2018 JAMB Chemistry Past Questions — Part 2

Questions 45–87

45. 45.

    The end products of burning a candle in the atmosphere are water and

    1. A.carbon (II) oxide
    2. B.sulphur (IV) oxide
    3. C.carbon (IV) oxide
    4. D.sulphur (VI) oxide

    Show answer

    Answer: C. carbon (IV) oxide

    AI explanation

    Analysis

    The question asks for the products formed when a candle burns in the atmosphere. A candle is primarily made of paraffin wax, which is a hydrocarbon (a compound consisting of carbon and hydrogen).

    Concept

    Combustion is a chemical reaction between a substance (the fuel) and oxygen. When a hydrocarbon burns completely in the presence of sufficient oxygen (as found in the atmosphere), it undergoes complete combustion.

    The general word equation for the complete combustion of a hydrocarbon is: Hydrocarbon + Oxygen → Carbon (IV) Oxide + Water

    • Carbon (IV) Oxide ($CO_2$): Produced from the oxidation of the carbon atoms in the wax.
    • Water ($H_2O$): Produced from the oxidation of the hydrogen atoms in the wax.

    Step-by-Step Solution

    1. Identify the fuel: A candle is a hydrocarbon ($C_nH_{2n+2}$).
    2. Identify the reactant: The candle burns in the atmosphere, which contains oxygen ($O_2$).
    3. Determine the reaction type: Burning in the atmosphere implies an excess of oxygen, leading to complete combustion.
    4. Analyze the products: • The carbon in the wax reacts with oxygen to form Carbon (IV) oxide ($CO_2$). • The hydrogen in the wax reacts with oxygen to form Water ($H_2O$).
    5. Evaluate the options: • A: Carbon (II) oxide ($CO$) is a product of incomplete combustion (usually when oxygen is limited). • B & D: Sulphur oxides are produced only if the fuel contains sulphur impurities, which is not the primary reaction for a candle. • C: Carbon (IV) oxide ($CO_2$) is the correct product of complete combustion.

    Conclusion

    The complete combustion of the hydrocarbon wax in a candle yields water and carbon (IV) oxide.

    Exam Tip: Remember that "Carbon (IV) oxide" is the IUPAC name for Carbon dioxide ($CO_2$). Always look for the "complete combustion" products ($CO_2$ and $H_2O$) when dealing with hydrocarbons in exam questions.

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46. 46.

    The salt that reacts with dilute hydrochloric acid to produce a pungent smelling gas which decolourizes acidified purple potassium tetraoxomanganate (VII) solution is

    1. A.Na$_2$SO$_4$
    2. B.Na$_2$SO$_3$
    3. C.Na$_2$S
    4. D.Na$_2$CO$_3$

    Show answer

    Answer: B. Na$_2$SO$_3$

    AI explanation

    Analysis

    To solve this question, we need to identify which of the given salts reacts with dilute hydrochloric acid (HCl) to produce a gas that:

    1. Has a pungent smell.
    2. Decolourizes acidified purple potassium tetraoxomanganate (VII) ($KMnO_4$).

    Concept Explanation

    • Na$_2$SO$_4$ (Sodium tetraoxosulphate VI): Does not react with dilute HCl to produce a gas.
    • Na$_2$SO$_3$ (Sodium trioxosulphate IV): Reacts with dilute HCl to produce sulphur (IV) oxide ($SO_2$) gas. $SO_2$ is a colourless gas with a characteristic pungent, choking smell (like burnt matches). It acts as a reducing agent, which causes the purple $KMnO_4$ to turn colourless.
    • Na$_2$S (Sodium sulphide): Reacts with dilute HCl to produce hydrogen sulphide ($H_2S$) gas. While $H_2S$ has a smell, it smells like rotten eggs, not a pungent choking smell, and it produces a yellow precipitate of sulphur when reacting with $KMnO_4$.
    • Na$_2$CO$_3$ (Sodium trioxocarbonate IV): Reacts with dilute HCl to produce carbon (IV) oxide ($CO_2$). $CO_2$ is odourless and does not decolourize acidified $KMnO_4$.

    Step-by-Step Solution

    1. Reaction: When sodium trioxosulphate (IV) reacts with hydrochloric acid: $$Na_2SO_3(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H_2O(l) + SO_2(g)$$
    2. Gas Identification: The gas produced is $SO_2$.
    3. Chemical Test: $SO_2$ is a strong reducing agent. When bubbled through acidified $KMnO_4$, it reduces the manganese from the +7 oxidation state (purple) to the +2 oxidation state (colourless): $$5SO_2 + 2MnO_4^- + 2H_2O \rightarrow 5SO_4^{2-} + 2Mn^{2+} + 4H^+$$ This confirms the decolourisation process.

    Conclusion

    The salt that satisfies both conditions (pungent smell and decolourisation of $KMnO_4$) is sodium trioxosulphate (IV).

    Exam Tip

    Whenever you see "decolourizes acidified $KMnO_4$," immediately think of reducing agents. Common reducing agents in your syllabus include $SO_2$, $H_2S$, and $Fe^{2+}$ salts. If the question mentions a "pungent smell," $SO_2$ is almost always the intended answer!

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47. 47.

    A quantity of electricity liberates 3.6g of Silver from its salt. What mass of aluminium Will be liberated from its salt by the same quantity of electricity? [Al = 27, Ag = 108].

    1. A.2.7g
    2. B.1.2g
    3. C.0.9g
    4. D.0.3g

    Show answer

    Answer: D. 0.3g

    AI explanation

    To solve this problem, we use Faraday’s Laws of Electrolysis. The key principle here is that the same quantity of electricity (charge, $Q$) will liberate masses of different elements in proportion to their equivalent masses.

    1. Analyze the question

    • Given: • Mass of Silver ($Ag$) liberated = $3.6\text{ g}$ • Molar mass of $Ag = 108\text{ g/mol}$ • Molar mass of $Al = 27\text{ g/mol}$
    • Required: Mass of Aluminium ($Al$) liberated by the same quantity of electricity.

    2. Explain the concept

    The relationship between the mass ($m$) of a substance deposited and the quantity of electricity ($Q$) is given by:

    $m = \frac{M \times Q}{n \times F}$

    Where:

    • $M$ = Molar mass
    • $n$ = Number of electrons transferred (valency)
    • $F$ = Faraday’s constant ($96500\text{ C/mol}$)

    Since the quantity of electricity ($Q$) is the same for both, we can use the ratio:

    $\frac{m_1}{E_1} = \frac{m_2}{E_2}$

    Where $E$ is the equivalent mass ($E = \frac{\text{Molar Mass}}{\text{Valency}}$).

    3. Show the solution

    Step 1: Determine the valency ($n$) of the metals.

    • Silver ($Ag$) is in Group 1, so its valency is $1$.
    • Aluminium ($Al$) is in Group 3, so its valency is $3$.

    Step 2: Calculate the equivalent mass for each.

    • Equivalent mass of $Ag = \frac{108}{1} = 108$
    • Equivalent mass of $Al = \frac{27}{3} = 9$

    Step 3: Use the ratio formula.

    $\frac{\text{Mass of } Ag}{\text{Equivalent mass of } Ag} = \frac{\text{Mass of } Al}{\text{Equivalent mass of } Al}$

    $\frac{3.6}{108} = \frac{\text{Mass of } Al}{9}$

    Step 4: Solve for the Mass of Aluminium.

    $\text{Mass of } Al = \frac{3.6 \times 9}{108}$

    $\text{Mass of } Al = \frac{32.4}{108}$

    $\text{Mass of } Al = 0.3\text{ g}$

    4. Identify the answer

    Comparing our result to the options provided, the correct answer is D.

    5. Exam Tip

    Always remember that the equivalent mass is simply the Molar Mass divided by the valency. If you ever forget the formula, just remember: "Mass deposited is directly proportional to the equivalent mass." Also, always check the valency of the metal—common ones like $Al^{3+}$, $Mg^{2+}$, and $Ag^+$ are frequently tested in JAMB!

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48. 48.

    Which of the following metals cannot replace hydrogen from water or steam?

    1. A.Sodium
    2. B.Magnesium
    3. C.Iron
    4. D.Copper

    Show answer

    Answer: D. Copper

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49. 49.

    2KClO$_3$(s) → 2KCl(s) + 3O$_2$(g) The importance of the catalyst in the reaction above is that

    1. A.heating may not be required before the reaction takes place
    2. B.the reaction is controllable even at a high temperature
    3. C.the reaction produces large quantity of oxygen
    4. D.the reaction takes place more rapidly at a lower temperature

    Show answer

    Answer: D. the reaction takes place more rapidly at a lower temperature

    AI explanation

    Analysis of the Question

    The question asks for the specific role of a catalyst in the thermal decomposition of potassium trioxochlorate(V) ($KClO_3$). In this reaction, manganese(IV) oxide ($MnO_2$) is typically added as a catalyst.

    Explanation of the Concept

    A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed in the process. It achieves this by providing an alternative reaction pathway with a lower activation energy.

    • Activation Energy: This is the minimum amount of energy required for reactant particles to collide effectively and form products.
    • Effect of Catalyst: By lowering the activation energy, more reactant molecules possess sufficient energy to react at a given temperature. Consequently, the reaction proceeds much faster than it would without the catalyst.

    Step-by-Step Solution

    1. Examine the reaction: $2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)$ is a thermal decomposition reaction. Normally, this reaction requires very high temperatures to break the bonds in $KClO_3$.
    2. Role of $MnO_2$: When $MnO_2$ is added, it acts as a catalyst. It does not change the amount of oxygen produced (that is determined by the stoichiometry of the equation), nor does it make the reaction "controllable" in the sense of safety.
    3. Impact on Temperature: Because the catalyst lowers the activation energy, the reaction can occur at a significantly lower temperature than would be required if the catalyst were absent. This makes the process more efficient and safer to manage in a laboratory setting.
    4. Evaluate Options: • A: Heating is still required; the catalyst just makes it easier. • B: While it helps, the primary scientific definition of the catalyst's role here is the speed and energy requirement. • C: The quantity of oxygen is determined by the moles of $KClO_3$ used, not the catalyst. • D: This correctly describes the kinetic effect of the catalyst: it speeds up the reaction at a lower temperature than would otherwise be needed.

    Conclusion

    The catalyst allows the decomposition to occur at a faster rate at a lower temperature, saving energy and reducing the risk associated with extreme heating.

    Exam Tip

    Whenever you see a question about a catalyst, look for keywords like "lowers activation energy" or "increases reaction rate." Remember: a catalyst never changes the yield (amount) of the product; it only changes how fast you get there!

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50. 50.

    To what temperature must a gas at 273k be heated in order to double both its volume and pressure?

    1. A.298K
    2. B.546K
    3. C.819K
    4. D.1092K

    Show answer

    Answer: D. 1092K

    AI explanation

    Analysis

    This question tests your understanding of the General Gas Law, which combines Boyle’s Law, Charles’s Law, and the Pressure Law. The formula is:

    $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

    Concept

    The General Gas Law states that for a fixed mass of gas, the ratio of the product of pressure and volume to the temperature remains constant.

    • $P_1, V_1, T_1$ are the initial pressure, volume, and temperature.
    • $P_2, V_2, T_2$ are the final pressure, volume, and temperature.

    Step-by-Step Solution

    1. Identify the given values: • Initial temperature ($T_1$) = $273\text{ K}$ • Initial pressure = $P_1$ • Initial volume = $V_1$

    2. Identify the final conditions: • The pressure is doubled: $P_2 = 2P_1$ • The volume is doubled: $V_2 = 2V_1$ • Final temperature ($T_2$) = ?

    3. Substitute these into the formula: $$\frac{P_1 \times V_1}{273} = \frac{(2P_1) \times (2V_1)}{T_2}$$

    4. Simplify the equation: $$\frac{P_1V_1}{273} = \frac{4P_1V_1}{T_2}$$

    5. Cancel out $P_1$ and $V_1$ from both sides: $$\frac{1}{273} = \frac{4}{T_2}$$

    6. Solve for $T_2$: $$T_2 = 273 \times 4$$ $$T_2 = 1092\text{ K}$$

    Conclusion

    By doubling both the pressure and the volume, the temperature must increase by a factor of $2 \times 2 = 4$. Multiplying the initial temperature of $273\text{ K}$ by $4$ gives $1092\text{ K}$.

    Exam Tip: When dealing with gas law problems, always ensure your temperature is in Kelvin. If a question gives you temperature in Celsius, add $273$ to it before plugging it into your formula. Also, remember that if both $P$ and $V$ increase, the temperature must increase significantly to maintain the balance!

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51. 51.

    2SO$_2$ $_{(g) }$+ O$_2$ $_{(g) }$ ↔ 2SO$_3$$_{(g) }$ ΔH = -395.7kJmol$^{-1}$ In the equation, an increase in temperature will shift the equilibrium position to the

    1. A.left and equilibrium constant decreases
    2. B.left and equilibrium constant increases
    3. C.right and equilibrium constant increases
    4. D.right and equilibrium constant decreases

    Show answer

    Answer: A. left and equilibrium constant decreases

    AI explanation

    Analysis

    This question tests your understanding of Le Chatelier’s Principle and the relationship between temperature and the equilibrium constant ($K$) in reversible reactions.

    Concept Explanation

    1. Le Chatelier’s Principle: This principle states that if a system at equilibrium is subjected to a change in temperature, pressure, or concentration, the system will adjust itself to counteract that change.
    2. Effect of Temperature on Equilibrium Position: • The reaction provided is exothermic because the enthalpy change ($\Delta H$) is negative ($-395.7 \text{ kJmol}^{-1}$). • In an exothermic reaction, heat is treated as a "product." • Increasing the temperature adds more heat to the system. According to Le Chatelier’s Principle, the system will shift to the left (the reactant side) to absorb the excess heat.
    3. Effect of Temperature on Equilibrium Constant ($K$): • For exothermic reactions, increasing the temperature decreases the value of the equilibrium constant ($K$). • Conversely, for endothermic reactions, increasing the temperature increases the value of $K$.

    Step-by-Step Solution

    • Step 1: Identify the nature of the reaction. Since $\Delta H$ is negative, the reaction is exothermic.
    • Step 2: Determine the shift. Increasing the temperature of an exothermic reaction favors the backward reaction (the reverse direction). Therefore, the equilibrium shifts to the left.
    • Step 3: Determine the effect on $K$. Because the reaction is exothermic, an increase in temperature causes the equilibrium constant ($K$) to decrease.

    Combining these two findings, an increase in temperature shifts the equilibrium to the left and causes the equilibrium constant to decrease.

    Conclusion

    The correct option is A.

    Exam Tip

    Remember the acronym "MEX": More heat (temperature) in an EXothermic reaction shifts the equilibrium to the left (the side of the reactants). Also, always associate "Exothermic" with "Decrease" in $K$ when temperature rises.

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52. 52.

    The Sulphide which is insoluble in dilute hydrochloric acid is

    1. A.FeS
    2. B.CuS
    3. C.ZnS
    4. D.Na2S

    Show answer

    Answer: B. CuS

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53. 53.

    Aluminium does not react with either dilute or concentrated trioxonitrate (V) acid because

    1. A.it is insoluble in either dilute or concentrated trioxonitrate (V) acid
    2. B.it is lower than hydrogen in the electrochemical series
    3. C.the reaction with trioxonitrate (V) acid has a light activation energy
    4. D.an insoluble oxide is formed on its surface which renders it passive

    Show answer

    Answer: D. an insoluble oxide is formed on its surface which renders it passive

    AI explanation

    Analysis of the Question

    The question asks why aluminium, a reactive metal, remains unreactive when placed in trioxonitrate(V) acid ($HNO_3$), regardless of whether the acid is dilute or concentrated.

    Concept Explanation: Passivity

    In chemistry, passivity refers to a phenomenon where a metal becomes "passive" or unreactive despite its position in the reactivity series. This happens when a thin, tough, and non-porous layer of an oxide forms on the surface of the metal. This layer acts as a protective barrier, preventing the underlying metal from coming into contact with the acid, thereby stopping any further chemical reaction.

    In the case of aluminium, when it is exposed to concentrated trioxonitrate(V) acid, a thin, tenacious layer of aluminium oxide ($Al_2O_3$) forms on its surface. This layer is insoluble in the acid and adheres tightly to the metal, effectively "passivating" it.

    Step-by-Step Solution

    1. Evaluate the Reactivity: Aluminium is high up in the electrochemical series (above hydrogen), so it should theoretically react with acids to displace hydrogen. However, the observation shows it does not.
    2. Analyze the Role of $HNO_3$: Trioxonitrate(V) acid is a strong oxidizing agent. When aluminium comes into contact with it, the acid oxidizes the surface of the aluminium to form aluminium oxide.
    3. Understand the Protective Layer: Unlike the oxides of some other metals, the aluminium oxide layer is very stable, non-porous, and insoluble in the acid.
    4. Conclusion: Because this oxide layer covers the entire surface of the aluminium, the acid molecules cannot reach the metal atoms beneath the layer to continue the reaction. This state is known as passivation.

    Evaluating the Options

    • A: Incorrect. Aluminium is not inherently "insoluble" in the sense of a physical mixture; it is the chemical reaction that is prevented by the oxide layer.
    • B: Incorrect. Aluminium is actually above hydrogen in the electrochemical series.
    • C: Incorrect. Activation energy is not the limiting factor here; the physical barrier (the oxide layer) is.
    • D: Correct. The formation of an insoluble oxide layer on the surface renders the metal passive.

    Exam Tip

    Whenever you see a question about why Aluminium, Iron, or Chromium do not react with concentrated trioxonitrate(V) acid, always look for the keyword "passivity" or "oxide layer." This is a classic JAMB Chemistry question that tests your understanding of metal surface chemistry!

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54. 54.

    On which of the following is the solubility of a gaseous substance dependent?  I. Nature of solvent  II. Nature of solute  III. Temperature  IV. Pressure

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

    Show answer

    Answer: A. I, II, III and IV

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55. 55.

    According to Charles' law, the volume of a gas becomes zero at

    1. A.0°C
    2. B.-100°C
    3. C.-273°C
    4. D.-373°C

    Show answer

    Answer: C. -273°C

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56. 56.

    The boiling of fat and aqueous caustic soda is referred to as

    1. A.hydrolysis
    2. B.esterification
    3. C.acidification
    4. D.saponification

    Show answer

    Answer: D. saponification

    AI explanation

    Analysis

    The question asks for the specific name of the chemical reaction that occurs when fat (a triglyceride) is heated with aqueous caustic soda (sodium hydroxide, NaOH).

    Concept Explanation

    This process is a classic reaction in organic chemistry involving esters.

    • Fats and Oils: These are naturally occurring esters formed from glycerol (a trihydric alcohol) and long-chain fatty acids.
    • Caustic Soda (NaOH): This is a strong base.
    • The Reaction: When an ester is heated with an alkali (base), the ester bond is broken. The fatty acid part reacts with the sodium from the base to form a sodium salt of a fatty acid, which is commonly known as soap. The glycerol is also released as a byproduct.

    This reaction is the fundamental process used in the commercial and domestic production of soap.

    Step-by-Step Solution

    1. Identify the reactants: Fat (ester) + NaOH (alkali).
    2. Identify the products: Soap (sodium salt of fatty acid) + Glycerol.
    3. Define the terms: • Hydrolysis: A general term for breaking down a compound using water. • Esterification: The reverse process; making an ester from an alcohol and an acid. • Acidification: The process of making a solution acidic. • Saponification: Specifically refers to the alkaline hydrolysis of fats or oils to produce soap.

    Since the reaction specifically produces soap from fat and caustic soda, the term "saponification" (from the Latin sapo, meaning soap) is the correct technical term.

    Conclusion

    The process of boiling fat with aqueous caustic soda to produce soap is known as saponification.

    Exam Tip: Remember the mnemonic "Saponification = Soap-making." Whenever you see "fat/oil" + "alkali" (like NaOH or KOH), always look for "saponification" in your options.

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57. 57.

    When air which contains the gases Oxygen, nitrogen, carbondioxide, water vapour and the rare gases, is passed through alkaline pyrogallol and then over quicklime, the only gases left are;

    1. A.nitrogen and carbondioxide
    2. B.the rare gases
    3. C.nitrogen and oxygen
    4. D.nitrogen and the rare gases

    Show answer

    Answer: D. nitrogen and the rare gases

    AI explanation

    Analysis of the Question

    This question tests your knowledge of the composition of air and the chemical properties of specific reagents used to absorb gases. To solve this, we must track what happens to each component of air as it passes through the two chemical agents.

    Concept Explanation

    Air is a mixture of gases. To isolate specific components, we use "scrubbers" or chemical reagents that react with and remove specific gases:

    1. Alkaline Pyrogallol: This is a strong reducing agent used specifically to absorb Oxygen ($O_2$). When air passes through it, all the oxygen is removed.
    2. Quicklime (Calcium Oxide, $CaO$): This is a basic oxide. It is used to dry gases and absorb acidic gases. Specifically, it reacts with Carbon dioxide ($CO_2$) and Water vapour ($H_2O$).

    Step-by-Step Solution

    • Step 1: Identify the components of air. The air contains: Nitrogen ($N_2$), Oxygen ($O_2$), Carbon dioxide ($CO_2$), Water vapour ($H_2O$), and Rare gases (Noble gases like Argon, Neon, etc.).

    • Step 2: Pass the air through Alkaline Pyrogallol. • The alkaline pyrogallol absorbs the Oxygen. • Remaining gases: Nitrogen, Carbon dioxide, Water vapour, and Rare gases.

    • Step 3: Pass the remaining gases over Quicklime ($CaO$). • The quicklime absorbs the Carbon dioxide (an acidic oxide) and the Water vapour (drying agent). • Remaining gases: Nitrogen and the Rare gases.

    Conclusion

    After the oxygen, carbon dioxide, and water vapour have been removed by the respective reagents, only the chemically inert gases (Nitrogen and the Rare gases) remain.

    Exam Tip

    Remember the "Gas Scrubbing" sequence for air:

    • KOH/NaOH solution: Removes $CO_2$.
    • Alkaline Pyrogallol: Removes $O_2$.
    • Conc. $H_2SO_4$ or Quicklime: Removes $H_2O$ (Water vapour).
    • Nitrogen and Rare gases are generally unreactive under these conditions, so they pass through unaffected.

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58. 58.

    The type of bonding in [Cu(NH$_3$)$^4$]$^{2+}$ is

    1. A.coordinate
    2. B.electrovalent
    3. C.metallic
    4. D.covalent

    Show answer

    Answer: A. coordinate

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59. 59.

    The collision theory explains reaction rates in terms of

    1. A.size of the product
    2. B.frequency of collision of the reactants
    3. C.size of reactant
    4. D.frequency of collision of the products

    Show answer

    Answer: B. frequency of collision of the reactants

    AI explanation

    Analysis

    The question asks for the fundamental basis of the collision theory in chemical kinetics. This theory is used to explain how chemical reactions occur and why reaction rates differ under various conditions.

    Concept Explanation

    The collision theory states that for a chemical reaction to occur between reactant particles, three conditions must be met:

    1. Collision: The reactant particles must collide with each other.
    2. Orientation: The particles must collide with the correct spatial orientation.
    3. Activation Energy: The particles must collide with sufficient kinetic energy (at least equal to the activation energy) to break existing bonds and form new ones.

    The rate of reaction is directly proportional to the number of effective collisions occurring per unit time. Therefore, the more frequently reactant particles collide with the necessary energy and orientation, the faster the reaction proceeds.

    Step-by-Step Solution

    1. Analyze the reactants vs. products: Collision theory focuses on the interaction between reactants. Products are the result of the reaction, so their collision frequency does not determine the rate of the forward reaction. This eliminates options A and D.
    2. Evaluate the role of size: While the size of particles can influence how often they collide (steric factor), the theory is primarily defined by the frequency and energy of the collisions of the reactants, not just their physical size. This makes option C less accurate than B.
    3. Focus on frequency: The core principle is that the rate of reaction is dependent on the frequency of effective collisions between reactant molecules. As the frequency of these collisions increases (e.g., by increasing concentration, pressure, or temperature), the reaction rate increases.

    Conclusion

    The collision theory explains that the rate of a reaction is determined by how often the reactant particles collide with enough energy to react. Thus, it is defined by the frequency of collision of the reactants.

    Exam Tip: Whenever you see a question about "Collision Theory," always look for keywords like "effective collisions," "activation energy," and "frequency of collisions." Remember: No collision, no reaction!

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60. 60.

    H$_2$ S$_{(g)}$ + Cl$_{2(g)}$ → 2HCl$_{(g)}$ + S$_{(g)}$ In the reaction above, the substance that is reduced is

    1. A.H$_2$S
    2. B.S
    3. C.HCl
    4. D.Cl$_2$

    Show answer

    Answer: D. Cl$_2$

    AI explanation

    To determine which substance is reduced in the reaction, we need to analyze the changes in the oxidation states of the elements involved.

    1. Analyze the concept

    In chemistry, reduction is defined as:

    • A decrease in oxidation state.
    • The gain of electrons.
    • The removal of oxygen or the addition of hydrogen.

    We use the acronym OIL RIG: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

    2. Step-by-step solution

    Let's assign oxidation numbers to each element in the reaction: H$_2$S$_{(g)}$ + Cl$_{2(g)}$ → 2HCl$_{(g)}$ + S$_{(s)}$

    • In H$_2$S: Hydrogen (H) is +1, so Sulfur (S) must be -2 (since the total charge of the molecule is 0).
    • In Cl$_2$: This is an element in its standard state, so its oxidation number is 0.
    • In HCl: Hydrogen (H) is +1, so Chlorine (Cl) must be -1.
    • In S: This is an element in its standard state, so its oxidation number is 0.

    Now, let's track the changes:

    • Sulfur (S): Changes from -2 (in H$_2$S) to 0 (in S). An increase in oxidation number indicates oxidation.
    • Chlorine (Cl): Changes from 0 (in Cl$_2$) to -1 (in HCl). A decrease in oxidation number indicates reduction.

    3. Conclusion

    Since the oxidation state of Chlorine decreased from 0 to -1, Chlorine (Cl$_2$) is the substance that gained electrons and was reduced.

    Exam Tip

    Always remember that in a reaction between a non-metal element (like Cl$_2$) and a compound containing a non-metal (like H$_2$S), the element that is "free" on the reactant side (Cl$_2$) is almost always the one being reduced, as it is gaining electrons to form an ion.

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61. 61.

    The number of electrons in the valence shell of an element of atomic number 14 is?

    1. A.1
    2. B.2
    3. C.3
    4. D.4

    Show answer

    Answer: D. 4

    AI explanation

    1. Analyze the question

    The question asks for the number of electrons in the outermost shell (valence shell) of an element with an atomic number of 14.

    2. Explain the concept

    To find the valence electrons, we must determine the electronic configuration of the element. The electronic configuration shows how electrons are distributed in the shells (K, L, M, N...).

    • The first shell (K) holds a maximum of 2 electrons.
    • The second shell (L) holds a maximum of 8 electrons.
    • The third shell (M) holds the remaining electrons.

    The valence shell is the outermost shell of an atom, and the electrons residing in this shell are called valence electrons.

    3. Show the solution

    • Step 1: Identify the atomic number. The atomic number is 14. This means there are 14 electrons to distribute.
    • Step 2: Distribute the electrons into shells: • 1st shell (K): 2 electrons • 2nd shell (L): 8 electrons • Total used so far: $2 + 8 = 10$ electrons. • Remaining electrons: $14 - 10 = 4$ electrons.
    • Step 3: Place the remaining electrons in the 3rd shell (M): • 3rd shell (M): 4 electrons.
    • Step 4: The electronic configuration is 2, 8, 4. Since the outermost shell is the 3rd shell, it contains 4 electrons.

    4. Identify the answer

    The number of electrons in the valence shell is 4. This corresponds to option D.

    5. Tip

    To quickly find the valence electrons for elements in the main groups (Groups 1–8), look at the last digit of the electronic configuration. For example, if an element has 15 electrons (2, 8, 5), it has 5 valence electrons and belongs to Group 5. Remember, the valence electrons determine the chemical properties and the group number of the element!

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62. 62.

    In the removal of water hardness, permutit is regenerated with a strong solution of

    1. A.NaOH
    2. B.Na$_3$SO$_4$
    3. C.Na$_3$CO$_3$
    4. D.NaCl

    Show answer

    Answer: D. NaCl

    AI explanation

    Analysis

    The question asks for the chemical substance used to regenerate "permutit" (also known as zeolite) after it has been used to soften hard water.

    Concept Explanation

    Permutit (Zeolite) is a naturally occurring or synthetic hydrated sodium aluminosilicate (represented as $Na_2Ze$).

    1. How it works: When hard water (containing $Ca^{2+}$ or $Mg^{2+}$ ions) passes through a bed of permutit, the sodium ions ($Na^+$) in the permutit are exchanged for the calcium or magnesium ions in the water. This removes the hardness: $Na_2Ze(s) + Ca^{2+}(aq) \rightarrow CaZe(s) + 2Na^+(aq)$
    2. The Regeneration Process: Eventually, the permutit becomes "exhausted" (all the $Na^+$ ions are replaced by $Ca^{2+}$ or $Mg^{2+}$). To reuse it, we must displace the calcium and magnesium ions and replace them with sodium ions again. This is done by flushing the exhausted permutit with a concentrated solution of sodium chloride (brine). $CaZe(s) + 2NaCl(aq) \rightarrow Na_2Ze(s) + CaCl_2(aq)$

    Step-by-Step Solution

    • Step 1: Identify the exhausted material. The exhausted permutit is in the form of Calcium-zeolite ($CaZe$).
    • Step 2: Identify the goal. We need to restore it to Sodium-zeolite ($Na_2Ze$).
    • Step 3: Select the reagent. We need a source of concentrated sodium ions ($Na^+$) that will force the equilibrium to shift back to the sodium form. A concentrated solution of common salt (NaCl) is the standard industrial reagent for this ion-exchange regeneration.

    Conclusion

    Comparing this to your options: A. NaOH (Sodium hydroxide) - Not used for this purpose. B. $Na_3SO_4$ - Incorrect formula and not used for regeneration. C. $Na_3CO_3$ - Incorrect formula (Sodium carbonate is $Na_2CO_3$). D. NaCl (Sodium chloride) - This is the correct reagent used for regeneration.

    Exam Tip: Remember that "Permutit" and "Zeolite" are often used interchangeably in JAMB questions. Always associate them with Ion-Exchange and Brine (NaCl) for regeneration. Don't get confused by the chemical formulas; just look for the sodium salt that provides the $Na^+$ ions needed to "recharge" the zeolite.

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63. 63.

    Which of the following gases may not be dried with concentrated sulphuric acid?

    1. A.HCl$_{(g)}$
    2. B.NH$_3$
    3. C.Cl$_2$
    4. D.SO$_2$

    Show answer

    Answer: B. NH$_3$

    AI explanation

    Analysis

    The question asks us to identify which of the listed gases cannot be dried using concentrated tetraoxosulphate(VI) acid ($H_2SO_4$). To answer this, we must understand the chemical property of concentrated $H_2SO_4$ as a drying agent.

    Concept

    Concentrated $H_2SO_4$ is a powerful acidic drying agent. In chemistry, when choosing a drying agent, the rule is: the drying agent must not react chemically with the gas being dried.

    • If a gas is acidic, it can be dried using acidic drying agents like concentrated $H_2SO_4$ or phosphorus(V) oxide ($P_2O_5$).
    • If a gas is basic (alkaline), it will react with an acidic drying agent (neutralization reaction) and therefore cannot be dried using it.

    Step-by-Step Solution

    1. Analyze the nature of the gases: • HCl (Hydrogen chloride): This is an acidic gas. It does not react with $H_2SO_4$. • NH$_3$ (Ammonia): This is a basic (alkaline) gas. • Cl$_2$ (Chlorine): This is a neutral/acidic gas. It does not react with $H_2SO_4$. • SO$_2$ (Sulphur dioxide): This is an acidic gas. It does not react with $H_2SO_4$.

    2. Evaluate the reaction with $H_2SO_4$: • Since Ammonia ($NH_3$) is a base, it reacts readily with concentrated $H_2SO_4$ (an acid) to form a salt (Ammonium tetraoxosulphate(VI)): $$2NH_{3(g)} + H_2SO_{4(conc)} \rightarrow (NH_4)_2SO_{4(s)}$$ • Because the gas is consumed in this reaction, it cannot be dried using this method.

    Conclusion

    Because ammonia is basic, it undergoes a neutralization reaction with concentrated sulphuric acid. Therefore, it is the gas that cannot be dried using this substance.

    Correct Option: B

    Exam Tip

    Remember the "Acid-Base" rule for drying gases:

    • Acidic gases (e.g., $CO_2$, $SO_2$, $HCl$) are dried with acidic agents (conc. $H_2SO_4$).
    • Basic gases (e.g., $NH_3$) are dried with basic agents (e.g., Calcium oxide, $CaO$).
    • Neutral gases (e.g., $H_2$, $O_2$, $N_2$) can generally be dried with either.

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64. 64.

    During the electrolysis of copper II sulphate between platinum electrodes, if litmus solution is added to the anode compartment

    1. A.the litmus turns blue but no gas is evolved
    2. B.the litmus turns blue and oxygen is evolved
    3. C.the litmus turns blue and hydrogen is evolved
    4. D.the litmus turns red and oxygen is evolved

    Show answer

    Answer: D. the litmus turns red and oxygen is evolved

    AI explanation

    Analysis

    To solve this, we need to understand the electrolysis of aqueous copper(II) tetraoxosulphate(VI) ($CuSO_4$) using inert electrodes (platinum).

    Concept

    In the electrolysis of $CuSO_4(aq)$:

    1. Ions present: $Cu^{2+}$, $SO_4^{2-}$, $H^+$, and $OH^-$ (from water).
    2. At the Cathode (Negative Electrode): $Cu^{2+}$ ions are preferentially discharged over $H^+$ ions because copper is lower in the electrochemical series. • Equation: $Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}$
    3. At the Anode (Positive Electrode): $OH^-$ ions are preferentially discharged over $SO_4^{2-}$ ions. • Equation: $4OH^-_{(aq)} \rightarrow 2H_2O_{(l)} + O_{2(g)} + 4e^-$
    4. The Resulting Solution: As $OH^-$ ions are removed from the solution to form oxygen gas, the concentration of $H^+$ ions increases relative to $OH^-$ ions. This makes the solution around the anode acidic.

    Step-by-Step Solution

    • Step 1: Identify the product at the anode. As established, $OH^-$ ions are discharged to produce oxygen gas ($O_2$).
    • Step 2: Determine the effect on the solution. The removal of $OH^-$ ions leaves behind an excess of $H^+$ ions, which increases the acidity of the electrolyte near the anode.
    • Step 3: Apply the litmus test. Litmus solution is an indicator that turns red in an acidic environment and blue in an alkaline environment. Since the anode compartment becomes acidic, the litmus will turn red.
    • Step 4: Combine the findings. The litmus turns red, and oxygen gas is evolved at the anode.

    Conclusion

    Comparing our findings to the options provided:

    • A, B, and C are incorrect because the solution becomes acidic (turning litmus red), not alkaline.
    • Option D correctly identifies that the litmus turns red due to the acidity and oxygen gas is evolved.

    Exam Tip

    Remember the "Anode-Acidic" rule for inert electrodes: When you electrolyze a salt of an oxyacid (like $SO_4^{2-}$ or $NO_3^-$) using inert electrodes, the anode compartment almost always becomes acidic because $OH^-$ is discharged, leaving $H^+$ ions behind.

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65. 65.

    In the preparation of oxygen by heating KCIO$_3$ , in the presence of MnO$_2$ only moderate heat is needed because the catalyst acts by $_2$

    1. A.lowering the pressure of the reaction
    2. B.increasing the surface area of the reaction
    3. C.increasing the rate of the reaction
    4. D.increasing the energy barrier of the reaction

    Show answer

    Answer: C. increasing the rate of the reaction

    AI explanation

    Analysis of the Question

    The question asks why the addition of Manganese(IV) oxide ($MnO_2$) allows the decomposition of Potassium trioxochlorate(V) ($KClO_3$) to occur at a lower temperature than it would otherwise. This is a classic question about the role of catalysts in chemical reactions.

    Concept Explanation

    A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed in the process.

    In chemical kinetics, for a reaction to occur, reactant particles must collide with a minimum amount of energy known as the activation energy ($E_a$).

    • Without a catalyst, the reaction requires a high amount of heat to provide enough particles with the energy to overcome this high activation energy barrier.
    • When a catalyst like $MnO_2$ is added, it provides an alternative reaction pathway that has a lower activation energy.
    • Because the energy barrier is lowered, more reactant molecules have sufficient energy to react at a lower temperature, thereby increasing the overall rate of the reaction.

    Step-by-Step Solution

    1. Identify the reaction: The decomposition of $KClO_3$ is: $2KClO_3(s) \xrightarrow{MnO_2, \Delta} 2KCl(s) + 3O_2(g)$
    2. Evaluate the options: • A (Lowering pressure): Catalysts do not affect the pressure of the system. • B (Increasing surface area): While surface area affects reaction rates in heterogeneous systems, it is not the mechanism by which a catalyst functions. • C (Increasing the rate of reaction): This is the definition of what a catalyst does. By providing an alternative pathway with a lower activation energy, the reaction proceeds much faster at a given temperature (or at a lower temperature). • D (Increasing the energy barrier): This is incorrect. A catalyst decreases the energy barrier (activation energy), it never increases it.

    Conclusion

    The presence of $MnO_2$ acts as a catalyst, which speeds up the decomposition of $KClO_3$ by providing an alternative pathway with a lower activation energy. Therefore, the reaction proceeds more rapidly (and at lower temperatures) than it would without the catalyst.

    Exam Tip: Always remember that a catalyst lowers the activation energy but does not change the enthalpy change ($\Delta H$) of the reaction. If you see an energy profile diagram in your exam, the "hump" (activation energy) is always smaller when a catalyst is present!

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66. 66.

    To what volume must 300cm$^3$ of 0.60M sodium hydroxide solution be diluted to give a 0.40M solution?

    1. A.450cm$^3$
    2. B.300cm$^3$
    3. C.200cm$^3$
    4. D.150cm$^3$

    Show answer

    Answer: A. 450cm$^3$

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67. 67.

    Which of the following is an example of a chemical change?

    1. A.dissolution of salt in water
    2. B.rusting of iron
    3. C.melting of ice
    4. D.separating a mixture by distillation

    Show answer

    Answer: B. rusting of iron

    AI explanation

    Analysis

    The question asks us to identify a chemical change from the given options. To do this, we must distinguish between physical changes and chemical changes.

    Concept: Physical vs. Chemical Change

    • Physical Change: This is a change in which no new substance is formed. The chemical composition of the material remains the same. Examples include changes in state (melting, freezing, boiling), dissolving (in most cases), and physical separation techniques (filtration, distillation).
    • Chemical Change: This is a change that results in the formation of one or more new substances with entirely different chemical properties. These changes are usually irreversible and often involve the absorption or release of energy (heat, light, or gas production).

    Step-by-Step Evaluation

    1. A. Dissolution of salt in water: When salt dissolves in water, it forms a solution. You can recover the salt simply by evaporating the water. No new chemical substance is created; it is a physical process.
    2. B. Rusting of iron: Rusting occurs when iron reacts with oxygen and moisture in the air to form a new substance called hydrated iron(III) oxide (rust). This is a chemical reaction that changes the composition of the iron, and it cannot be reversed by simple physical means.
    3. C. Melting of ice: This is a change of state from solid to liquid. The chemical formula remains $H_2O$. It is a classic example of a physical change.
    4. D. Separating a mixture by distillation: Distillation uses the difference in boiling points to separate components of a mixture. The substances involved do not change their chemical identity during this process.

    Conclusion

    Since rusting of iron involves the creation of a new substance through a chemical reaction, it is the only chemical change listed.

    Exam Tip

    Whenever you see a question about physical vs. chemical changes, ask yourself: "Can I easily get the original substance back using a physical method like heating or filtering?" If the answer is yes, it is likely a physical change. If the substance has permanently transformed into something else (like burning wood or rusting metal), it is a chemical change.

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68. 68.

    The reaction between an organic acid and an alcohol in the presence of an acid catalyst is known as;

    1. A.saponification
    2. B.dehydration
    3. C.esterification
    4. D.hydrolysis

    Show answer

    Answer: C. esterification

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69. 69.

    If the molecular mass of tetraoxosulphate (VI) acid is 98, calculate its vapour density.

    1. A.196
    2. B.49
    3. C.106
    4. D.82

    Show answer

    Answer: B. 49

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70. 70.

    The radio isotope used in industrial radiography for the rapid checking of faults in welds and casting is?

    1. A.carbon-14
    2. B.phosphorus-32
    3. C.cobalt-60
    4. D.iodine-131

    Show answer

    Answer: C. cobalt-60

    AI explanation

    Analysis

    The question asks for the specific radioisotope used in industrial radiography. Industrial radiography is a non-destructive testing method used to inspect the internal structure of materials, such as metal welds and castings, to detect cracks, voids, or other structural defects.

    Concept Explanation

    Radioisotopes are unstable isotopes of elements that emit radiation (alpha, beta, or gamma rays) as they decay. In industrial radiography, we require a source that emits gamma rays because they have high penetrating power, allowing them to pass through thick metal objects and expose photographic film on the other side.

    • Carbon-14: Primarily used in radiocarbon dating of organic materials (archaeology).
    • Phosphorus-32: Used in medical and agricultural research as a tracer to study plant metabolism or treat certain blood disorders.
    • Cobalt-60: A powerful emitter of gamma rays. Its high energy and long half-life make it the standard choice for industrial radiography to inspect heavy-duty welds and steel castings.
    • Iodine-131: Primarily used in medicine for diagnosing and treating thyroid conditions.

    Step-by-Step Solution

    1. Identify the requirement: We need a radioisotope capable of penetrating dense materials like steel welds.
    2. Evaluate the options: • A (Carbon-14) is a weak beta emitter, unsuitable for penetrating metal. • B (Phosphorus-32) is also a beta emitter, unsuitable for industrial imaging. • C (Cobalt-60) is a well-known gamma-ray source used extensively in industry for "taking X-rays" of metal parts. • D (Iodine-131) is a gamma emitter, but it is specifically used for biological/medical applications, not industrial structural testing.
    3. Conclusion: Cobalt-60 is the correct isotope for industrial radiography.

    Exam Tip

    When you see questions about radioisotopes, remember this "cheat sheet":

    • Carbon-14: Dating old things (fossils).
    • Cobalt-60: Checking metal welds (industrial).
    • Iodine-131: Thyroid treatment (medical).
    • Phosphorus-32: Plant/Blood research (tracer).

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71. 71.

    A correct electrochemical series can be obtained from Na, Ca, Al, Mg, Zn, Fe, Pb, H, Cu, Hg, Ag, Au by interchanging

    1. A.Al and Mg
    2. B.Zn and Fe
    3. C.Zn and Pb
    4. D.Pb and H

    Show answer

    Answer: A. Al and Mg

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72. 72.

    The general formula of alkanones is

    1. A.RCHO
    2. B.R$_2$CO
    3. C.RCOOH
    4. D.RCOOR

    Show answer

    Answer: B. R$_2$CO

    AI explanation

    Analysis

    The question asks for the general formula of alkanones. In organic chemistry, alkanones are more commonly known as ketones. To identify the correct formula, we need to look at the functional group characteristic of ketones.

    Concept Explanation

    • Ketones (Alkanones): These are organic compounds characterized by a carbonyl group ($C=O$) bonded to two carbon atoms. The general structure is $R-C(=O)-R'$, where $R$ and $R'$ are alkyl groups.
    • Functional Group: The carbonyl group ($C=O$) is located within the carbon chain, not at the end.
    • General Formula: Since a ketone requires the carbonyl carbon to be attached to two other carbon-containing groups (alkyl groups), the simplest representation is $R_2CO$.

    Let's evaluate the other options provided:

    • A. RCHO: This is the general formula for alkanals (aldehydes), where the carbonyl group is at the end of the chain.
    • C. RCOOH: This is the general formula for alkanoic acids (carboxylic acids).
    • D. RCOOR: This is the general formula for alkanoates (esters).

    Step-by-Step Solution

    1. Identify the homologous series: Alkanones = Ketones.
    2. Recall the structure of a ketone: $R-C(=O)-R$.
    3. Combine the components: Two $R$ groups attached to a $CO$ (carbonyl) unit.
    4. This matches the expression $R_2CO$.

    Conclusion

    The general formula for alkanones is $R_2CO$.

    Exam Tip: To avoid confusion in the exam hall, remember the "position" of the functional group:

    • Aldehydes (Alkanals): The 'H' is at the end ($R-CH=O$).
    • Ketones (Alkanones): The 'O' is sandwiched between two 'R' groups ($R-C=O-R$).
    • Acids (Alkanoic acids): They have an extra 'OH' ($R-COOH$).

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73. 73.

    What is the concentration of a solution containing 2g of NaOH in 100cm3 of solution? [Na = 23, O =16, H = 1]

    1. A.0.40 moldm$^{-3}$
    2. B.0.50 moldm$^{-3}$
    3. C.0.05 moldm$^{-3}$
    4. D.0.30 moldm$^{-3}$

    Show answer

    Answer: B. 0.50 moldm$^{-3}$

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74. 74.

    Sieving is a technique used to separate mixtures containing solid particles of

    1. A.small sizes
    2. B.large sizes
    3. C.different sizes
    4. D.the same size

    Show answer

    Answer: C. different sizes

    AI explanation

    Analysis

    The question asks for the principle behind sieving as a separation technique. We need to identify the specific property of the solid particles that allows this method to work effectively.

    Concept Explanation

    Sieving is a mechanical method used to separate a mixture of solids. It relies on the difference in particle size.

    A sieve (or mesh) acts as a barrier with holes of a specific diameter. When a mixture is passed through the sieve:

    • Particles smaller than the holes pass through (the filtrate).
    • Particles larger than the holes are retained on the sieve (the residue).

    If all particles were the same size, they would either all pass through or all be retained, making separation impossible. Therefore, the mixture must contain particles of varying sizes for the process to be effective.

    Step-by-Step Solution

    1. Identify the goal: Separate solid particles from a mixture.
    2. Examine the mechanism: The sieve acts as a filter that discriminates based on physical dimensions.
    3. Evaluate the options: • A & B: Simply having small or large particles doesn't guarantee separation; you need a comparison between them. • D: If particles are the same size, they cannot be separated by a sieve. • C: Because the sieve allows some particles to pass while blocking others, the mixture must contain particles of different sizes.

    Conclusion

    Sieving is only effective when the components of the mixture have different particle sizes, allowing the sieve to act as a size-selective barrier.

    Exam Tip: Think of a kitchen sieve used for garri or flour. You are trying to separate the fine powder (small particles) from the coarse chaff or lumps (large particles). This is only possible because the mixture contains particles of different sizes!

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75. 75.

    Which of the following pairs of substances will react further with oxygen to form a higher oxide?

    1. A.CO$_2$ and H$_2$O
    2. B.NO and H$_2$O
    3. C.CO and CO$_2$
    4. D.SO$_2$ and NO

    Show answer

    Answer: D. SO$_2$ and NO

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76. 76.

    What volume of oxygen will remain after reacting 8cm$^3$ of hydrogen gas with 20cm$^3$ of oxygen gas?

    1. A.10cm$^3$
    2. B.12cm$^3$
    3. C.14cm$^3$
    4. D.16cm$^3$

    Show answer

    Answer: D. 16cm$^3$

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77. 77.

    When large hydrocarbon molecules are heated at high temperature in the presence of a catalyst to give smaller molecules, the process is known as

    1. A.disintegration
    2. B.polymerization
    3. C.cracking
    4. D.degradation

    Show answer

    Answer: C. cracking

    AI explanation

    Analysis

    The question asks for the specific term used to describe the industrial process of breaking down large, heavy hydrocarbon molecules into smaller, more useful molecules (like petrol or ethene) using high heat and a catalyst.

    Concept Explanation

    In petroleum chemistry, the demand for shorter-chain hydrocarbons (which are used as fuel for cars) is much higher than the supply obtained from fractional distillation of crude oil. To bridge this gap, refineries use a process called cracking.

    • Cracking involves heating long-chain alkanes (which have high boiling points and are less valuable) to high temperatures, often in the presence of a catalyst (like silica or alumina).
    • This process breaks the carbon-carbon bonds, resulting in smaller, more volatile, and more valuable molecules like alkanes and alkenes.

    Let's look at why the other options are incorrect:

    • Disintegration: This is a general term for breaking apart, not a specific chemical process in petroleum refining.
    • Polymerization: This is the exact opposite of cracking. It involves joining many small monomer molecules together to form a large polymer chain.
    • Degradation: This usually refers to the breakdown of materials due to environmental factors (like weathering or biological decay), not a controlled industrial chemical process.

    Step-by-Step Solution

    1. Identify the input: Large hydrocarbon molecules (long-chain alkanes).
    2. Identify the conditions: High temperature and a catalyst.
    3. Identify the output: Smaller molecules.
    4. Match with the process: The chemical process defined by breaking large molecules into smaller ones under these specific conditions is cracking.

    Conclusion

    The process described is the standard definition of cracking in organic chemistry.

    Exam Tip: Always remember the "Cracking vs. Polymerization" pair. Cracking = Breaking (Large to Small). Polymerization = Building (Small to Large). This simple association will help you answer similar questions in your JAMB exam easily.

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78. 78.

    A substance that is used as a ripening agent for fruits is

    1. A.ethene
    2. B.propane
    3. C.methane
    4. D.butane

    Show answer

    Answer: A. ethene

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79. 79.

    An element used in the production of matches is

    1. A.nitrogen
    2. B.aluminium
    3. C.copper
    4. D.sulphur

    Show answer

    Answer: D. sulphur

    AI explanation

    Analysis

    The question asks for an element commonly used in the manufacturing of matches. We need to look at the chemical composition of match heads and the striking surface to identify which of the listed elements is essential.

    Concept Explanation

    Modern safety matches are designed to ignite only when struck against a specially prepared surface. The chemical process involves:

    1. Match Head: Typically contains an oxidizing agent (like potassium chlorate, $KClO_3$), a fuel (like sulphur or antimony trisulphide), and binding materials.
    2. Striking Surface: Contains red phosphorus and powdered glass.

    When the match is struck, the friction generates enough heat to convert a small amount of red phosphorus into white phosphorus, which ignites the potassium chlorate, providing the oxygen needed to burn the sulphur and the wooden stick.

    Step-by-Step Solution

    1. Evaluate Nitrogen (A): Nitrogen is an inert gas used in food packaging and the production of ammonia. It is not a component of match heads.
    2. Evaluate Aluminium (B): Aluminium is a structural metal used in construction and aircraft manufacturing. It is not used in the chemical composition of matches.
    3. Evaluate Copper (C): Copper is a transition metal used primarily for electrical wiring and plumbing. It plays no role in the combustion process of a match.
    4. Evaluate Sulphur (D): Sulphur is a non-metal that acts as a fuel in the match head. Its low ignition temperature makes it ideal for catching fire easily once the initial spark is produced.

    Conclusion

    Based on the chemical composition of matches, sulphur is the element listed that is essential for the ignition process.

    Exam Tip: When dealing with questions about industrial chemistry, always remember the "Big Three" for matches: Potassium chlorate (oxidizer), Red Phosphorus (initiator), and Sulphur (fuel). If you see any of these in an exam, you are on the right track!

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80. 80.

    The ionic radii of metals are usually

    1. A.greater than their atomic radii
    2. B.unaffected by the charge on the ion
    3. C.the same as their atomic radii
    4. D.less than their atomic radii

    Show answer

    Answer: D. less than their atomic radii

    AI explanation

    Analysis

    This question tests your understanding of atomic structure and the relationship between neutral atoms and their corresponding ions. Specifically, it asks how the size of a metal atom changes when it loses electrons to become a cation.

    Concept Explanation

    To understand this, we look at two main factors:

    1. Electron-Electron Repulsion: In a neutral metal atom, the number of electrons equals the number of protons. When a metal forms an ion, it loses its outermost valence electrons.
    2. Effective Nuclear Charge: When electrons are removed, the remaining electrons are held more tightly by the nucleus because the ratio of protons (positive charge) to electrons (negative charge) increases. Furthermore, the loss of an entire outer electron shell significantly reduces the "cloud" of the atom.

    Because the remaining electrons are pulled closer to the nucleus and the outermost shell is removed, the resulting ion is always smaller than the neutral atom from which it was formed.

    Step-by-Step Solution

    1. Identify the process: Metals lose electrons to form positively charged ions (cations). For example, a Sodium atom ($Na$) becomes a Sodium ion ($Na^+$).
    2. Compare the structure: • Neutral Atom ($Na$): Has 11 protons and 11 electrons. The electrons occupy three shells. • Cation ($Na^+$): Has 11 protons but only 10 electrons. The loss of the valence electron means the third shell is now empty, leaving only two shells.
    3. Evaluate the size: With fewer electrons and one less shell, the electrostatic attraction from the 11 protons pulls the remaining electron cloud much closer to the nucleus.
    4. Conclusion: The ionic radius of a metal is always smaller than its atomic radius.

    Identifying the Answer

    Comparing our conclusion to the options provided:

    • A is incorrect because ions are smaller, not larger.
    • B is incorrect because the charge directly influences the pull of the nucleus.
    • C is incorrect because the loss of electrons changes the size.
    • D is correct because the ionic radius is indeed less than the atomic radius.

    Exam Tip

    Remember the "Shell Loss Rule": When a metal becomes an ion, it loses its outermost shell. Think of it like peeling an orange—if you remove the outer layer, the remaining fruit is smaller! Conversely, non-metals (which gain electrons) form ions that are larger than their atomic radii because the added electrons increase repulsion.

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81. 81.

    What is the PH of 0.00 1 moldm$^3$ solution of the sodium hydroxide

    1. A.14
    2. B.13
    3. C.12
    4. D.11

    Show answer

    Answer: D. 11

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82. 82.

    Which of these alloys contains copper?

    1. A.solder
    2. B.steel
    3. C.permallory
    4. D.bronze

    Show answer

    Answer: D. bronze

    AI explanation

    1. Analyze the question

    The question asks you to identify which of the listed alloys is composed of copper as one of its primary metallic components.

    2. Explain the concept

    An alloy is a mixture of two or more elements, where at least one of the elements is a metal. To answer this, we need to look at the composition of each option provided:

    • Solder: Primarily composed of tin and lead. It is commonly used in electronics for joining components.
    • Steel: Primarily composed of iron and carbon. It is a fundamental material in construction and engineering.
    • Permalloy: Primarily composed of nickel and iron. It is known for its high magnetic permeability.
    • Bronze: Primarily composed of copper and tin. It is a classic alloy used historically for tools, weapons, and statues.

    3. Show the solution

    Let’s break down the composition of each option:

    • A. Solder: Sn (Tin) + Pb (Lead) — No copper.
    • B. Steel: Fe (Iron) + C (Carbon) — No copper.
    • C. Permalloy: Ni (Nickel) + Fe (Iron) — No copper.
    • D. Bronze: Cu (Copper) + Sn (Tin) — Contains copper.

    Since the question specifically asks for an alloy that contains copper, we can clearly see that Bronze is the correct choice.

    4. Identify the answer

    The correct option is D.

    5. Exam Tip

    To remember common alloys, use a simple mnemonic or grouping:

    • Brass = Copper + Zinc (Think "C-Z")
    • Bronze = Copper + Tin (Think "C-T")
    • Solder = Lead + Tin (Think "L-T")

    Always remember that "Bronze" and "Brass" both start with copper, but they differ by their second metal (Tin vs. Zinc).

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83. 83.

    The solubility of the solids that dissolves in a given solvent with the liberation of heat will

    1. A.increase with an increase in temperature
    2. B.decrease with an increase in temperature
    3. C.decrease with a decrease in temperature
    4. D.not be affected by changes in temperature

    Show answer

    Answer: B. decrease with an increase in temperature

    AI explanation

    Analysis

    This question tests your understanding of Le Chatelier’s Principle as it applies to the solubility of solids in liquids. Specifically, it focuses on how temperature changes affect a system at equilibrium when the dissolution process is exothermic (liberates heat).

    Concept Explanation

    According to Le Chatelier’s Principle, if a system at equilibrium is subjected to a change in temperature, the system will adjust itself to counteract that change.

    The dissolution process can be represented as: Solid + Solvent ⇌ Solution + Heat

    • Exothermic process: Heat is released (liberated) during dissolution.
    • Increasing the temperature: You are adding more heat to the system. To counteract this, the system will shift the equilibrium to the left (the side that absorbs heat) to consume the excess heat.
    • Result: Shifting to the left means the solid remains undissolved, effectively decreasing the solubility.

    Conversely, if the process were endothermic (absorbing heat), increasing the temperature would shift the equilibrium to the right, increasing solubility.

    Step-by-Step Solution

    1. Identify the process: The question states the solid dissolves with the "liberation of heat." This means the reaction is exothermic ($\Delta H < 0$).
    2. Apply Le Chatelier’s Principle: Increasing the temperature adds heat to the system.
    3. Predict the shift: The system tries to remove the added heat by favoring the reverse reaction (the endothermic direction).
    4. Conclusion: Because the reverse reaction is favored, the amount of solid that can remain dissolved decreases as the temperature rises.

    Identifying the Answer

    Based on the principle explained above, increasing the temperature causes the solubility of a substance that dissolves exothermically to decrease.

    Exam Tip: Remember the "Heat Rule":

    • Exothermic (Heat out): Increase Temp = Solubility Decreases.
    • Endothermic (Heat in): Increase Temp = Solubility Increases. Think of heat as a "product" in exothermic reactions; adding more product pushes the reaction backward!

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84. 84.

    Beryllium and Aluminium have similar properties because they

    1. A.are both metals
    2. B.belong to the same group
    3. C.belong to the same period
    4. D.are positioned diagonally to each other

    Show answer

    Answer: D. are positioned diagonally to each other

    AI explanation

    Analysis of the Question

    The question asks for the reason behind the chemical similarities between Beryllium (Be) and Aluminium (Al). This is a classic topic in Periodic Table trends known as the "Diagonal Relationship."

    Concept Explanation: The Diagonal Relationship

    In the Periodic Table, certain elements in the second period show a striking similarity to the elements located diagonally below them in the third period. This happens because of a balance between two factors:

    1. Atomic/Ionic Size: As you move across a period, size decreases. As you move down a group, size increases. These two opposing trends result in Beryllium and Aluminium having similar ionic radii.
    2. Charge-to-Size Ratio (Ionic Potential): Because their sizes are similar and their charges are comparable, they exert a similar polarizing power on anions. This leads to them forming compounds with similar covalent character and acidic/amphoteric properties.

    Specifically:

    • Beryllium (Group 2, Period 2) is diagonally related to Aluminium (Group 13, Period 3).
    • Lithium (Group 1, Period 2) is diagonally related to Magnesium (Group 2, Period 3).
    • Boron (Group 13, Period 2) is diagonally related to Silicon (Group 14, Period 3).

    Step-by-Step Solution

    1. Evaluate Option A: While both are metals, many other metals (like Sodium and Calcium) are not similar to each other. Being a metal is not the specific reason for their unique chemical resemblance.
    2. Evaluate Option B: Beryllium is in Group 2, and Aluminium is in Group 13. They do not belong to the same group.
    3. Evaluate Option C: Beryllium is in Period 2, and Aluminium is in Period 3. They do not belong to the same period.
    4. Evaluate Option D: Beryllium (atomic number 4) is located at the top-left of the p-block area, and Aluminium (atomic number 13) is located diagonally to its bottom-right. This diagonal placement is the standard chemical explanation for their shared properties, such as their oxides being amphoteric.

    Conclusion

    The similarity in the properties of Beryllium and Aluminium is a direct result of their diagonal positioning in the Periodic Table, which balances their ionic charges and sizes.

    Exam Tip

    To remember diagonal relationships, just think of the "staircase" pattern starting from Lithium. Remember the mnemonic: Li-Mg, Be-Al, B-Si. These pairs are famous in JAMB Chemistry for having similar chemical behaviors, such as forming covalent bonds and having amphoteric oxides!

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85. 85.

    Calculate the pH of 0.05 moldm$^{-3}$ H$_2$SO$_4$

    1. A.1.30
    2. B.1.12
    3. C.1.00
    4. D.0.30

    Show answer

    Answer: C. 1.00

    AI explanation

    Analysis

    To calculate the pH of a solution, we need to determine the concentration of hydrogen ions $[H^+]$ in the solution. Sulfuric acid ($H_2SO_4$) is a strong diprotic acid, meaning it dissociates completely in water to release two hydrogen ions per molecule.

    Concept

    The formula for pH is:

    $\text{pH} = -\log_{10}[H^+]$

    Since $H_2SO_4$ is a diprotic acid, the dissociation equation is:

    $H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$

    This means the concentration of hydrogen ions $[H^+]$ is twice the concentration of the acid:

    $[H^+] = 2 \times [\text{Concentration of } H_2SO_4]$

    Step-by-Step Solution

    1. Identify the concentration of the acid: $[\text{Acid}] = 0.05 \text{ mol dm}^{-3}$

    2. Calculate the concentration of hydrogen ions $[H^+]$: Since each mole of $H_2SO_4$ produces 2 moles of $H^+$, $[H^+] = 2 \times 0.05 \text{ mol dm}^{-3}$ $[H^+] = 0.1 \text{ mol dm}^{-3}$

    3. Calculate the pH: $\text{pH} = -\log_{10}(0.1)$ Recall that $0.1$ can be written as $10^{-1}$. $\text{pH} = -\log_{10}(10^{-1})$ Using the logarithm rule $\log(a^b) = b \log(a)$: $\text{pH} = -(-1) \times \log_{10}(10)$ Since $\log_{10}(10) = 1$: $\text{pH} = 1 \times 1 = 1.00$

    Conclusion

    The calculated pH is 1.00. Comparing this to the given options, it matches option C.

    Exam Tip

    Always check if the acid is monoprotic (like $HCl$, which gives $1H^+$) or diprotic (like $H_2SO_4$, which gives $2H^+$). A common mistake students make is forgetting to multiply the concentration by 2 for sulfuric acid, which would lead to an incorrect answer of 1.30 (Option A). Don't fall for that trap!

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86. 86.

    A basic postulate of the kinetic theory of gases is that the molecules of a gas move in straight lines between collisions. This implies that

    1. A.collisions are perfectly elastic
    2. B.forces of repulsion exist
    3. C.forces of repulsion and attraction are in equilibrium
    4. D.collisions are inelastic

    Show answer

    Answer: C. forces of repulsion and attraction are in equilibrium

    AI explanation

    Analysis of the Question

    This question tests your understanding of the Kinetic Molecular Theory of Gases. You are asked to identify what the movement of gas molecules in straight lines between collisions implies regarding the forces acting between them.

    Explanation of the Concept

    The Kinetic Molecular Theory makes several assumptions about ideal gases. One of the core postulates is that gas molecules are in constant, random, straight-line motion.

    • Why straight lines? If there were significant forces of attraction or repulsion between gas molecules, these forces would pull or push the molecules off their straight-line paths as they moved.
    • The Implication: Because they travel in straight lines until they hit something (the walls of the container or another molecule), it implies that there are no significant intermolecular forces (attraction or repulsion) acting on the molecules while they are in motion.
    • The Equilibrium State: In a gas, the molecules are far apart. The assumption is that the forces of attraction and repulsion are negligible or, more specifically, that these forces are in a state of balance (equilibrium) such that they do not influence the path of the molecules.

    Step-by-Step Solution

    1. Evaluate the motion: The fact that molecules move in straight lines means they are not being deflected by neighboring molecules.
    2. Analyze the forces: If there were strong forces of attraction, molecules would "clump" or curve toward each other. If there were strong forces of repulsion, they would veer away from each other.
    3. Connect to the theory: The kinetic theory assumes that for an ideal gas, the intermolecular forces are negligible. This is mathematically represented by the idea that the forces of attraction and repulsion are effectively in equilibrium, resulting in a net force of zero on the molecules as they travel.
    4. Review the options: • A: While collisions are assumed to be elastic, this is a separate postulate and does not explain the straight-line motion. • B: If forces of repulsion existed, the molecules would not move in straight lines. • C: This correctly explains why there is no net force to deflect the molecules from their straight-line path. • D: Kinetic theory assumes collisions are elastic, not inelastic.

    Conclusion

    The straight-line motion of gas molecules is a direct consequence of the absence of net intermolecular forces, meaning the forces of attraction and repulsion are in equilibrium.

    Exam Tip: Remember the acronym "V.E.R.T." for Kinetic Theory:

    • Volume of particles is negligible.
    • Elastic collisions.
    • Random, straight-line motion.
    • Temperature is proportional to kinetic energy.
    • No intermolecular forces (Attraction/Repulsion are in equilibrium).

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87. 87.

    Which of these sources of water may likely contain the least concentration of Ca$^{2+}$ and Mg$^{2+}$?

    1. A.Spring water
    2. B.River water
    3. C.Tap water
    4. D.Sea water

    Show answer

    Answer: D. Sea water

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