All practice questions

Question 8401Question →

Match each thermodynamic condition or combination on the left with its corresponding reaction spontaneity description on the right.

Click a left item, then click its matching right item

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Answer

\Delta G < 0

matches with 'The reaction is spontaneous under the given conditions';

\Delta G = 0

matches with 'The system has reached dynamic equilibrium';

\Delta H < 0

and

\Delta S > 0

matches with 'The reaction is spontaneous at all temperatures';

\Delta H > 0

and

\Delta S < 0

matches with 'The reaction is non-spontaneous at all temperatures'.

Matching

\Delta G < 0

to spontaneity under specified conditions,

\Delta G = 0

to dynamic equilibrium,

\Delta H < 0

and

\Delta S > 0

to spontaneity at all temperatures, and

\Delta H > 0

and

\Delta S < 0

to non-spontaneity at all temperatures follows directly from the Gibbs-Helmholtz relation

\Delta G = \Delta H - T\Delta S

.

Step-by-Step Solution

1

Recall the fundamental thermodynamic criterion for spontaneity involving Gibbs free energy change

(\Delta G)

.

A reaction is spontaneous when

\Delta G < 0

, non-spontaneous when

\Delta G > 0

, and at dynamic equilibrium when

\Delta G = 0

.

Gibbs free energy combines enthalpy and entropy factors to determine direction of feasible chemical changes.

2

Analyze the Gibbs-Helmholtz equation

\Delta G = \Delta H - T\Delta S

for sign combinations.

If

\Delta H < 0

and

\Delta S > 0

,

\Delta G = (-\text{value}) - T(+\text{value})

, which is always negative at any absolute temperature

T > 0\text{ K}

.

Exothermic enthalpy releases energy while positive entropy increases disorder, driving spontaneity unconditionally.

3

Evaluate the opposing sign combination where

\Delta H > 0

and

\Delta S < 0

.

\Delta G = (+\text{value}) - T(-\text{value}) = +\text{value} + T(\text{value})

, which is always positive.

Endothermic process with decreasing entropy is thermodynamically unfavorable at all temperatures.

Key Concept

Gibbs Free Energy Equation and Reaction Spontaneity

Question 8402Question →

During the industrial extraction of iron in a blast furnace, limestone ( $\text{CaCO}_3$ ) is added to remove silica impurities ( $\text{SiO}_2$ ). Arrange the following stages of slag formation and separation in their correct chronological sequence from first to last.

Drag items to arrange them in the correct order

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Answer

The correct chronological sequence is: Thermal decomposition of limestone (

\text{CaCO}_3

) into basic calcium oxide (

\text{CaO}

) and carbon(IV) oxide (

\text{CO}_2

)

\rightarrow

Acid-base reaction between basic calcium oxide (

\text{CaO}

) flux and acidic silicon(IV) oxide (

\text{SiO}_2

) impurity

\rightarrow

Formation of molten calcium trioxosilicate(IV) (

\text{CaSiO}_3

)

\rightarrow

Collection of molten slag as a protective layer floating above the denser liquid pig iron at the hearth.

Limestone (

\text{CaCO}_3

) decomposes thermally under high temperatures into calcium oxide (

\text{CaO}

) and carbon(IV) oxide (

\text{CO}_2

). Next, the basic

\text{CaO}

flux reacts with acidic silica (

\text{SiO}_2

) impurities in an acid-base neutralization to form molten calcium trioxosilicate(IV) (

\text{CaSiO}_3

, slag). Finally, the molten slag collects at the hearth and floats on top of the denser molten pig iron, forming a protective layer.

Step-by-Step Solution

1

Identify the initial thermal reaction of limestone in the furnace.

Limestone (

\text{CaCO}_3

) decomposes at around

800^\circ\text{C}-1000^\circ\text{C}

to yield calcium oxide (

\text{CaO}

) and carbon(IV) oxide (

\text{CO}_2

).

Calcium oxide (

\text{CaO}

) must first be synthesized to act as a basic flux.

2

Determine the chemical interaction between the flux and raw ore impurities.

Basic calcium oxide (

\text{CaO}

) reacts with acidic silicon(IV) oxide (

\text{SiO}_2

).

Silica is the main acidic impurity in hematite ore, requiring neutralization by the basic flux.

3

Identify the chemical product formed from this reaction.

Molten calcium trioxosilicate(IV) (

\text{CaSiO}_3

, slag) is formed via the reaction

\text{CaO(s)} + \text{SiO}_2\text{(s)} \rightarrow \text{CaSiO}_3\text{(l)}

.

Neutralization of silica forms the molten compound known as slag.

4

Describe the physical separation of slag at the hearth.

The molten slag drains down to the hearth and floats above the denser liquid iron layer.

Density differences allow slag to float on molten iron, preventing its re-oxidation by blast air.

Key Concept

Slag Formation and Impurity Removal in the Blast Furnace

Question 8403Question →

Which of the following represents the correct sequential procedure for separating two immiscible liquids using a separating funnel?

Drag items to arrange them in the correct order

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Answer

The correct procedural order begins with pouring the mixture into the closed funnel, shaking and venting pressure, allowing the layers to settle undisturbed, and finally removing the stopper to drain the lower layer.

The correct laboratory sequence requires securing the tap before filling, shaking with periodic pressure venting, resting the funnel to allow gravitational separation into distinct layers based on density, and finally removing the top stopper before draining the denser lower layer through the stopcock.

Step-by-Step Solution

1

Ensure the stopcock is closed and pour the liquid mixture into the funnel.

The mixture is safely contained within the separating funnel.

Prevents accidental loss of liquid before separation begins.

2

Stopper the funnel, invert it, and shake gently while opening the stopcock to release vapor pressure.

Vapor pressure is released safely without building up dangerous pressure inside the stoppered glass vessel.

Shaking ensures proper contact between liquid phases, while venting prevents pressure buildup.

3

Mount the funnel upright in a ring stand and allow it to stand undisturbed.

Two distinct phase layers separated by a visible interface form due to differences in density and immiscibility.

Gravity forces the denser liquid to the bottom layer and the lighter liquid to the top layer.

4

Unstopper the top of the funnel and turn the stopcock to drain the bottom layer into a container.

The lower denser layer is collected separately, completing the physical separation.

Removing the stopper prevents a negative pressure vacuum that would prevent the liquid from flowing out.

Key Concept

Standard procedural steps for separating immiscible liquids using a separating funnel

Question 8404Question →

Match each water treatment chemical or industrial effluent contaminant on the left with its correct chemical action or environmental impact on the right.

Click a left item, then click its matching right item

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Answer

The correct pairings connect calcium hydroxide and sodium carbonate to the precipitation of hardness-causing cations; non-biodegradable detergents to persistent foam formation that inhibits aeration; high BOD effluent to microbial oxygen depletion; and heavy metal ions to trophic bioaccumulation and neurological toxicity.

The correct pairings accurately link chemical reagents (

Ca(OH)_2, Na_2CO_3

) to precipitation of hardness-causing metal ions, non-biodegradable synthetic detergents to persistent surface foam formation, high BOD organic waste to microbial depletion of dissolved oxygen, and heavy metal ions (

Pb^{2+}, Hg^{2+}

) to biological biomagnification.

Step-by-Step Solution

1

Analyze the chemical function of lime and soda ash in water treatment

Calcium hydroxide (

Ca(OH)_2

) converts soluble

Ca(HCO_3)_2

into insoluble

CaCO_3

, and sodium carbonate (

Na_2CO_3

) precipitates

Ca^{2+}

and

Mg^{2+}

ions as insoluble carbonates (

CaCO_3

and

MgCO_3

).

Chemical precipitation using lime-soda softening removes both temporary and permanent water hardness.

2

Evaluate the environmental impact of branched alkylbenzene sulfonate detergents

Because soil and aquatic microbes lack enzymes to degrade branched-chain alkylbenzene sulfonates, these surfactants accumulate as stable surface foams.

Surface foam restricts light transmission required for photosynthesis and reduces re-aeration at the water-air boundary.

3

Relate Biochemical Oxygen Demand (BOD) to organic pollution effects

High BOD signifies extensive organic pollution, fueling exponential growth of decomposer bacteria that rapidly absorb dissolved oxygen.

Aerobic respiration by decomposers lowers the dissolved oxygen concentration, creating hypoxic aquatic conditions.

4

Assess the biological toxicity of heavy metal effluents

Heavy metal ions such as

Pb^{2+}

and

Hg^{2+}

are non-biodegradable toxins that accumulate in fatty tissues and biomagnify up food chains.

Inhibition of key metabolic enzymes by heavy metals causes chronic neurological and physiological damage.

Key Concept

Chemical Methods of Water Softening, Effluent Pollution Mechanisms, and Ecological Impacts

Question 8405Question →

But-2-ene exhibits geometric (cis-trans) isomerism because each carbon atom involved in the double bond is bonded to two non-identical groups, whereas but-1-ene does not exhibit geometric isomerism.

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because geometric isomerism requires restricted rotation about the

C=C

bond together with two distinct groups attached to each double-bonded carbon atom—a condition satisfied by but-2-ene but not by but-1-ene.

The statement correctly describes the structural rule for geometric isomerism in alkenes. But-2-ene meets the condition because both double-bonded carbons carry two non-identical groups (

-H

and

-CH_3

), whereas but-1-ene fails the condition because its terminal carbon carries two identical hydrogen atoms.

Step-by-Step Solution

1

Identify the structural requirements for geometric (cis-trans) isomerism.

Geometric isomerism in alkenes requires a rigid

C=C

double bond where each of the two unsaturated carbon atoms is attached to two non-identical substituents.

If either carbon atom of the double bond carries two identical groups, swapping those groups produces an identical molecule rather than a distinct stereoisomer.

2

Analyze the substituent groups attached to the double-bonded carbons in but-2-ene (

CH_3-CH=CH-CH_3

).

Carbon-2 is attached to

-H

and

-CH_3

, and Carbon-3 is also attached to

-H

and

-CH_3

.

Since both double-bonded carbons have two different groups attached, but-2-ene exists as two stereoisomers: cis-but-2-ene and trans-but-2-ene.

3

Analyze the substituent groups attached to the double-bonded carbons in but-1-ene (

CH_2=CH-CH_2-CH_3

).

Carbon-1 is attached to two identical hydrogen atoms (

-H

and

-H

).

The presence of two identical hydrogen atoms on Carbon-1 prevents the formation of cis-trans isomers for but-1-ene.

Key Concept

Structural Criteria for Geometric (Cis-Trans) Isomerism

Question 8406Question →

What volume of chlorine gas, measured at STP, is required for the complete reaction with a given mass of iron metal?

Fill in the blanks below

When

11.2\text{ g}

of iron reacts completely with excess dry chlorine gas according to the balanced equation:

2Fe(s) + 3Cl_2(g) \rightarrow 2FeCl_3(s)

the volume of chlorine gas consumed at STP is

\text{dm}^3

.
[Relative atomic mass:

\text{Fe} = 56

; Molar gas volume at STP

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

]

Show answer & explanation

Answer

The volume of chlorine gas consumed at STP is

6.72\text{ dm}^3

.

To find the volume of

Cl_2

gas consumed at STP, first determine the amount of

Fe

in moles:

\text{Moles of } Fe = \frac{11.2\text{ g}}{56\text{ g mol}^{-1}} = 0.20\text{ mol}

. From the balanced equation

2Fe(s) + 3Cl_2(g) \rightarrow 2FeCl_3(s)

,

2\text{ moles of } Fe

react with

3\text{ moles of } Cl_2

. Therefore,

0.20\text{ mol of } Fe

requires

0.20 \times \frac{3}{2} = 0.30\text{ mol of } Cl_2

. At STP,

1\text{ mole of gas}

occupies

22.4\text{ dm}^3

, so the volume of

Cl_2

is

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

.

Step-by-Step Solution

1

Calculate the number of moles of iron reacted

\text{Moles of } Fe = \frac{11.2\text{ g}}{56\text{ g mol}^{-1}} = 0.20\text{ mol}

Convert the mass of iron to moles using its relative atomic mass.

2

Determine the moles of chlorine gas (

Cl_2

) required using the stoichiometric mole ratio

\text{Moles of } Cl_2 = 0.20\text{ mol } Fe \times \frac{3\text{ mol } Cl_2}{2\text{ mol } Fe} = 0.30\text{ mol}

The balanced chemical equation shows that

2\text{ moles of } Fe

react with

3\text{ moles of } Cl_2

(a

2:3

mole ratio).

3

Calculate the volume of

Cl_2

gas consumed at STP

\text{Volume of } Cl_2 = 0.30\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 6.72\text{ dm}^3

Multiply the number of moles of

Cl_2

by the molar volume of a gas at STP.

Key Concept

Mass-Volume Stoichiometric Calculations at STP

Estimated Time:2m 0s

Question 8407Question →

What is the oxidation state of the central iron atom in the hexacyanoferrate(III) complex ion, $[Fe(CN)_6]^{3-}$ ?

+3

+2

+6

-3

Show answer & explanation

Answer: $+3$

Answer

The oxidation state of the central iron atom in

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

is

+3

.

In the complex ion

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

, six anionic cyanide ligands (

CN^-

), each with a

-1

charge, bond to the central iron atom. The sum of the oxidation state of iron (

x

) and the total ligand charge (

6 \times -1 = -6

) equals the overall charge of the complex ion (

-3

). Solving

x - 6 = -3

gives

x = +3

.

Step-by-Step Solution

1

Identify the charge of the ligands and the net charge of the complex ion.

Each cyanide ligand (

CN^-

) carries a

-1

charge. The overall complex ion has a net charge of

-3

.

Ligand charges must sum with the central metal oxidation state to equal the overall ion charge.

2

Set up an algebraic equation for the oxidation state of iron (

x

).

x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3

.

Solving the linear algebraic equation yields the precise oxidation state of the central atom.

Key Concept

Determination of central metal oxidation state in complex ions

Estimated Time:45s

Question 8408Question →

Read the passage below:

The clerk counted the crumpled notes for the third time, his fingers trembling under the weight of the magistrate's cold gaze across the polished mahogany desk. Outside, rain beat a relentless rhythm against the iron shutters. 'Well, Ezekiel?' the magistrate murmured, his voice smooth, like a silk ribbon drawn slowly over a sharpened blade. 'Does the ledger balance, or must we consult the constable's memory?' Ezekiel swallowed hard, laying the stack down without a word.

What is the author's tone toward the magistrate, and what overarching mood does the excerpt establish for the reader?

The tone is critical and sinister, while the mood is tense and ominous.

The tone is tense and ominous, while the mood is critical and sinister.

The tone is sympathetic and gentle, while the mood is calm and reflective.

The tone is humorous and satirical, while the mood is lighthearted and cheerful.

Show answer & explanation

Answer: The tone is critical and sinister, while the mood is tense and ominous.

Answer

The author's tone toward the magistrate is critical and sinister, while the overall mood created for the reader is tense and ominous.

The correct option accurately distinguishes between tone (the author's critical, sinister attitude toward the threatening magistrate) and mood (the tense, ominous feeling created by Ezekiel's fear and the dark setting).

Step-by-Step Solution

1

Analyze the literary devices and description of the magistrate to identify the author's tone.

The comparison of the magistrate's voice to 'a silk ribbon drawn slowly over a sharpened blade' shows a critical attitude that emphasizes disguised cruelty and menace.

Tone reflects the writer's attitude toward a character or subject.

2

Analyze the atmosphere, sensory details, and character reactions to determine the mood.

The trembling clerk, the relentless rain beating against iron shutters, and the implicit threat of arrest build a heavy, frightening atmosphere.

Mood is the emotional feeling or atmosphere generated in the reader by the text.

3

Synthesize tone and mood to select the matching interpretation.

The writer adopts a critical/sinister tone toward the magistrate, creating a tense/ominous mood.

Differentiating authorial perspective (tone) from ambient emotion (mood) ensures correct literary analysis.

Key Concept

Distinguishing between Tone and Mood in Unseen Prose

Estimated Time:1m 30s

Question 8409Question →

A fixed mass of gas is sealed inside a rigid vessel. Complete the following statement by calculating the final pressure after temperature change.

Fill in the blanks below

A gas sealed in a rigid container exerts a pressure of

240\text{ kPa}

at

127^\circ\text{C}

. When cooled at constant volume to

-73^\circ\text{C}

, the final pressure of the gas is

\text{kPa}

.

Show answer & explanation

Answer

120

According to Gay-Lussac's Pressure Law, for a given mass of gas at constant volume, pressure is directly proportional to absolute temperature (

P \propto T

). Converting temperatures to Kelvin gives

T_1 = 400\text{ K}

and

T_2 = 200\text{ K}

. Substituting into

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

yields

P_2 = 240 \times \frac{200}{400} = 120\text{ kPa}

.

Step-by-Step Solution

1

Convert initial and final temperatures from Celsius to Kelvin using

T = t + 273

.

T_1 = 127 + 273 = 400\text{ K}

and

T_2 = -73 + 273 = 200\text{ K}

.

Gas law calculations require thermodynamic absolute temperature in Kelvin.

2

Apply Pressure Law formula

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

at constant volume.

\frac{240}{400} = \frac{P_2}{200}

.

Pressure of a fixed mass of gas is directly proportional to absolute temperature when volume remains constant.

3

Solve for final pressure

P_2

.

P_2 = 240 \times \frac{200}{400} = 240 \times 0.5 = 120\text{ kPa}

.

Halving absolute temperature halves the pressure exerted by gas molecules.

Key Concept

Pressure Law (Gay-Lussac's Law of Temperature-Pressure)

Estimated Time:1m 30s

Question 8410Question →

An agricultural soil contaminated by acid mine drainage displays an abnormally low pH of $4.0$ , causing heavy metal ions such as $\text{Pb}^{2+}$ and $\text{Cd}^{2+}$ to remain highly soluble and toxic to crops. Which chemical substance is most appropriate to treat the soil to reduce acidity and precipitate these toxic metal ions?

Calcium carbonate (

\text{CaCO}_3

)

Ammonium tetraoxosulfate(VI) ((

\text{NH}_4)_2\text{SO}_4

)

Sodium chloride (

\text{NaCl}

)

Trioxonitrate(V) acid (

\text{HNO}_3

)

Show answer & explanation

Answer: Calcium carbonate ( $\text{CaCO}_3$ )

Answer

Calcium carbonate (

\text{CaCO}_3

) is the most suitable chemical agent for treating acidic, heavy-metal-polluted soil.

Calcium carbonate (

\text{CaCO}_3

) acts as a basic liming material. When applied to acidic soil, it reacts with hydrogen ions to raise the soil pH toward neutral levels. As the pH increases, heavy metal cations such as

\text{Pb}^{2+}

and

\text{Cd}^{2+}

form insoluble hydroxides and carbonates, immobilizing them and preventing plant absorption or groundwater leaching.

Step-by-Step Solution

1

Identify the cause of heavy metal mobility in soil.

Low soil pH (acidic conditions) keeps heavy metal ions in soluble, bioavailable forms like

\text{Pb}^{2+}

and

\text{Cd}^{2+}

.

Solubility of heavy metal cations increases significantly under acidic conditions.

2

Determine the required chemical treatment to remediate soil acidity.

Adding a basic neutralizing agent (liming) increases soil pH.

Neutralization lowers

[\text{H}^+]

concentration in the soil solution.

3

Evaluate the effect of raising soil pH on dissolved metal cations.

Heavy metal ions react with carbonate and hydroxide ions to form insoluble precipitates such as

\text{Pb(OH)}_2

and

\text{CdCO}_3

.

Raising pH decreases the solubility of heavy metals, immobilizing them in soil matrix.

Key Concept

Soil Liming and Heavy Metal Immobilization

Question 8411Question →

Match each aqueous salt solution to its characteristic effect on litmus paper at $25^\circ\text{C}$ .

Click a left item, then click its matching right item

Show answer & explanation

Answer

Ammonium chloride solution (

NH_4Cl(aq)

) matches with 'Turns blue litmus paper red (

pH < 7

)'; Sodium ethanoate solution (

CH_3COONa(aq)

) matches with 'Turns red litmus paper blue (

pH > 7

)'; Sodium chloride solution (

NaCl(aq)

) matches with 'Has no effect on either red or blue litmus paper (

pH = 7

)'.

Salt hydrolysis determines the acidity or alkalinity of an aqueous salt solution based on the strengths of the parent acid and base. Ammonium chloride (

NH_4Cl

) yields acidic solutions (

pH < 7

) turning blue litmus red due to

NH_4^+

cation hydrolysis. Sodium ethanoate (

CH_3COONa

) produces alkaline solutions (

pH > 7

) turning red litmus blue due to

CH_3COO^-

anion hydrolysis. Sodium chloride (

NaCl

) consists of spectator ions from a strong acid and strong base, undergoing no hydrolysis and remaining neutral (

pH = 7

).

Step-by-Step Solution

1

Analyze the parent acid and base for Ammonium chloride (

NH_4Cl

).

NH_4Cl

forms from

HCl

(strong acid) and

NH_3

(weak base). Cation hydrolysis occurs:

NH_4^+(aq) + H_2O(l) \rightleftharpoons NH_3(aq) + H_3O^+(aq)

. Excess

H_3O^+

turns blue litmus red.

Salts of strong acids and weak bases yield acidic solutions.

2

Analyze the parent acid and base for Sodium ethanoate (

CH_3COONa

).

CH_3COONa

forms from

CH_3COOH

(weak acid) and

NaOH

(strong base). Anion hydrolysis occurs:

CH_3COO^-(aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)

. Excess

OH^-

turns red litmus blue.

Salts of weak acids and strong bases yield alkaline solutions.

3

Analyze the parent acid and base for Sodium chloride (

NaCl

).

NaCl

forms from

HCl

(strong acid) and

NaOH

(strong base). Neither ion undergoes hydrolysis. The solution remains neutral (

pH = 7

) and does not change litmus color.

Salts of strong acids and strong bases do not undergo hydrolysis.

Key Concept

Salt Hydrolysis and Solution Acidity/Alkalinity

Estimated Time:45s

Question 8412Question →

Match each chemical system in Column A with its correct outcome and rationale based on the electrochemical series in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Potassium with zinc sulfate matches spontaneous displacement due to a more negative reduction potential; Copper with hydrochloric acid matches no reaction due to a positive reduction potential relative to hydrogen; Fluorine with sodium chloride matches spontaneous halide oxidation due to a higher reduction potential; Silver with gold(III) nitrate matches spontaneous metal displacement because silver acts as a stronger reducing agent than gold.

Each chemical system correctly pairs with its electrochemical behavior: Potassium displaces Zinc because it possesses a more negative reduction potential; Copper does not react with dilute acid because its reduction potential is positive relative to hydrogen; Fluorine displaces chloride ions because it has a higher reduction potential and thus greater oxidizing power; Silver displaces gold ions because silver has a lower reduction potential than gold, making it the stronger reducing agent.

Step-by-Step Solution

1

Analyze position of Potassium and Zinc in the electrochemical series.

Potassium (

E^\circ = -2.93\text{ V}

) has a more negative reduction potential than Zinc (

E^\circ = -0.76\text{ V}

), making it a stronger reducing agent.

Metals with more negative standard reduction potentials spontaneously displace ions of metals below them in the series.

2

Evaluate the reactivity of Copper in non-oxidizing acid.

Copper (

E^\circ = +0.34\text{ V}

) lies below Hydrogen (

E^\circ = 0.00\text{ V}

) in the electrochemical series.

Metals with positive reduction potentials cannot spontaneously reduce hydrogen ions to evolve

\text{H}_2

gas.

3

Compare oxidizing strengths of Fluorine and Chlorine.

Fluorine (

E^\circ = +2.87\text{ V}

) has a higher reduction potential than Chlorine (

E^\circ = +1.36\text{ V}

).

A halogen with a higher reduction potential acts as a stronger oxidizing agent and displaces halide ions with lower reduction potentials.

4

Determine feasibility of displacement between Silver and Gold ions.

Silver (

E^\circ = +0.80\text{ V}

) is more easily oxidized than Gold (

E^\circ = +1.50\text{ V}

).

The metal with the smaller reduction potential acts as the reducing agent, resulting in a positive standard cell potential (

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

).

Key Concept

Electrochemical Series and Reaction Spontaneity

Estimated Time:1m 30s

Question 8413Question →

Which of the following transition metal compounds is used as the catalyst in the industrial manufacture of tetraoxosulfate(VI) acid by the Contact process?

Vanadium(V) oxide (

V_2O_5

)

Finely divided iron (

Fe

)

Finely divided nickel (

Ni

)

Manganese(IV) oxide (

MnO_2

)

Show answer & explanation

Answer: Vanadium(V) oxide ( $V_2O_5$ )

Answer

Vanadium(V) oxide (

V_2O_5

) is the catalyst used in the Contact process.

Vanadium(V) oxide (

V_2O_5

) is the primary catalyst used in the Contact process to catalyze the reversible oxidation of

SO_2

to

SO_3

due to the ability of vanadium to vary its oxidation state between +5 and +4 during the catalytic cycle.

Step-by-Step Solution

1

Identify the key chemical transformation in the Contact process

The oxidation of sulfur(IV) oxide gas to sulfur(VI) oxide gas (

2SO_2 + O_2 \rightleftharpoons 2SO_3

).

This exothermic reversible step requires a catalyst to achieve an optimal reaction rate at reasonable temperature.

2

Recall the industrial catalyst specifically employed for this conversion

Vanadium(V) oxide (

V_2O_5

) is used as the preferred catalyst at around 450 °C.

Transition metals and their compounds exhibit variable oxidation states, allowing

V_2O_5

to facilitate electron transfer steps efficiently.

Key Concept

Catalytic behavior of transition metal compounds in industrial processes

Estimated Time:45s

Question 8414Question →

When excess hydrogen sulfide gas (

H_2S

) reacts with sulfur(IV) oxide (

SO_2

) according to the balanced chemical equation:

2H_2S(g) + SO_2(g) \rightarrow 3S(s) + 2H_2O(l)

What mass of elemental sulfur is precipitated when

5.6\text{ dm}^3

of

SO_2

gas at STP reacts completely?
(Molar volume of gas at STP =

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

; Relative atomic mass:

S = 32

)

24.0\text{ g}

8.0\text{ g}

22.4\text{ g}

16.0\text{ g}

Show answer & explanation

Answer: $24.0\text{ g}$

Answer

The mass of elemental sulfur precipitated is

24.0\text{ g}

.

First, find the moles of

SO_2

at STP:

\frac{5.6\text{ dm}^3}{22.4\text{ dm}^3\text{ mol}^{-1}} = 0.25\text{ mol}

. According to the balanced equation,

1\text{ mole of } SO_2

produces

3\text{ moles of } S

. Thus,

0.25\text{ mol of } SO_2

produces

0.75\text{ mol of } S

. Multiplying by the relative atomic mass of sulfur (

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

) gives

0.75 \times 32 = 24.0\text{ g}

.

Step-by-Step Solution

1

Calculate the number of moles of

SO_2

gas reacting at STP.

\text{Moles of } SO_2 = \frac{\text{Volume at STP}}{\text{Molar volume at STP}} = \frac{5.6\text{ dm}^3}{22.4\text{ dm}^3\text{ mol}^{-1}} = 0.25\text{ mol}

.

Gas volume at STP is converted to moles using the standard molar volume of

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

.

2

Determine the mole ratio between

SO_2

and

S

from the balanced reaction equation.

From

2H_2S(g) + SO_2(g) \rightarrow 3S(s) + 2H_2O(l)

,

1\text{ mol of } SO_2

yields

3\text{ mol of } S

.

Stoichiometric coefficients give the exact molar equivalence between reactants and products.

3

Calculate the moles of

S

formed.

\text{Moles of } S = 0.25\text{ mol } SO_2 \times 3 = 0.75\text{ mol } S.

Multiplying moles of

SO_2

by the stoichiometric factor of 3 gives the moles of produced sulfur.

4

Convert the moles of sulfur to mass in grams.

\text{Mass of } S = 0.75\text{ mol} \times 32\text{ g mol}^{-1} = 24.0\text{ g}.

Mass is obtained by multiplying number of moles by relative atomic mass.

Key Concept

Gas stoichiometry at STP and redox reaction between hydrogen sulfide and sulfur(IV) oxide

Estimated Time:1m 30s

Question 8415Question →

Hydrogen gas reacts with oxygen gas to form water according to the balanced chemical equation:

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

If

4.0\text{ moles}

of hydrogen gas (

\text{H}_2

) are mixed with

3.0\text{ moles}

of oxygen gas (

\text{O}_2

) and allowed to react completely, which of the following correctly identifies the limiting reactant and the amount of excess reactant remaining?

\text{H}_2

is the limiting reactant, and

1.0\text{ mole}

of

\text{O}_2

remains unreacted.

\text{O}_2

is the limiting reactant, and

1.0\text{ mole}

of

\text{H}_2

remains unreacted.

\text{H}_2

is the limiting reactant, and

2.0\text{ moles}

of

\text{O}_2

remain unreacted.

\text{O}_2

is the limiting reactant, and

2.0\text{ moles}

of

\text{H}_2

remain unreacted.

Show answer & explanation

Answer: $\text{H}_2$ is the limiting reactant, and $1.0\text{ mole}$ of $\text{O}_2$ remains unreacted.

Answer

\text{H}_2

is the limiting reactant, and

1.0\text{ mole}

of

\text{O}_2

remains unreacted.

The balanced chemical equation indicates a

2:1

mole ratio between

\text{H}_2

and

\text{O}_2

. Reacting

4.0\text{ moles}

of

\text{H}_2

requires

2.0\text{ moles}

of

\text{O}_2

. Since

3.0\text{ moles}

of

\text{O}_2

are present,

\text{H}_2

is completely consumed first (limiting reactant), leaving

1.0\text{ mole}

of

\text{O}_2

unreacted as excess.

Step-by-Step Solution

1

Determine the mole ratio from the balanced equation

The mole ratio of

\text{H}_2

to

\text{O}_2

is

2 : 1

.

The stoichiometric coefficients in

2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}

dictate that

2\text{ moles}

of

\text{H}_2

consume

1\text{ mole}

of

\text{O}_2

.

2

Calculate the required moles of oxygen for the given hydrogen

\text{Required moles of } \text{O}_2 = 4.0\text{ moles } \text{H}_2 \times \frac{1\text{ mole } \text{O}_2}{2\text{ moles } \text{H}_2} = 2.0\text{ moles } \text{O}_2

.

This determines how much oxygen is actually needed to completely react with all

4.0\text{ moles}

of hydrogen.

3

Identify the limiting reactant and excess amount

Available

\text{O}_2 = 3.0\text{ moles}

, which is greater than the required

2.0\text{ moles}

. Therefore,

\text{H}_2

is the limiting reactant and excess

\text{O}_2 = 3.0 - 2.0 = 1.0\text{ mole}

.

The reactant that runs out first (

\text{H}_2

) limits the reaction, leaving unreacted excess oxygen.

Key Concept

Limiting Reactant Determination

Question 8416Question →

A $160\text{ cm}^3$ sample of an unknown gas $Z$ diffuses through a porous partition in $30\text{ seconds}$ . Under identical conditions of temperature and pressure, an $80\text{ cm}^3$ sample of sulfur(IV) oxide ( $SO_2$ ) gas diffuses through the same partition in $20\text{ seconds}$ . What is the relative molecular mass of gas $Z$ ? [Relative atomic masses: $S = 32$ , $O = 16$ ]

16\text{ g/mol}

36\text{ g/mol}

48\text{ g/mol}

85.3\text{ g/mol}

Show answer & explanation

Answer: $36\text{ g/mol}$

Answer

The relative molecular mass of gas Z is

36\text{ g/mol}

.

The rate of diffusion for gas Z is

160/30 = 16/3\text{ cm}^3/\text{s}

and for

SO_2

is

80/20 = 4\text{ cm}^3/\text{s}

. According to Graham's law,

(r_Z / r_{SO_2})^2 = M_{SO_2} / M_Z

. Substituting the values gives

(4/3)^2 = 16/9 = 64 / M_Z

, yielding

M_Z = 36\text{ g/mol}

.

Step-by-Step Solution

1

Calculate the molar mass of sulfur(IV) oxide (

SO_2

)

M_{SO_2} = 32 + (2 \times 16) = 64\text{ g/mol}

Molar mass of the reference gas is required for Graham's Law calculation.

2

Determine the rates of diffusion for gas

Z

and

SO_2

r_Z = \frac{160\text{ cm}^3}{30\text{ s}} = \frac{16}{3}\text{ cm}^3/\text{s}

and

r_{SO_2} = \frac{80\text{ cm}^3}{20\text{ s}} = 4\text{ cm}^3/\text{s}

Rate of diffusion is defined as volume of gas diffused per unit time (

r = V/t

).

3

Calculate the ratio of the diffusion rates

\frac{r_Z}{r_{SO_2}} = \frac{16/3}{4} = \frac{4}{3}

Comparing the two rates simplifies substitution into Graham's Law equation.

4

Apply Graham's Law of Diffusion to solve for the unknown molar mass (

M_Z

)

\frac{r_Z}{r_{SO_2}} = \sqrt{\frac{M_{SO_2}}{M_Z}} \implies \left(\frac{4}{3}\right)^2 = \frac{64}{M_Z} \implies \frac{16}{9} = \frac{64}{M_Z} \implies M_Z = \frac{64 \times 9}{16} = 36\text{ g/mol}

According to Graham's Law, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

Key Concept

Graham's Law of Diffusion relates gas diffusion rates to their molar masses:

r_1 / r_2 = \sqrt{M_2 / M_1}

.

Question 8417Question →

During solvent extraction in a laboratory, an organic solute dissolved in an aqueous solution is extracted using an organic solvent in a separating funnel. Which of the following conditions is essential for this separation method to work effectively?

The solute must be significantly more soluble in the added organic solvent, and the two liquid solvents must be immiscible.

The organic solvent must have a higher boiling point than water so that it settles rapidly at the bottom of the funnel.

The organic solvent must be completely miscible with water so that a single homogenous solution is formed.

The solute must undergo a chemical reaction with the organic solvent to form an insoluble solid precipitate.

Show answer & explanation

Answer: The solute must be significantly more soluble in the added organic solvent, and the two liquid solvents must be immiscible.

Answer

The solute must be significantly more soluble in the added organic solvent, and the two liquid solvents must be immiscible.

For solvent extraction to be effective, the added solvent must not mix with the original solvent (they must be immiscible to form two distinct layers), and the target solute must have a much higher solubility in the extracting solvent than in the original solvent.

Step-by-Step Solution

1

Identify the key requirement for solvent extraction using a separating funnel.

Two liquid layers must form (immiscibility), and the solute must prefer one solvent over the other (differential solubility).

Solvent extraction works by partitioning a solute between two immiscible liquids based on its relative solubility in each.

Key Concept

Solvent Extraction and Immiscibility Principles

Estimated Time:45s

Question 8418Question →

When iron filings are heated in a stream of dry chlorine gas, compound $X$ is formed. Conversely, when iron filings react with dilute hydrochloric acid, compound $Y$ is produced. What are the correct IUPAC names of compounds $X$ and $Y$ respectively?

Iron(III) chloride and iron(II) chloride

Iron(II) chloride and iron(III) chloride

Iron(III) chloride and iron(III) chloride

Iron(II) chloride and iron(II) chloride

Show answer & explanation

Answer: Iron(III) chloride and iron(II) chloride

Answer

Iron(III) chloride and iron(II) chloride

Dry chlorine gas acts as a powerful oxidizing agent that oxidizes iron directly to iron(III) chloride (

\text{FeCl}_3

). In contrast, reacting iron with dilute hydrochloric acid generates iron(II) chloride (

\text{FeCl}_2

) because the evolved hydrogen gas acts as a reducing agent, maintaining iron in the

+2

oxidation state.

Step-by-Step Solution

1

Analyze the reaction of iron with dry chlorine gas

2\text{Fe}_{(s)} + 3\text{Cl}_{2(g)} \rightarrow 2\text{FeCl}_{3(s)}

Chlorine gas is a powerful oxidizing agent capable of taking iron from oxidation state 0 to +3, forming iron(III) chloride.

2

Analyze the reaction of iron with dilute hydrochloric acid

\text{Fe}_{(s)} + 2\text{HCl}_{(aq)} \rightarrow \text{FeCl}_{2(aq)} + \text{H}_{2(g)}

Dilute hydrochloric acid oxidizes iron to iron(II) chloride (

+2

state). The hydrogen gas (

\text{H}_2

) evolved during the reaction acts as a reducing agent, preventing any further oxidation to iron(III).

3

Match the products to compound

X

and compound

Y

Compound

X

is iron(III) chloride and compound

Y

is iron(II) chloride.

Sequential identification following the given stem conditions.

Key Concept

Variable oxidation states of iron and differential oxidizing strength of chlorine versus hydrogen ions.

Estimated Time:1m 0s

Question 8419Question →

Consider the reversible endothermic gas-phase reaction represented by the thermochemical equation below:

PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \quad \Delta H = +92.5\text{ kJ mol}^{-1}

Which of the following conditions will shift the position of equilibrium to the right to favor the yield of gaseous products?

Increasing the temperature and decreasing the total pressure

Adding a suitable catalyst to the reaction vessel

Increasing the total pressure while lowering the temperature

Continuous removal of phosphorus pentachloride reactant gas

Show answer & explanation

Answer: Increasing the temperature and decreasing the total pressure

Answer

Increasing the temperature and decreasing the total pressure will shift the position of equilibrium to the right.

The forward reaction is endothermic (

\Delta H > 0

), so increasing the temperature drives the reaction forward (to the right) to absorb excess thermal energy. Additionally, the decomposition converts 1 mole of gaseous reactant into 2 moles of gaseous products; decreasing the pressure shifts the equilibrium toward the right side, which contains more gas moles.

Step-by-Step Solution

1

Analyze the temperature effect using the enthalpy sign

Since

\Delta H = +92.5\text{ kJ mol}^{-1}

(endothermic), heat acts as a reactant. Increasing the system temperature shifts the equilibrium to the right to absorb the extra heat.

Le Chatelier's principle states that a system at equilibrium responds to stress by shifting in the direction that minimizes that stress.

2

Analyze the pressure effect by counting gaseous stoichiometric coefficients

The left side has 1 mole of gas (

PCl_5

), while the right side has 2 moles of gas (

PCl_3 + Cl_2

). Decreasing the total pressure causes the equilibrium to shift toward the side with a greater number of gas moles (to the right).

Lowering pressure causes the system to shift toward producing more gas particles to counteract the pressure decrease.

3

Combine both conditions to determine the overall shift

Both an increase in temperature and a decrease in pressure independently shift the equilibrium position to favor product formation.

Both stresses act synergistically to drive the equilibrium to the right.

Key Concept

Le Chatelier's Principle and Equilibrium Shifts

Estimated Time:1m 15s

Question 8420Question →

The table below shows the solubility of anhydrous copper(II) tetraoxosulfate(VI), $\text{CuSO}_4$ , in water at different temperatures:

Temperature ( $^\circ\text{C}$ )	Solubility ( $\text{g}$ of $\text{CuSO}_4$ per $100\text{ g}$ of $\text{H}_2\text{O}$ )
$20$	$20.0$
$40$	$29.0$
$80$	$55.0$

A saturated solution of copper(II) tetraoxosulfate(VI) in $200.0\text{ g}$ of water at $80^\circ\text{C}$ is cooled to $20^\circ\text{C}$ . Given that the molar mass of $\text{CuSO}_4 = 160\text{ g mol}^{-1}$ and $\text{H}_2\text{O} = 18\text{ g mol}^{-1}$ , what is the exact mass of hydrated copper(II) tetraoxosulfate(VI) pentahydrate crystals, $\text{CuSO}_4\cdot 5\text{H}_2\text{O}$ , that will deposit from the solution?

123.24\text{ g}

109.38\text{ g}

70.00\text{ g}

44.80\text{ g}

Show answer & explanation

Answer: $123.24\text{ g}$

Answer

The mass of copper(II) tetraoxosulfate(VI) pentahydrate crystals deposited is

123.24\text{ g}

.

When a hydrated salt crystallizes, it removes both solute and water of crystallization from the saturated solution. Taking into account that

m

grams of

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

contains

0.64m

grams of

\text{CuSO}_4

and

0.36m

grams of

\text{H}_2\text{O}

, setting up the solubility ratio at

20^\circ\text{C}

as

(110.0 - 0.64m) / (200.0 - 0.36m) = 0.20

gives

m = 123.24\text{ g}

.

Step-by-Step Solution

1

Calculate the initial mass of dissolved anhydrous

\text{CuSO}_4

at

80^\circ\text{C}

.

In

200.0\text{ g}

of water, mass of dissolved

\text{CuSO}_4 = 2 \times 55.0\text{ g} = 110.0\text{ g}

.

Solubility at

80^\circ\text{C}

is

55.0\text{ g}

per

100\text{ g}

of water.

2

Determine the molar masses of the anhydrous salt, water, and hydrate.

Molar mass of

\text{CuSO}_4 = 160\text{ g mol}^{-1}

;

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

;

\text{CuSO}_4\cdot 5\text{H}_2\text{O} = 160 + 5(18) = 250\text{ g mol}^{-1}

.

Needed to establish mass fractions of solute and solvent in the crystals.

3

Express the mass fractions of anhydrous salt and water in

m

grams of

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

crystals.

Anhydrous

\text{CuSO}_4

fraction

= \frac{160}{250}m = 0.64m

; Water fraction

= \frac{90}{250}m = 0.36m

.

As crystals form, they take away both anhydrous salt and water from the solution.

4

Set up the solubility saturation equation at

20^\circ\text{C}

.

\frac{110.0 - 0.64m}{200.0 - 0.36m} = \frac{20.0}{100.0} = 0.20

.

At

20^\circ\text{C}

, the remaining solution must remain saturated.

5

Solve for

m

.

110.0 - 0.64m = 0.20(200.0 - 0.36m) \implies 110.0 - 0.64m = 40.0 - 0.072m \implies 70.0 = 0.568m \implies m = 123.24\text{ g}

.

Isolating

m

gives the exact mass of hydrated crystals deposited.

Key Concept

Crystallization of Hydrated Salts from Saturated Solutions

Answer

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