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JAMB UTME - Chemistry - 2019

Question 1 Report

The electronic configuration of element Z is 1s $^{2}$ 2s $^{2}$ 2p $^{6}$ 3s $^{2}$ 3p $^{1}$ . What is the formula of the compound formed between Z and tetraoxosulphate (VI) ion.

Z

_{3}

(SO

_{4}

)

_{2}

ZSO

_{4}

Z

_{2}

SO

_{4}

Z

_{2}

(SO

_{4}

)

_{3}

Answer Details

Z = 1s $^{2}$ 2s $^{2}$ 2p $^{6}$ 3s $^{2}$ 3p $^{1}$
$∴$ We have Z $^{3 +}$ and SO $_{4}^{2 -}$
The reaction : Z $^{3 +}$ + SO $_{4}^{2 -}$ $\to$ Z $_{2}$ (SO $_{4}$ ) $_{3}$ .

Read lesson note on Atomic Structure And Bonding (JAMB)

Atomic Structure And Bonding

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Question 2 Report

How many alkoxyalkanes can be obtained from the molecular formula C $_{4}$ H $_{10}$ O?

4

2

1

3

Chemistry Of Carbon Compounds (Nigeria Only)

Acids, Bases And Salts (Nigeria Only)

Chemical Equilibra

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Question 4 Report

The IUPAC name of the compound CF $_{3}$ CHBrCl is

2-chloro-2-bromo-1,1,1-trifluoroethane

2-bromo-2-chloro-1,1,1-trifluoroethane

1-bromo-1-chloro-2,2,2-trifluoroethane

1,1,1-trifluoro-2-bromo-2-chloroethane

Periodic Chemistry (Ghana Only)

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Question 5 Report

A certain hydrocarbon on complete combustion at s.t.p produced 89.6dm $^{3}$ of CO $_{2}$ and 54g of water. The hydrocarbon should be

C

_{6}

H

_{6}

C

_{4}

H

_{10}

C

_{5}

H

_{10}

C

_{4}

H

_{6}

Chemistry Of Carbon Compounds (Nigeria Only)

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Question 6 Report

Hydrocarbons which will react with Tollen's reagent conform to the general formula

C

_{n}

H

_{2 n}

C

_{n}

H

_{2 n + 2}

C

_{n}

H

_{2 n - 2}

C

_{n}

H

_{2 n + 1}

OH

Chemistry Of Carbon Compounds (Nigeria Only)

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Question 12 Report

The cost of discharging 6.0g of a divalent metal, X from its salt is ₦12.00. What is the cost of discharging 9.0g of a trivalent metal, Y from its salt under the same condition?

[X = 63, Y = 27, 1F = 96,500C]

N12.00

N27.00

N63.00

N95.00

Answer Details

For X: X $^{2 +}$ + 2e $^{-}$ $\to$ X
2F = 63g
xF = 6g
x = $\frac{6 \times 2}{63} = \frac{4}{21} F$
$\frac{4}{21}$ F = N12.00
1F = $\frac{12}{\frac{4}{21}}$
= N63.00
1F is equivalent to N63.00.
For Y: Y $^{3 +}$ + 3e $^{-}$ $\to$ Y
3F = 27g
xF = 9g
x = $\frac{3 \times 9}{27}$
= 1F
1F = N63.00

Read lesson note on Chemical Combination (JAMB)

Chemical Combination

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Question 13 Report

At 27°C, 58.5g of sodium chloride is present in 250cm $^{3}$ of a solution. The solubility of sodium chloride at this temperature is?

(molar mass of sodium chloride = 111.0gmol $^{- 1}$ )

2.0 moldm

^{- 3}

0.25 moldm

^{- 3}

1.0 moldm

^{- 3}

0.5 moldm

^{- 3}

Answer Details

Given the Mass of the salt = 58.5g
Volume = 250 cm $^{3}$ = 0.25 dm $^{3}$
Mass concentration = $\frac{M a s s}{V o l u m e}$
= $\frac{58.5}{0.25}$ = 234 gdm $^{- 3}$
Solubility (in moldm $^{- 3}$ = $\frac{234}{111}$
= 2.11 moldm $^{- 3}$
$≊$ 2.0 moldm $^{- 3}$

Read lesson note on Solubility Of Substances (Ghana Only) (WAEC)

Solubility Of Substances (Ghana Only)

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Question 15 Report

Which of the following is a set of neutral oxides?

N

_{2}

O, H

_{2}

O, CO

N

_{2}

O, SO

_{2}

, CO

N

_{2}

O, Na

_{2}

O, ZnO

N

_{2}

O, SO

_{3}

, NO

Oxidation And Reduction

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Question 16 Report

By what amount must the temperature of 200cm $^{3}$ of Nitrogen at 27°C be increased to double the pressure if the final volume is 150cm $^{3}$ (Assume ideality)

150°C

177°C

75°C

120°C

Answer Details

Using the ideal gas law and equation:
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$\frac{P_{1} \times 200 {cm}^{3}}{300 K} = \frac{2 P \times 150 {cm}^{3}}{T_{2}}$
Cross multiply:
$T_{2} = \frac{300 \times 150 \times 2 P}{200 \times P}$
$= 450 K$ or $177^{\circ} C$
Don't forget to convert to $^{\circ} C$

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

Kinetic Theory Of Matter And Gas Laws

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Question 17 Report

Methane is prepared in the laboratory by heating a mixture of sodium ethanoate with soda lime. The chemical constituent(s) of soda lime is/are

Ca(OH)

_{2}

and NaOH

Na

_{2}

CO

_{3}

NaCl

NaOH

Chemical Combination

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Question 19 Report

For the general equation of the nature

XP + yQ ⇌ mR + nS, the expression for the equilibrium constant is

k[P]

^{x}

[Q]

^{y}

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

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

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

Answer Details

The expression for the equilibrium constant for the general equation XP + yQ ⇌ mR + nS is: Kc = [R]m[S]n / [P]x[Q]y where Kc is the equilibrium constant, [R] and [S] are the concentrations of the products, and [P] and [Q] are the concentrations of the reactants, all raised to the stoichiometric coefficients (m, n, x, y) in the balanced equation. This equation is known as the equilibrium constant expression and it represents the ratio of the concentrations of the products and reactants at equilibrium for a particular chemical reaction. The equilibrium constant is a measure of how far a reaction proceeds towards completion, with a larger value indicating a greater extent of reaction. The equilibrium constant expression is derived from the law of mass action, which states that the rate of a chemical reaction is proportional to the product of the concentrations of the reactants raised to their stoichiometric coefficients. At equilibrium, the rates of the forward and reverse reactions are equal, and the equilibrium constant expression represents the ratio of the rate constants for these two reactions. Therefore, the correct expression for the equilibrium constant for the general equation XP + yQ ⇌ mR + nS is Kc = [R]m[S]n / [P]x[Q]y.

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

Read lesson note on Chemical Equilibra (JAMB)

Chemical Kinetics And Equilibrium Systems (Ghana Only)

Chemical Equilibra

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Question 21 Report

Which of the following could not be alkane?

C

_{4}

H

_{10}

C

_{5}

H

_{12}

C

_{7}

H

_{14}

C

_{8}

H

_{18}

Chemistry Of Carbon Compounds (Nigeria Only)

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Question 22 Report

When chlorine water is exposed to bright sunlight, the following products are formed

HCl

_{(a q)}

and Cl

_{2}

HOCl

_{(a q)}

and Cl

_{2}

HCl

_{(a q)}

and O

_{2}

HOCl

_{(a q)}

and O

_{2}

Chemistry Of Carbon Compounds (Ghana Only)

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Question 24 Report

Which two gases can be used for the demonstration of the fountain experiment?

SO

_{3}

and NH

_{3}

SO

_{2}

and NH

_{3}

HCl and NH

_{3}

SO

_{2}

and HCl

Kinetic Theory Of Matter And Gas Laws

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Question 27 Report

A cell shorthand notation can be written as A / A $^{+}$ // B $^{2 +}$ /B. The double slash in the notation represents the

electrolyte

electrodes

salt bridge

solid - liquid interface

Electrolysis

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Question 28 Report

X is a substance which liberates CO $_{2}$ on treatment with concentrated H $_{2}$ SO $_{4}$ . A warm solution of X can decolorize acidified KMnO $_{4}$ . X is

HCOOH

NaHCO

_{3}

FeCO

_{3}

H

_{2}

C

_{2}

O

_{4}

Answer Details

It should be noted that for X to liberate CO $_{2}$ , X must be a carbonate or an oxalate. Since X decolorizes KMnO $_{4}$ , X must be an oxalate.
Therefore, X is H $_{2}$ C $_{2}$ O $_{4}$ .

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

Periodic Chemistry (Ghana Only)

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Question 29 Report

The oxidation state(s) of nitrogen in ammonium nitrite is/are

-3, +3

+5

-3, +5

-3

Answer Details

Ammonium nitrite = NH $_{4}$ NO $_{2}$
NH $_{4}^{+}$ : Let the oxidation number of Nitrogen = x
x + 4 = 1 $⟹$ x = 1 - 4
x = -3
NO $_{2}^{-}$ : x - 4 = -1
x = -1 + 4 $⟹$ x = +3.
The oxidation numbers for Nitrogen in Ammonium Nitrite = -3, +3.

Read lesson note on Oxidation And Reduction (JAMB)

Oxidation And Reduction

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Question 30 Report

The heat of formation of ethene, C $_{2}$ H $_{4}$ is 50 kJmol $^{- 1}$ , and that of ethane, C $_{2}$ H $_{6}$ is -82kJmol $^{- 1}$ . Calculate the heat evolved in the process:

C $_{2}$ H $_{4}$ + H $_{2}$ $\to$ C $_{2}$ H $_{6}$

-122kJ

-132kJ

-88kJ

-60kJ

Energy Changes

Stoichiometry And Chemical Reactions

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Question 31 Report

The IUPAC nomenclature of the compound

H $_{3}$ C - CH(CH $_{3}$ ) - CH(CH $_{3}$ ) - CH $_{2}$ - CH $_{3}$

3,4- dimethylhexane

2,3- dimethyl pentane

2- ethyl hexane

2,3- dimethylhexane

Chemistry Of Carbon Compounds (Nigeria Only)

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Question 33 Report

SO $_{3}$ is not directly dissolved in water in the industrial preparation of H $_{2}$ SO $_{4}$ by the contact process because

SO

_{3}

is an acidic gas

Acid is usually added to water and not water to acid

The reaction between SO

_{3}

and water is violently exothermic

SO

_{3}

does not dissolve readily in water

Acids, Bases And Salts

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Question 37 Report

The reaction $2 H_{2} S + S O_{2} \to 3 S + 2 H_{2} O$ is

not a redox reaction because there is no oxidant in the reaction

not a redox reaction because there is no reductant in the reaction

a redox reaction in which H

_{2}

S is the oxidant and SO

_{2}

is the reductant

a redox reaction in which SO

_{2}

is the oxidant and H

_{2}

S is the reductant

View Answer

Question 39 Report

200cm $^{3}$ of 0.50mol/dm $^{3}$ solution of calcium hydrogen trioxocarbonate (IV) is heated. The maximum weight of solid precipitated is

10g

25g

20g

15g

Answer Details

To solve this problem, we need to use the concept of stoichiometry and the solubility product constant (Ksp) of calcium hydrogen trioxocarbonate (IV). First, we need to write the balanced equation for the reaction that occurs when the solution of calcium hydrogen trioxocarbonate (IV) is heated: Ca(HCO3)2(s) → CaCO3(s) + H2O(g) + CO2(g) From the balanced equation, we can see that 1 mole of calcium hydrogen trioxocarbonate (IV) produces 1 mole of calcium carbonate. Therefore, we need to determine the number of moles of calcium hydrogen trioxocarbonate (IV) in the solution: Number of moles = concentration x volume Number of moles = 0.50 mol/dm³ x 0.2 dm³ Number of moles = 0.1 mol Since 1 mole of calcium hydrogen trioxocarbonate (IV) produces 1 mole of calcium carbonate, the number of moles of calcium carbonate produced will also be 0.1 mol. Next, we need to use the solubility product constant (Ksp) of calcium carbonate to determine the maximum amount of solid that can be precipitated: Ksp = [Ca²⁺][CO3²⁻] Ksp = 3.3 x 10⁻⁹ (at 25°C) At the maximum amount of solid precipitated, all the calcium carbonate formed will have precipitated, and the concentration of calcium ions and carbonate ions will be equal. Therefore, we can assume that the concentration of calcium ions and carbonate ions is both x. Substituting into the Ksp expression: Ksp = x² 3.3 x 10⁻⁹ = x² x = 5.74 x 10⁻⁵ mol/dm³ The mass of calcium carbonate precipitated can now be calculated: Mass = number of moles x molar mass Mass = 0.1 mol x 100.1 g/mol Mass = 10.01 g Therefore, the maximum weight of solid precipitated is approximately 10 g. Note that this calculation assumes that all the calcium carbonate precipitated as a solid, which may not always be the case in a real-world experiment. Additionally, this calculation does not take into account any losses due to filtration or other experimental errors.

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

Kinetic Theory Of Matter And Gas Laws

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Question 40 Report

The two ions responsible for hardness in water are

Ca

^{2 +}

and / or Mg

^{2 +}

K

^{+}

and / or Mg

^{2 +}

Ca

^{2 +}

and / or Li

^{+}

Ca

^{2 +}

and / or Na

^{+}

View Answer

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