JAMB UTME - Chemistry - 2002

Chemical Kinetics And Equilibrium Systems (Ghana Only)

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Question 6 Report

Three drops of a 1.0 mol dm-3 solution of NaOH are added to 20 cm3 of a solution of pH 8.4. The pH of the resulting solution will be

Close to that of pure water

Unaltered

Greater than 8.4 Correct answer

Less than 8.4

Answer Details

When NaOH is added to a solution, it reacts with water to form hydroxide ions (OH-) and sodium ions (Na+). The OH- ions can then react with hydrogen ions (H+) in the solution, causing the pH to increase. In this case, three drops of a 1.0 mol dm-3 solution of NaOH are added to a solution of pH 8.4. The amount of NaOH added is very small compared to the volume of the solution, so we can assume that the amount of water in the solution remains relatively constant. The reaction between NaOH and water can be written as follows: NaOH + H2O → Na+ + OH- + H2O The OH- ions produced by the reaction will react with the H+ ions in the solution: H+ + OH- → H2O This will cause the concentration of H+ ions in the solution to decrease, and the pH to increase. The exact change in pH will depend on the amount of H+ ions in the original solution and the amount of OH- ions produced by the reaction with NaOH. However, since the pH of the original solution is already quite high at 8.4, the addition of a small amount of NaOH will not cause a significant increase in pH. Therefore, the pH of the resulting solution will be greater than 8.4, but not by a significant amount. Therefore, the correct answer is "Greater than 8.4".

Read lesson note on Acids, Bases And Salts (Ghana Only) (WAEC)

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Question 7 Report

Which of the following decreases when a given mass of gas is compressed to half its initial volume?

Average intermolecular distance Correct answer

Frequency of collisions

Number of molecules present

Atomic radius of each particle

Answer Details

When a given mass of gas is compressed to half its initial volume, the average intermolecular distance decreases. This is because when the volume of the gas is decreased, the molecules are brought closer together, which leads to a decrease in the average distance between them. The frequency of collisions between the gas particles will increase as they are closer together, and the number of molecules present in the gas will remain the same, as the mass of the gas has not changed. The atomic radius of each particle also remains the same, as compression does not affect the size of individual gas particles.

Read lesson note on Kinetic Theory Of Matter And Gas Laws (JAMB)

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Question 8 Report

The gas that gives brown colouration in brown ring test is

NO Correct answer

CO2

NO2

Answer Details

The gas that gives brown coloration in the brown ring test is Nitric Oxide (NO). When FeSO4 is added to a solution containing nitrate ions and concentrated H2SO4 is carefully added down the side of the test tube, a brown ring appears at the junction of the two liquids, indicating the presence of nitrate ions. This is because NO reacts with iron(II) to form a complex ion, [Fe(NO)(H2O)5]2+, which gives the brown coloration at the interface of the two layers. The other options in this question, CO and CO2, do not produce a brown ring in this test.

Read lesson note on Kinetic Theory Of Matter And Gas Laws (JAMB)

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Question 9 Report

Which of the following chloride would exhibit the least ionic character?

MgCl2

CaCl2

LiCl

AlCl3 Correct answer

Answer Details

Read lesson note on Chemical Combination (JAMB)

Chemical Combination

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Question 10 Report

A fixed mass of gas has a volume of 92cm3 at 3 oC. When will be its volume at 18 oC if the pressure remains constant

15.3cm3

87.3cm3

552.0cm3

97.0cm3 Correct answer

Answer Details

This is a problem about the ideal gas law, which relates the pressure, volume, temperature, and amount of a gas. The equation for the ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin. To solve this problem, we are given the initial volume V1, initial temperature T1, and final temperature T2. We are also told that the pressure remains constant, which means that P1 = P2. Since the amount of gas is fixed, we can assume that n1 = n2. Using the ideal gas law, we can write: P1V1 = n1RT1 P2V2 = n2RT2 Since n1 = n2, we can simplify to: P1V1/T1 = P2V2/T2 We can solve for V2 by rearranging the equation: V2 = V1(T2/T1) Plugging in the given values, we get: V2 = 92 cm^3 * (291 K / 276 K) ≈ 97.0 cm^3 Therefore, the answer is 97.0 cm^3.

Read lesson note on Kinetic Theory Of Matter And Gas Laws (JAMB)

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Question 11 Report

Ethanol can easily be produced by

fermentation of starch Correct answer

distillation of starch solution

catalytic oxidation of methane

destructive distillation of wood

Answer Details

Ethanol can easily be produced by fermentation of starch. Fermentation is a process that involves the breakdown of sugar by microorganisms such as yeast or bacteria in the absence of oxygen. In the case of ethanol production, the sugar source can be derived from various agricultural crops such as corn, sugarcane, or even from waste products like fruit peels or bread. During the fermentation process, enzymes produced by the microorganisms break down the complex sugar molecules into simpler compounds, including ethanol and carbon dioxide. The ethanol produced can be purified and concentrated through distillation to obtain the final product. This process is relatively simple and can be carried out on a large scale, making it a popular method for ethanol production. The other options listed, such as the distillation of starch solution, catalytic oxidation of methane, or destructive distillation of wood, are not as commonly used for ethanol production.

Read lesson note on Organic Compounds (JAMB)

Organic Compounds

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Question 12 Report

Which of these pairs are synthetic and natural macromolecules respectively?

Haemoglobin and nylon, creatine and polyethylene

Nylon and polyethlene, creatine and Haemoglobin

Nylon and creatine, polyethylene and haemoglobin Correct answer

Polyethylene and creatine, nylon and haemoglobin

Answer Details

Read lesson note on Organic Compounds (JAMB)

Organic Compounds

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Question 13 Report

Ordinary glass is manufactured from silica, CaCO3 and

Na2CO3 Correct answer

K2CO3

K2SO4

NaHCO3

Answer Details

Read lesson note on Chemistry, Industry And The Environment (WAEC)

Read lesson note on Chemical Industry And Environment (WAEC)

Chemistry, Industry And The Environment

Chemical Industry And Environment

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Question 14 Report

A compound contains 31.91% potassium, 28.93% chlorine and the rest oxygen. What is the chemical formula of the compound?

[K = 39, Cl = 35.5, O = 16]

KClO4

KClO

KClO2

KClO3 Correct answer

Answer Details

To find the chemical formula of the compound, we first need to determine the ratio of the elements in it. From the given percentages, we can assume that we have 100 grams of the compound, and then convert the percentages to grams: - Potassium (K) = 31.91 grams - Chlorine (Cl) = 28.93 grams - Oxygen (O) = 100 - 31.91 - 28.93 = 39.16 grams Next, we need to determine the ratio of the elements in terms of their moles. To do this, we need to divide each element's mass by its molar mass: - K: 31.91 g / 39.1 g/mol = 0.815 mol - Cl: 28.93 g / 35.5 g/mol = 0.816 mol - O: 39.16 g / 16 g/mol = 2.45 mol We can see that the ratio of K to Cl to O is approximately 1:1:3. Therefore, the chemical formula of the compound is KClO3, option (D). Each formula unit of KClO3 contains one potassium atom (K), one chlorine atom (Cl), and three oxygen atoms (O).

Read lesson note on Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

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Question 15 Report

Iron is often galvanized in order to

make it more malleable

remove the impurities in it

protect it against corrosion Correct answer

render it passive

Answer Details

Iron is often galvanized in order to protect it against corrosion. Galvanizing is a process of coating a layer of zinc on the surface of iron or steel to prevent the formation of rust. Zinc is more reactive than iron and when it is exposed to air and water, it forms a layer of zinc oxide which acts as a protective layer for the underlying iron. This prevents the iron from rusting and increases its lifespan. Galvanizing is commonly used in outdoor applications where the iron or steel is exposed to the elements and is susceptible to corrosion.

Read lesson note on Metals And Their Compounds (JAMB)

Metals And Their Compounds

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Question 16 Report

The substance least considered as a source of environmental pollution is

Lead compounds

Organophosphorous compounds

Silicate minerals Correct answer

Uranium

Answer Details

All of the listed substances can potentially cause environmental pollution, but the substance least considered as a source of environmental pollution is silicate minerals. Silicate minerals are naturally occurring minerals that make up a large portion of the Earth's crust and are typically not harmful to the environment. On the other hand, lead compounds, organophosphorus compounds, and uranium are all known to be harmful to the environment and can cause pollution. Lead can contaminate soil and water, causing damage to wildlife and humans who come into contact with it. Organophosphorus compounds are often used as pesticides and can have negative effects on the health of humans and animals, as well as on the environment. Uranium is a radioactive element that can cause contamination of soil, water, and air, leading to harmful effects on living organisms. In summary, while all of the listed substances can potentially cause environmental pollution, silicate minerals are the least considered as a source of environmental pollution due to their natural occurrence and low likelihood of causing harm.

Read lesson note on Environmental Pollution (JAMB)

Environmental Pollution

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Question 17 Report

The solubility of a salt of molar mass 101g at 20oC is 0.34 mol dm-3. If 3.40g of the salt is dissolved completely in 250cm3 of water in a beaker, the resulting solution is

A suspension

Saturated

Unsaturated Correct answer

Supersaturated

Answer Details

To determine whether the resulting solution is a suspension, saturated, unsaturated, or supersaturated, we need to compare the amount of salt dissolved in water with its maximum solubility at that temperature. First, we need to calculate the maximum amount of salt that can dissolve in 250 cm³ of water at 20°C using the given solubility: 0.34 mol dm⁻³ = 0.34 mol / 1000 cm³ Maximum amount of salt that can dissolve in 250 cm³ of water = 0.34 x 250 / 1000 = 0.085 mol Next, we need to compare the maximum amount of salt that can dissolve in water with the actual amount of salt dissolved in 250 cm³ of water: Amount of salt dissolved in 250 cm³ of water = 3.40 g / 101 g/mol = 0.0337 mol Since the actual amount of salt dissolved in water (0.0337 mol) is less than the maximum amount of salt that can dissolve in water (0.085 mol), the resulting solution is unsaturated. Therefore, the answer is Unsaturated.

Read lesson note on Solubility (JAMB)

Solubility

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Question 18 Report

A major factor coonsidered in selecting a suitable method for preparing a simple salt is its

crystalline form

melting point

reactivity with dilute acids

solubility in water Correct answer

Answer Details

One major factor that is considered when selecting a suitable method for preparing a simple salt is its solubility in water. This is because the salt needs to dissolve in water in order to form a solution that can be used for various purposes, such as in chemistry experiments or in food preparation. If the salt is not soluble in water, then it may not be suitable for these applications. Other factors, such as its crystalline form, melting point, and reactivity with dilute acids, may also be important considerations depending on the specific application or use of the salt, but solubility in water is a key factor that must be considered in almost all cases.

Read lesson note on Acids, Bases And Salts (Ghana Only) (WAEC)

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Question 19 Report

The boiling of fat and aqueous caustic soda is referred to as

hydrolysis

esterification

acidification

saponification Correct answer

Answer Details

The boiling of fat and aqueous caustic soda is referred to as saponification. Saponification is a process in which fat or oil is mixed with a strong base, usually sodium hydroxide (caustic soda), to produce soap. During saponification, the ester bonds of the fat molecules are broken down by the strong base, releasing the fatty acid and glycerol molecules. The fatty acid then reacts with the base to form the soap molecule, which is a salt of the fatty acid. In simpler terms, saponification is a chemical reaction that turns fat into soap by boiling it with a strong base like caustic soda. This process is used to make soap in large quantities for commercial and industrial purposes.

Read lesson note on Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

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Question 20 Report

In the reaction E+F ↔ G+H, the backward reaction is favoured if the concentration of

F is increased

E is increased

E is reduced Correct answer

G is reduced

Answer Details

Read lesson note on Chemical Equilibria (JAMB)

Chemical Equilibria

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Question 21 Report

\( \mathrm{PCl_5(g) \rightarrow PCl_3(g) + Cl_2(g)} \)

In the reaction above, an increase in pressure will

decelerate the reaction

Increase the yield of

P C l_{3}

increase the yield of

P C l_{5}

Correct answer

accelerate the reaction

Answer Details

Read lesson note on Chemical Equilibria (JAMB)

Chemical Equilibria

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Question 22 Report

If the relative molecular mass of an element is not a whole number, it can be deduced that the element is

naturally radioactive

abundant in nature

a transition metal

an isotopic mixture Correct answer

Answer Details

If the relative molecular mass of an element is not a whole number, it can be deduced that the element is an isotopic mixture. This is because the relative molecular mass of an element is calculated by taking the weighted average of the masses of its isotopes, where the weighting factor is the abundance of each isotope in nature. If an element has isotopes with different masses and abundances, then the weighted average will not be a whole number. Therefore, the element must be an isotopic mixture.

Read lesson note on Atomic Structure And Bonding (JAMB)

Atomic Structure And Bonding

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Question 23 Report

The chromatography separation of ink is based on the ability of the components to

React with solvent

React with each other

Dissolve in each other in the column

Move at different speeds in the column Correct answer

Answer Details

The chromatography separation of ink is based on the ability of the components to move at different speeds in the column. Chromatography is a technique for separating mixtures based on the differences in their physical and/or chemical properties. In ink chromatography, the mixture of colored components in the ink is separated by passing it through a stationary phase (paper or column) and a mobile phase (solvent). The colored components in the ink mixture have different degrees of solubility in the mobile phase and different affinities for the stationary phase. As a result, they will move through the stationary phase at different rates, leading to their separation. The more soluble and less attracted to the stationary phase a component is, the faster it will move through the column. Conversely, the less soluble and more attracted to the stationary phase a component is, the slower it will move. Therefore, ink chromatography separates the components of the ink mixture based on their relative speeds of movement.

Read lesson note on Separation Of Mixtures And Purification Of Chemical Substances (JAMB)

Separation Of Mixtures And Purification Of Chemical Substances

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Question 24 Report

When H2S is passed into a solution of iron (III) chloride, the solution turns

Pale green Correct answer

Pale red

Brown

Colourless

Answer Details

When H2S is passed into a solution of iron (III) chloride, the solution turns brown. This is due to the formation of a precipitate of iron (III) sulfide (Fe2S3). The reaction is represented as follows: 2FeCl3(aq) + 3H2S(g) → Fe2S3(s) + 6HCl(aq) The brown precipitate of Fe2S3 indicates the presence of iron (III) ions in the solution. Therefore, the correct answer is "Brown".

Read lesson note on Acids, Bases And Salts (Ghana Only) (WAEC)

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Question 25 Report

The most important use of hydrogen is in the

manufacture of ammonia Correct answer

manufacture of methyl alcohol

manufacture of ethyl alcohol

hydrogenation of oils

Answer Details

The most important use of hydrogen is in the manufacture of ammonia. Ammonia is a vital chemical used in the production of fertilizers, which are essential for modern agriculture. Hydrogen gas is combined with nitrogen gas to produce ammonia in a process called the Haber process. This process is highly dependent on the availability of hydrogen gas, making it the most important use of hydrogen. While hydrogenation of oils, manufacture of methyl alcohol, and ethyl alcohol are also important uses of hydrogen, they are not as critical as the production of ammonia for the global economy.

Read lesson note on Rates Of Chemical Reaction (JAMB)

Read lesson note on Metals And Their Compounds (JAMB)

Read lesson note on Chemical Combination (JAMB)

Rates Of Chemical Reaction

Metals And Their Compounds

Chemical Combination

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Question 26 Report

The number of isomers formed by C6H14 is

5 Correct answer

Answer Details

The number of isomers formed by C6H14 is 5. Isomers are molecules with the same molecular formula but different structural arrangements of their atoms. C6H14 can form five isomers: n-hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, and 2,3-dimethylbutane. The first three are straight-chain isomers, while the last two are branched-chain isomers. These isomers have different physical and chemical properties, such as boiling points, melting points, and reactivities, which make them useful in various industrial applications.

Read lesson note on Chemistry Of Carbon Compounds (Nigeria Only) (WAEC)

Chemistry Of Carbon Compounds (Nigeria Only)

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Question 27 Report

If 0.75 mole of cyclopropane and 0.66 mole of oxygen are mixed in vessel with a total pressure of 0.7 atmosphere, what is the partial pressure of oxygen in the mixture

0.44 atmosphere

0.22 atmosphere

0.55 atmosphere

0.33 atmosphere Correct answer

Answer Details

The question involves finding the partial pressure of oxygen in a mixture of cyclopropane and oxygen. We are given the total pressure of the mixture, as well as the moles of each component. To solve this problem, we need to use the mole fraction of oxygen. The mole fraction is the number of moles of oxygen divided by the total number of moles in the mixture. First, we need to calculate the total number of moles in the mixture: Total moles = moles of cyclopropane + moles of oxygen Total moles = 0.75 + 0.66 Total moles = 1.41 Next, we can calculate the mole fraction of oxygen: Mole fraction of oxygen = moles of oxygen / total moles Mole fraction of oxygen = 0.66 / 1.41 Mole fraction of oxygen = 0.4681 Finally, we can calculate the partial pressure of oxygen using Dalton's law of partial pressures, which states that the total pressure of a gas mixture is equal to the sum of the partial pressures of the individual gases in the mixture: Partial pressure of oxygen = mole fraction of oxygen x total pressure Partial pressure of oxygen = 0.4681 x 0.7 Partial pressure of oxygen = 0.3277 or approximately 0.33 atmosphere. Therefore, the correct answer is, 0.33 atmosphere.

Read lesson note on Kinetic Theory Of Matter And Gas Laws (JAMB)

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Question 28 Report

The reaction of an alkene with hydrogen in the presence of catalyst is

a substitution reaction

an oxidation reaction

a nucleophilic reaction

an addition reaction Correct answer

Answer Details

The reaction of an alkene with hydrogen in the presence of a catalyst is an addition reaction. This is because the double bond of the alkene is broken, and the hydrogen atoms are added to the molecule in a process called hydrogenation. No atoms are lost or gained, and the reaction is typically exothermic, releasing energy.

Read lesson note on Periodic Chemistry (Ghana Only) (WAEC)

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Question 29 Report

What amount of mercury would be liberated if the same quantity of electricity that liberates 0.65g of zinc is applied?

2.01g

1.00g

4.02g Correct answer

8.04g

Answer Details

Zn2+ + 2e- → Zn
2F will be liberated 65g of Zn, if 0.65gm of Zn is liberated, the quantity of electricity used
X = (0.65/65) * 2
= 0.02F
∴ Hg+ + e- → Hg1
i.e 1F → 201gm of Hg
∴ 0.02 will liberated
Y = (0.02/1) * 201
= 4.02g

Read lesson note on Electrolysis (JAMB)

Electrolysis

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Question 30 Report

Which of the following statement is true of a proton?

The mass of a proton is 1840 times the mass of electron

the total mass of the protons in a particular nucleus is always half the nuclear size

The mass of a proton is one - twelfth the molar mass of carbon

The mass of a proton is 1.0008g Correct answer

Answer Details

Read lesson note on Atomic Structure And Bonding (JAMB)

Atomic Structure And Bonding

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Question 31 Report

25cm3 of a 0.2 mol dm3 solution of Na2CO3 requires 20 cm3 of a solution of HCl for neutralization. The concentration of the HCl solution is

0.5 mol dm3 Correct answer

0.6 mol dm3

0.2 mol dm3

0.4 mol dm3

Answer Details

The question describes a neutralization reaction between a solution of Na2CO3 and HCl. The volume of HCl required for neutralization is given as 20 cm3. From the balanced equation for the reaction, we know that 1 mole of Na2CO3 reacts with 2 moles of HCl. Therefore, the moles of HCl required for neutralization can be calculated as follows: moles of HCl = (moles of Na2CO3) x 2 moles of Na2CO3 = concentration x volume / 1000 Substituting the given values, we get: moles of Na2CO3 = 0.2 x 25 / 1000 = 0.005 moles of HCl = 0.005 x 2 = 0.01 The concentration of HCl solution can then be calculated by dividing the moles of HCl by the volume of HCl solution used: concentration of HCl = moles of HCl / volume of HCl solution concentration of HCl = 0.01 / 20 / 1000 concentration of HCl = 0.5 mol dm^-3 Therefore, the concentration of the HCl solution is 0.5 mol dm^-3.

Read lesson note on Stoichiometry And Chemical Reactions (WAEC)

Stoichiometry And Chemical Reactions

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Question 32 Report

The processes which return carbon (IV) oxide to the atmosphere include

respiration, decay and combustion Correct answer

photosynthesis, respiration and transpiration

ozone depletion, combustion and decay

photosynthesis, decay and respiration

Answer Details

The processes which return carbon (IV) oxide to the atmosphere include respiration, decay, and combustion. Respiration is the process by which living organisms, including plants and animals, release carbon dioxide into the atmosphere as they break down food molecules to release energy. Decay is the breakdown of organic matter by microorganisms, which also releases carbon dioxide. Combustion, on the other hand, is the burning of fuels, such as coal and oil, which releases carbon dioxide into the atmosphere as a byproduct of the reaction. While other processes, such as photosynthesis, transpiration, and ozone depletion, may play important roles in the carbon cycle, they do not directly return carbon dioxide to the atmosphere in the same way that respiration, decay, and combustion do.

Read lesson note on Chemistry, Industry And The Environment (WAEC)

Read lesson note on Chemical Industry And Environment (WAEC)

Chemistry, Industry And The Environment

Chemical Industry And Environment

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Question 33 Report

The formula CH2 for ethanoic acid is regarded as its

Molecular formula

General formula

Empirical formula Correct answer

Structural formula

Answer Details

Read lesson note on Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

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Question 34 Report

146C → X + B

X in the equation above represents

126C

125B

147N Correct answer

136C

Answer Details

146C → X + B
→ X + 0-1e
→ 147X

Read lesson note on Structure Of The Atom (WAEC)

Structure Of The Atom

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Question 35 Report

The major product of the dehydration of the compound above is

A Correct answer

Answer Details

Read lesson note on Chemistry, Industry And The Environment (WAEC)

Chemistry, Industry And The Environment

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Question 36 Report

The product of the electrolysis of dilute sodium hydroxide using platinum electrodes are

Hydrogen and oxygen gases Correct answer

Water and hydrogen gas

Water and sodium metal

Sodium metal and oxygen gas

Answer Details

Read lesson note on Chemistry And Industry (JAMB)

Read lesson note on Electrolysis (JAMB)

Chemistry And Industry

Electrolysis

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Question 37 Report

In the diagram above, the curve that represents the production of oxgyen gas from the decomposition of KCIO3 in the presence of MnO2 catalyst is

P Correct answer

Answer Details

Read lesson note on Kinetic Theory Of Matter And Gas Laws (JAMB)

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Question 38 Report

The compound above is

glycol

primary alkanol

secondary alkanol Correct answer

tertiary alkanol

Answer Details

Read lesson note on Chemistry Of Carbon Compounds (Nigeria Only) (WAEC)

Chemistry Of Carbon Compounds (Nigeria Only)

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Question 39 Report

A suitable reagent for distinguishing between ethanoic acid ethanol is

bromine water

Fehling's solution

sodium hydrogentrioxocarbonate (IV) Correct answer

ammoniacal silver trioxonitrate (V)

Answer Details

Read lesson note on Acids, Bases And Salts (JAMB)

Acids, Bases And Salts

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Question 40 Report

A red precipitate of copper (I) carbide is formed when ammonium solution of copper (I) chloride is introduced into

CH3CH2CH2CH3

CH2 = CH - CH2CH3

CH3CH2 - C = CH Correct answer

CH3 - C = C - CH3

Answer Details

Read lesson note on Periodic Chemistry (Ghana Only) (WAEC)

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Question 41 Report

Hydrogen is readily released when dilute hydrochloric acid reacts with

Na Correct answer

Answer Details

Hydrogen is readily released when dilute hydrochloric acid reacts with sodium (Na). When an acid reacts with a metal, it typically produces hydrogen gas and a salt. In the case of dilute hydrochloric acid and sodium, the reaction produces sodium chloride (NaCl) and hydrogen gas (H2). The reaction between dilute hydrochloric acid and sodium can be represented by the following equation: 2HCl(aq) + 2Na(s) → 2NaCl(aq) + H2(g) In this reaction, the hydrogen ions (H+) from the acid react with the metal atoms from the sodium (Na) to form hydrogen gas (H2). The sodium ions (Na+) combine with the chloride ions (Cl-) from the acid to form sodium chloride (NaCl), which dissolves in the water. The other options listed, such as Cu, Au, and Ag, do not react with hydrochloric acid to produce hydrogen gas. These metals have different chemical properties and reactivities compared to sodium, and would react differently with hydrochloric acid.

Read lesson note on Periodic Chemistry (Ghana Only) (WAEC)

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Question 42 Report

When a salt loses its water of crystallization to the atmosphere on exposure, the process is said to be

Efflorescence Correct answer

Deliquescence

Effervescence

Fluorescence

Answer Details

The process of a salt losing its water of crystallization to the atmosphere on exposure is known as efflorescence. This happens when a hydrate salt is exposed to a dry atmosphere, causing the water molecules in the crystal lattice to evaporate, leaving behind the anhydrous salt. This process can be reversed by adding water to the anhydrous salt to form the hydrate again. The other options are not relevant to this process: deliquescence refers to a process where a substance absorbs water from the atmosphere and dissolves in it; effervescence refers to the escape of gas from a liquid or a solid due to a chemical reaction, and fluorescence is the emission of light by a substance that has absorbed light or other electromagnetic radiation.

Read lesson note on Acids, Bases And Salts (Ghana Only) (WAEC)

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Question 43 Report

which of the following polymers is suitable for packaging and electrical insulation?

polyamide

polyethene

polycarbonate

polystyrene Correct answer

Answer Details

Read lesson note on Basic Biochemistry And Synthetic Polymers (WAEC)

Read lesson note on Organic Compounds (JAMB)

Basic Biochemistry And Synthetic Polymers

Organic Compounds

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Question 44 Report

Which of the following gives a precipitate when treated with NaOH solution?

AlCl3 Correct answer

NH4CL

CH3COONa

Na2CO3

Answer Details

When a substance reacts with NaOH solution to produce a solid or insoluble product, it is said to give a precipitate. We can use this knowledge to determine which of the given options will give a precipitate when treated with NaOH solution. Option A, AlCl3, will give a precipitate when treated with NaOH solution. This is because AlCl3 is an acidic salt and will react with the basic NaOH to produce a gelatinous precipitate of aluminium hydroxide (Al(OH)3), which is insoluble in water. Option B, NH4Cl, will not give a precipitate when treated with NaOH solution because NH4Cl is a salt of a strong acid (HCl) and a weak base (NH3), and it will not react with NaOH to form a precipitate. Option C, CH3COONa, will not give a precipitate when treated with NaOH solution because it is a salt of a weak acid (acetic acid) and a strong base (NaOH). Therefore, it will not react with NaOH to form a precipitate. Option D, Na2CO3, will give a precipitate when treated with NaOH solution. This is because Na2CO3 is a basic salt and will react with the acidic NaOH to produce a white precipitate of sodium carbonate (Na2CO3), which is insoluble in water. Therefore, the option that gives a precipitate when treated with NaOH solution is option A, AlCl3, and option D, Na2CO3. The answer is not uniquely defined.

Read lesson note on Acids, Bases And Salts (JAMB)

Read lesson note on Acids, Bases And Salts (Ghana Only) (WAEC)

Acids, Bases And Salts

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Question 45 Report

Steam changes the colour of anhydrous cobalt (II) chloride from

blue to pink Correct answer

white to red

white to green

blue to white

Answer Details

Anhydrous cobalt (II) chloride is blue in color, but when steam is passed over it, the blue color disappears and the compound turns pink. Therefore, the answer is "blue to pink". This is due to the fact that the cobalt (II) chloride reacts with the water in the steam to form pink hexahydrate cobalt (II) chloride.

Read lesson note on Periodic Chemistry (Ghana Only) (WAEC)