Methanoic acid mixes with water in all proportions and has about the same boiling point as water. Which of the following methods would you adopt to obtain pure water from a mixture of Sand, water and methanoic acid?

C. Neutralization with sodium hydroxide followed by distillation

B. Filtration followed by distillation

C. Neutralization with sodium hydroxide followed by distillation

D. Neutralization with sodium hydroxide followed by filtration

3H$_{2(g)}$+ N$_{2(g)}$⇔ 2NH$_{3(g)}$ ; H= -ve In the reaction above, lowering of temperature will

A. have no effect on the equilibrium position

B. increase the rate of both the forward and reverse reactions equally

Suitable reagents for the laboratory preparation of nitrogen are

A. sodium dioxonitrate(III) and ammonium chloride

B. sodium trioxonitrate(V) and ammonium chloride

C. sodium chloride and ammonium trioxonitrate(V)

D. sodium chloride and ammonium di-ozonitrate(III)

The presence of ammonia gas in a desiccator can exclusively be removed by

D. concentrated tetraoxosulphate (VI) acid

(I). 3CuO(s) + 2NH3(g) -----> 3Cu(s) + 3H2O(l) + N2(g) (II). 2NH3(g) + 3Cl2(g) -----> 6HCl(g) + N2(g) (III). 4NH3(g) + 3O2(g) -----> 6H2O(l) + N2(g) The reactions represented by the equations above demonstrate the

C. reducing properties of ammonia

B. acidic properties of ammonia

C. reducing properties of ammonia

D. oxidizing properties of ammonia

A sample of hard water contains some calcium sulphate and calcium hydrogen carbonate. The total hardness may therefore be removed by

B. adding excess calcium hydroxide

C. adding a calculated amount of calcium hydroxide

The situation obtained when a perfect gas expands into a vacuum is

D. ΔH is zero and T Δ S is positive

A. ΔH is positive and TΔ S is zero

B. ΔH is positive and T Δ S is negative

C. ΔH is negative and TΔ S is zero

D. ΔH is zero and T Δ S is positive

The choice of method for extracting a metal from its ores depends on the

B. position of the metal in the electrochemical series

D. position of the metal in the periodic table

The conductivity of an acid solution depends on the

A. amount of ions present and their mobilities

B. mobilities of the ions present

In the upper atmosphere, the ultra-violet light breaks off a free chlorine atom from chlorofluorocarbon molecule. The effect of this is that the free chlorine atom will

A. be very reactive and will attack ozone

B. be non-reactive and will not attack ozone

C. not change the level of ozone in the atmosphere

D. increase the level of ozone in the atmosphere

Which of the following statements is correct about the periodic table?

B. The valence electrons of the elements in the same period increase progressively across the period

A. Elements in the same period have the same number of valence electrons

B. The valence electrons of the elements in the same period increase progressively across the period

C. Elements in the same group have the same number of electron shells

D. The non-metallic Properties of the elements tend to decrease across each period

mE + nF -----> pG + qH In the equation shown, the equilibrium constant is given by?

C. $\frac{[G]^p[H]^q}{[E]^m[F]^n}$

A. $\frac{[E]^m[F]}{[G]^p[H]^q}$

C. $\frac{[G]^p[H]^q}{[E]^m[F]^n}$

In the laboratory preparation of oxygen, the gas cannot be collected by downward displacement of air because

B. the density of air is nearly the same as that of oxygen

A. the density of oxygen is greater than that of air

B. the density of air is nearly the same as that of oxygen

C. oxygen is a component of air

Elements P, Q, R, S have 6, 11, 15, 17 electrons respectively, therefore,

D. Q will form an electrovalent bond with S

A. P will form an electrovalent bond with R

B. Q will form a covalent bond with S

C. R will form an electrovalent bond with S

D. Q will form an electrovalent bond with S

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JAMB 2018 · Part 1 of 2

2018 JAMB Chemistry Past Questions

The 2018 JAMB Chemistry past questions with answers: 87 objective questions from the 2018 paper in the DeeepSolve question bank, each with its answer behind "Show answer". This is part 1 of 2: questions 1 to 44. 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 1–44

1.
Which of the following are mixtures? I. Petroleum II. Rubber latex III. Vulcanizer's solution IV. Carbon sulphide
1. A.I, II and III
2. B.I, II and IV
3. C.I and II only
4. D.I and IV
Show answer
Answer: A. I, II and III
AI explanation
To solve this question, we need to distinguish between pure substances (elements and compounds) and mixtures (physical combinations of substances that can be separated by physical means).
1. Analyze the components
I. Petroleum (Crude Oil): This is a complex mixture of various hydrocarbons (alkanes, cycloalkanes, and aromatic hydrocarbons) along with small amounts of sulfur, nitrogen, and oxygen compounds. It is separated into useful fractions (like petrol, kerosene, and diesel) through fractional distillation. It is a mixture.
II. Rubber Latex: This is a colloidal suspension of rubber particles in water, along with proteins, alkaloids, and resins. Because it contains various substances suspended in a liquid medium, it is a mixture.
III. Vulcanizer's Solution: This is a solution of rubber in a solvent like benzene or carbon disulfide. Since it involves a solute dissolved in a solvent, it is a mixture.
IV. Carbon Disulphide ( $CS_2$ ): This is a chemical compound consisting of one carbon atom and two sulfur atoms bonded together in a fixed ratio. Because it has a constant chemical composition and properties, it is a pure compound, not a mixture.
2. Evaluate the options
I, II, and III are all mixtures.
IV is a pure compound.
Therefore, the group that consists of mixtures is I, II, and III.
3. Conclusion
Comparing our analysis to the provided options, we find that Option A correctly identifies the mixtures.
Exam Tip
When you see questions about mixtures in Chemistry, remember:
Mixtures are usually "solutions," "suspensions," "colloids," or "crude" substances.
Compounds usually have a specific chemical formula (like $H_2O$ , $NaCl$ , or $CS_2$ ) and are formed by chemical bonding, not just physical mixing.
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2.
Methanoic acid mixes with water in all proportions and has about the same boiling point as water. Which of the following methods would you adopt to obtain pure water from a mixture of Sand, water and methanoic acid?
1. A.Extractive distillation
2. B.Filtration followed by distillation
3. C.Neutralization with sodium hydroxide followed by distillation
4. D.Neutralization with sodium hydroxide followed by filtration
Show answer
Answer: C. Neutralization with sodium hydroxide followed by distillation
AI explanation
1. Analyze the Question
We have a mixture containing three components: sand (insoluble solid), water (liquid), and methanoic acid (liquid). We need to obtain pure water. The challenge is that methanoic acid and water have similar boiling points, meaning simple distillation will not effectively separate them because they will vaporize together.
2. Explain the Concept
Filtration: This is used to separate insoluble solids from liquids. Sand is insoluble in water, so filtration will easily remove the sand.
Neutralization: Methanoic acid ( $HCOOH$ ) is an acid. If we add a base like sodium hydroxide ( $NaOH$ ), a neutralization reaction occurs: $HCOOH + NaOH \rightarrow HCOONa + H_2O$ This reaction produces sodium methanoate (a salt) and water.
Distillation: Once the acid is converted into a salt, the salt has a very high boiling point compared to water. Therefore, when you heat the mixture, only the water will vaporize, leaving the salt behind. This allows for the collection of pure water.
3. Show the Solution
Step 1: Filtration. First, pass the mixture through a filter paper. The sand will be trapped as residue, leaving a mixture of water and methanoic acid in the filtrate.
Step 2: Neutralization. Add sodium hydroxide ( $NaOH$ ) to the filtrate. The $NaOH$ reacts with the methanoic acid to form sodium methanoate ( $HCOONa$ ), which is a non-volatile salt dissolved in the water.
Step 3: Distillation. Now, perform distillation on the solution. Since sodium methanoate does not vaporize at the boiling point of water, pure water will distill over, leaving the salt in the flask.
4. Identify the Answer
Comparing our steps to the options:
Option A (Extractive distillation) is unnecessary and complex.
Option B (Filtration followed by distillation) would fail because the acid would still distill with the water.
Option C (Neutralization with sodium hydroxide followed by distillation) correctly addresses the need to remove the sand (implied in the process) and convert the acid into a non-volatile salt to allow for effective distillation.
Option D (Neutralization followed by filtration) is incorrect because the salt produced is soluble in water and would not be removed by filtration.
5. Exam Tip
When dealing with mixtures of liquids that have similar boiling points, look for a way to change the chemical properties of one component (like turning an acid into a salt) to make it non-volatile. This makes the subsequent distillation process much more effective!
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3.
3H $_{2(g)}$ + N $_{2(g)}$ ⇔ 2NH $_{3(g)}$ ; H= -ve In the reaction above, lowering of temperature will
1. A.have no effect on the equilibrium position
2. B.increase the rate of both the forward and reverse reactions equally
3. C.favour the forward reaction
4. D.favour the reverse reaction
Show answer
Answer: C. favour the forward reaction
AI explanation
Analysis
The question asks about the effect of temperature change on a chemical equilibrium system. Specifically, we are looking at the Haber process reaction: $3H_{2(g)} + N_{2(g)} \rightleftharpoons 2NH_{3(g)}$ The enthalpy change ( $\Delta H$ ) is negative, which means the reaction is exothermic (it releases heat).
Concept: Le Chatelier’s Principle
Le Chatelier’s Principle states that if a dynamic equilibrium is disturbed by changing the conditions (such as temperature, pressure, or concentration), the position of equilibrium shifts to counteract the change.
For temperature:
Exothermic reactions ( $\Delta H < 0$ ): Heat is treated as a product. Increasing the temperature shifts the equilibrium to the left (reverse), while lowering the temperature shifts the equilibrium to the right (forward) to produce more heat.
Endothermic reactions ( $\Delta H > 0$ ): Heat is treated as a reactant. Increasing the temperature shifts the equilibrium to the right (forward), while lowering the temperature shifts it to the left (reverse).
Step-by-Step Solution
Identify the nature of the reaction: The equation shows $\Delta H = -ve$ , confirming it is an exothermic reaction.
Treat heat as a component: We can rewrite the equation as: $3H_{2(g)} + N_{2(g)} \rightleftharpoons 2NH_{3(g)} + \text{Heat}$
Apply the change: The question states that the temperature is lowered.
Determine the shift: According to Le Chatelier’s Principle, the system will try to replace the heat that was removed. To do this, the equilibrium will shift in the direction that produces heat, which is the forward reaction.
Conclusion: By lowering the temperature, the system favours the production of ammonia ( $NH_3$ ), which is the forward reaction.
Evaluation of Options
A is incorrect because temperature changes always affect the equilibrium position of exothermic/endothermic reactions.
B is incorrect because while temperature affects rates, the equilibrium position specifically shifts to favour one side.
C is correct because lowering the temperature of an exothermic reaction shifts the equilibrium to the right (forward).
D is incorrect because that would happen if the temperature were increased.
Exam Tip
Think of an exothermic reaction as a "heater." If you turn down the temperature (the environment gets cold), the "heater" works harder to produce more heat to compensate. Therefore, it pushes the reaction forward.
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4.
If 1 litre of 2.2M sulphuric acid is poured into a bucket containing 10 litres of water and the resulting solution mixed thoroughly, the resulting sulphuric acid concentration will be
1. A.2.2M
2. B.1.1M
3. C.0.2M
4. D.0.11M
Show answer
Answer: C. 0.2M
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5.
Which of the following produces relatively few ions in solution?
1. A.NaOH
2. B.Ca(OH) $_2$
3. C.KOH
4. D.AI(OH) $_3$
Show answer
Answer: D. AI(OH) $_3$
AI explanation
Analysis
The question asks us to identify which of the given substances is a weak electrolyte (produces relatively few ions in solution). To solve this, we must distinguish between strong bases (which dissociate completely) and weak bases (which dissociate only partially).
Concept Explanation
Strong Bases: These are compounds that dissociate completely into their constituent ions when dissolved in water. Examples include Group 1 hydroxides (like NaOH and KOH) and some Group 2 hydroxides. Because they dissociate fully, they produce a high concentration of ions, making them strong electrolytes.
Weak Bases: These are compounds that do not dissociate completely in water. They exist in an equilibrium where only a small fraction of the molecules break apart into ions. Consequently, they produce relatively few ions in solution. Aluminum hydroxide, $\text{Al(OH)}_3$ , is a classic example of an insoluble or very weakly dissociating base.
Step-by-Step Solution
Evaluate NaOH (Option A): Sodium hydroxide is a Group 1 hydroxide. It is a strong base and dissociates completely: $\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-$ . It produces many ions.
Evaluate Ca(OH) $_2$ (Option B): Calcium hydroxide is a strong base (alkali). Although its solubility is limited, the portion that does dissolve dissociates completely into ions.
Evaluate KOH (Option C): Potassium hydroxide is also a Group 1 hydroxide. Like NaOH, it is a strong base that dissociates completely: $\text{KOH} \rightarrow \text{K}^+ + \text{OH}^-$ .
Evaluate Al(OH) $_3$ (Option D): Aluminum hydroxide is an amphoteric hydroxide. It is practically insoluble in water and does not dissociate significantly into ions. Therefore, it produces very few ions in solution compared to the strong bases listed.
Conclusion
Since $\text{Al(OH)}_3$ is a weak, poorly soluble base, it produces the fewest ions among the options provided.
Exam Tip
Remember the "Group 1 Rule": Hydroxides of Group 1 elements (Li, Na, K, Rb, Cs) are generally strong bases. If you see them in an exam question asking for "strong electrolytes," they are usually your best bet. Conversely, transition metal hydroxides (like $\text{Al(OH)}_3$ , $\text{Fe(OH)}_3$ , or $\text{Cu(OH)}_2$ ) are typically weak or insoluble.
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6.
A balanced chemical equation obeys the law of
1. A.conservation of mass
2. B.definite proportions
3. C.multiple proportions
4. D.conservation of energy
Show answer
Answer: A. conservation of mass
AI explanation
Analysis
The question asks for the fundamental scientific principle that necessitates balancing a chemical equation. In chemistry, we represent reactions using equations, and for these to be valid, they must account for every atom present before and after the reaction.
Concept Explanation
The Law of Conservation of Mass (formulated by Antoine Lavoisier) states that matter is neither created nor destroyed in a chemical reaction.
When you write a chemical equation, you are essentially creating a mathematical balance sheet. Because atoms cannot simply vanish or appear out of thin air, the total mass of the reactants must equal the total mass of the products. Balancing an equation ensures that the number of atoms of each element is the same on both sides of the arrow, which satisfies this law.
Option B (Definite Proportions) refers to the fact that a chemical compound always contains the same elements in the same proportion by mass.
Option C (Multiple Proportions) deals with two elements forming more than one compound.
Option D (Conservation of Energy) relates to the First Law of Thermodynamics, which is important in thermochemistry but is not the primary reason for balancing the stoichiometry of an equation.
Step-by-Step Solution
Examine the requirement: A balanced equation requires that the number of atoms of each element on the left (reactants) equals the number of atoms on the right (products).
Apply the principle: Since atoms have mass, having an equal number of atoms on both sides ensures that the total mass remains constant throughout the reaction.
Conclusion: This direct correspondence between the number of atoms and the mass of the substances confirms that the equation is governed by the Law of Conservation of Mass.
Exam Tip
Whenever you see a question about "balancing equations," immediately associate it with "conservation of mass." Think of it like a scale: if you put 10g of reactants on one side, you must have 10g of products on the other side. If the numbers don't match, the equation isn't "balanced," and the law is being violated.
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7.
Suitable reagents for the laboratory preparation of nitrogen are
1. A.sodium dioxonitrate(III) and ammonium chloride
2. B.sodium trioxonitrate(V) and ammonium chloride
3. C.sodium chloride and ammonium trioxonitrate(V)
4. D.sodium chloride and ammonium di-ozonitrate(III)
Show answer
Answer: A. sodium dioxonitrate(III) and ammonium chloride
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8.
The presence of ammonia gas in a desiccator can exclusively be removed by
1. A.phosphorus (V) oxide
2. B.fused calcium chloride
3. C.calcium oxide
4. D.concentrated tetraoxosulphate (VI) acid
Show answer
Answer: C. calcium oxide
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9.
(I). 3CuO(s) + 2NH3(g) -----> 3Cu(s) + 3H2O(l) + N2(g) (II). 2NH3(g) + 3Cl2(g) -----> 6HCl(g) + N2(g) (III). 4NH3(g) + 3O2(g) -----> 6H2O(l) + N2(g) The reactions represented by the equations above demonstrate the
1. A.basic properties of ammonia
2. B.acidic properties of ammonia
3. C.reducing properties of ammonia
4. D.oxidizing properties of ammonia
Show answer
Answer: C. reducing properties of ammonia
AI explanation
Analysis of the Question
To determine what these reactions demonstrate, we need to examine the role of ammonia ( $NH_3$ ) in each chemical equation. Specifically, we need to look at whether ammonia is gaining or losing electrons (or oxygen/hydrogen) in these processes.
Concept Explanation
Reducing Agent: A substance that causes reduction in another substance by donating electrons, adding hydrogen, or removing oxygen. In the process, the reducing agent itself gets oxidized (its oxidation state increases).
Oxidizing Agent: A substance that causes oxidation in another substance by accepting electrons, removing hydrogen, or adding oxygen. In the process, the oxidizing agent itself gets reduced.
Let's look at the oxidation state of Nitrogen in $NH_3$ : In $NH_3$ , the oxidation state of Nitrogen is -3. In the products ( $N_2$ ), the oxidation state of Nitrogen is 0. Since the oxidation state increases from -3 to 0, Nitrogen is being oxidized. Therefore, ammonia acts as a reducing agent.
Step-by-Step Solution
Let's analyze the reactions:
Reaction (I): $3CuO + 2NH_3 \rightarrow 3Cu + 3H_2O + N_2$ • Copper (II) oxide ( $CuO$ ) loses oxygen to become Copper ( $Cu$ ). This is reduction. • Ammonia ( $NH_3$ ) removes the oxygen from $CuO$ . Therefore, ammonia is the reducing agent.
Reaction (II): $2NH_3 + 3Cl_2 \rightarrow 6HCl + N_2$ • Chlorine ( $Cl_2$ ) gains hydrogen (or loses electrons) to become $HCl$ . • Ammonia provides the hydrogen and is oxidized to $N_2$ . Ammonia is the reducing agent.
Reaction (III): $4NH_3 + 3O_2 \rightarrow 6H_2O + 2N_2$ • Oxygen ( $O_2$ ) is reduced to water. • Ammonia is oxidized to nitrogen gas. Ammonia is the reducing agent.
In all three equations, ammonia is removing oxygen from a metal oxide, reacting with a non-metal (chlorine), or reacting with oxygen. In every case, ammonia is being oxidized, which confirms its role as a reducing agent.
Conclusion
Since ammonia is causing the other substances to be reduced while it is being oxidized, these reactions demonstrate the reducing properties of ammonia.
Exam Tip
Whenever you see ammonia reacting with metal oxides or halogens in JAMB chemistry, always look for the formation of Nitrogen gas ( $N_2$ ). The conversion of $NH_3$ to $N_2$ is a classic indicator that ammonia is acting as a reducing agent!
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10.
A sample of hard water contains some calcium sulphate and calcium hydrogen carbonate. The total hardness may therefore be removed by
1. A.boiling the water
2. B.adding excess calcium hydroxide
3. C.adding a calculated amount of calcium hydroxide
4. D.adding sodium carbonate
Show answer
Answer: D. adding sodium carbonate
AI explanation
1. Analyze the Question
The question asks for a method to remove the total hardness of water that contains both calcium sulphate ( $CaSO_4$ ) and calcium hydrogen carbonate ( $Ca(HCO_3)_2$ ).
2. Explain the Concept
Temporary Hardness: Caused by calcium hydrogen carbonate ( $Ca(HCO_3)_2$ ) or magnesium hydrogen carbonate. This can be removed by boiling.
Permanent Hardness: Caused by the presence of calcium sulphate ( $CaSO_4$ ) or magnesium sulphate. Boiling cannot remove this.
Total Hardness: The sum of both temporary and permanent hardness. To remove total hardness, you need a method that precipitates both the calcium ions ( $Ca^{2+}$ ) and the sulphate/carbonate ions effectively.
Chemical Precipitation: Adding sodium carbonate ( $Na_2CO_3$ ), also known as washing soda, is the standard industrial and laboratory method for removing total hardness. It reacts with both types of hardness to form insoluble calcium carbonate ( $CaCO_3$ ), which can be filtered out.
3. Show the Solution
Reaction with Temporary Hardness: $Ca(HCO_3)_2 + Na_2CO_3 \rightarrow CaCO_3(s) + 2NaHCO_3$
Reaction with Permanent Hardness: $CaSO_4 + Na_2CO_3 \rightarrow CaCO_3(s) + Na_2SO_4$
In both cases, the calcium ions are removed from the water as a solid precipitate ( $CaCO_3$ ), effectively softening the water. While calcium hydroxide (Option C) can remove temporary hardness and some permanent hardness (if magnesium is present), sodium carbonate is the universal reagent used to treat total hardness because it works efficiently for both types of calcium salts present in the sample.
4. Identify the Answer
Adding sodium carbonate is the most effective way to remove both temporary and permanent hardness simultaneously.
5. Exam Tip
Remember the "Carbonate Rule": If you see "hardness of water" in a JAMB question, look for sodium carbonate ( $Na_2CO_3$ ) or washing soda. It is the "magic" chemical that precipitates calcium and magnesium ions regardless of whether the hardness is temporary or permanent.
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11.
In the reaction between sodium hydroxide and sulphuric acid solutions, what volume of 0.5 molar sodium hydroxide would exactly neutralise 10cm3 of 1.25 molar sulphuric acid?
1. A.5cm $^3$
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 given a neutralization reaction between a base (Sodium hydroxide, NaOH) and an acid (Sulphuric acid, $H_2SO_4$ ). We need to find the volume of the base required to neutralize a specific volume and concentration of the acid.
2. Explain the Concept
To solve this, we use 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
First, we must write the balanced chemical equation:
$2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O$
From the equation:
$n_a = 1$ (coefficient of $H_2SO_4$ )
$n_b = 2$ (coefficient of $NaOH$ )
3. Show the Solution
Given data:
$C_a = 1.25 \, \text{mol/dm}^3$
$V_a = 10 \, \text{cm}^3$
$C_b = 0.5 \, \text{mol/dm}^3$
$V_b = ?$
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:
$12.5 = \frac{0.5 \times V_b}{2}$
Multiply by 2 to clear the fraction:
$25 = 0.5 \times V_b$
Solve for $V_b$ :
$V_b = \frac{25}{0.5}$
$V_b = 50 \, \text{cm}^3$
4. Identify the Answer
The volume of sodium hydroxide required is 50 cm $^3$ . This corresponds to option D.
5. Tip
Always remember to balance your chemical equation first! Many students forget the coefficients ( $n_a$ and $n_b$ ), which leads to incorrect answers. In neutralization reactions, the stoichiometry (the ratio of moles) is just as important as the concentration and volume.
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12.
What mass of Cu would be produced by the cathodic reduction of Cu $^{2+}$ when 1.60A of current passes through a solution of CuSO $_4$ for 1 hour. (F=96500Cmol $^{-1}$ , Cu=64)
1. A.7.64g
2. B.3.82g
3. C.1.91g
4. D.0.96g
Show answer
Answer: C. 1.91g
AI explanation
1. Analyze the question
We are asked to calculate the mass of Copper (Cu) deposited during electrolysis.
Current ( $I$ ): 1.60 A
Time ( $t$ ): 1 hour
Faraday's Constant ( $F$ ): 96,500 C/mol
Molar mass of Copper ( $M$ ): 64 g/mol
Reaction: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ (This tells us the number of moles of electrons, $n$ , is 2).
2. Explain the concept
To solve this, we use Faraday’s First Law of Electrolysis, which states that the mass of a substance deposited is proportional to the quantity of electricity passed. The formula is:
$m = \frac{M \times I \times t}{n \times F}$
Where:
$m$ = mass in grams
$M$ = molar mass
$I$ = current in Amperes
$t$ = time in seconds
$n$ = number of electrons transferred
$F$ = Faraday's constant
3. Show the solution
Step 1: Convert time to seconds. Since the formula requires time in seconds: $t = 1 \text{ hour} = 60 \times 60 = 3600 \text{ seconds}$
Step 2: Calculate the quantity of electricity ( $Q$ ). $Q = I \times t$ $Q = 1.60 \text{ A} \times 3600 \text{ s} = 5760 \text{ Coulombs}$
Step 3: Calculate the mass ( $m$ ) using the formula. $m = \frac{64 \times 5760}{2 \times 96500}$ $m = \frac{368640}{193000}$ $m \approx 1.91 \text{ g}$
4. Identify the answer
Comparing our result to the given options, the correct value is 1.91g.
Correct Option: C
5. Tip
Always remember to convert time from hours or minutes to seconds before plugging it into the Faraday formula. Also, check the valency of the metal ion (the charge) to determine the value of $n$ ; for $\text{Cu}^{2+}$ , $n=2$ , but for $\text{Ag}^+$ , $n=1$ . This is a common place where students lose marks!
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13.
The situation obtained when a perfect gas expands into a vacuum is
1. A.ΔH is positive and TΔ S is zero
2. B.ΔH is positive and T Δ S is negative
3. C.ΔH is negative and TΔ S is zero
4. D.ΔH is zero and T Δ S is positive
Show answer
Answer: D. ΔH is zero and T Δ S is positive
AI explanation
Analysis
This question tests your understanding of Thermodynamics, specifically the expansion of an ideal (perfect) gas into a vacuum. This process is known as Joule expansion or free expansion.
Concept Explanation
To solve this, we need to look at the First Law of Thermodynamics and the nature of an ideal gas:
Work Done ( $w$ ): When a gas expands into a vacuum, there is no external pressure ( $P_{ext} = 0$ ). Since work done $w = -P_{ext} \Delta V$ , it follows that $w = 0$ .
Internal Energy ( $\Delta U$ ): For an ideal gas, internal energy depends only on temperature. Since there is no work done and no heat exchange ( $q = 0$ ) with the surroundings (the system is isolated), the temperature remains constant. Therefore, $\Delta U = 0$ .
Enthalpy ( $\Delta H$ ): Enthalpy is defined as $H = U + PV$ . For an ideal gas, $PV = nRT$ . Since $T$ is constant and $U$ is constant, $\Delta H$ must also be zero.
Entropy ( $\Delta S$ ): Entropy is a measure of disorder. When a gas expands into a larger volume, the number of possible microstates increases, meaning the system becomes more disordered. Therefore, the entropy change ( $\Delta S$ ) is positive. Consequently, the term $T\Delta S$ is positive.
Step-by-Step Solution
Step 1: Identify the conditions of free expansion. Because the gas expands into a vacuum, no work is done ( $w = 0$ ) and no heat is exchanged ( $q = 0$ ).
Step 2: Apply the First Law of Thermodynamics ( $\Delta U = q + w$ ). Since $q=0$ and $w=0$ , then $\Delta U = 0$ .
Step 3: Relate $\Delta U$ to $\Delta H$ . For an ideal gas, $\Delta H = \Delta U + \Delta(PV)$ . Since $\Delta U = 0$ and $\Delta(PV) = \Delta(nRT) = 0$ (because $T$ is constant), then $\Delta H = 0$ .
Step 4: Determine the entropy change. Expansion into a vacuum is a spontaneous, irreversible process that increases the volume available to the gas molecules. This leads to an increase in entropy ( $\Delta S > 0$ ). Thus, $T\Delta S > 0$ .
Conclusion
Comparing our findings ( $\Delta H = 0$ and $T\Delta S > 0$ ) with the given options, we find that they match option D.
Exam Tip: Always remember that for an ideal gas, the internal energy and enthalpy are functions of temperature only. If the temperature doesn't change during an expansion, $\Delta U$ and $\Delta H$ will always be zero!
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14.
The Consecutive members of an alkane homologous series differ by
1. A.CH
2. B.CH $_2$
3. C.CH $_3$
4. D.C $_n$ H $_n$
Show answer
Answer: B. CH $_2$
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15.
The choice of method for extracting a metal from its ores depends on the
1. A.strength of the core
2. B.position of the metal in the electrochemical series
3. C.source of the core
4. D.position of the metal in the periodic table
Show answer
Answer: B. position of the metal in the electrochemical series
AI explanation
Analysis
The question asks for the primary factor that determines how a metal is extracted from its ore. In metallurgy, different metals have different affinities for oxygen and other non-metals, which dictates the energy required to isolate them.
Concept Explanation
The extraction of metals is based on their reactivity. The more reactive a metal is, the more strongly it holds onto its ore (usually oxides, sulfides, or carbonates), and the more energy is required to reduce it to its metallic state.
The Electrochemical Series (or Reactivity Series) ranks metals based on their tendency to lose electrons (oxidation).
Highly reactive metals (e.g., Potassium, Sodium, Calcium): These are at the top of the series. They hold onto their ores so strongly that they cannot be reduced by carbon or hydrogen. They must be extracted via electrolysis of their molten salts.
Moderately reactive metals (e.g., Zinc, Iron, Lead): These are in the middle of the series. They can be extracted by reduction with carbon or carbon monoxide in a blast furnace.
Less reactive metals (e.g., Copper, Silver, Gold): These are at the bottom of the series. They are often found in their native state or can be extracted by simple thermal decomposition or roasting.
Step-by-Step Solution
Evaluate the options: • A (Strength of the core): "Core" is not a standard metallurgical term for ore properties; this is irrelevant. • B (Position of the metal in the electrochemical series): This determines the metal's reactivity, which directly dictates whether electrolysis, chemical reduction, or thermal decomposition is needed. • C (Source of the core): While the location of the ore matters for mining, it does not dictate the chemical method of extraction. • D (Position of the metal in the periodic table): While reactivity trends exist in the periodic table, the electrochemical series is the specific tool used to predict the ease of reduction and the required extraction method.
Conclusion: The position of a metal in the electrochemical series provides the most accurate guide to its reactivity, which is the deciding factor for the extraction method.
Exam Tip
Remember the "Reactivity Ladder":
Top (High reactivity): Electrolysis (needs lots of electricity).
Middle (Moderate reactivity): Reduction with Carbon (needs heat).
Bottom (Low reactivity): Heat alone or naturally occurring (needs little to no effort).
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16.
The conductivity of an acid solution depends on the
1. A.amount of ions present and their mobilities
2. B.mobilities of the ions present
3. C.amount of ions present
4. D.temperature of the solution
Show answer
Answer: A. amount of ions present and their mobilities
AI explanation
Analysis
The question asks for the fundamental factors that determine how well an acidic solution conducts electricity. In chemistry, conductivity in an electrolyte (like an acid) is essentially the movement of charged particles (ions) through the solution.
Concept Explanation
To understand conductivity, think of it as "traffic flow" for electricity:
The Number of Ions: The more ions (charge carriers) you have in the solution, the more current can be carried. This is why strong acids (which ionize completely) conduct electricity better than weak acids (which only partially ionize).
The Mobility of Ions: Even if you have many ions, they must be able to move freely through the solvent to reach the electrodes. Mobility depends on the size of the ion, the viscosity of the solvent, and the temperature.
Conductivity ( $\kappa$ ) is mathematically related to the concentration of ions ( $c$ ) and their molar ionic conductivities ( $\lambda$ ), which represent how fast those specific ions move:
$\kappa = \sum c_i \cdot z_i \cdot F \cdot u_i$
(Where $c$ is concentration, $z$ is charge, $F$ is Faraday's constant, and $u$ is mobility).
Step-by-Step Solution
Evaluate Option C: While the amount of ions is important, it is not the only factor. If you have many ions but they are "sluggish" or trapped, conductivity will be low.
Evaluate Option B: Similarly, if you have very fast ions but only a tiny amount of them, the total conductivity will be low.
Evaluate Option A: This option combines both essential requirements. Conductivity is a product of the number of charge carriers (amount of ions) and how fast they can travel (their mobilities).
Evaluate Option D: While temperature does affect mobility, it is a secondary factor that influences the mobility mentioned in Option A, rather than being the primary definition of conductivity itself.
Conclusion
The conductivity of an acid solution is defined by the total concentration of ions available to carry the charge and the speed at which those specific ions can navigate the solution. Therefore, both factors must be considered.
Exam Tip
When you see questions about "conductivity," always look for the "two-part" answer: Quantity (how many) and Quality/Speed (how fast). This logic applies to both Chemistry (electrolytes) and Physics (current electricity).
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17.
In the upper atmosphere, the ultra-violet light breaks off a free chlorine atom from chlorofluorocarbon molecule. The effect of this is that the free chlorine atom will
1. A.be very reactive and will attack ozone
2. B.be non-reactive and will not attack ozone
3. C.not change the level of ozone in the atmosphere
4. D.increase the level of ozone in the atmosphere
Show answer
Answer: A. be very reactive and will attack ozone
AI explanation
Analysis
This question tests your understanding of environmental chemistry, specifically the depletion of the ozone layer by chlorofluorocarbons (CFCs). You need to understand the catalytic role of chlorine atoms in the stratosphere.
Concept Explanation
Chlorofluorocarbons (CFCs) are stable compounds at ground level. However, when they rise into the upper atmosphere (stratosphere), they are exposed to high-energy ultraviolet (UV) radiation. This radiation causes the CFC molecules to undergo photodissociation, which releases a highly reactive free chlorine atom ( $Cl\cdot$ ).
Once released, this chlorine atom acts as a catalyst in a cycle that destroys ozone ( $O_3$ ) molecules. The reaction typically follows this path:
$Cl + O_3 \rightarrow ClO + O_2$
$ClO + O \rightarrow Cl + O_2$
Because the chlorine atom is regenerated at the end of the cycle, a single chlorine atom can destroy thousands of ozone molecules, leading to the thinning of the ozone layer.
Step-by-Step Solution
Identify the trigger: UV light breaks the C-Cl bond in CFCs, creating a free radical chlorine atom.
Assess the nature of the atom: Free radicals are atoms or molecules with an unpaired electron, making them extremely chemically reactive.
Determine the interaction: Because the chlorine atom is highly reactive, it seeks to stabilize itself by reacting with nearby molecules. In the stratosphere, it encounters ozone ( $O_3$ ) and reacts with it to form chlorine monoxide ( $ClO$ ) and oxygen gas ( $O_2$ ).
Evaluate the options: • Option A correctly identifies that the chlorine atom is reactive and targets ozone. • Option B is incorrect because the atom is highly reactive. • Option C is incorrect because the reaction significantly reduces ozone levels. • Option D is incorrect because the process destroys ozone, it does not increase it.
Conclusion
The free chlorine atom is a highly reactive radical that initiates the catalytic destruction of the ozone layer. Therefore, it will attack ozone molecules.
Exam Tip: Whenever you see questions about CFCs and the atmosphere, remember the "Ozone Hole." The key mechanism is that CFCs provide the chlorine, and UV light provides the energy to start the chain reaction that destroys ozone.
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18.
The elements in the periodic table are listed in order of increasing
1. A.relative atomic number
2. B.atomic mass
3. C.relative isotopic mass
4. D.nuclear charge
Show answer
Answer: D. nuclear charge
AI explanation
Analysis
The question asks for the fundamental property that determines the arrangement of elements in the modern Periodic Table.
Concept Explanation
In the early days of chemistry, scientists like Dmitri Mendeleev arranged elements based on their atomic mass. However, this led to several inconsistencies (such as the placement of Tellurium and Iodine).
In 1913, Henry Moseley discovered that the properties of elements are a periodic function of their atomic number (the number of protons in the nucleus). Since the number of protons is equal to the number of electrons in a neutral atom, and the number of protons defines the nuclear charge, the modern Periodic Table is arranged by increasing atomic number, which corresponds directly to the increasing positive charge of the nucleus.
Step-by-Step Solution
Evaluate Option A: "Relative atomic number" is not a standard term; we simply use "atomic number."
Evaluate Option B: Atomic mass was the old method, which was replaced because it does not account for isotopes correctly.
Evaluate Option C: Relative isotopic mass refers to specific isotopes, not the elements themselves as they appear in the table.
Evaluate Option D: The atomic number of an element is defined by the number of protons in its nucleus. Since each proton carries a positive charge, the atomic number is effectively the nuclear charge. As you move from one element to the next in the periodic table, you are adding one proton, thereby increasing the nuclear charge by one unit.
Exam Tip
Always remember: Mendeleev = Atomic Mass, but Modern Periodic Law = Atomic Number (Nuclear Charge). If you see "Atomic Number" and "Nuclear Charge" as options, they are essentially the same thing, but "Nuclear Charge" is the more fundamental physical description.
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19.
How many atoms are present in 6.0g of magnesium? [Mg = 24, N.A = 6.02 x 10 $^{23}$ mol]
1. A.1.20 x 10 $^{22}$
2. B.2.4 x 10 $^{22}$
3. C.1.51 x 10 $^{23}$
4. D.3.02 x 10 $^{22}$
Show answer
Answer: C. 1.51 x 10 $^{23}$
AI explanation
Analysis
To find the number of atoms in a given mass of an element, we need to relate the mass to the number of moles, and then use Avogadro's constant ( $N_A$ ) to convert those moles into the actual number of atoms.
Concept
Calculate Moles ( $n$ ): The number of moles is calculated by dividing the given mass ( $m$ ) by the molar mass ( $M$ ) of the element. $ $n = \frac{m}{M}$ $
Calculate Atoms ( $N$ ): The number of atoms is the number of moles multiplied by Avogadro's constant ( $N_A$ ). $ $N = n \times N_A$ $
Step-by-Step Solution
Step 1: Identify the given values
Mass of Magnesium ( $m$ ) = 6.0 g
Molar Mass of Magnesium ( $M$ ) = 24 g/mol
Avogadro's constant ( $N_A$ ) = $6.02 \times 10^{23} \text{ mol}^{-1}$
Step 2: Calculate the number of moles of Magnesium
$n = \frac{6.0 \text{ g}}{24 \text{ g/mol}}$
$n = 0.25 \text{ moles}$
Step 3: Calculate the number of atoms
$N = 0.25 \times (6.02 \times 10^{23})$
$N = \frac{1}{4} \times 6.02 \times 10^{23}$
$N = 1.505 \times 10^{23}$
Rounding this to three significant figures, we get $1.51 \times 10^{23}$ .
Conclusion
Comparing our result to the given options, we can see that it matches option C.
Exam Tip
When dealing with mole calculations, always remember the "Mole Bridge": Mass $\leftrightarrow$ Moles $\leftrightarrow$ Number of Particles. If you are moving from mass to particles, you must pass through the "Mole" station first. Also, $0.25$ is the same as $1/4$ ; multiplying by $1/4$ is often much faster than using decimals during a timed exam!
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20.
An element X forms the following compounds with chlorine; XCl $_4$ , XCl $_3$ , XCl $_2$ . This illustrates the
1. A.law of multiple proportions
2. B.law of chemical proportions
3. C.law of simple proportions
4. D.law of conservation of mass
Show answer
Answer: A. law of multiple proportions
AI explanation
Analysis of the Question
The question asks us to identify the chemical law demonstrated when a single element (X) combines with another element (chlorine) to form more than one compound with different ratios of the elements.
Concept Explanation
The Law of Multiple Proportions states that when two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers.
In this scenario:
We have element X and element Cl.
They form three distinct compounds: XCl $_2$ , XCl $_3$ , and XCl $_4$ .
If we fix the mass of X at 1 mole, the ratios of the mass of chlorine in these compounds are 2:3:4.
Because these are small whole numbers, this is a classic illustration of the Law of Multiple Proportions.
Step-by-Step Solution
Identify the components: We have one element (X) reacting with another (Cl).
Observe the pattern: The element X is forming multiple compounds with the same second element (Cl).
Apply the definition: Whenever you see an element forming different compounds with another element (like CO and CO $_2$ , or in this case, XCl $_2$ , XCl $_3$ , and XCl $_4$ ), it is the definition of the Law of Multiple Proportions.
Evaluate the options: • Law of Multiple Proportions: Fits perfectly as it deals with multiple compounds formed by the same two elements. • Law of Chemical Proportions: This is a general term, not a specific scientific law. • Law of Simple Proportions: This is often confused with the Law of Definite Proportions, but it is not the standard name for this phenomenon. • Law of Conservation of Mass: This states that matter is neither created nor destroyed in a chemical reaction; it does not describe the formation of multiple compounds.
Conclusion
The correct option is A.
Exam Tip
Whenever you see a question involving two elements forming different compounds (e.g., $XY$ , $XY_2$ , $XY_3$ ), immediately look for "Multiple Proportions" in the options. It is a favorite topic for JAMB examiners!
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21.
The hydrogen ion concentration of a sample of orange juice is 2.0 X 10 $^{-11}$ moldm $^{-3}$ . What is its p $^{OH}$ ? [log102 = 0.3010]
1. A.3.30
2. B.13.00
3. C.10.70
4. D.11.30
Show answer
Answer: A. 3.30
AI explanation
1. Analyze the question
We are given the hydrogen ion concentration $[H^+]$ of a solution and asked to calculate the $pOH$ .
Given: $[H^+] = 2.0 \times 10^{-11} \, \text{mol dm}^{-3}$
Given: $\log_{10} 2 = 0.3010$
2. Explain the concept
To solve this, we use two fundamental relationships in acid-base chemistry:
pH formula: $pH = -\log_{10}[H^+]$
Relationship between pH and pOH: $pH + pOH = 14.00$ (at 25°C)
3. Show the solution
Step 1: Calculate the pH $pH = -\log_{10}(2.0 \times 10^{-11})$
Using the laws of logarithms ( $\log(a \times b) = \log a + \log b$ ): $pH = -(\log_{10} 2 + \log_{10} 10^{-11})$ $pH = -(0.3010 + (-11))$ $pH = -(0.3010 - 11)$ $pH = -(-10.699)$ $pH = 10.699 \approx 10.70$
Step 2: Calculate the pOH Using the relationship $pH + pOH = 14$ : $10.70 + pOH = 14.00$ $pOH = 14.00 - 10.70$ $pOH = 3.30$
4. Identify the answer
Comparing our result to the given options, the correct value is 3.30. This matches option A.
5. Tip
Always check if the question asks for pH or pOH. A common mistake in JAMB exams is calculating the pH and stopping there, forgetting that the question specifically asked for the pOH. Remember: $pH + pOH$ is always 14.00!
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22.
The knowledge of half-life can be used to
1. A.Create an element
2. B.Detect an element
3. C.to combine elements
4. D.Irradiate an element
Show answer
Answer: B. Detect an element
AI explanation
Analysis of the Question
This question tests your understanding of Radioactivity and the practical applications of Half-life in Chemistry and Physics. You are asked to identify the purpose of knowing the half-life of a radioactive substance.
Concept Explanation
The half-life ( $t_{1/2}$ ) of a radioactive isotope is the time required for exactly half of the radioactive nuclei in a sample to decay.
Because every radioactive isotope has a unique and constant half-life, this property acts like a "fingerprint." Scientists use this specific rate of decay to:
Identify/Detect elements: By measuring the rate at which a sample decays, scientists can determine which isotope is present.
Radiometric Dating: Determining the age of ancient artifacts (like carbon dating) or geological formations.
Medical Diagnosis: Choosing isotopes with the right half-life to track processes in the body without causing long-term radiation damage.
Step-by-Step Solution
Option A (Create an element): Creating elements (transmutation) is done through nuclear bombardment (e.g., in a particle accelerator), not by knowing a half-life.
Option B (Detect an element): Since each radioactive isotope decays at a specific, predictable rate (its half-life), scientists can measure the decay activity over time to identify what the element is. This is a standard method in analytical chemistry and nuclear physics.
Option C (Combine elements): Combining elements is a chemical reaction process involving valence electrons, which is unrelated to the nuclear decay process described by half-life.
Option D (Irradiate an element): Irradiation involves exposing a substance to radiation. While you need to know the half-life to manage the dosage, the half-life itself is a property of the material, not a tool used to perform the irradiation.
Conclusion
The most accurate application of the knowledge of half-life among the given options is the identification or detection of radioactive elements based on their unique decay rates.
Exam Tip: Whenever you see questions about "half-life," always associate it with time and identification. If you are ever asked about carbon dating in an exam, remember that it relies entirely on the half-life of Carbon-14!
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23.
The constituent common to duralumin and alnico is
1. A.Co
2. B.Mn
3. C.Al
4. D.Mg
Show answer
Answer: C. Al
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24.
In electrovalency, the oxidation number of the participating metal is always
1. A.decreasing
2. B.increasing
3. C.constant
4. D.neutral
Show answer
Answer: B. increasing
AI explanation
Analysis of the Question
This question tests your understanding of electrovalency (also known as ionic bonding) and the behavior of metal atoms during chemical reactions. You need to identify what happens to the oxidation state of a metal when it forms an ionic bond.
Concept Explanation
Electrovalency occurs when a metal atom transfers one or more electrons to a non-metal atom.
Metals have low ionization energies, meaning they easily lose their valence electrons to achieve a stable octet configuration (like a noble gas).
When a neutral metal atom (which has an oxidation number of 0) loses electrons, it becomes a positively charged ion (a cation).
Because the metal is losing negatively charged electrons, its oxidation number changes from 0 to a positive value (e.g., $Na \rightarrow Na^+ + e^-$ ).
In terms of oxidation numbers, moving from 0 to a positive value is mathematically an increase.
Step-by-Step Solution
Identify the starting state: A neutral metal atom in its elemental state has an oxidation number of 0.
Identify the process: In electrovalency, the metal loses electrons to form a cation.
Determine the change: When an atom loses electrons, it undergoes oxidation. Oxidation is defined as an increase in the oxidation number.
Evaluate the options: • Since the metal goes from 0 to a positive charge (e.g., $+1, +2, +3$ ), the oxidation number is increasing.
Conclusion
This aligns with the chemical principle that metals act as reducing agents by losing electrons, which results in an increase in their oxidation state.
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). Since metals lose electrons in electrovalency, they are always oxidized, meaning their oxidation number increases.
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25.
In the shown experiment (Fig. 1) the litmus paper will initially
1. A.be bleached
2. B.turn green
3. C.turn red
4. D.turn blue
Show answer
Answer: B. turn green
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26.
Diamond is a bad conductor of electricity because its bonding electrons are used in
1. A.crystal lattice formation
2. B.covalent bond formation
3. C.metallic bond formation
4. D.coordinate bond formation
Show answer
Answer: B. covalent bond formation
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27.
Which of the following statements is correct about the periodic table?
1. A.Elements in the same period have the same number of valence electrons
2. B.The valence electrons of the elements in the same period increase progressively across the period
3. C.Elements in the same group have the same number of electron shells
4. D.The non-metallic Properties of the elements tend to decrease across each period
Show answer
Answer: B. The valence electrons of the elements in the same period increase progressively across the period
AI explanation
To solve this question, let’s break down the fundamental organization of the Periodic Table.
1. Analyze the Question
The question asks us to identify the correct statement regarding the trends and properties of elements as they are arranged in the Periodic Table.
2. Explain the Concepts
Periods: These are the horizontal rows. Elements in the same period have the same number of electron shells. As you move from left to right across a period, the atomic number increases by one, meaning one proton and one electron are added to the atom. This results in the number of valence electrons increasing by one for each subsequent element.
Groups: These are the vertical columns. Elements in the same group have the same number of valence electrons, which is why they share similar chemical properties.
Metallic/Non-metallic Trends: As you move across a period from left to right, the metallic character decreases and the non-metallic character increases because the atoms become smaller and hold onto their electrons more tightly (higher electronegativity).
3. Evaluate the Options
Option A: Incorrect. Elements in the same group have the same number of valence electrons, not the same period.
Option B: Correct. As you move across a period (e.g., from Lithium to Neon), each element adds one electron to the valence shell. Therefore, the number of valence electrons increases progressively.
Option C: Incorrect. Elements in the same period have the same number of electron shells, not the same group.
Option D: Incorrect. Non-metallic properties actually increase across a period from left to right, they do not decrease.
4. Conclusion
Based on the periodic trends, the number of valence electrons increases by one as you move from one element to the next across a period. This confirms that Option B is the accurate statement.
Exam Tip
Remember the "Group vs. Period" rule:
Group = Grouping of valence electrons (same number).
Period = Physical shells (same number of shells). If you get confused, just look at the first two elements of Period 2: Lithium (Group 1, 1 valence electron) and Beryllium (Group 2, 2 valence electrons). This proves the number increases across the period!
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28.
The alkanoic acid found in human sweat is
1. A.CH $_3$ -COOH
2. B.CH $_3$ —CH $_2$ —COOH
3. C.H-COOH
4. D.CH $_3$
Show answer
Answer: B. CH $_3$ —CH $_2$ —COOH
AI explanation
Analysis of the Question
The question asks for the specific alkanoic acid (carboxylic acid) that is present in human sweat. This is a factual recall question based on the properties of organic compounds commonly found in biological systems.
Concept Explanation
Alkanoic acids (also known as carboxylic acids) have the general formula C $_n$ H $_{2n+1}$ COOH.
Methanoic acid (H-COOH): Found in ant stings.
Ethanoic acid (CH $_3$ COOH): Commonly known as acetic acid (vinegar).
Propanoic acid (CH $_3$ CH $_2$ COOH): This is a three-carbon acid. It is a metabolic byproduct and is known to be one of the organic acids present in human sweat, contributing to its characteristic odor.
Step-by-Step Solution
Evaluate Option A (CH $_3$ COOH): This is ethanoic acid. While it is a common organic acid, it is not the primary acid associated with the composition of human sweat.
Evaluate Option B (CH $_3$ CH $_2$ COOH): This is propanoic acid. Scientific literature identifies propanoic acid as one of the volatile fatty acids present in human sweat.
Evaluate Option C (H-COOH): This is methanoic acid (formic acid), which is primarily associated with insect stings.
Evaluate Option D (CH $_3$ ): This is a methyl group, not an acid. It is an alkyl radical and does not represent an alkanoic acid.
Correct Answer
The correct option is B.
Exam Tip
When dealing with organic chemistry questions about "common sources" of compounds:
Ants/Stings $\rightarrow$ Methanoic acid (Formic acid).
Vinegar $\rightarrow$ Ethanoic acid (Acetic acid).
Sweat/Body Odor $\rightarrow$ Propanoic acid.
Rancid Butter $\rightarrow$ Butanoic acid.
Memorizing these common associations will save you time during your JAMB exam!
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29.
Which of the following separation techniques can be employed in obtaining solvent from its solution?
1. A.Evaporation
2. B.Distillation
3. C.Filtration
4. D.Precipitation
Show answer
Answer: B. Distillation
AI explanation
Analysis
The question asks for a technique used to recover a solvent (the liquid part of a solution) from a solution. In a solution, the solute is dissolved in the solvent, and we need to separate them such that the solvent is collected rather than lost to the atmosphere.
Concept Explanation
To choose the right technique, we must understand what happens in each process:
Evaporation: This involves heating a solution until the solvent turns into vapor and escapes into the atmosphere. While this leaves the solute behind, the solvent is lost.
Distillation: This is a process of heating a liquid to form vapor and then cooling the vapor back into a liquid (condensation). This allows you to collect the pure solvent in a separate container (the distillate).
Filtration: This is used to separate an insoluble solid from a liquid. It cannot separate a dissolved solute from a solvent.
Precipitation: This is a chemical process where a dissolved substance is converted into an insoluble solid (precipitate). It is not a standard physical separation technique for recovering a solvent.
Step-by-Step Solution
Identify the goal: We need to recover the solvent.
Evaluate the options: • Evaporation loses the solvent to the air. • Filtration only works for mixtures, not solutions. • Precipitation changes the chemical state of the solute.
Analyze Distillation: In simple distillation, the solution is boiled. The solvent (which has a lower boiling point than the solute) vaporizes. The vapor travels through a condenser, where it is cooled and turns back into a liquid. This liquid is then collected. This is the standard laboratory method for obtaining a pure solvent from a solution.
Conclusion
Distillation is the only technique listed that allows for the recovery of the solvent in liquid form.
Exam Tip: Always remember the difference between Evaporation and Distillation: If the question asks to recover the solute (like salt from saltwater), use evaporation. If the question asks to recover the solvent (like pure water from saltwater), use distillation.
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30.
mE + nF -----> pG + qH In the equation shown, the equilibrium constant is given by?
1. A. $\frac{[E]^m[F]}{[G]^p[H]^q}$
2. B. $\frac{[E][F]}{[G][H]}$
3. C. $\frac{[G]^p[H]^q}{[E]^m[F]^n}$
4. D. $\frac{[G][H]}{[E][F]}$
Show answer
Answer: C. $\frac{[G]^p[H]^q}{[E]^m[F]^n}$
AI explanation
Analysis of the Question
This question tests your understanding of the Law of Mass Action as it applies to chemical equilibrium. You are asked to derive the expression for the equilibrium constant ( $K_c$ ) for a general reversible chemical reaction.
Concept Explanation
The equilibrium constant ( $K_c$ ) for a reversible reaction is defined as the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants, with each concentration raised to the power of its stoichiometric coefficient from the balanced chemical equation.
For a general reaction:
$aA + bB \rightleftharpoons cC + dD$
The equilibrium constant expression is:
$K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$
[ ] represents the molar concentration of the species.
The letters a, b, c, and d are the stoichiometric coefficients (the numbers in front of the chemical formulas).
Step-by-Step Solution
Identify the products and reactants: In the equation $mE + nF \rightleftharpoons pG + qH$ , the products are G and H, and the reactants are E and F.
Identify the coefficients: The coefficients are p and q for the products, and m and n for the reactants.
Apply the formula: • Place the products in the numerator: $[G]^p [H]^q$ • Place the reactants in the denominator: $[E]^m [F]^n$
Combine them: $ $K_c = \frac{[G]^p [H]^q}{[E]^m [F]^n}$ $
Comparing this result to your options, it matches option C.
Exam Tip
Always remember: "Products over Reactants." A common mistake students make is flipping the fraction or forgetting to raise the concentrations to the power of their coefficients. If you see a coefficient in the equation, it must appear as an exponent in your $K_c$ expression!
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31.
The refreshing and characteristic taste of soda water and other soft drinks is as a result of the presence of
1. A.carbon(IV)oxide
2. B.carbon(ll)oxide
3. C.soda
4. D.glucose
Show answer
Answer: A. carbon(IV)oxide
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32.
If one of the following oxides is heated with hydrogen or carbon using a bunsen burner. it is not reduced to the metal, Which one is it?
1. A.lead oxide
2. B.Magnesium oxide
3. C.Copper oxide
4. D.Tin oxide
Show answer
Answer: B. Magnesium oxide
AI explanation
Analysis
The question asks which of the listed metal oxides cannot be reduced to its metal form when heated with common reducing agents like hydrogen ( $H_2$ ) or carbon ( $C$ ). This is a test of your knowledge of the Reactivity Series of Metals.
Concept: The Reactivity Series
In chemistry, the ability of a metal oxide to be reduced by carbon or hydrogen depends on the position of the metal in the reactivity series:
Low/Moderate Reactivity: Metals like Copper, Lead, and Tin are relatively low in the reactivity series. Their oxides are easily reduced by carbon or hydrogen because the carbon or hydrogen has a higher affinity for oxygen than these metals do.
High Reactivity: Metals like Potassium, Sodium, Calcium, and Magnesium are very high in the reactivity series. They have a very strong affinity for oxygen. Because they form very stable oxides, carbon and hydrogen are not strong enough reducing agents to "pull" the oxygen away from them.
Step-by-Step Solution
Copper oxide ( $CuO$ ): Copper is low in the reactivity series. It is easily reduced by hydrogen: $CuO + H_2 \rightarrow Cu + H_2O$ .
Lead oxide ( $PbO$ ): Lead is moderately reactive. It can be reduced by carbon: $PbO + C \rightarrow Pb + CO$ .
Tin oxide ( $SnO_2$ ): Tin is also moderately reactive and can be reduced by carbon or hydrogen.
Magnesium oxide ( $MgO$ ): Magnesium is very high in the reactivity series (it is an alkaline earth metal). Its bond with oxygen is extremely strong. Heating $MgO$ with carbon or hydrogen will not result in the formation of magnesium metal because magnesium's affinity for oxygen is much greater than that of carbon or hydrogen.
Conclusion
Since Magnesium is a highly reactive metal, its oxide is stable and cannot be reduced by simple heating with carbon or hydrogen in a laboratory setting.
Exam Tip
Remember the mnemonic for the reactivity series: "Please Stop Calling Me A Careless Zebra Instead Try Learning How Copper Saves Gold." (Potassium, Sodium, Calcium, Magnesium, Aluminium, Carbon, Zinc, Iron, Tin, Lead, Hydrogen, Copper, Silver, Gold). Any metal above Carbon/Hydrogen in this list generally cannot be reduced by them!
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33.
The lUPAC name for
1. A.1-methyl pentane
2. B.3-methylbutane
3. C.2-methylbutane
4. D.1-dimethyl propane
Show answer
Answer: B. 3-methylbutane
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34.
In the laboratory preparation of oxygen, the gas cannot be collected by downward displacement of air because
1. A.the density of oxygen is greater than that of air
2. B.the density of air is nearly the same as that of oxygen
3. C.oxygen is a component of air
4. D.air is an impure substance
Show answer
Answer: B. the density of air is nearly the same as that of oxygen
AI explanation
Analysis
The question asks why oxygen gas, when prepared in the laboratory, is typically collected over water rather than by the downward displacement of air. We need to compare the physical properties of oxygen gas and air.
Concept Explanation
In the laboratory, gases are collected based on their physical properties:
Upward displacement of air: Used for gases that are less dense than air (e.g., Ammonia, Hydrogen).
Downward displacement of air: Used for gases that are denser than air (e.g., Chlorine, Hydrogen Chloride).
Over water: Used for gases that are sparingly soluble in water (e.g., Oxygen, Hydrogen).
When we talk about "downward displacement of air," we are essentially saying the gas is heavy enough to push the air out of the top of the collection jar. For this to be effective, there must be a significant difference in density between the gas being collected and the air.
Step-by-Step Solution
Density of Oxygen: The molar mass of oxygen ( $O_2$ ) is approximately $16 \times 2 = 32 \text{ g/mol}$ .
Density of Air: Air is a mixture of gases (mostly Nitrogen and Oxygen). The average molar mass of air is approximately $28.8 \text{ g/mol}$ .
Comparison: While oxygen is slightly denser than air, the difference is very small. Because their densities are so close, oxygen does not displace air efficiently. Instead of pushing the air out, the oxygen tends to mix with the air due to diffusion.
Conclusion: Because the densities are nearly identical, the gas would not be collected in a pure state using the downward displacement of air method.
Identifying the Answer
Looking at the options provided:
A is incorrect because, while oxygen is slightly denser, the difference is not significant enough to facilitate clean displacement.
B is correct because the similarity in density causes the gases to mix rather than separate, making the method ineffective.
C and D are true statements about air, but they do not explain the physical limitation of the collection method.
Exam Tip
When you see questions about gas collection, always remember the "Density Rule":
If Molar Mass of gas > 29 (approx. density of air), it can be collected by downward displacement of air.
If Molar Mass of gas < 29, it is collected by upward displacement of air.
If the Molar Mass is very close to 29 (like Oxygen at 32), the displacement method is inefficient, so we use water collection instead!
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35.
Calculate the percentage composition of oxygen in calcium trioxocarbonate(IV) [Ca=40, C=12, O=16]
1. A.16
2. B.48
3. C.40
4. D.12
Show answer
Answer: B. 48
AI explanation
1. Analyze the Question
We need to find the percentage by mass of oxygen in calcium trioxocarbonate(IV), which is commonly known as calcium carbonate ( $CaCO_3$ ). We are given the relative atomic masses: $Ca = 40$ , $C = 12$ , and $O = 16$ .
2. Explain the Concept
To calculate the percentage composition of an element in a compound, we use the formula:
$\text{Percentage composition} = \left( \frac{\text{Total mass of the element in the compound}}{\text{Molar mass of the compound}} \right) \times 100\%$
3. Show the Solution
Step 1: Calculate the molar mass of $CaCO_3$
$Ca = 40$
$C = 12$
$O = 16 \times 3 = 48$ (since there are 3 atoms of oxygen)
$\text{Molar mass} = 40 + 12 + 48 = 100 \text{ g/mol}$
Step 2: Calculate the percentage of oxygen The total mass of oxygen in the compound is $48$ .
$\text{Percentage of Oxygen} = \left( \frac{48}{100} \right) \times 100\%$
$\text{Percentage of Oxygen} = 48\%$
Note: While the question asks for the percentage composition, the options provided (16, 48, 40, 12) represent the numerical value of the mass contribution of oxygen in the formula unit. Based on the calculation, 48 is the correct value.
4. Identify the Answer
Comparing our result to the options provided, the correct value is 48.
5. Exam Tip
When calculating percentage composition, always double-check the number of atoms of the element in the chemical formula. A common mistake is forgetting to multiply the atomic mass of oxygen by its subscript (in this case, 3). Also, remember that the sum of the percentages of all elements in a compound must always equal 100%.
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36.This question needs a diagram, table or passage that we don't have yet, so it isn't shown.
37.
According to the Kinetic Theory an increase in temperature causes the kinetic energy of particles to
1. A.decrease
2. B.increase
3. C.be zero
4. D.remain constant
Show answer
Answer: B. increase
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38.
Which of the following is used to power steam engines?
1. A.lubricating oil
2. B.coal
3. C.bitumen
4. D.diesel oil
Show answer
Answer: B. coal
AI explanation
Analysis
The question asks for the primary fuel source historically and technically used to generate steam in steam engines. To answer this, we must look at the process of steam generation and the energy sources derived from petroleum and coal.
Concept Explanation
A steam engine works by heating water in a boiler to produce high-pressure steam. This steam then expands to push a piston or turn a turbine, converting thermal energy into mechanical work.
Coal: This is a solid fossil fuel composed mainly of carbon. When burned, it releases a significant amount of heat energy, making it the traditional and most efficient fuel for heating large quantities of water in boilers for steam engines (such as steam locomotives).
Lubricating oil: This is used to reduce friction between moving parts of an engine, not as a fuel source.
Bitumen: This is a thick, sticky form of petroleum used primarily for road surfacing and roofing. It is not suitable as a fuel for steam engines.
Diesel oil: This is used in internal combustion engines (diesel engines), where the fuel is burned inside the cylinder to create pressure, rather than to heat water for steam.
Step-by-Step Solution
Identify the requirement: We need a fuel capable of sustained, high-heat combustion to boil water.
Evaluate the options: • A (Lubricating oil): Incorrect. It is a lubricant, not a fuel. • B (Coal): Correct. Historically, coal was the primary fuel used to fire the boilers of steam engines. • C (Bitumen): Incorrect. It is too viscous and heavy for this purpose. • D (Diesel oil): Incorrect. While it is a fuel, it is specifically designed for internal combustion engines, not external combustion steam engines.
Conclusion: Coal provides the thermal energy required to convert water into steam in a boiler system.
Exam Tip
When you see questions about fossil fuels in Chemistry, remember their primary uses:
Coal: Heating and electricity generation (steam).
Bitumen: Road construction.
Diesel/Petrol: Internal combustion engines.
Kerosene: Domestic heating and lighting.
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39.
An aqueous solution of a metal salt, M. gives a white precipitate with NaOH which dissolves in excess NaOH. With aqueous ammonia, the solution of M also gives a white precipitate which dissolves in excess ammonia Therefore the cation in M is
1. A.Zn
2. B.Ca
3. C.Al
4. D.Pb
Show answer
Answer: A. Zn
AI explanation
Analysis of the Question
This question tests your knowledge of qualitative analysis, specifically the reactions of metal cations with sodium hydroxide (NaOH) and aqueous ammonia ( $NH_3$ ). We need to identify a cation that forms a white precipitate with both reagents and, crucially, is soluble in the excess of both reagents.
Concept Explanation
In qualitative analysis, we use reagents like NaOH and $NH_3$ to identify metal ions based on the solubility of their hydroxides.
Reaction with NaOH: Many metal ions form white precipitates (hydroxides) with NaOH. However, only a few are amphoteric (meaning they dissolve in excess NaOH to form complex ions). Common examples include $Zn^{2+}$ , $Al^{3+}$ , and $Pb^{2+}$ .
Reaction with Aqueous Ammonia: When $NH_3$ is added, it acts as a weak base. Most metal hydroxides precipitate. The key differentiator here is the ability to form a soluble ammine complex. • $Al^{3+}$ : Forms a white precipitate that is insoluble in excess ammonia. • $Pb^{2+}$ : Forms a white precipitate that is insoluble in excess ammonia. • $Zn^{2+}$ : Forms a white precipitate ( $Zn(OH)_2$ ) that dissolves in excess ammonia to form a colorless complex ion, $[Zn(NH_3)_4]^{2+}$ .
Step-by-Step Solution
Step 1: Identify the cation that forms a white precipitate with NaOH. All the options ( $Zn^{2+}$ , $Ca^{2+}$ , $Al^{3+}$ , $Pb^{2+}$ ) form white precipitates with NaOH.
Step 2: Check solubility in excess NaOH. $Zn^{2+}$ , $Al^{3+}$ , and $Pb^{2+}$ are soluble in excess NaOH. $Ca^{2+}$ is not.
Step 3: Check solubility in excess aqueous ammonia. • $Al^{3+}$ : Precipitate remains (insoluble). • $Pb^{2+}$ : Precipitate remains (insoluble). • $Zn^{2+}$ : The precipitate dissolves in excess ammonia due to the formation of the tetraamminezinc(II) ion: $Zn(OH)_{2(s)} + 4NH_{3(aq)} \rightarrow [Zn(NH_3)_4]^{2+}_{(aq)} + 2OH^-_{(aq)}$
Since only $Zn^{2+}$ satisfies both conditions (dissolving in excess NaOH and excess ammonia), it is the correct cation.
Conclusion
The cation that exhibits these specific amphoteric and complex-forming properties is Zinc ( $Zn^{2+}$ ).
Exam Tip
Remember the "Zinc-Copper-Silver" rule for ammonia: These three metals form soluble complexes with excess ammonia. If you see a question about a precipitate dissolving in excess ammonia, always look for these transition metals!
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40.
The periodic classification is an arrangement of the elements
1. A.atomic weights
2. B.isotopic weights
3. C.molecular weights
4. D.atomic numbers
Show answer
Answer: D. atomic numbers
AI explanation
1. Analyze the question
The question asks for the fundamental property used to arrange elements in the modern Periodic Table. You need to distinguish between the historical method (Mendeleev's approach) and the modern method (Moseley's approach).
2. Explain the concept
Historical Context: Early chemists like Dmitri Mendeleev arranged elements based on increasing atomic weights (or atomic masses). However, this led to several inconsistencies where elements did not fit perfectly into groups with similar chemical properties.
Modern Periodic Law: In 1913, Henry Moseley discovered that the properties of elements are a periodic function of their atomic numbers (the number of protons in the nucleus). This discovery corrected the flaws in the earlier tables and is the basis for the modern Periodic Table we use today.
3. Show the solution
Option A (Atomic weights): This was used in the early versions of the periodic table but was found to be inaccurate for predicting chemical behavior.
Option B (Isotopic weights): Isotopes of the same element have different weights but identical chemical properties. Arranging by isotopic weights would be chaotic and unscientific.
Option C (Molecular weights): Molecular weights apply to compounds or molecules, not individual elements in the periodic table.
Option D (Atomic numbers): The modern Periodic Table is strictly organized by increasing atomic number ( $Z$ ). This ensures that elements with the same number of valence electrons (which determine chemical reactivity) fall into the same vertical groups.
4. Identify the answer
The correct basis for the periodic classification of elements is the atomic number.
5. Exam Tip
Memory Trick: Remember "Moseley = Modern = Number."
Moseley (the scientist) gave us the Modern table based on the atomic Number.
If you see a question about the original or early periodic table, the answer is Atomic Weight. If it refers to the modern periodic table, the answer is always Atomic Number.
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41.
The derivative of benzene that can be used in making explosives is
1. A.aniline
2. B.xylene
3. C.phenol
4. D.toluene
Show answer
Answer: D. toluene
AI explanation
1. Analyze the Question
The question asks for a specific derivative of benzene that is commonly used in the production of explosives. We need to look at the chemical properties of the options provided to see which one is a precursor to a well-known explosive.
2. Explain the Concept
Benzene derivatives are aromatic compounds where one or more hydrogen atoms on the benzene ring are replaced by other functional groups.
In the context of explosives, the most famous derivative is Trinitrotoluene (TNT). TNT is produced by the nitration of toluene (methylbenzene). The methyl group ( $–CH_3$ ) on the benzene ring makes the ring more reactive toward electrophilic substitution (like nitration), allowing for the addition of three nitro ( $–NO_2$ ) groups to form the explosive compound.
3. Show the Solution
Aniline ( $C_6H_5NH_2$ ): Primarily used in the dye and rubber industries.
Xylene ( $C_6H_4(CH_3)_2$ ): Used as a solvent and in the production of polymers like polyester.
Phenol ( $C_6H_5OH$ ): Used in the production of plastics, resins, and as an antiseptic. While it can be nitrated to form picric acid (which is an explosive), it is not the primary industrial precursor associated with the term "derivative of benzene" in the context of standard JAMB chemistry questions regarding TNT.
Toluene ( $C_6H_5CH_3$ ): This is the direct precursor to Trinitrotoluene (TNT). The process involves reacting toluene with concentrated nitric acid in the presence of concentrated sulfuric acid.
Because TNT is one of the most widely recognized explosives derived from a benzene derivative, toluene is the correct answer.
4. Identify the Answer
The correct option is D.
5. Exam Tip
When you see questions about benzene derivatives and explosives, always look for Toluene. Remember the acronym TNT (Trinitrotoluene)—the "T" stands for Toluene! This will help you quickly identify the correct answer in your exam.
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42.
A given amount of gas occupies 10.0dm $^3$ at 4 atm and 273°C. The number of moles of the gas present is? [Molar volume of gas at s.t.p = 22.4 dm $^3$ ]
1. A.0.89 mol
2. B.1.90 mol
3. C.3.80 mol
4. D.5.70 mol
Show answer
Answer: A. 0.89 mol
AI explanation
Analysis
To solve this problem, we need to determine the number of moles ( $n$ ) of a gas given its volume ( $V$ ), pressure ( $P$ ), and temperature ( $T$ ). We will use the Ideal Gas Law equation:
$PV = nRT$
Concept
The Ideal Gas Law relates the pressure, volume, temperature, and the number of moles of a gas.
Pressure ( $P$ ): 4 atm
Volume ( $V$ ): 10.0 dm³
Temperature ( $T$ ): Must be in Kelvin. $T(K) = 273 + 273 = 546 \text{ K}$
Gas Constant ( $R$ ): Since pressure is in atm and volume is in dm³, we use $R = 0.0821 \text{ dm}^3\text{atm K}^{-1}\text{mol}^{-1}$
Alternatively, we can use the relationship between the given state and Standard Temperature and Pressure (STP) using the Combined Gas Law, but the Ideal Gas Law is more direct here.
Solution
Step 1: Convert temperature to Kelvin
$T = 273^\circ\text{C} + 273 = 546 \text{ K}$
Step 2: Apply the Ideal Gas Law formula
$PV = nRT$
$n = \frac{PV}{RT}$
Step 3: Substitute the values
$n = \frac{4 \times 10}{0.0821 \times 546}$
$n = \frac{40}{44.8266}$
$n \approx 0.892 \text{ mol}$
Alternative Method (Using STP): At STP ( $P = 1 \text{ atm}, T = 273 \text{ K}$ ), 1 mole occupies 22.4 dm³. Using the gas law $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ :
$\frac{4 \times 10}{546} = \frac{1 \times V_{stp}}{273}$
$V_{stp} = \frac{40 \times 273}{546} = \frac{40}{2} = 20 \text{ dm}^3$
Since 1 mole = 22.4 dm³, then $n = \frac{20}{22.4} \approx 0.893 \text{ mol}$ .
Both methods confirm the result is approximately 0.89 mol.
Conclusion
The calculated number of moles is 0.89 mol, which matches option A.
Exam Tip
Always remember to convert Celsius to Kelvin by adding 273 before plugging your temperature into any gas law equation. Also, check your units—if the volume is in dm³ and pressure in atm, ensure your gas constant ( $R$ ) matches those units!
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43.
Which of the compounds is composed of Al, Si, O and H?
1. A.Epson salt
2. B.Limestone
3. C.Clay
4. D.Urea
Show answer
Answer: C. Clay
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44.
Elements P, Q, R, S have 6, 11, 15, 17 electrons respectively, therefore,
1. A.P will form an electrovalent bond with R
2. B.Q will form a covalent bond with S
3. C.R will form an electrovalent bond with S
4. D.Q will form an electrovalent bond with S
Show answer
Answer: D. Q will form an electrovalent bond with S
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2018 JAMB Chemistry Past Questions

Questions 1–44

1. Analyze the components

2. Evaluate the options

3. Conclusion

Exam Tip

1. Analyze the Question

2. Explain the Concept

3. Show the Solution

4. Identify the Answer

5. Exam Tip

Analysis

Concept: Le Chatelier’s Principle

Step-by-Step Solution

Evaluation of Options

Exam Tip

Analysis

Concept Explanation

Step-by-Step Solution

Conclusion

Exam Tip

Analysis

Concept Explanation

Step-by-Step Solution

Exam Tip

Analysis of the Question

Concept Explanation

Step-by-Step Solution

Conclusion

Exam Tip

1. Analyze the Question

2. Explain the Concept

3. Show the Solution

4. Identify the Answer

5. 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

Step-by-Step Solution

Conclusion

Analysis

Concept Explanation

Step-by-Step Solution

Exam Tip

Analysis

Concept Explanation

Step-by-Step Solution

Conclusion

Exam Tip

Analysis

Concept Explanation

Step-by-Step Solution

Conclusion

Analysis

Concept Explanation

Step-by-Step Solution

Exam Tip

Analysis

Concept

Step-by-Step Solution

Conclusion

Exam Tip

Analysis of the Question

Concept Explanation

Step-by-Step Solution

Conclusion

Exam Tip

1. Analyze the question

2. Explain the concept

3. Show the solution

4. Identify the answer