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JAMB UTME - Chemistry - 2019

Question 1 Report

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

Calcium

Beryllium

Magnesium

Strontium

Answer Details

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.

Read lesson note on Metals And Their Compounds (JAMB)

Read lesson note on Organic Compounds (JAMB)

Metals And Their Compounds

Organic Compounds

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Question 3 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 4 Report

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

Answer Details

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

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Question 5 Report

The following are isoelectronic ions except

Na

^{+}

Mg

^{2 +}

Si

^{2 +}

O

^{2 +}

Answer Details

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.

Read lesson note on Acids, Bases And Salts (Nigeria Only) (WAEC)

Read lesson note on Atomic Structure And Bonding (JAMB)

Acids, Bases And Salts (Nigeria Only)

Atomic Structure And Bonding

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Question 6 Report

The emission of two successive beta particles from the nucleus \( {}^{32}_{15}\mathrm{P} \) will produce

_{14}^{30}

Si

_{13}^{32}

Al

_{17}^{32}

Cl

_{16}^{32}

S

Answer Details

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

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Question 8 Report

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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Question 11 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 12 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 14 Report

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

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Question 17 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 20 Report

The hybridization in the compound \( \mathrm{CH_3-CH_2-C\equiv H} \) is

sp

^{3}

and sp

sp

sp

^{2}

sp

^{3}

and sp

^{2}

Answer Details

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

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Question 21 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 23 Report

The reaction \(2H_2S + SO_2 \rightarrow 3S + 2H_2O\) 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

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Question 28 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 30 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 32 Report

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}

Answer Details

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

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Question 37 Report

Consider the reaction

\( \mathrm{A(s) + 2B(g) \rightarrow 2C(aq) + D(g)} \)

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

Answer Details

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.

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

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

Read lesson note on 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 39 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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