WAEC SSCE - Chemistry - 2021 (Essay)
Question 1 Report
(a)i. With the aid of an equation, explain briefly why aluminum metal is not affected by air.
(ii) In the extraction of aluminum from bauxite, state the:
- substance used for purifying the ore;
- composition of the mixture electrolyzed.
(b) ZnO is an amphoteric oxide. Write equations to illustrate this statement.
(c)i) List three uses of sodium trioxocarbonate(IV).
(ii) Explain briefly why a solution of trioxonitrate(V) acid turns yellowish on storage for some time.
(ii) Describe briefly how trioxonitrate(V) ions could be tested for in the laboratory.
(d) Write balanced chemical equations for the preparation of hydrogen chloride.
(i) using concentrated H\(_{2}\)SO\(_{4}\):
(ii) by direct combination of its constituent elements.
(iii) State one use of hydrogen chloride.
(a) (i) Aluminum metal is not affected by air because it is covered with a thin layer of aluminum oxide (Al2O3) that protects the metal from further oxidation. The equation for the formation of aluminum oxide is: 2Al + 3O2 → 2Al2O3 The aluminum oxide layer is very stable and prevents oxygen in the air from reaching the aluminum metal, thus protecting it from further reaction. (ii) In the extraction of aluminum from bauxite: - The substance used for purifying the ore is concentrated sodium hydroxide (NaOH) solution, which is used to dissolve the aluminum oxide (Al2O3) present in bauxite, leaving behind impurities. - The composition of the mixture electrolyzed is a molten mixture of aluminum oxide (Al2O3) and cryolite (Na3AlF6), which is used as a solvent to lower the melting point of Al2O3 and make it easier to extract aluminum by electrolysis. (b) ZnO is an amphoteric oxide, which means it can react with both acids and bases. The equations to illustrate this statement are: ZnO + 2HCl → ZnCl2 + H2O ZnO + 2NaOH + H2O → Na2Zn(OH)4 In the first equation, ZnO reacts with hydrochloric acid (HCl) to form zinc chloride (ZnCl2) and water (H2O). In the second equation, ZnO reacts with sodium hydroxide (NaOH) and water (H2O) to form sodium zincate (Na2Zn(OH)4). (c) (i) Three uses of sodium trioxocarbonate(IV) (sodium carbonate, Na2CO3) are: - It is used in the production of glass, where it acts as a flux to lower the melting point of silica (SiO2) and help it fuse into glass. - It is used as a water softener in laundry detergents, where it reacts with hard water ions (such as calcium and magnesium) to form insoluble precipitates that can be removed from clothing. - It is used as a food additive, where it acts as a leavening agent in baked goods to produce carbon dioxide (CO2) gas and make the dough rise. (ii) A solution of trioxonitrate(V) acid (HNO3) turns yellowish on storage for some time because it slowly decomposes to form nitrogen dioxide (NO2) gas, which is a brownish color. The equation for this decomposition is: 4HNO3 → 2H2O + 4NO2 + O2 (iii) Trioxonitrate(V) ions (NO3-) can be tested for in the laboratory using the Brown Ring Test. In this test, a solution of the sample containing NO3- ions is mixed with freshly prepared iron(II) sulfate (FeSO4) solution and concentrated sulfuric acid (H2SO4). A brown ring will form at the junction of the two layers if NO3- ions are present, due to the formation of a complex between Fe2+ ions, NO3- ions, and H2O molecules. (d) (i) The balanced chemical equation for the preparation of hydrogen chloride (HCl) using concentrated sulfuric acid (H2SO4) is: NaCl + H2SO4 → NaHSO4 + HCl (ii) The balanced chemical equation for the preparation of hydrogen chloride (HCl) by direct combination of its constituent elements is: H2 + Cl2 → 2HCl (iii) One use
Answer Details
(a) (i) Aluminum metal is not affected by air because it is covered with a thin layer of aluminum oxide (Al2O3) that protects the metal from further oxidation. The equation for the formation of aluminum oxide is: 2Al + 3O2 → 2Al2O3 The aluminum oxide layer is very stable and prevents oxygen in the air from reaching the aluminum metal, thus protecting it from further reaction. (ii) In the extraction of aluminum from bauxite: - The substance used for purifying the ore is concentrated sodium hydroxide (NaOH) solution, which is used to dissolve the aluminum oxide (Al2O3) present in bauxite, leaving behind impurities. - The composition of the mixture electrolyzed is a molten mixture of aluminum oxide (Al2O3) and cryolite (Na3AlF6), which is used as a solvent to lower the melting point of Al2O3 and make it easier to extract aluminum by electrolysis. (b) ZnO is an amphoteric oxide, which means it can react with both acids and bases. The equations to illustrate this statement are: ZnO + 2HCl → ZnCl2 + H2O ZnO + 2NaOH + H2O → Na2Zn(OH)4 In the first equation, ZnO reacts with hydrochloric acid (HCl) to form zinc chloride (ZnCl2) and water (H2O). In the second equation, ZnO reacts with sodium hydroxide (NaOH) and water (H2O) to form sodium zincate (Na2Zn(OH)4). (c) (i) Three uses of sodium trioxocarbonate(IV) (sodium carbonate, Na2CO3) are: - It is used in the production of glass, where it acts as a flux to lower the melting point of silica (SiO2) and help it fuse into glass. - It is used as a water softener in laundry detergents, where it reacts with hard water ions (such as calcium and magnesium) to form insoluble precipitates that can be removed from clothing. - It is used as a food additive, where it acts as a leavening agent in baked goods to produce carbon dioxide (CO2) gas and make the dough rise. (ii) A solution of trioxonitrate(V) acid (HNO3) turns yellowish on storage for some time because it slowly decomposes to form nitrogen dioxide (NO2) gas, which is a brownish color. The equation for this decomposition is: 4HNO3 → 2H2O + 4NO2 + O2 (iii) Trioxonitrate(V) ions (NO3-) can be tested for in the laboratory using the Brown Ring Test. In this test, a solution of the sample containing NO3- ions is mixed with freshly prepared iron(II) sulfate (FeSO4) solution and concentrated sulfuric acid (H2SO4). A brown ring will form at the junction of the two layers if NO3- ions are present, due to the formation of a complex between Fe2+ ions, NO3- ions, and H2O molecules. (d) (i) The balanced chemical equation for the preparation of hydrogen chloride (HCl) using concentrated sulfuric acid (H2SO4) is: NaCl + H2SO4 → NaHSO4 + HCl (ii) The balanced chemical equation for the preparation of hydrogen chloride (HCl) by direct combination of its constituent elements is: H2 + Cl2 → 2HCl (iii) One use
Question 2 Report
(a)i. State two characteristics of a homologous series.
(ii) Explain briefly why there are differences in the reaction of ethane and ethene.
(b) When crystals of sodium chloride were warmed with concentrated tetraoxosulphate( VI) acid, a gas was evolved.
(i) Name the gas.
(ii) State two physical properties of the gas.
(iii) Write a balanced chemical equation for the reaction.
(c)i. What are hydrocarbons?
(ii) State two natural sources of hydrocarbons.
(iii) A hydrocarbon contains 83% of carbon by mass. Calculate its empirical formula. [H=1.0, C=12.0]
(d) Draw and label a diagram of a set-up that could be used to electroplate a copper ornament with silver
(a)(i) Characteristics of a homologous series
- All members have the same functional group and can be represented by the same general formula.
- Successive members differ by a \(-CH_2-\) unit, that is, by 14 in relative molecular mass.
- They have similar chemical properties and show a gradual change in physical properties such as boiling point.
(a)(ii) Ethane is a saturated hydrocarbon containing only carbon-carbon single bonds. It undergoes mainly substitution reactions. Ethene is unsaturated and contains a carbon-carbon double bond, \(C=C\), which is reactive; it therefore readily undergoes addition reactions.
(b)(i) The gas evolved is hydrogen chloride, \(HCl\).
(b)(ii) Hydrogen chloride is a colourless gas with a pungent, choking smell. It is denser than air and is very soluble in water.
(b)(iii)
(c)(i) Hydrocarbons are organic compounds that contain carbon and hydrogen only.
(c)(ii) Natural sources of hydrocarbons include crude oil (petroleum) and natural gas.
(c)(iii) Empirical formula
Assume 100 g of the hydrocarbon.
| Element | Mass (g) | Moles | Mole ratio |
|---|---|---|---|
| C | 83 | \(\frac{83}{12}=6.92\) | \(\frac{6.92}{6.92}=1\) |
| H | \(100-83=17\) | \(\frac{17}{1}=17\) | \(\frac{17}{6.92}=2.46\approx2.5\) |
The simplest ratio is \(\mathrm{C:H}=1:2.5\). Multiplying throughout by 2 gives \(2:5\).
(d) Electroplating a copper ornament with silver
The silver electrode is connected to the positive terminal and is the anode. The copper ornament is connected to the negative terminal and is the cathode. Aqueous silver trioxonitrate(V), \(\mathrm{AgNO_3(aq)}\), is used as the electrolyte. Silver dissolves from the anode and is deposited on the copper ornament.
Answer Details
(a)(i) Characteristics of a homologous series
- All members have the same functional group and can be represented by the same general formula.
- Successive members differ by a \(-CH_2-\) unit, that is, by 14 in relative molecular mass.
- They have similar chemical properties and show a gradual change in physical properties such as boiling point.
(a)(ii) Ethane is a saturated hydrocarbon containing only carbon-carbon single bonds. It undergoes mainly substitution reactions. Ethene is unsaturated and contains a carbon-carbon double bond, \(C=C\), which is reactive; it therefore readily undergoes addition reactions.
(b)(i) The gas evolved is hydrogen chloride, \(HCl\).
(b)(ii) Hydrogen chloride is a colourless gas with a pungent, choking smell. It is denser than air and is very soluble in water.
(b)(iii)
(c)(i) Hydrocarbons are organic compounds that contain carbon and hydrogen only.
(c)(ii) Natural sources of hydrocarbons include crude oil (petroleum) and natural gas.
(c)(iii) Empirical formula
Assume 100 g of the hydrocarbon.
| Element | Mass (g) | Moles | Mole ratio |
|---|---|---|---|
| C | 83 | \(\frac{83}{12}=6.92\) | \(\frac{6.92}{6.92}=1\) |
| H | \(100-83=17\) | \(\frac{17}{1}=17\) | \(\frac{17}{6.92}=2.46\approx2.5\) |
The simplest ratio is \(\mathrm{C:H}=1:2.5\). Multiplying throughout by 2 gives \(2:5\).
(d) Electroplating a copper ornament with silver
The silver electrode is connected to the positive terminal and is the anode. The copper ornament is connected to the negative terminal and is the cathode. Aqueous silver trioxonitrate(V), \(\mathrm{AgNO_3(aq)}\), is used as the electrolyte. Silver dissolves from the anode and is deposited on the copper ornament.
Question 3 Report
(a) Distinguish between molecular formula and structural formula
(b) List three factors that determine the ionization energy of an atom.
(c) State the two conditions necessary for the establishment of a chemical equilibrium
(d) Consider the following table
| Element | A | B | C |
| Ionization energy KJ mol\(^{-1}\) | 619 | 518 | 594 |
(d)(i) State which of the elements is the strongest reducing agent.
(ii) Give a reason for the answer stated in (d)(i)
(e) State Graham's law of diffusion
(f) Consider the following salts: \(\mathrm{Mg(NO_3)_2}\), \(\mathrm{CaCO_3}\), \(\mathrm{Na_2SO_4}\). State which of the salts is/are:
(i) readily soluble in water:
(ii) insoluble in water.
(g) Classily each of the following products as addition polymer or condensation polymer:
(i) protein:
(ii) perspex:
(iii) nylon.
(h) Define atomic radius.
(i) Explain briefly why ethanol has a higher boiling point than propane even though they both have comparable molar masses.
(j) State three significance of the pH value in everyday life.
(a) Molecular formula and structural formula
A molecular formula shows the actual number of atoms of each element present in one molecule of a compound (for example ethanol \( C_2H_6O \)). A structural formula shows how those atoms are arranged and joined together by bonds (for example \( CH_3CH_2OH \)).
(b) Three factors that determine ionization energy
- The atomic radius (distance of the outermost electron from the nucleus).
- The nuclear charge (number of protons).
- The shielding (screening) effect of the inner electrons.
(c) Two conditions for chemical equilibrium
- The reaction must be reversible.
- The system must be closed (no reactant or product enters or leaves) and kept at constant temperature.
(d) Ionization energies: A = 619, B = 518, C = 594 \( \text{kJ mol}^{-1} \).
| Element | A | B | C |
|---|---|---|---|
| Ionization energy / kJ mol\(^{-1}\) | 619 | 518 | 594 |
(i) The strongest reducing agent is B.
(ii) B has the lowest ionization energy, so it loses (donates) its outermost electron most easily; a good reducing agent is one that readily gives up electrons.
(e) Graham's law of diffusion
At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or relative molecular mass): \( \text{rate} \propto \dfrac{1}{\sqrt{M}} \).
(f) (i) Readily soluble in water: \( Mg(NO_3)_2 \) and \( Na_2SO_4 \) (all nitrates and all sodium salts are soluble). (ii) Insoluble in water: \( CaCO_3 \).
(g) (i) Protein: condensation polymer. (ii) Perspex: addition polymer. (iii) Nylon: condensation polymer.
(h) Atomic radius
Atomic radius is half the distance between the nuclei of two identical atoms joined by a single covalent bond (half the internuclear distance).
(i) Ethanol versus propane
Ethanol molecules contain an \( -OH \) group and form hydrogen bonds between neighbouring molecules. These hydrogen bonds are much stronger than the weak van der Waals (dispersion) forces between propane molecules, so more energy is needed to separate ethanol molecules, giving it the higher boiling point even though the molar masses are comparable.
(j) Three significances of pH in everyday life
- Control of soil pH so that crops grow well (agriculture).
- Maintaining the correct pH of blood and other body fluids for good health (medicine).
- Quality control in the manufacture of foods, drinks, drugs and cosmetics, and in the treatment of drinking water.
Answer Details
(a) Molecular formula and structural formula
A molecular formula shows the actual number of atoms of each element present in one molecule of a compound (for example ethanol \( C_2H_6O \)). A structural formula shows how those atoms are arranged and joined together by bonds (for example \( CH_3CH_2OH \)).
(b) Three factors that determine ionization energy
- The atomic radius (distance of the outermost electron from the nucleus).
- The nuclear charge (number of protons).
- The shielding (screening) effect of the inner electrons.
(c) Two conditions for chemical equilibrium
- The reaction must be reversible.
- The system must be closed (no reactant or product enters or leaves) and kept at constant temperature.
(d) Ionization energies: A = 619, B = 518, C = 594 \( \text{kJ mol}^{-1} \).
| Element | A | B | C |
|---|---|---|---|
| Ionization energy / kJ mol\(^{-1}\) | 619 | 518 | 594 |
(i) The strongest reducing agent is B.
(ii) B has the lowest ionization energy, so it loses (donates) its outermost electron most easily; a good reducing agent is one that readily gives up electrons.
(e) Graham's law of diffusion
At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or relative molecular mass): \( \text{rate} \propto \dfrac{1}{\sqrt{M}} \).
(f) (i) Readily soluble in water: \( Mg(NO_3)_2 \) and \( Na_2SO_4 \) (all nitrates and all sodium salts are soluble). (ii) Insoluble in water: \( CaCO_3 \).
(g) (i) Protein: condensation polymer. (ii) Perspex: addition polymer. (iii) Nylon: condensation polymer.
(h) Atomic radius
Atomic radius is half the distance between the nuclei of two identical atoms joined by a single covalent bond (half the internuclear distance).
(i) Ethanol versus propane
Ethanol molecules contain an \( -OH \) group and form hydrogen bonds between neighbouring molecules. These hydrogen bonds are much stronger than the weak van der Waals (dispersion) forces between propane molecules, so more energy is needed to separate ethanol molecules, giving it the higher boiling point even though the molar masses are comparable.
(j) Three significances of pH in everyday life
- Control of soil pH so that crops grow well (agriculture).
- Maintaining the correct pH of blood and other body fluids for good health (medicine).
- Quality control in the manufacture of foods, drinks, drugs and cosmetics, and in the treatment of drinking water.
Question 4 Report
(a) What is the structure of:
(i) graphite:
(ii) diamond
(ii) Explain briefly why diamond is hard and a non-conductor of electricity while graphite is soft and an electrical conductor,
(b)i. State what is achieved at each of the following stages in the purification of town water supply:
- aeration;
- screening
- sedimentation.
(ii) Name two substances responsible for hardness in water.
(iii) State two methods for the removal of hardness in water.
(iv) Give one disadvantage of hard water
(c)i). Describe briefly the extraction of tin from its ore.
(ii) Write a balanced chemical equation for the reaction.
(iii) Write an equation for the reaction of tin with:
- oxygen;
- chlorine.
(a)(i) Structures
- Graphite: a giant covalent structure made of flat layers of carbon atoms. Within each layer every carbon is covalently bonded to three others in a hexagonal (honeycomb) arrangement, leaving one delocalized electron per atom. The layers are held together only by weak van der Waals forces.
- Diamond: a giant covalent structure in which every carbon atom is covalently bonded tetrahedrally to four other carbon atoms, forming a rigid three-dimensional network.
(a)(iii) Hardness and conductivity
Diamond is hard because its carbon atoms are held in a rigid three-dimensional network of strong covalent bonds, and it is a non-conductor because all four valence electrons of each carbon are used in bonding, leaving no free electrons to carry charge. Graphite is soft because its layers are held by weak forces and can slide over one another, and it conducts electricity because each carbon uses only three electrons in bonding, leaving one delocalized electron per atom free to move and carry charge.
(b)(i) Water purification stages
- Aeration: water is sprayed into air to add dissolved oxygen and drive off dissolved gases and unpleasant odours; it also oxidizes dissolved iron and manganese.
- Screening: large floating and suspended solids (leaves, sticks, debris) are removed by passing water through screens/gratings.
- Sedimentation: the water is left to stand (often with a coagulant such as alum) so that fine suspended particles settle out to the bottom.
(b)(ii) Substances causing hardness
Dissolved calcium and magnesium salts, that is, the hydrogencarbonates and sulphates of calcium and magnesium (\(Ca^{2+}\) and \(Mg^{2+}\) ions).
(b)(iii) Methods of removing hardness
Boiling (removes temporary hardness), adding washing soda (\(Na_2CO_3\)), ion exchange, or distillation.
(b)(iv) One disadvantage of hard water
It wastes soap by forming an insoluble scum, and it deposits scale (fur) in kettles, boilers and pipes.
(c)(i) Extraction of tin
Tin is obtained from its ore cassiterite, \(SnO_2\). The ore is concentrated by washing and froth flotation to remove earthy impurities, then roasted in air to remove sulphur and arsenic. The concentrated oxide is then reduced with carbon (coke) in a reverberatory or blast furnace, and the molten tin is run off and refined.
(c)(ii) Equation for the reduction
\[SnO_{2(s)} + 2C_{(s)} \to Sn_{(l)} + 2CO_{(g)}\](c)(iii) Reactions of tin
\[\text{With oxygen: } Sn_{(s)} + O_{2(g)} \to SnO_{2(s)}\]\[\text{With chlorine: } Sn_{(s)} + 2Cl_{2(g)} \to SnCl_{4(l)}\]Answer Details
(a)(i) Structures
- Graphite: a giant covalent structure made of flat layers of carbon atoms. Within each layer every carbon is covalently bonded to three others in a hexagonal (honeycomb) arrangement, leaving one delocalized electron per atom. The layers are held together only by weak van der Waals forces.
- Diamond: a giant covalent structure in which every carbon atom is covalently bonded tetrahedrally to four other carbon atoms, forming a rigid three-dimensional network.
(a)(iii) Hardness and conductivity
Diamond is hard because its carbon atoms are held in a rigid three-dimensional network of strong covalent bonds, and it is a non-conductor because all four valence electrons of each carbon are used in bonding, leaving no free electrons to carry charge. Graphite is soft because its layers are held by weak forces and can slide over one another, and it conducts electricity because each carbon uses only three electrons in bonding, leaving one delocalized electron per atom free to move and carry charge.
(b)(i) Water purification stages
- Aeration: water is sprayed into air to add dissolved oxygen and drive off dissolved gases and unpleasant odours; it also oxidizes dissolved iron and manganese.
- Screening: large floating and suspended solids (leaves, sticks, debris) are removed by passing water through screens/gratings.
- Sedimentation: the water is left to stand (often with a coagulant such as alum) so that fine suspended particles settle out to the bottom.
(b)(ii) Substances causing hardness
Dissolved calcium and magnesium salts, that is, the hydrogencarbonates and sulphates of calcium and magnesium (\(Ca^{2+}\) and \(Mg^{2+}\) ions).
(b)(iii) Methods of removing hardness
Boiling (removes temporary hardness), adding washing soda (\(Na_2CO_3\)), ion exchange, or distillation.
(b)(iv) One disadvantage of hard water
It wastes soap by forming an insoluble scum, and it deposits scale (fur) in kettles, boilers and pipes.
(c)(i) Extraction of tin
Tin is obtained from its ore cassiterite, \(SnO_2\). The ore is concentrated by washing and froth flotation to remove earthy impurities, then roasted in air to remove sulphur and arsenic. The concentrated oxide is then reduced with carbon (coke) in a reverberatory or blast furnace, and the molten tin is run off and refined.
(c)(ii) Equation for the reduction
\[SnO_{2(s)} + 2C_{(s)} \to Sn_{(l)} + 2CO_{(g)}\](c)(iii) Reactions of tin
\[\text{With oxygen: } Sn_{(s)} + O_{2(g)} \to SnO_{2(s)}\]\[\text{With chlorine: } Sn_{(s)} + 2Cl_{2(g)} \to SnCl_{4(l)}\]Question 5 Report
(a) In an experiment, 25.0 cm\(^{3}\) of H\(_{2}\)SO\(_{4}\) completely neutralized 24.0 cm\(^{3}\) of a 0.1 50 mol dm\(^{-3}\) aqueous KOH using a suitable indicator.
(i) Write a balanced chemical equation for the reaction.
(ii) Calculate the concentration of the acid solution.
(b)i. A burning magnesium ribbon was placed in a gas jar containing Carbon (IV) oxide.
- Write an equation for the reaction.
- Explain briefly why the magnesium ribbon burns in carbon(lV) oxide although the gas does not support combustion.
- Calculate the percentage mass of nitrogen in magnesium trioxonitrate (V). |N = 140. O = 16.0. Mg = 24.01]
(c) Consider the following organic compound: CH\(_{3}\)CH\(_{2}\)CH = CHCOOH.
(i) State two chemical reactions which could he sed te identify the compound.
(ii) What would be observed in each of the reactions stated in c(i)
(d) Describe briefly how soap is manufactured using pellets of sodium hydroxide and vegetable oil
(e) Define the term electronegativity
(a)(i) Equation for the neutralization
\[H_2SO_{4(aq)} + 2KOH_{(aq)} \to K_2SO_{4(aq)} + 2H_2O_{(l)}\](a)(ii) Concentration of the acid
Moles of KOH \(= \dfrac{24.0}{1000} \times 0.150 = 3.6 \times 10^{-3}\,mol\).
From the equation, 2 mol KOH react with 1 mol \(H_2SO_4\), so moles of acid \(= \dfrac{3.6 \times 10^{-3}}{2} = 1.8 \times 10^{-3}\,mol\).
\[[H_2SO_4] = \frac{1.8 \times 10^{-3}}{25.0/1000} = 0.072\,mol\,dm^{-3}\](b) Magnesium burning in carbon(IV) oxide
Equation:
\[2Mg_{(s)} + CO_{2(g)} \to 2MgO_{(s)} + C_{(s)}\]Explanation: magnesium is a very reactive metal and a powerful reducing agent. The heat of the burning ribbon is enough to decompose carbon(IV) oxide; magnesium removes the oxygen from the \(CO_2\), reducing it to carbon (seen as black specks) while itself being oxidized to white magnesium oxide. So the ribbon continues to burn even though \(CO_2\) does not normally support combustion.
Percentage mass of nitrogen in magnesium trioxonitrate(V), \(Mg(NO_3)_2\):
\[M = 24 + 2(14 + 48) = 148\]\[\%N = \frac{2 \times 14}{148} \times 100 = 18.92\%\](c) The compound \(CH_3CH_2CH=CHCOOH\)
This compound has both a carbon-carbon double bond (unsaturation) and a carboxylic acid (\(-COOH\)) group.
- (i) Reaction 1: add bromine water. Reaction 2: add sodium trioxocarbonate(IV) (\(Na_2CO_3\)) solution.
- (ii) With bromine water: the red-brown bromine water is decolourized (turns colourless), confirming the \(C=C\) double bond. With \(Na_2CO_3\): brisk effervescence occurs, giving off a colourless gas (\(CO_2\)) that turns limewater milky, confirming the acidic \(-COOH\) group.
(d) Manufacture of soap
Vegetable oil is boiled with concentrated sodium hydroxide (caustic soda) solution. The oil is hydrolyzed (saponified) to give soap (the sodium salt of the fatty acid) and glycerol. Common salt (sodium chloride) is then added to the mixture to salt out (precipitate) the soap, which floats to the top; it is filtered off, pressed and moulded.
(e) Electronegativity
Electronegativity is the relative tendency (power) of an atom to attract the shared pair of electrons in a covalent bond towards itself.
Answer Details
(a)(i) Equation for the neutralization
\[H_2SO_{4(aq)} + 2KOH_{(aq)} \to K_2SO_{4(aq)} + 2H_2O_{(l)}\](a)(ii) Concentration of the acid
Moles of KOH \(= \dfrac{24.0}{1000} \times 0.150 = 3.6 \times 10^{-3}\,mol\).
From the equation, 2 mol KOH react with 1 mol \(H_2SO_4\), so moles of acid \(= \dfrac{3.6 \times 10^{-3}}{2} = 1.8 \times 10^{-3}\,mol\).
\[[H_2SO_4] = \frac{1.8 \times 10^{-3}}{25.0/1000} = 0.072\,mol\,dm^{-3}\](b) Magnesium burning in carbon(IV) oxide
Equation:
\[2Mg_{(s)} + CO_{2(g)} \to 2MgO_{(s)} + C_{(s)}\]Explanation: magnesium is a very reactive metal and a powerful reducing agent. The heat of the burning ribbon is enough to decompose carbon(IV) oxide; magnesium removes the oxygen from the \(CO_2\), reducing it to carbon (seen as black specks) while itself being oxidized to white magnesium oxide. So the ribbon continues to burn even though \(CO_2\) does not normally support combustion.
Percentage mass of nitrogen in magnesium trioxonitrate(V), \(Mg(NO_3)_2\):
\[M = 24 + 2(14 + 48) = 148\]\[\%N = \frac{2 \times 14}{148} \times 100 = 18.92\%\](c) The compound \(CH_3CH_2CH=CHCOOH\)
This compound has both a carbon-carbon double bond (unsaturation) and a carboxylic acid (\(-COOH\)) group.
- (i) Reaction 1: add bromine water. Reaction 2: add sodium trioxocarbonate(IV) (\(Na_2CO_3\)) solution.
- (ii) With bromine water: the red-brown bromine water is decolourized (turns colourless), confirming the \(C=C\) double bond. With \(Na_2CO_3\): brisk effervescence occurs, giving off a colourless gas (\(CO_2\)) that turns limewater milky, confirming the acidic \(-COOH\) group.
(d) Manufacture of soap
Vegetable oil is boiled with concentrated sodium hydroxide (caustic soda) solution. The oil is hydrolyzed (saponified) to give soap (the sodium salt of the fatty acid) and glycerol. Common salt (sodium chloride) is then added to the mixture to salt out (precipitate) the soap, which floats to the top; it is filtered off, pressed and moulded.
(e) Electronegativity
Electronegativity is the relative tendency (power) of an atom to attract the shared pair of electrons in a covalent bond towards itself.
