All practice questions

Question 8321Question →

Copper(II) oxide ( $\text{CuO}$ ) is a black solid compound formed when copper metal is strongly heated in air. Given the relative atomic masses of copper ( $\text{Cu} = 64$ ) and oxygen ( $\text{O} = 16$ ), what is the percentage by mass of copper in pure copper(II) oxide?

Show answer & explanation

Answer: 80

Answer

The percentage by mass of copper in pure copper(II) oxide (

\text{CuO}

) is

80\%

.

The molar mass of copper(II) oxide (

\text{CuO}

) is

64 + 16 = 80\text{ g/mol}

. The relative mass contributed by copper is

64\text{ g/mol}

. Dividing

64

by

80

and multiplying by

100

gives

80\%

.

Step-by-Step Solution

1

Calculate the molar mass of copper(II) oxide (

\text{CuO}

).

Molar mass of

\text{CuO} = 64 + 16 = 80\text{ g/mol}

.

The molar mass of a binary compound is the sum of the relative atomic masses of its constituent elements.

2

Calculate the mass percentage of copper in the compound.

\text{Percentage of Cu} = \left(\frac{64}{80}\right) \times 100\% = 80\%

.

The mass percentage is found by dividing the mass contributed by copper by the total molar mass of the compound and multiplying by 100.

Key Concept

Percentage composition by mass of an element in a copper compound

Question 8322Question →

A fine suspension of micro-solid particles in a liquid medium passes through filter paper very slowly because the particles clog the pores. Which physical technique utilizes high-speed spinning to force the denser solid particles to settle rapidly at the bottom of the container?

Centrifugation

Sublimation

Fractional distillation

Magnetization

Show answer & explanation

Answer: Centrifugation

Answer

Centrifugation is the technique that uses high-speed rotation and centrifugal force to rapidly separate suspended solids from liquids based on density differences.

Centrifugation is specifically designed for separating fine solid suspensions in liquids when standard filtration is too slow or ineffective. Spinning the mixture at high speed exerts centrifugal force, driving denser suspended particles rapidly to the bottom.

Step-by-Step Solution

1

Analyze the physical properties of the mixture described in the stem.

The mixture consists of micro-solid particles suspended in a liquid medium that clog filter paper.

Identifying that filtration is ineffective due to pore clogging narrows down the required physical separation method.

2

Evaluate the mechanism of centrifugation.

High-speed spinning creates outward centrifugal force that forces heavier suspended particles to settle to the bottom as a pellet, leaving clear liquid supernatant.

Centrifugation relies on density differences between suspended solid particles and the liquid phase without heat or chemical modification.

Key Concept

Centrifugation and separation of suspensions

Estimated Time:1m 0s

Question 8323Question →

Given the standard reduction potentials ( $E^\circ$ ) below, arrange the metals in order of increasing reducing strength (from weakest reducing agent to strongest reducing agent):

\begin{aligned} \text{Ag}^+_{(\text{aq})} + \text{e}^- &\rightarrow \text{Ag}_{(\text{s})} \quad E^\circ = +0.80\text{ V} \\ \text{Cu}^{2+}_{(\text{aq})} + 2\text{e}^- &\rightarrow \text{Cu}_{(\text{s})} \quad E^\circ = +0.34\text{ V} \\ \text{Fe}^{2+}_{(\text{aq})} + 2\text{e}^- &\rightarrow \text{Fe}_{(\text{s})} \quad E^\circ = -0.44\text{ V} \\ \text{Zn}^{2+}_{(\text{aq})} + 2\text{e}^- &\rightarrow \text{Zn}_{(\text{s})} \quad E^\circ = -0.76\text{ V} \end{aligned}

Which sequence represents the correct order of increasing reducing strength?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order of increasing reducing strength is Silver (

\text{Ag}

), Copper (

\text{Cu}

), Iron (

\text{Fe}

), and Zinc (

\text{Zn}

).

Reducing power increases as standard reduction potential (

E^\circ

) becomes more negative, because metals with negative reduction potentials readily lose electrons. Silver has the most positive reduction potential (

+0.80\text{ V}

), followed by copper (

+0.34\text{ V}

), iron (

-0.44\text{ V}

), and zinc (

-0.76\text{ V}

). Therefore, the order from weakest to strongest reducing agent is Silver, Copper, Iron, and Zinc.

Step-by-Step Solution

1

Understand the relationship between standard reduction potential (

E^\circ

) and reducing strength.

A species with a more negative standard reduction potential has a greater tendency to undergo oxidation (lose electrons) and is therefore a stronger reducing agent. A species with a more positive

E^\circ

is a weaker reducing agent.

Reducing strength is inversely proportional to standard reduction potential.

2

Compare the given

E^\circ

values.

E^\circ(\text{Ag}^+/\text{Ag}) = +0.80\text{ V} > E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\text{ V} > E^\circ(\text{Fe}^{2+}/\text{Fe}) = -0.44\text{ V} > E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.76\text{ V}

Listing potentials from most positive to most negative orders the elements from weakest to strongest reducing agent.

3

Arrange the metals in increasing order of reducing power.

Silver (

\text{Ag}

) < Copper (

\text{Cu}

) < Iron (

\text{Fe}

) < Zinc (

\text{Zn}

)

Silver has the highest

E^\circ

and is the weakest reducing agent, while zinc has the lowest

E^\circ

and is the strongest reducing agent.

Key Concept

Reducing Strength and Standard Electrode Potentials

Question 8324Question →

Ionic compounds are generally brittle because applying a mechanical stress causes layers of ions to shift, bringing ions of identical charge into alignment and causing strong electrostatic repulsion.

True

False

Show answer & explanation

Answer: True

Answer

True. Applying mechanical stress causes layers of an ionic lattice to shift, bringing like-charged ions into direct alignment; the immediate electrostatic repulsion forces the crystal planes apart, rendering ionic solids brittle.

The statement is true because mechanical stress displaces adjacent rows in an ionic lattice, shifting like-charged ions into alignment. The powerful electrostatic repulsion generated between identical charges forces the crystal layers apart, causing the ionic solid to shatter.

Step-by-Step Solution

1

Identify the arrangement of particles in a solid ionic lattice.

An ionic crystal consists of alternating positive cations and negative anions organized in a regular three-dimensional array held by electrostatic attraction.

Determining the initial alternating charge pattern is necessary to understand how movement alters interionic forces.

2

Analyze the structural shift caused by an applied mechanical force.

The force causes one layer of ions to slide past another by one atomic position.

Mechanical impact displaces crystal planes relative to each other.

3

Evaluate the net electrostatic forces after displacement.

Ions of identical charge (

+

and

+

, or

-

and

-

) are brought into direct alignment, creating powerful repulsive forces that shatter the lattice along cleavage planes.

Electrostatic repulsion between like charges overcomes binding attraction, explaining the characteristic brittleness of ionic solids.

Key Concept

Brittleness and Mechanical Cleavage of Ionic Lattice Structures

Question 8325Question →

Complete the statement describing the redox reaction and observation when hydrogen sulfide gas acts as a reducing agent with iron(III) chloride solution.

Fill in the blanks below

When hydrogen sulfide gas (

H_2S

) is bubbled through a reddish-brown aqueous solution of iron(III) chloride (

FeCl_3

), the solution changes color to due to the reduction of

Fe^{3+}

to

Fe^{2+}

ions, while the hydrogen sulfide is oxidized to produce a yellow precipitate of elemental .

Show answer & explanation

Answer

The solution turns green (or pale green) due to the formation of

Fe^{2+}

ions, and a yellow precipitate of sulfur (or sulphur) is deposited.

In the chemical reaction

2Fe^{3+}_{(aq)} + H_2S_{(g)} \rightarrow 2Fe^{2+}_{(aq)} + S_{(s)} + 2H^{+}_{(aq)}

, iron(III) chloride acts as an oxidizing agent and is reduced to green iron(II) ions. Simultaneously, hydrogen sulfide acts as a reducing agent and is oxidized to elemental sulfur, which forms a yellow precipitate.

Step-by-Step Solution

1

Analyze the oxidation state changes in the reaction between

Fe^{3+}

ions and

H_2S

.

Iron is reduced from

+3

in

Fe^{3+}

to

+2

in

Fe^{2+}

. Sulfur is oxidized from

-2

in

H_2S

to

0

in elemental sulfur (

S

).

Iron(III) ions act as an oxidizing agent and undergo reduction, while hydrogen sulfide acts as a reducing agent and undergoes oxidation.

2

Relate the chemical species formed to their characteristic physical observations.

Aqueous

Fe^{3+}

ions (reddish-brown/yellow-brown) are converted into aqueous

Fe^{2+}

ions (green). The oxidation of hydrogen sulfide yields elemental sulfur, which appears as a pale yellow solid precipitate.

Observing distinct color changes and solid deposition is the primary laboratory method for identifying redox species.

Key Concept

Redox testing of hydrogen sulfide as a reducing agent with iron(III) salts

Estimated Time:1m 0s

Question 8326Question →

An aqueous solution of aluminium chloride ( $AlCl_3$ ) is acidic because the chloride ion ( $Cl^-$ ) undergoes anion hydrolysis in water to produce hydrogen ions ( $H^+$ ).

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Chloride ions do not undergo salt hydrolysis; the acidity of an aluminium chloride solution arises from cation hydrolysis of the hydrated aluminium ion.

The statement is false because chloride ions are spectator ions in aqueous solution and do not hydrolyze. The acidic nature of an aluminium chloride solution is instead driven by cation hydrolysis, in which the high charge density of the hydrated aluminium ion causes it to act as a Brønsted-Lowry acid by donating protons to water.

Step-by-Step Solution

1

Analyze the dissolution of

AlCl_3

in aqueous medium.

AlCl_3(s) + 6H_2O(l) \rightarrow [Al(H_2O)_6]^{3+}(aq) + 3Cl^-(aq)

Aluminium chloride dissociates fully in water to yield hydrated aluminium cations and chloride anions.

2

Evaluate the hydrolysis potential of the chloride anion (

Cl^-

).

Cl^-(aq) + H_2O(l) \rightarrow \text{No Reaction}

Chloride is the conjugate base of the strong acid

HCl

; hence, it is a spectator ion with negligible basicity and cannot undergo hydrolysis.

3

Evaluate the hydrolysis reaction of the hexaaquaaluminium(III) cation (

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

).

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

The high charge density of

Al^{3+}

weakens O-H bonds in coordinated water molecules, enabling proton transfer to free water molecules and increasing

[H_3O^+]

.

Key Concept

Cation Hydrolysis of Polyvalent Metal Ions

Question 8327Question →

Under identical conditions of temperature and pressure, a $100\text{ cm}^3$ sample of nitrogen(II) oxide ( $NO$ ) diffuses through a porous container in $20\text{ seconds}$ . If $100\text{ cm}^3$ of an unknown oxide of nitrogen diffuses through the exact same porous container in $35\text{ seconds}$ , what is the molecular formula of the unknown oxide of nitrogen? $[N = 14, O = 16]$

N_2O_4

NO_2

N_2O

N_2O_5

Show answer & explanation

Answer: $N_2O_4$

Answer

The molecular formula of the unknown oxide is

N_2O_4

.

The correct answer is derived using Graham's Law of Diffusion (

\frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}}

). Given that nitrogen(II) oxide (

NO

) has a molar mass of

30\text{ g/mol}

and diffuses in

20\text{ s}

, an oxide diffusing in

35\text{ s}

must have a molar mass of

30 \times (35/20)^2 = 91.875 \approx 92\text{ g/mol}

. The formula matching this molar mass is dinitrogen tetroxide (

N_2O_4

).

Step-by-Step Solution

1

Calculate the molar mass of the reference gas, nitrogen(II) oxide (

NO

).

M_1(NO) = 14 + 16 = 30\text{ g/mol}

.

Graham's law requires the molar mass of the reference gas to determine the unknown gas mass.

2

Apply Graham's Law of Diffusion in terms of time taken for equal volumes of gas to diffuse.

\frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}}

, where

t_1 = 20\text{ s}

,

t_2 = 35\text{ s}

, and

M_1 = 30\text{ g/mol}

.

Rate of diffusion is inversely proportional to diffusion time for fixed volume, so

\frac{\text{Rate}_1}{\text{Rate}_2} = \frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}}

.

3

Solve for the unknown molar mass

M_2

.

\frac{35}{20} = 1.75 \implies (1.75)^2 = \frac{M_2}{30} \implies 3.0625 = \frac{M_2}{30} \implies M_2 = 91.875 \approx 92\text{ g/mol}

.

Squaring the time ratio isolates the molar mass ratio.

4

Identify the formula of the nitrogen oxide with a molar mass of

92\text{ g/mol}

.

N_2O_4

:

2(14) + 4(16) = 28 + 64 = 92\text{ g/mol}

.

Matching the calculated molar mass to the chemical formula of oxides of nitrogen.

Key Concept

Graham's Law of Diffusion applied to Oxides of Nitrogen

Estimated Time:2m 0s

Question 8328Question →

When solid ammonium chloride ( $NH_4Cl$ ) is dissolved in water, the resulting solution exhibits an acidic pH. Which species undergoes hydrolysis to cause this acidity?

Ammonium ion (

NH_4^+

)

Chloride ion (

Cl^-

)

Both ammonium (

NH_4^+

) and chloride (

Cl^-

) ions

Un-ionized ammonium chloride molecule (

NH_4Cl

)

Show answer & explanation

Answer: Ammonium ion ( $NH_4^+$ )

Answer

The ammonium ion (

NH_4^+

) undergoes cation hydrolysis to form hydronium ions (

H_3O^+

), making the solution acidic.

Ammonium chloride (

NH_4Cl

) completely dissociates in water into

NH_4^+

and

Cl^-

. Since

NH_4^+

is the conjugate acid of the weak base

NH_3

, it reacts with water molecules (cation hydrolysis) to donate a proton, producing excess

H_3O^+

ions which lower the pH below 7.

Step-by-Step Solution

1

Identify the parent acid and base of the salt.

Ammonium chloride (

NH_4Cl

) is formed from the weak base ammonia (

NH_3

) and the strong acid hydrochloric acid (

HCl

).

Salts composed of a weak base and strong acid form acidic aqueous solutions due to cation hydrolysis.

2

Determine which ion hydrolyzes in aqueous solution.

The ammonium ion (

NH_4^+

) reacts with water:

NH_4^+_{(aq)} + H_2O_{(l)} \rightleftharpoons NH_{3(aq)} + H_3O^+_{(aq)}

.

Conjugate acids of weak bases are strong enough to donate protons to water molecules, generating hydronium ions.

Key Concept

Salt Hydrolysis of Weak Base and Strong Acid Salts

Question 8329Question →

Match each distillation term or phenomenon on the left with its correct chemical definition or description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Azeotropic mixture matches with a constant-boiling liquid mixture; Distillate matches with the purified liquid collected after vapor condensation; Refluxing matches with the continuous return of condensed vapor back down the column; Theoretical plate matches with a hypothetical zone where liquid and vapor reach equilibrium.

Each distillation concept is correctly matched based on foundational principles: an azeotropic mixture maintains constant composition at its boiling point; distillate refers to the condensed liquid product; refluxing is the downward flow of condensate inside the column; and a theoretical plate quantifies column separation performance.

Step-by-Step Solution

1

Define an azeotropic mixture.

Identify that azeotropes boil at a fixed temperature with identical liquid and vapor compositions.

Azeotropes act like pure substances during boiling and cannot be separated further by simple or fractional distillation.

2

Define distillate.

Identify distillate as the final condensed liquid collected in the receiving flask.

Vapors leaving the still head are condensed into liquid distillate.

3

Define refluxing in fractional distillation.

Identify refluxing as returning condensed liquid back down the fractionating column.

This establishes a temperature gradient and continuous equilibrium steps along the column.

4

Define a theoretical plate.

Identify a theoretical plate as a stage representing one complete vaporization-condensation cycle.

More theoretical plates in a column correspond to higher separation efficiency for liquids with close boiling points.

Key Concept

Distillation Terminology and Theoretical Principles

Question 8330Question →

A $5.00\text{ g}$ sample of impure zinc granules reacts completely with excess dilute tetraoxosulfate(VI) acid to produce $1.12\text{ dm}^3$ of hydrogen gas at s.t.p. What is the percentage purity of the zinc sample? $[\text{Zn} = 65, \text{Molar volume of gas at s.t.p.} = 22.4\text{ dm}^3\text{ mol}^{-1}]$

$30.0\%

$60.7\%

$65.0\%

$35.0\%

Show answer & explanation

Answer: $65.0\%

Answer

65.0\%

The balanced chemical equation

\text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2

shows a

1:1

mole ratio between zinc and hydrogen gas. At s.t.p.,

1.12\text{ dm}^3

of

\text{H}_2

represents

\frac{1.12}{22.4} = 0.05\text{ mol}

. Thus,

0.05\text{ mol}

of pure zinc was present, corresponding to

0.05 \times 65 = 3.25\text{ g}

. Dividing the pure zinc mass by the total sample mass (

5.00\text{ g}

) gives

\frac{3.25}{5.00} \times 100\% = 65.0\%

.

Step-by-Step Solution

1

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

\text{Moles of H}_2 = \frac{1.12\text{ dm}^3}{22.4\text{ dm}^3\text{ mol}^{-1}} = 0.05\text{ mol}

The reaction produces

1\text{ mol}

of

\text{H}_2

for every

1\text{ mol}

of pure zinc reacted:

\text{Zn (s)} + \text{H}_2\text{SO}_4\text{ (aq)} \rightarrow \text{ZnSO}_4\text{ (aq)} + \text{H}_2\text{ (g)}

.

2

Determine the mass of pure zinc that reacted.

\text{Mass of pure Zn} = 0.05\text{ mol} \times 65\text{ g mol}^{-1} = 3.25\text{ g}

Molar mass converts the stoichiometric mole amount to pure mass of the element.

3

Calculate the percentage purity of the zinc sample.

\text{Percentage Purity} = \left( \frac{3.25\text{ g}}{5.00\text{ g}} \right) \times 100\% = 65.0\%

Percentage purity is the ratio of pure substance mass to total sample mass expressed as a percentage.

Key Concept

Calculating percentage purity from stoichiometric gas yields at s.t.p.

Estimated Time:1m 30s

Question 8331Question →

When an unknown alkanol is warmed with acidified potassium heptaoxodichromate(VI) ( $K_2Cr_2O_7$ ) solution, the orange solution turns green and an alkanone is produced. Which of the following compounds undergoes this reaction?

Propan-2-ol

Propan-1-ol

2-Methylpropan-2-ol

Ethanol

Show answer & explanation

Answer: Propan-2-ol

Answer

Propan-2-ol is a secondary alkanol that undergoes oxidation to form an alkanone (propanone), changing the color of acidified potassium heptaoxodichromate(VI) from orange to green.

Propan-2-ol is a secondary alkanol (

CH_3-CH(OH)-CH_3

). Upon oxidation with acidified

K_2Cr_2O_7

, the orange dichromate(VI) ions (

Cr_2O_7^{2-}

) are reduced to green chromium(III) ions (

Cr^{3+}

), and the secondary alcohol is converted into propanone (

CH_3COCH_3

), which belongs to the alkanone family.

Step-by-Step Solution

1

Classify the given alkanols by structural type (primary, secondary, or tertiary).

Propan-1-ol and ethanol are primary alkanols; propan-2-ol is a secondary alkanol; 2-methylpropan-2-ol is a tertiary alkanol.

The reaction outcome of alkanol oxidation depends strictly on the classification of the hydroxyl-bearing carbon atom.

2

Determine the oxidation product for each classification type.

Primary alkanols oxidize to alkanals (and further to alkanoic acids); secondary alkanols oxidize to alkanones; tertiary alkanols resist mild oxidation.

Secondary alkanols have one hydrogen atom on the hydroxyl carbon, which allows dehydrogenation to form a carbonyl double bond (

C=O

) bounded by two alkyl groups (ketone/alkanone).

3

Match the specified product (alkanone) to the correct compound.

Propan-2-ol oxidizes to propanone (

CH_3COCH_3

), which is an alkanone.

Only the secondary alkanol propan-2-ol yields an alkanone.

Key Concept

Oxidation of Alkanols (Primary, Secondary, and Tertiary Classification)

Estimated Time:1m 0s

Question 8332Question →

A $14.3\text{ g}$ sample of hydrated sodium trioxocarbonate(IV), $\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}$ , was heated strongly in a crucible to constant mass. After complete heating, the mass of the remaining anhydrous residue was found to be $5.3\text{ g}$ . Given the relative atomic masses $(\text{Na} = 23, \text{C} = 12, \text{O} = 16, \text{H} = 1)$ , what is the value of $x$ in the formula of the hydrated salt?

10

7

5

1

Show answer & explanation

Answer: $10$

Answer

The value of

x

is

10

, giving the formula

\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}

.

Heating to constant mass completely removes the water of crystallization (

9.0\text{ g}

of

\text{H}_2\text{O}

, equal to

0.50\text{ mol}

). The remaining

5.3\text{ g}

of anhydrous

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

equals

0.05\text{ mol}

. Dividing

0.50\text{ mol}

by

0.05\text{ mol}

yields a mole ratio of

10

, which means

x = 10

.

Step-by-Step Solution

1

Calculate the mass of water lost during heating

\text{Mass of H}_2\text{O} = 14.3\text{ g} - 5.3\text{ g} = 9.0\text{ g}

The loss in mass upon heating to constant mass corresponds directly to the driven-off water of crystallization.

2

Determine the molar masses of anhydrous

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

and

\text{H}_2\text{O}

\text{Molar mass of Na}_2\text{CO}_3 = (2 \times 23) + 12 + (3 \times 16) = 106\text{ g/mol}

;

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

Molar masses are required to convert the given masses into mole quantities.

3

Calculate the number of moles of anhydrous salt and water

\text{Moles of Na}_2\text{CO}_3 = \frac{5.3\text{ g}}{106\text{ g/mol}} = 0.05\text{ mol}

;

\text{Moles of H}_2\text{O} = \frac{9.0\text{ g}}{18\text{ g/mol}} = 0.50\text{ mol}

Moles equal mass divided by molar mass.

4

Calculate the mole ratio to find

x

x = \frac{\text{Moles of H}_2\text{O}}{\text{Moles of Na}_2\text{CO}_3} = \frac{0.50}{0.05} = 10

The coefficient

x

represents the stoichiometric ratio of moles of water of crystallization per mole of anhydrous salt.

Key Concept

Determination of Water of Crystallization in Hydrated Salts

Question 8333Question →

Under identical conditions of temperature and pressure, a $100\text{ cm}^3$ sample of an unknown gas $Q$ diffuses through a porous plug in $80\text{ seconds}$ , whereas an equal volume of oxygen gas ( $O_2$ ) diffuses through the same plug in $40\text{ seconds}$ . What is the relative molecular mass of gas $Q$ ?
( $O = 16.0$ )

128

64

16

8

Show answer & explanation

Answer: 128

Answer

The relative molecular mass of gas

Q

is 128.

According to Graham's Law of diffusion, the time

t

required for a fixed volume of gas to diffuse is directly proportional to the square root of its molar mass

M

. Therefore,

\frac{t_Q}{t_{O2}} = \sqrt{\frac{M_Q}{M_{O2}}}

. Substituting

t_Q = 80\text{ s}

,

t_{O2} = 40\text{ s}

, and

M_{O2} = 32

gives

\frac{80}{40} = \sqrt{\frac{M_Q}{32}}

, so

2 = \sqrt{\frac{M_Q}{32}}

. Squaring both sides gives

4 = \frac{M_Q}{32}

, leading to

M_Q = 128

. Hence, the correct answer is 128.

Step-by-Step Solution

1

Calculate the molar mass of oxygen gas (

O_2

)

M_{O2} = 2 \times 16.0 = 32\text{ g/mol}

Oxygen exists as a diatomic gas, so its molecular mass is twice its atomic mass.

2

Set up Graham's Law relating diffusion time and molar mass for equal volumes

\frac{t_Q}{t_{O2}} = \sqrt{\frac{M_Q}{M_{O2}}}

The rate of diffusion is inversely proportional to the square root of molar mass, meaning diffusion time for a fixed volume is directly proportional to the square root of molar mass.

3

Substitute the given times (

t_Q = 80\text{ s}

,

t_{O2} = 40\text{ s}

) into the equation

\frac{80}{40} = \sqrt{\frac{M_Q}{32}} \implies 2 = \sqrt{\frac{M_Q}{32}}

Simplifying the time ratio yields a factor of 2.

4

Square both sides of the equation to solve for

M_Q

2^2 = \frac{M_Q}{32} \implies 4 = \frac{M_Q}{32} \implies M_Q = 4 \times 32 = 128

Squaring removes the radical, leaving a simple linear equation to calculate the unknown molar mass.

Key Concept

Graham's Law of Diffusion (Time-Molar Mass Relationship)

Question 8334Question →

A liquid pharmaceutical preparation shows a distinct illuminated path when a beam of light passes through it in a dark room (Tyndall effect) and shows no settling of particles upon standing for several weeks. However, when poured through standard filter paper, it leaves no residue. Which of the following correctly classifies this liquid system and explains its observed behavior?

It is a true solution because its solute particles are fully dissolved, homogeneous, and smaller than

1\text{ nm}

.

It is a colloidal system because its dispersed particle diameters (

1\text{ nm}

to

100\text{ nm}

) are large enough to scatter light but small enough to pass through standard filter paper pores.

It is a suspension because its dispersed phase particles are kept suspended solely by thermal convection currents within the solvent.

It is a true solution because its particles undergo rapid sedimentation only when exposed to atmospheric air.

Show answer & explanation

Answer: It is a colloidal system because its dispersed particle diameters ( $1\text{ nm}$ to $100\text{ nm}$ ) are large enough to scatter light but small enough to pass through standard filter paper pores.

Answer

The liquid preparation is a colloidal system because its dispersed particles (diameters between

1\text{ nm}

and

100\text{ nm}

) are sufficiently large to cause the Tyndall effect by scattering light, yet small enough to pass unhindered through the pores of standard filter paper without settling.

The correct response identifies the mixture as a colloidal system. Colloidal particles range in diameter from

1\text{ nm}

to

100\text{ nm}

. This intermediate particle size makes them small enough to pass through the microscopic pores of standard laboratory filter paper while remaining large enough to scatter incident light beams (Tyndall effect) and stay suspended indefinitely via Brownian motion.

Step-by-Step Solution

1

Analyze the light scattering observation.

The observation of a visible light path (Tyndall effect) rules out true solutions, which have particle sizes smaller than

1\text{ nm}

and do not scatter light.

Particles must be comparable in size to the wavelength of light (

1\text{ nm}

to

100\text{ nm}

) to scatter light effectively.

2

Analyze the filtration and stability observations.

Passage through standard filter paper without leaving a residue and absence of settling rules out suspensions.

Suspension particles exceed

100\text{ nm}

(or

0.1\text{ }\mu\text{m}

), causing them to be trapped by filter paper pores and to settle under gravity over time.

3

Synthesize particle size characteristics to classify the system.

The mixture exhibits properties unique to a colloidal dispersion.

Colloids uniquely combine stability against gravity, ability to pass through ordinary filter paper, and distinct scattering of light.

Key Concept

Physical distinction between true solutions, colloids, and suspensions based on particle size, filtration capability, and light scattering (Tyndall effect).

Question 8335Question →

A solid mixture containing insoluble calcium trioxocarbonate(IV) ( $\text{CaCO}_3$ ) and soluble hydrated magnesium tetraoxosulfate(VI) ( $\text{MgSO}_4 \cdot 7\text{H}_2\text{O}$ ) is thoroughly stirred in distilled water and filtered. To obtain pure hydrated magnesium tetraoxosulfate(VI) crystals from the filtrate without decomposing the salt or losing its water of crystallization, which procedure must be performed?

Evaporating the filtrate completely to dryness over a Bunsen flame

Concentrating the filtrate by gentle heating to saturation, followed by slow cooling

Fractionally distilling the filtrate to collect the dissolved salt as the first distillate

Filtering the filtrate a second time through a finer filter paper at room temperature

Show answer & explanation

Answer: Concentrating the filtrate by gentle heating to saturation, followed by slow cooling

Answer

Concentrating the filtrate by gentle heating to saturation, followed by slow cooling to allow crystal formation.

Concentrating the filtrate until it becomes saturated and then allowing it to cool slowly yields pure crystals of hydrated magnesium tetraoxosulfate(VI) (

\text{MgSO}_4 \cdot 7\text{H}_2\text{O}

) while preserving its water of crystallization.

Step-by-Step Solution

1

Analyze the components of the filtrate

The filtrate contains dissolved hydrated magnesium tetraoxosulfate(VI) (

\text{MgSO}_4 \cdot 7\text{H}_2\text{O}

) in water after insoluble

\text{CaCO}_3

is removed by filtration.

Filtration separates the insoluble solid residue (

\text{CaCO}_3

) from the soluble salt solution.

2

Determine the appropriate recovery method for hydrated salts

Crystallization is required instead of evaporation to dryness.

Evaporating to dryness thermally decomposes hydrated salts or drives off essential water of crystallization, destroying the crystalline structure.

3

Apply the crystallization procedure

Partially evaporate the solvent to prepare a hot saturated solution, then allow it to cool slowly to precipitate pure hydrated crystals.

Solubility of the salt decreases as the saturated solution cools, prompting pure crystals to separate out.

Key Concept

Crystallization vs Evaporation to Dryness for Hydrated Salts

Estimated Time:1m 0s

Question 8336Question →

A fixed mass of oxygen gas contained in a rigid metal vessel exerts a pressure of $1.20\text{ atm}$ at a temperature of $27^\circ\text{C}$ . If the volume of the vessel remains constant, at what absolute temperature in Kelvin (K) will the gas exert a pressure of $1.80\text{ atm}$ ?

Show answer & explanation

Answer: 450

Answer

The final absolute temperature of the gas is

450\text{ K}

.

According to Gay-Lussac's Pressure Law, for a fixed mass of gas at constant volume,

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

. Converting the initial temperature

27^\circ\text{C}

to Kelvin gives

T_1 = 27 + 273 = 300\text{ K}

. Substituting

P_1 = 1.20\text{ atm}

,

P_2 = 1.80\text{ atm}

, and

T_1 = 300\text{ K}

gives

T_2 = \frac{1.80 \times 300}{1.20} = 450\text{ K}

.

Step-by-Step Solution

1

Convert the given initial temperature from Celsius to Kelvin.

T_1 = 27 + 273 = 300\text{ K}

Gas laws require absolute temperature in Kelvin for proportional relationship calculations.

2

State the Pressure Law (Gay-Lussac's Law) equation for a fixed volume of gas.

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

The pressure of a fixed mass of gas is directly proportional to its absolute temperature at constant volume.

3

Substitute the given values into the Pressure Law equation.

\frac{1.20\text{ atm}}{300\text{ K}} = \frac{1.80\text{ atm}}{T_2}

Insert

P_1 = 1.20\text{ atm}

,

T_1 = 300\text{ K}

, and

P_2 = 1.80\text{ atm}

.

4

Rearrange and solve for

T_2

.

T_2 = \frac{1.80 \times 300}{1.20} = 450\text{ K}

Cross-multiplying yields the final absolute temperature.

Key Concept

Pressure Law (Gay-Lussac's Law)

Question 8337Question →

An industrial electroplating plant produces wastewater contaminated with dilute tetraoxosulfate(VI) acid ( $H_2SO_4$ ) and toxic copper(II) ions ( $Cu^{2+}$ ). Prior to releasing this effluent into municipal drainage, which of the following chemical treatments is most suitable to simultaneously neutralize the acidity and precipitate the heavy metal ions as an insoluble sludge?

Treatment with calcium hydroxide,

Ca(OH)_2

, to form insoluble calcium sulfate and copper(II) hydroxide

Addition of concentrated trioxonitrate(V) acid,

HNO_3

, to convert the copper ions into soluble copper(II) trioxonitrate(V)

Boiling the wastewater to remove the permanent hardness caused by dissolved copper(II) tetraoxosulfate(VI)

Chlorination of the wastewater to oxidize synthetic non-biodegradable polymers and dissolved heavy metals into harmless gases

Show answer & explanation

Answer: Treatment with calcium hydroxide, $Ca(OH)_2$ , to form insoluble calcium sulfate and copper(II) hydroxide

Answer

Treatment with calcium hydroxide,

Ca(OH)_2

, to form insoluble calcium sulfate and copper(II) hydroxide

Treatment with calcium hydroxide (slaked lime) provides hydroxide ions (

OH^-

) that neutralize free

H^+

ions from the acidic effluent and react with dissolved copper(II) ions (

Cu^{2+}

) to form insoluble copper(II) hydroxide,

Cu(OH)_2

, which precipitates out of solution as removable sludge.

Step-by-Step Solution

1

Analyze the chemical nature of the industrial effluent contaminants.

The effluent contains free hydrogen ions (

H^+

) from

H_2SO_4

(causing low pH) and soluble toxic

Cu^{2+}

ions.

Effective treatment must address both acidity and heavy metal toxicity.

2

Evaluate the chemical reaction between slaked lime,

Ca(OH)_2

, and the effluent components.

Acid neutralization:

H_2SO_4(aq) + Ca(OH)_2(s) \rightarrow CaSO_4(s/aq) + 2H_2O(l)

. Heavy metal precipitation:

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

.

Calcium hydroxide neutralizes

H^+

ions and precipitates

Cu^{2+}

as insoluble

Cu(OH)_2

sludge.

3

Confirm the physical removal mechanism of the precipitate.

The solid precipitate settles out as sludge during sedimentation and can be separated by filtration.

Precipitation converts dissolved pollutants into solids that can be easily removed prior to discharge.

Key Concept

Industrial Effluent Remediation and Heavy Metal Precipitation

Question 8338Question →

During the laboratory preparation of oxygen gas, manganese(IV) oxide ( $MnO_{2(s)}$ ) is added to hydrogen peroxide solution ( $H_2O_{2(aq)}$ ) to increase the speed of decomposition. Which of the following best explains how manganese(IV) oxide increases the rate of this reaction?

It provides an alternative reaction pathway with a lower activation energy.

It shifts the equilibrium position to yield a higher volume of oxygen gas.

It increases the average kinetic energy of the hydrogen peroxide molecules.

It increases the total concentration of the reacting hydrogen peroxide molecules.

Show answer & explanation

Answer: It provides an alternative reaction pathway with a lower activation energy.

Answer

Manganese(IV) oxide acts as a catalyst, increasing the rate of reaction by providing an alternative reaction pathway with a lower activation energy.

Adding a catalyst such as manganese(IV) oxide provides an alternative reaction pathway with a lower activation energy (

E_a

). As a result, a larger fraction of colliding reactant molecules possess energy equal to or greater than the activation energy threshold, leading to an increased frequency of effective collisions.

Step-by-Step Solution

1

Identify the role of manganese(IV) oxide in the decomposition of hydrogen peroxide.

Manganese(IV) oxide is a catalyst because it speeds up the reaction without being consumed.

Recognizing the function of the added chemical substance is the essential first step.

2

Apply Collision Theory to explain catalyst activity.

Catalysts provide a lower activation energy route.

Lowering activation energy allows a greater proportion of molecular collisions to have sufficient energy to react per unit time.

Key Concept

Effect of catalysts on reaction rate and activation energy

Question 8339Question →

A rigid steel cylinder contains a gas at a pressure of $1.20\text{ atm}$ and a temperature of $27^\circ\text{C}$ . If the cylinder is heated to $127^\circ\text{C}$ while keeping the volume constant, what is the final pressure of the gas?

1.60\text{ atm}

5.64\text{ atm}

0.90\text{ atm}

2.20\text{ atm}

Show answer & explanation

Answer: $1.60\text{ atm}$

Answer

The final pressure of the gas inside the cylinder is

1.60\text{ atm}

.

According to the Pressure Law, for a fixed mass of gas at constant volume, pressure is directly proportional to absolute temperature (

P_1/T_1 = P_2/T_2

). Converting temperatures to Kelvin gives

T_1 = 300\text{ K}

and

T_2 = 400\text{ K}

. Solving for

P_2

yields

1.20 \times (400/300) = 1.60\text{ atm}

.

Step-by-Step Solution

1

Convert initial and final temperatures from degrees Celsius to Kelvin.

T_1 = 27 + 273 = 300\text{ K}

and

T_2 = 127 + 273 = 400\text{ K}

.

Gas law equations require absolute temperatures in Kelvin.

2

Apply the Pressure Law (Gay-Lussac's Law) for constant volume.

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

.

Pressure is directly proportional to absolute temperature when volume is constant.

3

Substitute the known values into the equation and calculate the final pressure.

P_2 = 1.20\text{ atm} \times \frac{400\text{ K}}{300\text{ K}} = 1.60\text{ atm}

.

Simplifying

\frac{400}{300}

to

\frac{4}{3}

gives

1.20 \times \frac{4}{3} = 1.60\text{ atm}

.

Key Concept

Pressure-Temperature relationship (Pressure Law) and conversion to absolute temperature scale.

Estimated Time:1m 0s

Question 8340Question →

A chemist needs to separate a dry solid mixture containing iron turnings, iodine crystals, and sodium chloride into its pure individual components. Arrange the following procedural steps in the correct order to achieve this separation.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of steps is: First, pass a magnet over the dry mixture to isolate iron turnings; second, heat the remaining dry mixture under a cooled inverted funnel to sublime iodine crystals; third, dissolve the residual solid in distilled water and filter; fourth, evaporate the clear filtrate to dryness to obtain pure sodium chloride.

The correct order respects the physical states and properties of each component. Performing magnetization first isolates dry iron turnings cleanly. Controlled heating sublimates volatile iodine next. Dissolving the residue in water followed by evaporation isolates pure sodium chloride last.

Step-by-Step Solution

1

Remove ferromagnetic material using magnetization.

Iron turnings are completely isolated from the dry mixture.

Magnetization exploits the magnetic susceptibility of iron while the sample is dry.

2

Sublime the volatile solid component by controlled heating.

Iodine vaporizes and deposits as pure crystals on the cool funnel surface, leaving solid sodium chloride.

Iodine readily sublimates upon gentle heating, whereas sodium chloride has a very high melting point.

3

Dissolve the non-volatile residue in water and filter.

Sodium chloride dissolves completely into the aqueous filtrate.

Sodium chloride is highly soluble in water.

4

Evaporate water from the clear filtrate.

Dry crystals of pure sodium chloride are recovered.

Water vaporizes off, leaving behind the non-volatile dissolved salt.

Key Concept

Sequential separation of solid mixtures by exploiting distinct physical properties (magnetism, volatility/sublimation, and aqueous solubility).

Estimated Time:1m 30s

Answer

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