The two ions responsible for hardness in water are

A. Ca$^{2+}$ and / or Mg$^{2+}$

Consider the reaction A$_{(s)}$ + 2B$_{(g)}$ → 2C$_{(aq)}$ + D$_{(g)}$ What will be the effect of a decrease in pressure on the reaction?

B. rate of forward reaction will increase

C. rate of backward reaction will increase

D. rates of forward and backward reaction are not affected

Which of the following statements does not show Rutherford's account of Nuclear Theory? An atom contains a region

A. which contains protons and neutrons

C. which is massive and can cause deflection of a few projectiles

D. which is very large and in which close to 98% of projectiles pass undeflected

The IUPAC name of the compound CF$_3$CHBrCl is

B. 2-bromo-2-chloro-1,1,1-trifluoroethane

A. 2-chloro-2-bromo-1,1,1-trifluoroethane

B. 2-bromo-2-chloro-1,1,1-trifluoroethane

C. 1-bromo-1-chloro-2,2,2-trifluoroethane

D. 1,1,1-trifluoro-2-bromo-2-chloroethane

For the general equation of the nature xP + yQ ⇌ mR + nS, the expression for the equilibrium constant is

D. $\frac{[R]^m[S]^n}{[P]^x[Q]^y}$

C. $\frac{[P]^x[Q]^y}{[R]^m[S]^n}$

D. $\frac{[R]^m[S]^n}{[P]^x[Q]^y}$

Hydrogen diffused through a porous plug

C. four times as fast as oxygen

SO$_3$ is not directly dissolved in water in the industrial preparation of H$_2$SO$_4$ by the contact process because

C. The reaction between SO$_3$ and water is violently exothermic

B. Acid is usually added to water and not water to acid

C. The reaction between SO$_3$ and water is violently exothermic

D. SO$_3$ does not dissolve readily in water

In the reaction: M + N → P ΔH = +Q kJ Which of the following would increase the concentration of the product?

B. Increasing the concentration of P

Which of the following pairs cannot be represented with a chemical formula?

D. caustic soda and washing soda

The shapes of water, ammonia, carbon (iv) oxide and methane are respectively

B. angular, pyramidal, linear and tetrahedral

A. linear, tetrahedral, angular and pyramidal

B. angular, pyramidal, linear and tetrahedral

C. pyramidal, linear, tetrahedral and angular

D. tetrahedral, angular, pyramidal and linear

Which of the following statements about catalyst is false?

D. catalysts do not alter the mechanism of the reaction and never appear in the rate low

A. biological catalysts are called enzymes

B. catalyst do not appear in the balanced equation

C. catalyst reduce the activation energy for a reaction

D. catalysts do not alter the mechanism of the reaction and never appear in the rate low

Which of the following reactions is an oxidation process?

B. 2O$^{2-}$ + 4e$^-$ $\to$ O$_2$

C. OCl$^-$ + H$_2$O + 2e$^-$ $\to$ 2OH$^-$ + Cl$^-$

Which of the following sets of operation will completely separate a mixture of sodium chloride, sand and iodine?

D. sublimation, addition of water, filtration and evaporation to dryness

A. addition of water, filtration, evaporation to dryness and sublimation

B. filtration, evaporation to dryness, addition of water and sublimation

C. sublimation, filtration, evaporation and addition of water

D. sublimation, addition of water, filtration and evaporation to dryness

JAMB 2019

2019 JAMB Chemistry Past Questions

The 2019 JAMB Chemistry past questions with answers: 80 objective questions from the 2019 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.

Practise this paper as a timed JAMB-style CBT All JAMB Chemistry years

Questions

1.
A secondary alkanol can be oxidized to give an?
1. A.alkanone
2. B.alkyne
3. C.alkanoic acid
4. D.alkanal
Show answer
Answer: A. alkanone
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2.
The two ions responsible for hardness in water are
1. A.Ca $^{2+}$ and / or Mg $^{2+}$
2. B.K $^+$ and / or Mg $^{2+}$
3. C.Ca $^{2+}$ and / or Li $^+$
4. D.Ca $^{2+}$ and / or Na $^+$
Show answer
Answer: A. Ca $^{2+}$ and / or Mg $^{2+}$
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3.
Consider the reaction A $_{(s)}$ + 2B $_{(g)}$ → 2C $_{(aq)}$ + D $_{(g)}$ What will be the effect of a decrease in pressure on the reaction?
1. A.the equilibrium will shift
2. B.rate of forward reaction will increase
3. C.rate of backward reaction will increase
4. D.rates of forward and backward reaction are not affected
Show answer
Answer: A. the equilibrium will shift
AI explanation
Analysis
This question tests your understanding of Le Chatelier’s Principle, specifically how pressure changes affect chemical equilibrium in gaseous systems.
Concept
Le Chatelier’s Principle states that if a dynamic equilibrium is disturbed by changing the conditions (such as pressure, temperature, or concentration), the position of the equilibrium shifts to counteract the change.
For pressure changes:
Increase in pressure: The equilibrium shifts toward the side with the fewer number of moles of gas.
Decrease in pressure: The equilibrium shifts toward the side with the greater number of moles of gas.
Note: Only gaseous species (g) are considered. Solids (s) and aqueous solutions (aq) do not contribute to pressure changes in this context.
Step-by-Step Solution
Step 1: Identify the gaseous moles on each side of the equation. Reaction: $A_{(s)} + 2B_{(g)} \rightleftharpoons 2C_{(aq)} + D_{(g)}$
Reactants side: • $A$ is a solid (ignore). • $2B$ is a gas (count the coefficient: 2 moles). • Total gaseous moles = 2.
Products side: • $2C$ is aqueous (ignore). • $D$ is a gas (coefficient is 1: 1 mole). • Total gaseous moles = 1.
Step 2: Analyze the effect of a decrease in pressure. When pressure is decreased, the system tries to increase the pressure by shifting toward the side with more moles of gas.
Since the reactant side has 2 moles of gas and the product side has 1 mole, the equilibrium will shift to the left (toward the reactants).
Step 3: Evaluate the options.
Option A states "the equilibrium will shift." This is a correct description of what happens when the pressure is changed in a system where the number of gaseous moles is unequal on both sides.
Options B and C are specific directions (forward or backward), but since the shift is to the left (backward), Option A is the most general and accurate statement provided in the options.
Conclusion
Because there are 2 moles of gas on the reactant side and only 1 mole of gas on the product side, a change in pressure will cause the equilibrium position to shift to re-establish balance. Therefore, the equilibrium will shift.
Exam Tip: Always remember to ignore solids ( $s$ ) and liquids ( $l$ ) when calculating the effect of pressure on equilibrium. Only count the coefficients of the species marked as gas ( $g$ ).
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4.
A synthetic rubber is not obtained from the polymerization of?
1. A.neoprene
2. B.isoprene
3. C.butandiene
4. D.hexane
Show answer
Answer: D. hexane
AI explanation
Analysis of the Question
The question asks us to identify which of the listed substances is not a monomer used in the production of synthetic rubber. To solve this, we need to understand the process of polymerization and the chemical structures required to form rubber-like polymers.
Concept Explanation
Polymerization: This is a chemical process where small molecules called monomers react together to form a long, chain-like molecule called a polymer.
Synthetic Rubber: Synthetic rubbers are elastomers (polymers with elastic properties) produced by the polymerization of unsaturated hydrocarbons, typically dienes (compounds with two double bonds).
The Requirement: For a substance to undergo addition polymerization to form rubber, it must contain a carbon-carbon double bond (alkene) or multiple double bonds (dienes).
Step-by-Step Solution
Analyze Neoprene (Option A): Neoprene is a synthetic rubber produced by the polymerization of chloroprene (2-chloro-1,3-butadiene). It is a well-known synthetic rubber.
Analyze Isoprene (Option B): Isoprene (2-methyl-1,3-butadiene) is the monomer unit of natural rubber. Synthetic versions of polyisoprene are also produced as synthetic rubbers.
Analyze Butadiene (Option C): 1,3-butadiene is a fundamental monomer used in the production of various synthetic rubbers, such as Styrene-Butadiene Rubber (SBR) and Polybutadiene.
Analyze Hexane (Option D): Hexane ( $C_6H_{14}$ ) is an alkane. Alkanes are saturated hydrocarbons, meaning they contain only single covalent bonds between carbon atoms. Because they lack the double bonds necessary for addition polymerization, they cannot act as monomers to form synthetic rubber.
Conclusion
Since hexane is a saturated alkane and lacks the double bonds required for polymerization, it cannot be used to produce synthetic rubber.
Exam Tip
When you see questions about polymerization, look for the suffix "-ene" or "-diene." These indicate the presence of double bonds, which are the "hooks" that allow molecules to link together. Saturated compounds ending in "-ane" (like hexane, propane, or ethane) generally do not undergo addition polymerization.
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5.
Which of the following gases contains the least number of atoms at s.t.p?
1. A.1 mole of butane
2. B.4 moles of chlorine
3. C.3 moles of ozone
4. D.7 moles of argon
Show answer
Answer: D. 7 moles of argon
AI explanation
Analysis
To determine which gas contains the least number of atoms, we need to calculate the total number of atoms in each sample.
Key Concept: The number of atoms in a substance is calculated as:
$\text{Total Atoms} = (\text{Number of moles}) \times (\text{Number of atoms per molecule}) \times (\text{Avogadro's constant, } N_A)$
Since $N_A$ is a constant ( $6.02 \times 10^{23}$ ), we can simply compare the product of (moles $\times$ atoms per molecule) for each option.
Step-by-Step Calculation
A. 1 mole of butane ( $C_4H_{10}$ )
Number of atoms per molecule = $4 \text{ (carbon)} + 10 \text{ (hydrogen)} = 14$ atoms.
Total = $1 \text{ mole} \times 14 = \mathbf{14 \text{ units}}$
B. 4 moles of chlorine ( $Cl_2$ )
Number of atoms per molecule = $2$ atoms.
Total = $4 \text{ moles} \times 2 = \mathbf{8 \text{ units}}$
C. 3 moles of ozone ( $O_3$ )
Number of atoms per molecule = $3$ atoms.
Total = $3 \text{ moles} \times 3 = \mathbf{9 \text{ units}}$
D. 7 moles of argon ( $Ar$ )
Argon is a noble gas and exists as a monatomic molecule.
Number of atoms per molecule = $1$ atom.
Total = $7 \text{ moles} \times 1 = \mathbf{7 \text{ units}}$
Conclusion
Comparing the results:
Butane: 14
Chlorine: 8
Ozone: 9
Argon: 7
The sample with the lowest value is 7 units (Argon). Therefore, 7 moles of argon contains the least number of atoms.
Exam Tip
Always check the atomicity of the gas! Many students forget that noble gases (like Argon, Helium, Neon) are monatomic (1 atom per molecule), while diatomic gases (like $Cl_2, O_2, N_2$ ) have 2 atoms per molecule. Misidentifying the atomicity is the most common trap in this type of JAMB question.
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6.
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
1. A.Ca(OH) $_2$ and NaOH
2. B.Na $_2$ CO $_3$
3. C.NaCl
4. D.NaOH
Show answer
Answer: A. Ca(OH) $_2$ and NaOH
AI explanation
Analysis
The question asks for the chemical composition of soda lime, which is a reagent used in the laboratory preparation of methane (an alkane) through a process called decarboxylation.
Concept Explanation
In organic chemistry, the preparation of methane from sodium ethanoate ( $CH_3COONa$ ) involves heating the salt with soda lime. The reaction is represented by the following equation:
$CH_3COONa + NaOH \xrightarrow{CaO, \Delta} CH_4(g) + Na_2CO_3(s)$
Soda lime is not a single chemical compound but a mixture.
Sodium hydroxide ( $NaOH$ ) is the active reagent that provides the hydroxide ion to remove the carboxyl group ( $–COONa$ ) from the sodium ethanoate.
Calcium hydroxide ( $Ca(OH)_2$ ) is added to the mixture to keep the sodium hydroxide in a solid, granular form and to prevent it from being too deliquescent (absorbing moisture from the air and turning into a liquid). It also helps to moderate the reaction.
Step-by-Step Solution
Identify the components: Soda lime is a mixture of two main substances: Sodium hydroxide ( $NaOH$ ) and Calcium hydroxide ( $Ca(OH)_2$ ).
Evaluate the options: • Option A lists both $Ca(OH)_2$ and $NaOH$ , which matches the definition of soda lime. • Option B ( $Na_2CO_3$ ) is a product of the reaction, not the reagent. • Option C ( $NaCl$ ) is common table salt and is not involved in this reaction. • Option D ( $NaOH$ ) is only one part of the mixture; soda lime specifically refers to the combination of $NaOH$ and $Ca(OH)_2$ .
Conclusion
The chemical constituents of soda lime are indeed a mixture of sodium hydroxide and calcium hydroxide.
Exam Tip: Remember that "soda" refers to the sodium hydroxide ( $NaOH$ ) and "lime" refers to the calcium hydroxide ( $Ca(OH)_2$ ). This mnemonic will help you identify the mixture easily in future exams.
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7.
Elements in the periodic table are arranged in the order of their
1. A.atomic numbers
2. B.metallic character
3. C.mass number
4. D.relative atomic mass
Show answer
Answer: A. atomic numbers
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8.
A cell shorthand notation can be written as A / A $^+$ // B $^{2+}$ /B. The double slash in the notation represents the
1. A.electrolyte
2. B.electrodes
3. C.salt bridge
4. D.solid - liquid interface
Show answer
Answer: C. salt bridge
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9.
What technique is suitable for separating a binary solution of potassium chloride and potassium trioxochlorate (V)?
1. A.Fractional crystallization
2. B.Fractional distillation
3. C.Filtration
4. D.Evaporation
Show answer
Answer: A. Fractional crystallization
AI explanation
Analysis
The question asks for the most effective method to separate a mixture of two salts: potassium chloride (KCl) and potassium trioxochlorate(V) (KClO₃). Both are solids that dissolve in water to form a binary solution.
Concept Explanation
To separate two dissolved salts from a solution, we rely on the principle of solubility. Different substances have different solubilities at different temperatures.
Fractional crystallization is a technique used to separate two or more solutes from a solution based on their different solubilities at varying temperatures.
As a hot, saturated solution of the mixture cools, the substance with the lower solubility at lower temperatures will crystallize out of the solution first, while the more soluble substance remains dissolved.
Step-by-Step Solution
Understand the components: We have a mixture of KCl and KClO₃. Both are soluble in water.
Evaluate the methods: • Fractional distillation: Used for separating liquids with different boiling points (e.g., ethanol and water). Not suitable for salts. • Filtration: Used to separate an insoluble solid from a liquid. Since both salts are dissolved, filtration will not separate them. • Evaporation: If you evaporate the water completely, you will get a mixture of both salts as a solid residue, not separated. • Fractional crystallization: Because KClO₃ and KCl have significantly different solubility curves (KClO₃ is much less soluble in cold water than KCl), cooling the hot saturated solution causes the KClO₃ to crystallize out first, effectively separating it from the KCl.
Conclusion: Fractional crystallization is the standard laboratory technique for separating mixtures of salts with different temperature-dependent solubilities.
Correct Answer
The correct option is A.
Exam Tip
Whenever you see a question about separating two dissolved salts, immediately think of solubility. If the question mentions "fractional," it is almost always referring to separating components based on a specific physical property (like solubility for crystallization or boiling point for distillation). Don't let the complex chemical names intimidate you—focus on the physical properties!
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10.
Which quantum divides shells into orbitals?
1. A.Spin
2. B.Principal
3. C.Azimuthal
4. D.Magnetic
Show answer
Answer: C. Azimuthal
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11.
How many alkoxyalkanes can be obtained from the molecular formula C $_4$ H $_{10}$ O?
1. A.4
2. B.2
3. C.1
4. D.3
Show answer
Answer: D. 3
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12.
Which of the following pollutants will lead to the depletion of ozone layer?
1. A.chlorofluorocarbon
2. B.carbon (ii) oxide
3. C.hydrogen sulphide
4. D.sulphur (iv) oxide
Show answer
Answer: A. chlorofluorocarbon
AI explanation
Analysis
The question asks us to identify a specific pollutant responsible for the depletion of the ozone layer. The ozone layer ( $O_3$ ) is a region in the stratosphere that protects the Earth from harmful ultraviolet (UV) radiation.
Concept Explanation
The depletion of the ozone layer is primarily caused by the release of chemicals that contain chlorine or bromine atoms. When these chemicals reach the stratosphere, they are broken down by UV radiation, releasing free radicals (such as chlorine atoms). These radicals act as catalysts in a chain reaction that breaks down ozone molecules into oxygen, effectively thinning the protective layer.
Chlorofluorocarbons (CFCs): These are synthetic compounds formerly used in refrigerators, air conditioners, and aerosol sprays. They are chemically stable at the Earth's surface but decompose in the stratosphere to release chlorine atoms, which destroy ozone.
Carbon(II) oxide ( $CO$ ): This is a toxic gas produced by incomplete combustion. It is a major air pollutant but does not contribute to ozone depletion.
Hydrogen sulphide ( $H_2S$ ): This is a gas often associated with volcanic activity and decaying organic matter. It contributes to acid rain and air pollution but is not the primary cause of ozone depletion.
Sulphur(IV) oxide ( $SO_2$ ): This is a major contributor to acid rain, formed by burning fossil fuels containing sulphur. Like the others, it does not cause ozone depletion.
Step-by-Step Solution
Evaluate the options: We are looking for a substance that releases ozone-depleting substances (ODS) like chlorine into the stratosphere.
Identify the culprit: CFCs are the most well-known class of chemicals that cause ozone depletion. When a CFC molecule (e.g., $CFCl_3$ ) is exposed to UV light, it releases a chlorine atom ( $Cl$ ).
The Reaction: The chlorine atom reacts with ozone ( $O_3$ ): $Cl + O_3 \rightarrow ClO + O_2$ This reaction destroys the ozone molecule.
Conclusion: Among the given options, only chlorofluorocarbons have the chemical properties required to reach the stratosphere and catalyze the destruction of the ozone layer.
Correct Answer
The correct option is A.
Exam Tip
When you see "Ozone Depletion" in a Chemistry or Biology exam, immediately look for CFCs (Chlorofluorocarbons). If you see "Acid Rain," look for Sulphur(IV) oxide ( $SO_2$ ) or Nitrogen(IV) oxide ( $NO_2$ ). Remembering these associations will save you time in the exam hall!
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13.
At 27°C, 58.5g of sodium chloride is present in 250cm $^3$ of a solution. The solubility of sodium chloride at this temperature is? (molar mass of sodium chloride = 111.0gmol $^{-1}$ )
1. A.2.0 moldm $^{-3}$
2. B.0.25 moldm $^{-3}$
3. C.1.0 moldm $^{-3}$
4. D.0.5 moldm $^{-3}$
Show answer
Answer: A. 2.0 moldm $^{-3}$
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14.
Which of the following statements does not show Rutherford's account of Nuclear Theory? An atom contains a region
1. A.which contains protons and neutrons
2. B.which is positively charged
3. C.which is massive and can cause deflection of a few projectiles
4. D.which is very large and in which close to 98% of projectiles pass undeflected
Show answer
Answer: A. which contains protons and neutrons
AI explanation
Analysis of the Question
This question tests your knowledge of the Rutherford Gold Foil Experiment (1911). You are asked to identify which statement is NOT part of Rutherford's original findings.
Concept Explanation
Ernest Rutherford’s experiment involved firing alpha particles (positively charged projectiles) at a thin sheet of gold foil. His observations led to the following conclusions:
The Nucleus is Positively Charged: Since some alpha particles were deflected, the center must be positively charged to repel the positive alpha particles.
The Nucleus is Massive: Because a few particles bounced back, the center must contain almost all the mass of the atom.
The Atom is Mostly Empty Space: Since most particles passed straight through, the atom is mostly empty space.
The Nucleus is Small: The fact that only a tiny fraction of particles were deflected indicated that the nucleus occupies a very small volume compared to the total size of the atom.
Step-by-Step Evaluation
Option A: Rutherford discovered the nucleus, but he did not identify the presence of neutrons. The neutron was discovered by James Chadwick in 1932, long after Rutherford's model. Therefore, Rutherford could not have stated that the nucleus contains neutrons.
Option B: This is correct; Rutherford concluded the center was positively charged.
Option C: This is correct; he concluded the center was dense (massive) enough to deflect particles.
Option D: This is correct; he observed that most of the atom is empty space, allowing particles to pass through undeflected.
Conclusion
Since Rutherford's model predates the discovery of the neutron, he could not have included it in his theory. Therefore, statement A is the one that does not describe his account.
Exam Tip: When dealing with atomic models (Dalton, Thomson, Rutherford, Bohr), always remember the timeline of discovery. If a question mentions "neutrons," it is almost certainly a distractor for questions about early atomic models like Rutherford's or Thomson's, as neutrons were the last of the three subatomic particles to be discovered.
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15.
X is a substance which liberates CO $_2$ on treatment with concentrated H $_2$ SO $_4$ . A warm solution of X can decolorize acidified KMnO $_4$ . X is
1. A.HCOOH
2. B.NaHCO $_3$
3. C.FeCO $_3$
4. D.H $_2$ C $_2$ O $_4$
Show answer
Answer: D. H $_2$ C $_2$ O $_4$
AI explanation
To solve this problem, let's break down the chemical properties of the substance X based on the two clues provided:
1. Analysis of the Clues
Clue 1: Liberates CO $_2$ on treatment with concentrated H $_2$ SO $_4$ . Many carbonates and oxalates react with concentrated sulfuric acid to release carbon dioxide (or a mixture of CO and CO $_2$ ).
Clue 2: A warm solution of X decolorizes acidified KMnO $_4$ . This is the "deciding" clue. Acidified potassium permanganate (KMnO $_4$ ) is a powerful oxidizing agent. For a substance to decolorize it, that substance must be a reducing agent. • HCOOH (Formic acid): Can be oxidized, but it is not the classic answer for this specific reaction profile in JAMB chemistry. • NaHCO $_3$ (Sodium bicarbonate): Does not act as a reducing agent; it cannot decolorize KMnO $_4$ . • FeCO $_3$ (Iron(II) carbonate): While Fe²⁺ can be oxidized to Fe³⁺, the reaction with H $_2$ SO $_4$ primarily focuses on the carbonate part, and it is not the standard laboratory reagent used to demonstrate this specific dual-property test. • H $_2$ C $_2$ O $_4$ (Oxalic acid): This is a classic reducing agent. When heated, it reacts with acidified KMnO $_4$ to produce CO $_2$ and water, effectively decolorizing the purple permanganate solution.
2. The Chemistry of Oxalic Acid (H $_2$ C $_2$ O $_4$ )
Reaction with H $_2$ SO $_4$ : When oxalic acid is treated with concentrated H $_2$ SO $_4$ , it undergoes dehydration to produce carbon monoxide (CO) and carbon dioxide (CO $_2$ ): $ $H_2C_2O_4 \xrightarrow{conc. H_2SO_4} CO + CO_2 + H_2O$ $
Reaction with KMnO $_4$ : Oxalic acid is a well-known reducing agent. In the presence of heat, it reduces the purple MnO $_4^-$ ion to the colorless Mn²⁺ ion: $ $5H_2C_2O_4 + 2MnO_4^- + 6H^+ \rightarrow 10CO_2 + 2Mn^{2+} + 8H_2O$ $
3. Conclusion
Comparing the options, H $_2$ C $_2$ O $_4$ (Oxalic acid) is the only substance that perfectly fits both criteria: it releases CO $_2$ upon dehydration with concentrated H $_2$ SO $_4$ and acts as a strong enough reducing agent to decolorize acidified KMnO $_4$ .
Exam Tip: Whenever you see "decolorizes acidified KMnO $_4$ " in a chemistry question, immediately look for a reducing agent (like Fe²⁺, SO $_2$ , H $_2$ S, or Oxalic acid). If you see "liberates CO $_2$ with acid," look for a carbonate or an oxalate. Oxalic acid is the intersection of these two properties!
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16.
If acidified Potassium Dichromate(VI) (K $_2$ Cr $_2$ O $_7$ ) acts as oxidizing agent, color changes from
1. A.yellow to green
2. B.orange to green
3. C.yellow to red
4. D.orange to red
Show answer
Answer: B. orange to green
AI explanation
Analysis
The question asks for the characteristic color change observed when acidified potassium dichromate(VI) ( $K_2Cr_2O_7$ ) functions as an oxidizing agent in a chemical reaction.
Concept Explanation
Potassium dichromate(VI) is a strong oxidizing agent commonly used in organic chemistry (for example, to test for primary and secondary alcohols or to oxidize aldehydes to carboxylic acids).
In its original state, the dichromate ion ( $Cr_2O_7^{2-}$ ) is orange. When it acts as an oxidizing agent, it undergoes a reduction reaction. The chromium atom is reduced from an oxidation state of +6 to +3. The resulting chromium(III) ion ( $Cr^{3+}$ ) in an aqueous solution is green.
The half-equation for this reduction is: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
Step-by-Step Solution
Identify the reactant: Acidified potassium dichromate(VI) contains the dichromate ion ( $Cr_2O_7^{2-}$ ), which is naturally orange.
Identify the process: As an oxidizing agent, it accepts electrons from the substance being oxidized.
Identify the product: During this reduction process, the chromium is converted into the chromium(III) ion ( $Cr^{3+}$ ).
Observe the color change: The solution transitions from the color of the dichromate ion (orange) to the color of the chromium(III) ion (green).
Conclusion
The color change is from orange to green. This matches option B.
Exam Tip: Don't confuse this with Potassium Chromate ( $K_2CrO_4$ ), which is yellow. Remember: Dichromate is Orange (like a sunset), and Chromate is Yellow (like a lemon).
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17.
Which of the following is the best starting material for the preparation of oxygen? Heating of trioxonitrate (v) with
1. A.Pb
2. B.Zn
3. C.Ca
4. D.Na
Show answer
Answer: D. Na
AI explanation
Analysis of the Question
The question asks for the best starting material (a metal) to heat with a trioxonitrate(V) salt to produce oxygen gas. In chemistry, the thermal decomposition of metal nitrates follows a specific pattern based on the reactivity series of the metals.
Concept Explanation
When metal trioxonitrates(V) (nitrates) are heated, they decompose. The products depend on the position of the metal in the reactivity series:
Group 1 Metals (e.g., Na, K): These decompose to form the metal trioxonitrite(III) (nitrite) and oxygen gas. • Equation: $2NaNO_3(s) \rightarrow 2NaNO_2(s) + O_2(g)$
Metals between Lead and Copper: These decompose to form the metal oxide, nitrogen(IV) oxide, and oxygen gas.
Noble Metals (e.g., Ag, Au): These decompose to form the metal, nitrogen(IV) oxide, and oxygen gas.
The question asks for the "best starting material" for the preparation of oxygen. While heating sodium trioxonitrate(V) ( $NaNO_3$ ) alone produces oxygen, the question specifically asks for the reaction of a trioxonitrate(V) with a metal.
When we heat a metal trioxonitrate(V) with a reactive metal like Sodium (Na), the sodium acts as a reducing agent, effectively pulling oxygen away from the nitrate to form the nitrite or oxide, thereby releasing oxygen gas efficiently.
Step-by-Step Solution
Identify the reaction type: This is a redox reaction where the metal reduces the nitrate salt.
Evaluate the options: • Pb, Zn, Ca: These metals are less reactive than Sodium. While they can participate in various chemical reactions, Sodium is the most reactive metal among the choices provided. • Na (Sodium): Sodium is highly reactive. When sodium trioxonitrate(V) is heated with sodium, the reaction is highly favorable for the liberation of oxygen because sodium has a very high affinity for oxygen, driving the decomposition process forward.
Conclusion: In the context of laboratory preparation and chemical reactivity, Sodium ( $Na$ ) is the most effective choice to facilitate the release of oxygen from the nitrate salt.
This aligns with the principle that more reactive metals are more effective at driving the decomposition of nitrates to release oxygen.
Exam Tip
Always remember the "Reactivity Series" (Potassium, Sodium, Calcium, Magnesium, Aluminum, Zinc, Iron, Lead, Copper, Mercury, Silver, Gold). Metals higher up the series are stronger reducing agents. If you see a question about "best starting material" or "best reducing agent," look for the metal highest on the reactivity series!
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18.
The electronic configuration of element Z is 1s $^2$ 2s $^2$ 2p $^6$ 3s $^2$ 3p $^1$ . What is the formula of the compound formed between Z and tetraoxosulphate (VI) ion.
1. A.Z $_3$ (SO $_4$ ) $_2$
2. B.ZSO $_4$
3. C.Z $_2$ SO $_4$
4. D.Z $_2$ (SO $_4$ ) $_3$
Show answer
Answer: D. Z $_2$ (SO $_4$ ) $_3$
AI explanation
1. Analyze the Question
We are given the electronic configuration of an element Z and asked to determine the chemical formula of the compound it forms when it reacts with the tetraoxosulphate(VI) ion ( $SO_4^{2-}$ ).
2. Explain the Concept
Determining Valency: The valency of an element is determined by the number of electrons in its outermost shell (valence shell). • If an element has 1, 2, or 3 electrons in its valence shell, it tends to lose them to become stable, forming a positive ion (cation) with a charge equal to the number of electrons lost.
Writing Chemical Formulas: To write the formula of an ionic compound, we use the "criss-cross" method: • Write the symbols of the cation and the anion side-by-side. • Write their respective charges (valencies) as superscripts. • Criss-cross the numbers to become the subscripts of the opposite element.
3. Show the Solution
Step 1: Determine the valency of Z. The electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^1$ .
The outermost shell is the 3rd shell ( $3s^2 3p^1$ ).
Total valence electrons = $2 + 1 = 3$ .
Since Z has 3 valence electrons, it will lose them to achieve stability, forming an ion with a charge of +3 ( $Z^{3+}$ ).
Step 2: Identify the charge of the anion. The tetraoxosulphate(VI) ion is $SO_4^{2-}$ , which has a charge of -2.
Step 3: Apply the criss-cross method.
Cation: $Z^{3+}$
Anion: $(SO_4)^{2-}$
By criss-crossing the charges:
The 2 from the sulphate ion becomes the subscript for Z.
The 3 from the Z ion becomes the subscript for the sulphate group.
This gives us: $Z_2(SO_4)_3$
4. Identify the Answer
Comparing our result to the given options: A. $Z_3(SO_4)_2$ B. $ZSO_4$ C. $Z_2SO_4$ D. $Z_2(SO_4)_3$
The correct option is D.
5. Exam Tip
When dealing with polyatomic ions like $SO_4^{2-}$ , $CO_3^{2-}$ , or $PO_4^{3-}$ , always remember to treat the entire group as a single unit. If the subscript you are criss-crossing is greater than 1, you must place the polyatomic ion in parentheses before adding the subscript (e.g., $(SO_4)_3$ ).
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19.
When ammonia and hydrogen ion bond together to form ammonium ion, the bond formed is called
1. A.electrovalent bond
2. B.compound bond
3. C.ionic bond
4. D.covalent bond
Show answer
Answer: D. covalent bond
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20.
When the end alkyl groups of ethyl ethanoate are interchanged, the compound formed is
1. A.propylmethanoate
2. B.ethylpropanoate
3. C.methyl ethanoate
4. D.methyl propanoate
Show answer
Answer: D. methyl propanoate
AI explanation
Analysis
To solve this, we need to understand the structure and naming convention of esters. An ester is formed from an alcohol and a carboxylic acid. Its general formula is RCOOR', where R comes from the acid and R' comes from the alcohol.
Concept: Ester Nomenclature
Ethyl ethanoate is formed from: • Ethanoic acid (the acid part, which provides the "ethanoate" suffix): $CH_3COOH$ • Ethanol (the alcohol part, which provides the "ethyl" prefix): $C_2H_5OH$
The structure of ethyl ethanoate is: $CH_3COOCH_2CH_3$ • The "ethanoate" part is $CH_3COO-$ (2 carbons) • The "ethyl" part is $-CH_2CH_3$ (2 carbons)
Step-by-Step Solution
Identify the alkyl groups: In ethyl ethanoate, the acid part has 2 carbons ( $CH_3$ ) and the alcohol part (the alkyl group) has 2 carbons ( $C_2H_5$ ).
Interchange the groups: The question asks us to interchange the "end alkyl groups." • Currently, we have an ethyl group attached to an ethanoate group. • If we look at the total number of carbons available in the molecule ( $CH_3COOCH_2CH_3$ has 4 carbons total), we need to rearrange them. • If we move one carbon from the alcohol side to the acid side, or vice versa, we change the identity of the ester. • Specifically, if we take the methyl group ( $CH_3$ ) from the acid side and swap it with the ethyl group ( $C_2H_5$ ) from the alcohol side, we get: • Acid part: Propanoic acid ( $CH_3CH_2COO-$ ) • Alcohol part: Methyl ( $CH_3-$ )
Name the new compound: Combining these gives Methyl propanoate ( $CH_3CH_2COOCH_3$ ).
Conclusion
By interchanging the alkyl groups (moving the carbon chain length between the acid and alcohol components), we transform ethyl ethanoate into methyl propanoate.
Exam Tip
When dealing with ester naming, remember: "Alkyl" comes from the alcohol (the part attached to the single oxygen), and "Alkanoate" comes from the acid (the part containing the carbonyl group $C=O$ ). If you are asked to interchange groups, simply shift the carbon count between these two sides while keeping the total number of carbons the same!
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21.
If the cost of electricity required to discharge 10g of an ion X $^{3+}$ is N20.00, how much would it cost to discharge 6g of ion Y $^{2+}$ ? [1 faraday = 96,500C, atomic masses are X = 27, Y = 24]
1. A.N10.00
2. B.N6.00
3. C.N20.00
4. D.N9.00
Show answer
Answer: D. N9.00
AI explanation
1. Analyze the Question
We need to find the cost of discharging 6g of ion $Y^{2+}$ , given the cost of discharging 10g of ion $X^{3+}$ . This is a Faraday’s Law of Electrolysis problem where the cost is directly proportional to the quantity of electricity (charge) required.
2. Explain the Concept
According to Faraday’s First Law of Electrolysis, the mass ( $m$ ) of a substance deposited is given by:
$m = \frac{M \times Q}{n \times F}$
Where:
$M$ = Molar mass
$Q$ = Quantity of electricity (Charge)
$n$ = Valency (number of electrons transferred)
$F$ = Faraday's constant
Since the cost is proportional to the quantity of electricity ( $Q$ ), we can rearrange the formula to find $Q$ :
$Q = \frac{m \times n \times F}{M}$
The cost is directly proportional to $Q$ . Therefore, the ratio of costs equals the ratio of charges:
$\frac{\text{Cost}_1}{\text{Cost}_2} = \frac{Q_1}{Q_2}$
3. Show the Solution
Step 1: Calculate the charge ( $Q_x$ ) for ion $X^{3+}$
Mass ( $m_x$ ) = 10g
Valency ( $n_x$ ) = 3
Molar mass ( $M_x$ ) = 27
$Q_x = \frac{10 \times 3 \times F}{27} = \frac{30F}{27} = \frac{10}{9}F$
Step 2: Calculate the charge ( $Q_y$ ) for ion $Y^{2+}$
Mass ( $m_y$ ) = 6g
Valency ( $n_y$ ) = 2
Molar mass ( $M_y$ ) = 24
$Q_y = \frac{6 \times 2 \times F}{24} = \frac{12F}{24} = 0.5F$
Step 3: Set up the proportion to find the cost Let the cost for $Y^{2+}$ be $C_y$ .
$\frac{\text{Cost}_x}{Q_x} = \frac{C_y}{Q_y}$
$\frac{20}{(10/9)F} = \frac{C_y}{0.5F}$
Cancel $F$ from both sides:
$\frac{20 \times 9}{10} = \frac{C_y}{0.5}$
$18 = \frac{C_y}{0.5}$
$C_y = 18 \times 0.5 = 9$
The cost to discharge 6g of $Y^{2+}$ is N9.00.
4. Identify the Answer
Comparing our result to the options provided, the correct choice is D.
5. Tip
When dealing with electrolysis cost problems, always remember that the charge required is proportional to the number of moles multiplied by the valency ( $n$ ). You can simplify your work by using the formula: $\text{Cost} \propto \frac{m \times n}{M}$ . This saves you from having to write out the Faraday constant every time!
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22.
An element Z contains 80% of $^{16} _8$ Z and 20% of $^{18} _8$ Z. Its relative atomic mass is
1. A.16.2
2. B.17.6
3. C.16.0
4. D.16.4
Show answer
Answer: D. 16.4
AI explanation
Analysis
The question asks for the relative atomic mass of an element that exists as a mixture of two isotopes. To find this, we calculate the weighted average of the mass numbers based on their percentage abundance in nature.
Concept
The relative atomic mass of an element is calculated using the formula:
$\text{Relative Atomic Mass} = \sum (\text{Isotope Mass} \times \text{Fractional Abundance})$
Where:
Isotope Mass is the mass number of each isotope.
Fractional Abundance is the percentage divided by 100.
Step-by-Step Solution
Identify the data provided: • Isotope 1: Mass = 16, Abundance = 80% (or 0.80) • Isotope 2: Mass = 18, Abundance = 20% (or 0.20)
Calculate the contribution of each isotope: • Contribution of Isotope 1: $16 \times 0.80 = 12.8$ • Contribution of Isotope 2: $18 \times 0.20 = 3.6$
Sum the contributions to find the relative atomic mass: • Relative Atomic Mass $= 12.8 + 3.6$ • Relative Atomic Mass $= 16.4$
Conclusion
Comparing our result to the options provided, the calculated value is 16.4.
Exam Tip: When calculating relative atomic mass from isotopes, your final answer should always fall between the mass numbers of the two isotopes. Since 16.4 is between 16 and 18, it is a logical result. If you get a number outside that range, you know you've made a calculation error!
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23.
Which of the following is a set of neutral oxides?
1. A.N $_2$ O, H $_2$ O, CO
2. B.N $_2$ O, SO $_2$ , CO
3. C.N $_2$ O, Na $_2$ O, ZnO
4. D.N $_2$ O, SO $_3$ , NO
Show answer
Answer: A. N $_2$ O, H $_2$ O, CO
AI explanation
Analysis of the Question
The question asks you to identify a group of oxides that are neutral. In chemistry, a neutral oxide is one that does not react with either acids or bases to form salts. Unlike acidic oxides (which react with bases) or basic oxides (which react with acids), neutral oxides remain unchanged in the presence of these substances.
Concept Explanation
To solve this, we need to categorize the common oxides:
Neutral Oxides: These are rare. The most common ones you must memorize for JAMB are Carbon(II) oxide (CO), Nitrogen(I) oxide (N $_2$ O), and Water (H $_2$ O). Nitrogen(II) oxide (NO) is also generally considered neutral.
Acidic Oxides: These are usually non-metal oxides like SO $_2$ , SO $_3$ , CO $_2$ , and P $_4$ O $_{10}$ .
Basic Oxides: These are usually metal oxides like Na $_2$ O, CaO, and MgO.
Amphoteric Oxides: These react with both acids and bases. A classic example is Zinc oxide (ZnO).
Step-by-Step Evaluation
Let's examine the options provided:
Option A: N $_2$ O, H $_2$ O, CO • N $_2$ O (Nitrous oxide) is neutral. • H $_2$ O (Water) is neutral. • CO (Carbon monoxide) is neutral. • This set consists entirely of neutral oxides.
Option B: N $_2$ O, SO $_2$ , CO • SO $_2$ is an acidic oxide (it reacts with water to form sulfurous acid). This makes the set incorrect.
Option C: N $_2$ O, Na $_2$ O, ZnO • Na $_2$ O is a basic oxide. • ZnO is an amphoteric oxide. This makes the set incorrect.
Option D: N $_2$ O, SO $_3$ , NO • SO $_3$ is a strongly acidic oxide (it reacts with water to form sulfuric acid). This makes the set incorrect.
Conclusion
Based on the classification of oxides, Option A is the only set where every compound is a neutral oxide.
Exam Tip
When you see a question about oxides, remember the acronym "CAN" for neutral oxides: CO, Aqua (H $_2$ O), and Nitrogen oxides (N $_2$ O and NO). If you see any other oxides in the list, check if they are metal oxides (usually basic) or non-metal oxides (usually acidic) to quickly eliminate the wrong options!
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24.
A solution X, on mixing with AgNO $_3$ solution gives a white precipitate soluble in aqueous NH $_3$ , a solution Y, when also added to X, also gives a white precipitate which is soluble when heated solutions X and Y respectively contain
1. A.Al $^{3+}$ , Cl $^-$
2. B.Hg $^+$ , SO $_4 ^{2-}$
3. C.Pb $^{2+}$ , Cl $^-$
4. D.Ag $^+$ , SO $_4 ^{2-}$
Show answer
Answer: C. Pb $^{2+}$ , Cl $^-$
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25.
The IUPAC name of the compound CF $_3$ CHBrCl is
1. A.2-chloro-2-bromo-1,1,1-trifluoroethane
2. B.2-bromo-2-chloro-1,1,1-trifluoroethane
3. C.1-bromo-1-chloro-2,2,2-trifluoroethane
4. D.1,1,1-trifluoro-2-bromo-2-chloroethane
Show answer
Answer: B. 2-bromo-2-chloro-1,1,1-trifluoroethane
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26.
A certain hydrocarbon on complete combustion at s.t.p produced 89.6dm $^3$ of CO $_2$ and 54g of water. The hydrocarbon should be?
1. A.C $_6$ H $_6$
2. B.C $_4$ H $_{10}$
3. C.C $_5$ H $_{10}$
4. D.C $_4$ H $_6$
Show answer
Answer: D. C $_4$ H $_6$
AI explanation
1. Analyze the question
We are given the volume of $CO_2$ produced and the mass of $H_2O$ produced from the combustion of a hydrocarbon ( $C_xH_y$ ). We need to determine the empirical formula (and subsequently the molecular formula) of the hydrocarbon.
2. Explain the concept
Combustion reaction: $C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O$
Mole calculation: • At s.t.p, 1 mole of any gas occupies $22.4\text{ dm}^3$ . • Moles of $CO_2 = \frac{\text{Volume}}{22.4} = \frac{89.6}{22.4} = 4\text{ moles}$ . • Moles of $H_2O = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{54}{18} = 3\text{ moles}$ .
Relationship: The carbon atoms in the hydrocarbon end up in $CO_2$ , and the hydrogen atoms end up in $H_2O$ .
3. Show the solution
Step 1: Find the moles of Carbon ( $C$ ) Since each mole of $CO_2$ contains 1 mole of $C$ : Moles of $C = 4\text{ moles}$ .
Step 2: Find the moles of Hydrogen ( $H$ ) Since each mole of $H_2O$ contains 2 moles of $H$ : Moles of $H = 3 \times 2 = 6\text{ moles}$ .
Step 3: Determine the ratio The ratio of $C:H$ is $4:6$ . This gives us the empirical formula $C_4H_6$ .
Step 4: Verify against options
A. $C_6H_6$ (Ratio 1:1)
B. $C_4H_{10}$ (Ratio 4:10)
C. $C_5H_{10}$ (Ratio 1:2)
D. $C_4H_6$ (Ratio 4:6)
The calculated ratio matches option D.
4. Identify the answer
The correct option is D.
5. Tip
When solving combustion problems, always remember:
Carbon atoms come only from $CO_2$ .
Hydrogen atoms come only from $H_2O$ (remember to multiply the moles of water by 2 because there are two hydrogen atoms in one water molecule!).
If you get a fraction, multiply to get the simplest whole-number ratio.
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27.
For the general equation of the nature xP + yQ ⇌ mR + nS, the expression for the equilibrium constant is
1. A.k[P] $^x$ [Q] $^y$
2. B. $\frac{m[R][S]}{x[P]y[Q]}$
3. C. $\frac{[P]^x[Q]^y}{[R]^m[S]^n}$
4. D. $\frac{[R]^m[S]^n}{[P]^x[Q]^y}$
Show answer
Answer: D. $\frac{[R]^m[S]^n}{[P]^x[Q]^y}$
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28.
Hydrogen diffused through a porous plug
1. A.thrice as fast as oxygen
2. B.at the same speed as oxygen
3. C.four times as fast as oxygen
4. D.twice as fast as oxygen
Show answer
Answer: C. four times as fast as oxygen
AI explanation
1. Analyze the question
The question asks for the relative rate of diffusion of hydrogen gas ( $H_2$ ) compared to oxygen gas ( $O_2$ ) through a porous plug. This is a classic application of Graham’s Law of Diffusion.
2. Explain the concept
Graham’s Law of Diffusion states that the rate of diffusion ( $r$ ) of a gas is inversely proportional to the square root of its molar mass ( $M$ ):
$r \propto \frac{1}{\sqrt{M}}$
When comparing two gases (Gas 1 and Gas 2), the formula is:
$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$
Molar mass of Hydrogen ( $H_2$ ): $1 \times 2 = 2 \text{ g/mol}$
Molar mass of Oxygen ( $O_2$ ): $16 \times 2 = 32 \text{ g/mol}$
3. Show the solution
Let $r_H$ be the rate of hydrogen and $r_O$ be the rate of oxygen.
Using the formula:
$\frac{r_H}{r_O} = \sqrt{\frac{M_O}{M_H}}$
Substitute the molar masses:
$\frac{r_H}{r_O} = \sqrt{\frac{32}{2}}$
$\frac{r_H}{r_O} = \sqrt{16}$
$\frac{r_H}{r_O} = 4$
This means $r_H = 4 \times r_O$ . Therefore, hydrogen diffuses four times as fast as oxygen.
4. Identify the answer
Comparing our result to the given options, the correct choice is C.
5. Tip
When solving diffusion problems, always remember the mnemonic: "Lighter gases move faster." Since hydrogen is much lighter than oxygen, it must diffuse faster. If you ever get a result where the heavier gas is moving faster, check your calculation—you likely flipped the molar masses in the square root!
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29.
The oxidation state(s) of nitrogen in ammonium nitrite is/are
1. A.-3, +3
2. B.+5
3. C.-3, +5
4. D.-3
Show answer
Answer: A. -3, +3
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30.
Hydrocarbons which will react with Tollen's reagent conform to the general formula
1. A.C $_n$ H $_{2n}$
2. B.C $_n$ H $_{2n + 2}$
3. C.C $_n$ H $_{2n - 2}$
4. D.C $_n$ H $_{2n + 1}$ OH
Show answer
Answer: C. C $_n$ H $_{2n - 2}$
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31.
The IUPAC nomenclature of the compound H $_3$ C - CH(CH $_3$ ) - CH(CH $_3$ ) - CH $_2$ - CH $_3$
1. A.3,4- dimethylhexane
2. B.2,3- dimethyl pentane
3. C.2- ethyl hexane
4. D.2,3- dimethylhexane
Show answer
Answer: B. 2,3- dimethyl pentane
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32.
Hydrogen bond is a sort of
1. A.van der waals force
2. B.dative bond
3. C.ionic bond
4. D.a covalent bond
Show answer
Answer: A. van der waals force
AI explanation
Analysis
The question asks for the classification of a hydrogen bond. To answer this, we need to understand the nature of intermolecular forces and how they differ from intramolecular bonds (like ionic or covalent bonds).
Concept Explanation
Intramolecular bonds: These are strong forces that hold atoms together within a molecule (e.g., covalent bonds, ionic bonds, and dative/coordinate covalent bonds).
Intermolecular forces: These are weaker forces of attraction that exist between molecules.
Hydrogen Bond: A hydrogen bond is a special, strong type of dipole-dipole attraction that occurs when a hydrogen atom is covalently bonded to a highly electronegative atom (specifically Nitrogen, Oxygen, or Fluorine). Because the hydrogen atom is partially positive and the electronegative atom is partially negative, they attract neighboring molecules.
Van der Waals forces: This is a general term used to describe the weak electrostatic attractions between molecules. In many chemistry textbooks and examination syllabi (including JAMB), hydrogen bonding is categorized as a specific, stronger type of van der Waals force because it is an intermolecular attraction rather than a chemical bond formed by the sharing or transfer of electrons.
Step-by-Step Solution
Evaluate the options: • B, C, and D: Dative, ionic, and covalent bonds are all intramolecular bonds. They involve the sharing or transfer of electrons to form a compound. A hydrogen bond does not create a new molecule; it only influences how existing molecules interact with each other. • A: Van der Waals forces are intermolecular forces. Since hydrogen bonding is an attraction between molecules, it falls under this classification in the context of general chemistry examinations.
Conclusion: While a hydrogen bond is significantly stronger than standard London dispersion forces, it is classified as an intermolecular force, which is grouped under the umbrella of van der Waals forces in this context.
Exam Tip
When you see questions about "bonds," always ask yourself: "Is this holding atoms together inside a molecule, or is it holding two separate molecules together?" If it's holding molecules together, it is an intermolecular force (Van der Waals or Hydrogen bond). If it's holding atoms together, it is a chemical bond (Ionic, Covalent, or Dative).
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33.
A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. If the molar mass of the compound is 180. Find the molecular formula. [H = 1, C = 12, O = 16]
1. A.C $_3$ H $_6$ O $_3$
2. B.C $_6$ H $_6$ O $_3$
3. C.C $_6$ H $_{12}$ O $_6$
4. D.CH $_2$ O
Show answer
Answer: C. C $_6$ H $_{12}$ O $_6$
AI explanation
To find the molecular formula of this compound, we follow a systematic two-part process: finding the empirical formula first, and then determining the molecular formula.
Step 1: Determine the Empirical Formula
Assume we have 100g of the compound. This means we have:
Carbon (C): 40.0g
Hydrogen (H): 6.7g
Oxygen (O): 53.3g
Next, divide each mass by the respective atomic mass to find the number of moles:
Moles of C: $40.0 / 12 = 3.33$
Moles of H: $6.7 / 1 = 6.7$
Moles of O: $53.3 / 16 = 3.33$
Now, divide each result by the smallest value (3.33) to get the simplest ratio:
C: $3.33 / 3.33 = 1$
H: $6.7 / 3.33 \approx 2$
O: $3.33 / 3.33 = 1$
The empirical formula is CH $_2$ O.
Step 2: Determine the Molecular Formula
The empirical formula mass of CH $_2$ O is: $(12 \times 1) + (1 \times 2) + (16 \times 1) = 12 + 2 + 16 = 30$
We use the formula: $\text{Molecular Formula} = (\text{Empirical Formula})_n$ Where $n = \frac{\text{Molar Mass}}{\text{Empirical Formula Mass}}$
$n = \frac{180}{30} = 6$
Now, multiply the subscripts in the empirical formula by $n$ (which is 6):
C: $1 \times 6 = 6$
H: $2 \times 6 = 12$
O: $1 \times 6 = 6$
The molecular formula is C $_6$ H $_{12}$ O $_6$ .
Conclusion
Comparing our result to the given options, the correct choice is C.
Exam Tip: Always remember that the empirical formula represents the simplest ratio, while the molecular formula represents the actual number of atoms in a molecule. If you are ever stuck, you can quickly check the options by calculating their molar masses; only the correct molecular formula will match the molar mass provided in the question (180).
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34.
SO $_3$ is not directly dissolved in water in the industrial preparation of H $_2$ SO $_4$ by the contact process because
1. A.SO $_3$ is an acidic gas
2. B.Acid is usually added to water and not water to acid
3. C.The reaction between SO $_3$ and water is violently exothermic
4. D.SO $_3$ does not dissolve readily in water
Show answer
Answer: C. The reaction between SO $_3$ and water is violently exothermic
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35.
By what amount must the temperature of 200cm $^3$ of Nitrogen at 27°C be increased to double the pressure if the final volume is 150cm $^3$ (Assume ideality)
1. A.150°C
2. B.177°C
3. C.75°C
4. D.120°C
Show answer
Answer: A. 150°C
AI explanation
Analysis
To solve this problem, we use the General Gas Law (Combined Gas Law), which relates pressure ( $P$ ), volume ( $V$ ), and temperature ( $T$ ) for a fixed mass of an ideal gas:
$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
Concept
Initial state: $V_1 = 200 \text{ cm}^3$ , $T_1 = 27^\circ\text{C}$ , $P_1 = P$
Final state: $V_2 = 150 \text{ cm}^3$ , $P_2 = 2P$ (pressure is doubled), $T_2 = ?$
Crucial Rule: Always convert temperature from Celsius to Kelvin by adding 273. • $T_1 = 27 + 273 = 300 \text{ K}$
Step-by-Step Solution
Set up the equation: $ $\frac{P \times 200}{300} = \frac{2P \times 150}{T_2}$ $
Simplify the equation: The pressure $P$ appears on both sides, so it cancels out: $ $\frac{200}{300} = \frac{300}{T_2}$ $ $ $\frac{2}{3} = \frac{300}{T_2}$ $
Solve for $T_2$ : $ $2 \times T_2 = 3 \times 300$ $ $ $2T_2 = 900$ $ $ $T_2 = 450 \text{ K}$ $
Convert back to Celsius: $ $T_2 (\text{in } ^\circ\text{C}) = 450 - 273 = 177^\circ\text{C}$ $
Calculate the increase in temperature: The question asks for the amount by which the temperature must be increased: $ $\text{Increase} = T_2 - T_1$ $ $ $\text{Increase} = 177^\circ\text{C} - 27^\circ\text{C} = 150^\circ\text{C}$ $
Conclusion
The temperature must be increased by 150°C to reach the final temperature of 177°C.
Exam Tip: Always check if the question asks for the final temperature or the increase in temperature. Many students stop at 177°C and choose option B, but the question specifically asks for the "amount... increased," which is $177 - 27 = 150$ .
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36.
Which of the following will give a precipitate with an aqueous solution of copper (I) chloride?
1. A.But-2-ene
2. B.But-1-ene
3. C.But-2-yne
4. D.But-1-yne
Show answer
Answer: D. But-1-yne
AI explanation
Analysis
The question asks us to identify which of the given hydrocarbons reacts with an aqueous solution of copper(I) chloride (specifically ammoniacal copper(I) chloride) to form a precipitate.
Concept Explanation
This is a classic test for terminal alkynes.
Terminal alkynes are hydrocarbons with a triple bond at the end of the carbon chain (e.g., $R-C \equiv C-H$ ).
The hydrogen atom attached to the triply-bonded carbon is acidic.
When a terminal alkyne reacts with ammoniacal copper(I) chloride (or ammoniacal silver nitrate), the acidic hydrogen is replaced by the metal, forming a metal acetylide precipitate.
Internal alkynes (where the triple bond is not at the end) and alkenes do not have this acidic hydrogen, so they do not react with these reagents to form a precipitate.
Step-by-Step Solution
Analyze the options: • A. But-2-ene ( $CH_3-CH=CH-CH_3$ ): This is an alkene. It does not react with copper(I) chloride. • B. But-1-ene ( $CH_2=CH-CH_2-CH_3$ ): This is an alkene. It does not react with copper(I) chloride. • C. But-2-yne ( $CH_3-C \equiv C-CH_3$ ): This is an internal alkyne. The triple bond is between the second and third carbons. There is no hydrogen atom attached to the triply-bonded carbons, so it cannot form a precipitate. • D. But-1-yne ( $CH \equiv C-CH_2-CH_3$ ): This is a terminal alkyne. The triple bond is at the first carbon, meaning there is an acidic hydrogen ( $H-C \equiv C-CH_2-CH_3$ ). This hydrogen reacts with the copper(I) ions to form a reddish-brown precipitate of copper(I) butynide.
Conclusion: Only But-1-yne possesses the terminal acidic hydrogen required to react with the reagent to form a precipitate.
Exam Tip
Remember the "Terminal Rule": If you see a question about forming precipitates with ammoniacal $AgNO_3$ (Tollens' reagent) or $CuCl$ , always look for the molecule that ends in "-1-yne". If it's an alkene or an internal alkyne (like but-2-yne), it will not react!
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37.
Elements X and Y have electronic configurations 1s $^2$ 2s $^2$ 2p $^4$ and 1s $^2$ 2s $^2$ 2p $^6$ 3s $^2$ 3p $^1$ respectively. When they combine, the formula of the compound formed is
1. A.XY
2. B.X $_2$ Y
3. C.Y $_2$ X $_3$
4. D.X $_2$ Y $_3$
Show answer
Answer: C. Y $_2$ X $_3$
AI explanation
1. Analyze the Question
We are given the electronic configurations of two elements, X and Y. We need to determine their valencies (combining powers) based on their valence electrons to find the chemical formula of the compound they form.
2. Explain the Concept
Element X: The configuration is $1s^2 2s^2 2p^4$ . The outermost shell is the second shell ( $n=2$ ), which contains $2 + 4 = 6$ electrons. To achieve a stable octet (8 electrons), it needs to gain 2 electrons. Therefore, its valency is -2.
Element Y: The configuration is $1s^2 2s^2 2p^6 3s^2 3p^1$ . The outermost shell is the third shell ( $n=3$ ), which contains $2 + 1 = 3$ electrons. To achieve stability, it will lose these 3 electrons. Therefore, its valency is +3.
3. Show the Solution
To find the formula of the compound, we use the "criss-cross" method:
Identify the charges (valencies): • X has a charge of -2 (it acts as an anion). • Y has a charge of +3 (it acts as a cation).
Write the symbols with their charges: • $Y^{3+}$ and $X^{2-}$
Criss-cross the charges to become subscripts: • The charge of Y (3) becomes the subscript of X. • The charge of X (2) becomes the subscript of Y.
Resulting formula: • $Y_2X_3$
4. Identify the Answer
Comparing our result ( $Y_2X_3$ ) with the given options: A. XY B. X $_2$ Y C. Y $_2$ X $_3$ D. X $_2$ Y $_3$
The correct option is C.
5. Exam Tip
When determining the formula of a compound, remember the "Octet Rule": Elements with 1, 2, or 3 valence electrons generally lose them to form positive ions (cations), while elements with 5, 6, or 7 valence electrons generally gain electrons to form negative ions (anions). Always write the metal (cation) first, followed by the non-metal (anion).
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38.
The IUPAC name for CH $_3$ CH $_2$ COOCH $_2$ CH $_3$ is
1. A.methylpropanoate
2. B.ethylethanoate
3. C.ethylpropanoate
4. D.methylethanoate
Show answer
Answer: C. ethylpropanoate
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39.
In the reaction: M + N → P ΔH = +Q kJ Which of the following would increase the concentration of the product?
1. A.Adding a suitable catalyst
2. B.Increasing the concentration of P
3. C.Increasing the temperature
4. D.Decreasing the temperature
Show answer
Answer: C. Increasing the temperature
AI explanation
Analysis
The reaction is given as: M + N ⇌ P with ΔH = +Q kJ.
The positive sign of ΔH (+Q) indicates that the reaction is endothermic (it absorbs heat).
According to Le Chatelier’s Principle, if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system will adjust itself to counteract that change.
Concept: Le Chatelier’s Principle
Temperature: For an endothermic reaction, heat acts like a reactant. Increasing the temperature shifts the equilibrium to the right (forward) to absorb the excess heat, thereby increasing the concentration of the product (P).
Concentration: Increasing the concentration of reactants (M or N) shifts the equilibrium to the right. Increasing the concentration of the product (P) would shift the equilibrium to the left (reverse).
Catalyst: A catalyst increases the rate of both forward and backward reactions equally. It helps the system reach equilibrium faster but does not change the position of the equilibrium or the concentration of the products.
Step-by-Step Solution
Option A: A catalyst only speeds up the reaction; it does not increase the yield of the product.
Option B: Increasing the concentration of P would cause the system to shift to the left to consume the excess P, decreasing the yield.
Option C: Since the reaction is endothermic (ΔH > 0), adding heat (increasing temperature) forces the reaction to move forward to use up the heat, thus producing more P.
Option D: Decreasing the temperature would shift the equilibrium to the left (the exothermic direction), decreasing the concentration of P.
Conclusion
Increasing the temperature is the correct way to shift an endothermic reaction toward the product side.
Exam Tip: Remember the mnemonic "Endo-Heat-In": For Endothermic reactions, adding Heat shifts the reaction In (forward) to the product side. Conversely, for Exothermic reactions, cooling the system increases the product yield.
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40.
Which of the following pairs cannot be represented with a chemical formula?
1. A.air and bronze
2. B.bronze and sodium chloride
3. C.copper and sodium chloride
4. D.caustic soda and washing soda
Show answer
Answer: A. air and bronze
AI explanation
Analysis of the Question
The question asks us to identify a pair of substances that cannot be represented by a single chemical formula. To answer this, we must distinguish between pure substances (elements and compounds) and mixtures.
Concept Explanation
Chemical Formula: A chemical formula (like $H_2O$ or $NaCl$ ) represents the fixed ratio of atoms in a pure substance (an element or a compound).
Pure Substances: These have a constant composition and can be represented by a formula. • Elements: e.g., Copper ( $Cu$ ), Oxygen ( $O_2$ ). • Compounds: e.g., Sodium chloride ( $NaCl$ ), Caustic soda ( $NaOH$ ), Washing soda ( $Na_2CO_3 \cdot 10H_2O$ ).
Mixtures: These consist of two or more substances physically combined in varying proportions. Because their composition is not fixed, they cannot be represented by a single chemical formula. • Air: A mixture of gases (nitrogen, oxygen, argon, carbon dioxide, etc.). • Bronze: An alloy (a solid mixture) of copper and tin.
Step-by-Step Solution
Analyze Option A: • Air: A mixture of gases. No fixed formula. • Bronze: An alloy (mixture) of copper and tin. No fixed formula. • Conclusion: Neither can be represented by a chemical formula.
Analyze Option B: • Bronze: Mixture (No formula). • Sodium chloride: Compound ( $NaCl$ ). Has a formula.
Analyze Option C: • Copper: Element ( $Cu$ ). Has a formula. • Sodium chloride: Compound ( $NaCl$ ). Has a formula.
Analyze Option D: • Caustic soda: Compound ( $NaOH$ ). Has a formula. • Washing soda: Compound ( $Na_2CO_3 \cdot 10H_2O$ ). Has a formula.
Since both air and bronze are mixtures with variable compositions, they are the only pair that cannot be represented by a chemical formula.
Final Answer
The correct option is A.
Exam Tip: Always remember that alloys (like bronze, brass, and steel) and solutions (like air or salt water) are mixtures. If you see an alloy or a mixture in a "chemical formula" question, that is almost certainly the one that cannot be represented by a formula!
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41.
Which of the following metals is the most essential in the regulation of blood volume, blood pressure and osmotic equilibrium?
1. A.Zinc
2. B.Manganese
3. C.Sodium
4. D.Iron
Show answer
Answer: C. Sodium
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42.
The shapes of water, ammonia, carbon (iv) oxide and methane are respectively
1. A.linear, tetrahedral, angular and pyramidal
2. B.angular, pyramidal, linear and tetrahedral
3. C.pyramidal, linear, tetrahedral and angular
4. D.tetrahedral, angular, pyramidal and linear
Show answer
Answer: B. angular, pyramidal, linear and tetrahedral
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43.
The cost of discharging 6.0g of a divalent metal, X from its salt is ₦12.00. What is the cost of discharging 9.0g of a trivalent metal, Y from its salt under the same condition? [X = 63, Y = 27, 1F = 96,500C]
1. A.N12.00
2. B.N27.00
3. C.N63.00
4. D.N95.00
Show answer
Answer: C. N63.00
AI explanation
To solve this problem, we use Faraday’s Laws of Electrolysis, which state that the mass ( $m$ ) of a substance deposited is proportional to the quantity of electricity ( $Q$ ) passed through the electrolyte.
1. Analyze the Question
Metal X (Divalent): Mass ( $m_x$ ) = 6.0g, Molar mass ( $M_x$ ) = 63, Valency ( $n_x$ ) = 2. Cost = ₦12.00.
Metal Y (Trivalent): Mass ( $m_y$ ) = 9.0g, Molar mass ( $M_y$ ) = 27, Valency ( $n_y$ ) = 3. Cost = ?
2. Explain the Concept
The quantity of electricity ( $Q$ ) required to deposit a mass ( $m$ ) is given by:
$Q = \frac{m \times n \times F}{M}$
Where:
$n$ = valency (number of electrons involved)
$F$ = Faraday's constant
$M$ = Molar mass
Since the cost is proportional to the quantity of electricity used, we can say:
$\text{Cost} \propto Q \implies \text{Cost} = k \times \left( \frac{m \times n}{M} \right)$
Where $k$ is a constant representing the cost per unit of electricity.
3. Show the Solution
Step 1: Calculate the quantity of electricity for Metal X ( $Q_x$ )
$Q_x = \frac{6.0 \times 2}{63} = \frac{12}{63} \approx 0.1905 \text{ Faradays}$
We know that $Q_x$ costs ₦12.00. Therefore, the cost per Faraday ( $k$ ) is:
$k = \frac{12}{12/63} = 12 \times \frac{63}{12} = 63 \text{ Naira per Faraday}$
Step 2: Calculate the quantity of electricity for Metal Y ( $Q_y$ )
$Q_y = \frac{9.0 \times 3}{27} = \frac{27}{27} = 1.0 \text{ Faraday}$
Step 3: Calculate the cost for Metal Y
$\text{Cost}_y = Q_y \times k$
$\text{Cost}_y = 1.0 \times 63 = 63.00$
4. Identify the Answer
The cost of discharging 9.0g of metal Y is ₦63.00. This corresponds to option C.
5. Exam Tip
When dealing with electrolysis cost problems, always remember the "Faraday equivalent" formula: $Q = \frac{m \times n}{M}$ . By calculating the number of Faradays required for each metal first, you simplify the problem into a basic ratio calculation, which saves time and reduces errors in the exam hall.
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44.
Which of the following could not be alkane?
1. A.C $_4$ H $_{10}$
2. B.C $_5$ H $_{12}$
3. C.C $_7$ H $_{14}$
4. D.C $_8$ H $_{18}$
Show answer
Answer: C. C $_7$ H $_{14}$
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45.
In the reaction between sodium hydroxide and tetraoxosulphate (VI) solutions, what volume of 0.5 molar sodium hydroxide would exactly neutralize 10cm $^3$ of 1.25 molar tetraoxosulphate (vi) acid?
1. A.25cm $^3$
2. B.10cm $^3$
3. C.20cm $^3$
4. D.50cm $^3$
Show answer
Answer: D. 50cm $^3$
AI explanation
1. Analyze the question
We are asked to find the volume of a base (NaOH) required to neutralize a specific volume and concentration of an acid ( $H_2SO_4$ ). This is a classic titration stoichiometry problem.
2. Explain the concept
To solve this, we use the balanced chemical equation for the reaction and the titration formula:
$\frac{C_a V_a}{n_a} = \frac{C_b V_b}{n_b}$
Where:
$C_a$ = Concentration of the acid
$V_a$ = Volume of the acid
$n_a$ = Number of moles of acid from the balanced equation
$C_b$ = Concentration of the base
$V_b$ = Volume of the base
$n_b$ = Number of moles of base from the balanced equation
Balanced Equation: $2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O$ From the equation, $n_a = 1$ and $n_b = 2$ .
3. Show the solution
Given:
$C_a = 1.25\text{ M}$
$V_a = 10\text{ cm}^3$
$C_b = 0.5\text{ M}$
$V_b = ?$
$n_a = 1$
$n_b = 2$
Step-by-step calculation: Substitute the values into the formula:
$\frac{1.25 \times 10}{1} = \frac{0.5 \times V_b}{2}$
Multiply both sides to clear the fraction:
$12.5 = \frac{0.5 \times V_b}{2}$
$12.5 \times 2 = 0.5 \times V_b$
$25 = 0.5 \times V_b$
Now, solve for $V_b$ :
$V_b = \frac{25}{0.5}$
$V_b = 50\text{ cm}^3$
4. Identify the answer
The calculated volume is $50\text{ cm}^3$ , which corresponds to option D.
5. Tip
Always write out the balanced chemical equation first! Many students forget the stoichiometry coefficients ( $n_a$ and $n_b$ ). If you forget to include the "2" for the NaOH in this reaction, you would get $25\text{ cm}^3$ (Option A), which is a common "distractor" answer designed to catch students who miss the mole ratio.
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46.
Which of the following statements about catalyst is false?
1. A.biological catalysts are called enzymes
2. B.catalyst do not appear in the balanced equation
3. C.catalyst reduce the activation energy for a reaction
4. D.catalysts do not alter the mechanism of the reaction and never appear in the rate low
Show answer
Answer: D. catalysts do not alter the mechanism of the reaction and never appear in the rate low
AI explanation
Analysis of the Question
This question tests your understanding of the fundamental properties of catalysts in chemical kinetics. We need to identify which of the four statements is scientifically incorrect.
Concept Explanation
A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the process. Here is how they work:
Activation Energy: Catalysts provide an alternative reaction pathway with a lower activation energy ( $E_a$ ), allowing more reactant molecules to have enough energy to collide effectively.
Reaction Mechanism: By providing this alternative pathway, catalysts do change the mechanism of the reaction.
Rate Law: Because the mechanism changes, the catalyst often appears in the rate law expression (its concentration affects the rate).
Balanced Equation: Catalysts are not consumed, so they do not appear as net reactants or products in the overall balanced chemical equation.
Enzymes: These are protein molecules that act as catalysts in biological systems.
Step-by-Step Evaluation
Option A: True. Enzymes are indeed biological catalysts that speed up biochemical reactions in living organisms.
Option B: True. Since a catalyst is regenerated during the reaction, it is not a net reactant or product, so it is omitted from the overall stoichiometric equation.
Option C: True. This is the primary function of a catalyst; by lowering the activation energy, it makes it easier for the reaction to proceed.
Option D: False. This statement claims catalysts do not alter the mechanism and never appear in the rate law. In reality, catalysts do alter the reaction mechanism (that is precisely how they lower the activation energy), and because they participate in the rate-determining step, they often appear in the rate law expression.
Conclusion
Since the question asks for the false statement, Option D is the correct choice because it contains incorrect information regarding the role of catalysts in reaction mechanisms and rate laws.
Exam Tip
When you see questions about catalysts, remember the "Alternative Pathway" rule: If a catalyst changes the speed of a reaction, it must be doing so by changing the "path" (mechanism) the molecules take to become products. If the path changes, the rate law (which describes the path) must also change.
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47.
Which of the following reactions is an oxidation process?
1. A.2H $^+$ + 2e $^-$ $\to$ H $_2$
2. B.2O $^{2-}$ + 4e $^-$ $\to$ O $_2$
3. C.OCl $^-$ + H $_2$ O + 2e $^-$ $\to$ 2OH $^-$ + Cl $^-$
4. D.Cl + e $^-$ $\to$ Cl $^-$
Show answer
Answer: B. 2O $^{2-}$ + 4e $^-$ $\to$ O $_2$
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48.
Which of the following sets of operation will completely separate a mixture of sodium chloride, sand and iodine?
1. A.addition of water, filtration, evaporation to dryness and sublimation
2. B.filtration, evaporation to dryness, addition of water and sublimation
3. C.sublimation, filtration, evaporation and addition of water
4. D.sublimation, addition of water, filtration and evaporation to dryness
Show answer
Answer: D. sublimation, addition of water, filtration and evaporation to dryness
AI explanation
To solve this problem, we must look at the physical properties of each component in the mixture: sodium chloride (salt), sand, and iodine.
1. Analyze the Components
Iodine ( $I_2$ ): A solid that undergoes sublimation (changes directly from solid to gas when heated).
Sodium Chloride ( $NaCl$ ): Highly soluble in water.
Sand ( $SiO_2$ ): Insoluble in water.
2. Explain the Concept
To separate a mixture, we must choose a sequence of operations where each step targets the unique physical property of one component without interfering with the others.
Sublimation: Since iodine sublimes, heating the mixture will cause the iodine to turn into vapor, leaving behind the salt and sand.
Addition of Water: Once the iodine is removed, we add water to the remaining mixture. The salt will dissolve, while the sand will remain as a solid.
Filtration: This process separates the insoluble sand from the salt solution (filtrate).
Evaporation to Dryness: Finally, heating the salt solution will evaporate the water, leaving the solid sodium chloride behind.
3. Step-by-Step Solution
Sublimation: Heat the mixture. Iodine vaporizes and can be collected, leaving salt and sand.
Addition of water: Add water to the remaining salt and sand. The salt dissolves to form a solution.
Filtration: Pour the mixture through filter paper. The sand stays on the paper (residue), and the salt solution passes through (filtrate).
Evaporation to dryness: Heat the salt solution to remove the water, leaving pure sodium chloride crystals.
4. Identify the Answer
Comparing our logical steps to the options provided:
Option A is incorrect because you cannot filter before removing the iodine.
Option B is incorrect for the same reason.
Option C is incorrect because adding water before sublimation would dissolve the salt while the iodine is still present, making the separation messy and inefficient.
Option D follows the correct logical sequence: Sublimation $\rightarrow$ Addition of water $\rightarrow$ Filtration $\rightarrow$ Evaporation to dryness.
Exam Tip
When dealing with separation of mixtures, always think about the "easiest" component to remove first. If a component sublimes (like Iodine, Ammonium Chloride, or Naphthalene), sublimation should almost always be your first step!
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49.
Which important nitrogen-containing compound is produced in Haber's process?
1. A.NH $_3$
2. B.NO $_2$
3. C.NO
4. D.HNO $_3$
Show answer
Answer: A. NH $_3$
AI explanation
Analysis
The question asks for the specific nitrogen-containing compound synthesized during the industrial process known as the Haber process.
Concept Explanation
The Haber process (also known as the Haber-Bosch process) is the primary industrial method used to produce ammonia ( $NH_3$ ).
In this process, nitrogen gas ( $N_2$ ) from the air is reacted with hydrogen gas ( $H_2$ ) under specific conditions:
Temperature: About 400°C – 450°C.
Pressure: High pressure, typically 200 atmospheres.
Catalyst: Finely divided iron.
The chemical equation for the reaction is:
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
Step-by-Step Solution
Identify the reactants: The process uses Nitrogen ( $N_2$ ) and Hydrogen ( $H_2$ ).
Identify the product: The reaction is designed to synthesize ammonia ( $NH_3$ ), which is a vital component in the production of fertilizers.
Evaluate the options: • A. $NH_3$ (Ammonia): This is the direct product of the Haber process. • B. $NO_2$ (Nitrogen dioxide): This is typically produced in the Ostwald process or through the reaction of concentrated nitric acid. • C. $NO$ (Nitrogen monoxide): This is an intermediate in the Ostwald process. • D. $HNO_3$ (Nitric acid): This is produced via the Ostwald process, often using ammonia (the product of the Haber process) as a starting material.
Conclusion
Since the Haber process is specifically defined as the industrial synthesis of ammonia, the correct compound is $NH_3$ .
Exam Tip: Always distinguish between the Haber process (for Ammonia, $NH_3$ ) and the Ostwald process (for Nitric acid, $HNO_3$ ). Students often confuse these two in exams!
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50.
A radioactive nucleus has a half-life of 20 years, starting with 100,000 particles, how many particles will be left exactly at the end of 40 years
1. A.75,000 particles
2. B.35,000 particles
3. C.25,000 particles
4. D.50,000 particles
Show answer
Answer: C. 25,000 particles
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51.
What volume of 0.100M sodium trioxonitrate (V) solution contains 5g of solute ? [Na = 23, N = 14, O = 16]
1. A.1.7 litres
2. B.0.420 litres
3. C.0.588 litres
4. D.2.35 litres
Show answer
Answer: C. 0.588 litres
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52.
Which process(es) is/are involved in the turning of starch iodide paper blue-black by chlorine gas?
1. A.chlorine attacks the starch to give the blue-black color
2. B.chlorine attacks the iodide ion to give the blue-black color
3. C.chlorine oxidizes the iodide ion to produce iodine which attacks the starch to give the blue-black color
4. D.iodine attacks the starch to give the blue-black color
Show answer
Answer: C. chlorine oxidizes the iodide ion to produce iodine which attacks the starch to give the blue-black color
AI explanation
Analysis
The question asks for the chemical mechanism behind the starch-iodide test, which is a classic qualitative test used to detect the presence of oxidizing agents like chlorine ( $Cl_2$ ).
Concept Explanation
To understand this, we need to look at two distinct chemical reactions occurring in sequence:
Redox Reaction: Chlorine is a strong oxidizing agent. When it comes into contact with iodide ions ( $I^-$ ) present in the starch-iodide paper, it displaces the iodine. The chlorine is reduced to chloride ions, and the iodide ions are oxidized to elemental iodine ( $I_2$ ). • Equation: $Cl_{2(g)} + 2I^-_{(aq)} \rightarrow 2Cl^-_{(aq)} + I_{2(s)}$
Complex Formation: Starch acts as an indicator for iodine. Elemental iodine ( $I_2$ ) reacts with the helical structure of the starch molecule to form a deep blue-black starch-iodine complex. This is a physical-chemical adsorption process.
Step-by-Step Solution
Step 1: Chlorine gas acts as an oxidizing agent. It reacts with the iodide ions in the paper.
Step 2: This reaction releases free iodine ( $I_2$ ).
Step 3: The released iodine then interacts with the starch molecules in the paper.
Step 4: The interaction between the iodine and the starch creates the characteristic blue-black color.
Looking at the options:
Option A is incorrect because chlorine does not react directly with starch to produce the color.
Option B is incomplete; it mentions the iodide ion but ignores the essential role of the starch.
Option C correctly identifies that chlorine oxidizes the iodide to iodine, which then reacts with the starch.
Option D is partially correct but misses the initial step of the chlorine gas acting as the oxidizing agent to liberate the iodine.
Conclusion
The process requires both the oxidation of iodide to iodine by the chlorine and the subsequent reaction of that iodine with the starch. Therefore, option C provides the most complete and accurate scientific explanation.
Exam Tip: Always remember that the starch-iodide test is specifically a test for oxidizing agents. If you see "starch-iodide" in a question, immediately think: Oxidation of iodide $\rightarrow$ liberation of iodine $\rightarrow$ blue-black complex with starch.
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53.
Burning magnesium ribbon in air removes which of the following (i) oxygen (ii) nitrogen (iii) argon and (iv) carbon(iv)oxide?
1. A.i and iv only
2. B.i, ii, iii and iv only
3. C.i only
4. D.i and ii only
Show answer
Answer: D. i and ii only
AI explanation
Analysis of the Question
The question asks us to identify which gases from the atmosphere are consumed or reacted when a magnesium ribbon is burned in air.
Concept Explanation
When magnesium ribbon is burned in air, it undergoes a combustion reaction. While we often simplify this by saying it reacts with oxygen, magnesium is a highly reactive metal. At the high temperatures produced during combustion, it is capable of reacting with other components of the atmosphere:
Oxygen ( $O_2$ ): Magnesium reacts vigorously with oxygen to form magnesium oxide ( $MgO$ ). $ $2Mg_{(s)} + O_2_{(g)} \rightarrow 2MgO_{(s)}$ $
Nitrogen ( $N_2$ ): Because magnesium is a strong reducing agent, it can react with atmospheric nitrogen to form magnesium nitride ( $Mg_3N_2$ ). This is a characteristic property of Group 2 metals. $ $3Mg_{(s)} + N_2_{(g)} \rightarrow Mg_3N_2_{(s)}$ $
Argon ( $Ar$ ): Argon is a noble gas. It is chemically inert (unreactive) under these conditions and does not participate in the reaction.
Carbon(IV) oxide ( $CO_2$ ): While magnesium can react with $CO_2$ (which is why you should never use a $CO_2$ fire extinguisher on a magnesium fire), in the context of standard atmospheric composition and typical JAMB chemistry questions, the primary focus is on the reaction with the two most abundant reactive gases: oxygen and nitrogen.
Step-by-Step Solution
Step 1: Identify the reactive components. Magnesium is reactive enough to combine with both oxygen and nitrogen when heated.
Step 2: Evaluate the options. Since magnesium reacts with both oxygen (i) and nitrogen (ii), we look for an option that includes both.
Conclusion
Burning magnesium ribbon in air removes both oxygen and nitrogen because it forms both magnesium oxide and magnesium nitride. Therefore, (i) and (ii) are the correct components.
Exam Tip: Always remember that magnesium is unique among common metals because it can "burn" in nitrogen. If you see a question about magnesium reacting in air, don't just look for oxygen—look for nitrogen as well!
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54.
The hybridization in the compound $CH_3 - CH_2 - C \equiv H$ is
1. A.sp $^3$ and sp
2. B.sp
3. C.sp $^2$
4. D.sp $^3$ and sp $^2$
Show answer
Answer: A. sp $^3$ and sp
AI explanation
To solve this problem, we need to look at the bonding around each carbon atom in the molecule.
Note: There appears to be a slight typo in your question. The structure $CH_3 - CH_2 - C \equiv H$ is chemically impossible because carbon must form four bonds. The correct structure for this series is $CH_3 - CH_2 - C \equiv CH$ (butyne).
1. Analyze the concept: Hybridization
Hybridization is determined by the number of sigma ( $\sigma$ ) bonds and lone pairs around a central atom:
$sp^3$ hybridization: Occurs when a carbon atom is bonded to 4 other atoms (single bonds only).
$sp^2$ hybridization: Occurs when a carbon atom is bonded to 3 other atoms (one double bond).
$sp$ hybridization: Occurs when a carbon atom is bonded to 2 other atoms (one triple bond or two double bonds).
2. Step-by-step breakdown
Let's analyze the structure $CH_3 - CH_2 - C \equiv CH$ :
Carbon 1 ( $CH_3$ ): This carbon is bonded to 3 hydrogens and 1 carbon (all single bonds). Since it has 4 sigma bonds, it is $sp^3$ hybridized.
Carbon 2 ( $CH_2$ ): This carbon is bonded to 2 hydrogens and 2 carbons (all single bonds). Since it has 4 sigma bonds, it is $sp^3$ hybridized.
Carbon 3 ( $-C \equiv$ ): This carbon is bonded to one carbon on the left and one carbon on the right via a triple bond. It is bonded to only 2 atoms. Therefore, it is $sp$ hybridized.
Carbon 4 ( $\equiv CH$ ): This carbon is bonded to one carbon and one hydrogen. It is bonded to only 2 atoms. Therefore, it is $sp$ hybridized.
3. Conclusion
The molecule contains carbon atoms that are either $sp^3$ hybridized or $sp$ hybridized. Looking at the options provided, the combination present in the chain is $sp^3$ and $sp$ .
Option A is the best fit among the choices provided.
Exam Tip: Always count the number of atoms attached to the carbon. If it's 4 atoms, it's $sp^3$ ; if it's 3, it's $sp^2$ ; if it's 2, it's $sp$ . Ignore the pi bonds (the second and third lines in double/triple bonds) when counting for hybridization!
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55.
Which of the following is a physical change?
1. A.Exposing white phosphorus to air
2. B.Burning kerosene
3. C.Dissolving calcium in water
4. D.Freezing ice cream
Show answer
Answer: D. Freezing ice cream
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56.
The heat of formation of ethene, C $_2$ H $_4$ is 50 kJmol $^{-1}$ , and that of ethane, C $_2$ H $_6$ is -82kJmol $^{-1}$ . Calculate the heat evolved in the process: C $_2$ H $_4$ + H $_2$ $\to$ C $_2$ H $_6$
1. A.-122kJ
2. B.-132kJ
3. C.-88kJ
4. D.-60kJ
Show answer
Answer: B. -132kJ
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57.
When chlorine water is exposed to bright sunlight, the following products are formed
1. A.HCl $_{(aq)}$ and Cl $_2$
2. B.HOCl $_{(aq)}$ and Cl $_2$
3. C.HCl $_{(aq)}$ and O $_2$
4. D.HOCl $_{(aq)}$ and O $_2$
Show answer
Answer: C. HCl $_{(aq)}$ and O $_2$
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58.
Sulphur exists in six forms in the solid state. This property is known as
1. A.Isomerism
2. B.Allotropy
3. C.Isotopy
4. D.Isomorphism
Show answer
Answer: B. Allotropy
AI explanation
Analysis
The question asks for the specific term used to describe the ability of an element, like sulphur, to exist in more than one physical form in the same state (solid).
Concept Explanation
Allotropy: This is the property whereby an element can exist in two or more different physical forms in the same physical state. These forms are called allotropes. Common examples include carbon (diamond and graphite) and sulphur (rhombic and monoclinic sulphur).
Isomerism: This refers to compounds that have the same molecular formula but different structural arrangements. This is common in organic chemistry (e.g., butane and isobutane).
Isotopy: This refers to atoms of the same element that have the same atomic number but different mass numbers (different number of neutrons).
Isomorphism: This is the phenomenon where two or more different substances crystallize in the same geometric form (e.g., potash alum and chrome alum).
Step-by-Step Solution
Identify the subject: The question focuses on "Sulphur" and its "six forms in the solid state."
Evaluate the definitions: Since sulphur is a single element and the question describes different physical forms of that same element, we are looking for the definition of allotropy.
Compare with options: • Isomerism applies to molecules/compounds. • Isotopy applies to atoms of the same element. • Isomorphism applies to different substances with the same crystal shape. • Allotropy is the only term that describes different forms of the same element.
Conclusion
The property of an element existing in different forms in the same physical state is known as Allotropy.
Exam Tip: To remember this easily, think of the "A" in Allotropy as standing for Another form of the same element. Don't confuse it with Isotopes, which are about the nucleus (protons/neutrons), whereas Allotropes are about the physical arrangement of the atoms.
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59.
Consider the reaction: A + 2B $_{(g)} \rightleftharpoons$ 2C + D $_{(g)}$ ( $\Delta$ H = +ve) What will be the effect of decrease in temperature on the reaction?
1. A.Rate of backward reaction will increase
2. B.The equilibrium constant will shift
3. C.Rates of forward and backward reactions are not affected
4. D.Rate of forward reaction will increase
Show answer
Answer: A. Rate of backward reaction will increase
AI explanation
Analysis of the Question
This question tests your understanding of Le Chatelier’s Principle, which states that if a dynamic equilibrium is disturbed by changing the conditions (temperature, pressure, or concentration), the position of equilibrium moves to counteract the change.
Concept Explanation
The reaction is given as: A + 2B $_{(g)} \rightleftharpoons$ 2C + D $_{(g)}$ ( $\Delta$ H = +ve)
$\Delta$ H = +ve: This indicates that the forward reaction is endothermic (it absorbs heat).
Reverse Reaction: Since the forward reaction is endothermic, the backward (reverse) reaction must be exothermic (it releases heat).
According to Le Chatelier’s Principle, when you decrease the temperature of a system at equilibrium, the system will try to produce more heat to compensate for the loss. It does this by favoring the exothermic direction.
Step-by-Step Solution
Identify the nature of the reaction: The forward reaction is endothermic ( $\Delta$ H > 0). Therefore, the backward reaction is exothermic ( $\Delta$ H < 0).
Apply the temperature change: You are decreasing the temperature.
Determine the shift: The system will favor the exothermic direction to generate heat. Since the backward reaction is exothermic, the equilibrium position will shift to the left (towards the reactants).
Analyze the rates: As the equilibrium shifts to the left, the rate of the backward reaction increases relative to the forward reaction until a new equilibrium is established.
Evaluating the Options
A. Rate of backward reaction will increase: This is correct because the system favors the exothermic (backward) direction to counteract the cooling.
B. The equilibrium constant will shift: While the equilibrium constant ( $K$ ) changes with temperature, "shifting" is a term usually reserved for the position of equilibrium, not the constant itself.
C. Rates of forward and backward reactions are not affected: This is incorrect; temperature changes always affect the rate of reaction.
D. Rate of forward reaction will increase: This is incorrect because the forward reaction is endothermic; decreasing the temperature will actually decrease the rate of the forward reaction.
Exam Tip
Remember the "Heat-Balance" rule:
If $\Delta$ H is positive (Endothermic), think of heat as a reactant.
If $\Delta$ H is negative (Exothermic), think of heat as a product.
Decrease temperature $\rightarrow$ System moves to produce heat.
Increase temperature $\rightarrow$ System moves to consume heat.
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60.
Which of the following represents the kind of bonding present in ammonium chloride?
1. A.ionic only
2. B.covalent only
3. C.ionic and covalent
4. D.dative covalent
Show answer
Answer: C. ionic and covalent
AI explanation
Analysis
To determine the type of bonding in ammonium chloride ( $NH_4Cl$ ), we need to look at the structure of the compound and how its constituent parts are held together.
Concept Explanation
Ammonium chloride is an ionic salt composed of two distinct ions:
The Ammonium Ion ( $NH_4^+$ ): This is a polyatomic cation. Within this ion, the nitrogen atom is bonded to four hydrogen atoms. These bonds are covalent (sharing of electrons). Furthermore, one of these bonds is specifically a dative (coordinate) covalent bond, where the nitrogen atom donates its lone pair of electrons to a hydrogen ion ( $H^+$ ).
The Chloride Ion ( $Cl^-$ ): This is a simple anion.
Because the compound consists of a positively charged ammonium ion ( $NH_4^+$ ) and a negatively charged chloride ion ( $Cl^-$ ), there is a strong electrostatic force of attraction between them. This electrostatic attraction is the definition of ionic bonding.
Step-by-Step Solution
Identify the internal structure: Inside the $NH_4^+$ ion, the atoms are held together by covalent bonds (including a dative bond).
Identify the external structure: The $NH_4^+$ cation and the $Cl^-$ anion are held together by electrostatic forces, which is an ionic bond.
Combine the findings: Since the compound contains both covalent bonds (within the ammonium ion) and ionic bonds (between the ammonium and chloride ions), the bonding is both ionic and covalent.
Conclusion
While ammonium chloride does contain dative covalent bonds, the question asks for the kind of bonding present in the compound as a whole. Because it is an ionic salt containing a polyatomic ion with covalent internal bonds, it is categorized as having both ionic and covalent bonding.
Exam Tip
Whenever you see a salt containing a polyatomic ion (like $NH_4^+$ , $SO_4^{2-}$ , $NO_3^-$ , or $CO_3^{2-}$ ), remember that the bond between the metal/cation and the polyatomic ion is ionic, but the bonds inside the polyatomic ion are always covalent.
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61.
Which of the following properties increases from left to right along the period but decreases down the group in the Periodic Table? i. Atomic Number ii. Ionization energy iii. Metallic character iv. Electron affinity
1. A.ii and iii only
2. B.ii and iv only
3. C.iii and iv only
4. D.i and ii only
Show answer
Answer: B. ii and iv only
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62.
The emission of two successive beta particles from the nucleus $^{32} _{15} P$ will produce
1. A. $^{30} _{14}$ Si
2. B. $^{32} _{13}$ Al
3. C. $^{32} _{17}$ Cl
4. D. $^{32} _{16}$ S
Show answer
Answer: C. $^{32} _{17}$ Cl
AI explanation
Analysis of the Question
This question tests your understanding of radioactive decay, specifically beta ( $\beta$ ) emission. You are asked to determine the resulting nucleus after a phosphorus-32 atom undergoes two successive beta decays.
Concept Explanation
A beta particle is essentially a high-speed electron ( $^{0}_{-1}e$ ). When a nucleus emits a beta particle:
Mass Number ( $A$ ): Remains unchanged because the mass of an electron is negligible.
Atomic Number ( $Z$ ): Increases by 1. This happens because a neutron in the nucleus converts into a proton and an electron (the beta particle), which is then ejected.
The Rule: If a nucleus $^{A}_{Z}X$ emits one beta particle: $^{A}_{Z}X \rightarrow ^{A}_{Z+1}Y + ^{0}_{-1}e$
Step-by-Step Solution
Step 1: Identify the initial nucleus. The starting material is $^{32}_{15}P$ .
Mass number ( $A$ ) = 32
Atomic number ( $Z$ ) = 15
Step 2: First beta emission. $^{32}_{15}P \rightarrow ^{32}_{15+1}X + ^{0}_{-1}e$ $^{32}_{15}P \rightarrow ^{32}_{16}S + ^{0}_{-1}e$ (The element with atomic number 16 is Sulfur, S).
Step 3: Second beta emission. Now, the new nucleus ( $^{32}_{16}S$ ) undergoes a second beta decay: $^{32}_{16}S \rightarrow ^{32}_{16+1}Y + ^{0}_{-1}e$ $^{32}_{16}S \rightarrow ^{32}_{17}Cl + ^{0}_{-1}e$ (The element with atomic number 17 is Chlorine, Cl).
Conclusion
After two successive beta emissions, the atomic number increases by 2 (from 15 to 17), while the mass number remains 32. The resulting nucleus is $^{32}_{17}Cl$ .
Comparing this to the options provided: A. $^{30} _{14}$ Si B. $^{32} _{13}$ Al C. $^{32} _{17}$ Cl D. $^{32} _{16}$ S
The correct option is C.
Exam Tip
Remember the "Beta-Plus-One" rule: Every time you see "beta decay," just add 1 to the atomic number and keep the mass number the same. If it says "two successive," just add 2 to the atomic number. Don't let the "32" confuse you—beta decay never changes the mass number!
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63.
Which of the following alkaline earth metals react more spontaneously with water?
1. A.Calcium
2. B.Beryllium
3. C.Magnesium
4. D.Strontium
Show answer
Answer: D. Strontium
AI explanation
Analysis
The question asks us to identify which of the listed Group 2 elements (alkaline earth metals) reacts most spontaneously with water.
Concept Explanation
In the Periodic Table, reactivity for metals is determined by how easily an atom can lose its valence electrons to form positive ions (cations).
For Group 2 elements (Beryllium, Magnesium, Calcium, Strontium, Barium, Radium), reactivity increases as you move down the group. This happens because:
Atomic Size: As you move down the group, the number of electron shells increases.
Shielding Effect: The valence electrons are further away from the nucleus and are shielded by more inner shells.
Ionization Energy: Because the valence electrons are further from the nucleus and more shielded, the force of attraction between the nucleus and the valence electrons decreases. Consequently, it requires less energy to remove these electrons, making the metal more reactive.
Step-by-Step Solution
Locate the elements in Group 2: • Beryllium (Be) - Period 2 • Magnesium (Mg) - Period 3 • Calcium (Ca) - Period 4 • Strontium (Sr) - Period 5
Apply the trend: Since reactivity increases down the group, we compare their positions: • Be is the least reactive (it barely reacts with water). • Mg reacts very slowly with cold water. • Ca reacts steadily with cold water. • Sr reacts more vigorously than Calcium.
Conclusion: Among the options provided, Strontium is the lowest in the group, meaning it has the lowest ionization energy and the highest tendency to lose electrons, making it the most spontaneous in its reaction with water.
Identification of Answer
Comparing the options, Strontium (Sr) is the most reactive metal listed. Therefore, it reacts more spontaneously with water than Calcium, Magnesium, or Beryllium.
Exam Tip
Remember the mnemonic for Group 2 reactivity: "The further down, the more the boom!" As you go down the group, the metals become "softer" and more reactive. If you ever forget the trend, just remember that Francium (at the bottom of Group 1) is the most reactive metal—the same logic applies to Group 2!
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64.
Na $_2$ CO $_3$ + 2HCl $\to$ 2NaCl + H $_2$ O + CO $_2$ The indicator most suitable for this reaction should have a pH equal to
1. A.5
2. B.7
3. C.3
4. D.9
Show answer
Answer: A. 5
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65.
How many electrons will be found in the nucleus of an atom with mass number 23 and 17 neutrons?
1. A.none
2. B.6
3. C.7
4. D.8
Show answer
Answer: A. none
AI explanation
Analysis
To solve this, we need to understand the structure of an atom and the subatomic particles that reside within it.
Concept Explanation
An atom consists of two main regions:
The Nucleus: Located at the center of the atom, it contains protons and neutrons. These are collectively known as nucleons.
The Extra-nuclear region: This is the space surrounding the nucleus where electrons orbit in shells or energy levels.
The mass number ( $A$ ) of an atom is defined as the sum of its protons ( $Z$ ) and neutrons ( $N$ ):
$A = Z + N$
Crucially, electrons are never found inside the nucleus. They are negatively charged particles that exist outside the nucleus.
Step-by-Step Solution
Identify the components: • Mass number ( $A$ ) = 23 • Number of neutrons ( $N$ ) = 17 • Number of protons ( $Z$ ) = $A - N = 23 - 17 = 6$
Locate the particles: • The 6 protons and 17 neutrons are located inside the nucleus. • The electrons (which would be 6 in a neutral atom) are located in the shells outside the nucleus.
Conclusion: Since electrons are not part of the nuclear structure, the number of electrons found in the nucleus is zero.
Exam Tip
Always remember the "Nucleus Rule": The nucleus is the home of Protons and Neutrons only. If a question asks for the number of electrons in the nucleus, the answer is always "none" or "zero," regardless of the mass number or atomic number provided. Don't let the numbers distract you!
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66.
Which of the following factors will speed up the rate of evolution of carbon (iv) oxide in the reaction below? 2HCl + CaCO $_3$ → CaCl $_2$ + H $_2$ O + CO $_2$
1. A.Increasing the concentration of the acid
2. B.Increasing the surface area of CaCO $_3$
3. C.A and B
4. D.Reducing the concentration of the acid
Show answer
Answer: C. A and B
AI explanation
To solve this problem, we need to look at the factors that influence the rate of a chemical reaction.
1. Analyze the Question
The reaction provided is: 2HCl + CaCO $_3$ → CaCl $_2$ + H $_2$ O + CO $_2$
We are looking for factors that will increase the speed (rate) at which carbon (IV) oxide (CO $_2$ ) is produced.
2. Explain the Concept
The rate of a chemical reaction depends on the frequency of effective collisions between reactant particles. Several factors can increase this frequency:
Concentration: Increasing the concentration of a reactant (like HCl) increases the number of particles per unit volume. This leads to more frequent collisions between the acid and the calcium carbonate.
Surface Area: When a solid reactant (like CaCO $_3$ ) is in a powdered form rather than large lumps, more of its particles are exposed to the acid. This increases the number of active sites for collisions, thereby speeding up the reaction.
Temperature: Increasing the temperature gives particles more kinetic energy, causing them to move faster and collide more often with sufficient energy to react.
3. Show the Solution
Option A (Increasing the concentration of the acid): By increasing the concentration of HCl, we provide more H $^+$ ions to react with the CaCO $_3$ . This will definitely increase the rate of CO $_2$ evolution.
Option B (Increasing the surface area of CaCO $_3$ ): By breaking the CaCO $_3$ into smaller pieces or using a powder, we increase the surface area available for the acid to attack. This also increases the rate of CO $_2$ evolution.
Option C (A and B): Since both A and B are scientifically proven ways to increase the reaction rate, this option covers both correct factors.
Option D (Reducing the concentration of the acid): Reducing the concentration would decrease the number of collisions, thereby slowing down the reaction. This is incorrect.
4. Identify the Answer
Since both increasing the concentration of the acid and increasing the surface area of the solid reactant will speed up the reaction, Option C is the correct choice.
5. Exam Tip
When you see a question about the "rate of reaction," always think of the Collision Theory. Ask yourself: "Does this change make the particles hit each other more often or with more energy?" If the answer is yes, the rate will increase!
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67.
Which two gases can be used for the demonstration of the fountain experiment?
1. A.SO $_4$ and NH $_3$
2. B.SO $_3$ and NH $_3$
3. C.HCl and NH $_3$
4. D.SO $_2$ and CO
Show answer
Answer: C. HCl and NH $_3$
AI explanation
Analysis
The fountain experiment is a classic chemistry demonstration used to show the high solubility of certain gases in water. For the experiment to work, the gas must be extremely soluble in water so that when a small amount of water is introduced into the gas-filled flask, it dissolves rapidly, creating a vacuum that sucks more water into the flask, forming a "fountain."
Concept
The principle relies on Henry's Law and the polar nature of molecules. Gases that are highly soluble in water (often due to their ability to ionize or form strong hydrogen bonds with water) are the best candidates.
Ammonia ( $NH_3$ ): Highly soluble in water because it reacts to form ammonium hydroxide ( $NH_4OH$ ).
Hydrogen Chloride ( $HCl$ ): Highly soluble in water because it dissociates completely into $H^+$ and $Cl^-$ ions.
Step-by-Step Solution
Evaluate Option A ( $SO_4$ and $NH_3$ ): $SO_4$ (sulfate) is an ion, not a gas. This option is chemically incorrect.
Evaluate Option B ( $SO_3$ and $NH_3$ ): While $NH_3$ is highly soluble, $SO_3$ is a gas that reacts with water to form sulfuric acid, but it is not the standard pair used in the classic fountain experiment demonstration.
Evaluate Option C ( $HCl$ and $NH_3$ ): Both $HCl$ and $NH_3$ are gases that are extremely soluble in water. In a laboratory setting, these two are the standard gases used to demonstrate the fountain effect because they create a dramatic pressure drop almost instantly.
Evaluate Option D ( $SO_2$ and $CO$ ): $CO$ (carbon monoxide) is only slightly soluble in water, so it cannot create the vacuum required for a fountain.
Conclusion
The gases $HCl$ and $NH_3$ are the most common examples used in chemistry textbooks and laboratory demonstrations to show high solubility in water.
Exam Tip
Whenever you see a question about the "fountain experiment," look for gases that are highly polar or highly soluble in water. Remember: $NH_3$ (basic) and $HCl$ (acidic) are the "star players" for this experiment!
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68.
200cm $^3$ of 0.50mol/dm $^3$ solution of calcium hydrogen trioxocarbonate (IV) is heated. The maximum weight of solid precipitated is
1. A.10g
2. B.25g
3. C.20g
4. D.15g
Show answer
Answer: A. 10g
AI explanation
1. Analyze the Question
We are given the volume ( $200\text{ cm}^3$ ) and concentration ( $0.50\text{ mol/dm}^3$ ) of a solution of calcium hydrogen trioxocarbonate(IV), $\text{Ca(HCO}_3)_2$ . We need to find the mass of the solid precipitate formed when this solution is heated.
2. Explain the Concept
When calcium hydrogen trioxocarbonate(IV) is heated, it undergoes thermal decomposition to form calcium trioxocarbonate(IV) (which is the solid precipitate), water, and carbon(IV) oxide gas.
The balanced chemical equation is:
$\text{Ca(HCO}_3)_2(aq) \xrightarrow{\Delta} \text{CaCO}_3(s) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$
From the equation, 1 mole of $\text{Ca(HCO}_3)_2$ produces 1 mole of $\text{CaCO}_3$ .
3. Show the Solution
Step 1: Calculate the number of moles of $\text{Ca(HCO}_3)_2$
Volume ( $V$ ) = $200\text{ cm}^3 = 0.2\text{ dm}^3$
Concentration ( $C$ ) = $0.50\text{ mol/dm}^3$
$\text{Moles} = C \times V = 0.50 \times 0.2 = 0.1\text{ moles}$
Step 2: Determine the moles of $\text{CaCO}_3$ produced Since the mole ratio of $\text{Ca(HCO}_3)_2$ to $\text{CaCO}_3$ is $1:1$ , $0.1\text{ moles}$ of $\text{Ca(HCO}_3)_2$ will produce $0.1\text{ moles}$ of $\text{CaCO}_3$ .
Step 3: Calculate the molar mass of $\text{CaCO}_3$
$\text{Ca} = 40, \text{C} = 12, \text{O} = 16$
$\text{Molar mass} = 40 + 12 + (3 \times 16) = 40 + 12 + 48 = 100\text{ g/mol}$
Step 4: Calculate the mass of the precipitate
$\text{Mass} = \text{moles} \times \text{molar mass}$
$\text{Mass} = 0.1\text{ mol} \times 100\text{ g/mol} = 10\text{ g}$
4. Identify the Answer
The maximum weight of the solid precipitated is $10\text{ g}$ . This matches option A.
5. Tip
Always remember to convert your volume from $\text{cm}^3$ to $\text{dm}^3$ (by dividing by 1000) before calculating moles. Also, ensure your chemical equation is balanced before using the mole ratio to solve for the product!
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69.
2-methylprop-1-ene is an isomer of?
1. A.3-methyl but-1-ene
2. B.2-methyl but-1-ene
3. C.but-2-ene
4. D.pent-2-ene
Show answer
Answer: C. but-2-ene
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70.
Which of the following does not support the fact that air is a mixture?
1. A.the constituents of air are in a fixed proportion by mass
2. B.it cannot be represented with a chemical formula
3. C.the constituents of air can be separated by physical by physical means
4. D.none of the above
Show answer
Answer: A. the constituents of air are in a fixed proportion by mass
AI explanation
Analysis
To answer this question, we must understand the fundamental difference between a mixture and a compound.
A mixture consists of two or more substances physically combined, where the components retain their individual properties and can be present in varying proportions.
A compound is a substance formed when two or more elements are chemically bonded in a fixed proportion by mass.
Concept Explanation
Air as a mixture: Air is a mixture of gases (nitrogen, oxygen, argon, carbon dioxide, etc.). Because it is a mixture, it does not have a fixed chemical formula, and its components can be separated by physical methods like fractional distillation of liquid air.
Fixed proportion by mass: This is a characteristic property of compounds (Law of Definite Proportions). In a mixture, the components can vary in concentration depending on the location or conditions (e.g., air near a busy road has more carbon monoxide than air in a forest). Therefore, saying air has a "fixed proportion by mass" contradicts the definition of a mixture.
Step-by-Step Solution
Evaluate Option B: "It cannot be represented with a chemical formula." This is a true statement about mixtures. Since air is a mixture, this supports the fact that it is a mixture.
Evaluate Option C: "The constituents of air can be separated by physical means." This is a classic property of mixtures (e.g., fractional distillation). This also supports the fact that air is a mixture.
Evaluate Option A: "The constituents of air are in a fixed proportion by mass." This statement describes a compound, not a mixture. Because air is a mixture, its components are not chemically bonded in a fixed ratio. Therefore, this statement does not support the fact that air is a mixture.
Conclusion
Since Option A describes a property of a compound rather than a mixture, it is the correct answer.
Exam Tip: Always remember the acronym "M-P-S" for mixtures: Mixtures have Physical separation and Separate properties (the components don't change their identity). If you see "fixed proportion" or "chemical formula," you are likely looking at a compound!
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71.
Which of the following conditions will most enhance the spontaneity of a reaction?
1. A. $\Delta$ H is negative and greater than T $\Delta$ S
2. B. $\Delta$ H is negative and $\Delta$ S = 0
3. C. $\Delta$ H is positive and less than T $\Delta$ S
4. D. $\Delta$ H is positive and equal to T $\Delta$ S
Show answer
Answer: A. $\Delta$ H is negative and greater than T $\Delta$ S
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72.
The reaction $2H_2 S + SO_2 \to 3S + 2H_2 O$ is
1. A.not a redox reaction because there is no oxidant in the reaction
2. B.not a redox reaction because there is no reductant in the reaction
3. C.a redox reaction in which H $_2$ S is the oxidant and SO $_2$ is the reductant
4. D.a redox reaction in which SO $_2$ is the oxidant and H $_2$ S is the reductant
Show answer
Answer: D. a redox reaction in which SO $_2$ is the oxidant and H $_2$ S is the reductant
AI explanation
To determine the nature of this reaction, we need to analyze the oxidation states of the elements involved. A redox reaction occurs when there is a change in the oxidation state of atoms during a chemical process.
1. Analyze the oxidation states
Let's look at the oxidation states of the elements in the reactants and products:
In $H_2S$ : Hydrogen is $+1$ . Since the molecule is neutral, $2(+1) + S = 0$ , so Sulfur ( $S$ ) is $-2$ .
In $SO_2$ : Oxygen is $-2$ . Since the molecule is neutral, $S + 2(-2) = 0$ , so Sulfur ( $S$ ) is $+4$ .
In $3S$ (elemental sulfur): The oxidation state of an element in its standard state is $0$ .
In $H_2O$ : Hydrogen is $+1$ and Oxygen is $-2$ .
2. Identify the changes
Sulfur in $H_2S$ : Changes from $-2$ to $0$ . This is an increase in oxidation number, which is oxidation. The substance that undergoes oxidation is the reductant (reducing agent). Therefore, $H_2S$ is the reductant.
Sulfur in $SO_2$ : Changes from $+4$ to $0$ . This is a decrease in oxidation number, which is reduction. The substance that undergoes reduction is the oxidant (oxidizing agent). Therefore, $SO_2$ is the oxidant.
3. Conclusion
Since both oxidation and reduction have occurred, this is definitely a redox reaction. Specifically, $H_2S$ loses electrons (is oxidized) and $SO_2$ gains electrons (is reduced).
Comparing this to your options:
Option D states: "a redox reaction in which $SO_2$ is the oxidant and $H_2S$ is the reductant." This matches our findings perfectly.
Exam Tip
Remember the mnemonic OIL RIG:
Oxidation Is Loss (of electrons/increase in oxidation number).
Reduction Is Gain (of electrons/decrease in oxidation number). The substance that gets oxidized is the reducing agent, and the substance that gets reduced is the oxidizing agent.
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73.
The molecular shape and bond angle of water are respectively
1. A.linear, 180°
2. B.bent, 109.5°
3. C.tetrahedral, 109.5°
4. D.bent, 105°
Show answer
Answer: D. bent, 105°
AI explanation
Analysis
The question asks for the molecular geometry (shape) and the specific bond angle of a water molecule ( $H_2O$ ). To determine this, we look at the hybridization and the electron pair arrangement of the oxygen atom.
Concept: VSEPR Theory
The Valence Shell Electron Pair Repulsion (VSEPR) theory helps us predict the shape of molecules based on the repulsion between electron pairs around a central atom.
Electron Configuration: Oxygen has 6 valence electrons. In $H_2O$ , it forms two single covalent bonds with two hydrogen atoms.
Electron Pairs: Oxygen uses 2 electrons for bonding (bonding pairs) and retains 4 electrons as 2 lone pairs.
Geometry: According to VSEPR, 4 electron pairs (2 bonding + 2 lone) arrange themselves in a tetrahedral electron geometry to minimize repulsion.
Molecular Shape: Because the two lone pairs exert a stronger repulsive force than the bonding pairs, they "push" the hydrogen atoms closer together. This distortion changes the shape from tetrahedral to bent (or V-shaped).
Bond Angle: While a perfect tetrahedron has an angle of 109.5°, the strong repulsion from the two lone pairs on the oxygen atom compresses the $H-O-H$ bond angle to approximately 104.5° to 105°.
Step-by-Step Solution
Step 1: Identify the central atom (Oxygen).
Step 2: Count the electron domains. There are 2 bonding pairs and 2 lone pairs, totaling 4 domains.
Step 3: Determine the shape. Due to the 2 lone pairs, the molecule is not linear or tetrahedral; it is bent.
Step 4: Determine the angle. The repulsion from the lone pairs reduces the ideal tetrahedral angle (109.5°) to approximately 105°.
Comparing this to the options provided:
A is incorrect (that is the shape of $CO_2$ ).
B is incorrect (109.5° is the angle for a perfect tetrahedron like $CH_4$ ).
C is incorrect (the molecule is not tetrahedral in shape).
D matches our derivation.
Exam Tip
Always remember the "Lone Pair Rule": Lone pairs take up more space than bonding pairs. Every time you have a lone pair on the central atom, the bond angle will be slightly smaller than the "ideal" angle for that geometry (e.g., $NH_3$ is 107°, $H_2O$ is 105°).
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74.
The combustion of carbon(II)oxide in oxygen can be represented by equation: 2CO + O $_2$ → 2CO $_2$ Calculate the volume of the resulting mixture at the end of the reaction, if 50cm $^3$ of carbon(ii)oxide was exploded in 100cm $^3$ of oxygen
1. A.75cm $^3$
2. B.125cm $^3$
3. C.100cm $^3$
4. D.50cm $^3$
Show answer
Answer: B. 125cm $^3$
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75.
What mass of magnesium would be obtained by passing a current of 2 amperes for 2 hours 30 mins, through molten magnesium chloride? [1 faraday = 96500C, Mg = 24]
1. A.1.12g
2. B.4.48g
3. C.1.80g
4. D.2.00g
Show answer
Answer: D. 2.00g
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76.
Consider the equation below: Cr $_2$ O $_7 ^{2-}$ + 6Fe $^{2+}$ + 14H $^{+}$ $\to$ 2Cr $^{3+}$ + 6Fe $^{3+}$ + 7H $_2$ O. The oxidation number of chromium changes from
1. A.+5 to +3
2. B.+6 to +3
3. C.-2 to +3
4. D.+7 to +3
Show answer
Answer: B. +6 to +3
AI explanation
1. Analyze the Question
We are asked to determine the change in the oxidation number of chromium (Cr) as it moves from the reactant side to the product side in the given redox equation: Cr $_2$ O $_7^{2-}$ + 6Fe $^{2+}$ + 14H $^{+}$ $\to$ 2Cr $^{3+}$ + 6Fe $^{3+}$ + 7H $_2$ O
2. Explain the Concept
To find the oxidation number of an element in a polyatomic ion or compound, we use these rules:
The sum of oxidation numbers in a neutral compound is 0.
The sum of oxidation numbers in an ion is equal to the charge of the ion.
Oxygen (O) usually has an oxidation number of -2.
3. Show the Solution
Step 1: Find the oxidation number of Cr in the reactant (Cr $_2$ O $_7^{2-}$ ) Let the oxidation number of Cr be $x$ . Since there are two Cr atoms and seven O atoms, and the total charge is -2: $2(x) + 7(-2) = -2$ $2x - 14 = -2$ $2x = -2 + 14$ $2x = 12$ $x = +6$ So, the oxidation number of Cr in the reactant is +6.
Step 2: Find the oxidation number of Cr in the product (Cr $^{3+}$ ) In the product, chromium exists as a simple ion (Cr $^{3+}$ ). The oxidation number of a monatomic ion is simply its charge. Therefore, the oxidation number of Cr in the product is +3.
Step 3: Determine the change The oxidation number changes from +6 to +3.
4. Identify the Answer
Comparing our result to the options provided: A. +5 to +3 B. +6 to +3 C. -2 to +3 D. +7 to +3
The correct option is B.
5. Tip
When calculating oxidation numbers for complex ions like dichromate (Cr $_2$ O $_7^{2-}$ ), always remember to set the equation equal to the charge of the ion (-2), not zero. If you set it to zero, you will get the wrong answer. Also, remember that in any redox reaction, the species that undergoes a decrease in oxidation number (like Cr here, from +6 to +3) is the one being reduced.
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77.
The velocity, V of a gas is related to its mass, M by (k = proportionality constant)
1. A.V = $\frac{k}{M}$
2. B.V = $\frac{k}{M^{\frac{1}{2}}}$
3. C.V = $kM^2$
4. D.V = $(\frac{k}{M})^{\frac{1}{2}}$
Show answer
Answer: B. V = $\frac{k}{M^{\frac{1}{2}}}$
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78.
Which of the following describes the chemical property of acids?
1. A.None of the above
2. B.Acid + XCO $_3$ $\to$ Salt + H $_2$ O + NH $_3$
3. C.Acid + NaOH $\to$ Normal salt + CO $_2$
4. D.Acid + X $_n$ CO $_3$ $\to$ Salt + H $_2$ O + CO $_2$
Show answer
Answer: D. Acid + X $_n$ CO $_3$ $\to$ Salt + H $_2$ O + CO $_2$
AI explanation
Analysis of the Question
The question asks for the general chemical property of acids when they react with carbonates. In chemistry, acids undergo specific neutralization and displacement reactions. We need to identify the correct balanced chemical equation that represents the reaction between an acid and a metal carbonate.
Concept Explanation
When an acid reacts with a metal carbonate (represented here as $X_nCO_3$ , where $X$ is a metal), the reaction follows a standard pattern: Acid + Metal Carbonate $\to$ Salt + Water + Carbon(IV) oxide
Acid: Provides the hydrogen ions ( $H^+$ ).
Metal Carbonate: Provides the carbonate ion ( $CO_3^{2-}$ ).
Products: The metal from the carbonate combines with the anion of the acid to form a salt. The hydrogen from the acid combines with the oxygen from the carbonate to form water ( $H_2O$ ), and the remaining carbon and oxygen form carbon(IV) oxide ( $CO_2$ ).
Step-by-Step Solution
Let’s evaluate the options provided:
Option B: $Acid + XCO_3 \to Salt + H_2O + NH_3$ . This is incorrect because the reaction between an acid and a carbonate produces $CO_2$ , not ammonia ( $NH_3$ ).
Option C: $Acid + NaOH \to Normal salt + CO_2$ . This is incorrect because $NaOH$ is a base (alkali), not a carbonate. The reaction between an acid and a base (neutralization) produces only salt and water ( $Acid + Base \to Salt + Water$ ).
Option D: $Acid + X_nCO_3 \to Salt + H_2O + CO_2$ . This perfectly matches the general chemical principle described above. The " $n$ " in $X_nCO_3$ represents the valency of the metal $X$ , which is standard notation for chemical formulas.
Conclusion
The reaction between an acid and a metal carbonate consistently yields a salt, water, and carbon(IV) oxide. Option D is the only equation that correctly represents these products.
Exam Tip: Whenever you see a reaction involving a carbonate ( $CO_3$ ) and an acid, always look for $CO_2$ in the products. If you see $NH_3$ or other gases, you can immediately rule those options out!
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79.
The part of the total energy of a system that accounts for the useful work done in a system is known as
1. A.Enthalpy change
2. B.None of the above
3. C.Gibb's free energy
4. D.Entropy
Show answer
Answer: C. Gibb's free energy
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80.
The following are isoelectronic ions except
1. A.Na $^+$
2. B.Mg $^{2+}$
3. C.Si $^{2+}$
4. D.O $^{2+}$
Show answer
Answer: C. Si $^{2+}$
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2019 JAMB Chemistry Past Questions

Questions

Analysis

Concept

Step-by-Step Solution

Conclusion

Analysis of the Question

Concept Explanation

Step-by-Step Solution

Conclusion

Exam Tip

Analysis

Step-by-Step Calculation

Conclusion

Exam Tip

Analysis

Concept Explanation

Step-by-Step Solution

Conclusion

Analysis

Concept Explanation

Step-by-Step Solution

Correct Answer

Exam Tip

Analysis

Concept Explanation

Step-by-Step Solution

Correct Answer

Exam Tip

Analysis of the Question

Concept Explanation

Step-by-Step Evaluation

Conclusion

1. Analysis of the Clues

2. The Chemistry of Oxalic Acid (H2_22​C2_22​O4_44​)

3. Conclusion

Analysis

Concept Explanation

Step-by-Step Solution

Conclusion

Analysis of the Question

Concept Explanation

Step-by-Step Solution

Exam Tip

1. Analyze the Question

2. Explain the Concept

3. Show the Solution

4. Identify the Answer

5. Exam Tip

Analysis

Concept: Ester Nomenclature

Step-by-Step Solution

Conclusion

Exam Tip

1. Analyze the Question

2. Explain the Concept

3. Show the Solution

4. Identify the Answer

5. Tip

Analysis

Concept

Step-by-Step Solution

Conclusion

Analysis of the Question

Concept Explanation

Step-by-Step Evaluation

Conclusion

Exam Tip

1. Analyze the question

2. Explain the concept

3. Show the solution

4. Identify the answer

5. Tip

1. Analyze the question

2. Explain the concept

3. Show the solution

4. Identify the answer

5. Tip

Analysis

Concept Explanation

2. The Chemistry of Oxalic Acid (H $_2$ C $_2$ O $_4$ )