All practice questions

Question 6861Question →

During a volumetric analysis experiment, $20.0\text{ cm}^3$ of a $0.050\text{ mol dm}^{-3}$ sodium trioxocarbonate(IV) ( $\text{Na}_2\text{CO}_3$ ) solution was completely neutralized by $25.0\text{ cm}^3$ of a tetraoxosulfate(VI) acid ( $\text{H}_2\text{SO}_4$ ) solution. What is the mass concentration of the tetraoxosulfate(VI) acid solution in $\text{g dm}^{-3}$ ? $[\text{H} = 1.0, \text{O} = 16.0, \text{S} = 32.0]$

3.92\text{ g dm}^{-3}

7.84\text{ g dm}^{-3}

1.96\text{ g dm}^{-3}

6.13\text{ g dm}^{-3}

Show answer & explanation

Answer: $3.92\text{ g dm}^{-3}$

Answer

The mass concentration of the tetraoxosulfate(VI) acid solution is

3.92\text{ g dm}^{-3}

.

The balanced chemical reaction shows a

1:1

stoichiometric ratio between

\text{Na}_2\text{CO}_3

and

\text{H}_2\text{SO}_4

. Substituting the given values into the titration equation yields a molar concentration of

0.040\text{ mol dm}^{-3}

. Multiplying this molarity by the molar mass of

\text{H}_2\text{SO}_4

(

98.0\text{ g mol}^{-3}

) correctly yields

3.92\text{ g dm}^{-3}

.

Step-by-Step Solution

1

Write the balanced chemical equation for the neutralization reaction.

\text{Na}_2\text{CO}_3(aq) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{Na}_2\text{SO}_4(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)

Determines the mole ratio between the acid (

n_a = 1

) and base (

n_b = 1

).

2

Apply the volumetric neutralization formula to find the molar concentration of the acid (

C_a

).

\frac{C_a V_a}{C_b V_b} = \frac{n_a}{n_b} \implies \frac{C_a \times 25.0}{0.050 \times 20.0} = \frac{1}{1} \implies C_a = 0.040\text{ mol dm}^{-3}

Calculates the molarity of the tetraoxosulfate(VI) acid solution.

3

Calculate the molar mass of tetraoxosulfate(VI) acid (

\text{H}_2\text{SO}_4

).

\text{Molar Mass} = 2(1.0) + 32.0 + 4(16.0) = 98.0\text{ g mol}^{-3}

Required to convert molar concentration to mass concentration.

4

Convert molar concentration to mass concentration.

\text{Mass concentration} = C_a \times \text{Molar Mass} = 0.040 \times 98.0 = 3.92\text{ g dm}^{-3}

Obtains the final concentration in grams per cubic decimetre.

Key Concept

Volumetric Analysis and Concentration Conversions

Question 6862Question →

Which of the following chemical species contains both covalent and dative (coordinate) covalent bonds?

Ammonium ion (

NH_4^+

)

Methane (

CH_4

)

Sodium chloride (

NaCl

)

Water (

H_2O

)

Show answer & explanation

Answer: Ammonium ion ( $NH_4^+$ )

Answer

Ammonium ion (

NH_4^+

) contains three covalent bonds and one dative (coordinate) covalent bond.

The ammonium ion (

NH_4^+

) is formed when a neutral ammonia molecule (

NH_3

), which has three single covalent bonds and one unshared lone pair on the nitrogen atom, donates its lone pair of electrons to an electron-deficient hydrogen ion (

H^+

). Therefore,

NH_4^+

contains both three standard covalent bonds and one dative bond.

Step-by-Step Solution

1

Analyze the electronic configuration and bonding of ammonia (

NH_3

)

In

NH_3

, the central nitrogen atom forms 3 covalent bonds with hydrogen atoms, sharing three of its valence electrons, leaving one unshared lone pair of electrons.

Understanding the starting structure is essential to see where the unshared electron pair originates.

2

Examine the reaction of ammonia with a hydrogen ion (

H^+

)

The hydrogen ion (

H^+

) has an empty

1s

orbital and requires two electrons to achieve a stable duplet configuration.

A dative bond requires an electron donor (species with a lone pair) and an electron acceptor (species with an empty orbital).

3

Determine the bond types in

NH_4^+

Nitrogen donates its lone pair to the incoming

H^+

ion forming a coordinate bond, while retaining the three original covalent bonds.

This confirms the co-existence of 3 ordinary covalent bonds and 1 dative bond within the ammonium ion.

Key Concept

Covalent and Coordinate (Dative) Bonding in Polyatomic Ions

Estimated Time:1m 0s

Question 6863Question →

A solution of a monoprotic acid $HX$ contains $3.65\text{ g dm}^{-3}$ of the acid. If $25.0\text{ cm}^3$ of this acid solution neutralizes $20.0\text{ cm}^3$ of a $0.125\text{ mol dm}^{-3}$ sodium hydroxide ( $\text{NaOH}$ ) solution, what is the molar mass of the acid $HX$ in $\text{g mol}^{-3}$ ?

Show answer & explanation

Answer: 36.5

Answer

The molar mass of the acid

HX

is

36.5\text{ g mol}^{-3}

.

The reaction of monoprotic acid

HX

with

\text{NaOH}

follows a 1:1 stoichiometric ratio. Applying the titration formula

\frac{C_a V_a}{C_b V_b} = 1

yields an acid concentration of

0.100\text{ mol dm}^{-3}

. Dividing the mass concentration of

3.65\text{ g dm}^{-3}

by this molarity gives a molar mass of

36.5\text{ g mol}^{-3}

.

Step-by-Step Solution

1

Determine the stoichiometric mole ratio between acid and base.

The mole ratio

n_a : n_b

for

HX

reacting with

\text{NaOH}

is

1 : 1

.

A monoprotic acid donates one proton per molecule to react with one mole of sodium hydroxide.

2

Calculate the molar concentration of the acid (

C_a

).

C_a = \frac{0.125\text{ mol dm}^{-3} \times 20.0\text{ cm}^3}{25.0\text{ cm}^3} = 0.100\text{ mol dm}^{-3}

.

Using the titration relation

\frac{C_a V_a}{C_b V_b} = \frac{n_a}{n_b}

allows determination of the acid molarity.

3

Calculate the molar mass of the acid.

\text{Molar mass} = \frac{\text{Mass concentration}}{\text{Molarity}} = \frac{3.65\text{ g dm}^{-3}}{0.100\text{ mol dm}^{-3}} = 36.5\text{ g mol}^{-3}

.

Molar mass is defined as mass of substance per mole.

Key Concept

Determination of molar mass using volumetric titration stoichiometry

Estimated Time:1m 30s

Question 6864Question →

Complete the following statement regarding the chemical test for an oxidizing agent using moist starch-iodide paper.

Fill in the blanks below

When chlorine gas (

Cl_2

) is brought into contact with moist starch-iodide paper, the paper turns because iodide ions (

I^-

) undergo to form free iodine (

I_2

).

Show answer & explanation

Answer

The starch-iodide paper turns blue-black (or blue) because iodide ions (

I^-

) undergo oxidation to form elemental iodine (

I_2

).

Moist starch-iodide paper is used to test for oxidizing agents such as chlorine (

Cl_2

). Chlorine oxidizes iodide ions (

I^-

) to elemental iodine (

I_2

) through the loss of electrons (oxidation). The liberated iodine subsequently reacts with starch on the test paper to form a characteristic blue-black complex.

Step-by-Step Solution

1

Identify the chemical role of chlorine gas (

Cl_2

) reacting with iodide ions (

I^-

)

Chlorine acts as an oxidizing agent by accepting electrons from iodide ions:

Cl_2 + 2I^- \rightarrow 2Cl^- + I_2

.

Chlorine has a stronger electron affinity and higher standard reduction potential than iodine.

2

Determine the type of redox process occurring at the iodide ion

The oxidation number of iodine increases from

-1

in

I^-

to

0

in

I_2

, which represents oxidation (loss of electrons).

Oxidation is defined as the loss of electrons or an increase in oxidation state.

3

Identify the visual observation produced on the indicator paper

The liberated elemental iodine (

I_2

) reacts with the starch embedded in the paper to form a deep blue-black starch-iodine complex.

Starch serves as a specific indicator that yields a blue-black coloration in the presence of free molecular iodine.

Key Concept

Laboratory test for oxidizing agents using moist starch-iodide paper

Estimated Time:1m 0s

Question 6865Question →

Match each chemical species in Column A with its correct bonding description in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: Hydronium ion (

H_3O^+

) matches with oxygen donating a lone pair to a proton; Ammonium ion (

NH_4^+

) matches with nitrogen donating a lone pair to a proton; Aluminum chloride dimer (

Al_2Cl_6

) matches with bridging chlorine atoms donating lone pairs to electron-deficient aluminum centers; Carbon monoxide (

CO

) matches with a triple bond containing two covalent bonds and one coordinate bond from oxygen to carbon.

Each chemical species correctly corresponds to its specific electron pair donor-acceptor behavior:

H_3O^+

relies on oxygen lone pair donation to

H^+

,

NH_4^+

relies on nitrogen lone pair donation to

H^+

,

Al_2Cl_6

uses bridging chlorine lone pairs to complete aluminum octets, and

CO

uses a dative bond from oxygen to carbon alongside two covalent bonds.

Step-by-Step Solution

1

Analyze the coordinate bonding in simple protonated ions (

H_3O^+

and

NH_4^+

).

In

H_3O^+

, oxygen in

H_2O

acts as the electron pair donor to

H^+

. In

NH_4^+

, nitrogen in

NH_3

acts as the electron pair donor to

H^+

.

Distinguishing between oxygen and nitrogen donor atoms correctly pairs the protonated species.

2

Examine the structure of the aluminum chloride dimer (

Al_2Cl_6

).

Monomeric

AlCl_3

is electron-deficient (6 valence electrons around aluminum). Dimerization occurs when a lone pair from a chlorine atom on one

AlCl_3

molecule is donated to the aluminum atom of another

AlCl_3

molecule, creating two coordinate bridges.

This bridge satisfies the octet rule for both aluminum atoms in

Al_2Cl_6

.

3

Determine the bonding in carbon monoxide (

CO

).

Carbon has 4 valence electrons and oxygen has 6. To complete octets for both atoms, oxygen shares two pairs covalently and donates one lone pair to form a coordinate bond, creating a triple bond (

C \equiv O

).

This establishes that oxygen is the donor and carbon is the acceptor in the dative component of the triple bond.

Key Concept

Coordinate (Dative) Covalent Bonding and Electron Pair Donation

Question 6866Question →

Consider the redox reaction between dichromate ions (

\text{Cr}_2\text{O}_7^{2-}

) and iron(II) ions (

\text{Fe}^{2+}

) in an acidic medium:

\text{Cr}_2\text{O}_7^{2-} + x\text{Fe}^{2+} + y\text{H}^+ \rightarrow 2\text{Cr}^{3+} + x\text{Fe}^{3+} + z\text{H}_2\text{O}

What is the stoichiometric coefficient

x

of

\text{Fe}^{2+}

when the ionic equation is completely balanced?

Show answer & explanation

Answer: 6

Answer

The stoichiometric coefficient x of Fe²⁺ in the balanced redox reaction is 6.

In the reduction half-reaction, dichromate (

\text{Cr}_2\text{O}_7^{2-}

) contains two Cr atoms in the +6 oxidation state converting to two

\text{Cr}^{3+}

ions in the +3 state, which consumes 6 electrons. In the oxidation half-reaction, each

\text{Fe}^{2+}

ion loses 1 electron to form

\text{Fe}^{3+}

. To balance charge transfer, 6

\text{Fe}^{2+}

ions are needed for every 1

\text{Cr}_2\text{O}_7^{2-}

ion, making the stoichiometric coefficient

x

equal to 6.

Step-by-Step Solution

1

Determine the oxidation state changes for Chromium and Iron.

Chromium changes from +6 in

\text{Cr}_2\text{O}_7^{2-}

to +3 in

\text{Cr}^{3+}

, requiring 3 electrons per Chromium atom (

6e^-

total for two Cr atoms). Iron changes from +2 in

\text{Fe}^{2+}

to +3 in

\text{Fe}^{3+}

, releasing

1e^-

per Iron atom.

Identifying the number of electrons transferred in each half-reaction is required to balance the overall redox equation.

2

Balance the electron gain and loss.

The oxidation half-reaction (

\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-

) must be multiplied by 6 to balance the 6 electrons required by the dichromate ion.

The total number of electrons lost by the reducing agent must equal the total number of electrons gained by the oxidizing agent.

3

Combine the half-reactions and read the coefficient

x

.

The balanced chemical equation is

\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}

, giving

x = 6

.

The coefficient

x

corresponds directly to the stoichiometric multiplier applied to

\text{Fe}^{2+}

.

Key Concept

Balancing Redox Equations via Half-Reactions in Acidic Medium

Question 6867Question →

In the redox reaction represented by the equation $H_2S_{(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)} + S_{(s)}$ , which substance acts as the reducing agent?

H_2S

, because sulfur increases its oxidation state from

-2

to

0

Cl_2

, because chlorine decreases its oxidation state from

0

to

-1

H_2S

, because hydrogen is reduced from

+1

to

0

Cl_2

, because it loses electrons to form

HCl

Show answer & explanation

Answer: $H_2S$ , because sulfur increases its oxidation state from $-2$ to $0$

Answer

Hydrogen sulfide (

H_2S

) is the reducing agent because sulfur is oxidized, increasing its oxidation number from

-2

in

H_2S

to

0

in elemental

S

.

Hydrogen sulfide (

H_2S

) acts as the reducing agent because sulfur undergoes oxidation. Its oxidation number increases from

-2

in

H_2S

to

0

in elemental sulfur (

S

), indicating that it loses electrons to reduce chlorine.

Step-by-Step Solution

1

Assign oxidation numbers to each element in the given reaction.

In

H_2S

:

H = +1, S = -2

. In

Cl_2

:

Cl = 0

. In

HCl

:

H = +1, Cl = -1

. In elemental

S

:

S = 0

.

Determining oxidation states allows tracking of electron loss and gain.

2

Identify which species loses electrons (is oxidized).

Sulfur changes from

-2

to

0

, representing an increase in oxidation number (loss of electrons).

The reactant containing the element that undergoes oxidation is the reducing agent.

Key Concept

Oxidizing and Reducing Agents via Oxidation State Tracking

Question 6868Question →

A galvanic cell is constructed under standard conditions using the following two reduction half-reactions:

\text{Fe}^{3+}_{\text{(aq)}} + \text{e}^- \rightarrow \text{Fe}^{2+}_{\text{(aq)}}, \quad E^\circ = +0.77\text{ V}

\text{Sn}^{4+}_{\text{(aq)}} + 2\text{e}^- \rightarrow \text{Sn}^{2+}_{\text{(aq)}}, \quad E^\circ = +0.15\text{ V}

What is the standard cell potential ( $E^\circ_{\text{cell}}$ ) in volts for the spontaneous overall reaction?

Show answer & explanation

Answer: 0.62

Answer

The standard cell potential for the spontaneous reaction is +0.62 V.

In a spontaneous galvanic cell, reduction takes place at the cathode, which is the electrode with the higher standard reduction potential (+0.77 V for

\text{Fe}^{3+}/\text{Fe}^{2+}

). Oxidation occurs at the anode (+0.15 V for

\text{Sn}^{4+}/\text{Sn}^{2+}

). Substituting these values into

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

yields

+0.77\text{ V} - (+0.15\text{ V}) = +0.62\text{ V}

.

Step-by-Step Solution

1

Identify the cathode and anode based on standard reduction potentials.

Cathode: Fe3+/Fe2+ half-cell (E° = +0.77 V); Anode: Sn4+/Sn2+ half-cell (E° = +0.15 V).

In a spontaneous galvanic cell, reduction occurs at the electrode with the more positive standard reduction potential.

2

Apply the standard electromotive force equation.

E°cell = E°cathode - E°anode = +0.77 V - (+0.15 V) = +0.62 V

The cell potential measures the overall potential difference between the reduction half-cell and oxidation half-cell.

Key Concept

Standard Cell Potential Calculation

Estimated Time:1m 30s

Question 6869Question →

An aqueous solution of iron(III) chloride, $FeCl_3$ , exhibits an acidic pH because the $Fe^{3+}$ ion undergoes cation hydrolysis in water.

True

False

Show answer & explanation

Answer: True

Answer

True

The statement is correct because iron(III) chloride (

FeCl_3

) is a salt of a weak base (

Fe(OH)_3

) and a strong acid (

HCl

). The

Fe^{3+}

cation undergoes hydrolysis in water to liberate

H^+

ions, resulting in an acidic solution.

Step-by-Step Solution

1

Identify the parent acid and base of iron(III) chloride (

FeCl_3

).

Parent base is iron(III) hydroxide (

Fe(OH)_3

, weak) and parent acid is hydrochloric acid (

HCl

, strong).

Salts composed of weak base cations and strong acid anions produce acidic solutions upon hydrolysis.

2

Determine which ion undergoes salt hydrolysis.

Fe^{3+}

reacts with water molecules, while

Cl^-

remains an unreactive spectator ion.

Anions of strong acids do not hydrolyze, but cations of weak bases undergo cation hydrolysis to yield excess

H^+

ions.

3

Conclude the acid-base nature of the resulting solution.

The concentration of

H^+

increases, resulting in a solution pH less than 7.

Since cation hydrolysis releases

H^+

ions, the statement that

FeCl_3

forms an acidic solution is True.

Key Concept

Cation hydrolysis of salts formed from a weak base and a strong acid

Question 6870Question →

Consider the system of simultaneous equations $x - 3y = 2$ and $x^2 - 2xy - 4y^2 = 19$ . What is the sum of the $x$ -values of the real solution pairs $(x, y)$ ?

28

8

20

-20

Show answer & explanation

Answer: $28$

Answer

The sum of the

x

-values of the real solution pairs is

28

.

Rearranging the linear equation gives

x = 3y + 2

. Substituting this into

x^2 - 2xy - 4y^2 = 19

results in

(3y + 2)^2 - 2(3y + 2)y - 4y^2 = 19

, which simplifies to

y^2 - 8y + 15 = 0

. The roots are

y = 5

and

y = 3

. Substituting these back into

x = 3y + 2

yields

x = 17

and

x = 11

. Their sum is

17 + 11 = 28

.

Step-by-Step Solution

1

Express

x

in terms of

y

using the linear equation.

x = 3y + 2

Isolation of

x

allows direct substitution into the quadratic equation.

2

Substitute

x = 3y + 2

into the quadratic equation

x^2 - 2xy - 4y^2 = 19

.

(3y + 2)^2 - 2(3y + 2)y - 4y^2 = 19

This reduces the non-linear system to a single quadratic equation in terms of

y

.

3

Expand and simplify the quadratic equation.

(9y^2 + 12y + 4) - (6y^2 + 4y) - 4y^2 = 19 \implies -y^2 + 8y + 4 = 19 \implies y^2 - 8y + 15 = 0

Putting the quadratic expression into standard form

ay^2 + by + c = 0

facilitates finding its roots.

4

Solve the quadratic equation

y^2 - 8y + 15 = 0

for

y

.

(y - 5)(y - 3) = 0 \implies y_1 = 5, y_2 = 3

Factoring determines the ordinate values for the solution pairs.

5

Calculate the corresponding

x

-values and find their sum.

For

y_1 = 5

:

x_1 = 3(5) + 2 = 17

.
For

y_2 = 3

:

x_2 = 3(3) + 2 = 11

.
Sum =

17 + 11 = 28

.

Plugging the

y

-values back into

x = 3y + 2

gives the abscissas, which are added to answer the question.

Key Concept

Solving simultaneous linear and quadratic equations by algebraic substitution

Estimated Time:2m 0s

Question 6871Question →

If $x - 1$ , $x + 2$ , and $3x$ are three consecutive terms of a geometric progression (G.P.) with positive terms, what is the common ratio of the progression?

2

4

3

\frac{1}{2}

Show answer & explanation

Answer: $2$

Answer

The common ratio of the progression is

2

.

For any three consecutive terms in a G.P., the square of the middle term equals the product of the first and third terms. Solving

(x+2)^2 = (x-1)(3x)

gives

2x^2 - 7x - 4 = 0

, which yields

x = 4

for positive terms. Substituting

x = 4

gives the terms

3, 6, 12

, which have a common ratio of

6 \div 3 = 2

.

Step-by-Step Solution

1

Set up the condition for consecutive terms in a Geometric Progression.

(x + 2)^2 = (x - 1)(3x)

For three consecutive terms

a, b, c

in G.P., the middle term squared equals the product of the outer terms (

b^2 = ac

).

2

Expand both sides and rearrange into a quadratic equation.

x^2 + 4x + 4 = 3x^2 - 3x \implies 2x^2 - 7x - 4 = 0

Expanding allows gathering all terms on one side to solve for

x

.

3

Factorize the quadratic equation to find

x

.

(2x + 1)(x - 4) = 0 \implies x = 4

(since terms are positive,

x > 1

).

The solution

x = -\frac{1}{2}

gives negative terms, so

x = 4

is chosen.

4

Find the consecutive terms and calculate the common ratio

r

.

Terms are

4 - 1 = 3

,

4 + 2 = 6

, and

3(4) = 12

. Common ratio

r = \frac{6}{3} = 2

.

Dividing the second term by the first term gives the common ratio

r

.

Key Concept

Geometric Progression Consecutive Terms Property (

b^2 = ac

)

Estimated Time:1m 30s

Question 6872Question →

According to Lenz's law, what occurs at the near face of a stationary solenoid when the north pole of a bar magnet is moved rapidly towards it?

A magnetic north pole is induced at the near face to oppose the approaching magnet.

A magnetic south pole is induced at the near face to attract the approaching magnet.

A magnetic south pole is induced at the near face to repel the approaching magnet.

No magnetic polarity is produced because induction generates only electric voltage without magnetic fields.

Show answer & explanation

Answer: A magnetic north pole is induced at the near face to oppose the approaching magnet.

Answer

A magnetic north pole is induced at the near face to oppose the approaching magnet.

Lenz's law states that the direction of an induced current is always such that its magnetic effect opposes the motion or change causing it. When a north pole approaches the solenoid face, the induced current flows counter-clockwise (viewed from the magnet) to form an induced magnetic north pole at that face, thereby exerting a repulsive force that opposes the motion.

Step-by-Step Solution

1

Identify the cause of change in magnetic flux.

The approaching north pole increases magnetic flux linking the solenoid.

Electromagnetic induction occurs whenever magnetic flux linked with a conductor changes.

2

Apply Lenz's law to determine the polarity of the induced field.

The induced current must produce a magnetic field that opposes the increase in flux caused by the approaching north pole.

Lenz's law states that the induced effect always opposes the cause producing it.

3

Deduce the required magnetic pole at the near face.

A north pole must be set up at the near face because like magnetic poles repel each other, opposing the inward motion.

Repulsion between like poles (North against North) exerts a retarding force on the approaching magnet.

Key Concept

Lenz's Law of Electromagnetic Induction

Question 6873Question →

The following trial balance extract was taken from the books of a sole trader as at 31st December 2024:

Account	Debit ( $\text{₦}$ )	Credit ( $\text{₦}$ )
Motor Vehicles (at cost)	$2,400,000$
Provision for Depreciation on Motor Vehicles (1st January 2024)	$600,000$

Additional information:
- On 1st July 2024, a new delivery van costing $\text{₦}800,000$ was purchased on credit. This transaction was mistakenly recorded in the Purchases Journal.
- Depreciation is to be charged at $20\%$ per annum using the reducing balance method, calculated pro-rata for assets acquired during the year.

What is the total depreciation expense for motor vehicles to be charged to the Profit and Loss Account for the year ended 31st December 2024?

\text{₦}440,000

\text{₦}360,000

\text{₦}520,000

\text{₦}640,000

Show answer & explanation

Answer: $\text{₦}440,000$

Answer

The total depreciation expense to be charged to the Profit and Loss Account is

\text{₦}440,000

.

The total depreciation expense of

\text{₦}440,000

combines

\text{₦}360,000

(reducing balance depreciation on existing motor vehicles based on net book value of

\text{₦}1,800,000

) and

\text{₦}80,000

(pro-rata depreciation for 6 months on the newly capitalized delivery van of

\text{₦}800,000

).

Step-by-Step Solution

1

Adjust motor vehicles cost for capital expenditure misclassification

Corrected Motor Vehicles Cost =

\text{₦}2,400,000 + \text{₦}800,000 = \text{₦}3,200,000

Purchasing a fixed asset is a capital expenditure. Entering it in the Purchases Journal was an error that must be corrected by adding

\text{₦}800,000

to Motor Vehicles.

2

Calculate depreciation on existing motor vehicles using the reducing balance method

Net Book Value of existing vehicles =

\text{₦}2,400,000 - \text{₦}600,000 = \text{₦}1,800,000

.
Depreciation =

20\% \times \text{₦}1,800,000 = \text{₦}360,000

The reducing balance method computes depreciation as a percentage of the Net Book Value (Cost minus Accumulated Depreciation at the start of the year).

3

Calculate pro-rata depreciation for the new vehicle acquired on 1st July 2024

Pro-rata Depreciation =

20\% \times \text{₦}800,000 \times \frac{6}{12} = \text{₦}80,000

The asset was owned for 6 months (July to December), so depreciation is apportioned for half a year.

4

Sum the total depreciation expense for the year

Total Depreciation =

\text{₦}360,000 + \text{₦}80,000 = \text{₦}440,000

Both charges combine to form the total depreciation expense debited to the Profit and Loss Account.

Key Concept

Adjustments for Depreciation of Fixed Assets using Reducing Balance Method and Pro-Rata Apportionment

Question 6874Question →

What is the simplified form of the expression $\sqrt{50} - \sqrt{18} + \sqrt{8}$ ?

4\sqrt{2}

2\sqrt{10}

6\sqrt{2}

2\sqrt{2}

Show answer & explanation

Answer: $4\sqrt{2}$

Answer

The simplified form of the expression is

4\sqrt{2}

.

Simplifying each radical into basic surd form yields

\sqrt{50} = 5\sqrt{2}

,

\sqrt{18} = 3\sqrt{2}

, and

\sqrt{8} = 2\sqrt{2}

. Combining these like terms gives

(5 - 3 + 2)\sqrt{2} = 4\sqrt{2}

.

Step-by-Step Solution

1

Simplify each surd term into basic surd form

\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}

,

\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}

, and

\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}

Converting each surd to have a common radicand allows for addition and subtraction of like terms.

2

Substitute the simplified surds back into the expression and combine like terms

5\sqrt{2} - 3\sqrt{2} + 2\sqrt{2} = (5 - 3 + 2)\sqrt{2} = 4\sqrt{2}

Surds with identical radicands can be combined by operating on their coefficients.

Key Concept

Simplification and Addition/Subtraction of Like Surds

Question 6875Question →

Simplify the expression $\frac{3\frac{1}{3} - 1\frac{1}{4}}{\frac{2}{3} \text{ of } 2\frac{1}{2} + 1\frac{1}{6}}$ and express the answer correct to three significant figures. What is the resulting value?

0.735

0.852

1.07

0.736

Show answer & explanation

Answer: $0.735$

Answer

The simplified value correct to three significant figures is

0.735

.

Simplifying the numerator yields

\frac{25}{12}

. For the denominator, applying BODMAS gives

\frac{2}{3} \times \frac{5}{2} = \frac{5}{3}

, and adding

\frac{7}{6}

results in

\frac{17}{6}

. Dividing the numerator by the denominator gives

\frac{25}{34} = 0.73529...

. Rounding to three significant figures gives

0.735

.

Step-by-Step Solution

1

Convert mixed fractions to improper fractions and simplify the numerator.

3\frac{1}{3} - 1\frac{1}{4} = \frac{10}{3} - \frac{5}{4} = \frac{40 - 15}{12} = \frac{25}{12}

Find a common denominator (

12

) to subtract the fractions.

2

Evaluate the denominator following BODMAS rules (performing 'of' before addition).

\frac{2}{3} \text{ of } 2\frac{1}{2} = \frac{2}{3} \times \frac{5}{2} = \frac{5}{3}

. Then add

1\frac{1}{6}

:

\frac{5}{3} + \frac{7}{6} = \frac{10 + 7}{6} = \frac{17}{6}

.

'of' signifies multiplication and takes precedence over addition.

3

Divide the numerator by the denominator.

\frac{\frac{25}{12}}{\frac{17}{6}} = \frac{25}{12} \times \frac{6}{17} = \frac{25}{2 \times 17} = \frac{25}{34}

Dividing by a fraction is equivalent to multiplying by its reciprocal.

4

Convert the resulting fraction to a decimal and round to three significant figures.

\frac{25}{34} = 0.735294... \approx 0.735

The first non-zero digit is

7

. The first three significant figures are

7, 3, 5

, followed by

2

, so we round down.

Key Concept

Compound Fraction Operations and Approximation

Question 6876Question →

Which of the following materials is classified as non-biodegradable because soil micro-organisms cannot decompose it?

Starch

Polyethene

Cellulose

Protein

Show answer & explanation

Answer: Polyethene

Answer

Polyethene is non-biodegradable.

Polyethene is a synthetic addition polymer. Because soil micro-organisms lack enzymes capable of breaking down its continuous carbon-carbon single bonds, it remains intact in the environment for long periods, leading to persistent pollution.

Step-by-Step Solution

1

Classify the given substances into natural polymers and synthetic addition polymers.

Starch, cellulose, and proteins are natural polymers produced by living organisms, whereas polyethene is a synthetic addition polymer.

Biological decomposers possess specific enzymes capable of hydrolyzing the chemical linkages in natural polymers.

2

Determine biodegradability based on resistance to microbial action.

Polyethene contains a stable, non-polar carbon-carbon backbone that soil micro-organisms cannot digest.

Materials that resist natural biological breakdown by bacteria and fungi are classified as non-biodegradable.

Key Concept

Biodegradable vs Non-Biodegradable Polymers

Question 6877Question →

A $1.5\text{ mol}$ sample of carbon dioxide gas ( $\text{CO}_2$ ) is held in a container of volume $0.60\text{ dm}^3$ at a temperature of $300\text{ K}$ under a pressure of $50\text{ atm}$ . What is the compressibility factor ( $Z$ ) for the gas under these conditions? (Take $R = 0.082\text{ atm dm}^3\text{ mol}^{-1}\text{ K}^{-1}$ and express your answer to two decimal places.)

Show answer & explanation

Answer: 0.81

Answer

The compressibility factor (Z) for carbon dioxide under these conditions is 0.81.

The compressibility factor

Z

is defined as

Z = \frac{PV}{nRT}

. Substituting the given values yields

PV = 50 \times 0.60 = 30.0

and

nRT = 1.5 \times 0.082 \times 300 = 36.9

. Dividing

30.0

by

36.9

yields approximately

0.81

, which indicates negative deviation from ideality due to intermolecular attractive forces.

Step-by-Step Solution

1

Identify the given variables and the formula for the compressibility factor.

P = 50\text{ atm}

,

V = 0.60\text{ dm}^3

,

n = 1.5\text{ mol}

,

T = 300\text{ K}

,

R = 0.082\text{ atm dm}^3\text{ mol}^{-1}\text{ K}^{-1}

. Formula:

Z = \frac{PV}{nRT}

.

The compressibility factor

Z

measures the deviation of a real gas from ideal gas behavior (

Z = 1

for an ideal gas).

2

Calculate the actual pressure-volume product (

PV

).

PV = 50\text{ atm} \times 0.60\text{ dm}^3 = 30.0\text{ atm dm}^3

.

This represents the numerator in the compressibility ratio.

3

Calculate the theoretical ideal pressure-volume product (

nRT

).

nRT = 1.5\text{ mol} \times 0.082\text{ atm dm}^3\text{ mol}^{-1}\text{ K}^{-1} \times 300\text{ K} = 36.9\text{ atm dm}^3

.

This represents the expected

PV

value if the gas behaved ideally.

4

Compute the value of

Z

and round to two decimal places.

Z = \frac{30.0}{36.9} \approx 0.8130 \rightarrow 0.81

.

Dividing the observed

PV

by

nRT

gives

Z < 1

, indicating that intermolecular attractive forces dominate under these conditions.

Key Concept

Compressibility factor Z of real gases

Question 6878Question →

Which of the following powers is exercised exclusively by the five permanent members of the United Nations Security Council?

Casting a veto to block substantive resolutions

Casting votes on non-procedural draft resolutions

Serving two-year terms on the Security Council

Sponsoring motions in the General Assembly

Show answer & explanation

Answer: Casting a veto to block substantive resolutions

Answer

Casting a veto to block substantive resolutions

The power of veto, which enables a single member to block the adoption of any substantive resolution, is granted exclusively to the five permanent members of the United Nations Security Council.

Step-by-Step Solution

1

Identify the structure of the UN Security Council

The Security Council consists of 5 permanent members (China, France, Russia, United Kingdom, United States) and 10 non-permanent members.

Understanding member classification clarifies distinct voting privileges.

2

Distinguish powers held exclusively by permanent members

Only permanent members possess veto power, allowing any single one of them to prevent the adoption of a substantive resolution even if it receives the required total votes.

Non-permanent members vote on all issues but do not possess veto power.

Key Concept

Privileges of UN Security Council Permanent Members

Question 6879Question →

In a nation operating a unitary system of government, the national parliament enacts a law granting regional assemblies administrative authority over primary healthcare and local transport. A decade later, the national parliament passes a subsequent statute revoking these granted powers and dissolving the regional assemblies without requiring regional consent. Which of the following constitutional principles legally justifies the national parliament's action?

The central legislature retains supreme legal sovereignty, allowing delegated powers to be altered or revoked by ordinary legislative action.

The regional assemblies exercise constitutionally entrenched residual powers that require a dual-referendum process before revocation.

The legal sovereign automatically forfeits legislative supremacy once authority is devolved to sub-national administrative units.

Sub-national regions hold sovereign statehood rights, leaving the central parliament with override authority only during formal national emergencies.

Show answer & explanation

Answer: The central legislature retains supreme legal sovereignty, allowing delegated powers to be altered or revoked by ordinary legislative action.

Answer

The central legislature retains supreme legal sovereignty, allowing delegated powers to be altered or revoked by ordinary legislative action.

In a unitary system of government, supreme legal power (sovereignty) is vested in a single central authority. While a unitary government may grant powers to lower levels of administration through a process known as devolution or decentralization, these sub-national bodies exist at the discretion of the central legislature. Because these powers are statutory rather than constitutional, the national parliament retains legal supremacy and can modify, reduce, or completely revoke devolved powers by ordinary legislative enactment at any time.

Step-by-Step Solution

1

Analyze the constitutional framework described in the scenario

The central legislature creates regional assemblies via ordinary statute rather than a rigid constitutional division of powers.

Identifying whether power sharing stems from statutory delegation or constitutional entrenchment distinguishes unitary systems from federations.

2

Apply the core defining characteristic of a unitary system of government

In a unitary state, legal sovereignty is indivisible and concentrated entirely in the central authority.

Sub-national bodies in a unitary system are subordinate creations of the central parliament and possess no independent constitutional status.

3

Evaluate the legal validity of the parliament's revoking statute

Because parliament is legally sovereign and the regional powers were merely devolved by statute, parliament can repeal the statute and abolish the regional assemblies unilaterally.

Statutory delegation (devolution) preserves central legislative supremacy, granting parliament the power to give and take back authority.

Key Concept

Unitary Sovereignty vs. Statutory Devolution

Question 6880Question →

A large-scale retail enterprise operating in a major city occupies a floor space exceeding 5,000 square meters, combining the full food range of a supermarket with the extensive general merchandise array of a discount store under a single roof with centralized checkout counters. Which retail format is described, and what is its fundamental operational strategy compared to traditional department stores?

Hypermarket, relying on high stock turnover, low profit margins per item, and self-service to reduce overhead costs.

Supermarket, relying primarily on wholesale distributors to finance inventory stock and manage individual department operations.

Department store, utilizing decentralized purchasing by individual section managers to maximize internal economies of scale.

Variety store, offering single-price standardized lines across independent urban kiosk units to eliminate central warehousing overheads.

Show answer & explanation

Answer: Hypermarket, relying on high stock turnover, low profit margins per item, and self-service to reduce overhead costs.

Answer

Hypermarket, relying on high stock turnover, low profit margins per item, and self-service to reduce overhead costs.

A hypermarket is a giant retail establishment (typically over 5,000 square meters) that merges supermarket food retailing with department store non-food goods. Its business model hinges on low profit margins per unit, direct manufacturer purchasing, self-service customer flow, and high stock turnover to generate overall profitability.

Step-by-Step Solution

1

Analyze the structural characteristics given in the scenario.

Floor space exceeding 5,000 square meters combining full supermarket food lines and general discount merchandise under one roof with centralized checkouts defines a hypermarket.

Hypermarkets are distinguished from supermarkets and department stores by their vast physical size, suburban locations, and integration of groceries with durable non-food goods.

2

Evaluate the financial and operational mechanics of hypermarkets.

Hypermarkets operate on low gross profit margins compensated by extremely high volume stock turnover and reduced floor staff overhead via complete self-service.

By buying directly from manufacturers in huge quantities, hypermarkets secure maximum trade discounts and pass cost savings to consumers to drive high sales volume.

Key Concept

Hypermarkets and Large-Scale Retailing Operational Mechanics

Estimated Time:1m 30s

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