All practice questions

Question 3941Question →

While Fra Luca Pacioli is widely celebrated in accounting history, his 1494 work *Particularis de Computis et Scripturis* did not invent double-entry bookkeeping but rather codified existing mercantile practices. Which of the following historical conditions was the primary driver behind the original development of double-entry recording among Italian merchants prior to Pacioli's publication?

The expansion of long-distance commercial credit and merchant partnerships, which demanded a systematic method to track bilateral debt relationships and dual-aspect transactions

The passage of statutory joint-stock legislation requiring commercial enterprises to publish audited balance sheets for public investors

The establishment of chartered professional accounting bodies that instituted standardized rules for municipal tax auditing

The implementation of legal frameworks that merged a merchant's private assets with business obligations to streamline estate accounting

Show answer & explanation

Answer: The expansion of long-distance commercial credit and merchant partnerships, which demanded a systematic method to track bilateral debt relationships and dual-aspect transactions

Answer

The primary driver behind the development of double-entry bookkeeping in medieval Italy was the expansion of long-distance commercial credit and merchant partnerships, which required a systematic method to record bilateral debts and dual-aspect financial transactions.

Double-entry bookkeeping developed in Northern Italy during the 13th–14th centuries as commercial trade expanded rapidly. Merchants engaged in complex credit transactions, agency arrangements, and international trade routes, which necessitated a comprehensive method of tracking dual-aspect financial events (every transaction involves giving and receiving value). Pacioli's 1494 work documented these already established commercial practices.

Step-by-Step Solution

1

Analyze the historical context of pre-Pacioli accounting development

Pacioli did not invent double-entry bookkeeping; he documented practices developed by Italian merchants in trading hubs like Venice, Genoa, and Florence during the 13th and 14th centuries.

Understanding the distinction between inventing a system and codifying existing practice is key to accounting history.

2

Identify the socio-economic factors driving merchant bookkeeping needs

The Growth of international trade, credit transactions, and joint merchant ventures required tracking dynamic debtor/creditor positions and dual impact (debit and credit) for every transaction.

Primitive single-entry record-keeping could only track basic cash flow and owed amounts, making it insufficient for multi-party commercial credit networks.

3

Evaluate alternative distractors against historical timelines

Joint-stock acts and professional accounting bodies belong to the 19th-century Industrial Revolution era, while merging personal and business assets violates the business entity principle upon which accounting relies.

Distinguishing between medieval record-keeping origins and 19th-century industrial accounting evolution prevents historical misclassification.

Key Concept

Origins and Evolution of Double-Entry Bookkeeping

Estimated Time:2m 0s

Question 3942Question →

In differentiating the quadratic function $f(x) = 2x^2 + 5x$ from first principles, which of the following expressions represents the fully simplified difference quotient $\frac{f(x+h) - f(x)}{h}$ prior to evaluating the limit as $h \to 0$ ?

4x + 2h + 5

4x + h + 5

4x - 5 + 2h

2x + 2h + 5

Show answer & explanation

Answer: $4x + 2h + 5$

Answer

The fully simplified difference quotient is

4x + 2h + 5

.

Expanding

f(x+h) = 2(x+h)^2 + 5(x+h)

gives

2x^2 + 4xh + 2h^2 + 5x + 5h

. Subtracting

f(x) = 2x^2 + 5x

leaves

4xh + 2h^2 + 5h

. Dividing this expression term-by-term by

h

yields

4x + 2h + 5

.

Step-by-Step Solution

1

Evaluate

f(x+h)

for the function

f(x) = 2x^2 + 5x

.

f(x+h) = 2(x+h)^2 + 5(x+h) = 2(x^2 + 2xh + h^2) + 5x + 5h = 2x^2 + 4xh + 2h^2 + 5x + 5h

To apply first principles, we must replace every instance of

x

with

(x+h)

and expand completely.

2

Form the numerator of the difference quotient,

f(x+h) - f(x)

.

f(x+h) - f(x) = (2x^2 + 4xh + 2h^2 + 5x + 5h) - (2x^2 + 5x) = 4xh + 2h^2 + 5h

Subtracting

f(x)

cancels out the terms

2x^2

and

5x

that do not contain

h

.

3

Divide the result by

h

to get the difference quotient.

\frac{f(x+h) - f(x)}{h} = \frac{4xh + 2h^2 + 5h}{h} = 4x + 2h + 5

Factoring out

h

from the numerator and canceling it with the denominator simplifies the quotient.

Key Concept

Differentiation from First Principles (Difference Quotient)

Estimated Time:1m 30s

Question 3943Question →

In an X-ray tube, which of the following adjustments will increase the penetrating power (hardness) of the emitted X-rays?

Increasing the accelerating potential difference between the anode and cathode

Increasing the current passing through the filament heater

Decreasing the potential difference across the X-ray tube

Increasing the surface area of the target anode

Show answer & explanation

Answer: Increasing the accelerating potential difference between the anode and cathode

Answer

Increasing the accelerating potential difference between the anode and cathode increases the penetrating power of X-rays.

Increasing the accelerating potential difference between the anode and cathode increases the energy transferred to the colliding electrons. According to the Duane-Hunt law (

e V = h f_{\text{max}}

), higher potential difference leads to higher frequency X-ray photons, which increases their penetrating power (hardness).

Step-by-Step Solution

1

Identify the factor controlling electron kinetic energy

Maximum kinetic energy of electrons is

E_{\text{max}} = e V

, directly proportional to the tube voltage

V

.

Electrons gain energy from the electric field established by the potential difference.

2

Relate electron energy to X-ray photon energy and penetrating power

Higher kinetic energy produces X-ray photons of higher maximum frequency (

f_{\text{max}} = \frac{e V}{h}

) and shorter minimum wavelength (hard X-rays).

Hard X-rays with higher frequencies have greater energy and higher penetrating power.

Key Concept

X-ray Hardness vs. Intensity Control

Estimated Time:45s

Question 3944Question →

A flat circular coil consisting of $100$ turns and having a cross-sectional area of $4.0 \times 10^{-3}\text{ m}^2$ is positioned perpendicular to a uniform magnetic field of $0.50\text{ T}$ . If the direction of the magnetic field is completely reversed in a time interval of $0.02\text{ s}$ , what is the magnitude of the average electromotive force (e.m.f.) induced in the coil?

10\text{ V}

20\text{ V}

0.20\text{ V}

40\text{ V}

Show answer & explanation

Answer: $20\text{ V}$

Answer

20\text{ V}

According to Faraday's law of electromagnetic induction, the magnitude of induced electromotive force equals the rate of change of magnetic flux linkage. Because the magnetic field direction is completely reversed, the change in field magnitude is

\Delta B = 0.50 - (-0.50) = 1.0\text{ T}

. Multiplying by the area (

4.0 \times 10^{-3}\text{ m}^2

) and turn count (

100

), then dividing by the duration (

0.02\text{ s}

) gives

20\text{ V}

.

Step-by-Step Solution

1

Determine the change in magnetic flux density (

\Delta B

)

\Delta B = B_f - B_i = -0.50\text{ T} - 0.50\text{ T} = -1.0\text{ T}

, so

|\Delta B| = 1.0\text{ T}

Reversing the magnetic field direction changes the sign of the flux density relative to the coil area.

2

Calculate the total change in magnetic flux linkage (

\Delta \Phi_{total}

)

\Delta \Phi_{total} = N \cdot A \cdot |\Delta B| = 100 \times (4.0 \times 10^{-3}\text{ m}^2) \times 1.0\text{ T} = 0.40\text{ Wb-turns}

The total magnetic flux linkage is proportional to the number of turns, the surface area, and the net field change.

3

Apply Faraday's law of electromagnetic induction to find the induced e.m.f. (

\mathcal{E}

)

\mathcal{E} = \frac{\Delta \Phi_{total}}{\Delta t} = \frac{0.40\text{ Wb-turns}}{0.02\text{ s}} = 20\text{ V}

The induced e.m.f. magnitude is equal to the rate of change of magnetic flux linkage.

Key Concept

Faraday's Law of Electromagnetic Induction

Question 3945Question →

A progressive transverse wave with a frequency of $50\text{ Hz}$ travels through a medium with a wave speed of $300\text{ m/s}$ . If the wave crosses a boundary into a second medium where its speed decreases to $150\text{ m/s}$ , what is the frequency of the wave in the second medium?

50\text{ Hz}

25\text{ Hz}

100\text{ Hz}

75\text{ Hz}

Show answer & explanation

Answer: $50\text{ Hz}$

Answer

The frequency of the wave in the second medium is

50\text{ Hz}

.

The frequency of a wave is determined entirely by the vibrating source that creates it. When a wave passes from one medium to another, its speed and wavelength change, but its frequency remains constant. Therefore, the frequency remains

50\text{ Hz}

.

Step-by-Step Solution

1

Identify the given physical parameters.

Initial frequency

f_1 = 50\text{ Hz}

, initial speed

v_1 = 300\text{ m/s}

, new speed

v_2 = 150\text{ m/s}

.

Recognize which wave properties are given and which property is being asked for.

2

Apply the fundamental principle of wave propagation across different media boundaries.

Frequency

f

depends only on the source of vibration and remains constant during refraction/transmission across media boundaries.

While wave speed

v

and wavelength

\lambda

change when entering a new medium, frequency

f

remains unaffected.

3

Determine the frequency in the second medium.

f_2 = f_1 = 50\text{ Hz}

.

Since frequency is invariant across media boundaries,

f_2

must equal

50\text{ Hz}

.

Key Concept

Invariance of wave frequency across media boundaries

Estimated Time:45s

Question 3946Question →

A body is projected vertically upwards from the ground with an initial velocity $u$ . It passes a point at a height of $40\text{ m}$ above the ground at $t = 2\text{ s}$ while ascending and again at $t = 4\text{ s}$ while descending. Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the initial speed of projection $u$ of the body?

20\text{ m/s}

30\text{ m/s}

40\text{ m/s}

60\text{ m/s}

Show answer & explanation

Answer: $30\text{ m/s}$

Answer

The initial speed of projection of the body is

30\text{ m/s}

.

The position of a body thrown vertically upwards is given by

h = ut - \frac{1}{2}gt^2

. Rearranging this equation into standard quadratic form gives

t^2 - \left(\frac{2u}{g}\right)t + \frac{2h}{g} = 0

. The roots

t_1

and

t_2

correspond to the times the body reaches height

h

. By Vieta's formulas, the sum of the times is

t_1 + t_2 = \frac{2u}{g}

. Substituting

t_1 = 2\text{ s}

,

t_2 = 4\text{ s}

, and

g = 10\text{ m/s}^2

gives

6 = \frac{2u}{10}

, which solves to

u = 30\text{ m/s}

. Alternatively, the time to reach maximum height is the midpoint

t_{\text{top}} = \frac{t_1 + t_2}{2} = 3\text{ s}

, and at maximum height

v = 0 = u - gt_{\text{top}}

, yielding

u = 10 \times 3 = 30\text{ m/s}

.

Step-by-Step Solution

1

Set up the vertical motion displacement equation

h = ut - \frac{1}{2}gt^2

The equation describes the vertical position

h

at any time

t

for a projectile launched from ground level with initial speed

u

.

2

Rearrange the equation into standard quadratic form for

t

\frac{1}{2}gt^2 - ut + h = 0 \implies t^2 - \left(\frac{2u}{g}\right)t + \frac{2h}{g} = 0

The two solutions

t_1

and

t_2

represent the times at which the body reaches the specific height

h

.

3

Apply Vieta's formulas for the sum of roots of the quadratic equation

t_1 + t_2 = \frac{2u}{g}

The sum of the roots of a quadratic equation

t^2 - Bt + C = 0

is equal to the coefficient

B

.

4

Substitute given values to solve for

u

2 + 4 = \frac{2u}{10} \implies 6 = \frac{u}{5} \implies u = 30\text{ m/s}

Given

t_1 = 2\text{ s}

,

t_2 = 4\text{ s}

, and

g = 10\text{ m/s}^2

, direct substitution yields the initial velocity.

Key Concept

Vertical Motion under Gravity and Time Symmetry

Estimated Time:1m 0s

Question 3947Question →

The presence of a soluble non-volatile impurity in a pure liquid sample depresses its boiling point and causes it to boil over a narrow temperature range.

True

False

Show answer & explanation

Answer: False

Answer

False. Soluble non-volatile impurities elevate the boiling point of liquids and broaden the temperature range over which boiling occurs.

Soluble non-volatile impurities elevate the boiling point of liquids rather than depressing it, and cause the liquid to boil over a broad range of temperatures instead of a narrow, sharp temperature point.

Step-by-Step Solution

1

Analyze how non-volatile impurities affect the vapor pressure of a liquid.

Solute molecules occupy surface area, reducing the rate of solvent evaporation and lowering the vapor pressure.

Lower vapor pressure means the liquid cannot boil at its normal boiling point.

2

Determine the effect on boiling point and boiling range.

The liquid must be heated to a higher temperature to reach atmospheric pressure, causing boiling point elevation and a non-sharp boiling range.

A sharp, fixed boiling point is a criterion of purity; impure liquids boil over a range of elevated temperatures.

Key Concept

Effect of Soluble Impurities on Boiling Point and Purity Criteria

Question 3948Question →

Monochromatic light of frequency $9.0 \times 10^{14}\text{ Hz}$ is incident on a clean potassium emitter surface having a threshold frequency of $5.0 \times 10^{14}\text{ Hz}$ . Given that Planck's constant $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$ and the elementary charge $e = 1.6 \times 10^{-19}\text{ C}$ , what is the stopping potential in volts required to completely arrest the photoelectric current?

Show answer & explanation

Answer: 1.65

Answer

The stopping potential required to arrest the emitted photoelectrons is

1.65\text{ V}

.

Applying Einstein's photoelectric equation

K_{\max} = h f - h f_0 = h(f - f_0)

gives a maximum kinetic energy of

2.64 \times 10^{-19}\text{ J}

. Dividing this by the electronic charge

e = 1.6 \times 10^{-19}\text{ C}

yields the stopping potential

V_s = 1.65\text{ V}

.

Step-by-Step Solution

1

Determine the net energy available for photoelectron kinetic energy using Einstein's photoelectric equation

K_{\max} = h(f - f_0) = 6.6 \times 10^{-34}\text{ J}\cdot\text{s} \times (9.0 \times 10^{14} - 5.0 \times 10^{14})\text{ Hz} = 2.64 \times 10^{-19}\text{ J}

Photoelectron emission occurs only when photon energy

hf

exceeds the work function

W_0 = h f_0

, with the excess energy appearing as maximum kinetic energy.

2

Express stopping potential in terms of electron charge and maximum kinetic energy

V_s = \frac{K_{\max}}{e} = \frac{2.64 \times 10^{-19}\text{ J}}{1.6 \times 10^{-19}\text{ C}} = 1.65\text{ V}

The stopping potential

V_s

does work

e V_s

equal to the maximum kinetic energy

K_{\max}

of the fastest photoelectrons to bring them to rest.

Key Concept

Einstein's Photoelectric Equation and Stopping Potential Relationship

Question 3949Question →

A steady electric current of $2.5\,\text{A}$ flows through a conductor for $4.0\,\text{minutes}$ . What is the total electric charge that passes through any cross-section of the conductor during this period?

10\,\text{C}

96\,\text{C}

600\,\text{C}

150\,\text{C}

Show answer & explanation

Answer: $600\,\text{C}$

Answer

600\,\text{C}

Electric current (

I

) is defined as the rate of charge flow (

Q

) per unit time (

t

), expressed as

Q = I \times t

. Converting

4.0\,\text{minutes}

into seconds gives

4.0 \times 60 = 240\,\text{s}

. Multiplying the current of

2.5\,\text{A}

by

240\,\text{s}

yields

600\,\text{C}

.

Step-by-Step Solution

1

Convert the time duration from minutes to seconds

t = 4.0\,\text{minutes} = 4.0 \times 60\,\text{s} = 240\,\text{seconds}

The standard SI unit for time in electromagnetism formulas is seconds.

2

Apply the formula relating electric charge, current, and time (

Q = I \times t

)

Q = 2.5\,\text{A} \times 240\,\text{s} = 600\,\text{C}

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

I = \frac{Q}{t}

).

Key Concept

Relationship between Electric Current, Charge, and Time

Question 3950Question →

A resistance thermometer has a resistance of $4.0\,\Omega$ at the ice point ( $0^\circ\text{C}$ ) and $6.0\,\Omega$ at the steam point ( $100^\circ\text{C}$ ). When immersed in a liquid bath, its resistance is measured to be $5.2\,\Omega$ . What is the temperature of the bath?

60.0^\circ\text{C}

260.0^\circ\text{C}

20.0^\circ\text{C}

40.0^\circ\text{C}

Show answer & explanation

Answer: $60.0^\circ\text{C}$

Answer

60.0^\circ\text{C}

The temperature of the bath is found by calculating the fraction of resistance change relative to the total change between the ice point and steam point:

\theta = \frac{5.2 - 4.0}{6.0 - 4.0} \times 100^\circ\text{C} = 60.0^\circ\text{C}

.

Step-by-Step Solution

1

Identify the given thermometric parameters

R_0 = 4.0\,\Omega

,

R_{100} = 6.0\,\Omega

, and

R_\theta = 5.2\,\Omega

These are the measured resistance values corresponding to the lower fixed point, upper fixed point, and unknown temperature.

2

Apply the linear temperature interpolation formula for a resistance thermometer

\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C}

Temperature on the Celsius scale is proportional to the relative change in the thermometric property between fixed points.

3

Substitute the values and compute the temperature

\theta = \frac{5.2 - 4.0}{6.0 - 4.0} \times 100^\circ\text{C} = \frac{1.2}{2.0} \times 100^\circ\text{C} = 60.0^\circ\text{C}

Evaluating the expression yields the exact temperature of the liquid bath.

Key Concept

Linear interpolation on temperature scales using thermometric properties

Question 3951Question →

Which of the following gases is evolved when dilute hydrochloric acid ( $HCl$ ) reacts with calcium trioxocarbonate(IV) ( $CaCO_3$ )?

Carbon(IV) oxide (

CO_2

)

Hydrogen (

H_2

)

Chlorine (

Cl_2

)

Carbon(II) oxide (

CO

)

Show answer & explanation

Answer: Carbon(IV) oxide ( $CO_2$ )

Answer

Carbon(IV) oxide (

CO_2

)

A characteristic chemical property of acids is that they react with trioxocarbonate(IV) salts (such as

CaCO_3

) to produce a salt, water, and carbon(IV) oxide gas (

CO_2

).

Step-by-Step Solution

1

Identify the reactants and the class of reaction

Dilute hydrochloric acid (

HCl

) reacts with calcium trioxocarbonate(IV) (

CaCO_3

), which is an acid-carbonate reaction.

Acids characteristically react with trioxocarbonates to form salt, water, and carbon dioxide.

2

Write the balanced chemical equation

2HCl(aq) + CaCO_3(s) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)

The equation shows that carbon(IV) oxide gas (

CO_2

) is liberated during the chemical process.

Key Concept

Action of acids on trioxocarbonate(IV) salts

Estimated Time:45s

Question 3952Question →

Two $12\text{ }\mu\text{F}$ capacitors are connected in series, and this combination is placed in parallel with a third capacitor of capacitance $6\text{ }\mu\text{F}$ . What is the total equivalent capacitance of the arrangement?

3.0\text{ }\mu\text{F}

4.8\text{ }\mu\text{F}

12.0\text{ }\mu\text{F}

30.0\text{ }\mu\text{F}

Show answer & explanation

Answer: $12.0\text{ }\mu\text{F}$

Answer

The total equivalent capacitance of the network is

12.0\text{ }\mu\text{F}

.

Combining the two identical

12\text{ }\mu\text{F}

capacitors in series gives an equivalent capacitance of

6.0\text{ }\mu\text{F}

. Adding this in parallel with the

6.0\text{ }\mu\text{F}

capacitor yields

6.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 12.0\text{ }\mu\text{F}

.

Step-by-Step Solution

1

Calculate the equivalent capacitance (

C_s

) of the two

12\text{ }\mu\text{F}

capacitors connected in series.

C_s = \frac{12 \times 12}{12 + 12} = 6.0\text{ }\mu\text{F}

Capacitors in series combine reciprocally:

\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}

.

2

Calculate the total equivalent capacitance (

C_{eq}

) by adding

C_s

to the parallel

6.0\text{ }\mu\text{F}

capacitor.

C_{eq} = C_s + C_3 = 6.0\text{ }\mu\text{F} + 6.0\text{ }\mu\text{F} = 12.0\text{ }\mu\text{F}

Capacitors in parallel add directly:

C_{eq} = C_{series} + C_{parallel}

.

Key Concept

Equivalent capacitance of series-parallel capacitor networks

Question 3953Question →

Under fixed laboratory conditions of temperature and pressure, Gas A has a density of $0.09\text{ g/dm}^3$ while Gas B has a density of $1.44\text{ g/dm}^3$ . What is the ratio of the rate of diffusion of Gas A to the rate of diffusion of Gas B?

4.0

16.0

0.25

0.0625

Show answer & explanation

Answer: $4.0$

Answer

The ratio of the rate of diffusion of Gas A to Gas B is

4.0

.

Graham's Law states that the rate of diffusion of a gas is inversely proportional to the square root of its density or molar mass. Calculating

\sqrt{\frac{1.44}{0.09}} = \sqrt{16} = 4.0

correctly gives the relative rate ratio, showing that the lighter gas diffuses 4 times faster than the denser gas.

Step-by-Step Solution

1

State Graham's Law of Diffusion in terms of gas densities.

\frac{r_A}{r_B} = \sqrt{\frac{d_B}{d_A}}

Graham's law dictates that the rate of diffusion (

r

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

d

) at constant temperature and pressure.

2

Substitute the given density values into the equation.

\frac{r_A}{r_B} = \sqrt{\frac{1.44\text{ g/dm}^3}{0.09\text{ g/dm}^3}}

Gas B's density (

d_B = 1.44

) goes into the numerator and Gas A's density (

d_A = 0.09

) into the denominator due to the inverse relationship.

3

Simplify the fraction inside the square root and calculate the final square root.

\frac{r_A}{r_B} = \sqrt{16} = 4.0

Dividing

1.44

by

0.09

gives

16

, and taking the square root of

16

yields

4.0

.

Key Concept

Graham's Law relating gas diffusion rates to density

Estimated Time:1m 15s

Question 3954Question →

In dramatic structure, when a playwright juxtaposes a secondary character with the main character to accentuate specific moral qualities, behavioral traits, or flaws of the protagonist—such as Banquo's moral restraint highlighting Macbeth's unchecked ambition—which character type does this contrasting character represent?

Foil

Antagonist

Flat character

Tragic hero

Show answer & explanation

Answer: Foil

Answer

The contrasting character represents a dramatic foil.

A dramatic foil is a character whose values, decisions, or traits directly contrast with those of another character (usually the protagonist), thereby illuminating and sharpening the audience's understanding of the primary character's nature.

Step-by-Step Solution

1

Analyze the functional relationship between the characters described in the stem.

Banquo's moral choices serve to illuminate Macbeth's moral corruption by comparison.

Dramatic function relies on literary technique, not just character opposition.

2

Match the comparative function to standard dramatic character definitions.

The literary device where one character underscores the distinctive traits of another through contrast is a foil.

Distinguishes structural contrast (foil) from active plot opposition (antagonist).

Key Concept

Dramatic Foil and Characterization Functions

Question 3955Question →

If the solubility of sodium hydroxide ( $\text{NaOH}$ ) in water at $25^\circ\text{C}$ is $0.50\text{ mol/dm}^3$ , what is its concentration in $\text{g/dm}^3$ ? $[\text{Na} = 23, \text{O} = 16, \text{H} = 1]$

20.0\text{ g/dm}^3

40.0\text{ g/dm}^3

80.0\text{ g/dm}^3

0.0125\text{ g/dm}^3

Show answer & explanation

Answer: $20.0\text{ g/dm}^3$

Answer

The concentration of sodium hydroxide in

\text{g/dm}^3

is

20.0\text{ g/dm}^3

.

To convert concentration from

\text{mol/dm}^3

to

\text{g/dm}^3

, multiply the molar concentration by the molar mass of the solute. For

\text{NaOH}

, the molar mass is

23 + 16 + 1 = 40\text{ g/mol}

. Multiplying

0.50\text{ mol/dm}^3

by

40\text{ g/mol}

yields

20.0\text{ g/dm}^3

.

Step-by-Step Solution

1

Calculate the molar mass of sodium hydroxide (

\text{NaOH}

).

\text{Molar mass of NaOH} = 23 + 16 + 1 = 40\text{ g/mol}

.

Molar mass is required to convert concentration from moles per cubic decimetre to grams per cubic decimetre.

2

Convert molar solubility to mass concentration using the formula:

\text{Concentration in g/dm}^3 = \text{Molar concentration in mol/dm}^3 \times \text{Molar Mass}

.

\text{Concentration in g/dm}^3 = 0.50\text{ mol/dm}^3 \times 40\text{ g/mol} = 20.0\text{ g/dm}^3

.

Multiplying the amount of substance per volume by mass per mole yields mass per unit volume.

Key Concept

Conversion between molarity (mol/dm³) and mass concentration (g/dm³)

Question 3956Question →

Match each trade classification in Column A with its correct operational description in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Wholesale Trade matches with buying merchandise in bulk from manufacturers and selling in smaller quantities to distributors; Retail Trade matches with selling items in small units directly to final consumers; Entrepôt Trade matches with importing commodities from one foreign country to re-export them to another nation; Export Trade matches with selling goods produced locally to buyers in foreign markets.

Each trade classification correctly pairs with its fundamental commercial role: Wholesale trade deals in bulk distribution to retailers; Retail trade supplies the final consumer directly; Entrepôt trade handles re-exported transit goods; and Export trade sells domestic products abroad.

Step-by-Step Solution

1

Analyze Home Trade classifications

Home trade consists of wholesale trade (bulk purchasing from manufacturers to break bulk) and retail trade (selling small units directly to ultimate consumers).

Home trade occurs within the geographical boundaries of a single country and is divided based on scale and customer type.

2

Analyze Foreign Trade classifications

Foreign trade consists of import, export (selling home-made goods abroad), and entrepôt trade (re-exporting imported goods to another country).

Foreign trade crosses national borders and is categorized by the direction of physical trade flows.

Key Concept

Classification of Trade into Home Trade (Wholesale, Retail) and Foreign Trade (Import, Export, Entrepôt)

Question 3957Question →

A $13.90\text{ g}$ sample of hydrated iron(II) tetraoxosulfate(VI), $\text{FeSO}_4 \cdot x\text{H}_2\text{O}$ , was heated strongly until all water of crystallization was driven off. The mass of the remaining anhydrous salt was $7.60\text{ g}$ . What is the value of $x$ ? $[\text{Fe} = 56, \text{S} = 32, \text{O} = 16, \text{H} = 1]$

Show answer & explanation

Answer: 7

Answer

The value of

x

in the hydrated salt formula

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

is 7.

Heating the sample drives off all water of crystallization, leaving only anhydrous

\text{FeSO}_4

. The mass of water is

13.90\text{ g} - 7.60\text{ g} = 6.30\text{ g}

. Converting both components to moles gives

0.05\text{ mol}

of

\text{FeSO}_4

and

0.35\text{ mol}

of

\text{H}_2\text{O}

. The ratio

\frac{0.35}{0.05} = 7

, yielding

x = 7

.

Step-by-Step Solution

1

Find the mass of water lost

Mass of

\text{H}_2\text{O} = 13.90\text{ g} - 7.60\text{ g} = 6.30\text{ g}

The loss in mass upon heating represents the driven-off water of crystallization.

2

Calculate molar masses of anhydrous salt and water

Molar mass of

\text{FeSO}_4 = 152\text{ g/mol}

, Molar mass of

\text{H}_2\text{O} = 18\text{ g/mol}

Molar masses are required to convert the measured masses into mole quantities.

3

Calculate the amount in moles of both components

Moles of

\text{FeSO}_4 = \frac{7.60}{152} = 0.05\text{ mol}

; Moles of

\text{H}_2\text{O} = \frac{6.30}{18} = 0.35\text{ mol}

Stoichiometric coefficient

x

represents the mole ratio between water and anhydrous salt.

4

Determine the mole ratio

x = \frac{0.35\text{ mol}}{0.05\text{ mol}} = 7

Dividing the moles of water of crystallization by the moles of anhydrous salt yields the integer coefficient

x

.

Key Concept

Determining the formula of a hydrated salt from gravimetric data

Estimated Time:1m 30s

Question 3958Question →

A retail merchant in Lagos intends to stock goods from a manufacturing firm in Ibadan. Arrange the following commercial documents in the correct chronological sequence of their issuance, from the initiation of the transaction to its final settlement.

Drag items to arrange them in the correct order

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Answer

The correct chronological order of commercial documents in home trade is: Letter of Inquiry, Quotation, Order, Invoice, and Receipt.

A home trade transaction begins when the prospective buyer sends a Letter of Inquiry to ask about prices and terms. The seller replies with a Quotation containing unit prices, discounts, and delivery terms. After accepting the terms, the buyer sends an Order for the specified goods. Upon dispatching the goods, the seller issues an Invoice outlining the total debt incurred. Finally, when the buyer pays the bill, the seller provides a Receipt as conclusive evidence of payment.

Step-by-Step Solution

1

Identify the document that initiates trade contact.

The Letter of Inquiry is placed first.

Trade transactions begin when a buyer requests information regarding goods, prices, and terms.

2

Identify the document sent by the seller in response to the inquiry.

The Quotation is placed second.

The seller replies to the inquiry by providing specific prices, discounts, and delivery conditions.

3

Identify the buyer's formal purchase commitment.

The Order is placed third.

After evaluating the terms stated in the quotation, the buyer formally requests the goods.

4

Identify the document billing the buyer upon or following dispatch.

The Invoice is placed fourth.

The seller issues an invoice to summarize the goods supplied and state the total debt due for settlement.

5

Identify the final document confirming payment.

The Receipt is placed fifth.

Once the buyer settles the invoice, the seller issues a receipt as proof of payment.

Key Concept

Chronological Sequence of Commercial Documents in Home Trade

Question 3959Question →

A weak monobasic acid has a concentration of $0.20\text{ mol dm}^{-3}$ in aqueous solution. At equilibrium, the hydrogen ion concentration, $[H^+]$ , is measured to be $1.0 \times 10^{-3}\text{ mol dm}^{-3}$ . What is the percentage degree of ionization of the acid in this solution?

Show answer & explanation

Answer: 0.5

Answer

The percentage degree of ionization of the weak acid is 0.5%.

The percentage degree of ionization measures the fraction of acid molecules that ionize in water, expressed as a percentage. By substituting

[H^+] = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

and initial concentration

C = 0.20\text{ mol dm}^{-3}

into

\text{Percentage ionization} = \left(\frac{[H^+]}{C}\right) \times 100\%

, we obtain

\left(\frac{0.001}{0.20}\right) \times 100\% = 0.5\%

.

Step-by-Step Solution

1

Identify the relationship between degree of ionization (\alpha), hydrogen ion concentration (

[H^+]

), and initial acid concentration (

C

).

\alpha = \frac{[H^+]}{C}

For a weak monobasic acid ionizing according to

HA \rightleftharpoons H^+ + A^-

, the concentration of ionized hydrogen ions equals

\alpha C

.

2

Calculate the fractional degree of ionization (\alpha).

\alpha = \frac{1.0 \times 10^{-3}\text{ mol dm}^{-3}}{0.20\text{ mol dm}^{-3}} = 0.005

Dividing the equilibrium hydrogen ion concentration by the initial acid concentration gives the fraction of acid molecules that ionized.

3

Convert the fractional degree of ionization into a percentage.

\text{Percentage ionization} = 0.005 \times 100\% = 0.5\%

Multiplying the decimal fraction by 100 converts the degree of ionization into percentage form.

Key Concept

Degree of Ionization of Weak Acids

Estimated Time:1m 30s

Question 3960Question →

When potassium ethanoate, $CH_3COOK$ , is dissolved in distilled water, the resulting solution has a pH greater than 7. Which ion is responsible for this alkalinity by undergoing hydrolysis?

The ethanoate ion,

CH_3COO^-

The potassium ion,

K^+

The hydrogen ion,

H^+

The hydroxide ion,

OH^-

Show answer & explanation

Answer: The ethanoate ion, $CH_3COO^-$

Answer

The ethanoate ion,

CH_3COO^-

The ethanoate ion (

CH_3COO^-

) is the conjugate base of the weak acid ethanoic acid (

CH_3COOH

). When potassium ethanoate dissolves in water,

CH_3COO^-

hydrolyzes by accepting a proton from water to form

CH_3COOH

and releasing

OH^-

ions into the solution:

CH_3COO^-(aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)

. The accumulation of free hydroxide ions makes the solution alkaline (

pH > 7

).

Step-by-Step Solution

1

Identify the ions formed when potassium ethanoate dissolves in water.

Potassium ethanoate dissociates completely into potassium ions (

K^+

) and ethanoate ions (

CH_3COO^-

).

Soluble ionic salts completely dissociate into their constituent cations and anions in aqueous solution.

2

Determine which ion reacts with water (hydrolyzes).

Potassium (

K^+

) is derived from a strong base (

KOH

) and does not hydrolyze. Ethanoate (

CH_3COO^-

) is derived from a weak acid (

CH_3COOH

) and reacts with water.

Anions derived from weak acids act as weak conjugate bases and undergo anion hydrolysis.

3

Write the hydrolysis equation to identify the species giving rise to alkalinity.

CH_3COO^-(aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)

Hydrolysis of ethanoate generates free

OH^-

ions, making the solution basic (

pH > 7

).

Key Concept

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

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