All 31 questions from the West African Examinations Council (WAEC) Chemistry 2011 Theory paper, with the correct answer and a full explanation for each. Free, no signup needed.
PART B SECTION I (FOR ALL CANDIDATES)
1(a) The table below gives the volume/pressure data for a particular sample of a gas at a given temperature.
Volume/dm³ (V): 4.00, 2.00, 1.00 | Pressure/atm (P): 1.00, 2.00, 4.00
(i) Deduce a mathematical relationship between volume (V) and pressure (P). (ii) Name the law that can be deduced from the data. (iii) Calculate the pressure of the gas when the volume is 3.20 dm³.
Model answer
(i) As the volume decreases, the pressure increases - i.e. there is an inverse relationship between the two: V∝1/P, i.e. V=K/P, so K=PV.
(ii) The table portrays Boyle's law.
(iii) P1=1atm, V1=4.00dm³, V2=3.20dm³. Using P1V1=P2V2: 1 x 4 = P2 x 3.20, so P2 = 4/3.20 = 1.25 atm.
1(b)(iii) A standard galvanic cell constructed with Ag⁺(aq)/Ag(s) and Zn²⁺(aq)/Zn(s) couple is discharged until 3.3g of Ag forms. I. Write the overall cell reaction and standard cell potential. II. How many moles of electrons flowed through the circuit during the discharge? III. How many coulombs of charges flowed through the circuit?
Ag⁺(aq)+e⁻⇌Ag(s); E°=+0.80V
Zn²⁺(aq)+2e⁻⇌Zn(s); E°=-0.76V [Ag=108]
Model answer
I. Overall cell reaction: Zn+2Ag⁺→Zn²⁺+2Ag. Cell notation: Zn/Zn²⁺//Ag⁺/Ag. E°=E°red(right)-E°red(left) = (+0.80V)-(-0.76V) = +1.56V.
II. Ag⁺+e⁻→Ag(s): 1 mole of Ag (108g) requires 1 mole of electrons. Since mass produced = 3.3g, moles of electrons = (1/108) x 3.3 = 0.03 moles of electrons.
III. Amount of electricity: 1e⁻ = 1F = 96500C. Thus 0.03e⁻ = 0.03F = 0.03 x 96500 = 2895 C.
1(c)(i) Define each of the following terms: I. Activation energy; II. Exothermic reaction.
Model answer
I. Activation energy is the minimum energy required for a chemical reaction to occur.
II. An exothermic reaction is one during which heat is released to the surrounding.
2. The following table gives the atomic numbers of elements V, W, X, Y and Z.
Element: V, W, X, Y, Z | Atomic number: 11, 16, 18, 19, 24
(a) Which of the elements: (i) belong(s) to group 1? (ii) is/are noble gas(es)? (iii) form(s) coloured compound(s)? (iv) form(s) an anion? (v) react(s) with water to liberate hydrogen? (vi) react(s) with water to form alkaline solution?
Model answer
(i) V & Y belong to group 1 (11V=2,8,1), (19Y=2,8,8,1).
(ii) X is a noble gas (18X=2,8,8).
(iii) Z forms coloured compounds (since it is a transition element).
(iv) W forms an anion.
(v) V and Y react with water to liberate hydrogen.
(vi) V and Y react with water to form alkaline solution (V is sodium, Y is potassium): 2Na(s)+2H2O(l)→2NaOH(aq)+H2(g); 2K(s)+2H2O(l)→2KOH(aq)+H2(g).
2(c)(i) Write the formula of the compound formed between element V and element W. (ii) State the type of bond formed in 2(c)(i) above. Explain your answer.
Model answer
(i) 11V=2,8,1 (loses one electron, combining power 1); 16W=2,8,6 (gains two electrons, combining power 2). Formula: V2W.
(ii) Ionic bond/electrovalent bond - the union involves a transfer of electrons.
2(d)(i) What is a covalent compound? (ii) Give two factors that influence covalent bonding. (iii) State the type of bond that exists in each of the following substances: MgO, NH3 and Fe. (iv) What are intermolecular forces?
Model answer
(i) A covalent compound is a compound formed when the participating elements donate electrons equally for sharing in order to attain the octet.
(ii) Electronegativity difference and ionization energy.
(iii) MgO - electrovalent/ionic bonding; NH3 - covalent bond/hydrogen bonding; Fe - metallic bond.
(iv) Intermolecular forces are the forces that hold the molecules of a substance together and affect the properties of the substance.
3(b) The table below gives the solubility of salt Z at various temperatures.
Temperature(°C): 0,10,20,30,40,50,60 | Solubility(moldm⁻³): 0.13,0.21,0.31,0.45,0.63,0.85,1.10
(i) Plot a graph of solubility against temperature. (ii) From the graph determine the solubility of salt Z at 35°C. (iii) If 100cm³ of the saturated solution is cooled from 55°C to 35°C, calculate the mass of salt Z that would crystallize out. [Molar mass of salt Z = 100 g mol⁻¹]
Model answer
(ii) The solubility of Z at 35°C is about 0.525 moldm⁻³ (read from the graph).
(iii) Mole at 35°C = (0.525x100/1000) = 0.0525 mol. Mole at 55°C = (0.65x100/1000) = 0.0965 mol. Amount crystallized = 0.0965-0.0525 = 0.044 mol. Mass = 0.044 x 100 = 4.4 g.
3(c) Write a balanced equation to illustrate the reaction of Al2O3 with dilute I. HCl; II. NaOH. (ii) What is the name given to an oxide that exhibits both acidic and basic properties? (iii) Give one metallic oxide which exhibits these properties.
Model answer
I. Al2O3(s)+6HCl(aq)→2AlCl3(aq)+3H2O(l). II. Al2O3+2NaOH+3H2O→2NaAl(OH)4.
(ii) Amphoteric oxide.
(iii) Aluminium oxide (Al2O3); Zinc oxide (ZnO) are examples of amphoteric oxides.
4a(i) What is a functional group? (ii) State the functional group in each of the following compounds: I. CH3CH2CH(CH3)OH, II. CH3CH2CH2COOH.
Model answer
(i) A functional group refers to an atom, a group of atoms or a bond that is responsible for the chemical properties of organic compounds.
(ii) I. CH3CH2CH(CH3)OH has the hydroxyl functional group (-OH). II. CH3CH2CH2COOH has the carboxylic/alkanoic acid functional group (-COOH).
4b(i) Complete the following equations: I. Ester + H2O (H+, heat) → ...; II. Ester + H2O (NaOH, heat) → ...
Model answer
I. Acid-catalysed hydrolysis of the ester gives the corresponding alcohol and carboxylic acid, e.g. CH3CH2CH2OH + CH3COOH.
II. Base-catalysed hydrolysis (saponification) gives the sodium salt of the acid and the alcohol, e.g. CH3CH2CH2CH2COONa + CH3OH.
4d(i) Give the reagents required for the following conversions: I. CH2=CH2 to CH2CH2OH; II. CH2CH2OH to CH3COOH; III. CH3COOH to CH3COOCH2CH3.
Model answer
I. H2O and H2SO4 (catalyst) for hydration of ethene to ethanol.
II. Acidified KMnO4/K2Cr2O7 (oxidizing agent) to oxidize the alcohol to the acid.
III. Concentrated H2SO4 and ethanol (esterification, using excess alcohol and acid catalyst).
4e Consider the given unsaturated hydroxy-acid structure. (i) State what would be observed when the compound is treated with each of the following reagents: I. cold NaHCO3(aq); II. hot solution of I2 in NaOH(aq); III. bromine water. (ii) State the functional group responsible for the observations in (i) above.
Model answer
(i) I. Effervescence of CO2 is observed. II. A yellow precipitate is formed. III. The bromine water is decolorized.
(ii) I. Carboxylic group (-COOH) - a group oxidizable to methylcarbonyl group. III. The double carbon-to-carbon bond (C=C) is responsible for decolorizing bromine water.
7(a) A compound X reacts with excess HNO3(aq) to give Carbon (IV) oxide and another compound Y. A solution of Y reacts with NaOH(aq) to form a white precipitate which is soluble in excess NaOH(aq). Identify X and Y.
7(b)(i) Write a balanced equation to illustrate the reducing property of ammonia in its reaction with CuO. (ii) Explain why it is not advisable to heat ammonium dioxonitrate (III) directly. (iii) Give two uses of nitrogen.
Model answer
(i) 3CuO+2NH3→3Cu+3H2O+N2.
(ii) It is not advisable to heat ammonium dioxonitrate (III), NH4NO2, directly because the compound decomposes explosively.
(iii) It is used for the manufacture of NH3; it serves as an important diluent of air; nitrogen in liquid form serves as a cooling agent (any two).
7(c) Give the reason why: (i) dilute H2SO4 is not suitable for the preparation of CO2(g) from CaCO3(s); (ii) concentrated H2SO4 cannot be used to dry ammonia gas.
Model answer
(i) CaCO3 reacts with H2SO4 to form CaSO4, which is insoluble in water. The insoluble layer of CaSO4 covers the CaCO3, preventing further reaction.
(ii) Concentrated H2SO4 cannot be used to dry ammonia gas because the two react: H2SO4+2NH3→(NH4)2SO4.
7(d) State two (i) physical properties (ii) chemical properties of metals.
Model answer
(i) Physical properties: metals are malleable; they are ductile; they are good conductors of heat and electricity; they are lustrous (any two).
(ii) Chemical properties: they ionize by losing electrons; the majority of them displace hydrogen from acids (any two).
8a(i) Explain the term half-life. (ii) Two radioactive elements P and Q have half-lives of 1200 seconds and 3600 seconds respectively. I. Which of the elements is more stable? II. Give a reason for your answer.
Model answer
(i) The half-life of a radioactive material is the time taken for the total amount present originally to decay to half.
(ii) I. Q is more stable. II. Q is more stable because it has a longer half-life.
8a(ii - continued) Draw the energy profile diagram for the reaction H2(g)+I2(g)→2HI(g); ΔH=-13kJmol⁻¹. If the concentration of HI(g) increases from 0.000 to 0.002 moldm⁻³ in 80 seconds, what is the rate of reaction?
Model answer
The reaction is exothermic (ΔH=-13kJmol⁻¹), so the energy profile shows products at a lower energy level than reactants.
Rate = change in concentration / time taken = 0.002/80 = 2.5x10⁻⁵ moldm⁻³s⁻¹.
8b(i) Give one use of each of the following compounds: I. NaHCO3; II. CaSO4; III. CaCO3. (ii) State a drying agent that can be used for each of the following gases: I. SO2; II. HCl; III. NH3.
Model answer
(i) NaHCO3 is used for manufacturing baking powder. CaSO4 is used for making plaster of Paris. CaCO3 is used in the manufacture of glass.
(ii) SO2 can be dried with conc. H2SO4. HCl can be dried with conc. H2SO4. NH3 can be dried with CaO.
8c(i) Write an equation for the complete combustion of carbon in oxygen. (ii) Calculate the number of moles of carbon (IV) oxide produced from the complete combustion of 2.5g of carbon [C=12.0, O=16.0]. (iii) Mention one use of I. Carbon (II) oxide II. Carbon (IV) oxide.
Model answer
(i) C(s)+O2(g)→CO2(g).
(ii) Mole of carbon = 2.5/12 = 0.2083 mol. From the equation, 1 mole of carbon gives 1 mole of CO2, so 0.2083 mol of carbon gives 0.2083 mol of CO2.
(iii) Carbon (II) oxide (CO) is used for the extraction of metals; Carbon (IV) oxide (CO2) is used as a fire extinguisher.
8d An industrial raw material has the following composition by mass: iron=28.1%; chlorine=35.7%; water of crystallization=36.2%. Calculate the formula for the material. [H=1.00, O=16.0, Cl=35.5, Fe=56.0]
Model answer
Fe: 28.1/56 = 0.5018. Cl: 35.7/35.5 = 1.006. H2O: 36.2/18 = 2.011.
Dividing by the smallest (0.5018): Fe:Cl:H2O = 1 : 2 : 4.
Thus, the formula is FeCl2.4H2O.