WAEC SSCE - Chemistry - 2009 (Essay)
Question 1 Report
(a)(1) Define each of the following terms: I. deliquescence; II. efflorescence.
(ii) Give one example I substance that undergoes each of the processes in (a)(i).
(b) Outline how three colourless gases suspected to be ethane, ethene and ethyne could be distinguished in the laboratory.
(c)(i) What is water crystallization?
(ii) Hydrated calcium chloride (CaCl\(_2\) = 111, xH\(_2\)O) contains 49.32% water of cystallization. Calm the value of x. [ (CaCl\(_2\) = 111, H\(_2\)O = 18 ]
(d) Name one calcium compound used (i) in the manufactur cement; (ii) as a dessicant; (iii) in the production of plaster of Paris (POP);
(iv) to neutralise acidic soils
(e) A concentrated solution of sodium chloride was electrolysed using graphite electrodes.
(i) State the ions present in the solution
(ii) Name the products at the I. anode, II. cathode.
(iii) Give the by-product of the electrolysis.
(a)(i) Definitions
- I. Deliquescence: the process by which a substance absorbs moisture (water vapour) from the atmosphere and dissolves in it to form a solution.
- II. Efflorescence: the process by which a hydrated crystalline substance loses part or all of its water of crystallization to the atmosphere and becomes powdery.
(a)(ii) Examples
- Deliquescence: sodium hydroxide (\(NaOH\)) or calcium chloride (\(CaCl_2\)).
- Efflorescence: washing soda, \(Na_2CO_3\cdot 10H_2O\) (or \(Na_2SO_4\cdot 10H_2O\)).
(b) Distinguishing ethane, ethene and ethyne
Pass each gas separately through bromine water: ethane does not decolourize it (no reaction), while ethene and ethyne both decolourize the red-brown bromine water. To separate ethene from ethyne, pass each of the two remaining gases through ammoniacal silver trioxonitrate(V) solution (Tollens' reagent): ethyne gives a white precipitate (silver dicarbide/acetylide), while ethene gives no precipitate. Thus ethane gives no change with bromine water, ethene decolourizes bromine water but gives no precipitate, and ethyne decolourizes bromine water and gives a white precipitate.
(c)(i) Water of crystallization
Water of crystallization is the definite/fixed number of water molecules chemically combined with each formula unit of a substance in its crystalline (hydrated) form.
(c)(ii) Value of x in \(CaCl_2\cdot xH_2O\)
Water is 49.32% of the hydrate, so anhydrous \(CaCl_2\) is \(100 - 49.32 = 50.68\%\).
\[x = \frac{49.32/18}{50.68/111} = \frac{2.740}{0.4566} = 6\]Therefore \(x = 6\); the hydrate is \(CaCl_2\cdot 6H_2O\).
(d) Calcium compounds
- (i) Manufacture of cement: calcium trioxocarbonate(IV), \(CaCO_3\) (limestone).
- (ii) As a desiccant: anhydrous calcium chloride, \(CaCl_2\).
- (iii) Production of plaster of Paris: calcium tetraoxosulphate(VI), \(CaSO_4\cdot\tfrac{1}{2}H_2O\) (from gypsum).
- (iv) To neutralise acidic soils: calcium oxide (quicklime, \(CaO\)) or calcium hydroxide (slaked lime).
(e) Electrolysis of concentrated sodium chloride solution (graphite electrodes)
- (i) Ions present: \(Na^+\), \(H^+\), \(Cl^-\) and \(OH^-\) (from the salt and from water).
- (ii) Products: I. Anode: chlorine gas (\(Cl_2\)); II. Cathode: hydrogen gas (\(H_2\)).
- (iii) By-product: sodium hydroxide (\(NaOH\)) solution.
Answer Details
(a)(i) Definitions
- I. Deliquescence: the process by which a substance absorbs moisture (water vapour) from the atmosphere and dissolves in it to form a solution.
- II. Efflorescence: the process by which a hydrated crystalline substance loses part or all of its water of crystallization to the atmosphere and becomes powdery.
(a)(ii) Examples
- Deliquescence: sodium hydroxide (\(NaOH\)) or calcium chloride (\(CaCl_2\)).
- Efflorescence: washing soda, \(Na_2CO_3\cdot 10H_2O\) (or \(Na_2SO_4\cdot 10H_2O\)).
(b) Distinguishing ethane, ethene and ethyne
Pass each gas separately through bromine water: ethane does not decolourize it (no reaction), while ethene and ethyne both decolourize the red-brown bromine water. To separate ethene from ethyne, pass each of the two remaining gases through ammoniacal silver trioxonitrate(V) solution (Tollens' reagent): ethyne gives a white precipitate (silver dicarbide/acetylide), while ethene gives no precipitate. Thus ethane gives no change with bromine water, ethene decolourizes bromine water but gives no precipitate, and ethyne decolourizes bromine water and gives a white precipitate.
(c)(i) Water of crystallization
Water of crystallization is the definite/fixed number of water molecules chemically combined with each formula unit of a substance in its crystalline (hydrated) form.
(c)(ii) Value of x in \(CaCl_2\cdot xH_2O\)
Water is 49.32% of the hydrate, so anhydrous \(CaCl_2\) is \(100 - 49.32 = 50.68\%\).
\[x = \frac{49.32/18}{50.68/111} = \frac{2.740}{0.4566} = 6\]Therefore \(x = 6\); the hydrate is \(CaCl_2\cdot 6H_2O\).
(d) Calcium compounds
- (i) Manufacture of cement: calcium trioxocarbonate(IV), \(CaCO_3\) (limestone).
- (ii) As a desiccant: anhydrous calcium chloride, \(CaCl_2\).
- (iii) Production of plaster of Paris: calcium tetraoxosulphate(VI), \(CaSO_4\cdot\tfrac{1}{2}H_2O\) (from gypsum).
- (iv) To neutralise acidic soils: calcium oxide (quicklime, \(CaO\)) or calcium hydroxide (slaked lime).
(e) Electrolysis of concentrated sodium chloride solution (graphite electrodes)
- (i) Ions present: \(Na^+\), \(H^+\), \(Cl^-\) and \(OH^-\) (from the salt and from water).
- (ii) Products: I. Anode: chlorine gas (\(Cl_2\)); II. Cathode: hydrogen gas (\(H_2\)).
- (iii) By-product: sodium hydroxide (\(NaOH\)) solution.
Question 2 Report
(a) State Gay Lussac's Law.
(b) Carbon (II) oxide reacted with oxygen to form carbon (IV) oxide in a see tube.
(i) Write a balanced equation for the reaction.
(ii) If 40 cm\(^3\) of the carbon (II) oxide were mixed with cm\(^3\) of oxygen,
I. calculate the volume of carbon (IV) oxide produced
II. which reactant is in excess and how much?
III. what was the total volume of the gaseous mixture at the end of the reaction?
(c) Consider the following oxides: CaO, SiO\(_2\), CO, NO\(_2\) and ZnO. Which of the oxide(s)
(i) is an acidic oxide that is insoluble in water?
(ii) reacts with water to give alkaline solution?
(iii) is amphoteric?
(iv) is neutral?
(v) is/are gaseous at room temperature?
(d) Explain why
(i) colourless concentrated trioxonitrate (V) acid turns yellow,
(ii) dilute trioxonitrate (V) acid does not liberate hydrogen when it reacts with magnesium.
(e) Write a chemical equation for the thermal decomposition of (i) Cu(NO\(_3\))\(_{2(g)}\)
(ii) NH\(_4\)NO\(_{3(g)}\)
Answer Details
None
Question 3 Report
(a)(i) Define polymerization.
(ii) Mention two types of polymerization.
(iii) Give one example of I. naturally occurring polymer, II. synthetic polymer.
(b)(i) Describe how ethyne could be prepared in the laboratory.
(ii) Write an equation for the I. polymerization of ethyne II. complete reaction of ethyne with hydrogen bromide.
(iii) Name each of the products formed in (b)(ii) II.
(c)(i) Define each of the following terms: I. pollution; Il. pollutant.
(ii) Mention three types of pollution.
(iii) Give two ways by which pollution can be prevented.
(a)(i) Polymerization
Polymerization is the chemical process by which many small molecules (monomers) combine together to form one very large molecule (a polymer).
(a)(ii) Two types of polymerization
Addition polymerization and condensation polymerization.
(a)(iii) Examples
- I. Naturally occurring polymer: rubber (also starch, cellulose or protein).
- II. Synthetic polymer: polythene (also polyvinyl chloride, PVC, or nylon).
(b)(i) Laboratory preparation of ethyne
Place a few lumps of calcium dicarbide (calcium carbide) in a flask and add water drop by drop. Ethyne gas is evolved rapidly and is collected over water.
\[CaC_{2(s)} + 2H_2O_{(l)} \to Ca(OH)_{2(aq)} + C_2H_{2(g)}\](b)(ii) Equations
- I. Polymerization of ethyne (to benzene):
- II. Complete reaction of ethyne with hydrogen bromide:
(b)(iii) Name of the product
The product of the complete reaction with hydrogen bromide is 1,1-dibromoethane (\(CH_3CHBr_2\)).
(c)(i) Definitions
- I. Pollution: the introduction of harmful substances (pollutants) into the environment, causing damage to living things and their surroundings.
- II. Pollutant: the harmful substance that causes pollution.
(c)(ii) Three types of pollution
Air pollution, water pollution and land (soil) pollution. (Noise pollution is also acceptable.)
(c)(iii) Two ways of preventing pollution
- Proper treatment and safe disposal of industrial and domestic wastes (and recycling).
- Fitting factories and vehicles with filters/catalytic converters to reduce harmful emissions, and afforestation.
Answer Details
(a)(i) Polymerization
Polymerization is the chemical process by which many small molecules (monomers) combine together to form one very large molecule (a polymer).
(a)(ii) Two types of polymerization
Addition polymerization and condensation polymerization.
(a)(iii) Examples
- I. Naturally occurring polymer: rubber (also starch, cellulose or protein).
- II. Synthetic polymer: polythene (also polyvinyl chloride, PVC, or nylon).
(b)(i) Laboratory preparation of ethyne
Place a few lumps of calcium dicarbide (calcium carbide) in a flask and add water drop by drop. Ethyne gas is evolved rapidly and is collected over water.
\[CaC_{2(s)} + 2H_2O_{(l)} \to Ca(OH)_{2(aq)} + C_2H_{2(g)}\](b)(ii) Equations
- I. Polymerization of ethyne (to benzene):
- II. Complete reaction of ethyne with hydrogen bromide:
(b)(iii) Name of the product
The product of the complete reaction with hydrogen bromide is 1,1-dibromoethane (\(CH_3CHBr_2\)).
(c)(i) Definitions
- I. Pollution: the introduction of harmful substances (pollutants) into the environment, causing damage to living things and their surroundings.
- II. Pollutant: the harmful substance that causes pollution.
(c)(ii) Three types of pollution
Air pollution, water pollution and land (soil) pollution. (Noise pollution is also acceptable.)
(c)(iii) Two ways of preventing pollution
- Proper treatment and safe disposal of industrial and domestic wastes (and recycling).
- Fitting factories and vehicles with filters/catalytic converters to reduce harmful emissions, and afforestation.
Question 4 Report
(a)(i) Outline a suitable procedure for the preparation of ZnCl\(_2\) crystals stating from zinc granules.
(ii) Write a balanced equation for the reaction in (a)(i).
(b) Considei' the following solutions: Na\(_2\)SO\(_{4(aq)}\), CH3COOK\(_{(aq)}\), Pb(NO\(_3\))\(_{2(aq)}\) and MgCl\(_2\). Which of them has/have a pH
(i) greater than 7;
(ii) equal to 7;
(iii) less than 7?
(c)(i) Determine the oxidation number of Mn in I. MnO\(_2\); II. KMnO\(_4\).
(ii) State one laboratory use of each of the .compounds in (c)(i).
(d) Explain why oxidation and reduction processes are complementary.
(e) Consider the reaction represented by the equation: Fe\(_{(s)}\) + 2Ag\(^+_{(aq)}\) \(\to\) Fe\(^{2+}_{(aq)}\) + 2Ag\(_{(s)}\)
(i) Write a balanced ionic half equations for the reaction.
(ii) Which of the species is I. oxidized; II. reduced?
(iii) State the change in the oxidation number of silver during the reaction.
(a)(i) Preparation of ZnCl2 crystals from zinc granules
Add excess zinc granules to warm dilute hydrochloric acid in a beaker. The zinc reacts, giving off hydrogen gas (effervescence). When effervescence stops (all the acid is used up), filter off the excess unreacted zinc. Evaporate the filtrate over a water bath to the point of crystallization, allow it to cool so that crystals of zinc chloride form, then filter off and dry the crystals between filter papers.
(a)(ii) Equation
\[Zn_{(s)} + 2HCl_{(aq)} \to ZnCl_{2(aq)} + H_{2(g)}\](b) pH of the solutions
\(Na_2SO_4\) is the salt of a strong acid and strong base; \(CH_3COOK\) is the salt of a weak acid and strong base; \(Pb(NO_3)_2\) and \(MgCl_2\) are salts of strong acids and weak bases.
- (i) pH greater than 7: \(CH_3COOK\).
- (ii) pH equal to 7: \(Na_2SO_4\).
- (iii) pH less than 7: \(Pb(NO_3)_2\) and \(MgCl_2\).
(c)(i) Oxidation number of Mn
- I. In \(MnO_2\): \(Mn + 2(-2) = 0 \Rightarrow Mn = +4\).
- II. In \(KMnO_4\): \(+1 + Mn + 4(-2) = 0 \Rightarrow Mn = +7\).
(c)(ii) One laboratory use of each
- \(MnO_2\): used as a catalyst (e.g. in the decomposition of hydrogen peroxide) and as an oxidizing agent in the preparation of chlorine.
- \(KMnO_4\): used as an oxidizing agent in volumetric analysis (redox titrations), where it acts as its own indicator.
(d) Why oxidation and reduction are complementary
Oxidation is the loss of electrons and reduction is the gain of electrons. The electrons lost by the species being oxidized are exactly those gained by the species being reduced. Therefore the two processes must occur together and simultaneously (in equal number of electrons); one cannot take place without the other.
(e) The reaction \(Fe + 2Ag^+ \to Fe^{2+} + 2Ag\)
- (i) Half equations: \(Fe \to Fe^{2+} + 2e^-\) (oxidation); \(Ag^+ + e^- \to Ag\) (reduction).
- (ii) I. Oxidized: iron (\(Fe\)); II. Reduced: silver ion (\(Ag^+\)).
- (iii) The oxidation number of silver changes from \(+1\) to \(0\) (a decrease of 1).
Answer Details
(a)(i) Preparation of ZnCl2 crystals from zinc granules
Add excess zinc granules to warm dilute hydrochloric acid in a beaker. The zinc reacts, giving off hydrogen gas (effervescence). When effervescence stops (all the acid is used up), filter off the excess unreacted zinc. Evaporate the filtrate over a water bath to the point of crystallization, allow it to cool so that crystals of zinc chloride form, then filter off and dry the crystals between filter papers.
(a)(ii) Equation
\[Zn_{(s)} + 2HCl_{(aq)} \to ZnCl_{2(aq)} + H_{2(g)}\](b) pH of the solutions
\(Na_2SO_4\) is the salt of a strong acid and strong base; \(CH_3COOK\) is the salt of a weak acid and strong base; \(Pb(NO_3)_2\) and \(MgCl_2\) are salts of strong acids and weak bases.
- (i) pH greater than 7: \(CH_3COOK\).
- (ii) pH equal to 7: \(Na_2SO_4\).
- (iii) pH less than 7: \(Pb(NO_3)_2\) and \(MgCl_2\).
(c)(i) Oxidation number of Mn
- I. In \(MnO_2\): \(Mn + 2(-2) = 0 \Rightarrow Mn = +4\).
- II. In \(KMnO_4\): \(+1 + Mn + 4(-2) = 0 \Rightarrow Mn = +7\).
(c)(ii) One laboratory use of each
- \(MnO_2\): used as a catalyst (e.g. in the decomposition of hydrogen peroxide) and as an oxidizing agent in the preparation of chlorine.
- \(KMnO_4\): used as an oxidizing agent in volumetric analysis (redox titrations), where it acts as its own indicator.
(d) Why oxidation and reduction are complementary
Oxidation is the loss of electrons and reduction is the gain of electrons. The electrons lost by the species being oxidized are exactly those gained by the species being reduced. Therefore the two processes must occur together and simultaneously (in equal number of electrons); one cannot take place without the other.
(e) The reaction \(Fe + 2Ag^+ \to Fe^{2+} + 2Ag\)
- (i) Half equations: \(Fe \to Fe^{2+} + 2e^-\) (oxidation); \(Ag^+ + e^- \to Ag\) (reduction).
- (ii) I. Oxidized: iron (\(Fe\)); II. Reduced: silver ion (\(Ag^+\)).
- (iii) The oxidation number of silver changes from \(+1\) to \(0\) (a decrease of 1).
Question 5 Report
(a) Atoms of four non-metallic elements in the same group of the periodic table are arranged in order of increasing atomic radius as \(R < T < W < X\). Which of the elements
(i) would readily lose electron(s) from the outermost shell;
(ii) is most electronegative;
(iii) would T displace from aqueous solution;
(iv) is at the top of the group?
(b) The following table shows the electronic configuration of two elements Y and Z.
| Element | Electronic Configuration |
| Y | \(1s^2\ 2s^2\ 2p^5\) |
| Z | \(1s^2\ 2s^2\ 2p^5\ 3s^1\) |
(i) Name the I. group to which Y belongs II. period to which Z belongs.
(ii) What is the number of protons present in an atom of Z?
(iii) How many unpaired electrons are in an atom of Y?
(iv) Write the formula of the compound formed between Y and Z
(c) Name the type of bond(s) that exist(s) in each of the following compounds.
(i) \(\mathrm{CaCl}_2\); (ii) \(\mathrm{NH}_4\mathrm{Cl}\); \(\mathrm{CCl}_4\)
(d) Describe how the conductance of a molar solution of ammonia compares to that of sodium hydroxide solution.
(e) State the type of reaction illustrated by each of the following equations:
(i) \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}_{(/)} + \mathrm{CH}_3\mathrm{COOH}_{(aq)} \rightleftharpoons \mathrm{CH}_3\mathrm{COOC}_2\mathrm{H}_5{}_{(/)} + \mathrm{H}_2\mathrm{O}_{(/)}\)
(ii) \(\mathrm{H}+_{Y^+_{(aq)}} + \mathrm{OH}^+_{(aq)} \to \mathrm{H}_2\mathrm{O}\)
(f) Determine the volume of the residual gas when \(20.0\text{cm}^3\) of hydrogen was sparked with \(15.0\text{cm}^3\) of oxygen and the resulting mixture cooled to room temperature.
(a) Four non-metals of the same group with increasing atomic radius \( R < T < W < X \), so R is at the top (smallest) and X at the bottom (largest).
- (i) Would most readily lose electrons from the outermost shell: X (largest atom, weakest hold on outer electrons, least electronegative).
- (ii) Most electronegative: R (smallest atom, at the top).
- (iii) For non-metals reactivity decreases down the group, so T can displace those below it, namely W and X, from aqueous solution.
- (iv) At the top of the group: R.
(b) Y: \( 1s^2\,2s^2\,2p^5 \) (fluorine). Z: \( 1s^2\,2s^2\,2p^6\,3s^1 \) (sodium), giving 11 electrons in the neutral atom.
| Element | Electronic configuration |
|---|---|
| Y | \( 1s^2\,2s^2\,2p^5 \) |
| Z | \( 1s^2\,2s^2\,2p^6\,3s^1 \) |
- (i) I. Y belongs to Group VII (the halogens). II. Z is in Period 3.
- (ii) Number of protons in an atom of Z \( = 11 \).
- (iii) In Y the \( 2p^5 \) sub-shell has one unpaired electron, so there is 1 unpaired electron.
- (iv) Y forms \( Y^- \) and Z forms \( Z^+ \), so the compound is ZY (that is \( NaF \)).
(c) Bond types
- (i) \( CaCl_2 \): ionic (electrovalent) bond.
- (ii) \( NH_4Cl \): ionic bond (between \( NH_4^+ \) and \( Cl^- \)) together with covalent and coordinate (dative) bonds within the ammonium ion.
- (iii) \( CCl_4 \): covalent bond.
(d) Conductance of ammonia compared with sodium hydroxide
Ammonia solution is a weak base that is only slightly ionized, so it produces few ions and conducts electricity poorly. Sodium hydroxide is a strong base that is fully ionized, producing many ions, so at the same molar concentration it conducts electricity much better than ammonia.
(e) Type of reaction
- (i) \( CH_3CH_2OH + CH_3COOH \rightleftharpoons CH_3COOC_2H_5 + H_2O \): esterification (a reversible condensation reaction).
- (ii) \( H^+ + OH^- \rightarrow H_2O \): neutralization.
(f) Volume of residual gas
\[ 2H_2 + O_2 \rightarrow 2H_2O \]\( 20.0\ cm^3 \) of \( H_2 \) reacts with \( \dfrac{20.0}{2} = 10.0\ cm^3 \) of \( O_2 \). Oxygen supplied is \( 15.0\ cm^3 \), so oxygen left over \( = 15.0 - 10.0 = 5.0\ cm^3 \). The water formed condenses to a liquid on cooling.
Residual gas = 5.0 cm\(^3\) of oxygen.
Answer Details
(a) Four non-metals of the same group with increasing atomic radius \( R < T < W < X \), so R is at the top (smallest) and X at the bottom (largest).
- (i) Would most readily lose electrons from the outermost shell: X (largest atom, weakest hold on outer electrons, least electronegative).
- (ii) Most electronegative: R (smallest atom, at the top).
- (iii) For non-metals reactivity decreases down the group, so T can displace those below it, namely W and X, from aqueous solution.
- (iv) At the top of the group: R.
(b) Y: \( 1s^2\,2s^2\,2p^5 \) (fluorine). Z: \( 1s^2\,2s^2\,2p^6\,3s^1 \) (sodium), giving 11 electrons in the neutral atom.
| Element | Electronic configuration |
|---|---|
| Y | \( 1s^2\,2s^2\,2p^5 \) |
| Z | \( 1s^2\,2s^2\,2p^6\,3s^1 \) |
- (i) I. Y belongs to Group VII (the halogens). II. Z is in Period 3.
- (ii) Number of protons in an atom of Z \( = 11 \).
- (iii) In Y the \( 2p^5 \) sub-shell has one unpaired electron, so there is 1 unpaired electron.
- (iv) Y forms \( Y^- \) and Z forms \( Z^+ \), so the compound is ZY (that is \( NaF \)).
(c) Bond types
- (i) \( CaCl_2 \): ionic (electrovalent) bond.
- (ii) \( NH_4Cl \): ionic bond (between \( NH_4^+ \) and \( Cl^- \)) together with covalent and coordinate (dative) bonds within the ammonium ion.
- (iii) \( CCl_4 \): covalent bond.
(d) Conductance of ammonia compared with sodium hydroxide
Ammonia solution is a weak base that is only slightly ionized, so it produces few ions and conducts electricity poorly. Sodium hydroxide is a strong base that is fully ionized, producing many ions, so at the same molar concentration it conducts electricity much better than ammonia.
(e) Type of reaction
- (i) \( CH_3CH_2OH + CH_3COOH \rightleftharpoons CH_3COOC_2H_5 + H_2O \): esterification (a reversible condensation reaction).
- (ii) \( H^+ + OH^- \rightarrow H_2O \): neutralization.
(f) Volume of residual gas
\[ 2H_2 + O_2 \rightarrow 2H_2O \]\( 20.0\ cm^3 \) of \( H_2 \) reacts with \( \dfrac{20.0}{2} = 10.0\ cm^3 \) of \( O_2 \). Oxygen supplied is \( 15.0\ cm^3 \), so oxygen left over \( = 15.0 - 10.0 = 5.0\ cm^3 \). The water formed condenses to a liquid on cooling.
Residual gas = 5.0 cm\(^3\) of oxygen.
Question 6 Report
(a) (i) State Boyle's law.
(ii) Give the mathematical expression of Boyle's law.
(iii) Sketch the graphical representation of Boyle's law.
(b) A given mass of gas occupied 500 cm3 at 30°C and 6.5 x 104 Nm\(^{-2}\). Calculate the volume of the gas at s.t.p. [ Standard pressure = 1.01 x 105 Nm\(^{-2}\)]
(c) Explain why an inflated balloon expands in warm water.
(d)(i) What is meant by the term vapour density? (ii) Determine the vapour density of SO\(_{2}\). [ S = 32, O = 16 ]
(e)(i) Which of the following gases, SO\(_{2}\) and NH\(_{3}\) can be collected by I. upward delivery; II. downward delivery?
(ii) What would be the nature of the solution formed by dissolving each of the following substances in water separately. I. SO\(_{2}\) II. NH\(_{3}\)
(iii) State the property exhibited by SO\(_{2}\) and NH\(_{3}\) in the following equations:
I. 2H\(_{2}\)S\(_{(aq)}\) + SO\(_{2(g)}\) \(\to\) 3S\(_{(s)}\) + 2H\(_{2}\)O\(_{(/)}\)
II. 3CuO\(_{(s)}\) + 2NH\(_{3(g)}\) \(\to\) 3Cu\(_{(s)}\) + 3H\(_{2}\)O\(_{(/)}\) + N\(_{2(s)}\).
(f) State two alloys of tin.
(ii) Give one use of each of the alloys in (f)(i).
(a)(i) Boyle's law states that, at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure.
(ii) \[V\propto\frac{1}{P}\] Hence, \[PV=k\] where \(k\) is a constant for a fixed mass of gas at constant temperature.
(iii) The graph of pressure against volume is a rectangular hyperbola.
(b) Using the combined gas law:
\[\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\]
\[V_2=\frac{P_1V_1T_2}{P_2T_1}\]
\[V_2=\frac{(6.5\times10^4)(500)(273)}{(1.01\times10^5)(273+30)}\]
\[V_2=289.8\ \text{cm}^3\]
\[\boxed{V_2\approx290\ \text{cm}^3}\]
(c) In warm water, the air particles inside the balloon gain kinetic energy and move more rapidly. They collide more frequently and forcefully with the wall of the balloon, causing the balloon to expand until the internal and external pressures balance.
(d)(i) Vapour density is the ratio of the mass of a given volume of a gas or vapour to the mass of an equal volume of hydrogen at the same temperature and pressure.
(d)(ii)
\[M_r(\mathrm{SO_2})=32+(2\times16)=64\]
\[\text{Vapour density}=\frac{M_r}{2}=\frac{64}{2}=\boxed{32}\]
(e)(i)
- Upward delivery: \(\mathrm{NH_3}\).
- Downward delivery: \(\mathrm{SO_2}\).
(e)(ii)
- \(\mathrm{SO_2}\) dissolves in water to form an acidic solution.
- \(\mathrm{NH_3}\) dissolves in water to form a basic or alkaline solution.
(e)(iii)
- \(\mathrm{SO_2}\) acts as an oxidising agent.
- \(\mathrm{NH_3}\) acts as a reducing agent.
(f)(i) Two alloys of tin are:
- Bronze, an alloy of copper and tin.
- Soft solder, an alloy of tin and lead.
(f)(ii)
- Bronze is used for making coins, medals and ornaments.
- Soft solder is used for joining metals, for example in electrical connections and plumbing.
Answer Details
(a)(i) Boyle's law states that, at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure.
(ii) \[V\propto\frac{1}{P}\] Hence, \[PV=k\] where \(k\) is a constant for a fixed mass of gas at constant temperature.
(iii) The graph of pressure against volume is a rectangular hyperbola.
(b) Using the combined gas law:
\[\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\]
\[V_2=\frac{P_1V_1T_2}{P_2T_1}\]
\[V_2=\frac{(6.5\times10^4)(500)(273)}{(1.01\times10^5)(273+30)}\]
\[V_2=289.8\ \text{cm}^3\]
\[\boxed{V_2\approx290\ \text{cm}^3}\]
(c) In warm water, the air particles inside the balloon gain kinetic energy and move more rapidly. They collide more frequently and forcefully with the wall of the balloon, causing the balloon to expand until the internal and external pressures balance.
(d)(i) Vapour density is the ratio of the mass of a given volume of a gas or vapour to the mass of an equal volume of hydrogen at the same temperature and pressure.
(d)(ii)
\[M_r(\mathrm{SO_2})=32+(2\times16)=64\]
\[\text{Vapour density}=\frac{M_r}{2}=\frac{64}{2}=\boxed{32}\]
(e)(i)
- Upward delivery: \(\mathrm{NH_3}\).
- Downward delivery: \(\mathrm{SO_2}\).
(e)(ii)
- \(\mathrm{SO_2}\) dissolves in water to form an acidic solution.
- \(\mathrm{NH_3}\) dissolves in water to form a basic or alkaline solution.
(e)(iii)
- \(\mathrm{SO_2}\) acts as an oxidising agent.
- \(\mathrm{NH_3}\) acts as a reducing agent.
(f)(i) Two alloys of tin are:
- Bronze, an alloy of copper and tin.
- Soft solder, an alloy of tin and lead.
(f)(ii)
- Bronze is used for making coins, medals and ornaments.
- Soft solder is used for joining metals, for example in electrical connections and plumbing.
