All practice questions

Question 8161Question →

A uniform horizontal beam $AB$ of length $4.0\text{ m}$ and mass $10\text{ kg}$ is hinged smoothly to a vertical wall at end $A$ . It is held horizontally in static equilibrium by a light cable attached to end $B$ and anchored to the wall above $A$ , making an angle of $30^\circ$ with the beam. A mass of $5\text{ kg}$ is suspended from the beam at a distance of $3.0\text{ m}$ from hinge $A$ . Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the tension in the cable?

175\text{ N}

87.5\text{ N}

275\text{ N}

17.5\text{ N}

Show answer & explanation

Answer: $175\text{ N}$

Answer

The tension in the cable is

175\text{ N}

.

Applying the principle of moments about the hinge at end A, the clockwise moments due to the beam's center of mass (

100\text{ N}

at

2.0\text{ m}

) and the suspended load (

50\text{ N}

at

3.0\text{ m}

) are balanced by the counterclockwise moment of the cable tension (

T \sin 30^\circ

at

4.0\text{ m}

). Solving

(100 \times 2.0) + (50 \times 3.0) = 2.0 T

yields

T = 175\text{ N}

.

Step-by-Step Solution

1

Calculate the downward gravitational forces (weights) acting on the system.

Weight of beam

W_{\text{beam}} = m_{\text{beam}} g = 10\text{ kg} \times 10\text{ m/s}^2 = 100\text{ N}

acting at

2.0\text{ m}

from

A

. Weight of load

W_{\text{load}} = m_{\text{load}} g = 5\text{ kg} \times 10\text{ m/s}^2 = 50\text{ N}

acting at

3.0\text{ m}

from

A

.

Forces causing clockwise moments must be expressed in force units (newtons) and located at their respective lines of action.

2

Formulate the equilibrium condition using the Principle of Moments about hinge

A

.

\sum \tau_A = 0 \implies (W_{\text{beam}} \times 2.0\text{ m}) + (W_{\text{load}} \times 3.0\text{ m}) = T \sin(30^\circ) \times 4.0\text{ m}

The hinge

A

eliminates reaction forces at the hinge from the moment equation.

3

Substitute numerical values and solve for tension

T

.

(100 \times 2.0) + (50 \times 3.0) = T \times 0.5 \times 4.0 \implies 200 + 150 = 2.0 T \implies 350 = 2.0 T \implies T = 175\text{ N}$.

Perpendicular distance from

A

to line of action of tension is

4.0 \sin 30^\circ = 2.0\text{ m}

.

Key Concept

Equilibrium of rigid bodies and Principle of Moments under non-perpendicular forces

Estimated Time:2m 0s

Question 8162Question →

A satellite of mass $500\text{ kg}$ orbits a spherical planet of radius $R = 6.0 \times 10^6\text{ m}$ with surface gravitational acceleration $g = 10\text{ m/s}^2$ . The satellite is transferred from an initial circular orbit of radius $2R$ to a higher circular orbit of radius $3R$ . What is the minimum energy required, in megajoules ( $\text{MJ}$ ), to perform this transfer?

Show answer & explanation

Answer: 2500

Answer

The minimum energy required to perform the orbital transfer is

2500\text{ MJ}

.

The minimum energy needed to move a satellite between circular orbits is equal to the change in its total mechanical energy (

E = -\frac{GMm}{2r}

). Expressing

GM

as

gR^2

, the energy difference between radii

2R

and

3R

simplifies to

\Delta E = \frac{gRm}{12}

, which evaluates to

2500\text{ MJ}

.

Step-by-Step Solution

1

Relate surface acceleration due to gravity to planet mass and radius.

GM = gR^2

At the planet's surface (

r = R

), gravitational acceleration is

g = \frac{GM}{R^2}

.

2

Formulate the total mechanical energy equation for a circular orbit.

E = -\frac{GMm}{2r} = -\frac{gR^2 m}{2r}

Total energy is kinetic energy

\frac{GMm}{2r}

plus gravitational potential energy

-\frac{GMm}{r}

.

3

Calculate initial and final total energies.

E_1 = -\frac{gRm}{4}

and

E_2 = -\frac{gRm}{6}

Substitute the orbit radii

r_1 = 2R

and

r_2 = 3R

into the total energy equation.

4

Determine the net work required for the transfer.

\Delta E = E_2 - E_1 = \frac{gRm}{12}

The energy required equals the difference in total mechanical energy between the final and initial orbits.

5

Substitute given numerical values and convert joules to megajoules.

\Delta E = \frac{10 \times (6.0 \times 10^6) \times 500}{12} = 2.5 \times 10^9\text{ J} = 2500\text{ MJ}

Dividing

2.5 \times 10^9\text{ J}

by

10^6

converts the value to megajoules.

Key Concept

Total Mechanical Energy of a Satellite in Circular Orbit and Orbital Transfer Energy

Estimated Time:3m 0s

Question 8163Question →

During the disproportionation reaction of white phosphorus (

P_4

) in an alkaline aqueous medium, phosphorus reacts according to the equation:

P_4 + 3OH^- + 3H_2O \rightarrow PH_3 + 3H_2PO_2^-

What are the oxidation numbers of phosphorus in

P_4

,

PH_3

, and

H_2PO_2^-

respectively, and what is the systematic IUPAC name of the

H_2PO_2^-

ion?

0

,

-3

, and

+1

; Dihydrogendioxophosphate(I) ion

+4

,

-3

, and

+1

; Dihydrogendioxophosphate(III) ion

0

,

+3

, and

+3

; Dihydrogentrioxophosphate(III) ion

-4

,

+3

, and

+5

; Dioxophosphate(V) ion

Show answer & explanation

Answer: $0$ , $-3$ , and $+1$ ; Dihydrogendioxophosphate(I) ion

Answer

The oxidation numbers of phosphorus in

P_4

,

PH_3

, and

H_2PO_2^-

are

0

,

-3

, and

+1

respectively, and the systematic IUPAC name of

H_2PO_2^-

is the Dihydrogendioxophosphate(I) ion.

The option stating '

0

,

-3

, and

+1

; Dihydrogendioxophosphate(I) ion' is correct because uncombined phosphorus (

P_4

) has an oxidation number of

0

. In phosphine (

PH_3

), phosphorus has an oxidation state of

-3

because hydrogen is

+1

. In

H_2PO_2^-

, solving

2(+1) + x + 2(-2) = -1

gives

x = +1

. Following IUPAC rules for oxoanions with hydrogen ligands, two hydrogens ('Dihydrogen'), two oxygens ('dioxo'), and phosphorus(I) in an anion ('phosphate(I)') combine to form the name Dihydrogendioxophosphate(I) ion.

Step-by-Step Solution

1

Determine the oxidation state of phosphorus in

P_4

.

Oxidation state =

0

By definition, any element in its free, uncombined elemental state has an oxidation number of zero.

2

Determine the oxidation state of phosphorus in

PH_3

.

Oxidation state =

-3

Hydrogen bonded to non-metals has an oxidation state of

+1

. Setting up the equation:

x + 3(+1) = 0 \implies x = -3

.

3

Determine the oxidation state of phosphorus in

H_2PO_2^-

.

Oxidation state =

+1

Hydrogen is

+1

and oxygen is

-2

. Setting up the ion charge equation:

2(+1) + x + 2(-2) = -1 \implies +2 + x - 4 = -1 \implies x = +1

.

4

Derive the systematic IUPAC name for

H_2PO_2^-

.

Dihydrogendioxophosphate(I) ion

The polyatomic ion contains 2 hydrogen atoms ('Dihydrogen'), 2 oxygen ligands ('dioxo'), the central phosphorus atom in an anion suffix ('phosphate'), and Roman numeral '(I)' indicating its

+1

oxidation state.

Key Concept

Calculation of oxidation states in polyatomic ions and application of inorganic IUPAC nomenclature rules.

Estimated Time:2m 0s

Question 8164Question →

Smoke consists of fine solid particles dispersed in a gaseous medium. Which of the following colloidal classifications correctly describes smoke?

Aerosol

Emulsion

Foam

Sol

Show answer & explanation

Answer: Aerosol

Answer

Smoke is classified as an aerosol because it consists of solid particles dispersed in a gas.

Smoke consists of microscopic solid carbon particles suspended in air (a gas). Any colloidal dispersion where a solid or liquid is dispersed in a gas is categorized as an aerosol.

Step-by-Step Solution

1

Identify the dispersed phase and the dispersion medium of smoke.

Dispersed phase = solid (carbon/ash particles), Dispersion medium = gas (air).

Classification of colloids depends on the physical states of the dispersed phase and dispersion medium.

2

Match the phase combination (solid in gas) to the standard colloidal nomenclature.

Solid dispersed in a gas is termed a solid aerosol.

Colloids with a gaseous dispersion medium are broadly categorized as aerosols.

Key Concept

Classification of Colloidal Systems based on Dispersed Phase and Dispersion Medium

Question 8165Question →

But-1-ene and but-2-ene share the molecular formula $C_4H_8$ but differ in the location of their carbon-carbon double bond. Which type of structural isomerism do these two compounds exhibit?

Chain isomerism

Positional isomerism

Functional group isomerism

Geometric isomerism

Show answer & explanation

Answer: Positional isomerism

Answer

Positional isomerism

Positional isomerism occurs when compounds with the same carbon skeleton and the same functional group differ only in the location of that functional group on the chain. But-1-ene and but-2-ene both have a straight four-carbon chain, but the double bond starts at position 1 and position 2 respectively.

Step-by-Step Solution

1

Examine the structures of both molecules.

Both but-1-ene (

CH_2=CH-CH_2-CH_3

) and but-2-ene (

CH_3-CH=CH-CH_3

) have a straight four-carbon chain and an alkene functional group.

Comparing carbon chain structure and functional group identity helps determine isomer type.

2

Identify the structural difference between the two molecules.

The double bond is located between carbon-1 and carbon-2 in but-1-ene, but between carbon-2 and carbon-3 in but-2-ene.

Compounds with the same carbon framework that differ only in the location of the functional group are classified as positional isomers.

Key Concept

Positional Isomerism in Alkenes

Question 8166Question →

An acid salt is produced when the replaceable hydrogen ions of a polybasic acid are only partially replaced by a metallic or ammonium ion. Which of the following chemical equations represents the preparation of an acid salt?

NaOH + H_2SO_4 \rightarrow NaHSO_4 + H_2O

2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O

NH_3 + HCl \rightarrow NH_4Cl

CuO + H_2SO_4 \rightarrow CuSO_4 + H_2O

Show answer & explanation

Answer: $NaOH + H_2SO_4 \rightarrow NaHSO_4 + H_2O$

Answer

The equation

NaOH + H_2SO_4 \rightarrow NaHSO_4 + H_2O

represents the preparation of sodium hydrogensulfate (

NaHSO_4

), an acid salt.

The equation producing sodium hydrogensulfate (

NaHSO_4

) represents partial neutralization of dibasic tetraoxosulfate(VI) acid (

H_2SO_4

), leaving a replaceable hydrogen ion within the salt structure.

Step-by-Step Solution

1

Define an acid salt in terms of neutralization.

Acid salts are formed when polybasic acids undergo partial neutralization, retaining at least one replaceable hydrogen ion in the anion.

Monobasic acids (like

HCl

) cannot form acid salts, while dibasic acids (like

H_2SO_4

) can form both acid salts and normal salts depending on stoichiometry.

2

Analyze the stoichiometry and products of each reaction equation.

Reacting 1 mole of

NaOH

with 1 mole of

H_2SO_4

replaces only one of the two hydrogen atoms in

H_2SO_4

, giving

NaHSO_4

.

Because

NaHSO_4

contains ionizable hydrogen (

H^+

), it forms an acidic aqueous solution, satisfying the definition of an acid salt.

Key Concept

Classification and Synthesis of Acid Salts

Estimated Time:1m 0s

Question 8167Question →

Concentrated trioxonitrate(V) acid can be safely stored and transported in containers constructed from aluminium metal. Which of the following chemical phenomena accounts for this behavior?

Formation of an impervious, protective surface oxide film

High standard reduction potential preventing acid attack

Precipitation of an insoluble trioxonitrate(V) salt lining the container

Neutralization of the acid by the basic nature of aluminium metal

Show answer & explanation

Answer: Formation of an impervious, protective surface oxide film

Answer

Formation of an impervious, protective surface oxide film

Concentrated trioxonitrate(V) acid is a powerful oxidizing agent that renders aluminium passive by forming a thin, dense, and non-porous oxide coating over its surface. This continuous protective barrier prevents the acid from contacting the bulk metal underneath.

Step-by-Step Solution

1

Examine the chemical action of concentrated trioxonitrate(V) acid on aluminium metal.

Concentrated trioxonitrate(V) acid acts as a powerful oxidizing agent.

Upon contact with aluminium, it immediately oxidizes the metal surface.

2

Determine the physical consequence of the surface oxidation.

A thin, tough, and impermeable oxide layer covers the surface of the metal.

This process, termed passivity, shields the underlying bulk aluminium from undergoing further chemical attack by the acid.

Key Concept

Aluminium oxide passivity

Question 8168Question →

In the Solvay process for the industrial manufacture of sodium trioxocarbonate(IV), ammonia is an expensive reagent that must be recovered and reused to make the process economically viable. Which compound is reacted with ammonium chloride in the recovery tower to regenerate ammonia gas?

Calcium hydroxide

Calcium chloride

Carbon(IV) oxide

Sodium chloride

Show answer & explanation

Answer: Calcium hydroxide

Answer

Calcium hydroxide

Calcium hydroxide is a strong base that reacts with ammonium chloride in the Solvay process recovery plant according to the equation

2NH_4Cl + Ca(OH)_2 \rightarrow CaCl_2 + 2H_2O + 2NH_3

. This step regenerates ammonia gas for continuous recycling, making the Solvay process economically efficient.

Step-by-Step Solution

1

Identify the byproduct formed when ammonia reacts during the carbonating phase of the Solvay process.

Ammonium chloride (

NH_4Cl

) is formed along with sodium hydrogentrioxocarbonate(IV) (

NaHCO_3

).

Ammonia absorbs carbon(IV) oxide and reacts with brine to form ammonium chloride in solution.

2

Determine the reagent added to ammonium chloride in the recovery tower to liberate free ammonia gas.

Calcium hydroxide (

Ca(OH)_2

), obtained from slaking quicklime (

CaO

), is added.

Ammonium salts react with strong bases like calcium hydroxide upon heating to yield ammonia gas, water, and a calcium salt.

3

Write the balanced chemical equation for the recovery step.

2NH_4Cl(aq) + Ca(OH)_2(aq) \rightarrow CaCl_2(aq) + 2H_2O(l) + 2NH_3(g)

This reaction regenerates ammonia gas, which is recycled back into the ammoniating tower.

Key Concept

Ammonia recovery in the industrial Solvay process using calcium hydroxide

Question 8169Question →

Consider the redox reaction taking place in an acidic medium:

\text{MnO}_4^-(\text{aq}) + \text{H}_2\text{O}_2(\text{aq}) + \text{H}^+(\text{aq}) \rightarrow \text{Mn}^{2+}(\text{aq}) + \text{O}_2(\text{g}) + \text{H}_2\text{O}(\text{l})

When this chemical equation is balanced using the smallest whole-number coefficients, what is the stoichiometric coefficient of

\text{H}^+(\text{aq})

?

16

6

8

3

Show answer & explanation

Answer: 6

Answer

6

The balanced net redox equation is

2\text{MnO}_4^-(\text{aq}) + 5\text{H}_2\text{O}_2(\text{aq}) + 6\text{H}^+(\text{aq}) \rightarrow 2\text{Mn}^{2+}(\text{aq}) + 5\text{O}_2(\text{g}) + 8\text{H}_2\text{O}(\text{l})

. Thus, the stoichiometric coefficient of

\text{H}^+(\text{aq})

is 6.

Step-by-Step Solution

1

Write and balance the reduction half-reaction

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

Manganese is reduced from oxidation state +7 in

\text{MnO}_4^-

to +2 in

\text{Mn}^{2+}

, requiring 5 electrons. Charge and mass are balanced using

\text{H}^+

and

\text{H}_2\text{O}

.

2

Write and balance the oxidation half-reaction

\text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2\text{e}^-

Oxygen in

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

is oxidized from -1 to 0 in

\text{O}_2

, releasing 2 electrons per molecule.

3

Equalize electron loss and gain

Multiply reduction half-reaction by 2 and oxidation half-reaction by 5:

2\text{MnO}_4^- + 16\text{H}^+ + 10\text{e}^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}

5\text{H}_2\text{O}_2 \rightarrow 5\text{O}_2 + 10\text{H}^+ + 10\text{e}^-

The least common multiple of 5 and 2 electrons transferred is 10.

4

Combine the half-reactions and simplify redundant species

2\text{MnO}_4^- + 5\text{H}_2\text{O}_2 + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{O}_2 + 8\text{H}_2\text{O}

Subtracting

10\text{H}^+

and

10\text{e}^-

from both sides yields a net coefficient of 6 for

\text{H}^+

on the reactant side.

Key Concept

Ion-Electron Method for Balancing Redox Equations in Acidic Medium

Question 8170Question →

A sample of gas in a flexible container occupies a volume of $4.00\text{ dm}^3$ at $27^\circ\text{C}$ and a pressure of $1.50\text{ atm}$ . What will be the volume of the gas if the temperature is increased to $127^\circ\text{C}$ and the pressure is reduced to $1.00\text{ atm}$ ?

28.2\text{ dm}^3

8.00\text{ dm}^3

4.50\text{ dm}^3

6.00\text{ dm}^3

Show answer & explanation

Answer: $8.00\text{ dm}^3$

Answer

The final volume of the gas is

8.00\text{ dm}^3

.

Applying the General Gas Law

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

requires converting all temperatures to Kelvin:

T_1 = 27 + 273 = 300\text{ K}

and

T_2 = 127 + 273 = 400\text{ K}

. Rearranging for

V_2

gives

V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} = \frac{1.50 \times 4.00 \times 400}{1.00 \times 300} = 8.00\text{ dm}^3

. Thus, the option stating

8.00\text{ dm}^3

is correct.

Step-by-Step Solution

1

Convert temperatures from degrees Celsius to Kelvin

T_1 = 27 + 273 = 300\text{ K}

and

T_2 = 127 + 273 = 400\text{ K}

Gas laws require absolute temperature units in Kelvin for linear proportional relationships.

2

State the General Gas Law equation and rearrange for final volume (

V_2

)

V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}

The General Gas Law relates pressure, volume, and temperature for a fixed amount of gas.

3

Substitute the given values into the rearranged equation

V_2 = \frac{1.50\text{ atm} \times 4.00\text{ dm}^3 \times 400\text{ K}}{1.00\text{ atm} \times 300\text{ K}}

Plugging in

P_1 = 1.50\text{ atm}

,

V_1 = 4.00\text{ dm}^3

,

T_1 = 300\text{ K}

,

P_2 = 1.00\text{ atm}

, and

T_2 = 400\text{ K}

.

4

Evaluate the mathematical calculation

V_2 = \frac{2400}{300} = 8.00\text{ dm}^3

Simplifying the fraction gives the final gas volume.

Key Concept

General Gas Law (

P_1 V_1 / T_1 = P_2 V_2 / T_2

)

Question 8171Question →

Match each chemical species or atmospheric component to its primary environmental function or effect regarding global warming and ozone layer preservation:

Click a left item, then click its matching right item

Show answer & explanation

Answer

Chlorofluorocarbons (CFCs) match with releasing chlorine free radicals that break down stratospheric ozone; Carbon(IV) oxide matches with absorbing outgoing thermal infrared radiation in the troposphere; Stratospheric ozone matches with filtering out harmful solar ultraviolet radiation.

Chlorofluorocarbons release chlorine free radicals that catalyze the breakdown of ozone molecules in the stratosphere. Carbon(IV) oxide is a major greenhouse gas that absorbs infrared heat radiation in the troposphere. Stratospheric ozone shields the Earth by absorbing harmful solar ultraviolet rays.

Step-by-Step Solution

1

Identify the primary mechanism of Chlorofluorocarbons (CFCs)

CFCs diffuse to the stratosphere where UV radiation breaks them down to form chlorine radicals, causing ozone layer depletion.

Connecting CFCs to ozone destruction avoids confusing greenhouse heat-trapping with catalytic chemical breakdown.

2

Identify the primary mechanism of Carbon(IV) oxide

Carbon(IV) oxide absorbs infrared (heat) radiation emitted from Earth's surface, preventing thermal escape.

This establishes Carbon(IV) oxide as a principal greenhouse gas driving global warming.

3

Identify the protective role of stratospheric ozone

Stratospheric ozone absorbs short-wavelength UV rays from the Sun.

Protective ozone acts as a radiation shield rather than a thermal insulator.

Key Concept

Distinction between global warming mechanisms (infrared absorption by greenhouse gases) and ozone depletion mechanisms (ultraviolet photolysis releasing chlorine radicals).

Question 8172Question →

In an industrial furnace operation, producer gas is synthesized by passing air over red-hot coke. Which of the following pairs correctly identifies the main combustible constituent and the primary non-combustible diluent present in the resulting gas mixture?

Carbon(II) oxide and nitrogen

Carbon(II) oxide and hydrogen

Carbon(IV) oxide and nitrogen

Carbon(IV) oxide and hydrogen

Show answer & explanation

Answer: Carbon(II) oxide and nitrogen

Answer

The main combustible constituent of producer gas is carbon(II) oxide (

\text{CO}

) and the primary non-combustible diluent is nitrogen (

\text{N}_2

).

When air is passed over red-hot coke, oxygen reacts with carbon to form carbon(II) oxide (

\text{CO}

), which is the primary combustible fuel component. The unreactive nitrogen gas (

\text{N}_2

) present in air passes through the bed of coke without reacting, forming the main non-combustible diluent (approximately two-thirds of the total volume).

Step-by-Step Solution

1

Identify the reactants involved in the production of producer gas.

Air (containing

\text{O}_2

and

\text{N}_2

) is passed over incandescent coke (

\text{C}

).

Producer gas manufacture relies on passing air over red-hot carbon.

2

Determine the chemical reaction and resulting gaseous products.

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

, while atmospheric

\text{N}_2

remains unreacted.

Incomplete oxidation of coke produces

\text{CO}

, while

\text{N}_2

from air dilutes the product mixture.

3

Classify the role of each component in the product gas mixture.

\text{CO}

is combustible and serves as the fuel, whereas

\text{N}_2

is inert/non-combustible.

\text{CO}

can undergo further oxidation to

\text{CO}_2

releasing energy, whereas

\text{N}_2

does not burn.

Key Concept

Composition and industrial production of producer gas

Estimated Time:1m 0s

Question 8173Question →

In the carbon monoxide molecule ( $CO$ ), a triple covalent linkage holds the carbon and oxygen atoms together. Which statement correctly describes the nature of the bonding pairs between the two atoms and identifies the electron pair donor for the dative (coordinate) bond?

There are three shared electron pairs in total, with oxygen donating the lone pair required for the coordinate bond.

There are three shared electron pairs in total, with carbon donating the lone pair required for the coordinate bond.

There are two shared electron pairs in total, with oxygen donating the lone pair required for the coordinate bond.

There are four shared electron pairs in total, with both atoms contributing one lone pair each.

Show answer & explanation

Answer: There are three shared electron pairs in total, with oxygen donating the lone pair required for the coordinate bond.

Answer

The carbon monoxide molecule features three shared pairs of electrons forming a triple bond, where oxygen serves as the donor atom contributing the lone pair for the coordinate covalent bond.

In carbon monoxide (

CO

), carbon contributes 2 electrons and oxygen contributes 2 electrons to form two standard single covalent bonds. To allow carbon to attain a stable octet (8 valence electrons), oxygen donates both electrons from one of its lone pairs to form a third bond (dative/coordinate bond). Thus, there are 3 shared electron pairs (a triple bond), and oxygen is the donor.

Step-by-Step Solution

1

Determine valence electron counts for Carbon and Oxygen

Carbon (Group 14) has 4 valence electrons; Oxygen (Group 16) has 6 valence electrons.

Establishing valence electrons determines how many electrons are shared to complete octets.

2

Analyze standard covalent sharing between Carbon and Oxygen

Sharing 2 electrons from Carbon and 2 electrons from Oxygen forms 2 ordinary covalent bonds. This gives Oxygen 8 valence electrons, but leaves Carbon with only 6 valence electrons.

Two standard covalent bonds are insufficient to fulfill the octet rule for Carbon.

3

Identify the dative (coordinate) bond formation

Oxygen donates one of its remaining lone pairs into Carbon's vacant orbital to form a 3rd shared bond (coordinate bond), completing Carbon's octet.

This establishes a total of 3 shared pairs (a triple bond) with Oxygen acting as the donor atom.

Key Concept

Covalent and Dative (Coordinate) Bonding in Carbon Monoxide

Estimated Time:1m 30s

Question 8174Question →

Liquefied air is a mixture containing predominantly nitrogen (boiling point $-196^\circ\text{C}$ ), argon (boiling point $-186^\circ\text{C}$ ), and oxygen (boiling point $-183^\circ\text{C}$ ). When liquid air undergoes fractional distillation to separate its components industrially, which gas is collected first at the top of the fractionating column, and why?

Nitrogen, because it has the lowest boiling point and boils off first as the temperature rises.

Oxygen, because its higher boiling point enables it to vaporize more rapidly under column thermal gradient.

Oxygen, because components with higher boiling points condense at the top of the column and exit first.

Argon, because having an intermediate boiling point allows it to displace nitrogen and oxygen at the top fraction.

Show answer & explanation

Answer: Nitrogen, because it has the lowest boiling point and boils off first as the temperature rises.

Answer

Nitrogen, because it has the lowest boiling point and boils off first as the temperature rises.

In fractional distillation of liquid air, nitrogen has the lowest boiling point (

-196^\circ\text{C}

) compared to argon (

-186^\circ\text{C}

) and oxygen (

-183^\circ\text{C}

). As the temperature of the liquid air is slowly increased, nitrogen reaches its boiling point first, vaporizes, rises to the top of the column, and distills over as the first fraction.

Step-by-Step Solution

1

Compare the boiling points of all components in liquid air

Nitrogen:

-196^\circ\text{C}

, Argon:

-186^\circ\text{C}

, Oxygen:

-183^\circ\text{C}

.

The boiling point determines the volatility of each component in a miscible mixture.

2

Identify which component is the most volatile

Nitrogen has the lowest numerical boiling point (

-196^\circ\text{C}

), meaning it requires the least heat to convert from liquid to gas.

Lower boiling point indicates weaker intermolecular forces and higher vapor pressure at a given temperature.

3

Relate boiling order to fractional distillation output

As temperature increases from

-200^\circ\text{C}

, nitrogen reaches its boiling point first, vaporizes, rises to the top of the column, and is collected as the first fraction.

During fractional distillation of liquid air, components with lower boiling points vaporize first and emerge at the top of the fractionating column.

Key Concept

Order of separation in fractional distillation based on boiling points

Estimated Time:1m 30s

Question 8175Question →

An element $M$ forms a stable electrovalent chloride with the formula $MCl_2$ . If the dipositive cation $M^{2+}$ has the electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^6$ , which of the following statements correctly explains the electrical conductivity and lattice properties of $MCl_2$ ?

It does not conduct electricity in the solid state because its ions are held in fixed positions, but conducts in the molten state due to mobile

M^{2+}

and

Cl^-

ions.

It conducts electricity in both the solid and molten states because delocalized valence electrons move freely through the ionic crystal lattice.

It has a low melting point and conducts electricity in the solid state because

M^{2+}

cations migrate easily between lattice nodes.

It conducts electricity when molten because the neutral compound dissociates into free electrons and uncharged elemental atoms.

Show answer & explanation

Answer: It does not conduct electricity in the solid state because its ions are held in fixed positions, but conducts in the molten state due to mobile $M^{2+}$ and $Cl^-$ ions.

Answer

The compound does not conduct electricity in the solid state because its ions are held in fixed positions within the lattice, but conducts in the molten state due to mobile

M^{2+}

and

Cl^-

ions.

In giant electrovalent (ionic) lattices like

CaCl_2

, ions are immobilized in fixed positions in the solid state, making the solid a non-conductor. Upon melting, the electrostatic lattice forces are overcome, producing free mobile

M^{2+}

and

Cl^-

ions that conduct electricity.

Step-by-Step Solution

1

Identify the element and ion structure

The ion

M^{2+}

has 18 electrons (

1s^2 2s^2 2p^6 3s^2 3p^6

), corresponding to a neutral calcium atom (

Ca

, atomic number 20). The compound formed is calcium chloride (

CaCl_2

).

Determining the electronic structure confirms

MCl_2

is a typical giant ionic (electrovalent) lattice.

2

Analyze solid-state properties

In the solid state,

M^{2+}

cations and

Cl^-

anions are locked in a rigid three-dimensional crystal lattice by strong omnidirectional electrostatic forces of attraction. Because the ions cannot move, the solid is an electrical insulator.

Conduction of electricity requires mobile charge carriers.

3

Analyze molten-state properties

When heated to its high melting point, thermal energy overcomes the lattice energy, allowing the

M^{2+}

and

Cl^-

ions to move freely and carry electrical current.

Liquid ionic compounds conduct electricity via migration of ions toward oppositely charged electrodes.

Key Concept

Ionic Lattice Properties and Conduction Mechanism

Question 8176Question →

Three isomeric alkanes have the molecular formula $C_5H_{12}$ : pentane, 2-methylbutane, and 2,2-dimethylpropane. Which of the following statements correctly accounts for the trend in their boiling points?

Pentane has the highest boiling point because its straight-chain structure provides a larger surface area for intermolecular van der Waals forces.

2,2-Dimethylpropane has the highest boiling point because its compact spherical shape maximizes molecular interactions.

2-Methylbutane has a higher boiling point than pentane because branching increases the strength of intermolecular forces.

All three isomers have identical boiling points because they possess the same molecular formula and molar mass.

Show answer & explanation

Answer: Pentane has the highest boiling point because its straight-chain structure provides a larger surface area for intermolecular van der Waals forces.

Answer

Pentane has the highest boiling point because its straight-chain structure provides a larger surface area for intermolecular van der Waals forces.

Pentane possesses an unbranched, straight-chain hydrocarbon structure. This spatial arrangement allows adjacent molecules to lie close together with maximum surface contact. As a result, intermolecular van der Waals forces are strongest in pentane, requiring the highest temperature to transition from liquid to gas.

Step-by-Step Solution

1

Analyze the molecular structures of the three

C_5H_{12}

isomers.

Pentane is unbranched (straight-chain), 2-methylbutane is monobranched, and 2,2-dimethylpropane is highly branched (spherical).

Structural branching determines molecular shape and the overall contact area available between molecules.

2

Relate molecular shape to intermolecular forces.

Alkanes are non-polar and held together by weak London dispersion (van der Waals) forces, which scale with molecular surface contact area.

Greater surface contact area leads to stronger attractive forces that require more thermal energy to overcome.

3

Determine the boiling point trend based on surface area.

Pentane has the largest surface area of contact, giving it the strongest intermolecular forces and the highest boiling point, while 2,2-dimethylpropane has the lowest.

Increased branching compacts the molecule into a sphere, minimizing contact area and lowering the boiling point.

Key Concept

Effect of structural branching on alkane boiling points and intermolecular forces

Estimated Time:1m 0s

Question 8177Question →

Two aqueous solutions, one of ethanoic acid ( $\text{CH}_3\text{COOH}$ ) and the other of hydrochloric acid ( $\text{HCl}$ ), have the exact same concentration of $0.10\text{ mol dm}^{-3}$ . Which of the following statements correctly accounts for why hydrochloric acid is classified as a stronger acid than ethanoic acid?

Hydrochloric acid completely ionizes in aqueous solution, whereas ethanoic acid only partially ionizes.

Hydrochloric acid has a higher molar concentration of acid molecules in the solution than ethanoic acid.

Ethanoic acid undergoes complete dissociation in water while hydrochloric acid forms a reversible equilibrium.

Ethanoic acid produces a higher hydrogen ion concentration because it contains four hydrogen atoms per molecule.

Show answer & explanation

Answer: Hydrochloric acid completely ionizes in aqueous solution, whereas ethanoic acid only partially ionizes.

Answer

Hydrochloric acid completely ionizes in aqueous solution, whereas ethanoic acid only partially ionizes.

The strength of an acid is defined by its degree of ionization in aqueous solution. Hydrochloric acid (

\text{HCl}

) is a strong acid because it ionizes completely in water to yield hydrogen ions. In contrast, ethanoic acid (

\text{CH}_3\text{COOH}

) is a weak acid that ionizes only partially, setting up a dynamic equilibrium mixture of molecules and ions.

Step-by-Step Solution

1

Define acid strength in terms of degree of ionization.

Acid strength is determined by the extent to which an acid dissociates into hydrogen ions (

\text{H}^+

) in water, independent of solution concentration.

Strong acids dissociate completely, whereas weak acids dissociate only partially.

2

Compare the ionization behavior of hydrochloric acid and ethanoic acid.

Hydrochloric acid (

\text{HCl}

) is a strong monobasic acid that ionizes virtually

100\%

in water. Ethanoic acid (

\text{CH}_3\text{COOH}

) is a weak monobasic acid that ionizes only slightly, establishing a dynamic equilibrium.

Because both solutions have equal molar concentrations, the completely ionized strong acid yields a substantially greater concentration of

\text{H}^+

ions.

Key Concept

Relative Strength and Ionization of Acids and Bases

Question 8178Question →

Consider three Period 3 elements: sodium ( $Na$ ), magnesium ( $Mg$ ), and aluminium ( $Al$ ). As one moves from $Na$ to $Al$ across the period, there is a notable increase in both melting point and electrical conductivity per mole of metal. Which of the following best accounts for this observed trend in metallic bond strength and physical properties?

The ionic radius increases from

Na^+

to

Al^{3+}

, enabling larger cations to form a tighter lattice that conducts heat and electricity more effectively.

The number of delocalized valence electrons contributed per atom increases while cationic radius decreases, increasing electrostatic attraction and mobile charge density.

Aluminium transitions from metallic bonding to directional covalent giant lattice bonding, requiring significantly higher heat to break.

Sodium contributes more delocalized electrons per atom into the lattice than aluminium because sodium possesses a lower first ionization energy.

Show answer & explanation

Answer: The number of delocalized valence electrons contributed per atom increases while cationic radius decreases, increasing electrostatic attraction and mobile charge density.

Answer

The trend is best explained by the increase in the number of delocalized valence electrons contributed per atom combined with a smaller cationic radius, which increases electrostatic attraction and mobile charge carrier density.

Metallic bonding consists of electrostatic attractions between fixed positive metal cations and a surrounding delocalized sea of valence electrons. Moving from sodium to aluminium, each atom donates more valence electrons (

Na = 1e^-

,

Mg = 2e^-

,

Al = 3e^-

) into the electron sea while the ionic radius decreases (

Na^+ > Mg^{2+} > Al^{3+}

). The combination of higher cationic charge, smaller ionic radius, and greater electron density increases the electrostatic attraction, raising both melting points and electrical conductivity.

Step-by-Step Solution

1

Analyze the structural factors determining metallic bond strength.

Metallic bond strength depends directly on two main factors: (1) the charge on the metal cation (number of delocalized electrons donated per atom) and (2) the cationic radius (distance between cations and delocalized electrons).

Strength of electrostatic attraction follows Coulomb's law:

F \propto \frac{q_1 q_2}{r^2}

.

2

Compare valence electron contributions across Period 3 metals.

Sodium (

[Ne]3s^1

) donates 1 electron per atom (

Na^+

), Magnesium (

[Ne]3s^2

) donates 2 electrons per atom (

Mg^{2+}

), and Aluminium (

[Ne]3s^3

) donates 3 electrons per atom (

Al^{3+}

).

Higher delocalized electron count yields greater mobile charge density for electrical conductivity.

3

Compare cationic radii across the period.

Ionic radii decrease across the period:

Na^+ (102\text{ pm}) > Mg^{2+} (72\text{ pm}) > Al^{3+} (54\text{ pm})

.

Smaller cations allow delocalized electrons to approach closer to positively charged nuclei, dramatically strengthening electrostatic attraction.

Key Concept

Factors affecting metallic bond strength and properties (charge density and delocalized electron count)

Question 8179Question →

A student aims to recover pure hydrated zinc tetraoxosulfate(VI) crystals ( $\text{ZnSO}_4 \cdot 7\text{H}_2\text{O}$ ) from a mixture containing dissolved zinc tetraoxosulfate(VI) and insoluble fine sand. After filtering off the sand, why must the filtrate be concentrated by gentle heating to saturation and allowed to cool, rather than evaporated directly to complete dryness?

Evaporating to complete dryness destroys the water of crystallization, producing an anhydrous powder or decomposed residue instead of hydrated crystals.

Direct evaporation to dryness causes the dissolved salt to distill over along with the water vapor.

Heating the solution causes the salt to react chemically with water to form an insoluble metal oxide precipitate.

Crystallization is only applicable when the solute has a higher boiling point than the solvent, whereas evaporation requires lower boiling points.

Show answer & explanation

Answer: Evaporating to complete dryness destroys the water of crystallization, producing an anhydrous powder or decomposed residue instead of hydrated crystals.

Answer

Evaporating to complete dryness destroys the water of crystallization, producing an anhydrous powder or decomposed residue instead of hydrated crystals.

Hydrated salts contain water of crystallization chemically integrated into their crystal lattice structure. Heating a solution of such a salt to complete dryness drives off this essential water, leaving behind an anhydrous powder or causing thermal decomposition rather than yielding the intended hydrated crystals.

Step-by-Step Solution

1

Identify the structural requirements of the target product.

The target product is hydrated zinc tetraoxosulfate(VI),

\text{ZnSO}_4 \cdot 7\text{H}_2\text{O}

, which requires chemically bound water of crystallization to maintain its crystalline lattice.

Retaining water of crystallization necessitates gentle thermal treatment rather than harsh heating.

2

Analyze the impact of evaporating to complete dryness.

Heating a hydrated salt solution to complete dryness removes all water molecules, converting the hydrated salt into an amorphous anhydrous powder (

\text{ZnSO}_4

) or decomposing it.

High thermal energy breaks the coordination bonds holding water molecules in the crystal structure.

3

Evaluate the correct procedural approach.

Evaporating the filtrate only to the point of saturation and allowing it to cool slowly enables pure hydrated crystals to form as solubility decreases.

Crystallization preserves the stoichiometric water of crystallization and excludes soluble impurities.

Key Concept

Distinction between evaporation to dryness and crystallization for hydrated salts

Question 8180Question →

A

25.0\text{ g}

sample of limestone containing

80.0\%

calcium trioxocarbonate(IV) by mass is strongly heated until decomposition is complete according to the equation:

CaCO_3(s) \xrightarrow{\Delta} CaO(s) + CO_2(g)

What volume of carbon(IV) oxide gas, measured at room temperature and pressure (RTP), is liberated in this reaction?
(

M(CaCO_3) = 100\text{ g mol}^{-1}

; Molar volume of gas at RTP

= 24.0\text{ dm}^3\text{ mol}^{-1}

)

4.80\text{ dm}^3

4.48\text{ dm}^3

6.00\text{ dm}^3

5.60\text{ dm}^3

Show answer & explanation

Answer: $4.80\text{ dm}^3$

Answer

The volume of carbon(IV) oxide gas liberated at RTP is

4.80\text{ dm}^3

.

The mass of active

CaCO_3

is

80.0\%

of

25.0\text{ g}

, which equals

20.0\text{ g}

. Dividing by the molar mass (

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

) gives

0.20\text{ mol}

of

CaCO_3

. By stoichiometry,

0.20\text{ mol}

of

CO_2

is evolved. Multiplying by the molar volume at RTP (

24.0\text{ dm}^3\text{ mol}^{-1}

) yields

4.80\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the mass of pure calcium trioxocarbonate(IV) (

CaCO_3

) present in the limestone sample.

Mass of pure

CaCO_3 = \frac{80.0}{100} \times 25.0\text{ g} = 20.0\text{ g}

.

Impurities in the limestone do not produce

CO_2

gas upon heating.

2

Determine the amount in moles of pure

CaCO_3

decomposed.

Moles of

CaCO_3 = \frac{20.0\text{ g}}{100\text{ g mol}^{-1}} = 0.20\text{ mol}

.

Converting mass to amount in moles allows stoichiometric evaluation using balanced chemical equations.

3

Use the stoichiometric ratio from the balanced chemical equation to find the moles of

CO_2

produced.

Since

1\text{ mol } CaCO_3 \rightarrow 1\text{ mol } CO_2

, moles of

CO_2 = 0.20\text{ mol}

.

The thermal decomposition ratio between

CaCO_3

and

CO_2

is

1:1

.

4

Calculate the volume of

CO_2

gas at room temperature and pressure (RTP).

Volume of

CO_2 = 0.20\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 4.80\text{ dm}^3

.

Molar gas volume at RTP is

24.0\text{ dm}^3\text{ mol}^{-1}

.

Key Concept

Thermal decomposition of calcium carbonate and percentage purity stoichiometry at non-STP conditions

Answer

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