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Question 6621Question →

Arrange the following structural phases of dramatic plot development in their correct sequential order from the beginning of a play to its final outcome:

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct dramatic sequence begins with the Exposition, followed by the Inciting Incident, Rising Action, Climax, and finishes with the Denouement.

In classical dramatic structure (Freytag's Pyramid), a play begins with the Exposition to establish context, moves into the Inciting Incident which launches the conflict, escalates through the Rising Action, hits its peak at the Climax, and resolves during the Denouement.

Step-by-Step Solution

1

Identify the initial dramatic setup.

Exposition is established as the starting point where background information and characters are introduced.

Dramatic structure requires background context before action occurs.

2

Determine the trigger for dramatic tension.

Inciting Incident acts as the second phase.

A specific catalyst must disrupt the initial equilibrium to begin the central conflict.

3

Trace the escalation of conflicts.

Rising Action follows the catalyst.

Complications build upon the central conflict to heighten dramatic suspense.

4

Locate the turning point of maximum tension.

Climax is placed after the rising action.

The plot reaches a critical peak where the protagonist faces the central problem directly.

5

Conclude the structural trajectory.

Denouement forms the final phase.

Following the climax, the narrative unravels toward resolution and closure.

Key Concept

Freytag's Pyramid / Dramatic Structure

Question 6622Question →

A sample of pure methane ( $\text{CH}_4$ ) contains $3.00\text{ g}$ of carbon. When this sample undergoes complete combustion in excess oxygen gas, all the hydrogen present is converted into water vapor ( $\text{H}_2\text{O}$ ). Based on the Law of Definite Proportions, what is the total mass (in grams) of water vapor produced? [Atomic masses: $\text{H} = 1$ , $\text{C} = 12$ , $\text{O} = 16$ ]

Show answer & explanation

Answer: 9

Answer

The total mass of water vapor produced is

9.00\text{ g}

.

According to the Law of Definite Proportions, a chemical compound always contains its component elements in a fixed ratio by mass. In methane (

\text{CH}_4

), the ratio of mass of carbon to hydrogen is

12 : 4

(

3 : 1

). Therefore,

3.00\text{ g}

of carbon is combined with

1.00\text{ g}

of hydrogen. When methane undergoes complete combustion, all

1.00\text{ g}

of hydrogen is converted into water (

\text{H}_2\text{O}

). Since hydrogen makes up

\frac{2}{18}

of the mass of water,

1.00\text{ g}

of hydrogen yields

1.00 \times \frac{18}{2} = 9.00\text{ g}

of water.

Step-by-Step Solution

1

Calculate the mass of hydrogen present in the methane sample using the Law of Definite Proportions.

The mass of hydrogen in the sample is

1.00\text{ g}

.

In

\text{CH}_4

, the mass ratio of carbon to hydrogen is

12 : 4 = 3 : 1

. Given

3.00\text{ g}

of carbon, the mass of hydrogen is

\frac{3.00\text{ g}}{3} = 1.00\text{ g}

.

2

Determine the mass fraction of hydrogen in water (

\text{H}_2\text{O}

).

Hydrogen accounts for

\frac{2}{18}

of the total mass of water.

The molar mass of

\text{H}_2\text{O}

is

2(1) + 16 = 18\text{ g/mol}

, of which

2\text{ g}

is hydrogen.

3

Calculate the total mass of water vapor formed from the hydrogen.

The total mass of water produced is

9.00\text{ g}

.

All

1.00\text{ g}

of hydrogen from methane is converted into water. Mass of

\text{H}_2\text{O} = 1.00\text{ g} \times \frac{18}{2} = 9.00\text{ g}

.

Key Concept

Law of Definite Proportions (Constant Composition)

Question 6623Question →

Match each poetic extract below with the primary sound device it exemplifies.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The extract with repeated initial /k/ sounds matches Alliteration; the line repeating internal long /uː/ vowel sounds matches Assonance; the line containing words that imitate natural sounds matches Onomatopoeia; and the line repeating terminal /k/ consonant sounds matches Consonance.

Each poetic extract is accurately paired with the sound device matching its dominant phonetic structure.

Step-by-Step Solution

1

Examine each poetic extract to identify recurring phonetic patterns.

Distinguish between initial consonant repetition, internal vowel repetition, imitative word choice, and terminal consonant repetition.

Identifying the location and type of repeating sound elements determines the specific literary sound device.

2

Pair each extract with its corresponding definition.

Match initial consonant repetition with Alliteration, internal vowel repetition with Assonance, imitative sound words with Onomatopoeia, and ending consonant repetition with Consonance.

Applying the correct definitions ensures precise literary categorization.

Key Concept

Sound Devices and Musicality in Poetry

Question 6624Question →

Match each physical property or behavioral phenomenon of metals on the left with its corresponding microscopic structural feature of metallic bonding on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

High thermal conductivity matches rapid kinetic energy transfer via mobile valence electrons; Malleability and ductility match sliding of cation layers due to non-directional bonding; Maintenance of electrical conductivity matches uninterrupted delocalized electron sea cohesion during lattice movement; Higher melting points of transition metals match the participation of unpaired

d

-orbital electrons alongside

s

-electrons in metallic bonding.

Each physical property directly corresponds to specific structural attributes of the delocalized electron sea model: thermal conduction relies on kinetic energy transport by mobile electrons, malleability depends on non-directional layer slipping, conductivity preservation relies on continuous electron sea mobility, and melting point strength in transition metals relies on additional

d

-electron contributions to bonding.

Step-by-Step Solution

1

Analyze thermal conductivity

Identify mobile electrons as the primary mechanism for heat transfer in metals.

Kinetic energy is rapidly dispersed by mobile valence electrons colliding with lattice ions and other electrons.

2

Analyze malleability and ductility

Relate mechanical deformation to non-directional electrostatic forces.

Planes of positive metal cations can slide past each other because delocalized electrons adjust continuously to cushion repulsive cation-cation forces.

3

Analyze electrical conductivity during deformation

Connect continuous conductivity to fluid electron sea nature.

Deforming a metal does not break discrete bonds or interrupt the delocalized sea of electrons carrying charge.

4

Analyze transition metal melting points

Evaluate electron contribution to bonding strength.

Transition metals draw upon both

(n-1)d

and

ns

electrons for metallic cohesion, strengthening the bond far beyond single valence

s

-electron systems.

Key Concept

Electron sea model, non-directional bonding, and structural origins of metallic physical properties

Question 6625Question →

Two capacitors with capacitances of $3\text{ }\mu\text{F}$ and $6\text{ }\mu\text{F}$ are connected in parallel. This combination is then connected in series with a $9\text{ }\mu\text{F}$ capacitor across a $12\text{ }\text{V}$ direct current source. What is the total electrical energy stored in the network?

3.24 \times 10^{-4}\text{ }\text{J}

6.48 \times 10^{-4}\text{ }\text{J}

7.92 \times 10^{-4}\text{ }\text{J}

1.30 \times 10^{-3}\text{ }\text{J}

Show answer & explanation

Answer: $3.24 \times 10^{-4}\text{ }\text{J}$

Answer

The total electrical energy stored in the network is

3.24 \times 10^{-4}\text{ }\text{J}

.

First, find the equivalent capacitance of the parallel branch (

3\text{ }\mu\text{F} + 6\text{ }\mu\text{F} = 9\text{ }\mu\text{F}

). Next, combine this with the

9\text{ }\mu\text{F}

capacitor in series, yielding

C_{eq} = \frac{9 \times 9}{9 + 9} = 4.5\text{ }\mu\text{F} = 4.5 \times 10^{-6}\text{ }\text{F}

. Finally, substituting into the stored energy formula

E = \frac{1}{2} C_{eq} V^2

gives

E = 0.5 \times 4.5 \times 10^{-6} \times 144 = 3.24 \times 10^{-4}\text{ }\text{J}

.

Step-by-Step Solution

1

Calculate equivalent capacitance of the parallel section

C_p = C_1 + C_2 = 3\text{ }\mu\text{F} + 6\text{ }\mu\text{F} = 9\text{ }\mu\text{F}

Capacitors in parallel add directly.

2

Calculate total equivalent capacitance of the network

\frac{1}{C_{eq}} = \frac{1}{C_p} + \frac{1}{C_3} = \frac{1}{9\text{ }\mu\text{F}} + \frac{1}{9\text{ }\mu\text{F}} = \frac{2}{9\text{ }\mu\text{F}} \implies C_{eq} = 4.5\text{ }\mu\text{F} = 4.5 \times 10^{-6}\text{ }\text{F}

Capacitors in series combine via reciprocals.

3

Calculate total stored electrical energy

E = \frac{1}{2} C_{eq} V^2 = \frac{1}{2} \times (4.5 \times 10^{-6}\text{ }\text{F}) \times (12\text{ }\text{V})^2 = 3.24 \times 10^{-4}\text{ }\text{J}

Energy stored in a capacitor network depends on equivalent capacitance and potential difference across it.

Key Concept

Equivalent capacitance of series-parallel combinations and energy stored in a capacitor

Question 6626Question →

Ethanol ( $C_2H_5OH$ ) and dimethyl ether ( $CH_3OCH_3$ ) are structural isomers with the same relative molecular mass ( $46\text{ g mol}^{-1}$ ). However, ethanol boils at $78.4^\circ\text{C}$ while dimethyl ether boils at $-24^\circ\text{C}$ . Which of the following statements correctly explains this difference in boiling points?

Ethanol molecules are held together by strong intermolecular hydrogen bonding, whereas dimethyl ether molecules experience weaker dipole-dipole interactions.

The intramolecular

O-H

covalent bond in ethanol requires more thermal energy to break than the

C-O

covalent bond in dimethyl ether during boiling.

Dimethyl ether has weaker London dispersion forces because its effective molecular weight in the liquid state is lower than that of ethanol.

Ethanol undergoes complete ionic dissociation in liquid phase, resulting in strong electrovalent attractions between ions.

Show answer & explanation

Answer: Ethanol molecules are held together by strong intermolecular hydrogen bonding, whereas dimethyl ether molecules experience weaker dipole-dipole interactions.

Answer

Ethanol molecules are held together by strong intermolecular hydrogen bonding, whereas dimethyl ether molecules experience weaker dipole-dipole interactions.

The correct answer identifies that ethanol can form strong intermolecular hydrogen bonds due to its hydroxyl group (

-OH

), while dimethyl ether lacks a hydrogen atom attached directly to an electronegative atom and relies only on weaker dipole-dipole and van der Waals forces. Consequently, ethanol requires significantly higher thermal energy to separate its molecules into the gas phase.

Step-by-Step Solution

1

Analyze the molecular structures of ethanol (

C_2H_5OH

) and dimethyl ether (

CH_3OCH_3

).

Ethanol contains a hydrogen atom directly bonded to oxygen (

-OH

), enabling hydrogen bond donation and acceptance. Dimethyl ether (

CH_3-O-CH_3

) has oxygen bonded only to carbon atoms, preventing hydrogen bond formation between its own molecules.

Hydrogen bonding requires a hydrogen atom attached to a highly electronegative atom (

N

,

O

, or

F

).

2

Compare the nature and strength of intermolecular forces in both compounds.

Ethanol exhibits strong intermolecular hydrogen bonding in addition to dipole-dipole and dispersion forces. Dimethyl ether exhibits only dipole-dipole forces and dispersion forces.

Hydrogen bonds are significantly stronger than permanent dipole-dipole attractions.

3

Relate intermolecular force strength to boiling point trends.

More thermal energy is needed to separate ethanol molecules during vaporization, giving ethanol a much higher boiling point (

78.4^\circ\text{C}

) compared to dimethyl ether (

-24^\circ\text{C}

).

Boiling point increases with stronger intermolecular forces of attraction.

Key Concept

Intermolecular Hydrogen Bonding vs. Dipole-Dipole Attractions

Question 6627Question →

Under the same conditions of temperature and pressure, how many times faster will helium gas ( $\text{He}$ ) diffuse compared to methane gas ( $\text{CH}_4$ )? [Relative atomic masses: $\text{H} = 1$ , $\text{C} = 12$ , $\text{He} = 4$ ]

2 times faster

4 times faster

0.5 times as fast

8 times faster

Show answer & explanation

Answer: 2 times faster

Answer

Helium gas diffuses 2 times faster than methane gas.

The rate of diffusion of a gas is inversely proportional to the square root of its molar mass (

r \propto \frac{1}{\sqrt{M}}

). Methane (

\text{CH}_4

) has a molar mass of

16\text{ g/mol}

and helium (

\text{He}

) has a molar mass of

4\text{ g/mol}

. Therefore, the ratio of the diffusion rate of helium to methane is

\sqrt{16/4} = \sqrt{4} = 2

.

Step-by-Step Solution

1

Calculate the molar masses of both gases.

\text{Molar mass of He} = 4\text{ g/mol}

;

\text{Molar mass of CH}_4 = 12 + (4 \times 1) = 16\text{ g/mol}

.

Molar masses are required inputs for Graham's Law.

2

Apply Graham's Law formula:

\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}

.

\frac{r_{\text{He}}}{r_{\text{CH}_4}} = \sqrt{\frac{16}{4}} = \sqrt{4} = 2

.

The rate of diffusion of a gas is inversely proportional to the square root of its molecular mass.

Key Concept

Graham's Law of Diffusion states that the rate of diffusion (

r

) of a gas is inversely proportional to the square root of its molar mass (

M

).

Question 6628Question →

Match each mode or specialized concept of transportation in commercial trade with its corresponding operational feature or economic advantage.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Pipeline Transportation matches continuous, automated movement of liquid and gaseous bulk cargoes; Tramp Steamer Shipping matches unscheduled ocean cargo freight operating without fixed routes; Containerization System matches standardized unit-load cargo facilitating rapid intermodal transfer; Inland Waterway Transport matches cost-effective haulage of heavy bulk goods along natural rivers and canals.

Each transport mode or concept uniquely addresses specific cargo characteristics, operational flexibility, and economic requirements in commerce.

Step-by-Step Solution

1

Analyze the operational mechanism of Pipeline Transportation

Identified as specialized infrastructure dedicated to fluids/gases moving continuously without vehicle return trips.

Pipelines eliminate the need for discrete vehicles and return freight runs for liquids and gases.

2

Distinguish Tramp Steamers from Ocean Liners

Tramp steamers operate flexibly on charter contracts without fixed timetables or regular routes.

Commercial shipping classifies vessel operations into fixed-route liners and flexible charter trampers.

3

Evaluate the core commercial advantage of Containerization

Enables seamless intermodal transfers using standard box dimensions with enhanced security against damage and theft.

Unitized cargo handling streamlines transshipment across different transport modes.

4

Examine the characteristics of Inland Waterway Transport

Provides economical haulage for bulky goods on rivers, though constrained by channel depth and natural geography.

River craft offer low fuel costs per ton-mile but depend on water levels and river network paths.

Key Concept

Modes of Transportation and Their Operational Functions in Commerce

Question 6629Question →

Match each type of economy of scale with its corresponding business operational context.

Click a left item, then click its matching right item

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Answer

Technical Economies match with unit cost reductions from specialized machinery; Financial Economies match with securing credit at lower interest rates; Economies of Localization match with regional industry clustering and shared infrastructure; Economies of Information match with industry-wide shared research and trade information.

Each type of economy is accurately paired by distinguishing firm-level operational benefits (Technical and Financial) from industry-wide environmental advantages resulting from industrial expansion (Localization and Information).

Step-by-Step Solution

1

Differentiate between internal (firm-level) and external (industry-level) economies of scale.

Technical and Financial economies originate within individual expanding firms, while Localization and Information economies accrue to all firms due to industry growth.

Correct classification requires separating internal operational expansion from industry-wide environmental benefits.

2

Pair internal economies to firm-specific operational mechanisms.

Technical economies link to capital equipment utilization; Financial economies link to credit access and collateral.

Internal economies reflect cost savings controlled directly by the firm's own growth decisions.

3

Pair external economies to industry-wide development mechanisms.

Localization links to geographical clustering; Information links to shared research and publications.

External economies are external to the firm but internal to the industry as a whole.

Key Concept

Internal versus External Economies of Scale

Question 6630Question →

A public limited company intends to raise long-term capital to finance the construction of a new factory branch. To avoid diluting the ownership and voting control of existing equity holders, which source of capital is most appropriate for the company to issue?

Debentures

Ordinary shares

Bank overdrafts

Trade credit

Show answer & explanation

Answer: Debentures

Answer

Debentures provide long-term loan capital without conferring voting rights, thereby funding major capital projects without diluting shareholder control.

Debentures are a primary source of long-term loan capital. Holders of debentures are creditors who receive fixed interest payments and do not possess voting rights at general meetings. Consequently, raising capital through debentures allows a company to fund major long-term capital investments without diluting the ownership equity or voting power of existing shareholders.

Step-by-Step Solution

1

Identify the financial requirement specified in the scenario

The company needs long-term capital to construct a fixed asset (a new factory branch) while preserving existing voting control.

Matching the duration of the finance source with the asset life (matching principle) and protecting ownership rights dictates the appropriate source of capital.

2

Evaluate short-term versus long-term sources of business capital

Bank overdrafts and trade credit are short-term sources used for working capital, so they are inappropriate for long-term factory construction.

Short-term liabilities must be repaid quickly and cannot fund long-term fixed assets.

3

Compare long-term capital options (equity vs. loan capital)

Ordinary shares grant voting rights and dilute ownership control, whereas debentures represent loan capital with no voting rights attached.

Debenture holders are creditors, not owners, so issuing debentures raises long-term funds without altering equity voting power.

Key Concept

Distinguishing between long-term loan capital (debentures) and equity capital (ordinary shares) regarding control and ownership implications.

Estimated Time:1m 0s

Question 6631Question →

The mass of a helium nucleus $^{4}_{2}\text{He}$ is $4.0015\text{ u}$ . If the mass of a proton is $1.0073\text{ u}$ and the mass of a neutron is $1.0087\text{ u}$ , what is the binding energy per nucleon of the helium nucleus? (Take $1\text{ u} = 931\text{ MeV}$ )

7.10\text{ MeV}

28.40\text{ MeV}

14.20\text{ MeV}

0.0305\text{ MeV}

Show answer & explanation

Answer: $7.10\text{ MeV}$

Answer

The binding energy per nucleon of the helium nucleus is

7.10\text{ MeV}

.

The correct response accurately calculates the mass defect of the helium-4 nucleus (

0.0305\text{ u}

), converts it to total binding energy (

28.40\text{ MeV}

), and divides by the mass number

A = 4

to obtain

7.10\text{ MeV}

per nucleon.

Step-by-Step Solution

1

Calculate the total mass of the individual constituent nucleons (2 protons and 2 neutrons).

Total constituent mass =

2(1.0073\text{ u}) + 2(1.0087\text{ u}) = 2.0146\text{ u} + 2.0174\text{ u} = 4.0320\text{ u}

.

Helium-4 consists of

Z = 2

protons and

N = A - Z = 4 - 2 = 2

neutrons.

2

Calculate the mass defect (

\Delta m

).

\Delta m = 4.0320\text{ u} - 4.0015\text{ u} = 0.0305\text{ u}

.

Mass defect is the difference between the total mass of individual nucleons and the actual nuclear mass.

3

Convert mass defect into total binding energy (

E_b

).

E_b = 0.0305\text{ u} \times 931\text{ MeV/u} = 28.3955\text{ MeV}

.

Applying mass-energy equivalence using the factor

1\text{ u} = 931\text{ MeV}

.

4

Divide total binding energy by the total number of nucleons (

A = 4

).

\text{Binding Energy per nucleon} = \frac{28.3955\text{ MeV}}{4} = 7.098875\text{ MeV} \approx 7.10\text{ MeV}.

Binding energy per nucleon measures average stability per nuclear particle.

Key Concept

Binding Energy per Nucleon and Mass Defect

Estimated Time:1m 30s

Question 6632Question →

A sample of hydrated copper(II) sulfate ( $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$ ) has a mass of $24.95\text{ g}$ . What is the total number of moles of oxygen atoms contained in this sample? $[\text{Relative atomic masses: } \text{Cu} = 63.5, \text{S} = 32.0, \text{O} = 16.0, \text{H} = 1.0]$

Show answer & explanation

Answer: 0.9

Answer

The total number of moles of oxygen atoms contained in the sample is

0.90\text{ mol}

.

The molar mass of hydrated copper(II) sulfate (

\text{CuSO}_4 \cdot 5\text{H}_2\text{O}

) is

249.5\text{ g/mol}

. Dividing

24.95\text{ g}

by

249.5\text{ g/mol}

gives

0.10\text{ mol}

of the compound. Since each mole of

\text{CuSO}_4 \cdot 5\text{H}_2\text{O}

contains

9\text{ moles}

of oxygen atoms (

4

from

\text{CuSO}_4

and

5

from

5\text{H}_2\text{O}

), the total quantity of oxygen atoms is

0.10 \times 9 = 0.90\text{ mol}

.

Step-by-Step Solution

1

Calculate the molar mass of

\text{CuSO}_4 \cdot 5\text{H}_2\text{O}

Molar mass =

249.5\text{ g/mol}

Sum the relative atomic masses of all atoms present in one formula unit of the hydrated compound.

2

Calculate the moles of the hydrated salt

Moles of compound =

0.10\text{ mol}

Divide the mass of the sample (

24.95\text{ g}

) by its molar mass (

249.5\text{ g/mol}

).

3

Determine the stoichiometric multiplier for oxygen atoms

9 moles of oxygen atoms per mole of compound

Each formula unit contains 4 oxygen atoms in the sulfate group and 5 oxygen atoms in the water of crystallization.

4

Calculate total moles of oxygen atoms

Moles of oxygen atoms =

0.90\text{ mol}

Multiply the moles of compound (

0.10\text{ mol}

) by the 9 moles of oxygen atoms per mole of compound.

Key Concept

Stoichiometric relationship of constituent atoms in a hydrated compound

Question 6633Question →

A constant horizontal force acts on a body of mass $10\text{ kg}$ , accelerating it from rest to a speed of $12\text{ m s}^{-1}$ in a time of $4\text{ s}$ along a smooth horizontal surface. What is the average power delivered by the force during this time interval?

Show answer & explanation

Answer: 180

Answer

The average power delivered by the force during the 4-second interval is

180\text{ W}

.

By the work-energy theorem, the total work done by the constant force equals the gain in kinetic energy:

W = \frac{1}{2} m v^2 = \frac{1}{2} \times 10 \times 144 = 720\text{ J}

. The average power is the rate at which work is performed over time:

P = \frac{W}{t} = \frac{720\text{ J}}{4\text{ s}} = 180\text{ W}

.

Step-by-Step Solution

1

Calculate the final kinetic energy acquired by the body.

E_k = \frac{1}{2} m v^2 = \frac{1}{2} (10\text{ kg})(12\text{ m s}^{-1})^2 = 720\text{ J}

.

Since the body starts from rest on a smooth surface, all work done by the net force goes into increasing its kinetic energy.

2

Divide the total work done by the elapsed time to find the average power.

P_{\text{avg}} = \frac{W}{t} = \frac{720\text{ J}}{4\text{ s}} = 180\text{ W}

.

Average power is defined as the rate of doing work over a given time interval.

Key Concept

Work-Energy Theorem and Average Power

Question 6634Question →

Match each prose narrative motif or character arc to the central theme it predominantly conveys in literary prose analysis.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The narrative arcs correspond to their respective central themes as follows: the pursuit of financial wealth at the expense of family aligns with the alienating impact of unbridled acquisitiveness; foreign administrative imposition aligns with the erosion of indigenous heritage under colonial encroachment; the youth's journey toward identity aligns with the quest for self-actualization; and the failed anti-corruption crusade aligns with the entrenched nature of socio-political corruption.

Each narrative scenario serves as a concrete manifestation of its paired theme: financial obsessive focus resulting in emotional distance reflects the alienating nature of material acquisitiveness; foreign law displacing tradition reflects cultural erosion; overcoming societal pressures reflects self-actualization; and a failed moral crusade against bribery reflects the systemic endurance of corruption.

Step-by-Step Solution

1

Analyze the narrative focus of each scenario presented on the left.

Identified core character motives and conflict trajectories: greed vs. isolation, tradition vs. foreign law, societal constraint vs. self-identity, and moral reform vs. systemic corruption.

Themes in prose are derived by analyzing how character actions and narrative conflicts reflect broader human or societal statements.

2

Map each narrative scenario to its corresponding abstract thematic statement on the right.

Formed accurate conceptual pairs based on explicit textual evidence and literary interpretation.

A central theme abstracts specific plot events into universal commentary about human experience or societal structures.

Key Concept

Thematic Interpretation in Prose Literature

Question 6635Question →

Read the poetic excerpt below:

"The iron gate creaks, the dry leaves rustle,
And heavy boots clatter through the noisy bustle."

Which sound device is predominantly used in the excerpt to evoke auditory imagery?

Assonance

Onomatopoeia

Alliteration

Refrain

Show answer & explanation

Answer: Onomatopoeia

Answer

Onomatopoeia

The correct answer is Onomatopoeia because the poet intentionally uses words such as 'creaks', 'rustle', and 'clatter', whose pronunciations directly imitate the acoustic sounds produced by the gate, leaves, and boots.

Step-by-Step Solution

1

Analyze the poetic excerpt for distinct auditory and sound-based words.

Identified words like 'creaks', 'rustle', and 'clatter'.

These specific words phonetically resemble the actual physical noises they depict.

2

Match the observed technique to literary definitions of sound devices.

Determined that using words that imitate their natural sounds is defined as onomatopoeia.

The acoustic quality of the words directly creates the sensory sound experience for the reader.

Key Concept

Onomatopoeia and Auditory Imagery

Estimated Time:1m 0s

Question 6636Question →

BlueWave Marine Plc has an authorized share capital of $2,000,000$ ordinary shares of $₦5$ each. The company issued $80\%$ of its authorized shares to the public and subsequently called up $₦4$ per share. All shareholders paid the call in full except for holders of $50,000$ shares who defaulted on the payment. What is the total paid-up capital of the company in Naira?

Show answer & explanation

Answer: 6200000

Answer

The total paid-up capital of the company is

₦6,200,000

.

Paid-up capital is the portion of called-up capital that shareholders have actually paid into the business. First,

80\%

of the

2,000,000

authorized shares equals

1,600,000

issued shares. Calling up

₦4

per share yields a called-up amount of

₦6,400,000

. Since holders of

50,000

shares defaulted on

₦4

per share, the unpaid calls in arrears total

₦200,000

. Subtracting

₦200,000

from

₦6,400,000

results in a total paid-up capital of

₦6,200,000

.

Step-by-Step Solution

1

Determine the number of issued shares

1,600,000

shares

The company issued

80\%

of its

2,000,000

authorized shares (

0.80 \times 2,000,000 = 1,600,000

).

2

Calculate total called-up capital

₦6,400,000

The directors requested

₦4

per share across all

1,600,000

issued shares (

1,600,000 \times ₦4 = ₦6,400,000

).

3

Calculate calls in arrears

₦200,000

Holders of

50,000

shares failed to pay the requested

₦4

per share (

50,000 \times ₦4 = ₦200,000

).

4

Compute net paid-up capital

₦6,200,000

Paid-up capital represents actual cash received, which equals Called-Up Capital minus Calls in Arrears (

₦6,400,000 - ₦200,000 = ₦6,200,000

).

Key Concept

Paid-Up Capital and Calls in Arrears

Question 6637Question →

Complete the following statement on poetic structure by providing the correct literary terms for the specified stanzaic forms.

Fill in the blanks below

In poetic form, an eight-line stanza is designated as an , whereas a three-line stanza structured with an interlocking rhyme scheme of aba bcb cdc is known as .

Show answer & explanation

Answer

The structural term for an eight-line stanza is an octave (or octet), and the three-line stanza form using an interlocking rhyme pattern is terza rima.

An eight-line stanza unit in poetry is termed an octave (or octet). Terza rima is an Italian verse structure composed of tercets that link together using an interlocking rhyme scheme (aba bcb cdc).

Step-by-Step Solution

1

Identify the standard structural term for a poetic stanza consisting of exactly eight lines.

An eight-line stanza, commonly found at the beginning of an Italian (Petrarchan) sonnet, is called an octave or octet.

Poetic terminology classifies stanzas by line count: couplet (2), tercet (3), quatrain (4), quintet (5), sestet (6), septet (7), and octave/octet (8).

2

Determine the poetic term for a three-line stanza with the interlocking rhyme pattern aba bcb cdc.

The form is identified as terza rima.

Terza rima is a specific verse arrangement composed of tercets woven together by carrying over rhymes from one stanza to the next (aba bcb cdc ded).

Key Concept

Stanzaic Forms and Interlocking Rhyme Patterns

Question 6638Question →

Complete the reduction half-reaction by identifying the correct coefficient for the electrons needed to balance the charge.

Fill in the blanks below

In the balanced reduction half-reaction in acidic medium: $\text{MnO}_4^- + 8\text{H}^+ + e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$, the missing coefficient for the electrons is .

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Answer

5

The total charge on the left side of the equation is

+7

(from one

\text{MnO}_4^-

ion carrying

-1

and eight

\text{H}^+

ions carrying

+8

). The total charge on the right side is

+2

(from one

\text{Mn}^{2+}

ion). To make both sides equal in charge, 5 electrons (each having a

-1

charge) must be added to the reactant side.

Step-by-Step Solution

1

Calculate the total ionic charge on the reactant side before adding electrons.

Net reactant charge =

(-1) + 8(+1) = +7

One permanganate ion contributes a charge of

-1

and eight hydrogen ions contribute

+8

.

2

Calculate the total ionic charge on the product side.

Net product charge =

+2

One manganese(II) ion contributes

+2

and four water molecules are neutral (

0

).

3

Determine the number of electrons required to balance the overall charge.

+7 + 5(-1) = +2

, so 5 electrons are required.

Electrons carry a

-1

charge, so adding

5e^-

to the reactant side lowers its charge from

+7

to

+2

to match the product side.

Key Concept

Balancing charge in half-reactions by adding electrons

Question 6639Question →

A mixture of $30\text{ cm}^3$ of carbon(II) oxide and $25\text{ cm}^3$ of oxygen was sparked at constant temperature and pressure to form carbon(IV) oxide. What is the total volume of the resulting gaseous mixture?

30\text{ cm}^3

40\text{ cm}^3

55\text{ cm}^3

10\text{ cm}^3

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Answer: $40\text{ cm}^3$

Answer

The total volume of the resulting gaseous mixture is

40\text{ cm}^3

.

According to the balanced equation

2\text{CO}_{(g)} + \text{O}_{2(g)} \rightarrow 2\text{CO}_{2(g)}

, two volumes of carbon(II) oxide react with one volume of oxygen to yield two volumes of carbon(IV) oxide.

30\text{ cm}^3

of carbon(II) oxide consumes

15\text{ cm}^3

of oxygen, leaving

10\text{ cm}^3

of excess oxygen unreacted. The reaction produces

30\text{ cm}^3

of carbon(IV) oxide gas. Adding the volume of carbon(IV) oxide produced to the remaining unreacted oxygen gives a total residual gaseous volume of

30\text{ cm}^3 + 10\text{ cm}^3 = 40\text{ cm}^3

.

Step-by-Step Solution

1

Write and balance the stoichiometric chemical equation for the reaction.

2\text{CO}_{(g)} + \text{O}_{2(g)} \rightarrow 2\text{CO}_{2(g)}

To establish the mole and volume combining ratios according to Gay-Lussac's Law.

2

Determine combining volume ratios and identify the limiting reactant and unreacted gas.

By volume ratio,

2\text{ cm}^3

of

\text{CO}

reacts with

1\text{ cm}^3

of

\text{O}_2

. Therefore,

30\text{ cm}^3

of

\text{CO}

requires

\frac{1}{2} \times 30\text{ cm}^3 = 15\text{ cm}^3

of

\text{O}_2

. Unreacted

\text{O}_2 = 25\text{ cm}^3 - 15\text{ cm}^3 = 10\text{ cm}^3

.

Carbon(II) oxide is completely consumed first, making it the limiting reactant.

3

Calculate the volume of gaseous product formed.

Since

2\text{ cm}^3

of

\text{CO}

produces

2\text{ cm}^3

of

\text{CO}_2

(a

1:1

ratio),

30\text{ cm}^3

of

\text{CO}

produces

30\text{ cm}^3

of

\text{CO}_2

.

Gay-Lussac's Law applies directly to gaseous reactants and products under uniform conditions.

4

Sum the volumes of all gases present after the reaction completes.

Total residual volume = Unreacted

\text{O}_2

+ Produced

\text{CO}_2 = 10\text{ cm}^3 + 30\text{ cm}^3 = 40\text{ cm}^3

.

The final mixture contains both the newly formed gaseous product and the remaining excess reactant.

Key Concept

Gay-Lussac's Law of Combining Volumes states that when gases react under constant temperature and pressure, their combining volumes and the volumes of any gaseous products bear a simple whole-number ratio to one another.

Question 6640Question →

A radioactive sample of an isotope has a half-life of $20\text{ minutes}$ . If the initial mass of the sample is $80\text{ g}$ , what mass of the isotope, in grams, will remain undecayed after $60\text{ minutes}$ ?

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Answer: 10

Answer

10 g of the isotope remains undecayed.

The correct answer is derived from the half-life formula

N_t = N_0 \times (1/2)^n

. Since

60\text{ minutes}

contains three

20\text{ minute}

half-life periods (

n = 3

), the remaining mass is

80\text{ g} \times (1/2)^3 = 80\text{ g} / 8 = 10\text{ g}

.

Step-by-Step Solution

1

Determine the number of elapsed half-lives

3 half-lives

Divide the total time elapsed (

60\text{ minutes}

) by the duration of one half-life (

20\text{ minutes}

).

2

Calculate the remaining mass after 3 half-lives

10 g

Apply the radioactive decay relation

N_t = N_0 \times \left(\frac{1}{2}\right)^n

, yielding

80 \times \left(\frac{1}{2}\right)^3 = 10\text{ g}

.

Key Concept

Radioactive Half-Life and Exponential Decay

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