All practice questions

Calculate the solubility of the salt in the given solution and complete the statement below.

Fill in the blanks below

If

5.85\text{ g}

of sodium chloride (

\text{NaCl}

) is dissolved in water to prepare

250\text{ cm}^3

of a saturated solution at

25^\circ\text{C}

, the solubility of

\text{NaCl}

is

\text{mol/dm}^3

.

[\text{Na} = 23.0, \text{Cl} = 35.5]

Show answer & explanation

Answer

The solubility of sodium chloride at

25^\circ\text{C}

is

0.4\text{ mol/dm}^3

.

The solubility in

\text{mol/dm}^3

is obtained by dividing the moles of solute by the solution volume in

\text{dm}^3

. Here,

5.85\text{ g}

of

\text{NaCl}

corresponds to

0.1\text{ mol}

(

5.85 / 58.5

), and

250\text{ cm}^3

equals

0.25\text{ dm}^3

. Dividing

0.1\text{ mol}

by

0.25\text{ dm}^3

gives

0.4\text{ mol/dm}^3

.

Step-by-Step Solution

1

Calculate the molar mass of sodium chloride (

\text{NaCl}

).

Molar mass

= 23.0 + 35.5 = 58.5\text{ g/mol}

.

The molar mass is required to convert mass in grams to amount in moles.

2

Determine the number of moles of

\text{NaCl}

present.

\text{Moles} = \frac{5.85\text{ g}}{58.5\text{ g/mol}} = 0.1\text{ mol}

.

Solubility in

\text{mol/dm}^3

requires the quantity of solute in moles.

3

Convert the volume of solution from

\text{cm}^3

to

\text{dm}^3

.

\text{Volume} = \frac{250\text{ cm}^3}{1000} = 0.25\text{ dm}^3

.

Concentration units are per cubic decimeter (

\text{dm}^3

).

4

Calculate the molar concentration (solubility).

\text{Solubility} = \frac{0.1\text{ mol}}{0.25\text{ dm}^3} = 0.4\text{ mol/dm}^3

.

Solubility is calculated as moles of solute divided by volume of solution in

\text{dm}^3

.

Key Concept

Solubility and Concentration Determination

Estimated Time:45s

Question 8502Question →

Which of the following indicators is most suitable for detecting the end point in a volumetric titration of ethanoic acid ( $\text{CH}_3\text{COOH}$ ) against sodium hydroxide ( $\text{NaOH}$ ) solution?

Phenolphthalein

Methyl orange

Methyl red

Screened methyl orange

Show answer & explanation

Answer: Phenolphthalein

Answer

Phenolphthalein is the most suitable indicator because the equivalence point of a weak acid (ethanoic acid) titrated against a strong base (sodium hydroxide) occurs in the basic pH range.

In the titration of ethanoic acid (a weak acid) with sodium hydroxide (a strong base), the salt formed at neutralization undergoes hydrolysis to yield an alkaline solution with a pH greater than 7. Phenolphthalein is the correct choice because its color change interval (pH 8.3–10.0) coincides with this basic equivalence point.

Step-by-Step Solution

1

Identify the nature of the acid and base involved in the titration.

Ethanoic acid (

\text{CH}_3\text{COOH}

) is a weak acid, while sodium hydroxide (

\text{NaOH}

) is a strong base.

The relative strengths of the acid and base determine the pH of the salt solution at the equivalence point.

2

Determine the expected pH at the equivalence point.

The resulting salt (sodium ethanoate) undergoes hydrolysis to produce hydroxide ions, giving an alkaline equivalence point with a pH between 8 and 9.

Anions of weak acids hydrolyze in water to form basic solutions.

3

Match the equivalence point pH with the transition range of an indicator.

Phenolphthalein has a pH transition range of 8.3–10.0.

A suitable indicator must undergo a sharp color change in the pH region corresponding to the equivalence point of the reaction.

Key Concept

Indicator Selection for Acid-Base Titrations

Question 8503Question →

When excess dry ammonia gas is passed over

24.0\text{ g}

of heated copper(II) oxide (

CuO

), the oxide is completely reduced to copper metal according to the equation:

2NH_3(g) + 3CuO(s) \rightarrow 3Cu(s) + N_2(g) + 3H_2O(l)

What is the volume of nitrogen gas, in

\text{dm}^3

, evolved at s.t.p.?
(

Cu = 64.0

,

O = 16.0

, molar volume of gas at s.t.p.

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

)

Show answer & explanation

Answer: 2.24

Answer

The volume of nitrogen gas evolved at s.t.p. is

2.24\text{ dm}^3

.

According to the balanced chemical equation

2NH_3(g) + 3CuO(s) \rightarrow 3Cu(s) + N_2(g) + 3H_2O(l)

, 3 moles (

240.0\text{ g}

) of copper(II) oxide produce 1 mole (

22.4\text{ dm}^3

at s.t.p.) of nitrogen gas. Hence,

24.0\text{ g}

of

CuO

yields

\frac{24.0}{240.0} \times 22.4\text{ dm}^3 = 2.24\text{ dm}^3

of nitrogen gas.

Step-by-Step Solution

1

Calculate the molar mass of

CuO

Molar mass of

CuO = 64.0 + 16.0 = 80.0\text{ g mol}^{-1}

Required to convert the given mass of reactant to moles.

2

Find the number of moles of

CuO

Moles of

CuO = \frac{24.0\text{ g}}{80.0\text{ g mol}^{-1}} = 0.30\text{ mol}

Determines the exact amount of copper(II) oxide reacting.

3

Apply stoichiometry to find moles of

N_2

gas produced

Moles of

N_2 = 0.30\text{ mol } CuO \times \frac{1\text{ mol } N_2}{3\text{ mol } CuO} = 0.10\text{ mol } N_2

The mole ratio between

CuO

and

N_2

in the balanced chemical equation is

3:1

.

4

Convert moles of

N_2

to volume at s.t.p.

Volume of

N_2 = 0.10\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 2.24\text{ dm}^3

1\text{ mole}

of any ideal gas occupies

22.4\text{ dm}^3

at s.t.p.

Key Concept

Reduction of metallic oxides by ammonia gas and gas stoichiometry at STP

Question 8504Question →

Complete the statements regarding the neutron activation of sodium and its subsequent radioactive decay by filling in the appropriate values.

Fill in the blanks below

When Sodium-23 (

^{23}_{11}\text{Na}

) is bombarded with a neutron, it forms radioactive Sodium-24 (

^{24}_{11}\text{Na}

), which subsequently undergoes beta decay with a half-life of

15\text{ hours}

. Starting with an initial mass of

0.80\text{ g}

of pure

^{24}_{11}\text{Na}

, after an elapsed time of

45\text{ hours}

, the mass of

^{24}_{11}\text{Na}

remaining is grams, and the element formed as the stable daughter product is .

Show answer & explanation

Answer

The remaining mass of Sodium-24 after 45 hours is 0.10 grams, and the stable daughter element formed is Magnesium.

In 45 hours, exactly three 15-hour half-lives elapse, reducing the initial 0.80 g sample to 0.10 g. During beta decay, the emission of a beta particle (

^{0}_{-1}\beta

) increases the atomic number of the nuclide from 11 to 12, converting Sodium to Magnesium.

Step-by-Step Solution

1

Calculate the total number of half-lives elapsed (

n

)

n = \frac{t}{t_{1/2}} = \frac{45\text{ hours}}{15\text{ hours}} = 3\text{ half-lives}

The total elapsed time divided by the half-life period yields the number of decay cycles.

2

Determine the remaining mass of Sodium-24 using exponential decay

N_t = N_0 \times \left(\frac{1}{2}\right)^n = 0.80\text{ g} \times \left(\frac{1}{2}\right)^3 = 0.80\text{ g} \times 0.125 = 0.10\text{ g}

With each half-life cycle, the active radioactive mass decreases by half.

3

Write the balanced nuclear reaction for the beta decay of Sodium-24 to identify the daughter element

^{24}_{11}\text{Na} \rightarrow ^{24}_{12}\text{Mg} + ^{0}_{-1}\beta

In beta particle emission (

^{0}_{-1}\beta

), a neutron converts to a proton, increasing the atomic number

Z

from 11 to 12 while leaving the mass number

A = 24

unchanged. Element 12 on the periodic table is Magnesium (Mg).

Key Concept

Radioactive decay law and nuclear equation balancing for beta decay

Question 8505Question →

An organic compound with the molecular formula $C_4H_{10}O$ forms four isomeric alkanols. Which of these alkanol isomers contains an asymmetric carbon atom and exhibits optical isomerism?

Butan-1-ol

Butan-2-ol

2-Methylpropan-1-ol

2-Methylpropan-2-ol

Show answer & explanation

Answer: Butan-2-ol

Answer

Butan-2-ol is the only isomeric alkanol of formula

C_4H_{10}O

containing a chiral carbon atom.

Butan-2-ol has a chiral carbon at position 2 (

CH_3-CH(OH)-CH_2-CH_3

), which is bonded to four different substituent groups:

-H

,

-OH

,

-CH_3

, and

-CH_2CH_3

. The presence of this asymmetric center causes optical activity.

Step-by-Step Solution

1

Identify the requirement for optical isomerism.

A molecule exhibits optical isomerism if it contains at least one chiral (asymmetric) carbon atom—a carbon atom bonded to four different groups or atoms.

Chirality causes non-superimposable mirror images (enantiomers) capable of rotating plane-polarized light.

2

Analyze the structural formulas of the four isomeric alkanols of

C_4H_{10}O

.

1) Butan-1-ol:

CH_3-CH_2-CH_2-CH_2OH

2) Butan-2-ol:

CH_3-CH(OH)-CH_2-CH_3

3) 2-Methylpropan-1-ol:

(CH_3)_2CH-CH_2OH

4) 2-Methylpropan-2-ol:

(CH_3)_3C-OH

Writing structural formulas reveals the substituent groups attached to each carbon atom.

3

Examine carbon-2 in butan-2-ol.

Carbon-2 is attached to

-H

,

-OH

,

-CH_3

, and

-CH_2CH_3

. All four substituents are distinct.

Since carbon-2 has four different attached groups, it is an asymmetric (chiral) carbon atom.

Key Concept

Optical isomerism and chirality in alkanols

Question 8506Question →

Match each chemical testing reagent or reaction system on the left with its corresponding diagnostic observation and redox transformation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Acidified potassium dichromate(VI) solution turns from orange to green (

Cr_2O_7^{2-} \rightarrow Cr^{3+}

); starch-potassium iodide paper turns blue-black (

I^- \rightarrow I_2

); iron(III) chloride solution turns from reddish-brown to pale green with yellow sulfur precipitate (

Fe^{3+} \rightarrow Fe^{2+}

); acidified potassium tetraoxomanganate(VII) turns from purple to colorless (

MnO_4^- \rightarrow Mn^{2+}

).

Each laboratory reagent exhibits a specific color change reflecting its underlying redox reaction: potassium dichromate(VI) changes from orange to green (

Cr_2O_7^{2-} \rightarrow Cr^{3+}

); starch-potassium iodide paper turns blue-black due to iodine liberation (

I^- \rightarrow I_2

); iron(III) chloride changes from reddish-brown to pale green with yellow sulfur precipitation (

Fe^{3+} \rightarrow Fe^{2+}

); and potassium tetraoxomanganate(VII) changes from purple to colorless (

MnO_4^- \rightarrow Mn^{2+}

).

Step-by-Step Solution

1

Determine the diagnostic color change for acidified potassium dichromate(VI).

Orange

Cr_2O_7^{2-}

ions are reduced to green

Cr^{3+}

ions when reacting with a reducing agent.

Chromium undergoes a reduction in oxidation state from

+6

to

+3

.

2

Analyze the chemistry of the starch-potassium iodide test paper.

Oxidizing agents convert

I^-

(iodide) into

I_2

(iodine), producing a characteristic blue-black complex with starch.

Iodide ions act as the reducing agent in the indicator paper and undergo oxidation.

3

Identify the reaction between iron(III) ions and hydrogen sulfide.

Reddish-brown

Fe^{3+}

is reduced to pale green

Fe^{2+}

, while sulfide (

S^{2-}

) is oxidized to free yellow sulfur (

S

).

H_2S

is a classic laboratory reducing agent for iron(III) salts.

4

Determine the diagnostic observation for acidified potassium tetraoxomanganate(VII).

Purple

MnO_4^-

ions are reduced to virtually colorless

Mn^{2+}

ions in acid medium.

Manganese undergoes reduction from

+7

to

+2

oxidation state.

Key Concept

Qualitative laboratory tests and color changes associated with common oxidizing and reducing agents.

Question 8507Question →

An organic compound $Z$ with the molecular formula $C_5H_{10}O$ gives a positive orange precipitate when treated with 2,4-dinitrophenylhydrazine. When warmed with acidified potassium heptaoxodichromate(VI) ( $K_2Cr_2O_7$ ), compound $Z$ is readily oxidized to a carboxylic acid containing five carbon atoms. Upon reduction with lithium tetrahydridoaluminate(III) ( $LiAlH_4$ ), compound $Z$ yields a primary alcohol that exhibits optical activity (chirality). What is the IUPAC name of compound $Z$ ?

2-methylbutanal

pentan-2-one

pentanal

3-methylbutanal

Show answer & explanation

Answer: 2-methylbutanal

Answer

2-methylbutanal

2-methylbutanal has the molecular formula

C_5H_{10}O

and contains an alkanal functional group. It reacts with 2,4-dinitrophenylhydrazine to form a precipitate. Because it is an alkanal, it undergoes oxidation with acidified potassium heptaoxodichromate(VI) to produce 2-methylbutanoic acid (a 5-carbon carboxylic acid). Upon reduction with lithium tetrahydridoaluminate(III), it forms 2-methylbutan-1-ol, which has an asymmetric carbon atom at position 2 bonded to four different substituent groups (

-H

,

-CH_3

,

-CH_2CH_3

,

-CH_2OH

), rendering the molecule chiral and optically active.

Step-by-Step Solution

1

Identify the functional group class from the 2,4-DNPH test.

A positive test with 2,4-dinitrophenylhydrazine confirms that compound

Z

is a carbonyl compound (either an alkanal or an alkanone).

Both alkanals and alkanones form colored hydrazone precipitates with 2,4-DNPH.

2

Distinguish between an alkanal and an alkanone using oxidation behavior.

Compound

Z

readily oxidizes with acidified

K_2Cr_2O_7

to a carboxylic acid with 5 carbons, proving

Z

is an alkanal (pentanal isomer).

Alkanals are easily oxidized to carboxylic acids with the same number of carbon atoms, whereas alkanones resist oxidation under mild conditions.

3

Analyze the reduction product for chirality.

Reduction of an alkanal with

LiAlH_4

yields a primary alcohol. Among 5-carbon alkanals (pentanal, 2-methylbutanal, 3-methylbutanal, 2,2-dimethylpropanal), only 2-methylbutanal reduces to 2-methylbutan-1-ol,

CH_3CH_2CH(CH_3)CH_2OH

.

In 2-methylbutan-1-ol, C-2 is bonded to four distinct groups:

-H

,

-CH_3

,

-CH_2CH_3

, and

-CH_2OH

, making it a chiral molecule capable of optical isomerism.

Key Concept

Distinction between alkanals and alkanones via oxidation, reduction of carbonyls to alcohols, and optical isomerism in branched primary alcohols.

Question 8508Question →

During the electrolysis of concentrated sodium chloride solution (brine) using inert platinum electrodes, chlorine gas is liberated at the anode instead of oxygen gas. Which factor primarily accounts for the preferential discharge of chloride ions over hydroxide ions in this process?

The high concentration of chloride ions relative to hydroxide ions in the solution

The lower position of chloride ions relative to hydroxide ions in the electrochemical series

The chemical affinity between the platinum electrode and chloride ions

The greater ionic mass of chloride ions compared to hydroxide ions

Show answer & explanation

Answer: The high concentration of chloride ions relative to hydroxide ions in the solution

Answer

The high concentration of chloride ions relative to hydroxide ions in the solution

Under standard conditions, hydroxide ions are discharged preferentially to chloride ions because hydroxide ions lie lower in the electrochemical series of anions. However, in concentrated brine, the concentration of chloride ions is vastly higher than that of hydroxide ions, causing concentration to become the decisive factor leading to chlorine gas evolution.

Step-by-Step Solution

1

Identify the ions present at the anode during electrolysis of concentrated sodium chloride solution.

The anions attracted to the anode are chloride ions (

Cl^-

) and hydroxide ions (

OH^-

).

Anions migrate to the positively charged anode.

2

Compare the factors governing preferential discharge of these anions.

Based on position in the electrochemical series,

OH^-

is discharged more easily than

Cl^-

. However, because the solution is concentrated,

Cl^-

ions far outnumber

OH^-

ions.

High concentration of a particular ion can overcome its disadvantage in electrochemical series position.

3

Deduce the discharged product and the dominant factor.

Chloride ions (

Cl^-

) are discharged to form chlorine gas (

Cl_2

) due to the concentration effect.

The concentration factor overrides the standard electrochemical series order in concentrated brine.

Key Concept

Effect of concentration on preferential discharge of ions during electrolysis

Question 8509Question →

Consider a standard galvanic cell constructed using cobalt and silver half-cells under standard conditions:

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

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

Which of the following statements correctly describes the operational mechanics of this cell?

Cobalt acts as the cathode and undergoes oxidation, leading to a decrease in electrode mass.

Electrons flow spontaneously from the cobalt electrode to the silver electrode through the external conductor.

The standard electromotive force (

E^\circ_{\text{cell}}

) generated by the cell is

+0.52\text{ V}

.

Nitrate ions (

\text{NO}_3^-

) from the salt bridge migrate into the silver half-cell to maintain electrical neutrality.

Show answer & explanation

Answer: Electrons flow spontaneously from the cobalt electrode to the silver electrode through the external conductor.

Answer

Electrons flow spontaneously from the cobalt electrode to the silver electrode through the external conductor.

In a galvanic cell, oxidation occurs at the electrode with the lower standard reduction potential (cobalt electrode, making it the anode). Reduction occurs at the electrode with the higher standard reduction potential (silver electrode, making it the cathode). Electrons are generated at the anode by oxidation and travel through the external circuit to the cathode.

Step-by-Step Solution

1

Identify the anode and cathode based on standard reduction potentials (

E^\circ

).

The half-cell with the more negative standard reduction potential (

E^\circ(\text{Co}^{2+}/\text{Co}) = -0.28\text{ V}

) undergoes oxidation at the anode. The half-cell with the more positive potential (

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

) undergoes reduction at the cathode.

Species with higher standard reduction potentials are more easily reduced, while those with lower reduction potentials are more easily oxidized.

2

Determine the direction of electron flow in the external circuit.

Oxidation at the cobalt anode releases electrons:

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

. These electrons flow through the wire toward the silver cathode.

Electrons always flow spontaneously from the anode (site of oxidation) to the cathode (site of reduction) in a galvanic cell.

3

Calculate the cell potential (

E^\circ_{\text{cell}}

) to verify consistency.

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = +0.80\text{ V} - (-0.28\text{ V}) = +1.08\text{ V}

.

A positive standard cell electromotive force confirms that the cell reaction is spontaneous in the stated direction.

Key Concept

Galvanic Cell Mechanics and Standard Electrode Potentials

Question 8510Question →

The solubility of a salt $Z$ (molar mass = $101.0\text{ g mol}^{-1}$ ) in water is $5.0\text{ mol dm}^{-3}$ at $80^\circ\text{C}$ and $202.0\text{ g dm}^{-3}$ at $25^\circ\text{C}$ . If $250\text{ cm}^3$ of a saturated solution of salt $Z$ is cooled from $80^\circ\text{C}$ to $25^\circ\text{C}$ , what mass of salt $Z$ will crystallize out of the solution?

75.75\text{ g}

303.00\text{ g}

50.50\text{ g}

0.75\text{ g}

Show answer & explanation

Answer: $75.75\text{ g}$

Answer

The mass of salt Z that will crystallize out of solution is

75.75\text{ g}

.

Converting the solubility at

80^\circ\text{C}

(

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

) into mass concentration yields

505.0\text{ g dm}^{-3}

. Subtracting the solubility at

25^\circ\text{C}

(

202.0\text{ g dm}^{-3}

) gives

303.0\text{ g dm}^{-3}

precipitated. Multiplying by the volume ratio (

250\text{ cm}^3 / 1000\text{ cm}^3 = 0.25

) yields

75.75\text{ g}

.

Step-by-Step Solution

1

Convert the solubility at

80^\circ\text{C}

from

\text{mol dm}^{-3}

to

\text{g dm}^{-3}

.

\text{Solubility at } 80^\circ\text{C} = 5.0\text{ mol dm}^{-3} \times 101.0\text{ g mol}^{-1} = 505.0\text{ g dm}^{-3}

.

Solubility values given in molarity must be multiplied by molar mass to obtain concentration in mass per unit volume.

2

Calculate the mass of solute precipitated per

\text{dm}^3

upon cooling to

25^\circ\text{C}

.

\Delta\text{Solubility} = 505.0\text{ g dm}^{-3} - 202.0\text{ g dm}^{-3} = 303.0\text{ g dm}^{-3}

.

The difference between solubilities at the higher and lower temperatures gives the mass of solute that cannot remain dissolved in

1\text{ dm}^3

of water.

3

Scale the precipitated mass to the given volume of

250\text{ cm}^3

.

\text{Mass crystallized} = 303.0\text{ g dm}^{-3} \times \frac{250\text{ cm}^3}{1000\text{ cm}^3\text{ dm}^{-3}} = 303.0\text{ g} \times 0.25 = 75.75\text{ g}

.

The solution volume is

250\text{ cm}^3

(

0.25\text{ dm}^3

), so the precipitated mass is proportional to this fractional volume.

Key Concept

Crystallization calculations from solubility temperature changes

Question 8511Question →

An organic compound $Z$ with the molecular formula $C_5H_{10}O$ gives a negative result with Tollen's reagent and does not form a yellow precipitate when warmed with iodine in sodium hydroxide solution. Upon reduction with lithium tetrahydridoaluminate(III) ( $LiAlH_4$ ), compound $Z$ yields a secondary alkanol. Which of the following is the correct IUPAC name of compound $Z$ ?

pentan-3-one

pentan-2-one

pentanal

3-methylbutan-2-one

Show answer & explanation

Answer: pentan-3-one

Answer

pentan-3-one

Pentan-3-one is an alkanone with the structure

CH_3CH_2COCH_2CH_3

. Because it is a ketone, it cannot be oxidized by mild oxidizing agents like Tollen's reagent. Furthermore, because its carbonyl carbon is bonded to two ethyl groups rather than a methyl group, it gives a negative triiodomethane (iodoform) test. Reduction of pentan-3-one using

LiAlH_4

yields pentan-3-ol, which is a secondary alkanol.

Step-by-Step Solution

1

Analyze the functional group class using Tollen's reagent test.

Compound

Z

gives a negative Tollen's test, confirming it is an alkanone (ketone) rather than an alkanal (aldehyde).

Alkanals are easily oxidized to alkanoic acids and reduce Tollen's reagent to metallic silver, whereas alkanones resist mild oxidation.

2

Evaluate the iodoform (triiodomethane) test requirement.

The absence of a yellow precipitate (

CHI_3

) rules out compounds containing a methyl ketone (

CH_3CO-

) group.

Only methyl ketones (

R-COCH_3

) or ethanal (

CH_3CHO

) undergo triiodomethane formation with iodine in sodium hydroxide.

3

Examine the reduction reaction and identify the specific isomer.

Among the

C_5H_{10}O

ketone isomers (pentan-2-one, pentan-3-one, and 3-methylbutan-2-one), only pentan-3-one lacks a methyl ketone group and reduces with

LiAlH_4

to give pentan-3-ol (a secondary alkanol).

Pentan-3-one has the structural formula

CH_3CH_2COCH_2CH_3

, fulfilling both qualitative test observations and structural reduction requirements.

Key Concept

Distinction tests for carbonyl compounds (Tollen's test and iodoform test) and reduction outcomes of alkanones

Estimated Time:2m 0s

Question 8512Question →

The standard reduction potentials for gallium and nickel half-reactions at

25^\circ\text{C}

are given below:

\text{Ga}^{3+}(aq) + 3e^- \rightarrow \text{Ga}(s) \quad E^\circ = -0.56\text{ V}

\text{Ni}^{2+}(aq) + 2e^- \rightarrow \text{Ni}(s) \quad E^\circ = -0.25\text{ V}

What is the standard electromotive force (

E^\circ_{\text{cell}}

) for the spontaneous reaction between these two electrochemical systems?

+0.31\text{ V}

-0.31\text{ V}

+0.37\text{ V}

+0.81\text{ V}

Show answer & explanation

Answer: $+0.31\text{ V}$

Answer

+0.31\text{ V}

For a redox reaction to be spontaneous under standard conditions, the cell potential (

E^\circ_{\text{cell}}

) must be positive. In the electrochemical series, species with higher reduction potentials undergo reduction (cathode). Comparing

-0.25\text{ V}

(nickel) and

-0.56\text{ V}

(gallium), nickel has the higher potential and acts as the cathode. Applying

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

yields

-0.25\text{ V} - (-0.56\text{ V}) = +0.31\text{ V}

.

Step-by-Step Solution

1

Identify cathode and anode based on standard reduction potentials for spontaneity

Cathode:

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

(

E^\circ = -0.25\text{ V}

), Anode:

\text{Ga}^{3+}/\text{Ga}

(

E^\circ = -0.56\text{ V}

)

The half-cell with the more positive (less negative) standard reduction potential undergoes reduction at the cathode in a spontaneous cell.

2

Calculate the standard cell potential using

E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}}

E^\circ_{\text{cell}} = -0.25\text{ V} - (-0.56\text{ V}) = +0.31\text{ V}

Standard electromotive force is the difference between the reduction potential of the cathode and the reduction potential of the anode.

Key Concept

Standard Cell Potential and Reaction Spontaneity

Estimated Time:1m 30s

Question 8513Question →

A sample containing $2.5\text{ moles}$ of ammonia gas ( $\text{NH}_3$ ) is held in a vessel at $400\text{ K}$ under a pressure of $50.0\text{ atm}$ . Under these conditions, the compressibility factor ( $Z$ ) of ammonia is $0.880$ . What is the actual volume occupied by the gas sample in $\text{dm}^3$ ? (Take $R = 0.0821\text{ dm}^3\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$ )

Show answer & explanation

Answer: 1.44

Answer

The actual volume occupied by the ammonia gas sample under the given conditions is

1.44\text{ dm}^3

.

The compressibility factor

Z

is defined as

Z = \frac{P V}{n R T} = \frac{V_{\text{real}}}{V_{\text{ideal}}}

. Substituting

n = 2.5\text{ mol}

,

T = 400\text{ K}

,

P = 50.0\text{ atm}

,

R = 0.0821\text{ dm}^3\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}

, and

Z = 0.880

into

V = \frac{Z \cdot n R T}{P}

gives

1.44\text{ dm}^3

.

Step-by-Step Solution

1

Identify the relationship between real volume and compressibility factor Z

Z = \frac{P V_{\text{actual}}}{n R T}

The compressibility factor quantifies the deviation of a real gas from ideal gas behavior.

2

Rearrange the compressibility formula to solve for actual volume (

V_{\text{actual}}

)

V_{\text{actual}} = \frac{Z \cdot n R T}{P}

Isolating

V_{\text{actual}}

allows direct calculation using the provided parameters.

3

Substitute the given values into the equation and compute the result

V_{\text{actual}} = \frac{0.880 \times 2.5 \times 0.0821 \times 400}{50.0} = 1.44496\text{ dm}^3

Performing the algebraic calculation yields the volume occupied by the real gas.

Key Concept

Compressibility Factor and Real Gas Deviation

Question 8514Question →

Match each salt listed in Column A with its appropriate laboratory preparation method from Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Lead(II) sulfate pairs with Precipitation (Double Decomposition); Sodium nitrate pairs with Titration of an acid with a soluble alkali; Copper(II) sulfate pairs with Action of dilute acid on an insoluble base; Anhydrous iron(III) chloride pairs with Direct combination of constituent elements.

Matching each salt to its preparation method requires analyzing solubility and chemical properties: insoluble salts like lead(II) sulfate are formed by precipitation/double decomposition; soluble sodium salts require neutralization by titration; soluble copper salts are synthesized using an insoluble oxide; and volatile anhydrous halides like iron(III) chloride are synthesized by direct combination of elements.

Step-by-Step Solution

1

Classify the salts by solubility in water.

Lead(II) sulfate is insoluble, while sodium nitrate, copper(II) sulfate, and iron(III) chloride are soluble.

The method of salt preparation depends primarily on whether the target salt is soluble or insoluble.

2

Determine the preparation method for the insoluble salt.

Lead(II) sulfate is prepared by precipitation (double decomposition) combining soluble aqueous reactants such as lead(II) nitrate and sodium sulfate.

Precipitation is the standard method for preparing insoluble salts.

3

Determine preparation methods for the soluble salts based on reactant nature.

Sodium nitrate requires titration because both sodium hydroxide and sodium nitrate are soluble; Copper(II) sulfate uses an insoluble base (copper(II) oxide) reacted with dilute acid; Anhydrous iron(III) chloride requires direct combination to avoid hydrolysis by water.

Group 1/ammonium soluble salts require titration, while anhydrous iron(III) chloride cannot be prepared by evaporation from aqueous solution due to hydration and hydrolysis.

Key Concept

Laboratory Preparation Methods of Soluble and Insoluble Salts

Question 8515Question →

During the electrolysis of an aqueous solution containing equal concentrations of cations using inert carbon electrodes, ions migrate to the cathode where discharge occurs based on their position in the electrochemical series. Arrange the following cations in increasing order of their ease of preferential discharge at the cathode (from the least easily discharged to the most easily discharged):

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence from least easily discharged to most easily discharged is Potassium ion (

K^+

), Zinc ion (

Zn^{2+}

), Hydrogen ion (

H^+

), and Copper(II) ion (

Cu^{2+}

).

In electrolysis at inert electrodes under equal concentrations, the primary factor determining preferential discharge of cations at the cathode is their position in the electrochemical series. Cations positioned lower in the series gain electrons more easily (have higher reduction potentials). Therefore, Potassium ion (

K^+

) is the hardest to discharge, followed by Zinc ion (

Zn^{2+}

), Hydrogen ion (

H^+

), and finally Copper(II) ion (

Cu^{2+}

), which is the most easily discharged.

Step-by-Step Solution

1

Recall the electrochemical series position for cations

The order of cations from top (most reactive / hardest to discharge) to bottom (least reactive / easiest to discharge) is

K^+

,

Zn^{2+}

,

H^+

,

Cu^{2+}

.

Ions lower in the electrochemical series have higher standard reduction potentials and accept electrons more readily at the cathode.

2

Arrange the cations in increasing order of ease of discharge

Potassium ion (

K^+

) < Zinc ion (

Zn^{2+}

) < Hydrogen ion (

H^+

) < Copper(II) ion (

Cu^{2+}

).

Ease of discharge increases down the electrochemical series.

Key Concept

Position of cations in the electrochemical series determines their relative ease of discharge at the cathode.

Question 8516Question →

Although fluorine and chlorine are both Group 17 halogens, hydrogen fluoride ( $HF$ ) exhibits intermolecular hydrogen bonding in the liquid state while hydrogen chloride ( $HCl$ ) does not. Which of the following statements best accounts for this difference?

Fluorine has a higher electronegativity and smaller atomic radius than chlorine, creating a highly polar

H-F

bond.

The intramolecular covalent bond in an

HF

molecule is weaker than the covalent bond in an

HCl

molecule.

Hydrogen fluoride has a greater molar mass than hydrogen chloride, generating stronger dispersion forces.

Chlorine is more electronegative than fluorine, causing

HCl

molecules to experience non-polar interactions.

Show answer & explanation

Answer: Fluorine has a higher electronegativity and smaller atomic radius than chlorine, creating a highly polar $H-F$ bond.

Answer

Fluorine's high electronegativity combined with its small atomic size allows for strong electrostatic attraction between the hydrogen atom of one molecule and the lone pair on the fluorine atom of an adjacent molecule.

Hydrogen bonding requires hydrogen to be covalently attached to a very small, highly electronegative atom (N, O, or F). Fluorine fulfills both conditions, giving the

H-F

bond a large dipole moment and high localized charge density that attracts adjacent

HF

molecules strongly.

Step-by-Step Solution

1

Identify the structural requirements for hydrogen bonding.

Hydrogen bonding occurs specifically when hydrogen is bonded directly to small, highly electronegative atoms (N, O, or F).

High electronegativity creates a strong dipole with high positive charge density on the hydrogen atom.

2

Compare fluorine and chlorine properties.

Fluorine has an electronegativity of 4.0 and a small atomic radius, whereas chlorine has an electronegativity of 3.0 and a larger atomic radius.

Although chlorine is electronegative, its larger atomic volume diffuses electron charge density, preventing effective hydrogen bond formation.

3

Select the statement that correctly attributes hydrogen bond formation to electronegativity and atomic size.

The statement emphasizing fluorine's higher electronegativity and smaller atomic size relative to chlorine provides the correct rationale.

It fulfills the fundamental chemical criteria required for hydrogen bond formation.

Key Concept

Intermolecular Forces and Hydrogen Bonding

Estimated Time:1m 0s

Question 8517Question →

Consider the reversible industrial reaction represented by the thermochemical equation below:

CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g) \quad \Delta H = -90\text{ kJ mol}^{-1}

What is the effect of adding a suitable catalyst to this equilibrium mixture?

It has no effect on the equilibrium position.

It shifts the equilibrium position to the right to increase methanol yield.

It shifts the equilibrium position to the left to favor the reactants.

It increases the yield of methanol by lowering the activation energy of the forward reaction only.

Show answer & explanation

Answer: It has no effect on the equilibrium position.

Answer

Adding a catalyst has no effect on the equilibrium position.

A catalyst provides an alternative pathway with lower activation energy for both the forward and reverse reactions. As a result, both rates are increased equally, allowing dynamic equilibrium to be reached faster without changing the relative amounts of reactants and products at equilibrium.

Step-by-Step Solution

1

Analyze the fundamental action of a catalyst in a chemical system.

A catalyst lowers the activation energy for both the forward and reverse reactions by providing an alternative reaction pathway.

Lowering the activation energy increases the rate of reaction in both directions by the exact same proportion.

2

Apply Le Chatelier's principle and equilibrium concepts to assess position shift.

The equilibrium position and the equilibrium concentrations of reactants and products remain unaffected.

Because both forward and reverse rates accelerate equally, the system reaches dynamic equilibrium faster without altering the final yield.

Key Concept

Effect of a catalyst on chemical equilibrium

Question 8518Question →

Consider the unbalanced redox reaction occurring in acidic solution:

\text{a AsO}_3^{3-}(\text{aq}) + \text{b MnO}_4^-(\text{aq}) + \text{c H}^+(\text{aq}) \rightarrow \text{d AsO}_4^{3-}(\text{aq}) + \text{e Mn}^{2+}(\text{aq}) + \text{f H}_2\text{O}(\text{l})

When this ionic equation is balanced using the smallest possible whole-number coefficients, what is the value of the coefficient

c

for

\text{H}^+

?

Show answer & explanation

Answer: 6

Answer

The coefficient c for hydrogen ions (H⁺) in the balanced redox equation is 6.

Balancing the oxidation half-reaction shows that each arsenite ion produces 2 electrons and 2 H⁺ ions. The reduction half-reaction shows that each permanganate ion consumes 5 electrons and 8 H⁺ ions. Multiplying the oxidation half-reaction by 5 and the reduction half-reaction by 2 balances the total electron transfer at 10 electrons. Combining the equations gives 16 H⁺ on the left and 10 H⁺ on the right, which simplifies to 6 H⁺ on the reactant side.

Step-by-Step Solution

1

Balance the oxidation half-reaction (arsenite to arsenate)

AsO₃³⁻ + H₂O → AsO₄³⁻ + 2H⁺ + 2e⁻

Arsenic changes oxidation state from +3 to +5, releasing 2 electrons. Oxygen is balanced with H₂O and hydrogen with H⁺.

2

Balance the reduction half-reaction (permanganate to manganese(II))

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Manganese changes oxidation state from +7 to +2, consuming 5 electrons in acidic medium.

3

Equalize the electrons transferred in both half-reactions

5(AsO₃³⁻ + H₂O → AsO₄³⁻ + 2H⁺ + 2e⁻) and 2(MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O)

The total number of electrons gained and lost must equal 10 electrons.

4

Combine the half-reactions and cancel common terms

5 AsO₃³⁻ + 2 MnO₄⁻ + 6 H⁺ → 5 AsO₄³⁻ + 2 Mn²⁺ + 3 H₂O

Subtracting 10 H⁺ and 5 H₂O from both sides leaves 6 H⁺ on the reactant side.

Key Concept

Balancing Ion-Electron Redox Half-Reactions in Acidic Medium

Question 8519Question →

Synthetic addition polymers, such as polyethene and polyvinyl chloride, are biodegradable because soil micro-organisms can readily break down their carbon-carbon backbone.

True

False

Show answer & explanation

Answer: False

Answer

False. Synthetic addition polymers possess strong, non-polar carbon-carbon backbones that micro-organisms cannot decompose naturally.

Synthetic addition polymers are non-biodegradable because their backbone consists of stable C-C single bonds that soil bacteria and fungi cannot enzymatically break down.

Step-by-Step Solution

1

Examine the chemical structure of synthetic addition polymers like polyethene.

The polymer backbone consists exclusively of non-polar carbon-carbon (C-C) single bonds without ester or amide linkages.

Microbial degradation requires specific functional groups that enzymes can recognize and hydrolyze.

2

Assess the action of soil micro-organisms on this chemical structure.

Micro-organisms do not possess enzymes capable of cleaving the unreactive C-C backbone of addition polymers.

Because the material cannot be decomposed biologically, it is categorized as non-biodegradable.

Key Concept

Distinction between non-biodegradable synthetic addition polymers and biodegradable natural/condensation polymers

Question 8520Question →

Consider the reversible gas-phase synthesis of methanol represented by the thermochemical equation below:

CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g) \quad \Delta H < 0

Match each applied stress condition on the left with its correct effect on the system at equilibrium on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Increasing total pressure shifts the equilibrium to the right; increasing temperature shifts the equilibrium to the left; adding a catalyst accelerates the rate of reaching equilibrium without shifting its position; adding an inert gas at constant volume has no effect on the equilibrium position.

Each stress causes an equilibrium adjustment governed strictly by Le Chatelier's principle. Pressure increases shift the position toward the side with fewer gas moles (the product side). Temperature increases favor the endothermic direction (the reverse reaction). Catalysts accelerate reaction rates equally in both directions without altering the equilibrium position, and inert gas additions at constant volume do not modify the partial pressures of the reacting components.

Step-by-Step Solution

1

Analyze the effect of volume reduction (pressure increase)

Reactants contain

1 + 2 = 3

gaseous moles, while products contain

1

gaseous mole. Increasing pressure shifts equilibrium towards fewer gaseous moles (to the right).

Le Chatelier's principle states that an increase in pressure shifts equilibrium toward the side with fewer moles of gas.

2

Analyze the effect of temperature increase

The forward reaction is exothermic (

\Delta H < 0

). Increasing temperature causes an equilibrium shift in the endothermic direction (to the left).

According to Le Chatelier's principle, adding thermal energy shifts equilibrium to favor heat absorption.

3

Analyze the effect of catalyst addition

A catalyst decreases activation energy for both directions equally.

Catalysts increase reaction rate and reduce equilibrium attainment time, but do not alter relative thermodynamic stabilities or equilibrium positions.

4

Analyze the effect of inert gas addition at constant volume

Concentrations and partial pressures of reacting gases remain constant.

Since total volume is fixed, molar concentrations of reacting species do not change, leaving equilibrium undisturbed.

Key Concept

Le Chatelier's Principle on Equilibrium Position and Rate of Reaction

Answer

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