All practice questions

Match each aqueous salt solution on the left to the corresponding hydrolyzing ion or hydrolysis behavior on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Ammonium chloride matches with

NH_4^+

cation hydrolysis producing

H_3O^+

; Sodium sulfide matches with

S^{2-}

anion hydrolysis producing

OH^-

; Iron(III) nitrate matches with

[Fe(H_2O)_6]^{3+}

hydrolysis producing

H_3O^+

; Potassium sulfate matches with neither ion undergoing hydrolysis.

Matching each salt depends on identifying which ion hydrolyzes. Ammonium chloride contains the weak acid cation

NH_4^+

which yields hydronium ions; sodium sulfide contains the weak acid conjugate base

S^{2-}

which generates hydroxide ions; iron(III) nitrate contains the hydrated cation

[Fe(H_2O)_6]^{3+}

which donates a proton to water; and potassium sulfate consists only of non-hydrolyzing spectator ions.

Step-by-Step Solution

1

Identify the parent acid and base for each salt to determine which ions undergo hydrolysis.

Salts derived from weak parents hydrolyze:

NH_4^+

comes from weak base

NH_3

,

S^{2-}

comes from weak acid

H_2S

,

[Fe(H_2O)_6]^{3+}

is a weak acidic complex ion, whereas

K^+

and

SO_4^{2-}

come from strong parent species.

Only ions derived from weak acids or weak bases are strong enough conjugate species to react significantly with water.

2

Write the hydrolysis equilibrium equations for the reactive species.

NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+

,

S^{2-} + H_2O \rightleftharpoons HS^- + OH^-

, and

[Fe(H_2O)_6]^{3+} + H_2O \rightleftharpoons [Fe(H_2O)_5(OH)]^{2+} + H_3O^+

.

Cation hydrolysis increases the concentration of hydronium ions (

H_3O^+

), while anion hydrolysis increases the concentration of hydroxide ions (

OH^-

).

3

Pair each salt solution with its correct hydrolyzing species and resulting ionic effect.

Each salt is uniquely matched to its hydrolysis behavior.

Matches strictly conform to Brønsted-Lowry acid-base and salt hydrolysis principles.

Key Concept

Salt Hydrolysis and Solution Acidity/Alkalinity

Question 8542Question →

An atom of an element $Q$ has a mass equal to $2.5$ times the mass of a single carbon-12 atom ( $^{12}\text{C}$ ). What is the relative atomic mass of element $Q$ on the carbon-12 scale?

2.5

14.4

30

60

Show answer & explanation

Answer: $30$

Answer

The relative atomic mass of element

Q

on the carbon-12 scale is 30.

On the carbon-12 scale,

1\text{ atomic mass unit (amu)}

is defined as

\frac{1}{12}

of the mass of a single carbon-12 atom (

^{12}\text{C}

). This gives a single

^{12}\text{C}

atom a relative mass of

12

. If an atom of element

Q

is

2.5

times as heavy as one

^{12}\text{C}

atom, its relative atomic mass is

2.5 \times 12 = 30

.

Step-by-Step Solution

1

Define atomic mass unit on the carbon-12 scale

1 atomic mass unit (amu) =

\frac{1}{12}

of the mass of one carbon-12 atom

The standard reference point for relative atomic masses is defined as

\frac{1}{12}\text{th}

the mass of a single

^{12}\text{C}

atom, which means one carbon-12 atom equals

12\text{ amu}

.

2

Calculate the atomic mass of element

Q

in amu

\text{Mass of } Q = 2.5 \times 12\text{ amu} = 30\text{ amu}

Since one atom of element

Q

has a mass

2.5

times that of a carbon-12 atom, its absolute atomic mass relative to

1\text{ amu}

is

2.5 \times 12

.

3

Determine the relative atomic mass

\text{RAM} = \frac{30\text{ amu}}{1\text{ amu}} = 30

Relative atomic mass is a dimensionless ratio comparing the average mass of an atom to

\frac{1}{12}\text{th}

the mass of carbon-12.

Key Concept

Relative Atomic Mass on the Carbon-12 Scale

Question 8543Question →

Match each chemical reaction or test involving carbonyl compounds in Column A with its corresponding characteristic visual outcome in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Warming ethanal with Fehling's solution yields a brick-red precipitate; warming propanone with Tollen's reagent yields no visible reaction; warming propanal with acidified

KMnO_4

causes decolorization of the purple solution; and warming propanone with iodine in sodium hydroxide solution produces a pale yellow precipitate.

Alkanals (ethanal and propanal) are reducing agents due to the carbonyl hydrogen atom; thus, ethanal reduces Fehling's solution to a brick-red copper(I) oxide precipitate, and propanal reduces purple acidified

KMnO_4

to a colorless

Mn^{2+}

solution. Alkanones (propanone) lack this hydrogen atom, so propanone shows no reaction with Tollen's reagent. However, because propanone has a

CH_3CO-

group, it responds to the triiodomethane test by forming a pale yellow precipitate.

Step-by-Step Solution

1

Analyze the oxidation reactions of alkanals.

Ethanal reduces Fehling's solution to form brick-red

Cu_2O

, while propanal reduces acidified

KMnO_4

, turning the purple solution colorless.

Alkanals possess a hydrogen atom bonded to the carbonyl carbon, enabling easy oxidation by mild and strong oxidizing agents.

2

Analyze the oxidation behavior of alkanones.

Propanone yields no visible reaction with Tollen's reagent.

Alkanones lack a hydrogen atom on the carbonyl carbon and are resistant to oxidation by mild oxidizing agents.

3

Identify the triiodomethane (iodoform) test reaction.

Propanone forms a pale yellow precipitate of triiodomethane (

CHI_3

).

Propanone contains the

CH_3C=O

group required for a positive triiodomethane test.

Key Concept

Distinction tests and oxidation properties of alkanals and alkanones

Estimated Time:1m 0s

Question 8544Question →

At a constant temperature, dinitrogen tetroxide gas,

\text{N}_2\text{O}_4(g)

, dissociates into nitrogen dioxide gas,

\text{NO}_2(g)

, in a closed vessel according to the balanced equation:

\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)

If the total equilibrium pressure of the system is

2.0\text{ atm}

and the equilibrium mole fraction of

\text{N}_2\text{O}_4(g)

is

0.20

, what is the numerical value and unit of the equilibrium constant,

K_p

, for this reaction?

6.4\text{ atm}

3.2\text{ atm}

4.0\text{ atm}

0.16\text{ atm}^{-1}

Show answer & explanation

Answer: $6.4\text{ atm}$

Answer

The numerical value of the equilibrium constant

K_p

is

6.4\text{ atm}

.

To find

K_p

, first calculate the partial pressure of each gas using Dalton's Law (

P_i = \chi_i P_{\text{total}}

). The mole fraction of

\text{NO}_2

is

1.00 - 0.20 = 0.80

. Thus,

P_{\text{N}_2\text{O}_4} = 0.20 \times 2.0\text{ atm} = 0.40\text{ atm}

and

P_{\text{NO}_2} = 0.80 \times 2.0\text{ atm} = 1.60\text{ atm}

. Substituting into

K_p = \frac{(P_{\text{NO}_2})^2}{P_{\text{N}_2\text{O}_4}}

gives

\frac{(1.60)^2}{0.40} = 6.4\text{ atm}

.

Step-by-Step Solution

1

Determine the equilibrium mole fractions of all gaseous species.

Mole fraction of

\text{N}_2\text{O}_4(g)

,

\chi_{\text{N}_2\text{O}_4} = 0.20

. Since the sum of mole fractions equals 1.00, the mole fraction of

\text{NO}_2(g)

is

\chi_{\text{NO}_2} = 1.00 - 0.20 = 0.80

.

The sum of all mole fractions in a mixture must equal 1.

2

Calculate the equilibrium partial pressures of each gas.

P_{\text{N}_2\text{O}_4} = 0.20 \times 2.0\text{ atm} = 0.40\text{ atm}

, and

P_{\text{NO}_2} = 0.80 \times 2.0\text{ atm} = 1.60\text{ atm}

.

Dalton's Law states that partial pressure equals mole fraction multiplied by total pressure (

P_i = \chi_i P_{\text{total}}

).

3

Write the

K_p

expression and calculate its value and unit.

K_p = \frac{(P_{\text{NO}_2})^2}{P_{\text{N}_2\text{O}_4}} = \frac{(1.60\text{ atm})^2}{0.40\text{ atm}} = \frac{2.56\text{ atm}^2}{0.40\text{ atm}} = 6.4\text{ atm}

.

The equilibrium constant

K_p

uses partial pressures raised to the power of their respective stoichiometric coefficients.

Key Concept

Equilibrium Constant in Terms of Partial Pressures (

K_p

)

Estimated Time:2m 30s

Question 8545Question →

A saturated solution of potassium chlorate ( $\text{KClO}_3$ ) at $20^\circ\text{C}$ contains $7.35\text{ g}$ of solute dissolved in $100\text{ g}$ of distilled water. What is the solubility of $\text{KClO}_3$ at $20^\circ\text{C}$ in $\text{mol/dm}^3$ ? [Molar masses: $\text{K} = 39, \text{Cl} = 35.5, \text{O} = 16$ ; density of water $= 1.00\text{ g/cm}^3$ ]

Show answer & explanation

Answer: 0.6

Answer

The solubility of potassium chlorate at

20^\circ\text{C}

is

0.6\text{ mol/dm}^3

.

To convert mass of salt in a given solvent volume into solubility in

\text{mol/dm}^3

, first calculate the molar mass of

\text{KClO}_3

(

122.5\text{ g/mol}

). The amount of salt in moles is

7.35 / 122.5 = 0.06\text{ mol}

. Since

100\text{ g}

of water equals

0.1\text{ dm}^3

, the concentration of the saturated solution is

0.06\text{ mol} / 0.1\text{ dm}^3 = 0.6\text{ mol/dm}^3

.

Step-by-Step Solution

1

Calculate the molar mass of

\text{KClO}_3

122.5\text{ g/mol}

Sum the relative atomic masses:

39 (\text{K}) + 35.5 (\text{Cl}) + 3 \times 16 (\text{O}) = 122.5\text{ g/mol}

.

2

Calculate the number of moles of solute

0.06\text{ mol}

Divide the given mass by the molar mass:

\frac{7.35\text{ g}}{122.5\text{ g/mol}} = 0.06\text{ mol}

.

3

Convert the mass of solvent to volume in

\text{dm}^3

0.1\text{ dm}^3

Water density is

1.00\text{ g/cm}^3

, so

100\text{ g} = 100\text{ cm}^3 = 0.1\text{ dm}^3

.

4

Determine the molar solubility

0.6\text{ mol/dm}^3

Divide moles of solute by volume of solvent in

\text{dm}^3

:

\frac{0.06\text{ mol}}{0.1\text{ dm}^3} = 0.6\text{ mol/dm}^3

.

Key Concept

Solubility in mol/dm³

Question 8546Question →

During the industrial separation of liquefied air by fractional distillation, the mixture is gradually warmed to separate its primary gas components. Given the boiling points of nitrogen ( $-196^\circ\text{C}$ ), argon ( $-186^\circ\text{C}$ ), and oxygen ( $-183^\circ\text{C}$ ), which of the following correctly indicates the sequence in which these gases boil off from first to last?

Nitrogen, Argon, Oxygen

Oxygen, Argon, Nitrogen

Nitrogen, Oxygen, Argon

Oxygen, Nitrogen, Argon

Show answer & explanation

Answer: Nitrogen, Argon, Oxygen

Answer

The gases distill off in the order of Nitrogen, Argon, and Oxygen.

During fractional distillation of liquid air, substances with lower boiling points vaporize first as the temperature rises. Because

-196^\circ\text{C}

is lower than

-186^\circ\text{C}

, which is in turn lower than

-183^\circ\text{C}

, nitrogen distills off first, followed by argon, and finally oxygen.

Step-by-Step Solution

1

Compare the boiling points of the three gases given in the problem statement.

Nitrogen (

-196^\circ\text{C}

), Argon (

-186^\circ\text{C}

), and Oxygen (

-183^\circ\text{C}

).

Lower boiling point values (more negative numbers) mean higher volatility.

2

Arrange the boiling points in order from lowest to highest.

-196^\circ\text{C} < -186^\circ\text{C} < -183^\circ\text{C}

.

As liquefied air warms up, the substance with the lowest boiling point reaches its boiling temperature first and vaporizes.

3

Match the arranged boiling points to their corresponding gases to determine distillation order.

Nitrogen boils off first, followed by Argon, and then Oxygen.

The fractional distillation column separates components in increasing order of boiling points.

Key Concept

Fractional Distillation of Liquefied Air

Question 8547Question →

A solution containing trivalent ions $M^{3+}$ of an unknown metal $M$ is electrolyzed using inert electrodes. If a steady current of $2.50\text{ A}$ passed for $5790\text{ s}$ deposits $1.35\text{ g}$ of metal $M$ at the cathode, what is the relative atomic mass of metal $M$ ? [ $1\text{ Faraday} = 96500\text{ C/mol}$ ]

Show answer & explanation

Answer: 27

Answer

The relative atomic mass of metal

M

is

27\text{ g/mol}

.

Using Faraday's first law (

Q = I \times t

), the total charge passed is

2.50 \times 5790 = 14475\text{ C}

. Dividing by Faraday's constant (

96500\text{ C/mol}

) gives

0.15\text{ mol}

of electrons. Since the metal ion is trivalent (

M^{3+}

), the half-reaction is

M^{3+} + 3e^- \rightarrow M

, meaning

0.15\text{ mol}

of electrons deposits

0.05\text{ mol}

of metal. Dividing the mass (

1.35\text{ g}

) by the amount in moles (

0.05\text{ mol}

) yields a relative atomic mass of

27\text{ g/mol}

.

Step-by-Step Solution

1

Calculate the total electric charge passed (

Q

).

Q = 2.50\text{ A} \times 5790\text{ s} = 14475\text{ C}

.

Electric charge is the product of current in amperes and time in seconds.

2

Calculate the moles of electrons transferred.

n_e = \frac{14475\text{ C}}{96500\text{ C/mol}} = 0.15\text{ mol } e^-

.

One mole of electrons carries a charge of 1 Faraday (

96500\text{ C}

).

3

Determine the amount (in moles) of metal

M

deposited.

From

M^{3+} + 3e^- \rightarrow M

,

3\text{ moles of } e^-

deposit

1\text{ mole of } M

. Thus,

n_M = \frac{0.15\text{ mol}}{3} = 0.05\text{ mol}

.

The reduction of a trivalent ion requires 3 moles of electrons per mole of metal deposited.

4

Calculate the relative atomic mass (

M_r

).

M_r = \frac{1.35\text{ g}}{0.05\text{ mol}} = 27\text{ g/mol}

.

Molar mass is calculated by dividing the mass of the substance by the amount in moles.

Key Concept

Faraday's Laws of Electrolysis and Quantitative Calculations

Question 8548Question →

Which of the following organic compounds acts as a weak base in an aqueous solution due to the presence of an unshared pair of electrons on its nitrogen atom?

Methylamine (

CH_3NH_2

)

Ethanamide (

CH_3CONH_2

)

Ethanoic acid (

CH_3COOH

)

Ethanol (

CH_3CH_2OH

)

Show answer & explanation

Answer: Methylamine ( $CH_3NH_2$ )

Answer

Methylamine (

CH_3NH_2

)

Methylamine (

CH_3NH_2

) is an aliphatic amine. The nitrogen atom retains a localized lone pair of electrons that readily accepts a proton from water to form a methylammonium ion and a hydroxide ion, demonstrating basic behavior in aqueous solution.

Step-by-Step Solution

1

Identify the functional groups present in each compound.

Methylamine (

CH_3NH_2

) is an amine; ethanamide (

CH_3CONH_2

) is an amide; ethanoic acid (

CH_3COOH

) is a carboxylic acid; ethanol (

CH_3CH_2OH

) is an alcohol.

Basic properties in organic nitrogen compounds depend on the availability of the nitrogen lone pair.

2

Evaluate the availability of the unshared pair of electrons on the nitrogen atom.

In primary aliphatic amines like methylamine, the lone pair on nitrogen is readily available to accept a proton (

H^+

). In amides like ethanamide, resonance delocalizes the nitrogen lone pair toward the carbonyl oxygen, removing its basic character.

Proton acceptance (Lewis/Brønsted-Lowry basicity) requires an available lone pair.

Key Concept

Basicity of Amines versus Amides

Question 8549Question →

What is the oxidation number of chlorine in potassium trioxochlorate(V), $\text{KClO}_3$ ?

+5

+3

+7

-1

Show answer & explanation

Answer: $+5$

Answer

The oxidation number of chlorine in

\text{KClO}_3

is

+5

.

In potassium trioxochlorate(V),

\text{KClO}_3

, potassium contributes

+1

and three oxygen atoms contribute a total of

-6

. For the compound to be electrically neutral, chlorine must have an oxidation state of

+5

.

Step-by-Step Solution

1

Assign known oxidation numbers to potassium and oxygen

Potassium (Group 1 metal) has an oxidation state of

+1

. Oxygen has an oxidation state of

-2

.

Standard rules assign

+1

to alkali metals and

-2

to oxygen in oxoacids/oxoanions.

2

Formulate an equation for the neutral molecule

\text{KClO}_3

(+1) + x + 3(-2) = 0

The sum of all oxidation numbers in a neutral chemical compound must equal zero.

3

Solve for the unknown oxidation state

x

of chlorine

+1 + x - 6 = 0 \implies x - 5 = 0 \implies x = +5

Solving the linear algebraic equation yields

+5

.

Key Concept

Oxidation state calculation of halogens in oxoacids and oxoacid salts

Question 8550Question →

What volume of nitrogen(IV) oxide gas ( $\text{NO}_2$ ), measured at s.t.p. in $\text{dm}^3$ , is evolved when $3.2\text{ g}$ of pure copper metal reacts completely with excess concentrated trioxonitrate(V) acid?

(Molar mass of $\text{Cu} = 64\text{ g mol}^{-1}$ ; Molar volume of gas at s.t.p. = $22.4\text{ dm}^3\text{ mol}^{-1}$ )

Show answer & explanation

Answer: 2.24

Answer

2.24 dm^3 of nitrogen(IV) oxide gas is produced at s.t.p.

When copper reacts with concentrated trioxonitrate(V) acid, the reaction follows the stoichiometry

\text{Cu} + 4\text{HNO}_3 \rightarrow \text{Cu(NO}_3)_2 + 2\text{NO}_2 + 2\text{H}_2\text{O}

. Thus,

1\text{ mole}

of copper metal (

64\text{ g}

) yields

2\text{ moles}

of

\text{NO}_2

gas (

44.8\text{ dm}^3

at s.t.p.). For

3.2\text{ g}

(

0.05\text{ moles}

) of copper, the volume of

\text{NO}_2

formed is

0.10\text{ moles} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 2.24\text{ dm}^3

.

Step-by-Step Solution

1

Write the balanced equation for the reaction of copper with concentrated trioxonitrate(V) acid

\text{Cu} + 4\text{HNO}_3 \rightarrow \text{Cu(NO}_3)_2 + 2\text{NO}_2 + 2\text{H}_2\text{O}

Concentrated trioxonitrate(V) acid acts as a strong oxidizing agent, converting copper to copper(II) ions and reducing itself to brown nitrogen(IV) oxide gas.

2

Calculate the moles of copper metal present in 3.2 g

3.2 / 64 = 0.05 mol

Number of moles is mass divided by relative molar mass.

3

Determine the amount of nitrogen(IV) oxide gas produced in moles

2 * 0.05 = 0.10 mol

The stoichiometric mole ratio between Cu and NO2 is 1:2.

4

Convert moles of nitrogen(IV) oxide gas into volume at s.t.p.

0.10 * 22.4 = 2.24 dm^3

One mole of any ideal gas occupies 22.4 dm^3 at standard temperature and pressure.

Key Concept

Stoichiometry of copper redox reaction with concentrated trioxonitrate(V) acid producing nitrogen(IV) oxide gas.

Question 8551Question →

Match each nitrogen-containing organic compound on the left with its correct structural classification on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Methylamine pairs with Primary aliphatic amine; Ethanamide pairs with Neutral organic amide; Phenylamine pairs with Primary aromatic amine; Dimethylamine pairs with Secondary aliphatic amine.

Each compound matches its unique chemical definition: Methylamine is a

1^\circ

aliphatic amine, Ethanamide is a neutral amide, Phenylamine is a

1^\circ

aromatic amine, and Dimethylamine is a

2^\circ

aliphatic amine.

Step-by-Step Solution

1

Examine the functional groups and substituents attached to nitrogen in each compound.

Methylamine (

CH_3NH_2

) and Phenylamine (

C_6H_5NH_2

) each have one organic group attached (

1^\circ

). Dimethylamine (

(CH_3)_2NH

) has two organic groups attached (

2^\circ

). Ethanamide (

CH_3CONH_2

) has a carbonyl group (

C=O

) linked directly to nitrogen.

The number of alkyl/aryl groups determines amine degree (

1^\circ, 2^\circ, 3^\circ

), while a carbonyl-nitrogen bond defines an amide.

2

Classify by aliphatic, aromatic, or neutral amide characteristics.

Methylamine contains an alkyl group (

1^\circ

aliphatic amine). Phenylamine contains a benzene ring (

1^\circ

aromatic amine). Dimethylamine has two alkyl groups (

2^\circ

aliphatic amine). Ethanamide is an amide and exhibits neutral aqueous behavior due to lone pair resonance delocalization.

Structure and electronic delocalization determine both classification and relative basicity.

Key Concept

Classification of amines (primary, secondary, aromatic, aliphatic) and amides.

Question 8552Question →

An unknown disaccharide $X$ with the molecular formula $\text{C}_{12}\text{H}_{22}\text{O}_{11}$ does not reduce Fehling's solution. Upon acid-catalyzed hydrolysis, $X$ yields an equimolar mixture of two isomeric hexoses, $Y$ and $Z$ . Compound $Y$ rapidly produces a deep red color when heated with Seliwanoff's reagent, whereas compound $Z$ is oxidized by bromine water to form a monocarboxylic acid. Which of the following correctly identifies disaccharide $X$ and explains why it fails to react with Fehling's solution prior to hydrolysis?

Sucrose; because the glycosidic linkage involves the anomeric carbons of both glucose and fructose, eliminating free hemiacetal/hemiketal groups.

Maltose; because the glycosidic bond links carbon-1 of one glucose unit to carbon-4 of another, masking both reducing ends.

Lactose; because both monosaccharide units exist strictly in locked cyclic ring forms that cannot open to form aldehydes.

Sucrose; because fructose contains a ketone group which must first be hydrolyzed into an aldehyde group before Fehling's reduction can occur.

Show answer & explanation

Answer: Sucrose; because the glycosidic linkage involves the anomeric carbons of both glucose and fructose, eliminating free hemiacetal/hemiketal groups.

Answer

Sucrose; because the glycosidic linkage involves the anomeric carbons of both glucose and fructose, eliminating free hemiacetal/hemiketal groups.

Seliwanoff's test specifically identifies ketohexoses such as fructose through rapid dehydration to hydroxymethylfurfural and reaction with resorcinol. Bromine water selectively oxidizes aldoses like glucose to aldonic acids without oxidizing ketoses. A disaccharide yielding glucose and fructose upon hydrolysis is sucrose. Sucrose is a non-reducing sugar because its glycosidic bond connects C-1 of glucose and C-2 of fructose, locking both anomeric carbon atoms and preventing ring opening to form reactive carbonyl groups.

Step-by-Step Solution

1

Analyze the chemical test results of the hydrolysis products

Y

and

Z

.

Compound

Y

gives a positive Seliwanoff's test (rapid red color), which is characteristic of a ketose (fructose). Compound

Z

is oxidized by bromine water (a mild oxidizing agent), which selectively oxidizes aldoses (glucose) to aldonic acids.

Seliwanoff's reagent differentiates ketoses from aldoses, while bromine water differentiates aldoses from ketoses.

2

Identify the disaccharide

X

based on its hydrolysis products.

Disaccharide

X

hydrolyzes into glucose and fructose, identifying

X

as sucrose.

Sucrose (

\text{C}_{12}\text{H}_{22}\text{O}_{11}

) is composed of one glucose unit and one fructose unit.

3

Determine the structural basis for the non-reducing nature of sucrose.

In sucrose, the glycosidic bond connects

\text{C-1}

(

\alpha

-anomeric carbon of glucose) to

\text{C-2}

(

\beta

-anomeric carbon of fructose).

Because both potential reducing centers (anomeric carbons) are locked in the glycosidic bond, sucrose lacks a free hemiacetal or hemiketal group, rendering it a non-reducing sugar that does not reduce Fehling's reagent.

Key Concept

Glycosidic bond linkages and reducing vs. non-reducing carbohydrate behavior

Question 8553Question →

When chlorine gas is bubbled into a hot, concentrated solution of potassium hydroxide (

\text{KOH}

), it undergoes disproportionation according to the chemical equation:

3\text{Cl}_2\text{(g)} + 6\text{KOH(aq)} \rightarrow 5\text{KCl(aq)} + \text{KClO}_3\text{(aq)} + 3\text{H}_2\text{O(l)}

What mass of potassium trioxochlorate(V) (

\text{KClO}_3

) is produced when

6.72\text{ dm}^3

of chlorine gas measured at s.t.p. reacts completely?
[Molar volume of gas at s.t.p. =

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

; Molar masses:

\text{K} = 39\text{ g/mol}

,

\text{Cl} = 35.5\text{ g/mol}

,

\text{O} = 16\text{ g/mol}

]

12.25\text{ g}

36.75\text{ g}

11.43\text{ g}

9.05\text{ g}

Show answer & explanation

Answer: $12.25\text{ g}$

Answer

The correct mass of potassium trioxochlorate(V) produced is

12.25\text{ g}

.

The option specifying

12.25\text{ g}

is correct because

6.72\text{ dm}^3

of

\text{Cl}_2

gas at s.t.p. corresponds to

0.30\text{ mol}

. Based on the 3:1 stoichiometric ratio from the balanced equation (

3\text{Cl}_2 \rightarrow 1\text{KClO}_3

),

0.10\text{ mol}

of

\text{KClO}_3

is produced. Multiplying

0.10\text{ mol}

by the molar mass of

\text{KClO}_3

(

122.5\text{ g/mol}

) yields

12.25\text{ g}

.

Step-by-Step Solution

1

Calculate the amount of chlorine gas in moles at s.t.p.

n(\text{Cl}_2) = \frac{6.72\text{ dm}^3}{22.4\text{ dm}^3\text{mol}^{-1}} = 0.30\text{ mol}

At s.t.p.,

1\text{ mole}

of any ideal gas occupies

22.4\text{ dm}^3

.

2

Determine the moles of potassium trioxochlorate(V) (

\text{KClO}_3

) formed using stoichiometric ratios.

n(\text{KClO}_3) = \frac{1}{3} \times 0.30\text{ mol} = 0.10\text{ mol}

From the balanced equation,

3\text{ moles}

of

\text{Cl}_2

produce

1\text{ mole}

of

\text{KClO}_3

.

3

Calculate the molar mass of

\text{KClO}_3

.

\text{Molar Mass} = 39 + 35.5 + (3 \times 16) = 122.5\text{ g/mol}

Summing the atomic masses of one potassium, one chlorine, and three oxygen atoms.

4

Calculate the mass of

\text{KClO}_3

produced.

\text{Mass} = 0.10\text{ mol} \times 122.5\text{ g/mol} = 12.25\text{ g}

Mass is obtained by multiplying the number of moles by the molar mass.

Key Concept

Disproportionation reactions of halogens in hot concentrated alkalis and gas stoichiometry at s.t.p.

Question 8554Question →

An underground steel pipeline carrying natural gas is buried in moist soil. To protect the iron in the steel from corrosion, blocks of another metal are electrically connected to the pipeline at regular intervals. Given the standard reduction potentials below:

\text{Fe}^{2+}_{(\text{aq})} + 2e^- \rightarrow \text{Fe}_{(\text{s})}, \quad E^\circ = -0.44\text{ V}

\text{Cu}^{2+}_{(\text{aq})} + 2e^- \rightarrow \text{Cu}_{(\text{s})}, \quad E^\circ = +0.34\text{ V}

\text{Mg}^{2+}_{(\text{aq})} + 2e^- \rightarrow \text{Mg}_{(\text{s})}, \quad E^\circ = -2.37\text{ V}

\text{Ag}^{+}_{(\text{aq})} + e^- \rightarrow \text{Ag}_{(\text{s})}, \quad E^\circ = +0.80\text{ V}

Which metal can be attached to the pipeline to serve as an effective sacrificial anode, and what chemical change occurs to it during protection?

Magnesium, because it is oxidized preferentially due to its more negative reduction potential.

Copper, because its positive reduction potential allows it to supply electrons to the iron structure.

Silver, because metals with higher reduction potentials readily undergo sacrificial oxidation.

Magnesium, because it undergoes reduction at the cathode while keeping the iron oxidized.

Show answer & explanation

Answer: Magnesium, because it is oxidized preferentially due to its more negative reduction potential.

Answer

Magnesium serves as an effective sacrificial anode because it has a more negative standard reduction potential than iron, meaning it oxidizes more readily and supplies electrons to protect the iron pipeline.

Magnesium has a more negative standard reduction potential (-2.37 V) than iron (-0.44 V). In an electrochemical cell formed in moist soil, magnesium oxidizes preferentially (

\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-

), supplying electrons to the steel pipe. This makes the iron pipeline cathodic and prevents iron from oxidizing into rust.

Step-by-Step Solution

1

Compare the standard reduction potentials (

E^\circ

) of iron and the candidate metals.

E^\circ(\text{Mg}^{2+}/\text{Mg}) = -2.37\text{ V}

,

E^\circ(\text{Fe}^{2+}/\text{Fe}) = -0.44\text{ V}

,

E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\text{ V}

, and

E^\circ(\text{Ag}^{+}/\text{Ag}) = +0.80\text{ V}

.

A metal acts as a sacrificial anode only if it is more easily oxidized (more electropositive) than iron, which corresponds to having a more negative standard reduction potential.

2

Identify which metal will oxidize preferentially over iron.

Magnesium has the most negative reduction potential (-2.37 V < -0.44 V), so it oxidizes preferentially:

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

.

The metal with the lower (more negative) reduction potential readily loses electrons to the iron structure, making the iron the cathode in the electrochemical cell.

3

Determine the chemical process occurring at the sacrificial block.

Oxidation takes place at the sacrificial magnesium block (the anode), sacrificing the block while preserving the steel pipeline.

By definition, oxidation occurs at the anode (

\text{anode} = \text{oxidation}

).

Key Concept

Sacrificial Anodic Protection

Question 8555Question →

The standard reduction potentials for two half-reactions at

25^\circ\text{C}

are given below:

\text{Co}^{2+}(aq) + 2e^- \rightarrow \text{Co}(s) \quad E^\circ = -0.28\text{ V}

\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) \quad E^\circ = +0.34\text{ V}

What is the standard electromotive force ( $E^\circ_{\text{cell}}$ ) for the spontaneous reaction that occurs when these two half-cells are connected under standard conditions?

+0.62\text{ V}

+0.06\text{ V}

-0.62\text{ V}

-0.06\text{ V}

Show answer & explanation

Answer: $+0.62\text{ V}$

Answer

The standard electromotive force for the spontaneous cell reaction is

+0.62\text{ V}

.

For a spontaneous electrochemical cell, reduction occurs at the electrode with the more positive reduction potential (cathode:

\text{Cu}^{2+}/\text{Cu}

at

+0.34\text{ V}

), and oxidation occurs at the electrode with the more negative reduction potential (anode:

\text{Co}^{2+}/\text{Co}

at

-0.28\text{ V}

). Subtracting the anode potential from the cathode potential yields

E^\circ_{\text{cell}} = +0.34\text{ V} - (-0.28\text{ V}) = +0.62\text{ V}

.

Step-by-Step Solution

1

Identify the cathode and anode based on standard reduction potentials.

Copper half-cell (

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

) has the higher reduction potential and acts as the cathode (reduction). Cobalt half-cell (

E^\circ = -0.28\text{ V}

) has the lower reduction potential and acts as the anode (oxidation).

A species with a higher reduction potential is more easily reduced, driving spontaneous electron flow from anode to cathode.

2

Calculate the standard cell electromotive force (

E^\circ_{\text{cell}}

).

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = +0.34\text{ V} - (-0.28\text{ V}) = +0.34\text{ V} + 0.28\text{ V} = +0.62\text{ V}

.

The cell EMF for a spontaneous reaction must be positive.

Key Concept

Standard Cell Potential and Reaction Spontaneity

Estimated Time:1m 0s

Question 8556Question →

When concentrated tetraoxosulfate(VI) acid ( $H_2SO_4$ ) is added to sucrose crystals, a black mass of carbon and steam are produced. What property of the acid does this reaction demonstrate?

Dehydrating action

Oxidizing action

Reducing action

Precipitating action

Show answer & explanation

Answer: Dehydrating action

Answer

Dehydrating action

Concentrated tetraoxosulfate(VI) acid has a very high affinity for water. When added to carbohydrates such as sucrose (

C_{12}H_{22}O_{11}

), it removes hydrogen and oxygen atoms in a 2:1 ratio as water molecules, leaving behind a charred black mass of elemental carbon.

Step-by-Step Solution

1

Analyze the chemical transformation in the given reaction

Sucrose (

C_{12}H_{22}O_{11}

) reacts with concentrated

H_2SO_4

to yield elemental carbon (

12C

) and water vapor (

11H_2O

).

The acid extracts hydrogen and oxygen in the exact ratio of water from the carbohydrate molecule.

2

Identify the specific property corresponding to removing elements of water

This process is classified as dehydration.

Dehydration refers specifically to the removal of chemically combined water or its constituent elements from a substance.

Key Concept

Dehydrating action of concentrated tetraoxosulfate(VI) acid

Estimated Time:45s

Question 8557Question →

A waste management facility evaluates four industrial waste polymers: polyethene, starch, polyvinyl chloride (PVC), and nylon-6,6. Which of the following chemical statements correctly differentiates their biodegradation mechanisms and incineration hazards?

Starch and nylon-6,6 undergo microbial hydrolysis due to polar bonds in their main chains, while PVC releases toxic hydrogen chloride gas upon incineration.

Polyethene degrades rapidly in soil because synthetic addition polymers contain accessible carbon-carbon double bonds along their main chain.

Polyethene and PVC undergo spontaneous bacterial decomposition because saturated aliphatic carbon backbones are easily cleaved by soil enzymes.

Starch and nylon-6,6 resist microbial breakdown because condensation polymerization creates non-polar, hydrophobic polymer chains.

Show answer & explanation

Answer: Starch and nylon-6,6 undergo microbial hydrolysis due to polar bonds in their main chains, while PVC releases toxic hydrogen chloride gas upon incineration.

Answer

Starch and nylon-6,6 undergo microbial hydrolysis due to polar bonds in their main chains, while PVC releases toxic hydrogen chloride gas upon incineration.

Natural polymers like starch and condensation polymers like nylon-6,6 contain polar hydrolyzable linkages (ester, amide, glycosidic bonds) that soil micro-organisms readily break down. Additionally, incinerating chlorinated synthetic polymers such as PVC releases toxic and corrosive hydrogen chloride (

HCl

) gas into the environment.

Step-by-Step Solution

1

Analyze the backbone structure of addition polymers versus condensation/natural polymers.

Polyethene and PVC are synthetic addition polymers possessing saturated, non-polar all-carbon (

C-C

) backbones that lack site-specific functional groups for enzymatic cleavage.

Soil micro-organisms produce hydrolytic enzymes that target polar linkages rather than non-polar carbon-carbon single bonds.

2

Examine the biodegradability of starch and nylon-6,6.

Starch (a polysaccharide) and nylon-6,6 (a polyamide) contain polar linkages (glycosidic and amide bonds) capable of undergoing enzymatic hydrolysis.

Polar functional groups interact with water and microbial enzymes, enabling biodegradation into simpler monomers.

3

Evaluate the incineration behavior of organochlorine polymers like PVC.

Thermal decomposition of polyvinyl chloride (

[CH_2-CHCl]_n

) releases volatile gaseous hydrogen chloride (

HCl

).

Chlorine atoms bound to the polymer backbone undergo elimination as

HCl(g)

at elevated temperatures during combustion.

Key Concept

Chemical basis of polymer biodegradability and thermal incineration products

Question 8558Question →

A laboratory technician reacts separate samples of two copper-based alloys, $X$ and $Y$ , with excess concentrated trioxonitrate(V) acid ( $HNO_3$ ). Sample $X$ dissolves completely to yield a clear blue solution with no residue, whereas Sample $Y$ yields a blue solution containing an insoluble white solid. Which of the following correctly identifies Alloys $X$ and $Y$ and the constituent element responsible for the insoluble white solid?

Alloy

X

is brass, Alloy

Y

is bronze, and tin forms the insoluble solid.

Alloy

X

is bronze, Alloy

Y

is brass, and zinc forms the insoluble solid.

Alloy

X

is brass, Alloy

Y

is bronze, and copper forms the insoluble solid.

Alloy

X

is duralumin, Alloy

Y

is brass, and aluminium forms the insoluble solid.

Show answer & explanation

Answer: Alloy $X$ is brass, Alloy $Y$ is bronze, and tin forms the insoluble solid.

Answer

Alloy

X

is brass, Alloy

Y

is bronze, and tin forms the insoluble solid.

Brass (

Cu + Zn

) dissolves completely in concentrated trioxonitrate(V) acid to produce soluble copper(II) trioxonitrate(V) and zinc trioxonitrate(V). Bronze (

Cu + Sn

) contains tin, which is oxidized by concentrated trioxonitrate(V) acid to form insoluble metastannic acid (

H_2SnO_3

), creating a white precipitate in the blue copper solution.

Step-by-Step Solution

1

Identify the elemental compositions of the copper alloys brass and bronze.

Brass is an alloy of copper (

Cu

) and zinc (

Zn

). Bronze is an alloy of copper (

Cu

) and tin (

Sn

).

Knowing constituent elements is essential to predict chemical reactions with concentrated acid.

2

Analyze the chemical reaction of Brass (Alloy

X

) with concentrated

HNO_3

.

Both

Cu

and

Zn

react to form soluble metal trioxonitrate(V) salts:

Cu(NO_3)_2

(blue) and

Zn(NO_3)_2

(colorless). The mixture forms a completely clear blue solution.

All nitrate salts of copper and zinc are soluble in aqueous media.

3

Analyze the chemical reaction of Bronze (Alloy

Y

) with concentrated

HNO_3

.

Copper dissolves to form soluble blue

Cu(NO_3)_2

, while tin (

Sn

) is oxidized by concentrated

HNO_3

to form hydrated tin(IV) oxide (

SnO_2 \cdot xH_2O

or metastannic acid,

H_2SnO_3

), which precipitates as an insoluble white solid.

Tin exhibits anomalous behavior with concentrated nitric acid compared to zinc, forming an insoluble oxide rather than a soluble nitrate salt.

Key Concept

Chemical differentiation of brass and bronze based on constituent metal reactivity with concentrated trioxonitrate(V) acid

Estimated Time:1m 30s

Question 8559Question →

In the industrial manufacture of tetraoxosulfate(VI) acid via the Contact Process, several crucial chemical and physical steps are carried out in a specific sequence to maximize yield and prevent efficiency loss. Arrange the following steps of the Contact Process in the correct sequential order from first to last.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of the Contact Process is: Combustion/roasting to produce

SO_2

gas

\rightarrow

Purification of

SO_2

gas to remove catalyst poisons

\rightarrow

Catalytic conversion of

SO_2

to

SO_3

over

V_2O_5

catalyst

\rightarrow

Absorption of

SO_3

gas in concentrated

H_2SO_4

to form oleum

\rightarrow

Controlled dilution of oleum with water to yield

H_2SO_4

.

The Contact Process must follow a rigorous order: first generating raw

SO_2

gas, purifying it to prevent catalyst poisoning by impurities like

As_2O_3

, catalytically oxidizing

SO_2

to

SO_3

over

V_2O_5

, absorbing

SO_3

in 98%

H_2SO_4

to form oleum (

H_2S_2O_7

), and finally diluting oleum with water to produce concentrated

H_2SO_4

.

Step-by-Step Solution

1

Identify the initial feedstock generation stage.

Combustion of sulfur or roasting of sulfide ores produces

SO_2

gas.

Sulfur(IV) oxide is the essential chemical precursor required for the process.

2

Determine the necessary gas purification stage prior to catalysis.

Passing the

SO_2

and air mixture through scrubbers and precipitators removes dust particles and arsenic(III) oxide (

As_2O_3

).

Arsenic compounds act as catalyst poisons, permanently deactivating the vanadium(V) oxide catalyst if not removed first.

3

Identify the catalytic oxidation step.

Purified

SO_2

reacts with

O_2

over a

V_2O_5

catalyst at 450 °C and 1–2 atm to form

SO_3

.

This exothermic equilibrium reaction converts sulfur(IV) oxide to sulfur(VI) oxide.

4

Determine the absorption stage for sulfur(VI) oxide.

SO_3

gas is absorbed into concentrated (98%)

H_2SO_4

to form oleum (

H_2S_2O_7

).

Direct hydration of

SO_3

with water is extremely exothermic and produces a fine acid fog that cannot be easily condensed industrially.

5

Identify the final dilution/hydration step.

Oleum (

H_2S_2O_7

) is diluted with a calculated volume of water to form concentrated

H_2SO_4

.

Reacting oleum with water yields high-purity tetraoxosulfate(VI) acid safely and efficiently.

Key Concept

Sequential chemical and industrial stages of the Contact Process for tetraoxosulfate(VI) acid production

Estimated Time:2m 0s

Question 8560Question →

An allele whose phenotype is expressed in an organism only when two copies of it are present, and whose effect is masked in the presence of a contrasting allele, is best described as which of the following?

A recessive allele

A dominant allele

A codominant allele

A continuous allele

Show answer & explanation

Answer: A recessive allele

Answer

A recessive allele

The correct option is 'A recessive allele' because recessive traits require two identical recessive alleles (homozygous state) to be observed phenotypically. In a heterozygous individual, the dominant allele overrides and masks the recessive trait.

Step-by-Step Solution

1

Analyze the definition given in the stem.

The stem describes an allele that requires two copies (homozygous condition) to manifest phenotypically and is masked when paired with a different allele.

By definition in classical Mendelian genetics, an allele that is concealed in the heterozygous condition is recessive.

2

Evaluate the choices against basic genetics terminology.

A recessive allele fits this exact criterion, whereas a dominant allele masks others and a codominant allele expresses alongside others.

Distinguishing between dominant, recessive, and codominant expression is fundamental to genetic analysis.

Key Concept

Recessive Allele Expression

Estimated Time:45s

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept