JAMB UTME - Chemistry - 2017

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Question 2 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 3 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 4 Report

X(g) + 3Y(g) ---- 2z(g) H = +ve. if the reaction above takes place at room temperature, the G will be

negative

zero

positive

indeterminate

Answer Details

$\begin{array}{cc} Enthalpy Change [ΔH] & Entropy Change [ΔS] & Gibbs free Energy[ΔG] \\ Positive & Positive & depends on T, may be + or - \\ Negative & Positive & always negative \\ Negative & Negative & depends on T, may be + or - \\ Positive & Negative & always positive \end{array}$

?G= ?H ? T?S.
To determine whether ?G will be positive or negative, the value of ?H(change in enthalpy) and ?S (change in entropy) must be given. Likewise the temperature.

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

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

Chemical Kinetics And Equilibrium Systems (Ghana Only)

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Question 5 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 6 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 7 Report

The process that occurs when two equivalent forms of a compound are in equilibrium is

Isotopy

Resonance

Isomerism

Reforming

Answer Details

Resonance involves two forms of a compound.
Isomerism involves two or more forms of an element.
Reforming involves the rearrangement of molecule.

This deals with a two forms of a molecule where the chemical connectivity is the same but the electrons are distributed differently around the structure.

Read lesson note on Chemical Equilibria (JAMB)

Chemical Equilibria

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Question 8 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 9 Report

The IUPAC nomenclature of the structure above is

3-methybut-3-ene

2-methylbut-1-ene

2-ethylprop-1-ene

2-methylbut-2-ene

Answer Details

CH2CH2 - CCH3CH2
Start the numbering from the terminal carbon.

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

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Question 10 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)

Acids, Bases And Salts (Ghana Only)

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Question 11 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 12 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 13 Report

The densities of two gases, X and Y are 0.5gdm-3 and 2.0gdm-3 respectively. What is the rate of diffusion of X relative to Y?

0.1

0.5

2.0

4.0

Answer Details

The rate of dimension of a gas inversely proportional to the square root of its molecular mass or its density, which is Graham's Law of diffusion of gas.

R ∝ $\frac{1}{\sqrt{M m}}$ or R ∝ $\frac{1}{\sqrt{D}}$

Dx = 0.5gdm-3, Dy = 2gdm-3

R= $\frac{K}{\sqrt{D}}$

R $\sqrt{D}$ = k

R1 $\sqrt{D_{1}}$ = R1 $\sqrt{D_{2}}$

Rx $\sqrt{D_{x}}$ = Ry $\sqrt{D_{y}}$

$\frac{R_{x}}{R_{y}}$ = $\frac{\sqrt{D_{y}}}{\sqrt{D_{x}}}$

= $\frac{\sqrt{2}}{\sqrt{0.5}}$

= 2.0

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

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Question 14 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 15 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)

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Question 16 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 17 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)

Acids, Bases And Salts (Ghana Only)

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Question 18 Report

$ {}^{226}_{88}\mathrm{Ra} \rightarrow {}^{x}_{86}\mathrm{Rn} $ + alpha particle

226

220

227

222

Answer Details

The given equation represents the alpha decay of ^226Ra into an unknown nuclide ^86Rn and an alpha particle. During alpha decay, an atomic nucleus emits an alpha particle, which consists of two protons and two neutrons bound together, resulting in a decrease of two in the atomic number and a decrease of four in the mass number. To determine the atomic number and mass number of the unknown nuclide ^86Rn, we can subtract the atomic and mass numbers of the alpha particle from those of ^226Ra: Atomic number: 88 (Ra) - 2 (alpha particle) = 86 (Rn) Mass number: 226 (Ra) - 4 (alpha particle) = 222 (Rn) Therefore, the unknown nuclide produced in the alpha decay of ^226Ra is ^222Rn, and the correct option is (D) 222.

Read lesson note on Chemical Combination (JAMB)

Read lesson note on Atomic Structure And Bonding (JAMB)

Chemical Combination

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Question 19 Report

The shape of ammonia molecules is

trigonal planar

octahedral

square planar

tetrahedral

Answer Details

The shape of ammonia (NH3) molecules is tetrahedral. The ammonia molecule has one nitrogen atom and three hydrogen atoms. The nitrogen atom is at the center of the molecule, with the three hydrogen atoms arranged around it. The shape of the molecule is determined by the arrangement of the electrons in the outermost energy level of the nitrogen atom, which is called the valence shell. In the valence shell of the nitrogen atom, there are four pairs of electrons, one of which is a lone pair, while the other three are bonded to hydrogen atoms. These four pairs of electrons repel each other, and try to get as far away from each other as possible. This results in a tetrahedral arrangement of the atoms around the nitrogen atom. A tetrahedron is a three-dimensional shape with four triangular faces, and it looks like a pyramid with a triangular base. In the case of ammonia, the three hydrogen atoms form a triangular base, and the lone pair of electrons sits at the top of the pyramid, creating a tetrahedral shape. Therefore, the shape of ammonia molecules is tetrahedral.

Read lesson note on Atomic Structure And Bonding (JAMB)

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Question 20 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 21 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 22 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 23 Report

A given mass of gas occupies 2dm3 at 300k. At what temperature will its volume be doubled, keeping the pressure constant?

400k

480k

550k

600k

Answer Details

To solve this problem, we can use the combined gas law, which states that the product of pressure and volume divided by temperature is constant, as long as the amount of gas and the pressure remain constant. In this case, the pressure is constant and the initial conditions of the gas are: - Volume (V1) = 2 dm3 - Temperature (T1) = 300 K Let's call the final temperature T2, and the final volume V2 = 2V1 = 4 dm3, since we want the volume to be doubled. Using the combined gas law, we can set up the following equation: P * V1 / T1 = P * V2 / T2 Simplifying the equation, we get: V1 / T1 = V2 / T2 Substituting the values we know, we get: 2 / 300 = 4 / T2 Cross-multiplying and solving for T2, we get: T2 = 4 * 300 / 2 = 600 K Therefore, the temperature at which the gas will double its volume, keeping the pressure constant, is 600 K. Option (d) is the correct answer.

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

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Question 24 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 25 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 26 Report

Ethene is prepared industrially by

Reforming

Polymerization

Distillation

Cracking

Answer Details

Ethene, also known as ethylene, is prepared industrially by a process called cracking. Cracking involves breaking down larger hydrocarbon molecules, typically obtained from crude oil or natural gas, into smaller ones. This is usually done by heating the hydrocarbons to very high temperatures (up to 850°C) in the presence of a catalyst. The cracking process breaks the long chains of hydrocarbons into shorter ones, including the production of ethene. This is because the high temperature and catalyst cause the chemical bonds in the hydrocarbon molecules to break apart, leading to the formation of smaller molecules. These smaller molecules, including ethene, can then be separated from the other products and used in various applications such as the production of plastics. Therefore, out of the given options, the industrial preparation of ethene is accomplished by the process of cracking.

Read lesson note on Chemical Industry And Environment (WAEC)

Chemical Industry And Environment

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Question 27 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 28 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

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Air

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Question 29 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 30 Report

Due to the high reactivity of sodium, it is usually stored under

water

mercury

paraffin

phenol

Answer Details

Na is kept under kerosene (paraffin) to avoid reactivity with air.
Paraffin is also known as Kerosene.
Na(sodium) is kept in kerosene to prevent it from coming in contact with oxygen and moisture. If this happens, it will react with the moisture present in air and form sodium hydroxide.

Read lesson note on Metals And Their Compounds (JAMB)

Metals And Their Compounds

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Question 31 Report

The constituent common to duralumin and alnico is

Co

Mn

Al

Mg

Answer Details

The constituent common to duralumin and alnico is aluminum (Al). Duralumin is a type of aluminum alloy that typically contains around 4% copper and small amounts of other metals such as magnesium and manganese. It is known for its high strength-to-weight ratio and was commonly used in aircraft construction in the early 20th century. Alnico, on the other hand, is a type of permanent magnet alloy that contains aluminum, nickel, and cobalt (hence the name "alnico"). It also often contains small amounts of other metals such as copper and iron. Alnico magnets are known for their high magnetic strength and resistance to demagnetization. Although duralumin and alnico are very different in their properties and applications, they both contain aluminum as a common constituent. Aluminum is a versatile metal that is widely used in many different alloys due to its light weight, strength, and corrosion resistance.

Read lesson note on Atomic Structure And Bonding (JAMB)

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Question 32 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

Acids, Bases And Salts (Ghana Only)

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Question 33 Report

When few drops of concentrated trioxonitrate(V) acid is added to an unknown sample and wanned an intense yellow colouration is observed. The likely functional group present in the sample is

NH3—C—C=O

CHO

CO

CNH2

Answer Details

Xanthopreitic test for the presence of protein, when conc nitric acid is added to the drop, an intense yellow colouration is observed.

It contains all the functional group of protein which includes the amino, alkanol and the carboxylic group. Adding few drops of conc HNO3 to a protein, gives an intense yellow colouration.

It is called Xanthopreitic test.

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Acids, Bases And Salts

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Question 34 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 35 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 36 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 37 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)

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Question 38 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.

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Electrolysis

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Question 39 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

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Electrolysis

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Question 40 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)