WAEC SSCE - Chemistry - 2010 (Essay)
Question 1 Report
(a (i) Explain why water is referred to as a universal solvent.
(ii) Give one chemical test for water. H\(_2\)SO\(_4\)
(b) A current of 1.25A was passed through an electrolytic cell containing dil. for 40 minutes.
(i) Write (ii) Calculate the volume of gas produced at the anode at s.t.p. (1F = 96,500 C, Molar Volume of gas at s.t.p. = 22.4 dm\(^3\) mol\(^{-3}\) ]
(C) Consider the reaction represented by the following equation: Au\(_{(s)}\) + Cl\(_{2(g)}\) \(\to\) AuCl\(_{3(s)}\)
(I) Balance the equation
(ii) If 1.2509 of Au and 1.744g of Cl\(_2\) were mixed
I. determine which of the reactants is in excess II. calculate the excess amount [Au = 197.0, CI = 35.5]
(a) Water
- (i) Water is called a universal solvent because it dissolves a very large number of substances (many ionic and polar solids, liquids and gases), more than any other common solvent, owing to its polar nature.
- (ii) Chemical test for water: add the liquid to white anhydrous copper(II) sulphate, which turns blue; or to blue anhydrous cobalt(II) chloride paper, which turns pink. (To show it is pure water, it also boils at 100 degrees Celsius.)
(b) Electrolysis of dilute H\(_2\)SO\(_4\)
- (i) At the anode (oxidation), oxygen is liberated: \[2H_2O_{(l)} \to O_{2(g)} + 4H^{+}_{(aq)} + 4e^{-}\]
- (ii) Quantity of charge \(Q = It = 1.25 \times (40\times60) = 1.25 \times 2400 = 3000\ \text{C}\). Moles of electrons \(= \dfrac{3000}{96500} = 0.03109\) mol. From the equation, 4 moles of electrons give 1 mole of O\(_2\): \[n(O_2) = \frac{0.03109}{4} = 7.77\times10^{-3}\ \text{mol}.\] Volume at s.t.p. \(= 7.77\times10^{-3} \times 22.4 = 0.174\ \text{dm}^3\) (about 174 cm\(^3\)).
(c) Au and Cl\(_2\)
- (I) \[2Au + 3Cl_2 \to 2AuCl_3\]
- (ii) I. Moles of Au \(= \dfrac{1.250}{197.0} = 6.35\times10^{-3}\) mol; moles of Cl\(_2\) \(= \dfrac{1.744}{71.0} = 2.456\times10^{-2}\) mol. From the equation, 2 mol Au need 3 mol Cl\(_2\), so 6.35 x 10\(^{-3}\) mol Au need \(\dfrac{3}{2}\times6.35\times10^{-3} = 9.52\times10^{-3}\) mol Cl\(_2\). Since 2.456 x 10\(^{-2}\) mol Cl\(_2\) is available (more than 9.52 x 10\(^{-3}\)), chlorine is in excess (gold is the limiting reactant). II. Excess Cl\(_2\) \(= 2.456\times10^{-2} - 9.52\times10^{-3} = 1.50\times10^{-2}\) mol, i.e. mass \(= 1.50\times10^{-2} \times 71.0 = 1.07\ \text{g}\) of chlorine.
Answer Details
(a) Water
- (i) Water is called a universal solvent because it dissolves a very large number of substances (many ionic and polar solids, liquids and gases), more than any other common solvent, owing to its polar nature.
- (ii) Chemical test for water: add the liquid to white anhydrous copper(II) sulphate, which turns blue; or to blue anhydrous cobalt(II) chloride paper, which turns pink. (To show it is pure water, it also boils at 100 degrees Celsius.)
(b) Electrolysis of dilute H\(_2\)SO\(_4\)
- (i) At the anode (oxidation), oxygen is liberated: \[2H_2O_{(l)} \to O_{2(g)} + 4H^{+}_{(aq)} + 4e^{-}\]
- (ii) Quantity of charge \(Q = It = 1.25 \times (40\times60) = 1.25 \times 2400 = 3000\ \text{C}\). Moles of electrons \(= \dfrac{3000}{96500} = 0.03109\) mol. From the equation, 4 moles of electrons give 1 mole of O\(_2\): \[n(O_2) = \frac{0.03109}{4} = 7.77\times10^{-3}\ \text{mol}.\] Volume at s.t.p. \(= 7.77\times10^{-3} \times 22.4 = 0.174\ \text{dm}^3\) (about 174 cm\(^3\)).
(c) Au and Cl\(_2\)
- (I) \[2Au + 3Cl_2 \to 2AuCl_3\]
- (ii) I. Moles of Au \(= \dfrac{1.250}{197.0} = 6.35\times10^{-3}\) mol; moles of Cl\(_2\) \(= \dfrac{1.744}{71.0} = 2.456\times10^{-2}\) mol. From the equation, 2 mol Au need 3 mol Cl\(_2\), so 6.35 x 10\(^{-3}\) mol Au need \(\dfrac{3}{2}\times6.35\times10^{-3} = 9.52\times10^{-3}\) mol Cl\(_2\). Since 2.456 x 10\(^{-2}\) mol Cl\(_2\) is available (more than 9.52 x 10\(^{-3}\)), chlorine is in excess (gold is the limiting reactant). II. Excess Cl\(_2\) \(= 2.456\times10^{-2} - 9.52\times10^{-3} = 1.50\times10^{-2}\) mol, i.e. mass \(= 1.50\times10^{-2} \times 71.0 = 1.07\ \text{g}\) of chlorine.
Question 2 Report
(a)(i) Give the two reasons why soda lime is used instead of caustic soda in the preparation of methane.
(ii) List two physical properties of methane.
(iii) A hydrocarbon with a vapour density of 29 contains 82.76% carbon and 17.24% hydrogen. Determine the: I. empirical formula; II. molecular formula of the hydrocarbon. [ H = 1.00 C = 12.00 ]
(b)(i) What is meant by the term isomerism?
(ii) Draw the structures of the two isomers of the compound with the molecular formula C\(_2\)H\(_6\)O.
(iii) Give the name of each of the isomers in (b)(ii).
(iv) State the major difference between the isomers.
(c) Give three deductions that could be made from the qualitative and quantitative analysis of a given organic compound.
(d) Give one chemical test to distinguish between propene and propane.
(a)(i) Soda lime is used instead of caustic soda because:
- Soda lime does not attack the glass apparatus on heating, whereas caustic soda attacks glass.
- Soda lime is not deliquescent, whereas caustic soda absorbs moisture from the air and becomes difficult to handle.
(ii) Methane is:
- a colourless, odourless gas at room temperature;
- slightly soluble in water and less dense than air.
(iii)
Vapour density = 29.
Therefore, molar mass = \(2 \times 29 = 58\).
| Element | Percentage composition | Relative atomic mass | Mole ratio |
|---|---|---|---|
| C | 82.76 | 12 | \(82.76/12 = 6.90\) |
| H | 17.24 | 1 | \(17.24/1 = 17.24\) |
Dividing by the smaller value, 6.90:
\[\text{C : H} = 1 : 2.5 = 2 : 5\]
I. Empirical formula = \(\mathrm{C_2H_5}\).
Empirical formula mass = \((2 \times 12) + (5 \times 1) = 29\).
\[n=\frac{58}{29}=2\]
II. Molecular formula = \((\mathrm{C_2H_5})_2 = \boxed{\mathrm{C_4H_{10}}}\).
(b)(i) Isomerism is the existence of two or more compounds having the same molecular formula but different structural formulae, or different arrangements of atoms.
(ii) and (iii) The two isomers of \(\mathrm{C_2H_6O}\) are shown below.
(iv) The isomers have different functional groups and belong to different homologous series. Ethanol is an alkanol containing the hydroxyl group, \(\mathrm{-OH}\), while methoxymethane is an ether containing the \(\mathrm{-O-}\) linkage.
(c) Qualitative and quantitative analysis of an organic compound can be used to deduce:
- the types of elements present in the compound;
- the percentage composition, and hence the relative numbers of atoms, of the elements present;
- the empirical formula and, where the molar mass is known, the molecular formula of the compound.
(d) Pass each gas through bromine water. Propene decolourises the reddish-brown bromine water because it is unsaturated, whereas propane produces no colour change.
Answer Details
(a)(i) Soda lime is used instead of caustic soda because:
- Soda lime does not attack the glass apparatus on heating, whereas caustic soda attacks glass.
- Soda lime is not deliquescent, whereas caustic soda absorbs moisture from the air and becomes difficult to handle.
(ii) Methane is:
- a colourless, odourless gas at room temperature;
- slightly soluble in water and less dense than air.
(iii)
Vapour density = 29.
Therefore, molar mass = \(2 \times 29 = 58\).
| Element | Percentage composition | Relative atomic mass | Mole ratio |
|---|---|---|---|
| C | 82.76 | 12 | \(82.76/12 = 6.90\) |
| H | 17.24 | 1 | \(17.24/1 = 17.24\) |
Dividing by the smaller value, 6.90:
\[\text{C : H} = 1 : 2.5 = 2 : 5\]
I. Empirical formula = \(\mathrm{C_2H_5}\).
Empirical formula mass = \((2 \times 12) + (5 \times 1) = 29\).
\[n=\frac{58}{29}=2\]
II. Molecular formula = \((\mathrm{C_2H_5})_2 = \boxed{\mathrm{C_4H_{10}}}\).
(b)(i) Isomerism is the existence of two or more compounds having the same molecular formula but different structural formulae, or different arrangements of atoms.
(ii) and (iii) The two isomers of \(\mathrm{C_2H_6O}\) are shown below.
(iv) The isomers have different functional groups and belong to different homologous series. Ethanol is an alkanol containing the hydroxyl group, \(\mathrm{-OH}\), while methoxymethane is an ether containing the \(\mathrm{-O-}\) linkage.
(c) Qualitative and quantitative analysis of an organic compound can be used to deduce:
- the types of elements present in the compound;
- the percentage composition, and hence the relative numbers of atoms, of the elements present;
- the empirical formula and, where the molar mass is known, the molecular formula of the compound.
(d) Pass each gas through bromine water. Propene decolourises the reddish-brown bromine water because it is unsaturated, whereas propane produces no colour change.
Question 3 Report
(a) (i) Define a base according to Arrhenius concept.
(ii) Give one example of an Arrhenius base.
(iii) Identify each of the following substances in aqueous solutions as strong electrolyte, non-electrolyte or weak electrolyte.
I. C\(_{12}\)H\(_{22}\)O\(_{11}\) II. NH\(_3\) Ill. NaOH
(iv) Write a balanced equation to represent the reaction between CH\(_3\)COOH and KOH.
(b) Calculate the volume of 0.500 mol dm\(^{-3}\) HCI required to neutralize 20.00 cm\(^3\) of 0.300 mol dm\(^{-3}\) NaOH.
(c) Give the IUPAC name of each of the following salts: (ii) NaOCI; (iii) Mg(HCO\(_3\))\(_2\).
(d)(i) Define the term standard solution.
(ii) Consider the following compounds: NaOH and Na\(_2\)CO\(_3\). Which of the compounds is suitable for the preparation of a standard alkaline solution? Give reason for your answer
(iii) Fe completely reacted with dilute HCI.
I. Write an equation for the reaction
II. If 3.08g of Fe completely reacted with 50.0 cm\(^3\) of 2.20 mol dm\(^{-3}\) HCI, calculate the relative atomic mass of the metal.
(a) Acids, bases and electrolytes
- (i) According to Arrhenius, a base is a substance that dissociates in water to produce hydroxide ions (OH\(^-\)) as the only negative ions.
- (ii) Example: sodium hydroxide, NaOH (or KOH).
- (iii) I. C\(_{12}\)H\(_{22}\)O\(_{11}\) (sugar): non-electrolyte. II. NH\(_3\): weak electrolyte. III. NaOH: strong electrolyte.
- (iv) \[CH_3COOH + KOH \to CH_3COOK + H_2O\]
(b) Volume of HCl to neutralize 20.00 cm\(^3\) of 0.300 mol dm\(^{-3}\) NaOH (reaction 1:1). Moles NaOH \(= \dfrac{0.300 \times 20.00}{1000} = 6.00\times10^{-3}\) mol. Moles HCl needed = same = \(6.00\times10^{-3}\) mol. \[V = \frac{6.00\times10^{-3}}{0.500} = 0.0120\ \text{dm}^3 = 12.0\ \text{cm}^3.\]
(c) IUPAC names
- NaOCl: sodium chlorate(I) (sodium oxochlorate(I)).
- Mg(HCO\(_3\))\(_2\): magnesium hydrogentrioxocarbonate(IV) (magnesium hydrogencarbonate).
(d) Standard solution and iron
- (i) A standard solution is a solution whose concentration is accurately known.
- (ii) Na\(_2\)CO\(_3\) is the suitable one. Reason: sodium carbonate can be obtained pure, is stable and does not absorb moisture or carbon dioxide from the air, so a known mass gives an accurately known concentration; NaOH is deliquescent and absorbs CO\(_2\), so it cannot be weighed out accurately as a primary standard.
- (iii) I. \[Fe + 2HCl \to FeCl_2 + H_2\uparrow\] II. Moles of HCl \(= \dfrac{2.20 \times 50.0}{1000} = 0.110\) mol. From the equation Fe : HCl = 1 : 2, so moles of Fe \(= \dfrac{0.110}{2} = 0.0550\) mol. \[\text{Relative atomic mass} = \frac{\text{mass}}{\text{moles}} = \frac{3.08}{0.0550} = 56.0.\] So the relative atomic mass of the metal is 56.
Answer Details
(a) Acids, bases and electrolytes
- (i) According to Arrhenius, a base is a substance that dissociates in water to produce hydroxide ions (OH\(^-\)) as the only negative ions.
- (ii) Example: sodium hydroxide, NaOH (or KOH).
- (iii) I. C\(_{12}\)H\(_{22}\)O\(_{11}\) (sugar): non-electrolyte. II. NH\(_3\): weak electrolyte. III. NaOH: strong electrolyte.
- (iv) \[CH_3COOH + KOH \to CH_3COOK + H_2O\]
(b) Volume of HCl to neutralize 20.00 cm\(^3\) of 0.300 mol dm\(^{-3}\) NaOH (reaction 1:1). Moles NaOH \(= \dfrac{0.300 \times 20.00}{1000} = 6.00\times10^{-3}\) mol. Moles HCl needed = same = \(6.00\times10^{-3}\) mol. \[V = \frac{6.00\times10^{-3}}{0.500} = 0.0120\ \text{dm}^3 = 12.0\ \text{cm}^3.\]
(c) IUPAC names
- NaOCl: sodium chlorate(I) (sodium oxochlorate(I)).
- Mg(HCO\(_3\))\(_2\): magnesium hydrogentrioxocarbonate(IV) (magnesium hydrogencarbonate).
(d) Standard solution and iron
- (i) A standard solution is a solution whose concentration is accurately known.
- (ii) Na\(_2\)CO\(_3\) is the suitable one. Reason: sodium carbonate can be obtained pure, is stable and does not absorb moisture or carbon dioxide from the air, so a known mass gives an accurately known concentration; NaOH is deliquescent and absorbs CO\(_2\), so it cannot be weighed out accurately as a primary standard.
- (iii) I. \[Fe + 2HCl \to FeCl_2 + H_2\uparrow\] II. Moles of HCl \(= \dfrac{2.20 \times 50.0}{1000} = 0.110\) mol. From the equation Fe : HCl = 1 : 2, so moles of Fe \(= \dfrac{0.110}{2} = 0.0550\) mol. \[\text{Relative atomic mass} = \frac{\text{mass}}{\text{moles}} = \frac{3.08}{0.0550} = 56.0.\] So the relative atomic mass of the metal is 56.
Question 4 Report
(a) (i) Define saturated solution.
(ii) Tha solubility of KNO\(_3\) at 20°C was 3.00 mol dm\(^{-3}\) If 67.0g of KNO\(_3\) was added to 250 cm\(^{-3}\) of water and stirred at 20°C, determine whether the solution formed was saturated or not at that temperature. [ KNO\(_3\) = 101.0 ]
(b) (i) Distinguish between dative bond and covalent bond.
(ii) Explain why sugar and common salt do not conduct electricity in the solid state.
(iii) State the type of intermolecular forces present in: I. hydrogen fluoride; II argon.
(iv) Consider the compounds with the following structures. S - H ---- N and O - H ---- N In which of the compounds is the hydrogen bend stronger? Give reason for your answer.
(c) (i) State Dalton's Law of Partial Pressure.
(ii) If 200cm of carbon (IV) oxide were collected over water at 18°C and 700 mmHg, determine the volume of the dry gas at s.t.p. [ standard vapour pressure of water at 18°C = 15 mmHg]
(a) Saturated solution
- (i) A saturated solution is one that contains the maximum amount of dissolved solute that it can hold at a given temperature, in the presence of undissolved solute (any further solute added will not dissolve).
- (ii) Solubility of KNO\(_3\) at 20 degrees Celsius \(= 3.00\ \text{mol dm}^{-3}\). Amount added: molar mass of KNO\(_3 = 101.0\), so \[n = \frac{67.0}{101.0} = 0.663\ \text{mol}\ \text{in}\ 250\ \text{cm}^3\ (0.250\ \text{dm}^3).\] Concentration formed \[= \frac{0.663}{0.250} = 2.65\ \text{mol dm}^{-3}.\] Since \(2.65\ \text{mol dm}^{-3} < 3.00\ \text{mol dm}^{-3}\) (the solubility), the solution is unsaturated, i.e. it is not saturated at 20 degrees Celsius.
(b) Bonding
- (i) In an ordinary covalent bond each of the two bonded atoms contributes one electron to the shared pair, whereas in a dative (co-ordinate) bond both electrons of the shared pair are supplied by only one of the atoms.
- (ii) Sugar and common salt do not conduct in the solid state because sugar is a covalent (molecular) solid with no ions or free electrons, and although common salt is ionic, in the solid its ions are held rigidly in the lattice and are not free to move; conduction needs mobile charge carriers.
- (iii) I. Hydrogen fluoride: hydrogen bonding (plus weaker dipole-dipole/van der Waals). II. Argon: weak van der Waals (dispersion) forces only.
- (iv) The hydrogen bond is stronger in O-H----N. Reason: oxygen is more electronegative than sulphur, so the O-H bond is more polar and the hydrogen carries a greater partial positive charge, giving a stronger attraction to the nitrogen lone pair.
(c) Gas laws
- (i) Dalton's law of partial pressures: in a mixture of gases that do not react, the total pressure is equal to the sum of the partial pressures of the individual gases, each partial pressure being the pressure the gas would exert if it alone occupied the whole volume at the same temperature.
- (ii) Pressure of dry CO\(_2\) \(= 700 - 15 = 685\ \text{mmHg}\); \(V_1 = 200\ \text{cm}^3\), \(T_1 = 18 + 273 = 291\ \text{K}\). At s.t.p. \(P_2 = 760\ \text{mmHg}\), \(T_2 = 273\ \text{K}\). Using \(\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}\): \[V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{685 \times 200 \times 273}{291 \times 760} = 169.1\ \text{cm}^3.\] The volume of the dry gas at s.t.p. is about 169 cm\(^3\).
Answer Details
(a) Saturated solution
- (i) A saturated solution is one that contains the maximum amount of dissolved solute that it can hold at a given temperature, in the presence of undissolved solute (any further solute added will not dissolve).
- (ii) Solubility of KNO\(_3\) at 20 degrees Celsius \(= 3.00\ \text{mol dm}^{-3}\). Amount added: molar mass of KNO\(_3 = 101.0\), so \[n = \frac{67.0}{101.0} = 0.663\ \text{mol}\ \text{in}\ 250\ \text{cm}^3\ (0.250\ \text{dm}^3).\] Concentration formed \[= \frac{0.663}{0.250} = 2.65\ \text{mol dm}^{-3}.\] Since \(2.65\ \text{mol dm}^{-3} < 3.00\ \text{mol dm}^{-3}\) (the solubility), the solution is unsaturated, i.e. it is not saturated at 20 degrees Celsius.
(b) Bonding
- (i) In an ordinary covalent bond each of the two bonded atoms contributes one electron to the shared pair, whereas in a dative (co-ordinate) bond both electrons of the shared pair are supplied by only one of the atoms.
- (ii) Sugar and common salt do not conduct in the solid state because sugar is a covalent (molecular) solid with no ions or free electrons, and although common salt is ionic, in the solid its ions are held rigidly in the lattice and are not free to move; conduction needs mobile charge carriers.
- (iii) I. Hydrogen fluoride: hydrogen bonding (plus weaker dipole-dipole/van der Waals). II. Argon: weak van der Waals (dispersion) forces only.
- (iv) The hydrogen bond is stronger in O-H----N. Reason: oxygen is more electronegative than sulphur, so the O-H bond is more polar and the hydrogen carries a greater partial positive charge, giving a stronger attraction to the nitrogen lone pair.
(c) Gas laws
- (i) Dalton's law of partial pressures: in a mixture of gases that do not react, the total pressure is equal to the sum of the partial pressures of the individual gases, each partial pressure being the pressure the gas would exert if it alone occupied the whole volume at the same temperature.
- (ii) Pressure of dry CO\(_2\) \(= 700 - 15 = 685\ \text{mmHg}\); \(V_1 = 200\ \text{cm}^3\), \(T_1 = 18 + 273 = 291\ \text{K}\). At s.t.p. \(P_2 = 760\ \text{mmHg}\), \(T_2 = 273\ \text{K}\). Using \(\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}\): \[V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{685 \times 200 \times 273}{291 \times 760} = 169.1\ \text{cm}^3.\] The volume of the dry gas at s.t.p. is about 169 cm\(^3\).
Question 5 Report
(a) i) Give the name and nature of the radiations that are emitted during radioactivity.
ii) State two differences between chemical reaction and nuclear reaction.
iii) Balance the following nuclear reactions and identify X and Y.
I. \(^{212}_{84} PO\) \(\to\) \(^{208}_{82} Pb\) + X
II \(^{137}_{55}Cs\) \(\to\) \(^{137}_{55}Ba\) + Y
(b) The electron configuration of an element X is: 1s\(^2\) 2s\(^2\) 2p\(^6\) 3s\(^2\) 3p\(^5\)
(i) Deduce the atomic number of X.
(ii) To what group does X belong?
(iii) Give two properties of the group to which the element X belong.
(iv) Identify element X by name.
(v) Write a balanced equation to represent the reaction between the element X and hot concentrated NaOH.
(c)(i) Explain why: I. graphite is used as a lubricant; II. diamond is used as an industrial cutting tool.
(ii) Write an equation to represent the reaction between aqueous HCI and NH\(_3\) solution.
(iii) Name the type of reaction represented by the equation.
(a) Radioactivity
- (i) Radiations emitted: alpha (\(\alpha\)) radiation, positively charged helium nuclei; beta (\(\beta\)) radiation, negatively charged fast-moving electrons; gamma (\(\gamma\)) radiation, high-energy electromagnetic waves with no charge and no mass.
- (ii) Two differences between chemical and nuclear reactions: a chemical reaction involves only the outer (valence) electrons and the atomic nucleus is unchanged, whereas a nuclear reaction involves the nucleus and the element is changed into a different element; the energy change in a nuclear reaction is enormously greater than in a chemical reaction. (Also: chemical reactions can be affected by temperature/pressure/catalyst; nuclear reactions cannot.)
- (iii) Balancing:
- I. \(^{212}_{84}Po \to\ ^{208}_{82}Pb +\ ^{4}_{2}X\). Mass number: \(212-208 = 4\); atomic number: \(84-82 = 2\). So X is an alpha particle (\(^{4}_{2}He\)).
- II. \(^{137}_{55}Cs \to\ ^{137}_{56}Ba +\ ^{0}_{-1}Y\). Mass number unchanged (137); atomic number rises by 1 (55 to 56). So Y is a beta particle (\(^{0}_{-1}e\), an electron).
(b) Element X: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}\)
- (i) Atomic number: total electrons \(= 2+2+6+2+5 = 17\).
- (ii) Group: 7 valence electrons, so Group VII (17), the halogens.
- (iii) Two properties of the group: they are reactive non-metals existing as diatomic molecules; they are strong oxidizing agents whose reactivity decreases down the group; they form ions of charge \(-1\) (and coloured, with characteristic states, e.g. Cl\(_2\) is a greenish-yellow gas).
- (iv) Identity: X is chlorine.
- (v) Reaction with hot concentrated NaOH: \[3Cl_2 + 6NaOH \to 5NaCl + NaClO_3 + 3H_2O\]
(c) Carbon and neutralization
- (i) I. Graphite as a lubricant: graphite has a layered structure in which the flat layers are held together by weak forces, so the layers slide easily over one another, giving a slippery, lubricating action. II. Diamond as a cutting tool: in diamond every carbon atom is covalently bonded to four others in a rigid three-dimensional giant structure, making it extremely hard, so it can cut and grind other materials.
- (ii) \[HCl_{(aq)} + NH_3{(aq)} \to NH_4Cl_{(aq)}\]
- (iii) Type of reaction: neutralization (an acid-base reaction; also a combination reaction).
Answer Details
(a) Radioactivity
- (i) Radiations emitted: alpha (\(\alpha\)) radiation, positively charged helium nuclei; beta (\(\beta\)) radiation, negatively charged fast-moving electrons; gamma (\(\gamma\)) radiation, high-energy electromagnetic waves with no charge and no mass.
- (ii) Two differences between chemical and nuclear reactions: a chemical reaction involves only the outer (valence) electrons and the atomic nucleus is unchanged, whereas a nuclear reaction involves the nucleus and the element is changed into a different element; the energy change in a nuclear reaction is enormously greater than in a chemical reaction. (Also: chemical reactions can be affected by temperature/pressure/catalyst; nuclear reactions cannot.)
- (iii) Balancing:
- I. \(^{212}_{84}Po \to\ ^{208}_{82}Pb +\ ^{4}_{2}X\). Mass number: \(212-208 = 4\); atomic number: \(84-82 = 2\). So X is an alpha particle (\(^{4}_{2}He\)).
- II. \(^{137}_{55}Cs \to\ ^{137}_{56}Ba +\ ^{0}_{-1}Y\). Mass number unchanged (137); atomic number rises by 1 (55 to 56). So Y is a beta particle (\(^{0}_{-1}e\), an electron).
(b) Element X: \(1s^{2}2s^{2}2p^{6}3s^{2}3p^{5}\)
- (i) Atomic number: total electrons \(= 2+2+6+2+5 = 17\).
- (ii) Group: 7 valence electrons, so Group VII (17), the halogens.
- (iii) Two properties of the group: they are reactive non-metals existing as diatomic molecules; they are strong oxidizing agents whose reactivity decreases down the group; they form ions of charge \(-1\) (and coloured, with characteristic states, e.g. Cl\(_2\) is a greenish-yellow gas).
- (iv) Identity: X is chlorine.
- (v) Reaction with hot concentrated NaOH: \[3Cl_2 + 6NaOH \to 5NaCl + NaClO_3 + 3H_2O\]
(c) Carbon and neutralization
- (i) I. Graphite as a lubricant: graphite has a layered structure in which the flat layers are held together by weak forces, so the layers slide easily over one another, giving a slippery, lubricating action. II. Diamond as a cutting tool: in diamond every carbon atom is covalently bonded to four others in a rigid three-dimensional giant structure, making it extremely hard, so it can cut and grind other materials.
- (ii) \[HCl_{(aq)} + NH_3{(aq)} \to NH_4Cl_{(aq)}\]
- (iii) Type of reaction: neutralization (an acid-base reaction; also a combination reaction).
Question 6 Report
(a) Name one product of destructive distillation of coal that is (i) solid (ii) liquid; (iii) gas.
(b) (i) What is the major component of synthetic gas? (ii) Give one reason why synthetic gas is not a major source of air pollution.
(c) (i) Write a balanced chemical equation for the complete combustion of carbon. (ii) State two: I. physical; II. chemical properties of the product in (c)(i).
(d) (i) Name two allotropes of carbon that are I. crystalline, II. amorphous. (ii) State one use of each of the allotropes named in (d)(i).
(e) (i) By means of balanced chemical equations only, outline the process of manufacture of H\(_2\)SO\(_4\), by contact process.
(ii) State the function of H\(_2\)SO\(_4\) in each of the following reaction equations:
I. C\(_2\)H\(_5\)OH\(_{2(l)}\) \(\to\) C\(_2\)H\(_{4(g)}\);
II. Pb(NO\(_{3(g)}\) + H\(_2\)SO\(_{4(g)}\) —> PbSO\(_{4(s)}\) + 2HNO\(_{3(aq)}\)
(a) Products of destructive distillation of coal
- (i) Solid: coke.
- (ii) Liquid: coal tar (also ammoniacal liquor).
- (iii) Gas: coal gas.
(b) Synthetic (producer/water) gas
- (i) Major component: carbon monoxide, CO (together with hydrogen).
- (ii) Reason it is not a major source of air pollution: it burns completely and cleanly to give carbon dioxide and water vapour, producing no soot/smoke and containing no sulphur, so it does not release smoke or sulphur oxides.
(c) Complete combustion of carbon
- (i) \[C + O_2 \to CO_2\]
- (ii) I. Physical properties of CO\(_2\): colourless, odourless gas; denser than air; fairly soluble in water. II. Chemical properties: it is an acidic oxide, turning moist blue litmus red and turning lime water milky; it does not support ordinary combustion (it neither burns nor supports burning).
(d) Allotropes of carbon
- (i) I. Crystalline: diamond and graphite. II. Amorphous: charcoal and coke (also lampblack/soot).
- (ii) Uses: diamond, for cutting/drilling tools (and jewellery); graphite, as a lubricant and as electrodes; charcoal, as a fuel and adsorbent (decolouriser); coke, as a fuel and a reducing agent in the blast furnace.
(e) Contact process and roles of H\(_2\)SO\(_4\)
- (i) \[S + O_2 \to SO_2\] \[2SO_2 + O_2 \rightleftharpoons 2SO_3\] \[SO_3 + H_2SO_4 \to H_2S_2O_7\] \[H_2S_2O_7 + H_2O \to 2H_2SO_4\]
- (ii) I. In \(C_2H_5OH \to C_2H_4 + H_2O\), concentrated H\(_2\)SO\(_4\) acts as a dehydrating agent (it removes water from the ethanol). II. In \(Pb(NO_3)_2 + H_2SO_4 \to PbSO_4 + 2HNO_3\), H\(_2\)SO\(_4\) acts as a non-volatile (less volatile) acid that displaces the more volatile nitric acid from its salt (an acid in double decomposition).
Answer Details
(a) Products of destructive distillation of coal
- (i) Solid: coke.
- (ii) Liquid: coal tar (also ammoniacal liquor).
- (iii) Gas: coal gas.
(b) Synthetic (producer/water) gas
- (i) Major component: carbon monoxide, CO (together with hydrogen).
- (ii) Reason it is not a major source of air pollution: it burns completely and cleanly to give carbon dioxide and water vapour, producing no soot/smoke and containing no sulphur, so it does not release smoke or sulphur oxides.
(c) Complete combustion of carbon
- (i) \[C + O_2 \to CO_2\]
- (ii) I. Physical properties of CO\(_2\): colourless, odourless gas; denser than air; fairly soluble in water. II. Chemical properties: it is an acidic oxide, turning moist blue litmus red and turning lime water milky; it does not support ordinary combustion (it neither burns nor supports burning).
(d) Allotropes of carbon
- (i) I. Crystalline: diamond and graphite. II. Amorphous: charcoal and coke (also lampblack/soot).
- (ii) Uses: diamond, for cutting/drilling tools (and jewellery); graphite, as a lubricant and as electrodes; charcoal, as a fuel and adsorbent (decolouriser); coke, as a fuel and a reducing agent in the blast furnace.
(e) Contact process and roles of H\(_2\)SO\(_4\)
- (i) \[S + O_2 \to SO_2\] \[2SO_2 + O_2 \rightleftharpoons 2SO_3\] \[SO_3 + H_2SO_4 \to H_2S_2O_7\] \[H_2S_2O_7 + H_2O \to 2H_2SO_4\]
- (ii) I. In \(C_2H_5OH \to C_2H_4 + H_2O\), concentrated H\(_2\)SO\(_4\) acts as a dehydrating agent (it removes water from the ethanol). II. In \(Pb(NO_3)_2 + H_2SO_4 \to PbSO_4 + 2HNO_3\), H\(_2\)SO\(_4\) acts as a non-volatile (less volatile) acid that displaces the more volatile nitric acid from its salt (an acid in double decomposition).
