Chemistry 2024 Jamb Examination

Chemistry 2024 Jamb Examination
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Question 121

The percentage of carbon(IV) oxide in air is
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Carbon(IV) oxide, also known as carbon dioxide (COâ‚‚), is a minor constituent of the Earth's atmosphere. It makes up approximately 0.03% of the air by volume.
This concentration can vary slightly due to factors such as human activities and natural processes like respiration and volcanic eruptions.
Therefore, the correct percentage of COâ‚‚ in the air is 0.03%.
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Question 122

An example of a physical change is
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A physical change involves a change in the physical state of a substance without altering its chemical composition.
Boiling of water is a physical change because the water molecules remain the same, only changing from liquid to vapour.
The other options involve chemical reactions where new substances are formed.
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Question 123

Boyle's law can be expressed mathematically as
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Boyle’s law states that the pressure of a given mass of gas is inversely proportional to its volume at constant temperature.
Mathematically, this is expressed as \( PV = k \), where \( P \) is pressure, \( V \) is volume, and \( k \) is a constant.
Therefore, when pressure increases, volume decreases, and vice versa, provided temperature remains constant.
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Question 124

What is the vapour density of 560 cm³ of a gas that weighs 0.4 g at s.t.p?
[Molar volume of gas at s.t.p = 22.4 dm³]
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Given:
Volume of gas = 560 cm³ = 0.56 dm³
Mass = 0.4 g
Molar volume = 22.4 dm³

Molar mass \( M = \dfrac{22.4}{0.56} \times 0.4 = 16.0 \; \text{g mol}^{-1} \)
Vapour density (VD) = \( \dfrac{M}{2} = \dfrac{16}{2} = 8.0 \)
Thus, the vapour density of the gas is 8.0.
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Question 125

When the subsidiary quantum number (l) equals 1, the shape of the orbital is
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The subsidiary quantum number, \( l \), determines the shape of the orbital.
When \( l = 0 \), the orbital is an s-orbital which is spherical in shape.
When \( l = 1 \), it corresponds to a p-orbital which has a dumb-bell shape.
Therefore, for \( l = 1 \), the orbital is dumb-bell shaped.
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Question 126

Esterification reaction is analogous to
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Esterification involves the reaction of a carboxylic acid with an alcohol to form an ester and water.
It is a condensation reaction because two molecules combine with the elimination of a small molecule, usually water.
Hence, esterification is analogous to a condensation reaction.
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Question 127

The IUPAC nomenclature of \( \text{CH}_3\text{CH}_2\text{C(CH}_3\text{)=C(CH}_3\text{)}_2 \) is
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The compound \( \text{CH}_3\text{CH}_2\text{C(CH}_3\text{)=C(CH}_3\text{)}_2 \) contains five carbon atoms in the main chain, giving the parent name pentene.
The double bond starts at carbon 2, and two methyl groups are attached at carbons 2 and 3.
Therefore, its IUPAC name is 2,3-dimethylpent-2-ene.
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Question 128

\( \text{H}_2\text{SO}_4 + \text{C}_2\text{H}_5\text{OH} \xrightarrow{170^\circ \text{C}} \text{C}_2\text{H}_4 + \text{H}_2\text{O} \)
The reaction above illustrates
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Concentrated sulphuric acid at high temperature acts as a dehydrating agent on ethanol.
The reaction removes a molecule of water from ethanol to form ethene (Câ‚‚Hâ‚„).
This process is known as a dehydration reaction because water is eliminated from the reactant.
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Question 129

H2S(g) + Cl2(g) → 2HCl(g) + S(s)

What is the change in oxidation state of sulphur from reactant to product?
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In hydrogen sulphide, H2S, hydrogen has oxidation state +1 each.
Using the algebraic method: \(2(+1) + x = 0\) where \(x\) is the oxidation state of sulphur.
Therefore \(x = -2\).
The product S(s) is elemental sulphur with oxidation state 0.
Hence the change in oxidation state of sulphur is from \(-2\) to \(0\).
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Question 130

If a stable neutral atom has a mass number of 31, the number of electrons and neutrons respectively are
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Mass number (A) = protons (Z) + neutrons (N).
The common stable nuclide with mass number 31 is phosphorus-31, which has atomic number \(Z = 15\).
For a neutral atom, electrons = protons = 15.
Neutrons = \(A - Z = 31 - 15 = 16\).
Therefore the numbers of electrons and neutrons respectively are 15 and 16.
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