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JAMB 2012

2012 JAMB Chemistry Past Questions

The 2012 JAMB Chemistry past questions with answers: 45 objective questions from the 2012 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 JAMB-style CBT.

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

Questions

  1. 1.

    Geometric isomerism can exist in

    1. A.hex-3-ene
    2. B.hexane
    3. C.prop-l-ene
    4. D.3-methyl but -l-ene
    Show answer

    Answer: A. hex-3-ene

    AI explanation

    1. Analyze the Question

    The question asks us to identify which of the given organic compounds exhibits geometric isomerism (also known as cis-trans isomerism).

    2. Explain the Concept

    For a molecule to exhibit geometric isomerism, it must satisfy two main conditions:

    • Restricted Rotation: There must be a double bond (C=C) or a ring structure that prevents the atoms from rotating freely.
    • Different Groups: Each carbon atom involved in the double bond must be attached to two different groups.

    If one of the carbons in the double bond is attached to two identical groups (e.g., two hydrogen atoms), geometric isomerism is impossible because swapping those identical groups does not create a new spatial arrangement.

    3. Show the Solution

    Let’s examine each option:

    • A. Hex-3-ene (CH3CH2CH=CHCH2CH3CH_3CH_2CH=CHCH_2CH_3): The double bond is between carbon 3 and carbon 4. • Carbon 3 is attached to a hydrogen atom and an ethyl group (−CH2CH3-CH_2CH_3). • Carbon 4 is also attached to a hydrogen atom and an ethyl group (−CH2CH3-CH_2CH_3). Since both carbons have two different groups attached, this molecule can exist as cis-hex-3-ene or trans-hex-3-ene. This satisfies the condition.
    • B. Hexane (CH3CH2CH2CH2CH2CH3CH_3CH_2CH_2CH_2CH_2CH_3): This is an alkane with only single bonds. There is free rotation around all carbon-carbon bonds, so it cannot exhibit geometric isomerism.
    • C. Prop-1-ene (CH2=CHCH3CH_2=CHCH_3): The double bond is at the end (carbon 1). Carbon 1 is attached to two identical hydrogen atoms. Because of these identical atoms, it cannot show geometric isomerism.
    • D. 3-methylbut-1-ene (CH2=CHCH(CH3)2CH_2=CHCH(CH_3)_2): Again, the double bond is at the end (carbon 1). Carbon 1 is attached to two identical hydrogen atoms, which prevents geometric isomerism.
    4. Identify the Answer

    Based on the analysis, hex-3-ene is the only compound where the groups attached to the double-bonded carbons are distinct enough to allow for cis and trans configurations.

    5. Exam Tip

    The "Terminal Double Bond" Rule: In JAMB Chemistry, if you see an alkene with the double bond at the very beginning of the chain (like prop-1-ene or but-1-ene), it cannot exhibit geometric isomerism because the terminal carbon will always have two hydrogen atoms attached to it. Look for double bonds located towards the middle of the carbon chain!

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  2. 2.

    2 - methyl propan -2- ol is an example of a

    1. A.primary alkanol
    2. B.secondary alkanol
    3. C.tertiary alkanol
    4. D.quartenary alkanol
    Show answer

    Answer: C. tertiary alkanol

    AI explanation

    1. Analyze the question

    The question asks us to classify 2-methylpropan-2-ol based on its structure as an alkanol (alcohol).

    2. Explain the concept

    Alkanols are classified as primary, secondary, or tertiary based on the number of alkyl groups attached to the carbon atom that carries the hydroxyl (-OH) group:

    • Primary (1∘1^\circ) alkanol: The carbon atom bonded to the -OH group is attached to only one other carbon atom (or no other carbon atoms, as in methanol).
    • Secondary (2∘2^\circ) alkanol: The carbon atom bonded to the -OH group is attached to two other carbon atoms.
    • Tertiary (3∘3^\circ) alkanol: The carbon atom bonded to the -OH group is attached to three other carbon atoms.
    3. Show the solution

    Let’s draw the structure of 2-methylpropan-2-ol:

    1. Parent chain: "Propan" means 3 carbons in the main chain.
    2. Position of -OH: "2-ol" means the -OH group is on the second carbon.
    3. Substituent: "2-methyl" means a methyl group (−CH3-CH_3) is also on the second carbon.

    The structure looks like this:

            CH3
        |
  CH3 - C - CH3
        |
        OH
    • Identify the carbon atom bonded to the -OH group (the central carbon).
    • Count how many other carbon atoms are directly attached to this central carbon.
    • We can see there are three methyl groups (−CH3-CH_3) attached to that central carbon.

    Since the carbon bearing the -OH group is attached to three other carbon atoms, it is a tertiary alkanol.

    4. Identify the answer

    Comparing our finding to the options provided: A. Primary alkanol B. Secondary alkanol C. Tertiary alkanol D. Quaternary alkanol (Note: This is not a standard classification for alcohols).

    The correct option is C.

    5. Tip

    Memory Trick: Think of the number of carbons attached to the "functional carbon."

    • 1 carbon attached = Primary (1st)
    • 2 carbons attached = Secondary (2nd)
    • 3 carbons attached = Tertiary (3rd)

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  3. 3.

    A metal M displaces zinc from ZnCl solution This shows that

    1. A.electrons flow from zinc to M
    2. B.M is more electropositive than zinc
    3. C.M is more electonegative than zinc
    4. D.zinc is more electropositive than M
    Show answer

    Answer: B. M is more electropositive than zinc

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  4. 4.

    What number of moles of Cu 2+ will be deposited by 360 coulombs of electricity? [f = 96500 C mol -1]

    1. A.5.36 x 10 -4 mole
    2. B.1. 87 x 10 10 -3 mole
    3. C.9. 35 x 10 -4 mole
    4. D.3. 73 x 10 -3
    Show answer

    Answer: B. 1. 87 x 10 10 -3 mole

    AI explanation

    1. Analyze the Question

    We are asked to calculate the number of moles of Copper ions (Cu2+Cu^{2+}) deposited when a specific quantity of electricity (360 Coulombs) is passed through an electrolytic cell. We are given the Faraday constant (F=96500 C mol−1F = 96500 \text{ C mol}^{-1}).

    2. Explain the Concept

    To solve this, we use Faraday’s First Law of Electrolysis, which relates the quantity of electricity (QQ) to the amount of substance deposited.

    The reaction for the deposition of copper is:

    Cu(aq)2++2e−→Cu(s)Cu^{2+}_{(aq)} + 2e^- \rightarrow Cu_{(s)}

    From this equation, we see that 2 moles of electrons are required to deposit 1 mole of copper. The quantity of electricity (QQ) is related to the number of moles of electrons (nen_e) by the formula:

    Q=ne×FQ = n_e \times F

    Where:

    • QQ = Quantity of electricity in Coulombs (C)
    • nen_e = Number of moles of electrons
    • FF = Faraday's constant (96500 C mol−196500 \text{ C mol}^{-1})
    3. Show the Solution

    Step 1: Calculate the number of moles of electrons (nen_e) used.

    ne=QFn_e = \frac{Q}{F}
    ne=36096500n_e = \frac{360}{96500}
    ne≈0.0037305 moles of electronsn_e \approx 0.0037305 \text{ moles of electrons}

    Step 2: Relate moles of electrons to moles of Copper. From the half-equation Cu2++2e−→CuCu^{2+} + 2e^- \rightarrow Cu, we know that:

    2 moles of e− produce 1 mole of Cu2 \text{ moles of } e^- \text{ produce } 1 \text{ mole of } Cu

    Therefore, the moles of CuCu deposited is half the moles of electrons:

    Moles of Cu=ne2\text{Moles of } Cu = \frac{n_e}{2}
    Moles of Cu=0.00373052\text{Moles of } Cu = \frac{0.0037305}{2}
    Moles of Cu≈0.001865 moles\text{Moles of } Cu \approx 0.001865 \text{ moles}

    Step 3: Convert to standard form.

    0.001865=1.865×10−3 moles0.001865 = 1.865 \times 10^{-3} \text{ moles}

    Rounding to three significant figures, we get 1.87×10−3 mole1.87 \times 10^{-3} \text{ mole}.

    4. Identify the Answer

    Comparing our result with the given options, it matches Option B. (Note: The text in your prompt had a slight typo "10 10 -3", but it clearly refers to 10−310^{-3}).

    5. Exam Tip

    Always check the valency of the metal ion! If the question asked for Ag+Ag^+ (Silver), you would divide by 1 because it only needs 1 mole of electrons. For Cu2+Cu^{2+}, always remember to divide by 2. A quick way to remember: Moles of substance = Qn×F\frac{Q}{n \times F}, where nn is the charge on the ion.

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  5. 5.

    l . Evaporation. ll. Sublimation. lll. Diffusion. IV. Brownian motion. Which of the above can correctly be listed as evidences for the particulate nature of matter?

    1. A.I and lll only
    2. B.ll and lV only
    3. C.l, ll and lll only
    4. D.l, ll, lll and lV
    Show answer

    Answer: D. l, ll, lll and lV

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  6. 6.

    During a titration experiment, 0.05 moles of carbon (lV) oxide is liberated. What is the volume of gas liberated?

    1. A.22.40 dm 3
    2. B.11.20 dm 3
    3. C.2.24 dm 3
    4. D.1.12 dm 3
    Show answer

    Answer: D. 1.12 dm 3

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  7. 7.

    I. It has a varied composition from one place to another. ll. it constituents can be separated by physical means lll. It contains unreactive noble gases which of the above shows that air is a mixture?

    1. A.I and ll only
    2. B.ll and lll only
    3. C.l and lll only
    4. D.l,ll and lll
    Show answer

    Answer: A. I and ll only

    AI explanation

    To determine why air is classified as a mixture, let's analyze the properties provided in the question:

    1. Analyze the Statements
    • Statement I: "It has a varied composition from one place to another." A pure substance (like a compound) has a fixed, definite composition. A mixture, however, does not have a fixed ratio of components. For example, the amount of water vapor, dust, or pollutants in the air varies depending on your location (e.g., near the sea vs. in a dry desert). This is a characteristic property of a mixture.
    • Statement II: "Its constituents can be separated by physical means." This is the fundamental definition of a mixture. The components of air (nitrogen, oxygen, argon, etc.) can be separated using physical processes like fractional distillation of liquid air. Since no chemical reaction is required to separate them, this confirms air is a mixture.
    • Statement III: "It contains unreactive noble gases." While it is true that air contains noble gases like argon and neon, this statement describes the composition of air, not the nature of it being a mixture. Many compounds also contain various elements, but that doesn't make them mixtures. Therefore, this statement does not prove that air is a mixture.
    2. Conclusion

    Statements I and II are the specific characteristics that define air as a mixture. Statement III is merely a factual observation about what is found inside the air, but it does not serve as evidence for the classification of "mixture."

    3. Identify the Answer

    Based on the analysis, the correct options are I and II.

    Exam Tip: Always remember the "Golden Rule" for mixtures: A mixture is a physical combination of substances that can be separated by physical methods and does not have a fixed chemical formula. If you see a question asking for proof of a mixture, look for "physical separation" or "variable composition."

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  8. 8.

    The number of hydroxonium ions produced by one molecule of an acid in aqueous solution is its

    1. A.basicity
    2. B.acid strength
    3. C.pH
    4. D.concentration
    Show answer

    Answer: A. basicity

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  9. 9.

    Which of the following is used to study the arrangement of particles in crystal lattices?

    1. A.Alpha-particles
    2. B.Beta-particles
    3. C.Gamma-rays
    4. D.X-rays
    Show answer

    Answer: D. X-rays

    AI explanation

    Analysis

    The question asks for the specific type of radiation or particle used in scientific research to determine the internal structure and arrangement of atoms or ions within a crystal lattice. This process is known as X-ray crystallography.

    Concept Explanation

    When X-rays are directed at a crystalline solid, they interact with the electrons of the atoms in the crystal. Because the wavelength of X-rays is comparable to the spacing between atoms in a crystal lattice, the rays undergo diffraction.

    By analyzing the pattern of the diffracted X-rays (a technique called X-ray diffraction or XRD), scientists can calculate the precise positions of atoms, the bond lengths, and the bond angles within the crystal structure. This is the primary method used to determine the molecular structure of complex substances, including proteins and minerals.

    Step-by-Step Solution
    1. Evaluate Alpha-particles (A): These are helium nuclei. They have low penetrating power and are easily stopped by a sheet of paper or skin. They are not suitable for probing the internal lattice of crystals.
    2. Evaluate Beta-particles (B): These are high-energy electrons. While they can be used in some specialized electron microscopy, they are not the standard tool for studying crystal lattice arrangements in the way described.
    3. Evaluate Gamma-rays (C): These have very high energy and very short wavelengths. While they can penetrate matter, they are generally too energetic and difficult to control for the precise diffraction patterns needed to map crystal lattices.
    4. Evaluate X-rays (D): X-rays have the ideal wavelength (in the range of 10−1010^{-10} meters) to match the inter-atomic distances in crystals. When they hit the crystal, they produce a diffraction pattern that acts as a "fingerprint" of the atomic arrangement.
    Conclusion

    X-rays are the standard tool used in laboratories worldwide for determining the arrangement of particles in crystal lattices due to their specific wavelength and ability to diffract off atomic planes.

    Exam Tip: Whenever you see a question about "crystal structure," "atomic spacing," or "diffraction patterns," immediately associate it with X-rays. This is a classic JAMB Chemistry question that tests your knowledge of the applications of electromagnetic radiation.

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  10. 10.

    Which of the following will produce a solution with pH less than 7 at equivalent point?

    1. A.HNO 3 +NaOH
    2. B.H 2SO 4 + KOH
    3. C.HCL +Mg (OH) 2
    4. D.HNO 3 + KOH
    Show answer

    Answer: C. HCL +Mg (OH) 2

    AI explanation

    Analysis

    The question asks us to identify which combination of acid and base will result in a solution with a pH less than 7 at the equivalence point.

    A pH less than 7 indicates an acidic solution. In acid-base titrations, the pH at the equivalence point depends on the strength of the acid and the base involved:

    1. Strong Acid + Strong Base: Produces a neutral salt (pH = 7).
    2. Strong Acid + Weak Base: Produces an acidic salt (pH < 7).
    3. Weak Acid + Strong Base: Produces a basic salt (pH > 7).
    Concept Explanation

    To solve this, we must identify the strength of the reactants in each option:

    • Strong Acids: HClHCl, HNO3HNO_3, H2SO4H_2SO_4.
    • Strong Bases: NaOHNaOH, KOHKOH.
    • Weak Bases: Mg(OH)2Mg(OH)_2 (Magnesium hydroxide is sparingly soluble and acts as a weak base).

    When a strong acid reacts with a weak base, the resulting salt undergoes hydrolysis. The cation from the weak base reacts with water to release hydrogen ions (H+H^+), making the solution acidic.

    Step-by-Step Solution

    Let's evaluate the options:

    • A. HNO3HNO_3 (Strong Acid) + NaOHNaOH (Strong Base): These react to form a neutral salt (NaNO3NaNO_3). The pH is 7.
    • B. H2SO4H_2SO_4 (Strong Acid) + KOHKOH (Strong Base): These react to form a neutral salt (K2SO4K_2SO_4). The pH is 7.
    • C. HClHCl (Strong Acid) + Mg(OH)2Mg(OH)_2 (Weak Base): The reaction is: 2HCl+Mg(OH)2→MgCl2+2H2O2HCl + Mg(OH)_2 \rightarrow MgCl_2 + 2H_2O Because Mg(OH)2Mg(OH)_2 is a weak base, the Mg2+Mg^{2+} ions undergo hydrolysis in water to produce H+H^+ ions, resulting in a solution with a pH < 7.
    • D. HNO3HNO_3 (Strong Acid) + KOHKOH (Strong Base): These react to form a neutral salt (KNO3KNO_3). The pH is 7.
    Conclusion

    Option C is the only combination involving a weak base, which leads to an acidic equivalence point.

    Exam Tip

    Always remember the "Stronger Wins" rule for salt hydrolysis:

    • If the acid is stronger than the base, the salt is acidic (pH < 7).
    • If the base is stronger than the acid, the salt is basic (pH > 7).
    • If both are strong (or both are weak), the salt is neutral (pH = 7).

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  11. 11.

    Commercial bleaching can be carried out using

    1. A.sulphur (IV) oxide and ammonia
    2. B.hydrogen sulphide and chlorine
    3. C.chlorine and sulphur (IV) oxide
    4. D.ammonia and chlorine
    Show answer

    Answer: C. chlorine and sulphur (IV) oxide

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  12. 12.

    The saponification of an alkanoate to produce soap and alkanol involves

    1. A.dehydration
    2. B.esterification
    3. C.hydrolysis
    4. D.oxidation
    Show answer

    Answer: C. hydrolysis

    AI explanation

    Analysis of the Question

    The question asks for the chemical process involved when an alkanoate (specifically an ester, like a fat or oil) reacts to produce soap and an alkanol (alcohol).

    Concept Explanation

    Saponification is the alkaline hydrolysis of an ester (typically a triglyceride, which is a fat or oil).

    In this process:

    1. Reactants: An ester (alkanoate) reacts with a strong base (like Sodium Hydroxide, NaOH, or Potassium Hydroxide, KOH).
    2. Reaction: The base breaks the ester bonds.
    3. Products: The reaction yields a salt of a fatty acid (which is soap) and an alcohol (alkanol, usually glycerol).

    The term "hydrolysis" comes from "hydro" (water) and "lysis" (splitting). In saponification, the ester is split by the action of the hydroxide ions in the presence of water.

    Step-by-Step Solution
    1. Identify the reaction type: Saponification is defined as the alkaline hydrolysis of fats and oils.
    2. Examine the options: • A. Dehydration: This is the removal of water molecules from a compound. This is not the primary mechanism of saponification. • B. Esterification: This is the reverse of saponification. It is the reaction between an alkanoic acid and an alkanol to form an ester and water. • C. Hydrolysis: This is the chemical breakdown of a compound due to reaction with water (or hydroxide ions). Since saponification involves breaking the ester bond to form soap and alcohol, it is a form of hydrolysis. • D. Oxidation: This involves the loss of electrons or an increase in oxidation state. This is not the mechanism for soap formation.
    Conclusion

    Since saponification is specifically the alkaline hydrolysis of an ester, the correct process is hydrolysis.

    Exam Tip

    Think of Saponification as the "breaking down" of an ester. Whenever you see "splitting" or "breaking" a complex molecule using water or a base, think Hydrolysis. Remember: Esterification builds the ester; Hydrolysis (Saponification) breaks it down!

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  13. 13.

    An increase in the pressure exerted on gas at a constant temperature results in

    1. A.a decrease in the number of effective collisions
    2. B.a decrease in volume
    3. C.an increase in the average intermolecular distance
    4. D.an increase in volume
    Show answer

    Answer: B. a decrease in volume

    AI explanation

    Analysis

    The question asks for the effect of increasing pressure on a gas when the temperature is kept constant. This is a classic application of the Gas Laws, specifically Boyle's Law.

    Concept: Boyle's Law

    Boyle's Law states that for a fixed mass of an ideal gas at a constant temperature, the pressure (PP) of the gas is inversely proportional to its volume (VV).

    Mathematically, this is expressed as:

    P∝1VorPV=kP \propto \frac{1}{V} \quad \text{or} \quad PV = k

    Where:

    • PP = Pressure
    • VV = Volume
    • kk = Constant

    This means that if you increase the pressure on a gas, the particles are forced closer together, which causes the volume to decrease. Conversely, if you decrease the pressure, the volume increases.

    Step-by-Step Solution
    1. Identify the variables: We are dealing with Pressure (PP) and Volume (VV) at a constant Temperature (TT).
    2. Apply the relationship: According to Boyle's Law, PP and VV have an inverse relationship.
    3. Analyze the change: The question states that pressure is increased.
    4. Determine the result: Because the relationship is inverse, an increase in pressure must lead to a decrease in volume.
    5. Evaluate the options: • A: Incorrect. Increasing pressure actually increases the frequency of collisions because particles are closer together. • B: Correct. This aligns with Boyle's Law. • C: Incorrect. Increasing pressure forces particles closer together, so the intermolecular distance decreases, not increases. • D: Incorrect. This describes a decrease in pressure, not an increase.
    Conclusion

    When you compress a gas (increase pressure) at a constant temperature, the gas molecules occupy a smaller space, resulting in a decrease in volume.

    Exam Tip

    Whenever you see a gas law question, remember the "P-V-T" rule:

    • Boyle's Law: Temperature is constant (P∝1/VP \propto 1/V).
    • Charles's Law: Pressure is constant (V∝TV \propto T).
    • Pressure Law: Volume is constant (P∝TP \propto T).
    • Memory Trick: Think of a balloon. If you squeeze it (increase pressure), it gets smaller (decrease volume).

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  14. 14.

    Carbohydrates are large molecules with the molecular formula Cx_x(H2_2O)y_y. In which of the following pairs is x not equal to y?

    1. A.glucose and starch
    2. B.maltose and starch
    3. C.sucrose and fructose
    4. D.galactose and starch
    Show answer

    Answer: B. maltose and starch

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  15. 15.

    The highest rate of production of carbon (lV) oxide can be achieved using

    1. A.0.05 mol -3HCI and 5g powdered CaCO3
    2. B.0.05 mol -3HCI and 5g lump CaCO3
    3. C.0.10 mol -3HCI and 5g powdered CaCO3
    4. D.0.025 mol -3HCI and 5g powdered CaCO3
    Show answer

    Answer: C. 0.10 mol -3HCI and 5g powdered CaCO3

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  16. 16.

    2Na2O2(s)+2H2O(1)Na_2O_{2(s)} + 2H_2O_{(1)}→ 4NaOH(s)+O2(g)NaOH_{(s)} +O_{2(g)} The substance that is oxidized in the reaction above is

    1. A.Na2O2(s)Na_2O_{2(s)}
    2. B.NaOH(aq)
    3. C.H2O(1)
    4. D.O2(g)
    Show answer

    Answer: A. Na2O2(s)Na_2O_{2(s)}

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  17. 17.

    Which of the following metals is the least reactive?

    1. A.Pb
    2. B.Sn
    3. C.Hg
    4. D.Au
    Show answer

    Answer: D. Au

    AI explanation

    Analysis

    The question asks us to identify the least reactive metal among the given options. Reactivity in metals is determined by their ability to lose electrons to form positive ions (cations). The more easily a metal loses electrons, the more reactive it is.

    Concept: The Reactivity Series

    In Chemistry, the Reactivity Series (or Activity Series) is a list of metals arranged in descending order of their reactivity. Metals at the top of the series are highly reactive (e.g., Potassium, Sodium), while those at the bottom are very unreactive (noble metals).

    The order of the metals provided in the options, from most reactive to least reactive, is:

    1. Sn (Tin)
    2. Pb (Lead)
    3. Hg (Mercury)
    4. Au (Gold)
    Step-by-Step Solution
    1. Evaluate Sn (Tin): Tin is a moderately reactive metal that reacts slowly with acids and is often used in plating to prevent corrosion.
    2. Evaluate Pb (Lead): Lead is less reactive than Tin. It sits below Tin in the reactivity series.
    3. Evaluate Hg (Mercury): Mercury is a transition metal that is significantly less reactive than both Lead and Tin. It does not react with dilute acids to displace hydrogen.
    4. Evaluate Au (Gold): Gold is a noble metal. It is extremely resistant to chemical attack, oxidation, and corrosion. It is found at the very bottom of the reactivity series, making it the least reactive among the options provided.
    Conclusion

    Comparing the positions of these metals in the reactivity series, Gold (Au) is the least reactive. It is so unreactive that it is often found in its native (pure) state in nature.

    Exam Tip: Remember the mnemonic for the reactivity series: "Please Stop Calling Me A Careless Zebra Instead Try Learning How Copper Saves Gold." (Potassium, Sodium, Calcium, Magnesium, Aluminium, Carbon, Zinc, Iron, Tin, Lead, Hydrogen, Copper, Silver, Gold). The further to the right (or bottom) the metal is, the less reactive it becomes.

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  18. 18.

    How many atoms are present in 6.0g of magnesium? [Mg = 24, NA = 6.02 x 1023 mol−1mol^{-1}]

    1. A.1.20 x 1022
    2. B.2.41 x 1022
    3. C.1.51 x 1023
    4. D.3.02 x 1023
    Show answer

    Answer: C. 1.51 x 1023

    AI explanation

    Analysis

    To find the number of atoms in a given mass of an element, we need to relate the mass to the number of moles, and then use Avogadro's constant (NAN_A) to convert moles into the number of atoms.

    Concept
    1. Calculate Moles (nn): The number of moles is calculated by dividing the given mass (mm) by the molar mass (MM). $n=mMn = \frac{m}{M}$
    2. Calculate Atoms (NN): The number of atoms is the product of the number of moles and Avogadro's constant (NAN_A). $N=n×NAN = n \times N_A$
    Step-by-Step Solution

    Step 1: Identify the given values

    • Mass of Magnesium (mm) = 6.0 g
    • Molar mass of Magnesium (MM) = 24 g/mol
    • Avogadro's constant (NAN_A) = 6.02×1023 mol−16.02 \times 10^{23} \text{ mol}^{-1}

    Step 2: Calculate the number of moles of Magnesium

    n=6.0 g24 g/moln = \frac{6.0 \text{ g}}{24 \text{ g/mol}}
    n=0.25 moln = 0.25 \text{ mol}

    Step 3: Calculate the number of atoms

    N=0.25×(6.02×1023)N = 0.25 \times (6.02 \times 10^{23})
    N=14×6.02×1023N = \frac{1}{4} \times 6.02 \times 10^{23}
    N=1.505×1023N = 1.505 \times 10^{23}

    Rounding to two decimal places as presented in the options, we get 1.51×10231.51 \times 10^{23}.

    Conclusion

    Comparing our result to the options provided, the calculated value matches option C.

    Exam Tip

    When dealing with mole calculations, always check your units. If you are given mass in grams and molar mass in g/mol, your result will be in moles. If you ever get stuck, remember the "Mole Map": Mass →\rightarrow Moles →\rightarrow Particles (Atoms/Molecules). Dividing by molar mass takes you to moles; multiplying by Avogadro's number takes you to particles.

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  19. 19.

    The electronic configuration of Mg 2+ is

    1. A.ls 22s 22P 63s23P2
    2. B.Is22s22P6
    3. C.ls22s22p63s2
    4. D.ls22s22P4
    Show answer

    Answer: B. Is22s22P6

    AI explanation

    Analysis

    To find the electronic configuration of an ion like Mg2+\text{Mg}^{2+}, we must first determine the configuration of the neutral atom and then account for the loss of electrons.

    Concept
    1. Atomic Number: Magnesium (Mg\text{Mg}) has an atomic number of 12. This means a neutral magnesium atom has 12 protons and 12 electrons.
    2. Neutral Configuration: Following the Aufbau principle, we fill the orbitals in order of increasing energy: 1s→2s→2p→3s1s \rightarrow 2s \rightarrow 2p \rightarrow 3s. • For neutral Mg\text{Mg} (12 electrons): 1s22s22p63s21s^2 2s^2 2p^6 3s^2.
    3. Ion Formation: The Mg2+\text{Mg}^{2+} ion indicates that the neutral atom has lost 2 electrons. • We remove the electrons from the outermost shell (the highest energy level), which is the 3s3s orbital.
    Step-by-Step Solution
    1. Start with neutral Magnesium: 1s22s22p63s21s^2 2s^2 2p^6 3s^2 (Total = 12 electrons)
    2. Remove 2 electrons: Since the 3s3s orbital contains 2 electrons, removing them leaves us with: 1s22s22p61s^2 2s^2 2p^6 (Total = 10 electrons)

    Comparing this result to your options:

    • Option A: 1s22s22p63s23p21s^2 2s^2 2p^6 3s^2 3p^2 (Incorrect, this is Silicon)
    • Option B: 1s22s22p61s^2 2s^2 2p^6 (Correct)
    • Option C: 1s22s22p63s21s^2 2s^2 2p^6 3s^2 (Incorrect, this is neutral Magnesium)
    • Option D: 1s22s22p41s^2 2s^2 2p^4 (Incorrect)
    Exam Tip

    When dealing with cations (positive ions), always remove electrons from the shell with the highest principal quantum number (nn) first. For transition metals, remember that the 4s4s electrons are removed before the 3d3d electrons!

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  20. 20.

    Chlorination of water for town supply is carried out to

    1. A.make the water colourless
    2. B.remove germs from the water
    3. C.make the water tastedful
    4. D.remove odour from the water
    Show answer

    Answer: B. remove germs from the water

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  21. 21.

    The isomers of C3H8O ARE

    1. A.1 - propanol and 2 - propanol
    2. B.1 - propanol and 1 - propanol
    3. C.2 - propanol and 2 - propanone
    4. D.2 - propanol and 1 - propanol
    Show answer

    Answer: A. 1 - propanol and 2 - propanol

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  22. 22.

    The chemicals used to soften hard water involves the addition of

    1. A.insoluble sodium compounds which from soluble solutions of calcium and magesium
    2. B.solube soudium compounds which from soluble solutions of calcium and magnesium ions
    3. C.solube sodium compounds which from insoluble precipitates of calcium and magnesium ions
    4. D.insoluble precipitates of calcium and magnesium ions
    Show answer

    Answer: C. solube sodium compounds which from insoluble precipitates of calcium and magnesium ions

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  23. 23.

    The oxidation number of boron in NaBH 4 is

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

    Answer: D. +3

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  24. 24.

    If the elements X and Y have atomic numbers 11 and 17 respectively, what type of bond can they form?

    1. A.Dative
    2. B.Covalent
    3. C.Ionic
    4. D.Metallic
    Show answer

    Answer: C. Ionic

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  25. 25.

    The solubilites of different solutes in a given solvent can be compared by

    1. A.plotting their solubility curves on separete axes
    2. B.plotting their solubility curves on the same axes
    3. C.plotting some of the solubility curves on the x-axis and others on the y-axis
    4. D.plotting their solubility curves on the x-axis only
    Show answer

    Answer: B. plotting their solubility curves on the same axes

    AI explanation

    Analysis

    The question asks for the standard scientific method used to compare the solubilities of different substances (solutes) in a specific solvent at varying temperatures.

    Concept Explanation

    A solubility curve is a graphical representation of how the solubility of a solute changes with temperature. The y-axis typically represents the solubility (usually in grams per 100g of solvent), and the x-axis represents the temperature (in degrees Celsius).

    To compare two or more different solutes effectively, you must be able to see their solubility values side-by-side at the same temperature points. If you plot them on separate graphs (separate axes), you lose the visual reference needed to see which solute is more or less soluble than the other at a specific temperature.

    Step-by-Step Solution
    1. Understand the Goal: We want to compare the solubility of Substance A vs. Substance B.
    2. Standardize the Variables: For a valid comparison, both substances must be measured against the same independent variable (temperature) and the same dependent variable (solubility).
    3. Graphical Representation: By plotting both curves on the same set of axes (same x-axis for temperature and same y-axis for solubility), the curves will intersect or run parallel. This allows you to immediately identify: • Which substance is more soluble at a specific temperature. • How the rate of solubility change differs between the substances.
    4. Evaluate Options: • Option A is incorrect because separate axes make visual comparison difficult. • Option B is correct because it provides a common reference frame for both solutes. • Options C and D are mathematically incorrect and do not follow standard graphing conventions for solubility.
    Conclusion

    Exam Tip: When you see a solubility curve question in your JAMB exam, remember that the "steeper" the curve, the more sensitive the solubility of that substance is to temperature changes. If two curves intersect, that point of intersection represents the temperature at which both substances have the exact same solubility.

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  26. 26.

    Potassium trioxochlorate (V)has a solubility of 1.5 mol dm -3 at 45 oC.On cooling this solution to a temperature of 20 oC, the solubility was found to be 0.5 mol dm -3. What mass of KCIO 3 was crystalized out? [K = 39, Cl = 35.5 O =16]

    1. A.1.00g
    2. B.10.00g
    3. C.12.25g
    4. D.122.50g
    Show answer

    Answer: D. 122.50g

    AI explanation

    1. Analyze the Question

    We are given the solubility of Potassium trioxochlorate (V) (KClO3KClO_3) at two different temperatures. We need to find the mass of the salt that crystallizes out when the solution is cooled from 45∘C45^\circ C to 20∘C20^\circ C.

    2. Explain the Concept
    • Solubility: This is the amount of solute that dissolves in a given volume of solvent (usually 1dm31 dm^3 or 1000cm31000 cm^3) at a specific temperature.
    • Crystallization: When a saturated solution is cooled, its solubility decreases. The difference between the initial solubility and the final solubility represents the amount of solute that can no longer stay dissolved and therefore forms crystals.
    • Molar Mass Calculation: To convert moles to grams, we use the formula: $Mass=Number of moles×Molar Mass\text{Mass} = \text{Number of moles} \times \text{Molar Mass}$
    3. Show the Solution

    Step 1: Calculate the molar mass of KClO3KClO_3

    • K=39K = 39
    • Cl=35.5Cl = 35.5
    • O=16×3=48O = 16 \times 3 = 48
    • Molar Mass=39+35.5+48=122.5 g/mol\text{Molar Mass} = 39 + 35.5 + 48 = 122.5 \, g/mol

    Step 2: Calculate the number of moles that crystallized out

    • Solubility at 45∘C=1.5 mol/dm345^\circ C = 1.5 \, mol/dm^3
    • Solubility at 20∘C=0.5 mol/dm320^\circ C = 0.5 \, mol/dm^3
    • Moles crystallized=1.5−0.5=1.0 mol\text{Moles crystallized} = 1.5 - 0.5 = 1.0 \, mol

    Step 3: Convert the moles to mass

    • Mass=moles×molar mass\text{Mass} = \text{moles} \times \text{molar mass}
    • Mass=1.0 mol×122.5 g/mol\text{Mass} = 1.0 \, mol \times 122.5 \, g/mol
    • Mass=122.5 g\text{Mass} = 122.5 \, g
    4. Identify the Answer

    Comparing our result to the given options, the correct value is 122.50g.

    This matches option D.

    5. Exam Tip

    When dealing with solubility questions, always check the units carefully. If the question gives you solubility in mol/dm3mol/dm^3 but asks for mass in grams, you must calculate the molar mass of the compound and multiply it by the change in concentration. Don't forget to sum the atomic masses correctly!

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  27. 27.

    50 cm3_3 of gas was collected over water at 10ºC and 765 mm Hg. Calculate the volume of the gas at s.t.p. if the saturated vapour pressure of water at 10ºC is 5mm Hg

    1. A.49.19 cm 3
    2. B.48.87 cm 3
    3. C.48.55 cm 3
    4. D.48.23 cm 3
    Show answer

    Answer: D. 48.23 cm 3

    AI explanation

    To solve this problem, we use the Combined Gas Law, while accounting for the pressure of water vapor since the gas was collected over water.

    1. Analyze the question
    • Initial Volume (V1V_1): 50 cm350\text{ cm}^3
    • Initial Temperature (T1T_1): 10∘C=10+273=283 K10^\circ\text{C} = 10 + 273 = 283\text{ K}
    • Initial Pressure (P1P_1): The gas is collected over water, so the total pressure includes the pressure of the gas and the saturated vapour pressure (SVP) of water. • Pgas=Ptotal−Pwater vaporP_{\text{gas}} = P_{\text{total}} - P_{\text{water vapor}} • P1=765 mm Hg−5 mm Hg=760 mm HgP_1 = 765\text{ mm Hg} - 5\text{ mm Hg} = 760\text{ mm Hg}
    • S.T.P. Conditions: • Standard Pressure (P2P_2): 760 mm Hg760\text{ mm Hg} • Standard Temperature (T2T_2): 273 K273\text{ K}
    • Target: Find V2V_2 (Volume at S.T.P.)
    2. Explain the concept

    The Combined Gas Law is expressed as:

    P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

    When a gas is collected over water, it is "moist." To find the volume of the "dry" gas, we must subtract the vapour pressure of the water from the total atmospheric pressure.

    3. Show the solution

    Rearrange the formula to solve for V2V_2:

    V2=P1×V1×T2T1×P2V_2 = \frac{P_1 \times V_1 \times T_2}{T_1 \times P_2}

    Substitute the values:

    V2=760×50×273283×760V_2 = \frac{760 \times 50 \times 273}{283 \times 760}

    Notice that 760760 appears in both the numerator and the denominator, so they cancel out:

    V2=50×273283V_2 = \frac{50 \times 273}{283}
    V2=13650283V_2 = \frac{13650}{283}
    V2≈48.233 cm3V_2 \approx 48.233\text{ cm}^3
    4. Identify the answer

    Rounding to two decimal places, we get 48.23 cm348.23\text{ cm}^3. This matches option D.

    5. Exam Tip

    Always remember that whenever you see "collected over water," you must subtract the S.V.P. from the total pressure before plugging the values into your gas law equations. Also, ensure your temperature is always converted to Kelvin by adding 273273.

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  28. 28.

    2H2(g) + O2(g) → 2H2 O(g) In the reaction above, what volume of hydrogen would be left over when 300 cm3 of oxygen and 1000 cm3 of hydrogen are exploded in a sealed tube?

    1. A.200 cm3
    2. B.400 cm3
    3. C.600 cm3
    4. D.700 cm3
    Show answer

    Answer: B. 400 cm3

    AI explanation

    Analysis

    To solve this problem, we use Gay-Lussac’s Law of Combining Volumes, which states that gases react in simple whole-number ratios by volume under the same conditions of temperature and pressure.

    Concept

    The balanced chemical equation is: 2H₂(g) + O₂(g) → 2H₂O(g)

    From the stoichiometry of the equation:

    • 2 volumes of Hydrogen (H₂) react with 1 volume of Oxygen (O₂).
    • This means for every 1 cm³ of O₂, you need 2 cm³ of H₂.
    Step-by-Step Solution
    1. Determine the required volume of Hydrogen: We are given 300 cm³ of Oxygen. According to the ratio (2:1), the volume of Hydrogen needed to react completely with 300 cm³ of Oxygen is: Volume of H₂ needed = 2 × Volume of O₂ Volume of H₂ needed = 2 × 300 cm³ = 600 cm³
    1. Calculate the volume of Hydrogen left over: We started with 1000 cm³ of Hydrogen. Volume of H₂ left = (Initial volume of H₂) - (Volume of H₂ used) Volume of H₂ left = 1000 cm³ - 600 cm³ = 400 cm³
    Conclusion

    Since 600 cm³ of Hydrogen is consumed to react with the 300 cm³ of Oxygen, 400 cm³ of Hydrogen remains unreacted.

    Exam Tip: Always check the stoichiometry (the coefficients in the balanced equation) first. If you are given two volumes, one of them is usually the "limiting reactant." In this case, Oxygen was the limiting reactant because it ran out first, leaving the excess Hydrogen behind.

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  29. 29.

    Mineral acids are usually added to commercial hydrogen peroxide to

    1. A.oxidize it
    2. B.decompose it
    3. C.minimize its decomposition
    4. D.reduce it to water and oxygen
    Show answer

    Answer: C. minimize its decomposition

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  30. 30.

    Which of the following gases has a characteristic pungent smell, turns red litmus paper blue and forms dense white fumes with hydrogen chloride gas?

    1. A.N2
    2. B.N2O
    3. C.CI2
    4. D.NH3
    Show answer

    Answer: D. NH3

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  31. 31.

    A major factor considered in selecting a suitable method for preparing a simple salt is its

    1. A.Crystalline from
    2. B.melting point
    3. C.reactivity with dilute acids
    4. D.solubility in water
    Show answer

    Answer: D. solubility in water

    AI explanation

    Analysis

    The question asks for the primary criterion used to decide how to prepare a simple salt in the laboratory. In chemistry, the preparation method (e.g., titration, precipitation, or reaction of an acid with a metal/base) depends heavily on whether the salt can dissolve in water or if it exists as an insoluble solid.

    Concept Explanation

    The preparation of salts is governed by their solubility rules.

    1. Soluble Salts: If a salt is soluble in water, we often use methods like titration (for neutralization) or the reaction of an acid with a metal, metal oxide, or metal carbonate.
    2. Insoluble Salts: If a salt is insoluble in water, it is typically prepared by precipitation (double decomposition), where two soluble salt solutions are mixed to form the insoluble salt as a precipitate.

    Because the physical state of the salt in water determines the separation technique (like filtration vs. evaporation), solubility is the most critical factor.

    Step-by-Step Solution
    • Step 1: Evaluate the options. Melting point (B) and crystalline form (A) are physical properties, but they do not dictate the chemical synthesis route. Reactivity with acids (C) is a property of the reactants, not the salt being formed.
    • Step 2: Consider the preparation of a salt like Barium Sulphate (BaSO4BaSO_4). Since it is insoluble, we must use precipitation. Consider a salt like Sodium Chloride (NaClNaCl). Since it is soluble, we use titration or evaporation.
    • Step 3: Conclude that the decision to use filtration (for insoluble salts) or evaporation/crystallization (for soluble salts) is entirely dependent on the solubility of the salt in water.
    Conclusion

    This aligns with standard chemistry principles taught in the Nigerian secondary school curriculum, where solubility is the primary guide for salt preparation methods.

    Exam Tip

    Remember the "Solubility Rule" mnemonic: "All Nitrates are soluble; all Sodium, Potassium, and Ammonium salts are soluble." If you are asked about salt preparation in an exam, always check if the salt is soluble or insoluble first—that will lead you to the correct method immediately!

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  32. 32.

    An increase in entropy can best be illustrated by

    1. A.mixing of gases
    2. B.freezing of water
    3. C.the condensation of vapour
    4. D.solidifying candle wax
    Show answer

    Answer: A. mixing of gases

    AI explanation

    1. Analyze the Question

    The question asks for an example of an increase in entropy. In chemistry, entropy is a measure of the degree of disorder, randomness, or chaos in a system. An increase in entropy means the system is moving from a more ordered state to a more disordered state.

    2. Explain the Concept
    • Entropy (ΔS\Delta S): Think of entropy as the "messiness" of particles. • Solids have low entropy because particles are fixed in a rigid structure (highly ordered). • Liquids have higher entropy because particles can move past each other. • Gases have the highest entropy because particles move randomly at high speeds and occupy all available space (highly disordered).
    • The Rule: Processes that increase the freedom of movement or the volume available to particles result in an increase in entropy. Conversely, processes like freezing or condensation (moving from gas/liquid to solid) decrease entropy because they increase order.
    3. Show the Solution

    Let's evaluate the options:

    • A. Mixing of gases: When two different gases are mixed, the particles of each gas spread out into a larger volume and become more randomly distributed. This increase in spatial randomness represents a significant increase in entropy.
    • B. Freezing of water: Water molecules move from a liquid state (more random) to a solid ice state (ordered crystal lattice). This is a decrease in entropy.
    • C. The condensation of vapour: Vapour (gas) molecules lose kinetic energy and come together to form a liquid. This transition from high disorder to lower disorder is a decrease in entropy.
    • D. Solidifying candle wax: Similar to freezing, the wax molecules lose energy and settle into a more ordered solid structure. This is a decrease in entropy.
    4. Identify the Answer

    Since mixing gases increases the randomness and disorder of the system, it is the correct illustration of an increase in entropy.

    5. Exam Tip

    Remember the "Disorder Rule":

    • Solid →\rightarrow Liquid →\rightarrow Gas = Entropy Increases (ΔS>0\Delta S > 0)
    • Gas →\rightarrow Liquid →\rightarrow Solid = Entropy Decreases (ΔS<0\Delta S < 0)
    • Mixing or Dissolving usually Increases entropy because the particles are spreading out!

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  33. 33.

    Ethanol reacts with concentrated tetraoxosulphate (VI) acid at a temperature above 1700^0C to form

    1. A.ethanone
    2. B.ethene
    3. C.ethyne
    4. D.ethanal
    Show answer

    Answer: B. ethene

    AI explanation

    Analysis

    The question asks for the product formed when ethanol (C2H5OHC_2H_5OH) undergoes a reaction with concentrated tetraoxosulphate(VI) acid (H2SO4H_2SO_4) at a specific temperature (above 170°C). This is a classic organic chemistry reaction involving the dehydration of an alcohol.

    Concept: Dehydration of Alcohols

    Concentrated tetraoxosulphate(VI) acid acts as a dehydrating agent in this reaction. The outcome of the reaction between ethanol and concentrated H2SO4H_2SO_4 depends heavily on the temperature:

    1. At 140°C: The reaction produces an ether (ethoxyethane).
    2. At 170°C and above: The reaction undergoes intramolecular dehydration, where a molecule of water is removed from a single molecule of ethanol, resulting in the formation of an alkene (ethene).
    Step-by-Step Solution
    1. Identify the reactants: Ethanol (CH3CH2OHCH_3CH_2OH) and concentrated H2SO4H_2SO_4.
    2. Identify the process: Since the temperature is above 170°C, the concentrated H2SO4H_2SO_4 removes a water molecule (H2OH_2O) from the ethanol molecule.
    3. Write the chemical equation: $CH3CH2OH→conc.H2SO4,>170∘CCH2=CH2+H2OCH_3CH_2OH \xrightarrow{conc. H_2SO_4, >170^\circ C} CH_2=CH_2 + H_2O$
    4. Analyze the product: The product CH2=CH2CH_2=CH_2 is ethene, which is the simplest member of the alkene homologous series.
    Conclusion

    Comparing this to your options:

    • A. Ethanone (a ketone, not formed here)
    • B. Ethene (the correct alkene product)
    • C. Ethyne (an alkyne, not formed here)
    • D. Ethanal (an aldehyde, formed by oxidation, not dehydration)
    Exam Tip

    Remember the "Temperature Rule" for ethanol and H2SO4H_2SO_4:

    • 140°C = Ether (Think: "Two alcohols join to make an ether").
    • 170°C = Ethene (Think: "High heat breaks it down to an alkene").

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  34. 34.

    A compound contains 40.0% C, 6.7% H 53.3% O. If the molecular mass of the compound is 180, its molecular formula is [C =12, H =1,O]

    1. A.CH2O
    2. B.C3H6O3
    3. C.C6H6O3
    4. D.C6H12O6
    Show answer

    Answer: D. C6H12O6

    AI explanation

    To find the molecular formula of the compound, we follow these steps:

    1. Determine the Empirical Formula

    First, we assume a 100g sample of the compound. This means we have:

    • Carbon (C): 40.0g
    • Hydrogen (H): 6.7g
    • Oxygen (O): 53.3g

    Next, divide each mass by its relative atomic mass to find the number of moles:

    • Moles of C: 40.0/12=3.3340.0 / 12 = 3.33
    • Moles of H: 6.7/1=6.76.7 / 1 = 6.7
    • Moles of O: 53.3/16=3.3353.3 / 16 = 3.33

    Now, divide each value by the smallest number of moles (3.33) to get the simplest ratio:

    • C: 3.33/3.33=13.33 / 3.33 = 1
    • H: 6.7/3.33≈26.7 / 3.33 \approx 2
    • O: 3.33/3.33=13.33 / 3.33 = 1

    The empirical formula is CH2OCH_2O.

    2. Calculate the Empirical Formula Mass

    Using the atomic masses (C=12,H=1,O=16C=12, H=1, O=16):

    • Empirical Mass = 12+(2×1)+16=3012 + (2 \times 1) + 16 = 30
    3. Find the Molecular Formula

    The relationship between the molecular formula and the empirical formula is: Molecular Formula=(Empirical Formula)n\text{Molecular Formula} = (\text{Empirical Formula})_n Where n=Molecular MassEmpirical Formula Massn = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}}

    • n=180/30=6n = 180 / 30 = 6

    Now, multiply the empirical formula (CH2OCH_2O) by n=6n=6:

    • C(1×6)H(2×6)O(1×6)=C6H12O6C_{(1 \times 6)} H_{(2 \times 6)} O_{(1 \times 6)} = \mathbf{C_6H_{12}O_6}
    Conclusion

    Comparing our result to the options provided, the correct molecular formula is C6H12O6C_6H_{12}O_6.

    Exam Tip: Always remember that the empirical formula represents the simplest ratio of atoms, while the molecular formula represents the actual number of atoms in one molecule. If you are ever stuck, you can quickly check the options by calculating their molar masses—the one that equals 180 is likely your answer!

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  35. 35.

    Which of the following pollutants is associated with brain damage?

    1. A.Carbon (ll) oxide
    2. B.Radioactive fallout
    3. C.Biodegradable waste
    4. D.Sulphus (lV) oxide
    Show answer

    Answer: A. Carbon (ll) oxide

    AI explanation

    Analysis

    The question asks us to identify which of the listed pollutants is specifically linked to causing brain damage in humans. We need to evaluate the physiological effects of each pollutant listed.

    Concept Explanation
    • Carbon (II) oxide (CO): A colorless, odorless gas produced by incomplete combustion. It binds to hemoglobin in the blood much more strongly than oxygen, forming carboxyhemoglobin. This prevents oxygen from reaching vital organs, including the brain, leading to hypoxia (oxygen starvation), which can cause brain damage or death.
    • Radioactive fallout: While dangerous and capable of causing radiation sickness, cancer, and genetic mutations, it is not the primary pollutant associated with "brain damage" in the context of standard environmental chemistry questions.
    • Biodegradable waste: These are organic materials that decompose naturally. While they can cause water pollution (eutrophication), they are not direct neurotoxins.
    • Sulphur (IV) oxide (SO2SO_2): A major air pollutant that causes acid rain and respiratory issues (like asthma and bronchitis). It is primarily an irritant to the lungs and eyes.
    Step-by-Step Solution
    1. Evaluate Option D: Sulphur (IV) oxide is primarily a respiratory irritant. It does not target the brain directly.
    2. Evaluate Option C: Biodegradable waste affects ecosystems but does not have a direct toxicological link to brain damage.
    3. Evaluate Option B: Radioactive fallout is a severe health hazard, but in the context of common environmental pollutants taught in the Nigerian secondary school curriculum, it is categorized under radiation hazards rather than the specific pollutant-induced brain damage associated with chemical asphyxiants.
    4. Evaluate Option A: Carbon (II) oxide is a chemical asphyxiant. By depriving the brain of oxygen, it is a well-documented cause of brain damage, cognitive impairment, and neurological deficits.
    Conclusion

    Carbon (II) oxide is the correct answer because it directly interferes with oxygen transport to the brain, leading to severe neurological consequences.

    Exam Tip: Always remember that Carbon (II) oxide (CO) is a "silent killer." Whenever you see questions about CO, associate it with asphyxiation and oxygen deprivation.

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  36. 36.

    Group VII elements are

    1. A.monoatmic
    2. B.good oxidizing agents
    3. C.highly electropositive
    4. D.electron donors
    Show answer

    Answer: B. good oxidizing agents

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  37. 37.

    Which of the following compounds will burn with a brick-red colour in a non-luminous Bunsen flame?

    1. A.LiCI
    2. B.NaCl
    3. C.CaCl2
    4. D.MgCl2
    Show answer

    Answer: C. CaCl2

    AI explanation

    Analysis

    The question asks for the identification of a specific metal salt based on the characteristic flame colour it produces when heated in a non-luminous Bunsen flame. This is a classic test in qualitative analysis known as the Flame Test.

    Concept: Flame Test

    The flame test is used to detect the presence of certain metallic ions. When these metal salts are heated, the electrons in the metal atoms absorb energy and move to higher energy levels. As they return to their ground state, they emit light at specific wavelengths, which appear as distinct colours to the human eye.

    Here are the characteristic flame colours for the metals in your options:

    • Lithium (Li⁺): Produces a crimson (deep red) flame.
    • Sodium (Na⁺): Produces a golden-yellow flame.
    • Calcium (Ca²⁺): Produces a brick-red flame.
    • Magnesium (Mg²⁺): Does not produce a characteristic flame colour (it is often considered non-flame-active in standard school laboratory tests).
    Step-by-Step Solution
    1. Examine the options: We are looking for the compound that contains the metal ion responsible for a "brick-red" flame.
    2. Evaluate the metals: • LiCl contains Lithium (Crimson). • NaCl contains Sodium (Golden-yellow). • CaCl₂ contains Calcium (Brick-red). • MgCl₂ contains Magnesium (No characteristic colour).
    3. Match the observation: Since Calcium ions (Ca²⁺) are known to impart a brick-red colour to a non-luminous flame, the compound CaCl₂ is the correct choice.
    Conclusion
    Exam Tip

    To remember these easily for your JAMB or WASSCE exams, use this mnemonic:

    • Sodium = Sunny yellow
    • Calcium = Chili red (Brick-red)
    • Potassium = Purple (Lilac)
    • Lithium = Lovely crimson

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  38. 38.

    A piece of filter paper moistened with lead (ll) ethanoate solution turns black when the paper is dropped into a gas likely to be

    1. A.sulphur (VI) oxide
    2. B.hydrogen chloride
    3. C.sulphur (VI) oxide
    4. D.hydrogen sulphide
    Show answer

    Answer: D. hydrogen sulphide

    AI explanation

    Analysis

    The question asks us to identify a gas that reacts with lead (II) ethanoate to produce a black precipitate. This is a classic qualitative analysis test used in chemistry to detect the presence of specific gases.

    Concept

    Lead (II) ethanoate, also known as lead (II) acetate, has the chemical formula Pb(CH3COO)2\text{Pb(CH}_3\text{COO)}_2. When this compound comes into contact with hydrogen sulphide (H2S\text{H}_2\text{S}) gas, a double decomposition reaction occurs.

    The lead ions (Pb2+\text{Pb}^{2+}) react with the sulphide ions (S2−\text{S}^{2-}) from the gas to form lead (II) sulphide (PbS\text{PbS}), which is a black solid. The chemical equation for this reaction is:

    Pb(CH3COO)2(aq)+H2S(g)→PbS(s)+2CH3COOH(aq)\text{Pb(CH}_3\text{COO)}_2(aq) + \text{H}_2\text{S}(g) \rightarrow \text{PbS}(s) + 2\text{CH}_3\text{COOH}(aq)

    The formation of this black lead (II) sulphide precipitate is the standard confirmatory test for the presence of hydrogen sulphide gas.

    Step-by-Step Solution
    1. Identify the reagent: Lead (II) ethanoate is used as a test paper for gases.
    2. Analyze the reaction: When the gas reacts with the lead ions, it forms a metal sulphide.
    3. Evaluate the product: Metal sulphides, particularly lead (II) sulphide (PbS\text{PbS}), are characteristically black in color.
    4. Match with options: • Sulphur (VI) oxide (SO3\text{SO}_3) and hydrogen chloride (HCl\text{HCl}) do not form black precipitates with lead (II) ethanoate. • Hydrogen sulphide (H2S\text{H}_2\text{S}) is the specific gas that reacts to form the black lead (II) sulphide precipitate.
    Conclusion

    The gas that turns lead (II) ethanoate paper black is hydrogen sulphide. This matches option D in your list. (Note: Option A and C are identical in your provided list; both are incorrect).

    Exam Tip

    Whenever you see "lead (II) ethanoate" or "lead (II) acetate" in a question, immediately associate it with the detection of sulphide ions (S2−\text{S}^{2-}). Think of the "blackening" as the formation of "black lead sulphide." This is a very common question in JAMB and WASSCE practical chemistry!

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  39. 39.

    The purest form of iron which contains only about 0.1% carbon is

    1. A.pig iron
    2. B.wrought iron
    3. C.cast iron
    4. D.iron pyrite
    Show answer

    Answer: B. wrought iron

    AI explanation

    Analysis

    The question asks for the purest form of iron among the given options, specifically identifying the one with a very low carbon content (approximately 0.1%).

    Concept Explanation

    Iron is extracted from its ores and processed into different forms based on the amount of carbon and other impurities present. The properties of iron change significantly depending on its carbon content:

    • Pig Iron: This is the crude form of iron obtained directly from the blast furnace. It contains about 3% to 5% carbon, along with other impurities like silicon, sulfur, and phosphorus. It is brittle and hard.
    • Wrought Iron: This is the purest form of commercial iron. It is produced by refining pig iron to remove almost all impurities. It contains less than 0.2% carbon (typically around 0.1%). Because of its high purity, it is malleable, ductile, and tough.
    • Cast Iron: This is made by remelting pig iron with scrap iron and coke. It contains about 2% to 4% carbon. It is very hard but brittle.
    • Iron Pyrite (FeS2FeS_2): This is an ore of iron, commonly known as "fool's gold." It is a mineral, not a processed form of metallic iron.
    Step-by-Step Solution
    1. Evaluate Pig Iron: Contains 3–5% carbon. This is high, so it is not the purest form.
    2. Evaluate Wrought Iron: Contains approximately 0.1% carbon. This matches the description in the question perfectly.
    3. Evaluate Cast Iron: Contains 2–4% carbon. This is too high to be considered the purest form.
    4. Evaluate Iron Pyrite: This is a chemical compound (sulfide ore), not a metallic form of iron used in industry.

    Comparing these, wrought iron is the form that contains the least amount of carbon, making it the purest form of iron listed.

    Conclusion

    Exam Tip: Remember the "Carbon Rule": The higher the carbon content, the harder and more brittle the iron. The lower the carbon content, the softer, more malleable, and purer the iron. Wrought iron is the "purest" and "softest" of the three main industrial types.

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  40. 40.

    An example of oxidation - reduction enzyme is

    1. A.amylase
    2. B.protease
    3. C.lipase
    4. D.dehydrogenase
    Show answer

    Answer: D. dehydrogenase

    AI explanation

    Analysis

    The question asks for an example of an enzyme that catalyzes oxidation-reduction (redox) reactions. In biochemistry, enzymes are classified based on the type of reaction they catalyze.

    Concept Explanation

    Enzymes are biological catalysts that speed up chemical reactions in living organisms. They are categorized into six major classes by the International Union of Biochemistry and Molecular Biology (IUBMB):

    1. Oxidoreductases: These catalyze oxidation-reduction reactions (the transfer of electrons or hydrogen atoms from one molecule to another). Examples include dehydrogenases, oxidases, and peroxidases.
    2. Transferases: These transfer functional groups (e.g., methyl or phosphate groups).
    3. Hydrolases: These catalyze the breaking of chemical bonds by the addition of water (hydrolysis). Examples include amylase, protease, and lipase.
    4. Lyases: These break bonds by means other than hydrolysis or oxidation.
    5. Isomerases: These catalyze structural changes within a single molecule.
    6. Ligases: These catalyze the joining of two molecules.
    Step-by-Step Solution
    • Amylase (Option A): This is a hydrolase that breaks down starch into sugars.
    • Protease (Option B): This is a hydrolase that breaks down proteins into peptides or amino acids.
    • Lipase (Option C): This is a hydrolase that breaks down fats (lipids) into fatty acids and glycerol.
    • Dehydrogenase (Option D): This enzyme removes hydrogen atoms from a substrate, which is a fundamental process in biological oxidation. Therefore, it belongs to the class of oxidoreductases.

    Since the question specifically asks for an oxidation-reduction enzyme, dehydrogenase is the correct classification.

    Exam Tip

    Remember the suffix "-ase" usually denotes an enzyme. To distinguish them:

    • If it breaks bonds using water, it is a Hydrolase (Amylase, Protease, Lipase).
    • If it involves moving electrons or hydrogen, it is an Oxidoreductase (Dehydrogenase).

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  41. 41.

    CO(g) + H2_2 O(g)_{(g)} → CO2(g)_{2(g)} + H2(g)_{2(g)} Calculate the standard heat change of the reaction above, if the standard enthalpies of formation of CO2(g)_{2(g)}, H2_2O(g)_{(g)} and CO(g)_{(g)} in kJmol−1^{-1} are - 394, -242 and - 110 respectively.

    1. A.+ 262 KJ mol -1
    2. B.- 262 KJ mol -1
    3. C.+ 42 KJ mol -1
    4. D.- 42 KJ mol -1
    Show answer

    Answer: D. - 42 KJ mol -1

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  42. 42.

    A hydrogen atom which has lost an electron contains

    1. A.one proton only
    2. B.one neutron only
    3. C.one proton and one neutron
    4. D.one proton, one electron and one neutron
    Show answer

    Answer: A. one proton only

    AI explanation

    Analysis

    To solve this, we need to look at the atomic structure of a neutral hydrogen atom and understand what happens when it becomes an ion.

    Concept: Atomic Structure of Hydrogen
    1. Neutral Hydrogen (11H^1_1H): A standard hydrogen atom consists of 1 proton in the nucleus and 1 electron orbiting the nucleus. Notably, the most common isotope of hydrogen (Protium) contains zero neutrons.
    2. Ionization: When an atom "loses an electron," it becomes a positively charged ion (a cation). In the case of hydrogen, if it loses its only electron, it is left with only its nucleus.
    Step-by-Step Solution
    • Step 1: Identify the components of a neutral hydrogen atom: 1 proton, 0 neutrons, and 1 electron.
    • Step 2: Apply the change described in the question: The atom loses its 1 electron.
    • Step 3: Calculate the remaining particles: • Protons: 1 (remains unchanged) • Electrons: 1 - 1 = 0 • Neutrons: 0 (remains unchanged)
    • Step 4: Conclusion: After losing its electron, the hydrogen atom is left with only the 1 proton that constitutes its nucleus. This is effectively a hydrogen ion (H+H^+).
    Identifying the Answer

    Comparing our result to the options provided:

    • A. one proton only — Matches our finding.
    • B. one neutron only — Incorrect (Hydrogen has no neutrons).
    • C. one proton and one neutron — Incorrect (Hydrogen has no neutrons).
    • D. one proton, one electron and one neutron — Incorrect (This describes a different element or isotope).
    Exam Tip

    Always remember the "Hydrogen Exception": Hydrogen is the only element in the periodic table that typically has no neutrons in its nucleus. When you see questions about hydrogen, keep in mind that its mass number (1) is equal to its atomic number (1), confirming the absence of neutrons (Mass=Protons+Neutrons→1=1+0Mass = Protons + Neutrons \rightarrow 1 = 1 + 0).

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  43. 43.

    A common characteristic between zinc and the other transition elements is the ability to

    1. A.have variable oxidation states
    2. B.from complex ions
    3. C.act as a catalyst
    4. D.from coloured ions
    Show answer

    Answer: B. from complex ions

    AI explanation

    1. Analyze the question

    The question asks for a property that zinc shares with other transition elements. To answer this, we must recall the definition of transition elements and the unique electronic configuration of zinc.

    2. Explain the concept
    • Transition Elements: By definition, a transition element is an element that has an incomplete dd-subshell in its neutral state or in at least one of its common oxidation states.
    • Zinc (ZnZn): Zinc has an electronic configuration of [Ar]3d104s2[Ar] 3d^{10} 4s^2. Even in its common oxidation state (Zn2+Zn^{2+}), its configuration is [Ar]3d10[Ar] 3d^{10}. Because its dd-subshell is completely filled, zinc is technically not a transition element according to the strict IUPAC definition.
    • Comparison: • Variable oxidation states: Most transition metals exhibit multiple oxidation states (e.g., Fe2+,Fe3+Fe^{2+}, Fe^{3+}). Zinc only exhibits +2+2. • Coloured ions: Transition metals form coloured ions due to d−dd-d transitions. Because zinc has a full dd-subshell, it cannot undergo d−dd-d transitions, so its ions are colourless. • Catalytic activity: Transition metals act as catalysts due to their variable oxidation states and ability to provide surface area. Zinc is generally not considered a good catalyst compared to metals like Fe,Ni,Fe, Ni, or VV. • Complex ion formation: Both transition metals and zinc have the ability to form complex ions (coordination compounds) by donating lone pairs of electrons into empty orbitals (e.g., [Zn(NH3)4]2+[Zn(NH_3)_4]^{2+}).
    3. Show the solution
    • Option A: Incorrect. Zinc only shows a +2+2 oxidation state.
    • Option C: Incorrect. Zinc does not possess the variable oxidation states required for typical transition metal catalysis.
    • Option D: Incorrect. Zinc ions (Zn2+Zn^{2+}) are colourless because the 3d3d subshell is full.
    • Option B: Correct. Despite not being a "true" transition metal, zinc shares the ability to form stable complex ions with transition metals because it has empty 4s4s and 4p4p orbitals available for coordinate bonding.
    4. Identify the answer

    The correct option is B.

    5. Exam Tip

    When you see a question about zinc in the context of transition metals, always remember the "Zinc Exception": Zinc is the "odd one out" because it lacks variable oxidation states, coloured ions, and catalytic properties. However, it behaves like a transition metal specifically in its ability to form complex ions.

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  44. 44.

    Alkanals can be distinguished from alkanones by the reaction with

    1. A.sudan lll stain
    2. B.starch iodide paper
    3. C.lithium tetrahydrido aluminate (lll)
    4. D.fehling's solution
    Show answer

    Answer: D. fehling's solution

    AI explanation

    Analysis

    The question asks for a chemical reagent that can differentiate between alkanals (aldehydes) and alkanones (ketones). Both groups contain the carbonyl functional group (C=OC=O), but they differ in their reactivity due to the position of this group.

    Concept Explanation
    • Alkanals (Aldehydes): These have the carbonyl group at the end of the carbon chain. Because of this position, they are easily oxidized to carboxylic acids.
    • Alkanones (Ketones): These have the carbonyl group in the middle of the carbon chain. They are resistant to mild oxidation.

    To distinguish them, we use mild oxidizing agents. When an alkanal is heated with a mild oxidizing agent, it gets oxidized, causing a visible color change. An alkanone will not react under the same conditions, so no color change occurs.

    Step-by-Step Solution
    1. Evaluate the options: • A. Sudan III stain: This is used in biology to test for the presence of fats/lipids, not for distinguishing carbonyl compounds. • B. Starch iodide paper: This is typically used to test for the presence of oxidizing agents (like chlorine or peroxides), not specifically to distinguish aldehydes from ketones. • C. Lithium tetrahydrido aluminate (III) (LiAlH4LiAlH_4): This is a strong reducing agent. It will reduce both alkanals and alkanones to their respective alcohols, so it cannot be used to distinguish them. • D. Fehling's solution: This is a mild oxidizing agent containing copper(II) ions. • When heated with an alkanal, the blue Cu2+Cu^{2+} ions are reduced to a reddish-brown precipitate of copper(I) oxide (Cu2OCu_2O). • When heated with an alkanone, there is no reaction, and the solution remains blue.
    1. Conclusion: Because Fehling's solution reacts with alkanals but not with alkanones, it is the perfect reagent for distinguishing between the two.
    Exam Tip

    Remember the "Tollens' and Fehling's" rule: Aldehydes are "easy" to oxidize. They will react with mild oxidizing agents like Fehling's solution and Tollens' reagent (silver mirror test). Ketones are "stubborn" and will only react with very strong oxidizing agents under harsh conditions.

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  45. 45.

    Which of the following methods can be used to obtain pure water from a mixture of sand, water and methanoic acid?

    1. A.netralization with NaOHfoolowed by filtration
    2. B.neutralization with NaOH followed by distillation
    3. C.fractional distillation
    4. D.filtration followed by distillation
    Show answer

    Answer: B. neutralization with NaOH followed by distillation

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