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Question 8361Question →

Match each chemical species or substance involved in water chemistry with its primary role or behavior regarding water hardness.

Click a left item, then click its matching right item

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Answer

Magnesium hydrogencarbonate causes temporary hardness that decomposes on boiling; magnesium tetraoxosulfate(VI) causes permanent hardness unaffected by boiling; sodium trioxocarbonate(IV) removes all hardness via chemical precipitation; sodium permutit removes all hardness via ion exchange.

Magnesium hydrogencarbonate is a soluble hydrogencarbonate salt causing temporary hardness removed by boiling. Magnesium tetraoxosulfate(VI) is a soluble sulfate salt causing permanent hardness that cannot be removed by boiling. Sodium trioxocarbonate(IV) removes all hardness types by precipitating calcium/magnesium ions as insoluble carbonates. Permutit softened water by exchanging hardness cations for soluble sodium ions.

Step-by-Step Solution

1

Identify the cause of temporary hardness.

Magnesium hydrogencarbonate,

\text{Mg(HCO}_3)_2

, dissolves in water to cause temporary hardness. Upon boiling, hydrogencarbonate ions decompose to insoluble carbonate precipitates:

\text{Mg(HCO}_3)_2(aq) \rightarrow \text{MgCO}_3(s) + \text{H}_2\text{O}(l) + \text{CO}_2(g)

.

Temporary hardness is specifically caused by hydrogen carbonate salts of calcium and magnesium, which are thermally unstable.

2

Identify the cause of permanent hardness.

Magnesium tetraoxosulfate(VI),

\text{MgSO}_4

, causes permanent hardness because sulfate salts do not undergo thermal decomposition upon boiling.

Permanent hardness is caused by soluble tetraoxosulfate(VI) or chloride salts of calcium and magnesium.

3

Analyze the action of washing soda (

\text{Na}_2\text{CO}_3

).

Adding washing soda introduces

\text{CO}_3^{2-}

ions, which combine with dissolved

\text{Mg}^{2+}

or

\text{Ca}^{2+}

ions to precipitate them as

\text{MgCO}_3

or

\text{CaCO}_3

.

Precipitation of divalent metallic cations as insoluble carbonates removes both temporary and permanent hardness.

4

Analyze the action of sodium permutit (zeolite).

Permutit acts as an ion exchanger:

\text{Na}_2\text{Z}(s) + \text{Mg}^{2+}(aq) \rightarrow \text{MgZ}(s) + 2\text{Na}^+(aq)

.

Hardness-causing divalent cations are bound to the zeolite matrix while harmless sodium ions are released into solution.

Key Concept

Classification of water hardness causes (hydrogencarbonate vs sulfate salts) and chemical removal mechanisms (boiling, precipitation, ion exchange).

Question 8362Question →

A gas mixture containing $0.20\text{ mol}$ of oxygen ( $\text{O}_2$ ) and $0.30\text{ mol}$ of nitrogen ( $\text{N}_2$ ) is collected over water at $27^\circ\text{C}$ . The total pressure of the moist gas mixture is $775\text{ mmHg}$ and the saturated vapor pressure of water at $27^\circ\text{C}$ is $25\text{ mmHg}$ . If $0.50\text{ mol}$ of dry argon ( $\text{Ar}$ ) gas is subsequently added to the mixture at the same temperature while maintaining the total system pressure at $775\text{ mmHg}$ , what is the final partial pressure of nitrogen gas in the moist mixture in $\text{mmHg}$ ?

Show answer & explanation

Answer: 225

Answer

The final partial pressure of nitrogen gas in the moist mixture is

225\text{ mmHg}

.

When a gas is collected over water, the total pressure measured includes the saturated vapor pressure of water (

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

). Subtracting

P_{\text{H}_2\text{O}} = 25\text{ mmHg}

from

P_{\text{total}} = 775\text{ mmHg}

yields the total pressure of the dry gases (

P_{\text{dry}} = 750\text{ mmHg}

). Adding

0.50\text{ mol}

of argon increases the total dry gas quantity to

1.00\text{ mol}

(

0.20 + 0.30 + 0.50

). The mole fraction of nitrogen in this dry mixture is

\frac{0.30}{1.00} = 0.30

. Finally, multiplying this mole fraction by

P_{\text{dry}}

gives the partial pressure of nitrogen as

0.30 \times 750\text{ mmHg} = 225\text{ mmHg}

.

Step-by-Step Solution

1

Calculate the total pressure exerted by the dry gas mixture.

P_{\text{dry}} = 750\text{ mmHg}

According to Dalton's Law of Partial Pressures, the total pressure of a gas collected over water is

P_{\text{total}} = P_{\text{dry}} + P_{\text{H}_2\text{O}}

. Therefore,

P_{\text{dry}} = 775\text{ mmHg} - 25\text{ mmHg} = 750\text{ mmHg}

.

2

Calculate total moles of dry gas after argon addition.

n_{\text{total, dry}} = 1.00\text{ mol}

Sum the moles of oxygen (

0.20\text{ mol}

), nitrogen (

0.30\text{ mol}

), and added argon (

0.50\text{ mol}

):

0.20 + 0.30 + 0.50 = 1.00\text{ mol}

.

3

Determine the mole fraction of nitrogen gas in the dry mixture.

\chi_{\text{N}_2} = 0.30

The mole fraction is the ratio of the moles of nitrogen to the total moles of dry gas:

\frac{0.30\text{ mol}}{1.00\text{ mol}} = 0.30

.

4

Calculate the partial pressure of nitrogen gas.

P_{\text{N}_2} = 225\text{ mmHg}

The partial pressure of an individual gas component in a dry mixture is given by

P_{\text{N}_2} = \chi_{\text{N}_2} \times P_{\text{dry}} = 0.30 \times 750\text{ mmHg} = 225\text{ mmHg}

.

Key Concept

Dalton's Law of Partial Pressures with Vapor Pressure Correction and Mole Fraction Calculation

Question 8363Question →

Which of the following changes will increase the rate of reaction between dilute hydrochloric acid and zinc dust primarily by increasing the number of reactant particles per unit volume, thereby increasing the frequency of effective collisions without altering the average kinetic energy of the molecules?

Increasing the concentration of the hydrochloric acid solution

Increasing the temperature of the reaction mixture

Adding a catalyst that shifts the equilibrium position toward the products

Replacing zinc dust with larger zinc granules of equal mass

Show answer & explanation

Answer: Increasing the concentration of the hydrochloric acid solution

Answer

Increasing the concentration of the hydrochloric acid solution

Increasing the concentration of hydrochloric acid increases the number of hydrogen ions (

H^+

) present per unit volume of solution. This increases the frequency of collisions between reactant particles without affecting their average kinetic energy or altering the activation energy of the reaction pathway.

Step-by-Step Solution

1

Analyze how concentration influences particle collision mechanics according to collision theory.

A higher concentration contains a greater number of solute particles (

H^+

ions) per unit volume.

With more reactant particles crowded in the same volume, collision frequency increases.

2

Evaluate the energy condition specified in the stem.

Concentration changes do not alter particle speed or average kinetic energy, fulfilling the exact requirement.

Average kinetic energy is strictly dependent on temperature.

Key Concept

Effect of concentration on reaction rate via collision frequency

Question 8364Question →

When potassium sulfite ( $K_2SO_3$ ) is dissolved in distilled water at $25^\circ\text{C}$ , the solution turns red litmus paper blue. Which of the following net ionic equations correctly represents the hydrolysis reaction responsible for the basic nature of this solution?

SO_3^{2-}(aq) + H_2O(l) \rightleftharpoons HSO_3^-(aq) + OH^-(aq)

K^+(aq) + H_2O(l) \rightleftharpoons KOH(aq) + H^+(aq)

SO_3^{2-}(aq) + 2H_2O(l) \rightleftharpoons H_2SO_3(aq) + 2OH^-(aq)

2K^+(aq) + SO_3^{2-}(aq) + H_2O(l) \rightleftharpoons 2KOH(aq) + SO_2(g)

Show answer & explanation

Answer: $SO_3^{2-}(aq) + H_2O(l) \rightleftharpoons HSO_3^-(aq) + OH^-(aq)$

Answer

The net ionic equation representing the hydrolysis of potassium sulfite is

SO_3^{2-}(aq) + H_2O(l) \rightleftharpoons HSO_3^-(aq) + OH^-(aq)

, which generates hydroxide ions and causes the solution to be alkaline.

Potassium sulfite (

K_2SO_3

) dissociates in water into

K^+

and

SO_3^{2-}

ions. Because

KOH

is a strong base, the

K^+

ion does not react with water. However, sulfurous acid (

H_2SO_3

) is a weak acid, meaning its conjugate anion (

SO_3^{2-}

) acts as a Bronsted-Lowry base. It hydrolyzes by removing a proton from water to form

HSO_3^-

and

OH^-

. The excess hydroxide ions render the solution alkaline, turning red litmus paper blue.

Step-by-Step Solution

1

Identify the parent acid and parent base of the salt

K_2SO_3

.

Potassium sulfite (

K_2SO_3

) is formed from a strong base (

KOH

) and a weak diprotic acid (

H_2SO_3

).

Determining the strength of parent species dictates which ion undergoes hydrolysis in aqueous solution.

2

Determine which constituent ion undergoes hydrolysis in water.

Potassium ions (

K^+

) do not hydrolyze because they are cations of a strong base. Sulfite ions (

SO_3^{2-}

), being conjugate bases of a weak acid, undergo anion hydrolysis.

Conjugate bases of weak acids are sufficiently strong to abstract protons from water molecules.

3

Write the net ionic proton-transfer equation for anion hydrolysis.

SO_3^{2-}(aq) + H_2O(l) \rightleftharpoons HSO_3^-(aq) + OH^-(aq)

The sulfite anion abstracts one proton (

H^+

) from water, yielding hydrogen sulfite (

HSO_3^-

) and releasing a free hydroxide ion (

OH^-

), which increases solution pH above 7.

Key Concept

Salt Hydrolysis of Weak Acid-Strong Base Salts

Estimated Time:1m 30s

Question 8365Question →

Atmospheric pollutants contribute to environmental degradation through distinct chemical and physical processes. Which of the following statements correctly contrasts the mechanism of carbon(IV) oxide ( $\text{CO}_2$ ) in global warming with that of chlorofluorocarbons ( $\text{CFCs}$ ) in ozone layer depletion?

Carbon(IV) oxide absorbs re-radiated terrestrial infrared radiation in the troposphere, whereas chlorofluorocarbons undergo photolysis in the stratosphere to release free radicals that catalytically destroy ozone.

Carbon(IV) oxide decomposes under solar ultraviolet radiation in the stratosphere to destroy ozone, whereas chlorofluorocarbons absorb infrared radiation exclusively in the thermosphere.

Carbon(IV) oxide acts as a homogeneous catalyst for ozone destruction in the troposphere, whereas chlorofluorocarbons reflect incoming solar radiation back into space.

Carbon(IV) oxide traps high-energy ultraviolet radiation near the Earth's surface, whereas chlorofluorocarbons absorb outgoing terrestrial infrared radiation to cause stratospheric cooling.

Show answer & explanation

Answer: Carbon(IV) oxide absorbs re-radiated terrestrial infrared radiation in the troposphere, whereas chlorofluorocarbons undergo photolysis in the stratosphere to release free radicals that catalytically destroy ozone.

Answer

Carbon(IV) oxide absorbs re-radiated terrestrial infrared radiation in the troposphere, whereas chlorofluorocarbons undergo photolysis in the stratosphere to release free radicals that catalytically destroy ozone.

The statement accurately distinguishes the physical warming mechanism in the troposphere (absorption of outgoing long-wave thermal infrared radiation by carbon(IV) oxide) from the photochemical catalytic mechanism in the stratosphere (solar UV photolysis of chlorofluorocarbons yielding free chlorine radicals that decompose ozone).

Step-by-Step Solution

1

Analyze the atmospheric role and mechanism of carbon(IV) oxide (CO₂)

CO₂ is a major greenhouse gas located mainly in the troposphere. It allows short-wave solar radiation to pass through but absorbs longer-wave terrestrial infrared (heat) radiation, preventing its loss into space and driving global warming.

Understanding the physical interaction of greenhouse gases with thermal radiation clarifies how global warming occurs.

2

Analyze the atmospheric role and photolysis mechanism of chlorofluorocarbons (CFCs)

CFCs diffuse unchanged into the stratosphere, where high-energy solar UV radiation photolyzes C-Cl bonds to generate reactive chlorine free radicals (Cl•).

UV radiation provides the activation energy needed for homolytic cleavage of chlorofluorocarbons in the upper atmosphere.

3

Examine the catalytic ozone destruction cycle caused by chlorine radicals

The chlorine radical reacts with stratospheric ozone: Cl• + O₃ → ClO• + O₂ followed by ClO• + O• → Cl• + O₂, regenerating the chlorine radical for further destruction.

A single chlorine radical can destroy thousands of ozone molecules via a catalytic propagation cycle.

4

Synthesize the contrast between the two atmospheric phenomena

Global warming involves tropospheric trapping of thermal infrared radiation by gases like CO₂, whereas ozone depletion involves stratospheric catalytic destruction of O₃ by chlorine radicals produced via photolysis of CFCs.

Distinguishing between tropospheric thermal trapping and stratospheric photochemical radical reactions avoids misattributing pollutant functions.

Key Concept

Greenhouse Effect vs Stratospheric Ozone Depletion Mechanisms

Question 8366Question →

A laboratory mixture contains soluble potassium trioxonitrate(V) ( $\text{KNO}_3$ ) contaminated with insoluble clay particles. What is the correct sequential order of operational steps required to obtain pure, dry crystals of $\text{KNO}_3$ from this mixture?

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Answer

The correct sequential order of operational steps is: First, dissolve the impure mixture in warm distilled water with continuous stirring. Second, filter the mixture using filter paper and a glass funnel. Third, heat the clear filtrate gently in an evaporating dish until it reaches its saturation point. Fourth, allow the hot saturated filtrate to cool down slowly, then filter and dry the formed crystals between filter papers.

To separate and purify a soluble salt mixed with an insoluble solid, dissolution must occur first so that the salt enters the liquid phase while the insoluble component remains solid. Filtration then isolates the solid residue from the dissolved salt filtrate. The filtrate is heated only until it becomes saturated; subsequent slow cooling lowers solute solubility and allows pure crystals to precipitate cleanly without thermal degradation.

Step-by-Step Solution

1

Dissolve the mixture in distilled water.

A aqueous mixture containing dissolved

\text{KNO}_3

and suspended insoluble clay is formed.

Solubility differences allow the soluble salt to dissolve while insoluble impurities remain solid.

2

Filter the mixture through filter paper.

Insoluble clay is retained on the paper as residue, yielding clear aqueous

\text{KNO}_3

filtrate.

Filtration separates undissolved solid matter from liquid solution based on particle size.

3

Evaporate water until the solution is saturated.

A hot saturated solution of

\text{KNO}_3

is produced.

Evaporating solvent increases solute concentration until the crystallization point is achieved.

4

Cool the saturated solution slowly and dry the resulting precipitate.

Pure

\text{KNO}_3

crystals form and are isolated dry.

Solubility decreases upon cooling, causing the dissolved salt to crystallize out of solution cleanly.

Key Concept

Sequential purification of soluble salts from insoluble solids using dissolution, filtration, concentration to saturation, and crystallization.

Estimated Time:1m 30s

Question 8367Question →

A sample of argon gas occupies a volume of $400\text{ cm}^3$ at a temperature of $-23^\circ\text{C}$ . If the pressure remains constant and the gas expands to a volume of $600\text{ cm}^3$ , what is its final temperature in degrees Celsius ( $^\circ\text{C}$ )?

Show answer & explanation

Answer: 102

Answer

102 °C

Converting the initial temperature to Kelvin gives

T_1 = -23 + 273 = 250\text{ K}

. Applying Charles's Law

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

gives

T_2 = \frac{600 \times 250}{400} = 375\text{ K}

. Subtracting

273

converts this back to degrees Celsius:

375 - 273 = 102^\circ\text{C}

.

Step-by-Step Solution

1

Convert initial temperature from Celsius to Kelvin

T_1 = 250\text{ K}

Gas laws require temperature to be expressed on the absolute (Kelvin) scale.

2

Calculate final absolute temperature using Charles's Law

T_2 = \frac{600\text{ cm}^3 \times 250\text{ K}}{400\text{ cm}^3} = 375\text{ K}

At constant pressure, volume is directly proportional to absolute temperature (

V_1/T_1 = V_2/T_2

).

3

Convert final temperature back to degrees Celsius

t_2 = 375 - 273 = 102^\circ\text{C}

The question explicitly requests the final answer in degrees Celsius.

Key Concept

Charles's Law states that the volume of a fixed mass of gas is directly proportional to its absolute temperature at constant pressure (

V \propto T

). Temperatures must always be converted to Kelvin (

K = ^\circ\text{C} + 273

) prior to calculation.

Question 8368Question →

Match each chemical species with its corresponding central atom hybridization state and molecular geometry as predicted by Valence Shell Electron Pair Repulsion (VSEPR) theory.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Chlorate ion (

ClO_3^-

) matches

sp^3

hybridization with a trigonal pyramidal shape; Xenon difluoride (

XeF_2

) matches

sp^3d

hybridization with a linear shape; Tetrachloroiodate ion (

ICl_4^-

) matches

sp^3d^2

hybridization with a square planar shape; Sulfur dioxide (

SO_2

) matches

sp^2

hybridization with a bent shape.

Each chemical species is accurately paired by identifying the total number of electron domains around its central atom:

ClO_3^-

has 4 domains (

sp^3

, trigonal pyramidal shape),

XeF_2

has 5 domains (

sp^3d

, linear shape),

ICl_4^-

has 6 domains (

sp^3d^2

, square planar shape), and

SO_2

has 3 domains (

sp^2

, bent shape).

Step-by-Step Solution

1

Determine the valence electron count and steric number (bonding domains + lone pairs) for the central atom of each species.

Chlorate ion (

ClO_3^-

): 3 bonds + 1 lone pair = steric number 4; Xenon difluoride (

XeF_2

): 2 bonds + 3 lone pairs = steric number 5; Tetrachloroiodate ion (

ICl_4^-

): 4 bonds + 2 lone pairs = steric number 6; Sulfur dioxide (

SO_2

): 2 double-bond domains + 1 lone pair = steric number 3.

The total number of electron domains determines both the hybridization of atomic orbitals and the underlying electron pair geometry.

2

Assign the hybridization corresponding to each steric number.

Steric number 4 corresponds to

sp^3

; steric number 5 corresponds to

sp^3d

; steric number 6 corresponds to

sp^3d^2

; steric number 3 corresponds to

sp^2

.

Hybrid orbital set size equals the number of electron domains.

3

Apply VSEPR theory rules to deduce the molecular geometry (accounting only for atomic positions).

ClO_3^-

is trigonal pyramidal (

sp^3

);

XeF_2

is linear (

sp^3d

);

ICl_4^-

is square planar (

sp^3d^2

);

SO_2

is bent (

sp^2

).

Non-bonding lone pairs cause greater repulsion and define the non-spherical molecular geometry relative to electron geometry.

Key Concept

VSEPR Theory and Hybridization of Molecular Species and Polyatomic Ions

Question 8369Question →

Rhombic sulfur ( $\alpha$ -sulfur) and monoclinic sulfur ( $\beta$ -sulfur) are two well-known crystalline allotropes of sulfur. Which of the following statements correctly describes the stability and thermal transformation relationship between these two allotropes at standard atmospheric pressure?

Above

96^\circ\text{C}

, monoclinic sulfur is the stable crystalline form, whereas below

96^\circ\text{C}

, rhombic sulfur is the stable crystalline form.

Above

96^\circ\text{C}

, rhombic sulfur is the stable crystalline form, whereas below

96^\circ\text{C}

, monoclinic sulfur is the stable crystalline form.

Rhombic sulfur and monoclinic sulfur are chemical isomers that possess identical crystal lattices but different chemical formulas.

Monoclinic sulfur is an amorphous allotrope obtained by pouring boiling sulfur into cold water.

Show answer & explanation

Answer: Above $96^\circ\text{C}$ , monoclinic sulfur is the stable crystalline form, whereas below $96^\circ\text{C}$ , rhombic sulfur is the stable crystalline form.

Answer

Above

96^\circ\text{C}

, monoclinic sulfur is the stable crystalline form, whereas below

96^\circ\text{C}

, rhombic sulfur is the stable crystalline form.

The transition temperature for the two main crystalline allotropes of sulfur is

96^\circ\text{C}

. Below this temperature, rhombic sulfur (

\alpha

-sulfur) is the thermodynamically stable form. When heated above

96^\circ\text{C}

, rhombic sulfur slowly transforms into monoclinic sulfur (

\beta

-sulfur), which remains stable up to its melting point of

119^\circ\text{C}

.

Step-by-Step Solution

1

Identify the nature of the allotropes of sulfur.

Both rhombic sulfur (

\alpha

-sulfur) and monoclinic sulfur (

\beta

-sulfur) are crystalline allotropes made up of crown-shaped

S_8

molecules.

Allotropes are different structural forms of the same element in the same physical state.

2

Determine the transition temperature between rhombic and monoclinic sulfur.

The transition temperature is

96^\circ\text{C}

. At this temperature, both allotropes coexist in equilibrium.

Temperature determines which crystal lattice structure is thermodynamically stable.

3

Evaluate stability ranges above and below the transition temperature.

Rhombic sulfur is stable below

96^\circ\text{C}

, while monoclinic sulfur is stable between

96^\circ\text{C}

and its melting point (

119^\circ\text{C}

).

Heating rhombic sulfur above

96^\circ\text{C}

slowly converts it to monoclinic sulfur needles.

Key Concept

Transition temperature and allotropy of crystalline sulfur

Estimated Time:1m 0s

Question 8370Question →

In the industrial biotechnology process of ethanol production from sugar molasses, fermentation is carried out using yeast. Which of the following best explains why the fermentation vessel must be kept anaerobic (oxygen-free) and maintained at a temperature below $40^\circ\text{C}$ ?

To prevent the microbial oxidation of ethanol into ethanoic acid and protect the yeast enzymes from heat denaturation.

To shift the position of dynamic equilibrium toward the products by altering reaction enthalpy.

To convert the glucose substrate into a concentrated alkanoic acid via an esterification mechanism.

To ensure that the ethanol produced undergoes immediate polymerization into synthetic addition polymers.

Show answer & explanation

Answer: To prevent the microbial oxidation of ethanol into ethanoic acid and protect the yeast enzymes from heat denaturation.

Answer

The fermentation vessel must be kept anaerobic to prevent ethanol from being oxidized into ethanoic acid, and maintained below 40°C to prevent thermal denaturation of yeast enzymes.

The option identifying microbial oxidation prevention and enzyme protection is correct because yeast enzymes (zymase) denature and lose catalytic activity at elevated temperatures above 40°C. Furthermore, anaerobic conditions prevent aerobic bacteria from further oxidizing the formed ethanol into ethanoic acid.

Step-by-Step Solution

1

Analyze the biological role of yeast enzymes during industrial fermentation.

Yeast secretes enzymes such as invertase and zymase, which hydrolyze sucrose and ferment glucose into ethanol (

C_2H_5OH

) and carbon(IV) oxide (

CO_2

).

Enzymes are protein molecules that require an optimal temperature range (typically 25°C - 37°C); temperatures above 40°C cause irreversible denaturation.

2

Determine the necessity of anaerobic (oxygen-free) conditions in ethanol synthesis.

In the presence of atmospheric oxygen, aerobic microorganisms oxidize ethanol into ethanoic acid (

CH_3COOH

), spoiling the alcoholic yield.

Anaerobic respiration ensures glucose is converted to ethanol rather than carbon dioxide and water or ethanoic acid.

Key Concept

Industrial Biotechnology and Enzymatic Fermentation Conditions

Estimated Time:1m 0s

Question 8371Question →

Match each atmospheric phenomenon or chemical process on the left with its precise molecular mechanism or cause on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Tropospheric thermal radiation trapping matches vibrational absorption of terrestrial infrared radiation; Stratospheric ozone catalytic destruction matches the chlorine radical propagation reaction cycle; Chlorofluorocarbon (CFC) photolysis matches homolytic cleavage of carbon-chlorine bonds by ultraviolet light; Natural stratospheric ozone layer formation matches solar UV decomposition of

\text{O}_2

into atomic oxygen followed by reaction with

\text{O}_2

.

Each atmospheric process is accurately paired with its chemical mechanism: tropospheric heat trapping is driven by infrared absorption by greenhouse gases; stratospheric ozone catalytic breakdown occurs through free-radical propagation steps involving atomic chlorine; CFC photolysis involves solar UV cleavage of C-Cl bonds; and natural ozone formation requires UV dissociation of diatomic oxygen into atomic oxygen.

Step-by-Step Solution

1

Analyze the process of tropospheric thermal radiation trapping

Identify that global warming / greenhouse effect involves absorption and re-emission of terrestrial infrared (heat) radiation by gases like

\text{CO}_2

and

\text{CH}_4

.

Greenhouse gases selectively interact with outgoing longwave thermal radiation in the troposphere.

2

Analyze stratospheric ozone depletion and CFC reactions

Distinguish between CFC photolysis (UV cleavage of C-Cl bonds releasing Cl radicals) and the catalytic propagation cycle (

\text{Cl} + \text{O}_3 \rightarrow \text{ClO} + \text{O}_2

).

Photolysis produces reactive radicals, whereas propagation steps directly decompose ozone repeatedly.

3

Analyze the formation mechanism of the ozone layer

Identify that natural stratospheric ozone is created when UV light dissociates

\text{O}_2

into oxygen atoms, which combine with

\text{O}_2

to form

\text{O}_3

.

This dynamic photochemical cycle maintains the ozone shield in the stratosphere.

Key Concept

Greenhouse Effect, Global Warming, and Ozone Layer Depletion

Question 8372Question →

The specific heat capacity of a physical body increases proportionally with its mass, whereas its total heat capacity remains constant for any quantity of a given material.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Specific heat capacity is an intensive property independent of mass, whereas heat capacity is an extensive property that depends directly on mass.

The statement is false because specific heat capacity (

c

) is an intensive property that remains constant regardless of mass, while heat capacity (

C

) is an extensive property that increases directly with the mass of the object.

Step-by-Step Solution

1

Recall the mathematical definitions of heat capacity and specific heat capacity

Heat capacity

C = \frac{Q}{\Delta T} = mc

, whereas specific heat capacity

c = \frac{Q}{m\Delta T}

.

Establishing the functional dependence of both thermal quantities on mass.

2

Compare intensive and extensive classifications

Specific heat capacity

c

is normalized per unit mass (

\text{J kg}^{-1}\text{ K}^{-1}

) and is independent of mass. Heat capacity

C

(

\text{J K}^{-1}

) scales directly with mass

m

.

The statement falsely reverses which quantity depends on mass.

Key Concept

Distinguishing between extensive (heat capacity) and intensive (specific heat capacity) thermal properties

Estimated Time:1m 0s

Question 8373Question →

A $4.99\text{ g}$ sample of hydrated copper(II) tetraoxosulfate(VI), $\text{CuSO}_4 \cdot x\text{H}_2\text{O}$ , is heated to constant mass, yielding $3.19\text{ g}$ of anhydrous $\text{CuSO}_4$ . Separately, a salt whose saturated solution vapor pressure is greater than the partial pressure of water vapor in the surrounding atmosphere loses its water of crystallization to become an anhydrous powder upon exposure to air. $[\text{Cu} = 63.5, \text{S} = 32, \text{O} = 16, \text{H} = 1]$ . What is the value of $x$ in the hydrated salt, and what term describes the atmospheric behavior of the second salt?

x = 5

, and the second salt undergoes efflorescence

x = 5

, and the second salt undergoes deliquescence

x = 7

, and the second salt undergoes efflorescence

x = 7

, and the second salt undergoes hygroscopy

Show answer & explanation

Answer: $x = 5$ , and the second salt undergoes efflorescence

Answer

The value of

x

is 5, and the atmospheric behavior described is efflorescence.

Calculating the moles of anhydrous

\text{CuSO}_4

(

3.19\text{ g} / 159.5\text{ g/mol} = 0.020\text{ mol}

) and water (

1.80\text{ g} / 18\text{ g/mol} = 0.100\text{ mol}

) yields a mole ratio

x = 0.100 / 0.020 = 5

. Additionally, when a hydrate's vapor pressure is higher than the atmospheric partial pressure of water vapor, it loses water of crystallization to the surrounding air, which is defined as efflorescence.

Step-by-Step Solution

1

Calculate the molar masses of anhydrous

\text{CuSO}_4

and

\text{H}_2\text{O}

.

\text{Molar mass of CuSO}_4 = 63.5 + 32 + (4 \times 16) = 159.5\text{ g/mol}

;

\text{Molar mass of H}_2\text{O} = (2 \times 1) + 16 = 18\text{ g/mol}

.

Molar masses are needed to convert the experimental masses into mole quantities.

2

Determine the mass and moles of anhydrous

\text{CuSO}_4

and water of crystallization lost.

\text{Mass of CuSO}_4 = 3.19\text{ g} \Rightarrow n(\text{CuSO}_4) = \frac{3.19}{159.5} = 0.020\text{ mol}

.

\text{Mass of H}_2\text{O} = 4.99\text{ g} - 3.19\text{ g} = 1.80\text{ g} \Rightarrow n(\text{H}_2\text{O}) = \frac{1.80}{18} = 0.100\text{ mol}

.

Subtracting the anhydrous mass from the total hydrated mass gives the mass of lost water.

3

Calculate the stoichiometric integer ratio

x

.

x = \frac{n(\text{H}_2\text{O})}{n(\text{CuSO}_4)} = \frac{0.100}{0.020} = 5

.

The value of

x

is the mole ratio of water to anhydrous salt.

4

Identify the atmospheric phenomenon based on vapor pressure conditions.

The phenomenon is efflorescence.

When the vapor pressure of a hydrated salt's water of crystallization exceeds the atmospheric vapor pressure, the salt spontaneously loses water to the atmosphere, becoming anhydrous or lower-hydrated powder.

Key Concept

Water of crystallization stoichiometry and vapor pressure behavior in efflorescence vs deliquescence

Question 8374Question →

Match each class of alkanol listed on the left with its characteristic oxidation behavior on the right when reacted with acidified potassium heptaoxodichromate(VI) solution.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Primary alkanols pair with oxidation to an alkanal and then an alkanoic acid; secondary alkanols pair with oxidation to an alkanone; tertiary alkanols pair with resistance to oxidation under mild conditions.

Primary alkanols possess two alpha-hydrogens and oxidize in two steps to form alkanals and then alkanoic acids. Secondary alkanols possess one alpha-hydrogen and oxidize to form alkanones. Tertiary alkanols lack alpha-hydrogens entirely, rendering them resistant to oxidation under mild conditions.

Step-by-Step Solution

1

Examine the structural environment of primary alkanols (

R-CH_2OH

)

Primary alkanols have two

\alpha

-hydrogen atoms attached to the carbon holding the

-OH

group, permitting two sequential oxidation steps.

Oxidation requires the removal of hydrogen from the hydroxyl-bearing carbon atom.

2

Examine the structural environment of secondary alkanols (

R_2CHOH

)

Secondary alkanols have only one

\alpha

-hydrogen atom, yielding an alkanone (

R_2C=O

).

Alkanones resist further oxidation under mild conditions because no additional

\alpha

-hydrogens are available.

3

Examine the structural environment of tertiary alkanols (

R_3COH

)

Tertiary alkanols possess zero

\alpha

-hydrogen atoms on the hydroxyl-bearing carbon atom.

Without an

\alpha

-hydrogen, oxidation cannot proceed without breaking carbon-carbon bonds.

Key Concept

Classification and oxidation products of alkanols

Question 8375Question →

The standard reduction potentials for two half-cell reactions at

25^\circ\text{C}

are given below:

\text{Mn}^{2+}(aq) + 2e^- \rightarrow \text{Mn}(s) \quad E^\circ = -1.18\text{ V}

\text{Pb}^{2+}(aq) + 2e^- \rightarrow \text{Pb}(s) \quad E^\circ = -0.13\text{ V}

Calculate the standard cell potential ( $E^\circ_{\text{cell}}$ ), in volts, for the spontaneous electrochemical reaction between these two half-cells.

Show answer & explanation

Answer: 1.05

Answer

The standard cell potential for the spontaneous reaction is

+1.05\text{ V}

.

For a spontaneous redox reaction in a galvanic cell, the half-cell with the higher standard reduction potential acts as the cathode, and the one with the lower standard reduction potential acts as the anode. Lead(II) ions (

\text{Pb}^{2+}

,

E^\circ = -0.13\text{ V}

) have a higher reduction potential than manganese(II) ions (

\text{Mn}^{2+}

,

E^\circ = -1.18\text{ V}

). Substituting these values into

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

yields

E^\circ_{\text{cell}} = -0.13\text{ V} - (-1.18\text{ V}) = +1.05\text{ V}

.

Step-by-Step Solution

1

Identify cathode and anode roles for a spontaneous galvanic cell.

Cathode (reduction):

\text{Pb}^{2+}(aq) + 2e^- \rightarrow \text{Pb}(s)

with

E^\circ = -0.13\text{ V}

. Anode (oxidation):

\text{Mn}(s) \rightarrow \text{Mn}^{2+}(aq) + 2e^-

with

E^\circ = -1.18\text{ V}

.

For a spontaneous process (

E^\circ_{\text{cell}} > 0

), the species with the more positive reduction potential acts as the oxidizing agent and undergoes reduction at the cathode.

2

Use the cell potential formula

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

.

E^\circ_{\text{cell}} = -0.13\text{ V} - (-1.18\text{ V})

Subtracting the standard reduction potential of the anode from that of the cathode gives the overall electromotive force of the cell.

3

Perform the subtraction to find the final numerical answer.

E^\circ_{\text{cell}} = +1.05\text{ V}

-0.13 + 1.18 = 1.05

, confirming a positive standard cell potential for the spontaneous reaction.

Key Concept

Standard Cell Potential and Reaction Spontaneity

Question 8376Question →

Which property of transition metals primarily enables them to function as efficient catalysts in industrial chemical processes?

Ability to exhibit variable oxidation states and offer vacant d-orbitals

High electrical and thermal conductivity

Low melting points and low densities relative to s-block metals

Ability to shift the position of dynamic equilibrium to increase product yield

Show answer & explanation

Answer: Ability to exhibit variable oxidation states and offer vacant d-orbitals

Answer

Ability to exhibit variable oxidation states and offer vacant d-orbitals

Transition elements possess unfilled or partially filled d-orbitals and readily change oxidation states. These characteristics allow them to bind reactants temporarily or form unstable transition-state intermediates, offering reaction mechanisms with lower activation energy.

Step-by-Step Solution

1

Recall the fundamental feature of transition metals responsible for catalytic activity.

Transition metals have partially filled d-orbitals and can easily exchange electrons to adopt multiple oxidation states.

This allows them to form temporary bonds with reactants (adsorption) or form reactive intermediate species, which significantly lowers activation energy.

Key Concept

Catalytic behavior of transition metals due to variable oxidation states and available d-orbitals

Question 8377Question →

The solubility of a salt $Y$ (molar mass = $100\text{ g/mol}$ ) in water is $3.5\text{ mol/dm}^3$ at $70^\circ\text{C}$ and $1.5\text{ mol/dm}^3$ at $25^\circ\text{C}$ . What mass of salt $Y$ will crystallize out when $500\text{ cm}^3$ of its saturated solution is cooled from $70^\circ\text{C}$ to $25^\circ\text{C}$ ?

100.0\text{ g}

175.0\text{ g}

200.0\text{ g}

1.0\text{ g}

Show answer & explanation

Answer: $100.0\text{ g}$

Answer

The mass of salt

Y

that crystallizes out of solution is

100.0\text{ g}

.

Cooling a saturated solution reduces its capacity to retain solute. The difference in concentration between

70^\circ\text{C}

(

3.5\text{ mol/dm}^3

) and

25^\circ\text{C}

(

1.5\text{ mol/dm}^3

) is

2.0\text{ mol/dm}^3

. Scaling this concentration difference for

500\text{ cm}^3

(

0.5\text{ dm}^3

) gives

1.0\text{ mole}

of salt

Y

. Converting moles to mass yields

1.0\text{ mole} \times 100\text{ g/mol} = 100.0\text{ g}

.

Step-by-Step Solution

1

Calculate the difference in solubility between

70^\circ\text{C}

and

25^\circ\text{C}

in

\text{mol/dm}^3

.

\Delta S = 3.5\text{ mol/dm}^3 - 1.5\text{ mol/dm}^3 = 2.0\text{ mol/dm}^3

Solubility decreases as temperature drops, causing solute to precipitate.

2

Convert solution volume from

\text{cm}^3

to

\text{dm}^3

and determine the moles of solute precipitated.

V = \frac{500\text{ cm}^3}{1000} = 0.5\text{ dm}^3

;

\text{Moles precipitated} = 2.0\text{ mol/dm}^3 \times 0.5\text{ dm}^3 = 1.0\text{ mole}

Solubility is given per cubic decimeter, so volume scaling is necessary.

3

Convert the precipitated moles into mass in grams using molar mass.

\text{Mass} = 1.0\text{ mole} \times 100\text{ g/mol} = 100.0\text{ g}

Mass equals number of moles multiplied by molar mass.

Key Concept

Crystallization calculations based on temperature-dependent solubility curves

Question 8378Question →

Match each type of solution state with its corresponding characteristic physical state and behavior at a given temperature.

Click a left item, then click its matching right item

Show answer & explanation

Answer

An unsaturated solution matches with holding less solute than the maximum limit and being able to dissolve more solute. A saturated solution matches with holding the maximum solute in dynamic equilibrium with undissolved solute. A supersaturated solution matches with holding excess dissolved solute beyond normal solubility and rapidly crystallizing upon seeding.

An unsaturated solution can dissolve more solute; a saturated solution maintains dynamic equilibrium with maximum dissolved solute; a supersaturated solution holds solute beyond normal solubility limits and crystallizes when seeded.

Step-by-Step Solution

1

Identify the characteristic of an unsaturated solution.

It has not reached maximum capacity, so adding solute leads to further dissolution.

Solvent molecules are available to interact with and dissolve additional solute particles at that temperature.

2

Identify the characteristic of a saturated solution.

It exists in dynamic equilibrium with undissolved solid.

At saturation, the solvent holds the maximum possible concentration of solute at that specific temperature.

3

Identify the characteristic of a supersaturated solution.

It contains dissolved solute in excess of the saturation concentration and is metastable/unstable.

Disturbing the solution with a seed crystal initiates rapid crystallization to relieve the unstable excess concentration.

Key Concept

Saturation States of Solutions

Question 8379Question →

For the endothermic synthesis reaction $\text{X}_2(g) + \text{Y}_2(g) \rightarrow 2\text{XY}(g)$ , the potential energy of the reactants is $+40\text{ kJ mol}^{-1}$ and the potential energy of the products is $+80\text{ kJ mol}^{-1}$ . If the activation energy for the forward reaction is $60\text{ kJ mol}^{-1}$ , what is the activation energy for the reverse reaction?

20\text{ kJ mol}^{-1}

40\text{ kJ mol}^{-1}

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

140\text{ kJ mol}^{-1}

Show answer & explanation

Answer: $20\text{ kJ mol}^{-1}$

Answer

20\text{ kJ mol}^{-1}

The peak energy of the transition state is obtained by adding the forward activation energy (

60\text{ kJ mol}^{-1}

) to the reactant energy (

40\text{ kJ mol}^{-1}

), giving

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

. Subtracting the product energy level (

80\text{ kJ mol}^{-1}

) from this peak value yields an activation energy for the reverse reaction of

20\text{ kJ mol}^{-1}

.

Step-by-Step Solution

1

Calculate the potential energy of the transition state (activated complex).

\text{Energy of transition state} = \text{Energy of reactants} + E_{a,\text{forward}} = 40\text{ kJ mol}^{-1} + 60\text{ kJ mol}^{-1} = 100\text{ kJ mol}^{-1}

.

The forward activation energy is the energy barrier that reactants must overcome to reach the peak transition state.

2

Calculate the activation energy for the reverse reaction.

E_{a,\text{reverse}} = \text{Energy of transition state} - \text{Energy of products} = 100\text{ kJ mol}^{-1} - 80\text{ kJ mol}^{-1} = 20\text{ kJ mol}^{-1}

.

The reverse activation energy is the energy difference between the product level and the activated complex peak.

Key Concept

Activation energy relationship in energy profile diagrams

Estimated Time:1m 0s

Question 8380Question →

An aqueous solution of $0.01\text{ mol dm}^{-3}$ ethanoic acid ( $\text{CH}_3\text{COOH}$ ) exhibits a higher degree of ionization ( $\alpha$ ) than a $0.10\text{ mol dm}^{-3}$ ethanoic acid solution at the same temperature. Which of the following statements correctly explains this observation?

Dilution shifts the ionization equilibrium forward to favor the formation of free ions according to Ostwald's dilution law.

Diluting a weak acid increases its acid dissociation constant (

K_a

), causing it to ionize more completely.

Ethanoic acid converts into a strong acid upon dilution because its concentration decreases significantly.

Concentrated ethanoic acid solutions have a higher percentage ionization because more acid molecules are present per unit volume.

Show answer & explanation

Answer: Dilution shifts the ionization equilibrium forward to favor the formation of free ions according to Ostwald's dilution law.

Answer

Dilution shifts the ionization equilibrium forward to favor the formation of free ions according to Ostwald's dilution law.

Dilution increases the volume of the solvent relative to solute molecules, which shifts the position of the dynamic ionization equilibrium toward the side with a greater number of individual particles (ions) to maintain equilibrium, thereby increasing the degree of ionization.

Step-by-Step Solution

1

Write the ionization equilibrium expression for ethanoic acid

\text{CH}_3\text{COOH(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{CH}_3\text{COO}^-(\text{aq}) + \text{H}_3\text{O}^+(\text{aq})

Ethanoic acid is a weak monobasic acid that ionizes partially in aqueous medium.

2

Apply Ostwald's Dilution Law equation for weak electrolytes

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

, where

\alpha

is degree of ionization,

K_a

is acid dissociation constant, and

C

is molar concentration.

As concentration

C

decreases (dilution),

\alpha

increases, shifting equilibrium toward ion formation.

Key Concept

Ostwald's Dilution Law and Extent of Ionization of Weak Acids

Estimated Time:1m 0s

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