JAMB UTME - Chemistry - 2017

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Question 2 Report

The constituent of air necessary in the rusting process are

O2 and H2O

Ar and CO2

CO2 and H2O

O2 and CO2

Answer Details

The constituent of air includes O2, CO2, N2 and Noble gases. While for a rusting process to take place, the presence of O2, H2O and CO2 is important.

The constituent of air includes O2, CO2, N2 and Noble gases and for rusting process to take place.

The presence of O2 and CO2 as a constituent of air is indispensable

Read lesson note on Air (JAMB)

Air

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Question 3 Report

Calculate the amount in moles of silver deposited when 9650C of electricty is passed through a solution of silver salt [= 96500 Cmol-1]

0.05

10.80

10.00

0.10

Answer Details

The amount of silver deposited can be calculated using Faraday's law of electrolysis, which states that the amount of a substance deposited at an electrode during electrolysis is proportional to the number of moles of electric charge passed through the solution. We can start by using the formula: moles of substance = charge passed / Faraday's constant Where Faraday's constant is equal to 96500 Cmol^-1. Plugging in the given values: moles of silver = 9650 C / 96500 Cmol^-1 moles of silver = 0.1 So, the amount of silver deposited is 0.1 moles.

Read lesson note on Electrolysis (JAMB)

Electrolysis

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Question 4 Report

The oxidation number of iodine in KIO3 is

+7

+3

+5

+6

Answer Details

The oxidation number of an atom is a measure of the number of electrons that it has gained or lost in a chemical reaction. To determine the oxidation number of iodine in KIO3, we need to use the fact that the sum of the oxidation numbers of all the atoms in a compound is equal to the overall charge of the compound. In KIO3, potassium (K) has an oxidation number of +1 because it belongs to the alkali metal group and it always has a +1 oxidation state. Oxygen (O) usually has an oxidation number of -2, unless it is combined with a more electronegative atom or a peroxide, which is not the case in KIO3. Let's assume that iodine (I) has an oxidation number of x. Since KIO3 is a neutral compound, the sum of the oxidation numbers of all its atoms must be zero. Therefore, we can write the following equation: (+1) + x + 3(-2) = 0 Simplifying the equation, we get: x = +5 Therefore, the oxidation number of iodine in KIO3 is +5.

Read lesson note on Oxidation And Reduction (JAMB)

Oxidation And Reduction

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Question 5 Report

According to Charle's law, the volume of a gas becomes zero at

−100°c

−273°c

−373°c

0°c

Answer Details

According to Charles's law, the volume of a gas would theoretically become zero at -273°C or -459.4°F or 0 Kelvin, which is known as absolute zero. Charles's law is a gas law that describes the relationship between the volume and temperature of a gas, while the pressure and amount of gas are kept constant. The law states that at a constant pressure, the volume of a gas is directly proportional to its absolute temperature. This means that as the temperature of a gas decreases, its volume will also decrease proportionally. However, according to the laws of thermodynamics, it is impossible to cool a gas down to absolute zero or to completely eliminate its volume. This is because at absolute zero, the gas molecules would have zero kinetic energy, and therefore they would not be moving. This is a state that cannot be reached due to the third law of thermodynamics. Therefore, Charles's law is only valid within a certain temperature range and it cannot be applied to gases at extremely low temperatures or high pressures.

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

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Question 6 Report

The reddish–brown rust on ion roofing sheets consists of

Fe3 + (H2O)6

FeO.H2O

Fe2O3.3H2O

Fe3O4.2H22O

Answer Details

The reddish-brown rust on iron roofing sheets is mainly composed of Fe2O3.3H2O, which is a compound of iron and oxygen with water molecules. Iron roofing sheets, when exposed to air and moisture, undergo a chemical reaction. Oxygen in the air reacts with the iron to form a new compound called iron oxide (Fe2O3). As moisture (water) is also present, the iron oxide reacts with the water to form hydrated iron oxide (Fe2O3.3H2O), which is commonly known as rust. So, the reddish-brown rust on iron roofing sheets is mainly composed of Fe2O3.3H2O, which is a combination of iron, oxygen, and water molecules.

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

Acids, Bases And Salts

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Question 7 Report

If 100cm3 of oxygen pass through a porous plug is 50 seconds, the time taken for the same volume of hydrogen to pass through the same porous plug is? [O = 16, H = 1]

10.0s

12.5s

17.7s

32.0s

Answer Details

The rate at which a gas diffuses through a porous plug is inversely proportional to the square root of its molecular weight. The molecular weight of oxygen (O2) is 32 (16 x 2), while the molecular weight of hydrogen (H2) is 2 (1 x 2). Therefore, the square root of the ratio of the molecular weights of hydrogen and oxygen is: sqrt(2/32) = 0.25 This means that hydrogen diffuses through a porous plug 0.25 times as fast as oxygen. If it takes 50 seconds for 100cm3 of oxygen to pass through the porous plug, then it will take 0.25 times longer for the same volume of hydrogen to pass through the same porous plug. Thus, the time taken for the same volume of hydrogen to pass through the same porous plug is: 50 seconds x 0.25 = 12.5 seconds Therefore, the correct answer is 12.5s.

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

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Question 8 Report

Temporary hard water is formed when rain water containing dissolved carbon(IV) oxide flows over deposits of

CaCO3

Na2CO3

Na2SO4

CaSO4

Answer Details

Temporary hard water is formed when rainwater, which is naturally soft, comes into contact with deposits of calcium carbonate (CaCO3) in the ground. As the rainwater seeps through the soil, it dissolves carbon dioxide (CO2) to form carbonic acid (H2CO3). This weak acid reacts with the calcium carbonate to form calcium bicarbonate (Ca(HCO3)2), which is soluble in water. As a result, the water that flows over these deposits contains dissolved calcium bicarbonate, which gives it temporary hardness. This temporary hardness can be removed by boiling the water, which causes the calcium bicarbonate to decompose back into calcium carbonate and carbon dioxide. The calcium carbonate then forms a white precipitate, which can be removed by filtration or settling. So, the correct answer to the given question is CaCO3.

Read lesson note on Oxidation And Reduction (JAMB)

Oxidation And Reduction

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Question 9 Report

Tin is unaffected by air at ordinary temperature due to its

Low melting point

Weak electropositive character

High boiling point

White lustrous appearance

Answer Details

Tin is unaffected by air at ordinary temperature because of its high reactivity, which means it doesn't easily react or corrode with other elements such as oxygen, the main component of air. Tin has a high reactivity, which helps it resist corrosion and oxidation. This makes tin a great material for products that need to be resistant to corrosion, such as food containers and metal pipes.

Read lesson note on Chemical Combination (JAMB)

Chemical Combination

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Question 10 Report

The general formula of alkanones is

RCHO

R2CO

RCOOH

RCOOR

Answer Details

The general formula of alkanones is R2CO, which represents a functional group that consists of a carbon atom double-bonded to an oxygen atom (C=O) and attached to two alkyl (or aryl) groups represented by the "R" symbol. In simpler terms, alkanones are a type of organic compound that contain a carbon atom double-bonded to an oxygen atom and two additional carbon-containing groups. The "R" groups can be any combination of alkyl or aryl groups, which are made up of carbon and hydrogen atoms arranged in a specific way. The general formula R2CO applies to all alkanones, regardless of the specific alkyl or aryl groups attached to the central carbon atom. By knowing this formula, we can easily recognize an alkanone when we see one and predict its chemical properties based on the functional group it contains.

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

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Question 11 Report

The alkyl group is represented by the general formula

CnH2n

CnH2n − 2

CnH2n + 1

CnH2n + 2

Answer Details

The alkyl group is represented by the general formula CnH2n+1, where "n" represents the number of carbon atoms in the group. An alkyl group is a type of organic molecule that consists of a chain of carbon atoms, each of which is bonded to two hydrogen atoms. The formula CnH2n+1 represents this structure, where "n" is the number of carbon atoms in the chain. The number of hydrogen atoms in the chain is always two more than the number of carbon atoms, hence the formula CnH2n+1.

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

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Question 12 Report

The salt formed from a weak acid and a strong base hydrolyzes in water to form

A saturated solution

an acidic solution

a buffer solution

an alkaline solution

Answer Details

This is a solution formed from the hydrolyses of an alkali in water.

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

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Question 13 Report

An isomer of C5H12 is

2–ethyl butane

butane

2–methyl butane

2–methyl propane

Answer Details

C5H12 is the molecular formula for pentane, which has five carbon atoms and 12 hydrogen atoms. An isomer is a molecule that has the same molecular formula as another molecule, but with a different arrangement of atoms. Therefore, we need to look for molecules with the molecular formula C5H12 that have a different arrangement of atoms than pentane. Let's consider the options: - Butane (C4H10) has a different molecular formula than pentane (C5H12), so it cannot be an isomer of pentane. - 2-Methyl propane (C4H10) also has a different molecular formula than pentane, so it cannot be an isomer of pentane. - 2-Methyl butane (C5H12) has the same molecular formula as pentane, but with a different arrangement of atoms. Specifically, the carbon atoms in 2-methyl butane are arranged in a branched chain, while the carbon atoms in pentane are arranged in a straight chain. Therefore, 2-methyl butane is an isomer of pentane. - 2-Ethyl butane (C6H14) has a different molecular formula than pentane, so it cannot be an isomer of pentane. Therefore, the isomer of C5H12 is 2-methyl butane.

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

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Question 14 Report

In the laboratory preparation of trioxonitrate (V) acid the nitrogen(iv) oxide formed as a by-product is removed by

further heating

adding concentrated H2SO4

cooling the acid solution with cold water

bubbling air through the acid solution

Answer Details

In the laboratory preparation of trioxonitrate (V) acid, nitrogen(iv) oxide is formed as a by-product. Nitrogen(iv) oxide is a harmful gas that can cause respiratory problems, so it needs to be removed from the reaction mixture. One way to remove nitrogen(iv) oxide is by bubbling air through the acid solution. When air is bubbled through the solution, the nitrogen(iv) oxide reacts with the oxygen in the air to form nitrogen dioxide, which is a brown gas. Nitrogen dioxide is not as harmful as nitrogen(iv) oxide and can be safely vented out of the laboratory. Therefore, bubbling air through the acid solution is the method used to remove nitrogen(iv) oxide formed as a by-product in the laboratory preparation of trioxonitrate (V) acid.

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

Acids, Bases And Salts

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Question 15 Report

In the diagram above. X is

SO3

SO2

S

H2S

Answer Details

The setup represents the production of sulfur dioxide. And the cylinder marked X is SO2

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 16 Report

An oxide XO2 has a vapour density of 32. What is the atomic mass of X?

20

32

14

12

Answer Details

Molecular mass = vapour density X2

Mm of XO2 = x + 16(2) = x + 32

vapour density = 32

∴ x + 32 = 32 × 2

x + 32 = 64

x = 64 - 32

x = 32

∴ the relative molecular mass of X is 32
Relative molecular mass = vapour density × 2

Read lesson note on Atomic Structure And Bonding (JAMB)

Atomic Structure And Bonding

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Question 17 Report

The reaction of halogens in the presence of sunlight is an example of

oxidation reaction

addition reaction

hydrogenation reaction

substitution reaction

Answer Details

The reaction of halogens in the presence of sunlight is an example of a substitution reaction. In this type of chemical reaction, one atom or group of atoms in a molecule is replaced by another atom or group of atoms. In the case of halogens, such as chlorine or bromine, they can react with certain organic compounds by breaking the carbon-hydrogen bond and replacing the hydrogen atom with a halogen atom. This reaction is usually initiated by the energy from sunlight, which provides the necessary activation energy to break the bond. For example, in the presence of sunlight, chlorine gas can react with methane gas to form chloromethane and hydrogen chloride: CH4 + Cl2 -> CH3Cl + HCl In this reaction, one of the hydrogen atoms in methane is replaced by a chlorine atom, forming chloromethane, while hydrogen chloride is also produced as a byproduct. Overall, the reaction of halogens in the presence of sunlight is an example of a substitution reaction, where one atom or group of atoms is substituted for another.

Read lesson note on Oxidation And Reduction (JAMB)

Oxidation And Reduction

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Question 18 Report

Calculate the mass of copper deposited when a current of 0.5 ampere was passed through a solution of copper(II) chloride for 45 minutes in an electrolytic cell. [Cu = 64, F = 96500Cmol-1]

0.300g

0.250g

0.2242g

0.448g

Answer Details

M =

\frac{Molar mass × Quantity of Electricity}{96500 × no of charge}

= $\frac{M m I T}{96500 n}$

Copper II Chloride = CuCl2

CuCl2 → Cu2+ + 2Cl2

Mass of compound deposited = $\frac{Molar mass × Quantity of Electricity}{96500 × no of charge}$

Q = IT

I = 0.5A

T = 45 × 60

T = 2700s

Q = 0.5 × 2700

= 1350c

Molarmass = 64gmol-1

no of charge = + 2

Mass = $\frac{64 \times 1350}{96500 \times 2}$

Mass = 0.448g

Read lesson note on Electrolysis (JAMB)

Electrolysis

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Question 19 Report

The ideal gas laws and equations are true for all gases at

low pressures and lower temperatures

low temperatures and high pressures

high pressures and high temperatures

low pressure and high temperatures

Answer Details

The ideal gas laws and equations are most accurate for gases under conditions of low pressure and high temperature. At low pressure, the gas molecules are far apart and experience fewer intermolecular forces, allowing them to behave more ideally. At high temperature, the gas molecules have more kinetic energy, which helps to overcome any intermolecular forces that may be present. Under these conditions, the volume of a gas is directly proportional to its temperature and inversely proportional to its pressure, as described by the ideal gas law. However, as pressure and/or temperature increase, the behavior of real gases becomes more complex and may deviate from ideal gas behavior.

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

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Question 20 Report

The mass of silver deposited when a current of 10A is passed through a solution of silver salt for 4830s is – (Ag = 108 F = 96500(mol-1)

54.0g

27.0g

13.5g

108.0g

Answer Details

To calculate the mass of silver deposited, we can use Faraday's laws of electrolysis, which states that the amount of substance produced at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. The formula to calculate the mass of substance produced is: Mass (g) = (Current × Time × Atomic mass) / (Faraday's constant × Valency) Here, we have a current of 10A passing through a solution of silver salt for 4830s, and we know that the atomic mass of silver is 108 g/mol and the valency is 1 (since silver has a charge of +1). The value of Faraday's constant is 96500 C/mol. Substituting the values in the formula, we get: Mass (g) = (10A × 4830s × 108 g/mol) / (96500 C/mol × 1) Mass (g) = 54.0g Therefore, the mass of silver deposited is 54.0g, which is option (A) in the given options.

Read lesson note on Electrolysis (JAMB)

Electrolysis

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Question 21 Report

A sample of orange juice is found to have a PH of 3.80. What is the concentration of the hydroxide ion in the juice?

1.6

\times

10-4

6.3

\times

10-11

6.3

\times

10-4

1.6

\times

10-11

Answer Details

The pH scale measures the acidity or basicity of a solution. The pH is defined as the negative logarithm (base 10) of the concentration of the hydronium ion (H3O+), which is given by the expression: pH = -log[H3O+] To calculate the concentration of the hydroxide ion (OH-) in the juice, we need to use the equation for the ion product of water: Kw = [H3O+][OH-] = 1.0 × 10-14 where Kw is the ion product constant of water. At 25°C, Kw has a value of 1.0 × 10-14. To solve for [OH-], we need to rearrange the equation: [OH-] = Kw / [H3O+] Now we can substitute the pH of the orange juice to calculate the concentration of hydronium ion [H3O+]. pH = -log[H3O+] 3.80 = -log[H3O+] [H3O+] = 10^(-3.80) [H3O+] = 1.58 × 10^(-4) mol/L Substituting this value into the ion product equation: [OH-] = Kw / [H3O+] [OH-] = 1.0 × 10^-14 / (1.58 × 10^-4) [OH-] = 6.33 × 10^-11 mol/L Therefore, the concentration of the hydroxide ion in the orange juice is 6.33 × 10^-11 mol/L.

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

Acids, Bases And Salts

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Question 22 Report

Water for town supply is chlorinate to make it free from

bad colour

bacteria

temporary hardness

permanent hardness

Answer Details

Water for town supply is chlorinated to make it free from bacteria. Chlorine is added to water in small amounts to kill harmful bacteria and other microorganisms that may be present in the water. These microorganisms can cause illnesses such as typhoid, cholera, and dysentery if ingested. Chlorination is an effective and inexpensive method to disinfect water and protect public health. Chlorine works by breaking down the cell walls of microorganisms, which leads to their death. Chlorine also helps to control the growth of algae and other microorganisms in the water supply, which can cause taste and odor problems. Chlorination does not remove other impurities from the water such as bad color, temporary hardness, or permanent hardness. These impurities may require additional treatment methods such as filtration or water softening to be removed.

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Question 23 Report

The carbon atoms on ethane are

sp2 hybridized

sp3 hybridized

sp4 hybridized

sp hybridized

Answer Details

The carbon atoms in ethane are sp3 hybridized. Hybridization is a concept in chemistry that describes how the orbitals of an atom combine to form new hybrid orbitals. In the case of carbon, which has four valence electrons, the four orbitals (s, px, py, and pz) combine to form four hybrid orbitals that have equal energy and are oriented in a tetrahedral arrangement around the carbon atom. In ethane, each carbon atom is bonded to three other atoms: two other carbon atoms and three hydrogen atoms. Each of the carbon-hydrogen bonds is formed by overlapping the sp3 hybrid orbitals on the carbon atom with the 1s orbitals on the hydrogen atoms. The carbon-carbon bond is formed by overlapping two of the sp3 hybrid orbitals on each carbon atom. Overall, the sp3 hybridization of the carbon atoms in ethane allows for the formation of strong and stable covalent bonds between the carbon and hydrogen atoms, as well as between the two carbon atoms.

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

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Question 24 Report

The shape of the S-orbital is

elliptical

spiral

circular

spherical

Answer Details

The shape of the S-orbital is spherical. This means that it has a symmetrical, 3-dimensional shape that resembles a ball or a globe. It is not flat or elongated, but rather is evenly rounded in all directions. This is the most basic type of orbital and is used to describe the electrons in the outermost energy level of an atom.

Read lesson note on Atomic Structure And Bonding (JAMB)

Atomic Structure And Bonding

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Question 25 Report

Aluminium hydroxide is used in the dyeing industry as a

dye

dispersant

salt

mordant

Answer Details

Aluminium hydroxide is used in the dyeing industry as a mordant. It combines with a dye and thereby fixes it in a material.

Read lesson note on Chemical Kinetics And Equilibrium Systems (Ghana Only) (WAEC)

Read lesson note on Chemical Bonds (Ghana Only) (WAEC)

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

Chemical Kinetics And Equilibrium Systems (Ghana Only)

Chemical Bonds (Ghana Only)

Periodic Chemistry (Ghana Only)

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Question 26 Report

When ΔH is negative, a reaction is said to be

endothermic

exothermic

reversible

ionic

Answer Details

When ΔH is negative, a reaction is exothermic. In simpler terms, exothermic reactions release energy into their surroundings, which is often felt as heat. When ΔH is negative, it means that the enthalpy change of the reaction is negative, indicating that energy is being released. This can be seen as a decrease in temperature in the surrounding environment or a release of heat or light. Examples of exothermic reactions include combustion reactions, where a fuel combines with oxygen to release heat and light, or the reaction between an acid and a base to form a salt and water. In both cases, energy is released into the surrounding environment, and the enthalpy change is negative (ΔH < 0).

Read lesson note on Energy Changes (JAMB)

Energy Changes

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Question 27 Report

The gas that can be collected by downward displacement of air is

chlorine

sulphur (IV) oxide

carbon (IV) oxide

ammonia

Answer Details

Upward delivery works well for hydrogen and ammonia, which are both less densed than air. Sometimes, they are collected over water.

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 28 Report

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

k [P]x [Q]y

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

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

\frac{m [R] n [S]}{x [P] y [Q]}

Answer Details

Expression for equilibrium constant

k = $\frac{concentration of product}{concentration of reactant}$

Read lesson note on Chemical Equilibria (JAMB)

Chemical Equilibria

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Question 29 Report

CH2S(g) + O2(g) → 2Cn + SO2(g)

What is the change in the oxidation number of copper in the reaction?

0 to + 2

0 to + 1

C + 1 to 0

+ 2 to + 1

Answer Details

In the reactant;

Cu2S

2 Cu - 2(1) = 0

2 Cu = 2

Cu = $\frac{2}{2}$

Cu = +1

In the product, Cu

Cu = O

The oxidation number of Cu in Cu2S and Cu respectively is +1 and 0 respectively

Read lesson note on Oxidation And Reduction (JAMB)

Oxidation And Reduction

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Question 30 Report

The enzyme used in the hydrolysis of starch to dextrin and maltose is

amylase

diastatse

invertase

zymase

Answer Details

This is any amylase or a mixture of amylases that converts starch to dextrin and maltose.

Read lesson note on Structure Of The Atom (Ghana Only) (WAEC)

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

Structure Of The Atom (Ghana Only)

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Question 31 Report

The enzyme used in the hydrolysis of starch to dextrin and maltose is

Zymase

Sucrase

Amylase

Lactase

Answer Details

The enzyme used in the hydrolysis of starch to dextrin and maltose is called amylase. Starch is a complex carbohydrate found in many foods, including grains, potatoes, and vegetables. It consists of long chains of glucose molecules that are linked together. To digest starch, our bodies use the enzyme amylase, which breaks down the long chains into smaller molecules of dextrin and maltose. These smaller molecules are more easily absorbed into the bloodstream and used as energy by the body. Amylase is produced in the salivary glands and pancreas and is released into the digestive system when we eat. In the mouth, amylase begins breaking down starch as we chew our food, and then continues its work in the small intestine where it is released by the pancreas. In summary, amylase is the enzyme responsible for breaking down starch into smaller molecules of dextrin and maltose during digestion.

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Question 32 Report

The furring of kettles is caused by the presence in water of

calcium hydrogentrioxocarbonate (IV)

calcium trioxocarbonate (IV)

calcium tetraoxosulphate (VI)

calcium hydroxide

Answer Details

Furring of kettles is caused by the temporary hardness in water. Temporary hardness in water is caused by calcium and magnesium trioxocarbonate (IV)
CaCO3 causes the furring of kettles

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

Acids, Bases And Salts

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Question 33 Report

Calculate the amount in moles of a gas which occupies 10.5 dm3 at 6 atm and 30oC [R = 082 atm dm3 K-1 mol-1]

2.536

1.623

4.736

0.394

Answer Details

For an ideal gas PV = nRT

Amount in moles = n

Volume v = 10.5dm3

Pressure P = 6atm

Temperature T = 30°C + 273 = 303k

R, Gas constant = 0.082 atmdm3k-1 mol

Recall from ideal gas equation

pv = nRT

n = $\frac{R V}{R T}$

n = $\frac{6 \times 105}{0.082 \times 303}$

n= 2.536mol

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

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Question 34 Report

Ca(OH)2(s) + 2NH4Cl(g) → CaCl2(s) + 2H2O2(l) + X. In the reaction above X is

NO2

NH3

N2O

H2O

Answer Details

In the given chemical equation, the reactants are solid calcium hydroxide (Ca(OH)2) and gaseous ammonium chloride (NH4Cl). When they react, they form solid calcium chloride (CaCl2), liquid water (H2O), and X, an unknown product. To determine the identity of X, we need to balance the chemical equation first. The balanced chemical equation is: Ca(OH)2(s) + 2NH4Cl(g) → CaCl2(s) + 2H2O(l) + 2NH3(g) As per the balanced chemical equation, the product X is 2 molecules of ammonia (NH3). Therefore, the answer is (b) NH3.

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

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Question 35 Report

In the diagram above, Y is

fused CaO

H2O

NaOH

Concentrated H2SO4

Answer Details

In the preparation of sulphur dioxide by the action of dilute acids on sulphates and bisulphites. conc H2SO4 helps to release SO2 from the mixture.
The setup represents the production of sulphur dioxide

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 36 Report

In the laboratory preparation of ethyl ethanoate, the water present in the mixture is removed using a solution of

an hydrous CaCl4

concentrated NaCO4

dilute NaOH

concentrated H4SO4

Answer Details

This is used in the preparation of ethyl ethanoate to remove the water present since it serves as a dehydrated agent.

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

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Question 37 Report

The tincture of iodine means iodine dissolved in

ethanol

bromine chloride

chlorine water

water

Answer Details

It is also called weak iodine solution. Tincture solutions are characterized by the presence of alcohol.

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

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Question 38 Report

Incomplete oxidation of ethanol yields

CH3COOH

CH3COCH3

CH3CH2OCH2CH3

CH2CHO

Answer Details

C2H5OH → CH3CHO → CH3COOH
Ethanol.....oxidation Ethanal........Ethanoic Acid
Primary alcohol oxidises to aldehyde and later to carboxylic acid.
Secondary alcohol oxidises to ketones.
Ethanol is an example of primary alcohol and primary alcohol can be oxidised to aldehyde and carboxylic acid. Wherein, incomplete oxidation of primary alcohol yields aldehyde also known as alkaline while complete oxidation of primary alcohol yields carboxylic acid.

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