JAMB UTME - Chemistry - 2019

Wo Idahun

Ajụjụ 2 Ripọtì

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

_{2}

X

_{3}

X

_{2}

Y

_{3}

Wo Idahun

Ajụjụ 3 Ripọtì

A synthetic rubber is obtained from the polymerization of

neoprene

isoprene

butandiene

propylene

Basic Biochemistry And Synthetic Polymers

Wo Idahun

Ajụjụ 4 Ripọtì

Which of the following is the best starting material for the preparation of oxygen? Heating of trioxonitrate (v) with

Pb

Zn

Ca

Na

Chemical Combination

Wo Idahun

Ajụjụ 5 Ripọtì

The following are isoelectronic ions except

Na

^{+}

Mg

^{2 +}

Si

^{2 +}

O

^{2 +}

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Acids, Bases And Salts (Nigeria Only) (WAEC)

Gụọ akwụkwọ nkuzi na Atomic Structure And Bonding (JAMB)

Acids, Bases And Salts (Nigeria Only)

Wo Idahun

Ajụjụ 6 Ripọtì

Consider the equation below:

Cr $_{2}$ O $_{7}^{2 -}$ + 6Fe $^{2 +}$ + 14H $^{+}$ $\to$ 2Cr $^{3 +}$ + 6Fe $^{3 +}$ + 7H $_{2}$ O.

The oxidation number of chromium changes from

+5 to +3

+6 to +3

-2 to +3

+7 to +3

Akọwa Nkọwa

Cr $_{2}$ O $_{7}^{2 -}$ + 6Fe $^{2 +}$ + 14H $^{+}$ $\to$ 2Cr $^{3 +}$ + 6Fe $^{3 +}$ + 7H $_{2}$ O
The oxidation of Cr in Cr $_{2}$ O $_{7}^{2 -}$ :
Let the oxidation of Cr = x;
2x + (-2 x 7) = -2 $⟹$ 2x - 14 = -2
2x = 12 ; x = +6
Hence, the change in oxidation of Cr = +6 to +3

Gụọ akwụkwọ nkuzi na Oxidation And Reduction (JAMB)

Oxidation And Reduction

Wo Idahun

Ajụjụ 7 Ripọtì

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

sp

^{3}

and sp

sp

^{2}

sp

^{3}

and sp

^{2}

Akọwa Nkọwa

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

Wo Idahun

Ajụjụ 8 Ripọtì

What technique is suitable for separating a binary solution of potassium chloride and potassium trioxochlorate (V)?

Fractional crystallization

Fractional distillation

Filtration

Evaporation

Separation Of Mixtures And Purification Of Chemical Substances

Wo Idahun

Ajụjụ 9 Ripọtì

Which important nitrogen-containing compound is produced in Haber's process?

NH

_{3}

NO

_{2}

NO

HNO

_{3}

Wo Idahun

Ajụjụ 10 Ripọtì

Which quantum divides shells into orbitals?

Spin

Principal

Azimuthal

Magnetic

Wo Idahun

Ajụjụ 11 Ripọtì

The shapes of water, ammonia, carbon (iv) oxide and methane are respectively

linear, tetrahedral, angular and pyramidal

angular, pyramidal, linear and tetrahedral

pyramidal, linear, tetrahedral and angular

tetrahedral, angular, pyramidal and linear

Wo Idahun

Ajụjụ 12 Ripọtì

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

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Kinetic Theory Of Matter And Gas Laws (JAMB)

Wo Idahun

Ajụjụ 13 Ripọtì

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

Akọwa Nkọwa

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

Wo Idahun

Ajụjụ 14 Ripọtì

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}}

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Kinetic Theory Of Matter And Gas Laws (JAMB)

Wo Idahun

Ajụjụ 15 Ripọtì

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}}

Akọwa Nkọwa

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.

Gụọ akwụkwọ nkuzi na Chemical Kinetics And Equilibrium Systems (Ghana Only) (WAEC)

Gụọ akwụkwọ nkuzi na Chemical Equilibra (JAMB)

Chemical Kinetics And Equilibrium Systems (Ghana Only)

Chemical Equilibra

Wo Idahun

Ajụjụ 16 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Separation Of Mixtures And Purification Of Chemical Substances (JAMB)

Gụọ akwụkwọ nkuzi na Atomic Structure And Bonding (JAMB)

Separation Of Mixtures And Purification Of Chemical Substances

Wo Idahun

Ajụjụ 17 Ripọtì

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

Wo Idahun

Ajụjụ 18 Ripọtì

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

Wo Idahun

Ajụjụ 19 Ripọtì

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

Wo Idahun

Ajụjụ 20 Ripọtì

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)

Wo Idahun

Ajụjụ 21 Ripọtì

When the end alkyl groups of ethyl ethanoate are interchanged, the compound formed is

propylmethanoate

ethylpropanoate

methyl ethanoate

methyl propanoate

Wo Idahun

Ajụjụ 22 Ripọtì

Elements in the periodic table are arranged in the order of their

atomic numbers

metallic character

mass number

relative atomic mass

Wo Idahun

Ajụjụ 23 Ripọtì

Which of the following represents the kind of bonding present in ammonium chloride?

ionic only

covalent only

ionic and dative covalent

dative covalent

Wo Idahun

Ajụjụ 24 Ripọtì

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

Wo Idahun

Ajụjụ 25 Ripọtì

Which of the following sets of operation will completely separate a mixture of sodium chloride, sand and iodine?

addition of water, filtration, evaporation to dryness and sublimation

filtration, evaporation to dryness, addition of water and sublimation

sublimation, filtration, evaporation and addition of water

sublimation, addition of water, filtration and evaporation to dryness

Separation Of Mixtures And Purification Of Chemical Substances

Wo Idahun

Ajụjụ 26 Ripọtì

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

Wo Idahun

Ajụjụ 27 Ripọtì

Which of the following alkaline metals react more quickly spontaneously with water?

Calcium

Beryllium

Magnesium

Strontium

Akọwa Nkọwa

The alkaline earth metals (Be, Mg, Ca, Sr, Ba, and Ra) are the second most reactive metals in the periodic table, and, like the Group 1 metals, have increasing reactivity in the higher periods. Beryllium (Be) is the only alkaline earth metal that does not react with water or steam, even if metal is heated to red heat. Additionally, beryllium has a resistant outer oxide layer that lowers its reactivity at lower temperatures.

Magnesium shows insignificant reaction with water, but burns vigorously with steam or water vapor to produce white magnesium oxide and hydrogen gas:

Mg(s) + 2 H2O(l) ⟶ Mg(OH)2(s) + H2(g)

A metal reacting with cold water will produce metal hydroxide. However, if a metal reacts with steam, like magnesium, metal oxide is produced as a result of metal hydroxides splitting upon heating.

The hydroxides of calcium, strontium and barium are only slightly water-soluble but produce sufficient hydroxide ions to make the environment basic, giving a general equation of:

M(s) + 2 H2O(l) ⟶ M(OH)2(aq) + H2(g)

Order of reactivity	Metal	Reactions with water or steam
most reactive	potassium (K)	very vigorous reaction with cold water
↑	sodium (Na)	vigorous reaction with cold water
↓	calcium (Ca)	less vigorous reaction with cold water
least reactive	magnesium (Mg)	slow reaction with cold water, vigorous with steam

If metals react with cold water, hydroxides are produced.
Metals that react with steam form oxides.
Hydrogen is always produced when a metal reacts with cold water or steam.

Gụọ akwụkwọ nkuzi na Metals And Their Compounds (JAMB)

Gụọ akwụkwọ nkuzi na Organic Compounds (JAMB)

Metals And Their Compounds

Organic Compounds

Wo Idahun

Ajụjụ 28 Ripọtì

Which of the following is a physical change?

Exposing white phosphorus to air

Burning kerosene

Dissolving calcium in water

Freezing ice cream

Wo Idahun

Ajụjụ 29 Ripọtì

Which of the following metals is the most essential in the regulation of blood volume, blood pressure and osmotic equilibrium?

Zinc

Manganese

Sodium

Iron

Metals And Their Compounds

Wo Idahun

Ajụjụ 30 Ripọtì

Which of the following gases contains the least number of atoms at s.t.p?

1 mole of butane

4 moles of chlorine

3 moles of ozone

7 moles of argon

Wo Idahun

Ajụjụ 31 Ripọtì

A secondary alkanol can be oxidized to give an

alkanone

alkyne

alkanoic acid

alkanal