JAMB UTME - Chemistry - 2017

View Answer

Question 10 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)

View Answer

Question 11 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

View Answer

Question 12 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)

View Answer

Question 13 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)

View Answer

Question 14 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)

View Answer

Question 15 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)

View Answer

Question 16 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

View Answer

Question 17 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

View Answer

Question 18 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

View Answer

Question 19 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)

View Answer

Question 20 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)

View Answer

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

View Answer

Question 22 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)

View Answer

Question 23 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

View Answer

Question 24 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)

View Answer

Question 25 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)

View Answer

Question 26 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

View Answer

Question 27 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

View Answer

Question 28 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

View Answer

Question 29 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)

View Answer

Question 30 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)

View Answer

Question 31 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

View Answer

Question 32 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)

View Answer

Question 33 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

View Answer

Question 34 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)

View Answer

Question 35 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)

View Answer

Question 36 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)

View Answer

Question 37 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)

View Answer

Question 38 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)