Chemistry 2019 Jamb Examination
Chemistry 2019 Jamb Examination
Question 61
The oxidation state(s) of nitrogen in ammonium nitrite is/are
Ammonium nitrite has the formula NH₄NO₂.
In the ammonium ion (NH₄⁺), nitrogen has an oxidation state of -3, while in the nitrite ion (NO₂⁻), nitrogen has an oxidation state of +3.
Therefore, the two oxidation states of nitrogen present in ammonium nitrite are -3 and +3.
In the ammonium ion (NH₄⁺), nitrogen has an oxidation state of -3, while in the nitrite ion (NO₂⁻), nitrogen has an oxidation state of +3.
Therefore, the two oxidation states of nitrogen present in ammonium nitrite are -3 and +3.
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Question 62
Hydrogen bond is a sort of
A hydrogen bond arises when a hydrogen atom covalently bonded to a highly electronegative atom such as oxygen, nitrogen, or fluorine is attracted to another electronegative atom nearby.
This type of bond is essentially a weak form of a covalent interaction, often referred to as a dipole–dipole attraction, but it has partial covalent character.
Thus, a hydrogen bond is a form of a covalent bond.
This type of bond is essentially a weak form of a covalent interaction, often referred to as a dipole–dipole attraction, but it has partial covalent character.
Thus, a hydrogen bond is a form of a covalent bond.
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Question 63
Methane is prepared in the laboratory by heating a mixture of sodium ethanoate with soda lime. The chemical constituent(s) of soda lime is/are
Soda lime is a mixture of calcium hydroxide (Ca(OH)₂) and sodium hydroxide (NaOH).
It acts as a strong base and removes carbon dioxide during the laboratory preparation of methane from sodium ethanoate.
The mixture enhances the decarboxylation process, leading to the release of methane gas.
It acts as a strong base and removes carbon dioxide during the laboratory preparation of methane from sodium ethanoate.
The mixture enhances the decarboxylation process, leading to the release of methane gas.
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Question 64
Which of the following sets of operation will completely separate a mixture of sodium chloride, sand and iodine?
Sodium chloride dissolves in water, sand is insoluble, and iodine sublimes on heating.
Adding water dissolves sodium chloride, leaving sand, which is removed by filtration.
The filtrate is then evaporated to dryness to recover sodium chloride, and the residue (sand) is heated to sublime iodine.
Thus, the correct order of separation steps is: addition of water → filtration → evaporation to dryness → sublimation.
Adding water dissolves sodium chloride, leaving sand, which is removed by filtration.
The filtrate is then evaporated to dryness to recover sodium chloride, and the residue (sand) is heated to sublime iodine.
Thus, the correct order of separation steps is: addition of water → filtration → evaporation to dryness → sublimation.
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Question 65
200cm³ of 0.50 mol/dm³ solution of calcium hydrogen trioxocarbonate (IV) is heated. The maximum weight of solid precipitated is
The thermal decomposition of calcium hydrogen trioxocarbonate (IV) produces calcium carbonate as a solid:\[ \text{Ca(HCO₃)₂} \rightarrow \text{CaCO₃ (s)} + \text{CO₂ (g)} + \text{H₂O (l)} \]Moles of Ca(HCO₃)₂ = 0.50 × (200 ÷ 1000) = 0.1 mol.
Moles of CaCO₃ produced = 0.1 mol.
Molar mass of CaCO₃ = 100 g mol⁻¹.
Mass = 0.1 × 100 = 10 g.
Hence, the maximum solid formed is 10 g.
Moles of CaCO₃ produced = 0.1 mol.
Molar mass of CaCO₃ = 100 g mol⁻¹.
Mass = 0.1 × 100 = 10 g.
Hence, the maximum solid formed is 10 g.
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Question 66
Which of the following conditions will most enhance the spontaneity of a reaction?
A spontaneous reaction occurs when the Gibbs free energy change \(ΔG\) is negative.\[ΔG = ΔH - TΔS\]When \(ΔH\) is negative (exothermic) and \(TΔS\) is positive or smaller than the magnitude of \(ΔH\), \(ΔG\) becomes negative.
Thus, spontaneity is enhanced when \(ΔH\) is negative and its magnitude is greater than \(TΔS\).
Thus, spontaneity is enhanced when \(ΔH\) is negative and its magnitude is greater than \(TΔS\).
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Question 67
X is a substance which liberates CO₂ on treatment with concentrated H₂SO₄. A warm solution of X can decolorize acidified KMnO₄. X is
Both formic acid (HCOOH) and oxalic acid (H₂C₂O₄) liberate CO₂ when treated with concentrated H₂SO₄.
However, only oxalic acid acts as a strong reducing agent capable of decolorising acidified potassium permanganate (KMnO₄) solution.
Hence, the substance X must be oxalic acid (H₂C₂O₄).
However, only oxalic acid acts as a strong reducing agent capable of decolorising acidified potassium permanganate (KMnO₄) solution.
Hence, the substance X must be oxalic acid (H₂C₂O₄).
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Question 68
The velocity, V of a gas is related to its mass, M by (k = proportionality constant)
According to Graham’s law of diffusion, the rate (or velocity) of diffusion of a gas is inversely proportional to the square root of its molar mass:\[ V \propto \dfrac{1}{\sqrt{M}} \quad \Rightarrow \quad V = kM^{-1/2} \]Therefore, the correct mathematical relationship is \(V = kM^{-1/2}\), corresponding to option B.
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Question 69
The emission of two successive beta particles from the nucleus \(^{32}_{15}P\) will produce
Each β⁻ emission increases the atomic number by one while leaving the mass number unchanged.\[^{32}_{15}P \xrightarrow{β⁻} ^{32}_{16}S \xrightarrow{β⁻} ^{32}_{17}Cl\]After two successive β⁻ emissions, the new nucleus is \(^{32}_{17}Cl\).
Hence, the final product is chlorine-32.
Hence, the final product is chlorine-32.
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Question 70
Hydrogen diffused through a porous plug
From Graham’s law, rate of diffusion \(r \propto \dfrac{1}{\sqrt{M}}\).\[ \dfrac{r_H}{r_O} = \sqrt{\dfrac{M_O}{M_H}} = \sqrt{\dfrac{32}{2}} = 4 \]Therefore, hydrogen diffuses four times as fast as oxygen through a porous plug.