WAEC SSCE - Chemistry - 2016 (Essay)
Question 1 Report
(a)(i) Name a suitable drying agent for the preparation of carbon (IV) oxide in the laboratory.
(ii) Using one chemical test, distinguish between carbon (II) oxide and carbon (IV) oxide.
(b)(i) Describe briefly how oxygen and nitrogen could be obtained separately from air on an industrial scale
(ii) State how a lighted splint can be used to distinguish between samples of oxygen and nitrogen.
(c)(i) Give one reason why bauxite is usually preferred as the ore for the extraction of aluminium.
(ii). List two main impurities. usually present in bauxite.
(iii) State the function of sodium hydroxide solution in the extraction of aluminium from its ore.
(iv) Explain briefly why it is difficult to extract aluminium by chemical reduction of aluminium oxide
(v) Write an equation for the reaction of aluminium oxide with aqueous sodium hydroxide.
(d) (i) The melting and boiling points of sodium chloride are 801 °C and 141.3 °C respectively. Explain briefly why sodium chloride does not conduct electricity at 25°C but does so between 801 °C and 1413 °C.
(ii) State the reason why sodium metal is stored under paraffin oil in the laboratory.
(e)(i) State what would be observed when aqueous sodium trioxocarbonate(IV) is added to a solution containing iron (III) ions
(ii) Write a balanced equation for the reaction in (e)(i).
Answer Details
None
Question 2 Report
(a) Consider the following atoms: \(^R_T\)X; \(^S_T\)X.
(i) State the phenomenon exhibited by the two atoms.
(ii) What is the difference between the atoms?
(iii) Give two examples of elements that exhibit the phenomenon stated in (ai)
(iv) lf T is 17, write the electron configuration of the element
(b)(i) State two differences between metals and-non-metals with respect to their:
I. physical properties;
2. chemical properties.
(ii) Give one example of each for the following compounds:
I. an amphoteric oxide;
II. a hydride which evolves hydrogen when reacted with water;
Ill. a trioxocarbonate (IV) salt which is readily decomposed on heating;
IV. a chloride salt which is readily hydrolyzed in water.
(c)(i) State three characteristic properties of transition metals.
(ii) Write the electron configuration of \(_{30}Zn\)
(iii.) Explain briefly why zinc is not considered as a typical transition element.
(d) Consider the reaction represented by the following equation:- Na\(_2\)CO\(_{3(aq)}\) + MgCl\(_{2(aq)}\) ----> 2NaCl\(_{(aq)}\) + MgCO\(_{3(aq)}\). Calculate the mass of sodium trioxocarbonate (IV) needed to produce 3.36 of magnesium trioxocarbonate (IV). [C = 12.0, O = 16.0, Na = 23.0, Mg = 24:0 ]
(a) The atoms \(^R_TX\) and \(^S_TX\) have the same atomic number T but different mass numbers R and S.
- (i) The phenomenon exhibited is isotopy (they are isotopes).
- (ii) The difference is that they have different numbers of neutrons (hence different mass numbers), while having the same number of protons.
- (iii) Two examples of elements exhibiting isotopy: chlorine (\(^{35}Cl\) and \(^{37}Cl\)) and carbon (\(^{12}C\) and \(^{14}C\)). (Hydrogen and oxygen are also acceptable.)
- (iv) If T = 17, the electron configuration is 2, 8, 7.
(b)(i) Metals compared with non-metals:
| Metals | Non-metals | |
|---|---|---|
| Physical | Shiny/lustrous, malleable and ductile, good conductors of heat and electricity, mostly solids with high melting points. | Dull, brittle (if solid), poor conductors (except graphite), many are gases or low-melting solids. |
| Chemical | Form basic oxides; are electropositive, lose electrons to form positive ions (reducing agents). | Form acidic oxides; are electronegative, gain electrons to form negative ions (oxidizing agents). |
(ii) Examples:
- I. Amphoteric oxide: zinc oxide, \(ZnO\) (or \(Al_2O_3\), PbO).
- II. Hydride evolving hydrogen with water: sodium hydride, \(NaH\) (\(NaH + H_2O \rightarrow NaOH + H_2\)).
- III. Trioxocarbonate(IV) salt readily decomposed on heating: copper(II) trioxocarbonate(IV), \(CuCO_3\) (or \(ZnCO_3\)).
- IV. Chloride salt readily hydrolysed in water: aluminium chloride, \(AlCl_3\) (or \(FeCl_3\)).
(c)(i) Three characteristic properties of transition metals: they show variable oxidation states, form coloured ions/compounds, and act as catalysts (they also form complex ions).
(ii) Electron configuration of \(_{30}Zn\): 2, 8, 18, 2 (\(1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^2\)).
(iii) Zinc is not regarded as a typical transition element because its 3d subshell is completely filled (3d^{10}); it shows only one oxidation state (+2) and forms mainly white/colourless compounds, so it lacks the characteristic transition-metal properties.
(d) \(Na_2CO_3 + MgCl_2 \rightarrow 2NaCl + MgCO_3\). To produce 3.36 g of \(MgCO_3\):
Molar mass of \(MgCO_3 = 24 + 12 + (3 \times 16) = 84\,g\,mol^{-1}\)
\[\text{moles of } MgCO_3 = \frac{3.36}{84} = 0.04\,mol\]
From the equation, 1 mole \(Na_2CO_3\) gives 1 mole \(MgCO_3\), so moles of \(Na_2CO_3 = 0.04\,mol\).
Molar mass of \(Na_2CO_3 = (2 \times 23) + 12 + (3 \times 16) = 106\,g\,mol^{-1}\)
\[\text{mass of } Na_2CO_3 = 0.04 \times 106 = 4.24\,g\]
Answer Details
(a) The atoms \(^R_TX\) and \(^S_TX\) have the same atomic number T but different mass numbers R and S.
- (i) The phenomenon exhibited is isotopy (they are isotopes).
- (ii) The difference is that they have different numbers of neutrons (hence different mass numbers), while having the same number of protons.
- (iii) Two examples of elements exhibiting isotopy: chlorine (\(^{35}Cl\) and \(^{37}Cl\)) and carbon (\(^{12}C\) and \(^{14}C\)). (Hydrogen and oxygen are also acceptable.)
- (iv) If T = 17, the electron configuration is 2, 8, 7.
(b)(i) Metals compared with non-metals:
| Metals | Non-metals | |
|---|---|---|
| Physical | Shiny/lustrous, malleable and ductile, good conductors of heat and electricity, mostly solids with high melting points. | Dull, brittle (if solid), poor conductors (except graphite), many are gases or low-melting solids. |
| Chemical | Form basic oxides; are electropositive, lose electrons to form positive ions (reducing agents). | Form acidic oxides; are electronegative, gain electrons to form negative ions (oxidizing agents). |
(ii) Examples:
- I. Amphoteric oxide: zinc oxide, \(ZnO\) (or \(Al_2O_3\), PbO).
- II. Hydride evolving hydrogen with water: sodium hydride, \(NaH\) (\(NaH + H_2O \rightarrow NaOH + H_2\)).
- III. Trioxocarbonate(IV) salt readily decomposed on heating: copper(II) trioxocarbonate(IV), \(CuCO_3\) (or \(ZnCO_3\)).
- IV. Chloride salt readily hydrolysed in water: aluminium chloride, \(AlCl_3\) (or \(FeCl_3\)).
(c)(i) Three characteristic properties of transition metals: they show variable oxidation states, form coloured ions/compounds, and act as catalysts (they also form complex ions).
(ii) Electron configuration of \(_{30}Zn\): 2, 8, 18, 2 (\(1s^2\,2s^2\,2p^6\,3s^2\,3p^6\,3d^{10}\,4s^2\)).
(iii) Zinc is not regarded as a typical transition element because its 3d subshell is completely filled (3d^{10}); it shows only one oxidation state (+2) and forms mainly white/colourless compounds, so it lacks the characteristic transition-metal properties.
(d) \(Na_2CO_3 + MgCl_2 \rightarrow 2NaCl + MgCO_3\). To produce 3.36 g of \(MgCO_3\):
Molar mass of \(MgCO_3 = 24 + 12 + (3 \times 16) = 84\,g\,mol^{-1}\)
\[\text{moles of } MgCO_3 = \frac{3.36}{84} = 0.04\,mol\]
From the equation, 1 mole \(Na_2CO_3\) gives 1 mole \(MgCO_3\), so moles of \(Na_2CO_3 = 0.04\,mol\).
Molar mass of \(Na_2CO_3 = (2 \times 23) + 12 + (3 \times 16) = 106\,g\,mol^{-1}\)
\[\text{mass of } Na_2CO_3 = 0.04 \times 106 = 4.24\,g\]
Question 3 Report
(a) What are nucleons?
(b) State Graham's law of diffusion.
(c) Explain briefly why aluminium does not corrode easily.
(d) State three examples of periodic properties.
(e) State two reasons why real gases deviate from ideal gas behaviour.
(f) List three uses of fractional distillation in industry.
(g) What factors determine the selective discharge of ions at the electrodes during electrolysis?
(h) State the type of reaction represented by each of the following equations:
(i) C\(_2\)H\(_6\) + Br\(_2\) ---> C\(_2\)H\(_5\)Br + HBr;
(ii) C\(_2\)H\(_4\) + .Br\(_2\) ---> CH\(_4\)Br\(_2\)
(i) Name the products formed when butane burns in limited supply of air.
(j) List three methods of separating a solid from a liquid.
(a) Nucleons are the particles found in the nucleus of an atom, that is, the protons and neutrons.
(b) 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 of its molar mass).
\[\text{rate} \propto \frac{1}{\sqrt{\rho}}\]
(c) Aluminium does not corrode easily because it forms a thin, tough, coherent and impervious layer of aluminium oxide on its surface. This oxide layer sticks firmly to the metal and prevents air and moisture from reaching the metal underneath, protecting it from further attack.
(d) Three periodic properties: atomic radius, ionization energy and electronegativity (electron affinity and metallic/non-metallic character are also acceptable).
(e) Two reasons real gases deviate from ideal behaviour:
- The molecules of a real gas have an appreciable (finite) volume, which is not negligible compared with the volume of the container, especially at high pressure.
- There are intermolecular forces of attraction between real gas molecules, especially at high pressure and low temperature.
(f) Three industrial uses of fractional distillation: refining of crude petroleum into its various fractions; separation of liquid air into oxygen and nitrogen; separation of ethanol from a water-ethanol mixture (rectification of spirits).
(g) Factors that determine selective discharge of ions during electrolysis: the position of the ion in the electrochemical (discharge) series, the concentration of the ion in the electrolyte, and the nature of the electrode used.
(h) Types of reaction:
- (i) \(C_2H_6 + Br_2 \rightarrow C_2H_5Br + HBr\) is a substitution reaction.
- (ii) \(C_2H_4 + Br_2 \rightarrow C_2H_4Br_2\) is an addition reaction.
(i) When butane burns in a limited supply of air, incomplete combustion occurs, giving carbon(II) oxide (and carbon/soot) and water vapour.
(j) Three methods of separating a solid from a liquid: filtration, evaporation and decantation (centrifugation and crystallization are also acceptable).
Answer Details
(a) Nucleons are the particles found in the nucleus of an atom, that is, the protons and neutrons.
(b) 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 of its molar mass).
\[\text{rate} \propto \frac{1}{\sqrt{\rho}}\]
(c) Aluminium does not corrode easily because it forms a thin, tough, coherent and impervious layer of aluminium oxide on its surface. This oxide layer sticks firmly to the metal and prevents air and moisture from reaching the metal underneath, protecting it from further attack.
(d) Three periodic properties: atomic radius, ionization energy and electronegativity (electron affinity and metallic/non-metallic character are also acceptable).
(e) Two reasons real gases deviate from ideal behaviour:
- The molecules of a real gas have an appreciable (finite) volume, which is not negligible compared with the volume of the container, especially at high pressure.
- There are intermolecular forces of attraction between real gas molecules, especially at high pressure and low temperature.
(f) Three industrial uses of fractional distillation: refining of crude petroleum into its various fractions; separation of liquid air into oxygen and nitrogen; separation of ethanol from a water-ethanol mixture (rectification of spirits).
(g) Factors that determine selective discharge of ions during electrolysis: the position of the ion in the electrochemical (discharge) series, the concentration of the ion in the electrolyte, and the nature of the electrode used.
(h) Types of reaction:
- (i) \(C_2H_6 + Br_2 \rightarrow C_2H_5Br + HBr\) is a substitution reaction.
- (ii) \(C_2H_4 + Br_2 \rightarrow C_2H_4Br_2\) is an addition reaction.
(i) When butane burns in a limited supply of air, incomplete combustion occurs, giving carbon(II) oxide (and carbon/soot) and water vapour.
(j) Three methods of separating a solid from a liquid: filtration, evaporation and decantation (centrifugation and crystallization are also acceptable).
Question 4 Report
(a) (i) Draw the structures of the isomers of the alkene with molecular formurat C\(_4\)H\(_8\)
(ii) State the class of alkanols to which each of the following compounds belongs:
I. CH\(_3\)C(CH\(_3\))\(_2\)OH;
II. CH\(_3\)CH(CH\(_3\))CH\(_2\)OH;
III. CH\(_3\)CH\(_2\)CH(CH\(_3\))OH.
(b) (i) Write the formulae of the products formed in the following reactions:
I. CH\(_3\)CH\(_2\)COOH \(\frac{K_{(s)}}{}\)
II. CH\(_3\)CH\(_2\)COOH. \(\frac{C_4H_6OH, heat}{Conc.H_2SO_4}\)
III. CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)OH \(\frac{H^+/KMnO_4}{(excess)}\)
(ii) Name the major product(s) of each of the reactions in (b)(i).
(c) A gaseous hydrocarbon R of mass 7.0 g occupies a volume of 2.24 dm\(^3\) at s. t.p. If the percentage composition by mass of hydrogen is 14.3, determine its:
(i) empirical formula;
(ii) molecular formula. [ H = 1.00, C = 12.0, Molar volume of gas at s.t.p, = 22.4 dm\(^3\) ]
(d) Define structural isomerism.
(a)(i) Structures of the isomers of the alkene C\(_4\)H\(_8\)
The alkene of molecular formula C\(_4\)H\(_8\) exists as the following isomers (three structural isomers, of which but-2-ene shows an additional pair of geometric isomers):
Thus the structural isomers are but-1-ene, but-2-ene and 2-methylprop-1-ene; but-2-ene occurs as cis- and trans-forms.
(a)(ii) Class of alkanol
| Compound | Structure | Class |
| I | CH\(_3\)C(CH\(_3\))\(_2\)OH (the \(-\)OH carbon is joined to three other carbons) | Tertiary (3°) alkanol |
| II | CH\(_3\)CH(CH\(_3\))CH\(_2\)OH (the \(-\)OH carbon is joined to one other carbon) | Primary (1°) alkanol |
| III | CH\(_3\)CH\(_2\)CH(CH\(_3\))OH (the \(-\)OH carbon is joined to two other carbons) | Secondary (2°) alkanol |
(b)(i) Formulae of the products
I. Propanoic acid with potassium metal:
\[2CH_3CH_2COOH + 2K \rightarrow 2CH_3CH_2COOK + H_2\]Products: CH\(_3\)CH\(_2\)COOK and H\(_2\).
II. Propanoic acid with butanol (C\(_4\)H\(_9\)OH) and concentrated H\(_2\)SO\(_4\), heat (esterification):
\[CH_3CH_2COOH + C_4H_9OH \xrightarrow[\Delta]{conc.\,H_2SO_4} CH_3CH_2COOC_4H_9 + H_2O\]Products: CH\(_3\)CH\(_2\)COOC\(_4\)H\(_9\) and H\(_2\)O.
III. Butan-1-ol oxidised by acidified KMnO\(_4\) (excess) — the primary alkanol is oxidised fully to the carboxylic acid:
\[CH_3CH_2CH_2CH_2OH \xrightarrow{H^+/KMnO_4\,(excess)} CH_3CH_2CH_2COOH\]Product: CH\(_3\)CH\(_2\)CH\(_2\)COOH.
(b)(ii) Names of the major product(s)
- I. Potassium propanoate (and hydrogen gas)
- II. Butyl propanoate (and water)
- III. Butanoic acid
(c) Empirical and molecular formula of hydrocarbon R
Given: mass of R = 7.0 g, volume = 2.24 dm\(^3\) at s.t.p., % H by mass = 14.3, molar volume = 22.4 dm\(^3\), H = 1.00, C = 12.0.
(i) Empirical formula
Since R is a hydrocarbon, % carbon = \(100 - 14.3 = 85.7\%\).
| Element | C | H |
| % by mass | 85.7 | 14.3 |
| Moles = %÷A\(_r\) | \(\frac{85.7}{12.0}=7.14\) | \(\frac{14.3}{1.0}=14.3\) |
| Divide by smaller (7.14) | \(\frac{7.14}{7.14}=1\) | \(\frac{14.3}{7.14}=2\) |
\[\therefore \text{Empirical formula} = CH_2\]
(ii) Molecular formula
Number of moles of R:
\[n = \frac{\text{volume at s.t.p.}}{\text{molar volume}} = \frac{2.24}{22.4} = 0.1\ \text{mol}\]Molar mass:
\[M_r = \frac{\text{mass}}{\text{moles}} = \frac{7.0}{0.1} = 70\ \text{g mol}^{-1}\]Let the molecular formula be \((CH_2)_n\). The empirical formula mass \(= 12.0 + 2(1.0) = 14\).
\[14n = 70 \quad\Rightarrow\quad n = \frac{70}{14} = 5\]\[\therefore \text{Molecular formula} = C_5H_{10}\](d) Structural isomerism
Structural isomerism is the phenomenon in which two or more compounds have the same molecular formula but different structural (arrangement of atoms) formulae, and hence different properties.
Answer Details
(a)(i) Structures of the isomers of the alkene C\(_4\)H\(_8\)
The alkene of molecular formula C\(_4\)H\(_8\) exists as the following isomers (three structural isomers, of which but-2-ene shows an additional pair of geometric isomers):
Thus the structural isomers are but-1-ene, but-2-ene and 2-methylprop-1-ene; but-2-ene occurs as cis- and trans-forms.
(a)(ii) Class of alkanol
| Compound | Structure | Class |
| I | CH\(_3\)C(CH\(_3\))\(_2\)OH (the \(-\)OH carbon is joined to three other carbons) | Tertiary (3°) alkanol |
| II | CH\(_3\)CH(CH\(_3\))CH\(_2\)OH (the \(-\)OH carbon is joined to one other carbon) | Primary (1°) alkanol |
| III | CH\(_3\)CH\(_2\)CH(CH\(_3\))OH (the \(-\)OH carbon is joined to two other carbons) | Secondary (2°) alkanol |
(b)(i) Formulae of the products
I. Propanoic acid with potassium metal:
\[2CH_3CH_2COOH + 2K \rightarrow 2CH_3CH_2COOK + H_2\]Products: CH\(_3\)CH\(_2\)COOK and H\(_2\).
II. Propanoic acid with butanol (C\(_4\)H\(_9\)OH) and concentrated H\(_2\)SO\(_4\), heat (esterification):
\[CH_3CH_2COOH + C_4H_9OH \xrightarrow[\Delta]{conc.\,H_2SO_4} CH_3CH_2COOC_4H_9 + H_2O\]Products: CH\(_3\)CH\(_2\)COOC\(_4\)H\(_9\) and H\(_2\)O.
III. Butan-1-ol oxidised by acidified KMnO\(_4\) (excess) — the primary alkanol is oxidised fully to the carboxylic acid:
\[CH_3CH_2CH_2CH_2OH \xrightarrow{H^+/KMnO_4\,(excess)} CH_3CH_2CH_2COOH\]Product: CH\(_3\)CH\(_2\)CH\(_2\)COOH.
(b)(ii) Names of the major product(s)
- I. Potassium propanoate (and hydrogen gas)
- II. Butyl propanoate (and water)
- III. Butanoic acid
(c) Empirical and molecular formula of hydrocarbon R
Given: mass of R = 7.0 g, volume = 2.24 dm\(^3\) at s.t.p., % H by mass = 14.3, molar volume = 22.4 dm\(^3\), H = 1.00, C = 12.0.
(i) Empirical formula
Since R is a hydrocarbon, % carbon = \(100 - 14.3 = 85.7\%\).
| Element | C | H |
| % by mass | 85.7 | 14.3 |
| Moles = %÷A\(_r\) | \(\frac{85.7}{12.0}=7.14\) | \(\frac{14.3}{1.0}=14.3\) |
| Divide by smaller (7.14) | \(\frac{7.14}{7.14}=1\) | \(\frac{14.3}{7.14}=2\) |
\[\therefore \text{Empirical formula} = CH_2\]
(ii) Molecular formula
Number of moles of R:
\[n = \frac{\text{volume at s.t.p.}}{\text{molar volume}} = \frac{2.24}{22.4} = 0.1\ \text{mol}\]Molar mass:
\[M_r = \frac{\text{mass}}{\text{moles}} = \frac{7.0}{0.1} = 70\ \text{g mol}^{-1}\]Let the molecular formula be \((CH_2)_n\). The empirical formula mass \(= 12.0 + 2(1.0) = 14\).
\[14n = 70 \quad\Rightarrow\quad n = \frac{70}{14} = 5\]\[\therefore \text{Molecular formula} = C_5H_{10}\](d) Structural isomerism
Structural isomerism is the phenomenon in which two or more compounds have the same molecular formula but different structural (arrangement of atoms) formulae, and hence different properties.
Question 5 Report
(a)(i) Determine the oxidation number of sulphur in Na\(_2\)S\(_2\)O\(_3\)
(ii) Name the allotropes of sulphur.
(iii) State two ways in which the structure of graphite and diamond are similar.
(b)(i) Name two green-house gases.
(ii) State one effect of an increased level of green-house gases on the environment.
(iii) State one source from which nitrogen (I) oxide is released into the environment.
(iv) Write a chemical equation to show the effect of heat on each of the following compounds: I. KNO\(_{3(s)}\) II. AgNO\(_{3(s)}\)
(C)(i)Describe briefly how pure crystals of calcium chloride could be obtained from a solution of calcium chloride
ii) Explain briefly each of the following observations:
I. ammonia gas is highly soluble in water;
II. boiling ploint of chlorine is lower than that of iodine
(d) Consider the reaction represented by the following equation:- 2NaCl + H\(_2\)SO\(_{4(s)}\) \(\to\) Na\(_2\)SO\(_{4(s)}\) + 2HCl\(_{(g)}\)
Calculate the volume of HCl gas that can be obtained at s.t.p. from 5.85 g of sodium. chloride. [ Na = 23.0, Cl = 35.5, Molar volume of gas at s.t.p. = 22.4 dm\(^3\)]
(a)(i) Oxidation number of sulphur in \(Na_2S_2O_3\). Let it be \(x\); Na is \(+1\), O is \(-2\), and the compound is neutral:
\[2(+1) + 2x + 3(-2) = 0\]
\[2 + 2x - 6 = 0 \;\Rightarrow\; 2x = 4 \;\Rightarrow\; x = +2\]
Sulphur has an average oxidation number of +2.
(ii) The allotropes of sulphur are rhombic sulphur and monoclinic sulphur (plastic sulphur is also sometimes listed).
(iii) Two ways in which graphite and diamond are similar: both are allotropes of carbon (made up of carbon atoms only), and both are giant covalent (macromolecular) structures with very high melting points.
(b)(i) Two greenhouse gases: carbon(IV) oxide (\(CO_2\)) and methane (\(CH_4\)). (Water vapour and CFCs are also acceptable.)
(ii) One effect of an increased level of greenhouse gases: global warming (a rise in the average atmospheric temperature, leading to melting of ice caps and rising sea levels).
(iii) One source of nitrogen(I) oxide, \(N_2O\): bacterial action on nitrogenous fertilizers in the soil (or from vehicle exhausts).
(iv) Effect of heat:
I. \[2KNO_3 \rightarrow 2KNO_2 + O_2\]
II. \[2AgNO_3 \rightarrow 2Ag + 2NO_2 + O_2\]
(c)(i) To obtain pure crystals of calcium chloride from its solution: evaporate the solution carefully to concentrate it to the point of crystallization, then allow it to cool so that crystals form; filter off the crystals and dry them (calcium chloride is deliquescent, so drying should be done quickly in a desiccator).
(ii) I. Ammonia is highly soluble in water because ammonia molecules are polar and form hydrogen bonds with water (and react with water to form ammonium hydroxide).
II. The boiling point of chlorine is lower than that of iodine because chlorine molecules are smaller and lighter, so the van der Waals (intermolecular) forces between them are weaker than the stronger van der Waals forces between the larger iodine molecules; less energy is needed to separate chlorine molecules.
(d) \(2NaCl + H_2SO_4 \rightarrow Na_2SO_4 + 2HCl\). From 5.85 g of NaCl:
Molar mass of \(NaCl = 23 + 35.5 = 58.5\,g\,mol^{-1}\)
\[\text{moles of } NaCl = \frac{5.85}{58.5} = 0.1\,mol\]
From the equation, 2 moles NaCl give 2 moles HCl (1:1), so moles of \(HCl = 0.1\,mol\).
\[\text{volume at s.t.p.} = 0.1 \times 22.4 = 2.24\,dm^3\]
Answer Details
(a)(i) Oxidation number of sulphur in \(Na_2S_2O_3\). Let it be \(x\); Na is \(+1\), O is \(-2\), and the compound is neutral:
\[2(+1) + 2x + 3(-2) = 0\]
\[2 + 2x - 6 = 0 \;\Rightarrow\; 2x = 4 \;\Rightarrow\; x = +2\]
Sulphur has an average oxidation number of +2.
(ii) The allotropes of sulphur are rhombic sulphur and monoclinic sulphur (plastic sulphur is also sometimes listed).
(iii) Two ways in which graphite and diamond are similar: both are allotropes of carbon (made up of carbon atoms only), and both are giant covalent (macromolecular) structures with very high melting points.
(b)(i) Two greenhouse gases: carbon(IV) oxide (\(CO_2\)) and methane (\(CH_4\)). (Water vapour and CFCs are also acceptable.)
(ii) One effect of an increased level of greenhouse gases: global warming (a rise in the average atmospheric temperature, leading to melting of ice caps and rising sea levels).
(iii) One source of nitrogen(I) oxide, \(N_2O\): bacterial action on nitrogenous fertilizers in the soil (or from vehicle exhausts).
(iv) Effect of heat:
I. \[2KNO_3 \rightarrow 2KNO_2 + O_2\]
II. \[2AgNO_3 \rightarrow 2Ag + 2NO_2 + O_2\]
(c)(i) To obtain pure crystals of calcium chloride from its solution: evaporate the solution carefully to concentrate it to the point of crystallization, then allow it to cool so that crystals form; filter off the crystals and dry them (calcium chloride is deliquescent, so drying should be done quickly in a desiccator).
(ii) I. Ammonia is highly soluble in water because ammonia molecules are polar and form hydrogen bonds with water (and react with water to form ammonium hydroxide).
II. The boiling point of chlorine is lower than that of iodine because chlorine molecules are smaller and lighter, so the van der Waals (intermolecular) forces between them are weaker than the stronger van der Waals forces between the larger iodine molecules; less energy is needed to separate chlorine molecules.
(d) \(2NaCl + H_2SO_4 \rightarrow Na_2SO_4 + 2HCl\). From 5.85 g of NaCl:
Molar mass of \(NaCl = 23 + 35.5 = 58.5\,g\,mol^{-1}\)
\[\text{moles of } NaCl = \frac{5.85}{58.5} = 0.1\,mol\]
From the equation, 2 moles NaCl give 2 moles HCl (1:1), so moles of \(HCl = 0.1\,mol\).
\[\text{volume at s.t.p.} = 0.1 \times 22.4 = 2.24\,dm^3\]
