WAEC SSCE - Chemistry - 1995 (Essay)
Pergunta 1 Relatório
(a)(i) Explain the term pH
(ii) If sodium hydroxide solution were added to a solution of a strong acid, what would happen to the pH of the solution?
(b) Give one example of each of the following:
(i) acidic oxide
(ii) acid salt.
(a)(i) The pH of a solution is a measure of its acidity or alkalinity; it is the negative logarithm to base ten of the hydrogen ion concentration, \(pH = -\log[H^+]\).
(ii) Adding sodium hydroxide to a strong acid neutralizes some of the H+ ions, so the pH increases (the solution becomes less acidic and moves toward, then above, 7).
(b)(i) Acidic oxide: carbon(IV) oxide, CO2 (or SO2, SO3).
(ii) Acid salt: sodium hydrogen trioxocarbonate(IV), NaHCO3 (or sodium hydrogen tetraoxosulphate(VI), NaHSO4).
Detalhes da Resposta
(a)(i) The pH of a solution is a measure of its acidity or alkalinity; it is the negative logarithm to base ten of the hydrogen ion concentration, \(pH = -\log[H^+]\).
(ii) Adding sodium hydroxide to a strong acid neutralizes some of the H+ ions, so the pH increases (the solution becomes less acidic and moves toward, then above, 7).
(b)(i) Acidic oxide: carbon(IV) oxide, CO2 (or SO2, SO3).
(ii) Acid salt: sodium hydrogen trioxocarbonate(IV), NaHCO3 (or sodium hydrogen tetraoxosulphate(VI), NaHSO4).
Pergunta 2 Relatório
(a)(i) Explain what is meant by saturated solution
(ii) Describe in outline, a suitable procedure for preparing a saturated solution of sodium trioxonitrate(V) at 30°C.
(ii) State two techniques that can be used to recover crystals of sodium trioxonitrate(V) from its saturated solution.
(b) 1.0dm\(^3\) of an aqueous solution at 90°C contains 404g of potassium trioxonitrate(V) and 245g of potassium trioxochlorate (V).
(i) Determine which of the two salts will separate out when the solution is cooled to 60°C. N = 14. O = 16, CI = 35.5, K = 39; Solubility of KNO\(_3\) in water at 60\(^o\)C = 5.14 mol.dm\(^{-3}\), Solubility of KCIO\(_3\) in water at 60°C = 1.61 mol.dm\(^{-3}\)
(ii) Calculate the mass of salt that will separate out at 60°C
(c)(i) List two salts which cause hardness of water.
(ii) Explain why temporary hardness of water result in the furring of kettle.
(a)(i) A saturated solution is a solution that contains the maximum amount of dissolved solute it can hold at a given temperature, in the presence of (and in equilibrium with) undissolved solute.
(a)(ii) Preparing a saturated solution of \(NaNO_3\) at 30 C
- Warm some distilled water and keep it at 30 C in a thermostatic bath.
- Add sodium trioxonitrate(V) a little at a time, stirring, until no more will dissolve and some solid remains undissolved at 30 C.
- Filter off the excess undissolved solid; the filtrate is the saturated solution at 30 C.
(a)(iii) Two techniques to recover the crystals: evaporation of the solvent (evaporate to the point of crystallisation and allow to cool) and cooling/crystallisation of the hot saturated solution.
(b) Molar masses: \(KNO_3 = 39+14+48 = 101\ \text{g mol}^{-1}\); \(KClO_3 = 39+35.5+48 = 122.5\ \text{g mol}^{-1}\).
Amounts present in \(1.0\ dm^3\):
\[n(KNO_3) = \frac{404}{101} = 4.0\ \text{mol dm}^{-3}\]\[n(KClO_3) = \frac{245}{122.5} = 2.0\ \text{mol dm}^{-3}\](i) At 60 C the solubility of \(KNO_3\) is \(5.14\ \text{mol dm}^{-3}\); since \(4.0 < 5.14\), all the \(KNO_3\) stays in solution. The solubility of \(KClO_3\) is only \(1.61\ \text{mol dm}^{-3}\); since \(2.0 > 1.61\), the excess crystallises. Therefore potassium trioxochlorate(V), \(KClO_3\), separates out.
(ii) Amount separating \(= 2.0 - 1.61 = 0.39\ \text{mol}\).
\[\text{mass} = 0.39 \times 122.5 = 47.8\ \text{g}\](c)(i) Two salts causing hardness: calcium hydrogentrioxocarbonate(IV), \(Ca(HCO_3)_2\), and calcium tetraoxosulphate(VI), \(CaSO_4\) (magnesium salts also cause hardness).
(c)(ii) Temporary hardness furs a kettle because, on boiling, the dissolved \(Ca(HCO_3)_2\) decomposes to insoluble calcium trioxocarbonate(IV), which deposits as a hard scale (fur) on the kettle: \[Ca(HCO_3)_2 \xrightarrow{\text{heat}} CaCO_3 + H_2O + CO_2\]
Detalhes da Resposta
(a)(i) A saturated solution is a solution that contains the maximum amount of dissolved solute it can hold at a given temperature, in the presence of (and in equilibrium with) undissolved solute.
(a)(ii) Preparing a saturated solution of \(NaNO_3\) at 30 C
- Warm some distilled water and keep it at 30 C in a thermostatic bath.
- Add sodium trioxonitrate(V) a little at a time, stirring, until no more will dissolve and some solid remains undissolved at 30 C.
- Filter off the excess undissolved solid; the filtrate is the saturated solution at 30 C.
(a)(iii) Two techniques to recover the crystals: evaporation of the solvent (evaporate to the point of crystallisation and allow to cool) and cooling/crystallisation of the hot saturated solution.
(b) Molar masses: \(KNO_3 = 39+14+48 = 101\ \text{g mol}^{-1}\); \(KClO_3 = 39+35.5+48 = 122.5\ \text{g mol}^{-1}\).
Amounts present in \(1.0\ dm^3\):
\[n(KNO_3) = \frac{404}{101} = 4.0\ \text{mol dm}^{-3}\]\[n(KClO_3) = \frac{245}{122.5} = 2.0\ \text{mol dm}^{-3}\](i) At 60 C the solubility of \(KNO_3\) is \(5.14\ \text{mol dm}^{-3}\); since \(4.0 < 5.14\), all the \(KNO_3\) stays in solution. The solubility of \(KClO_3\) is only \(1.61\ \text{mol dm}^{-3}\); since \(2.0 > 1.61\), the excess crystallises. Therefore potassium trioxochlorate(V), \(KClO_3\), separates out.
(ii) Amount separating \(= 2.0 - 1.61 = 0.39\ \text{mol}\).
\[\text{mass} = 0.39 \times 122.5 = 47.8\ \text{g}\](c)(i) Two salts causing hardness: calcium hydrogentrioxocarbonate(IV), \(Ca(HCO_3)_2\), and calcium tetraoxosulphate(VI), \(CaSO_4\) (magnesium salts also cause hardness).
(c)(ii) Temporary hardness furs a kettle because, on boiling, the dissolved \(Ca(HCO_3)_2\) decomposes to insoluble calcium trioxocarbonate(IV), which deposits as a hard scale (fur) on the kettle: \[Ca(HCO_3)_2 \xrightarrow{\text{heat}} CaCO_3 + H_2O + CO_2\]
Pergunta 3 Relatório
What name is given to each of the following? The
(a) irregular random movement of smoke particles in air.
(b) existence of an element in various forms in the same physical-state
(c) disintegration of atomic nuclei, accompanied with radiation emission.
(d) conversion of a solid difitrunto vapour without melting.
(a) The irregular random movement of smoke particles in air - Brownian motion.
(b) The existence of an element in various forms in the same physical state - allotropy.
(c) The disintegration of atomic nuclei accompanied by the emission of radiation - radioactivity.
(d) The conversion of a solid directly into vapour without melting - sublimation.
Detalhes da Resposta
(a) The irregular random movement of smoke particles in air - Brownian motion.
(b) The existence of an element in various forms in the same physical state - allotropy.
(c) The disintegration of atomic nuclei accompanied by the emission of radiation - radioactivity.
(d) The conversion of a solid directly into vapour without melting - sublimation.
Pergunta 4 Relatório
(a) List three properties of a system that is in a state of chemical equilibrium.
(b) Consider reaction represented by the following equation: 3H\(_{2(g)}\) + N\(_{2(g)}\) \(\rightleftharpoons\) 2NH\(_{3(g)}\); H = 92KJ
(i) Explain the effect of increasing the temperature of the reaction on the yield of ammonia
(ii) Uses of energy profile diagram to illustrate the effect of a positive catalyst on the rate of either the forward reaction or the reverse reaction.
(c) In the extraction of aluminium from bauxite:
(i) outline the procedure used for purifying the ore;
(ii) write equation for the reaction at each electrode, during the electrolysis of the pure alumina;
(iii) state the function of molten cryolite in the electrolytic cell for the extraction.
(a) Three properties of a system in a state of chemical equilibrium
- The change is dynamic: the forward and reverse reactions continue simultaneously and the rate of the forward reaction equals the rate of the reverse reaction.
- The macroscopic properties remain constant with time: the concentrations of reactants and products (and observable properties such as colour and pressure) do not change once equilibrium is reached.
- It is reached only in a closed system, and at equilibrium the free-energy change of the system is zero (\(\Delta G = 0\)).
(b) \(3H_{2(g)} + N_{2(g)} \rightleftharpoons 2NH_{3(g)};\ \Delta H = -92\ \text{kJ}\) (the forward reaction is exothermic).
(i) Effect of increasing the temperature on the yield of ammonia
Since the forward reaction is exothermic, by Le Chatelier's principle an increase in temperature favours the endothermic (reverse) direction. The equilibrium therefore shifts to the left, so the yield of ammonia decreases as the temperature is raised.
(ii) Effect of a positive catalyst (energy profile diagram)
A positive catalyst provides an alternative reaction pathway of lower activation energy. On the energy profile below, the peak of the catalysed curve (green, dashed) is lower than that of the uncatalysed curve (red, solid), while the energy levels of the reactants and of the products are unchanged. Because the activation-energy barrier is lowered by the same amount for both directions, the catalyst speeds up the forward and the reverse reactions equally; equilibrium is reached faster, but the position of equilibrium (and hence the yield) is unchanged.
(c) Extraction of aluminium from bauxite
(i) Purifying the ore (Bayer process): the powdered bauxite is dissolved, under pressure, in hot concentrated sodium hydroxide solution. The amphoteric aluminium oxide dissolves as sodium aluminate, while insoluble impurities such as \(Fe_2O_3\) and \(SiO_2\) do not dissolve and are filtered off:
\[Al_2O_{3(s)} + 2NaOH_{(aq)} + 3H_2O_{(l)} \rightarrow 2NaAl(OH)_{4(aq)}\]
The filtered sodium aluminate solution is then diluted and seeded so that aluminium hydroxide precipitates out:
\[NaAl(OH)_{4(aq)} \rightarrow Al(OH)_{3(s)} + NaOH_{(aq)}\]
The aluminium hydroxide is filtered off, washed, dried and heated strongly (calcined) to give pure alumina:
\[2Al(OH)_{3(s)} \xrightarrow{\text{heat}} Al_2O_{3(s)} + 3H_2O_{(g)}\]
(ii) Electrode reactions during the electrolysis of the pure molten alumina:
\[\text{Cathode (reduction): } Al^{3+} + 3e^- \rightarrow Al\]
\[\text{Anode (oxidation): } 2O^{2-} \rightarrow O_2 + 4e^-\]
(iii) Function of molten cryolite: it acts as a solvent (flux) for the alumina and lowers the melting point of the electrolyte from about 2045 °C to about 950 °C, so that electrolysis can be carried out at a much lower temperature while also increasing the electrical conductivity of the melt.
Detalhes da Resposta
(a) Three properties of a system in a state of chemical equilibrium
- The change is dynamic: the forward and reverse reactions continue simultaneously and the rate of the forward reaction equals the rate of the reverse reaction.
- The macroscopic properties remain constant with time: the concentrations of reactants and products (and observable properties such as colour and pressure) do not change once equilibrium is reached.
- It is reached only in a closed system, and at equilibrium the free-energy change of the system is zero (\(\Delta G = 0\)).
(b) \(3H_{2(g)} + N_{2(g)} \rightleftharpoons 2NH_{3(g)};\ \Delta H = -92\ \text{kJ}\) (the forward reaction is exothermic).
(i) Effect of increasing the temperature on the yield of ammonia
Since the forward reaction is exothermic, by Le Chatelier's principle an increase in temperature favours the endothermic (reverse) direction. The equilibrium therefore shifts to the left, so the yield of ammonia decreases as the temperature is raised.
(ii) Effect of a positive catalyst (energy profile diagram)
A positive catalyst provides an alternative reaction pathway of lower activation energy. On the energy profile below, the peak of the catalysed curve (green, dashed) is lower than that of the uncatalysed curve (red, solid), while the energy levels of the reactants and of the products are unchanged. Because the activation-energy barrier is lowered by the same amount for both directions, the catalyst speeds up the forward and the reverse reactions equally; equilibrium is reached faster, but the position of equilibrium (and hence the yield) is unchanged.
(c) Extraction of aluminium from bauxite
(i) Purifying the ore (Bayer process): the powdered bauxite is dissolved, under pressure, in hot concentrated sodium hydroxide solution. The amphoteric aluminium oxide dissolves as sodium aluminate, while insoluble impurities such as \(Fe_2O_3\) and \(SiO_2\) do not dissolve and are filtered off:
\[Al_2O_{3(s)} + 2NaOH_{(aq)} + 3H_2O_{(l)} \rightarrow 2NaAl(OH)_{4(aq)}\]
The filtered sodium aluminate solution is then diluted and seeded so that aluminium hydroxide precipitates out:
\[NaAl(OH)_{4(aq)} \rightarrow Al(OH)_{3(s)} + NaOH_{(aq)}\]
The aluminium hydroxide is filtered off, washed, dried and heated strongly (calcined) to give pure alumina:
\[2Al(OH)_{3(s)} \xrightarrow{\text{heat}} Al_2O_{3(s)} + 3H_2O_{(g)}\]
(ii) Electrode reactions during the electrolysis of the pure molten alumina:
\[\text{Cathode (reduction): } Al^{3+} + 3e^- \rightarrow Al\]
\[\text{Anode (oxidation): } 2O^{2-} \rightarrow O_2 + 4e^-\]
(iii) Function of molten cryolite: it acts as a solvent (flux) for the alumina and lowers the melting point of the electrolyte from about 2045 °C to about 950 °C, so that electrolysis can be carried out at a much lower temperature while also increasing the electrical conductivity of the melt.
Pergunta 5 Relatório
(a)(i) Sketch a graph to illustrate Charles' law.
(ii) A gas occupies 500 cm\(^{ 3}\) at 2TC. calculate its volume at 40°C constant pressure.
(b) List two gases that are used as refrigerant
(a)(i) Charles' law states that at constant pressure the volume of a fixed mass of gas is directly proportional to its absolute (kelvin) temperature. A graph of volume against absolute temperature is therefore a straight line passing through the origin:
The line is straight and, when produced, passes through the origin (0 K, 0 cm\(^3\)), showing \(V \propto T\) at constant pressure. If the same volume is plotted against temperature in \(^{\circ}\)C, the straight line is displaced and cuts the temperature axis at \(-273\ ^{\circ}\text{C}\) (absolute zero).
(a)(ii) Convert both temperatures to kelvin:
\[T_1 = 27 + 273 = 300\ \text{K}, \qquad T_2 = 40 + 273 = 313\ \text{K}, \qquad V_1 = 500\ \text{cm}^3\]At constant pressure, Charles' law gives:
\[\frac{V_1}{T_1} = \frac{V_2}{T_2} \quad\Rightarrow\quad \frac{500}{300} = \frac{V_2}{313}\] \[V_2 = \frac{500 \times 313}{300} = 521.7 \approx 522\ \text{cm}^3\]The gas occupies 522 cm\(^3\) at 40 \(^{\circ}\)C.
(b) Two gases used as refrigerants:
- Ammonia (NH\(_3\))
- Carbon dioxide (CO\(_2\))
Detalhes da Resposta
(a)(i) Charles' law states that at constant pressure the volume of a fixed mass of gas is directly proportional to its absolute (kelvin) temperature. A graph of volume against absolute temperature is therefore a straight line passing through the origin:
The line is straight and, when produced, passes through the origin (0 K, 0 cm\(^3\)), showing \(V \propto T\) at constant pressure. If the same volume is plotted against temperature in \(^{\circ}\)C, the straight line is displaced and cuts the temperature axis at \(-273\ ^{\circ}\text{C}\) (absolute zero).
(a)(ii) Convert both temperatures to kelvin:
\[T_1 = 27 + 273 = 300\ \text{K}, \qquad T_2 = 40 + 273 = 313\ \text{K}, \qquad V_1 = 500\ \text{cm}^3\]At constant pressure, Charles' law gives:
\[\frac{V_1}{T_1} = \frac{V_2}{T_2} \quad\Rightarrow\quad \frac{500}{300} = \frac{V_2}{313}\] \[V_2 = \frac{500 \times 313}{300} = 521.7 \approx 522\ \text{cm}^3\]The gas occupies 522 cm\(^3\) at 40 \(^{\circ}\)C.
(b) Two gases used as refrigerants:
- Ammonia (NH\(_3\))
- Carbon dioxide (CO\(_2\))
Pergunta 6 Relatório
(a) Mention one process apart from respiration, which increases the amount of carbon (IV) oxide in the atmosphere.
(b)(i) State one use of sodium hydrogentrioxocarbonate (IV).
(ii) Write an equation to show the action of heat on sodium hydrogentrioxocarbonate(IV).
(a) Process (other than respiration) that increases atmospheric \(CO_2\)
Combustion of fuels (the burning of wood, coal, petroleum and other fossil fuels). The decay of dead organic matter by bacteria is also acceptable.
(b)(i) Use of sodium hydrogentrioxocarbonate(IV), \(NaHCO_3\)
It is used as baking powder in bread and cake making (it releases \(CO_2\) which makes the dough rise). It is also used as an antacid to relieve stomach acidity and as a component of dry-powder fire extinguishers.
(b)(ii) Action of heat on \(NaHCO_3\)
On heating it decomposes to give sodium trioxocarbonate(IV), water and carbon(IV) oxide:
\[2NaHCO_3 \xrightarrow{\text{heat}} Na_2CO_3 + H_2O + CO_2\]Detalhes da Resposta
(a) Process (other than respiration) that increases atmospheric \(CO_2\)
Combustion of fuels (the burning of wood, coal, petroleum and other fossil fuels). The decay of dead organic matter by bacteria is also acceptable.
(b)(i) Use of sodium hydrogentrioxocarbonate(IV), \(NaHCO_3\)
It is used as baking powder in bread and cake making (it releases \(CO_2\) which makes the dough rise). It is also used as an antacid to relieve stomach acidity and as a component of dry-powder fire extinguishers.
(b)(ii) Action of heat on \(NaHCO_3\)
On heating it decomposes to give sodium trioxocarbonate(IV), water and carbon(IV) oxide:
\[2NaHCO_3 \xrightarrow{\text{heat}} Na_2CO_3 + H_2O + CO_2\]Pergunta 7 Relatório
(a) Name the device used for producing an electric current from a chemical
(b) Copy and complete the table below.
| Electrolyte | Product at the anode (carbon) | Product at the cathode (carbon) |
| Dilute \( \mathrm{NaCl}_{(aq)} \) Concentrated \( \mathrm{NaCI}_{(aq)} \) |
(a) The device used for producing an electric current from a chemical reaction is an electrochemical cell (also called a galvanic or voltaic cell, or simply a chemical cell).
(b) Completed table
Reasoning:
- In dilute \( NaCl_{(aq)} \) the solution is mostly water. At the anode, \( OH^- \) is discharged in preference to the dilute \( Cl^- \), giving oxygen. At the cathode, \( H^+ \) is discharged in preference to \( Na^+ \) (sodium is high in the electrochemical series), giving hydrogen.
- In concentrated \( NaCl_{(aq)} \) the high concentration of \( Cl^- \) causes chloride to be discharged at the anode instead of \( OH^- \), giving chlorine. Hydrogen is still liberated at the cathode.
| Electrolyte | Product at the anode (carbon) | Product at the cathode (carbon) |
|---|---|---|
| Dilute \( NaCl_{(aq)} \) | Oxygen, \( O_2 \) | Hydrogen, \( H_2 \) |
| Concentrated \( NaCl_{(aq)} \) | Chlorine, \( Cl_2 \) | Hydrogen, \( H_2 \) |
Electrode half-equations: at the cathode \( 2H^+ + 2e^- \rightarrow H_2 \); at the dilute anode \( 4OH^- \rightarrow O_2 + 2H_2O + 4e^- \); at the concentrated anode \( 2Cl^- \rightarrow Cl_2 + 2e^- \).
Detalhes da Resposta
(a) The device used for producing an electric current from a chemical reaction is an electrochemical cell (also called a galvanic or voltaic cell, or simply a chemical cell).
(b) Completed table
Reasoning:
- In dilute \( NaCl_{(aq)} \) the solution is mostly water. At the anode, \( OH^- \) is discharged in preference to the dilute \( Cl^- \), giving oxygen. At the cathode, \( H^+ \) is discharged in preference to \( Na^+ \) (sodium is high in the electrochemical series), giving hydrogen.
- In concentrated \( NaCl_{(aq)} \) the high concentration of \( Cl^- \) causes chloride to be discharged at the anode instead of \( OH^- \), giving chlorine. Hydrogen is still liberated at the cathode.
| Electrolyte | Product at the anode (carbon) | Product at the cathode (carbon) |
|---|---|---|
| Dilute \( NaCl_{(aq)} \) | Oxygen, \( O_2 \) | Hydrogen, \( H_2 \) |
| Concentrated \( NaCl_{(aq)} \) | Chlorine, \( Cl_2 \) | Hydrogen, \( H_2 \) |
Electrode half-equations: at the cathode \( 2H^+ + 2e^- \rightarrow H_2 \); at the dilute anode \( 4OH^- \rightarrow O_2 + 2H_2O + 4e^- \); at the concentrated anode \( 2Cl^- \rightarrow Cl_2 + 2e^- \).
Pergunta 8 Relatório
(a) State the type of reaction involved in the conversion of:
(i) proteins.to amino acids:
(ii) ethanol to ethene;
(iii) benzene to bromobenzene
(b) Write an equation to show that ethene reasts with hydrogen in the presence of finely divided nickel.
(a)
- (i) Proteins to amino acids: hydrolysis.
- (ii) Ethanol to ethene: dehydration (an elimination reaction).
- (iii) Benzene to bromobenzene: substitution (halogenation).
(b) Ethene reacts with hydrogen over finely divided nickel to give ethane:
\[C_2H_4 + H_2 \xrightarrow{Ni} C_2H_6\]
Detalhes da Resposta
(a)
- (i) Proteins to amino acids: hydrolysis.
- (ii) Ethanol to ethene: dehydration (an elimination reaction).
- (iii) Benzene to bromobenzene: substitution (halogenation).
(b) Ethene reacts with hydrogen over finely divided nickel to give ethane:
\[C_2H_4 + H_2 \xrightarrow{Ni} C_2H_6\]
Pergunta 9 Relatório
(a) List two differences between solids and liquids.
(b) The graph below is the heating curve for a solid X. Use the graph to answer Questions (i) — (iii) below.
(i) What is the melting point of X?
(ii) If the vapour of X is cooled, at what temperature will it start to condense?
(iii) (I) As X is heated, state what happens to the: I. frequency of collision of molecules of X;
(II) value of the entropy of the system.
(a) Two differences between solids and liquids
| Solids | Liquids |
|---|---|
| Have a definite (fixed) shape of their own. | Have no definite shape; they take the shape of the container. |
| Particles are closely packed and held in fixed positions, so solids are almost incompressible and do not flow. | Particles are more loosely held and can slide over one another, so liquids flow. |
(b) The heating curve for solid X is shown below. Each horizontal (flat) portion is a change of state, where the added heat is used to break the forces between particles instead of raising the temperature.
(i) Melting point of X
The melting point is the temperature of the first flat portion of the curve, where the solid is turning into liquid at constant temperature.
\[ \text{Melting point of X} = 60\,^{\circ}\text{C} \](ii) Temperature at which the vapour of X starts to condense
Condensation is the reverse of boiling, so the vapour condenses at the boiling point of X. This is the temperature of the second flat portion of the curve.
\[ \text{Condensation temperature} = 210\,^{\circ}\text{C} \](iii) As X is heated:
(I) Frequency of collision of molecules of X: it increases. Heating gives the molecules more kinetic energy, so they move faster and collide with one another more often.
(II) Value of the entropy of the system: it increases. As X changes from solid to liquid to gas the particles become more disordered and more freely arranged, so the entropy of the system rises.
Detalhes da Resposta
(a) Two differences between solids and liquids
| Solids | Liquids |
|---|---|
| Have a definite (fixed) shape of their own. | Have no definite shape; they take the shape of the container. |
| Particles are closely packed and held in fixed positions, so solids are almost incompressible and do not flow. | Particles are more loosely held and can slide over one another, so liquids flow. |
(b) The heating curve for solid X is shown below. Each horizontal (flat) portion is a change of state, where the added heat is used to break the forces between particles instead of raising the temperature.
(i) Melting point of X
The melting point is the temperature of the first flat portion of the curve, where the solid is turning into liquid at constant temperature.
\[ \text{Melting point of X} = 60\,^{\circ}\text{C} \](ii) Temperature at which the vapour of X starts to condense
Condensation is the reverse of boiling, so the vapour condenses at the boiling point of X. This is the temperature of the second flat portion of the curve.
\[ \text{Condensation temperature} = 210\,^{\circ}\text{C} \](iii) As X is heated:
(I) Frequency of collision of molecules of X: it increases. Heating gives the molecules more kinetic energy, so they move faster and collide with one another more often.
(II) Value of the entropy of the system: it increases. As X changes from solid to liquid to gas the particles become more disordered and more freely arranged, so the entropy of the system rises.
Pergunta 10 Relatório
(a) State three reasons why air is classified as a mixture.
(b) List two methods that can be used to separate a mixture of iodine crystals and iron filings.
(a) Air is classified as a mixture because:
- Its components can be separated by physical means (fractional distillation of liquid air).
- Its components retain their own individual properties.
- Its composition is variable (the proportions are not fixed) and the components are not combined in any fixed ratio, and no chemical reaction or energy change occurs when they are together.
(b) A mixture of iodine crystals and iron filings can be separated by:
- Using a magnet to remove the iron filings (magnetic separation).
- Sublimation - warming the mixture so the iodine sublimes and is collected, leaving the iron filings behind.
Detalhes da Resposta
(a) Air is classified as a mixture because:
- Its components can be separated by physical means (fractional distillation of liquid air).
- Its components retain their own individual properties.
- Its composition is variable (the proportions are not fixed) and the components are not combined in any fixed ratio, and no chemical reaction or energy change occurs when they are together.
(b) A mixture of iodine crystals and iron filings can be separated by:
- Using a magnet to remove the iron filings (magnetic separation).
- Sublimation - warming the mixture so the iodine sublimes and is collected, leaving the iron filings behind.
Pergunta 11 Relatório
(a)(i) List the three types of particles present in atoms.
(ii) name the element which does not contain all the three particles in its atom. Mention the particle that is not present.
(b) Give the reason why:
(i) the relative atomic masses of some elements are not whole number;
(ii) relative atomic masses are used instead of the actual masses of atoms in grams;
(iii) metals are good conductors of electricity.
(c)(i) Name the type of bond present in the oxonium ion,
(ii) State one effect of the existence of intermolecular hydrogen bonding on the physical properties of ethanol.
(d)(i) Explain what is meant by water of crystallization.
(ii) When 5.0g of a compound Y was heated to constant mass, 1.8g of water vapour was given off. Determine the number of molecules of water of crystallization in one molecule of Y, given that the molar mass of its anhydrous form is 160g. [H = 1, 0 = 16]
(a)(i) Three types of particles in atoms
- Protons (positively charged, in the nucleus)
- Neutrons (uncharged, in the nucleus)
- Electrons (negatively charged, in the shells)
(a)(ii) The element is hydrogen (the ordinary isotope, \(^{1}_{1}H\)). The particle absent is the neutron (its atom has 1 proton and 1 electron only).
(b) Reasons
- (i) The relative atomic masses of some elements are not whole numbers because such elements exist as a mixture of isotopes; the relative atomic mass is the weighted average of the masses of the isotopes according to their relative abundances.
- (ii) Relative atomic masses are used instead of actual masses in grams because the actual mass of a single atom is extremely small (of the order of \(10^{-23}\) g) and inconvenient to handle, whereas relative values are simple comparative numbers.
- (iii) Metals are good conductors of electricity because they contain mobile (delocalised) free electrons that drift through the metallic lattice and carry charge.
(c)(i) The oxonium ion \(H_3O^+\) contains ordinary covalent bonds and one coordinate (dative) covalent bond (the oxygen lone pair is donated to the extra proton).
(c)(ii) Intermolecular hydrogen bonding raises the boiling point of ethanol (it boils much higher than expected for its molecular mass because extra energy is needed to break the hydrogen bonds). It also makes ethanol fully miscible with water.
(d)(i) Water of crystallisation is the fixed number of water molecules chemically combined with one formula unit of a salt in its crystal structure.
(d)(ii) Mass of anhydrous salt \(= 5.0 - 1.8 = 3.2\) g.
\[n(\text{anhydrous}) = \frac{3.2}{160} = 0.02\ \text{mol}\]\[n(H_2O) = \frac{1.8}{18} = 0.10\ \text{mol}\]\[\text{ratio } H_2O : \text{salt} = \frac{0.10}{0.02} = 5\]Therefore there are 5 molecules of water of crystallisation in one molecule of Y.
Detalhes da Resposta
(a)(i) Three types of particles in atoms
- Protons (positively charged, in the nucleus)
- Neutrons (uncharged, in the nucleus)
- Electrons (negatively charged, in the shells)
(a)(ii) The element is hydrogen (the ordinary isotope, \(^{1}_{1}H\)). The particle absent is the neutron (its atom has 1 proton and 1 electron only).
(b) Reasons
- (i) The relative atomic masses of some elements are not whole numbers because such elements exist as a mixture of isotopes; the relative atomic mass is the weighted average of the masses of the isotopes according to their relative abundances.
- (ii) Relative atomic masses are used instead of actual masses in grams because the actual mass of a single atom is extremely small (of the order of \(10^{-23}\) g) and inconvenient to handle, whereas relative values are simple comparative numbers.
- (iii) Metals are good conductors of electricity because they contain mobile (delocalised) free electrons that drift through the metallic lattice and carry charge.
(c)(i) The oxonium ion \(H_3O^+\) contains ordinary covalent bonds and one coordinate (dative) covalent bond (the oxygen lone pair is donated to the extra proton).
(c)(ii) Intermolecular hydrogen bonding raises the boiling point of ethanol (it boils much higher than expected for its molecular mass because extra energy is needed to break the hydrogen bonds). It also makes ethanol fully miscible with water.
(d)(i) Water of crystallisation is the fixed number of water molecules chemically combined with one formula unit of a salt in its crystal structure.
(d)(ii) Mass of anhydrous salt \(= 5.0 - 1.8 = 3.2\) g.
\[n(\text{anhydrous}) = \frac{3.2}{160} = 0.02\ \text{mol}\]\[n(H_2O) = \frac{1.8}{18} = 0.10\ \text{mol}\]\[\text{ratio } H_2O : \text{salt} = \frac{0.10}{0.02} = 5\]Therefore there are 5 molecules of water of crystallisation in one molecule of Y.
Pergunta 12 Relatório
(a) What is the change in oxidation state of chromium in the reaction represented by the following equation?
3SO\(_2\) + Cr\(_2\)O\(^2_{-7}\) + 2H\(^+\) -> 3SO\(_4^{2-}\) + 2Cr\(^{3+}\) + H\(_2\)O
(b) Use the half equations given below to deduce the equation for the reaction between iron(II) ions and heptaoxodichromate (VI) ions in acidic solution.
Fe\(^{2+}\) --> Fe\(^{3+}\) + e\(^-\)
Cr\(_2\)O\(^{2-}_7\) + 14H\(^+\) + 6e\(^-\) ----> 2Cr\(^{3+}\) + 7H\(_2\)O.
(a) Change in oxidation state of chromium
In the heptaoxodichromate(VI) ion \(Cr_2O_7^{2-}\), each chromium has an oxidation number of +6 (oxygen is \(-2\): \(2x + 7(-2) = -2\Rightarrow x = +6\)).
In the product \(Cr^{3+}\), chromium is +3.
The oxidation state therefore decreases from +6 to +3, a change of 3 units. Chromium is reduced (each Cr gains 3 electrons).
(b) Deriving the overall equation
To cancel electrons, multiply the iron half-equation by 6 so that 6 electrons are supplied to match the 6 electrons gained by the dichromate:
\[6Fe^{2+} \to 6Fe^{3+} + 6e^-\]\[Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O\]Adding and cancelling the electrons gives the overall equation:
\[6Fe^{2+} + Cr_2O_7^{2-} + 14H^+ \to 6Fe^{3+} + 2Cr^{3+} + 7H_2O\]Detalhes da Resposta
(a) Change in oxidation state of chromium
In the heptaoxodichromate(VI) ion \(Cr_2O_7^{2-}\), each chromium has an oxidation number of +6 (oxygen is \(-2\): \(2x + 7(-2) = -2\Rightarrow x = +6\)).
In the product \(Cr^{3+}\), chromium is +3.
The oxidation state therefore decreases from +6 to +3, a change of 3 units. Chromium is reduced (each Cr gains 3 electrons).
(b) Deriving the overall equation
To cancel electrons, multiply the iron half-equation by 6 so that 6 electrons are supplied to match the 6 electrons gained by the dichromate:
\[6Fe^{2+} \to 6Fe^{3+} + 6e^-\]\[Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O\]Adding and cancelling the electrons gives the overall equation:
\[6Fe^{2+} + Cr_2O_7^{2-} + 14H^+ \to 6Fe^{3+} + 2Cr^{3+} + 7H_2O\]Pergunta 13 Relatório
(a)(i) What is isomerism?
(ii) Name the alkanol that is isomeric with methoxymethane (CH\(_3\)OCH\(_3\)).
(b)(i) Outline the laboratory:preparation of ethylethanoate. (Diagrams not required)
(ii) Write the structural formula of ethylethancqte
(iii) State two physical properties of ethylethanoate.
(c) When gas oil which consists of larger hydrocarbons was subjected to high temperature and pressure, the following reaction occurred.
C\(_{17}\)H\(_{36(l)}\) \(\to\) 3C\(_2\)H\(_{4(g)}\) + C\(_3\)H\(_{6(g)}\) + Q\(_{(l)}\)
(i) What name is given to the process indicated above?
(ii) State the importance of the process to the petroleum industry.
(iii) Find the formula of the product which Q represents in the equation above.
(iv) Mention one type of chemical industry that utilizes ethene as raw material.
(d) Consider the following compounds: CH\(_3\) — (CH\(_2\))\(_2\) —CH\(_3\); C\(_6\)H\(_5\) —CH = CH\(_2\); CH=C — CH\(_3\). State which of them:
(i) is used as a domestic fuel;
(ii) is an aromatic compound,
(iii) participates in such situation but not addition reactions;
(iv) would react with two moles of hydrogen per mole.
(a)(i) Isomerism is the existence of two or more compounds having the same molecular formula but different structural arrangements of their atoms (and hence different properties).
(a)(ii) Methoxymethane \(CH_3OCH_3\) has molecular formula \(C_2H_6O\). The isomeric alkanol is ethanol, \(C_2H_5OH\).
(b)(i) Laboratory preparation of ethyl ethanoate
- Warm a mixture of ethanoic acid and ethanol together with a little concentrated tetraoxosulphate(VI) acid (which acts as catalyst and dehydrating agent) under reflux.
- Esterification occurs: \[CH_3COOH + C_2H_5OH \underset{\Delta}{\overset{H_2SO_4}{\rightleftharpoons}} CH_3COOC_2H_5 + H_2O\]
- Distil off the ester and purify it by shaking with sodium trioxocarbonate(IV) solution (to remove acid) followed by drying and redistillation.
(b)(ii) Structural formula: \(CH_3-CO-O-CH_2-CH_3\) (i.e. \(CH_3COOC_2H_5\)).
(b)(iii) Two physical properties: it is a colourless liquid with a sweet, fruity smell; it is only slightly soluble in water but volatile.
(c) \(C_{17}H_{36} \to 3C_2H_4 + C_3H_6 + Q\)
- (i) The process is cracking (thermal/catalytic cracking).
- (ii) It is important because it breaks large, less useful hydrocarbon molecules into smaller, more useful ones (petrol and gaseous alkenes), increasing the yield of petrol and supplying alkenes as raw materials.
- (iii) Balancing carbon: \(17 = (3\times2) + 3 + C_Q \Rightarrow C_Q = 8\). Balancing hydrogen: \(36 = (3\times4) + 6 + H_Q \Rightarrow H_Q = 18\). Therefore Q is \(C_8H_{18}\) (octane).
- (iv) The plastics (polymer) industry, which polymerises ethene to poly(ethene). The manufacture of ethanol is also acceptable.
(d) The compounds are butane \(CH_3(CH_2)_2CH_3\), phenylethene \(C_6H_5CH=CH_2\), and propyne \(CH\equiv C-CH_3\).
- (i) Domestic fuel: butane, \(CH_3(CH_2)_2CH_3\).
- (ii) Aromatic compound: phenylethene (styrene), \(C_6H_5CH=CH_2\).
- (iii) Undergoes substitution but not addition: butane (a saturated alkane).
- (iv) Reacts with two moles of hydrogen per mole: propyne, \(CH\equiv C-CH_3\) (the triple bond adds \(2H_2\)).
Detalhes da Resposta
(a)(i) Isomerism is the existence of two or more compounds having the same molecular formula but different structural arrangements of their atoms (and hence different properties).
(a)(ii) Methoxymethane \(CH_3OCH_3\) has molecular formula \(C_2H_6O\). The isomeric alkanol is ethanol, \(C_2H_5OH\).
(b)(i) Laboratory preparation of ethyl ethanoate
- Warm a mixture of ethanoic acid and ethanol together with a little concentrated tetraoxosulphate(VI) acid (which acts as catalyst and dehydrating agent) under reflux.
- Esterification occurs: \[CH_3COOH + C_2H_5OH \underset{\Delta}{\overset{H_2SO_4}{\rightleftharpoons}} CH_3COOC_2H_5 + H_2O\]
- Distil off the ester and purify it by shaking with sodium trioxocarbonate(IV) solution (to remove acid) followed by drying and redistillation.
(b)(ii) Structural formula: \(CH_3-CO-O-CH_2-CH_3\) (i.e. \(CH_3COOC_2H_5\)).
(b)(iii) Two physical properties: it is a colourless liquid with a sweet, fruity smell; it is only slightly soluble in water but volatile.
(c) \(C_{17}H_{36} \to 3C_2H_4 + C_3H_6 + Q\)
- (i) The process is cracking (thermal/catalytic cracking).
- (ii) It is important because it breaks large, less useful hydrocarbon molecules into smaller, more useful ones (petrol and gaseous alkenes), increasing the yield of petrol and supplying alkenes as raw materials.
- (iii) Balancing carbon: \(17 = (3\times2) + 3 + C_Q \Rightarrow C_Q = 8\). Balancing hydrogen: \(36 = (3\times4) + 6 + H_Q \Rightarrow H_Q = 18\). Therefore Q is \(C_8H_{18}\) (octane).
- (iv) The plastics (polymer) industry, which polymerises ethene to poly(ethene). The manufacture of ethanol is also acceptable.
(d) The compounds are butane \(CH_3(CH_2)_2CH_3\), phenylethene \(C_6H_5CH=CH_2\), and propyne \(CH\equiv C-CH_3\).
- (i) Domestic fuel: butane, \(CH_3(CH_2)_2CH_3\).
- (ii) Aromatic compound: phenylethene (styrene), \(C_6H_5CH=CH_2\).
- (iii) Undergoes substitution but not addition: butane (a saturated alkane).
- (iv) Reacts with two moles of hydrogen per mole: propyne, \(CH\equiv C-CH_3\) (the triple bond adds \(2H_2\)).
Pergunta 14 Relatório
(a) Mention the chemical substance manufactured starting from each of the foirownc sets of materials:
(i) sugar and yeast;
(ii) ammonia, air and water;
(iii) vegetable oil and caustic alkali.
(b) State one air pollutant generated during the manufacture of fertilizers.
(a) Substances manufactured
- (i) Sugar and yeast: Ethanol (ethyl alcohol). The yeast supplies the enzyme zymase which ferments the sugar: \[C_6H_{12}O_{6} \xrightarrow{\text{zymase}} 2C_2H_5OH + 2CO_2\]
- (ii) Ammonia, air and water: Trioxonitrate(V) acid (nitric acid, \(HNO_3\)). In the Ostwald process ammonia is catalytically oxidised to nitrogen(II) oxide, further oxidised by air to nitrogen(IV) oxide, then dissolved in water: \[4NH_3 + 5O_2 \to 4NO + 6H_2O\]\[2NO + O_2 \to 2NO_2\]\[3NO_2 + H_2O \to 2HNO_3 + NO\]
- (iii) Vegetable oil and caustic alkali: Soap (by saponification, the alkaline hydrolysis of the oil).
(b) Air pollutant from fertilizer manufacture
One example is sulphur(IV) oxide, \(SO_2\) (released where sulphuric acid is used to make ammonium tetraoxosulphate(VI) fertilizer). Oxides of nitrogen or escaped ammonia are also acceptable.
Detalhes da Resposta
(a) Substances manufactured
- (i) Sugar and yeast: Ethanol (ethyl alcohol). The yeast supplies the enzyme zymase which ferments the sugar: \[C_6H_{12}O_{6} \xrightarrow{\text{zymase}} 2C_2H_5OH + 2CO_2\]
- (ii) Ammonia, air and water: Trioxonitrate(V) acid (nitric acid, \(HNO_3\)). In the Ostwald process ammonia is catalytically oxidised to nitrogen(II) oxide, further oxidised by air to nitrogen(IV) oxide, then dissolved in water: \[4NH_3 + 5O_2 \to 4NO + 6H_2O\]\[2NO + O_2 \to 2NO_2\]\[3NO_2 + H_2O \to 2HNO_3 + NO\]
- (iii) Vegetable oil and caustic alkali: Soap (by saponification, the alkaline hydrolysis of the oil).
(b) Air pollutant from fertilizer manufacture
One example is sulphur(IV) oxide, \(SO_2\) (released where sulphuric acid is used to make ammonium tetraoxosulphate(VI) fertilizer). Oxides of nitrogen or escaped ammonia are also acceptable.
