1 Answers Chapter 1 Test Yourself 1.1 (pg. 8) 1. Chemistry is one of the science fields that studies the structure, properties, composition and interactions of matter. 2. Preservative, vitamin, pesticide, paint, detergent (any other acceptable answers). 3. Technological development in food preservation (any other acceptable answers). 4. (a) Researcher, doctor, pharmacist (any other acceptable answers). (b) Biotechnology researcher, farmer, biotechnology engineer (any other acceptable answers). Brain Teaser (pg. 10) Add ice to the distilled water and stir the mixture using a thermometer until 10 °C of temperature is obtained. Test Yourself 1.2 (pg. 11) 1. Scientific method is a systematic method used by scientists and researchers when solving problems in science. 2. Scientific method helps chemists to solve problems through systematic scientific steps. 3. (Student’s suitable answer) Hypothesis: The addition of sugar will increase the rate of melting of ice Manipulated variable: Mass of sugar used Responding variable: Rate of melting of ice Fixed variable: Surrounding temperature, type of sugar used, mass of ice Procedure: 1. Weigh 200 g ice and put in a small pail. 2. Put the small pail in a shaded place for 5 minutes. 3. After 5 minutes, use a measuring cylinder to measure the volume of water formed in the pail. 4. Record the reading in a table. 5. Repeat steps 1 to 4 by adding 10 g, 30 g and 50 g sugar. Brain Teaser (pg. 16) No. There are specific methods to get rid of different types of chemical wastes in the laboratory. Chemical wastes that are thrown directly into the sink or rubbish bin could pollute the environment and endanger human health. Test Yourself 1.3 (pg. 18) 1. – Do not play in the laboratory. – Wear protective equipment when carrying out experiments. – Do not bring food or drinks into the laboratory. (any other acceptable answers). 2. (a) A specially designed equipment to carry out experiments that release fumes that are toxic, flammable or pungent. (b) Used to wash and clean the body when accidents occur to the body. This equipment is also used to put out fire that occurs on the body. (c) Used to protect the body and clothes from spillage of chemicals such as acid, alkali and organic solvent. 3. Solid residue such as glass and rubber must be thrown in special containers. 4. Oxygen gas – put a glowing wooden splinter into the test tube; the glowing wooden splinter is rekindled. Hydrogen gas – put a lighted wooden splinter at the mouth of the test tube; a ‘pop’ sound is heard. 5. Repeat the titration until the difference of two volume values does not exceed 0.05 cm3 .
2 Achievement Test 1 (pg. 20) 1. (a) Salt, sugar, paint, preservative, detergent (any other acceptable answers). (b) Salt – as flavouring agent or preservative. Sugar – as flavouring agent or preservative Paint – used to paint buildings, wood or metal as protection Preservative – to preserve food so that it can be kept for a longer period Detergent – used as cleaning agent to remove dirt from clothes (any other acceptable answers). 2. (a) Agricultural industry, cosmetics industry, medical industry (any other acceptable answers). (b) Agriculture – provides source of food and income to the country Cosmetics – provides source of income to the country Medicine – provides new medicines to the people and improve their quality of life (any other acceptable answers). 3. Safety glasses – used to prevent dust or chemicals from splashing into the eyes Laboratory coat – used to protect the body and clothes from chemical spills such as acid, alkali and organic solvent Gloves – used when handling chemicals to protect the hands from injury, chemicals or infections (any other acceptable answers) 4. Making observations ➞ Making an inference ➞ Identifying the problem ➞ Making a hypothesis ➞ Identifying variables ➞ Controlling the variables ➞ Planning the experiment ➞ Collecting data ➞ Interpreting data ➞ Making a conclusion ➞ Writing a report. 5. Precautionary steps to be taken: 1. Inform the teacher or laboratory assistant immediately about the accident. 2. Mark the spillage area as a restricted zone. 3. Sprinkle sulphur powder to cover the spillage 4. Contact the Fire and Rescue Department for further action. 6. (a) To give a different surface area for the cloth. (b) The larger the surface area of the cloth, the faster the cloth will dry. (c) Cloth with larger surface area will dry faster than cloth with smaller surface area. (d) Manipulated variable: Surface area of cloth Responding variable: Rate of drying of cloth Fixed variable: Surrounding temperature, type of cloth used, size of cloth (e) Surface area of cloth (cm2 ) 25 50 100 Time taken for the cloth to dry (minutes) Enrichment Corner (pg. 21) 1. (a) Used gloves, broken conical flask, bromine, concentrated acid (b) Gloves and conical flask – must be thrown into a special container Bromine – must be stored in a covered container away from light and heat source Concentrated acid – must be put in a labelled container and disposed 2. (Student’s suitable answer) Hypothesis: Soil with acidic pH is not suitable to plant vegetables Method to control: Manipulated variable: Type of soil // neutral soil, acidic soil and alkaline soil Responding variable: Rate of growth of vegetables in soil of different pH Fixed variable: Type of vegetable, same light intensity
1 Answers Chapter 2 Activity 2.2 (pg. 26 – 27) Discussion: 1. (a) Naphthalene, C10H8 is a flammable substance. (b) Water bath method helps to distribute heat evenly. This ensures even heating. 2. (a) To ensure even cooling (b) To distribute heat evenly (c) Supercooling will happen 3. During melting, heat energy that is absorbed by the naphthalene particles C10H8 is used to overcome attraction force between the particles so that the solid changes to liquid. During cooling, heat energy that is released to the surroundings is balanced by the heat energy released when particles attract each other to form a solid. Test Yourself 2.1 (pg. 29) 1. Atom 2. (a) Evaporation (b) Water molecules more freely and faster 3. (a) Time (minutes) 50 43 Temperature (°C) (b) Heat energy that is absorbed by the lauric acid particles, C12H24O2 is used to overcome attraction force between the particles until the solid changes to liquid. Brain Teaser (pg. 29) The following table shows the mass and charge of a proton, neutron and electron. Subatomic particle Proton Neutron Electron Mass (g) 1.67 × 10–24 1.67 × 10–24 9.11 × 10–28 Charge (C) +1.60 × 10–19 0 –1.60 × 10–19 A proton is positively-charged. An electron is negatively-charged. Because the charge of a proton and electron is of the same magnitude, the relative charge of a proton is +1 whereas the relative charge of an electron is –1. The mass of a proton and neutron is the same. To determine the relative mass of a proton and neutron, the following formula is used. Mass of neutron Mass of proton = 1.67 × 10–24 1.67 × 10–24 = 1
2 To determine the relative mass of an electron, the following formula is used. Mass of electron Mass of proton = 9.11 × 10–28 1.67 × 10–24 = 0.0005 (equal is to 0) Hence, the relative charge and relative mass of the subatomic particles are as follows. Subatomic particle Proton Neutron Electron Relative charge +1 0 –1 Relative mass 1 1 0 Test Yourself 2.2 (pg. 32) 1. (a) Electron (b) Proton and neutron (c) Subatomic particle Proton Neutron Electron Relative charge –1 +1 Neutral Relative mass 0 1 1 2. (a) James Chadwick (b) Shell Nucleus that contains protons and neutrons Electron Brain Teaser (pg. 32) No. Each element has a specific proton number. Activity 2.5 (pg. 34) 1. (a) False (b) Nucleon number = 11 + 12 = 23 (c) Proton number = 7 (d) Number of electrons received = 2 (e) Nucleon number = 4 + 5 = 9 (f) Number of neutrons = 40 – 20 = 20 (g) Number of protons = 3 3. BOHRIUM Test Yourself 2.3 (pg. 36) 1. 23 2. 39 19Z 3. Number of protons = 11; Number of neutrons = 12; Number of electrons = 10
3 4. (a) 2.8 (b) X (c) 10 p 10 n Electron Nucleus that contains protons and neutrons Test Yourself 2.4 (pg. 39) 1. Isotopes are atoms of the same element with the same proton number / same number of protons but with different nucleon number / number of neutrons. 2. Atoms W and X This is because the atom W and X have the same number of protons / same number of protons but different nucleon number / number of neutrons. 3. Similarity: • This atom is from the same element, that is oxygen • The number of protons in the nucleus of this atom is the same • The chemical properties of these atoms are the same Differences: • The nucleon number of all three atoms is different. The nucleon number for oxygen-16 is 16, the nucleon number for oxygen-17 is 17 and the nucleon number for oxygen-18 is 18. • The number of neutrons in the nucleus of all three atoms is different. Oxygen-16 has 8 neutrons, oxygen-17 has 9 neutrons and oxygen-18 has 10 neutrons. • The physical properties of all three atoms are different. 4. Relative atomic mass of magnesium = (79.0 × 24) + (10.0 × 25) + (11.0 × 26) 100 = 24.32 5. (a) Cobalt-60 (b) Positive effect: • Cancer cells can be killed Negative effect: • Immunity system is weakened • Hair loss Achievement Test 2 (pg. 40 – 41) 1. (a) Physical state Solid Liquid Gas Substance D, E B, C A (b) Substance C (c) Substance B changes from gas to liquid. The kinetic energy of substance B particles decreases. The attraction force between the particles become stronger.
4 2. (b) Time (minutes) P Q R S Melting point Temperature (°C) (b) (c) Water bath (d) Thermometer Boiling tube Water Lauric acid, C12H24O2 Heat 3. Alcohol evaporates when it is rubbed onto the skin. Heat is absorbed from the skin by alcohol particles causing Chen Ling to feel cold. 4. (a) Proton (b) Neutron (c) X: 2.1 Y: 2.7 Z: 2.8.3 (d) Nucleon number of Z atom = 13 + 14 = 27 5. y = nucleon number of y B isotope Relative atomic mass of boron = 10.81 (80.0 × 11) + (20.0 × y) 100 = 10.81 880 + 20y = 1081 y = 1 081 – 880 20 y = 10.05 Hence, the nucleon number of y B isotope = 10
5 6. (a) Number of protons = 78 Number of neutrons = 195 – 78 = 117 (b) Charge of platinum ion = 78 – 74 = +4 7. (Suitable student’s answer) Reasonable • The dosage of iodine-131 that is given for treatment is small • The thyroid gland that secretes excess hormone can be destroyed • Problems caused by hyperthyroidism such as rapid pulse rate, excessive sweating, drastic weight loss and others can be overcome Unreasonable • The thyroid gland is destroyed when undergoing hyperthyroidism treatment using iodine-131 • The thyroid gland plays an important role in producing hormones for body metabolism • The patient needs to take thyroid pills after undergoing hyperthyroidism treatment Enrichment corner (pg. 41) 1. (Student’s suitable answer) Tie a transparent plastic bag at the shoot or stem using a string. Make sure that the transparent plastic bag is tied tightly. Water vapour produced by the transpiration of plant will accumulate in the plastic bag and condense to form water.
1 Answers Chapter 3 Brain Teaser (pg. 44) The problem is because not all elements can combine with hydrogen or displace hydrogen. Hydrogen exists as a diatomic molecule and is difficult to exist as monoatomic. Brain Teaser (pg. 45) The RAM of magnesium is 24. Activity 3.2 (pg. 46 – 47) A. Discussion: 1. 12 2. Washer 3. Average mass of one atom of an element compared to 1 12 mass of one atom of carbon-12 B. Discussion: 1. Average mass of one molecule compared to 1 12 mass of one atom of carbon-12 2. (Student’s suitable answer) 3. Relative molecular mass is the sum of all relative atomic masses of all atoms that form a molecule 4. (Student’s suitable answer using the following relationship: RAM of molecule W = RAM of A + RAM of B + RAM of C) Activity 3.3 (pg. 48) 1. 2 5. 92 9. 294 2. 48 6. 58 10. 242 3. 28 7. 135 4. 17 8. 99 Test Yourself 3.1 (pg. 49) 1. Relative atomic mass is the average mass of one atom of an element compared to 1 12 mass of one atom of carbon-12 2. (a) 12 lithium atoms (b) 27 helium atoms 3. (a) 16 (c) 98 (b) 148 (d) 46 Activity 3.5 (pg. 51 – 52) 1. (a) 6.02 × 1022 atoms (b) 2.107 × 1024 atoms 2. (a) 7.224 × 1023 molecules (b) 4.816 × 1023 molecules 3. (a) 1.806 × 1024 formula units (b) 1.505 × 1023 formula units 4. (a) 10 mol (c) 0.15 mol (b) 0.5 mol (d) 6 mol 5. (a) 30 mol (b) 3.612 × 1025 ions 6. (a) 1.204 × 1023 ethene molecules (b) 4.816 × 1023 H atoms (c) 7.224 × 1023 atoms Activity 3.6 (pg. 53 – 54) 1. (a) 22.4 g (b) 61.6 g 2. (a) 0.5 mol (b) 0.06 mol
2 3. 6.6 g 4. 55 g mol–1 Activity 3.7 (pg. 55) 1. 13.44 dm3 at STP; 14.4 dm3 at room conditions 2. (a) 0.002 mol (b) 1.75 mol 3. 12 dm3 Activity 3.9 (pg. 58) 2. (a) (i) 1.505 × 1023 atoms (ii) 7.224 × 1023 atoms (b) (i) 4.48 dm3 (ii) 0.448 dm3 (c) 3.2 g (d) 2.4682 × 1023 methane molecules; 6.56 g (e) 2.2 g Test Yourself 3.2 (pg. 58) 1. (a) 207 g mol–1 (c) 85 g mol–1 (b) 119.5 g mol–1 (d) 160 g mol–1 2. 4.816 × 1024 molecules 3. 8.5 g 4. 0.15 mol 5. 11.2 dm3 6. 72 g mol–1 7. Yes. 4 g of hydrogen gas contains 2 mol of hydrogen gas, that is 1.204 × 1024 H2 molecules. 14 g of nitrogen gas contains 0.5 mol nitrogen gas, that is 3.01 × 1023 N2 molecules. Brain Teaser (pg. 60) The molecular formula of a compound is a multiple of its empirical formula. Molecular formula = (Empirical formula)n The molecular formula of some compounds is the same as its empirical formula because, the value of the multiplication n = 1. The molecular formula of some compounds is different from its empirical formula because the value of the multiplication n = 2, 3, 4, ... Brain Teaser(pg. 61) C3 H7 Activity 3.11 (pg. 61) 1. KBr 2. SnCl4 3. YI2 4. C7 H14O2 Activity 3.12 (pg. 62) Discussion 1. To remove the oxide layer on the surface of the magnesium tape 2. Magnesium oxide 3. – Step 6 aims to prevent the loss of white fumes – Step 7 aims to allow air to enter the crucible so that oxygen in air can react with magnesium tape – Step 11 aims to ensure that the magnesium is completely burned 4. The results obtained is not accurate and could affect the determination of the empirical formula.
3 Activity 3.13 (pg. 63 – 64) Discussion: 1. To produce hydrogen gas 2. To remove air in the glass tube 3. To prevent air from entering the apparatus because oxygen in air will oxidize the hot copper back to copper(II) oxide 4. To ensure that all copper(II) oxide are completely changed to copper Activity 3.14 (pg. 66) 1. C2 H4 O2 2. C8 H18 3. 16.25 g 4. Urea because the percentage by mass of nitrogen is the highest, that is 46.67% % nitrogen in ammonium nitrate = 2(14) 14 + 4(1) + 14 + 3(16) × 100% = 35 % % nitrogen in urea = 2(14) 12 + 16 + 2[14 + 2(1)] × 100% = 46.67 % % nitrogen in calcium nitrate = 2(14) 40 + 2[14 + 3(16)] × 100% = 17.07 % Activity 3.16 (pg. 69) 1. (a) Calcium chloride (c) Magnesium nitrate (e) Sodium nitrate (b) Potassium bromide (d) Zinc carbonate (f) Ammonium chloride 2. (a) Nitrogen monoxide (c) Sulphur trioxide (e) Boron trifluoride (b) Carbon dioxide (d) Carbon tetrachloride (f) Carbon disulphide 3. Dinitrogen trioxide Test Yourself 3.3 (pg. 69) 1. Empirical formula is the formula that shows the simplest ratio of the number of atoms of each element in a substance. Molecular formula is the formula that shows the actual number of atoms of each element in a molecule of a substance. 2. C4 H5 N2 O 3. CaCO3 ; NaF 4. (a) Molecular formula P2 O4 ; Empirical formula PO2 (b) Diphosphorus tetraoxide Activity 3.17 (pg. 70) 1. (a) N2 (g) + 3H2 (g) → 2NH3 (g) (b) 2Na(s) + 2H2 O(l) → 2NaOH(aq) + H2 (g) (c) CuCO3 (s) ∆ → CuO(s) + CO2 (g) (d) 4Al(s)) + 3O2 (g) → 2Al2 O3 (s) 2. (a) 2KI(aq) + Br2 (aq) → I2 (s) + 2KBr(aq) (b) Zn(s) + 2AgNO3 (aq) → Zn(NO3 )2 (aq) + 2Ag(s) (c) C3 H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2 O(l) (d) 2AgNO3 (s) ∆ → 2Ag(s) + 2NO2 (s) + O2 (g)
4 Activity 3.18 (pg. 71) Discussion: 1. Activity (a) Reactant and Product (b) Physical state (c) Chemical formula A Reactant: Copper(II) carbonate Solid CuCO3 Product: Copper(II) oxide Carbon dioxide Solid Gas CuO CO2 B Reactant: Hydrogen chloride Ammonia Gas Gas HCl NH3 Product: Ammonium chloride Solid NH4 Cl C Reactant: Potassium iodide Lead(II) nitrate Aqueous Aqueous KI Pb(NO3 )2 Product: Potassium nitrate Lead(II) iodide Aqueous Solid KNO3 PbI2 2. Reaction A CuCO3 (s) → CuO(s) + CO2 (g) Reaction B HCl(g) + NH3 (g) → NH4 Cl(s) Reaction C 2KI(aq) + Pb(NO3 )2 (aq) → 2KNO3 (aq) + PbI2 (s) Activity 3.20 (pg. 73 – 74) 1. 5 g 2. 32.5 g 3. 8 g 4. 3.612 × 1023 ammonia molecules Activity 3.21 (pg. 74) 1. Volume of oxygen (dm3 ) 1 5 10 20 50 Mass of potassium chlorate(V) required (g) 3.403 17.014 34.028 68.056 170.139 2. Mass of potassium chlorate(V) (kg) 0.25 0.5 1 1.5 2 Volume of oxygen (dm3 ) 73.469 146.939 293.877 440.816 587.755 Test Yourself 3.4 (pg. 74) 1. (a) Cu(s) + 2AgNO3 (aq) → Cu(NO3 )2 (aq) + 2Ag(s) (b) Zn(s) + Cl2 (g) → ZnCl2 (s) 2. (a) Water and oxygen gas (b) 10.08 dm3
5 Achievement Test 3 (pg. 76 – 77) 1. Molar mass is the mass of one mole of substance in gram Molar volume is the volume of one mole of gas 2. Number of particles in a substance = Avogadro’s constant, NA × Number of moles of substance (or number of moles of substance = Number of particles in a substance Avogadro’s constant, NA ) 3. Mass of one nitrogen atom is 14 times larger than 1 12 mass of one carbon-12 atom 4. (a) C3 H4 O3 (b) 176 5. Al(OH)3 and MgCO3 6. C6 H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2 O(l) 7. (a) 278 g mol–1 (b) 20.14% 8. YO2 9. Q, R, P 10. Molecular formula is more suitable to be used in an equation because a molecular formula shows the actual number of each type of atom in a molecular compound. Hence, a stoichiometric equation that uses molecular formula can give the actual mole ratio of reactants involved. Hence, it can be used to solve various numerical reaction problems. Enrichment corner (pg. 77) 1. 3H2 O (g) + 2Fe(s) → Fe2 O3 (s) + 3H2 (g) Mass of steam = 100 56 × 3 2 × 18 = 48.2 g 2. Total CO2 = 42 × 24 hours × 18 days × 5 people = 90720 g Mass of LiOH required = 90720 44 × 2 × 24 = 98967.27 g Number of absorption tubes = 98967.27 750 = 131.96 ≈ 132 of tubes
1 Answers Chapter 4 Brain Teaser (pg. 81) The elements in the Periodic Table of Elements are arranged in order of increasing proton number, from 1 to 118. Test Yourself 4.1 (pg. 82) 1. (a) Antoine Lavoisier (b) John Newlands (c) Johann W. Dobereiner 2. Difference: Dobereiner grouped three elements having the same average mass and named them triads. All three elements in a triad were found to have the same chemical properties. His classification showed the relationship between the chemical properties of elements with atomic mass. Newlands arranged elements in order of increasing atomic mass. He arranged seven elements in a row because he found that the chemical and physical properties of the first element repeats for every eighth element. He names this arrangement as Octave Law. Similarity: Both relate atomic mass with chemical properties. Test Yourself 4.2 (pg. 84) 1. Mg, Cu, F 2. (a) 2.8.8.1; Group 1 (b) 2.4; Group 14 (c) 2.8.7; Group 17 (d) 2.8.8; Group 18 3. C Carbon 2.4 Li Lithium 2.1 Brain Teaser (pg. 85) Density = Mass Volume The size of atoms increases when going down Group 18. The increase in size increases the atomic mass. When going down Group 18, the increase in atomic mass of the element is faster than the increase in volume. Hence, the density of elements increases when going down Group 18. Test Yourself 4.3 (pg. 87) 1. 2 2. Octet 3. The neon atom has a stable octet arrangement of electrons
2 4. The size of atoms increases when going down the group because the number of filled electron shells increases. The melting point and boiling point of argon are higher than those of helium. The size of the atom increases when going down the group because the number of filled electron shells increases / atomic size of argon is larger compared to helium atom. The attraction force between argon atoms are stronger compared to helium atoms. More heat energy is needed to overcome the attraction force between argon atoms. Experiment 4,1 (pg. 88 – 90) Discussion 1. (a) 2Li(s) + 2H2 O(l) → 2LiOH(aq) + H2 (g) 2Na(s) + 2H2 O(l) → 2NaOH(aq) + H2 (g) 2K(s) + 2H2 O(l) → 2KOH(aq) + H2 (g) (b) 4Li(s) + O2 (g) → 2Li2 O(s) 4Na(s) + O2 (g) → 2Na2 O(s) 4K(s) + O2 (g) → 2K2 O(s) (c) 2Li(s) + Cl2 (g) → 2LiCl(s) 2Na(s) + Cl2 (g) → 2NaCl(s) 2K(s) + Cl2 (g) → 2KCl(s) 2. With water, oxygen and chlorine: Li, Na, K Test Yourself 4.4 (pg. 92) 1. Lithium and rubidium (any Group 1 element) 2. (a) • The melting point and boiling point of X are the highest, followed with Y and Z. • The size of atom Z is the largest, followed with Y and X (b) 4X(s) + O2 (g) → 2X2 O(s) (c) Reactivity in increasing order: X, Y, Z Metal Z is the most reactive, followed with metals Y and X. The reactivity of Group 1 metals increases when going down the group because the size of atoms increases. Hence it is easier for the atom to donate the valence electron that is further away from the nucleus. Test Yourself 4.5 (pg. 96) 1. • The melting point and boiling point increases • The colour becomes darker • The density increases 2. The reactivity of Group 17 elements depends on the ability of the atom to receive an electron easily. The size of the fluorine atom is smaller than the size of the iodine atom. Hence the nuclear attraction towards electrons in the fluorine atom is stronger compared to iodine. Hence it is easier for fluorine to receive an electron compared to iodine. 3. • Disinfectant contains iodine. • Drinking water contains chlorine. 4. Astatine is rare radioactive element because it is chemically unstable. Astatine can be produced form the radioactive decay of uranium, neptunium and plutonium. Hence, astatine is not used in the laboratory. Experiment 4.2 (pg. 97 – 98) Discussion 1. Basic oxide: Sodium oxide, Na2 O, magnesium oxide, MgO Amphoteric oxide: Aluminium oxide, Al2 O3 Acidic oxide: Silicon(IV) oxide, SiO2 , sulphur dioxide, SO2 2. MgO(s) + 2HNO3 (aq) → Mg(NO3 )2 (aq) + H2 O(l)
3 3. Al2 O3 (s) + 2NaOH(aq) → 2NaAlO2 (aq) + H2 O(l) 4. Element with basic oxide: Na and Mg Element with acidic oxide: Si, P, S and Cl Test Yourself 4.6 (pg. 100) 1. These elements have three electron shells. Sodium, Na (2.8.1), magnesium, Mg (2.8.2), aluminium, Al (2.8.3), silicon, Si (2.8.4), phosphorus, P (2.8.5), sulphur, S (2.8.6), chlorine, Cl (2.8.7) and argon, Ar (2.8.8). 2. Silicon has high melting point and boiling point. Chlorine has low melting point and boiling point. (Student’s suitable answer) 3. (a) Chlorine, Cl has smaller atomic size than aluminium. This is because when going across Period 3 from left to right, the number of shells is the same but the number of protons in the atom increases. This causes the nuclear attraction towards the electron to increase and causes the atomic radius to decrease. Hence, the size of the atom decreases. (b) Aluminium, Al (c) Ne, C, Cl, Al, Na Test Yourself 4.7 (pg. 104) 1. (a) Transition elements form coloured ions (b) Have different oxidation numbers / form complex ions 2. Haber Process – iron filings / Contact Process – vanadium(V) oxide Achievement Test 4 (pg. 106 – 107) 1. According to increase in proton number 2. (a) Group 17 (b) Period 3 3. Neon 4. This element is a shiny grey, soft metal. Group 1 elements react vigorously with oxygen, water and chlorine gas. 5. Aluminium 6. (a) Electron arrangement of X = 2.2 Electron arrangement of Y = 2.8.7 (b) Chemical properties of X Chemical properties of Y Forms basic oxide Forms acidic oxide Reacts with oxygen to form metal oxide Reacts with metal to form metal halide (c) • The reactivity of an element that is in the same group as Element X depends on the tendency of the atom to lose two electrons • When going down the group, the increasing size of the atom causes the distance between the nucleus and the valence electrons to increase • The nuclear attraction towards the valence electrons become weaker • It is easier for the atom to lose electrons. Hence, the reactivity of the element increases when going down the group. • The reactivity of an element that is in the same group as Element Y depends on the tendency of the atom to receive or pull one electron. • When going down the group, the increasing size of the atom causes the distance between the nucleus and the valence electrons to increase. • The nuclear attraction towards the valence electrons become weaker • It is harder for the atom to receive or attract electrons. Hence, the reactivity of the element decreases when going down the group.
4 7. 2Na(s) + Cl2 (g) → 2NaCl(s) 8. (a) Group 13 (b) Period 3 (c) G is a metal 9. (a) Metal: Na, Mg, Al Metalloid: Si Non-metal: P, S, Cl, Ar (b) The atomic radius decreases from Na to Cl. This is because as the number of electrons increases, the number of protons also increases. Hence the nuclear charge increases and its attraction towards the electrons also increases causing the electronegativity to increase. Hence, the size of the atom decreases as the atomic radius to decreases. (c) Argon (d) 2Na(s) + 2H2 O(l) → 2NaOH(aq) + H2 (g) 10. (a) Green (b) Brown Enrichment corner (pg. 107) 1. Silicon is used in the manufacture of microchips because it is a half metal or known as metalloid. A metalloid shows properties of a metal and non-metal as it is a weak electrical conductor with insulator properties. Hence, silicon does not allow large current to pass through it compared to metals that will cause the microchip to become hot and then burn. Metals such as lithium cannot be used because lithium is a very reactive metal and is not suitable to make semiconductors that require a composition of metal and non-metal.
1 Answers Chapter 5 Brain Teaser (pg. 110) The electron arrangement in the inner shell has achieved a stable duplet or octet electron arrangement. Hence, the electrons in the inner shell are not transferred or shared with other atoms. Thus, electrons in the inner shells are not involved in chemical bonds. Test Yourself 5.1 (pg. 111) 1. A chemical bond means a bond formed between atoms that transfer or share electrons to form a compound. 2. Ionic bond and covalent bond 3. Noble gas atoms have stable duplet or octet electron arrangement. Noble gas atoms do not need to transfer or share electrons with other atoms. 4. No. To achieve a stable octet electron arrangement, sodium, Na has to donate one valence electron. Test Yourself 5.2 (pg. 114) 1. (a) Al3+ and F– (b) Al → Al3+ + 3e– F + e– → F– (c) 2.8.3 Aluminium, Al + + + 2.7 Fluorine, F Al Al F 2.8 Fluoride ion, F– F 2.7 Fluorine, F F 2.7 Fluorine, F 2.8 Aluminium ion, Al3+ 3+ – F 2.8 Fluoride ion, F– F 2.8 Fluoride ion, F– F – – 2. (a) KCl (b) • The electron arrangement of potassium atom, K is 2.8.8.1. The potassium atom donates one electron to achieve a stable octet electron arrangement and K+ ion is formed. • The donated electron is transferred to a chlorine atom. • The electron arrangement of chlorine atom, Cl is 2.8.7. The chlorine atom receives one electron to form Cl– ion with a stable octet electron arrangement. • The potassium ion, K+ and chloride ion, Cl– of opposite charges are attracted to each other by strong electrostatic attraction force. • Potassium chloride compound, KCl is formed.
2 Test Yourself 5.3 (pg. 117) 1. Single bond, double bond, triple bond 2. A covalent bond is formed when non-metal atoms share electrons to achieve a stable duplet or octet electron arrangement. 3. 2.6 Oxygen atom 1 Hydrogen atom 2 H + O 2 2.8 Water molecule, H2 O 2 H O H 4. • Yes • The electron arrangement of carbon, C is 2.4. Carbon atom has four valence electrons. Carbon atom needs four electrons to achieve a stable octet electron arrangement, that is 2.8. Thus, carbon atom contributes four electrons to be shared. • The electron arrangement of hydrogen, H is one. Hydrogen atom needs one electron to achieve a stable duplet electron arrangement. • Thus, one carbon atom shares four pairs of electrons with four hydrogen atoms. Four single covalent bonds are formed and a covalent compound with formula CH4 is formed. 5. Only valence electrons are involved in the formation of ionic bonds and covalent bonds. Ionic bond involves electron transfer, whereas covalent bond involves sharing of electrons. Ionic bond involves a metal and non-metal element, whereas covalent bond involves non-metal elements only. Brain Teaser (pg. 118) Dry curly hair forms hydrogen bonds with each other in the hair structure. When curly hair gets wet, the protein molecules form hydrogen bonds with water molecules. Water molecules then form hydrogen bonds with other hair protein molecules and stick together. This will cause wet, curly hair to appear straight. Test Yourself 5.4 (pg. 119) 1. The attraction force between hydrogen atom, H bonded to highly electronegative atoms, such as nitrogen, N, oxygen, O or fluorine, F with nitrogen atoms, N, oxygen, O or fluorine, F in other molecules. 2. Hydrogen fluoride molecule, HF consists of one hydrogen atom, H and one fluorine atom, F. Fluorine atom has high electronegativity. The attraction force is formed between the hydrogen atom that is bonded to the fluorine atom in one molecule with a fluorine atom in another hydrogen fluoride, HF molecule. A strong hydrogen bond is formed. A lot of heat energy is required to overcome the attractive van der Waals forces between the hydrogen fluoride molecules in addition to the hydrogen bonds. Hence, hydrogen fluoride has high boiling point and exists as liquid at room temperature. 3. No. Even though chlorine atom has high electronegativity, its size is too large. 4. Cellulose in paper has oxygen atom bonded to hydrogen atom. Water molecules will form hydrogen bonds with cellulose in paper. Thus, paper stick together when wet. Test Yourself 5.5 (pg. 121) 1. Dative bond or coordinate bond is a type of covalent bond where the shared pair of electrons originate from one atom only. 2. Hydrogen ion, H+ does not have any electron in its shell. The nitrogen atom in ammonia has a pair of electrons that can be shared with the hydrogen ion. The shared electrons form a dative bond.
3 3. Can. The nitrogen atom in ammonia has a pair of electrons that can be shared with the boron atom in boron trifluoride through the formation of a dative bond so that the boron atom can have a stable octet electron arrangement. Test Yourself 5.6 (pg. 122) 1. Mobile electrons that are not owned by any atom or ion. 2. The valence electron of metal atoms can be easily donated and delocalised in solid state. Positively charged metal ions are formed. All delocalised valence electrons can move freely within the metal structure to form an electron sea. The electrostatic attraction between the electron sea and the positively charged metal ions form metallic bonds. 3. Aluminium metal has valence electrons that are easily delocalised to form an electron sea. These delocalised electrons can move freely and carry charges from the negative terminal to the positive terminal and conduct electricity. Experiment 5.1 (pg. 123 – 124) Discussion: 1. Lead(II) bromide, PbBr2 and magnesium chloride, MgCl2 are ionic compounds whereas naphthalene C10H8 is a covalent compound. 2. Sodium chloride can conduct electricity in molten and aqueous forms, dissolve in water but cannot dissolve in organic solvents. Sodium chloride has high melting point and boiling point. Test Yourself 5.7 (pg. 130) 1. Ionic compounds have higher melting and boiling points than covalent compounds. 2. Simple molecules and giant covalent compounds are formed through the sharing of electrons between atoms. 3. (a) Magnesium hydroxide is an ionic compound and can dissolve in water (b) No (c) In the solid form, magnesium ions and hydroxide ions cannot move freely because they are bonded by strong electrostatic forces (ionic bond). 4. (a) The melting point and boiling point of diamond is higher than methane. This is because diamonds have strong covalent bonds. A lot of energy is needed to overcome these forces to melt and boil diamonds. On the other hand, methane molecules are held by weak Van der Waals forces. Very little energy is needed to overcome these forces to melt and boil methane (b) Diamond cannot conduct electricity. Diamond is a covalent compound. The molecules in covalent compounds are neutral and do not carry any charges. Achievement Test 5 (pg. 132 – 133) 1. The attraction force formed when non-metal atoms share electrons to achieve a stable duplet or octet electron arrangement. 2. Silicon dioxide is a giant covalent compound. Hence a lot of heat energy is needed to overcome the strong covalent bonds in the compound. 3. (a) G and E / G and H (b) DE2 (c) H2 has low melting point and boiling point. The Van der Waals forces between the molecules are very weak. Very little energy is needed to overcome these weak attraction force. 4. (a) Q: 2.8.2 R: 2.6
4 (b) (i) Ionic bond (ii) The electron arrangement of atom Q is 2.8.2. Atom Q has 2 valence electrons. In order to achieve a stable octet electron arrangement, atom Q donates 2 valence electrons to form Q2+ ion. The electrons are transferred to atom R. The electron arrangement of atom R is 2.6. Atom R has 6 valence electrons. In order to achieve a stable octet electron arrangement, atom R receives 2 electrons to form R2– ion. The Q2+ and R2– ions are attracted to each other by strong electrostatic attraction force. Ionic bond is formed and ionic compound S with formula QR is formed. 5. (a) Element J (b) The electron arrangement of atom J has 1 valence electron. In order to achieve a stable octet electron arrangement, atom J donates 1 valence electrons to form J+ ion. The electrons are transferred to atom K. The electron arrangement of atom K is 2.7. Atom K has 7 valence electrons. In order to achieve a stable octet electron arrangement, atom K receives 1 electron to form K– ion. The J+ and K– ions are attracted to each other by strong electrostatic attraction force. Ionic bond is formed and a white solid ionic compound T with formula JK is formed. 6. The electron arrangement of atom D is 2.8.7. Atom D has 7 valence electrons. Atom D needs one more electron to achieve a stable octet electron arrangement. Since the formula of the compound is ED3 , this means that one E atom shares three pairs of electrons with three D atoms. Thus, the electron arrangement of atom E could be 2.5 or 2.8.5. 7. There are hydrogen bonds between water molecules. Hydrogen bonds are strong attraction force that require a lot of heat energy to overcome them. Hence, the boiling point of water is high, and water exists as a liquid at room temperature. Hydrogen chloride does not form hydrogen bonds between molecules. Hence, very little heat energy is needed to overcome the weak attractive Van der Waals forces. The boiling point of hydrogen chloride is lower and HCl exists as a gas at room temperature. 8. The valence electrons of copper can be easily donated and delocalised in solid state. All the delocalised electrons can move freely within the metal structure and form an electron sea. These freely moving electrons carry electrical charges from the negative terminal to the positive terminal and conduct electricity in electrical wires. 9. The white solid is an ionic compound. Ionic compounds have different charges. When placed in water, the positive ions are attracted to the partially negatively-charged oxygen atom in water molecules whereas the negative ions are attracted to the partially positively-charged hydrogen atom in the water molecule. The attraction force between the water molecules and these ions are strong enough to overcome the electrostatic attraction force between the ions in the solid ionic compound. Thus, this compound can dissolve in water. When this white solid is melted, it can conduct electricity. This is because the ions in the molten form can move freely. These freely moving electrons can carry electrical charges from the positive to the negative terminal. Enrichment Corner (pg. 133) 1. A polynucleotide is a covalent compound that has hydrogen atoms bonded to nitrogen atoms. The nitrogen atom is small and has high electronegativity. An attraction force is formed between the hydrogen atom and the oxygen atom of another polynucleotide, causing many hydrogen bonds to be formed between the two polynucleotides. Hence, polynucleotides spiral together to form DNA.
1 Answers Chapter 6 Brain Teaser (pg. 137) No. Magnesium, Mg reacts with hydrochloric acid, HCl to produce magnesium chloride, MgCl2 and hydrogen gas, H2 but not water. A base is a substance that reacts with an acid to produce salt and water only. Experiment 6.1 (pg. 139 – 140) Interpreting Data: 1. The blue litmus paper changes to red. 2. Solid oxalic acid, C2 H2 O4 does not show acidic properties but solid oxalic acid, C2 H2 O4 that has been dissolved in water shows acidic properties. 3. Water is required by an acid to show acidic properties. Discussion: 1. Hydrogen ion, H+ 2. Without water, solid oxalic acid, C2 H2 O4 exists in the form of molecules. Hydrogen ions, H+ are not present. Thus, solid oxalic acid, C2 H2 O4 does not show acidic properties. The colour of the blue litmus paper remains unchanged. When solid oxalic acid, C2 H2 O4 is dissolved in water, the oxalic acid molecules, C2 H2 O4 ionise to produce hydrogen ions, H+ . Hydrogen ions, H+ cause oxalic acid, C2 H2 O4 to exhibit acidic properties. Hence the blue litmus paper changes to red. 3. An acid is a substance that changes blue litmus paper to red when the blue litmus paper is dipped in an acid that has been dissolved in water. Experiment 6.2 (pg. 141) Interpreting Data: 1. Sodium hydroxide pellets, NaOH do not show alkaline properties but the resulting sodium hydroxide solution show alkaline properties. 2. Red litmus paper must be moistened to detect alkaline properties. Discussion: 1. Hydroxide ion, OH– 2. Without water, the hydroxide ions, OH– in the sodium hydroxide pellets, NaOH cannot move freely as they are bonded in its lattice structure. Thus, it does not show alkaline properties. The colour of the red litmus paper remains unchanged. When sodium hydroxide pellets, NaOH are dissolved in water, it will ionise and hydroxide ions, OH– will move freely in water. The presence of mobile hydroxide ions, OH– causes the sodium hydroxide solution, NaOH to show alkaline properties. Hence the red litmus paper changes to blue. 3. An alkali is a substance that changes red litmus paper to blue when the red litmus paper is dipped in an alkali that has been dissolved in water. Test Yourself 6.1 (pg. 142) 1. (a) Chemical that ionises in water to produce hydrogen ions, H+ (b) Chemical that ionises in water to produce hydroxide ions, OH– 2. Carbonic acid is a diprotic acid because when it ionises in water, one acid molecule produces two hydrogen ions, H+ . 3. (a) Washing powder was not dissolved in water before the experiment was carried out. (b) Put 2 cm3 distilled water into a test tube containing washing powder and shake. Then, put in a red litmus paper. The washing powder only shows its alkaline properties in the presence of water because washing powder that is dissolved in water can ionise to produce hydroxide ions, OH– .
2 Experiment 6.3 (pg. 146) Interpreting Data: 1. When the concentration of hydrochloric acid, HCl decreases, pH value increases. 2. When the concentration of hydrochloric acid, HCl decreases, the concentration of hydrogen ions, H+ also decreases. 3. When the concentration of hydrogen ions, H+ decreases, pH value increases. Discussion: 1. (a) The concentration of hydrogen ions, H+ decreases. (b) pH value increases. (c) The acidity of the aqueous solution decreases 2. When the concentration of hydrogen ions, H+ increases, pH value decreases and the acidity of the aqueous solution increases. Experiment 6.4 (pg. 147) Interpreting Data: 1. When the concentration of sodium hydroxide solution, NaOH decreases, pH value decreases. 2. (a) The concentration of hydroxide ions, OH– decreases. (b) pH value decreases. (c) The alkalinity of the sodium hydroxide solution, NaOH decreases. 3. When the concentration of hydroxide ions, OH– increases, pH value increases and the alkalinity of the sodium hydroxide solution, NaOH increases. Test Yourself 6.2 (pg. 148) 1. pH = – log [H+ ] 2. pH = – log [H+ ] = – log [0.001] = – (–3) = 3 3. Ca(OH) → Ca2+ + 2OH– 0.05 mol dm–3 0.10 mol dm–3 pOH = –log [0.1] = – (–1) = 1.0 pOH of calcium hydroxide solution, Ca(OH)2 = 1.0 pH of calcium hydroxide solution, Ca(OH)2 = 14.0 – 1.0 = 13.0 Test Yourself 6.3 (pg. 151) 1. (a) A strong acid is an acid that ionises completely in water to produce high concentration of hydrogen ions, H+ (b) A weak acid is an acid that ionises partially in water to produce low concentration of hydrogen ions, H+ (c) A strong alkali is an alkali that ionises completely in water to produce high concentration of hydroxide ions, OH– (d) A weak alkali is an alkali that ionises partially in water to produce low concentration of hydroxide ions, OH– 2. Ammonia, NH3 is a weak alkali whereas potassium hydroxide, KOH is a strong alkali. The concentration of hydroxide ions, OH– in ammonia, NH3 is lower. Hence, the pH value of ammonia, NH3 is lower than the pH value of potassium hydroxide, KOH. 3. Nitric acid HNO3 is a strong acid that ionises completely in water to produce high concentration of hydrogen ions, H+ whereas oxalic acid, C2 H2 O4 is a weak acid that ionises partially in water to produce low concentration of hydrogen ions, H+ . The higher the concentration of H+ ions, the lower the pH value.
3 Activity 6.4 (pg. 152 – 155) A. Reaction between acid and base Discussion: 1. Copper(II) oxide powder dissolves in sulphuric acid, H2 SO4 . 2. Blue colour 3. CuO(s) + H2 SO4 (aq) → CuSO4 (aq) + H2 O(l) 4. Acid + Base → Salt + Water B. Reaction between acid and reactive metal Discussion: 1. Effervescence occurs. A ‘pop’ sound is produced when a lighted wooden splinter is put at the mouth of the test tube. 2. Hydrogen gas 3. 2HCl(aq) + Zn(s) → ZnCl2 (aq) + H2 (g) 4. Acid + reactive metal → Salt + Hydrogen gas C. Reaction between acid and metal carbonate Discussion: 1. To ensure that all the nitric acid, HNO3 has completely reacted with the marble chips, CaCO3 . 2. Filter the mixture in the test tube 3. (a) Calcium nitrate, Ca(NO3 )2 (b) Carbon dioxide gas, CO2 4. 2HNO3 (aq) + CaCO3 (s) → Ca(NO3 )2 (aq) + H2 O(l) + CO2 (g) 5. Acid + metal carbonate → salt + carbon dioxide gas + water Test Yourself 6.4 (pg. 158) 1. (a) 2HCl(aq) + Ba(OH)2 (aq) → BaCl2 (aq) + 2H2 O(l) (b) 2HCl(aq) + Mg(s) → MgCl2 (aq) + H2 (g) (c) 2HCl(aq) + ZnCO3 (s) → ZnCl2 (aq) + CO2 (g) + H2 O(l) 2. (a) Alkali + acid → salt + water (b) Alkali + ammonium salt → salt + water + ammonia gas (c) Alkali + metal ion → insoluble metal hydroxide + cation from alkali Activity 6.6 (pg. 161) 1. Concentration = 6 g 0.2 dm3 = 30 g dm–3 2. Concentration = 0.4 mol 2 dm3 = 0.2 mol dm–3 3. Concentration = 0.5 mol dm–3 × [2(1) + 32 + 4(16)] g mol–1 = 49 g dm–3 4. Molarity = 1.989 g dm–3 (23 + 35.5) g mol–1 = 0.034 mol dm–3 5. Number of moles, n = 0.2 mol dm–3 × 2.5 dm3 = 0.5 mol NaOH
4 6. Number of moles, n = 0.1 mol dm–3 × 250 cm3 1000 = 0.025 mol Ba(OH)2 Ba(OH)2 → Ba2+ + 2OH– Since 1 mole Ba(OH)2 produces 2 moles OH– 0.025 mol Ba(OH)2 produces 0.05 moles OH– Test Yourself 6.5 (pg. 161) 1. The mass of solute in 1 dm3 solution 2. g dm–3 and mol dm–3 3. Molarity = 0.03 mol 1.2 dm3 = 0.025 mol dm–3 4. Concentration = 2 mol dm–3 × [2(1) + 32 + 4(16)] g mol–1 = 196 g dm–3 5. Concentration = 1.9 g 0.1 dm3 = 19 g dm–3 Concentration = Molarity × molar mass 19 g dm–3 = 0.2 mol dm–3 × molar mass Molar mass = 1.9 g dm–3 0.2 mol dm–3 = 95 g mol–1 Relative mass of MgY2 = relative atomic mass of Mg + relative atomic mass of Y 95 = 24 + 2n 2n = 95 – 24 = 71 n = 71 2 = 35.5 Relative atomic mass of element Y is 35.5 Activity 6.7 (pg. 162 – 163) Discussion: 1. To ensure that all the sodium carbonate residue, Na2 CO3 stuck to the wall of the beaker and filter funnel has been transferred to the volumetric flask. 2. So that the prepared standard solution has the exact concentration as the concentration of the solution that would be prepared. 3. Make sure that the eye level is at the observed meniscus level. 4. To prevent evaporation of water from happening that will change the concentration of the prepared standard solution. Activity 6.8 (pg. 165 – 166) Discussion: 1. M1 V1 = M2 V2 (1.0)(V1) = (0.2) (100) V1 = 20 cm3 2. 20 cm3 pipette
5 3. A beaker cannot accurately measure the volume of desired standard solution. 4. No. A pipette has been accurately calibrated and the volume of the solution that comes out of the pipette is as shown on the pipette label. Brain Teaser (pg. 165) If the pipette is rinsed with distilled water, the remaining water droplets in the pipette will dilute the solution Activity 6.9 (pg. 167) 1. 2 mol dm–3 × V1 = 0.1 mol dm–3 × 50 cm3 V1 = 0.1 mol dm–3 × 50 cm3 2 mol dm–3 = 2.5 cm3 2. 0.5 mol dm–3 × 50 cm3 = M2 × 80 cm3 M2 = 0.5 mol dm–3 × 50 cm3 80 cm3 = 0.3125 mol dm–3 3. 1.2 mol dm–3 × 50 cm3 = 0.5 mol dm–3 × V2 1.2 mol dm–3 × 50 cm3 0.5 mol dm–3 = V2 V2 = 120 cm3 4. M1 × 50 cm3 = 0.2 mol dm–3 × 250 cm3 M1 = 0.2 mol dm–3 × 250 cm3 50 cm3 = 1.0 mol dm–3 Test Yourself 6.6 (pg. 167) 1. A solution whose exact concentration is known. 2. 0.15 mol dm–3 × V1 = 0.018 mol dm–3 × 500 cm3 V1 = 0.018 mol dm–3 × 500 cm3 0.15 mol dm–3 = 60 cm3 Hence, X = 60 3. 1.5 mol dm–3 × 25 cm3 = M2 × 150 cm3 1.5 mol dm–3 × 25 cm3 150 cm3 = M2 M2 = 0.25 mol dm–3 4. 0.2 mol dm–3 × 50 cm3 = 0.025 mol dm–3 × V2 0.2 mol dm–3 × 50 cm3 0.025 mol dm–3 = V2 V2 = 400 cm3 Volume of distilled water = 400 cm3 – 50 cm3 = 350 cm3
6 Activity 6.10 (pg. 168) 1. (a) 2HCl(aq) + Ba(OH)2 (aq) → BaCl2 (aq) + 2H2 O(l) H+ (aq) + OH– (aq) → H2 O(l) (b) H2 SO4 (aq) + 2KOH(aq) → K2 SO4 (aq) + 2H2 O(l) H+ (aq) + OH– (aq) → H2 O(l) (c) HNO3 (aq) + NaOH(aq) → NaNO3 (aq) + H2 O(l) H+ (aq) + OH– (aq) → H2 O(l) Activity 6.13 (pg. 170 – 171) Interpreting data: 2. H+ (aq) + OH– (aq) → H2 O(l) Discussion: 1. So that the colour change in the solution in the conical flask can be clearly observed. 2. If the conical flask is rinsed with potassium hydroxide solution, KOH before the titration, the actual volume of KOH in the conical flask will exceed 25 cm3 . More nitric acid, HNO3 will be required to neutralise the potassium hydroxide solution, KOH. The calculation to determine the concentration of the potassium hydroxide solution, KOH is inaccurate. 3. A point where the colour of the solution in the conical flask changes from pink to colourless when nitric acid, HNO3 is titrated into the conical flask containing potassium hydroxide solution, KOH. Activity 6.14 (pg. 173) 1. H2 SO4 + 2NaOH → Na2 SO4 + 2H2 O 0.1(Va ) 0.2(25) = 1 Va = 1 2 × 0.2(25) 0.1 = 25 cm3 2. 0.125(Va ) 1(25) = 1 2 Va = 1 2 × 1(25) 0.125 = 100 cm3 Total volume of solution = 100 cm3 + 25 cm3 = 125 cm3 3. 2HNO3 + Ca(OH)2 → Ca(NO3 )2 + 2H2 O Ma (50) 0.25(50) = 1 2 Ma = 2 × 0.25(50) 50 = 0.50 mol dm–3 4. H2 SO4 + 2KOH → K2 SO4 + 2H2 O 0.5(15) Mb (20) = 1 2 2 × (0.5)(15) = 1 × (Mb )(20) Mb = 0.75 mol dm–3
7 Test Yourself 6.7 (pg. 174) 1. Reaction between an acid and an alkali to form salt and water only. 2. Yellow to orange 3. HNO3 + NH3 → NH4 NO3 1(Va ) 0.75(50) = 1 1 Va = 0.75(50) = 37.5 cm3 4. 2HCl + Ba(OH)2 → BaCl2 + 2H2 O 0.05(Va ) 0.1(25) = 2 1 Va = 2 × 0.1(25) 0.05 = 100 cm3 5. Ma (20) 0.1(10) = 1 1 Ma = 1 2 × 0.1(10) 20 = 0.05 mol dm–3 6. 2HCl + CuO → CuCl2 + H2 O Number of moles of CuO = 6 g (64 + 16) g mol–1 = 0.075 mol Based on the chemical equation, 2 mol HCl reacts with 1 mol CuO 0.15 mol HCl reacts with 0.075 mol CuO 0.15 mol = Ma (500) 1000 Ma = 0.3 mol dm–3 Test Yourself 6.8 (pg. 178) 1. Ionic compound formed when hydrogen ion, H+ in an acid is replaced with a metal ion or ammonium ion, NH4 + . 2. • Different crystals have different geometrical shapes • Has specific geometrical shape such as cube, cuboid, rhombus and prism • Has smooth surface, straight sides and sharp corners • Has a fixed angle between two adjacent surfaces 3. (a) Ammonium phosphate is used as fertilizer. (b) Sodium hydrogen carbonate is used to neutralize excess hydrochloric acid in the stomach. Brain Teaser (pg. 180) Insoluble salt. Even though lead(II) chloride, PbCl2 and lead(II) iodide, PbI2 salts dissolve in hot water, the solubility of a salt refers to the solubility of the salt in water at room temperature and atmospheric pressure.
8 Activity 6.19 (pg. 181 – 182) Discussion: 1. To determine the end point so that the volume of hydrochloric acid, HCl, V cm3 that is needed to neutralize potassium hydroxide solution, KOH can be determined. 2. To prepare pure salt solution. 3. The crystals formed are those of a soluble salt. If the crystals are rinsed with a lot of distilled water, the water-soluble crystals will dissolve in distilled water. 4. HCl(aq) + KOH (aq) → KCl(aq) + H2 O(l) 5. Sodium nitrate salt and ammonium nitrate salt Activity 6.20 (pg. 182 – 183) Discussion: 1. To ensure that all the nitric acid, HNO3 has completely reacted. 2. The first filtration is to remove excess copper(II) oxide. The second filtration is to obtain copper(II) nitrate, Cu(NO3 )2 crystals. 3. CuO(s) + 2HNO3 (aq) → Cu(NO3 )2 (aq) + H2 O(l). 4. Yes. The reaction between nitric acid, HNO3 and copper(II) oxide, CuO produces salt and water. Activity 6.21 (pg. 184) Discussion: 1. Yes. To ensure that all sulphuric acid, H2 SO4 has completely reacted with zinc, Zn. 2. Zn(s) + H2 SO4 (aq) → ZnSO4 (aq) + H2 (g). 3. Copper is not a reactive metal and is less electropositive than hydrogen in the electrochemical series. Copper cannot react with dilute acid. Activity 6.23 (pg. 185 – 186) Discussion: 1. To remove dissolved impurities. 2. Crystal growth. 3. No. In order to purify a salt, the salt must first be removed from undissolved impurities by first dissolving the salt in water. Activity 6.24 (pg. 186 – 187) Discussion: 1. BaCl2 (aq) + Na2 SO4 (aq) → 2NaCl(aq) + BaSO4 (s) 2. Ba2+(aq) + SO4 2–(aq) → BaSO4 (s) 3. To remove sodium chloride, NaCl that might be stuck to the surface of barium sulphate, BaSO4 . 4. Not suitable. Initially, the reaction between barium carbonate, BaCO3 and sulphuric acid, H2 SO4 produces barium sulphate, BaSO4 , carbon dioxide and water. However, barium sulphate, BaSO4 is not soluble and covers the surface of barium carbonate, BaCO3 . This prevents sulphuric acid from further reacting with barium carbonate in the inner part. 5. • Copper carbonate salt Copper(II) nitrate aqueous solution and sodium carbonate aqueous solution are needed • Calcium sulphate salt • Calcium chloride aqueous solution and magnesium sulphate aqueous solution are needed Activity 6.25 (pg. 187) 1. (a) Ag+ (aq) + Cl– (aq) → AgCl(s) (b) Pb2+(aq) + CrO4 2–(aq) → PbCrO4 (s) (c) Cu2+(aq) + CO3 2–(aq) → CuCO3 (s)
9 Experiment 6.6 (pg. 188 – 189) Discussion: 1. So that the height of the precipitate formed is not affected by the space in the test tube. 2. From test tubes 1 to 4, more iodide ions, I– react with lead(II) ions, Pb2+. Thus, more lead(II) iodide precipitate, PbI2 is formed and the higher is the height of the precipitate. In test tube 5, the height of the precipitate is maximum because all the lead(II) ions, Pb2+ have reacted completely with iodide ions, I– to form lead(II) iodide precipitate, PbI2 . From test tubes 6 to 8, the height of the precipitate does not change because all the lead(II) ions, Pb2+ have reacted completely with iodide ions, I– . Brain Teaser (pg. 190) 2Ag+ (aq) + CO3 2–(aq) → Ag2 CO3 (s) Test Yourself 6.9 (pg. 190) 1. Soluble salts: NaNO3 , NaCl, Pb(NO3 )2 , MgSO4 , K2 CO3 , (NH4 )2 SO4 , ZnCl2 Insoluble salts: BaSO4 , CaCO3 , AgCl, PbI2 , BaCrO4 2. Calcium nitrate, sodium sulphate Ca2+(aq) + SO4 2–(aq) → CaSO4 (s) 3. Pipeclay triangle Evaporating dish Zinc nitrate solution, Zn(NO3 )2 Distilled water Wash bottle Filter paper Zinc nitrate crystals, Zn(NO3 )2 Zinc nitrate solution, Zn(NO3 )2 Zinc nitrate solution, Zn(NO3 ) 2 Glass rod Nitric acid, HNO3 Spatula Excess zinc powder Filter tunnel Filter paper Excess zinc powder residue Zinc nitrate solution, Zn(NO3 ) 2 Zinc nitrate crystals, Zn(NO3 )2 Heat
10 Activity 6.26 (pg. 191 – 193) Interpreting data: 1. Gas test Observation Inference A A glowing wooden splinter relights when it is put in the test tube Oxygen gas, O2 is released B A ‘pop’ sound is produced when a lighted wooden splinter is put near the mouth of the test tube Hydrogen gas, H2 is released C Limewater turns cloudy Carbon dioxide gas, CO2 is released D Moist red litmus paper turns blue A Pungent smell Ammonia gas, NH3 is released E Moist blue litmus paper turns red and then white A pungent greenish yellow gas is produced Chlorine gas, Cl2 is released F White fumes form when a glass rod dipped in concentrated ammonia is put near the mouth of the test tube Hydrogen chloride gas, HCl is released G The purple colour of acidified potassium manganate(VII) solution, KMnO4 is decolourised Sulphur dioxide gas, SO2 is released H A brown, pungent gas is produced Moist blue litmus paper turns red Nitrogen dioxide gas, NO2 is released 2. An acidic or alkaline gas only shows acidic and alkaline properties in the presence of water. Hence litmus paper must be moistened. 3. Gas Chemical test Method Observation Oxygen gas, O2 Put a glowing wooden splinter in the test tube The glowing wooden splinter is rekindled Hydrogen gas, H2 Put a lighted wooden splinter is near the mouth of the test tube A ‘pop’ sound is produced Carbon dioxide gas, CO2 Pass the colourless gas through limewater Limewater turns cloudy Ammonia gas, NH3 Put a moist red litmus paper near the mouth of the test tube The moist red litmus paper turns blue. Chlorine gas, Cl2 Put a moist blue litmus paper near the mouth of the test tube The moist blue litmus paper turns red and is then bleached. Hydrogen chloride gas, HCl Put a glass rod dipped in concentrated ammonia, NH3 near the mouth of the test tube White fumes are formed Sulphur dioxide gas, SO2 Pass the colourless gas through acidified potassium manganate(VII), KMnO4 solution The purple colour of acidified potassium manganate(VII) solution, KMnO4 is decolourised Nitrogen dioxide gas, NO2 Put a moist blue litmus paper near the mouth of the test tube The moist blue litmus paper turns red Discussion: 1. White fumes are produced. 2. Ammonium chloride gas, NH4 Cl 3. Moist blue litmus paper turns red
11 Experiment 6.7 (pg. 194 – 195) Interpreting Data 1. To detect presence of carbon dioxide gas, CO2 2. Carbon dioxide gas, CO2 3. Sodium carbonate, Na2 CO3 Discussion 1. CaCO3 (s) → CaO(s) + CO2 (g) ZnCO3 (s) → ZnO(s) + CO2 (g) PbCO3 (s) → PbO(s) + CO2 (g) CuCO3 (s) → CuO(s) + CO2 (g) 2. Carbonate salt → metal oxide + carbon dioxide gas 3. Potassium carbonate (K2 CO3 ) Experiment 6.8 (pg. 195 – 196) Interpreting Data: 1. Sodium nitrate, NaNO3 2. Nitrogen dioxide gas, NO2 3. Glowing wooden splinter is rekindled. Oxygen gas, O2 is released. Discussion: 1. • 2Mg(NO3 )2 (s) → 2MgO(s) + 4NO2 (g) + O2 (g) • 2Zn(NO3 )2 (s) → 2ZnO(s) + 4NO2 (g) + O2 (g) • 2Pb(NO3 )2 (s) → 2PbO(s) + 4NO2 (g) + O2 (g) • 2Cu(NO3 )2 (s) → 2CuO(s) + 4NO2 (g) + O2 (g) 2. Nitrate salt → metal oxide + oxygen gas + nitrogen dioxide gas Test Yourself 6.10 (pg. 197) 1. Sulphur dioxide gas, SO2 2. Brown gas is released. A black residue is formed. 3. Analysis of anion present in salt Y sample: • Colourless gas that turns limewater cloudy is carbon dioxide gas, CO2 • Carbonate salt produces carbon dioxide gas when heated strongly • Carbonate ion, CO3 2– present in salt sample Y Analysis of cation present in salt Y sample: • The heated residue that is yellow when hot, white when cold is zinc oxide, ZnO • Zinc ion, Zn2+ present in salt sample Y Hence, salt Y is zinc carbonate, ZnCO3 Activity 6.27 (pg. 198 – 199) Interpreting Data: 1. (a) Iron(II) sulphate, FeSO4 , copper(II) carbonate, CuCO3 , copper(II) chloride, CuCl2 (b) Iron(III) chloride, FeCl3 (c) Copper(II) sulphate, CuSO4 , copper(II) nitrate, Cu(NO3 )2 (d) Ammonium nitrate, NH4 NO3 , potassium nitrate, KNO3 , sodium chloride, NaCl, calcium carbonate, CaCO3 , calcium nitrate, Ca(NO3 )2 , magnesium sulphate, MgSO4 , magnesium carbonate, MgCO3 , zinc sulphate, ZnSO4 , zinc chloride, ZnCl2 , lead(II) nitrate, Pb(NO3 )2 , lead(II) chloride, PbCl2 , lead(II) sulphate, PbSO4 2. (a) Copper(II) carbonate, CuCO3 (b) Copper(II) chloride, CuCl2 (c) Iron(II) sulphate, FeSO4
12 Discussion: 1. No. There are many various salts with the same colour. 2. Colourless 3. Soluble salts: Iron(II) sulphate, FeSO4 , copper(II) chloride, CuCl2 , Iron(III) chloride, FeCl3 , copper(II) sulphate, CuSO4 , copper(II) nitrate, Cu(NO3 ) 2 , ammonium nitrate, NH4 NO3 , potassium nitrate, KNO3 , sodium chloride, NaCl, calcium nitrate, Ca(NO3 ) 2 , magnesium sulphate, MgSO4 , zinc sulphate, ZnSO4 , zinc chloride, ZnCl2 , lead(II) nitrate, Pb(NO3 ) 2 Insoluble salts: calcium carbonate, CaCO3 , magnesium carbonate, MgCO3 , lead (II) chloride, PbCl2 , lead(II) sulphate, PbSO4 , copper(II) carbonate, CuCO3 4. The salt that remains as solid in distilled water after the salt is put in a test tube containing distilled water. Experiment 6.9 (pg. 202 – 203) Interpreting data: 1. Carbon dioxide gas, CO2 2. (a) Silver chloride, AgCl (b) Ag+ (aq) + Cl– (aq) → AgCl(s) 3. (a) Barium sulphate, BaSO4 (b) Ba2+(aq) + SO4 2– (aq) → BaSO4 (s) 4. (a) CO3 2– (b) Cl– (c) SO4 2– (d) NO3 – Discussion: 1. To remove any carbonate ions, CO3 2– that might be present in the solution 2. No. Without acid, the white precipitate formed could be silver carbonate, Ag2 CO3 or silver chloride, AgCl. Experiment 6.10 (pg. 205 – 207) Interpreting data: Based on Part A experiment: 1. (a) Iron(II) ion, Fe2+ (b) Iron(III) ion, Fe3+ (c) Copper(II) ion, Cu2+ (d) Calcium(II) ion, Ca2+, magnesium ion, Mg2+, aluminium ion, Al3+, zinc ion, Zn2+, lead(II) ion, Pb2+ 2. Ammonium nitrate, NH4 NO3 . Ammonia gas, NH3 . 3. (a) Zinc ion, Zn2+, aluminium ion, Al3+, lead(II) ion, Pb2+ (b) Calcium(II) ion, Ca2+, magnesium ion, Mg2+ Based on Part B experiment: 1. (a) Iron(II) ion, Fe2+ (b) Iron(III) ion, Fe3+ (c) Copper(II) ion, Cu2+ (d) Magnesium ion, Mg2+, aluminium ion, Al3+, zinc ion, Zn2+, lead(II) ion, Pb2+ 2. Calcium nitrate, Ca(NO3 )2 , ammonium nitrate, NH4 NO3 . 3. (a) Zinc ion, Zn2+, aluminium ion, Al3+, lead(II) ion, Pb2+ (b) Magnesium ion, Mg2+, aluminium ion, Al3+, lead(II) ion, Pb2+ 4. Copper(II) ion, Cu2+ Discussion: 1. Zinc ion, Zn2+ 2. Zn2+(aq) + 2OH– (aq) → Zn(OH)2 (s)
13 Experiment 6.11 (pg. 210 – 211) Discussion: 1. Pb2+(aq) + 2I– (aq) → PbI2 (s) 2. Sodium hydroxide solution, NaOH Procedure: 1. Pour 2 cm3 salt solution into a test tube. 2. Add 2 cm3 sodium hydroxide solution, NaOH and shake. 3. Heat the mixture. 4. Put a moist red litmus paper at the mouth of the test tube. Observation: The moist red litmus paper turns blue. Hence, ammonium ion, NH4 + is present. 3. A white precipitate is formed. When heated, the white precipitate dissolves in hot water to form a colourless solution. When cooled, the white precipitate forms again. Test Yourself 6.11 (pg. 214) 1. Iron(II) ion, Fe2+ 2. The gas produced is sulphur dioxide, SO2 . 3. Calcium ion, Ca2+ 4. Procedure: 1. Pour 2 cm3 salt solutions S1, S2 and S3 into three separate test tubes. 2. Add sodium hydroxide solution, NaOH to each test tube slowly until in excess. 3. Record the observations. 4. Repeat steps 1 to 3 using ammonia solution, NH3 to replace sodium hydroxide solution, NaOH. Observation: • The white precipitate dissolves in both alkali solutions to form a colourless solution. Hence, the salt solution contains zinc ions, Zn2+. • The white precipitate does not dissolve in excess of both alkali solutions. Hence, the salt solution could contain lead(II) ions, Pb2+ or aluminium ion, Al3+. Procedure: 1. Pour 2 cm3 salt solution of an unknown cation into a test tube. 2. Add some potassium iodide solution, KI. Shake the mixture until well mixed. 3. Heat the mixture. Record the observations. 4. Let the test tube cool in running water. Record the observation. 5 Repeat steps 1 to 4 for each solution of unknown cation. Observation: • A yellow precipitate is formed. The yellow precipitate dissolves when heated to produce a colourless solution. The yellow precipitate forms again when cooled. Hence the salt solution contains lead(II) ions, Pb2+. • No precipitate is formed. The salt solution contains aluminium ions, Al3+. Achievement Test 6 (pg. 216 – 217) 1. (a) (i) Diprotic acid (ii) Monoprotic acid (b) Acid P ionises in water to produce two hydrogen ions, H+ per molecule whereas acid Q ionises in water to produce one hydrogen ion, H+ per molecule. (c) The pH value of acid P is lower than acid Q. Acid P is a strong acid whereas acid Q is a weak acid. The concentration of hydrogen ions, H+ in acid P is higher. Thus, the pH value is lower.
14 (d) (i) H2 SO4 + Zn → ZnSO4 + H2 (ii) n = (0.1(10) 1000 = 0.001 mol 1 mol HCl produces 1 mol H2 0.001 mol H2 is produced Volume = 0.001 × 24 = 0.024 dm3 (e) (0.1)(V1 ) = (0.05)(100) V1 = 50 cm3 Procedure: 1. Using a pipette, transfer 50 cm3 solution of acid P into a 100 cm3 volumetric flask. 2. Add distilled water until the level of solution is near to the calibration mark. 3. Add distilled water carefully with a dropper until the meniscus level of the solution aligns with the calibration mark on the volumetric flask. 4. Put the stopper on the mouth of the volumetric flask and invert it several times until the solution is thoroughly mixed. 2. (a) Average volume of sulphuric acid = 25.10 + 25.20 + 25.30 3 = 25.20 cm3 (b) H2 SO4 + 2NaOH → Na2 SO4 + 2H2 O (c) (0.5)(25.20) Mb (25.0) = 1 2 Mb = 1.008 mol dm–3 3. (a) (i) CuO (ii) NO2 (iii) O2 (iv) Cu(NO3 )2 (b) 2Cu(NO3 )2 (s) → 2CuO(s) + 4NO2 (g) + O2 (g) (c) Procedure: 1. 2 cm3 of blue solution Y is added into the test tube. 2. 2 cm3 of dilute sulphuric acid, H2 SO4 is added into the test tube followed by 2 cm3 iron(II) sulphate solution, FeSO4 . 3. The mixture is shaked. 4. A few drop of concentrated sulphuric acid, H2 SO4 are dripped slowly down the wall of the tilted test tube. Observation: Brown ring is formed. Nitrate ion, NO3 – is present. (d) Nitric acid, HNO3 Procedure: 1. Pour 20 cm3 2 mol dm–3 of nitric acid, HNO3 into a beaker. Heat the acid. 2. Using a spatula, add the black solid P gradually into the acid. Stir the mixture with a glass rod. 3. Keep on adding the black solid P until some of the solid is no longer dissolved. 4. Filter the excess black solid from the mixture. 5. Pour the filtrate into an evaporating dish and heat the filtrate until a saturated salt solution is obtained. 6. Let the saturated solution cool until salt crystals form.
15 7. Filter the contents in the evaporating dish to obtain the salt crystals. Rinse the crystals with some distilled water. 8. Dry the salt crystals between two pieces of filter paper. Enrichment Corner (pg. 217) 1. Procedure: Thread Glass rod Glass rod 400 cm3 hot water 300 g copper(II) sulphate powder, CuSO4 Excess copper(II) sulphate powder, CuSO4 Hot saturated copper(II) sulphate solution, CuSO4 Hot saturated copper(II) sulphate solution, CuSO4 Hot saturated copper(II) sulphate solution, CuSO4 Copper(II) sulphate powder, CuSO4 Copper(II) sulphate crystal, CuSO4 Copper(II) sulphate crystal, CuSO4 Nail varnish 1. Dissolve 300 g copper(II) sulphate, CuSO4 in 400 cm3 hot water. 2. Stir the mixture with a glass rod. 3. Filter the mixture to remove excess solid copper(II) sulphate, CuSO4 solid. 4. Let the saturated solution cool. 5. Tie a small copper(II) sulphate crystal, CuSO4 with a thread. 6. Hang the crystal in the solution. Leave aside for a week. 7. Remove the large crystal formed. Then filter the copper(II) sulphate solution, CuSO4 to remove any impurity in the copper(II) sulphate solution, CuSO4 . 8. Repeat steps 6 and 7 as many times to obtain the desired crystal size. 9. Paint the copper(II) sulphate, CuSO4 crystal with colourless nail varnish so that it looks shiny.
1 Answers Chapter 7 Activity 7.1 (pg. 222 – 224) A. Reaction between zinc, Zn and sulphuric acid, H2 SO4 Discussion: 1. Effervescence/ Gas bubbles are formed. The volume of the gas formed can be measured. 2. Hydrogen gas. 3. Zn(s) + H2 SO4 (aq) → ZnSO4 (aq) + H2 (g) 4. Effervescence stopped. No gas bubbles are formed. B. Reaction between nitric acid, HNO3 and marble chips, CaCO3 Discussion: 1. Reduction in the mass of marble, CaCO3 (the weighing scale reading decreases). 2. Loss of carbon dioxide gas produced. CaCO3 (s) + 2HNO3 (aq) → Ca(NO3 )2 (aq) + H2 O(l) + CO2 (g) 3. Effervescence stopped/ No gas bubbles are formed / No change in mass. C. Reaction between potassium iodide solution, KI and lead(II) nitrate solution, Pb(NO3 )2 Discussion: 1. Yellow precipitate 2. Pb(NO3 )2 (aq) + 2KI(aq) → PbI2 (s) + 2KNO3 (aq) Activity 7.3 (pg. 229) 1. 0 1 2 3 4 6 7 20 40 50 30 10 5 Time (minute) Volume of gas (cm3 ) 2. (a) (22 – 0) cm3 (1 – 0) min = 22 cm3 min–1 (b) (48 – 45) cm3 (5 – 4) min = 3 cm3 min–1 (c) 50 cm3 6 min = 8.33 cm3 min–1 3. (a) 10.62 cm3 min–1 (b) 4.17 cm3 min–1
2 Test Yourself 7.1 (pg. 229) 1. Rate of reaction is the change in the quantity of a reactant per unit time or the change in the quantity of product per unit time. 2. (a) Slow (b) Fast (c) Slow (d) Fast 3. (a) The increase in the volume of carbon dioxide gas, CO2 produced or the reduction in the mass of calcium carbonate, CaCO3 . (b) Formation of sulphur precipitate, S. (c) Reduction in gas pressure. (d) Reduction in the electrical conductivity or change in pH value of the neutralization reaction. 4. (a) P: 50 cm3 60 s = 0.83 cm3 s–1 Q: 50 cm3 95 s = 0.53 cm3 s–1 R: 50 cm3 20 s = 2.50 cm3 s–1 (b) R, P, Q. The higher the rate of reaction, the more reactive the metal. Experiment 7.2 (pg. 232 – 233) Discussion: 1. Formation of an insoluble solid in water. 2. Sulphur 3. • The total volume of the solution is maintained. • The height of the solution in the conical flask is maintained (use the same conical flask). • Use a white paper with the same ‘X’ mark. 4. The time taken for a fixed mass of sulphur to be formed. Experiment 7.3 (pg. 234) Discussion: 1. S2 O3 2–(aq) + 2H+ (aq) → S(s) + SO2 (g) + H2 O(l) 2. Yes. Thiosulphate ions, S2 O3 2– react with hydrogen ions, H+ present in all acid solutions. Experiment 7.4 (pg. 235) Discussion: 1. The release of oxygen gas can rekindle a glowing wooden splinter. The time for oxygen gas to rekindle a glowing wooden splinter. 2. The faster the glowing wooden splinter is rekindled, the higher the rate of reaction. Activity 7.4 (pg. 238) 1. First method: The experiment is carried out using 1.0 g marble powder, CaCO3 and 50 cm3 of 0.5 mol dm–3 hydrochloric acid, HCl at room temperature. Manipulated variable : Size of marble chips, CaCO3 Responding variable : Rate of reaction Fixed variable : Mass of marble chips, CaCO3 , volume and concentration of hydrochloric acid, HCl Second method: The experiment is carried out using 1.0 g marble powder, CaCO3 and 50 cm3 of 0.5 mol dm–3 hydrochloric acid, HCl at 50 °C. Manipulated variable : Temperature of reaction Responding variable : Rate of reaction Fixed variable : Mass of marble chips, CaCO3 , volume and concentration of hydrochloric acid, HCl
3 2. (a) 2H2 O2 (aq) → 2H2 O(l) + O2 (g) (b) Problem statement: What substance is the most effective catalyst for the decomposition of hydrogen peroxide, H2 O2 ? Hypothesis: The most effective catalyst will decompose hydrogen peroxide, H2 O2 in the shortest time Variables: Manipulated variable : Type of catalyst Responding variable : Rate of reaction Fixed variable : Mass of catalyst, volume and concentration of hydrogen peroxide, H2 O2 and temperature Delivery tube Oxygen gas, O2 Basin Water Conical flask Measuring cylinder Hydrogen peroxide, H2O2 Catalyst Procedure: 1. Set up the apparatus as shown in the diagram. 2. Put 0.1 g manganese(IV) oxide powder, MnO2 into 50 cm3 10-volume hydrogen peroxide solution, H2 O2 . 3. Start the stop watch immediately. 4. Record the volume of gas collected in the measuring cylinder every 10 seconds. 5. Repeat steps 1 to 4 using copper(II) oxide, CuO and iron, Fe. (c) Manganese(IV) oxide powder, MnO2 because the time taken for the reaction to complete is the shortest, that is 40 seconds. Test Yourself 7.2 (pg. 239) 1. (a) Concentration of reactant solution, size of reactants, temperature and presence of catalyst. (b) (i) Use powdered zinc to replace zinc granules Manipulated variable : Size of zinc granules Fixed variable : Temperature, volume and concentration of sulphuric acid, mass of zinc granules (ii) Increasing the temperature of sulphuric acid Manipulated variable : Temperature of sulphuric acid Fixed variable : Volume and concentration of sulphuric acid, mass and size of zinc granules (iii) Increasing the concentration of sulphuric acid Manipulated variable : Concentration of sulphuric acid Fixed variable : Temperature, volume of sulphuric acid, mass and size of zinc granules (iv) Adding a catalyst Manipulated variable : Presence of catalyst Fixed variable : Temperature, volume and concentration of sulphuric acid, mass and surface area of zinc granules
4 2. (a) Time for the effervescence to stop or time for all the marble, CaCO3 to fully dissolve. (b) Concentration of hydrochloric acid, reaction temperature and size of marble, CaCO3 . (c) Experiment IV has the highest rate of reaction because the concentration of hydrochloric acid is higher, the temperature is higher and the size of marble is smaller. Test Yourself 7.3 (pg. 242) 1. (a) smaller (b) low (c) pressure/temperature (d) exposed 2. (a) (i) The higher the temperature, the lower the percentage of ammonia produced (ii) The higher the pressure, the higher the percentage of ammonia produced (b) Low temperature (350 °C) • The percentage of products is high; rate of reaction is low. High temperature (550 °C) • The percentage of products is low; rate of reaction is high. Adding a catalyst can increase the rate of reaction, enabling the reaction to occur at low temperature. 3. (a) Excess unburnt hydrocarbon, carbon monoxide, CO and nitrogen oxide, NO (b) In the car engine, there is • high temperature • high pressure Both factors can increase the rate of reaction. Test Yourself 7.4 (pg. 248) 1. (a) 7 (b) 7 (c) 3 (d) 3 2. Effective collision • Has minimum energy to overcome the activation energy. • Collide at the correct orientation to break and form chemical bonds. 3. • Catalyst provides an alternative reaction pathway. • Alternative reaction pathway has lower activation energy. • More reactant particles can overcome the low activation energy. • The frequency of effective collisions between the particles increases, hence the rate of reaction increases. 4. (a) Activation energy is the minimum energy that reactant particles must overcome in order for a reaction to occur. (b) Ea Reaction pathway Energy Reactants Products (c) (i) Exothermic reaction releases heat to the surroundings. (ii) Endothermic reaction absorbs heat from the surroundings. Achievement Test 7 (pg. 250 – 251) 1. (a) g (mass); cm3 (volume); mol dm–3 (concentration) (b) g s–1; cm3 s–1; mol dm–3 s–1 2. (d), (c), (b), (a), (e)
5 3. (a) Change in the murkiness of milk. The time taken for the ‘ ’ mark to disappear from view. (b) Manipulated variable : Temperature Fixed variable : Type of catalyst, volume and concentration of milk, size of boiling tubes, paper marked ‘ ’ (c) (i) Rate of reaction is inversely proportional with time, that is rate = 1 time (ii) Temperature (°C) 15.0 25.0 35.0 45.0 55.0 65.0 Time taken for the ‘ ’ mark to disappear from view (minute) 12.0 7.0 2.5 4.0 7.0 19.0 1 time (minute–1) 0.083 0.143 0.400 0.250 0.143 0.053 (iii) 0 Temperature (°C) Rate of reaction (minute–1) 15 25 35 45 55 65 (iv) Rate of reaction increases and then decreases with the increase in temperature. Initially, the rate of reaction increases because (i) presence of catalyst increases the rate of reaction (ii) increase of temperature increases the rate of reaction Finally, the rate of reaction decreases because (i) enzymes function optimally at 37 °C (body temperature). (ii) enzymes get denatured at high temperature and lose their ability to increase the rate of reaction. 4. (a) Reactant particles must collide for a reaction to occur. Rate of reaction depends on the frequency of effective collision. (b) (i) NH4 NO3 (aq) → N2 O(g) + 2H2 O(l) + heat (ii) Reaction pathway N2 O + 2H2 O NH4 NO3 Ea Energy (c) + +
6 Enrichment corner (pg. 251) 1. 1.0 mol dm–3 hydrochloric acid, HCl reacts faster because the concentration of acid is higher • The higher the concentration of the acid, the more hydrogen ions, H+ there are per unit volume. • The frequency of collisions between hydrogen ions, H+ and calcium carbonate, CaCO3 increases • The frequency of effective collisions between the particles increases • The rate of reaction increases 2. Sn(s) + 2HCl(aq) → SnCl2 (aq) + H2 (g) Hydrochloric acid, HCI Delivery tube Hydrogen gas, H2 Burette Basin Water Conical flask Tin granules, Sn Hydrochloric acid, HCI Delivery tube Hydrogen gas, H2 Burette Basin Water Conical flask Tin powder, Sn Experiment I • 100 cm3 0.04 mol dm–3 hydrochloric acid • 10 g tin granules • Room temperature Experiment II • 100 cm3 0.04 mol dm–3 hydrochloric acid • 10 g tin powder • Room temperature Procedure: 1. Put 10 g tin granules into 100 cm3 of 0.04 mol dm–3 hydrochloric acid. 2. Collect the gas produced and measure its volume every 30 seconds. 3. Repeat steps 1 to 2 using tin powder to replace tin granules. 4. Plot a graph of volume of gas against time. Volume of gas (cm3 ) 0 Time (minute) Experiment I Experiment II 0.5 1.0 1.5 2.0 2.5 3.0 3.5 Explanation: • Gradient of tangent = rate of reaction • Gradient of tangent at t = 0 (initial rate of reaction) • Gradient of tangent at t = 0 is steeper in Experiment II compared to Experiment I • The rate of reaction is higher when tin powder is used • The smaller the size of the particles, the higher the rate of reaction
1 Answers Chapter 8 Experiment 8.2 (pg. 255 – 256) Discussion: 1. To remove the oxide layer from the surface of the metal. 2. The rate of corrosion of stainless steel is lower compared to iron. 3. Copper block 4. The smaller the diameter of the dent, the harder the substance. Test Yourself 8.1 (pg. 255 – 256) 1. (a) When a force is applied, the layers of atoms slide over each other. Pure metal is soft, that is ductile and malleable. (b) Foreign atoms (carbon atom) disturb the orderly arrangement of atoms in pure iron. The layers of atoms are harder to move or slide over each other. Thus the addition of carbon strengthens the structure of iron and makes iron stronger and harder. 2. (a) To make gold harder and stronger (b) % gold = 18 24 × 100 = 75%; % copper = 6 24 × 100 = 25% 3. (a) Washing machines are in contact with water. Water and oxygen cause metals to rust. Stainless steel is resistant to corrosion. (b) Electrical cables are hung on tall poles. Aluminium alloy has low density. (c) Pure gold is soft. Gold alloy is harder and stronger. Test Yourself 8.2 (pg. 262) 1. (a) Fused silica glass (b) Sodium ion 2. Mass of silica, SiO2 = 80 100 × 1000 g = 800 g Mass of boron oxide, B2 O3 = 15 100 × 1000 g = 150 g Mass of alumina, Al2 O3 = 5 100 × 1000 g = 50 g 3. Advantages: • The transparent property enables the food in the container to be visible • Can be easily cleaned and does not leave any traces • High heat resistance Disadvantages: • Heavy because lead glass has high density • Risk of lead poisoning, as lead can dissolve and diffuse into food Test Yourself 8.3 (pg. 265) 1. (a) metal; organic (b) Carbon, titanium carbide, silicon carbide 2. Can • Hard and strong, that is, not easily broken • Chemically inert, that is, safe to be used • Has low expansion coefficient or resistant to heat, that is resistant to thermal shock
2 Cannot • Expensive to produce • Not transparent 3. Iron(II) oxide. Iron(II) compound contains Fe2+ ions that give it a green colour. Test Yourself 8.4 (pg. 271) 1. (a) Combination of two or more non-homogeneous substances (b) Not suitable. Concrete is resistant to compression but is brittle due to its weak stretching strength (c) The stretching weakness of concrete can be strengthened by inserting steel rods in concrete. The steel rods are not brittle because it can withstand stretching. (d) To build bridges and buildings 2. (a) Matrix substance: plastic; Strengthening substance: fibre glass (b) • Has high strength-weight ratio • Strong and durable (resistant towards compression and stretching forces) • Does not pollute the water that it stores (chemically inert) • Low building cost 3. (a) • Protective jacket • Sheath layer • Core (b) To transmit information and data in the form of light reflection (c) Transmission of high definition cable TV needs • High capacity • High speed • High band width Fibre optics use light that can send information faster than copper wire. Fibre optics have wider band with compared to copper wire. 4. (a) Glass, silver chloride and copper(I) chloride (b) Silver chloride (c) • to make building windows • to make camera lenses Achievement Test 8 (pg. 273 – 274) 1. (a) % carbon that must be removed = (4.0 – 0.8) 4.0 × 100 = 80% (b) (i) % chromium = 18%, % nickel = 8%, % carbon = 1% (ii) Stainless steel does not rust when it is in contact with water and oxygen 2. (a) Silica, SiO2 ; soda (Na2 CO3 ); lead(II) oxide (PbO) (b) Advantages: • Transparent / high refractive index Disadvantages: • High density / break easily (brittle) / low resistance to scratching / does not absorb UV rays (c) Polycarbonate lens: • Light • Does not break easily when dropped • Many attractive / unique lens design • Eye safety – protection from UV rays • High clarity 3. (a) Silicon dioxide and aluminium oxide (b) Fe3+ ion (c) Manufacture of ceramics and bricks
3 4. (a) Heat insulator / light / resistant to corrosion (b) Heat insulator / electrical insulator 5. (a) Yttrium barium copper oxide ceramic, YBCO (b) When temperature drops, the electrical resistance of metals that are not superconductors also drops but the resistance does not disappear even though at the lowest temperature 0 K When temperature drops, the electrical resistance of superconductor X also drops and the resistance disappears at a very low critical temperature (4 K) – close to absolute zero, 0 K. (c) Cool the substance in liquid nitrogen or helium Enrichment Corner (pg. 274) 1. Silicon carbide has a giant structure and all the atoms are bonded together by strong covalent bonds. Hence, a lot of energy is required to overcome the strong bonds.