WAEC SSCE - Chemistry - 2005 (Essay)
Question 1 Report
(a)(i) List three observable changes that take place when a dilute solution of copper (II) chloride is electrolysed using platinum electrodes. Stage III
(ii) Calculate the quantity of electricity used during electrolysis when a current of 0.21 ampere flows for 2 hours.
(iii) State what is meant by the term preferential discharge of ions in electrolysis
(iv) Give one factor which influences the preferentia! discharge of ions during electrolysis.
(v) State one Q difference between-a conductor and an electrolyte.
(b) Consider the reaction represented by the equation below: Na\(_2\)S\(_2\)O\(_{3(aq)}\) + 2HCI\(_{(aq)}\) \(\to\) 2NaCI\(_{(aq)}\) + H\(_2\)O\(_{(l)}\) + SO\(_{2(g)}\) + S\(_{(s)}\)
(i) List two factors that can affect the rate of this reaction.
(i) Which of the products can be readily used to measure the rate of the reaction. Give a reason for your answer.
(iii) Name two instruments that can be used to measure factors in (b)(i) above.
(c)(i) State the reasons for regarding rusting and burning as oxidation processes.
(ii) I. Write the balanced half equations for the following redox reaction: Mg\(_{(s)}\) + Fe\(^{2+}_{(aq)}\) ----> Mg\(^{2+}_{(aq)}\) + Fe\(_{(s)}\)
II. Which of the reactants is the oxidizing agent?
Ill. State the change in the oxidation number of the oxidizing agent.
(d)(i) State one ore from which each of the following metals can be extracted. I. Tin II. Iron (ii) List two uses of copper (iii) Name one alloy of tin.
Answer Details
None
Question 2 Report
(i) Write the structure of 2—chloro-2—methylpropane.
(ii) Consider the compound X represented by the structure below:
I. State the functional group in X; II. Give the IUPAC name of X; Ill. State the homologous series to which X belongs. IV. Give the names of two compounds from which X is formed. V. State one physical characteristt of X.
(b)(i) List two products obtained from fractional distillation of petroleum;
(ii) State one use of each product in (b)(i) above,
(iii) Mention one disadvantage of crude oil production
(c)(i) Mention the monomer of protein (ii) A compound has an empirical formula of CHO\(_2\) and its molar mass is 90. Deduce the molecular formula of the compound. [H = 1, C = 12, O = 16].
(d) Use the reaction scheme below to answer questions (i).— (iv).
(i) State the reagents needed for stages II and V;
(ii) Nanie the product Q of reaction in stage IV;
(iii) State the conditions required for stage III;
(iv) Give the names of the processes in stages I, II, Ill and V respectively.
(i) Structure of 2-chloro-2-methylpropane: a central carbon carries three methyl groups and one chlorine atom, (CH3)3C-Cl. Displayed:
CH3
|
CH3 - C - Cl
|
CH3
(ii) Compound X is CH3COOC2H5.
- I. Functional group: the ester (alkanoate) group, -COO-.
- II. IUPAC name: ethyl ethanoate.
- III. Homologous series: the alkanoates (esters).
- IV. Two compounds from which X is formed: ethanoic acid and ethanol.
- V. One physical characteristic: X has a sweet, fruity, pleasant smell (it is a colourless volatile liquid).
(b)(i) Two products of the fractional distillation of petroleum: petrol (gasoline) and kerosene.
(ii) One use of each:
- Petrol: fuel for car and vehicle (petrol) engines.
- Kerosene: fuel for heating and lighting, and as aviation (jet) fuel.
(iii) One disadvantage of crude oil production: it causes oil spillage and environmental pollution.
(c)(i) Monomer of protein: the amino acid.
(c)(ii) Molecular formula from empirical formula CHO2, molar mass 90. Empirical formula mass \(= 12 + 1 + 2(16) = 45\).
\[ n = \frac{90}{45} = 2 \]Molecular formula \(= (CHO_2)_2 = \) C2H2O4.
(d) Reaction scheme (ethanol / ethene route):
- (i) Reagents: Stage II requires concentrated H2SO4; Stage V requires hydrogen gas.
- (ii) Product Q of the reaction in Stage IV is 1,2-dibromoethane.
- (iii) Conditions for Stage III: high temperature, high pressure and a catalyst.
- (iv) Names of the processes: Stage I - fermentation; Stage II - dehydration; Stage III - polymerisation; Stage V - hydrogenation.
Answer Details
(i) Structure of 2-chloro-2-methylpropane: a central carbon carries three methyl groups and one chlorine atom, (CH3)3C-Cl. Displayed:
CH3
|
CH3 - C - Cl
|
CH3
(ii) Compound X is CH3COOC2H5.
- I. Functional group: the ester (alkanoate) group, -COO-.
- II. IUPAC name: ethyl ethanoate.
- III. Homologous series: the alkanoates (esters).
- IV. Two compounds from which X is formed: ethanoic acid and ethanol.
- V. One physical characteristic: X has a sweet, fruity, pleasant smell (it is a colourless volatile liquid).
(b)(i) Two products of the fractional distillation of petroleum: petrol (gasoline) and kerosene.
(ii) One use of each:
- Petrol: fuel for car and vehicle (petrol) engines.
- Kerosene: fuel for heating and lighting, and as aviation (jet) fuel.
(iii) One disadvantage of crude oil production: it causes oil spillage and environmental pollution.
(c)(i) Monomer of protein: the amino acid.
(c)(ii) Molecular formula from empirical formula CHO2, molar mass 90. Empirical formula mass \(= 12 + 1 + 2(16) = 45\).
\[ n = \frac{90}{45} = 2 \]Molecular formula \(= (CHO_2)_2 = \) C2H2O4.
(d) Reaction scheme (ethanol / ethene route):
- (i) Reagents: Stage II requires concentrated H2SO4; Stage V requires hydrogen gas.
- (ii) Product Q of the reaction in Stage IV is 1,2-dibromoethane.
- (iii) Conditions for Stage III: high temperature, high pressure and a catalyst.
- (iv) Names of the processes: Stage I - fermentation; Stage II - dehydration; Stage III - polymerisation; Stage V - hydrogenation.
Question 3 Report
(a)(i) State Graham's law of diffusion
(ii) If \(100\ \text{cm}^3\) of oxygen diffused in 4 seconds and \(50\text{cm}^3\) of gas Y diffused in 3 seconds, calculate the relative molecular mass of gas Y. (0 = 16)
(b) Consider the following equilibrium reaction: \(X + 2Y_{(g)} \rightleftharpoons XY_{2(9)}\) \(\Delta H = -52\text{KJ mol}^{-1}\)
(i) State what happens to the yield of \(XY_2\) when the temperature is increased
(ii) Explain the effect of decrease in pressure on the equilibrium position.
(iii) State the effect of a catalyst on the I. position of equilibrium II. activation energy
(c)(i) State the differences between the solubilities of solids and gases in liquids.
(ii) Name the physical-properties used it choosing separation techniques for the following mixtures:
I. kerosene and petrol II. calcium trioxocarbonate (IV) and potassium chloride. III. ammonium chloride and sodium chloride.
(d)(i) State a method of preparing each of the following salts:
| Acid | Basicity |
| \(H_3PO_4\) | |
| \(CH_3COOH\) | |
| \(HNO_2\) |
(iii) State the difference between anhydrous and hydrated salts.
(a)(i) Graham's law of diffusion
At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or relative molecular mass).
\[ \text{rate} \propto \frac{1}{\sqrt{M}} \](a)(ii) Relative molecular mass of gas Y
Rate of oxygen \( = \dfrac{100}{4} = 25\ \text{cm}^3\text{s}^{-1} \); rate of Y \( = \dfrac{50}{3} = 16.67\ \text{cm}^3\text{s}^{-1} \). Relative molecular mass of \( O_2 = 2 \times 16 = 32 \).
\[ \frac{R_{O_2}}{R_Y} = \sqrt{\frac{M_Y}{M_{O_2}}} \]\[ \frac{25}{16.67} = \sqrt{\frac{M_Y}{32}} \;\Rightarrow\; 1.5 = \sqrt{\frac{M_Y}{32}} \]Squaring both sides: \( 2.25 = \dfrac{M_Y}{32} \), so \( M_Y = 2.25 \times 32 = 72 \).
Relative molecular mass of gas Y = 72.
(b) \( X + 2Y_{(g)} \rightleftharpoons XY_{2(g)} \), \( \Delta H = -52\ \text{kJ mol}^{-1} \) (the forward reaction is exothermic).
(i) An increase in temperature favours the endothermic (backward) direction, so the yield of \( XY_2 \) decreases.
(ii) A decrease in pressure shifts the equilibrium towards the side with the greater number of gas molecules. The left side has 3 gas molecules \( (X + 2Y) \) while the right has 1 \( (XY_2) \); the position of equilibrium therefore moves backward (to the left) and the yield of \( XY_2 \) decreases.
(iii) A catalyst: I. has no effect on the position of equilibrium; II. lowers the activation energy of both the forward and backward reactions equally.
(c)(i) Solubility of solids compared with gases in liquids
- The solubility of most solids increases with a rise in temperature, whereas the solubility of gases decreases with a rise in temperature.
- Pressure has almost no effect on the solubility of solids, but the solubility of gases increases markedly with an increase in pressure.
(c)(ii) Physical property used for separation
- I. Kerosene and petrol: difference in boiling points (fractional distillation).
- II. Calcium trioxocarbonate(IV) and potassium chloride: difference in solubility in water (\( CaCO_3 \) insoluble, \( KCl \) soluble); dissolve, then filter.
- III. Ammonium chloride and sodium chloride: difference in the ability to sublime (\( NH_4Cl \) sublimes); sublimation.
(d)(ii) Basicity of the acids
Basicity is the number of replaceable hydrogen ions produced per molecule of the acid on ionization.
| Acid | Basicity |
|---|---|
| \( H_3PO_4 \) | 3 |
| \( CH_3COOH \) | 1 |
| \( HNO_2 \) | 1 |
Only the hydrogen of the \( -COOH \) group in ethanoic acid is replaceable, so its basicity is 1.
(d)(iii) Anhydrous and hydrated salts
An anhydrous salt contains no water of crystallization, while a hydrated salt contains a fixed number of molecules of water of crystallization chemically combined with it (for example \( CuSO_4 \) is anhydrous while \( CuSO_4 \cdot 5H_2O \) is hydrated).
Answer Details
(a)(i) Graham's law of diffusion
At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or relative molecular mass).
\[ \text{rate} \propto \frac{1}{\sqrt{M}} \](a)(ii) Relative molecular mass of gas Y
Rate of oxygen \( = \dfrac{100}{4} = 25\ \text{cm}^3\text{s}^{-1} \); rate of Y \( = \dfrac{50}{3} = 16.67\ \text{cm}^3\text{s}^{-1} \). Relative molecular mass of \( O_2 = 2 \times 16 = 32 \).
\[ \frac{R_{O_2}}{R_Y} = \sqrt{\frac{M_Y}{M_{O_2}}} \]\[ \frac{25}{16.67} = \sqrt{\frac{M_Y}{32}} \;\Rightarrow\; 1.5 = \sqrt{\frac{M_Y}{32}} \]Squaring both sides: \( 2.25 = \dfrac{M_Y}{32} \), so \( M_Y = 2.25 \times 32 = 72 \).
Relative molecular mass of gas Y = 72.
(b) \( X + 2Y_{(g)} \rightleftharpoons XY_{2(g)} \), \( \Delta H = -52\ \text{kJ mol}^{-1} \) (the forward reaction is exothermic).
(i) An increase in temperature favours the endothermic (backward) direction, so the yield of \( XY_2 \) decreases.
(ii) A decrease in pressure shifts the equilibrium towards the side with the greater number of gas molecules. The left side has 3 gas molecules \( (X + 2Y) \) while the right has 1 \( (XY_2) \); the position of equilibrium therefore moves backward (to the left) and the yield of \( XY_2 \) decreases.
(iii) A catalyst: I. has no effect on the position of equilibrium; II. lowers the activation energy of both the forward and backward reactions equally.
(c)(i) Solubility of solids compared with gases in liquids
- The solubility of most solids increases with a rise in temperature, whereas the solubility of gases decreases with a rise in temperature.
- Pressure has almost no effect on the solubility of solids, but the solubility of gases increases markedly with an increase in pressure.
(c)(ii) Physical property used for separation
- I. Kerosene and petrol: difference in boiling points (fractional distillation).
- II. Calcium trioxocarbonate(IV) and potassium chloride: difference in solubility in water (\( CaCO_3 \) insoluble, \( KCl \) soluble); dissolve, then filter.
- III. Ammonium chloride and sodium chloride: difference in the ability to sublime (\( NH_4Cl \) sublimes); sublimation.
(d)(ii) Basicity of the acids
Basicity is the number of replaceable hydrogen ions produced per molecule of the acid on ionization.
| Acid | Basicity |
|---|---|
| \( H_3PO_4 \) | 3 |
| \( CH_3COOH \) | 1 |
| \( HNO_2 \) | 1 |
Only the hydrogen of the \( -COOH \) group in ethanoic acid is replaceable, so its basicity is 1.
(d)(iii) Anhydrous and hydrated salts
An anhydrous salt contains no water of crystallization, while a hydrated salt contains a fixed number of molecules of water of crystallization chemically combined with it (for example \( CuSO_4 \) is anhydrous while \( CuSO_4 \cdot 5H_2O \) is hydrated).
Question 4 Report
(i) Name two amorphous forms of carbon
(ii) State the reason why graphite is a lubricant but diamond is not.
(iii) Draw and label a diagram for the laboratory preparation of a dry sample of carbon (IV) oxide.
(b)(i) Give one example of the following: I. Soil pollutant; II. Water pollutant; III. Air pollutant.
(ii) State the major use of sulphur (IV) oxide in a chemical industry.
(c)(i) Explain in terms of the kinetic theory why petrol is volatile
(ii) State two criteria for determining the purity of a substance.
(iii) Mention one use of each of the following gases: I. Krypton; II. Argon
(d)(i) When zinc metal was added to aqueous copper (I) tetraoxosulphate (VI), the solution turned colourless. I. Name the compound in the colourless solution. II. Write the ionic equation for the reaction. Ill. State what would be observed when a few drops of sodium hydroxide solution is added to a portion of the colourless solution.
(ii) Calculate the volume of CO\(_2\) produced when 5.3g of Na\(_2\)CO\(_3\) reacted with excess HNO\(_{3(aq)}\) + Na\(_2\)CO\(_3\) + 2HNO\(_{3(aq)}\) \(\to\) 2NaNO\(_{3(aq)}\) + CO\(_{2(g)}\) + H\(_2\)O\(_{(l)}\) [H = 1, C = 12, N = 14, O = 16, Na = 23, 1 mole of a gas occupies 22.4 dm\(^3\) at s.t.p.]
(a)(i) Two amorphous forms of carbon: wood charcoal and soot (lamp-black). (Coke and carbon black are also acceptable.)
(ii) Why graphite is a lubricant but diamond is not: in graphite the carbon atoms are arranged in flat parallel layers held together by weak forces, so the layers slide easily over one another, giving graphite its slippery, lubricating property. In diamond every carbon atom is joined to four others by strong covalent bonds in a rigid three-dimensional lattice with no layers to slide, so diamond is hard and cannot act as a lubricant.
(iii) Laboratory preparation of a dry sample of carbon(IV) oxide: marble chips (calcium trioxocarbonate(IV)) are placed in a flask and dilute hydrochloric acid is run onto them through a thistle funnel whose stem dips below the acid. The carbon(IV) oxide evolved is passed first through water to remove any hydrogen chloride gas, then through concentrated tetraoxosulphate(VI) acid to dry it, and the dry gas is finally collected in a gas jar by upward delivery (downward displacement of air), since carbon(IV) oxide is denser than air.
\( \text{CaCO}_{3(s)} + 2\text{HCl}_{(aq)} \rightarrow \text{CaCl}_{2(aq)} + \text{H}_2\text{O}_{(l)} + \text{CO}_{2(g)} \)
(b)(i) I. Soil pollutant: pesticides (or excess fertilizers, polythene, oil spillage). II. Water pollutant: sewage/domestic waste (or factory effluent, crude oil). III. Air pollutant: sulphur(IV) oxide (or carbon(II) oxide, oxides of nitrogen, chlorofluorocarbons).
(ii) The major industrial use of sulphur(IV) oxide is in the manufacture of tetraoxosulphate(VI) acid, \( \text{H}_2\text{SO}_4 \), by the Contact process.
(c)(i) The molecules of petrol are held together by only very weak intermolecular forces of attraction. At room temperature the molecules already possess enough kinetic energy to overcome these weak forces, so they escape readily from the liquid surface into the vapour state; hence petrol is volatile.
(ii) Two criteria for purity: a pure substance has a sharp, constant melting point and a fixed, constant boiling point. (A fixed density or refractive index is also acceptable.)
(iii) I. Krypton: used in photographic flash lamps and fluorescent (fluorescent tube) lighting. II. Argon: used to provide an inert atmosphere in gas-filled electric light bulbs and in welding.
(d)(i) I. The colourless solution contains zinc tetraoxosulphate(VI), ZnSO₄.
II. Ionic equation: \( \text{Zn}_{(s)} + \text{Cu}^{2+}_{(aq)} \rightarrow \text{Zn}^{2+}_{(aq)} + \text{Cu}_{(s)} \).
III. On adding a few drops of sodium hydroxide solution, a white gelatinous precipitate of zinc hydroxide, \( \text{Zn(OH)}_2 \), is formed (which redissolves in excess sodium hydroxide).
(d)(ii) Volume of CO₂ produced from 5.3 g of Na₂CO₃:
\[ \text{Na}_2\text{CO}_3 + 2\text{HNO}_{3(aq)} \rightarrow 2\text{NaNO}_{3(aq)} + \text{CO}_{2(g)} + \text{H}_2\text{O}_{(l)} \]
Molar mass of \( \text{Na}_2\text{CO}_3 = (2\times23) + 12 + (3\times16) = 106 \ \text{g mol}^{-1} \).
Amount of \( \text{Na}_2\text{CO}_3 = \dfrac{5.3}{106} = 0.05 \ \text{mol} \).
From the equation, 1 mol \( \text{Na}_2\text{CO}_3 \) gives 1 mol \( \text{CO}_2 \), so amount of \( \text{CO}_2 = 0.05 \ \text{mol} \).
Volume at s.t.p. \( = 0.05 \times 22.4 = \mathbf{1.12 \ dm^3} \).
Answer Details
(a)(i) Two amorphous forms of carbon: wood charcoal and soot (lamp-black). (Coke and carbon black are also acceptable.)
(ii) Why graphite is a lubricant but diamond is not: in graphite the carbon atoms are arranged in flat parallel layers held together by weak forces, so the layers slide easily over one another, giving graphite its slippery, lubricating property. In diamond every carbon atom is joined to four others by strong covalent bonds in a rigid three-dimensional lattice with no layers to slide, so diamond is hard and cannot act as a lubricant.
(iii) Laboratory preparation of a dry sample of carbon(IV) oxide: marble chips (calcium trioxocarbonate(IV)) are placed in a flask and dilute hydrochloric acid is run onto them through a thistle funnel whose stem dips below the acid. The carbon(IV) oxide evolved is passed first through water to remove any hydrogen chloride gas, then through concentrated tetraoxosulphate(VI) acid to dry it, and the dry gas is finally collected in a gas jar by upward delivery (downward displacement of air), since carbon(IV) oxide is denser than air.
\( \text{CaCO}_{3(s)} + 2\text{HCl}_{(aq)} \rightarrow \text{CaCl}_{2(aq)} + \text{H}_2\text{O}_{(l)} + \text{CO}_{2(g)} \)
(b)(i) I. Soil pollutant: pesticides (or excess fertilizers, polythene, oil spillage). II. Water pollutant: sewage/domestic waste (or factory effluent, crude oil). III. Air pollutant: sulphur(IV) oxide (or carbon(II) oxide, oxides of nitrogen, chlorofluorocarbons).
(ii) The major industrial use of sulphur(IV) oxide is in the manufacture of tetraoxosulphate(VI) acid, \( \text{H}_2\text{SO}_4 \), by the Contact process.
(c)(i) The molecules of petrol are held together by only very weak intermolecular forces of attraction. At room temperature the molecules already possess enough kinetic energy to overcome these weak forces, so they escape readily from the liquid surface into the vapour state; hence petrol is volatile.
(ii) Two criteria for purity: a pure substance has a sharp, constant melting point and a fixed, constant boiling point. (A fixed density or refractive index is also acceptable.)
(iii) I. Krypton: used in photographic flash lamps and fluorescent (fluorescent tube) lighting. II. Argon: used to provide an inert atmosphere in gas-filled electric light bulbs and in welding.
(d)(i) I. The colourless solution contains zinc tetraoxosulphate(VI), ZnSO₄.
II. Ionic equation: \( \text{Zn}_{(s)} + \text{Cu}^{2+}_{(aq)} \rightarrow \text{Zn}^{2+}_{(aq)} + \text{Cu}_{(s)} \).
III. On adding a few drops of sodium hydroxide solution, a white gelatinous precipitate of zinc hydroxide, \( \text{Zn(OH)}_2 \), is formed (which redissolves in excess sodium hydroxide).
(d)(ii) Volume of CO₂ produced from 5.3 g of Na₂CO₃:
\[ \text{Na}_2\text{CO}_3 + 2\text{HNO}_{3(aq)} \rightarrow 2\text{NaNO}_{3(aq)} + \text{CO}_{2(g)} + \text{H}_2\text{O}_{(l)} \]
Molar mass of \( \text{Na}_2\text{CO}_3 = (2\times23) + 12 + (3\times16) = 106 \ \text{g mol}^{-1} \).
Amount of \( \text{Na}_2\text{CO}_3 = \dfrac{5.3}{106} = 0.05 \ \text{mol} \).
From the equation, 1 mol \( \text{Na}_2\text{CO}_3 \) gives 1 mol \( \text{CO}_2 \), so amount of \( \text{CO}_2 = 0.05 \ \text{mol} \).
Volume at s.t.p. \( = 0.05 \times 22.4 = \mathbf{1.12 \ dm^3} \).
Question 5 Report
(a) The electronic configurations of atoms of elements A, B, C and D are given as follows: A. Is\(^2\)2s\(^2\)2p\(^2\); B. 1s\(^2\)2s\(^2\)2sp\(^1\) ; C. 1s\(^2\)2s\(^2\) 2p\(^1\) ; D. 1s\(^2\) 2s\(^2\)
(I) Arrange the elements in order of increasing atomic size, giving reasons
(ii) State which of the elements I. is divalent II. contains atoms with two unpaired electrons in the grouped state. Ill, readily loses one electron from its atom during chemical bonding IV. belongs to group Ill in the Periodic Table.
(b)(i) State one difference between electrovalent and covalent bonds.
(ii) Name two other bonds apart from the ones in (b)(i) above which bind atoms and molecules together.
(iii) State two characteristics of a covalent compound.
(c)(i) What is isotopy?
(ii) Illustrate with suitable example
(iii) Two isotopes of Z with mass numbers 18 and 20 are in the ratio 1:2 Determine the relative atomic mass of Z.
(d)(i) Which of the following elements: calcium, fluorine, iodine neon, magnesium and helium are I. halogens II. noble gases Ill. alkaline earth metals.
(ii) Write a balanced equation for the bombardment of \(^7_3Li\) with protons to produce \(^8_4\beta\) and \(\gamma\)-rays
(iii) State one use of radioactive isotopes.
Question 6 Report
(a)(i) State the two types of hardness in water.
(ii) Name a salt that causes each type of hardness.
(ii) Write a balanced equation for the removal of each type of hardness.
(iv) State one effect of hard water on soap.
(b)(i) State whether the pH of each of the following is less than, equal to, or greater than 7.
I. Glucose solution II. Chlroine water III. Lime water IV. Sour milk
(ii) Give the difference between the following compounds: I. an acidic oxide and an amphoteric oxide; II. concentrated acid and a dilute acid; Ill. a normal salt and an acid salt
(c)(i) Iron reacts with H\(_2\)SO\(_4\) according to the equation: Fe\(_{(s)}\) + H\(_2\)SO\(_{4(aq)}\) ---> FeSO\(_{4(aq)}\) + H\(_{2(g)}\)
Calculate the mass of FeSO\(_4\) that would be produced by 0.5 mole of Fe. [H = 1, S = 32, Fe = 56]
(ii) List two allotropes of sulphur
(d)(i) State what would La observed when a damp starch-iodide paper is dropped into a gas jar of chloride
(ii) Explain your ansv.er in (d)(i) above.
(iii) State the products formed when ammonia reacts with excess chlorine.
(a) Hardness of water
- (i) Temporary hardness and permanent hardness.
- (ii) Temporary hardness: calcium hydrogen trioxocarbonate (IV), Ca(HCO3)2. Permanent hardness: calcium tetraoxosulphate (VI), CaSO4.
- (iii) Removal of temporary hardness (by boiling):
Ca(HCO3)2(aq) → CaCO3(s) + H2O(l) + CO2(g)
Removal of permanent hardness (by adding washing soda):
CaSO4(aq) + Na2CO3(aq) → CaCO3(s) + Na2SO4(aq) - (iv) Hard water forms an insoluble scum with soap and does not readily form a lather, so it wastes soap.
(b)(i) pH values
- I. Glucose solution: equal to 7.
- II. Chlorine water: less than 7.
- III. Lime water: greater than 7.
- IV. Sour milk: less than 7.
(b)(ii) Differences
- I. An acidic oxide reacts with bases only to form salt and water, while an amphoteric oxide reacts with both acids and bases to form salt and water.
- II. A concentrated acid contains little or no water (a high proportion of acid), while a dilute acid contains a large amount of water (a low proportion of acid).
- III. In a normal salt all the ionizable hydrogen of the acid has been replaced by a metal, while in an acid salt only part of the ionizable hydrogen has been replaced (it still contains replaceable hydrogen).
(c)(i) Mass of FeSO4 from 0.5 mole of Fe
From Fe + H2SO4 → FeSO4 + H2, the mole ratio Fe : FeSO4 is 1 : 1, so moles of FeSO4 = 0.5 mol.
Molar mass of FeSO4 = 56 + 32 + 64 = 152 g mol-1.
Mass = 0.5 × 152 = 76 g.
(c)(ii) Rhombic sulphur and monoclinic sulphur.
(d)(i) The damp starch-iodide paper turns blue-black.
(d)(ii) Chlorine is a stronger oxidizing agent than iodine; it oxidizes the iodide ions on the paper to iodine, and the liberated iodine turns the starch blue-black.
(d)(iii) With excess chlorine, ammonia gives nitrogen trichloride (NCl3) and hydrogen chloride: NH3 + 3Cl2 → NCl3 + 3HCl.
Answer Details
(a) Hardness of water
- (i) Temporary hardness and permanent hardness.
- (ii) Temporary hardness: calcium hydrogen trioxocarbonate (IV), Ca(HCO3)2. Permanent hardness: calcium tetraoxosulphate (VI), CaSO4.
- (iii) Removal of temporary hardness (by boiling):
Ca(HCO3)2(aq) → CaCO3(s) + H2O(l) + CO2(g)
Removal of permanent hardness (by adding washing soda):
CaSO4(aq) + Na2CO3(aq) → CaCO3(s) + Na2SO4(aq) - (iv) Hard water forms an insoluble scum with soap and does not readily form a lather, so it wastes soap.
(b)(i) pH values
- I. Glucose solution: equal to 7.
- II. Chlorine water: less than 7.
- III. Lime water: greater than 7.
- IV. Sour milk: less than 7.
(b)(ii) Differences
- I. An acidic oxide reacts with bases only to form salt and water, while an amphoteric oxide reacts with both acids and bases to form salt and water.
- II. A concentrated acid contains little or no water (a high proportion of acid), while a dilute acid contains a large amount of water (a low proportion of acid).
- III. In a normal salt all the ionizable hydrogen of the acid has been replaced by a metal, while in an acid salt only part of the ionizable hydrogen has been replaced (it still contains replaceable hydrogen).
(c)(i) Mass of FeSO4 from 0.5 mole of Fe
From Fe + H2SO4 → FeSO4 + H2, the mole ratio Fe : FeSO4 is 1 : 1, so moles of FeSO4 = 0.5 mol.
Molar mass of FeSO4 = 56 + 32 + 64 = 152 g mol-1.
Mass = 0.5 × 152 = 76 g.
(c)(ii) Rhombic sulphur and monoclinic sulphur.
(d)(i) The damp starch-iodide paper turns blue-black.
(d)(ii) Chlorine is a stronger oxidizing agent than iodine; it oxidizes the iodide ions on the paper to iodine, and the liberated iodine turns the starch blue-black.
(d)(iii) With excess chlorine, ammonia gives nitrogen trichloride (NCl3) and hydrogen chloride: NH3 + 3Cl2 → NCl3 + 3HCl.
