Chemistry Grade 10 Second Term Note PDF

CHEMISTRY SCHEME OF WORK & LESSON NOTES FOR SECOND TERM 2024/2025 SESSION GRADE 10 JANUARY 6, 2025 PRINCETON COLLEGE 9/33, OLATUNDE ONIMOLE STREET, SURULERE, LAGOS Week Topic 1 Revision...

CHEMISTRY SCHEME OF WORK & LESSON NOTES FOR SECOND TERM 2024/2025 SESSION GRADE 10 JANUARY 6, 2025 PRINCETON COLLEGE 9/33, OLATUNDE ONIMOLE STREET, SURULERE, LAGOS Week Topic 1 Revision of first term’s examination questions 2 The mole concept 3 Laws of Chemical combination 4 Chemical bonding 5 First Continuous Assessment Tests 6 Intermolecular forces 7 Mid-Term Break 8 Gas Laws 9 Second Continuous Assessment Tests 10 Gas Laws 11 Revision 12 & 13 Examination Week 1 Periods 1 and 2 Topic Revision of first term’s examination Learning Objectives At the end of this lesson, you should be able make corrections to the first term’s examination questions Reference material First term’s examination question paper Students’ activities The students respond to questions from the first term’s examination question papers as the teacher moderates the lesson Question 1 a The graph is the heating curve for a solid M Use the graph to answer the questions that follow. i What is the melting point of solid M? ii If the vapour of M is cooled, at what temperature will it start to condense? iii What is the boiling temperature of solid M? Question 4 a Explain why energy has to be supplied to turn a liquid into a gas. b Water is a colorless liquid at room temperature. A pure sample of water is slowly heated from 0 oC to 100 oC and its temperature is measured every minute. The results are represented on the graph below. i Name the change that occurs in the region D to E ii Sketch on the graph how the line would continue if water was heated to higher temperature. iii Complete the following table that compares the separation and movement of the particles in regions C to D and with those of E to F. C to D E to F separation (distance …………………… …………………… between particles) …………………… …………………… …………………… movement of particles random and slow ………………….... can particles move apart to …………………… …………………… fill any volume? …………………… …………………… Evaluation Question 3 The graph shows how the temperature of a substance changes as it is cooled over a period of 30 minutes. The substance is a gas at the start. Each letter on the graph may be used once, more than once or not at all. i Which letter, S. T, V, W, X,, Y or Z shows when I the particles in the substance have the most kinetic energy II the particles in the substance are furthest apart III the particles exist both as a gas and a liquid ii. Use the graph to estimate the freezing point of the substance. Assignment Question 5 In an experiment, a sample of solid Z was continually heated for 11 minutes. The graph shows how the temperature of the sample of solid Z changed during the first 9 minutes.  i What is the melting point of solid Z? ii The sample of solid Z began to boil at 9 minutes. It was boiled for 2 minutes. Use this information to sketch on the grid how the temperature of the sample solid Z changed between 9 minutes and 11 minutes. iii The sample of solid Z was continually heated between 2 minutes and 5 minutes. Explain, in terms of attractive forces, why there was no increase in temperature of the sample of solid Z between 2 minutes and 5 minutes. iv Describe how the motion of particles of sample Z changed from 0 minutes to 2 minutes. v A sample of Z was allowed to cool from 120 oC to 20 oC. The total time taken was 8 minutes. Starting from point X, sketch on the grid how the temperature of the sample of Z changed from 0 minutes to 8 minutes. Week 2 Period 1 Topic The mole concept Learning Objectives At the end of this lesson, you should be able to: o Define the mole o Relate the mole, mass and molar mass o Perform simple calculations involving the mole, mass and molar mass Reference books Extensive Chemistry for Senior Secondary Schools and Colleges. Revised Edition. Cambridge IGCSE Chemistry. Third Edition. Key words mole molar mass Method of Teaching The students do conversion of the number of moles of an element and compound to mass using simulation The mole A mole is 6.02 × 1023 particles. This numerical value is a constant known as Avogadro constant. The relationship between the mole, mass and molar mas number of moles = mass. molar mass n = m M The unit of the mole is mol, The unit of mass is gram (g), The unit of molar mass is gram per mol (g/mol) Illustration How many moles of carbon are there in 88 grams of CO2? 1 mol of CO2 contains 1mol of C molar mass of CO2 = 44 g/mol 44 g of CO2 contains 1 mol of C ∴ 88 g of CO2.will contain 88 g x 1mol 44 g = 2 mol ∴ There are 2 mol of C in 88 g of CO2 Evaluation Stevic Chemistry Workbook. Page 39, Questions 1, 2 and 3. Assignment Extensive Chemistry for Senior Secondary Schools and Colleges. Page 75, A Question 11 Week 3 Period 1 Topic The laws of chemical combination Learning Objectives At the end of this lesson, you should be able to: o State and verify the law of conservation of mass o State and verify the law of definite proportion Reference books Extensive Chemistry for Senior Secondary Schools and Colleges. Revised Edition. Cambridge IGCSE Chemistry. Third Edition. Key words conservation law proportion multiple Method of Teaching Laboratory Method The students make drawings of some of the laboratory apparatus on display. The laws of chemical combination ▪ The law of conservation of mass ▪ The law of definite proportions ▪ The law of multiple proportions The law of conservation of mass The law of conservation of mass states that matter is neither created nor destroyed in chemical reactions. In an experiment, 63.5 g of copper combines with 16 g of oxygen to give 79.5 g of cupric oxide (a black oxide of copper). This experiment is in agreement with the law of conservation of mass. The law of definite proportions (or the law of constant composition) The law of definite proportions states that in a given compound, the constituent elements are always combined in the same proportions by mass, regardless of the origin or mode of preparation of the compound. What this law means is that when elements react chemically, they combine in specific proportions, not in random proportions. A sample of pure water, whatever the source, always contains 88.9 % by mass of oxygen and 11.1 % by mass of hydrogen. Any one of the following methods may prepare the compound, copper (II) oxide ▪ Heating copper in oxygen ▪ Dissolving copper in trioxonitrate (V) acid and igniting the copper (II) nitrate formed. ▪ Dissolving copper in trioxonitrate (V) acid, precipitating copper (II) hydroxide, and strongly heating the copper (II) hydroxide. and in each case, the ratio, copper : oxygen by mass is always constant. 2.16 g of mercuric oxide gave on decomposition 0.16 g of oxygen. In another experiment, 16 g of mercury was obtained by the decomposition of 17.28 g of mercuric oxide. Show that these data conform to the law of definite proportions. Experiment 1 Mass of mercuric oxide = 2.16 g Mass of oxygen evolved from mercuric oxide = 0.16 g ∴ Mass of mercury in the compound = (2.16 - 0.16) g = 2.00 g Mass of mercury : mass of oxygen ratio = 2.00 : 0.16 Divide through by the smaller mass, 0.16 = 12.5 : 1  Multiply through by 2 to obtain whole numbers = 25: 2 Experiment 2 Mass of mercuric oxide = 17.28 g Mass of mercury in mercuric oxide = 16.00 g ∴ Mass of oxygen in the compound = (17.28 - 16.00) g = 1.28 g Mass of mercury: mass of oxygen ratio = 16.00: 1.28 Divide through by the smaller mass, 1.28 = 12.5: 1 Multiply through 2 to obtain whole numbers = 25: 2 In both cases, the ratio of the mass of mercury to the mass of oxygen is the same, 25 ; 2 thus conforming to the law of definite proportions. Evaluation Stevic Chemistry Workbook. Page 50, Questions 3, 5 and 11. Assignment Stevic Chemistry Workbook. Pages 52, Questions 1. Week 3 Period 2 Topic The laws of chemical combination Learning Objectives At the end of this lesson, you should be able to: o State and verify the law of multiple proportion Reference books Extensive Chemistry for Senior Secondary Schools and Colleges. Revised Edition. Cambridge IGCSE Chemistry. Third Edition. Key words conservation law proportion multiple Method of Teaching Laboratory Method The students make drawings of some of the laboratory apparatus on display. The law of multiple proportion The law of multiple proportions states that when two elements combine to form more than one compound, the several masses of one element that combine with a fixed mass of the other bear simple whole number ratio. Evaluation Stevic Chemistry Workbook. Page 50, Questions 1, 2 and 4. Assignment Stevic Chemistry Workbook. Page 53, Questions 3. Week 4 Period 1 Topic Chemical bonding Learning Objectives At the end of this lesson, you should be able to: o Mention the types of chemical bonds o Highlight the characteristics of compounds with specific chemical bonds Reference books Extensive Chemistry for Senior Secondary Schools and Colleges. Revised Edition. Cambridge IGCSE Chemistry. Third Edition. Key words matter property composition Method of Teaching Demonstration method The students in small groups, perform some physical and chemical changes in the Chemistry laboratory. Chemical bonds Chemical bonds refer to the force of attraction between atoms of elements so as to attain a stable (duplet or octet) structure. The electron structure of group zero or group eight elements account for their stability and inability to react with other elements. As a result, other elements try to possess this structure by bonding with other atoms. Types of chemical bonds Chemical bonds are divided into inter-atomic bonds and intermolecular bonds The inter-atomic bonds are either electrovalent (ionic) bonds or covalent bonds The intermolecular bonds include Van der Waals forces of attraction, dipole forces of attraction, and hydrogen bond. Electrovalent or ionic bonds Electrovalent bond is formed as a result of electron transfer from a metal to a non- metal. Electrovalent bonding in sodium chloride An electrovalent bond is formed when a sodium atom transfers one electron to a chlorine atom. Na+ and Cl- are formed. This can be represented schematically in the following way: The resulting ions, which have opposite charges, are attracted to one another by an electrovalent bond. Properties of electrovalent compounds ▪ Electrovalent compounds are composed of ions ▪ They are soluble in polar solvents like water ▪ They have high melting and boiling points. ▪ Electrovalent compounds in solution are electrolytes. Therefore, they are good conductors of electricity. Covalent bonds Covalent bond is formed as a result of electron sharing between two non-metals. Covalent bonding in hydrogen sulphide (H2S) A covalent bond is formed when two hydrogen atom share electrons with a sulphur atom. Each of the participating atoms donates an electron for sharing, whereby the central atom is sulphur, surrounded by two hydrogen atoms. Other examples of covalent compounds include, CO2, SO2, CH4, Cl2, N2 Properties of covalent compounds ▪ Covalent compounds have low melting and boiling points. ▪ They are non-electrolytes ▪ They are insoluble in polar solvents like water but soluble in organic solvents like benzene Dative covalent bonds or coordinate covalent bonds Dative covalent bond is formed when one of the participating atoms donates a lone pair of electrons for sharing. Dative covalent bonds are present in the molecules of ammonium ion (NH4+), hydroxonium ion (H3O+), Cu (II) tetraamine ion [Cu (NH3)4]2+. Evaluation Extensive Chemistry for Senior Secondary Schools and Colleges. Page 61, B Question 1 Assignment Extensive Chemistry for Senior Secondary Schools and Colleges. Page 62, B Question 3. Week 5 Continuous Assessment Tests Week 6 Period 1 Topic Intermolecular forces Learning Objectives At the end of this lesson, you should be able to: o Mention the types of chemical bonds o Highlight the characteristics of compounds with specific chemical bonds Reference books Extensive Chemistry for Senior Secondary Schools and Colleges. Revised Edition. Cambridge IGCSE Chemistry. Third Edition. Key words matter property composition Method of Teaching Demonstration method The students in small groups, perform some physical and chemical changes in the Chemistry laboratory. Van der Waals forces of attraction Van der Waals forces are found in gaseous molecules. They are relatively weak forces of attraction. This force is responsible for the liquid formation from gases. Dipole forces of attraction This force of attraction is one that occurs between molecules like water. It is a relatively weak force of attraction when compared with hydrogen bonds. Hydrogen bonds Hydrogen bonds are formed between hydrogen in a molecule and a highly electronegative element like fluorine or oxygen or nitrogen in another molecule. Hydrogen bonds are present between molecules of hydrogen fluoride or water or ammonia. Properties of molecules with hydrogen bonds ▪ Hydrogen bonds accounts for high boiling points in compounds like ethanol, water, and ethanoic acid. ▪ It is also responsible for the pairing of nitrogenous bases in deoxyribonucleic acid (DNA). ▪ Hydrogen bonds are also responsible for the solubility of compounds like ethanol and ethanoic acid in water. Metallic bonds Metallic bonding is a form of chemical combination that involves metals only. The metallic bond is formed when metal atoms form a crystal lattice of closely packed cations and the valence electrons from the metal atoms move freely within the lattice. Hence, the attraction between the cations and free moving electrons is the metallic bond. The strength of the metallic bond differs from one metal to another, and is responsible for the characteristics of metals such as  ▪ Malleability, the ability to be beaten into different shapes ▪ Conductivity, the ability to conduct heat and electricity ▪ Ductility, the ability to be drawn into wires ▪ Sonorous, the ability to produce sound ▪ Lustre, the ability to shine when polished Evaluation Extensive Chemistry for Senior Secondary Schools and Colleges. Page 60, A Question 4 and 9. Assignment Extensive Chemistry for Senior Secondary Schools and Colleges. Page 60, A Question 5 and 6. Week 7 Mid-Term Break Week 8 Period 1 Topic The Gas laws Learning Objectives At the end of this lesson, you should be able to: o State the gas laws o Perform calculations using the laws Reference books Extensive Chemistry for Senior Secondary Schools and Colleges. Revised Edition. Cambridge IGCSE Chemistry. Third Edition. Key words particle proton neutron electron valency Method of Teaching Demonstration method The students write the chemical symbols of the elements with proton numbers 1 – 30. The Gas laws The gas laws are laws proposed to explain the properties and behavior of gases. Boyle’s law Boyle’s law states that the volume, V of a given mass of a gas is inversely proportional to the pressure, P provided temperature is kept constant. Mathematically, Vα1 P V=K where K is a constant of proportionality P  K = PV ∴ P1V1 = P2V2 Illustration Given that 200 cm3 of a gas exerts a pressure of about 540 mm Hg. At what pressure will the volume be tripled? V1 = 200 cm3 P1 = 540 mm Hg V2 = (3 x 200) cm3, = 600 cm3 P2 = ? P1V1 = P2V2 P2 = P1V1 V2 P2 = 540 mm Hg x 200 cm3 600 cm3 P2 = 180 mm Hg Charles’ law Charles’ law states that the volume, V of a given mass of gas is directly proportional to the absolute temperature; T provided pressure is kept constant. Mathematically, VαT V = kT where K is a constant of proportionality K=V  T ∴ V1 = V2 T1 T2 Illustration At what temperature will a gas of 70 cm3 become 42 cm3 if the initial temperature is 300 K? V1 = 70 cm3 V2 = 42 cm3 T1 = 300 K T2 = ? V1 = V2 T1 T2 T2 = V2T1 V1 T2 = 42 cm3 x 300 K 70 cm3 T2 = 180 K General gas law The general gas law is a combination of Charles’ law and Boyle’s law. P1V1 = P2V2. T1 T2 At standard temperature and pressure (S.T.P,) the temperature is 273 K, while pressure is 760 mm Hg or 1 atm or 1.0125 x 105 Nm-2. Illustration If the pressure of a gas is 2.02 x 105 Nm-2 at 303 K when the volume is 5 dm3, what will be the final volume at S.T.P.? P1V1 = P2V2 T1 T2 V2 = P1V1T2 T1P2. V2 = 2.02 x 105 Nm-2 x 5 dm3 x 273 K 303 K x 1.0125 x 105 Nm-2 V2 = 8.99 dm3  9.0 dm3 Evaluation Stevic Chemistry Workbook. Page 68, Question 1. Assignment Extensive Chemistry for Senior Secondary Schools and Colleges. Page 74, A Question 3. Week 8 Period 2 Topic The Gas laws Learning Objectives At the end of this lesson, you should be able to: o State the gas laws o Perform calculations using the laws Reference books Extensive Chemistry for Senior Secondary Schools and Colleges. Revised Edition. Cambridge IGCSE Chemistry. Third Edition. Key words particle proton neutron electron valency Method of Teaching Demonstration method The students write the chemical symbols of the elements with proton numbers 1 – 30..Ideal gas equation The ideal gas equation is shown below PV = nRT. Where P is pressure in atm V is volume in dm3 n is number of moles of the gas R is the gas constant with value of 0.0821atm dm3 mol-1 K-1 T is temperature in Kelvin Dalton’s law of partial pressure Dalton’s law of partial pressure states that when two or more gases that are chemically unreactive are confined in a vessel, the pressures exerted by these gases are equal to the sum of their individual partial pressure. Mathematically, PT = PA + PB + PC +… PA = XA × PT XA = nA nA + nB + nC … XA + XB + XC = 1 Where PT is the total partial pressure PA is the partial pressure of gas A XA is the mole fraction of gas A nA is the number of moles of gas A Illustration Given that 4 g of Ne, 4 g of O2 and 2 g of H2 where kept in a cylinder. If they exert a total partial pressure of 2.1 atm, calculate i. the number of moles of each gas ii. the mole fraction of each gas iii. the partial pressure of Ne.  [Ar; Ne = 20, O = 16, H = 1] number of mole (n) = mass (m) molar mass (M) i nNe = 4 g,Ne = 0.2 mol 20 g mol-1 nO2 = 4g = 0.125 mol 32 g mol-1 nH2 = 2g = 1 mol 2 g mol-1 ii XNe = nNe nNe + nO2 + nH2; = 0.2 mol (0.2 + 0.125 +1) mol = 0.2 1.325 XNe = 0.151 XO2 = nO2 nNe + nO2 + nH2; = 0.125 mol (0.2 + 0.125 + 1) mol = 0.125 1.325 XO2 = 0.09 XH2 = nH2 nNe + nO2 + nH2; = 1 mol (0.2 + 0.125 + 1) mol = 1 1.325 XH2 = 0.755  iii PNe = XNe × PT PNe = 0.151 × 2.1atm PNe = 0.3171atm Graham’s law of diffusion Graham’s law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its vapour density or relative molecular mass. Illustration If it takes 30 seconds for 15 cm3 of hydrogen gas to diffuse through a porous partition, how long will it take equal volume of oxygen gas to diffuse through same medium under the same condition? [Ar: H = 1, O = 16] Gay Lussac’s law of combining volumes Gay Lusaac’s law of combining volumes states that when gases combine, they do so in volumes which bear a simple whole number ratio to each another and to the volume of the products if gaseous. N2(g) + 3H2(g) → 2NH3(g) The ratio of the reacting volume of nitrogen to hydrogen is 1: 3. Evaluation Stevic Chemistry Workbook. Page 73, Question 2. Assignment Extensive Chemistry for Senior Secondary Schools and Colleges. Page 76, A Question 30. Week 9 Period 2 Topic The Gas laws Learning Objectives At the end of this lesson, you should be able to: o State the gas laws o Perform calculations using the laws Reference books Extensive Chemistry for Senior Secondary Schools and Colleges. Revised Edition. Cambridge IGCSE Chemistry. Third Edition. Key words particle proton neutron electron valency Method of Teaching Demonstration method The students write the chemical symbols of the elements with proton numbers 1 – 30..Ideal gas equation The ideal gas equation is shown below PV = nRT. Where P is pressure in atm V is volume in dm3 n is number of moles of the gas R is the gas constant with value of 0.0821atm dm3 mol-1 K-1 T is temperature in Kelvin Dalton’s law of partial pressure Dalton’s law of partial pressure states that when two or more gases that are chemically unreactive are confined in a vessel, the pressures exerted by these gases are equal to the sum of their individual partial pressure. Mathematically, PT = PA + PB + PC +… PA = XA × PT XA = nA nA + nB + nC … XA + XB + XC = 1 Where PT is the total partial pressure PA is the partial pressure of gas A XA is the mole fraction of gas A nA is the number of moles of gas A Illustration Given that 4 g of Ne, 4 g of O2 and 2 g of H2 where kept in a cylinder. If they exert a total partial pressure of 2.1 atm, calculate iv. the number of moles of each gas v. the mole fraction of each gas vi. the partial pressure of Ne.  [Ar; Ne = 20, O = 16, H = 1] number of mole (n) = mass (m) molar mass (M) i nNe = 4 g,Ne = 0.2 mol 20 g mol-1 nO2 = 4g = 0.125 mol 32 g mol-1 nH2 = 2g = 1 mol 2 g mol-1 ii XNe = nNe nNe + nO2 + nH2; = 0.2 mol (0.2 + 0.125 +1) mol = 0.2 1.325 XNe = 0.151 XO2 = nO2 nNe + nO2 + nH2; = 0.125 mol (0.2 + 0.125 + 1) mol = 0.125 1.325 XO2 = 0.09 XH2 = nH2 nNe + nO2 + nH2; = 1 mol (0.2 + 0.125 + 1) mol = 1 1.325 XH2 = 0.755  iii PNe = XNe × PT PNe = 0.151 × 2.1atm PNe = 0.3171atm Graham’s law of diffusion Graham’s law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its vapour density or relative molecular mass. Illustration If it takes 30 seconds for 15 cm3 of hydrogen gas to diffuse through a porous partition, how long will it take equal volume of oxygen gas to diffuse through same medium under the same condition? [Ar: H = 1, O = 16] Gay Lussac’s law of combining volumes Gay Lusaac’s law of combining volumes states that when gases combine, they do so in volumes which bear a simple whole number ratio to each another and to the volume of the products if gaseous. N2(g) + 3H2(g) → 2NH3(g) The ratio of the reacting volume of nitrogen to hydrogen is 1: 3. Evaluation Stevic Chemistry Workbook. Page 73, Question 2. Assignment Extensive Chemistry for Senior Secondary Schools and Colleges. Page 76, A Question 30. Week 10 Period 1 Topic The Gas laws Learning Objectives At the end of this lesson, you should be able to: o State the gas laws o Perform calculations using the laws Reference books Extensive Chemistry for Senior Secondary Schools and Colleges. Revised Edition. Cambridge IGCSE Chemistry. Third Edition. Key words particle proton neutron electron valency Method of Teaching Demonstration method The students write the chemical symbols of the elements with proton numbers 1 – 30..Ideal gas equation The ideal gas equation is shown below PV = nRT. Where P is pressure in atm V is volume in dm3 n is number of moles of the gas R is the gas constant with value of 0.0821atm dm3 mol-1 K-1 T is temperature in Kelvin Dalton’s law of partial pressure Dalton’s law of partial pressure states that when two or more gases that are chemically unreactive are confined in a vessel, the pressures exerted by these gases are equal to the sum of their individual partial pressure. Mathematically, PT = PA + PB + PC +… PA = XA × PT XA = nA nA + nB + nC … XA + XB + XC = 1 Where PT is the total partial pressure PA is the partial pressure of gas A XA is the mole fraction of gas A nA is the number of moles of gas A Illustration Given that 4 g of Ne, 4 g of O2 and 2 g of H2 where kept in a cylinder. If they exert a total partial pressure of 2.1 atm, calculate vii. the number of moles of each gas viii. the mole fraction of each gas ix. the partial pressure of Ne.  [Ar; Ne = 20, O = 16, H = 1] number of mole (n) = mass (m) molar mass (M) i nNe = 4 g,Ne = 0.2 mol 20 g mol-1 nO2 = 4g = 0.125 mol 32 g mol-1 nH2 = 2g = 1 mol 2 g mol-1 ii XNe = nNe nNe + nO2 + nH2; = 0.2 mol (0.2 + 0.125 +1) mol = 0.2 1.325 XNe = 0.151 XO2 = nO2 nNe + nO2 + nH2; = 0.125 mol (0.2 + 0.125 + 1) mol = 0.125 1.325 XO2 = 0.09 XH2 = nH2 nNe + nO2 + nH2; = 1 mol (0.2 + 0.125 + 1) mol = 1 1.325 XH2 = 0.755  iii PNe = XNe × PT PNe = 0.151 × 2.1atm PNe = 0.3171atm Graham’s law of diffusion Graham’s law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its vapour density or relative molecular mass. Illustration If it takes 30 seconds for 15 cm3 of hydrogen gas to diffuse through a porous partition, how long will it take equal volume of oxygen gas to diffuse through same medium under the same condition? [Ar: H = 1, O = 16] Gay Lussac’s law of combining volumes Gay Lusaac’s law of combining volumes states that when gases combine, they do so in volumes which bear a simple whole number ratio to each another and to the volume of the products if gaseous. N2(g) + 3H2(g) → 2NH3(g) The ratio of the reacting volume of nitrogen to hydrogen is 1: 3. Evaluation Stevic Chemistry Workbook. Page 73, Question 2. Assignment Extensive Chemistry for Senior Secondary Schools and Colleges. Page 76, A Question 30.