All practice questions

Question 6701Question →

A sound wave traveling through air at a speed of $340\text{ m/s}$ has a wavelength of $0.85\text{ m}$ . Upon entering a second gaseous medium, its speed decreases to $272\text{ m/s}$ . What is the wavelength of the wave in the second medium?

0.68\text{ m}

1.06\text{ m}

0.85\text{ m}

0.54\text{ m}

Show answer & explanation

Answer: $0.68\text{ m}$

Answer

The wavelength of the wave in the second medium is

0.68\text{ m}

.

Because wave frequency is determined by the source, it stays constant at

400\text{ Hz}

across both media. Using

\lambda_2 = \frac{v_2}{f}

, the wavelength in the second medium is

\frac{272\text{ m/s}}{400\text{ Hz}} = 0.68\text{ m}

.

Step-by-Step Solution

1

Calculate the frequency of the wave in the first medium using the wave equation

v = f \lambda

.

f = \frac{v_1}{\lambda_1} = \frac{340\text{ m/s}}{0.85\text{ m}} = 400\text{ Hz}

.

The source determines the wave frequency, which remains unchanged when passing into a new medium.

2

Apply the wave equation with the constant frequency to find the new wavelength in the second medium.

\lambda_2 = \frac{v_2}{f} = \frac{272\text{ m/s}}{400\text{ Hz}} = 0.68\text{ m}

.

Wave speed changes in a new medium lead directly to proportional changes in wavelength since frequency is constant.

Key Concept

Invariance of wave frequency across media boundaries and application of the wave equation

v = f \lambda

.

Question 6702Question →

Consider the thermal decomposition of potassium trioxochlorate(V) represented by the balanced equation:

2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)

What mass of

KClO_3

is required to produce

6.72\text{ dm}^3

of oxygen gas measured at STP?

[K = 39.0,\text{ } Cl = 35.5,\text{ } O = 16.0;\text{ Molar volume of gas at STP } = 22.4\text{ dm}^3\text{ mol}^{-1}]

24.5\text{ g}

12.25\text{ g}

36.75\text{ g}

22.87\text{ g}

Show answer & explanation

Answer: $24.5\text{ g}$

Answer

24.5\text{ g}

of

KClO_3

is required.

According to the balanced chemical equation,

2\text{ moles}

of

KClO_3

(

245.0\text{ g}

) produce

3\text{ moles}

of

O_2

(

67.2\text{ dm}^3

at STP). By direct proportion,

6.72\text{ dm}^3

of

O_2

requires

\frac{6.72}{67.2} \times 245.0 = 24.5\text{ g}

of

KClO_3

.

Step-by-Step Solution

1

Calculate the molar mass of

KClO_3

and the total mass of 2 moles of

KClO_3

.

Molar mass of

KClO_3 = 39.0 + 35.5 + 3(16.0) = 122.5\text{ g mol}^{-1}

. Mass of

2\text{ moles} = 2 \times 122.5 = 245.0\text{ g}

.

The balanced chemical equation shows

2\text{ moles}

of

KClO_3

undergo decomposition.

2

Calculate the volume of 3 moles of

O_2

at STP.

Volume of

3\text{ moles of } O_2 = 3 \times 22.4\text{ dm}^3 = 67.2\text{ dm}^3

.

At STP,

1\text{ mole}

of any gas occupies

22.4\text{ dm}^3

.

3

Set up a proportion to find the mass of

KClO_3

needed to yield

6.72\text{ dm}^3

of

O_2

.

\text{Mass of } KClO_3 = \frac{6.72\text{ dm}^3}{67.2\text{ dm}^3} \times 245.0\text{ g} = 24.5\text{ g}.

Direct stoichiometric ratio relates mass of reactant to volume of gaseous product.

Key Concept

Mass-Volume Stoichiometry at STP

Question 6703Question →

A merchant receives a written, unconditional promise signed by a debtor, undertaking to pay a specified sum of money on demand or at a fixed future date to a named payee. Which instrument of credit does this document represent?

Bill of Exchange

Promissory Note

Letter of Credit

Credit Note

Show answer & explanation

Answer: Promissory Note

Answer

Promissory Note

A promissory note is defined legally as an unconditional promise in writing, made and signed by the debtor (maker), promising to pay on demand or at a fixed or determinable future time a sum certain in money to a specified person or bearer.

Step-by-Step Solution

1

Analyze the core features described in the document.

The document is an unconditional promise written and signed directly by the debtor to pay a specified sum.

Identifying who initiates the instrument (debtor vs creditor) and the nature of the obligation (promise vs order) determines the specific credit instrument.

2

Distinguish between an order to pay and a promise to pay.

An unconditional promise to pay issued by the debtor is a Promissory Note, whereas an unconditional order to pay issued by a creditor is a Bill of Exchange.

Understanding key legal definitions under commercial credit instruments ensures accurate classification.

Key Concept

Forms and Instruments of Credit

Estimated Time:1m 0s

Question 6704Question →

An industrial wastewater slurry contains suspended micro-fine latex particles, dissolved sodium chloride, and insoluble magnetite ( $Fe_3O_4$ ) powder. To efficiently recover pure magnetite powder, isolate the micro-fine latex particles, and obtain pure water from the mixture without chemical additives, which sequence of separation techniques should be applied?

Application of a magnetic field to extract magnetite, high-speed centrifugation to sediment latex particles, followed by distillation to recover pure water

High-speed centrifugation to separate dissolved sodium chloride, magnetization to remove latex particles, followed by evaporation to recover pure water

Gravity filtration to separate dissolved sodium chloride, sublimation to vaporize magnetite powder, followed by magnetization to isolate latex particles

Sublimation to remove suspended latex particles, gravity filtration to isolate magnetite powder, followed by centrifugation to recover pure water from salt

Show answer & explanation

Answer: Application of a magnetic field to extract magnetite, high-speed centrifugation to sediment latex particles, followed by distillation to recover pure water

Answer

Application of a magnetic field to extract magnetite, high-speed centrifugation to sediment latex particles, followed by distillation to recover pure water

Magnetite is ferromagnetic and can be selectively extracted using a magnetic field. Colloidal latex micro-particles do not settle under normal gravity due to thermal agitation (Brownian motion), requiring high-speed centrifugation to force sedimentation based on density. Distillation then vaporizes and condenses water to isolate it from dissolved non-volatile sodium chloride.

Step-by-Step Solution

1

Identify the physical properties of each component in the mixture

Magnetite (

Fe_3O_4

) is magnetic and insoluble; latex forms a micro-suspended colloid; sodium chloride forms a true aqueous solution; water is the liquid solvent.

Separation techniques must exploit differences in specific physical properties such as magnetism, particle mass/density, and boiling point.

2

Apply magnetization first

Insoluble magnetic

Fe_3O_4

is attracted and cleanly separated from the liquid slurry.

Magnetic separation isolates the strongly magnetic solid without altering or contaminating liquid phase components.

3

Apply high-speed centrifugation to the remaining mixture

Suspended colloidal latex particles sediment rapidly out of the liquid phase.

Colloidal particles remain in suspension under normal gravity due to Brownian motion, but high centrifugal force forces them to settle based on density differences.

4

Perform distillation on the remaining true solution

Water evaporates and condenses as pure distillate while sodium chloride remains in the distillation flask.

Distillation separates volatile solvent liquid from non-volatile dissolved solid solutes based on boiling point differences.

Key Concept

Sequential physical separation of complex mixtures exploiting magnetism, centrifugal sedimentation of colloids, and differential boiling points.

Estimated Time:2m 0s

Question 6705Question →

A saturated solution contains $12.0\text{ g}$ of sodium hydroxide ( $\text{NaOH}$ ) dissolved in $500\text{ cm}^3$ of solution at $25^\circ\text{C}$ . What is the solubility of sodium hydroxide in $\text{mol dm}^{-3}$ at this temperature?
[Molar mass of $\text{NaOH} = 40.0\text{ g mol}^{-1}$ ]

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

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

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

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

Show answer & explanation

Answer: $0.60\text{ mol dm}^{-3}$

Answer

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

To find the solubility in

\text{mol dm}^{-3}

, first determine the number of moles of

\text{NaOH}

by dividing the mass (

12.0\text{ g}

) by the molar mass (

40.0\text{ g mol}^{-1}

), yielding

0.30\text{ mol}

. Next, convert the volume to cubic decimeters (

500\text{ cm}^3 = 0.50\text{ dm}^3

). Finally, divide the amount in moles by the volume in

\text{dm}^3

to get

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

.

Step-by-Step Solution

1

Calculate the amount of

\text{NaOH}

in moles

\text{Moles} = \frac{12.0\text{ g}}{40.0\text{ g mol}^{-1}} = 0.30\text{ mol}

Solubility in

\text{mol dm}^{-3}

requires expressing the amount of solute in moles.

2

Convert the solution volume from

\text{cm}^3

to

\text{dm}^3

\text{Volume} = \frac{500\text{ cm}^3}{1000\text{ cm}^3\text{ dm}^{-3}} = 0.50\text{ dm}^3

Standard concentration units require volume expressed in cubic decimeters.

3

Divide the number of moles by the volume in

\text{dm}^3

\text{Solubility} = \frac{0.30\text{ mol}}{0.50\text{ dm}^3} = 0.60\text{ mol dm}^{-3}

Solubility is the maximum amount of solute in moles that dissolves per

\text{dm}^3

of solution at a given temperature.

Key Concept

Solubility calculation in mol/dm³

Question 6706Question →

A public limited company raises long-term corporate finance by issuing various financial instruments to investors. Which statement accurately describes the financial status and rights of debenture holders in a public limited company?

They are creditors of the company who receive fixed interest payments regardless of whether the company makes a profit.

They are co-owners of the company who hold voting rights at general meetings and receive variable dividends.

They receive dividends paid exclusively out of distributable profits and carry residual claims upon winding up.

They exercise internal governance authority and possess the legal power to appoint board directors.

Show answer & explanation

Answer: They are creditors of the company who receive fixed interest payments regardless of whether the company makes a profit.

Answer

Debenture holders are creditors of the public limited company who receive fixed interest payments regardless of profit.

In a public limited company, debenture holders are lenders (creditors) rather than owners. Because debentures represent borrowed capital, the company is legally required to pay the agreed fixed rate of interest periodically as an expense, regardless of whether it registers a profit or a loss.

Step-by-Step Solution

1

Identify the nature of debentures as a corporate security instrument.

Debentures represent long-term borrowed capital (debt) rather than equity capital.

Understanding whether an instrument is debt or equity determines the legal rights and financial status of the holder.

2

Analyze the financial return and legal status associated with debentures.

Holders are creditors entitled to fixed interest charged against revenue, rather than owners receiving profit dividends.

Interest on debt is an obligation payable prior to profit distribution and does not depend on profit availability.

Key Concept

Debentures as Loan Capital in Public Limited Companies

Question 6707Question →

A standard solution is prepared by dissolving $1.575\text{ g}$ of hydrated ethanedioic acid ( $\text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O}$ ) in distilled water to make $250.0\text{ cm}^3$ of solution. A $25.0\text{ cm}^3$ sample of this acid solution requires $25.0\text{ cm}^3$ of a $0.100\text{ mol dm}^{-3}$ sodium hydroxide ( $\text{NaOH}$ ) solution for complete neutralization. What is the integer value of $x$ in the formula of the hydrated acid? [Relative atomic masses: $\text{H} = 1$ , $\text{C} = 12$ , $\text{O} = 16$ ]

Show answer & explanation

Answer: 2

Answer

The integer value of

x

is 2.

The value of

x

is calculated as

2

. Based on the reaction stoichiometry,

25.0\text{ cm}^3

of

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

\text{NaOH}

contains

0.0025\text{ mol}

of base, which neutralizes

0.00125\text{ mol}

of the diprotic acid in the

25.0\text{ cm}^3

sample. The entire

250.0\text{ cm}^3

solution therefore contains

0.0125\text{ mol}

of acid. Dividing the mass (

1.575\text{ g}

) by

0.0125\text{ mol}

gives a molar mass of

126\text{ g mol}^{-1}

for

\text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O}

. Subtracting the molar mass of anhydrous

\text{H}_2\text{C}_2\text{O}_4

(

90\text{ g mol}^{-1}

) gives

36\text{ g mol}^{-1}

for water, which corresponds to

x = 36 / 18 = 2

.

Step-by-Step Solution

1

Determine the mole ratio from the balanced chemical neutralization equation.

\text{H}_2\text{C}_2\text{O}_4 \cdot x\text{H}_2\text{O} + 2\text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + (x+2)\text{H}_2\text{O}

. The stoichiometric ratio of acid to base is

1 : 2

.

Ethanedioic acid is a diprotic acid requiring two moles of hydroxide ions for complete neutralization per mole of acid.

2

Calculate the amount in moles of sodium hydroxide solution used in the titration.

\text{Moles of NaOH} = 0.100\text{ mol dm}^{-3} \times \frac{25.0}{1000}\text{ dm}^3 = 0.0025\text{ mol}

.

Number of moles is equal to molar concentration multiplied by volume in cubic decimeters.

3

Calculate the total moles of hydrated acid present in the

250.0\text{ cm}^3

volumetric flask.

\text{Moles in } 25.0\text{ cm}^3 \text{ aliquot} = \frac{0.0025}{2} = 0.00125\text{ mol}

. Total moles in

250.0\text{ cm}^3 = 0.00125 \times \frac{250.0}{25.0} = 0.0125\text{ mol}

.

Using the stoichiometric ratio (

n_a/n_b = 1/2

) and scaling up from the

25.0\text{ cm}^3

aliquot to the full

250.0\text{ cm}^3

solution volume.

4

Compute the molar mass of the hydrated acid and solve for

x

.

\text{Molar mass} = \frac{1.575\text{ g}}{0.0125\text{ mol}} = 126\text{ g mol}^{-1}

. Molar mass of anhydrous

\text{H}_2\text{C}_2\text{O}_4 = 2(1) + 2(12) + 4(16) = 90\text{ g mol}^{-1}

. Mass of

x\text{H}_2\text{O} = 126 - 90 = 36\text{ g mol}^{-1}

. Thus,

x = \frac{36}{18} = 2

.

Subtracting the molar mass of the anhydrous acid from the total molar mass gives the mass of the water of crystallization, which is divided by the molar mass of water (

18\text{ g mol}^{-1}

) to find

x

.

Key Concept

Volumetric Analysis and Water of Crystallization Determination

Question 6708Question →

A light rigid bar $PQ$ of length $5.0\text{ m}$ is hinged at end $P$ and held horizontally. An upward force of $40\text{ N}$ is applied at end $Q$ at an angle of $30^\circ$ to the bar. To keep the bar in horizontal equilibrium, a vertical downward force $F$ is applied at a distance of $2.0\text{ m}$ from $P$ . What is the magnitude of the force $F$ ?

20\text{ N}

50\text{ N}

86.6\text{ N}

100\text{ N}

Show answer & explanation

Answer: $50\text{ N}$

Answer

The magnitude of the force

F

required to maintain equilibrium is

50\text{ N}

.

The force of

40\text{ N}

applied at an angle of

30^\circ

to the bar has a perpendicular component of

40\sin 30^\circ = 20\text{ N}

. The counterclockwise moment about pivot

P

is

20\text{ N} \times 5.0\text{ m} = 100\text{ N}\cdot\text{m}

. For the bar to remain in equilibrium, the clockwise moment created by

F

must equal

100\text{ N}\cdot\text{m}

, giving

F \times 2.0\text{ m} = 100\text{ N}\cdot\text{m}

, which yields

F = 50\text{ N}

.

Step-by-Step Solution

1

Calculate the perpendicular component of the force applied at end

Q

.

F_{\perp} = 40\text{ N} \times \sin(30^\circ) = 40 \times 0.5 = 20\text{ N}

Only the force component perpendicular to the bar produces a moment about the pivot

P

.

2

Calculate the counterclockwise moment produced by the force at

Q

about pivot

P

.

\text{Moment}_{Q} = 20\text{ N} \times 5.0\text{ m} = 100\text{ N}\cdot\text{m}

Moment is defined as perpendicular force multiplied by distance from the pivot.

3

Apply the Principle of Moments about pivot

P

to find force

F

.

F \times 2.0\text{ m} = 100\text{ N}\cdot\text{m} \implies F = \frac{100}{2.0} = 50\text{ N}

For rotational equilibrium, total clockwise moment about

P

must equal total counterclockwise moment about

P

.

Key Concept

Principle of Moments and Rotational Equilibrium with Forces at an Angle

Question 6709Question →

A uniform cylindrical wire of length $2.0\,\text{m}$ and radius $1.0\,\text{mm}$ is connected to a direct-current source. When a potential difference of $12\,\text{V}$ is applied across its ends, a steady current of $4.0\,\text{A}$ flows through it. Taking $\pi \approx 3.142$ , what is the resistivity of the material of the wire?

4.71 \times 10^{-6}\,\Omega\cdot\text{m}

4.71 \times 10^{-3}\,\Omega\cdot\text{m}

1.89 \times 10^{-5}\,\Omega\cdot\text{m}

9.43 \times 10^{-6}\,\Omega\cdot\text{m}

Show answer & explanation

Answer: $4.71 \times 10^{-6}\,\Omega\cdot\text{m}$

Answer

The resistivity of the material of the wire is

4.71 \times 10^{-6}\,\Omega\cdot\text{m}

.

Using Ohm's law (

R = V/I

), the wire's resistance is

3.0\,\Omega

. Substituting this resistance, the wire length (

2.0\,\text{m}

), and the circular area (

A = \pi r^2 = 3.142 \times 10^{-6}\,\text{m}^2

) into

\rho = R A / L

yields

4.71 \times 10^{-6}\,\Omega\cdot\text{m}

.

Step-by-Step Solution

1

Calculate the resistance of the wire using Ohm's Law.

R = \frac{V}{I} = \frac{12\,\text{V}}{4.0\,\text{A}} = 3.0\,\Omega

Ohm's Law relates potential difference, current, and resistance.

2

Calculate the cross-sectional area of the wire from its radius.

A = \pi r^2 = 3.142 \times (1.0 \times 10^{-3}\,\text{m})^2 = 3.142 \times 10^{-6}\,\text{m}^2

The wire has a circular cross-section.

3

Rearrange the resistance formula

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

to solve for resistivity

\rho

.

\rho = \frac{R A}{L} = \frac{3.0\,\Omega \times 3.142 \times 10^{-6}\,\text{m}^2}{2.0\,\text{m}} = 4.713 \times 10^{-6}\,\Omega\cdot\text{m} \approx 4.71 \times 10^{-6}\,\Omega\cdot\text{m}

Resistivity is an intrinsic property derived from resistance, length, and cross-sectional area.

Key Concept

Relationship between resistance, potential difference, current, and material resistivity.

Question 6710Question →

A sample of well water forms a white precipitate when boiled and requires a large amount of soap to produce lather. Which soluble compound present in the water is responsible for this temporary hardness?

\text{Ca(HCO}_3)_2

\text{CaSO}_4

\text{MgCl}_2

\text{Na}_2\text{SO}_4

Show answer & explanation

Answer: $\text{Ca(HCO}_3)_2$

Answer

Calcium hydrogentrioxocarbonate(IV),

\text{Ca(HCO}_3)_2

, is responsible for temporary hardness.

Temporary water hardness is caused by dissolved hydrogentrioxocarbonate(IV) salts such as calcium hydrogentrioxocarbonate(IV),

\text{Ca(HCO}_3)_2

. Heating the water decomposes this soluble compound into insoluble calcium trioxocarbonate(IV), carbon dioxide gas, and water, thereby removing the calcium ions responsible for hardness.

Step-by-Step Solution

1

Identify the type of water hardness based on the behavior upon heating.

Because boiling removes the hardness by forming a precipitate, the water sample exhibits temporary hardness.

Temporary hardness is caused by hydrogentrioxocarbonate(IV) salts of calcium or magnesium, which undergo thermal decomposition.

2

Determine the chemical formula and reaction involved.

\text{Ca(HCO}_3)_2(aq) \xrightarrow{\Delta} \text{CaCO}_3(s) + \text{H}_2\text{O}(l) + \text{CO}_2(g)

Soluble calcium hydrogentrioxocarbonate(IV) breaks down on heating to yield insoluble calcium trioxocarbonate(IV), effectively precipitating out calcium ions.

Key Concept

Temporary hardness in water is caused by dissolved calcium and magnesium hydrogentrioxocarbonates(IV) and can be removed by boiling.

Question 6711Question →

Calculate the mass, in grams, of nitrogen contained in a $16.4\text{ g}$ sample of pure calcium trioxonitrate(V), $\text{Ca(NO}_3)_2$ . $[\text{Ca} = 40, \text{N} = 14, \text{O} = 16]$

Show answer & explanation

Answer: 2.8

Answer

The mass of nitrogen in the sample is

2.8\text{ g}

.

The molar mass of

\text{Ca(NO}_3)_2

is calculated as

40 + 2(14 + 3 \times 16) = 164\text{ g/mol}

. A

16.4\text{ g}

sample corresponds to

\frac{16.4}{164} = 0.1\text{ mol}

of

\text{Ca(NO}_3)_2

. Because each formula unit contains 2 nitrogen atoms,

0.1\text{ mol}

of compound yields

0.2\text{ mol}

of nitrogen. Multiplying

0.2\text{ mol}

by the molar mass of atomic nitrogen (

14\text{ g/mol}

) gives

2.8\text{ g}

.

Step-by-Step Solution

1

Calculate the molar mass of calcium trioxonitrate(V),

\text{Ca(NO}_3)_2

164\text{ g/mol}

Sum the relative atomic masses of all atoms present:

40 + 2(14 + 3 \times 16) = 164\text{ g/mol}

.

2

Calculate the number of moles of

\text{Ca(NO}_3)_2

present in

16.4\text{ g}

0.1\text{ mol}

Use the formula

\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{16.4\text{ g}}{164\text{ g/mol}} = 0.1\text{ mol}

.

3

Determine the number of moles of nitrogen atoms in

0.1\text{ mol}

of

\text{Ca(NO}_3)_2

0.2\text{ mol}

of N atoms

Each formula unit of

\text{Ca(NO}_3)_2

contains 2 nitrogen atoms.

4

Calculate the mass of the nitrogen atoms

2.8\text{ g}

Multiply the moles of nitrogen by its atomic mass:

0.2\text{ mol} \times 14\text{ g/mol} = 2.8\text{ g}

.

Key Concept

Mole Concept and Mass Composition of Compounds

Estimated Time:1m 30s

Question 6712Question →

A progressive wave traveling along a stretched string has a frequency of $250\text{ Hz}$ and a speed of $300\text{ m/s}$ . What is the minimum distance, in meters, between two points on the string that differ in phase by $\frac{\pi}{3}\text{ rad}$ ?

Show answer & explanation

Answer: 0.2

Answer

The minimum distance between the two points is

0.2\text{ m}

.

The wavelength is found using

\lambda = \frac{v}{f} = \frac{300}{250} = 1.2\text{ m}

. Substituting

\lambda = 1.2\text{ m}

and

\Delta \phi = \frac{\pi}{3}\text{ rad}

into the phase difference formula

\Delta \phi = \frac{2\pi \Delta x}{\lambda}

gives

\Delta x = \frac{(\pi / 3) \times 1.2}{2\pi} = 0.2\text{ m}

.

Step-by-Step Solution

1

Calculate the wavelength ($\lambda$) from wave speed ($v$) and frequency ($f$)

$\lambda = \frac{300}{250} = 1.2\text{ m}$

The fundamental wave equation relates wave speed, frequency, and wavelength as $v = f\lambda$.

2

Apply the phase difference formula to solve for spatial separation ($\Delta x$)

$\Delta x = \frac{\Delta \phi \cdot \lambda}{2\pi} = \frac{(\pi / 3) \cdot 1.2}{2\pi} = 0.2\text{ m}$

A full cycle of $2\pi\text{ radians}$ corresponds to a spatial displacement of one wavelength ($\lambda$).

Key Concept

Relationship between phase difference and spatial displacement

Question 6713Question →

Read the poetic excerpt below and answer the question that follows:

"Count not the gold that fills the merchant’s chest,
Nor praise the marble where the tyrants rest;
For iron rusts, and gilded crowns decay,
And silent dust consumes their proud display."

What is the prevailing tone of the speaker in the lines above?

Didactic and contemptuous

A gloomy feeling of despair and horror

The biographical poet's personal regret over poverty

Reverent and awestruck

Show answer & explanation

Answer: Didactic and contemptuous

Answer

Didactic and contemptuous

The correct option is 'Didactic and contemptuous' because the speaker uses instructive language ('Count not', 'Nor praise') to convey a moral lesson while expressing scorn toward material wealth and political power by emphasizing their inevitable decay.

Step-by-Step Solution

1

Analyze the speaker's stance and diction toward the subject matter.

The imperatives 'Count not' and 'Nor praise' show an instructive (didactic) stance aimed at teaching a moral lesson.

Tone is defined as the speaker's attitude toward the subject.

2

Evaluate the imagery and choice of words describing wealth and power.

Phrases like 'iron rusts', 'crowns decay', and 'silent dust consumes' dismiss earthly power as temporary and meaningless, establishing a contemptuous tone.

Diction and imagery objectively justify the specific emotional stance of the speaker.

Key Concept

Tone, Mood, and Persona in Poetry

Question 6714Question →

Read the prose extract below:

"Throughout the harsh Harmattan season, the dried riverbed became a winding highway of scorched earth and white pebbles."

Which literary device is employed in the phrase "the dried riverbed became a winding highway"?

Metaphor

Simile

Oxymoron

Assonance

Show answer & explanation

Answer: Metaphor

Answer

Metaphor

The correct answer is Metaphor because the author asserts directly that the dried riverbed was a winding highway, making an implicit comparison without using explicit connective words such as 'like' or 'as'.

Step-by-Step Solution

1

Analyze the targeted phrase in the prose extract

The phrase compares 'the dried riverbed' directly to 'a winding highway'.

Identifying the subject and what it is being compared to establishes the figurative framework.

2

Check for comparative connective markers

No comparative words such as 'like' or 'as' are present in the sentence.

The absence of 'like' or 'as' distinguishes a metaphor from a simile.

3

Determine the dominant literary device

Directly asserting that one thing is another creates a metaphor.

A metaphor makes an implicit, direct comparison between two unlike entities.

Key Concept

Literary Devices in Prose: Metaphor vs. Simile

Question 6715Question →

A manufacturing firm distributes its consumer goods directly to buyers by issuing illustrated catalogs and accepting orders primarily through postal or electronic communication, completely dispensing with physical storefronts. Which large-scale retail trading format is this enterprise employing?

Mail order business

Department store

Tied shop

Supermarket

Show answer & explanation

Answer: Mail order business

Answer

Mail order business

The mail order business is a form of large-scale retailing where trading is conducted entirely without physical shop premises, relying instead on descriptive catalogs, advertisements, and postal or courier delivery services.

Step-by-Step Solution

1

Identify the key operational characteristics in the question scenario

The business uses catalog distribution, remote ordering, and lacks physical shop premises.

Retail formats are classified based on whether they operate via physical storefronts or direct non-store communication channels.

2

Match the features to the appropriate large-scale retail format

Mail order retailing is specifically designed to eliminate physical stores by relying on catalogs and mail/postal delivery.

Department stores, tied shops, and supermarkets all depend on physical retail locations for customer visits.

Key Concept

Mail Order Retailing

Question 6716Question →

Match each physical property of metals on the left with the microscopic structural explanation on the right that best accounts for it.

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Answer

Thermal conductivity matches with rapid transfer of thermal energy by mobile delocalized electrons through the lattice. High metallic lustre matches with re-emission of absorbed light by oscillating free electrons on the metal surface. Ductility matches with sliding of positive cation layers past each other into wires without breaking metallic bonds.

Thermal conductivity is caused by mobile delocalized electrons rapidly transferring heat throughout the metal lattice. Metallic lustre occurs because surface free electrons absorb and re-emit light photons. Ductility is enabled by positive cation layers sliding past each other within the sea of delocalized electrons without disrupting the non-directional metallic bonding.

Step-by-Step Solution

1

Identify the cause of heat transport in metals

Thermal conductivity relies on delocalized valence electrons carrying thermal energy rapidly through the crystal lattice.

Free electrons gain kinetic energy when heated and collide with surrounding ions and electrons to distribute heat.

2

Identify the cause of optical reflection (lustre)

Metallic lustre is produced by surface delocalized electrons oscillating in response to incoming light waves and reflecting them.

The un-bound nature of free valence electrons allows immediate absorption and re-radiation of visible light photons.

3

Identify the mechanical feature enabling deformation into wires

Ductility relies on layers of cations sliding past one another while held together by non-directional metallic attraction.

Because metallic bonding is non-directional, moving cation layers do not cause repulsive strain that breaks the crystal structure.

Key Concept

Properties of metals in terms of the delocalized electron sea model

Question 6717Question →

Match each type of cooperative society with the primary operational focus or economic need it addresses for its members.

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Answer

Consumers' Cooperative Society matches with protection against artificial scarcity and inflated retail prices; Producers' Cooperative Society matches with elimination of middlemen markup on raw materials and collective marketing of outputs; Credit and Thrift Cooperative Society matches with encouragement of regular savings and micro-loans; Farmers' Cooperative Society matches with provision of affordable agricultural inputs and equipment hiring.

Each cooperative type specializes in solving specific financial or operational hurdles: consumer cooperatives eliminate retail middleman profits, producer cooperatives support manufacturing production and joint sales, credit and thrift societies provide micro-finance, and farmers' cooperatives provide agricultural inputs and machinery access.

Step-by-Step Solution

1

Identify the primary mandate of Consumers' Cooperative Societies.

Connect Consumers' Cooperative Society to retail price protection and direct bulk purchasing for domestic items.

Consumer cooperatives focus on making essential consumer goods affordable by eliminating retail distribution intermediaries.

2

Analyze the core functions of Producers' Cooperative Societies.

Match Producers' Cooperative Society to raw material sourcing and joint output sales.

Producer cooperatives assist small-scale manufacturing units with affordable raw materials and joint market access.

3

Examine the role of Credit and Thrift Cooperative Societies.

Link Credit and Thrift Cooperative Society to savings mobilization and accessible credit.

Credit and thrift societies encourage financial discipline and supply short-term loans without conventional banking hurdles.

4

Determine the objective of Farmers' Cooperative Societies.

Link Farmers' Cooperative Society to farm input supply and agricultural mechanization support.

Farmers' cooperatives specifically address agricultural productivity by reducing input procurement costs for growers.

Key Concept

Classification and functional roles of different types of cooperative societies.

Estimated Time:1m 30s

Question 6718Question →

Complete the statement below regarding the volumetric analysis of a mixture containing sodium carbonate and sodium hydrogen carbonate using double indicators.

Fill in the blanks below

During the titration of a mixture of sodium carbonate and sodium hydrogen carbonate against standard hydrochloric acid, the first equivalence point marking the half-neutralization of sodium carbonate to sodium hydrogen carbonate is indicated by , which turns from pink to at a pH range of approximately 8.3 to 10.0.

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Answer

The first equivalence point is detected using phenolphthalein indicator, which transitions from pink to colorless (or colourless).

Phenolphthalein is the correct indicator for the first equivalence point because its transition interval (pH 8.3–10.0) matches the pH of the sodium hydrogen carbonate solution formed when all carbonate ions are neutralized. Its color changes from pink in alkaline medium to colorless at the endpoint.

Step-by-Step Solution

1

Identify the chemical reaction occurring at the first equivalence point of the carbonate-bicarbonate titration.

At the first endpoint, all

\text{CO}_3^{2-}

ions are protonated to form

\text{HCO}_3^-

according to

\text{CO}_3^{2-} + \text{H}^+ \rightarrow \text{HCO}_3^-

. The pH at this stage is weakly alkaline (around pH 8.3).

Sodium hydrogen carbonate is an amphiprotic salt that forms a solution with a pH near 8.3, requiring an indicator with a pH transition range in the alkaline region.

2

Determine the appropriate indicator and its observed color change at this pH transition.

Phenolphthalein has a pH transition range of 8.3–10.0. In basic solution containing carbonate ions, it displays a pink color, and turns colorless once all carbonate ions are converted to hydrogen carbonate ions.

Methyl orange changes color at pH 3.1–4.4, which corresponds to the second equivalence point where hydrogen carbonate is completely converted to carbon dioxide and water.

Key Concept

Double Indicator Titration of Carbonate and Hydrogen Carbonate Mixtures

Question 6719Question →

Match each prose extract below to the dominant literary device or figurative element it demonstrates.

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Answer

Extract 1 ("The solitary lantern wept...") matches Personification. Extract 2 ("The village council hall became a buzzing hive...") matches Metaphor. Extract 3 ("His voice... was as smooth... as cool morning oil") matches Simile. Extract 4 ("...wrapped in a deafening silence") matches Oxymoron.

Each extract contains distinct stylistic markers: giving human actions like weeping to a lamp is personification; directly stating a hall is a beehive is a metaphor; comparing voice to oil using 'as' is a simile; and placing 'deafening' next to 'silence' creates an oxymoron.

Step-by-Step Solution

1

Analyze Extract 1 for human attributes given to non-human entities.

The lantern is described as weeping tears, which is a human action, identifying the device as Personification.

Attributing human qualities to non-living objects defines personification.

2

Analyze Extract 2 for direct figurative comparison.

The hall is directly equated to a buzzing hive, establishing a Metaphor.

Directly asserting that one thing is another without comparative signal words constitutes a metaphor.

3

Analyze Extract 3 for explicit comparison markers.

The comparison uses the marker "as... as", identifying it as a Simile.

Comparisons using "like" or "as" are similes.

4

Analyze Extract 4 for contradictory word combinations.

"Deafening silence" combines opposing concepts of loud sound and complete quiet, identifying it as an Oxymoron.

Two adjacent contradictory terms form an oxymoron.

Key Concept

Literary Devices and Figurative Language in Prose

Question 6720Question →

An organic compound contains $40.0\%$ carbon, $6.7\%$ hydrogen, and $53.3\%$ oxygen by mass. If the vapour density of the compound is $30$ , what is its molecular formula? $[C = 12, H = 1, O = 16]$

CH_2O

C_2H_4O_2

C_3H_6O_3

C_4H_8O_4

Show answer & explanation

Answer: $C_2H_4O_2$

Answer

The molecular formula of the compound is

C_2H_4O_2

.

Dividing mass percentage by relative atomic mass yields a

1:2:1

mole ratio for

\text{C}:\text{H}:\text{O}

, establishing an empirical formula of

CH_2O

(mass

= 30\text{ g/mol}

). Multiplying the vapour density (

30

) by

2

gives a molar mass of

60\text{ g/mol}

. The molecular formula multiplier

n = 60 / 30 = 2

, yielding

C_2H_4O_2

.

Step-by-Step Solution

1

Determine the mole ratio of each constituent element

Moles of

C = \frac{40.0}{12} = 3.33

, Moles of

H = \frac{6.7}{1} = 6.70

, Moles of

O = \frac{53.3}{16} = 3.33

Mass percentages are divided by their respective relative atomic masses to yield mole quantities.

2

Calculate the simplest whole-number ratio to obtain the empirical formula

Ratio

C : H : O = \frac{3.33}{3.33} : \frac{6.70}{3.33} : \frac{3.33}{3.33} = 1 : 2 : 1

. Empirical formula is

CH_2O

.

Dividing all mole values by the smallest value (

3.33

) converts the mole ratio into simple integers.

3

Compute the molar mass from the given vapour density

\text{Molar Mass} = 2 \times \text{Vapour Density} = 2 \times 30 = 60\text{ g/mol}

The molar mass of a volatile substance is equal to twice its vapour density.

4

Find the molecular formula multiplier

n

and determine the molecular formula

Empirical formula mass of

CH_2O = 12 + (2 \times 1) + 16 = 30\text{ g/mol}

.

n = \frac{60}{30} = 2

. Molecular formula =

(CH_2O)_2 = C_2H_4O_2

.

The molecular formula is obtained by multiplying the subscripts of the empirical formula by

n

, where

n = \frac{\text{Molar Mass}}{\text{Empirical Mass}}

.

Key Concept

Deriving Empirical and Molecular Formulae using Vapour Density

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