All practice questions

Question 6921Question →

A uniform cylindrical rod of length $20\text{ cm}$ floats vertically in liquid $X$ of density $800\text{ kg/m}^3$ with $15\text{ cm}$ of its length submerged. When transferred to liquid $Y$ , it floats vertically with $12\text{ cm}$ of its length submerged. What is the density of liquid $Y$ ?

640\text{ kg/m}^3

1000\text{ kg/m}^3

1333\text{ kg/m}^3

500\text{ kg/m}^3

Show answer & explanation

Answer: $1000\text{ kg/m}^3$

Answer

The density of liquid

Y

is

1000\text{ kg/m}^3

.

According to the Law of Flotation, a floating body displaces its own weight of liquid. Therefore,

h_X \rho_X = h_Y \rho_Y

. Substituting

15\text{ cm} \times 800\text{ kg/m}^3 = 12\text{ cm} \times \rho_Y

yields

\rho_Y = 1000\text{ kg/m}^3

.

Step-by-Step Solution

1

Apply the Law of Flotation for a floating body of uniform cross-sectional area

A

.

Weight of rod

W = \text{Upthrust} = A \cdot h_{\text{submerged}} \cdot \rho_{\text{liquid}} \cdot g

.

A floating body displaces a weight of fluid equal to its own total weight.

2

Equate the upthrust in liquid

X

to the upthrust in liquid

Y

.

A \cdot h_X \cdot \rho_X \cdot g = A \cdot h_Y \cdot \rho_Y \cdot g \implies h_X \cdot \rho_X = h_Y \cdot \rho_Y

.

Since the rod is identical and floating freely in both liquids, its weight

W

remains unchanged.

3

Substitute the known values (

h_X = 15\text{ cm}

,

\rho_X = 800\text{ kg/m}^3

,

h_Y = 12\text{ cm}

) and solve for

\rho_Y

.

\rho_Y = \frac{h_X \cdot \rho_X}{h_Y} = \frac{15 \times 800}{12} = 1000\text{ kg/m}^3

.

Rearranging the linear equation yields the density of liquid

Y

.

Key Concept

Law of Flotation and Hydrometer Principle

Estimated Time:1m 15s

Question 6922Question →

The glass manufacturing industry located in Ughelli, Delta State, was established in that area primarily due to the ready availability of which key raw material?

Silica sand

Limestone

Bauxite

Rock salt

Show answer & explanation

Answer: Silica sand

Answer

Silica sand

Silica sand (

SiO_2

) forms the main structural matrix of commercial glass. Chemical industries that rely on high volumes of heavy raw materials, such as the glass factory at Ughelli, are sited directly near silica sand deposits to reduce transport overheads.

Step-by-Step Solution

1

Identify the primary raw material required for glass manufacturing.

Glass is predominantly composed of silicon(IV) oxide (

SiO_2

), which is obtained from silica sand.

Chemical industries are sited close to heavy bulk raw materials to minimize transportation costs.

2

Relate the raw material to the specified industrial location in Nigeria.

Ughelli in Delta State has vast natural deposits of high-purity silica sand, making it an ideal site for glass production.

Proximity to raw materials is the key siting factor for bulk material chemical industries.

Key Concept

Industrial Siting Factors and Raw Material Allocation

Estimated Time:45s

Question 6923Question →

Calculate the solubility in $\text{mol/dm}^3$ of sodium nitrate ( $\text{NaNO}_3$ ) at $25^\circ\text{C}$ , if $17.0\text{ g}$ of the salt dissolves in $100.0\text{ g}$ of water to form a saturated solution. [Relative atomic masses: $\text{Na} = 23, \text{N} = 14, \text{O} = 16$ ; density of water $= 1.0\text{ g/cm}^3$ ]

Show answer & explanation

Answer: 2

Answer

The solubility of sodium nitrate at

25^\circ\text{C}

is

2.0\text{ mol/dm}^3

.

To find solubility in

\text{mol/dm}^3

, first convert

17.0\text{ g}

of

\text{NaNO}_3

into moles by dividing by its molar mass (

85.0\text{ g/mol}

), obtaining

0.20\text{ mol}

. Next, convert

100.0\text{ g}

of water into volume, which equals

0.100\text{ dm}^3

. Dividing

0.20\text{ mol}

by

0.100\text{ dm}^3

yields

2.0\text{ mol/dm}^3

.

Step-by-Step Solution

1

Determine the molar mass of the solute (

\text{NaNO}_3

)

Molar mass of

\text{NaNO}_3 = 23 + 14 + (3 \times 16) = 85.0\text{ g/mol}

Molar mass is required to convert mass of solute to amount in moles.

2

Calculate the amount of

\text{NaNO}_3

in moles

\text{Moles} = \frac{17.0\text{ g}}{85.0\text{ g/mol}} = 0.20\text{ mol}

Solubility in

\text{mol/dm}^3

measures the amount of solute in moles per cubic decimeter of solvent.

3

Convert the mass of solvent (water) to volume in

\text{dm}^3

100.0\text{ g of water} = 100.0\text{ cm}^3 = 0.100\text{ dm}^3

Density of water is

1.0\text{ g/cm}^3

, and

1000\text{ cm}^3 = 1\text{ dm}^3

.

4

Calculate the solubility in

\text{mol/dm}^3

\text{Solubility} = \frac{0.20\text{ mol}}{0.100\text{ dm}^3} = 2.0\text{ mol/dm}^3

Dividing the moles of solute by the volume of solvent in

\text{dm}^3

gives the concentration in

\text{mol/dm}^3

.

Key Concept

Solubility expressed in molar concentration (mol/dm³)

Question 6924Question →

In a nuclear fusion reaction, two light nuclei fuse together to form a heavier nucleus. The total mass of the reactants before fusion is $4.028\text{ u}$ , and the total mass of the products after fusion is $4.003\text{ u}$ . Given that $1\text{ u} = 931.5\text{ MeV}$ , what is the total energy released in this reaction in $\text{MeV}$ ?

Show answer & explanation

Answer: 23.2875

Answer

The total energy released in the reaction is 23.2875 MeV.

The energy released in nuclear fusion is calculated using the mass defect $\Delta m = m_{\text{reactants}} - m_{\text{products}}$. Subtracting

4.003\text{ u}

from

4.028\text{ u}

yields a mass defect of

0.025\text{ u}

. Multiplying this mass defect by

931.5\text{ MeV/u}

gives the total energy released as

23.2875\text{ MeV}

.

Step-by-Step Solution

1

Calculate the mass defect (\Delta m)

\Delta m = 4.028\text{ u} - 4.003\text{ u} = 0.025\text{ u}

Mass defect is the loss of mass during nuclear fusion that gets converted into energy.

2

Convert mass defect into energy released

E = 0.025\text{ u} \times 931.5\text{ MeV/u} = 23.2875\text{ MeV}

According to mass-energy equivalence, each atomic mass unit (u) of missing mass yields 931.5 MeV of energy.

Key Concept

Mass Defect and Energy Release in Nuclear Reactions

Question 6925Question →

A radioactive sample has a half-life of $12\text{ minutes}$ . If its initial activity is $96\text{ Bq}$ , what is the remaining activity of the sample, in $\text{Bq}$ , after an elapsed time of $36\text{ minutes}$ ?

Show answer & explanation

Answer: 12

Answer

The remaining activity of the radioactive sample after

36\text{ minutes}

is

12\text{ Bq}

.

In

36\text{ minutes}

, exactly

3

half-lives elapse (

36 / 12 = 3

). The remaining activity reduces to

(1/2)^3 = 1/8

of the initial value, giving

96\text{ Bq} / 8 = 12\text{ Bq}

.

Step-by-Step Solution

1

Calculate the number of half-lives that have elapsed

n = 3

Divide the total elapsed time (

36\text{ minutes}

) by the half-life duration (

12\text{ minutes}

).

2

Calculate the remaining activity of the isotope

A = 12\text{ Bq}

After

n = 3

half-lives, the remaining activity fraction is

\left(\frac{1}{2}\right)^3 = \frac{1}{8}

. Multiplying initial activity

96\text{ Bq}

by

\frac{1}{8}

yields

12\text{ Bq}

.

Key Concept

Radioactive Decay Law and Half-life

Question 6926Question →

Match each chemical species on the left with the correct oxidation state of its central atom on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Potassium trioxovanadate(V) (

KVO_3

) matches +5; Sodium thiosulfate (

Na_2S_2O_3

) matches +2; Potassium hexachloroplatinate(IV) (

K_2PtCl_6

) matches +4; Trioxophosphate(III) acid (

H_3PO_3

) matches +3.

Each chemical species is correctly paired with the oxidation state of its central atom calculated by setting the sum of all oxidation numbers equal to zero.

Step-by-Step Solution

1

Calculate oxidation state of Vanadium in

KVO_3

Potassium is +1, Oxygen is -2.

1 + V + 3(-2) = 0 \implies V = +5

.

The sum of oxidation numbers in a neutral compound equals zero.

2

Calculate oxidation state of Sulfur in

Na_2S_2O_3

Sodium is +1, Oxygen is -2.

2(1) + 2S + 3(-2) = 0 \implies 2S = 4 \implies S = +2

.

Average oxidation state calculation per sulfur atom.

3

Calculate oxidation state of Platinum in

K_2PtCl_6

Potassium is +1, Chlorine is -1.

2(1) + Pt + 6(-1) = 0 \implies Pt = +4

.

Halogens in binary complex salts take an oxidation number of -1.

4

Calculate oxidation state of Phosphorus in

H_3PO_3

Hydrogen is +1, Oxygen is -2.

3(1) + P + 3(-2) = 0 \implies P = +3

.

Hydrogen in covalent oxoacids takes an oxidation state of +1.

Key Concept

Oxidation number calculation using standard rules for alkali metals, halogens, hydrogen, and oxygen in neutral inorganic compounds.

Estimated Time:1m 30s

Question 6927Question →

Arrange the following organic functional groups in decreasing order of priority (highest priority first) for selection as the principal functional group suffix when naming polyfunctional compounds according to IUPAC nomenclature rules:

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence in decreasing order of IUPAC principal functional group priority is: Carboxylic acid (

-\text{COOH}

), Ester (

-\text{COOR}

), Aldehyde (

-\text{CHO}

), Alcohol (

-\text{OH}

), and Amine (

-\text{NH}_2

).

According to official IUPAC nomenclature seniority rules for principal functional groups: Carboxylic acids (

-\text{COOH}

) take top priority, followed by acid derivatives like Esters (

-\text{COOR}

), then Aldehydes (

-\text{CHO}

), Alcohols (

-\text{OH}

), and finally Amines (

-\text{NH}_2

).

Step-by-Step Solution

1

Identify the highest-priority functional group among carboxylic derivatives and oxygen/nitrogen species.

Carboxylic acids (

-\text{COOH}

) occupy the top hierarchy level among organic functional groups.

IUPAC nomenclature assigns the highest principal suffix priority to carboxylic acid functional groups over esters, carbonyls, alcohols, and amines.

2

Compare carboxylic acid derivatives to carbonyl and hydroxyl groups.

Ester (

-\text{COOR}

) ranks higher than Aldehyde (

-\text{CHO}

).

Carboxylic acid derivatives (esters, acyl halides, amides) take priority over aldehydes and ketones.

3

Determine priority among oxygen- and nitrogen-containing groups.

Aldehyde (

-\text{CHO}

) > Alcohol (

-\text{OH}

) > Amine (

-\text{NH}_2

).

Carbonyl groups rank higher than hydroxyl groups, which in turn rank higher than amino groups.

Key Concept

IUPAC Functional Group Seniority Hierarchy for Polyfunctional Nomenclature

Question 6928Question →

A weak monobasic acid, $\text{HA}$ , has an acid dissociation constant ( $K_a$ ) of $4.5 \times 10^{-5}\text{ mol dm}^{-3}$ at $25^\circ\text{C}$ . Calculate the percentage ionization of a $0.05\text{ mol dm}^{-3}$ solution of this acid.

Show answer & explanation

Answer: 3

Answer

The percentage ionization of the weak monobasic acid solution is 3.0%.

According to Ostwald's dilution law for weak monobasic acids,

K_a = \alpha^2 C

. Rearranging to solve for the degree of ionization gives

\alpha = \sqrt{K_a / C} = \sqrt{(4.5 \times 10^{-5}) / 0.05} = \sqrt{9.0 \times 10^{-4}} = 0.03

. Expressed as a percentage,

0.03 \times 100\% = 3.0\%

.

Step-by-Step Solution

1

Relate acid dissociation constant (

K_a

), initial molar concentration (

C

), and degree of ionization (

\alpha

)

K_a = \alpha^2 C

For a weak monobasic acid undergoing partial ionization (

\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-

), Ostwald's dilution law simplifies to

K_a = \frac{\alpha^2 C}{1 - \alpha} \approx \alpha^2 C

because

\alpha \ll 1

.

2

Substitute the given values into the simplified expression and calculate

\alpha

\alpha = \sqrt{\frac{4.5 \times 10^{-5}}{0.05}} = \sqrt{9.0 \times 10^{-4}} = 0.03

Rearranging the equation yields

\alpha = \sqrt{K_a / C}

. Dividing the acid dissociation constant by the molar concentration gives

9.0 \times 10^{-4}

, and taking the square root yields

0.03

.

3

Convert the fractional degree of ionization to percentage ionization

3.0%

Multiplying the fractional degree of ionization by 100% gives the percentage of acid molecules ionized in solution.

Key Concept

Ostwald's Dilution Law and Degree of Ionization of Weak Acids

Estimated Time:1m 30s

Question 6929Question →

Calculate the final absolute pressure of the oxygen gas inside the industrial storage vessel by completing the blank below.

Fill in the blanks below

An industrial storage vessel containing oxygen gas at a fixed volume exerts a pressure of

180\text{ kPa}

at a temperature of

27^\circ\text{C}

. If the temperature of the gas rises to

127^\circ\text{C}

, the new pressure exerted by the gas will be kPa.

Show answer & explanation

Answer

The final pressure exerted by the oxygen gas is

240\text{ kPa}

.

According to Gay-Lussac's Pressure Law, for a fixed mass of gas at constant volume,

\frac{P_1}{T_1} = \frac{P_2}{T_2}

where temperature must be in Kelvin. Converting the temperatures:

T_1 = 27 + 273 = 300\text{ K}

and

T_2 = 127 + 273 = 400\text{ K}

. Rearranging for

P_2

gives

P_2 = 180 \times \frac{400}{300} = 240\text{ kPa}

.

Step-by-Step Solution

1

Convert temperatures from Celsius to the Kelvin absolute scale

T_1 = 27^\circ\text{C} + 273 = 300\text{ K}

and

T_2 = 127^\circ\text{C} + 273 = 400\text{ K}

All gas law calculations require temperatures to be expressed in absolute units (Kelvin).

2

Apply the Pressure Law (Gay-Lussac's Law) mathematical formula

\frac{P_1}{T_1} = \frac{P_2}{T_2} \implies P_2 = P_1 \times \frac{T_2}{T_1}

At constant volume, the pressure of a given mass of gas is directly proportional to its absolute temperature.

3

Substitute the given numerical values into the formula and solve for

P_2

P_2 = 180\text{ kPa} \times \frac{400\text{ K}}{300\text{ K}} = 180 \times \frac{4}{3} = 240\text{ kPa}

Calculates the resulting pressure in kilopascals.

Key Concept

Pressure Law (Gay-Lussac's Law of Temperature-Pressure)

Question 6930Question →

At a specific temperature and pressure, $2.0\text{ moles}$ of a real gas occupy a volume of $18.0\text{ dm}^3$ . Under the exact same conditions, $2.0\text{ moles}$ of an ideal gas occupy a volume of $22.5\text{ dm}^3$ . What is the compressibility factor ( $Z$ ) of the real gas under these conditions?

Show answer & explanation

Answer: 0.8

Answer

0.8

The compressibility factor

Z

measures the deviation of a real gas from ideal gas behavior and is calculated as

Z = \frac{V_{\text{real}}}{V_{\text{ideal}}}

. Substituting the given values yields

Z = \frac{18.0}{22.5} = 0.80

. A value of

Z < 1

indicates that intermolecular attractive forces dominate, causing the real gas to occupy less volume than predicted by the ideal gas equation.

Step-by-Step Solution

1

Identify the formula for compressibility factor

Z = \frac{V_{\text{real}}}{V_{\text{ideal}}}

The compressibility factor

Z

quantifies deviation from ideality as the ratio of molar volume of a real gas to that of an ideal gas at the same temperature and pressure.

2

Substitute the given values into the equation

Z = \frac{18.0\text{ dm}^3}{22.5\text{ dm}^3}

Both volumes are measured under identical pressure, temperature, and mole count.

3

Calculate the numerical value

Z = 0.80

Dividing 18.0 by 22.5 gives 0.80, indicating

Z < 1

due to predominant intermolecular attraction.

Key Concept

Compressibility factor (Z = V_real / V_ideal) measuring real gas deviation

Question 6931Question →

Liquid air undergoes fractional distillation to isolate noble gases and other atmospheric components. Based on their boiling points, in which sequence are nitrogen, argon, and oxygen collected as vapors from first to last?

Nitrogen, Argon, Oxygen

Oxygen, Argon, Nitrogen

Argon, Nitrogen, Oxygen

Nitrogen, Oxygen, Argon

Show answer & explanation

Answer: Nitrogen, Argon, Oxygen

Answer

Nitrogen, Argon, Oxygen

During the fractional distillation of liquid air, components boil off in order of increasing boiling points (most volatile to least volatile). Nitrogen has the lowest boiling point (

-196^\circ\text{C}

), followed by Argon (

-186^\circ\text{C}

), and finally Oxygen (

-183^\circ\text{C}

). Thus, the collection sequence is Nitrogen, Argon, Oxygen.

Step-by-Step Solution

1

Identify the boiling points of the three components in liquid air

Nitrogen (

N_2

) =

-196^\circ\text{C}

, Argon (

Ar

) =

-186^\circ\text{C}

, Oxygen (

O_2

) =

-183^\circ\text{C}

Fractional distillation separates liquefied gases based on differences in their boiling points.

2

Arrange the gases in increasing order of boiling points (from most volatile to least volatile)

-196^\circ\text{C} < -186^\circ\text{C} < -183^\circ\text{C}

, corresponding to Nitrogen

\rightarrow

Argon

\rightarrow

Oxygen

The substance with the lowest boiling point vaporizes and boils off first during distillation.

Key Concept

Isolation of noble gases from liquid air by fractional distillation

Question 6932Question →

Match each partnership revaluation event on the left with its correct posting entry in the Revaluation Account on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Increases in asset values and decreases in liability values are credited to the Revaluation Account as gains, whereas decreases in asset values and increases in liability values are debited to the Revaluation Account as losses.

The Revaluation Account functions as a nominal account. Any transaction that increases the net assets of the partnership (such as an appreciation in asset value or a reduction in liability value) represents a revaluation gain and is posted to the credit side. Any transaction that reduces net assets (such as asset write-downs or additional liabilities) represents a revaluation loss and is posted to the debit side.

Step-by-Step Solution

1

Classify each revaluation transaction as a gain or a loss to the partnership.

Plant increase and creditors reduction are gains; premises decrease and expense provision are losses.

Revaluation gains increase the overall net asset value of the firm, while revaluation losses reduce it.

2

Apply nominal account double-entry rules to the Revaluation Account.

Gains are posted on the credit side, and losses are posted on the debit side.

The Revaluation Account is a nominal account prepared to record profit or loss on the revaluation of assets and liabilities.

Key Concept

Double Entry Rules for Partnership Revaluation Account

Question 6933Question →

During a fiscal year, the Ministry of Transportation of a state government recorded the following financial disbursements:

- Construction of a modern bus terminal: $₦85,000,000$
- Routine servicing and minor repairs of fleet vehicles: $₦4,200,000$
- Payment of monthly salaries and allowances to ministry staff: $₦38,500,000$
- Major structural rehabilitation and capacity expansion of existing railway bridges: $₦52,000,000$
- Purchase of new traffic monitoring equipment: $₦19,800,000$
- Refueling of operational vehicles and office stationery supplies: $₦3,500,000$

What is the total recurrent expenditure (in $₦$ ) of the Ministry for the fiscal year?

Show answer & explanation

Answer: 46200000

Answer

The total recurrent expenditure of the Ministry for the fiscal year is ₦46,200,000.

Recurrent expenditures are ongoing operational expenses essential for daily administration, personnel payments, and regular asset maintenance. Adding routine servicing (

₦4,200,000

), salaries (

₦38,500,000

), and refueling/stationery (

₦3,500,000

) gives

₦46,200,000

. Capital expenditures such as building new terminals, acquiring equipment, and major bridge expansions are excluded because they yield long-term physical assets.

Step-by-Step Solution

1

Classify each public sector expenditure item as either recurrent expenditure or capital expenditure.

Recurrent expenditures: Routine servicing (

₦4,200,000

), staff salaries (

₦38,500,000

), and refueling/stationery (

₦3,500,000

). Capital expenditures: Bus terminal construction (

₦85,000,000

), major bridge expansion (

₦52,000,000

), and equipment acquisition (

₦19,800,000

).

Recurrent expenditures represent operational costs consumed within the financial year, whereas capital expenditures create non-current assets or increase their long-term capacity/lifespan.

2

Sum the classified recurrent expenditure amounts to obtain the total.

Total Recurrent Expenditure =

₦4,200,000 + ₦38,500,000 + ₦3,500,000 = ₦46,200,000

.

Aggregating all operational and recurring personnel expenses provides the true total of public recurrent spending.

Key Concept

Public sector recurrent expenditure comprises regular operational running costs, staff salaries, administrative expenses, and routine maintenance of existing public infrastructure, whereas capital expenditure involves creating or expanding long-term physical assets.

Question 6934Question →

Under a consignment agreement, if the consignor pays a del-credere commission to the consignee, who bears the loss arising from bad debts on credit sales?

The consignee

The consignor

The credit customer's bank

Both the consignor and consignee equally

Show answer & explanation

Answer: The consignee

Answer

The consignee bears the financial loss arising from bad debts when a del-credere commission is received.

Del-credere commission is specifically paid by the consignor to compensate the consignee for taking on credit risk. Therefore, when del-credere commission is paid, the consignee must absorb any bad debt losses.

Step-by-Step Solution

1

Identify the purpose of a del-credere commission in consignment accounts.

Del-credere commission is an extra allowance paid to a consignee for taking on the risk of bad debts on credit sales.

By accepting this commission, the consignee guarantees payment of credit sales to the consignor.

2

Determine who absorbs the loss when credit customers default.

Because the consignee has received del-credere commission, any bad debt loss is debited to the consignee's own account rather than charged to the consignor.

The risk of non-payment shifts entirely from the consignor to the consignee.

Key Concept

Del-Credere Commission and Bad Debt Risk Allocation

Question 6935Question →

A manufacturing firm integrates its enterprise accounting system with an Electronic Data Interchange (EDI) network and an Electronic Funds Transfer (EFT) payment gateway to execute B2B vendor payments automatically upon invoice matching. To maintain internal financial control, which security mechanism specifically guarantees non-repudiation and verifies the authenticity of transmitted financial instructions before the accounts payable ledger is updated?

Digital signatures using public key cryptography to encrypt and verify the sender's identity and payload integrity

Periodic batch processing of payment files at off-peak hours to systematically prevent transaction interception

User password access controls restricted to individual accounting data entry workstations

Converting standardized EDI documents into unformatted plain text before network transmission

Show answer & explanation

Answer: Digital signatures using public key cryptography to encrypt and verify the sender's identity and payload integrity

Answer

Digital signatures using public key cryptography to encrypt and verify the sender's identity and payload integrity

Digital signatures utilizing public key infrastructure provide non-repudiation, ensuring that the party sending the electronic payment instruction cannot deny initiating the transaction, while simultaneously confirming message integrity before accounts payable updates occur.

Step-by-Step Solution

1

Identify the key objective of the control

The control must ensure non-repudiation and origin authenticity for automated electronic payment instructions.

In automated B2B accounting environments, transactions trigger funds transfer without manual approval, requiring cryptographically secure identification.

2

Evaluate technical security controls in EDI/EFT accounting workflows

Digital signatures use asymmetric encryption to bind the sender's private key to the transmitted data payload.

This guarantees that the instruction originated from the authorized partner and was not altered in transit.

Key Concept

Non-repudiation and Security Controls in EDI and EFT Accounting

Question 6936Question →

Bookkeeping involves interpreting financial data and preparing reports for management decision-making, whereas accounting is limited to the clerical task of recording daily financial transactions.

True

False

Show answer & explanation

Answer: False

Answer

False. Bookkeeping is restricted to the routine, clerical recording of transactions, whereas accounting involves interpreting, analyzing, and summarizing financial information for decision-making.

The statement is false because bookkeeping represents the recording phase of financial transactions, while accounting involves the interpretation, synthesis, and reporting of those records to support business decisions.

Step-by-Step Solution

1

Define the primary scope and function of bookkeeping.

Bookkeeping is the primary stage of the accounting cycle responsible for recording daily financial transactions in journals and ledgers.

Establishing the functional boundary of bookkeeping clarifies its routine, non-analytical nature.

2

Define the primary scope and function of accounting.

Accounting builds upon bookkeeping records to analyze, summarize, interpret, and communicate financial information to stakeholders.

Establishing the analytical nature of accounting highlights where higher-level decision-making reporting occurs.

3

Compare the given statement against the established definitions.

The statement misattributes analytical reporting to bookkeeping and routine record-keeping to accounting.

Since the responsibilities are reversed in the statement, the assertion is false.

Key Concept

Scope and Functional Distinctions Between Bookkeeping and Accounting

Question 6937Question →

The adjusted cash book of Tunde & Sons shows a debit balance of $\text{NGN } 12,000$ . Cheques issued to suppliers totaling $\text{NGN } 3,500$ have not yet been presented to the bank for payment. What is the balance as per the bank statement?

\text{NGN } 15,500

\text{NGN } 8,500

\text{NGN } 12,000

\text{NGN } 19,000

Show answer & explanation

Answer: $\text{NGN } 15,500$

Answer

\text{NGN } 15,500

Starting with a favourable (debit) adjusted cash book balance, unpresented cheques must be added to determine the bank statement balance because these cheques have reduced the cash book balance but have not yet been deducted from the bank account.

Step-by-Step Solution

1

Identify the starting adjusted cash book balance.

Adjusted Cash Book debit balance =

\text{NGN } 12,000

.

The reconciliation statement starts from the corrected cash book figure.

2

Add unpresented cheques to the adjusted cash book balance.

\text{NGN } 12,000 + \text{NGN } 3,500 = \text{NGN } 15,500

.

Unpresented cheques have been deducted in the cash book but not yet cleared by the bank, meaning the bank's record reflects a higher balance.

Key Concept

Preparation of Bank Reconciliation Statement from Adjusted Cash Book Balance

Question 6938Question →

At a constant temperature, doubling the volume of a sealed container holding a liquid and its saturated vapour will cause the saturated vapour pressure to be halved.

True

False

Show answer & explanation

Answer: False

Answer

The statement is false. Saturated vapour pressure depends exclusively on temperature and is independent of the container volume.

Saturated vapour pressure is entirely independent of the volume of the space occupied by the vapour. As long as liquid remains in the container to evaporate, expanding the volume simply causes more liquid to evaporate until the vapour pressure reaches its equilibrium saturation value at that temperature.

Step-by-Step Solution

1

Analyze the physical state of a saturated vapour in contact with its liquid.

A saturated vapour exists in dynamic equilibrium with its liquid phase, meaning the rate of evaporation equals the rate of condensation at that specific temperature.

Dynamic equilibrium determines the equilibrium vapour pressure above a liquid surface.

2

Examine the effect of increasing the container volume at constant temperature.

Expanding the volume temporarily reduces vapour concentration, causing the rate of evaporation to exceed condensation until the space is re-saturated.

Phase change allows the mass of the gas phase to change, unlike in closed ideal gas systems.

3

Conclude the value of the final vapour pressure.

The pressure returns to the exact same saturated vapour pressure (SVP) value as before the expansion.

SVP is an intrinsic property dependent only on temperature and the identity of the liquid, independent of volume.

Key Concept

Independence of Saturated Vapour Pressure from Volume

Question 6939Question →

Complete the statement below regarding the statutory responsibilities of consumer protection regulatory bodies in Nigeria.

Fill in the blanks below

While the is statutorily mandated to formulate quality standards for industrial manufactured products and administer MANCAP certification in Nigeria, the enforces market competition law and resolves broad consumer grievances across all commercial sectors.

Show answer & explanation

Answer

The first blank is filled by the Standards Organisation of Nigeria (SON) and the second blank is filled by the Federal Competition and Consumer Protection Commission (FCCPC).

The Standards Organisation of Nigeria (SON) is responsible for establishing technical quality specifications for manufactured goods and enforcing the Mandatory Conformity Assessment Programme (MANCAP). In contrast, the Federal Competition and Consumer Protection Commission (FCCPC) focuses on market regulation, curbing anti-competitive behavior, and securing redress for exploited consumers across all trade sectors.

Step-by-Step Solution

1

Identify the agency responsible for setting product quality standards and issuing MANCAP certification for manufactured goods in Nigeria.

The agency is the Standards Organisation of Nigeria (SON).

SON sets national standardization benchmarks, conducts factory quality audits, and enforces the Mandatory Conformity Assessment Programme (MANCAP) for locally manufactured products.

2

Identify the primary agency responsible for promoting market competition and general consumer dispute resolution across sectors.

The agency is the Federal Competition and Consumer Protection Commission (FCCPC).

FCCPC (which superseded the Consumer Protection Council) has statutory authority over market competition, prevention of anti-competitive practices, and addressing general consumer complaints.

Key Concept

Functions and statutory distinctions between Nigerian regulatory agencies (SON vs. FCCPC)

Question 6940Question →

Both Aves (birds) and Mammalia (mammals) maintain constant core body temperatures through endogenous heat production and complex physiological feedback loops. When exposed to heat stress, these homoiothermic classes utilize distinct evaporative cooling and integumentary mechanisms. Which of the following thermoregulatory adaptations is exclusive to mammals and absent in birds?

Secretion of watery fluid from specialized cutaneous sweat glands distributed across the integument to facilitate surface evaporative heat loss

Modulation of cardiac output and peripheral vasodilation utilizing a three-chambered cardiac septum to redirect warm blood away from core organs

Elimination of metabolic heat through dermal excretion of hypertonic nitrogenous wastes directly from cutaneous nephric pores

Suppression of cellular oxidative phosphorylation to reduce metabolic ATP yield and eliminate metabolic heat generation during locomotion

Show answer & explanation

Answer: Secretion of watery fluid from specialized cutaneous sweat glands distributed across the integument to facilitate surface evaporative heat loss

Answer

Secretion of watery fluid from specialized cutaneous sweat glands distributed across the integument to facilitate surface evaporative heat loss.

Mammals possess specialized sweat (sudoriferous) glands within the dermis that secrete watery fluid onto the epidermal surface, enabling evaporative heat dissipation during thermal stress. Birds completely lack sweat glands in their integument and instead rely primarily on respiratory evaporative cooling (panting and gular fluttering) along with behavioral heat dissipation.

Step-by-Step Solution

1

Analyze thermoregulatory integumentary structures in Aves versus Mammalia.

Identify that mammalian skin uniquely contains sweat (sudoriferous) glands, whereas avian skin lacks sweat glands completely.

Birds have dry skin devoid of epidermal glands except for the uropygial (preen) gland at the base of the tail.

2

Evaluate evaporative cooling strategies across homoiothermic classes under heat stress.

Mammals dissipate heat via cutaneous sweating and vasodilation, while birds dissipate heat via panting and gular fluttering (rapid vibration of the hyoid apparatus/gular throat pouch).

Feathers insulate bird skin, making cutaneous sweating ineffective, so avian heat loss is predominantly respiratory.

3

Rule out incorrect physiological assertions regarding cardiac and metabolic features.

Confirm that both classes have 4-chambered hearts and rely on aerobic respiration, ruling out options suggesting 3-chambered hearts or metabolic shutdown.

Complete double circulation with four chambers is required to sustain high metabolic rates in all homoiotherms.

Key Concept

Integumentary and physiological thermoregulation differences between Aves and Mammalia

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