WAEC SSCE - Chemistry - 2001 (Essay)
Tambaya 1 Rahoto
(a)(i) State three characteristics of a homologous series.
(ii) Give the name and structural formula of the second member of the alkyne series.
(iii) Write an equation to represent the combustion of ethane in excess oxygen.
(b) Name the type of reaction involved in the conversion of ethanol to
(i) ethene;
(ii) ethylethanoate;
(iii) chloroethane;
(iv) ethoxide
(v) ethanoic acid
(c) Consider the following compound.
(i) Write its IUPAC name.
(ii) Give its molecular formula and empirical formula
(iii) List the products of its H H 0 reaction with saturated Na\(_2\)CO\(_3\) solution.
(iv) State with reason whether its boiling point will be higher or lower than that of the corresponding alkane.
(d) A vegetable oil X was treated with activated charcoal and then with a gas Y in the presence of a catalyst in order to manufacture
(i) Identify Y.
(ii) State the function of the activated charcoal.
(iii) What is the catalyst used?
(iv) If a sample of X is heated with concentrated sodium hydroxide solution, list the products that will be obtained.
(a)(i) Three characteristics of a homologous series.
- All members are represented by the same general formula (e.g. alkanes \(C_nH_{2n+2}\)).
- Successive members differ from the next by a \(-CH_2-\) unit (a relative molecular mass difference of 14).
- Members have similar chemical properties because they contain the same functional group.
- There is a gradual change in physical properties (e.g. boiling point rises steadily) as molar mass increases.
(Any three of the above.)
(a)(ii) Second member of the alkyne series.
Name: propyne. Structural formula: \(CH_3-C{\equiv}CH\).
(a)(iii) Combustion of ethane in excess oxygen.
\[2C_2H_{6(g)} + 7O_{2(g)} \rightarrow 4CO_{2(g)} + 6H_2O_{(l)}\]
(b) Type of reaction in the conversion of ethanol to:
- ethene: dehydration (an elimination reaction);
- ethyl ethanoate: esterification (a condensation reaction);
- chloroethane: substitution (halogenation, e.g. with \(PCl_5\) or conc. \(HCl\));
- ethoxide: redox / displacement (sodium displaces hydrogen: \(2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2\));
- ethanoic acid: oxidation.
(c) The compound shown is \(CH_3-CH_2-COOH\) (an ethyl group joined to a carboxyl, \(-COOH\), group).
(i) IUPAC name: propanoic acid.
(ii) Molecular and empirical formula: Molecular formula \(= C_3H_6O_2\). The ratio \(C{:}H{:}O = 3{:}6{:}2\) has no common factor greater than 1, so the empirical formula is also \(C_3H_6O_2\).
(iii) Products of reaction with saturated \(Na_2CO_3\) solution: a carboxylic acid reacts with the carbonate to give a salt, water and carbon dioxide:
\[2CH_3CH_2COOH + Na_2CO_3 \rightarrow 2CH_3CH_2COONa + H_2O + CO_2\]
Products: sodium propanoate, water and carbon dioxide (the effervescence of \(CO_2\) confirms the \(-COOH\) group).
(iv) Boiling point compared with the corresponding alkane: The corresponding alkane (same carbon skeleton) is butane. Propanoic acid has a higher boiling point. This is because carboxylic acid molecules form strong intermolecular hydrogen bonds (they even pair up as dimers), whereas butane molecules are held only by weak van der Waals (dispersion) forces; more energy is needed to separate the acid molecules.
(d) Manufacture from vegetable oil X (production of margarine):
- Identity of Y: hydrogen (\(H_2\)); the process is hydrogenation of the unsaturated oil.
- Function of the activated charcoal: it acts as an adsorbent that purifies/decolourises (bleaches) the oil by adsorbing coloured impurities and odours.
- Catalyst used: finely divided nickel (Ni).
- Heating X with concentrated \(NaOH\) (saponification): the oil (an ester/triglyceride) is hydrolysed to give soap (the sodium salts of the fatty acids) and glycerol (propane-1,2,3-triol).
Bayanin Amsa
(a)(i) Three characteristics of a homologous series.
- All members are represented by the same general formula (e.g. alkanes \(C_nH_{2n+2}\)).
- Successive members differ from the next by a \(-CH_2-\) unit (a relative molecular mass difference of 14).
- Members have similar chemical properties because they contain the same functional group.
- There is a gradual change in physical properties (e.g. boiling point rises steadily) as molar mass increases.
(Any three of the above.)
(a)(ii) Second member of the alkyne series.
Name: propyne. Structural formula: \(CH_3-C{\equiv}CH\).
(a)(iii) Combustion of ethane in excess oxygen.
\[2C_2H_{6(g)} + 7O_{2(g)} \rightarrow 4CO_{2(g)} + 6H_2O_{(l)}\]
(b) Type of reaction in the conversion of ethanol to:
- ethene: dehydration (an elimination reaction);
- ethyl ethanoate: esterification (a condensation reaction);
- chloroethane: substitution (halogenation, e.g. with \(PCl_5\) or conc. \(HCl\));
- ethoxide: redox / displacement (sodium displaces hydrogen: \(2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2\));
- ethanoic acid: oxidation.
(c) The compound shown is \(CH_3-CH_2-COOH\) (an ethyl group joined to a carboxyl, \(-COOH\), group).
(i) IUPAC name: propanoic acid.
(ii) Molecular and empirical formula: Molecular formula \(= C_3H_6O_2\). The ratio \(C{:}H{:}O = 3{:}6{:}2\) has no common factor greater than 1, so the empirical formula is also \(C_3H_6O_2\).
(iii) Products of reaction with saturated \(Na_2CO_3\) solution: a carboxylic acid reacts with the carbonate to give a salt, water and carbon dioxide:
\[2CH_3CH_2COOH + Na_2CO_3 \rightarrow 2CH_3CH_2COONa + H_2O + CO_2\]
Products: sodium propanoate, water and carbon dioxide (the effervescence of \(CO_2\) confirms the \(-COOH\) group).
(iv) Boiling point compared with the corresponding alkane: The corresponding alkane (same carbon skeleton) is butane. Propanoic acid has a higher boiling point. This is because carboxylic acid molecules form strong intermolecular hydrogen bonds (they even pair up as dimers), whereas butane molecules are held only by weak van der Waals (dispersion) forces; more energy is needed to separate the acid molecules.
(d) Manufacture from vegetable oil X (production of margarine):
- Identity of Y: hydrogen (\(H_2\)); the process is hydrogenation of the unsaturated oil.
- Function of the activated charcoal: it acts as an adsorbent that purifies/decolourises (bleaches) the oil by adsorbing coloured impurities and odours.
- Catalyst used: finely divided nickel (Ni).
- Heating X with concentrated \(NaOH\) (saponification): the oil (an ester/triglyceride) is hydrolysed to give soap (the sodium salts of the fatty acids) and glycerol (propane-1,2,3-triol).
Tambaya 2 Rahoto
(a)(i) Draw and label a simple cell for the electrolytic purification of copper.
(ii) Write can equation for the reaction at each electrode in (a)(i) above.
(iii) State with reason whether the Daniell cell is an electrolytic cell or an electrochemical cell.
(iv) What is the function of MnO\(_2\) in the Laclanche cell?
(b) Consider the following equation: MnO\(^-_4\) + 8H\(^+\) + xe\(^-\) \(\to\) Mn\(^{2+}\) + yH\(_2\)O. State the
(i) values of x and y;
(ii) oxidation state of Mn in MnO\(^-_4\).
(c)(i) List three factors that affect selective discharge of ions during electrolysis
(ii) State Faraday's second law of electrolysis.
(iii) A voltameter containing silver trioxonitrate(V) solution was connected in series to another voltameter containing copper (II) tetraoxosulphate(VI) solution. When a current ri 0.200 ampere was passed through the solutions, 0.780g of silver was deposited. Calculate the
I. mass of copper that would be deposited in the copper voltameter
II. quantity of electricity used and the time of current flow. [Cu = 63.5 ; Ag = 108; 1F = 96500C]
(a)(i) Electrolytic purification of copper
Impure copper is made the anode and pure copper is the cathode. Both are immersed in acidified copper(II) sulfate solution and connected to a direct-current supply. Pure copper is deposited on the cathode, while insoluble impurities settle as anode mud.
(ii) Electrode reactions
Anode (oxidation): \[\mathrm{Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^-}\]
Cathode (reduction): \[\mathrm{Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}}\]
(iii) The Daniell cell is an electrochemical (galvanic) cell because it converts chemical energy from a spontaneous redox reaction into electrical energy.
(iv) In the Leclanché cell, \(\mathrm{MnO_2}\) acts as a depolariser. It removes hydrogen formed at the carbon electrode, thereby preventing polarisation and a fall in voltage.
(b)(i)
Balancing oxygen atoms gives \(y=4\), and balancing charge gives \(x=5\):
\[\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}\]
Therefore, \(\boxed{x=5}\) and \(\boxed{y=4}\).
(ii) Let the oxidation state of Mn be \(z\):
\[z+4(-2)=-1\]
\[z=+7\]
Hence, the oxidation state of Mn in \(\mathrm{MnO_4^-}\) is \(\boxed{+7}\).
(c)(i) Factors affecting selective discharge of ions
- Position of the ions in the electrochemical series.
- Concentration of the ions in the electrolyte.
- Nature of the electrode.
(ii) Faraday's second law of electrolysis
When the same quantity of electricity passes through different electrolytes, the masses of substances liberated or deposited are proportional to their chemical equivalent masses.
(iii) Calculation
Since the voltameters are connected in series, the same quantity of electricity passes through both solutions.
For silver:
\[\mathrm{Ag^+ + e^- \rightarrow Ag}\]
\[\text{Moles of Ag deposited}=\frac{0.780}{108}=7.222\times10^{-3}\ \text{mol}\]
One mole of Ag requires one mole of electrons. Therefore:
\[\text{Moles of electrons}=7.222\times10^{-3}\ \text{mol}\]
I. Mass of copper deposited
\[\mathrm{Cu^{2+}+2e^-\rightarrow Cu}\]
\[\text{Moles of Cu}=\frac{7.222\times10^{-3}}{2}=3.611\times10^{-3}\ \text{mol}\]
\[\text{Mass of Cu}=3.611\times10^{-3}\times63.5=0.229\ \text{g}\]
\[\boxed{\text{Mass of copper deposited}=0.229\ \text{g}}\]
II. Quantity of electricity and time of flow
\[Q=nF=(7.222\times10^{-3})(96500)=696.9\ \text{C}\]
\[\boxed{Q\approx697\ \text{C}}\]
Since \(Q=It\):
\[t=\frac{Q}{I}=\frac{696.9}{0.200}=3484.5\ \text{s}\]
\[\boxed{t\approx3.48\times10^3\ \text{s}=58.1\ \text{min}}\]
Bayanin Amsa
(a)(i) Electrolytic purification of copper
Impure copper is made the anode and pure copper is the cathode. Both are immersed in acidified copper(II) sulfate solution and connected to a direct-current supply. Pure copper is deposited on the cathode, while insoluble impurities settle as anode mud.
(ii) Electrode reactions
Anode (oxidation): \[\mathrm{Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^-}\]
Cathode (reduction): \[\mathrm{Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}}\]
(iii) The Daniell cell is an electrochemical (galvanic) cell because it converts chemical energy from a spontaneous redox reaction into electrical energy.
(iv) In the Leclanché cell, \(\mathrm{MnO_2}\) acts as a depolariser. It removes hydrogen formed at the carbon electrode, thereby preventing polarisation and a fall in voltage.
(b)(i)
Balancing oxygen atoms gives \(y=4\), and balancing charge gives \(x=5\):
\[\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}\]
Therefore, \(\boxed{x=5}\) and \(\boxed{y=4}\).
(ii) Let the oxidation state of Mn be \(z\):
\[z+4(-2)=-1\]
\[z=+7\]
Hence, the oxidation state of Mn in \(\mathrm{MnO_4^-}\) is \(\boxed{+7}\).
(c)(i) Factors affecting selective discharge of ions
- Position of the ions in the electrochemical series.
- Concentration of the ions in the electrolyte.
- Nature of the electrode.
(ii) Faraday's second law of electrolysis
When the same quantity of electricity passes through different electrolytes, the masses of substances liberated or deposited are proportional to their chemical equivalent masses.
(iii) Calculation
Since the voltameters are connected in series, the same quantity of electricity passes through both solutions.
For silver:
\[\mathrm{Ag^+ + e^- \rightarrow Ag}\]
\[\text{Moles of Ag deposited}=\frac{0.780}{108}=7.222\times10^{-3}\ \text{mol}\]
One mole of Ag requires one mole of electrons. Therefore:
\[\text{Moles of electrons}=7.222\times10^{-3}\ \text{mol}\]
I. Mass of copper deposited
\[\mathrm{Cu^{2+}+2e^-\rightarrow Cu}\]
\[\text{Moles of Cu}=\frac{7.222\times10^{-3}}{2}=3.611\times10^{-3}\ \text{mol}\]
\[\text{Mass of Cu}=3.611\times10^{-3}\times63.5=0.229\ \text{g}\]
\[\boxed{\text{Mass of copper deposited}=0.229\ \text{g}}\]
II. Quantity of electricity and time of flow
\[Q=nF=(7.222\times10^{-3})(96500)=696.9\ \text{C}\]
\[\boxed{Q\approx697\ \text{C}}\]
Since \(Q=It\):
\[t=\frac{Q}{I}=\frac{696.9}{0.200}=3484.5\ \text{s}\]
\[\boxed{t\approx3.48\times10^3\ \text{s}=58.1\ \text{min}}\]
Tambaya 3 Rahoto
(a)(i) Explain what is meant by acid anhydride and give one example
(ii) State three chemical properties of hydrochloric acid.
(b) Explain each of the following observations:
(i) Tetraoxosulphate (VI) acid can form two types of salts unlike trioxonitrate (V) acid.
(ii) Copper and iron react with concentrated H\(_2\)SO\(_4\) but only one of them reacts with the dilute acid.
(iii) On adding dilute H\(_2\)SO\(_4\) separately to zinc dust and zinc granules of the same mass, the dust produced more vigorous effervescence.
(c)(i) Define activation energy.
(ii) Sketch and label an energy profile diagram for the following reaction: A + B --> C + D; AH = xkJmol\(^{-1}\)
(iii) Explain why the heat of reaction of the mineral acids with sodium hydroxide is constant in value.
(d) Consider the reaction represented by the following equation:
Q\(_{(s)}\) \(\rightleftharpoons\) Q\(_{(l)}\) \(\Delta\) = xkJmol\(^{-1}\)
(i) State with reason which of Q\(_{(s)}\) and O\(_{(J)}\) has the higher entropy.
(ii) What will be the effect of decrease in temperature on the system at equilibrium?
(a)(i) An acid anhydride is an oxide which reacts with water to form an acid; it may be regarded as an acid with water removed. For example:
\[ SO_3 + H_2O \rightarrow H_2SO_4 \]
Therefore, sulphur(VI) oxide, \(SO_3\), is an acid anhydride.
(a)(ii) Three chemical properties of hydrochloric acid are:
- It reacts with metals above hydrogen in the reactivity series to form a chloride salt and hydrogen gas: \[ Zn + 2HCl \rightarrow ZnCl_2 + H_2 \]
- It neutralises bases or alkalis to form a chloride salt and water: \[ HCl + NaOH \rightarrow NaCl + H_2O \]
- It reacts with carbonates or hydrogencarbonates to form a chloride salt, water and carbon(IV) oxide: \[ Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2 \]
(b)(i) Tetraoxosulphate(VI) acid, \(H_2SO_4\), is dibasic: it has two ionisable hydrogen ions. It can therefore form:
- an acid salt, such as \(NaHSO_4\); and
- a normal salt, such as \(Na_2SO_4\).
Trioxonitrate(V) acid, \(HNO_3\), is monobasic because it has only one ionisable hydrogen ion. It forms only normal salts, such as \(NaNO_3\).
(b)(ii) Iron is above hydrogen in the reactivity series, so it reacts with dilute sulphuric acid and displaces hydrogen gas:
\[ Fe + H_2SO_4 \rightarrow FeSO_4 + H_2 \]
Copper is below hydrogen in the reactivity series, so it cannot displace hydrogen from dilute \(H_2SO_4\). However, hot concentrated sulphuric acid is an oxidising agent, so it reacts with both copper and iron. Thus, only iron reacts with the dilute acid.
(b)(iii) Zinc dust has a greater surface area than zinc granules of the same mass. More zinc particles are exposed to the acid, so collisions between zinc and acid particles occur more frequently. The reaction is therefore faster and hydrogen gas is evolved more vigorously.
(c)(i) Activation energy is the minimum energy which reacting particles must possess for a collision to result in a chemical reaction.
(c)(ii) The energy profile must show reactants, products, activation energy, and the enthalpy change. Since the question gives \(\Delta H=x\text{ kJ mol}^{-1}\) without stating whether \(x\) is positive or negative, the relative positions of products and reactants depend on its sign. The diagram below shows the endothermic case, where \(x\) is positive.
If \(x\) is negative, the reaction is exothermic and the products should be drawn below the reactants, with \(\Delta H=-x\text{ kJ mol}^{-1}\).
(c)(iii) Mineral acids such as hydrochloric acid, nitric acid and sulphuric acid are strong acids. In dilute solution, they ionise completely to produce \(H^+\) ions. Sodium hydroxide also ionises completely to produce \(OH^-\) ions. Therefore, the net ionic equation is always:
\[ H^+_{(aq)} + OH^-_{(aq)} \rightarrow H_2O_{(l)} \]
Since the same reaction—the formation of water—occurs each time, the heat of neutralisation is approximately constant, about \(-57\text{ kJ mol}^{-1}\) of water formed.
(d)(i) \(Q_{(l)}\) has the higher entropy. In a liquid, particles are less ordered and have more freedom of movement than particles in a solid, which are held in fixed positions. The symbol \(O_{(l)}\) in the question appears to be a typographical error for \(Q_{(l)}\).
(d)(ii) Melting, \(Q_{(s)} \rightleftharpoons Q_{(l)}\), is endothermic in the forward direction. A decrease in temperature favours the exothermic reverse reaction, freezing. The equilibrium therefore shifts to the left, producing more \(Q_{(s)}\).
Bayanin Amsa
(a)(i) An acid anhydride is an oxide which reacts with water to form an acid; it may be regarded as an acid with water removed. For example:
\[ SO_3 + H_2O \rightarrow H_2SO_4 \]
Therefore, sulphur(VI) oxide, \(SO_3\), is an acid anhydride.
(a)(ii) Three chemical properties of hydrochloric acid are:
- It reacts with metals above hydrogen in the reactivity series to form a chloride salt and hydrogen gas: \[ Zn + 2HCl \rightarrow ZnCl_2 + H_2 \]
- It neutralises bases or alkalis to form a chloride salt and water: \[ HCl + NaOH \rightarrow NaCl + H_2O \]
- It reacts with carbonates or hydrogencarbonates to form a chloride salt, water and carbon(IV) oxide: \[ Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2 \]
(b)(i) Tetraoxosulphate(VI) acid, \(H_2SO_4\), is dibasic: it has two ionisable hydrogen ions. It can therefore form:
- an acid salt, such as \(NaHSO_4\); and
- a normal salt, such as \(Na_2SO_4\).
Trioxonitrate(V) acid, \(HNO_3\), is monobasic because it has only one ionisable hydrogen ion. It forms only normal salts, such as \(NaNO_3\).
(b)(ii) Iron is above hydrogen in the reactivity series, so it reacts with dilute sulphuric acid and displaces hydrogen gas:
\[ Fe + H_2SO_4 \rightarrow FeSO_4 + H_2 \]
Copper is below hydrogen in the reactivity series, so it cannot displace hydrogen from dilute \(H_2SO_4\). However, hot concentrated sulphuric acid is an oxidising agent, so it reacts with both copper and iron. Thus, only iron reacts with the dilute acid.
(b)(iii) Zinc dust has a greater surface area than zinc granules of the same mass. More zinc particles are exposed to the acid, so collisions between zinc and acid particles occur more frequently. The reaction is therefore faster and hydrogen gas is evolved more vigorously.
(c)(i) Activation energy is the minimum energy which reacting particles must possess for a collision to result in a chemical reaction.
(c)(ii) The energy profile must show reactants, products, activation energy, and the enthalpy change. Since the question gives \(\Delta H=x\text{ kJ mol}^{-1}\) without stating whether \(x\) is positive or negative, the relative positions of products and reactants depend on its sign. The diagram below shows the endothermic case, where \(x\) is positive.
If \(x\) is negative, the reaction is exothermic and the products should be drawn below the reactants, with \(\Delta H=-x\text{ kJ mol}^{-1}\).
(c)(iii) Mineral acids such as hydrochloric acid, nitric acid and sulphuric acid are strong acids. In dilute solution, they ionise completely to produce \(H^+\) ions. Sodium hydroxide also ionises completely to produce \(OH^-\) ions. Therefore, the net ionic equation is always:
\[ H^+_{(aq)} + OH^-_{(aq)} \rightarrow H_2O_{(l)} \]
Since the same reaction—the formation of water—occurs each time, the heat of neutralisation is approximately constant, about \(-57\text{ kJ mol}^{-1}\) of water formed.
(d)(i) \(Q_{(l)}\) has the higher entropy. In a liquid, particles are less ordered and have more freedom of movement than particles in a solid, which are held in fixed positions. The symbol \(O_{(l)}\) in the question appears to be a typographical error for \(Q_{(l)}\).
(d)(ii) Melting, \(Q_{(s)} \rightleftharpoons Q_{(l)}\), is endothermic in the forward direction. A decrease in temperature favours the exothermic reverse reaction, freezing. The equilibrium therefore shifts to the left, producing more \(Q_{(s)}\).
Tambaya 4 Rahoto
(a) List two substances that can be used in the laboratory to
(i) dry hydrogen;
(ii) remove carbon (IV) oxide from a sample of air;
(iii) convert hot copper (II) oxide to copper;
(iv) prepare zinc chloride by the action of dilute HCI.
(b)(i) Name two alloys which contain lead.
(ii) State and explain what is observed on bubbling \( \mathrm{H_2S} \) into a solution of \( \mathrm{Pb(NO)_2} \).
(iii) A metal M exists as a silvery white solid at temperatures above 18°C and as a grey solid below 18°C.
I. name the phenomenon exhibited by M.
II. What term is used to describe the temperature given as 18°C in this case?
(c)(i) Write an equation for the action of heat on each of the following compounds:
I. \( \mathrm{AgNO_3} \)
II. \( \mathrm{(NH4)_2CO_3} \).
(ii) Copy and complete the table below
| Metal | Name of main ore | Method of extraction | One major use Haematite |
| — | Haematite | — | — |
| — | — | Electrolysis of molten oxide | — |
(a) Two substances for each purpose:
| Purpose | Substances |
|---|---|
| (i) Dry hydrogen | Concentrated H2SO4; fused (anhydrous) calcium chloride, CaCl2 |
| (ii) Remove CO2 from air | Sodium hydroxide solution (NaOH); soda lime |
| (iii) Convert hot CuO to copper | Hydrogen gas; carbon (II) oxide, CO (coke/carbon also acceptable) |
| (iv) Prepare ZnCl2 with dilute HCl | Zinc metal; zinc carbonate, ZnCO3 (zinc oxide also acceptable) |
(b)(i) Two alloys that contain lead: solder (tin and lead) and type metal (lead, tin and antimony).
(b)(ii) On bubbling H2S into a solution of Pb(NO3)2, a black precipitate is observed. This is lead(II) sulfide, PbS, which is insoluble and black:
\[\text{Pb(NO}_3)_2(aq) + \text{H}_2\text{S}(g) \rightarrow \text{PbS}(s) + 2\text{HNO}_3(aq)\](b)(iii) Metal M is tin.
- I. The phenomenon (existing in two different solid forms) is allotropy (the change from white to grey tin is known as tin pest).
- II. The temperature of 18 °C, at which one form changes into the other, is the transition temperature.
(c)(i) Action of heat:
I. On silver trioxonitrate(V):
\[2\text{AgNO}_3 \xrightarrow{\ \Delta\ } 2\text{Ag} + 2\text{NO}_2 + \text{O}_2\]II. On ammonium trioxocarbonate(IV):
\[(\text{NH}_4)_2\text{CO}_3 \xrightarrow{\ \Delta\ } 2\text{NH}_3 + \text{H}_2\text{O} + \text{CO}_2\](c)(ii) Completed table:
| Metal | Name of main ore | Method of extraction | One major use |
|---|---|---|---|
| Iron | Haematite | Reduction in the blast furnace (with coke / carbon (II) oxide) | Making steel for construction |
| Aluminium | Bauxite | Electrolysis of molten oxide (alumina) | Making aircraft bodies and cooking utensils |
Bayanin Amsa
(a) Two substances for each purpose:
| Purpose | Substances |
|---|---|
| (i) Dry hydrogen | Concentrated H2SO4; fused (anhydrous) calcium chloride, CaCl2 |
| (ii) Remove CO2 from air | Sodium hydroxide solution (NaOH); soda lime |
| (iii) Convert hot CuO to copper | Hydrogen gas; carbon (II) oxide, CO (coke/carbon also acceptable) |
| (iv) Prepare ZnCl2 with dilute HCl | Zinc metal; zinc carbonate, ZnCO3 (zinc oxide also acceptable) |
(b)(i) Two alloys that contain lead: solder (tin and lead) and type metal (lead, tin and antimony).
(b)(ii) On bubbling H2S into a solution of Pb(NO3)2, a black precipitate is observed. This is lead(II) sulfide, PbS, which is insoluble and black:
\[\text{Pb(NO}_3)_2(aq) + \text{H}_2\text{S}(g) \rightarrow \text{PbS}(s) + 2\text{HNO}_3(aq)\](b)(iii) Metal M is tin.
- I. The phenomenon (existing in two different solid forms) is allotropy (the change from white to grey tin is known as tin pest).
- II. The temperature of 18 °C, at which one form changes into the other, is the transition temperature.
(c)(i) Action of heat:
I. On silver trioxonitrate(V):
\[2\text{AgNO}_3 \xrightarrow{\ \Delta\ } 2\text{Ag} + 2\text{NO}_2 + \text{O}_2\]II. On ammonium trioxocarbonate(IV):
\[(\text{NH}_4)_2\text{CO}_3 \xrightarrow{\ \Delta\ } 2\text{NH}_3 + \text{H}_2\text{O} + \text{CO}_2\](c)(ii) Completed table:
| Metal | Name of main ore | Method of extraction | One major use |
|---|---|---|---|
| Iron | Haematite | Reduction in the blast furnace (with coke / carbon (II) oxide) | Making steel for construction |
| Aluminium | Bauxite | Electrolysis of molten oxide (alumina) | Making aircraft bodies and cooking utensils |
Tambaya 5 Rahoto
(a) State three characteristic properties of
(i) electrovalent compounds;
(ii) alpha particles
(iii) catalysts
(b)(i) Write the electronic configuration of silicon (atomic number 14) and state the group to which it belongs in the Periodic Table.
(ii) State the type of chemical bonding between silicon and oxygen in SiO\(_2\)
(iii) A chip used in a microcomputer contains 5.72 x 10\(^{-3}\)g of silicon, calculate the number of silicon atoms in the chip.
[Si = 28; Avogadro constant = 6.02 x 10\(^{23}\) mol\(^{-1}\)]
(c) An element X belongs to the same group as sodium but is more reactive.
(i) Suggest with reason whether X would be a reducing or oxidizing agent.
(ii) What would be a suitable method of storing X in the laboratory?
(iii) Describe briefly what would be observed if a small piece of X were dropped into a trough of cold water which had been coloured with red litmus.
(iv) Write an equation to show how the oxide of X would react with dilute HCI.
(v) Suggest the likely colour of the salts of X
(a) Characteristic properties
(i) Electrovalent (ionic) compounds:
- High melting and boiling points.
- Conduct electricity when molten or dissolved in water.
- Usually soluble in water but insoluble in organic solvents.
(ii) Alpha particles:
- Positively charged (helium nuclei, 2 protons and 2 neutrons).
- Deflected by electric and magnetic fields (toward the negative plate).
- Low penetrating power (stopped by a sheet of paper) but strongly ionising.
(iii) Catalysts:
- Alter the rate of a reaction without being used up.
- Needed only in small amounts and are chemically unchanged at the end.
- Are specific and can be poisoned by impurities.
(b) Silicon
(i) Electronic configuration of Si (Z = 14): \(1s^2\,2s^2\,2p^6\,3s^2\,3p^2\), i.e. 2, 8, 4. It belongs to Group IV (Group 14).
(ii) Bonding between silicon and oxygen in SiO2: covalent (a giant covalent structure).
(iii) Number of silicon atoms:
Moles of Si \( = \dfrac{5.72 \times 10^{-3}}{28} = 2.043 \times 10^{-4}\ \text{mol} \)
Number of atoms \( = 2.043 \times 10^{-4} \times 6.02 \times 10^{23} = \mathbf{1.23 \times 10^{20}\ atoms} \)
(c) Element X (same group as Na but more reactive)
(i) X is a reducing agent. Being more electropositive than sodium, it loses its outer electron even more readily, so it reduces other species while being oxidised itself.
(ii) Store X under paraffin oil (kerosene) to keep it away from air and moisture.
(iii) If dropped into cold water coloured with red litmus, X would float and dart about on the surface, melt into a shining ball, hiss and give off a gas (hydrogen) which may ignite; the red litmus turns blue as an alkaline hydroxide is formed.
(iv) \[ X_2O + 2HCl \to 2XCl + H_2O \]
(v) The salts of X are likely to be white (colourless), as for other Group I salts.
Bayanin Amsa
(a) Characteristic properties
(i) Electrovalent (ionic) compounds:
- High melting and boiling points.
- Conduct electricity when molten or dissolved in water.
- Usually soluble in water but insoluble in organic solvents.
(ii) Alpha particles:
- Positively charged (helium nuclei, 2 protons and 2 neutrons).
- Deflected by electric and magnetic fields (toward the negative plate).
- Low penetrating power (stopped by a sheet of paper) but strongly ionising.
(iii) Catalysts:
- Alter the rate of a reaction without being used up.
- Needed only in small amounts and are chemically unchanged at the end.
- Are specific and can be poisoned by impurities.
(b) Silicon
(i) Electronic configuration of Si (Z = 14): \(1s^2\,2s^2\,2p^6\,3s^2\,3p^2\), i.e. 2, 8, 4. It belongs to Group IV (Group 14).
(ii) Bonding between silicon and oxygen in SiO2: covalent (a giant covalent structure).
(iii) Number of silicon atoms:
Moles of Si \( = \dfrac{5.72 \times 10^{-3}}{28} = 2.043 \times 10^{-4}\ \text{mol} \)
Number of atoms \( = 2.043 \times 10^{-4} \times 6.02 \times 10^{23} = \mathbf{1.23 \times 10^{20}\ atoms} \)
(c) Element X (same group as Na but more reactive)
(i) X is a reducing agent. Being more electropositive than sodium, it loses its outer electron even more readily, so it reduces other species while being oxidised itself.
(ii) Store X under paraffin oil (kerosene) to keep it away from air and moisture.
(iii) If dropped into cold water coloured with red litmus, X would float and dart about on the surface, melt into a shining ball, hiss and give off a gas (hydrogen) which may ignite; the red litmus turns blue as an alkaline hydroxide is formed.
(iv) \[ X_2O + 2HCl \to 2XCl + H_2O \]
(v) The salts of X are likely to be white (colourless), as for other Group I salts.
Tambaya 6 Rahoto
(a) Describe briefly a suitable procedure for preparing a pure sample of MgSO\(_4\) starting from MgO.
(b)(i) Mention two sources of water pollution.
(ii) Explain why the sample of air collected in the process of boiling water is richer in oxygen than atmospheric air
(iii) Mention one substance used as coagulant in water treatment plants.
(c)(i) State two physical porperties of chlorine.
(ii) Write an equation to show how chlorine reacts with iron
(iii) Why is Chlorine preferred to sulphur (IV) oxide in the bleaching of cotton
(d) Bleaching powder reacts with dilute HCl according to the reaction below;
CaOCl\(_{2(s)}\) + 2HCI\(_{(aq)}\) -> CaCl\(_{2(aq)}\) + H\(_2\)O\(_{(l)}\) + Cl\(_{2(g)}\)
Calculate the mass of bleaching powder that will produce 400cm\(^3\) of chlorine at 25\(^o\)C and a pressure of 1.20 x 10\(^5\) NM\(^{-2}\). [O = 16.0; Cl = 35.5; Ca = 40.0;1 mole of gas occupies 22.4 dm\(^3\) at s.t.p; standard pressure = 1.01 x 10\(^6\) Nm\(^{-2}\)]
(a) Preparing pure MgSO4 from MgO
Warm some dilute tetraoxosulphate(VI) acid in a beaker and add magnesium oxide (a base) a little at a time, stirring, until no more dissolves (excess MgO ensures all the acid is used up). Filter to remove the unreacted MgO. Evaporate the filtrate until saturated (to the point of crystallisation), allow to cool so that MgSO4·7H2O crystallises, then filter off and dry the crystals between filter papers.
\[ MgO + H_2SO_4 \to MgSO_4 + H_2O \]
(b) Water
(i) Two sources of water pollution: industrial effluents/waste; sewage or domestic waste (also agricultural run-off, oil spillage).
(ii) Oxygen is more soluble in water than nitrogen, so the dissolved air has a higher proportion of oxygen than atmospheric air; when this air is expelled by boiling, it is richer in oxygen (about 33% O2) than ordinary air (about 21% O2).
(iii) Coagulant: alum (aluminium sulphate).
(c) Chlorine
(i) Two physical properties: it is a greenish-yellow gas with a choking, pungent smell (also: denser than air, moderately soluble in water, poisonous).
(ii) \[ 2Fe + 3Cl_2 \to 2FeCl_3 \]
(iii) Chlorine is preferred to sulphur(IV) oxide because chlorine bleaches by oxidation, giving a permanent effect, whereas SO2 bleaches by reduction, which is temporary (the colour returns on exposure to air).
(d) Mass of bleaching powder
\[ CaOCl_2 + 2HCl \to CaCl_2 + H_2O + Cl_2 \]
Convert 400 cm3 of Cl2 at 25°C (298 K) and \(1.20 \times 10^5\ \text{N m}^{-2}\) to s.t.p. (273 K, \(1.01 \times 10^5\ \text{N m}^{-2}\)) using \( \dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \):
\[ V_2 = \frac{1.20\times10^5 \times 0.400 \times 273}{298 \times 1.01\times10^5} = 0.435\ \text{dm}^3 \]
Moles of Cl2 \( = \dfrac{0.435}{22.4} = 1.94 \times 10^{-2}\ \text{mol} \)
From the equation, 1 mol CaOCl2 gives 1 mol Cl2, so moles of CaOCl2 \( = 1.94 \times 10^{-2}\ \text{mol} \).
Molar mass of CaOCl2 \( = 40 + 16 + 71 = 127\ \text{g mol}^{-1} \)
Mass \( = 1.94 \times 10^{-2} \times 127 = \mathbf{2.47\ g} \)
(The stated "standard pressure = 1.01 × 106" is taken as 1.01 × 105 N m-2, the true standard pressure.)
Bayanin Amsa
(a) Preparing pure MgSO4 from MgO
Warm some dilute tetraoxosulphate(VI) acid in a beaker and add magnesium oxide (a base) a little at a time, stirring, until no more dissolves (excess MgO ensures all the acid is used up). Filter to remove the unreacted MgO. Evaporate the filtrate until saturated (to the point of crystallisation), allow to cool so that MgSO4·7H2O crystallises, then filter off and dry the crystals between filter papers.
\[ MgO + H_2SO_4 \to MgSO_4 + H_2O \]
(b) Water
(i) Two sources of water pollution: industrial effluents/waste; sewage or domestic waste (also agricultural run-off, oil spillage).
(ii) Oxygen is more soluble in water than nitrogen, so the dissolved air has a higher proportion of oxygen than atmospheric air; when this air is expelled by boiling, it is richer in oxygen (about 33% O2) than ordinary air (about 21% O2).
(iii) Coagulant: alum (aluminium sulphate).
(c) Chlorine
(i) Two physical properties: it is a greenish-yellow gas with a choking, pungent smell (also: denser than air, moderately soluble in water, poisonous).
(ii) \[ 2Fe + 3Cl_2 \to 2FeCl_3 \]
(iii) Chlorine is preferred to sulphur(IV) oxide because chlorine bleaches by oxidation, giving a permanent effect, whereas SO2 bleaches by reduction, which is temporary (the colour returns on exposure to air).
(d) Mass of bleaching powder
\[ CaOCl_2 + 2HCl \to CaCl_2 + H_2O + Cl_2 \]
Convert 400 cm3 of Cl2 at 25°C (298 K) and \(1.20 \times 10^5\ \text{N m}^{-2}\) to s.t.p. (273 K, \(1.01 \times 10^5\ \text{N m}^{-2}\)) using \( \dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \):
\[ V_2 = \frac{1.20\times10^5 \times 0.400 \times 273}{298 \times 1.01\times10^5} = 0.435\ \text{dm}^3 \]
Moles of Cl2 \( = \dfrac{0.435}{22.4} = 1.94 \times 10^{-2}\ \text{mol} \)
From the equation, 1 mol CaOCl2 gives 1 mol Cl2, so moles of CaOCl2 \( = 1.94 \times 10^{-2}\ \text{mol} \).
Molar mass of CaOCl2 \( = 40 + 16 + 71 = 127\ \text{g mol}^{-1} \)
Mass \( = 1.94 \times 10^{-2} \times 127 = \mathbf{2.47\ g} \)
(The stated "standard pressure = 1.01 × 106" is taken as 1.01 × 105 N m-2, the true standard pressure.)
