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JAMB UTME - Chemistry - 2019

Ibeere 1 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

Ka ìwé ẹ̀kọ́ nípa Chemical Kinetics And Equilibrium Systems (Ghana Only) (WAEC)

Ka ìwé ẹ̀kọ́ nípa Electrolysis (JAMB)

Chemical Kinetics And Equilibrium Systems (Ghana Only)

Electrolysis

Ibeere 4 Ìròyìn

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

Awọn alaye Idahun

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 $^{\circ} C$

Ka ìwé ẹ̀kọ́ nípa Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

Ibeere 6 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 7 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.

Ka ìwé ẹ̀kọ́ nípa Kinetic Theory Of Matter And Gas Laws (JAMB)

Kinetic Theory Of Matter And Gas Laws

Ibeere 10 Ìròyìn

The following are isoelectronic ions except

Na

^{+}

Mg

^{2 +}

Si

^{2 +}

O

^{2 +}

Awọn alaye Idahun

Two or more ions are said to be isoelectronic if they have the same electronic structure and the same number of valence electrons.
Na $^{+}$ = 10 electrons = 2, 8
Mg $^{2 +}$ = 10 electrons = 2,8
O $^{2 -}$ = 10 electrons = 2,8
Si $^{2 +}$ = 12 electrons = 2,8,2
$⟹$ Si $^{2 +}$ is not isoelectronic with the rest.

Ka ìwé ẹ̀kọ́ nípa Acids, Bases And Salts (Nigeria Only) (WAEC)

Ka ìwé ẹ̀kọ́ nípa Atomic Structure And Bonding (JAMB)

Acids, Bases And Salts (Nigeria Only)

Atomic Structure And Bonding

Ibeere 13 Ìròyìn

A solution X, on mixing with AgNO $_{3}$ solution gives a white precipitate soluble in aqueous NH $_{3}$ , a solution Y, when also added to X, also gives a white precipitate which is soluble when heated solutions X and Y respectively contain

Al

^{3 +}

, Cl

^{-}

Hg

^{+}

, SO

_{4}^{2 -}

Pb

^{2 +}

, Cl

^{-}

Ag

^{+}

, SO

_{4}^{2 -}

Chemistry Of Carbon Compounds (Ghana Only)

Ibeere 15 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 16 Ìròyìn

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)

Ibeere 17 Ìròyìn

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

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Kinetic Theory Of Matter And Gas Laws (JAMB)

Ka ìwé ẹ̀kọ́ nípa Chemical Kinetics And Equilibrium Systems (Ghana Only) (WAEC)

Ka ìwé ẹ̀kọ́ nípa Periodic Chemistry (Ghana Only) (WAEC)

Kinetic Theory Of Matter And Gas Laws

Chemical Kinetics And Equilibrium Systems (Ghana Only)

Periodic Chemistry (Ghana Only)

Ibeere 18 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.

Ka ìwé ẹ̀kọ́ nípa Chemical Kinetics And Equilibrium Systems (Ghana Only) (WAEC)

Ka ìwé ẹ̀kọ́ nípa Chemical Equilibra (JAMB)

Chemical Kinetics And Equilibrium Systems (Ghana Only)

Chemical Equilibra

Ibeere 19 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 20 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 21 Ìròyìn

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

Ibeere 23 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 $_{4}$ .

Ka ìwé ẹ̀kọ́ nípa Periodic Chemistry (Ghana Only) (WAEC)

Periodic Chemistry (Ghana Only)

Ibeere 24 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 26 Ìròyìn

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

-3, +3

+5

-3, +5

-3

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Oxidation And Reduction (JAMB)

Oxidation And Reduction

Ibeere 27 Ìròyìn

The hybridization in the compound $C H_{3} - C H_{2} - C \equiv H$ is

sp

^{3}

and sp

sp

sp

^{2}

sp

^{3}

and sp

^{2}

Awọn alaye Idahun

The hybridization in a and b is sp $^{3}$ hybridization while in c and d is sp hybridization.

Ibeere 28 Ìròyìn

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

Ibeere 30 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 31 Ìròyìn

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)

Ibeere 33 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 34 Ìròyìn

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

Ibeere 35 Ìròyìn

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

methylpropanoate

ethylethanoate

ethylpropanoate

methylethanoate

Chemistry Of Carbon Compounds (Nigeria Only)

Ibeere 38 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 39 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

Ka ìwé ẹ̀kọ́ nípa Structure Of The Atom (WAEC)

Structure Of The Atom

Ibeere 40 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

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