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JAMB UTME - Chemistry - 2019

Ibeere 1 Ìròyìn

If the cost of electricity required to discharge 10g of an ion X $^{3 +}$ is N20.00, how much would it cost to discharge 6g of ion Y $^{2 +}$ ?

[1 faraday = 96,500C, atomic masses are X = 27, Y = 24]

N10.00

N6.00

N20.00

N9.00

Awọn alaye Idahun

X $^{3 +}$ + 3e $^{-}$ $\to$ X
3F = 27g
xF = 10g
$\frac{x}{3} = \frac{10}{27} ⟹ x = \frac{10}{9} F$
$\frac{10}{9}$ F $\equiv$ N20.00
1F is equivalent to x
$\frac{1}{\frac{10}{9}} = \frac{x}{20}$
$\frac{9}{10} = \frac{x}{20} ⟹ x = N 18.00$
1F is equivalent to N18.00.
Y $^{2 +}$ + 2e $^{-}$ $\to$ Y
2F = 24g
xF = 6g
x = $\frac{6 \times 2}{24} = \frac{1}{2} F$
1F = N18.00
$\frac{1}{2}$ F = $\frac{1}{2} \times N 18.00$
= N9.00

Ibeere 2 Ìròyìn

Which of the following reactions is an oxidation process?

2H

^{+}

+ 2e

^{-}

\to

H

_{2}

2O

^{2 -}

+ 4e

^{-}

\to

O

_{2}

OCl

^{-}

+ H

_{2}

O + 2e

^{-}

\to

2OH

^{-}

+ Cl

^{-}

Cl + e

^{-}

\to

Cl

^{-}

Oxidation And Reduction

Ibeere 5 Ìròyìn

An element Z contains 80% of $_{8}^{16}$ Z and 20% of $_{8}^{18}$ Z. Its relative atomic mass is

16.2

17.6

16.0

16.4

Awọn alaye Idahun

R.A.M of Z = $16 (\frac{80}{100}) + 18 (\frac{20}{100})$
= $12.8 + 3.6$
= 16.4

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

Read lesson note on Atomic Structure And Bonding (JAMB)

Separation Of Mixtures And Purification Of Chemical Substances

Atomic Structure And Bonding

Ibeere 6 Ìròyìn

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)

Ibeere 7 Ìròyìn

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

Ibeere 10 Ìròyìn

In the reaction between sodium hydroxide and tetraoxosulphate (VI) solutions, what volume of 0.5 molar sodium hydroxide would exactly neutralize 10cm $^{3}$ of 1.25 molar tetraoxosulphate (vi) acid?

25cm

^{3}

10cm

^{3}

20cm

^{3}

50cm

^{3}

Awọn alaye Idahun

Equation of reaction : 2NaOH + H $_{2}$ SO $_{4}$ → Na $_{2}$ SO $_{4}$ + 2H $_{2}$ O

Concentration of a base, CB = 0.5M

Volume of acid, V $_{A}$ = 10cm $^{3}$

Concentration of an acid, C $_{A}$ = 1.25M

Volume of base, V $_{B}$ = ?

Recall:
$\frac{C_{A} V_{A}}{C_{B} V_{B}} = \frac{n_{A}}{n_{B}}$ ... (1)
N.B: From the equation,
$\frac{n_{A}}{n_{B}} = \frac{1}{2}$

From (1)
$\frac{1.25 \times 10}{0.5 \times V_{B}} = \frac{1}{2}$
$\frac{12.5}{0.5 V_{B}} = \frac{1}{2}$
25 = 0.5V $_{B}$

VB = 50.0 cm $^{3}$

Read lesson note on Stoichiometry And Chemical Reactions (WAEC)

Stoichiometry And Chemical Reactions

Ibeere 11 Ìròyìn

Elements X and Y have electronic configurations 1s $^{2}$ 2s $^{2}$ 2p $^{4}$ and 1s $^{2}$ 2s $^{2}$ 2p $^{6}$ 3s $^{2}$ 3p $^{1}$ respectively. When they combine, the formula of the compound formed is

XY

X

_{2}

Y

Y

_{2}

X

_{3}

X

_{2}

Y

_{3}

Ibeere 12 Ìròyìn

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)

Ibeere 13 Ìròyìn

Consider the reaction: A + 2B $_{(g)} ⇌$ 2C + D $_{(g)}$ ( $Δ$ H = +ve)

What will be the effect of decrease in temperature on the reaction?

Rate of backward reaction will increase

The equilibrium constant will shift

Rates of forward and backward reactions are not affected

Rate of forward reaction will increase

Chemical Equilibra

Ibeere 14 Ìròyìn

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

Ibeere 15 Ìròyìn

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

Ibeere 17 Ìròyìn

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

Awọn alaye Idahun

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

Ibeere 19 Ìròyìn

What mass of magnesium would be obtained by passing a current of 2 amperes for 2 hours, through molten magnesium chloride?
[1 faraday = 96500C, Mg = 24]

1.12g

4.48g

1.80g

2.00g

Awọn alaye Idahun

Current (I) = 2A; Time (t) = 2 hours = 7200 secs
Q = It
= 2 x 7200 = 14400C
1 F = 96500C
x = 14400C
x = $\frac{14400}{96500}$
= 0.15F
Mg $^{2 +}$ + 2e $^{-}$ $\to$ Mg
2F $\to$ 24g
0.15F $\to$ x
2x = 24 x 0.15
x = $\frac{24 \times 0.15}{2}$
= 1.8g

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

Read lesson note on Electrolysis (JAMB)

Chemical Kinetics And Equilibrium Systems (Ghana Only)

Electrolysis

Ibeere 20 Ìròyìn

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}

Awọn alaye Idahun

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)

Ibeere 21 Ìròyìn

The combustion of carbon(ii)oxide in oxygen can be represented by equation.

2CO + O $_{2}$ ? 2CO $_{2}$

Calculate the volume of the resulting mixture at the end of the reaction if 50cm $_{3}$ of carbon(ii)oxide was exploded in 100cm $_{3}$ of oxygen

75cm

^{3}

125cm

^{3}

100cm

^{3}

50cm

^{3}

Kinetic Theory Of Matter And Gas Laws

Stoichiometry And Chemical Reactions

Ibeere 22 Ìròyìn

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)

Ibeere 23 Ìròyìn

The emission of two successive beta particles from the nucleus $_{15}^{32} P$ will produce

_{14}^{30}

Si

_{13}^{32}

Al

_{17}^{32}

Cl

_{16}^{32}

S

Awọn alaye Idahun

Explanation:

_{15}^{32}

P

\to

2

_{- 1}^{0}

e +

_{b}^{a}

X
32 = 2(0) + a
a = 32
15 = 2(-1) + b
b = 15 + 2
b = 17

∴

_{b}^{a}

X =

_{17}^{32}

Cl

Read lesson note on Structure Of The Atom (WAEC)

Structure Of The Atom

Ibeere 27 Ìròyìn

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

Ibeere 28 Ìròyìn

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

^{+}

Ibeere 29 Ìròyìn

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}}

Awọn alaye Idahun

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

Ibeere 30 Ìròyìn

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

Ibeere 32 Ìròyìn

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)

Ibeere 33 Ìròyìn

Na $_{2}$ CO $_{3}$ + 2HCl $\to$ 2NaCl + H $_{2}$ O + CO $_{2}$

The indicator most suitable for this reaction should have a pH equal to

5

7

3

9

Acids, Bases And Salts

Ibeere 37 Ìròyìn

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}

Awọn alaye Idahun

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

Ibeere 39 Ìròyìn

The velocity, V of a gas is related to its mass, M by (k = proportionality constant)

V =

\frac{k}{M}

V =

\frac{k}{M^{\frac{1}{2}}}

V =

k M^{2}

V =

(\frac{k}{M})^{\frac{1}{2}}

Awọn alaye Idahun

Recall:
V = $\sqrt{\frac{3 R T}{M}}$
$∴ V \propto \frac{1}{\sqrt{M}}$
$V = \frac{k}{\sqrt{M}}$
V = $\frac{k}{M^{\frac{1}{2}}}$

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

Kinetic Theory Of Matter And Gas Laws

Ibeere 40 Ìròyìn

If acidified Potassium Dichromate(VI) (K $_{2}$ Cr $_{2}$ O $_{7}$ ) acts as oxidizing agent, color changes from

yellow to green

orange to green

yellow to red

orange to red

Oxidation And Reduction

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