All practice questions

Question 8261Question →

Photochemical smog is primarily formed during cold, humid winter mornings when atmospheric sulfur dioxide ( $SO_2$ ) and carbon soot interact with trapped fog droplets.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Photochemical smog requires solar radiation, nitrogen oxides, and hydrocarbons under warm, dry conditions, whereas cold, humid conditions involving sulfur dioxide and soot produce industrial (sulfurous) smog.

The statement is false because it describes classical industrial (sulfurous) smog rather than photochemical smog. Photochemical smog develops in sunny, warm environments through solar UV-driven reactions of nitrogen oxides (

NO_x

) and unburnt hydrocarbons, yielding oxidants such as ground-level ozone (

O_3

) and peroxyacetyl nitrate (PAN).

Step-by-Step Solution

1

Examine the environmental conditions and chemical precursors cited in the statement.

The statement describes cold, humid winter mornings with sulfur dioxide (

SO_2

) and carbonaceous soot.

Identifying the atmospheric parameters is essential to distinguish between atmospheric smog classifications.

2

Compare industrial (sulfurous) smog and photochemical smog mechanisms.

Industrial smog originates from sulfur dioxide (

SO_2

) and particulates reacting in damp air, whereas photochemical smog requires sunlight-driven photochemical reactions involving nitrogen oxides (

NO_x

) and volatile organic compounds (VOCs).

Solar ultraviolet light is mandatory for initiating the photolysis of

NO_2

into

NO

and oxygen radicals, which leads to photochemical smog.

3

Evaluate the statement's validity.

The statement misidentifies industrial smog conditions as those of photochemical smog.

Since the stem describes conditions unique to industrial sulfurous smog, the claim regarding photochemical smog is incorrect.

Key Concept

Distinction between Photochemical Smog and Industrial (Sulfurous) Smog

Question 8262Question →

Marble monuments and historical stone structures located near industrial regions often undergo rapid deterioration due to acidic rainfall. Which primary air pollutant reacts with atmospheric oxygen and water to produce tetraoxosulfate(VI) acid ( $H_2SO_4$ ), causing this environmental degradation?

Sulfur(IV) oxide

Carbon(IV) oxide

Chlorofluorocarbon

Methane

Show answer & explanation

Answer: Sulfur(IV) oxide

Answer

Sulfur(IV) oxide is the primary air pollutant that dissolves and oxidizes in atmospheric moisture to form tetraoxosulfate(VI) acid (

H_2SO_4

), causing acid rain.

Sulfur(IV) oxide (

SO_2

) emitted from power plants and factories reacts with oxygen and water in the atmosphere to form tetraoxosulfate(VI) acid (

H_2SO_4

). This strong acid lowers rainfall pH and reacts with calcium carbonate (

CaCO_3

) in marble, causing structural weathering.

Step-by-Step Solution

1

Identify the chemical pollutant associated with tetraoxosulfate(VI) acid formation.

Sulfur(IV) oxide (

SO_2

) is produced during the combustion of sulfur-containing fossil fuels.

Atmospheric oxidation converts

SO_2

into

SO_3

, which then dissolves in water vapor to yield sulfuric acid (

H_2SO_4

).

2

Differentiate acid rain precursors from greenhouse gases and ozone-depleting substances.

Sulfur(IV) oxide is specifically an acid rain precursor, while carbon(IV) oxide and methane are greenhouse gases, and CFCs are ozone depleters.

Connecting specific environmental impacts to their corresponding atmospheric pollutants ensures accurate identification.

Key Concept

Acid Rain Formation from Sulfur Oxides

Estimated Time:45s

Question 8263Question →

When a $16.0\text{ g}$ sample of pure ammonium trioxonitrate(V), $NH_4NO_3$ , undergoes complete thermal decomposition, it produces dinitrogen monoxide gas, $N_2O$ , and water vapor. Assuming standard temperature and pressure (STP, where molar gas volume = $22.4\text{ dm}^3\text{mol}^{-1}$ ), what is the volume of the oxide of nitrogen collected, and how does its rate of diffusion compare to that of carbon(IV) oxide, $CO_2$ , under identical conditions? (Molar masses: $N = 14\text{ g/mol}$ , $O = 16\text{ g/mol}$ , $H = 1\text{ g/mol}$ , $C = 12\text{ g/mol}$ )

4.48\text{ dm}^3

, and its rate of diffusion is equal to that of

CO_2

4.80\text{ dm}^3

, and its rate of diffusion is equal to that of

CO_2

4.48\text{ dm}^3

, and it diffuses

\sqrt{2}

times faster than

CO_2

8.96\text{ dm}^3

, and its rate of diffusion is slower than that of

CO_2

Show answer & explanation

Answer: $4.48\text{ dm}^3$ , and its rate of diffusion is equal to that of $CO_2$

Answer

4.48\text{ dm}^3

, and its rate of diffusion is equal to that of

CO_2

Thermal decomposition of

16.0\text{ g}

(

0.20\text{ mol}

) of

NH_4NO_3

produces

0.20\text{ mol}

of

N_2O

. At STP,

0.20\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 4.48\text{ dm}^3

. Both

N_2O

and

CO_2

have a molar mass of

44\text{ g/mol}

, so by Graham's law of diffusion, their relative rates of diffusion are identical.

Step-by-Step Solution

1

Write the balanced chemical equation for the thermal decomposition of ammonium trioxonitrate(V)

NH_4NO_3(s) \rightarrow N_2O(g) + 2H_2O(g)

Establishes the stoichiometric mole ratio between

NH_4NO_3

and

N_2O

, which is

1 : 1

.

2

Calculate the molar mass and number of moles of

NH_4NO_3

reacted

Molar mass of

NH_4NO_3 = 2(14) + 4(1) + 3(16) = 80\text{ g/mol}

. Number of moles =

\frac{16.0\text{ g}}{80\text{ g/mol}} = 0.20\text{ mol}

.

Determines the mole quantity of reactant available.

3

Determine the volume of

N_2O

produced at STP

Moles of

N_2O = 0.20\text{ mol}

. Volume at STP =

0.20\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 4.48\text{ dm}^3

.

Applies standard molar volume of gas at STP.

4

Compare the rates of diffusion of

N_2O

and

CO_2

using Graham's law

Molar mass of

N_2O = 2(14) + 16 = 44\text{ g/mol}

. Molar mass of

CO_2 = 12 + 2(16) = 44\text{ g/mol}

. Ratio of diffusion rates =

\sqrt{\frac{44}{44}} = 1

.

Gases with equal molar masses diffuse at equal rates under identical conditions.

Key Concept

Thermal decomposition of nitrogen oxides, molar volume at STP, and Graham's law of diffusion

Estimated Time:2m 0s

Question 8264Question →

In the laboratory preparation of ethyne gas, water is added dropwise to solid calcium carbide ( $CaC_2$ ). What is the IUPAC name of the inorganic compound produced as a byproduct in this reaction?

Show answer & explanation

Answer: Calcium hydroxide; calcium hydroxide; Ca(OH)2; Ca(OH)₂

Answer

Calcium hydroxide

Hydrolysis of calcium carbide (

CaC_2

) produces ethyne (

C_2H_2

) as the desired hydrocarbon gas along with calcium hydroxide (

Ca(OH)_2

) as the inorganic byproduct.

Step-by-Step Solution

1

Write the balanced chemical equation for the hydrolysis of calcium carbide.

CaC_2(s) + 2H_2O(l) \rightarrow C_2H_2(g) + Ca(OH)_2(s)

Calcium carbide reacts exothemically with water to liberate ethyne gas and leave behind a solid residue of calcium hydroxide.

2

Identify the chemical name of the inorganic byproduct

Ca(OH)_2

.

Calcium hydroxide

The inorganic salt contains calcium ions (

Ca^{2+}

) and hydroxide ions (

OH^-

), giving the IUPAC name calcium hydroxide.

Key Concept

Laboratory preparation of ethyne by hydrolysis of calcium carbide

Question 8265Question →

How many of the acyclic structural isomers with the molecular formula $C_4H_8Cl_2$ possess at least one chiral carbon atom?

3

2

4

5

Show answer & explanation

Answer: 3

Answer

3 acyclic structural isomers of

C_4H_8Cl_2

possess at least one chiral carbon atom.

The correct response is 3. A chiral carbon atom is defined as a carbon atom bonded to four different atoms or functional groups. Among the nine constitutional acyclic isomers of dichlorobutane (

C_4H_8Cl_2

), only 1,2-dichlorobutane (at carbon-2), 1,3-dichlorobutane (at carbon-2), and 2,3-dichlorobutane (at carbon-2 and carbon-3) possess carbon atoms meeting this criterion.

Step-by-Step Solution

1

Identify all acyclic structural (constitutional) isomers of

C_4H_8Cl_2

.

There are 9 total acyclic structural isomers: 1,1-dichlorobutane, 1,2-dichlorobutane, 1,3-dichlorobutane, 1,4-dichlorobutane, 2,2-dichlorobutane, 2,3-dichlorobutane, 1,1-dichloro-2-methylpropane, 1,2-dichloro-2-methylpropane, and 1,3-dichloro-2-methylpropane.

Structural isomers must be systematically generated based on butane and methylpropane carbon skeletons.

2

Examine each structural isomer for the presence of a chiral (asymmetric) carbon atom.

A chiral carbon must be bonded to four completely different atoms or groups.

Chirality requires tetrahedral asymmetry around a carbon atom.

3

Determine which specific isomers contain chiral carbons.

1. 1,2-dichlorobutane: Carbon-2 is bonded to

-H

,

-Cl

,

-CH_2Cl

, and

-CH_2CH_3

(Chiral).
2. 1,3-dichlorobutane: Carbon-2 is bonded to

-H

,

-Cl

,

-CH_3

, and

-CH_2CH_2Cl

(Chiral).
3. 2,3-dichlorobutane: Carbon-2 and Carbon-3 are both bonded to

-H

,

-Cl

,

-CH_3

, and

-CH(Cl)CH_3

(Chiral).

All other structural isomers have identical groups attached to each carbon (e.g., two

-Cl

atoms on C-1 in 1,1-dichlorobutane, or two methyl groups on C-2 in methylpropane derivatives).

Key Concept

Chirality and Structural Isomerism in Haloalkanes

Question 8266Question →

Fill in the blanks to complete the chemical observation and oxidation state change during the test for sulfur(IV) oxide gas. What are the correct terms to fill in the blanks?

Fill in the blanks below

When

SO_2

gas is passed through an acidified solution of potassium heptaoxodichromate(VI),

K_2Cr_2O_7

, the solution turns from orange to because the dichromate ion is reduced, changing the oxidation state of chromium from

+6

to .

Show answer & explanation

Answer

The solution turns green because chromium is reduced from an oxidation state of +6 to +3.

Sulfur(IV) oxide (

SO_2

) is a strong reducing agent. When passed into an acidified solution of potassium heptaoxodichromate(VI), it reduces the orange dichromate ion (

Cr_2O_7^{2-}

, where

Cr

is in the

+6

oxidation state) to the green chromium(III) ion (

Cr^{3+}

, where

Cr

is in the

+3

oxidation state). Thus, the color turns green and the final oxidation state is

+3

.

Step-by-Step Solution

1

Identify the role of

SO_2

and

K_2Cr_2O_7

in the redox reaction.

SO_2

acts as a reducing agent and is oxidized to

SO_4^{2-}

, while

K_2Cr_2O_7

acts as an oxidizing agent.

Reducing agents cause the reduction of other species while being oxidized themselves.

2

Determine the color change of the acidified

K_2Cr_2O_7

solution.

The orange

Cr_2O_7^{2-}

ion is reduced to the green

Cr^{3+}

ion.

The formation of hydrated chromium(III) ions in solution imparts a distinct green color.

3

Determine the initial and final oxidation states of chromium.

In

Cr_2O_7^{2-}

, chromium has an oxidation state of

+6

. Upon reduction to

Cr^{3+}

, its oxidation state becomes

+3

.

The half-reaction is

Cr_2O_7^{2-} + 14H^+ + 6e^- \nrightarrow 2Cr^{3+} + 7H_2O

.

Key Concept

Laboratory identification test for reducing agents using acidified potassium heptaoxodichromate(VI)

Question 8267Question →

In the biological treatment of municipal and industrial wastewater via the activated sludge process, several sequential operations are performed to purify water and manage waste. Arrange the following steps of the process in the correct chronological order from first to last.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence is: (1) Screening and primary sedimentation to remove coarse solids and heavy inorganic grit, (2) Aeration in tanks containing aerobic bacteria to digest dissolved organic pollutants, (3) Secondary clarification in settling basins to separate the activated biological sludge flocs from the purified effluent, and (4) Disinfection of the clarified water using chlorine or ultraviolet light prior to discharge.

The logical sequence follows the engineering progression from coarse physical removal of solids, through microbial digestion of dissolved organics during aeration, solid-liquid separation of biomass in secondary clarifiers, and final pathogenic disinfection.

Step-by-Step Solution

1

Identify the preliminary physical separation stage.

Primary screening and sedimentation occurs first.

Large debris and heavy suspended solids must be removed physically before exposing effluent to microbial action.

2

Identify the biological oxidation phase.

Aeration with aerobic microorganisms follows primary sedimentation.

Microbial breakdown requires dissolved oxygen supplied in aeration tanks to convert soluble organic waste into biomass and gas.

3

Identify the biomass separation stage.

Secondary clarification follows aeration.

The biological flocs created during aeration need quiescent conditions to settle out as activated sludge, leaving clear liquid above.

4

Identify the final pathogen neutralization stage.

Disinfection is the final operational step.

Harmful bacterial or viral pathogens are inactivated right before environmental discharge to protect aquatic ecosystems and public health.

Key Concept

Sequential physical, biological, and chemical stages of industrial biotechnology in activated sludge wastewater treatment.

Question 8268Question →

Which organic gas is produced when ethanol ( $C_2H_5OH$ ) is heated with excess concentrated tetraoxosulfate(VI) acid ( $H_2SO_4$ ) at $170^\circ\text{C}$ ?

Ethene (

C_2H_4

)

Ethane (

C_2H_6

)

Ethyne (

C_2H_2

)

Ethoxyethane (

(C_2H_5)_2O

)

Show answer & explanation

Answer: Ethene ( $C_2H_4$ )

Answer

Ethene (

C_2H_4

)

When ethanol is heated with excess concentrated tetraoxosulfate(VI) acid at a high temperature (

170^\circ\text{C}

), the acid acts as a dehydrating agent, removing a molecule of water from ethanol to form ethene (

C_2H_4

).

Step-by-Step Solution

1

Identify the functional group and reagent conditions given in the reaction stem.

Ethanol (

C_2H_5OH

) is reacted with excess concentrated

H_2SO_4

at

170^\circ\text{C}

.

Concentrated tetraoxosulfate(VI) acid acts as a powerful dehydrating agent at high temperatures.

2

Determine the type of dehydration taking place under these specific temperature conditions.

Intramolecular dehydration occurs, removing one water molecule (

H_2O

) from a single ethanol molecule.

At

170^\circ\text{C}

with excess acid, the removal of

-OH

and a neighboring

-H

atom creates a carbon-carbon double bond.

3

Write the balanced chemical equation to confirm the product formula.

C_2H_5OH \xrightarrow{\text{conc. } H_2SO_4, 170^\circ\text{C}} C_2H_4 + H_2O

The organic gas evolved is ethene (

C_2H_4

), an alkene.

Key Concept

Laboratory preparation of alkenes via dehydration of alkanols

Estimated Time:45s

Question 8269Question →

Arrange the following chemical and electrochemical stages of the rusting of iron in chronological sequence, from initial anodic oxidation to the final formation of rust.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct chronological sequence is: Oxidation of iron metal to

\text{Fe}^{2+}

ions

\rightarrow

Reduction of dissolved oxygen to

\text{OH}^-

ions

\rightarrow

Precipitation of

\text{Fe(OH)}_2

\rightarrow

Oxidation of

\text{Fe(OH)}_2

to hydrated

\text{Fe}_2\text{O}_3\cdot x\text{H}_2\text{O}

.

Rusting is an electrochemical process. Iron metal initially oxidizes at anodic sites to produce

\text{Fe}^{2+}

ions and electrons. The released electrons migrate through the iron to cathodic regions, where dissolved atmospheric oxygen is reduced to

\text{OH}^-

ions. These ions react in the electrolyte solution to form insoluble green

\text{Fe(OH)}_2

, which is subsequently oxidized by dissolved oxygen to form reddish-brown hydrated iron(III) oxide (rust,

\text{Fe}_2\text{O}_3\cdot x\text{H}_2\text{O}

).

Step-by-Step Solution

1

Identify the initial anodic oxidation process

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

Rusting begins as an electrochemical reaction where iron metal acts as the anode and undergoes oxidation.

2

Identify the cathodic reduction process

\text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4\text{e}^- \rightarrow 4\text{OH}^-\text{(aq)}

Electrons released during anodic oxidation flow to cathodic sites where atmospheric oxygen dissolved in water is reduced.

3

Determine the ionic precipitation reaction

\text{Fe}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} \rightarrow \text{Fe(OH)}_2\text{(s)}

The formed cations and anions diffuse towards each other in the aqueous electrolyte layer, forming an insoluble precipitate.

4

Determine the final oxidation step to rust

4\text{Fe(OH)}_2\text{(s)} + \text{O}_2\text{(g)} + 2x\text{H}_2\text{O(l)} \rightarrow 2(\text{Fe}_2\text{O}_3\cdot x\text{H}_2\text{O})\text{(s)}

Dissolved oxygen further oxidizes iron(II) hydroxide to hydrated iron(III) oxide, commonly known as rust.

Key Concept

Electrochemical mechanism of iron rusting

Question 8270Question →

Carbon(IV) oxide ( $\text{CO}_2$ ) drives global warming primarily by absorbing high-energy solar ultraviolet radiation in the troposphere, whereas chlorofluorocarbons ( $\text{CFCs}$ ) destroy stratospheric ozone by absorbing long-wave infrared radiation emitted from the Earth's surface.

True

False

Show answer & explanation

Answer: False

Answer

The statement is false. Carbon(IV) oxide (

\text{CO}_2

) traps heat by absorbing outgoing terrestrial infrared (thermal) radiation, not incoming solar ultraviolet radiation. Meanwhile, chlorofluorocarbons (

\text{CFCs}

) cause ozone depletion when high-energy solar ultraviolet photolysis releases reactive chlorine free radicals (

\text{Cl}^\bullet

) in the stratosphere.

The statement is false because it reverses the radiation types and chemical mechanisms involved in global warming and ozone depletion. Carbon(IV) oxide (

\text{CO}_2

) traps terrestrial long-wave infrared radiation in the troposphere to produce the greenhouse effect. In contrast, ozone depletion by chlorofluorocarbons (

\text{CFCs}

) occurs in the stratosphere via the photolytic generation of chlorine free radicals (

\text{Cl}^\bullet

) induced by high-energy solar ultraviolet radiation.

Step-by-Step Solution

1

Analyze the mechanism and type of electromagnetic radiation associated with the greenhouse effect and global warming.

Greenhouse gases such as

\text{CO}_2

,

\text{CH}_4

, and

\text{N}_2\text{O}

allow short-wave solar radiation to pass through the atmosphere but absorb long-wave infrared (heat) radiation re-emitted by the Earth's surface.

Trapping terrestrial infrared radiation in the troposphere causes atmospheric warming.

2

Analyze the mechanism and type of electromagnetic radiation associated with ozone layer depletion.

Chlorofluorocarbons (

\text{CFCs}

) in the stratosphere absorb high-energy ultraviolet (UV) radiation, leading to homolytic fission of

\text{C}-\text{Cl}

bonds:

\text{CF}_2\text{Cl}_2 \xrightarrow{h\nu} \text{CF}_2\text{Cl}^\bullet + \text{Cl}^\bullet

. The resulting chlorine radical initiates catalytic ozone breakdown (

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

).

Ozone depletion is a photolytic radical-catalyzed chemical process triggered by solar UV light, not an infrared heat absorption phenomenon.

3

Compare the statement's claims against the established chemical mechanisms.

The statement incorrectly attributes UV absorption to

\text{CO}_2

's greenhouse mechanism and IR absorption to

\text{CFCs}

' ozone destruction mechanism.

Both radiation types and underlying physical/chemical mechanisms are interchanged.

Key Concept

Mechanistic Distinction Between the Greenhouse Effect (Infrared Trapping) and Stratospheric Ozone Depletion (Ultraviolet Photolysis)

Question 8271Question →

What is the relative molecular mass of hydrated sodium trioxocarbonate(IV) crystals, $Na_2CO_3 \cdot 10H_2O$ ? $[Na = 23, C = 12, O = 16, H = 1]$

286

106

124

268

Show answer & explanation

Answer: 286

Answer

The relative molecular mass of hydrated sodium trioxocarbonate(IV),

Na_2CO_3 \cdot 10H_2O

, is 286.

The relative molecular mass of a hydrated compound is calculated by taking the sum of the relative atomic masses of all constituent elements in both the anhydrous salt and the water molecules of crystallization. For

Na_2CO_3 \cdot 10H_2O

,

(23 \times 2) + 12 + (16 \times 3) + 10 \times [(1 \times 2) + 16] = 46 + 12 + 48 + 180 = 286

.

Step-by-Step Solution

1

Calculate the relative formula mass of anhydrous

Na_2CO_3

(23 \times 2) + 12 + (16 \times 3) = 46 + 12 + 48 = 106

Sum the relative atomic masses of all atoms present in the anhydrous salt portion.

2

Calculate the relative mass of the water of crystallization (

10H_2O

)

10 \times [(1 \times 2) + 16] = 10 \times 18 = 180

Multiply the relative molecular mass of one water molecule (18) by the coefficient 10.

3

Add the mass of the anhydrous salt to the mass of the water of crystallization

106 + 180 = 286

Combine both components to obtain the total relative molecular mass of the hydrated crystal.

Key Concept

Relative Molecular Mass of Hydrated Salts

Question 8272Question →

An oxide of nitrogen contains $30.4\%$ nitrogen and $69.6\%$ oxygen by mass. If its relative molar mass is $92\text{ g/mol}$ , what is the molecular formula of the compound? $[N = 14, O = 16]$

Show answer & explanation

Answer: N2O4; N₂O₄; $N_2O_4$

Answer

The molecular formula of the oxide is

N_2O_4

.

First, the empirical formula is derived by dividing the percentage composition by relative atomic masses (

30.4/14 = 2.17

for N and

69.6/16 = 4.35

for O), which gives a simple whole number ratio of

1:2

(

NO_2

). The empirical formula mass of

NO_2

is

46\text{ g/mol}

. Dividing the molar mass (

92\text{ g/mol}

) by the empirical mass (

46\text{ g/mol}

) yields an integer factor of

2

. Multiplying the empirical formula by

2

gives the molecular formula

N_2O_4

.

Step-by-Step Solution

1

Calculate the relative number of moles of each element.

Moles of

N = \frac{30.4}{14} = 2.17

; Moles of

O = \frac{69.6}{16} = 4.35

.

Dividing mass percentage by atomic mass gives the mole ratio.

2

Determine the simplest whole number mole ratio.

Ratio

N : O = \frac{2.17}{2.17} : \frac{4.35}{2.17} = 1 : 2

. Empirical formula =

NO_2

.

Dividing by the smallest mole value yields the empirical formula subscripts.

3

Calculate the empirical formula mass and the multiplier integer

n

.

Empirical formula mass =

14 + 2(16) = 46\text{ g/mol}

.

n = \frac{92}{46} = 2

.

The integer multiplier

n

is the ratio of molar mass to empirical formula mass.

4

Multiply empirical formula subscripts by

n

.

Molecular formula =

(NO_2)_2 = N_2O_4

.

Applying the integer multiplier gives the actual number of atoms in the molecule.

Key Concept

Determining empirical and molecular formulas from elemental percentage composition and relative molar mass.

Estimated Time:1m 30s

Question 8273Question →

Match each calcium-containing substance on the left with its correct chemical function or industrial preparation description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Calcium fluoride matches with serving as a flux in calcium extraction; Calcium oxide matches with calcination product of limestone used to dry ammonia; Calcium sulfate hemihydrate matches with partial dehydration product of gypsum; Calcium hydroxide matches with slaked lime used to detect carbon(IV) oxide.

Each calcium compound is correctly paired according to standard industrial practices and chemical properties: calcium fluoride lowers the electrolytic bath melting point; calcium oxide is a basic desiccant produced from limestone; calcium sulfate hemihydrate is formed by partially dehydrating gypsum; and slaked lime solution forms a precipitate with carbon(IV) oxide.

Step-by-Step Solution

1

Identify the industrial metallurgical role of calcium fluoride in calcium metal extraction.

Calcium fluoride acts as a flux in fused

\text{CaCl}_2

electrolysis to decrease the operating temperature.

Lowering the melting point improves electrical conductivity and reduces thermal energy consumption.

2

Analyze the industrial preparation and chemical nature of calcium oxide.

Thermal decomposition of

\text{CaCO}_3

yields basic

\text{CaO}

, which does not react with basic gases like

\text{NH}_3

.

Acidic drying agents such as concentrated

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

would react with ammonia, making basic quicklime the required choice.

3

Determine the formula and thermal origin of Plaster of Paris.

Controlled heating of gypsum yields calcium sulfate hemihydrate (

\text{CaSO}_4\cdot\frac{1}{2}\text{H}_2\text{O}

).

Heating at

120^\circ\text{C}

drives off part of the water of crystallization without causing complete dehydration to anhydrous anhydrite.

4

Relate calcium hydroxide to its slaking reaction and analytical application.

Slaking

\text{CaO}

produces

\text{Ca(OH)}_2

, whose aqueous solution forms an insoluble milky

\text{CaCO}_3

precipitate with

\text{CO}_2

.

The reaction of dissolved calcium hydroxide with carbon(IV) oxide produces insoluble calcium trioxocarbonate(IV).

Key Concept

Chemical properties, industrial preparations, and extraction roles of calcium and its major compounds.

Question 8274Question →

An electrovalent compound is insoluble in water when its lattice enthalpy is smaller in magnitude than the total hydration enthalpy of its constituent gaseous ions.

True

False

Show answer & explanation

Answer: False

Answer

False. An electrovalent compound is soluble in water when the magnitude of its hydration enthalpy exceeds its lattice enthalpy, allowing ion-water electrostatic attractions to overcome ionic crystal lattice forces.

The statement is false because for an electrovalent compound to dissolve in water, the hydration energy released when ions interact with water molecules must overcome the lattice energy holding the crystal together. If hydration enthalpy is greater in magnitude than lattice enthalpy, the compound is soluble rather than insoluble.

Step-by-Step Solution

1

Identify the enthalpy changes during the dissolution of an ionic solid.

Dissolution depends on two key thermodynamic quantities: lattice enthalpy (energy required to separate solid ions into gaseous ions) and hydration enthalpy (energy released when gaseous ions are solvated by water).

The overall enthalpy of solution is approximated by

\Delta H_{\text{solution}} = \Delta H_{\text{lattice}} + \Delta H_{\text{hydration}}

.

2

Evaluate the condition where

|\Delta H_{\text{lattice}}| < |\Delta H_{\text{hydration}}|

.

The energy released during ion hydration is greater than the energy required to break the ionic lattice.

This leads to an exothermic dissolution process (

\Delta H_{\text{solution}} < 0

), which strongly favors the compound dissolving in water.

3

Determine the truth value of the statement.

The statement asserts that such a compound is insoluble, which contradicts chemical thermodynamic principles.

Therefore, the given statement is false.

Key Concept

Lattice Enthalpy vs Hydration Enthalpy in Ionic Compound Solubility

Question 8275Question →

In water purification and industrial wastewater management, specific chemical and physical processes are used to eliminate target pollutants. Match each water treatment process on the left with its corresponding chemical function on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Coagulation matches with clumping fine suspended solids using coagulants such as potash alum; Aeration matches with expelling dissolved volatile gases and oxidising soluble iron compounds; Chlorination matches with destroying pathogenic microorganisms to sanitize water; Activated Carbon Filtration matches with adsorbing dissolved organic impurities, dyes, and unpleasant odours.

Each process corresponds directly to its functional role in water purification: Coagulation uses coagulants like alum to clump fine suspended matter; Aeration strips unpleasant volatile gases and oxidises soluble metals; Chlorination kills disease-causing microorganisms; Activated Carbon Filtration adsorbs dissolved organic impurities and odours.

Step-by-Step Solution

1

Identify the primary chemical mechanism of Coagulation

Coagulation uses salts like alum to neutralize particle charges, resulting in the aggregation of fine suspended matter into larger settled masses.

Alum provides trivalent cations (

Al^{3+}

) to destabilize colloidal suspensions.

2

Identify the primary chemical mechanism of Aeration

Aeration increases dissolved oxygen to oxidize dissolved ferrous iron (

Fe^{2+}

) to ferric iron (

Fe^{3+}

) and strips out foul-smelling gases like

H_2S

.

Physical gas exchange and oxidation improve water taste and clarity.

3

Identify the primary biological mechanism of Chlorination

Chlorination serves as the final disinfection stage to eliminate harmful biological pathogens.

Chlorine generates active oxidizing species (

HOCl / OCl^-

) that rupture bacterial cell walls.

4

Identify the primary physical mechanism of Activated Carbon Filtration

Activated carbon adsorbs non-polar organic contaminants, synthetic detergents, and residual pigments due to its extremely porous structure.

High internal surface area promotes strong Van der Waals forces to trap organic pollutants.

Key Concept

Municipal and Industrial Water Treatment Processes

Question 8276Question →

A $50.0\text{ cm}^3$ gaseous mixture consisting of equal volumes of carbon(II) oxide and carbon(IV) oxide is passed through an excess solution of concentrated potassium hydroxide. What volume of gas remains unabsorbed, and which oxide of carbon was removed from the mixture?

25.0\text{ cm}^3

, with carbon(IV) oxide being removed

25.0\text{ cm}^3

, with carbon(II) oxide being removed

50.0\text{ cm}^3

, with neither gas being removed

0.0\text{ cm}^3

, with both gases being removed

Show answer & explanation

Answer: $25.0\text{ cm}^3$ , with carbon(IV) oxide being removed

Answer

25.0\text{ cm}^3

of gas remains unabsorbed, with carbon(IV) oxide being the component removed by potassium hydroxide solution.

Carbon(IV) oxide (

\text{CO}_2

) is an acidic oxide that dissolves and reacts with concentrated alkali solutions such as potassium hydroxide to form potassium trioxocarbonate(IV) and water. Since the mixture contains

25.0\text{ cm}^3

of

\text{CO}_2

and

25.0\text{ cm}^3

of neutral carbon(II) oxide (

\text{CO}

), only

\text{CO}_2

is absorbed, leaving

25.0\text{ cm}^3

of

\text{CO}

gas remaining.

Step-by-Step Solution

1

Determine the initial volume of each gas component in the mixture.

The mixture has

50.0\text{ cm}^3

total volume with equal proportions, giving

25.0\text{ cm}^3

of

\text{CO}

and

25.0\text{ cm}^3

of

\text{CO}_2

.

Equal volume proportions mean each gas makes up half of the total volume.

2

Analyze the reaction of each gas with concentrated potassium hydroxide (

\text{KOH}

) solution.

\text{CO}_2

reacts via

2\text{KOH(aq)} + \text{CO}_2\text{(g)} \rightarrow \text{K}_2\text{CO}_3\text{(aq)} + \text{H}_2\text{O(l)}

, absorbing all

25.0\text{ cm}^3

of

\text{CO}_2

.

\text{CO}

does not react.

\text{CO}_2

is an acidic oxide which neutralizes alkali solutions, whereas

\text{CO}

is a neutral oxide.

3

Calculate the unabsorbed volume of gas.

25.0\text{ cm}^3

of

\text{CO}

remains as unabsorbed gas.

Subtracting the absorbed volume of

\text{CO}_2

(

25.0\text{ cm}^3

) from total volume (

50.0\text{ cm}^3

) leaves

25.0\text{ cm}^3

.

Key Concept

Chemical Behavior of Oxides of Carbon toward Alkalis

Question 8277Question →

Arrange the following sequential stages involved in the industrial isolation of pure nitrogen gas from atmospheric air via fractional distillation in the correct chronological order, from ambient air intake to nitrogen gas collection.

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Answer

The correct chronological order of the process is: 1. Removal of dust, moisture, and carbon(IV) oxide; 2. Compression and Joule-Thomson expansion to produce liquid air; 3. Feeding liquid air into a fractionating column and gradual warming; 4. Selective vaporization and collection of nitrogen gas at -196 °C.

The industrial preparation of nitrogen gas relies on fractional distillation of liquid air. Ambient air must first be purified of water vapor and carbon(IV) oxide to prevent cryogenic equipment blockages caused by solid ice formation. The clean air is compressed under high pressure (around 200 atmospheres) and subjected to rapid Joule-Thomson expansion, which cools it repeatedly until it condenses into liquid air at approximately

-200\text{ }^\circ\text{C}

. When liquid air is fed into a fractionating column and warmed gradually, nitrogen gas boils off first at

-196\text{ }^\circ\text{C}

(

77\text{ K}

) due to having a lower boiling point than argon (

-186\text{ }^\circ\text{C}

) and oxygen (

-183\text{ }^\circ\text{C}

).

Step-by-Step Solution

1

Identify the initial purification stage required before gas liquefaction.

Air is scrubbed to remove dust, water vapor, and

CO_2

.

Freezing points of water (

0\text{ }^\circ\text{C}

) and carbon(IV) oxide (

-78.5\text{ }^\circ\text{C}

) are much higher than liquefaction temperature, meaning they would solidify and clog cryogenic tubes if not removed first.

2

Determine the phase change process of purified gaseous air into liquid air.

Purified air is compressed to

\sim 200\text{ atm}

and allowed to expand through a fine nozzle.

The Joule-Thomson expansion causes progressive cooling until air liquefies around

-200\text{ }^\circ\text{C}

.

3

Analyze the fractionating column feed and thermal gradient.

Liquid air enters the fractionating column and is warmed slowly.

Gradual heating drives components with lower boiling points to vaporize first.

4

Compare boiling points to establish which component distills first.

Nitrogen boils off at

-196\text{ }^\circ\text{C}

, followed by argon (

-186\text{ }^\circ\text{C}

) and oxygen (

-183\text{ }^\circ\text{C}

).

Nitrogen has the lowest boiling point, so it boils off first as a gas at the top of the column.

Key Concept

Fractional Distillation of Liquid Air for Industrial Production of Nitrogen

Question 8278Question →

During a chemical reaction between zinc metal and dilute hydrochloric acid, the volume of hydrogen gas produced was recorded over time. At $10\text{ s}$ , the total volume of hydrogen gas collected was $15.0\text{ cm}^3$ , and at $30\text{ s}$ , the volume collected reached $45.0\text{ cm}^3$ . What is the average rate of hydrogen gas evolution in $\text{cm}^3\text{ s}^{-1}$ over this $20\text{-second}$ time interval?

Show answer & explanation

Answer: 1.5

Answer

The average rate of evolution of hydrogen gas over the time interval is

1.5\text{ cm}^3\text{ s}^{-1}

.

The average rate of reaction is calculated by dividing the volume of gas produced during the interval (

\Delta V = 45.0\text{ cm}^3 - 15.0\text{ cm}^3 = 30.0\text{ cm}^3

) by the elapsed time (

\Delta t = 30\text{ s} - 10\text{ s} = 20\text{ s}

), yielding

1.5\text{ cm}^3\text{ s}^{-1}

.

Step-by-Step Solution

1

Determine the volume of hydrogen gas evolved during the specified time interval.

\Delta V = 45.0\text{ cm}^3 - 15.0\text{ cm}^3 = 30.0\text{ cm}^3

The volume change represents the quantity of product formed specifically between

10\text{ s}

and

30\text{ s}

.

2

Determine the time elapsed over the interval.

\Delta t = 30\text{ s} - 10\text{ s} = 20\text{ s}

The reaction rate is measured over the duration between the two observations.

3

Compute the average rate of gas evolution.

\text{Rate} = \frac{\Delta V}{\Delta t} = \frac{30.0\text{ cm}^3}{20\text{ s}} = 1.5\text{ cm}^3\text{ s}^{-1}

The average rate of reaction with respect to gas volume is defined as the change in volume per unit time.

Key Concept

Average Rate of Reaction

Question 8279Question →

A sample of hydrated magnesium tetraoxosulfate(VI) has the formula $MgSO_4 \cdot xH_2O$ . If the relative molecular mass of the hydrated salt is $246$ , what is the value of $x$ ? $[Mg = 24, S = 32, O = 16, H = 1]$

5

6

7

10

Show answer & explanation

Answer: $7$

Answer

The value of

x

is

7

.

The relative formula mass of anhydrous

MgSO_4

is calculated as

24 + 32 + (4 \times 16) = 120

. Subtracting this from the total relative molecular mass of

246

leaves

126

for the water of crystallization (

xH_2O

). Since each

H_2O

molecule has a relative mass of

(2 \times 1) + 16 = 18

, dividing

126

by

18

gives

x = 7

.

Step-by-Step Solution

1

Calculate the relative formula mass of the anhydrous salt

MgSO_4

.

24 + 32 + (4 \times 16) = 120

Summing the relative atomic masses of magnesium, sulfur, and four oxygen atoms gives the mass of the anhydrous portion.

2

Calculate the relative molecular mass of a single water molecule

H_2O

.

(2 \times 1) + 16 = 18

Two hydrogen atoms and one oxygen atom combine to give a relative molecular mass of 18.

3

Set up the equation for the relative molecular mass of the hydrated salt and solve for

x

.

120 + 18x = 246 \implies 18x = 126 \implies x = 7

Subtracting the anhydrous mass of 120 from the total mass of 246 gives 126 for the water content, and dividing by 18 yields 7 molecules of water of crystallization.

Key Concept

Relative Molecular Mass of Hydrated Salts

Estimated Time:1m 30s

Question 8280Question →

Match each mixture separation scenario on the left with its corresponding distillation requirement or thermal principle on the right.

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Answer

The correct matches pair liquid mixtures with small boiling point differences to fractional distillation involving fractionating columns, non-volatile solute solutions to simple distillation, industrial liquefied gases to cryogenic fractional distillation, and widely separated boiling point liquids to simple distillation.

Each mixture scenario is accurately paired according to volatility differences, non-volatility of solid solutes, temperature threshold limits (

25^\circ\text{C}

difference rule), and specialized industrial gas liquefaction conditions.

Step-by-Step Solution

1

Analyze the thermal criteria for distillation methods based on boiling point differences.

Miscible liquids with boiling point differences less than

25^\circ\text{C}

require a fractionating column (fractional distillation), whereas differences exceeding

50^\circ\text{C}

allow single-stage simple distillation.

Fractionating columns create a temperature gradient allowing multiple vaporizations and condensations.

2

Evaluate the volatility of solutes in liquid solutions.

Dissolved solid salts are non-volatile and remain in the boiling flask while solvent vapor turns into pure distillate.

Non-volatile compounds do not contribute to the vapor phase pressure at distillation temperatures.

3

Examine industrial gas separation requirements.

Gaseous air components must first be liquefied cryogenically before undergoing fractional distillation to separate components with close boiling points like argon and nitrogen.

Distillation requires liquid-vapor equilibrium, which only exists for air at cryogenic temperatures.

Key Concept

Simple vs Fractional Distillation Operational Criteria

Answer

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