hubungan mol dengan bilangan zarah [NA = 6.02 x 1023] Mol = Number of particle Avogadro constant A) Complete the following table. Lengkapkan jadual dibawah Number of Mole of Substance Bilangan mol bahan Number of atom Bilangan atom Number of Mole of Substance Bilangan mol bahan Number of atom Bilangan atom 1.0 mol of magnesium 6.02 x 1023 0.001 mol 6.02 x 1020 0.1 mol of iron 6.02 x 1022 0.5 mol 3.01 x 1023 0.5 mol of argon 3.01 x 1023 0.1 mol 6.02 x 1022 0.01 mol of aluminium 6.02 x 1021 0.05 mol 3.01 x 1022 5.0 mol of sodium 3.01 x 1024 5 mol 3.01 x 1024 2.5 mol of lithium 1.505 x 1024 2.5 mol 1.505 x 1024 0.05 mol of copper 3.01 x 1022 1.5 mol 9.03 x 1023 0.01 mol of silver 6.02 x 1021 0.0015 mol 9.03 x 1020 0.1 mol of helium 6.02 x 1022 0.25 mol 1.505 x 1023 B) Complete the following table Lengkapkan jadual di bawah Number of Mole of Substance Bilangan mol bahan Number of molecules Bilangan molekul Number of atoms Bilangan atom 1.0 mole of water 1.0 mol air 6.02 x 1023 6.02 x 1023 x 3 = 1.806 x 1024 0.1 mole of ammonia 0.1 mol ammonia 6.02 x 1022 6.02 x 1023 x 4 = 2.408 x 1024 0.5 mole of sulphur dioxide 0.5 mol sulfur dioksida 3.01 x 1023 3.01 x 1023 x 3 = 9.03 x 1023
4 2.0 mole of carbon dioxide 2.0 mol karbon dioksida 1.204 x 1024 1.204 x 1024 x 3 =3.612 x 1024 0.02 mole of oxygen 0.02 mol oksigen 1.204 x 1022 1.204 x 1022 x 2 =2.408 x 1022 5.0 mole of hydrogen 5.0 mol hydrogen 3.01 x 1024 3.01 x 1024 x 2 = 6.02 x 1024 0.05 mole of methane 0.05 mol metana 3.01 x 1022 3.01 x 1022 x 5 = 1.505 x 1023 0.1 mole of chlorine 0.1 mol klorin 6.02 x 1022 6.02 x 1022 x 2 = 1.204 x 1022 0.05 mole of bromine 0.05 mol bromine 3.01 x 1022 3.01 x 1022 x 2 = 6.02 x 1023 0.1 mole of tetrachloromethane 0.1 mol tetraklorometana 6.02 x 1022 6.02 x 1022 x 5 = 3.01 x 1024 C) Calculate the number of cation and anion for the following substances Kirakan bilangan kation dan anion bagi bahan-bahan berikut Number of mole of substance Bilangan mol bahan Number of cations Bilangan kation Number of anions Bilangan anion 1.0 mole of magnesium oxide 1.0 mol magnesium oksida 6.02 x 1023 6.02 x 1023 1.0 mole of sodium oxide 1.0 mol natrium oksida 6.02 x 1023 x 2 = 1.204 x 1023 6.02 x 1023 0.1 mole of silver nitrate 0.1 mol argentum nitrat 6.02 x 1022 6.02 x 1022 0.1 mole of aluminium nitrate 0.1 mol aluminium nitrat 6.02 x 1022 6.02 x 1022 x 3 = 1.806 x 1022 0.02 mole of copper(II) sulphate 0.02 mol kuprum(II) sulfat 1.204 x 1022 1.204 x 1022 0.5 mole of lead(II) nitrate 0.5 mol plumbum(II) nitrat 3.01 x 1023 3.01 x 1023 x 2 = 6.02 x 1023 0.01 mole of zinc chloride 0.01 mol zink klorida 6.02 x 1021 6.02 x 1021 x 2 = 1.204 x 1022 2.0 mole of iron(II) chloride 2.0 mol ferum(II) klorida 1.204 x 1024 1.204 x 1024 x 2 = 2.408 x 1024 2.0 mole of calcium carbonate 2.0 mol kalsium karbonat 1.204 x 1024 1.204 x 1024 1.5 mole of sulphuric acid 1.5 mol asid sulfurik 9.03 x 1023 x 2 = 1.806 x 1024 9.03 x 1023 0.01 mole of potassium hydroxide 0.01 mol kalium hidroksida 6.02 x 1021 6.02 x 1021
5 2.1 The relationship between mole and mass Hubungan mol dengan jisim bahan Mol = Mass Molar mass Complete the following table Lengkapkan jadual di bawah Mass of substance Jisim bahan Number of Mole of Substance Bilangan mol bahan Number of Mole of Substance Bilangan mol bahan Mass of substance Jisim bahan 12 g of magnesium 12 g magnesium 12/24= 0.5 mol 1.0 mole of magnesium 1.0 mol magnesium 1.0 x 24 = 24 g 16 g of oxygen 16 g oksigen 16/(16 x 2) = 0.5 mol 0.5 mole of oxygen 0.5 mol oksigen 0.5 x 916 x2) = 16 g 6 g of water 6 g air 6/ (1 x 2 + 16) = 0.333 mol 0.01 mole of water 0.01 g air 0.01 x (2 x 1+16) = 0.18 g 10 g of chlorine 10 g klorin 10/(35.5 x 2) = 0.014 mol 1.2 mole of chlorine 1.2 mol klorin 1.2 x (35.5 x 2) = 85.2 g 15 g sodium chloride 15 g natrium klorida 15/ (23+ 16) = 0.256 mol 2.5 mole of sodium chloride 2.5 mol natrium klorida 2.5 x (23 + 35.5) = 146.25 g 20 g zink carbonate 20 g zink karbonat 20/(65+12+16 x 3) = 0.16 mol 0.05 mole of zink carbonate 0.05 mol zink karbonat 0.05 x (65+ 12+16 x 3) = 6.25 g 25 g silver nitrate 25 g argentum nitrat 25/ (108+14+16 x 3) = 0.147 mol 0.01 mole of silver nitrate 0.01 mol argentum nitrat 0.01 x (108+12+16 x 3) = 1.7 g 2.3 The relationship between mole and volume of gas Hubungan mol dengan Isipadu gas Mol = Volume of gas Molar volume Volume of gas at room temperature Isipadu gas pada suhu bilik Number of mole of gas Bilangan mol gas Number of mole of gas Bilangan mol gas Volume of gas at room condition Isipadu gas pada keadaan bilik 20 dm3 of hydrogen 20 dm3 hidrogen 20/24 =0.833 mol 0.01 mole of hydrogen 0.01 mol hydrogen 0.01 x 24 = 0.24 dm3 800 cm3 of oxygen 800 cm3 oksigen V=800/1000 =0.8 dm3 n=0.8/24 =0.033 mol 0.02 mole of oxygen 0.02 mol oksigen 0.02 x 24 = 0.48 dm3 1500 cm3 of nitrogen 1500 cm3 nitrogen V=1500/1000 =1.5 dm3 n=1.5/24 =0.0625 mol 0.05 mole of carbon monoxide 0.005 mol karbon monoksida 0.05 x 24 = 1.2 dm3
6 5 dm3 of carbon dioxide 5 dm3 karbon dioksida 5/24 = 0.208 mol 1.5 mole of steam 1.5 mol wap air 50 dm3 of sulfur dioxide 50 dm3 sulfur dioksida 50/24 =2.083 mol 1.0 mole of nitrogen 1.0 mol nitrogen 1.0 x 24 = 24 dm3 2000 cm3 ammonia 2000 cm3 ammonia V=2000/1000 =2.0 dm3 n=2.0/24 =0.083 mol 1.2 mole of carbon dioxide 1.2 mol karbon dioksida 1.2 x 24 = 28.8 dm3 1500 cm3 chlorine gas 1500 cm3 gas klorin V=00/1000 =1.5 dm3 n=1.5/24 =0.0625 mol 15 mole of methane 15 mol metana 15 x 24 = 360 dm3 Determine the number of mole of the following substance Tentukan bilangan mol bagi bahan-bahan berikut Quantity of substance Kuantiti Bahan Number of mole Bilangan mol 200 cm3 0.1 mol dm-3 hydrochloric acid 200 cm3 asid hidroklorik 0.1 mol dm-3 n=(0.1 x 200) / 1000 = 0.02 mol 25 cm3 0.5 mol dm-3 sulphuric acid 25 cm3 asid sulfurik 0.5 mol dm-3 n=(0.5 x 25) / 1000 = 0.0125 mol 20 cm3 0.5 mol dm-3 soadium chloride solution 20 cm3 larutan natrium klorida 0.5 mol dm-3 n=(0.5 x 20) / 1000 = 0.01 mol 250 cm3 0.01 mol dm-3 lead(II) nitrate solution 250 cm3 larutan plumbum(II) nitrat 0.01 mol dm3 n=(0.01 x 250) / 1000 = 0.025 mol 500 cm3 0.1 mol dm-3 zinc sulphate solution 500 cm3 larutan zink sulfat 0.1 mol dm-3 n=(0.1 x 500) / 1000 = 0.05 mol 1.5 dm3 0.2 mol dm-3 sulphuric acid solution 1.5 dm3 larutan asid sulfurik 0.2 mol dm-3 n=0.2 x 1.5 = 0.3 mol 2.5 dm3 0.25 mol dm-3 soadium hydroxide solution 2.5 dm3 larutan natrium hidroksida 0.25 mol dm-3 n=0.25 x 2.5 = 0.625 mol
7 UJIAN PENGESANAN 2 Name : ………………………………………… Class : ……………………… 1. Calculate the mass of 0.1 mol of ammonia. [ Relative atomic mass : N, 14 ; H,1] Mass of ammonia = 0.1 x (14 + 1 x 3) = 1.7 g 2. Calculate the number of mole of molecules present in 3.6 dm3 of nitrogen gas at room temperature and pressure [ 1 mol of gas = 24 dm3 at room temperature] Number of mole of molecules= 3.6 / 24 = 0.15 mol 3. Calculate the volume occupied by 0.5 mole of ethane gas at STP. [ 1 mol of gas = 22.4 dm3 at STP] Volume = 0.5 x 22.4 = 11.2 dm3 4. In a container containing 2.2g of carbon dioxide gas. Calculate the number of molecule and atom in the container. Number of mole= 2.2 / (12 + 16 x 2)= 0.05 mol Number of molecule = 0.05 x 6.02 x 1023 = 3.01x 1022 molecule Number of atom= 0.05 x 6.02 x 1023 x 3 = 9.03x 1022 atom 5. Calculate the mass of 18.8 dm3 of propane at standard temperature and pressure. [ 1 mol of gas = 22.4 dm3 at STP] Number of mole= 18.8 / 22.4 =0.839 mol Mass of propene = 0.839 x ( 12 x3 + 1 x 8) = 36.916 g 6. Calculate the mass of 9 x 1021 molecules of nitrogen dioxide gas. Number of mole = 9 x 1021 /6.02 x 1023 = 0.015 mol Mass of nitrogen dioxide = 0.015 x (14+16 x2) = 0.69 g
8 3.0 CHEMICAL EQUATION PERSAMAAN KIMIA Chemical Reaction For Group 1 Elements 1. Alkali metal + Water → Metal Hydroxide + Hydrogen gas Lithium + Water → Lithium hydroxide + Hydrogen 2 Li + 2 H2O → 2 LiOH + H2 2. Alkali metal + Oxygen → Metal Oxide Sodium + Oxygen → sodium Oxide 4 Na + O2 → 2Na2O 3. Metal oxide + Water → Metal hydroxide Sodium Oxide + Water → Sodium Hydroxide Na2O + H2O → 2NaOH 4. Alkali metal + Halogen → Metal Halide Lithium + Chlorine → Lithium Chloride 2Li + Cl2 → 2 LiCl Chemical Reaction For Group 17 Elements 1. Halogen + Water → Acid + Acid Chlorine + Water → Hydrochloric Acid + Hypochlorous Acid Cl2 + H2O → HCl + HOCl 2. Halogen + Iron → Iron(III) Halide Bromine + Iron → Iron(III) bromide 3Br2 + 2 Fe → 2FeBr3 3. Halogen + Sodium Hydroxide → Sodium Halide + Sodium halate + water Bromine + Sodium Hydroxide → Sodium bromide + Sodium Bromate + Water Br2 + 2NaOH → NaBr + NaOBr + H2O Chemical Reaction For Acids 1. Acid + Alkali → Salt + Water Hydrochloric acid + Sodium Hydroxide → Sodium Chloride + Water HCl + NaOH → NaCl + H2O 2. Acid + base → Salt + Water Sulphuric acid + Magmesium Oxide → Magnesium sulphate + Water H2SO4 + MgO → MgSO4 + H2O 3. Metal + Acid → Salt + hydrogen Magnesium + Hydrochloric Acid → Magnesium Chloride + hydrogen Mg + 2 HCl → MgCl2 + H2
9 4. Metal oxide + Acid → salt + Water Calcium Oxide + Sulphuric acid → Calcium sulphate + Water CaO + H2SO4 → CaSO4 + H2O 5. Metal Carbonate + Acid → Salt + Water + Carbon Dioxide Zinc Carbonate + Nitric Acid → Zinc nitrate + Water + Carbon dioxide ZnCO3 + 2HNO3 → Zn(NO3)2 + H2O + CO2 Chemical Reaction For Alkali 1. Acid + Alkali → Salt + Water Nitric acid + Potassium Hydroxide → Potassium nitrate + Water HNO3 + KOH → KNO3 + H2O 2. Ammonium Salt + Alkali → Ammonia + salt + Water Ammonium nitrate + Sodium hydroxide → Sodium nitrate + Ammonia + water NH4NO3 + NaOH → NaNO3 + NH3 + H2O Chemical Reaction For Salt 1. Action Of Heat Carbonate salt → Metal Oxide + Carbon Dioxide Calcium carbonate → Calcium oxide + Carbon Dioxide CaCO3 → CaO + CO2 Nitrate salt → Metal Oxide + Oxygen + Nitrogen Dioxide Lead (II) nitrate → Lead(II) oxide + Oxygen + Nitrogen dioxide 2Pb(NO3)2 → 2PbO + O2 + 4NO2 2. Precipitation Reaction / Double Decomposition Reaction Lead (II) Nitrate + Sodium Chloride → Sodium Nitrate + Lead (II) Chloride Pb(NO3)2 + 2NaCl → 2NaNO3 + PbCl2 3. Displacement Reaction Copper (II) Sulphate + Zinc → Zinc Sulphate + Copper CuSO4 + Zn → ZnSO4 + Cu Contact Process 1. Prepare sulphur dioxide Burning of sulphur in dry air to form sulphur dioxide Sulphur + Oxygen → sulphur dioxide S + O2 → SO2 Burning of metal sulphide to form sulphur dioxide and metal oxide Zinc sulphide + Oxigen → Sulphur dioxide + Zinc Oxide. 2ZnS + 3O2 → 2SO2 + 2ZnO
10 2. Conversion of Sulphur dioxide to sulphur trioxide. Sulphur dioxide + Oxygen → sulphur trioxide 2SO2 + O2 → 2SO3 3. Sulphur trioxide is dissolved in concentrated sulphuric acid to produce oleum. Sulphur trioxide + Sulphuric acid → Oleum SO3 + H2SO4 → H2S2O7 4. Oleum is diluted with water to produce concentrated sulphuric acid. Oleum + Water → Sulphuric acid H2S2O7 + H2O → 2H2SO4 Haber Process Nitrogen + Hydrogen → Ammonia N2 + 3H2 → 2 NH3 EXERCISE LATIHAN Write the balance CHEMICAL EQUATION for the following chemical reaction if the reactants used for the reactions are: 1. Potassium + Water ………………………………………………………………………………………......... . 2. Sodium + Water ………………………………………………………………………………………......... . 3. Sodium + Oxygen ………………………………………………………………………………………......... . 4. Lithium + Oxygen ……………………………………………………………………………………………. . 5. Potassium Oxide + Water ………………………………………………………………………………………......... . 6. Lithium Oxide + Water ……………………………………………………………………………………………. 7. Potassium + Bromine ………………………………………………………………………………………......... .
11 8. Lithium + Chlorine ………………………………………………………………………………………......... . 9. Sodium + Iodine ……………………………………………………………………………………………. . 10. Bromine + Water ………………………………………………………………………………………......... . 11. Iodine + Water ………………………………………………………………………………………......... . 12. Chlorine + Iron ………………………………………………………………………………………......... . 13. Iodine + Iron ………………………………………………………………………………………......... . 14. Chlorine + Sodium hydroxide ………………………………………………………………………………………......... . 15. Iodine + Sodium Hydroxide ……………………………………………………………………………………………. . 16. Hydrochloric Acid + Zinc Metal ………………………………………………………………………………………......... . 17. Hydrochloric Acid + Magnesium Metal ……………………………………………………………………………………………. . 18. Sulphuric Acid + Calcium Metal ……………………………………………………………………………………………. . 19. Sulphuric Acid + Copper (II) Oxide ………………………………………………………………………………………......... .
12 20. Hydrochloric Acid + Magnesium Oxide …………………………………………………………………………………………… 21. Nitric Acid + Zinc Oxide …………………………………………………………………………………………… 22. Nitric Acid + Magnesium Carbonate ………………………………………………………………………………………......... 23. Sulphuric Acid + Sodium Carbonate ………………………………………………………………………………………......... . 24. Hydrochloric Acid + Calcium carbonate ……………………………………………………………………………………………. 25. Sulphuric Acid + Potassium hydroxide ………………………………………………………………………………………......... . 26. Hydrochloric acid + Sodium hydroxide ……………………………………………………………………………………………. 27. Ethanoic Acid + sodium hydroxide …………………………………………………………………………………………… 28. Ethanoic Acid + Magnesium ………………………………………………………………………………………......... . 29. Ammonium Chloride + Sodium hydroxide ………………………………………………………………………………………......... . 30. Ammonium Nitrate + Sodium hydroxide …………………………………………………………………………………………… 31. Lead ( II ) Nitrate + Sodium Chloride ………………………………………………………………………………………......... . 32. Barium Nitrate + Zinc Sulpahte …………………………………………………………………………………………… 33. Lead (II) nitrate + PotassiumIodide …………………………………………………………………………………………… 34. Lead (II) nitrate + Potassium Dichromate (VI)
13 …………………………………………………………………………………………… 35. Silver nitrate + Potassium Chloride …………………………………………………………………………………………… 36. Zinc Nitrate + Calcium ………………………………………………………………………………………......... . 37. Copper (II ) Sulphate + Magnesium ………………………………………………………………………………………......... . 38. Silver nitrate + Copper ……………………………………………………………………………………………. . 39. Heat Solid of Copper Carbonate ………………………………………………………………………………………......... . 40. Heat solid of Ammonium Chloride …………………………………………………………………………………………… 41. Heat Solid of Magnesium Nitrate ……………………………………………………………………………………………. 42. Heat Solid of Lead (II) Nitrate …………………………………………………………………………………………… 43. Heat solid of calcium carbonate ……………………………………………………………………………………………
14 4.0 The relationship between mol and concentaration@ volume of solution Hubungan mol dengan kepekatan@ isipadu larutan 4.1 Concentration of solution Kepekatan larutan Concentration Kepekatan (gdm-3 ) = Mass of solute dissolved (g) Volume of solution ( dm3 ) Concentration Kepekatan (mol dm-3 ) = Mol of solute dissolved (g) Volume of solution ( dm) Dilution Pencairan M1V1 = M2V2 1. The concentration of a sodium hydroxide solution is 25.0 g dm-3 . Calculate the molarity of the solution. Kepekatan larutan natrium hidroksida adalah 25.0 g dm-3 . Kirakan kemolaran larutan ini. Molarity of sodium hydroxide = 25.0 / (23+16+1) = 0.625 mol dm-3 2. The molarity of copper (II) chloride solution is 0.01 mol dm-3 . Calculate the concentration of the solution in g dm-3 . Kemolaran larutan kuprum(II) klorida ialah 0.01 mol dm-3 .Kirakan kepekatan larutan dalam g dm-3 . Concentration of copper(II) chloride = 0.01 x (64+ 35.5 x2 ) = 1.35 g dm-3 3. Ammonia solution used in school laboratories usually has a concentration of 2.0 mol dm-3 . Calculate the number of mole of ammonia in 250 cm3 of the ammonia solution. Larutan ammonia yang biasa digunakan di makmal mempunyai kepekatan 2.0 mol dm-3. Kirakan bilangan mol ammonia yang terdapat di dalamm 250 cm3 larutan ammonia. Number of mole = (2.0 x 250)/1000= 0.5 mol 4. What is the mass of zinc chloride required to dissolve in water to prepare a 500 cm3 of solution with a concentration of 0.2 mol dm-3 .
15 Apakah jisim zink klorida yang perlu dilarutkan di dalam air untuk menyediakan 500 cm3 larutan berkepekatan 0.1 mol dm-3 . Number of mole = (0.2 x 500)/1000= 0.1 mol Mass of zinc chloride = 0.1 x (65 + 35.5 x 2) =13.6 g 4. Aminah need to to prepare 250 cm3 of sodium hydroxide solution with concentration 0.5 mol dm-3 . Calculate the mass of sodium hydroxide powder required to prepare the solution. Aminah dikehendaki menyediakan 250 cm3 larutan natrium hidroksida dengan kepekatan 0.5 mol dm-3. Kirakan jisim natrrium hidroksida yang diperlukan untuk menyediakan larutan ini. Number of mole = (0.5 x 250)/1000= 0.125 mol Mass of zinc chloride = 0.125 x (23+ 1+ 16) =5 g 5. Calculate the mass of copper(II) nitrate powder required to dissolve in water to prepare 2500 cm3 of solution with concentration 0.01 mol dm-3 . Kirakan jisim kuprum(II) nitrat yang perlu dilarutkan di dalam air untuk menyediakan 2.5 dm3 larutan dengan kepekatan 0.01 mol dm-3 . Number of mole = (0.01 x 2500)/1000= 0.025 mol Mass of zinc chloride = 0.025 x (64+ 2(14+16 x 3)) =4.7 g 6. What is the molarity of the lithium hydroxide solution produced when 950 cm3 of distilled water is added to 50 cm3 of lithium hydroxide solution of 1.0 mol dm-3 ? Apakah kemolaran larutan litium hidroksida yang terhasil apabila 950 cm3 air suling ditambah kepada 50 cm3 larutan litium hidroksida 1.0 mol dm-3 ? M1V1=M2V2 1.0 x 50 = M2 x (950 + 50) M2 = 0.05 mol dm-3 7. What is the volume of the potassium hydroxide solution 2.0 mol dm-3 is needed to prepare 250 cm3 of potassium hydroxide 0.5 mol dm-3 . Determine the volume of distilled water used. Apakah isipadu larutan kalium hidroksida 2.0 mol dm-3 yang diperlukan untuk menyediakan 250 cm3 larutan kalium hidroksida 0.5 mol dm-3 . Tentukan isipadu air suling yang digunakan. M1V1=M2V2 2.0 x V1 = 0.5 x 250 M2 = 62.5 cm3
16 8. The concentration of sodium hydroxide solution is 10.0 g dm-3 . Kepekatan larutan natrium hidroksida adalah 10.0 g dm-3 . a) What is the molarity of the solution? Apakah kemolaran larutan ini? molarity of sodium hydroxide = 10.0/ (23+16 +1) = 0.25 mol dm-3 b) What is the molarity of the solution produced when 200 cm3 of distilled water is added to 50 cm3 of this solution? Apakah kemolaran larutan yang terhasil apabila 200 cm3 air suling di tambah kepada 50 cm3 larutan ini? M1V1=M2V2 0.25 x 500 = M2 x 250 M2 = 0.05 mol dm-3 9. What is the volume of distilled water required to be added to 60 cm3 nitric acid of 2.0 mol dm-3 to produce a 0.25 mol dm-3 nitric acid? Berapakah isipadu air suling yang perlu ditambah kepada 60 cm3 asid nitrik untuk menghasilkan 0.25 mol dm-3 asid nitrik? M1V1=M2V2 2.0 x 60 = 0.25 x V2 V2 = 480 cm3 Volume needed to be added to 60 cm3 of nitric acid = 480-60 = 420 cm3 10. Distilled water is added into 20 cm3 of sodium hydroxide solution in a beaker to produce 250 cm3 of 0.2 mol dm-3 sodium hydroxide solution. What is the molarity of sodium hydroxide solution in the beaker. Air suling ditambah kepada 20 cm3 larutan natrium hidroksida untuk menghasilkan 250 cm3 larutan natrium hidroksida 0.2 mol dm-3 . Apakah kemolaran larutan natrium hidroksida? M1V1=M2V2 M1 x 20 = 0.2 x 250 V2 = 2.5 mol dm-3 11. What is the volume of 2.0 mol dm-3 of copper(II) nitrate solution is needed to prepare 100 cm3 of 0.2 mol dm-3 of copper(II) nitrate solution. Berapakah isipadu 2.0 mol dm-3 larutan kuprum(II) nitrat yang diperlukan untuk menyediakan 100 cm3 larutan kuprum(II) nitrat 0.2 mol dm-3 . M1V1=M2V2 M1 x 2.0 = 0.2 x 100 V2 = 10 cm3
17 4.2 Rumusan Konsep MOL a) Calculate the mass of NaOH used to prepare 250 cm3 0.5 mol dm-3 sodium hydroxide solution. Kirakan jisim NaOH yang digunakan untuk menyediakan 250 cm3 larutan natrium hidroksida 0.5 mol dm-3 Number of mole = (0.5 x 250)/1000= 0.125 mol Mass = 0.125 x (23 + 16+ 1) = 5.0 g b) Determine the number of particles found in 500 cm3 of suphur dioxide gas at room temperature. Tentukan bilangan zarah yang terdapat di dalam 500 cm3 gas sulfur dioksida pada suhu bilik. Volume in dm3 = 500 / 1000 =0.5 dm3 Number of mole = 0.5 / 24 = 0.0208 mol Number of particles = 0.0208 x 6.02 x 1023 = 1.252 x 1022 c) A balloon contains 6.03 x 1020 carbon dioxide molecules. Calculate the volume of carbon dioxide in the balloon. Sebuah belon mengandungi 6.03 x 1020 molekul karbon dioksida. Kirakan isipadu gas karbon dioksida di dalam belon tersebut. Number of mole = 6.02 x 1020 /6.02 x 1023 = 0.01 mol Volume of gas = 0.01 x 24 = 0.24 dm3 Number of Particles Bilangan zarah Mass Jisim MV/1000 Volume of gas Isipadu gas MOLE × NA ÷ NA ÷ Molar mass × Molar mass ÷ Molar volume × Molar volume
18 d) What is the number of hydrogen atoms in 36 g of water? Berapakah bilangan atom hidrogen dalam 36 g air? Number of mole = 36 / (1 x 2 + 16) = 2 mol Number of atoms= 2 x 6.02 x 1023 x 2 = 2.408 x 1024 e) What is the number of nitrogen atoms in 8.5 g of ammonia? Berapakah bilangan atom nitrogen dalam 8.5 g ammonia? Number of mole = 8.5 / (14 +1 x 3) = 0.5 mol Number of atoms= 0.5 x 6.02 x 1023 = 3.01 x 1023 f) A beaker contains 500 cm3 of 0.1 mol dm-3 copper(II) chloride solution. Sebuah bikar mengandungi 500 cm3 larutan kuprum(II) klorida 0.1 mol dm-3 . (i) What is the mass of copper(II) chloride found in the beaker? Berapakah jisim kuprum(II) klorida yang terdapat di dalam bikar tersebut? Number of mole = (0.1 x 500) / 1000=0.05 mol Mass of copper(II) chloride = 0.05 x ( 64 + 35.5 x 2) = 6.75 g (ii) Determine the number of cations found in the beaker. Tentukan bilangan kation yang terdapat di dalam bikar tersebut. Number of cations = 0.05 x 6.02 x 1023 = 3.01 x 1022 (iii) Calculate the total number of particles found in the beaker. Kirakan jumlah bilangan zarah yang terdapat di dalam bikar tersebut. Number of particles = 0.05 x 6.02 x 1023 x 3= 9.03 x 1022 g) A bottle contains 1.5 dm3 of chlorine gas at room temperature. Sebuah botol mengandungi gas klorin sebanyak 1.5 dm3 pada suhu bilik. (i) What kind of particles are contained in chlorine gas? Apakah jenis zarah yang terdapat di dalam gas klorin? Molecules (ii) How many chlorine particles are in the bottle? Berapakah bilangan zarah klorin di dalam botol tersebut? Number of moles = 1.5/24 =0.0625 mol Number of particles = 0.0625 x 6.02 x 1023 = 3.7625 x 1022 (iii) Calculate the number of chlorine atoms in the bottle. Kirakan bilangan atom klorin di dalam botol tersebut. Number of particles = 0.0625 x 6.02 x 1023 x 2= 7.525 x 1022 (iv) Determine the mass of chlorine contained in the bottle. Tentukan jisim klorin yang terdapat di dalam botol tersebut Mass of chlorine = 0.0625 x 24 =1.5 dm3
19 h) 8.0 g of magnesium nitrate is dissolved in water to produce a solution with a concentration of 0.02 mol dm-3 . Calculate the volume of the solution prepared. 8.0 g magnesium nitrat dilarutkan ke dalam air menghasilkan larutan berkepekatan 0.02 mol dm-3. Kirakan isipadu larutan yang disediakan. Number of mole = 8.0/( 24 +2(14+16 x 3)= 0.054 mol 0.054 = (0.02 x V )/ 1000 V = (0.054 x 1000)/ 0.02 = 2.7 cm3 i) 50 cm3 of sulphuric acid contains 0.002 mol of hydrogen ions. What is the molarity of sulphuric acid? 50 cm3 asid sulfurik mengandungi 0.002 mol ion hydrogen. Apakah kemolaran asid sulfurik? Number of mole= 0.002 / 2 = 0.001 mole 0.001 = (M x 50 )/ 1000 M = (0.001 x 1000) / 50 = 0.02 mol dm-3 j) 20.0 g of potassium hydroxide is dissolved in water to produce a solution with a concentration of 0.02 mol dm-3 . Calculate the volume of the solution prepared. 20.0 g kalium hidroksida telah dilarutkan di dalam air menghasilkan larutan dengan kepekatan 0.02 mol dm-3 . Kirakan isipadu larutan yang disediakan. Number of mole = 20 / (39+16+1) = 0.357 mol n= MV/1000 0.357 = (0.02 x V )/ 1000 V = (0.357 x 1000) / 0.02 = 17.850 cm3 /17.85 dm3 k) 8.0 g of zinc sulphate is dissolved in water to form 500 cm3 solution. Calculate the concentration of zinc sulphate solution in g dm-3 and mol dm-3 . 8.0 g zink sulfat larut di dalam air menghasilkan 500 cm3 larutan. Kirakan kepekatan larutan zink sulfat dalam g dm-3 dan mol dm-3 . Number of mole = 8.0/ (65 + 32 +16 x 4) = 0.05 mol n= MV/1000 0.05 = (M x 500 )/ 1000 M = (0.05 x 1000) / 500 = 0.1 mol dm-3 l) 4.0 g sodium hydroxide powder is dissolved in the water and made up to 150 cm3 of solution. Calculate the concentration of sodium hydroxide solution in g dm-3 and mol dm-3 ? 4.0 g natrium hidroksida larut di dalam air membentuk 250 cm3 larutan. Kirakan kepekatan larutan natrium hidroksida dalam g dm-3 dan mol dm-3 . Concentration = 4.0/ 0.15 =26.67 g dm-3 Molarity = 26.67 / ( 23+ 16+1) = 0.667 mol dm-3
20 m)What is the mass of sodium carbonate required to prepare a 200 cm3 solution with concentration 0.5 mol dm-3 ? Berapakah jisim natrium karbonat yang diperlukan untuk menyediakan 200 cm3 larutan dengan kepekatan 0.5 mol dm-3 ? Number of mole, n n=MV/1000 n= (0.5 x 200) / 1000= 0.1 mol Mass of sodium carbonate = 0.1 x (23 + 12 + 16 x 3) 8.3 g n) Calculate the number of mole of sodium hydroxide in 250 cm3 of 0.5 mol dm-3 sodium hydroxide solution. Kirakan bilangan mol natrium hidroksida dalam 250 cm3 larutan natrium hidroksida 0.5 mol dm-3 . Number of mole, n n=MV/1000 n= (0.5 x 250) / 1000= 0.125 mol o) Calculate the number of mole of sulphuric acid in 500 cm3 of 0.2 mol dm-3 of sulphuric acid solution. Kirakan bilangan mol asid sulfurik dalam 500 cm3 larutan asid sulfurik 0.2 mol dm-3 . Number of mole, n n=MV/1000 n= (0.2 x 500) / 1000= 0.1 mol
21 4.3 STOICHIOMETRY 1 1. 10.0 g of iron are displaced from iron(II) sulphate solution by zinc. Zink telah menyesarkan 10.0 g ferum daripada larutan ferum(II) sulfat. [RMM: Fe=56, Zn=65, S=32, O=16] a) Write a balanced chemical equation Tuliskan persamaan kimia seimbang Zn + FeSO4→ Fe + ZnSO4 b) Calculate the number of mole of iron displaced. Kirakan bilangan mol ferum yang telah disesarkan. Number of mole, n = 10.0/ 56= 0.1786 mol c) Calculate the number of mole of zinc sulphate produced. Kirakan bilangan mol zink sulfat yang terhasil. 1 mol Fe : 1 mol ZnSO4 0.1786 mol Fe : 0.1786 mol ZnSO4 d) Determine the mass of zinc sulphate formed Tentukan jisim zink sulfat yang terbentuk Mass of Zinc sulphate = 0.1786 x ( 65 + 32 + 16 x 4) = 28.755 g 2. Decomposition of calcium carbonate when heated will produce calcium oxide and carbon dioxide. Penguraian kalsium karbonat apabila dipanaskan akan menghasilkan kalsium oksida dan karbon dioksida. (RMM : Ca=40, C=12, O=16) a) Write a balance chemical equation. Tuliskan persamaan kimia seimbang CaCO3 → CaO + CO2 b) 20.0 g of calcium carbonate is decomposed when heated. Determine 20.0 g kalsium karbonat terurai apabila dipanaskan. Tentukan (i) The number of mole of calcium carbonate decomposed Bilangan mol kalsium karbonat yang terurai Number of mole = 20.0/ (40 + 12 + 16 x 3) =0.2 mol (ii) the volume of gas produced at room temperature Isipadu gas yang terbebas pada suhu bilik Volume of gas = 0.2 x 24 = 4.8 dm3 (iii) The mass of calcium oxide produced Jisim kalsium oksida yang terhasil Mass of calcium oxide = 0.2 x (40+16) = 11.2 g
22 3. Lead is extracted according to the following equation, Plumbum diekstrak mengikut persamaan berikut, C + 2PbO → CO2 + 2Pb [RMM: Pb= 207; O=16, C=12] a) Determine the number of moles of lead extracted from 0.2 mole of lead(II) oxide. Tentukan bilangan mol plumbum yang telah diekstrak daripada 0.2 mol plumbum(II) oksida. 2mol PbO : 2 mol of Pb 0.2 mol of PbO : 0.2 mol of Pb b) Calculate the number of mole of carbon required to extract 0.2 mole of Lead (II) oxide. Kirakan bilangan mol karbon yang diperlukan untuk mengekstrak 0.2 mol plumbum(II) oksida. 2mol PbO : 1 mol of C 0.2 mol of PbO : 0.1 mol of C c) Determine the mass of lead are produced if 22.3 g of lead(II) oxide is heated with excessive carbon. Tentukan jisim plumbum yang terhasil sekiranya 22.3 g plumbum(II) oksida dipanaskan dengan karbon berlebihan. Number of mole of PbO = 22.3 / (207 +16) = 0.1 mol 2mol PbO : 2 mol of Pb 0.1 mol of PbO : 0.1 mol of Pb mass of Pb = 0.1 x 207 = 20.7 g 4. Magnesium ribbon reacts with hydrochloric acid to produce salt and hydrogen gas. Pita magnesium bertindak balas dengan asid hidroklorik menghasilkan garam dan gas hidrogen. [RMM: Mg=24, Cl=35.5, H=1: Molar volume of gas at room condition= 24 dm3 mol-1 ] a) Write the balance chemical equation for this reaction. Tuliskan persamaan kimia seimbang bagi tindak balas ini. Mg + 2HCl → MgCl2 + H2 d) If excess magnesium ribbon is added to 50 cm3 of 2.0 mol dm-3 hydrochloric acid, calculate Jika pita magnesium berlebihan dimasukkan kedalam 50 cm3 asid hidroklorik 2.0 mol dm-3 , kirakan (i) The number of mole of salt formed. Bilangan mol garam yang terbentuk. Number of moles of HCl = (2.0 x 50) /1000 = 0.1 mol 2 mol of HCl : 1 mol of MgCl2 0.1 mol of HCl : 0.05 mol of MgCl2
23 (ii) The mass of salt formed. Jisim garam terbentuk. Mass of salt, MgCl2 = 0.05 x (24 + 35.5 x 2 ) = 4.75 g (iii)The volume of hydrogen gas evolved at room temperature Isipadu gas hidrogen yang terbebas pada suhu bilik. 2 mol of HCl : 1 mol of H2 0.1 mol of HCl : 0.05 mol of H2 Volume of gas = 0.05 x 24 = 1.2 dm3 (iv) The number of particles of hydrogen released Bilangan zarah gas hidrogen yang terbebas Number of hydrogen particles = 0.05 mol x 6.02 x 1023 = 3.01 x 1022 5. 15.0 g of zinc oxide powder is reacted with excess dilute nitric acid. 15.0 g serbuk zink oksida telah bertindak balas dengan asid nitrik cair berlebihan. [RMM Zn=65, O=16, N=15, H=1 a) Write a chemical equation for the reaction. Tuliskan persamaan kimia bagi tindak balas tersebut ZnO + 2HNO3 → Zn(NO3)2 + H2O b) Calculate the number of mole of zinc oxide used. Kirakan bilangan mol zink oksida yang digunakan. Number of mole= 15/ (65 + 16) =0.185 mol c) Calculate the mass of zinc nitrate formed. Kirakan jisim zink nitrat yang terbentuk. 1mol of HCl : 1 mol of Zn(NO3)2 0.185 mol of HCl : 0.185 mol of of Zn(NO3)2 Mass of Zn(NO3)2 = 0.185 x (65 + 2(14 + 16 x 3))= 34.965 g 6. Iron metal reacts with excess sulphuric acid to produce iron(II) sulphate and hydrogen gas. Logam ferum bertindak balas dengan asid sulfurik menghasilkan ferum(II) sulfat dan gas hidrogen. [RMM: Fe=56, S=32, O=16, H=1: Molar volume of gas at room condition= 24 dm3 mol-1 ] a) Write a balanced chemical equation for the reaction. Tuliskan persamaan kimia seimbang bagi tindak balas ini. Fe + H2SO4 → FeSO4 + H2 b) If 5.6 g of iron metal is used in the reaction, calculate Sekiranya 5.6 g logam ferum digunakan dalam tindak balas ini, kirakan (i) the mass of iron(II) sulphate formed. Jisim ferum(II) sulfat terbentuk. Number of mole = 5.6 / 56 = 0.1 mol 1 mol Fe : 1 mol FeSO4 0.1 mol Fe : 0.1 mol FeSO4 mass of FeSO4= 0.1 x (56 + 32+16x4) =13.6 g
24 (ii) the volume of hydrogen gas produced at room condition. Isipadu gas hidrogen terbebas. 2 mol Fe : 1 mol H2 0.1 mol Fe : 0.1 mol H2 Volume of hydrogen gas = 0.1 x 24 = 2.4 dm3 7. Propane burns in air as represented by the equation: Propana dibakar dalam udara diwakilkan persamaan berikut; C3H8 + 5O2 → 3CO2 + 4H2O If if 22 g of carbon dioxide is produced, Jika 22 g karbon dioksida terhasil, [RAM: C=12, O=16, H=1: Molar volume of gas at room condition= 24 dm3 mol-1 ] (i) Calculate the mass of propane burnt Kirakan jisim propana yang telah terbakar Number of mole of CO2= 22 / (12 +16 x 2) =0.5 mol 3mol of CO2 : 1 mol of C3H8 0.5mol of CO2 : 0.1667 mol of C3H8 Mass of C3H8= 0.1667 x (12 x 3 + 1 x 8) = 7.3348 g (ii) Determine the volume of oxygen gas needed Tentukan isipadu gas oksigen yang diperlukan 3mol of CO2 : 5 mol of O2 0.5 mol of CO2 : 0.8333 mol of O2 Volume of gas = 0.8333 x 24 = 20 dm3 8. 6.0 g of magnesium is completely burnt in excess oxygen. Calculate the mass of magnesium oxide formed. 6.0 g magnesium terbakar lengkap di dalam oksigen berlebihan. Kirakan jisim magnesium oksida yang terhasil. [RAM: Mg=24, O=16] 2Mg + O2 → 2MgO Number of mole of Mg = 6.0 /24 =0.25 mol 2 mol of Mg : 2 mol of MgO 0.25 mol of Mg : 0.25 mol of MgO Mass of MgO = 0.25 x (24+ 16) =10 g
25 9. Sodium bicarbonate decomposes when heated as represented by the equation; Natrium bikarbonat terurai apabila di panaskan seperti ditunjukkan dalam persama berikut; 2NaHCO3 → Na2CO3 + CO2 + H2O [RAM: H=1, C=12, O=16, Na=11; Molar volume of gas at room condition= 24 dm3 mol-1 ] a) If 2000 dm3 carbon dioxide gas released at room temperature, calculate the mass of sodium bicarbonate that decomposed. Sekiranya 2000 cm3 gas karbon dioksida terbebas, kirakan jisim natrium bikarbonat yang terurai. Volume of gas = 2000/1000 = 2 dm3 Number of mole of gas = 2/24 = 0.0833 mol 1mol CO2 : 2 mol of NaHCO3 0.0833 mol CO2 : 0.1667 mol of NaHCO3 Mass of NaHCO3 = 0.1667 x (23+1+12+ 16x 3) = 14.0 g b) What is the mass of sodium carbonate formed when 1.2 dm3 of carbon dioxide gas released at room temperature? Berapakah jisim natrium karbonat yangterbentuk apabila 1.2 dm3 gas karbon dioksida terbebas pada suhu bilik? Number of mole of gas = 1.2/24 = 0.05 mol 1mol CO2 : 1 mol of Na2CO3 0.05 mol CO2 : 0.05 mol of Na2CO3 Mass of Na2CO3 = 0.05 x (23 x 2 + 12 + 16 x 3) = 5.3 g c) What is the mass of sodium bicarbonate needed to produce 12g of sodium carbonate? Berapakah jisim natrium bikarbonat yang diperlukan untuk menghasilkan 12 g natrium karbonat? Number of mole of Na2CO3 = 12 / (23 x 2 + 12 + 16 x 3)=0.1132 mol 1 mol of Na2CO3 : 2 mol NaHCO3 0.1132 mol of Na2CO3 : 0.2264 mol NaHCO3 Mass of NaHCO3 = 0.2264 x (23+1+12+ 16x 3) = 19.02 g
26 10. Lead(II) carbonate reacts with hydrochloric acid to produce lead(II) chloride. Calculate the mass of lead(II) carbonate that had reacted with the excess hydrochloric acid to produce 5000 cm3 of carbon dioxide at room temperature and pressure. Plumbun(II) karbonat bertindak balas dengan asid hidroklorik untk menghasilkan plumbum(II) klorida. Kirakan jisim plumbum(II) karbonat yang telah bertindak balas dengan asid hidroklorik berlebihan menghasilkan 5000 cm3 gas karbon dioksida pada suhu dan tekanan bilik. PbCO3 + 2HCl → PbCl2 + CO2 + H2O Volume of gas = 5000/1000 = 5 dm3 Number of mole = 5/24 = 0.2083 mol 1 mol of CO2 : 1 mol of PbCO3 0.2083 mol of CO2 : 0.2083 mol of PbCO3 Mass of PbCO3 = 0.2083 x (207 + 12 + 16 x 3 ) =55.63 g 4.4 STOICHIOMETRY 2 ( Neutralization) MaVa = a MbVb b 1. 25.0 cm3 of 2.0 mol dm-3 potassium hydroxide solution is titrated with nitric acid from a burette. 22.50 cm3 of nitric acid is needed for complete neutralization. 25.0 cm3 larutan kalium hidroksida dititrat denga asid nitrik daripada buret. Di dapati 22.50 cm3 asid nitrik diperlukan untuk mencapai takat akhir penitratan. a) Write a chemical equation for the neutralization. Tuliskan persamaan kimia bagi peneutralan ini HNO3 + KOH → KNO3 + H2O b) Calculate the number of mole of potassium hydroxide used in this experiment. Kirakan bilangan mol larutan kalium hidroksida yang digunakan dalam eksperimen ini. Number of mole of KOH , n = 2.0 25 1000 = 0.05 mol c) Calculate the number of moles of sulphuric acid needed to exactly neutralize 25.0 cm3 of 2 mol dm-3 sodium hydroxide. Kirakan bilangan mol asid nitric yang diperlukan untuk meneutralkan 25.0 cm3 larutan kalium hidroksida 2 mol dm-3. 1mol of KOH : 1 mol of HNO3 0.05 mol of KOH : 0.05 mol of HNO3 d) Determine the molarity of nitric acid. Tentukan kemolaran asid nitric. Molarity of nitric acid = 0.05 x 1000 22.5 = 2.22 moldm-3
27 2. 20 cm3 of sulphuric acid solution can neutralize 25 cm3 of 0.2 mol dm-3 sodium hydroxide solution. What is the concentration of acid? 20 cm3 larutan asid sulfurik boleh meneutralkan 25 cm3 larutan natrium hidroksida 0.2 mol dm-3 . Berapakah kepekatan asid? H2SO4 + 2NaOH → Na2SO4 + 2H2O 20 0.2 25 = 1 2 Molarity of acid, Ma= Ma= ( ½) x (0.2 x 25) 20 = 0.125 mol dm-3 3. What is the volume of 0.1 mol dm-3 potassium hydroxide needed to neutralize 50 cm3 of 0.25 mol dm-3 hydrochloric acid? Berapakah isipadu kalium hidroksida 0.1 mol dm-3 yang diperlukan untuk meneutralkan 50 cm3 asid hidroklorik 0.25 mol dm-3 ? HCl + KOH → KCl + H2O 0.25 50 0.1 = 1 1 Volume of alkali, Vb= Vb= 1 x (0.25 x 50) 0.1 = 125 cm3 4. The reaction between phosphoric acid and sodium hydroxide is represented by the following equation: Tindak balas antara asid fosforik dan natrium hidroksida boleh diwakilkan oleh persamaan berikut; H3PO4 + 3NaOH → Na3PO4 + 3H2O If 20 cm3 of the acid neutralize 50 cm3 of 0.2 mol dm-3 of sodium hydroxide, Calculate the molarity of phosphoric acid. Sekiranya 20 cm3 asid meneutralkan 50 cm3 larutan natrium hidroksida 0.2 mol dm-3 , kirakan kemolaran asid fosforik. 20 0.2 50 = 1 3 Molarity of acid , Ma = Ma= ( 1 3 ) x (0.2 x 50) 20 = 0.1667 mol dm-3
28 5. 20 cm3 of hydrochloric acid reacts completely with 50 cm3 of 0.1 mol dm-3 of calcium hydroxide solution. What is the concentration of the hydrochloric acid? 20 cm3 asid hidroklorik meneutralkan 50 cm3 larutan kalsium hidroksida 0.1 mol dm-3 . Berapakah kepekatan asid hidroklorik? 2HCl + Ca(OH)2 → CaCl2 + 2H2O 20 0.1 50 = 2 1 Ma= (1/2) x (0.1 x 50) 20 = 0.125 mol dm-3 5. The reaction between phosphoric acid and potassium hydroxide is represented by the following equation: Tindak balas antara asid fosforik dan natrium hidroksida boleh diwakilkan oleh persamaan berikut; H3PO4 + 3KOH → K3PO4 + 3H2O What is the volume of 1.0 moldm-3 potassium hydroxide solution used to neutralize 25.0 cm3 1.0 mol dm-3 phosphoric acid. Berapakah isipadu larutan kalium hidroksida 1.0 mol dm-3 yang digunakan untuk meneutralkan 20 cm3 asid fosforik 0.5 mol dm-3 ? 1.0 25 1.0 = 1 3 Vb= 3 x (1.0 x 25) 1.0 = 75 cm3 6. What is the volume of 2.0 mol dm-3 of potassium hydroxide is needed to neutralise 50 cm3 of 0.5 mol dm-3 sulphuric acid? Berakah isipadu larutan kalium hidroksida 2.0 mol dm-3 yang diperlukan untuk meneutralkan 50 cm3 asid sulfurik 0.5 mol dm-3 ? H2SO4 + 2KOH → K2SO4 + 2H2O 0.5 50 2.0 = 1 2 Vb= 2 x (0.5 x 50) 2.0 = 25 cm3
29 Chapter 2: The structure of atom/ Bab 2: Struktur atom [Perak2022-Set02-01] 1 (a) Jadual 1 menunjukkan nama beberapa bahan kimia dengan formula kimia. Table 1 shows the names of some chemical substance with its chemical formula. Bahan kimia Chemical substance Formula kimia Chemical formula Iodin Iodine I2 Kuprum(II) sulfat Copper(II) sulphate CuSO4 Aluminium Aluminium Al Tetraklorometana Tetrachloromethane CCl4 Jadual / Table 1 Berdasarkan Jadual 1, Based on Table 1, (i) Nyatakan jenis zarah bagi tetraklorometana. State the type of particles of tetrachloromethane. Molekul / Molecule ..................................................................................................................... [1 markah][1 mark] (ii) Kelaskan bahan kimia dalam Jadual 1 kepada unsur dan sebatian. Classify the substances in Table 1 into elements and compounds. Unsur Element Sebatian Compound Iodin / Iodine / I2 Aluminium /Aluminium / Al Tetraklorometana / Tetrachloromethane / CCl4 Kuprum(II) sulfat / Copper(II) sulphate / CuSO4 [2 markah / marks] (b) (i) Namakan tiga zarah subatom dalam suatu atom. Name the three subatomic particles in an atom. Proton, neutron, electron Proton, neutron, electron ..................................................................................................................... [1 markah][1 mark]
30 (ii) Atom natrium mempunyai 12 neutron dan 11 proton. Berapakah bilangan elektron dalam satu atom natrium? Sodium atom has 12 neutrons and 11 protons. How many electrons are there in one sodium atom? 11 ..................................................................................................................... [1 markah][1 mark] [Pahang2022-01] 1. Rajah 1 menunjukkan lengkung pemanasan pada ais. Diagram 1 shows the heating curve on ice. Rajah 1 / Diagram 1 (a) Nyatakan jenis zarah dalam ais. State the type of particles in ice. Molekul / Molecule ….……………………………………………………………………………………………… [1 markah/1 mark] (b) Apakah maksud takat lebur? What is the meaning of melting point? Takat lebur ialah suhu (malar) apabila sesuatu bahan bertukar daripada keadaan pepejal menjadi cecair pada tekanan tertentu Melting point is (constant) temperature when a substance turns from solid to liquid at certain pressure. ….……………………………………………………………………………………………… ….……………………………………………………………………………………………… [1 markah/1 mark] (c) Nyatakan keadaan fizik bagi ais semasa proses peleburan. State the physical state of ice during the melting process. Campuran pepejal dan cecair Mixture of solid and liquid ….……………………………………………………………………………………………… [1 markah/1 mark] D C B 70 0 -20 Suhu /Temperature (℃) Masa/Time (min) A
31 (d) Mengapakah suhu kekal malar di antara B dan C walaupun pemanasan diteruskan? Why does the temperature remain constant between B and C even as heating continues? 1.Tenaga haba yang diserap 2.digunakan untuk mengatasi daya tarikan antara zarah 1.The heat energy is absorbed to overcome the attraction force between particles ….……………………………………………………………………………………………… ….……………………………………………………………………………………………… [2 markah/2 marks] [JohorPPD BatuPahat 2022-01] 1. Jadual 1 menunjukkan takat lebur dan takat didih bahan P, Q, R, S dan T. Table 1 shows the melting and boiling point of substance P, Q, R, S and T. BAHAN SUBSTANCE TAKAT LEBUR (°C) MELTING POINT (°C) TAKAT DIDIH (°C) BOILING POINT (°C) P -101.0 -35.0 Q -94.0 65.0 R 17.8 290.0 S 97.8 883.0 T 801.0 1413.0 Jadual 1 Table 1 a) Nyatakan satu bahan yang wujud dalam keadaan pepejal pada suhu bilik. Name a substance that exists in a solid state at room temperature. S atau T ..................................................................................................................... [1 markah] [1 mark] (b) (i) Bahan manakah yang akan melalui proses pembekuan apabila diletakkan dalam peti ais yang mempunyai suhu 2.0 °C? Which substance will go through freezing process when placed in a refrigerator that has a temperature of 2.0 °C? R ..................................................................................................................... [1 markah] [1 mark] (ii) Apakah perubahan keadaan fizik bagi bahan dalam b(i)? What is the change in physical state of substance in b(i)? Cecair kepada pepejal ..................................................................................................................... [1 markah] [1 mark]
32 c) Lakarkan graf pemanasan cecair Q. Labelkan takat didih Q pada graf. Sketch the heating graph of liquid Q. Label the boiling point of Q on the graph. [2 markah / 2 marks ] - Bentuk graf - Label takat didih Masa (s) Time (s) Suhu (°C) Masa (s) Time (s) Suhu (°C) Temperature (°C) Takat didih, 65°C
33 [Selangor2022-Set3-01] 1. Jadual 1 menunjukkan nombor proton dan nombor nukleon bagi tiga atom. Huruf yang digunakan bukan simbol sebenar bagi atom-atom itu. Gunakan huruf tersebut untuk menjawab soalan berikut. Table 1 shows the proton number and nucleon number of three atoms. The letters used are not the actual symbols of the atoms. Use the letters to answer the following questions. Atom Atom Nombor proton Proton number Nombor nukleon Nucleon number T 11 23 U 11 24 V 12 24 Jadual 1 Table 1 (a) Apakah yang dimaksudkan dengan nombor nukleon? What is meant by the nucleon number? Jumlah bilangan proton dan neutron di dalam nukleus sesuatu atom. The total number of protons and neutrons in the nucleus of an atom. ..................................................................................................................... [1 markah] [1 mark] (b) Nyatakan bilangan neutron bagi atom V. State the number of neutrons for atom V. 12 ..................................................................................................................... [1 markah] [1 mark] (c) Nyatakan sepasang isotop. Terangkan jawapan anda. State a pair of isotopes. Explain your answer. •T dan U •Atom-atom dari unsur yang sama dan mempunyai bilangan proton yang sama tetapi bilangan neutron yang berbeza. •T and U Atoms from the same element and have the same number of protons but different number of neutrons. ..................................................................................................................... ..................................................................................................................... [2 markah][2 marks] (d) Nyatakan satu kegunaan atom U dalam kehidupan harian. State one use of atom U in daily life. Untuk mengesan kebocoran paip bawah tanah. To detect the leakage of underground pipe. ..................................................................................................................... [1 markah] [1 mark]
34 [Melaka2022-03] FE 3. Rajah 2 menunjukkan susunan radas untuk menentukan formula empirik bagi magnesium oksida Diagram 2 shows the apparatus set-up to determine the empirical formula of magnesium oxide Rajah 2/ Diagram 2 (a) Semasa menjalankan eksperimen ini, mengapakah penutup mangkuk pijar perlu dibuka sekali sekala? When carrying out this experiment, why does the crucible lid need to be opened once a while? Untuk membenarkan udara/ oksigen masuk ke dalam mangkuk pijar// To allow air/ oxygen to enter the crucible ..................................................................................................................... [1 markah] [1 mark] (b) Keputusan eksperimen ditunjukkan dalam Jadual 2. The result of the experiment is shown in Table 2. Penerangan Description Jisim (g) Mass (g) Mangkuk pijar + penutup Crucible + lid 26.35 Mangkuk pijar + penutup + pita magnesium Crucible + lid + magnesium ribbon 28.75 Mangkuk pijar + penutup + magnesium oksida Crucible + lid + magnesium oxide 30.35 Jadual 2/ Table 2
35 Berdasarkan keputusan dalam Jadual 2, Based on the results in Table 2, (i) hitung jisim bagi: calculate the mass of: Magnesium : ............................................................................................ Magnesium Oksigen : ............................................................................................ Oxygen [2 markah/ marks] Magnesium : (28.75 – 26.35) g // 2.4 g Oksigen : (30.35 – 28.75) g // 1.6 g (ii) Hitung bilangan mol magnesium dan oksigen. Calculate the number of moles of magnesium and atom. [Jisim atom relatif: Mg = 24, O = 16] [Relative atomic mass: Mg = 24, O = 16] Magnesium : 2.4÷24// 0.1 Oksigen : 1.6 ÷ 16// 0.1 [2 markah/ marks] (iii) Tentukan formula empirik bagi magnesium oksida dalam eksperimen itu. Determine the empirical formula for the magnesium oxide in the experiment. MgO ..................................................................................................................... [1 markah] [1 mark]
36 [SPM2021-V2-03] FE 3. Rajah 2 menunjukkan susunan radas untuk menentukan formula empirik bagi oksida R. Diagram 2 shows the apparatus set-up to determine the empirical formula of oxide of R. Rajah 2 Diagram 2 (a) Gas hidrogen dapat dihasilkan melalui tindak balas magnesium dan asid. Nyatakan satu asid yang boleh digunakan. Hydrogen gas can be produced through the reaction of magnesium and an acid. State one acid that can be used. Asid hidroklorik/ hydrochloric acid/ HCl Sebarang asid kuat – HNO3/ H2SO4 Any strong acid ..................................................................................................................... [1 markah] [1 mark] (b) Cadangkan oksida R. Suggest an oxide of R. Kuprum //copper // Cu Fe// Sn// Pb// Ag ..................................................................................................................... [1 markah] [1 mark]
37 (c) Jadual 2 menunjukkan data yang diperoleh daripada eksperimen dalam Rajah 2. Table 2 shows the data obtained from the experiment in Diagram 2. Penerangan Description Jisim (g) Mass (g Jisim tiub pembakaran dan kertas asbestos Mass of combustion tube and asbestos paper 32.50 Jisim tiub pembakaran, kertas asbestos dan oksida R Mass of combustion tube, asbestos paper and oxide of R 37.30 Jisim tiub pembakaran, kertas asbestos dan R Mass of combustion tube, asbestos paper and R 36.34 Jadual 2 Table 2 Tentukan formula empirik bagi oksida R. Determine the empirical formula of oxide of R. [Jisim atom relatif: R = 64 ; O = 16] [Relative atomic mass : R = 64 ; O = 16] Unsur Element R Oksigen Oxygen Jisim (g) Mass (g) 36.34 - 32.50 = 3.84 37.30 - 36.34 = 0.96 Bilangan mol Number of mol 3.84/64 = 0.06 0.96/16 = 0.06 Nisbah mol Ratio of mol 1 1 Formula empirik: .............................................................................. Empirical formula Formula empiric/ Empirical formula : RO [4M] [4 markah] [4 marks] [Selangor2022-Set3-03] FE 3. Sebatian karbon P mengandungi 6.67% hidrogen, 40.00% karbon dan 53.33% oksigen. Carbon compound P contains 6.67% hydrogen, 40.00% carbon and 53.33% oxygen. (a) (i) Apakah maksud formula empirik? What is the meaning of empirical formula? Formula kimia yang menunjukkan nisbah mol teringkas bagi bilangan atom setiap unsur dalam suatu molekul sebatian. A chemical formula that shows the simplest mole ratio for the number of atoms of each element in a molecule of a compound. ..................................................................................................................... ..................................................................................................................... [1 markah] [1 mark]
38 (ii) Hitungkan formula empirik bagi sebatian P. [Jisim atom relatif: C = 12, H = 1, O = 16] Calculate the empirical formula of compound P. [.Relative atomic mass: C = 12, H=l,0=16] Atom Atom C H O Jisim, g Mass, g 40.00 6.67 53.33 Bilangan mol, mol Number of mole, mol 40.00 12 = 3.33 6.67 1 = 6.67 53.33 16 = 3.33 Nisbah mol teringkas Simplest mole ratio 3.33 3.33 = 1 6.67 3.33 = 2 3.33 3.33 = 1 Formula empirik = CH2O Empirical formula [3 markah] [3 marks] (b) Jisim molekul relatif bagi sebatian P ialah 180. Hitungkan formula molekul bagi sebatian P. The relative molecular mass of compound P is 180. Calculate the molecular formula of compound P. (CH2O)n = 180 [12 + 2(1) + 16]n = 180 30n = 180 n = 6 Formula molekul P = C6H12O6 Molecular formula P = C6H12O6 [2 markah] [2 marks] [2 markah/2marks]
39 [SPM2021-V1-06] Fe 6 Jadual 3 menunjukkan persamaan perkataan bagi dua tindak balas melibatkan logam X dan oksida logam Y. Formula empirik bagi oksida X dan oksida Y ditentukan melalui Kaedah I dan Kaedah II. Table 3 shows the word equations for two reactions involving metal X and metal oxide Y. The empirical formulae of X oxide and Y oxide are determined through Method I and Method II. Kaedah Method Persamaan perkataan Word equation I X + Oksigen → Oksida X X + Oxygen → X Oxide II Hidrogen + Oksida Y → Y + Air Hydrogen + Y Oxide → Y + Water Jadual 3 Table 3 (a) Apakah yang dimaksudkan dengan formula empirik? What is meant by empirical formula? Formula kimia yang menunjukkan nisbah paling ringkas bagi bilangan atom setiap jenis unsur dalam sesuatu sebatian. The chemical formula that shows the simplest ratio of the number of atoms of each element in a compound ..................................................................................................................... [1 Markah | 1 mark] (b) Cadangkan logam X dan logam Y. Terangkan mengapa anda memilih logam tersebut. Suggest metal X and metal Y. Explain why you choose the metal. Logam X/Metal X : Magnesium / Mg //Aluminium / Al // zink//Zn Penerangan/ Explanation : X lebih reaktif berbanding hidrogen // X is more reactive than hydrogen// X reaktif terhadap oksigen X is reactive towards oxygen Logam Y/ Metal Y : ................................................................ Kuprum //copper // Cu Fe// Sn// Pb// Ag
40 Penerangan/ Explanation : ................................................................ Y kurang reaktif daripada hidrogen // Y is less reactive than hydrogen [4 markah][4 marks] (c) (i) 1.08 g X bertindak balas dengan 0.96 g oksigen. Apakah formula empirik bagi oksida X? 1.08 g of X reacts with 0.96 g oxygen. What is the empirical formula of X oxide? [Jisim atom relatif: X = 27 , O = 16 ] [Relative atomic mass : X = 27 , O = 16 ] Atom Atom X O Jisim, g Mass, g 1.08/27 0.96/16 Bilangan mol, mol Number of mole, mol = 0.04 = 0.06 Nisbah mol teringkas Simplest mole ratio 2 3 X2O3 //Al2O3 [3 markah] [3 marks] (ii) Kaedah yang manakah lebih sesuai digunakan untuk menentukan formula empirik bagi oksida plumbum? Which method is suitable to be used to determine the empirical formula for lead oxide? Kaedah II Method II [1 markah] [1 mark]
41 [MRSM2022-02] FM 2 Aspirin digunakan untuk meredakan demam serta melegakan kesakitan ringan dan sederhana. Formula struktur aspirin ditunjukkan dalam Rajah 2. Aspirin is used to reduce fever and relieve mild and moderate pain. Structural formula of aspirin is shown in Diagram 2. Rajah 2 Diagram 2 (a) (i) Nyatakan maksud formula molekul. State the meaning of molecular formula. Chemical formula that shows the actual number of atoms of each element in a molecule/compound ..................................................................................................................... [1 markah] [1 mark] (ii) Tuliskan formula molekul bagi aspirin. Write the molecular formula of aspirin. C9H8O4 ..................................................................................................................... [1 markah] [1 mark] (iii) Nyatakan jenis zarah bagi aspirin. State the type of particle of aspirin. Molecule ..................................................................................................................... [1 markah] [1 mark]
42 (b) Persamaan kimia berikut mewakili pembakaran aluminium dengan oksigen. The following chemical equation represents the burning of aluminium with oxygen. 4Al + 3O2 → 2Al2O3 Berikan dua maklumat yang boleh ditafsirkan daripada persamaan kimia di atas. Give two information that can be interpreted from the chemical equation above. P1. Identify reactants and product P2. Mole ratio//quantity ratio of reactant and product Sample answer: 4 mol of aluminium reacts with 3 mol of oxygen to produce 2 mol of aluminium oxide // 4 aluminium atoms react with 3 oxvaen molecules to produce 2 units of aluminium ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... [2 markah] [2 marks]
43 [SPM2021-V1-03] k1 3. Rajah 2 menunjukkan carta alir bagi tindak balas bermula dengan unsur Y. Unsur Y terletak dalam Kumpulan 1 Jadual Berkala Unsur. Diagram 2 shows the flow chart for the reaction starting with element Y. Element Y is located in Group 1 of the Periodic Table of Elements Rajah 2 Diagram 2 (a) Nyatakan bilangan elektron valens bagi unsur Kumpulan I. State the number of valence electron of Group I elements. 1 ..................................................................................................................... [1 Markah | 1 mark] (b) Tulis persamaan kimia bagi Tindak balas I. Write the chemical equation for Reaction I. 4Y + O2 2Y2O 4Na + O2 2Na2O Reject untuk Li/ K ..................................................................................................................... [2 markah] [2 mark] (c) 0.5 mol unsur Y terbakar dalam oksigen seperti yang ditunjukkm dalam tindak balas I. Hitung jisim oksida Y yang terbenfuk. 0.5 mol of element Y is burnt in oxygen as show in Reaction I. Calculate the mass of Y oxide formed [Jisim molar oksida Y = 62 g mol-1] [Molar mass of oxide Y = 62 g mol-1]
44 Diberikan 0.5 mol Given Nisbah/ Ratio 4Y + O2 2Y2O 4 mol Y menghasilkan 2 mol Y2O 0.5 mol Y menghasilkan 0.5 X 2/4 = 0.25 mol Y2O // 4 mol of Y produced 2 mol of Y2O // 0.5 mol of Y produced 0.5 X 2/4 = 0.25 mol of Y2O Diberikan jisim 62 g mol-1 Given the mass is Jisim/ Mass Y2O = 0.25 x 62g = 15.5 g [2 markah] [2 mark] (d) Ramalkan nilai pH bagi larutan YOH yang terbentuk. Predict the pH value of YOH solution formed. 12//13//14 reject r > 12 ..................................................................................................................... [1 Markah | 1 mark]
45 [SPM2021-V2-04] k1 4. Rajah 3 menunjukkan susunan radas bagi tindak balas unsur X dengan gas oksigen. X berada dalam Kumpulan 1 Jadual Berkala Unsur. Diagram 3 shows the apparatus set-up for the reaction of element X with oxygen gas. X is in Group 1 in the Periodic Table of Elements. Rajah 3 Diagram 3 Berdasarkan Rajah 3, Based on Diagram 3, (a). Nyatakan nama bagi Kumpulan 1. State the name of Group 1. Logam Alkali// Alkali metal ..................................................................................................................... [1 markah] [1 mark] (b) Kalium berada di bawah X dalam Kumpulan 1 dan lebih reaktif berbanding X. Terangkan mengapa. Potassium is located below X in Group 1 and it is more reactive compare to X. Explain why. Daya tarikan antara nucleus terhadap electron valens atom kalium lebih lemah berbanding dengan atom X. The force attraction between nucleus to the valence electrons in potassium atom is weaker than X atom. Atom kalium mudah mendermakan electron valensnya Potassium atom easier to donate valens electron ..................................................................................................................... ..................................................................................................................... [2 markah] [2 marks]
46 (c) (i) Tulis persamaan kimia bagi tindak balas yang terlibat dalam Rajah 3. Write the chemical equation for the reaction involved in Diagram 3. 4X + O2 → 2X2O 4Na + O2 → 2Na2O Kenapa Na.. sebab jisim di soalan bawah adalah 23 ..................................................................................................................... [2 markah] [2 marks] (ii) Hitung jisim hasil tindak balas yang terbentuk apabila 0.5 mol unsur X terbakar dalam oksigen berlebihan. [Jisim atom relatif: X=23; O=16] Calculate the mass of the product formed when 0.5 mol of element X is burned in excess oxygen. [Relative atomic mass: X=23; O=16] Diberikan 0.5 mol Given Nisbah/ Ratio 4X + O2 → 2X2O 4 mol X menghasilkan 2 mol X2O 0.5 mol X menghasilkan 0.5 X 2/4 = 0.25 mol X2O // 4 mol of X produced 2 mol of X2O // 0.5 mol of X produced 0.5 X 2/4 = 0.25 mol of X2O Jisim/ Mass = 0.25 X [23 X 2 + 16] = 0.25 x 62 = 15.5 g [2 markah] [2 marks]
47 [Perlis2022-08] 8. Rajah 7 menunjukkan sebahagian daripada Jadual Berkala Unsur. A, B, C, D dan E bukan merupakan simbol sebenar unsur. Diagram 7 shows part of the Periodic Table of Element. A, B, C, D and E are not the actual symbols of the elements. Rajah 7 Diagram 7 Berdasarkan rajah 7, Based on diagram 7, (a) (i) Unsur manakah yang merupakan halogen? Which element is a halogen? E ..................................................................................................................... [1 markah][1 mark] (ii) Nyatakan dua unsur yang terletak dalam kumpulan yang sama. State two elements that are placed in the same group. A and D ..................................................................................................................... [1 markah][1 mark] (b) 0.1 mol wul besi bertindak balas dengan gas klorin berlebihan. 0.1 mol of iron wool reacts with excess chlorine gas. (i) Seimbangkan persamaan kimia bagi tindak balas itu. Balanced chemical equation for the reaction. ......... Fe + …….. Cl2 → ……. FeCl3 ......2... Fe + …3….. Cl2 → ……2. FeCl3 [1 markah][1 mark]
48 (ii) Hitung jisim hasil tindak balas yang terbentuk. Calculate the mass of the product formed. [Jisim atom relatif: Cl = 35.5, Fe = 56] [Relative atomic mass: Cl = 35.5, Fe = 56] 2 mole of Fe produce 2 mol FeCl3 0.1 mole of Fe produce 0.1 mol FeCl3 Mass of FeCl3 = 0.1 x [ 56 + (3 x 35.5) ] = 16.26 g [2 markah] [2 marks] (c) Jadual 3 menunjukkan keputusan apabila dua unsur Kumpulan 1 bertindak balas dengan oksigen. Table 3 shows the result when Group 1 elements react with oxygen. Unsur Element Pemerhatian Observation A Terbakar sangat perlahan dengan nyalaan merah. Burn slowly with red flame. D Terbakar sangat cergas dengan nyalaan ungu. Burn vigorously with purple flame. Jadual 3 Table 3 Terangkan perbezaan kereaktifan bagi pemerhatian antara unsur A dan unsur D. Explain reactivity differences in the observation between A and D elements. P1 : Saiz atom A lebih kecil/ Size A atom smaller// Saiz atom D lebih besar/ Size D atom bigger P2 Atom A sukar membebaskan satu elektron valens/ A atom harder/more difficult to release electron// Atom D mudah membebaskan satu elektron valens/ D atom easier to release one valence electron atau/or Daya tarikan antara nukleus dan elektron valens di dalam atom A lebih kuat/ Force of attraction between nucleus and valence electron in A atom stronger// Daya tarikan antara nukleus dan elektron valens di dalam atom D lebih lemah/
49 Force of attraction between nucleus and valence electron in D atom weaker ..................................................................................................................... ……............................................................................................................... [2 markah] [2 marks] (d) Oksida B bersifat amfoterik. Cadangkan dua bahan kimia yang sesuai untuk menunjukkan oksida logam B ialah oksida amfoterik. Berikan alasan anda. B oxide is amphoteric in properties. Suggest two chemical substances to show metal B oxide is amphoteric oxide. Give your reason. Natrium hidroksida & asid hidroklorik sodium hydroxide & Hydrochloric acid ● any acid and alkali accepted Amphoteric oxide reacts with both acid and alkali. ..................................................................................................................... ..................................................................................................................... ..................................................................................................................... [3 markah] [3 marks]