WAEC SSCE - Chemistry - 1989 (Essay)
Question 1 Report
(a) Write the formula of the oxide of nitrogen in which nitrogen has oxidation number of (i) +1;
(ii) +2;
(iii) +3;
(iv) +4;
(b) State which of the oxides in (a) above is/are: (i) acidic (ii) neutral.
(a) Oxides of nitrogen by oxidation number of nitrogen:
- (i) +1: N\(_2\)O (dinitrogen oxide).
- (ii) +2: NO (nitrogen(II) oxide / nitrogen monoxide).
- (iii) +3: N\(_2\)O\(_3\) (dinitrogen trioxide).
- (iv) +4: NO\(_2\) (nitrogen(IV) oxide / nitrogen dioxide).
(b) Classification:
- (i) Acidic: N\(_2\)O\(_3\) and NO\(_2\) (they react with water/alkalis to form acids or salts).
- (ii) Neutral: N\(_2\)O and NO (they are neither acidic nor basic).
Answer Details
(a) Oxides of nitrogen by oxidation number of nitrogen:
- (i) +1: N\(_2\)O (dinitrogen oxide).
- (ii) +2: NO (nitrogen(II) oxide / nitrogen monoxide).
- (iii) +3: N\(_2\)O\(_3\) (dinitrogen trioxide).
- (iv) +4: NO\(_2\) (nitrogen(IV) oxide / nitrogen dioxide).
(b) Classification:
- (i) Acidic: N\(_2\)O\(_3\) and NO\(_2\) (they react with water/alkalis to form acids or salts).
- (ii) Neutral: N\(_2\)O and NO (they are neither acidic nor basic).
Question 2 Report
(a) Name the gas evolved when dilute hydrochloric acid is added to each of the following solids and the mixture is warmed
(i) sodium trioxocarbonate (IV);
(ii) potassium trioxosulphate (IV);
(iii) iron (II) sulphide.
(b) Name the reagent(s) that you would use to identify the gas in (a)(ii) above.
(a) Gas evolved on warming each solid with dilute hydrochloric acid:
- (i) Sodium trioxocarbonate(IV), Na\(_2\)CO\(_3\): carbon dioxide, CO\(_2\). \[Na_2CO_3 + 2HCl \to 2NaCl + H_2O + CO_2\uparrow\]
- (ii) Potassium trioxosulphate(IV), K\(_2\)SO\(_3\): sulphur(IV) oxide (sulphur dioxide), SO\(_2\). \[K_2SO_3 + 2HCl \to 2KCl + H_2O + SO_2\uparrow\]
- (iii) Iron(II) sulphide, FeS: hydrogen sulphide, H\(_2\)S. \[FeS + 2HCl \to FeCl_2 + H_2S\uparrow\]
(b) Reagent to identify the gas in (a)(ii), SO\(_2\): acidified potassium tetraoxomanganate(VII) solution, which is decolourised (purple to colourless) by the sulphur dioxide; alternatively acidified potassium dichromate(VI), which turns from orange to green.
Answer Details
(a) Gas evolved on warming each solid with dilute hydrochloric acid:
- (i) Sodium trioxocarbonate(IV), Na\(_2\)CO\(_3\): carbon dioxide, CO\(_2\). \[Na_2CO_3 + 2HCl \to 2NaCl + H_2O + CO_2\uparrow\]
- (ii) Potassium trioxosulphate(IV), K\(_2\)SO\(_3\): sulphur(IV) oxide (sulphur dioxide), SO\(_2\). \[K_2SO_3 + 2HCl \to 2KCl + H_2O + SO_2\uparrow\]
- (iii) Iron(II) sulphide, FeS: hydrogen sulphide, H\(_2\)S. \[FeS + 2HCl \to FeCl_2 + H_2S\uparrow\]
(b) Reagent to identify the gas in (a)(ii), SO\(_2\): acidified potassium tetraoxomanganate(VII) solution, which is decolourised (purple to colourless) by the sulphur dioxide; alternatively acidified potassium dichromate(VI), which turns from orange to green.
Question 3 Report
Name the type of chemical process involved in the production of
(a) polythene from ethene
(b) ethene from kerosene fraction of petroleum.
(c) soap from vegetable oils
(d) margarine from vegetable oils
Type of chemical process involved:
- (a) Polythene from ethene: addition polymerization.
- (b) Ethene from the kerosene fraction of petroleum: cracking (thermal or catalytic cracking).
- (c) Soap from vegetable oils: saponification (alkaline hydrolysis of the oil).
- (d) Margarine from vegetable oils: hydrogenation (addition of hydrogen to the unsaturated oil).
Answer Details
Type of chemical process involved:
- (a) Polythene from ethene: addition polymerization.
- (b) Ethene from the kerosene fraction of petroleum: cracking (thermal or catalytic cracking).
- (c) Soap from vegetable oils: saponification (alkaline hydrolysis of the oil).
- (d) Margarine from vegetable oils: hydrogenation (addition of hydrogen to the unsaturated oil).
Question 4 Report
(a) Give two characteristic features of boiling
(b) What will be the effect of the following on the boiling point of water:
(i) addition of crystals of sodium chloride,
(ii) reduction of the atmospheric pressure?
(c) State two ways in which boiling differs from evaporation.
(a) Two characteristic features of boiling: it takes place at a definite, fixed temperature (the boiling point) for a given external pressure, and the temperature stays constant while boiling continues; bubbles of vapour form and rise throughout the whole (bulk) of the liquid.
(b) Effect on the boiling point of water:
- (i) Addition of sodium chloride crystals: the boiling point is raised (the dissolved salt causes an elevation of boiling point).
- (ii) Reduction of the atmospheric pressure: the boiling point is lowered (water boils at a lower temperature when the pressure above it is reduced).
(c) Two ways boiling differs from evaporation: boiling occurs only at a fixed temperature (the boiling point), whereas evaporation occurs at all temperatures below the boiling point; boiling takes place throughout the whole body of the liquid (with bubble formation), whereas evaporation takes place only at the surface of the liquid.
Answer Details
(a) Two characteristic features of boiling: it takes place at a definite, fixed temperature (the boiling point) for a given external pressure, and the temperature stays constant while boiling continues; bubbles of vapour form and rise throughout the whole (bulk) of the liquid.
(b) Effect on the boiling point of water:
- (i) Addition of sodium chloride crystals: the boiling point is raised (the dissolved salt causes an elevation of boiling point).
- (ii) Reduction of the atmospheric pressure: the boiling point is lowered (water boils at a lower temperature when the pressure above it is reduced).
(c) Two ways boiling differs from evaporation: boiling occurs only at a fixed temperature (the boiling point), whereas evaporation occurs at all temperatures below the boiling point; boiling takes place throughout the whole body of the liquid (with bubble formation), whereas evaporation takes place only at the surface of the liquid.
Question 5 Report
(a) Name the components of
(i) producer gas; (ii) water gas;
(b) Give the reason why water is a better fuel than producer gas.
(a) Components
- (i) Producer gas: carbon(II) oxide (CO) and nitrogen (N2), together with a small amount of carbon(IV) oxide (CO2). It is made by passing air over red-hot coke.
- (ii) Water gas: carbon(II) oxide (CO) and hydrogen (H2). It is made by passing steam over red-hot coke: \[C_{(s)} + H_2O_{(g)} \rightarrow CO_{(g)} + H_{2(g)}\]
(b) Why water gas is the better fuel
Water gas is a better fuel than producer gas because both of its components, carbon(II) oxide and hydrogen, are combustible and release heat when burnt. Producer gas, on the other hand, contains a large proportion (about half) of nitrogen, which is chemically inert and does not burn. This non-combustible nitrogen dilutes the fuel and carries away heat, so producer gas has a much lower calorific (heat) value than water gas.
Answer Details
(a) Components
- (i) Producer gas: carbon(II) oxide (CO) and nitrogen (N2), together with a small amount of carbon(IV) oxide (CO2). It is made by passing air over red-hot coke.
- (ii) Water gas: carbon(II) oxide (CO) and hydrogen (H2). It is made by passing steam over red-hot coke: \[C_{(s)} + H_2O_{(g)} \rightarrow CO_{(g)} + H_{2(g)}\]
(b) Why water gas is the better fuel
Water gas is a better fuel than producer gas because both of its components, carbon(II) oxide and hydrogen, are combustible and release heat when burnt. Producer gas, on the other hand, contains a large proportion (about half) of nitrogen, which is chemically inert and does not burn. This non-combustible nitrogen dilutes the fuel and carries away heat, so producer gas has a much lower calorific (heat) value than water gas.
Question 6 Report
(a) Write the chemical equation for the formation of named alkanoate.
(b)(i) What are the monomers of protein called?
(ii) Write the two functional groups present in the monomers named in (b)(i) above
(iii) State the type of reaction that leads to the formation of proteins from their monomers.
(a) Formation of an alkanoate (ester)
An alkanoate is formed when an alkanoic acid reacts with an alkanol in the presence of a little concentrated tetraoxosulphate(VI) acid as catalyst (esterification):
\[CH_3COOH + C_2H_5OH \underset{}{\overset{H_2SO_4}{\rightleftharpoons}} CH_3COOC_2H_5 + H_2O\]
(ethanoic acid + ethanol give ethyl ethanoate and water).
(b) Proteins
- (i) The monomers of protein are called amino acids (specifically 2-amino acids / alpha-amino acids).
- (ii) Two functional groups present in the monomers: the amino group (\(-NH_2\)) and the carboxyl group (\(-COOH\)).
- (iii) The reaction that joins the amino acids into proteins is condensation polymerization (forming peptide/amide linkages with the elimination of water molecules).
Answer Details
(a) Formation of an alkanoate (ester)
An alkanoate is formed when an alkanoic acid reacts with an alkanol in the presence of a little concentrated tetraoxosulphate(VI) acid as catalyst (esterification):
\[CH_3COOH + C_2H_5OH \underset{}{\overset{H_2SO_4}{\rightleftharpoons}} CH_3COOC_2H_5 + H_2O\]
(ethanoic acid + ethanol give ethyl ethanoate and water).
(b) Proteins
- (i) The monomers of protein are called amino acids (specifically 2-amino acids / alpha-amino acids).
- (ii) Two functional groups present in the monomers: the amino group (\(-NH_2\)) and the carboxyl group (\(-COOH\)).
- (iii) The reaction that joins the amino acids into proteins is condensation polymerization (forming peptide/amide linkages with the elimination of water molecules).
Question 7 Report
Consider the compounds represented as A and B below:
(a) What is the celationship between A and B? (
b) Name A and B.
(c) Will the chemical properties of A and B be the same? Give one reason for your answer.
Both A and B have the same molecular formula, C2H2Cl2, with one H and one Cl on each doubly bonded carbon. They differ only in how the two chlorine atoms are arranged about the rigid C=C double bond.
- A: the two Cl atoms are on opposite sides of the double bond (one up, one down).
- B: the two Cl atoms are on the same side of the double bond (both down).
(a) Relationship between A and B
They are geometric (cis-trans) isomers, a form of stereoisomerism. Restricted rotation about the C=C double bond fixes the positions of the substituents, so the two arrangements are distinct compounds with the same structural formula.
(b) Names
- A = trans-1,2-dichloroethene (Cl atoms on opposite sides).
- B = cis-1,2-dichloroethene (Cl atoms on the same side).
(c) Are the chemical properties the same?
Yes, A and B have essentially the same chemical properties. Reason: they possess the same functional group (a C=C double bond) and the same number and type of atoms and bonds, so they undergo the same chemical reactions. Their differences are mainly physical (for example melting point, boiling point and polarity: the cis form is polar while the trans form is non-polar), not chemical.
Answer Details
Both A and B have the same molecular formula, C2H2Cl2, with one H and one Cl on each doubly bonded carbon. They differ only in how the two chlorine atoms are arranged about the rigid C=C double bond.
- A: the two Cl atoms are on opposite sides of the double bond (one up, one down).
- B: the two Cl atoms are on the same side of the double bond (both down).
(a) Relationship between A and B
They are geometric (cis-trans) isomers, a form of stereoisomerism. Restricted rotation about the C=C double bond fixes the positions of the substituents, so the two arrangements are distinct compounds with the same structural formula.
(b) Names
- A = trans-1,2-dichloroethene (Cl atoms on opposite sides).
- B = cis-1,2-dichloroethene (Cl atoms on the same side).
(c) Are the chemical properties the same?
Yes, A and B have essentially the same chemical properties. Reason: they possess the same functional group (a C=C double bond) and the same number and type of atoms and bonds, so they undergo the same chemical reactions. Their differences are mainly physical (for example melting point, boiling point and polarity: the cis form is polar while the trans form is non-polar), not chemical.
Question 8 Report
(a) When is a sample of water said to be hard?
(b) State one difference between temporary and permanent hardness of water
(c) Give one method of removing hardness completely from water
(d) Name two local is used for the production of soap.
(a) A sample of water is said to be hard when it does not readily form a lather (foam) with soap, but instead forms an insoluble scum, because it contains dissolved calcium and magnesium salts.
(b) Temporary hardness is caused by dissolved hydrogen carbonates of calcium and magnesium and can be removed simply by boiling, whereas permanent hardness is caused by the sulphates and chlorides of calcium and magnesium and cannot be removed by boiling.
(c) Hardness can be completely removed by distillation (or by the ion-exchange/permutit method, which removes both temporary and permanent hardness).
(d) Two local materials used for making soap: a vegetable oil or animal fat (for example palm oil or palm-kernel oil) and wood ash (which supplies the alkali/caustic potash).
Answer Details
(a) A sample of water is said to be hard when it does not readily form a lather (foam) with soap, but instead forms an insoluble scum, because it contains dissolved calcium and magnesium salts.
(b) Temporary hardness is caused by dissolved hydrogen carbonates of calcium and magnesium and can be removed simply by boiling, whereas permanent hardness is caused by the sulphates and chlorides of calcium and magnesium and cannot be removed by boiling.
(c) Hardness can be completely removed by distillation (or by the ion-exchange/permutit method, which removes both temporary and permanent hardness).
(d) Two local materials used for making soap: a vegetable oil or animal fat (for example palm oil or palm-kernel oil) and wood ash (which supplies the alkali/caustic potash).
Question 9 Report
(a) State Le Chatelier's principle
(b) Use Le Chetelier's principle to deduce the conditions that favour a high yield of ammonia in the Haber process
(c) Give the chemical test for ammonia.
(d) State what would be observed when aqueous ammonia solution is added to:
(i) zinc chloride solution, (ii) copper (II) tetraoxosulohate (V) solution
(e) Explain why the H — N — H bond angle in ammonia is less than that of H — C — H in methane
(f) Give two uses of ammonia.
(a) Le Chatelier's principle
When a system at equilibrium is subjected to a change in conditions (concentration, temperature or pressure), the equilibrium shifts in the direction that tends to oppose or reduce the effect of that change.
(b) Conditions for a high yield of ammonia in the Haber process
The reaction is: \[N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \quad \Delta H = -ve\]
- High pressure (about 200 to 250 atm): the forward reaction produces fewer gas molecules (4 volumes to 2 volumes), so raising the pressure shifts the equilibrium to the right, increasing the yield of ammonia.
- Low temperature: the forward reaction is exothermic, so a lower temperature favours a higher yield. However, too low a temperature makes the reaction very slow, so a moderate (compromise) temperature of about 450 to 500 °C is used together with an iron catalyst to reach equilibrium quickly.
- Continuous removal of ammonia (by liquefaction) drives the equilibrium forward.
(c) Chemical test for ammonia
Ammonia turns moist red litmus paper blue, and it gives dense white fumes of ammonium chloride when a glass rod dipped in concentrated hydrochloric acid is held in it: \[NH_{3(g)} + HCl_{(g)} \rightarrow NH_4Cl_{(s)}\]
(d) Observations with aqueous ammonia
- (i) Zinc chloride solution: a white gelatinous precipitate of zinc hydroxide forms, which dissolves in excess ammonia to give a colourless solution.
- (ii) Copper(II) tetraoxosulphate(VI) solution: a pale blue precipitate of copper(II) hydroxide forms, which dissolves in excess ammonia to give a deep royal-blue solution.
(e) H-N-H bond angle less than H-C-H
In methane the carbon atom has four bonding pairs of electrons and no lone pair, so the equal bond-pair repulsions give the ideal tetrahedral angle of 109.5°. In ammonia the nitrogen atom has three bonding pairs and one lone pair. A lone pair repels more strongly than a bonding pair, so it compresses the bonding pairs closer together, reducing the H-N-H angle to about 107°.
(f) Two uses of ammonia
- Manufacture of nitrogenous fertilizers such as ammonium salts and urea.
- Manufacture of trioxonitrate(V) acid (nitric acid) in the Ostwald process. (It is also used as a refrigerant.)
Answer Details
(a) Le Chatelier's principle
When a system at equilibrium is subjected to a change in conditions (concentration, temperature or pressure), the equilibrium shifts in the direction that tends to oppose or reduce the effect of that change.
(b) Conditions for a high yield of ammonia in the Haber process
The reaction is: \[N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \quad \Delta H = -ve\]
- High pressure (about 200 to 250 atm): the forward reaction produces fewer gas molecules (4 volumes to 2 volumes), so raising the pressure shifts the equilibrium to the right, increasing the yield of ammonia.
- Low temperature: the forward reaction is exothermic, so a lower temperature favours a higher yield. However, too low a temperature makes the reaction very slow, so a moderate (compromise) temperature of about 450 to 500 °C is used together with an iron catalyst to reach equilibrium quickly.
- Continuous removal of ammonia (by liquefaction) drives the equilibrium forward.
(c) Chemical test for ammonia
Ammonia turns moist red litmus paper blue, and it gives dense white fumes of ammonium chloride when a glass rod dipped in concentrated hydrochloric acid is held in it: \[NH_{3(g)} + HCl_{(g)} \rightarrow NH_4Cl_{(s)}\]
(d) Observations with aqueous ammonia
- (i) Zinc chloride solution: a white gelatinous precipitate of zinc hydroxide forms, which dissolves in excess ammonia to give a colourless solution.
- (ii) Copper(II) tetraoxosulphate(VI) solution: a pale blue precipitate of copper(II) hydroxide forms, which dissolves in excess ammonia to give a deep royal-blue solution.
(e) H-N-H bond angle less than H-C-H
In methane the carbon atom has four bonding pairs of electrons and no lone pair, so the equal bond-pair repulsions give the ideal tetrahedral angle of 109.5°. In ammonia the nitrogen atom has three bonding pairs and one lone pair. A lone pair repels more strongly than a bonding pair, so it compresses the bonding pairs closer together, reducing the H-N-H angle to about 107°.
(f) Two uses of ammonia
- Manufacture of nitrogenous fertilizers such as ammonium salts and urea.
- Manufacture of trioxonitrate(V) acid (nitric acid) in the Ostwald process. (It is also used as a refrigerant.)
Question 10 Report
(a) The half-life of 56/25 Mn is 9.3 x 10\(^3\)S. What does the statement mean?
(b) Give the three types of radiation that are usually emitted by radioactive substances.
(a) The statement that the half-life of \(^{56}_{25}Mn\) is \(9.3\times10^{3}\) s means that the time required for one-half of the atoms (nuclei) present in any given sample of the radioactive manganese-56 to decay (disintegrate) is \(9.3\times10^{3}\) seconds. After each interval of \(9.3\times10^{3}\) s, the amount of the undecayed Mn-56 remaining is reduced to half of what it was at the start of that interval.
(b) The three types of radiation usually emitted by radioactive substances:
- Alpha (\(\alpha\)) radiation, positively charged helium nuclei.
- Beta (\(\beta\)) radiation, negatively charged fast-moving electrons.
- Gamma (\(\gamma\)) radiation, high-energy electromagnetic waves carrying no charge.
Answer Details
(a) The statement that the half-life of \(^{56}_{25}Mn\) is \(9.3\times10^{3}\) s means that the time required for one-half of the atoms (nuclei) present in any given sample of the radioactive manganese-56 to decay (disintegrate) is \(9.3\times10^{3}\) seconds. After each interval of \(9.3\times10^{3}\) s, the amount of the undecayed Mn-56 remaining is reduced to half of what it was at the start of that interval.
(b) The three types of radiation usually emitted by radioactive substances:
- Alpha (\(\alpha\)) radiation, positively charged helium nuclei.
- Beta (\(\beta\)) radiation, negatively charged fast-moving electrons.
- Gamma (\(\gamma\)) radiation, high-energy electromagnetic waves carrying no charge.
Question 11 Report
(a) Distinguish between a conductor and an electrolyte
(b)(i) State Faraday's first law of electrolysis
(ii) Describe how you would investigate Faraday's law of electrolysis, using copper (II) tetraoxosulphate (VI) solution and copper electrodes.
(c) 0.222g of a divalent metal is deposited when a current of 0.45 ampere is passed through a solution of its salt for 25 minutes using appropriate electrodes. Calculate the relative atomic mass of the metal. 1F = 96500C mol\(^{-1}\)
`(d) State two applications of electrolysis.
(a) Conductor vs electrolyte
A conductor is a substance that allows electricity to pass through it without undergoing any chemical change; it conducts by the movement of free electrons (for example, a metal). An electrolyte is a substance which, in the molten state or in aqueous solution, conducts electricity and is chemically decomposed by it; it conducts by the movement of ions.
(b)(i) Faraday's first law
The mass of a substance liberated or deposited at an electrode during electrolysis is directly proportional to the quantity of electricity (charge) passed through the electrolyte.
(ii) Investigation
- Weigh a clean, dry copper cathode and record its mass.
- Set up an electrolytic cell containing copper(II) tetraoxosulphate(VI) solution with two copper electrodes, connected in series with an ammeter, a rheostat and a switch.
- Pass a steady current I (read on the ammeter) for a measured time t (with a stopwatch); the charge is Q = It.
- Remove, wash, dry and re-weigh the cathode to find the mass of copper deposited.
- Repeat for different values of Q, then plot a graph of mass deposited against Q. A straight line through the origin confirms that mass is proportional to charge.
(c) Relative atomic mass
\[Q = It = 0.45 \times (25 \times 60) = 675\,C\]
For a divalent metal, M2+ + 2e- → M, so 2 × 96500 C deposit 1 mole of metal.
\[\text{moles of metal} = \frac{675}{2 \times 96500} = 3.497 \times 10^{-3}\,mol\]
\[\text{RAM} = \frac{0.222}{3.497 \times 10^{-3}} \approx 63.5\]
The relative atomic mass of the metal is about 63.5 (the metal is copper).
(d) Two applications of electrolysis
- Electroplating of metals (for example, coating with silver, chromium or nickel).
- Purification (refining) of metals such as copper; also the extraction of reactive metals such as aluminium and sodium.
Answer Details
(a) Conductor vs electrolyte
A conductor is a substance that allows electricity to pass through it without undergoing any chemical change; it conducts by the movement of free electrons (for example, a metal). An electrolyte is a substance which, in the molten state or in aqueous solution, conducts electricity and is chemically decomposed by it; it conducts by the movement of ions.
(b)(i) Faraday's first law
The mass of a substance liberated or deposited at an electrode during electrolysis is directly proportional to the quantity of electricity (charge) passed through the electrolyte.
(ii) Investigation
- Weigh a clean, dry copper cathode and record its mass.
- Set up an electrolytic cell containing copper(II) tetraoxosulphate(VI) solution with two copper electrodes, connected in series with an ammeter, a rheostat and a switch.
- Pass a steady current I (read on the ammeter) for a measured time t (with a stopwatch); the charge is Q = It.
- Remove, wash, dry and re-weigh the cathode to find the mass of copper deposited.
- Repeat for different values of Q, then plot a graph of mass deposited against Q. A straight line through the origin confirms that mass is proportional to charge.
(c) Relative atomic mass
\[Q = It = 0.45 \times (25 \times 60) = 675\,C\]
For a divalent metal, M2+ + 2e- → M, so 2 × 96500 C deposit 1 mole of metal.
\[\text{moles of metal} = \frac{675}{2 \times 96500} = 3.497 \times 10^{-3}\,mol\]
\[\text{RAM} = \frac{0.222}{3.497 \times 10^{-3}} \approx 63.5\]
The relative atomic mass of the metal is about 63.5 (the metal is copper).
(d) Two applications of electrolysis
- Electroplating of metals (for example, coating with silver, chromium or nickel).
- Purification (refining) of metals such as copper; also the extraction of reactive metals such as aluminium and sodium.
Question 12 Report
(a) Give one physical property of
(i) diamond;
(ii) graphite
(b) Give two uses of diamond
(c) evidence that shows that both graphite and diamond are allotropes of carbon.
(a) Physical property
- (i) Diamond: it is extremely hard (the hardest natural substance) and transparent; it does not conduct electricity.
- (ii) Graphite: it is soft, slippery and greyish-black (opaque); it conducts electricity.
(b) Two uses of diamond: for making cutting and drilling tools (e.g. glass cutters and rock drills) and as an abrasive for grinding and polishing; it is also used in jewellery.
(c) Evidence that graphite and diamond are allotropes of carbon: both substances are composed of carbon only, and when each is completely burnt in excess oxygen the only product formed is carbon dioxide (in the same proportion). Producing the same single product on combustion shows that both are simply different physical forms of the same element, carbon, which is what allotropy means.
Answer Details
(a) Physical property
- (i) Diamond: it is extremely hard (the hardest natural substance) and transparent; it does not conduct electricity.
- (ii) Graphite: it is soft, slippery and greyish-black (opaque); it conducts electricity.
(b) Two uses of diamond: for making cutting and drilling tools (e.g. glass cutters and rock drills) and as an abrasive for grinding and polishing; it is also used in jewellery.
(c) Evidence that graphite and diamond are allotropes of carbon: both substances are composed of carbon only, and when each is completely burnt in excess oxygen the only product formed is carbon dioxide (in the same proportion). Producing the same single product on combustion shows that both are simply different physical forms of the same element, carbon, which is what allotropy means.
Question 13 Report
(a) List three characteristics of a homologous series
(b) Give one example of;
(i) alkanes; (ii) alkynes.
(c) A hydrocarbon contains 7.7% by mass of hydrogen and 92.3% by mass of carbon. The relative molar mass of the compound is 78.
(i) Derive the empirical formula of the compound and hence its molecular formula.
(ii) Name the hydrocarbon and write its structural formula. (H=1, C=12)
(d) Two hydrocarbons, X and Y were treated separated with acidified potassium tetraoxomanganate (VII) solution. X decolorized the solution and Y did not. Which of X and Y will undergo
(i) substitution reaction only,
(ii) both addition and substitution reactions.
(iii) polymerization?
(e) If ethanol is to be converted into ethanoic acid
(i) What are the conditions required?
(ii) name the type of reaction that will be involved and write the equation
(a) Three characteristics of a homologous series
- All members can be represented by the same general formula.
- Consecutive members differ by a -CH2- group (a difference of 14 in relative molar mass).
- They have similar chemical properties because they contain the same functional group, and show a gradual change in physical properties as molar mass increases.
(b) Examples
- (i) Alkane: methane, CH4.
- (ii) Alkyne: ethyne, C2H2.
(c) Formula from composition
| Element | C | H |
|---|---|---|
| % mass | 92.3 | 7.7 |
| ÷ atomic mass | 92.3/12 = 7.69 | 7.7/1 = 7.7 |
| Ratio | 1 | 1 |
Empirical formula = CH (empirical mass = 12 + 1 = 13).
\[n = \frac{78}{13} = 6\]
Molecular formula = (CH)6 = C6H6.
(ii) The hydrocarbon is benzene. Its structural formula is a hexagonal ring of six carbon atoms with alternating single and double bonds, each carbon carrying one hydrogen atom (often drawn as a hexagon with an inscribed circle).
(d) X decolorized acidified KMnO4, so X is unsaturated (an alkene/alkyne); Y did not, so Y is saturated (an alkane).
- (i) Substitution only: Y.
- (ii) Both addition and substitution: X.
- (iii) Polymerization: X.
(e) Ethanol to ethanoic acid
- (i) Conditions: heat (reflux) the ethanol with an acidified oxidizing agent such as acidified potassium tetraoxomanganate(VII) or acidified potassium dichromate(VI).
- (ii) Type of reaction: oxidation. \[CH_3CH_2OH + 2[O] \rightarrow CH_3COOH + H_2O\]
Answer Details
(a) Three characteristics of a homologous series
- All members can be represented by the same general formula.
- Consecutive members differ by a -CH2- group (a difference of 14 in relative molar mass).
- They have similar chemical properties because they contain the same functional group, and show a gradual change in physical properties as molar mass increases.
(b) Examples
- (i) Alkane: methane, CH4.
- (ii) Alkyne: ethyne, C2H2.
(c) Formula from composition
| Element | C | H |
|---|---|---|
| % mass | 92.3 | 7.7 |
| ÷ atomic mass | 92.3/12 = 7.69 | 7.7/1 = 7.7 |
| Ratio | 1 | 1 |
Empirical formula = CH (empirical mass = 12 + 1 = 13).
\[n = \frac{78}{13} = 6\]
Molecular formula = (CH)6 = C6H6.
(ii) The hydrocarbon is benzene. Its structural formula is a hexagonal ring of six carbon atoms with alternating single and double bonds, each carbon carrying one hydrogen atom (often drawn as a hexagon with an inscribed circle).
(d) X decolorized acidified KMnO4, so X is unsaturated (an alkene/alkyne); Y did not, so Y is saturated (an alkane).
- (i) Substitution only: Y.
- (ii) Both addition and substitution: X.
- (iii) Polymerization: X.
(e) Ethanol to ethanoic acid
- (i) Conditions: heat (reflux) the ethanol with an acidified oxidizing agent such as acidified potassium tetraoxomanganate(VII) or acidified potassium dichromate(VI).
- (ii) Type of reaction: oxidation. \[CH_3CH_2OH + 2[O] \rightarrow CH_3COOH + H_2O\]
Question 14 Report
(a) Explain the differences in the reactions of zinc with dilute trioxonitrate (V) acid and zinc with dilute hydrochloric acid.
(b) Write equations to illustrate how ammonia gas can be converted into trioxonitrate (V) acid.
(c) Calculate the mass of sodium trioxonitrate (V) produced when 30.0g of pure sodium hydroxide reacts with 100cm\(^3\) of 1.00 M trioxonitrate (V) acid. (H =1, N = 14, 0 = 16, Na = 23)
(d) Write the equations for the decomposition by heat of:
(i) sodium trioxonitrate(V);
(ii) copper (II) trioxonitrate (V);
(iii) mercury (II) trioxonitrate (V);
(a) Zinc with dilute HNO3 vs dilute HCl
Dilute hydrochloric acid is a non-oxidizing acid. Because zinc lies above hydrogen in the activity series, it displaces hydrogen ions as hydrogen gas: \[Zn_{(s)} + 2HCl_{(aq)} \rightarrow ZnCl_{2(aq)} + H_{2(g)}\]
Trioxonitrate(V) acid (nitric acid) is a strong oxidizing acid. The nitrate ion, rather than the hydrogen ion, is reduced, so hydrogen gas is not evolved; instead oxides of nitrogen (or ammonium salt with very dilute acid) are formed, for example: \[3Zn + 8HNO_3 \rightarrow 3Zn(NO_3)_2 + 2NO + 4H_2O\]
(b) Ammonia to trioxonitrate(V) acid (Ostwald process)
\[4NH_3 + 5O_2 \xrightarrow{Pt} 4NO + 6H_2O\]
\[2NO + O_2 \rightarrow 2NO_2\]
\[4NO_2 + O_2 + 2H_2O \rightarrow 4HNO_3\]
(c) Mass of sodium trioxonitrate(V)
\[NaOH + HNO_3 \rightarrow NaNO_3 + H_2O\]
Moles of HNO3 = \(1.00 \times \frac{100}{1000} = 0.100\,mol\). Moles of NaOH = \(\frac{30}{40} = 0.75\,mol\).
The acid is the limiting reagent, so moles of NaNO3 = 0.100 mol. Molar mass of NaNO3 = 23 + 14 + 48 = 85 g mol-1.
\[\text{mass} = 0.100 \times 85 = 8.5\,g\]
(d) Decomposition by heat
- (i) \[2NaNO_3 \rightarrow 2NaNO_2 + O_2\]
- (ii) \[2Cu(NO_3)_2 \rightarrow 2CuO + 4NO_2 + O_2\]
- (iii) \[2Hg(NO_3)_2 \rightarrow 2Hg + 4NO_2 + O_2\]
Answer Details
(a) Zinc with dilute HNO3 vs dilute HCl
Dilute hydrochloric acid is a non-oxidizing acid. Because zinc lies above hydrogen in the activity series, it displaces hydrogen ions as hydrogen gas: \[Zn_{(s)} + 2HCl_{(aq)} \rightarrow ZnCl_{2(aq)} + H_{2(g)}\]
Trioxonitrate(V) acid (nitric acid) is a strong oxidizing acid. The nitrate ion, rather than the hydrogen ion, is reduced, so hydrogen gas is not evolved; instead oxides of nitrogen (or ammonium salt with very dilute acid) are formed, for example: \[3Zn + 8HNO_3 \rightarrow 3Zn(NO_3)_2 + 2NO + 4H_2O\]
(b) Ammonia to trioxonitrate(V) acid (Ostwald process)
\[4NH_3 + 5O_2 \xrightarrow{Pt} 4NO + 6H_2O\]
\[2NO + O_2 \rightarrow 2NO_2\]
\[4NO_2 + O_2 + 2H_2O \rightarrow 4HNO_3\]
(c) Mass of sodium trioxonitrate(V)
\[NaOH + HNO_3 \rightarrow NaNO_3 + H_2O\]
Moles of HNO3 = \(1.00 \times \frac{100}{1000} = 0.100\,mol\). Moles of NaOH = \(\frac{30}{40} = 0.75\,mol\).
The acid is the limiting reagent, so moles of NaNO3 = 0.100 mol. Molar mass of NaNO3 = 23 + 14 + 48 = 85 g mol-1.
\[\text{mass} = 0.100 \times 85 = 8.5\,g\]
(d) Decomposition by heat
- (i) \[2NaNO_3 \rightarrow 2NaNO_2 + O_2\]
- (ii) \[2Cu(NO_3)_2 \rightarrow 2CuO + 4NO_2 + O_2\]
- (iii) \[2Hg(NO_3)_2 \rightarrow 2Hg + 4NO_2 + O_2\]
