Chemistry 2024 Jamb Examination
Question 121
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
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
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
[Molar volume of gas at s.t.p = 22.4 dm³]
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 \( l = 0 \), 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
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 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
The reaction above illustrates
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
What is the change in oxidation state of sulphur from reactant to product?
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
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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\begin{array}{|l|r|}
\hline
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