All practice questions

Question 6721Question →

A group of small-scale cocoa farmers in Ondo State formed an enterprise to bulk-purchase agricultural inputs directly from manufacturers and collectively process and market their harvested crops. Which type of cooperative society best describes this business organization?

Producer Cooperative Society

Consumer Cooperative Society

Credit and Thrift Cooperative Society

Retail Cooperative Society

Show answer & explanation

Answer: Producer Cooperative Society

Answer

The business enterprise is a Producer Cooperative Society because it is organized by independent producers to jointly source production inputs and market their harvested output.

A Producer Cooperative Society is established by independent producers, such as farmers or artisanal craftsmen, to pool their resources for purchasing raw materials or production inputs at lower costs and collectively processing and selling their finished output to achieve economies of scale.

Step-by-Step Solution

1

Analyze the primary objective and membership of the business organization.

The members are cocoa farmers (producers) seeking to purchase production inputs collectively and market their farm produce.

Identifying the target membership and core function determines the specific classification of cooperative society.

2

Distinguish between the different classifications of cooperative societies based on their main functions.

Producer cooperatives focus on aiding members in production and crop marketing, whereas consumer cooperatives focus on retail goods distribution, and credit societies focus on micro-financing and savings.

Matching operational features to established economic definitions identifies the correct type of cooperative organization.

Key Concept

Classification and Functions of Producer Cooperative Societies

Estimated Time:1m 0s

Question 6722Question →

When a gas $X$ is bubbled into an acidified solution of potassium heptaoxodichromate(VI), $K_2Cr_2O_7$ , the solution changes color from orange to green, accompanied by the formation of a yellow precipitate. Which of the following correctly identifies gas $X$ and describes its role in the reaction?

Gas

X

is hydrogen sulfide (

H_2S

), and it acts as a reducing agent because the oxidation state of sulfur increases from

-2

to

0

.

Gas

X

is sulfur(IV) oxide (

SO_2

), and it acts as an oxidizing agent because it causes chromium to be reduced.

Gas

X

is chlorine (

Cl_2

), and it acts as a reducing agent because it is oxidized to chloride ions (

Cl^-

).

Gas

X

is carbon(II) oxide (

CO

), and it acts as an oxidizing agent by precipitating elemental sulfur from the acid solution.

Show answer & explanation

Answer: Gas $X$ is hydrogen sulfide ( $H_2S$ ), and it acts as a reducing agent because the oxidation state of sulfur increases from $-2$ to $0$ .

Answer

Gas

X

is hydrogen sulfide (

H_2S

), acting as a reducing agent as sulfur is oxidized from an oxidation state of

-2

to

0

.

Hydrogen sulfide (

H_2S

) acts as a reducing agent by donating electrons to dichromate ions (

Cr_2O_7^{2-}

), reducing chromium from oxidation state

+6

(orange) to

+3

(green). Simultaneously, sulfide ions (

S^{2-}

) in

H_2S

are oxidized to elemental sulfur (

S_{(s)}

), which appears as a characteristic yellow precipitate.

Step-by-Step Solution

1

Analyze the color change of acidified potassium heptaoxodichromate(VI),

K_2Cr_2O_7

.

The orange color is due to

Cr_2O_7^{2-}

ions (where Cr has an oxidation state of

+6

). The green color indicates reduction to

Cr^{3+}

ions (oxidation state

+3

).

Potassium heptaoxodichromate(VI) acts as an oxidizing agent when reduced from

+6

to

+3

.

2

Identify the role of gas

X

and the origin of the yellow precipitate.

Gas

X

causes the reduction of dichromate, so gas

X

must be a reducing agent. The yellow precipitate formed is insoluble elemental sulfur (

S_{(s)}

).

A yellow colloidal precipitate in wet redox tests of gases is characteristic of sulfide oxidation to sulfur.

3

Determine the oxidation state change for hydrogen sulfide (

H_2S

).

In

H_2S

, sulfur has an oxidation state of

-2

. In elemental sulfur (

S

), its oxidation state is

0

.

An increase in oxidation state (from

-2

to

0

) represents oxidation; the species undergoing oxidation is the reducing agent.

Key Concept

Laboratory identification of reducing agents using acidified potassium heptaoxodichromate(VI)

Question 6723Question →

Arrange the following steps in the correct chronological order to illustrate how water hardness is removed using the zeolite (permutit) ion-exchange method and how the column is subsequently regenerated.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence for removing water hardness and regenerating the ion-exchange bed is: first, passing hard water into the column; second, exchanging calcium and magnesium ions for sodium ions; third, collecting the softened water; and fourth, regenerating the exhausted column with concentrated brine.

Hard water softening via the permutit process begins by feeding water containing hardness-causing cations (

\text{Ca}^{2+}

,

\text{Mg}^{2+}

) into the zeolite column. As the water percolates, an ion exchange occurs where

\text{Ca}^{2+}

and

\text{Mg}^{2+}

bind to the matrix while

\text{Na}^+

is released. The softened water free of calcium and magnesium is then collected. Finally, once the column becomes exhausted, a concentrated brine solution (

\text{NaCl}

) is flushed through to regenerate the zeolite bed for repeated use.

Step-by-Step Solution

1

Introduce hard water containing calcium and magnesium ions into the column.

Untreated water makes contact with the active sodium zeolite matrix.

Ion exchange requires direct contact between the dissolved divalent metal ions and the exchanger bed.

2

Execute the cation exchange reaction.

Calcium and magnesium ions bind to the zeolite structure, releasing sodium ions into solution.

Sodium zeolite (

\text{Na}_2\text{Z}

) trades its

\text{Na}^+

ions for

\text{Ca}^{2+}

or

\text{Mg}^{2+}

due to structural affinity.

3

Collect the effluent water.

Water free of

\text{Ca}^{2+}

and

\text{Mg}^{2+}

ions is obtained.

Sodium salts left in solution do not react with soap to form scum or cause scale build-up.

4

Regenerate the exhausted exchanger with concentrated sodium chloride (brine).

The column is converted back to active sodium zeolite (

\text{Na}_2\text{Z}

).

High concentration of

\text{Na}^+

ions forces the reverse reaction, removing bound

\text{Ca}^{2+}

and

\text{Mg}^{2+}

ions.

Key Concept

Ion-Exchange Softening and Permutit Regeneration

Question 6724Question →

In commercial practice, various credit arrangements dictate when legal ownership of goods transfers and what rights sellers retain. Match each credit trading scheme on the left with its corresponding legal ownership feature and seller remedy on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Hire Purchase Scheme matches with ownership remaining with the vendor until the final installment is paid with repossession rights; Deferred Payment Scheme matches with ownership passing upon delivery while payment is postponed into future installments; Credit Sale Scheme matches with ownership passing immediately upon delivery restricting the vendor to legal debt recovery.

Hire Purchase keeps legal ownership with the seller until the final installment is settled, allowing the seller to repossess upon default. Deferred Payment transfers ownership immediately upon delivery while deferring installments to future dates. Credit Sale transfers ownership immediately at the time of delivery, meaning the seller cannot repossess the goods and can only sue for breach of contract to recover the debt.

Step-by-Step Solution

1

Analyze the legal timing of ownership transfer for Hire Purchase

In Hire Purchase, ownership stays with the seller until the last payment is completed. The seller maintains repossession rights.

The buyer acts as a hirer until exercising the option to buy upon full payment.

2

Analyze Deferred Payment characteristics

In Deferred Payment, title passes to the buyer at delivery, but financial settlement is postponed over periodic intervals.

It is an outright sale where only payment, not ownership, is delayed.

3

Analyze Credit Sale remedies and ownership

In Credit Sale, title transfers instantly upon delivery. The vendor cannot repossess the goods if default occurs.

Since ownership has transferred, the seller's legal remedy is limited to suing the debtor for the unpaid sum.

Key Concept

Legal timing of title transfer and seller repossession rights across Hire Purchase, Deferred Payment, and Credit Sale schemes

Question 6725Question →

Match each mixture or suspension in List I with the most appropriate physical separation technique based on component properties in List II.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings match: ammonium chloride and sodium nitrate with sublimation via gentle heating; cobalt powder and brass with magnetic separation; cellular organelles suspension with high-speed centrifugation; and iodine crystals and sand with iodine sublimation.

Each mixture is paired with the separation technique that targets a distinct physical property difference: sublimable volatile solids (

NH_4Cl

, iodine) separated by sublimation; ferromagnetic cobalt separated by magnetization; and fine organelle colloidal suspensions separated by high-speed centrifugation.

Step-by-Step Solution

1

Analyze the physical properties of the components in the mixture of ammonium chloride and sodium nitrate.

Ammonium chloride (

NH_4Cl

) sublimes readily when heated, while

NaNO_3

has a high melting point and does not sublime.

Difference in volatility and sublimability allows ammonium chloride to be isolated as a sublimates deposit.

2

Evaluate the magnetic properties of cobalt powder and brass.

Cobalt is ferromagnetic and strongly attracted to a magnetic field, while brass (alloy of Cu and Zn) is non-magnetic.

A magnetic separator can selectively withdraw cobalt from the non-magnetic brass powder.

3

Determine the physical behavior of fine organelle suspensions.

Microscopic cellular organelles suspended in liquid buffers experience Brownian motion and take extremely long periods to settle under normal gravity.

Applying high rotational speed (centrifugal force) accelerates sedimentation according to particle density and mass differences.

4

Examine the mixture of iodine crystals and silica sand.

Iodine transitions directly from solid to vapor upon heating (

I_{2(s)} \rightarrow I_{2(g)}

) and re-crystallizes on a cold condenser surface, whereas silica sand (

SiO_2

) is heat-stable and non-volatile.

Sublimation isolates iodine in high purity while sand remains in the heating container.

Key Concept

Physical Separation Techniques: Sublimation, Magnetization, and Centrifugation

Question 6726Question →

A uniform horizontal wooden beam $AB$ of length $4.0\text{ m}$ and mass $8.0\text{ kg}$ is supported on a pivot placed $1.0\text{ m}$ from end $A$ . A mass of $5.0\text{ kg}$ is suspended from end $B$ . What is the magnitude of the downward vertical force $F$ in newtons that must be applied at end $A$ to keep the beam in horizontal equilibrium? (Take $g = 10\text{ m/s}^2$ ).

Show answer & explanation

Answer: 230

Answer

The magnitude of the downward force required at end A to maintain horizontal equilibrium is 230 N.

Taking moments about the pivot, the downward force F at end A produces an anticlockwise moment of F × 1.0 m. This balances the clockwise moments produced by the weight of the beam (80 N × 1.0 m) and the load at end B (50 N × 3.0 m). Equating anticlockwise and clockwise moments gives F × 1.0 = 80 + 150 = 230 N.

Step-by-Step Solution

1

Determine the weights and distance of each force from the pivot point

Weight of beam = 80 N acting at 1.0 m right of pivot; weight at B = 50 N acting at 3.0 m right of pivot; force F acts at 1.0 m left of pivot.

The center of gravity of a uniform 4.0 m beam is at its midpoint (2.0 m from end A).

2

Set up the moment equilibrium equation about the pivot

F × 1.0 = (80 × 1.0) + (50 × 3.0)

For static rotational equilibrium, total anticlockwise moments equal total clockwise moments.

3

Solve for the force magnitude F

F = 230 N

Summing clockwise moments yields 80 + 150 = 230 N m, which divided by 1.0 m gives F = 230 N.

Key Concept

Principle of Moments and Static Equilibrium

Estimated Time:1m 30s

Question 6727Question →

A potential difference of $16\,\text{V}$ is applied across a uniform conductor of length $4.0\,\text{m}$ and cross-sectional area $1.5 \times 10^{-6}\,\text{m}^2$ . If the resistivity of the conductor material is $3.0 \times 10^{-7}\,\Omega\cdot\text{m}$ , what is the electric current, in amperes, flowing through the conductor?

Show answer & explanation

Answer: 20

Answer

The electric current flowing through the conductor is

20\,\text{A}

.

The electrical resistance of the wire is first determined using the formula

R = \frac{\rho L}{A} = \frac{(3.0 \times 10^{-7})(4.0)}{1.5 \times 10^{-6}} = 0.8\,\Omega

. Then, by applying Ohm's law (

I = \frac{V}{R}

), the current is computed as

I = \frac{16}{0.8} = 20\,\text{A}

.

Step-by-Step Solution

1

Calculate the electrical resistance of the conductor from its physical dimensions and resistivity.

R = 0.8\,\Omega

Substitute

\rho = 3.0 \times 10^{-7}\,\Omega\cdot\text{m}

,

L = 4.0\,\text{m}

, and

A = 1.5 \times 10^{-6}\,\text{m}^2

into

R = \frac{\rho L}{A}

.

2

Apply Ohm's law to calculate the current flowing through the conductor.

I = 20\,\text{A}

Substitute potential difference

V = 16\,\text{V}

and calculated resistance

R = 0.8\,\Omega

into

I = \frac{V}{R}

.

Key Concept

Relationship between resistivity, resistance, potential difference, and electric current

Question 6728Question →

A progressive transverse wave traveling in Medium 1 is described by the equation $y = 0.04 \sin(200\pi t - 5\pi x)$ , where $x$ and $y$ are in meters and $t$ is in seconds. If the wave enters Medium 2 where its propagation speed drops to $20\text{ m/s}$ , what is the wavelength of the wave in Medium 2?

0.40\text{ m}

0.20\text{ m}

0.10\text{ m}

0.80\text{ m}

Show answer & explanation

Answer: $0.20\text{ m}$

Answer

0.20\text{ m}

When a wave propagates from one medium into another, its frequency remains unchanged because frequency is determined solely by the periodic source. From the wave equation in Medium 1, the angular frequency

\omega = 200\pi\text{ rad/s}

, which corresponds to a frequency

f = \frac{\omega}{2\pi} = 100\text{ Hz}

. Using

v = f\lambda

in Medium 2 with

v = 20\text{ m/s}

, the new wavelength is

\lambda = \frac{20}{100} = 0.20\text{ m}

.

Step-by-Step Solution

1

Extract the angular frequency from the wave equation.

Comparing

y = 0.04 \sin(200\pi t - 5\pi x)

to

y = A \sin(\omega t - k x)

gives

\omega = 200\pi\text{ rad/s}

.

The coefficient of

t

inside the sine term represents the angular frequency

\omega

.

2

Calculate the wave frequency.

f = \frac{\omega}{2\pi} = \frac{200\pi}{2\pi} = 100\text{ Hz}

.

Frequency is related to angular frequency by

\omega = 2\pi f

.

3

Apply the boundary condition for wave refraction.

Frequency in Medium 2 is

f = 100\text{ Hz}

.

Frequency is determined strictly by the wave source and remains invariant when passing between different media.

4

Calculate the wavelength in Medium 2 using the wave speed equation.

\lambda_2 = \frac{v_2}{f} = \frac{20\text{ m/s}}{100\text{ Hz}} = 0.20\text{ m}.

The fundamental wave equation

v = f\lambda

allows finding wavelength from speed and frequency.

Key Concept

Frequency Invariance Across Media Boundaries

Question 6729Question →

Match each retail business format listed on the left with its defining operational characteristic on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Supermarket matches with self-service retail focused predominantly on foodstuffs and household consumables; Automatic Vending Machine matches with automated round-the-clock dispensing without human attendants; Tied House matches with selling goods exclusively from a single manufacturer under contract; Variety Store matches with stocking diverse low-cost general merchandise at uniform prices.

Each retail format is correctly matched to its specific structural and operational definition: Supermarket relates to self-service grocery and food retailing; Automatic Vending Machine relates to 24/7 automated unattended dispensing; Tied House represents exclusive contract-bound retailing for one producer; and Variety Store corresponds to selling broad low-cost merchandise at uniform price points.

Step-by-Step Solution

1

Analyze Supermarket operational features

Identified as a large retail outlet selling food and household goods primarily via self-service

Supermarkets specialize in food and fast-moving consumer goods arranged on open display shelves for customer selection.

2

Analyze Automatic Vending Machine features

Identified as a slot machine or automated dispenser operating continuously without attendants

Vending machines eliminate salespersons by using mechanical or electronic dispensing systems upon payment insertion.

3

Analyze Tied House contract structure

Identified as an exclusive retail arrangement tied to a single producer

Tied houses (common in trade and brewing/beverage industries) operate under exclusive supply agreements.

4

Analyze Variety Store characteristics

Identified as a retail shop offering diverse low-priced merchandise

Variety stores sell a wide range of small goods (stationery, toiletries, hardware) at low standard price points.

Key Concept

Operational and structural characteristics distinguishing small-scale and large-scale retail formats

Question 6730Question →

Match each poetic excerpt with the dominant tone, mood, or persona it demonstrates.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Each poetic line matches its corresponding literary feature: 'We weep to see...' corresponds to the melancholic tone regarding transience; 'Shall I compare thee...' corresponds to the admirative tone of romantic praise; 'Do not go gentle...' corresponds to the defiant persona urging resistance against death; and 'I wander'd lonely...' corresponds to the serene mood of peaceful reflection.

The pairings are justified by direct textual evidence: weeping over early departure conveys a melancholic tone; praising beauty beyond summer days expresses an admiring tone; calling to rage against dying light creates a defiant persona; and floating alone like a cloud establishes a serene mood.

Step-by-Step Solution

1

Examine the word choice and emotional attitude expressed by the speaker in each excerpt.

Identified clear evidence of sorrow, romantic admiration, resistance, and peaceful isolation.

Tone, mood, and persona are established through diction and speaker attitude.

2

Distinguish between tone (author/speaker attitude), mood (atmosphere felt by reader), and persona (the adopted voice).

Categorized each statement according to whether it describes speaker attitude, atmospheric feeling, or speaker role.

Accurate analysis requires separating the speaker's emotional stance from reader experience and voice construction.

3

Align each poetic line with its correct textual interpretation.

Matched all four pairs based on textual evidence.

Direct evidence from the excerpts validates the pairings.

Key Concept

Analyzing Tone, Mood, and Persona in Poetry

Question 6731Question →

Which of the following methods is most suitable for preparing lead(II) sulfate ( $PbSO_4$ ), an insoluble salt, in the laboratory?

Neutralization of dilute tetraoxosulfate(VI) acid with aqueous lead(II) nitrate solution by titration

Direct combination of lead metal and concentrated tetraoxosulfate(VI) acid

Double decomposition between aqueous lead(II) nitrate and aqueous sodium sulfate

Action of dilute tetraoxosulfate(VI) acid on insoluble lead(II) carbonate followed by evaporation to dryness

Show answer & explanation

Answer: Double decomposition between aqueous lead(II) nitrate and aqueous sodium sulfate

Answer

Double decomposition between aqueous lead(II) nitrate and aqueous sodium sulfate.

Insoluble salts such as lead(II) sulfate (

PbSO_4

) are prepared in the laboratory by double decomposition (precipitation), where two aqueous solutions containing soluble salts are mixed to precipitate the desired insoluble salt.

Step-by-Step Solution

1

Identify the solubility of the target salt

Lead(II) sulfate (

PbSO_4

) is insoluble in water.

The choice of preparation method depends primarily on whether the salt is soluble or insoluble.

2

Select the appropriate synthesis method for insoluble salts

Insoluble salts are best prepared by double decomposition (precipitation) using two soluble salts as starting materials.

Mixing solutions containing

Pb^{2+}

ions (e.g.,

Pb(NO_3)_2(aq)

) and

SO_4^{2-}

ions (e.g.,

Na_2SO_4(aq)

) results in the immediate precipitation of insoluble

PbSO_4(s)

.

3

Formulate the balanced chemical equation

Pb(NO_3)_2(aq) + Na_2SO_4(aq) \rightarrow PbSO_4(s) + 2NaNO_3(aq)

The insoluble lead(II) sulfate precipitates out and can easily be collected by filtration, washed with distilled water, and dried.

Key Concept

Preparation of insoluble salts by precipitation (double decomposition)

Question 6732Question →

During the municipal purification and treatment of river water for public supply, several chemical reagents are added at different stages of the process. Which of the following chemicals is correctly paired with its primary functional role in this water treatment scheme?

Potash alum — Coagulating fine colloidal particles to form settleable flocs

Chlorine — Removing permanent hardness caused by dissolved calcium tetraoxosulfate(VI)

Slaked lime — Sterilizing the water by destroying pathogenic microorganisms

Activated carbon — Precipitating dissolved iron and manganese ions out of solution

Show answer & explanation

Answer: Potash alum — Coagulating fine colloidal particles to form settleable flocs

Answer

Potash alum — Coagulating fine colloidal particles to form settleable flocs

Potash alum (aluminum potassium sulfate) serves as a coagulating agent. In raw river water, fine suspended clay particles carry negative electrical charges that keep them suspended. The addition of alum neutralizes these charges, causing the tiny particles to clump together into larger, heavier flocs that readily settle to the bottom during the sedimentation phase.

Step-by-Step Solution

1

Identify the primary purpose of each reagent in municipal water purification.

Potash alum

\text{KAl(SO}_4\text{)}_2\cdot 12\text{H}_2\text{O}

supplies

\text{Al}^{3+}

ions that neutralize negative charges on suspended colloidal clay particles.

This charge neutralization enables microscopic particles to aggregate (coagulate) into visible flocs.

2

Distinguish between physical particle coagulation, chemical softening, and disinfection.

Coagulation removes turbidity via alum, filtration removes flocs using sand beds, and chlorination destroys disease-causing micro-organisms.

Each chemical step serves a unique, non-interchangeable objective in town water supply treatment.

Key Concept

Chemical Reagent Functional Roles in Municipal Water Treatment

Estimated Time:1m 0s

Question 6733Question →

When acidified potassium heptaoxodichromate(VI), $K_2Cr_2O_7$ , oxidizes sulfur(IV) oxide gas, $SO_2$ , the sulfur species is converted into a polyatomic oxoanion. What is the systematic IUPAC name of the resulting oxoanion, and what is the net change in the oxidation state of the sulfur atom per atom during this redox reaction?

Tetraoxosulfate(VI) ion, with an increase of

+2

Tetraoxosulfate(VI) ion, with an increase of

+6

Trioxosulfate(IV) ion, with a decrease of

-2

Heptaoxodisulfate(VI) ion, with an increase of

+4

Show answer & explanation

Answer: Tetraoxosulfate(VI) ion, with an increase of $+2$

Answer

The resulting oxoanion is the tetraoxosulfate(VI) ion, and the net change in the oxidation state of sulfur is an increase of

+2

.

In the reaction between

SO_2

and

K_2Cr_2O_7

in acid, sulfur is oxidized from

+4

in

SO_2

to

+6

in the sulfate ion (

SO_4^{2-}

). The oxidation state increases by

+2

. The oxoanion

SO_4^{2-}

contains four oxygen atoms bonded to sulfur in the

+6

state, which systematically names it tetraoxosulfate(VI) ion.

Step-by-Step Solution

1

Determine the oxidation state of sulfur in the reactant sulfur(IV) oxide (

SO_2

).

In

SO_2

, oxygen has an oxidation state of

-2

. Let

x

be the oxidation state of sulfur:

x + 2(-2) = 0 \Rightarrow x = +4

.

Establishing the initial oxidation state is essential to calculate the oxidation state change.

2

Identify the chemical formula of the product oxoanion formed when

SO_2

is oxidized by acidified

K_2Cr_2O_7

.

Acidified dichromate oxidizes

SO_2

to sulfate ions,

SO_4^{2-}

.

Oxidation of sulfur(IV) oxide in aqueous acid yields the sulfate oxoanion.

3

Determine the oxidation state of sulfur in the product oxoanion (

SO_4^{2-}

) and calculate the net change.

In

SO_4^{2-}

, let

y

be the oxidation state of sulfur:

y + 4(-2) = -2 \Rightarrow y = +6

. The change in oxidation state is

+6 - (+4) = +2

(an increase of 2).

Determining the final state gives the quantitative change required by the problem.

4

Apply systematic IUPAC nomenclature rules to name the oxoanion

SO_4^{2-}

.

The ion has four oxygen atoms attached to central sulfur with oxidation state

+6

, giving the name tetraoxosulfate(VI) ion.

IUPAC oxoanion rules require specifying the number of oxygen atoms (tetraoxo-), the central element root (sulfate), and its Roman numeral oxidation state ((VI)).

Key Concept

Calculation of oxidation numbers in oxoanions and systematic IUPAC nomenclature of inorganic redox species.

Estimated Time:2m 0s

Question 6734Question →

In Ferdinand Oyono's prescribed novel *Houseboy*, why does Toundi Ondoua become a target of relentless hostility and persecution from the colonial authorities?

He possesses intimate knowledge of Madame's adulterous affair with Monsieur Moreau.

He is caught stealing church funds intended for Father Gilbert's mission project.

He publicly denounces Father Gilbert's religious teachings during Sunday mass.

He leads an organized labor strike among indigenous domestic workers in Dangan.

Show answer & explanation

Answer: He possesses intimate knowledge of Madame's adulterous affair with Monsieur Moreau.

Answer

Toundi Ondoua becomes a target of persecution because he possesses intimate knowledge of Madame's adulterous affair with Monsieur Moreau.

In Ferdinand Oyono's *Houseboy*, Toundi's downfall is primarily caused by his position as an eyewitness to Madame's extra-marital affair with Monsieur Moreau. The colonial rulers cannot tolerate an African servant holding knowledge that shatters their pretense of moral superiority, prompting Moreau and the Commandant to treat him with extreme brutality.

Step-by-Step Solution

1

Analyze Toundi's role in the Commandant's household in *Houseboy*.

Toundi serves as the houseboy and closely observes the private behavior and secret dealings of the white colonial master and his wife.

Understanding his proximity to the main characters reveals why his presence becomes threatening to them.

2

Identify the key conflict involving Madame and Monsieur Moreau.

Madame engages in an extra-marital affair with the prison director, Monsieur Moreau, while the Commandant is away on tour.

Toundi becomes the silent keeper of this scandalous secret.

3

Evaluate why the colonial figures turn brutally against Toundi.

Because Toundi knows about the affair, his presence constantly reminds Madame and Moreau of their moral hypocrisy, leading Moreau and the Commandant to scapegoat, arrest, and torture him.

In colonial narrative dynamics, an indigenous servant knowing a master's secret strips away the master's pretense of authority.

Key Concept

Analysis of character conflict and thematic exposure of colonial hypocrisy in prescribed African prose.

Question 6735Question →

An electronics retailer sells a refrigerator to a buyer who pays an initial deposit and agrees to settle the remaining balance in monthly installments. Under the terms of the agreement, legal ownership of the refrigerator passes to the buyer immediately upon delivery, and the seller retains only the right to sue for any unpaid debt in the event of default. Which credit transaction scheme is illustrated in this scenario?

Hire purchase scheme

Deferred payment scheme

Equipment leasing scheme

Factor credit arrangement

Show answer & explanation

Answer: Deferred payment scheme

Answer

The correct transaction scheme is the deferred payment scheme.

Under a deferred payment (or credit sale) scheme, legal title and ownership of the goods pass to the buyer immediately upon contract execution or delivery. If the buyer defaults on subsequent installment payments, the seller cannot repossess the goods because ownership has transferred; the seller's sole legal remedy is to sue the buyer for the unpaid debt balance.

Step-by-Step Solution

1

Identify the timing of legal ownership (title) transfer described in the scenario.

Ownership transfers to the buyer immediately upon delivery.

This key distinction separates deferred payment/credit sales from hire purchase agreements.

2

Identify the seller's legal remedy in case of buyer default.

The seller can only sue to recover the unpaid balance and cannot repossess the goods.

Because title has already passed to the buyer, the seller no longer owns the property and therefore lacks repossession rights.

3

Match these characteristics to the correct commercial credit scheme.

Deferred payment scheme.

Immediate transfer of title combined with debt recovery rights defines a deferred payment transaction.

Key Concept

Distinction between Hire Purchase and Deferred Payment Schemes

Estimated Time:1m 0s

Question 6736Question →

A steady electric current of $3.2\,\text{A}$ flows through a conductor for $5.0\,\text{minutes}$ . Given that the elementary charge is $1.6 \times 10^{-19}\,\text{C}$ , how many electrons pass through a cross-section of the conductor during this period?

1.0 \times 10^{20}

3.6 \times 10^{20}

6.0 \times 10^{21}

1.536 \times 10^{-16}

Show answer & explanation

Answer: $6.0 \times 10^{21}$

Answer

The number of electrons passing through the cross-section of the conductor is

6.0 \times 10^{21}

.

Electric current

I

is related to total electric charge

Q

and time

t

by

Q = I \times t

. Converting time into seconds gives

t = 5.0 \times 60 = 300\,\text{s}

. The total charge passed is

Q = 3.2\,\text{A} \times 300\,\text{s} = 960\,\text{C}

. Using the charge quantization formula

Q = n \cdot e

, the number of electrons

n

is given by

n = \frac{960\,\text{C}}{1.6 \times 10^{-19}\,\text{C}} = 6.0 \times 10^{21}

.

Step-by-Step Solution

1

Convert the time from minutes into seconds.

t = 5.0\,\text{minutes} = 5.0 \times 60\,\text{s} = 300\,\text{s}

SI units require time to be in seconds when calculating electric charge.

2

Calculate the total electric charge passing through the conductor.

Q = I \times t = 3.2\,\text{A} \times 300\,\text{s} = 960\,\text{C}

Electric current is defined as the rate of flow of charge (

I = Q/t

).

3

Determine the number of electrons using charge quantization.

n = \frac{Q}{e} = \frac{960\,\text{C}}{1.6 \times 10^{-19}\,\text{C}} = 6.0 \times 10^{21}

Total charge is equal to the number of carrier electrons multiplied by the elementary charge (

Q = n \cdot e

).

Key Concept

Quantization of Electric Charge and Current

Question 6737Question →

When

6.62\text{ g}

of lead(II) trioxonitrate(V) is completely decomposed by heating according to the balanced chemical equation:

2Pb(NO_3)_2(s) \rightarrow 2PbO(s) + 4NO_2(g) + O_2(g)

What is the total volume of gaseous products liberated at STP?
[

1\text{ mole of gas at STP} = 22.4\text{ dm}^3

; relative atomic masses:

Pb = 207

,

N = 14

,

O = 16

]

0.896\text{ dm}^3

1.12\text{ dm}^3

1.20\text{ dm}^3

0.224\text{ dm}^3

Show answer & explanation

Answer: $1.12\text{ dm}^3$

Answer

1.12\text{ dm}^3

The correct answer is

1.12\text{ dm}^3

. Decomposing

6.62\text{ g}

(

0.02\text{ mol}

) of

Pb(NO_3)_2

yields

0.04\text{ mol}

of

NO_2

and

0.01\text{ mol}

of

O_2

, totaling

0.05\text{ mol}

of gas. At STP,

0.05\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 1.12\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the molar mass of lead(II) trioxonitrate(V),

Pb(NO_3)_2

.

\text{Molar Mass} = 207 + 2 \times (14 + 3 \times 16) = 331\text{ g mol}^{-1}

.

Molar mass is needed to convert the given mass into moles of reactant.

2

Determine the amount (in moles) of

Pb(NO_3)_2

reacted.

n(Pb(NO_3)_2) = \frac{6.62\text{ g}}{331\text{ g mol}^{-1}} = 0.02\text{ mol}

.

Quantitative stoichiometric relations require knowing the exact mole quantity of the reactant.

3

Identify the total mole ratio of gaseous products to reactant from the balanced chemical equation.

2\text{ moles } Pb(NO_3)_2 \rightarrow 4\text{ moles } NO_2(g) + 1\text{ mole } O_2(g) = 5\text{ moles of total gas}

. Total gas mole ratio

= \frac{5}{2} = 2.5

.

Both

NO_2

and

O_2

are gases at STP, so both contribute to the total volume evolved.

4

Calculate the total moles and volume of gas liberated at STP.

\text{Total moles of gas} = 0.02 \times 2.5 = 0.05\text{ mol}

.

\text{Total volume at STP} = 0.05\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 1.12\text{ dm}^3

.

Multiplying total gaseous moles by the standard molar gas volume gives the total volume at STP.

Key Concept

Mass-Volume Stoichiometric Calculation for Reaction Systems Yielding Multiple Gaseous Products

Question 6738Question →

Match each Rochdale cooperative principle on the left with its correct operational description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Democratic Control matches with granting each member equal voting power of one vote; Patronage Dividend matches with distributing surplus profits based on transaction volume; Voluntary and Open Membership matches with allowing free entry and exit without discrimination; Limited Interest on Capital matches with capping financial returns on contributed equity.

Each Rochdale cooperative principle directly corresponds to its operational definition: Democratic Control enforces equal voting rights (one member, one vote); Patronage Dividend returns surplus based on member trading volume; Voluntary and Open Membership ensures non-discriminatory admission and resignation; and Limited Interest on Capital restricts financial yield to keep member service paramount.

Step-by-Step Solution

1

Identify the defining features of governance in cooperative societies.

Democratic Control aligns with 'one member, one vote' regardless of shareholding size.

Cooperatives prioritize member equality over capital ownership.

2

Determine how surplus profits are allocated to cooperative members.

Patronage Dividend aligns with surplus distribution based on transaction volume.

Rewards are given for member patronage rather than capital investment.

3

Analyze admission and membership rules.

Voluntary and Open Membership aligns with non-discriminatory, free entry and exit.

Rochdale principles mandate that membership must remain open to all who wish to use the service.

4

Evaluate rules regarding returns on member capital contributions.

Limited Interest on Capital aligns with capping returns on equity.

This prevents profit maximization from overriding mutual service delivery.

Key Concept

Rochdale Principles of Cooperation

Question 6739Question →

A hydrocarbon was analyzed and found to contain $85.7\%$ carbon and $14.3\%$ hydrogen by mass. What is the empirical formula of the hydrocarbon? [Relative atomic masses: $\text{C} = 12$ , $\text{H} = 1$ ]

Show answer & explanation

Answer: CH2; CH_2; CH₂

Answer

CH₂

By dividing the mass percentage of each element by its relative atomic mass (Carbon:

85.7/12 = 7.142

, Hydrogen:

14.3/1 = 14.3

), the mole ratio obtained is

1:2

after dividing by the smallest value (

7.142

). Hence, the empirical formula is

\text{CH}_2

.

Step-by-Step Solution

1

Calculate the mole ratio of each element by dividing their mass percentages by their respective relative atomic masses.

Moles of Carbon =

\frac{85.7}{12} = 7.142\text{ mol}

; Moles of Hydrogen =

\frac{14.3}{1} = 14.3\text{ mol}

Molar mass converts mass percentages into relative molar quantities.

2

Divide each mole value by the smallest number of moles to obtain the simplest whole-number ratio.

Carbon ratio =

\frac{7.142}{7.142} = 1

; Hydrogen ratio =

\frac{14.3}{7.142} = 2

Empirical formula represents the simplest whole-number ratio of atoms in a compound.

3

Write the chemical formula using the calculated whole-number ratio.

Empirical Formula =

\text{CH}_2

Combining the simplest ratio yields 1 atom of C for every 2 atoms of H.

Key Concept

Empirical Formula Calculation from Percentage Composition

Estimated Time:1m 30s

Question 6740Question →

A transverse progressive wave traveling along a stretched string is represented by the mathematical wave equation $y = 0.05 \sin(160\pi t - 8\pi x)$ , where $x$ and $y$ are measured in meters and $t$ is in seconds. What is the speed of propagation of the wave in $\text{m/s}$ ?

Show answer & explanation

Answer: 20

Answer

The speed of propagation of the wave is

20\text{ m/s}

.

Comparing the given equation

y = 0.05 \sin(160\pi t - 8\pi x)

with the standard wave equation

y = A \sin(\omega t - kx)

gives the angular frequency

\omega = 160\pi\text{ rad/s}

and wave number

k = 8\pi\text{ rad/m}

. Substituting these into

v = \frac{\omega}{k}

yields

v = \frac{160\pi}{8\pi} = 20\text{ m/s}

.

Step-by-Step Solution

1

Compare the given wave equation with the standard progressive wave equation.

The standard form is

y = A \sin(\omega t - kx)

. Comparing parameters yields

\omega = 160\pi\text{ rad/s}

and

k = 8\pi\text{ rad/m}

.

Matching coefficients allows direct extraction of angular frequency and wave number.

2

Calculate the wave speed using the relation between angular frequency and wave number.

v = \frac{\omega}{k} = \frac{160\pi\text{ rad/s}}{8\pi\text{ rad/m}} = 20\text{ m/s}

.

Wave speed is defined as the ratio of angular frequency to wave number (

v = \lambda f = \frac{\omega}{k}

).

Key Concept

Wave Equation Parameter Extraction and Wave Speed Calculation

Answer

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