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JAMB UTME - Chemistry - 2019

Question 1 Rapport

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

Détails de la réponse

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

Voir la réponse

Question 2 Rapport

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

Détails de la réponse

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

Lisez la note de cours sur Chemical Kinetics And Equilibrium Systems (Ghana Only) (WAEC)

Lisez la note de cours sur Electrolysis (JAMB)

Chemical Kinetics And Equilibrium Systems (Ghana Only)

Electrolysis

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Question 3 Rapport

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 Rapport

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)

Voir la réponse

Question 5 Rapport

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen. If the molar mass of the compound is 180. Find the molecular formula.

[H = 1, C = 12, O = 16]

C

_{3}

H

_{6}

O

_{3}

C

_{6}

H

_{6}

O

_{3}

C

_{6}

H

_{12}

O

_{6}

CH

_{2}

O

Chemistry Of Carbon Compounds (Nigeria Only)

Voir la réponse

Question 6 Rapport

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)

Voir la réponse

Question 8 Rapport

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}

Voir la réponse

Question 9 Rapport

Which of the following factors will speed up the rate of evolution of carbon (iv) oxide in the reaction below?

2HCl + CaCO $_{3}$ → CaCl $_{2}$ + H $_{2}$ O + CO $_{2}$

Increasing the concentration of the acid

Increasing the surface area of CaCO

_{3}

A and B

Reducing the concentration of the acid

Kinetic Theory Of Matter And Gas Laws

Oxidation And Reduction

Voir la réponse

Question 10 Rapport

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

Voir la réponse

Question 11 Rapport

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

Voir la réponse

Question 14 Rapport

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

Détails de la réponse

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

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Chemical Combination

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Question 15 Rapport

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

-3, +3

+5

-3, +5

-3

Détails de la réponse

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.

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Oxidation And Reduction

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Question 16 Rapport

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)

Voir la réponse

Question 18 Rapport

Which of the following conditions will most enhance the spontaneity of a reaction?

Δ

H is negative and greater than T

Δ

S

Δ

H is negative and

Δ

S = 0

Δ

H is positive and less than T

Δ

S

Δ

H is positive and equal to T

Δ

S

Kinetic Theory Of Matter And Gas Laws

Voir la réponse

Question 19 Rapport

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

Voir la réponse

Question 20 Rapport

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}

Détails de la réponse

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 $^{3}$

Lisez la note de cours sur Stoichiometry And Chemical Reactions (WAEC)

Stoichiometry And Chemical Reactions

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Question 21 Rapport

Consider the reaction

A $_{(s)}$ + 2B $_{(g)}$ → 2C $_{(a q)}$ + D $_{(g)}$

What will be the effect of a decrease in pressure on the reaction?

the equilibrium will shift

rate of forward reaction will increase

rate of backward reaction will increase

rates of forward and backward reaction are not affected

Détails de la réponse

Given: The equation below

A $_{(s)}$ + 2B $_{(g)}$ → 2C $_{(a q)}$ + D $_{(g)}$

Since we have a higher number of moles of gaseous species on the LHS, i.e 2, a decrease in pressure will favor the forward reaction.

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Lisez la note de cours sur Chemical Kinetics And Equilibrium Systems (Ghana Only) (WAEC)

Lisez la note de cours sur Periodic Chemistry (Ghana Only) (WAEC)

Kinetic Theory Of Matter And Gas Laws

Chemical Kinetics And Equilibrium Systems (Ghana Only)

Periodic Chemistry (Ghana Only)

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Question 24 Rapport

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

Détails de la réponse

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

Lisez la note de cours sur Separation Of Mixtures And Purification Of Chemical Substances (JAMB)

Lisez la note de cours sur Atomic Structure And Bonding (JAMB)

Separation Of Mixtures And Purification Of Chemical Substances

Atomic Structure And Bonding

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Question 27 Rapport

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}

Détails de la réponse

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 $_{4}$ .

Lisez la note de cours sur Periodic Chemistry (Ghana Only) (WAEC)

Periodic Chemistry (Ghana Only)

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Question 28 Rapport

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

^{+}

Voir la réponse

Question 30 Rapport

The IUPAC name for CH $_{3}$ CH $_{2}$ COOCH $_{2}$ CH $_{3}$ is

methylpropanoate

ethylethanoate

ethylpropanoate

methylethanoate

Chemistry Of Carbon Compounds (Nigeria Only)

Voir la réponse

Question 33 Rapport

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

Détails de la réponse

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

Lisez la note de cours sur Structure Of The Atom (WAEC)

Structure Of The Atom

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Question 34 Rapport

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

Détails de la réponse

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 $^{\circ} C$

Lisez la note de cours sur Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

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Question 35 Rapport

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}}

Détails de la réponse

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.

Lisez la note de cours sur Chemical Kinetics And Equilibrium Systems (Ghana Only) (WAEC)

Lisez la note de cours sur Chemical Equilibra (JAMB)

Chemical Kinetics And Equilibrium Systems (Ghana Only)

Chemical Equilibra

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Question 37 Rapport

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

Voir la réponse

Question 39 Rapport

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}}

Détails de la réponse

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

Lisez la note de cours sur Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

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