WAEC SSCE - Chemistry - 1991 (Essay)
Question 1 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 2 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 3 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
(a) Industrial process by which ethene is obtained from petroleum fractions: cracking (thermal or catalytic cracking).
(b) IUPAC name of the isomer shown
The structure is ClCH=CHCl, with the two chlorine atoms on opposite sides of the C=C double bond, so it is the trans isomer: (E)-1,2-dichloroethene (trans-1,2-dichloroethene).
(c) Reactions of benzene
(i) As an unsaturated hydrocarbon (addition reaction): benzene adds hydrogen in the presence of a nickel catalyst to give cyclohexane. \[\text{C}_6\text{H}_6 + 3\text{H}_2 \xrightarrow{\text{Ni},\ 180^\circ\text{C}} \text{C}_6\text{H}_{12}\]
(ii) As a saturated hydrocarbon (substitution reaction): benzene reacts with chlorine in the presence of anhydrous AlCl3 (or FeCl3) as catalyst, in the dark, to give chlorobenzene. \[\text{C}_6\text{H}_6 + \text{Cl}_2 \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{Cl} + \text{HCl}\]
Answer Details
(a) Industrial process by which ethene is obtained from petroleum fractions: cracking (thermal or catalytic cracking).
(b) IUPAC name of the isomer shown
The structure is ClCH=CHCl, with the two chlorine atoms on opposite sides of the C=C double bond, so it is the trans isomer: (E)-1,2-dichloroethene (trans-1,2-dichloroethene).
(c) Reactions of benzene
(i) As an unsaturated hydrocarbon (addition reaction): benzene adds hydrogen in the presence of a nickel catalyst to give cyclohexane. \[\text{C}_6\text{H}_6 + 3\text{H}_2 \xrightarrow{\text{Ni},\ 180^\circ\text{C}} \text{C}_6\text{H}_{12}\]
(ii) As a saturated hydrocarbon (substitution reaction): benzene reacts with chlorine in the presence of anhydrous AlCl3 (or FeCl3) as catalyst, in the dark, to give chlorobenzene. \[\text{C}_6\text{H}_6 + \text{Cl}_2 \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{Cl} + \text{HCl}\]
Question 4 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.
Question 5 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.
(a) Three characteristics of a catalyst
- It remains chemically unchanged in mass and composition at the end of the reaction.
- Only a small quantity is needed to catalyse a large amount of reactants.
- It is specific in action (a given catalyst speeds up a particular reaction), and it alters only the rate of the reaction, not the position of equilibrium or the products formed.
(b) A manufacturing process using each metal as catalyst
- (i) Iron: the Haber process for the manufacture of ammonia.
- (ii) Nickel: the hydrogenation of vegetable oils to make margarine.
- (iii) Platinum: the Ostwald process for the manufacture of trioxonitrate(V) acid (oxidation of ammonia). (Platinum, as platinum-rhodium, is also used in the Contact process.)
(c) An organic catalyst
An enzyme, for example zymase (or diastase / amylase).
Answer Details
(a) Three characteristics of a catalyst
- It remains chemically unchanged in mass and composition at the end of the reaction.
- Only a small quantity is needed to catalyse a large amount of reactants.
- It is specific in action (a given catalyst speeds up a particular reaction), and it alters only the rate of the reaction, not the position of equilibrium or the products formed.
(b) A manufacturing process using each metal as catalyst
- (i) Iron: the Haber process for the manufacture of ammonia.
- (ii) Nickel: the hydrogenation of vegetable oils to make margarine.
- (iii) Platinum: the Ostwald process for the manufacture of trioxonitrate(V) acid (oxidation of ammonia). (Platinum, as platinum-rhodium, is also used in the Contact process.)
(c) An organic catalyst
An enzyme, for example zymase (or diastase / amylase).
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) Gaseous hydrocarbons
- (i) Used for welding: ethyne (acetylene), C2H2 (its oxy-acetylene flame is very hot).
- (ii) Major raw material for the plastic industry: ethene (ethylene), C2H4.
(b) Structural formula and process
Structural formula of ethyne: \[H-C \equiv C-H\] Ethyne is converted to neoprene rubber by polymerization (the monomer chloroprene, made from ethyne, polymerizes to neoprene).
(c) Potatoes
- (i) On complete dehydration of the carbohydrate (starch), the main product is carbon (with water driven off).
- (ii) Converting potatoes to ethanol:
First, the starch is hydrolysed. Boiling the mashed potato with dilute acid, or using the enzyme diastase, breaks starch down to maltose, and maltase converts maltose to glucose: \[(C_6H_{10}O_5)_n + nH_2O \rightarrow n C_6H_{12}O_6\] Then the glucose is fermented by the enzyme zymase in yeast at about 30 to 35 °C: \[C_6H_{12}O_6 \xrightarrow{zymase} 2C_2H_5OH + 2CO_2\] - (iii) A fairly pure sample of ethanol is obtained from the fermented mixture by fractional distillation (see diagram note). The technique is called fractional distillation.
Answer Details
(a) Gaseous hydrocarbons
- (i) Used for welding: ethyne (acetylene), C2H2 (its oxy-acetylene flame is very hot).
- (ii) Major raw material for the plastic industry: ethene (ethylene), C2H4.
(b) Structural formula and process
Structural formula of ethyne: \[H-C \equiv C-H\] Ethyne is converted to neoprene rubber by polymerization (the monomer chloroprene, made from ethyne, polymerizes to neoprene).
(c) Potatoes
- (i) On complete dehydration of the carbohydrate (starch), the main product is carbon (with water driven off).
- (ii) Converting potatoes to ethanol:
First, the starch is hydrolysed. Boiling the mashed potato with dilute acid, or using the enzyme diastase, breaks starch down to maltose, and maltase converts maltose to glucose: \[(C_6H_{10}O_5)_n + nH_2O \rightarrow n C_6H_{12}O_6\] Then the glucose is fermented by the enzyme zymase in yeast at about 30 to 35 °C: \[C_6H_{12}O_6 \xrightarrow{zymase} 2C_2H_5OH + 2CO_2\] - (iii) A fairly pure sample of ethanol is obtained from the fermented mixture by fractional distillation (see diagram note). The technique is called 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.
One oxide in each case:
- (a) Can act as a reducing agent: carbon(II) oxide, CO.
- (b) Can be used as a refrigerant: sulphur(IV) oxide, SO2.
- (c) Is the anhydride of a strong acid: sulphur(VI) oxide, SO3 (the anhydride of tetraoxosulphate(VI) acid). \[SO_3 + H_2O \rightarrow H_2SO_4\]
- (d) Is yellow when hot and white when cold: zinc oxide, ZnO.
- (f) Is used as a pigment in paints: titanium(IV) oxide, TiO2 (a white pigment).
Answer Details
One oxide in each case:
- (a) Can act as a reducing agent: carbon(II) oxide, CO.
- (b) Can be used as a refrigerant: sulphur(IV) oxide, SO2.
- (c) Is the anhydride of a strong acid: sulphur(VI) oxide, SO3 (the anhydride of tetraoxosulphate(VI) acid). \[SO_3 + H_2O \rightarrow H_2SO_4\]
- (d) Is yellow when hot and white when cold: zinc oxide, ZnO.
- (f) Is used as a pigment in paints: titanium(IV) oxide, TiO2 (a white pigment).
Question 8 Report
Sketch a curve to show how the solubility of a gas varies with increasing temperature
Solubility of a gas against temperature
For a gas dissolved in water, solubility decreases as temperature increases. Heating gives gas particles more kinetic energy, so they escape from the liquid more easily.
Examination point: label the horizontal axis as temperature and the vertical axis as solubility, then draw a smooth downward-sloping curve.
Answer Details
Solubility of a gas against temperature
For a gas dissolved in water, solubility decreases as temperature increases. Heating gives gas particles more kinetic energy, so they escape from the liquid more easily.
Examination point: label the horizontal axis as temperature and the vertical axis as solubility, then draw a smooth downward-sloping curve.
Question 9 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 10 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 11 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 12 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 that affect the rate of a chemical reaction: temperature and concentration of the reactants. (Surface area, a catalyst and, for gases, pressure are also acceptable.)
(b) The data to be plotted are:
| Vol. of 2 mol dm-3 HCl used (cm3) | 5 | 15 | 25 | 35 | 45 |
|---|---|---|---|---|---|
| Vol. of H2 evolved (cm3) | 120 | 360 | 550 | 600 | 600 |
Plot volume of H2 (y-axis) against volume of acid (x-axis). The graph rises steeply and in direct proportion at first (about 24 cm3 of H2 per 1 cm3 of acid), then bends over and becomes a horizontal line (a plateau) at 600 cm3 once all the magnesium has reacted.
(c) From the graph:
- (i) At 50 cm3 of acid the curve is already on the plateau, so the volume of hydrogen is 600 cm3 (adding more acid gives no more gas because the magnesium is used up).
- (ii) To produce 480 cm3 of hydrogen, read on the rising (proportional) part of the graph. Since \(120\ \text{cm}^3\) H2 corresponds to \(5\ \text{cm}^3\) acid, the ratio is \(24\ \text{cm}^3\) H2 per cm3 acid: \[V_{acid} = \frac{480}{24} = \mathbf{20\ cm^3}\]
- (iii) The acid needed to dissolve the magnesium exactly is the volume at which the graph just reaches the plateau (600 cm3), which is read as 30 cm3.
(d) At 30 cm3 of acid, the acid is exactly enough to react with all 0.72 g of magnesium; this is the point where the maximum volume of hydrogen (600 cm3) is first produced. Below this volume the acid is the limiting reactant (less hydrogen). Above it, the magnesium is completely used up and is now the limiting reactant, so no extra hydrogen is formed and the graph stays flat.
(e) First find the moles of magnesium used:
\[n_{Mg} = \frac{0.72}{24} = 0.03\ \text{mol}\](i) From (c)(iii), 0.03 mol Mg needs 30 cm3 of the acid. For 1 mole:
\[V = \frac{30}{0.03} = 1000\ \text{cm}^3 = \mathbf{1\ dm^3}\](ii) From the plateau, 0.03 mol Mg liberates 600 cm3 of hydrogen. For 1 mole:
\[V_{H_2} = \frac{600}{0.03} = 20000\ \text{cm}^3 = \mathbf{20\ dm^3}\](iii) Deduce the mole ratios. The acid is 2 mol dm-3, so 1 dm3 contains 2 mol HCl. Therefore:
- 1 mol Mg reacts with 2 mol HCl (from (e)(i)).
- 1 mol Mg produces about 1 mol H2 (20 dm3 is close to the molar gas volume, from (e)(ii)).
This gives the ratio Mg : HCl : H2 = 1 : 2 : 1, so the equation is:
\[\text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g)\]Answer Details
(a) Two factors that affect the rate of a chemical reaction: temperature and concentration of the reactants. (Surface area, a catalyst and, for gases, pressure are also acceptable.)
(b) The data to be plotted are:
| Vol. of 2 mol dm-3 HCl used (cm3) | 5 | 15 | 25 | 35 | 45 |
|---|---|---|---|---|---|
| Vol. of H2 evolved (cm3) | 120 | 360 | 550 | 600 | 600 |
Plot volume of H2 (y-axis) against volume of acid (x-axis). The graph rises steeply and in direct proportion at first (about 24 cm3 of H2 per 1 cm3 of acid), then bends over and becomes a horizontal line (a plateau) at 600 cm3 once all the magnesium has reacted.
(c) From the graph:
- (i) At 50 cm3 of acid the curve is already on the plateau, so the volume of hydrogen is 600 cm3 (adding more acid gives no more gas because the magnesium is used up).
- (ii) To produce 480 cm3 of hydrogen, read on the rising (proportional) part of the graph. Since \(120\ \text{cm}^3\) H2 corresponds to \(5\ \text{cm}^3\) acid, the ratio is \(24\ \text{cm}^3\) H2 per cm3 acid: \[V_{acid} = \frac{480}{24} = \mathbf{20\ cm^3}\]
- (iii) The acid needed to dissolve the magnesium exactly is the volume at which the graph just reaches the plateau (600 cm3), which is read as 30 cm3.
(d) At 30 cm3 of acid, the acid is exactly enough to react with all 0.72 g of magnesium; this is the point where the maximum volume of hydrogen (600 cm3) is first produced. Below this volume the acid is the limiting reactant (less hydrogen). Above it, the magnesium is completely used up and is now the limiting reactant, so no extra hydrogen is formed and the graph stays flat.
(e) First find the moles of magnesium used:
\[n_{Mg} = \frac{0.72}{24} = 0.03\ \text{mol}\](i) From (c)(iii), 0.03 mol Mg needs 30 cm3 of the acid. For 1 mole:
\[V = \frac{30}{0.03} = 1000\ \text{cm}^3 = \mathbf{1\ dm^3}\](ii) From the plateau, 0.03 mol Mg liberates 600 cm3 of hydrogen. For 1 mole:
\[V_{H_2} = \frac{600}{0.03} = 20000\ \text{cm}^3 = \mathbf{20\ dm^3}\](iii) Deduce the mole ratios. The acid is 2 mol dm-3, so 1 dm3 contains 2 mol HCl. Therefore:
- 1 mol Mg reacts with 2 mol HCl (from (e)(i)).
- 1 mol Mg produces about 1 mol H2 (20 dm3 is close to the molar gas volume, from (e)(ii)).
This gives the ratio Mg : HCl : H2 = 1 : 2 : 1, so the equation is:
\[\text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g)\]Question 13 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 14 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.
