WAEC SSCE - Chemistry - 1997 (Essay)
Question 1 Report
(a) Give one example of a metal which:
(i) can displace hydrogen from cold water
(ii) at red-heat, reacts reversibly with steam
(iii) does not react with dilute hydrochloric acid
(b) When dry hydrogen was passed over lead (II) oxide, a greyish solid J was obtained.
(i) Identify J
(ii) What type of reaction was involved in the formation of J from lead (II) oxide?
(a)(i) Metal that displaces hydrogen from cold water: sodium (or potassium, calcium).
(ii) Metal that reacts reversibly with steam at red heat: iron (3Fe + 4H2O ⇌ Fe3O4 + 4H2).
(iii) Metal that does not react with dilute hydrochloric acid: copper (or silver, gold).
(b) Passing dry hydrogen over lead(II) oxide gives a greyish solid J.
(i) J is lead (Pb). (ii) The reaction is a reduction (redox); the hydrogen reduces PbO to Pb: PbO + H2 → Pb + H2O.
Answer Details
(a)(i) Metal that displaces hydrogen from cold water: sodium (or potassium, calcium).
(ii) Metal that reacts reversibly with steam at red heat: iron (3Fe + 4H2O ⇌ Fe3O4 + 4H2).
(iii) Metal that does not react with dilute hydrochloric acid: copper (or silver, gold).
(b) Passing dry hydrogen over lead(II) oxide gives a greyish solid J.
(i) J is lead (Pb). (ii) The reaction is a reduction (redox); the hydrogen reduces PbO to Pb: PbO + H2 → Pb + H2O.
Question 2 Report
(a)(i) State the two main processes involved in the manufacture of oxygen from air
(ii) Name the type of chemical bonding which exists between oxygen atoms in a molecule of oxygen
(b) What term is used to describe the relationship between oxygen and ozone (O\(_3\))?
(a)(i) The two main processes in obtaining oxygen from air are the liquefaction of air (compressing and cooling air until it liquefies) followed by the fractional distillation of the liquid air (nitrogen boils off first at -196°C, leaving oxygen).
(ii) The bonding between the two oxygen atoms in an O2 molecule is a covalent bond (a double covalent bond).
(b) The relationship between oxygen (O2) and ozone (O3) is allotropy; they are allotropes of the element oxygen.
Answer Details
(a)(i) The two main processes in obtaining oxygen from air are the liquefaction of air (compressing and cooling air until it liquefies) followed by the fractional distillation of the liquid air (nitrogen boils off first at -196°C, leaving oxygen).
(ii) The bonding between the two oxygen atoms in an O2 molecule is a covalent bond (a double covalent bond).
(b) The relationship between oxygen (O2) and ozone (O3) is allotropy; they are allotropes of the element oxygen.
Question 3 Report
(a) List two products obtained when crude oil is refined
(b)(i) What is the general formula for alkanols?
(ii) State the type of reaction involved in the conversion of ethanol to ethanoic acid
(iii) Write an equation to show how ethanoic acid reacts with sodium trioxocarbonate (IV).
(a) Two products from refining crude oil: petrol (gasoline) and kerosene (also diesel, bitumen, lubricating oil).
(b)(i) General formula for alkanols: CnH2n+1OH.
(ii) Converting ethanol to ethanoic acid is an oxidation reaction.
(iii) Ethanoic acid with sodium trioxocarbonate(IV):
2CH3COOH + Na2CO3 → 2CH3COONa + H2O + CO2
Answer Details
(a) Two products from refining crude oil: petrol (gasoline) and kerosene (also diesel, bitumen, lubricating oil).
(b)(i) General formula for alkanols: CnH2n+1OH.
(ii) Converting ethanol to ethanoic acid is an oxidation reaction.
(iii) Ethanoic acid with sodium trioxocarbonate(IV):
2CH3COOH + Na2CO3 → 2CH3COONa + H2O + CO2
Question 4 Report
Consider the reaction represented by the following Q equation: N\(_2\)O\(_{4(g)}\) \(\rightleftharpoons\) 2NO\(_{2(g)}\) \(\Delta\)H = +57.2KJmol\(^{-1}\)
(a) When is the reaction said to be at equilibrium?
(b) Mention two conditions that can favour the forward reaction
(c) Name the principle involved in (b) above.
(a) The reaction is at equilibrium when the rate of the forward reaction equals the rate of the reverse reaction, so that the concentrations of N2O4 and NO2 remain constant (a dynamic equilibrium).
(b) The forward reaction is endothermic (ΔH = +57.2 kJ mol-1) and produces more moles of gas (1 → 2). Two conditions that favour it are:
- Increasing the temperature.
- Decreasing the pressure (increasing the volume).
(Removing NO2 as it forms would also favour it.)
(c) The principle involved is Le Chatelier's principle.
Answer Details
(a) The reaction is at equilibrium when the rate of the forward reaction equals the rate of the reverse reaction, so that the concentrations of N2O4 and NO2 remain constant (a dynamic equilibrium).
(b) The forward reaction is endothermic (ΔH = +57.2 kJ mol-1) and produces more moles of gas (1 → 2). Two conditions that favour it are:
- Increasing the temperature.
- Decreasing the pressure (increasing the volume).
(Removing NO2 as it forms would also favour it.)
(c) The principle involved is Le Chatelier's principle.
Question 5 Report
(a) State Gay Lussac's law of combining volumes
(b) Hydrogen reacts with oxygen according to the following equation;
2H\(_{2(g)}\) + O\(_{2(g)}\) ---> 2H\(_2\)O\(_{(g)}\). If 50cm\(^3\) of hydrogen were sparked with 30cm\(^{3}\) of oxygen, calculate the volume of unused oxygen after cooling to the initial temperature and pressure.
(a) Gay-Lussac's law of combining volumes: When gases react, they do so in volumes that bear a simple whole-number ratio to one another and to the volume of any gaseous product, provided the temperature and pressure remain constant.
(b) 2H2 + O2 → 2H2O. Hydrogen and oxygen react in the volume ratio 2 : 1.
50 cm3 of H2 requires \(\dfrac{50}{2}=25\ \text{cm}^3\) of O2.
Oxygen supplied = 30 cm3; oxygen used = 25 cm3.
\[ \text{Unused oxygen}=30-25=5\ \text{cm}^3 \]Answer Details
(a) Gay-Lussac's law of combining volumes: When gases react, they do so in volumes that bear a simple whole-number ratio to one another and to the volume of any gaseous product, provided the temperature and pressure remain constant.
(b) 2H2 + O2 → 2H2O. Hydrogen and oxygen react in the volume ratio 2 : 1.
50 cm3 of H2 requires \(\dfrac{50}{2}=25\ \text{cm}^3\) of O2.
Oxygen supplied = 30 cm3; oxygen used = 25 cm3.
\[ \text{Unused oxygen}=30-25=5\ \text{cm}^3 \]Question 6 Report
(a) Mention three chemical properties of chlorine
(b) What type of reaction is represented by each of the following equations?
(i) Mg\(_{(s)}\) + 2HCI\(_{(aq)}\) --> Mg\(_2\)Cl\(_{2(aq)}\) + H\(_{2(g)}\)
(ii) Ag\(^+_{aq}\) + Cl\(^-\)\(_{(aq)}\) ----> AgCl\(_{(s)}\)
(a) Three chemical properties of chlorine:
- It is a strong oxidizing agent (e.g. oxidizes Fe2+ to Fe3+).
- It bleaches moist coloured materials (bleaching in the presence of water).
- It displaces less reactive halogens (Br-, I-) from solutions of their salts.
(It also reacts directly with metals and with hydrogen to form chlorides.)
(b) (i) Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g): a displacement reaction (redox).
(ii) Ag+(aq) + Cl-(aq) → AgCl(s): a precipitation (ionic double-decomposition) reaction.
Answer Details
(a) Three chemical properties of chlorine:
- It is a strong oxidizing agent (e.g. oxidizes Fe2+ to Fe3+).
- It bleaches moist coloured materials (bleaching in the presence of water).
- It displaces less reactive halogens (Br-, I-) from solutions of their salts.
(It also reacts directly with metals and with hydrogen to form chlorides.)
(b) (i) Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g): a displacement reaction (redox).
(ii) Ag+(aq) + Cl-(aq) → AgCl(s): a precipitation (ionic double-decomposition) reaction.
Question 7 Report
(a)(i) What is meant by cracking of petroleum fractions?
(ii) Write an equation for the laboratory preparation of ethene from ethanol.
(iii) Give one chemical test to distinguish between ethane and ethene.
(b)(i) Name the class of carbohydrates to which starch and cellulose belong.
(ii) What process is used for isolating ethanol from the other products of fermentation of sugar?
(iii) Name the organic product of the reaction between ethanol and sodium
(iv). Write the structural formula of 2-chloroethanol.
(c) State the reason why:
(i) benzene produces more soot than ethene on burning in excess air;
(ii) ethanoic acid has a higher boiling point than methanoic acid;
(iii) sodium chloride is used during the manufacture of soap.
(d) Give one use of: (i) ethyne (ii) coal (iii) carbon black
(a)(i) Cracking is the breaking down of large, long-chain hydrocarbon molecules from heavier petroleum fractions into smaller, more useful molecules (such as petrol and alkenes) using heat and/or a catalyst.
(ii) C2H5OH → C2H4 + H2O (heated with concentrated H2SO4 at about 170°C).
(iii) Add bromine water: ethene rapidly decolourizes the red-brown bromine water, while ethane does not.
(b)(i) Starch and cellulose are polysaccharides. (ii) Ethanol is isolated by fractional distillation. (iii) The organic product of ethanol + sodium is sodium ethoxide (C2H5ONa). (iv) 2-chloroethanol: ClCH2CH2OH.
(c)(i) Benzene produces more soot because it has a higher proportion of carbon (high carbon-to-hydrogen ratio), so it undergoes more incomplete combustion, releasing unburnt carbon. (ii) Ethanoic acid has a higher relative molecular mass and larger molecule than methanoic acid, giving stronger van der Waals forces (in addition to hydrogen bonding), so more energy is needed to boil it. (iii) Sodium chloride is added to precipitate (salt out) the soap, separating it from the glycerol and water.
(d)(i) Ethyne: oxy-acetylene welding/cutting. (ii) Coal: as a fuel (and a source of coke and coal tar). (iii) Carbon black: as a reinforcing filler in motor tyres (and a pigment in ink).
Answer Details
(a)(i) Cracking is the breaking down of large, long-chain hydrocarbon molecules from heavier petroleum fractions into smaller, more useful molecules (such as petrol and alkenes) using heat and/or a catalyst.
(ii) C2H5OH → C2H4 + H2O (heated with concentrated H2SO4 at about 170°C).
(iii) Add bromine water: ethene rapidly decolourizes the red-brown bromine water, while ethane does not.
(b)(i) Starch and cellulose are polysaccharides. (ii) Ethanol is isolated by fractional distillation. (iii) The organic product of ethanol + sodium is sodium ethoxide (C2H5ONa). (iv) 2-chloroethanol: ClCH2CH2OH.
(c)(i) Benzene produces more soot because it has a higher proportion of carbon (high carbon-to-hydrogen ratio), so it undergoes more incomplete combustion, releasing unburnt carbon. (ii) Ethanoic acid has a higher relative molecular mass and larger molecule than methanoic acid, giving stronger van der Waals forces (in addition to hydrogen bonding), so more energy is needed to boil it. (iii) Sodium chloride is added to precipitate (salt out) the soap, separating it from the glycerol and water.
(d)(i) Ethyne: oxy-acetylene welding/cutting. (ii) Coal: as a fuel (and a source of coke and coal tar). (iii) Carbon black: as a reinforcing filler in motor tyres (and a pigment in ink).
Question 8 Report
(a) List two industrial uses of concentrated tetraoxosulphate (VI) acid
(b) The diagram below represents the set-up for the electrolysis of dilute tetraoxosulphate (VI) acid. Use it to answer Questions (i) to (iii).
(i) Which letter on the diagram represents the battery?
(ii) Write an equation for the reaction occurring at the cathode
(iii) What product does V represent?
(a) Two industrial uses of concentrated tetraoxosulphate(VI) acid
- Manufacture of fertilizers such as ammonium tetraoxosulphate(VI) and calcium superphosphate.
- As the electrolyte in lead-acid accumulators (car batteries); also used in petroleum refining, and in the production of detergents, paints and dyes.
(b) Electrolysis of dilute tetraoxosulphate(VI) acid
Reading the set-up: the zig-zag component S is the rheostat (variable resistor), R is the cell drawn as two parallel plates, and T is the switch (key). The two inverted tubes V and U collect the gases given off at electrodes P and Q, which dip into the dilute H2SO4.
(i) The letter that represents the battery (cell) is R.
(ii) Reaction at the cathode. At the cathode, hydrogen ions are discharged (reduced) in preference to the poorly discharged tetraoxosulphate(VI) ions:
\[ 2H^{+}_{(aq)} + 2e^{-} \rightarrow H_{2(g)} \]
(iii) Product V. V is collected at the electrode joined to the negative terminal, i.e. the cathode, so V represents hydrogen gas. It is the more abundant gas: for every 1 volume of oxygen liberated at the anode, 2 volumes of hydrogen are collected here, since
\[ 2H_2O_{(l)} \rightarrow 2H_{2(g)} + O_{2(g)} \]
(overall) gives twice as much hydrogen as oxygen.
Answer Details
(a) Two industrial uses of concentrated tetraoxosulphate(VI) acid
- Manufacture of fertilizers such as ammonium tetraoxosulphate(VI) and calcium superphosphate.
- As the electrolyte in lead-acid accumulators (car batteries); also used in petroleum refining, and in the production of detergents, paints and dyes.
(b) Electrolysis of dilute tetraoxosulphate(VI) acid
Reading the set-up: the zig-zag component S is the rheostat (variable resistor), R is the cell drawn as two parallel plates, and T is the switch (key). The two inverted tubes V and U collect the gases given off at electrodes P and Q, which dip into the dilute H2SO4.
(i) The letter that represents the battery (cell) is R.
(ii) Reaction at the cathode. At the cathode, hydrogen ions are discharged (reduced) in preference to the poorly discharged tetraoxosulphate(VI) ions:
\[ 2H^{+}_{(aq)} + 2e^{-} \rightarrow H_{2(g)} \]
(iii) Product V. V is collected at the electrode joined to the negative terminal, i.e. the cathode, so V represents hydrogen gas. It is the more abundant gas: for every 1 volume of oxygen liberated at the anode, 2 volumes of hydrogen are collected here, since
\[ 2H_2O_{(l)} \rightarrow 2H_{2(g)} + O_{2(g)} \]
(overall) gives twice as much hydrogen as oxygen.
Question 9 Report
(a)(i) Define entropy
(ii) What term is used to describe a reaction in which heat is absorbed from the surrounding?
(b) State two conditions that can lead to ineffective collisions during a chemical reaction.
(a)(i) Entropy is a measure of the degree of disorder or randomness of a system.
(ii) A reaction in which heat is absorbed from the surroundings is described as endothermic.
(b) Two conditions that lead to ineffective collisions:
- The colliding particles possess energy less than the activation energy.
- The particles collide with the wrong (unfavourable) orientation.
Answer Details
(a)(i) Entropy is a measure of the degree of disorder or randomness of a system.
(ii) A reaction in which heat is absorbed from the surroundings is described as endothermic.
(b) Two conditions that lead to ineffective collisions:
- The colliding particles possess energy less than the activation energy.
- The particles collide with the wrong (unfavourable) orientation.
Question 10 Report
(a) State two properties of alkalis
(b)(i) Why is tetraoxosulphate (VI) acid able to produce two types of salt?
(ii) Write an equation for the reactior involved on heating sodium trioxosulphate (IV) with hydrochloric acid.
(a) Two properties of alkalis: they turn red litmus blue (pH greater than 7); they have a soapy/slippery feel. (They also neutralize acids to give a salt and water, and taste bitter.)
(b)(i) Tetraoxosulphate(VI) acid, H2SO4, produces two types of salt because it is dibasic: it contains two replaceable (ionizable) hydrogen atoms. When only one H is replaced by a metal it forms an acid salt (e.g. NaHSO4); when both are replaced it forms a normal salt (e.g. Na2SO4).
(ii) Heating sodium trioxosulphate(IV), Na2SO3, with hydrochloric acid:
Na2SO3(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + SO2(g)
Answer Details
(a) Two properties of alkalis: they turn red litmus blue (pH greater than 7); they have a soapy/slippery feel. (They also neutralize acids to give a salt and water, and taste bitter.)
(b)(i) Tetraoxosulphate(VI) acid, H2SO4, produces two types of salt because it is dibasic: it contains two replaceable (ionizable) hydrogen atoms. When only one H is replaced by a metal it forms an acid salt (e.g. NaHSO4); when both are replaced it forms a normal salt (e.g. Na2SO4).
(ii) Heating sodium trioxosulphate(IV), Na2SO3, with hydrochloric acid:
Na2SO3(s) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + SO2(g)
Question 11 Report
(a)(i) State two general methods of preparing soluble salts.
(ii) Mention three pieces of apparatus required for determining the solubility of a salt at a given temperature.
(b) The solubilities of two salts represented as K and L were determined at various temperatures. The results are shown in the table below:
| Temperature \( (^{o}\mathrm{C}) \) | 0 | 20 | 40 | 60 | 80 | 90 |
| Solubility of K \( (\mathrm{mol.\ dm}^{-3}) \) | 0.38 | 0.46 | 0.54 | 0.62 | 0.69 | 0.73 |
| Solubility of L \( (\mathrm{mol.\ dm}^{-3}) \) | 0.12 | 0.34 | 0.64 | 1.08 | 1.64 | 2.00 |
(i) Plot the solubility curves of K and L on the same graph. Use the curves to answer questions (ii) - (iv) below.
(ii) What is the solubility of K at 50°C?
(iii) At what temperature is the solubility of L equal to \(1.0\mathrm{mol.\ dm}^{-3}\)?
(iv) Over what temperature range is K more soluble that L?
(v) Given that the molar mass of L is 101g, determine whether a solution containing 3.4g of L per \(250\mathrm{cm}^{3}\) at 20°C is saturated or unsaturated.
(a)(i) General methods of preparing soluble salts
- Neutralisation of an acid by an alkali or base.
- Direct combination of the constituent elements.
(a)(ii) Apparatus required
A thermometer, a water bath and a glass stirrer.
(b)(i) Solubility curves
The solubility curves of K and L are plotted below on the same axes.
(b)(ii) From the curve for K, at 50°C:
Solubility of K = 0.58 mol dm−3.
(b)(iii) From the curve for L, a solubility of 1.0 mol dm−3 is obtained at approximately 57°C.
(b)(iv) K is more soluble than L from 0°C to about 31.5°C.
(b)(v)
Mass concentration of L in the solution is:
\[\frac{3.4\ \text{g}}{250\ \text{cm}^3}\times 1000 = 13.6\ \text{g dm}^{-3}\]
At 20°C, the solubility of L is 0.34 mol dm−3. In mass units:
\[0.34\times 101 = 34.34\ \text{g dm}^{-3}\]
Since 13.6 g dm−3 is less than 34.34 g dm−3, the solution is unsaturated.
Answer Details
(a)(i) General methods of preparing soluble salts
- Neutralisation of an acid by an alkali or base.
- Direct combination of the constituent elements.
(a)(ii) Apparatus required
A thermometer, a water bath and a glass stirrer.
(b)(i) Solubility curves
The solubility curves of K and L are plotted below on the same axes.
(b)(ii) From the curve for K, at 50°C:
Solubility of K = 0.58 mol dm−3.
(b)(iii) From the curve for L, a solubility of 1.0 mol dm−3 is obtained at approximately 57°C.
(b)(iv) K is more soluble than L from 0°C to about 31.5°C.
(b)(v)
Mass concentration of L in the solution is:
\[\frac{3.4\ \text{g}}{250\ \text{cm}^3}\times 1000 = 13.6\ \text{g dm}^{-3}\]
At 20°C, the solubility of L is 0.34 mol dm−3. In mass units:
\[0.34\times 101 = 34.34\ \text{g dm}^{-3}\]
Since 13.6 g dm−3 is less than 34.34 g dm−3, the solution is unsaturated.
Question 12 Report
(a) List two properties of alpha particles.
(b) X is an element which exists as an isotopic mixture containing 90% of \(^{39}_{19} X\) and 10% of \(^{41}_{19}X\).
(i) How many neutrons are present in the isotope \(^{41}_{19}X\)?
(ii) Calculate the mean relative atomic mass of X.
(a) Two properties of alpha particles: they are positively charged (each is a helium nucleus, He2+); they have low penetrating power (stopped by paper or the skin). (They are also relatively heavy and cause intense ionization.)
(b)(i) Neutrons in 4119X = mass number - atomic number = 41 - 19 = 22 neutrons.
(ii) Mean relative atomic mass
\[ \bar{A}=\frac{(90\times39)+(10\times41)}{100}=\frac{3510+410}{100}=39.2 \]Answer Details
(a) Two properties of alpha particles: they are positively charged (each is a helium nucleus, He2+); they have low penetrating power (stopped by paper or the skin). (They are also relatively heavy and cause intense ionization.)
(b)(i) Neutrons in 4119X = mass number - atomic number = 41 - 19 = 22 neutrons.
(ii) Mean relative atomic mass
\[ \bar{A}=\frac{(90\times39)+(10\times41)}{100}=\frac{3510+410}{100}=39.2 \]Question 13 Report
(a)(i) List two reactants for the laboratory preparation of ammonia
(ii) State three physical properties of ammonia
(iii) Describe in outline, the manufacture of ammonia by the Haber process.
(b) Write an equation in each case to show the:
(i) reaction between ammonia gas and heated copper (II) oxide
(ii) action of heat on ammonium trioxocarbonate (IV)
(c)(i) Which industrial process is used for convey ammonia to trioxonitrate (V) acid?
(ii) Give the reason why electropositive metals do not generally are off hydrogen with dilute trioxonitrate (V) acid
(d) Give one example in each case, to show how trioxonitrate (V) acid reacts generally with: (i) bases (ii) non-metals.
(a)(i) Two reactants for laboratory preparation of ammonia: an ammonium salt (e.g. ammonium chloride, NH4Cl) and a base (e.g. calcium hydroxide, slaked lime).
(ii) Three physical properties of ammonia: colourless gas; pungent (choking) smell; less dense than air; very soluble in water.
(iii) Haber process: Nitrogen (from air) and hydrogen (from natural gas/water) are mixed in the ratio 1 : 3 by volume and passed over a finely divided iron catalyst at about 450°C and a pressure of about 200-250 atm:
N2(g) + 3H2(g) ⇌ 2NH3(g)
The ammonia is cooled and liquefied, and the unreacted gases are recycled.
(b)(i) 3CuO + 2NH3 → 3Cu + N2 + 3H2O
(ii) (NH4)2CO3 → 2NH3 + H2O + CO2
(c)(i) Ammonia is converted to trioxonitrate(V) acid by the Ostwald process.
(ii) Electropositive metals do not liberate hydrogen with dilute HNO3 because HNO3 is a strong oxidizing agent: it oxidizes any hydrogen formed to water while it is itself reduced to oxides of nitrogen (NO or NO2), so oxides of nitrogen are evolved instead of hydrogen.
(d)(i) With bases (neutralization): HNO3 + NaOH → NaNO3 + H2O.
(ii) With non-metals (oxidation): C + 4HNO3 → CO2 + 4NO2 + 2H2O.
Answer Details
(a)(i) Two reactants for laboratory preparation of ammonia: an ammonium salt (e.g. ammonium chloride, NH4Cl) and a base (e.g. calcium hydroxide, slaked lime).
(ii) Three physical properties of ammonia: colourless gas; pungent (choking) smell; less dense than air; very soluble in water.
(iii) Haber process: Nitrogen (from air) and hydrogen (from natural gas/water) are mixed in the ratio 1 : 3 by volume and passed over a finely divided iron catalyst at about 450°C and a pressure of about 200-250 atm:
N2(g) + 3H2(g) ⇌ 2NH3(g)
The ammonia is cooled and liquefied, and the unreacted gases are recycled.
(b)(i) 3CuO + 2NH3 → 3Cu + N2 + 3H2O
(ii) (NH4)2CO3 → 2NH3 + H2O + CO2
(c)(i) Ammonia is converted to trioxonitrate(V) acid by the Ostwald process.
(ii) Electropositive metals do not liberate hydrogen with dilute HNO3 because HNO3 is a strong oxidizing agent: it oxidizes any hydrogen formed to water while it is itself reduced to oxides of nitrogen (NO or NO2), so oxides of nitrogen are evolved instead of hydrogen.
(d)(i) With bases (neutralization): HNO3 + NaOH → NaNO3 + H2O.
(ii) With non-metals (oxidation): C + 4HNO3 → CO2 + 4NO2 + 2H2O.
Question 14 Report
(a)(i) List three characteristic properties of transition metals
(ii) 0.45g of a metal M was deposited when a current of 1.8 amperes was passed for 12.5 minutes through a solution containing M\(^{2+}\). Calculate the relative atomic mass of M. [1 Faraday = 96500 C]
(iii) Give the reason why copper-plated iron corrodes easily when the surface is scratched.
(b)(i) State the law of definite proportions (constant composition).
(ii) Describe in outline, an experimental procedure for determining the proportion of oxygen in a given sample of copper(II) oxide.
(iii) Write an equation to show how copper (II) oxide can be obtained directly from copper (II) trioxonitrate (V)
(a)(i) Three characteristic properties of transition metals: they show variable oxidation states; they form coloured ions/compounds; they act as catalysts. (They also form complex ions.)
(ii) Current = 1.8 A, time = 12.5 min = 750 s, M2+ (so 2 electrons per atom).
\[ Q=1.8\times750=1350\ \text{C};\quad n(e^-)=\frac{1350}{96500}=0.01399\ \text{mol} \] \[ n(M)=\frac{0.01399}{2}=6.995\times10^{-3}\ \text{mol} \] \[ \text{Relative atomic mass}=\frac{0.45}{6.995\times10^{-3}}=64.3\approx 64 \](iii) When the copper coat is scratched, the more reactive (more electropositive) iron is exposed. Iron then acts as the anode and copper as the cathode in an electrochemical cell, so the iron loses electrons and corrodes faster than it would on its own.
(b)(i) Law of definite proportions: A given chemical compound always contains the same elements combined together in the same fixed proportion by mass, whatever its source or method of preparation.
(ii) Weigh a known mass of dry copper(II) oxide in a weighed boat/tube. Pass dry hydrogen over it while heating; the hydrogen reduces the oxide to copper (CuO + H2 → Cu + H2O). Heat to constant mass, cool in the stream of hydrogen, and reweigh. The loss in mass is the mass of oxygen, so the percentage of oxygen = (loss in mass / mass of CuO) × 100.
(iii) 2Cu(NO3)2 → 2CuO + 4NO2 + O2
Answer Details
(a)(i) Three characteristic properties of transition metals: they show variable oxidation states; they form coloured ions/compounds; they act as catalysts. (They also form complex ions.)
(ii) Current = 1.8 A, time = 12.5 min = 750 s, M2+ (so 2 electrons per atom).
\[ Q=1.8\times750=1350\ \text{C};\quad n(e^-)=\frac{1350}{96500}=0.01399\ \text{mol} \] \[ n(M)=\frac{0.01399}{2}=6.995\times10^{-3}\ \text{mol} \] \[ \text{Relative atomic mass}=\frac{0.45}{6.995\times10^{-3}}=64.3\approx 64 \](iii) When the copper coat is scratched, the more reactive (more electropositive) iron is exposed. Iron then acts as the anode and copper as the cathode in an electrochemical cell, so the iron loses electrons and corrodes faster than it would on its own.
(b)(i) Law of definite proportions: A given chemical compound always contains the same elements combined together in the same fixed proportion by mass, whatever its source or method of preparation.
(ii) Weigh a known mass of dry copper(II) oxide in a weighed boat/tube. Pass dry hydrogen over it while heating; the hydrogen reduces the oxide to copper (CuO + H2 → Cu + H2O). Heat to constant mass, cool in the stream of hydrogen, and reweigh. The loss in mass is the mass of oxygen, so the percentage of oxygen = (loss in mass / mass of CuO) × 100.
(iii) 2Cu(NO3)2 → 2CuO + 4NO2 + O2
