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WAEC (WASSCE) 2020

2020 WAEC (WASSCE) Chemistry Past Questions and Answers

The 2020 WAEC (WASSCE) Chemistry past questions with answers: 49 objective questions from the 2020 paper in the DeeepSolve question bank, each with its answer behind "Show answer". Try each question before you check, then practise the paper as a timed WAEC-style CBT.

Past questions © the examination body (JAMB/WAEC/NECO). DeeepSolve is not affiliated with them. Rights holders: see /copyright.

Questions

  1. 1.

    An example of a biodegradable pollutant is?

    1. A.plastic
    2. B.sewage
    3. C.carbon (II) oxide
    4. D.hydrogen sulphide
    Show answer

    Answer: B. sewage

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  2. 2.

    which of the following quantities is a molar quantity?

    1. A.Molarity
    2. B.Molar mass
    3. C.Mass concentration
    4. D.Molality
    Show answer

    Answer: B. Molar mass

    AI explanation

    Step 1: Analyze the question The question asks which quantity is a "molar quantity." In chemistry, a molar quantity is a property expressed per mole of a substance. In other words, it is divided by the amount of substance (n).

    Step 2: Explain the concept A molar quantity = (some property) ÷ (amount of substance in moles).

    Examples:

    • Molar mass = mass ÷ moles (g/mol)
    • Molar volume = volume ÷ moles (dm³/mol)
    • Molar enthalpy = enthalpy change ÷ moles (kJ/mol)

    Step 3: Check each option

    • A. Molarity is moles of solute per volume of solution (mol/dm³). It is a concentration, not a per-mole quantity. ✗
    • B. Molar mass is mass per mole (g/mol). It is divided by the amount of substance, so it is a true molar quantity. ✓
    • C. Mass concentration is mass per unit volume (g/dm³). It has no mole in its definition. ✗
    • D. Molality is moles of solute per kg of solvent (mol/kg). Again, a concentration measure. ✗

    Step 4: Answer The correct option is B. Molar mass.

    Tip: Molarity and molality are concentrations (moles per volume or per mass of solvent). A true "molar" quantity has the mole in the denominator, as in "per mole." Ask yourself, "Is it something divided by moles?" If yes, it's molar.

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  3. 3.

    The high solubility of ethanol in water is due to?

    1. A.its low boiling point.
    2. B.its low freezing point.
    3. C.its covalent nature
    4. D.hydrogen bonding
    Show answer

    Answer: D. hydrogen bonding

    AI explanation

    Step 1: Analyze the question We need the reason ethanol (C₂H₅OH) dissolves so well in water.

    Step 2: The concept "Like dissolves like." Solubility depends on how well the attractions between solute and solvent molecules compare with the attractions within each pure substance. Water is a polar molecule that forms hydrogen bonds. Hydrogen bonding occurs when hydrogen is attached to a highly electronegative atom (F, O, or N) and is attracted to a lone pair on another such atom.

    Step 3: Apply it to ethanol

    • Ethanol has an -OH (hydroxyl) group, so its structure is CH₃CH₂–O–H.
    • The O–H bond is highly polar, so the hydrogen is partially positive and the oxygen is partially negative.
    • The -OH group of ethanol forms hydrogen bonds with the water molecules (O–H···O).
    • These new ethanol-water hydrogen bonds are about as strong as the hydrogen bonds in pure water, so the two liquids mix freely (they are miscible).

    Step 4: Check the other options

    • A. Low boiling point: This describes volatility and has no bearing on solubility.
    • B. Low freezing point: This is also unrelated to dissolving.
    • C. Covalent nature: Many covalent compounds, such as hexane and methane, do not dissolve in water, so being covalent does not explain solubility.
    • D. Hydrogen bonding: This is the specific interaction between ethanol and water, so it is correct.

    Answer: D. Hydrogen bonding

    Exam tip: When you see a small molecule with an -OH or -NH₂ group dissolving in water (alcohols, small amines, sugars), think "hydrogen bonding." Remember that as the carbon chain gets longer, solubility drops because the non-polar part dominates.

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  4. 4.

    f 5.0 cm−3^-3 of 0.200 mol dm(^-3 Na Na_2COCO_3 wasdilutedto250cm(−3 was diluted to 250cm(^-3 solution, what would be the concentration of the resulting solution?

    1. A.0.004 mol dm(^-3$
    2. B.0.020 mol dm(^-3$
    3. C.0.200 mol dm(^-3$
    4. D.0.400 mol dm(^-3$
    Show answer

    Answer: A. 0.004 mol dm(^-3$

    AI explanation

    Analyze the question: A small volume (5.0 cm³) of 0.200 mol dm⁻³ Na₂CO₃ is diluted to a final volume of 250 cm³. We need the new concentration.

    Concept: When a solution is diluted, the amount (moles) of solute stays the same. Only the volume changes. This gives the dilution formula:

    C₁V₁ = C₂V₂

    Solution:

    • C₁ = 0.200 mol dm⁻³
    • V₁ = 5.0 cm³
    • V₂ = 250 cm³
    • C₂ = ?

    C₂ = (C₁ × V₁) / V₂ C₂ = (0.200 × 5.0) / 250 C₂ = 1.0 / 250 C₂ = 0.004 mol dm⁻³

    Check with moles: Moles of Na₂CO₃ = 0.200 × (5.0/1000) = 0.001 mol. Concentration in 250 cm³ = 0.001 / 0.250 = 0.004 mol dm⁻³ ✔

    Answer: A. 0.004 mol dm⁻³

    Tip: In dilution questions, the concentration drops by the same factor the volume rises. Here the volume increased 50 times (5 → 250), so the concentration falls 50 times: 0.200 ÷ 50 = 0.004. Also, the volumes can stay in cm³ because they cancel in C₁V₁ = C₂V₂.

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  5. 5.

    Which of the following processes is not exhibited by atoms in order to attain more stable electron configuration?

    1. A.Gaining of electrons
    2. B.Hybridization of orbitals
    3. C.Losing electrons
    4. D.Sharing electron
    Show answer

    Answer: B. Hybridization of orbitals

    AI explanation

    Analyze the question

    The question asks which process atoms do NOT use to reach a stable electron configuration (usually a full outer shell, like the noble gases).

    Explain the concept

    Atoms become stable by achieving an octet (or duplet for H and He). They do this through three main routes:

    1. Gaining electrons: non-metals like Cl and O form anions (ionic bonding). Example: Cl + e⁻ → Cl⁻
    2. Losing electrons: metals like Na and Mg form cations (ionic bonding). Example: Na → Na⁺ + e⁻
    3. Sharing electrons: non-metals form covalent bonds. Example: H₂, H₂O, CH₄

    Evaluate each option

    • A. Gaining of electrons: a way atoms become stable (e.g., Cl⁻).
    • B. Hybridization of orbitals: this is a theoretical model that describes the mixing of atomic orbitals (e.g., sp³ in methane) to explain molecular geometry and bond formation. It is not a process by which atoms gain, lose or share electrons to fill their outer shell. It happens as part of bonding but is not itself a way of attaining a noble gas configuration.
    • C. Losing electrons: a way atoms become stable (e.g., Na⁺).
    • D. Sharing electrons: a way atoms become stable (covalent bonding).

    Answer

    Option B, hybridization of orbitals, is the process not exhibited by atoms as a means of attaining a stable electron configuration.

    Tip

    When you see "attain stable configuration," think of the three electron-transfer routes: gain, lose, share. Anything else, such as hybridization, is a model for explaining shape and not a stabilizing electron process.

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  6. 6.

    The metallic bond in magnesium is stronger than that in calcium because magnesium has a

    1. A.larger atomic size.
    2. B.smaller atomic size
    3. C.greater number of valence electrons
    4. D.lower melting point
    Show answer

    Answer: B. smaller atomic size

    AI explanation

    Analyze the question: We are comparing the strength of metallic bonding in magnesium (Mg) and calcium (Ca). Both are Group 2 elements, so each has 2 valence electrons.

    Concept: Metallic bond strength depends on:

    1. The number of delocalised (valence) electrons contributed per atom.
    2. The charge on the metal cation.
    3. The size of the atom or ion. A smaller size means a higher charge density, so the positive nuclei attract the "sea" of delocalised electrons more strongly.

    Solution:

    • Mg and Ca are both in Group 2, so each has 2 valence electrons. This rules out option C, because the number of valence electrons is the same.
    • Mg (atomic number 12, 2,8,2) has 3 electron shells. Ca (atomic number 20, 2,8,8,2) has 4 shells.
    • So Mg has a smaller atomic (and ionic) radius than Ca.
    • With a smaller radius, the Mg²⁺ ions sit closer to the delocalised electrons. The attraction is stronger, so the metallic bond is stronger.
    • Option A is the opposite of the true situation, since Mg is smaller, not larger.
    • Option D is wrong because a lower melting point means a weaker metallic bond. Mg actually has a slightly higher melting point (about 650 °C) than... in fact Ca melts at about 842 °C, so Mg's melting point is not the reason, and "lower melting point" would not explain a stronger bond anyway.

    Note on the data: Ca has a higher melting point than Mg in real data, which is an exception to the simple trend. However, the question asks for the reason based on the standard WASSCE explanation, which is atomic size. Among the options given, only B fits the reasoning.

    Correct answer: B. Smaller atomic size

    Tip: For metals in the same group, remember "smaller size, stronger bond." For metals in the same period, remember "more valence electrons, stronger bond" (for example, Al is stronger than Mg, which is stronger than Na).

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  7. 7.

    Which of the following arrangement of elements is in decreasing order of electronegativity?

    1. A.Na, Mg, Al, Si, P
    2. B.Na, A1, Mg, P, Si
    3. C.P, Mg, Na, Si, Al
    4. D.P, Si, Al, Mg, Na
    Show answer

    Answer: D. P, Si, Al, Mg, Na

    AI explanation

    Analyze the question: We need the arrangement in which electronegativity goes from highest to lowest.

    Concept: Na, Mg, Al, Si and P are all in Period 3. Across a period from left to right, nuclear charge increases while the electrons are added to the same shell. The atomic radius decreases, so the nucleus attracts bonding electrons more strongly. This means electronegativity increases from left to right across a period.

    Order of the elements in Period 3: Na, Mg, Al, Si, P.

    So electronegativity increases in this order: Na < Mg < Al < Si < P

    Pauling values for checking:

    • Na = 0.93
    • Mg = 1.31
    • Al = 1.61
    • Si = 1.90
    • P = 2.19

    Decreasing order means we reverse the list: P > Si > Al > Mg > Na

    Checking the options:

    • A. Na, Mg, Al, Si, P: this is increasing order, so it is wrong.
    • B. Na, Al, Mg, P, Si: this is jumbled, so it is wrong.
    • C. P, Mg, Na, Si, Al: this is jumbled, so it is wrong.
    • D. P, Si, Al, Mg, Na: this is the correct decreasing order.

    Tip: For Period 3 questions, remember that electronegativity, ionization energy and non-metallic character all increase from left to right. Atomic radius and metallic character decrease. If a question asks for decreasing electronegativity in a period, simply read the period backwards, from the right-hand element to the left-hand one.

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  8. 8.

    The boiling point of pentane is higher than that of propane because?

    1. A.carbon-carbon single bonds are stronger than carbon-hydrogen bonds.
    2. B.Pentane has more covalent bonds to break
    3. C.Pentane does not burn easily as propane
    4. D.The intermolecular forces in pentane are stronger than those of propane
    Show answer

    Answer: D. The intermolecular forces in pentane are stronger than those of propane

    AI explanation

    Analyze the question: We are asked why pentane (C₅H₁₂) boils at a higher temperature (about 36 °C) than propane (C₃H₈, about -42 °C).

    Concept: Boiling is a physical change. When a liquid boils, the molecules separate from one another, but the bonds inside each molecule stay intact. So what matters is the strength of the intermolecular forces, not the covalent bonds within the molecules.

    Alkanes are non-polar, so the only forces between their molecules are weak van der Waals (London dispersion) forces. The strength of these forces increases with molecular size and surface area, because a larger electron cloud is more easily polarised.

    Solution:

    • Pentane has 5 carbon atoms and 12 hydrogen atoms, so it has a larger molar mass and more electrons than propane (3 carbons, 8 hydrogens).
    • A larger electron cloud means stronger London dispersion forces between pentane molecules.
    • More energy (a higher temperature) is needed to overcome these forces, so pentane boils at a higher temperature.

    Checking the options:

    • A. Wrong. Covalent bonds are not broken during boiling. Also, C-C bonds are not stronger than C-H bonds in general.
    • B. Wrong. Covalent bonds are not broken when a liquid boils, so the number of them does not matter.
    • C. Wrong. How easily a substance burns is a chemical property and has nothing to do with boiling point.
    • D. Correct. Pentane's intermolecular forces are stronger than propane's.

    Tip: In "boiling point" questions, remember that boiling or melting breaks intermolecular forces, not covalent bonds. Within a homologous series, a bigger molecule means a higher boiling point.

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  9. 9.

    Which of the following oxides has a giant covalent structure?

    1. A.Al2_2O3_3
    2. B.Na2_2O
    3. C.P4_4O10_10
    4. D.SiO2_2
    Show answer

    Answer: D. SiO2_2

    AI explanation

    Step 1: Analyze the question We need the oxide whose atoms are all joined by covalent bonds in a continuous, three-dimensional network (a giant covalent or macromolecular structure).

    Step 2: Concept: structure of oxides across Period 3

    • Na₂O, MgO (metals on the left): giant ionic lattices, basic oxides.
    • Al₂O₃: ionic with some covalent character, amphoteric oxide, still treated as a giant ionic lattice at WASSCE level.
    • SiO₂: giant covalent (macromolecular) structure.
    • P₄O₁₀, SO₂, SO₃ (non-metals on the right): simple molecular covalent structures, acidic oxides.

    Step 3: Check each option

    • A. Al₂O₃: giant ionic/amphoteric, so not the answer.
    • B. Na₂O: giant ionic lattice of Na⁺ and O²⁻, so not the answer.
    • C. P₄O₁₀: discrete P₄O₁₀ molecules held together by weak van der Waals forces. It is covalent but simple molecular, not giant.
    • D. SiO₂: each Si atom is bonded tetrahedrally to 4 oxygen atoms, and each O is shared between 2 Si atoms. This forms a continuous network like diamond, giving a very high melting point (about 1710 °C) and extreme hardness. This is a giant covalent structure.

    Step 4: Answer The correct option is D. SiO₂ (silicon(IV) oxide, the main component of sand and quartz).

    Exam tip: Remember the trend across Period 3: ionic → giant covalent → simple molecular. Na₂O and MgO are ionic, SiO₂ is the one giant covalent oxide, and P₄O₁₀, SO₂ and SO₃ are simple molecular. Whenever you see "giant covalent" with oxides, think of silicon, just as carbon is the giant covalent element (diamond and graphite).

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  10. 10.

    Group VIl elements in their combined states are called?

    1. A.Halogens
    2. B.Anions
    3. C.Halide
    4. D.Cations
    Show answer

    Answer: C. Halide

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  11. 11.

    When an ionic bond is broken, bonding electrons are

    1. A.shared between participating atoms
    2. B.gained by the most electropositive atom
    3. C.gained by the most electronegative atom
    4. D.lost by both participating atoms.
    Show answer

    Answer: C. gained by the most electronegative atom

    AI explanation

    Analyze the question: It asks what happens to the bonding electrons when an ionic bond is broken. This is about heterolytic fission (heterolysis).

    Concept:

    • In an ionic bond, electrons have already been transferred from the electropositive atom (metal) to the electronegative atom (non-metal). For example, in NaCl, the compound is Na⁺Cl⁻.
    • When the bond is broken (for example, when the compound dissolves or melts), the ions separate. The electron pair stays with the more electronegative atom, which keeps the negative charge as an anion (Cl⁻).
    • In a covalent bond, by contrast, electrons are shared, and they may split evenly if the bond breaks homolytically.

    Evaluating the options:

    • A. Shared between participating atoms: This describes a covalent bond, not an ionic one. ✗
    • B. Gained by the most electropositive atom: The electropositive atom tends to lose electrons, not gain them. ✗
    • C. Gained by the most electronegative atom: The electronegative atom attracts the bonding electrons and keeps them, forming the anion. ✓
    • D. Lost by both atoms: Both atoms cannot lose electrons, because the electrons have to go somewhere. ✗

    **Answer: C.

    Tip: Remember that the electronegative atom "wins" the electrons, both when the bond forms and when it breaks. Think of it as the atom that is "greedy" for electrons. Ionic bond breaking = heterolytic fission, where the electron pair goes to one atom, the more electronegative one.

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  12. 12.

    If 20 cm3^3 of sodium hydroxide was neutralized by 20 cm3^3 of 0.01 mol dm3^3 tetraoxosulphate(VI) acid, what is the concentration of the sodium hydroxide solution?

    1. A.0.010
    2. B.0.020
    3. C.0.100
    4. D.0.150
    Show answer

    Answer: B. 0.020

    AI explanation

    Step 1: Analyze the question We have a neutralization (titration) between NaOH (a base) and H₂SO₄ (tetraoxosulphate(VI) acid). We need the concentration of NaOH.

    Step 2: Balanced equation 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

    The mole ratio is NaOH : H₂SO₄ = 2 : 1. This ratio is the key point, because sulphuric acid is dibasic.

    Step 3: Use the titration formula (C_A × V_A) / (C_B × V_B) = n_A / n_B

    Here the acid is A and the base is B:

    • C_A = 0.01 mol/dm³, V_A = 20 cm³, n_A = 1
    • C_B = ?, V_B = 20 cm³, n_B = 2

    (0.01 × 20) / (C_B × 20) = 1/2

    C_B × 20 = 2 × (0.01 × 20) = 0.4

    C_B = 0.4 / 20 = 0.02 mol/dm³

    Check by moles:

    • Moles of H₂SO₄ = 0.01 × 20/1000 = 2 × 10⁻⁴ mol
    • Moles of NaOH = 2 × 2 × 10⁻⁴ = 4 × 10⁻⁴ mol
    • Concentration of NaOH = 4 × 10⁻⁴ / 0.020 dm³ = 0.02 mol/dm³ ✓

    Answer: B (0.020)

    Exam tip: Always write the balanced equation first. Many students pick 0.010 (option A) because they forget that H₂SO₄ gives two H⁺ ions per molecule, so it neutralizes twice as many moles of NaOH. When the volumes are equal, the base concentration is simply the acid concentration multiplied by 2 in this case.

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  13. 13.

    Which of the following statements is correet for a reaction at equilibrium?

    1. A.All reactions cease to occur
    2. B.The eaction has gone to completion
    3. C.The rates of the forward and backward reactions are equal
    4. D.The amount of product equals the amount of reactants
    Show answer

    Answer: C. The rates of the forward and backward reactions are equal

    AI explanation

    Step 1: Analyze the question You are asked which statement is true for a reaction at equilibrium. This is about dynamic chemical equilibrium in reversible reactions.

    Step 2: The concept In a reversible reaction, e.g. N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the reactants form products (forward reaction), and the products also break back down into reactants (backward reaction).

    At equilibrium:

    • The rate of the forward reaction = the rate of the backward reaction.
    • The reaction has not stopped. It is dynamic, meaning both reactions keep going at the same speed.
    • Because the rates are equal, the concentrations of reactants and products stay constant (not necessarily equal).

    Step 3: Check each option

    • A. All reactions cease to occur. Wrong. Both forward and backward reactions continue; equilibrium is dynamic, not static.
    • B. The reaction has gone to completion. Wrong. At equilibrium, reactants and products coexist. Completion would mean no reactants are left.
    • C. The rates of the forward and backward reactions are equal. Correct. This is the defining condition of equilibrium.
    • D. The amount of product equals the amount of reactants. Wrong. The amounts are constant, but they are usually not equal. Their ratio is determined by the equilibrium constant, Kc.

    Step 4: Answer

    Exam tip: Remember "Dynamic, not dead". At equilibrium, nothing appears to change, but both reactions are still happening at equal rates. Constant does not mean equal. WASSCE and JAMB often test this distinction.

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  14. 14.

    The most suitable process of obtaining water from an aqueous solution of sugar i

    1. A.Crystallization
    2. B.Distillation
    3. C.Filtration
    4. D.Decantation
    Show answer

    Answer: B. Distillation

    AI explanation

    Analyze the question: You have sugar dissolved in water (an aqueous solution). The goal is to get the water out of it, not the sugar.

    Concept: Sugar dissolves completely in water, so it forms a homogeneous mixture. Different separation methods suit different mixtures:

    • Crystallization: recovers the dissolved solid (sugar) by evaporating part of the solvent. The water is lost as vapour, not collected.
    • Distillation: heats the solution so the water boils off as vapour. The vapour is then cooled in a condenser and collected as pure liquid water. The sugar stays behind in the flask because it has a very high boiling point.
    • Filtration: separates an insoluble solid from a liquid. Dissolved sugar passes through filter paper, so this fails.
    • Decantation: pours off a liquid from a settled insoluble solid. It does not work for dissolved substances.

    Solution:

    1. Sugar solution → heat in a distillation flask.
    2. Water (boiling point 100 °C) vaporizes; sugar does not.
    3. Vapour passes through a condenser (cooled by running water) and turns back into liquid.
    4. Pure water collects in the receiver (the distillate), and sugar remains in the flask.

    Answer: Distillation, which is option B.

    Tip: Remember the pairing:

    • Want the solvent (water) from a solution → Distillation
    • Want the dissolved solid → Crystallization/Evaporation
    • Insoluble solid in liquid → Filtration/Decantation

    Keep practicing these separation techniques; they appear often in WASSCE, NECO and JAMB!

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  15. 15.

    Dilution factor is the?

    1. A.amount of distilled water that is added to the concentrated solution to dilute it
    2. B.Quantity of distilled water added to l dm3^3 of the concentrated solution to give the dilute solution
    3. C.number of times the volume of the concentrated solution is diluted to give the dilute solution
    4. D.act of diluting the concentrated solution to obtain the dilute solution
    Show answer

    Answer: C. number of times the volume of the concentrated solution is diluted to give the dilute solution

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  16. 16.

    Equal masses of calcium trioxocarbonate(iv) were added to dilute hydrochloric acid at the temperature specified. Under which of the following conditions would the reaction be slowest?

    1. A.Calcium trioxocarbonate ( iv ) chips a 20°C
    2. B.CalcIum trioxocarbonate (IV) chips at 40°C
    3. C.Calcium trioxocarbonate (IV) powder at 20'C
    4. D.Calcium trioxocarbonate (IV) powder at 40'C
    Show answer

    Answer: A. Calcium trioxocarbonate ( iv ) chips a 20°C

    AI explanation

    Analyze the question: Equal masses of CaCO₃ react with dilute HCl under different conditions of particle size (chips vs powder) and temperature (20°C vs 40°C). We need the condition that gives the slowest rate.

    Concept: Two factors affect the rate here:

    1. Surface area: Powder has a much larger surface area exposed to the acid than chips, so more collisions happen per second and the reaction is faster. Chips have less surface area, so the reaction is slower.
    2. Temperature: A higher temperature gives particles more kinetic energy. More collisions have enough energy to overcome the activation energy, so the reaction is faster. A lower temperature makes it slower.

    Equation: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g)

    Comparing the options:

    • A. Chips, 20°C: small surface area and low temperature, so slowest
    • B. Chips, 40°C: small surface area but high temperature, so moderate
    • C. Powder, 20°C: large surface area but low temperature, so moderate
    • D. Powder, 40°C: large surface area and high temperature, so fastest

    Option A has both the rate-lowering factors (larger lumps and lower temperature), so the reaction is slowest there.

    Answer: A

    Tip: For "slowest" or "fastest" questions, check each factor separately. The slowest case has the lowest temperature, lowest concentration and smallest surface area (big lumps). The fastest case has the opposite of each.

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  17. 17.

    Calcium chloride is an ionic compound. Which of the following statements account for its ionic character? I. Calcium has high ionization energy.  II. Calcium has low ionization energy. III. Chlorine has high electron affinity. IV. Chlorine has high Ionization energy.

    1. A.I and Il only
    2. B.I, II and IV only
    3. C.II, IIl and IV only
    4. D.I, II, III and IV
    Show answer

    Answer: C. II, IIl and IV only

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  18. 18.

    What is the mass of silver deposited when 24,125 C of electricity is passed through a solution or silver salt. [Ag = 108, IF = 96,500 C ]

    1. A.432g
    2. B.108g
    3. C.54g
    4. D.27g
    Show answer

    Answer: D. 27g

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  19. 19.

    An alkanol containing 60% carbon by mass would have a molecular formula [H=1.0. C-12.0, O= 16.0]

    1. A.CH3_3OH
    2. B.C2_2H5_5OH
    3. C.C3_3H7_7OH
    4. D.C4_4H9_9OH
    Show answer

    Answer: C. C3_3H7_7OH

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  20. 20.

    The type of isomerism exhibited by cis and trans isomers is?

    1. A.positional
    2. B.geometrical
    3. C.functional
    4. D.optical
    Show answer

    Answer: B. geometrical

    AI explanation

    Step 1: Analyze the question You are asked what type of isomerism is shown by cis and trans isomers.

    Step 2: The concept Isomers are compounds with the same molecular formula but different arrangements of atoms. The main types are:

    • Positional isomerism: the same functional group sits at different positions on the carbon chain (e.g., propan-1-ol and propan-2-ol).
    • Functional group isomerism: the compounds have different functional groups (e.g., ethanol and methoxymethane).
    • Optical isomerism: the compounds have a chiral carbon and rotate plane-polarized light differently.
    • Geometrical (cis-trans) isomerism: the compounds differ in the spatial arrangement of groups around a C=C double bond (or a ring), because rotation about the bond is restricted.

    Step 3: Apply it Take but-2-ene, CH₃CH=CHCH₃:

    • cis-but-2-ene: both CH₃ groups are on the same side of the double bond.
    • trans-but-2-ene: the CH₃ groups are on opposite sides of the double bond.

    The formula and the order of atoms are the same. Only the spatial arrangement differs, and that is due to the restricted rotation of the C=C bond. This is geometrical isomerism.

    Step 4: Answer The correct option is B. geometrical.

    Exam tip: "Cis/trans" means "geometry". Remember that geometrical isomerism needs two things: a C=C bond (restricted rotation), and two different groups on each carbon of the double bond. For example, but-1-ene cannot show it, because one carbon of its double bond has two identical hydrogen atoms.

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  21. 21.

    Which of the following substances is an example of a fine chemical?

    1. A.Sodium hydroxide
    2. B.Hydrochloric acid
    3. C.Ethylene
    4. D.Ammonia
    Show answer

    Answer: B. Hydrochloric acid

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  22. 22.

    Consider the reaction represented by the following equation: AgNO3_3 + NaCl → AgCl + NaNO3_3. The steps that could be taken to obtain pure dry sample of AgCI (s) from the mixture includes

    1. A.heating to saturation and drying
    2. B.filtering and evaporation to dryness
    3. C.filtering, washing and drying
    4. D.crystallizing and allowing to cool
    Show answer

    Answer: C. filtering, washing and drying

    AI explanation

    Analyze the question: When aqueous AgNO₃ reacts with aqueous NaCl, AgCl forms as an insoluble white precipitate, while NaNO₃ stays dissolved in the water. We need the steps to get pure, dry AgCl solid.

    Concept: This is a precipitation reaction, so the separation method depends on solubility:

    • AgCl is insoluble in water, so it is a solid precipitate.
    • NaNO₃ is soluble, so it remains in solution (the filtrate).

    Solution:

    1. Filter the mixture. The AgCl stays on the filter paper as the residue, and the NaNO₃ solution passes through as the filtrate.
    2. Wash the residue with distilled water. This removes any NaNO₃ solution still clinging to the AgCl, which makes it pure.
    3. Dry the residue, in an oven or between filter papers, to get dry AgCl.

    Checking the other options:

    • A. Heating to saturation is used for obtaining crystals of a dissolved solute, not for separating a precipitate.
    • B. Evaporating to dryness would leave NaNO₃ mixed in with the AgCl, so the sample would be impure.
    • D. Crystallizing and cooling applies to soluble salts, not to an insoluble precipitate.

    Answer: Option C, filtering, washing and drying.

    Tip: For an insoluble solid mixed with a soluble one, remember F-W-D: Filter, Wash, Dry. Use evaporation or crystallization only to recover the soluble substance from the filtrate.

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  23. 23.

    The bond between NH3_3 and H+^+ in NH4+^+ is?

    1. A.dative
    2. B.covalent
    3. C.hydrogen
    4. D.electrovalent
    Show answer

    Answer: A. dative

    AI explanation

    Step 1: Analyze the question The question asks what type of bond forms when an NH₃ molecule joins to an H⁺ ion to make the ammonium ion, NH₄⁺.

    Step 2: The concept

    • In NH₃, nitrogen has 5 valence electrons. Three are shared with hydrogen atoms, leaving one lone pair.
    • H⁺ has no electrons, only an empty 1s orbital.
    • When one atom supplies both electrons of the shared pair, the bond is called a dative covalent (coordinate) bond.

    Step 3: Working

    1. NH₃ has a lone pair on nitrogen.
    2. H⁺ has an empty orbital that can accept a pair of electrons.
    3. Nitrogen donates its lone pair to H⁺, forming N→H.

    H₃N: + H⁺ → [H₃N→H]⁺

    1. Once formed, all four N–H bonds in NH₄⁺ are identical. But the question asks specifically about the bond between NH₃ and H⁺, which is the one formed by donation of the lone pair.

    Step 4: Check the other options

    • B. Covalent: This is partly true, since a dative bond is a type of covalent bond. However, "dative" is the more specific and correct answer, because both electrons come from nitrogen.
    • C. Hydrogen: This is a weak intermolecular attraction (such as between H and N, O or F on different molecules). It is not the bond formed here.
    • D. Electrovalent: This involves the transfer of electrons to form ions, which does not happen here.

    Answer: A. dative

    Exam tip: Whenever you see NH₄⁺, H₃O⁺, or similar ions formed by adding H⁺ to a molecule with a lone pair (NH₃, H₂O), think dative/coordinate bond. The key phrase is "one atom supplies both electrons."

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  24. 24.

    The following molecules have double covalent bonds between two atoms except?

    1. A.oxygen
    2. B.carbon (IV)oxide
    3. C.ethene
    4. D.water
    Show answer

    Answer: D. water

    AI explanation

    Analyze the question: We need the molecule that does NOT contain a double covalent bond between two atoms.

    Concept: A double covalent bond means two atoms share two pairs of electrons (shown as =).

    Checking each option:

    • A. Oxygen (O₂): O=O. Each oxygen atom needs 2 more electrons, so they share two pairs. This is a double bond.
    • B. Carbon (IV) oxide (CO₂): O=C=O. Carbon forms a double bond with each oxygen atom. This has double bonds.
    • C. Ethene (C₂H₄): H₂C=CH₂. There is a C=C double bond. This has a double bond.
    • D. Water (H₂O): H–O–H. Oxygen shares one pair of electrons with each hydrogen atom, and hydrogen needs only one electron to fill its shell (duplet). So only single bonds exist, plus two lone pairs on oxygen.

    Answer: Water is the only molecule with no double bond, so the correct option is D.

    Tip: Hydrogen can only ever form a single bond because it needs just one more electron to complete its duplet. So any molecule where the bonds are all to hydrogen (like H₂O, NH₃, CH₄) will have only single bonds. Remember: O₂ is double-bonded, N₂ is triple-bonded, and CO₂ has two double bonds.

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  25. 25.

    Which of the following compounds has the lowest boiling point?

    1. A.C4_4H10_10
    2. B.CH3_3COOH
    3. C.H2_2O
    4. D.C2_2H5_5OH
    Show answer

    Answer: A. C4_4H10_10

    AI explanation

    Analyze the question: We need the compound with the weakest intermolecular forces, since that gives the lowest boiling point.

    Concept: Boiling point depends on the strength of the forces between molecules:

    • Hydrogen bonding is strong, so these compounds have high boiling points.
    • Van der Waals (London dispersion) forces are weak, so these compounds have low boiling points.

    Solution: Look at each option.

    | Option | Compound | Main intermolecular force | Boiling point | |---|---|---|---| | A | C₄H₁₀ (butane) | Weak van der Waals only (non-polar alkane) | about 0 °C | | B | CH₃COOH (ethanoic acid) | Strong H-bonding, and it forms dimers | about 118 °C | | C | H₂O (water) | Extensive H-bonding | 100 °C | | D | C₂H₅OH (ethanol) | H-bonding | about 78 °C |

    Butane is a non-polar hydrocarbon. It has no -OH group, so it cannot form hydrogen bonds. Its molecules are held together only by weak van der Waals forces, so little energy is needed to separate them.

    The other three compounds all contain an -OH group and form hydrogen bonds, so they boil at much higher temperatures.

    Answer: A. C₄H₁₀

    Exam tip: In boiling-point questions, check for hydrogen bonding first (-OH, -COOH, -NH₂, or H bonded to F, O or N). A compound with none of these, such as an alkane, will usually have the lowest boiling point. Order of strength: H-bonding > dipole-dipole > van der Waals.

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  26. 26.

    The initial volume and pressure of a given mass of gas is V and 3.P. What is its pressure if its volume is increased to  2V at constant temperature?

    1. A.3 P
    2. B.2P
    3. C.frac32frac{3}{2} P
    4. D.frac23frac{2}{3} P
    Show answer

    Answer: C. frac32frac{3}{2} P

    AI explanation

    Analyze the question

    A fixed mass of gas at constant temperature changes volume from V to 2V. We are asked for the new pressure. The initial pressure is written as "3.P", which means 3P (the "." is a typo-style separator, and the options confirm this).

    Concept: Boyle's Law

    At constant temperature, the pressure of a fixed mass of gas is inversely proportional to its volume:

    P₁V₁ = P₂V₂

    Solution

    • P₁ = 3P
    • V₁ = V
    • V₂ = 2V
    • P₂ = ?

    (3P)(V) = P₂(2V)

    P₂ = 3PV / 2V = (3/2)P

    Doubling the volume should halve the pressure, and half of 3P is 1.5P, which matches our result.

    Answer: Option C, (3/2)P.

    Tip: For Boyle's Law, if the volume is multiplied by a factor, the pressure is divided by the same factor. Here the volume is ×2, so the pressure is ÷2: 3P ÷ 2 = 3P/2. Also, watch out for option B (2P), which comes from wrongly treating the initial pressure as P instead of 3P.

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  27. 27.

    Which of the following hydroxides is not readily soluble in water?

    1. A.NH4_4OH
    2. B.Ca(OH)2_2
    3. C.NaOH
    4. D.KOH
    Show answer

    Answer: B. Ca(OH)2_2

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  28. 28.

    Which of the following statements about the solubility of a salt is correct?

    1. A.A salt whose solubility increases with temperature would not crystallize easily on cooling
    2. B.A salt whose solubility is independent of temperature would normally crystallize out on cooling
    3. C.Crystallization would be efficient in separating out a salt whose solubility increases considerably with temperature
    4. D.Solubility of a solid does not affect its crystallization
    Show answer

    Answer: C. Crystallization would be efficient in separating out a salt whose solubility increases considerably with temperature

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  29. 29.

    Which of the following compounds would release hydrogen when reacted with sodium metal?  I. CH3_3COOH II. CH3_3CH2_2OH III. CH3_3COOCH3_3

    1. A.I only
    2. B.I and II only
    3. C.Il and IIl only
    4. D.I and Ill only
    Show answer

    Answer: B. I and II only

    AI explanation

    Analyze the question: Sodium is a reactive metal. It releases hydrogen gas when it reacts with compounds that have an acidic hydrogen, meaning a hydrogen that can be replaced by the metal. In organic chemistry, these are the –OH groups in carboxylic acids and alcohols.

    Concept:

    • Carboxylic acids (–COOH) react with sodium to give a sodium salt and H₂.
    • Alcohols (–OH) react with sodium more slowly to give a sodium alkoxide and H₂.
    • Esters (–COO–R) have no hydrogen attached to oxygen, so they do not react with sodium to release hydrogen.

    Step-by-step solution:

    I. CH₃COOH (ethanoic acid) has an acidic –OH hydrogen: 2CH₃COOH + 2Na → 2CH₃COONa + H₂↑ It releases hydrogen.

    II. CH₃CH₂OH (ethanol) has a hydroxyl group: 2CH₃CH₂OH + 2Na → 2CH₃CH₂ONa + H₂↑ It releases hydrogen (sodium ethoxide forms).

    III. CH₃COOCH₃ (methyl ethanoate) is an ester. It has no –OH group and no replaceable hydrogen, so there is no reaction with sodium and no H₂.

    Answer: Only I and II release hydrogen, so the correct option is B.

    Tip: Remember "O–H releases H₂ with Na". If you see –OH (in an acid or an alcohol), sodium will liberate hydrogen. Esters, ethers, and alkanes will not.

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  30. 30.

    Which or the following substances is a polypeptide?

    1. A.Starch
    2. B.Glycogen
    3. C.Protein
    4. D.Fats
    Show answer

    Answer: C. Protein

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  31. 31.

    "Electrons always occupy the lowest empty energy level" is a statement of

    1. A.Aufbau Principle
    2. B.Hund's rule
    3. C.Pauli Exclusion Principle
    4. D.Periodic law
    Show answer

    Answer: A. Aufbau Principle

    AI explanation

    Step 1: Analyze the question The statement says electrons fill the lowest available energy level first. We need the principle or rule that describes this.

    Step 2: Review the concepts

    • Aufbau Principle: "Aufbau" is German for "building up". Electrons fill orbitals starting from the lowest energy level and move to higher ones. The order is 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p, and so on.
    • Hund's rule: Electrons occupy degenerate orbitals (orbitals of the same energy, such as the three 2p orbitals) singly first, with parallel spins, before they pair up.
    • Pauli Exclusion Principle: No two electrons in an atom can have the same set of four quantum numbers. An orbital holds a maximum of 2 electrons, and they must have opposite spins.
    • Periodic law: The properties of elements are a periodic function of their atomic numbers. It has nothing to do with how electrons fill orbitals.

    Step 3: Match the statement "Electrons always occupy the lowest empty energy level" describes filling orbitals in order of increasing energy. That is the Aufbau Principle.

    Step 4: Answer The correct option is A. Aufbau Principle.

    Exam tip: Link each rule to a keyword:

    • Aufbau = "build up", lowest energy first
    • Hund = "hundred buses": passengers sit singly in each seat before sharing
    • Pauli = "pairs have opposite spins"

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  32. 32.

    How many electrons does  3115{31}_{15}P3−^3- contain?

    1. A.12
    2. B.15
    3. C.16
    4. D.18
    Show answer

    Answer: D. 18

    AI explanation

    Analyze the question: We need the number of electrons in the phosphide ion, P³⁻, where the isotope is written as ³¹₁₅P.

    Concept:

    • The subscript (15) is the atomic number, which equals the number of protons.
    • The superscript (31) is the mass number (protons + neutrons).
    • In a neutral atom, electrons = protons.
    • A charge of 3− means the atom has gained 3 electrons.

    Solution:

    1. Protons = atomic number = 15
    2. Neutral P atom has 15 electrons
    3. P³⁻ gains 3 electrons: 15 + 3 = 18 electrons

    Check: 18 electrons is the same as argon's configuration (2, 8, 8), which is why phosphorus readily forms a 3− ion.

    The mass number 31 is not needed for the electron count. It only tells us there are 31 − 15 = 16 neutrons, which is the trap behind option C.

    Answer: D. 18.

    Tip: For ions, start with the atomic number, then add electrons for negative charges and subtract electrons for positive charges. Remember: anions gain, cations lose.

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  33. 33.

    Which of the following pairs of elements has the greatest difference in electronegativity?

    1. A.Na and F
    2. B.Na and CI
    3. C.Na and Br
    4. D.Na and I
    Show answer

    Answer: A. Na and F

    AI explanation

    Analyze the question: We need the pair with the largest electronegativity difference. Sodium is the same in every pair, so the answer depends on which halogen is most electronegative.

    Concept: Electronegativity is an atom's ability to attract shared electrons. Down a group, it decreases because atomic size and shielding increase. Across a period, it increases. The halogens (Group 17) therefore follow this order:

    F > Cl > Br > I

    Pauling values:

    | Element | Electronegativity | |---|---| | Na | 0.93 | | F | 3.98 | | Cl | 3.16 | | Br | 2.96 | | I | 2.66 |

    Differences with Na:

    • Na and F: 3.98 − 0.93 = 3.05
    • Na and Cl: 3.16 − 0.93 = 2.23
    • Na and Br: 2.96 − 0.93 = 2.03
    • Na and I: 2.66 − 0.93 = 1.73

    Answer: Na and F has the greatest difference, so the correct option is A. This is why NaF is the most ionic of these compounds.

    Tip: When one element is fixed in every pair, just pick the other element that is farthest away on the scale. Fluorine is the most electronegative element in the periodic table, so any pair with F and a metal will usually give the biggest difference.

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  34. 34.

    The oxidation state of chlorine in NaClO3_3 is?

    1. A.+1
    2. B.+3
    3. C.+5
    4. D.+6
    Show answer

    Answer: C. +5

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  35. 35.

    The IUPAC name of the compound represented by the structure below is ?

    1. A.2- chloro but-I ,3-diene.
    2. B.But-1,3-chlorodiene.
    3. C.2-chloro but-diene.
    4. D.3 -chloro but -1.3-diene.
    Show answer

    Answer: C. 2-chloro but-diene.

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  36. 36.

    When 250 cm3 of a saturated solution of CuSO4_4 at 30°C was evaporated to dryness, 5.0 g of the salt was obtained.. What is the solubility of the salt at 30°C? [CuSO4_4 = 160]

    1. A.0.031
    2. B.0.125
    3. C.0.640
    4. D.1.560
    Show answer

    Answer: B. 0.125

    AI explanation

    Analyze the question: A saturated solution of 250 cm³ gives 5.0 g of CuSO₄ on evaporation. We need the solubility, which in WASSCE chemistry is usually given in mol/dm³ (the options are small numbers, so they are molar values). Concept: Solubility (mol/dm³) = (mass of solute per dm³) ÷ molar mass Solution: Step 1: Find the mass in 1000 cm³ (1 dm³). 250 cm³ contains 5.0 g 1000 cm³ contains (5.0 × 1000) ÷ 250 = 20 g Step 2: Convert to moles using the molar mass of 160 g/mol. Moles = 20 ÷ 160 = 0.125 mol Solubility = 0.125 mol/dm³ Answer: B Tip: When the options are small decimals, the solubility is in mol/dm³. First scale the mass up to 1 dm³ (here ×4), then divide by the molar mass. Check the working with the shortcut: 5 g ÷ 160 = 0.03125 mol in 0.25 dm³, so ×4 = 0.125 mol/dm³. Option A (0.031) is the trap answer for students who forget to scale up to 1 dm³.

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  37. 37.

    Which of the following substances is not a reducing agent?

    1. A.C
    2. B.CO
    3. C.O2_2
    4. D.H2_2
    Show answer

    Answer: C. O2_2

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  38. 38.

    What quantity of electrons is lost when one mole of iron (II) ions is oxidized to iron (III) ?

    1. A.0 mole
    2. B.3 moles
    3. C.I mole
    4. D.2 moles
    Show answer

    Answer: C. I mole

    AI explanation

    Analyze the question We need the number of moles of electrons lost when 1 mole of Fe²⁺ is oxidized to Fe³⁺.

    Concept Oxidation is loss of electrons (remember OIL RIG: Oxidation Is Loss, Reduction Is Gain). The number of electrons lost per ion equals the change in charge (oxidation state).

    Solution The half-equation is:

    Fe²⁺ → Fe³⁺ + e⁻

    • Oxidation state changes from +2 to +3, which is an increase of 1.
    • So each Fe²⁺ ion loses 1 electron.
    • 1 mole of Fe²⁺ therefore loses 1 mole of electrons (6.02 × 10²³ electrons).

    Checking the options

    • A. 0 mole: wrong, because oxidation always involves electron loss.
    • B. 3 moles: wrong, because that would be Fe → Fe³⁺ (from the neutral atom).
    • C. 1 mole: correct (the option "I mole" is 1 mole).
    • D. 2 moles: wrong, because that would be Fe → Fe²⁺.

    Tip Count only the change in charge. Fe²⁺ → Fe³⁺ is a change of +1, so 1 electron is lost. Do not count from the neutral atom unless the question starts with Fe.

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  39. 39.

    A factor that is considered most important when siting a chemical industry is?

    1. A.nearness to other industrial establishments
    2. B.nearness to raw materials
    3. C.favourable climate conditions
    4. D.availability of storage facilities
    Show answer

    Answer: B. nearness to raw materials

    AI explanation

    Analyze the question: It asks which factor matters most when choosing a location (siting) for a chemical industry.

    Explain the concept: Industrial location depends on several factors: raw materials, power, labour, transport, market, water, and government policy. For chemical industries, the raw materials are usually bulky, heavy, or costly to transport. Some are also hazardous or perishable. Locating the plant near its raw materials therefore cuts transport costs and reduces losses, which makes production more economical.

    Solution, option by option:

    • A. Nearness to other industrial establishments: This can help, but it is not the main factor. It is a secondary advantage, not a necessity.
    • B. Nearness to raw materials: This is the most important factor. The raw materials are the foundation of production, and the cost of moving them heavily affects profit. Examples are cement plants sited near limestone, such as Ewekoro, and petrochemical plants near crude oil sources in the Niger Delta.
    • C. Favourable climate conditions: Climate matters little for most chemical processes, since they take place in controlled plants.
    • D. Availability of storage facilities: Useful, but these can be built anywhere, so this is not the deciding factor.

    Exam tip: When a question asks for the "most important" factor in siting an industry, think "raw materials first." This is especially true for industries whose raw materials are bulky and whose finished product weighs less (weight-losing industries). Remember it as: "No raw materials, no products."

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  40. 40.

    Which of the following solids would not decompose on heating?

    1. A.Ammonium chloride
    2. B.Lead (i) trioxonitrate (V)
    3. C.Potassium trioxocarbonate (IV)
    4. D.Sodium hydrogen trioxocarbonate (IV)
    Show answer

    Answer: C. Potassium trioxocarbonate (IV)

    AI explanation

    Analyzing the question: We need the solid that is thermally stable, meaning it does not break down when heated with a Bunsen burner.

    Concept: Thermal stability of salts depends on the metal present. Salts of very reactive metals (K, Na) are the most stable. Their carbonates resist decomposition under ordinary laboratory heating. The exception is that hydrogen carbonates decompose easily.

    Checking each option:

    • A. Ammonium chloride (NH₄Cl): It decomposes (sublimes and dissociates) on heating: NH₄Cl(s) ⇌ NH₃(g) + HCl(g) It decomposes.
    • B. Lead(II) trioxonitrate(V), Pb(NO₃)₂: Heavy-metal nitrates decompose, giving the oxide, nitrogen dioxide and oxygen: 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂ It decomposes.
    • C. Potassium trioxocarbonate(IV), K₂CO₃: Potassium is highly electropositive, so its carbonate is very stable. It does not decompose at ordinary laboratory temperatures. It does not decompose.
    • D. Sodium hydrogen trioxocarbonate(IV), NaHCO₃: It decomposes easily on heating: 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂ It decomposes.

    Answer: Option C.

    Exam tip: Remember that Group 1 carbonates (except lithium) are stable to heat. Hydrogen carbonates, nitrates, and ammonium salts almost always decompose. Think "K and Na carbonates are tough, everything else breaks down."

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  41. 41.

    What volume of oxygen at s.t.p is required to burn completely 7.5 dm3^3 of methane according to the following equation? CH4_4 g_g + 202_2 g_g  → CO2_2 g_g + H2_2Og_g

    1. A.3.75 dm3^3
    2. B.7.59 dm3^3
    3. C.15.0 dm3^3
    4. D.30.0dm3^3
    Show answer

    Answer: C. 15.0 dm3^3

    AI explanation

    Analyze the question: We need the volume of oxygen that burns 7.5 dm³ of methane completely. Both are gases at the same conditions, so we can use Gay-Lussac's law of combining volumes.

    Concept: At the same temperature and pressure, the volume ratio of gases equals the mole ratio in the balanced equation (Avogadro's law).

    Balanced equation: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

    (The question prints "20₂", which is meant to be 2O₂, and the equation as written is not fully balanced because the water needs a coefficient of 2. This does not affect the oxygen ratio.)

    Solution:

    • From the equation, 1 volume of CH₄ reacts with 2 volumes of O₂.
    • Volume of O₂ = 2 × 7.5 dm³ = 15.0 dm³

    Answer: 15.0 dm³, which is option C.

    Tip: For gas-volume questions, you do not need to convert to moles or use 22.4 dm³. If all the gases are at the same conditions, just read the coefficients as volume ratios. Also, always check that the equation is balanced first.

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  42. 42.

    Which of the following reactions are always exothermic? 1. Neutralization Il. Decomposition III. Combustion

    1. A.I and II only
    2. B.I and III only
    3. C.II and IIl only
    4. D.I, II and III
    Show answer

    Answer: B. I and III only

    AI explanation

    Analyze the question: We need to identify which reaction types always release heat (ΔH negative).

    Concept: An exothermic reaction releases heat to the surroundings. Let's check each type:

    • I. Neutralization: An acid reacting with a base to form salt and water always releases heat. For a strong acid and strong base, H⁺ + OH⁻ → H₂O has ΔH ≈ -57.3 kJ/mol. Always exothermic.
    • II. Decomposition: Breaking compounds into simpler substances usually requires energy (for example, CaCO₃ → CaO + CO₂ is endothermic, and so is the electrolysis of water). It is not always exothermic. Some decompositions (such as that of hydrogen peroxide) are exothermic, but that is not true in general.
    • III. Combustion: Burning a substance in oxygen always releases heat and light. Always exothermic.

    Solution: The reactions that are always exothermic are I (neutralization) and III (combustion).

    Correct answer: B (I and III only)

    Tip: Remember the pairing "Neutralization and Combustion = always release heat," while decomposition usually needs heat (think of "thermal decomposition," where you must heat the substance). Watch for the word "always" in exam questions, because it rules out types with exceptions.

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  43. 43.

    Which of the following pairs of properties of alkali metals decreases down the group?

    1. A.First ionization energy and reactivity
    2. B.melting point and atomic radius
    3. C.Reactivity and electronegativity
    4. D.First ionization energy and melting point
    Show answer

    Answer: D. First ionization energy and melting point

    AI explanation

    Analyze the question: We need the pair in which both properties decrease as we go down Group 1 (Li → Na → K → Rb → Cs).

    Concept: Going down the group, the number of electron shells increases, so the atomic radius increases. The outer electron is farther from the nucleus and better shielded, so it is held less strongly.

    Check each property:

    • Atomic radius: increases
    • First ionization energy: decreases (the outer electron is easier to remove)
    • Electronegativity: decreases (less attraction for bonding electrons)
    • Reactivity: increases (metals lose their outer electron more easily)
    • Melting point: decreases (metallic bonds weaken as the atoms get larger, e.g. Li 181°C, Na 98°C, K 63°C, Cs 29°C)

    Check each option:

    • A. First ionization energy decreases, but reactivity increases. ✗
    • B. Melting point decreases, but atomic radius increases. ✗
    • C. Reactivity increases, but electronegativity decreases. ✗
    • D. First ionization energy decreases and melting point decreases. ✓

    **Answer: D.

    Tip: For alkali metals going down, remember "Reactivity and radius go UP; almost everything else (ionization energy, electronegativity, melting point, density trend aside) goes DOWN." Any option pairing reactivity or radius with a decreasing property is a trap.

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  44. 44.

    The atomic number of an atom would be equal to its mass number if it

    1. A.has a totally filled valence shell
    2. B.has a high charge to mass ratio
    3. C.does not contain neutrons
    4. D.exhibits isotopy
    Show answer

    Answer: C. does not contain neutrons

    Get a step-by-step AI explanation in the app

  45. 45.

    How many moles of H2_2SO4 are there in 50 cm3^3 of 0.108 mol dm3^3 solution of the acid?

    1. A.5.4x 10
    2. B.5.4 x 10−3^{-3}
    3. C.5.4 x 10−2^{-2}
    4. D.5.4x 10−1^{-1}
    Show answer

    Answer: B. 5.4 x 10−3^{-3}

    AI explanation

    Analyze the question: You are asked for the number of moles of H₂SO₄ in 50 cm³ of a 0.108 mol/dm³ solution.

    Concept: Moles = concentration (mol/dm³) × volume (dm³)

    Since concentration is in mol/dm³, the volume must be converted from cm³ to dm³ first: 1 dm³ = 1000 cm³

    Solution:

    1. Convert the volume: 50 cm³ = 50 ÷ 1000 = 0.050 dm³
    1. Apply the formula: n = C × V n = 0.108 × 0.050 n = 0.0054 mol
    1. Write in standard form: 0.0054 = 5.4 × 10⁻³ mol

    Check against the options:

    • A. 5.4 × 10 = 54 (too large)
    • B. 5.4 × 10⁻³ = 0.0054 ✔
    • C. 5.4 × 10⁻² = 0.054 (this comes from a decimal slip)
    • D. 5.4 × 10⁻¹ = 0.54 (too large)

    Exam tip: Always convert cm³ to dm³ by dividing by 1000 before using n = C × V. A quick shortcut: moles = (C × V in cm³) ÷ 1000, so 0.108 × 50 = 5.4, then 5.4 ÷ 1000 = 5.4 × 10⁻³. Also, the 2 in H₂SO₄ doesn't matter here, because you are counting moles of the acid itself, not moles of H⁺ ions.

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  46. 46.

    Which of the following standard conditions is not correct about energy changes?

    1. A.Standard temperature is 298K
    2. B.B. Standard pressure is 1 atm
    3. C.Concentration of solution must be 1 mol/dm3^3
    4. D.ΔH at 298 K is the activation energy
    Show answer

    Answer: D. ΔH at 298 K is the activation energy

    AI explanation

    Analyzing the question: We need the statement that is NOT correct about the standard conditions used for energy changes (thermochemistry).

    Concept: Standard conditions for measuring enthalpy changes (ΔH°) are:

    • Temperature: 298 K (25 °C)
    • Pressure: 1 atm (101.325 kPa, about 100 kPa)
    • Concentration of solutions: 1 mol/dm³
    • Substances in their standard physical states

    Checking each option:

    • A. Standard temperature is 298 K. Correct. This is the standard temperature for thermochemical measurements.
    • B. Standard pressure is 1 atm. Correct. This is the standard pressure used in WASSCE chemistry.
    • C. Concentration of solution must be 1 mol/dm³. Correct. Solutions must have unit concentration.
    • D. ΔH at 298 K is the activation energy. Incorrect. ΔH is the enthalpy change, the difference between the energy of the products and the energy of the reactants (ΔH = H_products − H_reactants). Activation energy (Ea) is the minimum energy the reactants need for the reaction to occur, which is the energy barrier from the reactants up to the transition state. They are different quantities, and ΔH does not become the activation energy just because it is measured at 298 K.

    Answer: D

    Tip: Remember the standard conditions as "298, 1, 1": 298 K, 1 atm, 1 mol/dm³. Also keep in mind that ΔH is the overall energy difference, while activation energy is the energy hill the reactants must climb. Any option that mixes the two up is likely to be the wrong statement.

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  47. 47.

    Which or the following products could be formed during incomplete combustion of a hydrocarbon. i.Carbon ii. Hydrogen III. Carbon (Il) oxide

    1. A.I only
    2. B.I and II only
    3. C.I and III only
    4. D.II and III only
    Show answer

    Answer: C. I and III only

    AI explanation

    Analyze the question: We need to identify which products can form when a hydrocarbon burns in a limited supply of oxygen (incomplete combustion).

    Concept:

    • Complete combustion (plenty of oxygen): hydrocarbon + O₂ → CO₂ + H₂O
    • Incomplete combustion (limited oxygen): the carbon is not fully oxidised, so you get carbon monoxide (CO), and/or soot (carbon, C), along with water.

    Equations:

    • CH₄ + 3/2 O₂ → CO + 2H₂O (carbon(II) oxide formed)
    • CH₄ + O₂ → C + 2H₂O (carbon/soot formed)

    Check each statement:

    • i. Carbon: Yes. It appears as soot (a black deposit), as seen on a smoky kerosene lamp or a candle flame with limited air.
    • ii. Hydrogen: No. Hydrogen in the hydrocarbon is oxidised easily and ends up as water (H₂O), not as hydrogen gas.
    • iii. Carbon(II) oxide: Yes. It is the poisonous gas CO, produced in incomplete combustion (for example, from generator exhaust or charcoal burning in a closed room).

    So the correct statements are i and iii only.

    **Answer: C. I and III only.

    Exam tip: Remember that incomplete combustion gives C, CO, and H₂O, while complete combustion gives CO₂ and H₂O. Hydrogen gas is never a product of burning a hydrocarbon in oxygen. Also remember that CO is dangerous because it binds to haemoglobin, a point that often appears in follow-up questions.

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  48. 48.

    A balanced chemical equation is based on the law of?

    1. A.Periodicity
    2. B.Constant composition
    3. C.Multiple proportion
    4. D.Conservation of mass
    Show answer

    Answer: D. Conservation of mass

    Get a step-by-step AI explanation in the app

  49. 49.

    Which of the following metals does not react with water to produce hydrogen?

    1. A.Copper
    2. B.Potassium
    3. C.Sodium
    4. D.Zinc
    Show answer

    Answer: A. Copper

    AI explanation

    Step 1: Analyze the question We need the metal that does not react with water to give hydrogen gas. This is a question about the reactivity series.

    Step 2: The concept In the reactivity series, the order from most to least reactive is:

    K > Na > Ca > Mg > Al > Zn > Fe > Pb > H > Cu > Ag > Au

    • Metals above hydrogen can displace it from water or acid, in some form. Very reactive ones (K, Na, Ca) react with cold water. Mg reacts with hot water or steam. Zn and Fe react with steam.
    • Metals below hydrogen (Cu, Ag, Au) do not displace hydrogen from water at all.

    Step 3: Check each option

    • A. Copper: It is below hydrogen in the series, so it does not react with water or steam. ✔
    • B. Potassium: It reacts violently with cold water: 2K + 2H₂O → 2KOH + H₂
    • C. Sodium: It reacts vigorously with cold water: 2Na + 2H₂O → 2NaOH + H₂
    • D. Zinc: It does not react with cold water, but it reacts with steam: Zn + H₂O(g) → ZnO + H₂. So it does produce hydrogen under the right conditions.

    Step 4: Answer Copper is the metal that does not react with water or steam to produce hydrogen.

    Tip: Remember the reactivity series with the phrase "Please Stop Calling Me A Careless Zebra, Instead Try Learning How Copper Saves Gold" (K, Na, Ca, Mg, Al, C, Zn, Fe, Sn, Pb, H, Cu, Ag, Au). Anything below H (Cu, Ag, Au) will not give hydrogen with water or dilute acids. In "which does not react" questions, pick the metal lowest in the series.

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