WAEC SSCE - Chemistry - 1991 (Essay)
Question 1 Report
(a) Give the products of electrolysis of dilute copper (II) tetraoxosulphate(VI) solution using the following materials as electrodes:
(i) carbon rod;
(ii) copper rods
(b) For each of the process in (a) above;
(i) write the anodic half reaction;
(ii) state how electrolysis affects the pH of electrolyte
(a) Products of electrolysis of dilute copper(II) tetraoxosulphate(VI) solution
- (i) Carbon (inert) electrodes: copper is deposited at the cathode; oxygen gas is evolved at the anode. The blue colour fades and the solution becomes acidic (tetraoxosulphate(VI) acid remains).
- (ii) Copper electrodes: copper is deposited at the cathode; the copper anode dissolves (loses mass) so no gas is evolved. The concentration and colour of the solution stay essentially the same.
(b) For each process
(i) Anodic half reaction
- Carbon anode: \[2H_2O_{(l)} \rightarrow O_{2(g)} + 4H^+_{(aq)} + 4e^-\]
- Copper anode: \[Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^-\]
(ii) Effect on pH
- With carbon electrodes, oxygen is released and H+ ions accumulate in solution, so the pH decreases (the electrolyte becomes more acidic).
- With copper electrodes, copper dissolves at the anode at the same rate as it deposits at the cathode, so the Cu2+ concentration is unchanged and the pH remains essentially constant.
Answer Details
(a) Products of electrolysis of dilute copper(II) tetraoxosulphate(VI) solution
- (i) Carbon (inert) electrodes: copper is deposited at the cathode; oxygen gas is evolved at the anode. The blue colour fades and the solution becomes acidic (tetraoxosulphate(VI) acid remains).
- (ii) Copper electrodes: copper is deposited at the cathode; the copper anode dissolves (loses mass) so no gas is evolved. The concentration and colour of the solution stay essentially the same.
(b) For each process
(i) Anodic half reaction
- Carbon anode: \[2H_2O_{(l)} \rightarrow O_{2(g)} + 4H^+_{(aq)} + 4e^-\]
- Copper anode: \[Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^-\]
(ii) Effect on pH
- With carbon electrodes, oxygen is released and H+ ions accumulate in solution, so the pH decreases (the electrolyte becomes more acidic).
- With copper electrodes, copper dissolves at the anode at the same rate as it deposits at the cathode, so the Cu2+ concentration is unchanged and the pH remains essentially constant.
Question 2 Report
(a)(i) Give two uses of chlorine.
(ii) State the action of chlorine on moist blue litmus paper
(b) Draw a labelled diagram for the laboratory preparation of a dry sample of chlorine
(c) State the type of reaction involved between chlorine and (i) aqueous iron (II) chloride;
(ii) propane. Write an equation for each reaction and name the product formed in (c)(ii).
(d) Consider the reactions the following equations: Cl\(_{2(g)}\) + 2Br\(^-_{(aq)}\) \(\to\) 2Cl\(^-_{(aq)}\) + Br\(_{2(g)}\)
F\(_{2(g)}\) + 2Cl\(^-_{(aq)}\) ---> 2F\(^-_{(aq)}\) + Cl\(_{2(g)}\)
From the equations, arrange bromine, chlorine and fluorine in increasing order of oxidizing ability. Give the reason for your answer.
(a)(i) Two uses of chlorine are:
- sterilising drinking water and swimming-pool water;
- manufacture of bleaching agents and for bleaching wood pulp and textiles.
(a)(ii) Chlorine first turns moist blue litmus paper red and then bleaches it white.
(b) A dry sample of chlorine is prepared by warming manganese(IV) oxide with concentrated hydrochloric acid. The gas is washed with saturated sodium chloride solution to remove hydrogen chloride, dried with concentrated tetraoxosulphate(VI) acid, and collected by downward displacement of air.
\[\mathrm{MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+2H_2O(l)+Cl_2(g)}\]
(c)(i) Chlorine reacts with aqueous iron(II) chloride by a redox reaction. Iron(II) ions are oxidised to iron(III) ions.
\[\mathrm{Cl_2(aq)+2Fe^{2+}(aq)\rightarrow 2Cl^-(aq)+2Fe^{3+}(aq)}\]
Equivalently:
\[\mathrm{Cl_2(aq)+2FeCl_2(aq)\rightarrow 2FeCl_3(aq)}\]
(c)(ii) Chlorine reacts with propane in sunlight or ultraviolet light by a substitution reaction (free-radical substitution).
\[\mathrm{C_3H_8(g)+Cl_2(g)\xrightarrow{UV\ light}C_3H_7Cl(g)+HCl(g)}\]
The organic product is chloropropane, formed as a mixture of 1-chloropropane and 2-chloropropane.
(d) The increasing order of oxidising ability is:
\[\mathrm{Br_2 < Cl_2 < F_2}\]
Chlorine oxidises bromide ions to bromine, so chlorine is a stronger oxidising agent than bromine. Fluorine oxidises chloride ions to chlorine, so fluorine is a stronger oxidising agent than chlorine.
Answer Details
(a)(i) Two uses of chlorine are:
- sterilising drinking water and swimming-pool water;
- manufacture of bleaching agents and for bleaching wood pulp and textiles.
(a)(ii) Chlorine first turns moist blue litmus paper red and then bleaches it white.
(b) A dry sample of chlorine is prepared by warming manganese(IV) oxide with concentrated hydrochloric acid. The gas is washed with saturated sodium chloride solution to remove hydrogen chloride, dried with concentrated tetraoxosulphate(VI) acid, and collected by downward displacement of air.
\[\mathrm{MnO_2(s)+4HCl(aq)\rightarrow MnCl_2(aq)+2H_2O(l)+Cl_2(g)}\]
(c)(i) Chlorine reacts with aqueous iron(II) chloride by a redox reaction. Iron(II) ions are oxidised to iron(III) ions.
\[\mathrm{Cl_2(aq)+2Fe^{2+}(aq)\rightarrow 2Cl^-(aq)+2Fe^{3+}(aq)}\]
Equivalently:
\[\mathrm{Cl_2(aq)+2FeCl_2(aq)\rightarrow 2FeCl_3(aq)}\]
(c)(ii) Chlorine reacts with propane in sunlight or ultraviolet light by a substitution reaction (free-radical substitution).
\[\mathrm{C_3H_8(g)+Cl_2(g)\xrightarrow{UV\ light}C_3H_7Cl(g)+HCl(g)}\]
The organic product is chloropropane, formed as a mixture of 1-chloropropane and 2-chloropropane.
(d) The increasing order of oxidising ability is:
\[\mathrm{Br_2 < Cl_2 < F_2}\]
Chlorine oxidises bromide ions to bromine, so chlorine is a stronger oxidising agent than bromine. Fluorine oxidises chloride ions to chlorine, so fluorine is a stronger oxidising agent than chlorine.
Question 3 Report
(a)State three characteristics of a catalyst.
(b) Mention one manufacturing process in which each of the following metals is used as a catalyst:
(i) iron;
(ii) nickel;
(iii) platinum
(c) Give one example of an organic catalyst.
Answer Details
None
Question 4 Report
Sketch a curve to show how the solubility of a gas varies with increasing temperature
At constant pressure, the solubility of a gas in a liquid decreases as the temperature increases.
Answer Details
At constant pressure, the solubility of a gas in a liquid decreases as the temperature increases.
Question 5 Report
A colourless and odourless gas X burns in oxygen with a pale blue flame.
(a) Suggest two gases which X could be.
(b) Give one chemical test that could be used to confirm which of the two gases X is.
(a) Two possible gases
A colourless, odourless gas that burns in oxygen with a pale blue flame could be hydrogen (H2) or carbon(II) oxide (CO). Both are colourless and odourless, and both burn with a pale blue flame.
(b) Chemical test to distinguish them
Burn a sample of X and test the product of combustion:
- If X is hydrogen, it burns to form water only, which turns white anhydrous copper(II) tetraoxosulphate(VI) blue (or blue cobalt chloride paper pink) and gives no reaction with limewater.
- If X is carbon(II) oxide, it burns to form carbon(IV) oxide, which turns limewater milky. \[2CO + O_2 \rightarrow 2CO_2\]
Therefore, passing the combustion product into limewater identifies the gas: a milky precipitate confirms carbon(II) oxide, while the formation of water confirms hydrogen.
Answer Details
(a) Two possible gases
A colourless, odourless gas that burns in oxygen with a pale blue flame could be hydrogen (H2) or carbon(II) oxide (CO). Both are colourless and odourless, and both burn with a pale blue flame.
(b) Chemical test to distinguish them
Burn a sample of X and test the product of combustion:
- If X is hydrogen, it burns to form water only, which turns white anhydrous copper(II) tetraoxosulphate(VI) blue (or blue cobalt chloride paper pink) and gives no reaction with limewater.
- If X is carbon(II) oxide, it burns to form carbon(IV) oxide, which turns limewater milky. \[2CO + O_2 \rightarrow 2CO_2\]
Therefore, passing the combustion product into limewater identifies the gas: a milky precipitate confirms carbon(II) oxide, while the formation of water confirms hydrogen.
Question 6 Report
(a) Name one gaseous hydrocarbon which is
(i) used for welding.
(ii) a major raw material for the plastic industry.
(b) Write the structural formula of the hydrocarbon in (a)(i) above. Name the process by which it can be converted to neoprene rubber.
(c) Potatoes contain a high proportion of carbohydrate.
(i) Give the main product formed when potatoes are dehydrated completely
(ii) Describe how you would convert potatoes to ethanol. State the reactions involved the process and write equation for the final stage of the conversion.
(iii) Draw a labelled diagram of the apparatus you would use to obtain a sample of fairly pure ethanol from the product formed in (c)(ii) above. What is the name given to the technique?
(a)
(i) Acetylene (ethyne).
(ii) Ethene.
(b)
The structural formula of acetylene is:
H—C≡C—H
It is converted to neoprene rubber by polymerisation, after conversion to chloroprene (2-chlorobuta-1,3-diene).
(c)(i) Complete dehydration of potatoes gives mainly starch.
(c)(ii) Conversion of potatoes to ethanol
- Wash, crush and pressure-cook the potatoes to rupture the cells and release the starch granules.
- Add malt. The enzyme diastase in malt hydrolyses starch to maltose.
- Allow the mixture to cool, then add yeast. The enzyme maltase hydrolyses maltose to glucose.
- Keep the mixture warm (about 30–35°C) in the absence of air. The enzyme zymase in yeast ferments glucose to ethanol and carbon(IV) oxide.
The reactions are:
Starch hydrolysis:
\[\left(C_6H_{10}O_5\right)_n+\frac{n}{2}H_2O\xrightarrow{\text{diastase}}\frac{n}{2}C_{12}H_{22}O_{11}\]
Maltose hydrolysis:
\[C_{12}H_{22}O_{11}+H_2O\xrightarrow{\text{maltase}}2C_6H_{12}O_6\]
Final fermentation stage:
\[C_6H_{12}O_6\xrightarrow{\text{zymase}}2C_2H_5OH+2CO_2\]
(c)(iii) The fermented liquid is distilled using the apparatus shown below. Ethanol, with a boiling point of about 78°C, distils before most of the water.
The technique is fractional distillation.
Answer Details
(a)
(i) Acetylene (ethyne).
(ii) Ethene.
(b)
The structural formula of acetylene is:
H—C≡C—H
It is converted to neoprene rubber by polymerisation, after conversion to chloroprene (2-chlorobuta-1,3-diene).
(c)(i) Complete dehydration of potatoes gives mainly starch.
(c)(ii) Conversion of potatoes to ethanol
- Wash, crush and pressure-cook the potatoes to rupture the cells and release the starch granules.
- Add malt. The enzyme diastase in malt hydrolyses starch to maltose.
- Allow the mixture to cool, then add yeast. The enzyme maltase hydrolyses maltose to glucose.
- Keep the mixture warm (about 30–35°C) in the absence of air. The enzyme zymase in yeast ferments glucose to ethanol and carbon(IV) oxide.
The reactions are:
Starch hydrolysis:
\[\left(C_6H_{10}O_5\right)_n+\frac{n}{2}H_2O\xrightarrow{\text{diastase}}\frac{n}{2}C_{12}H_{22}O_{11}\]
Maltose hydrolysis:
\[C_{12}H_{22}O_{11}+H_2O\xrightarrow{\text{maltase}}2C_6H_{12}O_6\]
Final fermentation stage:
\[C_6H_{12}O_6\xrightarrow{\text{zymase}}2C_2H_5OH+2CO_2\]
(c)(iii) The fermented liquid is distilled using the apparatus shown below. Ethanol, with a boiling point of about 78°C, distils before most of the water.
The technique is fractional distillation.
Question 7 Report
Give one oxide in each case which;
(a) can act as a reducing agent
(b) can be used as a refrigerant
(c) is the anhydride of a strong acid;
(d) is yellow when hot and white then hot and white when cold;
(f) is usad as a pigment in paints.
Answer Details
None
Question 8 Report
(a) Name the industrial process by which ethene is obtained from petroleum fractions
(b) Give the I. U. P.A.C name of the isomer whose structure is shown below.
(c) Illustrate with an equation, one reaction in which benzene behaves as:
(i) unsaturated hydrocarbon
(ii) a saturated hydrocarbon
Answer Details
None
Question 9 Report
The set-up shown in the diagram below was used to separate a drop of universal indicator into the constituent dyes using ethyl ethanoate as the solvent.
(a) What name is given to the separate strated in the diagram?
(b) State: (i) how many components are resolved in the separation;
(ii) the material normally used in laborary as the adsorbent medium;
(iii) which of the labels the point of application of the indicator.
(a) Paper chromatography.
(b)
(i) Three components are resolved.
(ii) Chromatography paper (filter paper/cellulose) is the adsorbent medium normally used in the laboratory.
(iii) Label III marks the point of application of the universal indicator.
Answer Details
(a) Paper chromatography.
(b)
(i) Three components are resolved.
(ii) Chromatography paper (filter paper/cellulose) is the adsorbent medium normally used in the laboratory.
(iii) Label III marks the point of application of the universal indicator.
Question 10 Report
(a) State two factors which can affect the rate of a chemical reaction.
(b) 0.72g of magnesium was added to different volumes of 2 mol. per dm\(^{3}\) hydrochloric acid. The volume of liberated was as measured at room temperature and pressure. The result of the experiment was as tabulated
| vol. of 2 mol. per dm\(^{3}\) HCl used (cm\(^{3}\) | Vol. of H\(_2\) evolved in cm\(^{3}\) (to the nearest 10cm\(^{3}\)) |
| 5 | 120 |
| 15 | 360 |
| 25 | 550 |
| 35 | 600 |
| 45 | 600 |
Use the data in the table to plot a graph of the volume of hydrogen liberated against the volume of acid used.
(c) From the graph in (b) above, determine the volume of: (i) hydrogen that would be produced if 50 cm\(^{3}\) of the acid were added to 0.72g of magnesium.
(ii) the acid which must be added to 0.72 g of magnesium to produce 480 cm\(^{3}\) of hydrogen;
(iii) the acid needed exactly to dissolve 0.72 g of magnesium completely.
(d) Explain your answer to (c)(iii).
(e) From your answers to (c) above, deduce the: (i) volume of the acid which will dissolve 1 mole of magnesium completely. (Mg = 24)
(ii) volume of hydrogen that would be liberated if 1 mole of magnesium dissolves completely in the acid;
(iii) equation for the reaction between magnesium and hydrochloric acid. Show clearly how you arrived at you answers
(a) Two factors which affect the rate of a chemical reaction are:
- Concentration of the reactants.
- Temperature of the reacting mixture.
(b) The graph is plotted with volume of 2 mol dm−3 HCl on the horizontal axis and volume of hydrogen evolved on the vertical axis.
| Volume of HCl used / cm3 | 5 | 15 | 25 | 35 | 45 |
|---|---|---|---|---|---|
| Volume of H2 evolved / cm3 | 120 | 360 | 550 | 600 | 600 |
(c) Reading from the graph:
- For 50 cm3 of acid, the volume of hydrogen produced is 600 cm3.
- The volume of acid required to produce 480 cm3 of hydrogen is approximately 21 cm3.
- The volume of acid needed to dissolve the 0.72 g of magnesium completely is 35 cm3.
(d) At 35 cm3 of acid, the graph becomes horizontal at 600 cm3 of hydrogen. This shows that all the magnesium has been used up. Addition of more acid cannot produce more hydrogen because magnesium is now the limiting reactant.
(e)
Mass of magnesium used = 0.72 g.
\[n(\mathrm{Mg})=\frac{0.72}{24}=0.030\ \text{mol}\]
0.030 mol of Mg is completely dissolved by 35 cm3 of the acid.
\[\text{Volume for 1 mol Mg}=\frac{35}{0.030}=1166.7\ \text{cm}^3\]
Therefore, 1.17 × 103 cm3 of the acid will dissolve 1 mole of magnesium completely.
0.030 mol of Mg produces 600 cm3 of hydrogen.
\[\text{Volume of }\mathrm{H_2}\text{ for 1 mol Mg}=\frac{600}{0.030}=2.00\times10^4\ \text{cm}^3\]
Therefore, 2.00 × 104 cm3 of hydrogen is liberated when 1 mole of magnesium dissolves completely.
The balanced equation is:
\[\mathrm{Mg_{(s)}+2HCl_{(aq)}\rightarrow MgCl_{2(aq)}+H_{2(g)}}\]
Answer Details
(a) Two factors which affect the rate of a chemical reaction are:
- Concentration of the reactants.
- Temperature of the reacting mixture.
(b) The graph is plotted with volume of 2 mol dm−3 HCl on the horizontal axis and volume of hydrogen evolved on the vertical axis.
| Volume of HCl used / cm3 | 5 | 15 | 25 | 35 | 45 |
|---|---|---|---|---|---|
| Volume of H2 evolved / cm3 | 120 | 360 | 550 | 600 | 600 |
(c) Reading from the graph:
- For 50 cm3 of acid, the volume of hydrogen produced is 600 cm3.
- The volume of acid required to produce 480 cm3 of hydrogen is approximately 21 cm3.
- The volume of acid needed to dissolve the 0.72 g of magnesium completely is 35 cm3.
(d) At 35 cm3 of acid, the graph becomes horizontal at 600 cm3 of hydrogen. This shows that all the magnesium has been used up. Addition of more acid cannot produce more hydrogen because magnesium is now the limiting reactant.
(e)
Mass of magnesium used = 0.72 g.
\[n(\mathrm{Mg})=\frac{0.72}{24}=0.030\ \text{mol}\]
0.030 mol of Mg is completely dissolved by 35 cm3 of the acid.
\[\text{Volume for 1 mol Mg}=\frac{35}{0.030}=1166.7\ \text{cm}^3\]
Therefore, 1.17 × 103 cm3 of the acid will dissolve 1 mole of magnesium completely.
0.030 mol of Mg produces 600 cm3 of hydrogen.
\[\text{Volume of }\mathrm{H_2}\text{ for 1 mol Mg}=\frac{600}{0.030}=2.00\times10^4\ \text{cm}^3\]
Therefore, 2.00 × 104 cm3 of hydrogen is liberated when 1 mole of magnesium dissolves completely.
The balanced equation is:
\[\mathrm{Mg_{(s)}+2HCl_{(aq)}\rightarrow MgCl_{2(aq)}+H_{2(g)}}\]
Question 11 Report
(a) Give two reasons why fused alumina is mixed with cryolite in the electrolytic extraction of aluminium.
(b) Write an equation for the reaction of sodium hydroxide with bauxite.
(a) Why fused alumina is mixed with cryolite
- Cryolite dissolves the alumina and lowers its melting point from about 2050 °C to about 950 °C, greatly reducing the energy and cost of the extraction.
- The molten cryolite increases the electrical conductivity of the melt, allowing electrolysis to proceed efficiently.
(b) Reaction of sodium hydroxide with bauxite
The amphoteric alumina in bauxite dissolves in hot concentrated sodium hydroxide to form soluble sodium aluminate: \[Al_2O_{3(s)} + 2NaOH_{(aq)} \rightarrow 2NaAlO_{2(aq)} + H_2O_{(l)}\]
Answer Details
(a) Why fused alumina is mixed with cryolite
- Cryolite dissolves the alumina and lowers its melting point from about 2050 °C to about 950 °C, greatly reducing the energy and cost of the extraction.
- The molten cryolite increases the electrical conductivity of the melt, allowing electrolysis to proceed efficiently.
(b) Reaction of sodium hydroxide with bauxite
The amphoteric alumina in bauxite dissolves in hot concentrated sodium hydroxide to form soluble sodium aluminate: \[Al_2O_{3(s)} + 2NaOH_{(aq)} \rightarrow 2NaAlO_{2(aq)} + H_2O_{(l)}\]
Question 12 Report
(a) Give three differences between electrovalent compounds and covalent compounds
(b) List two physical properties of metals which can be accounted for by their structure
(c) Thorium (Th) metal undergoes a reaction represented by the following equation:
\(^{234}_{90}Th \to X + ^{234}_{91}Pa\)
(i) State the type of process involved in the reaction
(ii) Balance the equation equation and hence identify X.
(iii) Name one equipment which can be used to detect X.
(iv) Sketch a curveto show the mass of given quantity of thorium will change over a long period of time.
(d) Y is a moderately reactive divalent found naturally in the combined state as the trioxocarbonate (IV) salt, YCO\(_3\) is decomposed by strong heat, state the steps you would use in extracting Y from the ore. Write equation to show the chemical processes involved.
(a) Differences between electrovalent and covalent compounds
| Electrovalent (ionic) compounds | Covalent compounds |
|---|---|
| They consist of oppositely charged ions held by strong electrostatic forces. | They consist of molecules formed by sharing electrons. |
| They conduct electricity when molten or in aqueous solution because their ions are mobile. | They generally do not conduct electricity because they have no mobile ions or electrons. |
| They are generally soluble in water and have high melting and boiling points. | They are generally insoluble in water, soluble in organic solvents, and have relatively low melting and boiling points. |
(b)
- Metals are good conductors of electricity because they contain mobile, delocalised electrons.
- Metals are malleable and ductile because layers of positive metal ions can slide over one another while metallic bonding is maintained.
(c)
(i) The process is radioactive beta decay, involving emission of a beta particle.
(ii) The mass number is unchanged, while the atomic number increases from 90 to 91. Therefore,
\[{}^{234}_{90}\mathrm{Th}\ \longrightarrow\ {}^{0}_{-1}\mathrm{e}+{}^{234}_{91}\mathrm{Pa}\]
Thus, \(X\) is a beta particle, \({}^{0}_{-1}\mathrm{e}\).
(iii) A Geiger-Müller tube or Geiger-Müller counter can be used to detect the beta radiation.
(iv) The mass of thorium decreases exponentially with time, becoming half of its previous value in each successive half-life.
(d)
- Crush and concentrate the carbonate ore, \(\mathrm{YCO_3}\).
- Heat the concentrated ore strongly (calcination) to decompose the carbonate to the oxide:
\[\mathrm{YCO_3(s)\ \xrightarrow{\Delta}\ YO(s)+CO_2(g)}\] - Reduce the oxide with carbon in a furnace to obtain the metal:
\[\mathrm{YO(s)+C(s)\ \xrightarrow{\Delta}\ Y(s)+CO(g)}\] - Tap off the metal and purify it where necessary.
Answer Details
(a) Differences between electrovalent and covalent compounds
| Electrovalent (ionic) compounds | Covalent compounds |
|---|---|
| They consist of oppositely charged ions held by strong electrostatic forces. | They consist of molecules formed by sharing electrons. |
| They conduct electricity when molten or in aqueous solution because their ions are mobile. | They generally do not conduct electricity because they have no mobile ions or electrons. |
| They are generally soluble in water and have high melting and boiling points. | They are generally insoluble in water, soluble in organic solvents, and have relatively low melting and boiling points. |
(b)
- Metals are good conductors of electricity because they contain mobile, delocalised electrons.
- Metals are malleable and ductile because layers of positive metal ions can slide over one another while metallic bonding is maintained.
(c)
(i) The process is radioactive beta decay, involving emission of a beta particle.
(ii) The mass number is unchanged, while the atomic number increases from 90 to 91. Therefore,
\[{}^{234}_{90}\mathrm{Th}\ \longrightarrow\ {}^{0}_{-1}\mathrm{e}+{}^{234}_{91}\mathrm{Pa}\]
Thus, \(X\) is a beta particle, \({}^{0}_{-1}\mathrm{e}\).
(iii) A Geiger-Müller tube or Geiger-Müller counter can be used to detect the beta radiation.
(iv) The mass of thorium decreases exponentially with time, becoming half of its previous value in each successive half-life.
(d)
- Crush and concentrate the carbonate ore, \(\mathrm{YCO_3}\).
- Heat the concentrated ore strongly (calcination) to decompose the carbonate to the oxide:
\[\mathrm{YCO_3(s)\ \xrightarrow{\Delta}\ YO(s)+CO_2(g)}\] - Reduce the oxide with carbon in a furnace to obtain the metal:
\[\mathrm{YO(s)+C(s)\ \xrightarrow{\Delta}\ Y(s)+CO(g)}\] - Tap off the metal and purify it where necessary.
Question 13 Report
State the atoms represented as shown below:
(a) State the relationship between the two atoms.
(b) What is the difference between them?
(c) Give two examples of elements which exhibit the phenomenon illustrated above.
The diagram shows the same chemical symbol X written twice, each time with a subscript on the lower left and a superscript on the upper left:
- Left atom: \( {}^{q}_{p}\text{X} \)
- Right atom: \( {}^{r}_{p}\text{X} \)
In the standard nuclide notation \( {}^{A}_{Z}\text{X} \), the lower figure is the atomic number \(Z\) (number of protons) and the upper figure is the mass number \(A\) (protons plus neutrons). Reading the two atoms off the diagram:
| Atom | Atomic number (lower) | Mass number (upper) |
|---|---|---|
| Left | \(p\) | \(q\) |
| Right | \(p\) | \(r\) |
Both atoms carry the same lower figure \(p\), so both have the same number of protons and are therefore atoms of the same element X. They differ only in the upper figure (\(q\) against \(r\)).
(a) Relationship between the two atoms
They are isotopes of the same element. Isotopes are atoms of the same element that have the same atomic number but different mass numbers.
(b) The difference between them
They have different mass numbers (\(q\) for the left atom and \(r\) for the right atom). Since the mass number is protons plus neutrons and both have the same number of protons \(p\), the difference lies wholly in the number of neutrons.
Number of neutrons is given by \( N = A - Z \):
\[ N_{\text{left}} = q - p \qquad N_{\text{right}} = r - p \]Their number of protons (\(p\)) and, in the neutral state, their number of electrons are the same, so they have identical chemical properties and differ only in relative atomic mass and physical properties that depend on mass.
(c) Two examples of elements that exhibit this phenomenon (isotopy)
- Chlorine: \( {}^{35}_{17}\text{Cl} \) and \( {}^{37}_{17}\text{Cl} \)
- Carbon: \( {}^{12}_{6}\text{C} \) and \( {}^{14}_{6}\text{C} \)
(Hydrogen, \( {}^{1}_{1}\text{H} \), \( {}^{2}_{1}\text{H} \) and \( {}^{3}_{1}\text{H} \), and oxygen, \( {}^{16}_{8}\text{O} \) and \( {}^{18}_{8}\text{O} \), are also acceptable examples.)
Answer Details
The diagram shows the same chemical symbol X written twice, each time with a subscript on the lower left and a superscript on the upper left:
- Left atom: \( {}^{q}_{p}\text{X} \)
- Right atom: \( {}^{r}_{p}\text{X} \)
In the standard nuclide notation \( {}^{A}_{Z}\text{X} \), the lower figure is the atomic number \(Z\) (number of protons) and the upper figure is the mass number \(A\) (protons plus neutrons). Reading the two atoms off the diagram:
| Atom | Atomic number (lower) | Mass number (upper) |
|---|---|---|
| Left | \(p\) | \(q\) |
| Right | \(p\) | \(r\) |
Both atoms carry the same lower figure \(p\), so both have the same number of protons and are therefore atoms of the same element X. They differ only in the upper figure (\(q\) against \(r\)).
(a) Relationship between the two atoms
They are isotopes of the same element. Isotopes are atoms of the same element that have the same atomic number but different mass numbers.
(b) The difference between them
They have different mass numbers (\(q\) for the left atom and \(r\) for the right atom). Since the mass number is protons plus neutrons and both have the same number of protons \(p\), the difference lies wholly in the number of neutrons.
Number of neutrons is given by \( N = A - Z \):
\[ N_{\text{left}} = q - p \qquad N_{\text{right}} = r - p \]Their number of protons (\(p\)) and, in the neutral state, their number of electrons are the same, so they have identical chemical properties and differ only in relative atomic mass and physical properties that depend on mass.
(c) Two examples of elements that exhibit this phenomenon (isotopy)
- Chlorine: \( {}^{35}_{17}\text{Cl} \) and \( {}^{37}_{17}\text{Cl} \)
- Carbon: \( {}^{12}_{6}\text{C} \) and \( {}^{14}_{6}\text{C} \)
(Hydrogen, \( {}^{1}_{1}\text{H} \), \( {}^{2}_{1}\text{H} \) and \( {}^{3}_{1}\text{H} \), and oxygen, \( {}^{16}_{8}\text{O} \) and \( {}^{18}_{8}\text{O} \), are also acceptable examples.)
Question 14 Report
Methane is obtained when a powdered mixture of anhydrous sodium ethanoate and soda-lime is heated in a hard glass test tube.
(a) Write an equation for the reaction.
b) Explain briefly why soda-lime is preferred to sodium hydroxide for the preparation.
(a) Equation
The soda-lime supplies sodium hydroxide, which reacts with the anhydrous sodium ethanoate to release methane: \[CH_3COONa_{(s)} + NaOH_{(s)} \rightarrow CH_{4(g)} + Na_2CO_{3(s)}\]
(b) Why soda-lime is preferred to sodium hydroxide
Soda-lime (a mixture of sodium hydroxide and calcium oxide) is preferred because solid sodium hydroxide alone is strongly deliquescent and corrosive: when heated it melts, absorbs moisture and attacks (corrodes) the glass test tube, which may crack. Soda-lime is a dry, granular, easier-to-handle solid that does not fuse or attack the glass in the same way, so it gives a safer and more convenient preparation.
Answer Details
(a) Equation
The soda-lime supplies sodium hydroxide, which reacts with the anhydrous sodium ethanoate to release methane: \[CH_3COONa_{(s)} + NaOH_{(s)} \rightarrow CH_{4(g)} + Na_2CO_{3(s)}\]
(b) Why soda-lime is preferred to sodium hydroxide
Soda-lime (a mixture of sodium hydroxide and calcium oxide) is preferred because solid sodium hydroxide alone is strongly deliquescent and corrosive: when heated it melts, absorbs moisture and attacks (corrodes) the glass test tube, which may crack. Soda-lime is a dry, granular, easier-to-handle solid that does not fuse or attack the glass in the same way, so it gives a safer and more convenient preparation.
