WAEC SSCE - Chemistry - 2023 (Essay)
Question 1 Report
a. Starting with calcium chloride, describe briefly how a solid sample of calcium trioxocarbonate (IV) can be prepared in the laboratory.
b. With relevant equations outline the procedure for the purification of impure copper.
c. Copper reacts with concentrated trioxonitrate (V) acid: i. write a balanced chemical equation for the reaction;
ii. state what would be observed in the reaction,
iii. state why the copper is oxidized;
iv. an excess of copper is added to 25.0 \(cm^3\) of 16.0 mol \(dm^3\) \(HNO_3\). Calculate the volume of the gas formed at s.t.p. [H=1.0, N=14.0, O= 16.0, Cu=63.0; Molar volume of gas at .s.t.p.=22.4 \(dm^3\)]
di. Pure \(HNO_3\), is a colourless liquid but when exposed to air, it turns yellowish-brown in colour. Explain briefly this observation.
ii. Write a balanced equation for the laboratory preparation of hydrogen trioxonitrate (V) acid.
a. Water is added to the \(CaCl_2\): to form a solution. \(Na_2CO\), is then added to the solution to precipitate \(CaCO_3\),which is filtered, washed, and dried.
b. Electricity is passed through a solution of CuSO4 using an impure copper as the anode and pure copper as the cathode. During electrolysis,the anode loses mass as copper dissolves and the cathode gains mass as copper is deposited.
Anode: \(Cu_s → Cu^{2+}_{(aq)} + 2e^-\)
Cathode: \(Cu^{2+}_{(aq)} + 2e^- → Cu_{(s)}\)
ci. \(Cu_{(s)} + 4HNO_{3(aq)} →Cu(NO_3)_{2(aq)} + 2NO_{2(g)} + 2H_2O_{(l)}\)
ii. -blue solution
-(reddish)brown gas
-gas bubbles/effervescence
-metal dissolves/metal disappears
iii. Because copper loses electrons/oxidation number increases.
iv. Moles of acid=0.025x16= 0.4 mole
Mole of \(NO_2\) = moles of acid/2 =0.2 mole
Volume of gas (\(NO_2)\) at s.t.p. = 0.2 x 22.4 \(dm^{-3}\) = 4.48 mol \(dm^{-3}\)
di. Pure HNO3 undergoes decomposition to \(NO_2\)(brown gas). This gas dissolves in the rest of the acid leaving it with that colour.
ii. -2\(KNO_3 + H_2SO_4→K_2SO_4+2HNO_3\)
- \(NaNO_3 + H_2SO_4→NaHSO_4 + HNO_3\)
Answer Details
a. Water is added to the \(CaCl_2\): to form a solution. \(Na_2CO\), is then added to the solution to precipitate \(CaCO_3\),which is filtered, washed, and dried.
b. Electricity is passed through a solution of CuSO4 using an impure copper as the anode and pure copper as the cathode. During electrolysis,the anode loses mass as copper dissolves and the cathode gains mass as copper is deposited.
Anode: \(Cu_s → Cu^{2+}_{(aq)} + 2e^-\)
Cathode: \(Cu^{2+}_{(aq)} + 2e^- → Cu_{(s)}\)
ci. \(Cu_{(s)} + 4HNO_{3(aq)} →Cu(NO_3)_{2(aq)} + 2NO_{2(g)} + 2H_2O_{(l)}\)
ii. -blue solution
-(reddish)brown gas
-gas bubbles/effervescence
-metal dissolves/metal disappears
iii. Because copper loses electrons/oxidation number increases.
iv. Moles of acid=0.025x16= 0.4 mole
Mole of \(NO_2\) = moles of acid/2 =0.2 mole
Volume of gas (\(NO_2)\) at s.t.p. = 0.2 x 22.4 \(dm^{-3}\) = 4.48 mol \(dm^{-3}\)
di. Pure HNO3 undergoes decomposition to \(NO_2\)(brown gas). This gas dissolves in the rest of the acid leaving it with that colour.
ii. -2\(KNO_3 + H_2SO_4→K_2SO_4+2HNO_3\)
- \(NaNO_3 + H_2SO_4→NaHSO_4 + HNO_3\)
Question 2 Report
a. Describe how iron and aluminum reacts with each of the following substances: i. dilute \(H_2SO_4\); ii. dilute \(HNO_3\)
bi. Write an equation for the burning of sulphur in air.
ii. Name the catalyst used in the contact process.
iii. In the contact process, why is an excess of air used?
iv. Why is it necessary to cool the catalyst used in 5(b)(ii)?
v. Give a reason why the air used in the contact process needs to be as clean as possible.
vi. State two reasons why SO2 should not be discharged into the atmosphere.
ci. State the reagents and conditions used in the laboratory preparation of chlorine.
ii. State two uses of chlorine.
di. Name the drying agents for each of the following gases: 1. Hydrogen;
i2. sulphur (IV) oxide;
i3. Ammonia.
ii1. State the components of the following
Bronze; ii2. Brass.
ai. Iron dissolves readily in dilute \(H_2SO_4\) liberating hydrogen and forming \(FeSO_4\) while with Aluminum there is no reaction.
ii. With aluminum there is no reaction(1)while iron dissolves in dilute \(HNO_3\) liberating hydrogen and forming \(Fe(NO_3)_2\)
bi. \(S+O_2→SO_2\)
ii. Vanadium(V) oxide/ platinized asbestos.
iii. To favour the formation of more product/\(SO_3\).
iv. The reaction is exothermic/high temperature favours the formation of more reactants/backward reaction.
v. Catalysts are easily poisoned/easily damaged by dirt.
vi. -SO2 damages buildings
-damages living things
-causes acid rain
-irritates eyes, nose and throat
-causes lung and respiratory diseases eg coughing, bronchitis, etc
ci. Reagents
-manganese(IV)oxide/MnO2
-concentrated HCI
Conditions
-heating
Concentrated HCI(1)
Bleaching powder/CaOCl2/KMnO4crystals
Room temperature/heat is not required
ii.
-for bleaching wood pulp and textiles
-for the production of plastics
-for recovering of tin from its ore/extraction of tin
-as a germicide and disinfectant/insecticide/in swimming pools
-to sterilize/purification of water supply
-for commercial manufacture of HCI
-production of dyes and drugs
-manufacture of solvents eg \(CCL_4\), \(CHCI_3\)
di1. Hydrogen-fused calcium chloride/conc.tetraoxosulphate(VI)acid
i2. Sulphur(IV)oxide-concentrated tetraoxosulphate(VI)acid/ anhydrous Calcium chloride
i3. calcium oxide/ quick lime/silica gel
ii1. copper and tin
ii2. Copper and zinc
Answer Details
ai. Iron dissolves readily in dilute \(H_2SO_4\) liberating hydrogen and forming \(FeSO_4\) while with Aluminum there is no reaction.
ii. With aluminum there is no reaction(1)while iron dissolves in dilute \(HNO_3\) liberating hydrogen and forming \(Fe(NO_3)_2\)
bi. \(S+O_2→SO_2\)
ii. Vanadium(V) oxide/ platinized asbestos.
iii. To favour the formation of more product/\(SO_3\).
iv. The reaction is exothermic/high temperature favours the formation of more reactants/backward reaction.
v. Catalysts are easily poisoned/easily damaged by dirt.
vi. -SO2 damages buildings
-damages living things
-causes acid rain
-irritates eyes, nose and throat
-causes lung and respiratory diseases eg coughing, bronchitis, etc
ci. Reagents
-manganese(IV)oxide/MnO2
-concentrated HCI
Conditions
-heating
Concentrated HCI(1)
Bleaching powder/CaOCl2/KMnO4crystals
Room temperature/heat is not required
ii.
-for bleaching wood pulp and textiles
-for the production of plastics
-for recovering of tin from its ore/extraction of tin
-as a germicide and disinfectant/insecticide/in swimming pools
-to sterilize/purification of water supply
-for commercial manufacture of HCI
-production of dyes and drugs
-manufacture of solvents eg \(CCL_4\), \(CHCI_3\)
di1. Hydrogen-fused calcium chloride/conc.tetraoxosulphate(VI)acid
i2. Sulphur(IV)oxide-concentrated tetraoxosulphate(VI)acid/ anhydrous Calcium chloride
i3. calcium oxide/ quick lime/silica gel
ii1. copper and tin
ii2. Copper and zinc
Question 3 Report
a. In an experiment, 20.0 \(cm^3\) of a solution containing 4g/\(dm^3\) of sodium hydroxide was neutralized by 8.0 \(cm^3\) of dilute tetraoxosulphate (VI) acid: i. Write a balanced equation for the reaction; ii. calculate the concentration of the acid in mol/ \(dm^3\)
bi. State two postulates of the Kinetic theory of gases which real gases do not obey
ii. Explain briefly why real gases do not obey the postulates stated in 2(b)(i).
c. Consider the following compound:
Name the compound
ii. name the two structural isomers of the compound;
iii. state the chemical process involved in the preparation of the compound from starch,
iv. write the chemical equation for the steps involved in the process in 2(c)(ii).
v. name two enzymes involved in the process in 2(c)(iii)
d. Explain briefly the term structural isomerism.
ai. \(H_2SO_4 + 2NaOH → Na_2SO_4 + 2H_2O\)
ii. Concentration of NaOH(\(C_n)\)
\(C_n\) = 4/40= 0.1 mol/\(dm^3\)
\(\frac{C_AV_A}{C_nV_n} = \frac{1}{2}\)
\(C_A \times \frac{8}{0.1} \times 20 = \frac{1}{2}\)
\(C_A =0.1 \times 20 \times \frac{1}{2} \times 8 = 0.125 mol /dm^3\)
bi. -The volume of a gas is negligible compared to the total volume of the container
-There are no forces of attraction or repulsion between molecules of the gas/collision between molecules is elastic.
ii. The volume of a gas is not negligible because at high pressure the molecules occupy appreciable volume and at low temperature the
intermolecular forces of attraction and repulsion become significant.
ci. Glucose.
ii. -Fructose
-Galactose
iii. hydrolysis
iv. 2(\(C_6H_{10}O_5)_n + nH_2O → nC_{12}H_{12}O_{11}\)
\(C_{12}H_{12}O_{11} + H_2O → 2C_6H_12O_6\)
v. -maltase
-diastase
d. It Is the existence of compounds with the same molecular formula but different structural formulae/arrangement of the atoms/linkage of the atoms.
Answer Details
ai. \(H_2SO_4 + 2NaOH → Na_2SO_4 + 2H_2O\)
ii. Concentration of NaOH(\(C_n)\)
\(C_n\) = 4/40= 0.1 mol/\(dm^3\)
\(\frac{C_AV_A}{C_nV_n} = \frac{1}{2}\)
\(C_A \times \frac{8}{0.1} \times 20 = \frac{1}{2}\)
\(C_A =0.1 \times 20 \times \frac{1}{2} \times 8 = 0.125 mol /dm^3\)
bi. -The volume of a gas is negligible compared to the total volume of the container
-There are no forces of attraction or repulsion between molecules of the gas/collision between molecules is elastic.
ii. The volume of a gas is not negligible because at high pressure the molecules occupy appreciable volume and at low temperature the
intermolecular forces of attraction and repulsion become significant.
ci. Glucose.
ii. -Fructose
-Galactose
iii. hydrolysis
iv. 2(\(C_6H_{10}O_5)_n + nH_2O → nC_{12}H_{12}O_{11}\)
\(C_{12}H_{12}O_{11} + H_2O → 2C_6H_12O_6\)
v. -maltase
-diastase
d. It Is the existence of compounds with the same molecular formula but different structural formulae/arrangement of the atoms/linkage of the atoms.
Question 4 Report
a. A compound contains 52.2% C, 13.1 % H and Oxygen only. The vapour density of the compound is 23.
Determine i. its empirical formula; ii. its molecular formula, [H=1.0, C=12.0, O= 16.0]
iii1). The compound reacts with sodium metal to produce hydrogen gas and when warmed with acidified \(KMnO_{4(aq)}\) gives a solution that turns from purple to colorless. It also forms a sweet-smelling liquid when heated with ethanoic acid in the presence of concentrated \(H_2SO_4\)
name the functional group present in the compound; iii2). draw the structural formula of the compound.
b. Outline the chemical equations for the production of ethanol from cooked cassava.
ci. Explain briefly why a piece of aluminum does not react with water.
ii. How can a pure sample of aluminum chloride crystals be prepared from aluminium
d. Describe how water can be separated from aqueous \(CuSO_4\)
ai. %0 = 100 - (52.2+13.1)
=34.7
C H O
52.2/12, 13.1/1, 34.7/16
4.35, 13.1, 2.168
4.35/2.168, 13.1/2.168, 2.168/2.168
2.00, 6.04, 1
empirical formula = \(C_2H_6O\)
ii. 2 x vapour density = molecular mass
2x23=46
(\(C_2H_6O)_n\) = 46
[(12 x 2) + (1 x 6) + 16]n = 46
46n = 46
n = 1
molecular formula = \(C_2H_6O\)
iii1. Hydroxyl
iii2. 
b. 2(\(C_6H_{10}O_5)_n + nH_2O --diatase→ nC_{12}H_{22}O_{11}\)
\(C_{12}H_{22}O_{11} + H_2O --maltase→ C_6H_{12}O_6 + C_6H_{12}O_6\)
\(C_6H_{12}O_6 --zymase→ 2C_2H_5OH + 2CO_2\)
ci. The reaction starts but the metal produces a layer of an oxide/aluminum oxide layer on its surface which is impermeable(to water) and stops the reaction.
ii. Add dilute hydrochloric acid to excess aluminum filter off the excess metal leave the filtrate in a warm place/evaporate the filtrate to the point of crystallization/leave it in the sun.
d. The flask is connected to the condenser and the solution is heated. The water vaporizes and is converted to(liquid)water in the condenser.
Answer Details
ai. %0 = 100 - (52.2+13.1)
=34.7
C H O
52.2/12, 13.1/1, 34.7/16
4.35, 13.1, 2.168
4.35/2.168, 13.1/2.168, 2.168/2.168
2.00, 6.04, 1
empirical formula = \(C_2H_6O\)
ii. 2 x vapour density = molecular mass
2x23=46
(\(C_2H_6O)_n\) = 46
[(12 x 2) + (1 x 6) + 16]n = 46
46n = 46
n = 1
molecular formula = \(C_2H_6O\)
iii1. Hydroxyl
iii2.
b. 2(\(C_6H_{10}O_5)_n + nH_2O --diatase→ nC_{12}H_{22}O_{11}\)
\(C_{12}H_{22}O_{11} + H_2O --maltase→ C_6H_{12}O_6 + C_6H_{12}O_6\)
\(C_6H_{12}O_6 --zymase→ 2C_2H_5OH + 2CO_2\)
ci. The reaction starts but the metal produces a layer of an oxide/aluminum oxide layer on its surface which is impermeable(to water) and stops the reaction.
ii. Add dilute hydrochloric acid to excess aluminum filter off the excess metal leave the filtrate in a warm place/evaporate the filtrate to the point of crystallization/leave it in the sun.
d. The flask is connected to the condenser and the solution is heated. The water vaporizes and is converted to(liquid)water in the condenser.
Question 5 Report
a. What is a transition element?
b. Consider the electron configuration of the following elements: A=2, 8,6; B=2,8,2; C=2,8,1, D=2,8,8
State the element which forms a: i. doubly charged cation; ii. soluble trioxocarbonate (IV).
c. Explain briefly why there is a general increase in the first ionization energies of the elements across the period in the periodic table.
d. Give two examples of an aliphatic compound
e. Explain briefly why alkanols are stronger bases than water.
f. State the major raw materials used in the Solvay process
g. What is geometric isomerism?
h. Give a reason why water gas is a better fuel than producer gas.
i. Define the term heat of combustion.
ji. State Faraday's second law of electrolysis
ii. Calculate the amount of silver deposited when 10920 coulombs of electricity is passed through a solution of a silver salt.
[IF = 96500 C mol-1]
a. A transition element is one which has incompletely filled d-orbitals.
bi. B
ii. C
c. Across a period(from left to right) there is a gradual increase in the number of protons in the nucleus/effective nuclear-charge.
This increases the force of attraction between the nucleus and the electrons hence more energy is needed to remove the outermost electron.
d. -methane
-ethene
-ethyne
-ethanol
-ethanoic acid
e. -Alkanols are formed by replacing one hydrogen of water by an alkyl group
-Alkyl groups have a positive inductive effect/ are electron releasing/electron donating
-This increases the electron density of the oxygen
f. -NaCl
-CaCO3
-NH3
-C
g. Geometric isomerism is the existence of two compounds with the same molecular formula but differ in the arrangement of groups attached to the carbon-containing double bond
h. This is because constituents of water gases(CO and H2)are both combustible and give a higher calorific/heat value than producer gas
i. It is defined as the heat change/released when 1 mole of a substance is burnt in excess oxygen.
OR It is defined as the heat change/released when 1 mole of a substance is completely burnt in air
ji. When the same quantity of electricity is passed through different electrolytes, the relative number of moles of the elements discharged is inversely proportional to the charges on the ions of the element.
ii. 96500 C liberates 1.0 mol of Ag(s)
10920 C will liberate 10920/96500
=0.113 mole
Answer Details
a. A transition element is one which has incompletely filled d-orbitals.
bi. B
ii. C
c. Across a period(from left to right) there is a gradual increase in the number of protons in the nucleus/effective nuclear-charge.
This increases the force of attraction between the nucleus and the electrons hence more energy is needed to remove the outermost electron.
d. -methane
-ethene
-ethyne
-ethanol
-ethanoic acid
e. -Alkanols are formed by replacing one hydrogen of water by an alkyl group
-Alkyl groups have a positive inductive effect/ are electron releasing/electron donating
-This increases the electron density of the oxygen
f. -NaCl
-CaCO3
-NH3
-C
g. Geometric isomerism is the existence of two compounds with the same molecular formula but differ in the arrangement of groups attached to the carbon-containing double bond
h. This is because constituents of water gases(CO and H2)are both combustible and give a higher calorific/heat value than producer gas
i. It is defined as the heat change/released when 1 mole of a substance is burnt in excess oxygen.
OR It is defined as the heat change/released when 1 mole of a substance is completely burnt in air
ji. When the same quantity of electricity is passed through different electrolytes, the relative number of moles of the elements discharged is inversely proportional to the charges on the ions of the element.
ii. 96500 C liberates 1.0 mol of Ag(s)
10920 C will liberate 10920/96500
=0.113 mole
