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Question 1 Report
(a) Giving different examples, mention one metal in each case which produces hydrogen on reacting with
(i) dilute mineral acid
(ii) cold water;
(iii) steam;
(iv) hot, concentrated alkali.
(b) In an experiment, excess 0.50 mol dm\(^{-3}\) HCI was added to 1Og of granulated zinc in a beaker. Other conditions remaining constant, state how the reaction rate would be affected in each case, if the experiment was repeated using:
(i) 1.0 mol dm\(^{-3}\) HCI;
(ii) 8.0g of granulated zinc;
(iii) 10g of zinc dust;
(iv) a higher volume of 0.50 mol dm HCI;
(v) a reaction vessel dipped in crushed ice;
(vi) equal volumes of water and 0.50 mol dm\(^3\) HCI.
(c) Aluminium is extracted from its ore by electrolysis.
(i) Name the ore from which the metal is extracted.
(ii) State the role of molten cryolite in the extraction.
(iii) Describe in outline how the ore is purified before electrolysis
(iv) Calculate the current in amperes required to produce 18.0g of aluminium in 1.50 hours. [Al = 27.0; F = 96500C]
(d) Give the reason why
(i) aluminium, which is a reactive metal, is resistant to corrosion.
(ii) metals are generally good reducing agents.
Question 2 Report
(a) Explain in terms of the kinetic theory why a tyre should not be overinflated.
(b)The following results were obtained at room temperature in an experiment to verify one of the gas laws using a glass syringe:
| Pressure (P) of air in syringe (atm) | Volume (V) of air in syringe (\(cm^3\)) | \(\frac{I}{V}\) |
| 0.100 | 10.00 | 0.100 |
| 0.125 | 8.00 | 0.125 |
| 0.150 | 6.60 | 0.150 |
| 0.175 | 5.60 | 0.179 |
| 0.200 | 4.80 | 0.208 |
| 0.225 | 4.40 | 0.227 |
(i) Plot a graph of P against \(\frac{1}{v}\), using 1 cm to represent 0.01 atm on the vertical axis and 1cm to represent 0.02 unit on the horizontal axis.
(ii) Which of the gas laws is in agreement with the results?
(c) The flow chart below represents the stages involved in the manufacture of H\(_2\)SO\(_4\).
| S + O\(_2\) \(\to\) SO\(_2\) | SO\(_2\) +x \(\to\) SO\(_3\) | SO\(_3\) +Conc. H\(_2\)SO\(_4\) \(\to\) Y | Y +H\(_2\)O \(\to\) Conc H\(_2\)SO\(_4\) |
| stage I | stage II | stage III | stage IV |
(i) Name the process represented by the chart.
(ii) Identify reactant X and product Y.
(iii) What are the operating temperature and pressure at stage II?
(iv) Mention the stage which requires a catalyst and state the catalyst used.
(v) Give the reason why the SO\(_3\) produced in stage II is not dissolved directly in water to form the acid
(d) When K\(_4\)Cr\(_2\)C\(_7\) dissolves in water, the following equilibrium is established:
\(\mathrm{Cr_2O_{7(aq)}^{2-} + H_2O_{(l)} \to 2CrO_{4(aq)}^{2-} + 2H_{aq}}\)
(i) State the colour observed on adding a few drops of dilute H\(_2\)SO\(_4\) to the system.
(ii) Explain your answer in (d)(1).
(iii) What principle is applicable to this explanation?
According to the kinetic theory, the air inside a tyre consists of tiny particles in continuous rapid random motion. These particles constantly collide with the inner walls of the tyre, and the total force of these collisions per unit area is the pressure the gas exerts. When the tyre is overinflated, more particles are forced into the same fixed volume, so the frequency of collisions with the walls increases and the pressure rises sharply. In addition, as the vehicle moves, friction warms the trapped air; the particles gain kinetic energy and move faster, striking the walls harder and more often, which raises the pressure still further. If the internal pressure exceeds the strength of the tyre material, the tyre bursts. The tyre should therefore not be overinflated.
The recorded readings, with \(\tfrac{1}{V}\) computed to three significant figures, are:
| Pressure, \(P\) (atm) | Volume, \(V\) (cm\(^3\)) | \(\dfrac{1}{V}\) (cm\(^{-3}\)) | \(P\times V\) (atm cm\(^3\)) |
|---|---|---|---|
| 0.100 | 10.00 | 0.100 | 1.00 |
| 0.125 | 8.00 | 0.125 | 1.00 |
| 0.150 | 6.60 | 0.152 | 0.99 |
| 0.175 | 5.60 | 0.179 | 0.98 |
| 0.200 | 4.80 | 0.208 | 0.96 |
| 0.225 | 4.40 | 0.227 | 0.99 |
(i) Plotting \(P\) (vertical axis) against \(\tfrac{1}{V}\) (horizontal axis) gives a straight line passing through the origin:
The graph is a straight line through the origin, which shows that \(P\) is directly proportional to \(\tfrac{1}{V}\). The gradient of the line,
\[\text{slope}=\frac{\Delta P}{\Delta(1/V)}=\frac{0.225-0.100}{0.227-0.100}=\frac{0.125}{0.127}\approx 1.0\ \text{atm cm}^3,\]which equals the constant product \(P\times V\) in the last column of the table. Hence \(PV=\text{constant}\).
(ii) The result that agrees with the data is Boyle's Law, which states that at constant temperature the volume of a fixed mass of gas is inversely proportional to its pressure, i.e. \(P\propto\tfrac{1}{V}\), so \(PV=\text{constant}\).
(i) The process represented by the chart is the Contact Process.
(ii) Reactant X is oxygen, O\(_2\) (air); product Y is oleum (fuming sulphuric acid), H\(_2\)S\(_2\)O\(_7\).
(iii) At Stage II the operating temperature is about 450 °C (in the range 400-500 °C) and the pressure is about 1-2 atm (approximately atmospheric).
(iv) The stage that requires a catalyst is Stage II (the conversion of SO\(_2\) to SO\(_3\)), and the catalyst used is vanadium(V) oxide, V\(_2\)O\(_5\).
(v) The SO\(_3\) is not dissolved directly in water because the reaction is strongly exothermic; the large amount of heat released would vaporise the water and create a dense, choking mist (fog) of fine sulphuric acid droplets that is difficult to condense and would escape the plant. Instead SO\(_3\) is absorbed in concentrated H\(_2\)SO\(_4\) to form oleum, which is then safely diluted with water.
The equilibrium established is
\[\text{Cr}_2\text{O}_7^{2-}{}_{(aq)}+\text{H}_2\text{O}_{(l)}\rightleftharpoons 2\text{CrO}_4^{2-}{}_{(aq)}+2\text{H}^+{}_{(aq)}\]where \(\text{Cr}_2\text{O}_7^{2-}\) (dichromate) is orange and \(\text{CrO}_4^{2-}\) (chromate) is yellow.
(i) On adding a few drops of dilute H\(_2\)SO\(_4\), the colour changes from yellow to orange (the solution becomes orange).
(ii) Dilute H\(_2\)SO\(_4\) supplies H\(^+\) ions, increasing their concentration on the right-hand side of the equilibrium. The system responds by shifting the position of equilibrium to the left so as to remove some of the added H\(^+\). This produces more of the orange dichromate ion, \(\text{Cr}_2\text{O}_7^{2-}\), and less of the yellow chromate ion, so the solution turns orange.
(iii) The principle applicable to this explanation is Le Chatelier's Principle.
Answer Details
According to the kinetic theory, the air inside a tyre consists of tiny particles in continuous rapid random motion. These particles constantly collide with the inner walls of the tyre, and the total force of these collisions per unit area is the pressure the gas exerts. When the tyre is overinflated, more particles are forced into the same fixed volume, so the frequency of collisions with the walls increases and the pressure rises sharply. In addition, as the vehicle moves, friction warms the trapped air; the particles gain kinetic energy and move faster, striking the walls harder and more often, which raises the pressure still further. If the internal pressure exceeds the strength of the tyre material, the tyre bursts. The tyre should therefore not be overinflated.
The recorded readings, with \(\tfrac{1}{V}\) computed to three significant figures, are:
| Pressure, \(P\) (atm) | Volume, \(V\) (cm\(^3\)) | \(\dfrac{1}{V}\) (cm\(^{-3}\)) | \(P\times V\) (atm cm\(^3\)) |
|---|---|---|---|
| 0.100 | 10.00 | 0.100 | 1.00 |
| 0.125 | 8.00 | 0.125 | 1.00 |
| 0.150 | 6.60 | 0.152 | 0.99 |
| 0.175 | 5.60 | 0.179 | 0.98 |
| 0.200 | 4.80 | 0.208 | 0.96 |
| 0.225 | 4.40 | 0.227 | 0.99 |
(i) Plotting \(P\) (vertical axis) against \(\tfrac{1}{V}\) (horizontal axis) gives a straight line passing through the origin:
The graph is a straight line through the origin, which shows that \(P\) is directly proportional to \(\tfrac{1}{V}\). The gradient of the line,
\[\text{slope}=\frac{\Delta P}{\Delta(1/V)}=\frac{0.225-0.100}{0.227-0.100}=\frac{0.125}{0.127}\approx 1.0\ \text{atm cm}^3,\]which equals the constant product \(P\times V\) in the last column of the table. Hence \(PV=\text{constant}\).
(ii) The result that agrees with the data is Boyle's Law, which states that at constant temperature the volume of a fixed mass of gas is inversely proportional to its pressure, i.e. \(P\propto\tfrac{1}{V}\), so \(PV=\text{constant}\).
(i) The process represented by the chart is the Contact Process.
(ii) Reactant X is oxygen, O\(_2\) (air); product Y is oleum (fuming sulphuric acid), H\(_2\)S\(_2\)O\(_7\).
(iii) At Stage II the operating temperature is about 450 °C (in the range 400-500 °C) and the pressure is about 1-2 atm (approximately atmospheric).
(iv) The stage that requires a catalyst is Stage II (the conversion of SO\(_2\) to SO\(_3\)), and the catalyst used is vanadium(V) oxide, V\(_2\)O\(_5\).
(v) The SO\(_3\) is not dissolved directly in water because the reaction is strongly exothermic; the large amount of heat released would vaporise the water and create a dense, choking mist (fog) of fine sulphuric acid droplets that is difficult to condense and would escape the plant. Instead SO\(_3\) is absorbed in concentrated H\(_2\)SO\(_4\) to form oleum, which is then safely diluted with water.
The equilibrium established is
\[\text{Cr}_2\text{O}_7^{2-}{}_{(aq)}+\text{H}_2\text{O}_{(l)}\rightleftharpoons 2\text{CrO}_4^{2-}{}_{(aq)}+2\text{H}^+{}_{(aq)}\]where \(\text{Cr}_2\text{O}_7^{2-}\) (dichromate) is orange and \(\text{CrO}_4^{2-}\) (chromate) is yellow.
(i) On adding a few drops of dilute H\(_2\)SO\(_4\), the colour changes from yellow to orange (the solution becomes orange).
(ii) Dilute H\(_2\)SO\(_4\) supplies H\(^+\) ions, increasing their concentration on the right-hand side of the equilibrium. The system responds by shifting the position of equilibrium to the left so as to remove some of the added H\(^+\). This produces more of the orange dichromate ion, \(\text{Cr}_2\text{O}_7^{2-}\), and less of the yellow chromate ion, so the solution turns orange.
(iii) The principle applicable to this explanation is Le Chatelier's Principle.
Question 3 Report
(a) What term is used to describe each of the following processes?
(i) Alkaline hydrolysis of fats and oils;
(ii) The conversion of glucose into ethanol by enzymatic action;
(iii) Thermal decomposition of higher petroleum fractions into lower molecular mass hydrocarbons in the presence of catalyst.
(b)(i) Write the structure and IUPAC name for one alkanoic acid with the molecular formula C\(_4\)H\(_8\)0\(_2\).
(ii) Arrange the following compounds in order of increasing boiling point: Butane; Butanoic acid; Methylpropane.
(iii) Give an explanation for your answer in (b)(ii).
(c)(i) Ethanol was used for preparing a gas X which decolorized bromine water. Identify X and describe briefly its laboratory preparation.
(ii) Write an equation to show how ethanol reacts with sodium
(iii) Give the reagent and reaction conditions for the conversion of ethanol into C\(_2\)H\(_5\)COOC\(_2\)H\(_5\).
(d) State the type of recction involved in each of the conversions indicated below:
(i)C\(_6\)H\(_6\)C\(_6\)H\(_5\)CH\(_3\)
(ii) nC\(_2\)H\(_4\) \(\to\) (CH\(_2\) - CH\(_2\)),
(iii) CH\(_3\)CH\(_2\)CH(OH)CH\(_3\) -> CH\(_3\)CH\(_2\)CCH\(_3\)
(iv) (C\(_6\)H\(_{10}\)O\(_5\)) -> C\(_6\)H\(_{12}\)O\(_6\).
(iv)
Answer Details
None
Question 4 Report
a)(i) Give two uses of ammonia.
(ii) Name the process by which ammoniacal liquor can be obtained from coal and list two other products of the reaction
(iii) What type of reaction is involved in the conversion of ammoniacal liquor to (NH\(_4\))\(_2\)SO\(_4\) by dilute H\(_2\)SO\(_4\)?
(iv) Sketch and label an energy profile diagram to show the effect of presence of Pt/Rh on the reaction represented by the following equation: 4NH\(_3\) + 5O\(_2\) \(\to\) 6H\(_2\)O + 4NO; \(\Delta\)H = —907 kJmol\(^1\)
(b) Rock salt is an impure form of sodium chloride.
(i) Outline a suitable procedure for preparing a pure sample of sodium chloride from rock salt.
(ii) State two methods that can be used to prepare chlorine from rock salt. Write an appropriate equation in each case.
(c) Lead pigments were used in a water colour painting which turned black after prolonged exposure to an air pollutant. The original colour was restored by using H\(_2\)O\(_2\) which converted the black substance to a simple, white lead (II) salt.
(i) Which pollutant turned the painting black?
(ii) Write the formula of the black substance
(iii) What is the white salt?
(iv) State the role of H\(_2\)O in the restoration process.
(a)(i) Two uses of ammonia are:
(a)(ii) Ammoniacal liquor is obtained by the destructive distillation of coal.
Two other products are:
Coal gas is also produced.
(a)(iii) The reaction is a neutralisation (acid-base) reaction.
\(2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4\)
(a)(iv) Energy profile diagram for the reaction:
The reaction is exothermic because the products are at a lower energy level than the reactants. Pt/Rh lowers the activation energy but does not change the value of \(\Delta H\).
(b)(i) Procedure for preparing pure sodium chloride from rock salt:
(b)(ii) Two methods of preparing chlorine from rock salt are:
(c)(i) The air pollutant is hydrogen sulphide, \(H_2S\).
(c)(ii) The black substance is lead(II) sulphide, \(PbS\).
(c)(iii) The white salt is lead(II) tetraoxosulphate(VI), \(PbSO_4\).
(c)(iv) Hydrogen peroxide, \(H_2O_2\), acts as an oxidising agent. It oxidises black lead(II) sulphide to white lead(II) sulphate:
\(PbS + 4H_2O_2 \rightarrow PbSO_4 + 4H_2O\)
Answer Details
(a)(i) Two uses of ammonia are:
(a)(ii) Ammoniacal liquor is obtained by the destructive distillation of coal.
Two other products are:
Coal gas is also produced.
(a)(iii) The reaction is a neutralisation (acid-base) reaction.
\(2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4\)
(a)(iv) Energy profile diagram for the reaction:
The reaction is exothermic because the products are at a lower energy level than the reactants. Pt/Rh lowers the activation energy but does not change the value of \(\Delta H\).
(b)(i) Procedure for preparing pure sodium chloride from rock salt:
(b)(ii) Two methods of preparing chlorine from rock salt are:
(c)(i) The air pollutant is hydrogen sulphide, \(H_2S\).
(c)(ii) The black substance is lead(II) sulphide, \(PbS\).
(c)(iii) The white salt is lead(II) tetraoxosulphate(VI), \(PbSO_4\).
(c)(iv) Hydrogen peroxide, \(H_2O_2\), acts as an oxidising agent. It oxidises black lead(II) sulphide to white lead(II) sulphate:
\(PbS + 4H_2O_2 \rightarrow PbSO_4 + 4H_2O\)
Question 5 Report
(a)(i) State three methods of preparing salts, giving one example in each case of a salt so prepared.
(ii) What type of salt is each of the following? NaH\(_2\)PO\(_4\); (CH\(_3\)COO)\(_2\)Pb; KAI(SO\(_4\))\(_2\). 12H\(_2\)O.
(b)(i) Write an equation for the reaction between dilute HCI and a solution of AgNO\(_3\).
(ii) Explain why NaNO\(_3\) is preferred to AgNO\(_3\) in the preparation of oxygen by thermal decomposition of trioxonitrate (V) salts.
(iii) When silver wire was dipped into an aqueous solution of CuSO\(_4\), the wire remained intact but when the wire was replaced with zinc rod, the rod decreased in size. Give an explanation for this observation.
(c) When a sample of a crystalline salt X was exposed to air, there was a loss in mass.
(i) What phenomenon was exhibited by X?
(ii) Suggest two substances which X could be.
(iii) On heating 5.00 g of a fresh sample of X to constant mass, 1.80g was lost in the form of water vapour. Calculate the number of molecules of water of crystallization in one molecule of X. [H = 1.00; O = 16.00; Anhydrous form of X = 160 g mol\(^{-1}\)
Answer Details
None
Question 6 Report
(a) What is the shape of (i) p - orbital; (ii) a molecule of methane; (iii) a molecule of carbon (IV) oxide?
(b) Consider the following elements: Ne, S, CI, 0, Fe, Mg. State which of them
(i) exhibit(s) allotropy;
(ii) form(s) coloured ions;
(iii) is/are malleable;
(iv) consist(s) of molecules that are far apart at room temperature;
(v) form(s) hydrides by sharing electrons with hydrogen;
(vi) has/have complete outermost shell.
(c)(i) List three applications of radioactivity in different fields.
(ii) Explain clearly the difference between the following reactions involving electron loss from lead.
\(^{211} pb\) \(\to\) \(^{ 211}Bi\) + \(^0_{-1}\); Pb \(\to\) pb\(^{3+}\) _ 2e\(^-\)
(iii) Give one advantage and one disadvantage of nuclear power generation over the use of fossil fuels.
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