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JAMB UTME - Chemistry - 2019

Vraag 1 Verslag

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

Antwoorddetails

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.

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Kinetic Theory Of Matter And Gas Laws

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Vraag 5 Verslag

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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Vraag 6 Verslag

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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Vraag 8 Verslag

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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Vraag 9 Verslag

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}

Antwoorddetails

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

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Vraag 12 Verslag

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

Antwoorddetails

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 $^{\circ} C$

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Kinetic Theory Of Matter And Gas Laws

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Vraag 14 Verslag

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

Antwoorddetails

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

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Vraag 18 Verslag

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

-3, +3

+5

-3, +5

-3

Antwoorddetails

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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Vraag 19 Verslag

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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Vraag 20 Verslag

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)

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Vraag 23 Verslag

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}

Antwoorddetails

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 $_{4}$ .

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Periodic Chemistry (Ghana Only)

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Vraag 24 Verslag

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

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Vraag 25 Verslag

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

Antwoorddetails

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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Vraag 28 Verslag

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

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Vraag 29 Verslag

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

Antwoorddetails

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)

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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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Vraag 30 Verslag

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)

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Vraag 31 Verslag

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

Antwoorddetails

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

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Structure Of The Atom

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Vraag 33 Verslag

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

^{+}

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Vraag 34 Verslag

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

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Vraag 35 Verslag

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)

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Vraag 36 Verslag

The following are isoelectronic ions except

Na

^{+}

Mg

^{2 +}

Si

^{2 +}

O

^{2 +}

Antwoorddetails

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.

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Acids, Bases And Salts (Nigeria Only)

Atomic Structure And Bonding

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Vraag 38 Verslag

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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