All practice questions

Question 3921Question →

Which of the following best describes the fundamental attraction responsible for metallic bonding in solid metals?

Electrostatic attraction between positive metal cations and a sea of delocalized valence electrons

Mutual sharing of localized electron pairs between adjacent atoms

Electrostatic attraction between oppositely charged discrete ions formed by electron transfer

Weak intermolecular forces existing between neutral metal atoms

Show answer & explanation

Answer: Electrostatic attraction between positive metal cations and a sea of delocalized valence electrons

Answer

The metallic bond is defined as the strong electrostatic attraction between positively charged metal cations fixed in a lattice and a surrounding sea of mobile, delocalized valence electrons.

The correct option accurately defines metallic bonding as the electrostatic force of attraction binding positive metal ions to a fluid sea of delocalized valence electrons, which accounts for characteristic metallic properties like electrical conductivity and malleability.

Step-by-Step Solution

1

Identify the valence electron behavior in metals

Metal atoms lose their outer valence electrons to form positive cations, producing a shared pool of delocalized electrons free to move throughout the giant structure.

Low ionization energies in metals allow valence electrons to become detached easily from individual atoms.

2

Determine the nature of the attractive force holding the lattice together

Strong electrostatic attraction acts non-directionally between the positive ions and the mobile electron sea.

Opposite charges attract each other, forming a stable metallic lattice.

Key Concept

Nature of Metallic Bonding

Estimated Time:45s

Question 3922Question →

Three business partners incorporated a private limited company to process agricultural produce. One year after operations commenced, one partner decided to sell her equity stake to an outside investor without obtaining consent from the remaining members. Which regulation or characteristic of a private limited company prevents this transfer from occurring freely?

The statutory restriction on share transfer contained in the Articles of Association

The object clause set out in the Memorandum of Association

The statutory obligation to issue a public prospectus before trading shares

The unlimited liability clause specified in the partnership deed

Show answer & explanation

Answer: The statutory restriction on share transfer contained in the Articles of Association

Answer

The statutory restriction on share transfer contained in the Articles of Association

Private limited companies are legally required to restrict the right to transfer shares in order to preserve their private character. The specific internal rules and restrictions governing share transfers are documented within the Articles of Association.

Step-by-Step Solution

1

Identify the legal characteristics governing share mobility in a private limited company.

Unlike public limited companies whose shares are freely transferable on a stock exchange, a private limited company restricts the transfer of its shares to protect the interest of existing members.

This restriction ensures that outside parties cannot acquire ownership control without the consent of existing shareholders.

2

Determine which company formation document regulates internal share transfer procedures.

The Articles of Association regulates internal management, director powers, and shareholder rights, including share transfer restrictions.

The Memorandum of Association covers external governance and object clauses, whereas the Articles of Association handles internal regulations.

Key Concept

Share Transfer Restrictions and Articles of Association in Private Limited Companies

Estimated Time:1m 15s

Question 3923Question →

Match each vacuum flask component or surface feature on the left with its primary mechanism for controlling heat transfer on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Silvered inner surfaces match with radiation reflection; evacuated space matches with elimination of conduction and convection; cork stopper matches with prevention of convection and conduction at the opening; dull black casing matches with maximizing thermal radiation emission/absorption.

Each feature of the vacuum flask targets a specific heat transfer mode: silvering reflects infrared radiation; the vacuum eliminates particle-dependent transfer (conduction and convection); cork acts as an insulator preventing convection and conduction at the top; and black surfaces maximize thermal radiation emission/absorption.

Step-by-Step Solution

1

Analyze the silvered inner walls of a vacuum flask

Silver surfaces are good reflectors of heat rays (infrared waves).

Radiant heat travels via electromagnetic waves and is reflected by shiny metallic coatings, minimizing radiation heat loss.

2

Analyze the vacuum space between the glass walls

Conduction and convection cannot occur across a vacuum.

Both conduction (particle vibration/electron flow) and convection (fluid movement) strictly require a physical medium.

3

Analyze the cork/plastic stopper

Cork prevents hot air circulation and thermal conduction across the opening.

Cork is a poor conductor of heat and stops evaporative/convective air currents from leaving the container.

4

Analyze the dull black exterior

Dull black surfaces are efficient radiation emitters/absorbers.

According to radiation principles, black matte surfaces radiate energy much faster than polished surfaces.

Key Concept

Modes of Heat Transfer and Practical Applications in Thermal Insulation

Question 3924Question →

A

4.00\text{ g}

sample of an impure copper(II) oxide ore is heated in a stream of dry hydrogen gas until reduction is complete according to the equation:

\text{CuO}_{(s)} + \text{H}_{2(g)} \rightarrow \text{Cu}_{(s)} + \text{H}_2\text{O}_{(g)}

If

2.54\text{ g}

of pure copper metal is obtained, what is the percentage purity of the copper(II) oxide in the ore?

[\text{Relative atomic masses: Cu} = 63.5, \text{O} = 16.0]

79.5\%

63.5\%

36.5\%

20.5\%

Show answer & explanation

Answer: $79.5\%$

Answer

The percentage purity of the copper(II) oxide in the ore is

79.5\%

.

The option stating

79.5\%

is correct because

2.54\text{ g}

of copper corresponds to

0.04\text{ mol}

of

\text{Cu}

. According to the chemical equation,

0.04\text{ mol}

of

\text{Cu}

requires

0.04\text{ mol}

of pure

\text{CuO}

, which weighs

0.04 \times 79.5\text{ g mol}^{-1} = 3.18\text{ g}

. Dividing

3.18\text{ g}

of pure

\text{CuO}

by the total sample mass of

4.00\text{ g}

yields

79.5\%

.

Step-by-Step Solution

1

Calculate the molar mass of copper(II) oxide (CuO) and the moles of copper metal produced.

\text{Molar mass of CuO} = 63.5 + 16.0 = 79.5\text{ g mol}^{-1}

.

\text{Moles of Cu} = \frac{2.54\text{ g}}{63.5\text{ g mol}^{-1}} = 0.04\text{ mol}

.

Converting the given mass of product into moles allows stoichiometric ratio calculations.

2

Determine the moles and mass of pure copper(II) oxide in the sample.

From the equation,

1\text{ mol CuO} \rightarrow 1\text{ mol Cu}

. Moles of pure

\text{CuO} = 0.04\text{ mol}

. Mass of pure

\text{CuO} = 0.04\text{ mol} \times 79.5\text{ g mol}^{-1} = 3.18\text{ g}

.

Stoichiometry dictates that 1 mole of CuO produces 1 mole of Cu upon complete reduction.

3

Calculate the percentage purity of the copper(II) oxide sample.

\text{Percentage purity} = \frac{\text{Mass of pure CuO}}{\text{Total mass of sample}} \times 100\% = \frac{3.18\text{ g}}{4.00\text{ g}} \times 100\% = 79.5\%

.

Percentage purity is the ratio of pure reactive compound mass to total impure sample mass expressed as a percentage.

Key Concept

Determining percentage purity using stoichiometric reduction yields

Question 3925Question →

What is the pH of an aqueous solution prepared by dissolving $0.49\text{ g}$ of tetraoxosulfate(VI) acid ( $\text{H}_2\text{SO}_4$ ) in distilled water to make $500\text{ cm}^3$ of solution? [Molar mass of $\text{H}_2\text{SO}_4 = 98\text{ g mol}^{-1}$ , $\log_{10} 2 = 0.301$ ]

1.70

2.00

2.30

12.30

Show answer & explanation

Answer: $1.70$

Answer

The pH of the solution is

1.70

.

Dissolving

0.49\text{ g}

of

\text{H}_2\text{SO}_4

(molar mass

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

) in

0.5\text{ dm}^3

yields a

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

solution. Because tetraoxosulfate(VI) acid fully ionizes into

2\text{H}^+

and

\text{SO}_4^{2-}

, the hydrogen ion concentration

[\text{H}^+]

is

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

. Taking the negative logarithm base 10 gives

\text{pH} = -\log_{10}(0.02) = 1.70

.

Step-by-Step Solution

1

Calculate the amount of

\text{H}_2\text{SO}_4

in moles

\text{Moles of } \text{H}_2\text{SO}_4 = \frac{0.49\text{ g}}{98\text{ g mol}^{-1}} = 0.005\text{ mol}

Converting mass of solute to moles using molar mass.

2

Determine the molar concentration of the acid solution

\text{Volume} = 500\text{ cm}^3 = 0.5\text{ dm}^3

;

\text{Molarity} = \frac{0.005\text{ mol}}{0.5\text{ dm}^3} = 0.01\text{ mol dm}^{-3}

Molarity is defined as moles of solute per cubic decimetre of solution.

3

Calculate the hydrogen ion concentration

[\text{H}^+]

[\text{H}^+] = 2 \times 0.01\text{ mol dm}^{-3} = 0.02\text{ mol dm}^{-3} = 2.0 \times 10^{-2}\text{ mol dm}^{-3}

\text{H}_2\text{SO}_4

is a strong dibasic acid that ionizes completely to produce two

\text{H}^+

ions per molecule.

4

Calculate the pH of the solution

\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(2.0 \times 10^{-2}) = 2 - \log_{10} 2 = 2 - 0.301 = 1.699 \approx 1.70

Applying the logarithmic definition of pH.

Key Concept

pH Calculation of Dibasic Strong Acids

Question 3926Question →

Using the first principles of differentiation for the reciprocal function $f(x) = \frac{4}{x}$ , evaluate the limit of the difference quotient as $h \to 0$ : $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ . What is the simplified expression for the derivative $\frac{dy}{dx}$ ?

Show answer & explanation

Answer: -\frac{4}{x^2}; -4/x^2; -4 / x^2; -\frac{4}{x^{2}}

Answer

The derivative

\frac{dy}{dx}

is

-\frac{4}{x^2}

.

Substituting

f(x) = \frac{4}{x}

into the first principles limit formula yields

\lim_{h \to 0} \frac{4x - 4(x+h)}{h \cdot x(x+h)} = \lim_{h \to 0} \frac{-4h}{h \cdot x(x+h)}

. Canceling

h

gives

\lim_{h \to 0} -\frac{4}{x(x+h)} = -\frac{4}{x^2}

.

Step-by-Step Solution

1

Write the first principles formula and substitute

f(x) = \frac{4}{x}

.

\frac{dy}{dx} = \lim_{h \to 0} \frac{\frac{4}{x+h} - \frac{4}{x}}{h}

Apply the definition of differentiation from first principles.

2

Combine the fractions in the numerator using a common denominator

x(x+h)

.

\frac{4}{x+h} - \frac{4}{x} = \frac{4x - 4(x+h)}{x(x+h)} = \frac{4x - 4x - 4h}{x(x+h)} = \frac{-4h}{x(x+h)}

Simplify the numerator into a single fractional expression.

3

Divide the simplified numerator by

h

and cancel the common factor of

h

.

\frac{\frac{-4h}{x(x+h)}}{h} = \frac{-4h}{h \cdot x(x+h)} = -\frac{4}{x(x+h)}

Eliminate the indeterminate factor of

h

from the denominator.

4

Evaluate the limit as

h \to 0

.

\lim_{h \to 0} -\frac{4}{x(x+h)} = -\frac{4}{x(x+0)} = -\frac{4}{x^2}

Substitute

h = 0

into the simplified algebraic expression.

Key Concept

Differentiation from First Principles

Question 3927Question →

In the Kano Closed-Settlement Zone of Northern Nigeria, agricultural land is cropped continuously every year with intensive manure application, whereas in surrounding lower-density rural districts, farmers rely on bush fallowing with multi-year land rotation. Which of the following statements best explains this spatial variation in agricultural land-use intensity?

High population pressure creates severe land scarcity, compelling farmers to adopt labor-intensive soil conservation and continuous cultivation to maximize yield per unit area.

The establishment of commercial tree crop plantations such as cocoa and oil palm in dry northern zones demands permanent land clearance and continuous tillage.

Agricultural processing centers are situated at urban market locations to minimize the transport costs of weight-losing raw farm produce from distant fallow fields.

High rural-to-urban migration from dense settlement zones serves as a pull factor that increases rural land holdings and enables longer fallow periods.

Show answer & explanation

Answer: High population pressure creates severe land scarcity, compelling farmers to adopt labor-intensive soil conservation and continuous cultivation to maximize yield per unit area.

Answer

High population pressure creates severe land scarcity, compelling farmers to adopt labor-intensive soil conservation and continuous cultivation to maximize yield per unit area.

In agricultural geography, high rural population density leads to land fragmentation and scarcity. Farmers can no longer afford to leave land idle under traditional bush fallowing, forcing an intensification of land use through continuous cropping, heavy application of animal manure, and labor-intensive soil management as seen in the Kano Closed-Settlement Zone.

Step-by-Step Solution

1

Analyze the spatial contrast in land use intensity between the Kano Closed-Settlement Zone and surrounding rural areas.

Identified that the Kano Closed-Settlement Zone is characterized by extreme population density and fragmented land holdings compared to outlying districts.

Agricultural land-use intensity is strongly governed by man-land ratios and population pressure on land resources (Boserup's agricultural intensification theory).

2

Evaluate the relationship between land scarcity and fallow duration.

As population density increases, land scarcity prevents long fallow periods, forcing a transition from bush fallowing to permanent/continuous cultivation supported by organic fertilization (manuring).

Farmers must substitute land expansion with intensive labor and land management practices to sustain yields per unit of land.

3

Distinguish correct demographic and land-use principles from distractors involving crop belts, location economics, and migration dynamics.

Confirmed that land scarcity driven by high rural population density is the primary driver of continuous intensive cropping in semi-arid West African agricultural systems.

Other options misattribute cash crop climatic belts, industrial location principles, or migration push/pull dynamics.

Key Concept

Agricultural Land-Use Intensification and Population Density

Estimated Time:2m 0s

Question 3928Question →

An experimental effusion cell measures gas diffusion rates through a micro-porous membrane at fixed temperature and pressure. In a calibration run, $120\text{ cm}^3$ of neon gas ( $\text{Ne}$ , atomic mass $= 20\text{ g/mol}$ ) effuses through the membrane in $30\text{ seconds}$ . Calculate the volume (in $\text{cm}^3$ ) of sulfur trioxide gas ( $\text{SO}_3$ , atomic masses: $\text{S} = 32\text{ g/mol}$ , $\text{O} = 16\text{ g/mol}$ ) that will effuse through the exact same membrane in $50\text{ seconds}$ .

Show answer & explanation

Answer: 100; 100 cm3; 100cm3; 100 cm^3; 100 cm³

Answer

100 cm³

The effusion rate of neon is

r_{\text{Ne}} = \frac{120\text{ cm}^3}{30\text{ s}} = 4\text{ cm}^3/\text{s}

. Given

M_{\text{Ne}} = 20\text{ g/mol}

and

M_{\text{SO}_3} = 80\text{ g/mol}

, Graham's law dictates

\frac{r_{\text{Ne}}}{r_{\text{SO}_3}} = \sqrt{\frac{80}{20}} = 2

. Thus,

r_{\text{SO}_3} = \frac{4}{2} = 2\text{ cm}^3/\text{s}

. Over a period of

50\text{ seconds}

, the volume of

\text{SO}_3

effused is

2\text{ cm}^3/\text{s} \times 50\text{ s} = 100\text{ cm}^3

.

Step-by-Step Solution

1

Calculate the rate of effusion of neon gas (

r_{\text{Ne}}

)

r_{\text{Ne}} = \frac{120\text{ cm}^3}{30\text{ s}} = 4\text{ cm}^3/\text{s}

Effusion rate is defined as the volume of gas effusing per unit time.

2

Determine the molar mass of sulfur trioxide (

\text{SO}_3

)

M_{\text{SO}_3} = 32 + 3(16) = 80\text{ g/mol}

The molar mass is calculated from the constituent relative atomic masses of sulfur and oxygen.

3

Apply Graham's Law of Effusion to determine the effusion rate of sulfur trioxide (

r_{\text{SO}_3}

)

\frac{r_{\text{Ne}}}{r_{\text{SO}_3}} = \sqrt{\frac{M_{\text{SO}_3}}{M_{\text{Ne}}}} \implies \frac{4}{r_{\text{SO}_3}} = \sqrt{\frac{80}{20}} = \sqrt{4} = 2 \implies r_{\text{SO}_3} = 2\text{ cm}^3/\text{s}

Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

4

Compute the total volume of sulfur trioxide effused in

50\text{ seconds}

V_{\text{SO}_3} = r_{\text{SO}_3} \times t = 2\text{ cm}^3/\text{s} \times 50\text{ s} = 100\text{ cm}^3

Multiplying the calculated rate of effusion by the given time yields the total volume.

Key Concept

Graham's Law of Diffusion and Effusion

Question 3929Question →

A naturally occurring element $X$ consists of two stable isotopes, $^{63}X$ and $^{65}X$ . If the relative atomic mass of element $X$ is $63.6$ , what is the percentage abundance of the heavier isotope $^{65}X$ ?

30\%

70\%

60\%

50\%

Show answer & explanation

Answer: $30\%$

Answer

The percentage abundance of the heavier isotope

^{65}X

is

30\%

.

The relative atomic mass of an element with isotopes is calculated using the weighted average formula:

\text{RAM} = \sum (\text{mass} \times \text{fractional abundance})

. Setting

p

as the fraction of

^{65}X

gives

63.6 = 63(1 - p) + 65p = 63 + 2p

. Solving yields

2p = 0.6

, so

p = 0.30

, which corresponds to

30\%

.

Step-by-Step Solution

1

Set up the relative atomic mass weighted average equation

63.6 = \frac{(m_1 \times x_1) + (m_2 \times x_2)}{100}

Relative atomic mass is the weighted average of the atomic masses of naturally occurring isotopes based on their fractional abundances.

2

Express isotopic abundances in terms of a single variable

p

Let

p

be the fraction of

^{65}X

, so the fraction of

^{63}X

is

1 - p

.

The sum of the fractional abundances of all isotopes of an element must equal 1 (or

100\%

).

3

Substitute known values and solve for

p

63.6 = 63(1 - p) + 65p \implies 63.6 = 63 + 2p \implies 2p = 0.6 \implies p = 0.30

Algebraic expansion isolates the variable corresponding to the heavier isotope abundance.

4

Convert the fractional abundance to percentage

0.30 \times 100\% = 30\%

Multiplying the decimal fraction by 100 yields the percentage abundance.

Key Concept

Calculation of Relative Atomic Mass from Isotopic Abundance

Estimated Time:1m 30s

Question 3930Question →

An expanding commercial enterprise structured as a public limited company intends to raise long-term capital by inviting members of the general public to subscribe to its newly issued shares. Which official document is the company legally required to prepare, register, and issue to the public for this purpose?

Articles of Association

Prospectus

Memorandum of Association

Certificate of Incorporation

Show answer & explanation

Answer: Prospectus

Answer

The Prospectus is the statutory document issued by a public limited company to invite the public to subscribe to its shares or debentures.

A Prospectus is specifically designed and legally mandated to invite members of the general public to purchase or subscribe for shares or debentures in a public limited company.

Step-by-Step Solution

1

Identify the primary operational goal described in the question stem.

The company needs to invite the general public to purchase its newly issued shares.

Only public limited companies are permitted to invite public subscriptions for capital securities.

2

Evaluate the statutory function of company documents.

The Prospectus contains financial performance, director information, and share offering details designed to inform prospective public investors.

Company law mandates the publication and registration of a Prospectus before public offer of securities.

Key Concept

Public Share Offerings and the Role of a Prospectus

Question 3931Question →

A $6.95\text{ g}$ sample of hydrated iron(II) tetraoxosulfate(VI), $\text{FeSO}_4 \cdot x\text{H}_2\text{O}$ , is dissolved in dilute tetraoxosulfate(VI) acid and made up to $250\text{ cm}^3$ in a volumetric flask. A $25.0\text{ cm}^3$ portion of this solution requires $25.0\text{ cm}^3$ of $0.020\text{ mol dm}^{-3}$ acidified potassium tetraoxomanganate(VII), $\text{KMnO}_4$ , for complete titration. Given the relative atomic masses $\text{Fe} = 56$ , $\text{S} = 32$ , $\text{O} = 16$ , and $\text{H} = 1$ , what is the integer value of $x$ , the number of molecules of water of crystallization per formula unit?

Show answer & explanation

Answer: 7

Answer

The integer value of x is 7.

By using the titration volume and concentration of acidified potassium tetraoxomanganate(VII), the amount of iron(II) ions in the sample is calculated. Knowing the total mole amount of anhydrous iron(II) tetraoxosulfate(VI) allows determination of the mass of the anhydrous salt component (3.80 g). Subtracting this from the initial hydrated sample mass (6.95 g) gives the mass of water of crystallization (3.15 g). Dividing the moles of water (0.175 mol) by the moles of anhydrous salt (0.025 mol) yields exactly 7 water molecules of crystallization per formula unit.

Step-by-Step Solution

1

Calculate the moles of

\text{KMnO}_4

consumed in the titration

n(\text{KMnO}_4) = 0.020 \text{ mol dm}^{-3} \times 0.0250 \text{ dm}^3 = 0.00050 \text{ mol}

Concentration and volume of titrant are provided.

2

Determine moles of

\text{Fe}^{2+}

present in the

25.0\text{ cm}^3

aliquot using the redox reaction stoichiometry

n(\text{Fe}^{2+})_{25\text{cm}^3} = 5 \times 0.00050 \text{ mol} = 0.0025 \text{ mol}

The mole ratio of

\text{MnO}_4^-

to

\text{Fe}^{2+}

in acidic redox titration is

1:5

according to

\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}

.

3

Calculate the total moles of

\text{FeSO}_4

in the original

250\text{ cm}^3

volumetric flask

n(\text{FeSO}_4)_{\text{total}} = 0.0025 \text{ mol} \times \left(\frac{250\text{ cm}^3}{25.0\text{ cm}^3}\right) = 0.025 \text{ mol}

The aliquot represents one-tenth of the total solution volume.

4

Calculate the mass of anhydrous

\text{FeSO}_4

in the sample

Molar mass of

\text{FeSO}_4 = 56 + 32 + (4 \times 16) = 152 \text{ g mol}^{-1}

. Mass

= 0.025 \text{ mol} \times 152 \text{ g mol}^{-1} = 3.80 \text{ g}

.

Converting moles of anhydrous salt to mass using molar mass.

5

Determine the mass and moles of water of crystallization

Mass of

\text{H}_2\text{O} = 6.95 \text{ g} - 3.80 \text{ g} = 3.15 \text{ g}

. Moles of

\text{H}_2\text{O} = \frac{3.15 \text{ g}}{18 \text{ g mol}^{-1}} = 0.175 \text{ mol}

.

Subtracting anhydrous mass from initial mass gives water of crystallization mass, converted to moles using molar mass of

\text{H}_2\text{O} = 18 \text{ g mol}^{-1}

.

6

Compute the hydration coefficient

x = \frac{n(\text{H}_2\text{O})}{n(\text{FeSO}_4)}

x = \frac{0.175 \text{ mol}}{0.025 \text{ mol}} = 7

The coefficient

x

represents the mole ratio of water of crystallization to anhydrous salt.

Key Concept

Quantitative determination of water of crystallization in hydrated salts via redox volumetric analysis

Question 3932Question →

A $0.04\text{ mol dm}^{-3}$ aqueous solution of a weak monobasic acid, $\text{HA}$ , is $2.0\%$ ionized at $25^\circ\text{C}$ . If distilled water is added to this solution until its total volume is quadrupled, what is the degree of ionization of the acid in the diluted solution?

0.5\%

1.0\%

4.0\%

8.0\%

Show answer & explanation

Answer: $4.0\%$

Answer

The degree of ionization of the acid in the diluted solution is

4.0\%

.

By Ostwald's dilution law for weak monobasic acids,

K_a = \alpha^2 C

, which rearranges to

\alpha = \sqrt{\frac{K_a}{C}}

. Since

K_a

is constant at a fixed temperature, the degree of ionization

\alpha

is inversely proportional to the square root of the concentration (

\alpha \propto \frac{1}{\sqrt{C}}

). Quadrupling the volume decreases the concentration by a factor of 4 (

C_2 = \frac{C_1}{4}

), which increases the degree of ionization by a factor of

\sqrt{4} = 2

. Therefore, the new degree of ionization is

2.0\% \times 2 = 4.0\%

.

Step-by-Step Solution

1

Calculate the acid dissociation constant (

K_a

) using the initial concentration and degree of ionization.

Initial concentration

C_1 = 0.04\text{ mol dm}^{-3}

, initial degree of ionization

\alpha_1 = 2.0\% = 0.02

. Using

K_a \approx \alpha_1^2 C_1

, we get

K_a = (0.02)^2 \times 0.04 = 4.0 \times 10^{-4} \times 0.04 = 1.6 \times 10^{-5}\text{ mol dm}^{-3}

.

The value of

K_a

depends only on temperature and remains constant upon dilution.

2

Determine the new concentration (

C_2

) after quadrupling the solution volume.

C_2 = \frac{C_1}{4} = \frac{0.04\text{ mol dm}^{-3}}{4} = 0.01\text{ mol dm}^{-3}

.

Diluting a solution to 4 times its original volume reduces its molar concentration by a factor of 4.

3

Calculate the new degree of ionization (

\alpha_2

) in the diluted solution.

\alpha_2 = \sqrt{\frac{K_a}{C_2}} = \sqrt{\frac{1.6 \times 10^{-5}}{0.01}} = \sqrt{1.6 \times 10^{-3}} = \sqrt{16 \times 10^{-4}} = 4.0 \times 10^{-2} = 0.04 = 4.0\%$.

Applying Ostwald's dilution law to find the updated degree of dissociation at the lower concentration.

Key Concept

Ostwald's Dilution Law and the relationship between dilution, concentration, and degree of ionization of weak electrolytes.

Estimated Time:2m 30s

Question 3933Question →

Match each atmospheric pollutant or component in Column I with its corresponding chemical mode of action or environmental transformation mechanism in Column II.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Carbon monoxide matches with irreversible hemoglobin binding; Sulfur dioxide matches with oxidation to an acid anhydride causing limestone corrosion; Chlorofluorocarbons match with UV-induced photolysis producing chlorine radicals; Nitrogen dioxide matches with solar photolysis initiating photochemical smog.

Each atmospheric pollutant is accurately matched to its chemical mechanism: carbon monoxide forms carboxyhemoglobin in blood, sulfur dioxide forms acid anhydrides causing marble/limestone corrosion, chlorofluorocarbons generate ozone-depleting chlorine radicals under UV radiation, and nitrogen dioxide photolyzes in sunlight to drive photochemical smog formation.

Step-by-Step Solution

1

Analyze Carbon monoxide (

\text{CO}

)

It readily binds to blood hemoglobin to form carboxyhemoglobin.

\text{CO}

toxicological impact relies on inhibiting cellular respiration by reducing oxygen transport.

2

Analyze Sulfur dioxide (

\text{SO}_2

)

It oxidizes to

\text{SO}_3

(sulfur trioxide, an acid anhydride), forming sulfuric acid rain.

Acid rain dissolves building materials containing calcium carbonate (

\text{CaCO}_3

).

3

Analyze Chlorofluorocarbons (

\text{CFCs}

)

High-energy solar ultraviolet radiation breaks

\text{C-Cl}

bonds in

\text{CFCs}

, producing free chlorine radicals.

Chlorine radicals act as catalysts in the breakdown of stratospheric ozone molecules into oxygen gas.

4

Analyze Nitrogen dioxide (

\text{NO}_2

)

In the troposphere,

\text{NO}_2

absorbs near-UV sunlight, splitting into

\text{NO}

and atomic oxygen (

\text{O}

).

Atomic oxygen combines with

\text{O}_2

to form tropospheric ozone, a key ingredient in photochemical smog.

Key Concept

Chemical transformations and atmospheric impacts of primary air pollutants

Question 3934Question →

A $200\text{ cm}^3$ sample of a saturated potassium trioxonitrate(V) solution, $\text{KNO}_3$ , at $60^\circ\text{C}$ has a concentration of $1.5\text{ mol/dm}^3$ . The solution is cooled to $25^\circ\text{C}$ to form a supersaturated solution. When a seed crystal is added, excess solute crystallizes until a new saturated concentration of $0.5\text{ mol/dm}^3$ is reached at $25^\circ\text{C}$ . What mass of $\text{KNO}_3$ crystallizes out of the solution? [Molar mass of $\text{KNO}_3 = 101\text{ g/mol}$ ]

0.20\text{ g}

20.2\text{ g}

30.3\text{ g}

10.1\text{ g}

Show answer & explanation

Answer: $20.2\text{ g}$

Answer

20.2\text{ g}

of

\text{KNO}_3

crystallizes out of the solution.

Cooling a saturated solution creates a unstable supersaturated state. Adding a seed crystal induces rapid crystallization of the excess solute until saturation equilibrium is re-established at the lower temperature. The difference in concentration is

1.0\text{ mol/dm}^3

. For

0.200\text{ dm}^3

, this equals

0.20\text{ mol}

of

\text{KNO}_3

, which has a mass of

0.20\text{ mol} \times 101\text{ g/mol} = 20.2\text{ g}

.

Step-by-Step Solution

1

Convert the volume of the solution from

\text{cm}^3

to

\text{dm}^3

.

\text{Volume} = \frac{200\text{ cm}^3}{1000} = 0.200\text{ dm}^3

.

Concentrations are given in

\text{mol/dm}^3

, so volume must be in

\text{dm}^3

.

2

Calculate the moles of

\text{KNO}_3

dissolved initially at

60^\circ\text{C}

and remaining at

25^\circ\text{C}

.

\text{Initial moles} = 1.5\text{ mol/dm}^3 \times 0.200\text{ dm}^3 = 0.30\text{ mol}

.

\text{Final moles} = 0.5\text{ mol/dm}^3 \times 0.200\text{ dm}^3 = 0.10\text{ mol}

.

Determining the mole difference shows how much solute leaves the supersaturated state upon seeding.

3

Determine the amount of moles precipitated and convert to mass.

\text{Moles precipitated} = 0.30\text{ mol} - 0.10\text{ mol} = 0.20\text{ mol}

.

\text{Mass precipitated} = 0.20\text{ mol} \times 101\text{ g/mol} = 20.2\text{ g}

.

Multiplying the precipitated moles by the molar mass gives the required mass in grams.

Key Concept

Mass of solute crystallized from a supersaturated solution upon reaching saturation equilibrium

Question 3935Question →

In ammonium trioxonitrate(V), $NH_4NO_3$ , nitrogen exists in two distinct ionic environments. What are the respective oxidation numbers of the nitrogen atom in the cation and the nitrogen atom in the anion, and what is the systematic IUPAC name of the nitrogen-containing oxoanion formed when the anion is reduced by gaining two electrons?

-3

and

+5

; dioxonitrate(III) ion

+1

and

+5

; trioxonitrate(III) ion

0

and

+5

; dioxonitrate(III) ion

-3

and

+3

; trioxonitrate(V) ion

Show answer & explanation

Answer: $-3$ and $+5$ ; dioxonitrate(III) ion

Answer

The respective oxidation numbers are

-3

for nitrogen in the ammonium cation and

+5

for nitrogen in the trioxonitrate(V) anion. The reduced species formed upon gaining two electrons is the dioxonitrate(III) ion.

In the ionic compound

NH_4NO_3

, the nitrogen atom in the cation

NH_4^+

has an oxidation state of

-3

because

x + 4(+1) = +1

. The nitrogen atom in the anion

NO_3^-

has an oxidation state of

+5

because

y + 3(-2) = -1

. When

NO_3^-

is reduced by two electrons, the oxidation number of nitrogen decreases from

+5

to

+3

, converting

NO_3^-

to

NO_2^-

. In IUPAC nomenclature,

NO_2^-

is named the dioxonitrate(III) ion due to its two oxo ligands and nitrogen oxidation state of

+3

.

Step-by-Step Solution

1

Determine the oxidation state of nitrogen in the ammonium cation (

NH_4^+

).

Let the oxidation state of N be

x

. Since hydrogen is

+1

and the net charge is

+1

:

x + 4(+1) = +1 \implies x = -3

.

Ammonium is a polyatomic cation where the sum of oxidation numbers equals the net ionic charge.

2

Determine the oxidation state of nitrogen in the trioxonitrate(V) anion (

NO_3^-

).

Let the oxidation state of N be

y

. Since oxygen is

-2

and the net charge is

-1

:

y + 3(-2) = -1 \implies y = +5

.

Nitrate is a polyatomic anion where oxygen exhibits an oxidation number of

-2

.

3

Calculate the oxidation state of nitrogen after two-electron reduction of the anion and determine its IUPAC name.

Reduction by gaining 2 electrons decreases the oxidation number of nitrogen:

+5 - 2 = +3

. The resulting species containing nitrogen in the

+3

state with two oxygen atoms (

NO_2^-

) is systematically named the dioxonitrate(III) ion.

Reduction corresponds to a gain of electrons (decrease in oxidation number), and IUPAC rules for oxoanions specify the prefix for oxygen count ('dioxo-') followed by the central element and its oxidation state in Roman numerals.

Key Concept

Polyatomic Ion Oxidation Numbers and IUPAC Nomenclature of Oxoanions

Question 3936Question →

Calculate the mass number and atomic number of the final daughter nuclide in the given radioactive decay series.

Fill in the blanks below

When a nucleus of Thorium-232 (

^{232}_{90}\text{Th}

) undergoes a decay series emitting

6

alpha particles (

\alpha

) and

4

beta particles (

\beta^-

), the resulting stable daughter nuclide has a mass number of and an atomic number of .

Show answer & explanation

Answer

The final stable daughter nuclide has a mass number of 208 and an atomic number of 82.

Emission of 6 alpha particles reduces the mass number by

6 \times 4 = 24

(from 232 to 208) and the atomic number by

6 \times 2 = 12

. The subsequent emission of 4 beta particles increases the atomic number by

4 \times 1 = 4

, resulting in a final atomic number of

90 - 12 + 4 = 82

.

Step-by-Step Solution

1

Calculate the total change in mass number (

A

) caused by alpha and beta particle emissions.

Each alpha particle (

^{4}_{2}\text{He}

) reduces

A

by 4, while beta particles (

^{0}_{-1}\text{e}

) do not change

A

. Total decrease in

A = 6 \times 4 = 24

.

Alpha particles carry 4 atomic mass units, whereas beta particles have virtually zero mass number.

2

Determine the final mass number.

Final mass number =

232 - 24 = 208

.

Subtracting the lost nucleon mass from the initial mass number gives the mass number of the resulting nucleus.

3

Calculate the total change in atomic number (

Z

) caused by alpha and beta particle emissions.

Each alpha particle reduces

Z

by 2, while each beta particle increases

Z

by 1. Total change in

Z = (6 \times -2) + (4 \times +1) = -12 + 4 = -8

.

Alpha decay removes 2 protons, whereas beta minus decay converts a neutron into a proton, increasing nuclear charge by 1.

4

Determine the final atomic number.

Final atomic number =

90 - 8 = 82

.

Applying the net charge change to the initial atomic number of Thorium (90) yields 82, which corresponds to Lead (Pb).

Key Concept

Conservation of mass number and atomic number in decay series

Question 3937Question →

Complete the sentence below by filling in the blank with the appropriate idiomatic expression that best fits the context.

Fill in the blanks below

Despite explicit warnings from the regulatory agency, the investment directors continued to with clients' capital, ultimately precipitating the collapse of the fund.

Show answer & explanation

Answer

play fast and loose

The expression 'play fast and loose' is an established English idiom meaning to act irresponsibly, dishonestly, or recklessly with someone or something, which accurately matches the scenario of reckless handling of investment capital.

Step-by-Step Solution

1

Analyze the context and tone of the sentence

The context involves investment directors engaging in reckless, deceitful, and irresponsible financial actions with clients' funds despite regulatory warnings.

Identifying the context helps determine the precise meaning required for the missing idiomatic expression.

2

Identify the idiomatic phrase that matches this figurative meaning

The idiom 'play fast and loose' means to behave dishonestly, deceitfully, or irresponsibly, especially by taking improper liberties with rules or trust.

This exact phrase fits both the syntax of the infinitive 'to [idiom]' and the semantic requirement of reckless financial handling.

Key Concept

Idiomatic Expressions - Expressing Deceit and Recklessness

Question 3938Question →

During the public inquiry, the panel meticulously cross-examined the key witness regarding the submitted documentation. Which word nearest in meaning to 'truthfulness' appropriately completes the sentence below?

Fill in the blanks below

The members of the commission expressed serious reservations concerning the of the statements contained in the report.

Show answer & explanation

Answer

veracity (or truthfulness)

The word 'veracity' means conformity to facts, accuracy, or truthfulness. In the context of investigating official statements, questioning their veracity means doubting whether they are true.

Step-by-Step Solution

1

Analyze the context of the sentence to identify the required meaning of the blank.

The sentence discusses doubts surrounding whether the statements in a report are factual and honest.

The phrase 'reservations concerning the...' requires a noun denoting factual accuracy or truthfulness.

2

Identify a precise formal synonym for 'truthfulness' suitable for an administrative or academic register.

The noun 'veracity' signifies adherence to truth, accuracy, or correctness.

'Veracity' fits the formal register of the sentence and directly substitutes for 'truthfulness'.

Key Concept

Synonyms and Contextual Word Meaning

Estimated Time:1m 0s

Question 3939Question →

A mountain climber of mass $60\text{ kg}$ climbs a vertical height of $15\text{ m}$ in a time of $30\text{ s}$ . Taking the acceleration due to gravity $g = 10\text{ m s}^{-2}$ , calculate the average power expended by the climber in Watts.

Show answer & explanation

Answer: 300

Answer

The average power expended by the climber is

300\text{ W}

.

The work done in lifting a mass

m

through vertical height

h

is given by

W = mgh = 60 \times 10 \times 15 = 9000\text{ J}

. The average power is the rate of doing work,

P = \frac{W}{t} = \frac{9000\text{ J}}{30\text{ s}} = 300\text{ W}

.

Step-by-Step Solution

1

Identify the given values and formula for work done against gravity.

Mass

m = 60\text{ kg}

, vertical displacement

h = 15\text{ m}

, acceleration due to gravity

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

, and time

t = 30\text{ s}

. Work done formula:

W = mgh

.

The climber works against gravity to increase their potential energy by an amount equal to

mgh

.

2

Calculate the total work done.

W = 60\text{ kg} \times 10\text{ m s}^{-2} \times 15\text{ m} = 9000\text{ J}

.

Multiplying force (

mg

) by vertical distance (

h

) yields work done in Joules.

3

Calculate the average power.

P = \frac{W}{t} = \frac{9000\text{ J}}{30\text{ s}} = 300\text{ W}

.

Power is defined as work done divided by the time interval (

P = \frac{W}{t}

).

Key Concept

Power as the rate of doing work against gravity

Estimated Time:45s

Question 3940Question →

Match each commercial contract scenario described in Column I with its corresponding legal concept, mode of discharge, or legal remedy in Column II.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The scenarios correctly match as follows: the statutory export prohibition constitutes discharge by frustration due to supervening illegality; the honest but false representation of vehicle condition is innocent misrepresentation rendering the contract voidable; the partial project claim following wrongful repudiation is a remedy of quantum meruit; and the refusal to transfer unique real estate where damages are inadequate warrants specific performance.

Each legal situation aligns with established common law contract principles: supervening statutory prohibition causes discharge by frustration; false statements made without fraud constitute innocent misrepresentation; recovery for partial work executed before wrongful repudiation is sought via quantum meruit; and specific performance is decreed for land contracts due to the unique nature of real property.

Step-by-Step Solution

1

Analyze Scenario 1 (Statutory export prohibition)

Performance becomes legally impossible after contract formation due to external legal changes without fault of either party.

This satisfies the legal criteria for discharge by frustration (supervening illegality).

2

Analyze Scenario 2 (Believed true statement inducing contract)

An untrue statement of material fact was made without fraudulent intent or deceit.

This is an innocent misrepresentation, acting as a vitiating element that makes the contract voidable.

3

Analyze Scenario 3 (Partial performance interrupted by repudiation)

Work was performed before the contract was wrongfully terminated by the counterparty.

The aggrieved party can claim quantum meruit ('as much as he has earned') rather than suing purely for unliquidated breach damages.

4

Analyze Scenario 4 (Refusal to transfer unique land parcel)

Breach occurs over unique property where monetary damages cannot restore the injured buyer.

Courts grant the equitable remedy of specific performance to compel performance of the express contract.

Key Concept

Law of Contract: Discharge, Vitiating Elements, and Remedies for Breach

Answer

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