All practice questions

Question 8461Question →

Match each chemical compound on the left with its characteristic solubility trend in water as temperature increases on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Potassium trioxonitrate(V) matches with steep increase in solubility; Sodium chloride matches with nearly constant solubility; Calcium tetraoxosulfate(VI) matches with decreasing solubility as temperature rises.

Potassium trioxonitrate(V) shows a steep increase in solubility with rising temperature due to its endothermic nature. Sodium chloride exhibits minimal temperature sensitivity, keeping its curve nearly flat. Calcium tetraoxosulfate(VI) exhibits retrograde solubility, decreasing as temperature increases because its dissolution is exothermic.

Step-by-Step Solution

1

Identify the thermodynamic enthalpy change associated with dissolving each salt in water.

Potassium trioxonitrate(V) dissolution is endothermic, sodium chloride dissolution has a near-zero enthalpy change, and calcium tetraoxosulfate(VI) dissolution is exothermic.

Le Chatelier's principle determines how temperature affects solubility equilibria based on whether heat is absorbed or released.

2

Relate enthalpy of solution to the slope of the solubility curve.

Potassium trioxonitrate(V) has a steep positive curve, sodium chloride has a nearly horizontal curve, and calcium tetraoxosulfate(VI) has a negative curve.

Endothermic dissolution shifts right with heat (increasing solubility), whereas exothermic dissolution shifts left with heat (decreasing solubility).

Key Concept

Solubility curves represent how solute solubility varies with temperature based on whether the dissolution process is endothermic or exothermic.

Question 8462Question →

Which of the following alkenes with the molecular formula $C_5H_{10}$ exhibits geometric (cis-trans) isomerism?

Pent-2-ene

Pent-1-ene

2-Methylbut-2-ene

3-Methylbut-1-ene

Show answer & explanation

Answer: Pent-2-ene

Answer

Pent-2-ene is the only isomer listed that exhibits geometric (cis-trans) isomerism because each carbon of the double bond is attached to two distinct groups.

Pent-2-ene has a double bond between carbon-2 and carbon-3. Carbon-2 is attached to a hydrogen atom and a methyl group (

CH_3

), while carbon-3 is attached to a hydrogen atom and an ethyl group (

CH_2CH_3

). Because neither carbon atom of the double bond holds two identical groups, spatial restriction gives rise to distinct cis and trans stereoisomers.

Step-by-Step Solution

1

Recall the necessary structural condition for geometric (cis-trans) isomerism in alkenes.

For a molecule to show cis-trans isomerism around a double bond

C=C

, each of the two carbon atoms in the double bond must be attached to two different atoms or groups.

If either carbon in the double bond has two identical groups attached, rotating spatial arrangements results in identical molecules.

2

Examine the connectivity of each option at the

C=C

double bond.

Pent-2-ene has

C_2

bonded to

-H

and

-CH_3

, and

C_3

bonded to

-H

and

-CH_2CH_3

. Both carbons have two different groups.

This satisfies the criteria for both cis and trans geometric arrangements.

3

Check the remaining options for duplicate attached groups on double-bonded carbons.

Pent-1-ene and 3-methylbut-1-ene both have a terminal

=CH_2

(two H atoms on

C_1

). 2-Methylbut-2-ene has two methyl groups on

C_2

.

None of these three options satisfy the non-identical substituent requirement on both double-bonded carbons.

Key Concept

Geometric Isomerism Requirements in Alkenes

Estimated Time:1m 0s

Question 8463Question →

During the industrial separation of liquified air by fractional distillation, nitrogen gas is collected before oxygen at the top of the fractionating column. Which of the following explains why nitrogen is collected first?

Nitrogen has a lower boiling point than oxygen

Nitrogen has a higher boiling point than oxygen

Nitrogen is more dense than liquid oxygen

Nitrogen reacts with liquid oxygen to form a volatile oxide

Show answer & explanation

Answer: Nitrogen has a lower boiling point than oxygen

Answer

Nitrogen is collected first because it has a lower boiling point than oxygen.

Nitrogen has a lower boiling point (-196 °C) than oxygen (-183 °C). In fractional distillation, liquid mixtures are separated according to boiling point, so the component with the lowest boiling point vaporizes first and distills at the top of the column.

Step-by-Step Solution

1

Identify the principle governing industrial fractional distillation of liquid air.

Components are separated according to their boiling points, with the substance having the lowest boiling point vaporizing and distilling over first.

Fractional distillation depends on boiling point differences to separate liquid mixtures.

2

Compare the boiling points of nitrogen and oxygen.

Nitrogen has a boiling point of -196 °C (77 K) while oxygen has a boiling point of -183 °C (90 K).

Because -196 °C is lower than -183 °C, nitrogen vaporizes first at the lower temperature.

Key Concept

Fractional distillation of liquid air based on boiling points

Estimated Time:45s

Question 8464Question →

In an experiment investigating reaction kinetics, a strip of magnesium ribbon is reacted with excess dilute hydrochloric acid. If the ribbon is ground into a fine powder while keeping the total mass of magnesium, temperature, and acid concentration constant, which of the following best explains why the initial rate of reaction increases based on collision theory?

The activation energy of the reaction is lowered because smaller particles react faster.

The average kinetic energy of the reacting particles increases, causing more energetic collisions.

The greater exposed surface area increases the frequency of collisions between reactant particles per unit time.

A greater proportion of colliding particles possess energy equal to or greater than the activation energy.

Show answer & explanation

Answer: The greater exposed surface area increases the frequency of collisions between reactant particles per unit time.

Answer

The greater exposed surface area increases the frequency of collisions between reactant particles per unit time.

Increasing the surface area of a solid reactant by powdering it exposes a larger number of atoms to reactant ions in solution. This increases the total frequency of collisions occurring per unit time, resulting in a higher rate of effective collisions and thus a faster overall reaction rate.

Step-by-Step Solution

1

Analyze the physical change made to the solid reactant.

Grinding the magnesium ribbon into a powder increases its total surface area exposed to the aqueous hydrochloric acid solution.

Smaller particle size provides a greater number of exposed surface atoms per unit mass.

2

Apply collision theory principles to evaluate collision mechanics.

A higher number of exposed surface atoms increases the number of collisions per second (collision frequency) between magnesium atoms and hydrogen ions.

Collision theory states that reaction rate is directly proportional to collision frequency when kinetic energy and orientation parameters are constant.

3

Distinguish surface area effects from temperature and catalyst effects.

Neither activation energy nor particle kinetic energy changes, as temperature and reaction pathway remain identical.

Activation energy is lowered only by catalysts, and average kinetic energy is increased only by raising temperature.

Key Concept

Effect of Surface Area on Collision Frequency in Collision Theory

Question 8465Question →

Match each sulfur-related compound or allotrope on the left with its correct physical characteristic, qualitative test, or stability range on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Rhombic sulfur matches the octahedral allotrope stable up to 96°C; Hydrogen sulfide matches the rotten-egg smelling gas turning lead(II) ethanoate paper black; Sulfur(IV) oxide matches the choking gas turning acidified potassium dichromate(VI) green; Monoclinic sulfur matches the needle-shaped allotrope stable between 96°C and 119°C.

Rhombic sulfur (

\alpha

-sulfur) is octahedral and stable below

96^\circ\text{C}

. Monoclinic sulfur (

\beta

-sulfur) is needle-shaped and stable between

96^\circ\text{C}

and

119^\circ\text{C}

. Hydrogen sulfide (

H_2S

) is a gas with a rotten-egg odor that reacts with

Pb^{2+}

to form black insoluble

PbS

. Sulfur(IV) oxide (

SO_2

) has a pungent choking smell and reduces orange dichromate(VI) solutions to green

Cr^{3+}

ions.

Step-by-Step Solution

1

Analyze the crystal structures and thermal stability range of sulfur allotropes.

Rhombic sulfur is octahedral and stable below 96°C. Monoclinic sulfur is needle-shaped and stable between 96°C and 119°C.

96°C is the transition temperature at which the two crystalline allotropes exist in equilibrium.

2

Identify the qualitative gas tests and chemical properties of

H_2S

and

SO_2

.

H_2S

forms black

PbS

precipitate with lead(II) ethanoate paper.

SO_2

reduces orange dichromate(VI) to green

Cr^{3+}

ions.

Both gases are reducing agents, but

H_2S

forms insoluble sulfides while

SO_2

undergoes specific color-change redox reactions with dichromate ions.

Key Concept

Physical and chemical properties of sulfur allotropes, hydrogen sulfide, and sulfur(IV) oxide

Question 8466Question →

In the auto-reduction stage of copper extraction, copper(I) oxide (

\text{Cu}_2\text{O}

) reacts with copper(I) sulfide (

\text{Cu}_2\text{S}

) according to the following balanced equation:

\text{Cu}_2\text{S (s)} + 2\text{Cu}_2\text{O (s)} \rightarrow 6\text{Cu (s)} + \text{SO}_2\text{ (g)}

If

14.3\text{ g}

of copper(I) oxide reacts completely with excess copper(I) sulfide, what mass of metallic copper in grams is produced?

[\text{Cu} = 63.5, \text{O} = 16.0]

Show answer & explanation

Answer: 19.05

Answer

The mass of metallic copper produced is 19.05 g.

According to the balanced chemical equation, 2 moles of copper(I) oxide react with copper(I) sulfide to yield 6 moles of metallic copper, which simplifies to a 1:3 molar ratio. Given that the molar mass of Cu2O is 143 g/mol, 14.3 g represents 0.1 mol of Cu2O. Based on the 1:3 ratio, this produces 0.3 mol of copper metal, corresponding to 19.05 g of Cu.

Step-by-Step Solution

1

Calculate the molar mass of copper(I) oxide

Molar mass of Cu2O = 143.0 g/mol

Required to convert the given mass of reactant to moles.

2

Determine moles of Cu2O reacted

Moles of Cu2O = 14.3 g / 143.0 g/mol = 0.10 mol

Establishes the quantitative amount of Cu2O in the chemical system.

3

Determine moles of copper metal produced using stoichiometry

Moles of Cu = 0.10 mol * (6 / 2) = 0.30 mol

The balanced chemical equation shows 2 moles of Cu2O yield 6 moles of Cu metal.

4

Calculate the mass of copper metal produced

Mass of Cu = 0.30 mol * 63.5 g/mol = 19.05 g

Converts the stoichiometric amount of product moles into grams.

Key Concept

Auto-reduction in copper extraction and stoichiometric mass calculations

Question 8467Question →

Consider the gaseous reversible reaction $2\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g)$ taking place in a closed container at a fixed temperature. Once the system reaches dynamic equilibrium, which of the following statements correctly describes the microscopic and macroscopic behavior of the reaction?

The rate of formation of

\text{N}_2\text{O}_4

equals its rate of decomposition, keeping all macroscopic properties such as pressure and intensity of color constant.

Introducing a positive catalyst increases the net yield of

\text{N}_2\text{O}_4

at equilibrium by accelerating only the forward reaction rate.

The concentrations of

\text{NO}_2

and

\text{N}_2\text{O}_4

must become equal to one another for the system to achieve dynamic equilibrium.

All chemical reactions cease completely at equilibrium, causing the total mass of the system to stabilize permanently.

Show answer & explanation

Answer: The rate of formation of $\text{N}_2\text{O}_4$ equals its rate of decomposition, keeping all macroscopic properties such as pressure and intensity of color constant.

Answer

The rate of formation of

\text{N}_2\text{O}_4

equals its rate of decomposition, keeping all macroscopic properties such as pressure and intensity of color constant.

In a closed system at dynamic equilibrium, the forward reaction rate equals the reverse reaction rate. As a result of this balance at the microscopic level, no net change occurs in macroscopic properties such as concentration, total gas pressure, or intensity of color.

Step-by-Step Solution

1

Analyze the microscopic nature of dynamic equilibrium in reversible chemical reactions.

Dynamic equilibrium implies that both forward (

2\text{NO}_2 \rightarrow \text{N}_2\text{O}_4

) and reverse (

\text{N}_2\text{O}_4 \rightarrow 2\text{NO}_2

) reactions occur concurrently at identical rates.

At the molecular level, reactants continue forming products while products revert to reactants at the same speed.

2

Connect microscopic rate equality to observable macroscopic properties.

Because the forward and reverse rates are equal, the net concentrations of species, total system pressure, and color intensity stay strictly constant.

When rate of formation equals rate of consumption, observable macroscopic properties do not change with time.

3

Evaluate common misconceptions regarding equilibrium state features.

Equilibrium does not require equal concentrations, nor does chemical activity stop, nor does a catalyst alter the equilibrium position.

Concentration equality is not required by

K_c

, chemical activity continues dynamically, and catalysts alter activation energy for both directions equally.

Key Concept

Microscopic vs. Macroscopic Characteristics of Dynamic Equilibrium

Estimated Time:1m 30s

Question 8468Question →

Match each transition metal complex ion on the left with its corresponding structural, electronic, and magnetic characteristics on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

[Fe(CN)_6]^{3-}

matches Octahedral geometry with a

3d^5

low-spin central metal ion containing 1 unpaired electron;

[Ni(CN)_4]^{2-}

matches Square planar geometry with a

3d^8

central metal ion that is diamagnetic;

[Co(NH_3)_6]^{3+}

matches Octahedral geometry with a

3d^6

low-spin central metal ion that is diamagnetic;

[Cu(NH_3)_4]^{2+}

matches Square planar geometry with a

3d^9

central metal ion containing 1 unpaired electron.

Each complex ion's central metal ion exhibits a specific oxidation state, electronic configuration, coordination geometry, and spin state based on crystal field theory and ligand field strength.

[Fe(CN)_6]^{3-}

features

Fe^{3+}

(

3d^5

) in a low-spin octahedral state with 1 unpaired electron.

[Ni(CN)_4]^{2-}

features

Ni^{2+}

(

3d^8

) in a square planar diamagnetic configuration.

[Co(NH_3)_6]^{3+}

features

Co^{3+}

(

3d^6

) in a low-spin octahedral diamagnetic state.

[Cu(NH_3)_4]^{2+}

features

Cu^{2+}

(

3d^9

) in a square planar configuration with 1 unpaired electron.

Step-by-Step Solution

1

Determine the oxidation state and d-electron count of the central metal ion in each complex ion.

For

[Fe(CN)_6]^{3-}

,

Fe^{3+}

is

3d^5

. For

[Ni(CN)_4]^{2-}

,

Ni^{2+}

is

3d^8

. For

[Co(NH_3)_6]^{3+}

,

Co^{3+}

is

3d^6

. For

[Cu(NH_3)_4]^{2+}

,

Cu^{2+}

is

3d^9

.

Ligand charges (

CN^-

=

-1

,

NH_3

=

0

) determine the oxidation state of the central transition metal ion.

2

Analyze ligand strength, coordination geometry, and crystal field splitting to determine magnetic character.

[Fe(CN)_6]^{3-}

is octahedral low-spin (

t_{2g}^5

, 1 unpaired

e^-

).

[Ni(CN)_4]^{2-}

is square planar (

dsp^2

, diamagnetic).

[Co(NH_3)_6]^{3+}

is octahedral low-spin (

t_{2g}^6

, diamagnetic).

[Cu(NH_3)_4]^{2+}

is square planar (

3d^9

, 1 unpaired

e^-

).

Strong-field ligands (

CN^-

,

NH_3

) induce electron pairing in low-spin octahedral or square planar configurations.

3

Match each complex ion to its complete set of physical, electronic, and magnetic properties.

Each complex correctly aligns with its unique d-electron configuration, geometry, and spin state.

Verifies all coordination parameters systematically.

Key Concept

Electronic Configuration, Oxidation State, Geometry, and Magnetic Properties of Transition Metal Complexes

Question 8469Question →

Arrange the following chemical substances in order of increasing boiling point, starting from the substance with the lowest boiling point to the one with the highest boiling point based on the nature and relative strength of their intermolecular forces.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order of increasing boiling point is Methane (

CH_4

) < Hydrogen sulfide (

H_2S

) < Ammonia (

NH_3

) < Water (

H_2O

).

The sequence reflects the increasing magnitude of intermolecular forces: non-polar Methane (

CH_4

) relies solely on weak London dispersion forces (lowest boiling point). Polar Hydrogen sulfide (

H_2S

) has dipole-dipole interactions but lacks hydrogen bonding because sulfur is not electronegative enough. Ammonia (

NH_3

) undergoes hydrogen bonding due to nitrogen's high electronegativity. Water (

H_2O

) forms an extensive network of strong hydrogen bonds, resulting in the highest boiling point.

Step-by-Step Solution

1

Identify the type of intermolecular forces operating in each substance

CH_4

is non-polar (London dispersion forces only);

H_2S

is polar (dipole-dipole and dispersion forces);

NH_3

is polar with hydrogen bonding;

H_2O

is polar with strong, extensive hydrogen bonding.

Boiling point depends directly on the total magnitude of attraction between molecules in the liquid state.

2

Compare non-hydrogen-bonding substances (

CH_4

vs

H_2S

)

CH_4

has the weakest intermolecular forces (dispersion only), while

H_2S

has additional permanent dipole-dipole attractions.

Permanent dipole-dipole interactions in polar molecules generally create stronger attraction than non-polar dispersion forces of comparable size.

3

Compare hydrogen-bonding substances (

NH_3

vs

H_2O

)

Both

NH_3

and

H_2O

form hydrogen bonds, but

H_2O

forms up to four hydrogen bonds per molecule in a 3D network, whereas

NH_3

is limited by its single lone pair to fewer hydrogen bonds per molecule.

Oxygen is more electronegative than nitrogen, and water has an optimal 1:1 ratio of lone pairs to hydrogen atoms for maximum hydrogen-bonding capacity.

4

Synthesize the complete sequence from lowest to highest boiling point

CH_4

<

H_2S

<

NH_3

<

H_2O

Intermolecular attraction strength increases in the order: London dispersion forces < dipole-dipole interactions < moderate hydrogen bonding < extensive hydrogen bonding.

Key Concept

Relative strengths of intermolecular forces (London dispersion, dipole-dipole, and hydrogen bonding) and their effect on physical properties like boiling point.

Estimated Time:2m 0s

Question 8470Question →

A sample of crude naphthalene contaminated with non-volatile inorganic salts is subjected to gentle heating under a cool watch glass. The naphthalene deposits on the underside of the watch glass as pure crystals without melting first. Which fundamental property of naphthalene enables this method of purification?

A relatively high vapor pressure as a solid, enabling direct transition from solid to vapor phase upon heating

A large difference in boiling points between the components when dissolved in a liquid medium

Thermal instability causing chemical decomposition into gaseous elements followed by recombination

A dramatic increase in aqueous solubility at elevated temperatures compared to room temperature

Show answer & explanation

Answer: A relatively high vapor pressure as a solid, enabling direct transition from solid to vapor phase upon heating

Answer

A relatively high vapor pressure as a solid, enabling direct transition from solid to vapor phase upon heating

Sublimation relies on the physical property of certain solids (like naphthalene, iodine, camphor, and ammonium chloride) to exert a high vapor pressure below their melting point, transitioning directly from solid to vapor phase upon heating.

Step-by-Step Solution

1

Identify the separation technique described in the scenario

The process described (solid converting directly to gas and re-depositing on a cool surface without passing through a liquid state) is sublimation.

Sublimation occurs when a substance undergoes a direct phase transition from solid to gas upon heating.

2

Determine the physical property responsible for sublimation

Sublimable solids possess weak intermolecular van der Waals forces and a high vapor pressure in the solid state.

When heat energy is applied, molecules overcome lattice forces and escape directly into the vapor phase without forming a liquid.

Key Concept

Sublimation physical principle

Estimated Time:1m 0s

Question 8471Question →

Complete the statement by calculating the required volume of gas at STP and the mass of metal produced in the following reduction reaction.

Fill in the blanks below

When

16.0\text{ g}

of iron(III) oxide (

Fe_2O_3

) is completely reduced by carbon monoxide gas according to the equation:

Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)

the volume of carbon monoxide gas consumed at STP is , and the mass of iron metal produced is .

[Relative atomic masses:

\text{Fe} = 56

,

\text{O} = 16

; Molar volume of gas at STP

= 22.4\text{ dm}^3\text{ mol}^{-1}

]

Show answer & explanation

Answer

The volume of carbon monoxide consumed at STP is

6.72\text{ dm}^3

and the mass of iron metal produced is

11.2\text{ g}

.

Based on the stoichiometry of the reaction

Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)

,

1\text{ mol}

(

160\text{ g}

) of

Fe_2O_3

reacts with

3\text{ mol}

(

0.30\text{ mol}

for

16.0\text{ g}

) of

CO

gas, yielding a volume of

6.72\text{ dm}^3

at STP, and produces

2\text{ mol}

(

0.20\text{ mol}

for

16.0\text{ g}

) of

Fe

, corresponding to

11.2\text{ g}

.

Step-by-Step Solution

1

Calculate the molar mass of

Fe_2O_3

and determine the number of moles of

Fe_2O_3

present.

Molar mass of

Fe_2O_3 = (2 \times 56) + (3 \times 16) = 160\text{ g mol}^{-1}

. Moles of

Fe_2O_3 = \frac{16.0\text{ g}}{160\text{ g mol}^{-1}} = 0.10\text{ mol}

.

Converting given mass to moles is required for stoichiometric mole-ratio calculations.

2

Use the mole ratio from the balanced equation to find the moles and volume of

CO(g)

required at STP.

From the balanced equation,

1\text{ mol } Fe_2O_3

reacts with

3\text{ mol } CO

.
Moles of

CO = 3 \times 0.10\text{ mol} = 0.30\text{ mol}

.
Volume of

CO

at STP

= 0.30\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 6.72\text{ dm}^3

.

Gas volume at STP is obtained by multiplying moles of gas by the molar volume (

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

).

3

Use the mole ratio from the balanced equation to calculate the mass of iron (

Fe

) produced.

From the equation,

1\text{ mol } Fe_2O_3

produces

2\text{ mol } Fe

.
Moles of

Fe = 2 \times 0.10\text{ mol} = 0.20\text{ mol}

.
Mass of

Fe = 0.20\text{ mol} \times 56\text{ g mol}^{-1} = 11.2\text{ g}

.

Mass of product is calculated by multiplying its moles by its relative atomic mass.

Key Concept

Mass-Mass and Mass-Volume Stoichiometric Calculations

Question 8472Question →

During a paper chromatography experiment to analyze a natural dye extract, a component spot migrated $4.8\text{ cm}$ from the origin line, while the solvent front traveled a total distance of $12.0\text{ cm}$ from the origin line. What is the retention factor ( $R_f$ ) of this component?

0.40

2.50

0.60

0.80

Show answer & explanation

Answer: $0.40$

Answer

The retention factor (

R_f

) of the component is

0.40

.

The retention factor (

R_f

) is dimensionless and measures the relative mobility of a solute compared to the solvent front. It is calculated using the expression

R_f = \frac{\text{distance moved by solute}}{\text{distance moved by solvent front}}

. Substituting

4.8\text{ cm}

and

12.0\text{ cm}

yields

0.40

.

Step-by-Step Solution

1

Identify the given values from the chromatographic analysis.

Distance moved by solute spot (

d_{\text{solute}}) = 4.8\text{ cm}

; Distance moved by solvent front (

d_{\text{solvent}}) = 12.0\text{ cm}

.

These measurements are required to calculate the retention factor.

2

Apply the retention factor (

R_f

) formula.

R_f = \frac{d_{\text{solute}}}{d_{\text{solvent}}} = \frac{4.8\text{ cm}}{12.0\text{ cm}} = 0.40

.

The retention factor is defined as the ratio of solute displacement to solvent displacement relative to the origin line.

Key Concept

Retention factor (

R_f

) calculation in paper chromatography

Question 8473Question →

In the van der Waals equation of state for real gases, $\left(P + \frac{a n^2}{V^2}\right)(V - n b) = n R T$ , which physical property of real gas molecules is corrected by the parameter $a$ ?

Intermolecular attractive forces between the gas molecules

The finite volume occupied by the gas molecules themselves

The kinetic energy lost during inelastic molecular collisions

The increase in molecular speed under elevated pressure

Show answer & explanation

Answer: Intermolecular attractive forces between the gas molecules

Answer

The parameter

a

in the van der Waals equation accounts for the intermolecular attractive forces between gas molecules.

In ideal gas theory, intermolecular forces are assumed to be non-existent. However, real gas molecules exert weak attractive forces on one another. These cohesive forces pull inward on molecules approaching the container wall, reducing the measured pressure. The term

\frac{a n^2}{V^2}

is added to the measured pressure to account for these intermolecular attractive forces, where

a

is a characteristic constant for the specific gas.

Step-by-Step Solution

1

Analyze the kinetic theory assumption vs. real gas behavior

Ideal gases assume zero intermolecular forces, but real gas molecules attract each other.

Intermolecular attraction reduces the force of impact of molecules against the container walls, resulting in a lower real pressure than ideal pressure.

2

Identify the role of term

\frac{a n^2}{V^2}

in the van der Waals equation

The pressure correction term

P_{\text{ideal}} = P_{\text{real}} + \frac{a n^2}{V^2}

adds a pressure factor proportional to intermolecular attraction (

a

).

The constant

a

specifically measures the strength of attraction between molecules of a given real gas.

Key Concept

Van der Waals Equation and Corrections for Real Gases

Estimated Time:1m 0s

Question 8474Question →

The solubility product ( $K_{sp}$ ) of lead(II) chloride ( $\text{PbCl}_2$ ) at $25^\circ\text{C}$ is $3.2 \times 10^{-5}\text{ mol}^3\text{ dm}^{-9}$ . What is the concentration of chloride ions ( $\text{Cl}^-$ ) in $\text{mol dm}^{-3}$ in a saturated solution of lead(II) chloride at $25^\circ\text{C}$ ?

2.0 \times 10^{-2}\text{ mol dm}^{-3}

4.0 \times 10^{-2}\text{ mol dm}^{-3}

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

3.17 \times 10^{-2}\text{ mol dm}^{-3}

Show answer & explanation

Answer: $4.0 \times 10^{-2}\text{ mol dm}^{-3}$

Answer

The concentration of chloride ions in the saturated solution is

4.0 \times 10^{-2}\text{ mol dm}^{-3}

.

For the dissolution equilibrium

\text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{Cl}^-(aq)

, the solubility product expression is

K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2

. Letting

s

equal the molar solubility of

\text{PbCl}_2

,

[\text{Pb}^{2+}] = s

and

[\text{Cl}^-] = 2s

. Substituting into

K_{sp}

gives

K_{sp} = s(2s)^2 = 4s^3

. Given

K_{sp} = 3.2 \times 10^{-5}

, solving

4s^3 = 3.2 \times 10^{-5}

yields

s^3 = 8.0 \times 10^{-6}

and

s = 2.0 \times 10^{-2}\text{ mol dm}^{-3}

. The chloride ion concentration is

[\text{Cl}^-] = 2s = 4.0 \times 10^{-2}\text{ mol dm}^{-3}

.

Step-by-Step Solution

1

Write the solubility equilibrium equation and express

K_{sp}

in terms of molar solubility (

s

).

\text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{Cl}^-(aq)

, so

K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 = s(2s)^2 = 4s^3

.

Dissolution of one mole of

\text{PbCl}_2

produces one mole of

\text{Pb}^{2+}

ions and two moles of

\text{Cl}^-

ions.

2

Substitute the given

K_{sp}

value and calculate the molar solubility (

s

).

3.2 \times 10^{-5} = 4s^3 \implies s^3 = 8.0 \times 10^{-6} \implies s = 2.0 \times 10^{-2}\text{ mol dm}^{-3}

.

Dividing

K_{sp}

by 4 gives

s^3

, and taking the cube root yields

s

.

3

Determine the concentration of chloride ions (

[\text{Cl}^-]

).

[\text{Cl}^-] = 2s = 2 \times (2.0 \times 10^{-2}\text{ mol dm}^{-3}) = 4.0 \times 10^{-2}\text{ mol dm}^{-3}

.

Since two moles of chloride ions are released per mole of salt dissolved,

[\text{Cl}^-]

equals

2s

.

Key Concept

Solubility product (

K_{sp}

) calculation for

MX_2

type salts and stoichiometric determination of ion concentrations.

Question 8475Question →

Match each aluminium alloy or chemical substance in Column I with its primary industrial composition or application in Column II.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Duralumin matches with High-strength alloy of aluminium, copper, magnesium, and manganese used in aircraft structural frames; Magnalium matches with Lightweight corrosion-resistant alloy of aluminium and magnesium used in balance beams and aircraft parts; Alnico matches with Alloy of aluminium, nickel, cobalt, and iron utilized in manufacturing strong permanent magnets; Molten Cryolite matches with Molten solvent added to lower the operating temperature of alumina and improve electrical conductivity.

Each substance is correctly paired based on standard chemistry principles: Duralumin is an

Al-Cu-Mg-Mn

structural aircraft alloy, Magnalium is an

Al-Mg

lightweight alloy, Alnico is an

Al-Ni-Co-Fe

magnetic alloy, and molten cryolite (

Na_3AlF_6

) lowers the melting temperature of alumina during electrolysis.

Step-by-Step Solution

1

Identify the chemical composition and primary application of Duralumin.

Duralumin contains

Al

,

Cu

,

Mg

, and

Mn

, known for its structural strength in aircraft manufacture.

Copper adds tensile strength to aluminium while retaining low density.

2

Identify the composition and application of Magnalium.

Magnalium is an alloy of

Al

and

Mg

, valued for low density and high corrosion resistance.

Magnesium enhances hardness and lightness without increasing susceptibility to oxidation.

3

Determine the composition and use of Alnico.

Alnico consists of

Al

,

Ni

,

Co

, and

Fe

, used for permanent magnets.

Ferromagnetic elements combined with aluminium create high magnetic retentivity.

4

Determine the role of cryolite in the industrial extraction of aluminium.

Cryolite acts as an electrolytic solvent, lowering the melting point of

Al_2O_3

and increasing conductivity.

Pure alumina has an extremely high melting point (

2050^\circ\text{C}

); dissolving it in molten cryolite reduces energy consumption.

Key Concept

Industrial extraction of aluminium and compositions/applications of its major alloys

Question 8476Question →

In chemical kinetics and energetics, potential energy profile diagrams illustrate key energy values along a reaction pathway. Match each parameter of an energy profile diagram on the left with its correct chemical definition or significance on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Activation energy of forward reaction matches the energy difference between the activated complex peak and reactants. Transition state matches the high-energy, unstable species formed at the maximum potential energy point. Enthalpy change of reaction matches the difference between product potential energy and reactant potential energy. Action of a positive catalyst matches providing an alternate reaction path with a lower peak activation barrier.

The parameters are matched according to standard kinetic definitions: forward activation energy measures the height of the energy barrier from the reactants; the transition state is the highest energy activated complex; enthalpy change is the net potential energy difference between products and reactants; and a catalyst lowers the activation energy barrier.

Step-by-Step Solution

1

Determine the physical meaning of forward activation energy (

E_{a,\text{f}}

).

It is the minimum energy required to boost reactants to the transition state peak.

Reactant molecules must overcome this energy barrier to undergo effective collisions.

2

Identify the feature located at the highest point of the energy profile diagram.

The apex represents the transition state (activated complex).

This configuration has maximum potential energy and minimum stability along the reaction coordinate.

3

Relate total enthalpy change (

\Delta H

) to potential energy values on the diagram.

\Delta H

equals the potential energy of products minus potential energy of reactants.

Enthalpy change depends purely on the net difference between final and initial energy states.

4

Analyze how addition of a positive catalyst affects energy parameters.

It lowers the peak height by introducing a new reaction pathway.

Catalysts alter the mechanism and lower activation energy without changing initial or final energy levels.

Key Concept

Activation Energy and Energy Profile Diagrams

Question 8477Question →

In the Contact Process for the industrial manufacture of tetraoxosulfate(VI) acid, sulfur(IV) oxide gas reacts with oxygen gas according to the equation: $2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)$ . What volume of oxygen gas, measured in $\text{dm}^3$ at stp, is required to completely react with $56\text{ dm}^3$ of $SO_2$ gas at stp?

Show answer & explanation

Answer: 28

Answer

The volume of oxygen gas required at stp is 28 dm³.

According to the balanced chemical equation

2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)

, 2 volumes of

SO_2

gas require 1 volume of

O_2

gas for complete oxidation. Therefore,

56\text{ dm}^3

of

SO_2

requires half its volume in oxygen, which equals

28\text{ dm}^3

.

Step-by-Step Solution

1

Determine the stoichiometric ratio between

SO_2

and

O_2

from the balanced equation.

2 volumes of

SO_2(g)

react with 1 volume of

O_2(g)

.

By Gay-Lussac's Law of Combining Volumes, gases react in simple whole-number volume ratios under the same conditions of temperature and pressure.

2

Compute the required volume of

O_2

gas for

56\text{ dm}^3

of

SO_2

.

\text{Volume of } O_2 = \frac{56\text{ dm}^3}{2} = 28\text{ dm}^3

.

Since the ratio of

SO_2

to

O_2

is

2:1

, the volume of oxygen gas required is half the volume of sulfur(IV) oxide gas.

Key Concept

Stoichiometric Volume Calculations in Gas Reactions (Contact Process)

Question 8478Question →

Match each mixture separation requirement on the left with the most appropriate physical technique on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Separating insoluble calcium carbonate from water pairs with Filtration; recovering thermally stable sodium chloride pairs with Evaporation to dryness; obtaining pure hydrated copper(II) tetraoxosulfate(VI) crystals pairs with Crystallization.

Filtration is used for separating insoluble solids from liquids because the solid residue remains on the filter medium. Evaporation to dryness is appropriate for thermally stable soluble solids where heating does not cause chemical decomposition. Crystallization is necessary for thermally unstable or hydrated salts to preserve their crystalline structure and water of crystallization upon controlled cooling of a saturated solution.

Step-by-Step Solution

1

Analyze the solubility and physical state of the component in 'Separating insoluble calcium carbonate particles suspended in water'.

Calcium carbonate (

\text{CaCO}_3

) is an insoluble solid suspended in liquid water. The appropriate method to separate an insoluble solid from a liquid is filtration.

Filtration relies on particle size differences where the insoluble residue is trapped on filter paper while the liquid filtrate passes through.

2

Analyze the thermal stability of 'Recovering thermally stable anhydrous sodium chloride from an aqueous salt solution'.

Sodium chloride (

\text{NaCl}

) is a soluble, heat-stable solid that does not decompose upon heating. Evaporating to dryness removes all water leaving dry salt.

Evaporation to dryness is suitable when the solute does not decompose at high temperatures and water of crystallization is not required.

3

Analyze the composition of 'Obtaining pure hydrated copper(II) tetraoxosulfate(VI) crystals from an aqueous solution'.

Copper(II) tetraoxosulfate(VI) pentahydrate (

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

) contains water of crystallization and decomposes/dehydrates if heated to dryness. It requires gentle heating to form a saturated solution, followed by slow cooling.

Crystallization preserves the hydration structure and purity of thermally sensitive hydrated salts.

Key Concept

Selection of physical separation methods based on solid solubility, liquid interaction, and thermal stability of hydrated salts.

Question 8479Question →

Alloys are frequently used in structural engineering and manufacturing because they exhibit greater mechanical strength and hardness than pure metals. Which of the following structural factors best explains why introducing a secondary element increases the hardness of a metal?

Atoms of differing sizes disrupt the regular crystal lattice, hindering the movement of atomic layers past one another.

The constituent elements undergo a chemical reaction to form rigid giant covalent networks throughout the solid.

Delocalized valence electrons are completely removed, converting the metallic lattice into an ionic crystal structure.

The added atoms expand in volume to completely fill all interatomic spaces and stop thermal vibrations.

Show answer & explanation

Answer: Atoms of differing sizes disrupt the regular crystal lattice, hindering the movement of atomic layers past one another.

Answer

Atoms of differing sizes disrupt the regular crystal lattice, hindering the movement of atomic layers past one another.

In a pure metal, identical atoms form regular layers that slip over one another with relative ease. When a secondary element with a different atomic radius is introduced to form an alloy, the regular lattice arrangement is distorted. This disruption prevents atomic layers from sliding smoothly, significantly increasing the hardness and tensile strength of the alloy.

Step-by-Step Solution

1

Consider the structure of a pure metal.

In pure metals, identical atoms are arranged in uniform, orderly layers that easily slide over each other under shear stress, making the metal malleable and relatively soft.

Uniform atomic radii allow smooth plane slipping along lattice planes.

2

Analyze the structural effect of alloying.

Adding atoms of a different size introduces structural irregularities and distorts the host crystal lattice.

The distortion creates friction and resistance against the sliding of atomic layers, resulting in enhanced hardness and strength.

Key Concept

Lattice distortion and mechanical strengthening of alloys

Estimated Time:45s

Question 8480Question →

During the municipal treatment of river water containing dissolved iron(II) salts and unpleasant odors caused by dissolved hydrogen sulfide gas, raw water is initially sprayed into the air during an aeration process. What is the primary chemical purpose of this aeration step prior to coagulation?

To oxidize soluble iron(II) compounds into insoluble iron(III) hydroxide while expelling volatile dissolved gases

To destroy pathogenic bacteria and harmful microorganisms present in the raw water supply

To precipitate calcium and magnesium hydrogen trioxocarbonates(IV) causing temporary water hardness

To cause fine colloidal clay suspensions to aggregate into larger settleable flocs

Show answer & explanation

Answer: To oxidize soluble iron(II) compounds into insoluble iron(III) hydroxide while expelling volatile dissolved gases

Answer

Aeration oxidizes soluble iron(II) compounds into insoluble iron(III) hydroxide and expels volatile dissolved gases from the water.

Aeration introduces atmospheric oxygen into raw water, which oxidizes soluble iron(II) compounds into an insoluble iron(III) hydroxide precipitate (

Fe(OH)_3

). Additionally, the splashing action strips out volatile gases responsible for bad tastes and odors, such as hydrogen sulfide (

H_2S

).

Step-by-Step Solution

1

Identify the chemical changes occurring when air contacts raw water during aeration.

Dissolved atmospheric oxygen reacts with dissolved

Fe^{2+}

ions, while volatile gases like

H_2S

escape into the atmosphere.

Aeration increases dissolved oxygen concentration and promotes liquid-gas equilibrium exchange.

2

Determine the resulting chemical products formed.

Soluble iron(II) ions form insoluble iron(III) hydroxide,

Fe(OH)_3

, precipitate.

Insoluble precipitates formed during aeration can subsequently be settled and removed during sedimentation and filtration.

Key Concept

Aeration in Municipal Water Purification

Estimated Time:1m 0s

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept