WAEC SSCE - Chemistry - 1993 (Essay)
Question 1 Report
(a) State one air pollution that causes:
(i) blood poisoning
(ii) acid poisoning
(iii) blackening of the walls of buildings
(b) Mention one major chemical industry in each case which requires the following as raw materials:
(i) petrochemicals;
(ii) cellulose.
(a)
(i) Carbon(II) oxide (carbon monoxide) prevents the oxygenation of the blood.
(ii) Sulphur(IV) oxide dissolves in rain water in the atmosphere to form acid rain.
(iii) Hydrogen sulphide reacts with the lead content of paint to form black lead(II) sulphide, PbS.
(b)
(i) Petrochemical industry.
(ii) Paper industry.
Answer Details
(a)
(i) Carbon(II) oxide (carbon monoxide) prevents the oxygenation of the blood.
(ii) Sulphur(IV) oxide dissolves in rain water in the atmosphere to form acid rain.
(iii) Hydrogen sulphide reacts with the lead content of paint to form black lead(II) sulphide, PbS.
(b)
(i) Petrochemical industry.
(ii) Paper industry.
Question 2 Report
(a) Name the residue obtained on strongly heating the following:
(i) ZnCO\(_3\) in an open crucible;
(ii) CuSO\(_4\), 5H\(_2\)O and then allowing it to cool in a desiccator.
(b) State the colour changes observed on heating and cooling in each case in(a) above.
Question 3 Report
(a) If L is the Avogadro constant and E° is standard cell potential, state what X and Y stand for in the following expressions
(i) X = \(\frac{\text{Mass of L molecules of gas or vapour}}{\text{ Mass of L molecules of hydrogen}}\)
(ii) Y = -nFE°
(b) State two differences between a primary cell and a secondary cell.
(a)
- (i) \(X = \dfrac{\text{mass of L molecules of gas or vapour}}{\text{mass of L molecules of hydrogen}}\) is the relative density (vapour density) of the gas. Because L molecules is one mole, this ratio equals the molar mass of the gas divided by the molar mass of hydrogen, i.e. relative density \(= \tfrac{1}{2}\times\) molar mass.
- (ii) \(Y = -nFE^\circ\) is the standard change in Gibbs (free) energy of the cell reaction, \(\Delta G^\circ\), where n is the number of moles of electrons transferred and F is the Faraday constant.
(b) Two differences between a primary cell and a secondary cell:
| Primary cell | Secondary cell |
|---|---|
| Cannot be recharged; used once and discarded. | Can be recharged and used repeatedly. |
| The cell reaction is irreversible. | The cell reaction is reversible. |
| Example: Leclanche (dry) cell. | Example: lead-acid accumulator. |
Answer Details
(a)
- (i) \(X = \dfrac{\text{mass of L molecules of gas or vapour}}{\text{mass of L molecules of hydrogen}}\) is the relative density (vapour density) of the gas. Because L molecules is one mole, this ratio equals the molar mass of the gas divided by the molar mass of hydrogen, i.e. relative density \(= \tfrac{1}{2}\times\) molar mass.
- (ii) \(Y = -nFE^\circ\) is the standard change in Gibbs (free) energy of the cell reaction, \(\Delta G^\circ\), where n is the number of moles of electrons transferred and F is the Faraday constant.
(b) Two differences between a primary cell and a secondary cell:
| Primary cell | Secondary cell |
|---|---|
| Cannot be recharged; used once and discarded. | Can be recharged and used repeatedly. |
| The cell reaction is irreversible. | The cell reaction is reversible. |
| Example: Leclanche (dry) cell. | Example: lead-acid accumulator. |
Question 4 Report
State which of the following can exhibit geometric isomerism:
Give reason for your answer
The four compounds shown are:
- (a) \(CH_3-CH{=}CH-CH_3\) (but-2-ene);
- (b) \(CH_3-CH_2-CHBr-CH_2Br\) (1,2-dibromobutane);
- (c) \(HOOC-CH(CH_3)-CH_2-COOH\) (2-methylbutanedioic acid);
- (d) \(HOOC-CH{=}CH-COOH\) (butenedioic acid).
Compounds that exhibit geometric (cis-trans) isomerism: (a) and (d).
Reason. Geometric isomerism requires two conditions to be met together:
- a carbon-to-carbon double bond (\(C{=}C\)), because the double bond prevents (restricts) free rotation about that bond; and
- each of the two doubly-bonded carbon atoms must carry two different groups.
In (a) each doubly-bonded carbon carries an \(H\) and a \(CH_3\) group (two different groups), so cis- and trans-but-2-ene exist. In (d) each doubly-bonded carbon carries an \(H\) and a \(-COOH\) group, so the cis form (maleic acid) and the trans form (fumaric acid) exist. Both conditions are satisfied, so (a) and (d) show geometric isomerism.
Compounds (b) and (c) contain only single (C-C) bonds. Single bonds allow free rotation, so the atoms cannot be locked into fixed cis/trans positions; therefore (b) and (c) do not exhibit geometric isomerism.
Answer Details
The four compounds shown are:
- (a) \(CH_3-CH{=}CH-CH_3\) (but-2-ene);
- (b) \(CH_3-CH_2-CHBr-CH_2Br\) (1,2-dibromobutane);
- (c) \(HOOC-CH(CH_3)-CH_2-COOH\) (2-methylbutanedioic acid);
- (d) \(HOOC-CH{=}CH-COOH\) (butenedioic acid).
Compounds that exhibit geometric (cis-trans) isomerism: (a) and (d).
Reason. Geometric isomerism requires two conditions to be met together:
- a carbon-to-carbon double bond (\(C{=}C\)), because the double bond prevents (restricts) free rotation about that bond; and
- each of the two doubly-bonded carbon atoms must carry two different groups.
In (a) each doubly-bonded carbon carries an \(H\) and a \(CH_3\) group (two different groups), so cis- and trans-but-2-ene exist. In (d) each doubly-bonded carbon carries an \(H\) and a \(-COOH\) group, so the cis form (maleic acid) and the trans form (fumaric acid) exist. Both conditions are satisfied, so (a) and (d) show geometric isomerism.
Compounds (b) and (c) contain only single (C-C) bonds. Single bonds allow free rotation, so the atoms cannot be locked into fixed cis/trans positions; therefore (b) and (c) do not exhibit geometric isomerism.
Question 5 Report
(a) X and Y belong to the same period in the Periodic Table is a group I element while Y belongs to group VII. State which of the elements would
(i) be a good oxidizing agent
(ii) have the smaller atomic volume
(iii) have the higher ionization potential
(b) Explain your answer in (a)(i) above.
X is a Group I element (a metal, e.g. sodium) and Y is a Group VII element (a non-metal, e.g. chlorine); both are in the same period.
(a)
- (i) A good oxidizing agent: Y (the Group VII element).
- (ii) Smaller atomic volume / smaller atomic radius: Y (across a period the nuclear charge rises and the atom contracts).
- (iii) Higher ionization potential (ionization energy): Y.
(b) Explanation for (a)(i): Y (Group VII) has 7 electrons in its outermost shell and needs to gain only one electron to attain a stable octet. It therefore has a strong tendency to accept (gain) electrons from other species; a species that gains electrons is an oxidizing agent. X (Group I) instead loses its single valence electron easily and acts as a reducing agent. Hence Y is the good oxidizing agent.
Answer Details
X is a Group I element (a metal, e.g. sodium) and Y is a Group VII element (a non-metal, e.g. chlorine); both are in the same period.
(a)
- (i) A good oxidizing agent: Y (the Group VII element).
- (ii) Smaller atomic volume / smaller atomic radius: Y (across a period the nuclear charge rises and the atom contracts).
- (iii) Higher ionization potential (ionization energy): Y.
(b) Explanation for (a)(i): Y (Group VII) has 7 electrons in its outermost shell and needs to gain only one electron to attain a stable octet. It therefore has a strong tendency to accept (gain) electrons from other species; a species that gains electrons is an oxidizing agent. X (Group I) instead loses its single valence electron easily and acts as a reducing agent. Hence Y is the good oxidizing agent.
Question 6 Report
(a)(i) What is meant by the rate of a chemical reaction?
(ii) Explain in terms of the vision theory, the effect of temperature increase on reaction rate.
(b) When hydrogen peroxide is exposed to air, it decomposes
(i) Write an equation for the reaction.
(ii) Outline an experiment to illustrate that effect of a named catalyst on the rate of decomposition.
(iii) Sketch an energy profile diagram to show the effect of the catalyst on the reaction rate, given that the reaction is exothermic.
(c)(i) Explain why enthalpy data alone cannot be used to predict whether a reaction can occur spontaneously or not.
(a)(i) The rate of a chemical reaction is the change in concentration of a reactant or product per unit time.
(a)(ii) According to collision theory, reactant particles must collide effectively before products can form. On increasing the temperature, particles acquire greater kinetic energy. They move faster, collide more frequently, and a greater proportion of the collisions have energy equal to or greater than the activation energy. Hence, the number of effective collisions per second increases and the reaction rate increases.
(b)(i)
(b)(ii) Place equal volumes and concentrations of hydrogen peroxide solution in two identical conical flasks fitted with bungs and delivery tubes leading to gas syringes. Add the same measured mass of manganese(IV) oxide, \(\mathrm{MnO_2}\), to one flask and immediately replace the bung. Do not add \(\mathrm{MnO_2}\) to the other flask, which serves as the control. Record the volume of oxygen collected in each gas syringe at equal time intervals.
The flask containing \(\mathrm{MnO_2}\) produces oxygen much more rapidly, shown by a steeper volume of oxygen against time curve and by vigorous effervescence. The gas relights a glowing splint, confirming that it is oxygen. \(\mathrm{MnO_2}\) is a catalyst because it increases the rate without being used up chemically.
(b)(iii) The catalysed pathway has a lower activation energy, while the enthalpy change, \(\Delta H\), remains negative because the products are at a lower energy than the reactants.
(c)(i) Enthalpy change alone cannot predict spontaneity because spontaneity also depends on the entropy change and on temperature. At constant temperature and pressure, a reaction is spontaneous only when:
Thus, an endothermic reaction may be spontaneous if the increase in entropy is sufficiently large at a suitable temperature, whereas an exothermic reaction may be non-spontaneous if the entropy change is sufficiently unfavourable.
Answer Details
(a)(i) The rate of a chemical reaction is the change in concentration of a reactant or product per unit time.
(a)(ii) According to collision theory, reactant particles must collide effectively before products can form. On increasing the temperature, particles acquire greater kinetic energy. They move faster, collide more frequently, and a greater proportion of the collisions have energy equal to or greater than the activation energy. Hence, the number of effective collisions per second increases and the reaction rate increases.
(b)(i)
(b)(ii) Place equal volumes and concentrations of hydrogen peroxide solution in two identical conical flasks fitted with bungs and delivery tubes leading to gas syringes. Add the same measured mass of manganese(IV) oxide, \(\mathrm{MnO_2}\), to one flask and immediately replace the bung. Do not add \(\mathrm{MnO_2}\) to the other flask, which serves as the control. Record the volume of oxygen collected in each gas syringe at equal time intervals.
The flask containing \(\mathrm{MnO_2}\) produces oxygen much more rapidly, shown by a steeper volume of oxygen against time curve and by vigorous effervescence. The gas relights a glowing splint, confirming that it is oxygen. \(\mathrm{MnO_2}\) is a catalyst because it increases the rate without being used up chemically.
(b)(iii) The catalysed pathway has a lower activation energy, while the enthalpy change, \(\Delta H\), remains negative because the products are at a lower energy than the reactants.
(c)(i) Enthalpy change alone cannot predict spontaneity because spontaneity also depends on the entropy change and on temperature. At constant temperature and pressure, a reaction is spontaneous only when:
Thus, an endothermic reaction may be spontaneous if the increase in entropy is sufficiently large at a suitable temperature, whereas an exothermic reaction may be non-spontaneous if the entropy change is sufficiently unfavourable.
Question 7 Report
Mention the respective properties of the following allotropes of carbon that account for their uses as indicated:
(a) diamond used for drilling rocks;
(b) diamond used as jewels;
(c) graphite used as electrodes;
(d) graphite used for slowing down neutrons in nuclear reactors;
(e) wood charcoal used in gas masks.
- (a) Diamond for drilling rocks: diamond is extremely hard (the hardest natural substance) because each carbon atom is covalently bonded to four others in a rigid three-dimensional structure, so it can cut and drill hard rocks.
- (b) Diamond as jewels: diamond has a very high refractive index and disperses light strongly, so it sparkles brilliantly when cut, making it valuable as a gem.
- (c) Graphite as electrodes: graphite is a good conductor of electricity (it has delocalised/mobile electrons between its layers) and has a high melting point, so it is used as an inert electrode.
- (d) Graphite for slowing down neutrons in nuclear reactors: graphite is a good moderator; its carbon atoms slow down fast neutrons without absorbing them, and it withstands the high temperatures involved.
- (e) Wood charcoal in gas masks: wood charcoal is highly porous with a very large surface area, so it adsorbs (takes onto its surface) poisonous gases, purifying the air breathed.
Answer Details
- (a) Diamond for drilling rocks: diamond is extremely hard (the hardest natural substance) because each carbon atom is covalently bonded to four others in a rigid three-dimensional structure, so it can cut and drill hard rocks.
- (b) Diamond as jewels: diamond has a very high refractive index and disperses light strongly, so it sparkles brilliantly when cut, making it valuable as a gem.
- (c) Graphite as electrodes: graphite is a good conductor of electricity (it has delocalised/mobile electrons between its layers) and has a high melting point, so it is used as an inert electrode.
- (d) Graphite for slowing down neutrons in nuclear reactors: graphite is a good moderator; its carbon atoms slow down fast neutrons without absorbing them, and it withstands the high temperatures involved.
- (e) Wood charcoal in gas masks: wood charcoal is highly porous with a very large surface area, so it adsorbs (takes onto its surface) poisonous gases, purifying the air breathed.
Question 8 Report
(a)(i) Define the term polymerization.
(ii) List the three conditions required for the polymerization on of ethene.
(iii) State the property which is common to compounds that can be easily polymerized
(b) Write appropriate equations to show how the following can be obtained from propan-1-ol in the labouratory
(i) propene;
(ii) propylmethanoate. State the type of reaction involved in each case.
(c)(i) A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. Determine its molecular formula if its molar mass is 180 (H = 1, C = 12, O = 16)
(ii) Explain why ethanoic acid boils at a much higher temperature than butane even though their molar masses almost equal.
(a)(i) Polymerization: the chemical process in which many small molecules (monomers) join together to form a very large molecule (a polymer).
(a)(ii) Conditions for polymerization of ethene: high pressure, high temperature (moderately raised temperature), and a catalyst (e.g. a trace of oxygen or a Ziegler-Natta catalyst).
(a)(iii) The common property is that such compounds are unsaturated (they contain a carbon-carbon double bond, C=C).
(b) From propan-1-ol (CH3CH2CH2OH):
- (i) Propene, by dehydration with concentrated H2SO4 (or hot Al2O3): \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{CH}_3\text{CH}=\text{CH}_2 + \text{H}_2\text{O}\). Type: elimination (dehydration).
- (ii) Propyl methanoate, by esterification with methanoic acid: \(\text{HCOOH} + \text{C}_3\text{H}_7\text{OH} \rightleftharpoons \text{HCOOC}_3\text{H}_7 + \text{H}_2\text{O}\). Type: esterification (condensation).
(c)(i) Molecular formula from 40.0% C, 6.7% H, 53.3% O:
| Element | %/Ar | ratio |
|---|---|---|
| C | 40.0/12 = 3.33 | 1 |
| H | 6.7/1 = 6.7 | 2 |
| O | 53.3/16 = 3.33 | 1 |
Empirical formula = CH2O (mass = 30). \(n = \dfrac{180}{30} = 6\), so the molecular formula is C6H12O6.
(c)(ii) Ethanoic acid molecules contain O-H groups and form strong intermolecular hydrogen bonds (indeed dimers) between molecules, whereas butane, being non-polar, has only weak van der Waals forces. Much more energy is needed to overcome the hydrogen bonds, so ethanoic acid boils at a much higher temperature despite the similar molar mass.
Answer Details
(a)(i) Polymerization: the chemical process in which many small molecules (monomers) join together to form a very large molecule (a polymer).
(a)(ii) Conditions for polymerization of ethene: high pressure, high temperature (moderately raised temperature), and a catalyst (e.g. a trace of oxygen or a Ziegler-Natta catalyst).
(a)(iii) The common property is that such compounds are unsaturated (they contain a carbon-carbon double bond, C=C).
(b) From propan-1-ol (CH3CH2CH2OH):
- (i) Propene, by dehydration with concentrated H2SO4 (or hot Al2O3): \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{CH}_3\text{CH}=\text{CH}_2 + \text{H}_2\text{O}\). Type: elimination (dehydration).
- (ii) Propyl methanoate, by esterification with methanoic acid: \(\text{HCOOH} + \text{C}_3\text{H}_7\text{OH} \rightleftharpoons \text{HCOOC}_3\text{H}_7 + \text{H}_2\text{O}\). Type: esterification (condensation).
(c)(i) Molecular formula from 40.0% C, 6.7% H, 53.3% O:
| Element | %/Ar | ratio |
|---|---|---|
| C | 40.0/12 = 3.33 | 1 |
| H | 6.7/1 = 6.7 | 2 |
| O | 53.3/16 = 3.33 | 1 |
Empirical formula = CH2O (mass = 30). \(n = \dfrac{180}{30} = 6\), so the molecular formula is C6H12O6.
(c)(ii) Ethanoic acid molecules contain O-H groups and form strong intermolecular hydrogen bonds (indeed dimers) between molecules, whereas butane, being non-polar, has only weak van der Waals forces. Much more energy is needed to overcome the hydrogen bonds, so ethanoic acid boils at a much higher temperature despite the similar molar mass.
Question 9 Report
(a) Give one disadvantage of:
(i) hard water
(ii) soft water
(b) Explain why the degree of hardness in a sample of clear lime water is higher than in another sample has that been turned milky by carbon (IV) oxide.
(a) One disadvantage of:
- (i) Hard water: it wastes soap because it forms an insoluble scum with soap before it can lather (it can also fur/scale kettles and boilers).
- (ii) Soft water: it dissolves lead from lead pipes, giving water that can cause lead (plumbo) poisoning (soft water is more corrosive to pipes).
(b) Clear lime water contains dissolved calcium hydroxide, Ca(OH)2, which supplies Ca2+ ions responsible for hardness. When carbon(IV) oxide is bubbled through and the lime water turns milky, the Ca2+ ions are precipitated as insoluble calcium trioxocarbonate(IV):
\[ \text{Ca(OH)}_2 + \text{CO}_2 \to \text{CaCO}_3\downarrow + \text{H}_2\text{O} \]Removing Ca2+ from solution as the insoluble carbonate lowers the concentration of hardness-causing ions. Hence the milky sample has a lower degree of hardness than the clear lime water.
Answer Details
(a) One disadvantage of:
- (i) Hard water: it wastes soap because it forms an insoluble scum with soap before it can lather (it can also fur/scale kettles and boilers).
- (ii) Soft water: it dissolves lead from lead pipes, giving water that can cause lead (plumbo) poisoning (soft water is more corrosive to pipes).
(b) Clear lime water contains dissolved calcium hydroxide, Ca(OH)2, which supplies Ca2+ ions responsible for hardness. When carbon(IV) oxide is bubbled through and the lime water turns milky, the Ca2+ ions are precipitated as insoluble calcium trioxocarbonate(IV):
\[ \text{Ca(OH)}_2 + \text{CO}_2 \to \text{CaCO}_3\downarrow + \text{H}_2\text{O} \]Removing Ca2+ from solution as the insoluble carbonate lowers the concentration of hardness-causing ions. Hence the milky sample has a lower degree of hardness than the clear lime water.
Question 10 Report
(a)(i) List four characteristic properties of transition metals
(ii) Name two metals that can be extracted from their ore by electrolysis.
(b)(i) Determine the oxidation number of chromium in Cr\(_2\)O\(^{2-}_{7}\)
(ii) State the colour observed on adding a few drops of dilute tetraoxosulphate (VI) acid to the system representedby the following equation: Cr\(_2\)O\(^{2-}_{7(aq)}\) + H\(_2O_{(l)}\) \(\rightleftharpoons\) 2CrO\(^{2-}_{4(aq)}\) + 2H\(^+_{(aq)}\). Explain your answer.
(c)(i) State and explain what would be observed if hydrogen sulphide gas were bubbled into acidified K\(_2\)Cr\(_2\)O\(_7\). Write an equation for the reaction.
(ii) What precaution should be taken to avoid excessive exposure to hydrogen sulphide gas while it is being generated in the laboratory?
Answer Details
None
Question 11 Report
The compound whose formula is written below is a major component of a soft fatty substance:
(a) State the change that would be observed in the physical state of the fatty substance if hydrogen were bubbled through it for long time in the presence of finely divided nickel at about 180\(^o\)C
(c) Determine the amount (in mole) of hydrogen that would be consumed if one mole of the component reacted completely with hydrogen.
(c) State the product of the reaction of the fatty substance with hot concentrated sodium hydroxide solution
Question 12 Report
(a)(i) Distinguish between a strong acid and a concentrated acid.
(ii) What is meant by amphoteric oxide? Give one example.
(b)(i) Describe the manufacture of tetraoxosulphate (VI) acid by contact process.
(ii) Write one equation each to show the action of tetraoxosulphate (VI) acid respectively as a dehydrating agent and as an oxidizing agent.
(iii) Give the reason why tetraoxosulphate classified as a heavy chemical.
(a)(i) Strong acid vs concentrated acid: a strong acid is one that ionises (dissociates) completely into ions in aqueous solution, whereas a concentrated acid is one containing a large amount of acid (solute) in a small amount of water (little water). Strength refers to degree of ionisation; concentration refers to amount of solute per unit volume.
(a)(ii) Amphoteric oxide: an oxide that can react with both acids and bases to form a salt and water, e.g. aluminium oxide, Al2O3 (or ZnO, PbO).
(b)(i) Contact process for H2SO4:
- Sulphur (or a sulphide ore) is burnt in air to give sulphur(IV) oxide: \(\text{S} + \text{O}_2 \to \text{SO}_2\).
- The purified, dry SO2 and excess air are passed over a catalyst of vanadium(V) oxide (V2O5) at about 450 °C and 1-2 atm, oxidising it to sulphur(VI) oxide: \(2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3\).
- The SO3 is absorbed in concentrated H2SO4 to form oleum (fuming sulphuric acid): \(\text{SO}_3 + \text{H}_2\text{SO}_4 \to \text{H}_2\text{S}_2\text{O}_7\).
- The oleum is diluted carefully with water to give concentrated H2SO4: \(\text{H}_2\text{S}_2\text{O}_7 + \text{H}_2\text{O} \to 2\text{H}_2\text{SO}_4\).
(b)(ii) Equations:
- As a dehydrating agent (removing the elements of water from sugar): \(\text{C}_{12}\text{H}_{22}\text{O}_{11} \xrightarrow{\text{conc. H}_2\text{SO}_4} 12\text{C} + 11\text{H}_2\text{O}\).
- As an oxidizing agent (hot concentrated acid with copper): \(\text{Cu} + 2\text{H}_2\text{SO}_4 \to \text{CuSO}_4 + \text{SO}_2 + 2\text{H}_2\text{O}\).
(b)(iii) Tetraoxosulphate(VI) acid is classified as a heavy chemical because it is manufactured and used in very large (industrial) quantities and is relatively cheap and impure.
Answer Details
(a)(i) Strong acid vs concentrated acid: a strong acid is one that ionises (dissociates) completely into ions in aqueous solution, whereas a concentrated acid is one containing a large amount of acid (solute) in a small amount of water (little water). Strength refers to degree of ionisation; concentration refers to amount of solute per unit volume.
(a)(ii) Amphoteric oxide: an oxide that can react with both acids and bases to form a salt and water, e.g. aluminium oxide, Al2O3 (or ZnO, PbO).
(b)(i) Contact process for H2SO4:
- Sulphur (or a sulphide ore) is burnt in air to give sulphur(IV) oxide: \(\text{S} + \text{O}_2 \to \text{SO}_2\).
- The purified, dry SO2 and excess air are passed over a catalyst of vanadium(V) oxide (V2O5) at about 450 °C and 1-2 atm, oxidising it to sulphur(VI) oxide: \(2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3\).
- The SO3 is absorbed in concentrated H2SO4 to form oleum (fuming sulphuric acid): \(\text{SO}_3 + \text{H}_2\text{SO}_4 \to \text{H}_2\text{S}_2\text{O}_7\).
- The oleum is diluted carefully with water to give concentrated H2SO4: \(\text{H}_2\text{S}_2\text{O}_7 + \text{H}_2\text{O} \to 2\text{H}_2\text{SO}_4\).
(b)(ii) Equations:
- As a dehydrating agent (removing the elements of water from sugar): \(\text{C}_{12}\text{H}_{22}\text{O}_{11} \xrightarrow{\text{conc. H}_2\text{SO}_4} 12\text{C} + 11\text{H}_2\text{O}\).
- As an oxidizing agent (hot concentrated acid with copper): \(\text{Cu} + 2\text{H}_2\text{SO}_4 \to \text{CuSO}_4 + \text{SO}_2 + 2\text{H}_2\text{O}\).
(b)(iii) Tetraoxosulphate(VI) acid is classified as a heavy chemical because it is manufactured and used in very large (industrial) quantities and is relatively cheap and impure.
Question 13 Report
Classify each of the following as physical change or a chemical change:
(a) fractional distillation of liquefied air;
(b) cracking of petroleum fractions;
(c) conversion of rhombic sulphur to monoclinic sulphur;
(d) chromatographic separation of chlorophyll.
- (a) Fractional distillation of liquefied air: physical change (the components are only separated by boiling point; no new substance is formed).
- (b) Cracking of petroleum fractions: chemical change (large molecules are broken into new, smaller molecules).
- (c) Conversion of rhombic sulphur to monoclinic sulphur: physical change (only a change in the crystalline form/allotrope; the substance is still sulphur).
- (d) Chromatographic separation of chlorophyll: physical change (the pigments are merely separated; no new substance is formed).
Answer Details
- (a) Fractional distillation of liquefied air: physical change (the components are only separated by boiling point; no new substance is formed).
- (b) Cracking of petroleum fractions: chemical change (large molecules are broken into new, smaller molecules).
- (c) Conversion of rhombic sulphur to monoclinic sulphur: physical change (only a change in the crystalline form/allotrope; the substance is still sulphur).
- (d) Chromatographic separation of chlorophyll: physical change (the pigments are merely separated; no new substance is formed).
Question 14 Report
(a) State the phenomenon illustrated by the:
(i) spreading of the smell of hydrogen sulphide gas in the laboratory;
(ii) existence of atoms of the same element having different mass numbers
(b) The atomic number of an element is 17. It has different atoms containing 18 neutrons and 20 neutrons, with a relative abundance of 75% and 25% respectively. Calculate the relative atomic mass of the element.
(a)
(i) The spreading of the smell of hydrogen sulphide gas is called diffusion, which is the movement of particles from a region of higher concentration to a region of lower concentration.
(ii) Isotopy.
(b)
Atomic number = 17
Mass numbers of the atoms:
17 + 18 = 35
17 + 20 = 37
Relative atomic mass = (35 × 0.75) + (37 × 0.25)
= 26.25 + 9.25
= 35.5
Answer Details
(a)
(i) The spreading of the smell of hydrogen sulphide gas is called diffusion, which is the movement of particles from a region of higher concentration to a region of lower concentration.
(ii) Isotopy.
(b)
Atomic number = 17
Mass numbers of the atoms:
17 + 18 = 35
17 + 20 = 37
Relative atomic mass = (35 × 0.75) + (37 × 0.25)
= 26.25 + 9.25
= 35.5
