All practice questions

Question 8281Question →

A compound has an empirical formula of $CH_2$ and a relative molecular mass of $42$ . What is the molecular formula of the compound? $[C = 12, H = 1]$

C_3H_6

C_2H_4

C_4H_8

CH_2

Show answer & explanation

Answer: $C_3H_6$

Answer

The molecular formula of the compound is

C_3H_6

.

The empirical formula mass of

CH_2

is

14\text{ g/mol}

. Dividing the given relative molecular mass (

42

) by

14

gives an integer factor

n = 3

. Multiplying the empirical formula indices by

3

gives

C_3H_6

.

Step-by-Step Solution

1

Calculate the empirical formula mass of

CH_2

.

Empirical formula mass

= 12 + 2(1) = 14\text{ g/mol}

.

The empirical formula mass is required to find the integer ratio multiplier

n

.

2

Determine the integer multiplier

n

by dividing the relative molecular mass by the empirical formula mass.

n = \frac{42}{14} = 3

.

The integer multiplier indicates how many empirical units constitute the molecular formula.

3

Multiply the subscripts of the empirical formula

CH_2

by

n = 3

.

Molecular formula

= (CH_2)_3 = C_3H_6

.

Applying the integer multiplier yields the true formula of the compound.

Key Concept

Molecular Formula Derivation from Empirical Formula and Molar Mass

Question 8282Question →

Match each redox testing reagent or indicator on the left with its characteristic diagnostic observation on the right when reacting with an oxidizing or reducing agent.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Acidified potassium tetraoxomanganate(VII) turns from purple to colorless with reducing agents; starch-iodide paper turns blue-black with oxidizing agents; acidified potassium heptaoxodichromate(VI) turns from orange to green with reducing agents; and freshly prepared iron(II) sulfate changes from pale green to reddish-brown with oxidizing agents.

Each testing reagent displays a distinct diagnostic color change depending on whether it reacts with an oxidizing or reducing agent. Acidified

\text{KMnO}_4

changes from purple to colorless in the presence of a reducing agent due to reduction of

\text{MnO}_4^-

to

\text{Mn}^{2+}

. Starch-iodide paper turns blue-black in the presence of an oxidizing agent as iodide is oxidized to iodine. Acidified

\text{K}_2\text{Cr}_2\text{O}_7

changes from orange to green in the presence of a reducing agent as

\text{Cr}_2\text{O}_7^{2-}

is reduced to

\text{Cr}^{3+}

. Freshly prepared

\text{FeSO}_4

changes from pale green to reddish-brown when an oxidizing agent oxidizes

\text{Fe}^{2+}

to

\text{Fe}^{3+}

.

Step-by-Step Solution

1

Identify the role of acidified

\text{KMnO}_4

solution in redox testing.

Acidified

\text{KMnO}_4

contains manganese in the

+7

oxidation state (purple). When it oxidizes a reducing agent, manganese is reduced to

\text{Mn}^{2+}

(colorless).

This is the classic quantitative and qualitative test for reducing agents.

2

Determine the response of moist starch-iodide paper to oxidizing gases.

Oxidizing agents liberate free iodine (

I_2

) from iodide ions (

I^-

). Free iodine reacts with starch to yield a distinctive blue-black color.

This tests specifically for oxidizing agents such as chlorine or ozone.

3

Identify the color change associated with acidified

\text{K}_2\text{Cr}_2\text{O}_7

.

Dichromate ions (orange, chromium oxidation state

+6

) are reduced to

\text{Cr}^{3+}

ions (green, oxidation state

+3

) by reducing agents.

The reduction of dichromate(VI) to chromium(III) causes the orange-to-green color change.

4

Determine the oxidation behavior of iron(II) sulfate solution.

Iron(II) ions (pale green) act as a reducing agent and are oxidized to iron(III) ions (reddish-brown/yellow) by oxidizing agents.

The oxidation of

\text{Fe}^{2+}

to

\text{Fe}^{3+}

shifts the solution color from green to brown/yellow.

Key Concept

Laboratory Diagnostic Tests for Oxidizing and Reducing Agents

Question 8283Question →

Match each chemical mixture recovery requirement on the left with the most appropriate physical separation technique on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Insoluble chalk suspended in water is separated by filtration. Hydrated copper(II) tetraoxosulfate(VI) crystals require evaporation to saturation followed by cooling to avoid thermal decomposition. Dry sodium chloride is recovered by evaporation to complete dryness. Potassium trioxonitrate(V) is purified from sodium chloride impurity using fractional crystallization.

Each technique is uniquely suited to the physical properties of the mixture: filtration removes insoluble suspensions; evaporation to saturation followed by cooling yields hydrated crystals without thermal decomposition; direct evaporation to dryness efficiently recovers anhydrous thermally stable salts; fractional crystallization separates multiple soluble solutes using solubility-temperature gradients.

Step-by-Step Solution

1

Analyze the solubility and thermal stability of each solute/mixture component.

Chalk is insoluble; hydrated copper sulfate is heat-sensitive; sodium chloride is heat-stable; potassium nitrate and sodium chloride mixture contains two soluble salts with different solubility-temperature slopes.

Physical properties determine which technique yields pure product without thermal breakdown.

2

Match solid-liquid heterogeneous suspensions to filtration.

Chalk powder suspended in water matches filtration.

Filtration separates undissolved solid particles from liquid filtrates.

3

Distinguish between thermal evaporation to dryness and crystallization for soluble salts.

Thermally stable salt (NaCl) matches evaporation to complete dryness, while hydrated salt (CuSO4·5H2O) matches evaporation to saturation followed by cooling.

Evaporating hydrated salts to dryness destroys water of crystallization, yielding anhydrous powder rather than crystals.

4

Match multi-solute solution separation based on temperature-dependent solubility to fractional crystallization.

Separating KNO3 from NaCl impurity matches fractional crystallization.

Cooling a hot saturated solution of both salts causes KNO3 (steep solubility rise with temperature) to crystallize out first.

Key Concept

Selection of separation techniques (filtration, evaporation, crystallization, fractional crystallization) based on solubility and thermal stability.

Question 8284Question →

When a mechanical stress is applied to a solid metal, the material deforms without shattering. Which of the following structural features of metallic bonding is directly responsible for this malleability?

The ability of layers of metal cations to slide past each other without disrupting the electrostatic attraction to the delocalized electron sea

The rapid breaking and re-formation of rigid directional covalent bonds between adjacent metal atoms

The rigid positioning of delocalized electrons that locks positive metal cations into fixed lattice positions

The expansion of localized electron orbitals which absorbs mechanical force without shifting metal cations

Show answer & explanation

Answer: The ability of layers of metal cations to slide past each other without disrupting the electrostatic attraction to the delocalized electron sea

Answer

The ability of layers of metal cations to slide past each other without disrupting the electrostatic attraction to the delocalized electron sea

In metallic lattices, delocalized valence electrons move freely throughout the array of positive metal cations. When a mechanical force is applied, layers of cations slide past one another. The mobile electron sea adapts immediately to the shifted cations, maintaining the non-directional electrostatic attraction throughout the lattice so that the metal deforms (malleability) instead of fracturing.

Step-by-Step Solution

1

Identify the atomic-scale structure of a metallic lattice.

Solid metals consist of a giant lattice of positive metal cations immersed in a fluid sea of delocalized valence electrons.

Understanding the non-directional nature of metallic bonds is necessary to explain physical properties.

2

Analyze how applied mechanical force alters the lattice structure.

Under mechanical stress, planes of positive cations slide past one another.

Applied mechanical forces induce shear stress across crystal lattice planes.

3

Determine why the metallic structure deforms rather than breaking.

Because the delocalized electrons are mobile and non-directional, they adjust immediately to the shifted cation layers, maintaining attractive electrostatic forces throughout the lattice and preventing repulsive cleavage.

Non-directional electrostatic attraction preserves structural cohesion during deformation.

Key Concept

Metallic Bonding and Malleability

Question 8285Question →

Ammonium chloride ( $NH_4Cl$ ) is synthesized by reacting ammonia gas with hydrogen chloride gas. Which types of chemical bonding exist within solid ammonium chloride?

Covalent, coordinate (dative), and ionic bonds

Covalent, ionic, and metallic bonds

Ionic, coordinate, and hydrogen bonds

Covalent, metallic, and coordinate bonds

Show answer & explanation

Answer: Covalent, coordinate (dative), and ionic bonds

Answer

Covalent, coordinate (dative), and ionic bonds

Ammonium chloride (

NH_4Cl

) exhibits three distinct types of chemical bonds: (1) Covalent bonds between nitrogen and three hydrogen atoms in the original ammonia molecule, (2) A coordinate (dative) bond formed when nitrogen donates its lone pair to a hydrogen ion (

H^+

) to form the ammonium ion (

NH_4^+

), and (3) An ionic (electrovalent) bond between the

NH_4^+

cation and the

Cl^-

anion.

Step-by-Step Solution

1

Analyze the structure of the ammonia molecule (

NH_3

).

Nitrogen shares electron pairs with three hydrogen atoms, forming three polar covalent bonds and retaining one unshared lone pair of electrons.

Covalent bonding occurs when non-metal atoms share pairs of valence electrons.

2

Analyze the formation of the ammonium ion (

NH_4^+

).

Nitrogen donates its lone pair of electrons to an electron-deficient hydrogen ion (

H^+

), forming one coordinate (dative) covalent bond.

A coordinate bond is formed when one atom provides both electrons for the shared pair.

3

Analyze the interaction between the ammonium ion (

NH_4^+

) and the chloride ion (

Cl^-

).

Electrostatic attraction between the positively charged

NH_4^+

cation and negatively charged

Cl^-

anion forms an ionic bond.

Oppositely charged ions attract each other to form a stable crystal lattice.

Key Concept

Coexistence of covalent, dative, and ionic bonding in polyatomic salts

Estimated Time:1m 0s

Question 8286Question →

A student dilutes $100\text{ cm}^3$ of a $0.05\text{ mol dm}^{-3}$ nitric acid ( $\text{HNO}_3$ ) solution with distilled water to achieve a total volume of $500\text{ cm}^3$ at $25^\circ\text{C}$ . What is the pOH of the resulting diluted solution?

2.0

12.0

1.3

12.7

Show answer & explanation

Answer: $12.0$

Answer

The pOH of the resulting diluted solution is

12.0

.

Diluting

100\text{ cm}^3

of

0.05\text{ mol dm}^{-3}\text{ HNO}_3

to

500\text{ cm}^3

decreases the hydrogen ion concentration to

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

. The pH of this solution is

-\log_{10}(0.01) = 2.0

. Since

\text{pH} + \text{pOH} = 14

at

25^\circ\text{C}

, the pOH is

14 - 2.0 = 12.0

.

Step-by-Step Solution

1

Calculate the hydrogen ion concentration of the diluted solution using the dilution formula

C_1 V_1 = C_2 V_2

.

C_2 = \frac{0.05\text{ mol dm}^{-3} \times 100\text{ cm}^3}{500\text{ cm}^3} = 0.01\text{ mol dm}^{-3} = 1.0 \times 10^{-2}\text{ mol dm}^{-3}

.

Nitric acid is a strong monoprotic acid that fully dissociates in water, so

[\text{H}^+] = C_2

.

2

Determine the pH of the diluted solution.

\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.0 \times 10^{-2}) = 2.0

.

The negative logarithm of the hydrogen ion concentration defines the pH.

3

Calculate the pOH using the water ion product relationship at

25^\circ\text{C}

.

\text{pOH} = 14 - \text{pH} = 14 - 2.0 = 12.0

.

The sum of pH and pOH equals

14

at standard room temperature.

Key Concept

Dilution effect on hydrogen ion concentration and the relation

\text{pH} + \text{pOH} = 14

Estimated Time:1m 30s

Question 8287Question →

During the municipal treatment of water for public consumption, alum, $Al_2(SO_4)_3$ , is added to untreated river water. What is the primary chemical function of adding alum in this process?

To cause coagulation and sedimentation of fine suspended impurities

To kill disease-causing pathogenic microorganisms in the water

To eliminate permanent hardness caused by dissolved calcium salts

To neutralize alkaline industrial effluents and adjust pH

Show answer & explanation

Answer: To cause coagulation and sedimentation of fine suspended impurities

Answer

The primary function of adding alum in municipal water treatment is to cause coagulation and sedimentation of fine suspended impurities.

Alum (

Al_2(SO_4)_3

) serves as a coagulating agent during water treatment. It destabilizes colloidal particles suspended in water, allowing them to join together into heavy flocs that settle out easily during sedimentation.

Step-by-Step Solution

1

Identify the role of alum (

Al_2(SO_4)_3

) in water purification

Alum provides trivalent aluminum ions (

Al^{3+}

) which neutralize the negative surface charges on microscopic suspended particles.

Neutralizing these surface charges causes the tiny particles to clump together into larger, visible aggregates called flocs.

2

Distinguish coagulation from other treatment stages

The heavy flocs created by coagulation settle to the bottom of sedimentation tanks by gravity.

Disinfection (destroying pathogens) is performed later using chlorine gas or sodium hypochlorite, while softening removes dissolved calcium and magnesium ions.

Key Concept

Coagulation and Flocculation in Municipal Water Treatment

Question 8288Question →

Match each structural feature of the electron sea model on the left with the macroscopic metal property it directly accounts for on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

1. Movement of delocalized electrons toward a positive terminal matches High electrical conductivity. 2. Layers of cations sliding past each other matches Malleability and ductility. 3. Absorption and re-emission of light by free electrons matches Lustrous (shiny) appearance. 4. Strong non-directional electrostatic attraction matches High melting and boiling points.

Each structural feature in the electron sea model directly dictates a specific macroscopic behavior: delocalized electron motion provides electrical conductivity, cation layer flexibility allows deformation (malleability/ductility), light oscillation by surface electrons causes shiny luster, and extensive electrostatic forces produce high thermal melting thresholds.

Step-by-Step Solution

1

Relate electric charge transport to metallic conduction

Free electrons moving toward a positive potential corresponds to high electrical conductivity.

Electric current in solid metals consists of a net flow of delocalized valence electrons.

2

Analyze deformation behavior of metallic lattices under pressure

Sliding layers of cations buffered by the electron sea corresponds to malleability and ductility.

Metals deform without shattering because metallic bonding is non-directional.

3

Connect light interaction with free electron oscillations

Free surface electrons absorbing and re-emitting light photons corresponds to luster.

Unbound electrons respond dynamically to electromagnetic waves, reflecting light.

4

Examine thermal stability of the lattice bonding

Strong omnidirectional electrostatic attraction corresponds to high melting and boiling points.

Separating metallic particles requires inputting significant thermal energy to overcome electrostatic bonds.

Key Concept

Electron Sea Model of Metallic Bonding

Question 8289Question →

A solid mixture consists of insoluble silicon(IV) oxide ( $\text{SiO}_2$ ) and soluble potassium tetraoxosulfate(VI) ( $\text{K}_2\text{SO}_4$ ). After adding distilled water to the mixture and stirring thoroughly, which procedure should be used to separate the insoluble solid and obtain pure crystals of potassium tetraoxosulfate(VI)?

Filter the mixture to remove

\text{SiO}_2

, then heat the filtrate to concentrate it before allowing it to cool and crystallize.

Evaporate the entire mixture to complete dryness and collect the combined residue.

Perform fractional distillation on the suspension to collect

\text{K}_2\text{SO}_4

as the first distillate fraction.

Filter the mixture to collect

\text{K}_2\text{SO}_4

on the filter paper while

\text{SiO}_2

passes into the filtrate.

Show answer & explanation

Answer: Filter the mixture to remove $\text{SiO}_2$ , then heat the filtrate to concentrate it before allowing it to cool and crystallize.

Answer

Filtration of the mixture to remove the insoluble silicon(IV) oxide (

\text{SiO}_2

), followed by partial evaporation of the filtrate to concentrate it and subsequent cooling to allow potassium tetraoxosulfate(VI) (

\text{K}_2\text{SO}_4

) crystals to form.

Filtration successfully separates the insoluble silicon(IV) oxide (

\text{SiO}_2

) as the residue on the filter paper from the soluble potassium tetraoxosulfate(VI) (

\text{K}_2\text{SO}_4

) in the filtrate. Heating the filtrate to concentrate it and then letting it cool slowly yields pure crystals of the salt.

Step-by-Step Solution

1

Dissolve the mixture in water and filter the suspension.

Insoluble silicon(IV) oxide (

\text{SiO}_2

) is retained on the filter paper as the residue, while dissolved potassium tetraoxosulfate(VI) (

\text{K}_2\text{SO}_4

) passes through as the clear filtrate.

Filtration separates insoluble solids from liquids based on particle size and solubility.

2

Evaporate the filtrate partially until it reaches its crystallization point (saturated solution).

Excess solvent is driven off without decomposing the salt or trapping impurities.

Controlled evaporation concentrates the solution to saturation without heating to complete dryness.

3

Allow the hot saturated solution to cool slowly at room temperature.

Pure crystals of

\text{K}_2\text{SO}_4

form and deposit out of solution.

Solubility decreases with decreasing temperature, causing the solute to crystallize.

Key Concept

Separation of insoluble and soluble solid mixtures using filtration and crystallization

Estimated Time:1m 0s

Question 8290Question →

Three unlabelled liquid mixtures, $P$ , $Q$ , and $R$ , are tested in a laboratory. Mixture $P$ passes completely through both filter paper and semi-permeable membranes without scattering light. Mixture $Q$ passes through filter paper but is retained by semi-permeable membranes and scatters a narrow beam of light. Mixture $R$ leaves a solid residue on filter paper upon gravity filtration. Which mixture will undergo precipitation upon the addition of an electrolyte, and what physical property accounts for this behavior?

Mixture

P

, because its dissolved solute particles adsorb electrolyte ions to form insoluble precipitates.

Mixture

Q

, because its dispersed particles carry surface charges and have diameters between

1\text{ nm}

and

100\text{ nm}

.

Mixture

R

, because Brownian motion maintains its suspension until electrolyte ions increase its density.

Mixture

Q

, because its solute molecules are smaller than

1\text{ nm}

and easily aggregate into coarse solids.

Show answer & explanation

Answer: Mixture $Q$ , because its dispersed particles carry surface charges and have diameters between $1\text{ nm}$ and $100\text{ nm}$ .

Answer

Mixture

Q

, because its dispersed particles carry surface charges and have diameters between

1\text{ nm}

and

100\text{ nm}

.

The experimental observations identify Mixture

Q

as a colloid because it passes through ordinary filter paper, is retained by a semi-permeable membrane, and displays light scattering (Tyndall effect). Colloidal particles range in size between

1\text{ nm}

and

100\text{ nm}

and carry surface electrical charges. Addition of an electrolyte neutralizes these charges, leading to coagulation.

Step-by-Step Solution

1

Analyze the filtration and optical properties of each mixture to classify them.

Mixture

P

is a true solution (particle size

< 1\text{ nm}

). Mixture

Q

is a colloidal system (particle size

1\text{--}100\text{ nm}

, exhibits Tyndall scattering, retained by semi-permeable membrane). Mixture

R

is a suspension (particle size

> 100\text{ nm}

, retained by filter paper).

Differentiation among true solutions, colloids, and suspensions is based on particle size and membrane permeability.

2

Determine which system undergoes coagulation upon adding an electrolyte.

Colloidal particles (Mixture

Q

) carry electric charges that stabilize them. Adding an electrolyte supplies ions of opposite charge, neutralizing the colloidal particles and causing them to coagulate/precipitate.

True solutions do not coagulate on electrolyte addition, and suspensions settle naturally by gravity.

Key Concept

Classification and properties of true solutions, colloidal systems, and suspensions

Question 8291Question →

Consider the half-reaction representing the oxidation of thiosulfate ions to tetrathionate ions:

2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow \text{S}_4\text{O}_6^{2-}(\text{aq}) + n e^-

What is the number of electrons,

n

, required to balance the charge in this half-reaction?

Show answer & explanation

Answer: 2

Answer

The number of electrons required to balance the charge in the half-reaction is 2.

To balance a half-reaction, both atom counts and net electric charges must be equal on both sides of the equation. The reactant side contains 2 thiosulfate ions (

2\text{S}_2\text{O}_3^{2-}

), giving a net charge of

2 \times (-2) = -4

. The product side contains 1 tetrathionate ion (

\text{S}_4\text{O}_6^{2-}

), giving a net charge of

-2

. Adding 2 electrons (

2 e^-

) to the product side lowers its total charge to

-4

, equalizing the charge on both sides.

Step-by-Step Solution

1

Calculate the total charge of the reactant species.

Reactant charge = 2 * (-2) = -4.

There are 2 thiosulfate ions, each with an ionic charge of -2.

2

Calculate the net charge of the ionic product species.

Product charge (excluding electrons) = -2.

There is 1 tetrathionate ion with an ionic charge of -2.

3

Equate the overall charges on both sides to solve for the number of electrons n.

-4 = -2 - n, giving n = 2.

Adding 2 electrons (each carrying a -1 charge) to the product side brings the total product charge to -4, matching the reactant side.

Key Concept

Balancing electric charge in oxidation half-reactions

Question 8292Question →

A sealed rigid glass flask contains carbon dioxide gas at a temperature of $7^\circ\text{C}$ and a pressure of $1.4\text{ atm}$ . If the flask is heated to a final temperature of $77^\circ\text{C}$ at constant volume, what is the new pressure exerted by the gas?

1.75\text{ atm}

15.4\text{ atm}

1.12\text{ atm}

2.10\text{ atm}

Show answer & explanation

Answer: $1.75\text{ atm}$

Answer

The new pressure exerted by the gas is

1.75\text{ atm}

.

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

P_1/T_1 = P_2/T_2

). Converting the temperatures to Kelvin (

T_1 = 7 + 273 = 280\text{ K}

and

T_2 = 77 + 273 = 350\text{ K}

) and solving for

P_2

gives

P_2 = 1.4 \times \frac{350}{280} = 1.75\text{ atm}

.

Step-by-Step Solution

1

Convert initial and final temperatures from Celsius to Kelvin.

T_1 = 7 + 273 = 280\text{ K}

and

T_2 = 77 + 273 = 350\text{ K}

.

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

2

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

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

.

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

3

Substitute the values to calculate the final pressure

P_2

.

P_2 = 1.4 \times \frac{350}{280} = 1.4 \times 1.25 = 1.75\text{ atm}

.

Simplifying the fraction

\frac{350}{280} = \frac{5}{4} = 1.25

gives the final answer.

Key Concept

Pressure Law (Gay-Lussac's Law)

Question 8293Question →

Under identical conditions of temperature and pressure, a neutral oxide of nitrogen diffuses $1.21$ times faster than carbon(IV) oxide ( $CO_2$ ). Which of the following is a characteristic chemical property of this oxide of nitrogen? ( $\text{Relative atomic masses: } C = 12, N = 14, O = 16$ )

It reacts rapidly with atmospheric oxygen at room temperature to form reddish-brown fumes.

It relights a glowing wooden splint almost as vigorously as oxygen gas.

It dissolves readily in water at STP to produce an acidic solution of trioxonitrate(V) acid.

It readily condenses near room temperature into a pale yellow liquid undergoing reversible dimerization.

Show answer & explanation

Answer: It reacts rapidly with atmospheric oxygen at room temperature to form reddish-brown fumes.

Answer

The gas reacts rapidly with atmospheric oxygen at room temperature to form reddish-brown fumes.

Using Graham's law of diffusion, the molar mass of the unknown gas is calculated as

\frac{44}{1.21^2} \approx 30.0\text{ g mol}^{-1}

, which corresponds uniquely to nitrogen(II) oxide (

NO

). A defining chemical test for

NO

is its immediate oxidation by air to give reddish-brown fumes of nitrogen(IV) oxide (

NO_2

).

Step-by-Step Solution

1

Calculate the molar mass of carbon(IV) oxide (

CO_2

).

M(CO_2) = 12 + 2(16) = 44\text{ g mol}^{-1}

.

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

2

Apply Graham's law of diffusion to determine the molar mass of the unknown nitrogen oxide (

M_x

).

\frac{R_x}{R_{CO_2}} = \sqrt{\frac{M(CO_2)}{M_x}} \implies 1.21 = \sqrt{\frac{44}{M_x}} \implies (1.21)^2 = \frac{44}{M_x} \implies M_x = \frac{44}{1.4641} \approx 30.0\text{ g mol}^{-1}

.

The rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

3

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

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

.

For nitrogen(II) oxide (

NO

),

M(NO) = 14 + 16 = 30\text{ g mol}^{-1}

.

Matching the calculated molar mass to known oxides of nitrogen (

N_2O = 44

,

NO = 30

,

NO_2 = 46

).

4

Determine the key chemical property of nitrogen(II) oxide (

NO

).

Nitrogen(II) oxide (

NO

) is a colorless neutral gas that reacts spontaneously with oxygen in air to form brown nitrogen(IV) oxide (

2NO + O_2 \rightarrow 2NO_2

).

To select the correct property corresponding to

NO

.

Key Concept

Graham's Law of Diffusion and Chemical Properties of Oxides of Nitrogen

Estimated Time:2m 0s

Question 8294Question →

An industrial gas cylinder is equipped with a pressure safety relief valve calibrated to open when the internal gas pressure reaches $3.60\text{ atm}$ . At an ambient temperature of $27^\circ\text{C}$ , the enclosed gas exerts a pressure of $2.40\text{ atm}$ . Assuming the volume of the container remains strictly constant, what is the maximum temperature in degrees Celsius ( $^\circ\text{C}$ ) to which the cylinder can be heated before the safety valve opens?

40.5^\circ\text{C}

177^\circ\text{C}

200^\circ\text{C}

450^\circ\text{C}

Show answer & explanation

Answer: $177^\circ\text{C}$

Answer

The maximum temperature before the safety valve opens is

177^\circ\text{C}

.

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

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

). Converting

27^\circ\text{C}

to absolute temperature yields

T_1 = 300\text{ K}

. Substituting

P_1 = 2.40\text{ atm}

,

P_2 = 3.60\text{ atm}

, and

T_1 = 300\text{ K}

gives

T_2 = 450\text{ K}

. Converting

450\text{ K}

back to Celsius gives

450 - 273 = 177^\circ\text{C}

.

Step-by-Step Solution

1

Convert the initial temperature from Celsius to Kelvin

T_1 = 27^\circ\text{C} + 273 = 300\text{ K}

Gas laws require thermodynamic temperature expressed in absolute units (Kelvin).

2

Apply the Pressure Law (Gay-Lussac's Law) formula at constant volume

\frac{P_1}{T_1} = \frac{P_2}{T_2} \implies \frac{2.40}{300} = \frac{3.60}{T_2}

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

3

Solve for the final absolute temperature

T_2

T_2 = 300 \times \frac{3.60}{2.40} = 450\text{ K}

Cross-multiplying gives the threshold temperature in Kelvin.

4

Convert the absolute temperature back to degrees Celsius

t_2 = 450\text{ K} - 273 = 177^\circ\text{C}

The question specifically requests the maximum temperature in degrees Celsius.

Key Concept

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

Estimated Time:1m 30s

Question 8295Question →

At $25^\circ\text{C}$ , an aqueous solution of a weak monoacidic base has a concentration of $0.08\text{ mol dm}^{-3}$ and a degree of ionization ( $\alpha$ ) of $0.025$ ( $2.5\%$ ). What is the concentration of hydroxide ions, $[\text{OH}^-]$ , in the solution in $\text{mol dm}^{-3}$ ?

Show answer & explanation

Answer: 0.002

Answer

The concentration of hydroxide ions,

[\text{OH}^-]

, in the solution is

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

.

For a weak monoacidic base in aqueous solution, only a fraction (

\alpha

) of the base molecules ionize to form hydroxide ions. The concentration of hydroxide ions is given by

[\text{OH}^-] = C \alpha

. Substituting the concentration

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

and degree of ionization

0.025

yields

[\text{OH}^-] = 0.08 \times 0.025 = 0.002\text{ mol dm}^{-3}

.

Step-by-Step Solution

1

Identify the relationship between degree of ionization and hydroxide ion concentration for a weak monoacidic base

[\text{OH}^-] = C \cdot \alpha

A weak monoacidic base

\text{BOH}

ionizes partially according to

\text{BOH}_{(aq)} \rightleftharpoons \text{B}^+_{(aq)} + \text{OH}^-_{(aq)}

, so the concentration of produced hydroxide ions equals the initial concentration multiplied by the degree of ionization.

2

Substitute the given values into the equation

[\text{OH}^-] = 0.08\text{ mol dm}^{-3} \times 0.025

The initial concentration

C = 0.08\text{ mol dm}^{-3}

and the degree of ionization

\alpha = 2.5\% = 0.025

.

3

Perform the multiplication to find the final concentration

[\text{OH}^-] = 0.002\text{ mol dm}^{-3}

Multiplying

0.08

by

0.025

yields

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

(or

2.0 \times 10^{-3}\text{ mol dm}^{-3}

).

Key Concept

Ionization equilibrium of weak bases and calculation of hydroxide ion concentration

Estimated Time:1m 15s

Question 8296Question →

According to Hückel's rule, a planar cyclic conjugated compound exhibits aromatic stability if the number of delocalized $\pi$ -electrons in its conjugated ring system is equal to which expression, where $n$ is a non-negative integer?

4n + 2

4n

2n + 2

4n - 2

Show answer & explanation

Answer: $4n + 2$

Answer

The expression for the number of delocalized

\pi

-electrons in an aromatic compound is

4n + 2

.

Hückel's rule states that a cyclic, planar, fully conjugated molecule possesses aromatic stability when it contains

(4n + 2)

delocalized

\pi

-electrons, where

n

is an integer (

0, 1, 2, 3, \dots

). For example, benzene has

6

\pi

-electrons, satisfying the rule for

n = 1

.

Step-by-Step Solution

1

Recall the structural criteria for aromaticity.

An aromatic molecule must be cyclic, planar, fully conjugated, and possess a specific number of delocalized

\pi

-electrons.

These structural conditions allow complete ring delocalization of

\pi

-electrons.

2

Apply Hückel's rule to determine the required

\pi

-electron count.

The total number of delocalized

\pi

-electrons must equal

(4n + 2)

, where

n = 0, 1, 2, 3, \dots

This formula predicts closed-shell electronic stability for planar conjugated rings.

Key Concept

Hückel's Rule of Aromaticity

Estimated Time:45s

Question 8297Question →

Calculate the relative molecular mass of hydrated magnesium tetraoxosulfate(VI), $MgSO_4 \cdot 7H_2O$ . $[Mg = 24, S = 32, O = 16, H = 1]$

Show answer & explanation

Answer: 246

Answer

246

The relative molecular mass of

MgSO_4 \cdot 7H_2O

is calculated by summing the atomic masses of all constituent atoms. The anhydrous part

MgSO_4

contributes

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

. The water of crystallization

7H_2O

contributes

7 \times (2 + 16) = 126

. The total relative molecular mass is

120 + 126 = 246

.

Step-by-Step Solution

1

Calculate the mass of the anhydrous salt component,

MgSO_4

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

Sum the relative atomic masses of one magnesium atom, one sulfur atom, and four oxygen atoms.

2

Calculate the mass of the water of crystallization component,

7H_2O

7 \times (2 + 16) = 126

Multiply the molecular mass of one water molecule (18) by the coefficient 7.

3

Combine the mass of

MgSO_4

and

7H_2O

120 + 126 = 246

The total relative molecular mass of a hydrated salt is the sum of the anhydrous salt mass and the water of crystallization mass.

Key Concept

Relative Molecular Mass of Hydrated Salts

Question 8298Question →

Consider the standard reduction potentials ( $E^\circ$ ) at $25^\circ\text{C}$ for the following half-reactions:

\text{Ce}^{4+}(aq) + e^- \rightarrow \text{Ce}^{3+}(aq) \quad E^\circ = +1.61\text{ V}

\text{Br}_2(l) + 2e^- \rightarrow 2\text{Br}^-(aq) \quad E^\circ = +1.07\text{ V}

\text{Fe}^{3+}(aq) + e^- \rightarrow \text{Fe}^{2+}(aq) \quad E^\circ = +0.77\text{ V}

\text{I}_2(s) + 2e^- \rightarrow 2\text{I}^-(aq) \quad E^\circ = +0.54\text{ V}

\text{Sn}^{4+}(aq) + 2e^- \rightarrow \text{Sn}^{2+}(aq) \quad E^\circ = +0.15\text{ V}

Which of the following chemical species can spontaneously oxidize $\text{I}^-(aq)$ to $\text{I}_2(s)$ under standard conditions, but is UNABLE to oxidize $\text{Br}^-(aq)$ to $\text{Br}_2(l)$ ?

\text{Fe}^{3+}(aq)

\text{Ce}^{4+}(aq)

\text{Sn}^{4+}(aq)

\text{Sn}^{2+}(aq)

Show answer & explanation

Answer: $\text{Fe}^{3+}(aq)$

Answer

The species

\text{Fe}^{3+}(aq)

is the correct choice because its standard reduction potential (+0.77 V) lies strictly between the reduction potentials of

\text{I}_2/\text{I}^-

(+0.54 V) and

\text{Br}_2/\text{Br}^-

(+1.07 V).

To oxidize

\text{I}^-

(

E^\circ_{\text{red}} = +0.54\text{ V}

), an oxidizing agent must have a standard reduction potential greater than

+0.54\text{ V}

. To fail to oxidize

\text{Br}^-

(

E^\circ_{\text{red}} = +1.07\text{ V}

), its reduction potential must be less than

+1.07\text{ V}

. The species

\text{Fe}^{3+}(aq)

has

E^\circ = +0.77\text{ V}

, which satisfies

+0.54\text{ V} < +0.77\text{ V} < +1.07\text{ V}

. Thus,

E^\circ_{\text{cell}}(\text{Fe}^{3+}/\text{I}^-) = +0.77 - 0.54 = +0.23\text{ V} > 0

(spontaneous) and

E^\circ_{\text{cell}}(\text{Fe}^{3+}/\text{Br}^-) = +0.77 - 1.07 = -0.30\text{ V} < 0

(non-spontaneous).

Step-by-Step Solution

1

Determine the required condition for spontaneous oxidation of

\text{I}^-(aq)

to

\text{I}_2(s)

.

The oxidation half-reaction is

2\text{I}^-(aq) \rightarrow \text{I}_2(s) + 2e^-

with

E^\circ_{\text{ox}} = -0.54\text{ V}

. For

E^\circ_{\text{cell}} = E^\circ_{\text{red}}(\text{oxidant}) + E^\circ_{\text{ox}} > 0

, the oxidant must have

E^\circ_{\text{red}} > +0.54\text{ V}

.

A redox reaction is spontaneous if the cell potential

E^\circ_{\text{cell}}

is positive.

2

Determine the required condition for non-spontaneous oxidation of

\text{Br}^-(aq)

to

\text{Br}_2(l)

.

The oxidation half-reaction is

2\text{Br}^-(aq) \rightarrow \text{Br}_2(l) + 2e^-

with

E^\circ_{\text{ox}} = -1.07\text{ V}

. For

E^\circ_{\text{cell}} = E^\circ_{\text{red}}(\text{oxidant}) + E^\circ_{\text{ox}} < 0

, the oxidant must have

E^\circ_{\text{red}} < +1.07\text{ V}

.

An unfeasible (non-spontaneous) reaction corresponds to a negative overall standard cell potential.

3

Combine the potential boundary constraints and evaluate the options.

The standard reduction potential of the ideal oxidant must satisfy

+0.54\text{ V} < E^\circ_{\text{red}} < +1.07\text{ V}

. Among the choices,

\text{Fe}^{3+}(aq)

has

E^\circ_{\text{red}} = +0.77\text{ V}

, which falls squarely within this range.

Comparing standard electrode potentials directly establishes which species can act as selective oxidizing agents.

Key Concept

Predicting reaction spontaneity and selective oxidation using standard electrode potentials (

E^\circ

).

Estimated Time:2m 0s

Question 8299Question →

When excess copper turnings are reacted with dilute trioxonitrate(V) acid ( $\text{HNO}_3$ ), $1.12\text{ dm}^3$ of nitrogen(II) oxide ( $\text{NO}$ ) gas, measured at STP, is evolved. What mass of copper metal was oxidized during this reaction?

$(\text{Molar mass of Cu} = 64\text{ g mol}^{-1},\text{ molar volume of gas at STP} = 22.4\text{ dm}^3\text{ mol}^{-1})$

4.8\text{ g}

3.2\text{ g}

1.6\text{ g}

2.1\text{ g}

Show answer & explanation

Answer: $4.8\text{ g}$

Answer

4.8\text{ g}

of copper metal was oxidized.

The option specifying

4.8\text{ g}

is correct because the balanced chemical reaction of copper turnings with dilute trioxonitrate(V) acid is

3\text{Cu} + 8\text{HNO}_3 \rightarrow 3\text{Cu(NO}_3)_2 + 2\text{NO} + 4\text{H}_2\text{O}

. Converting

1.12\text{ dm}^3

of

\text{NO}

at STP gives

0.05\text{ mol}

. Multiplying by the stoichiometric ratio

\frac{3}{2}

gives

0.075\text{ mol}

of

\text{Cu}

, which equals

0.075 \times 64 = 4.8\text{ g}

.

Step-by-Step Solution

1

Write the balanced chemical equation for the reaction of copper with dilute trioxonitrate(V) acid.

3\text{Cu}(s) + 8\text{HNO}_3(aq) \rightarrow 3\text{Cu(NO}_3)_2(aq) + 2\text{NO}(g) + 4\text{H}_2\text{O}(l)

Dilute

\text{HNO}_3

acts as an oxidizing agent, reducing to nitrogen(II) oxide (

\text{NO}

) gas rather than hydrogen gas.

2

Calculate the moles of nitrogen(II) oxide (

\text{NO}

) gas produced at STP.

\text{Moles of NO} = \frac{1.12\text{ dm}^3}{22.4\text{ dm}^3\text{ mol}^{-1}} = 0.05\text{ mol}

At STP, one mole of any ideal gas occupies

22.4\text{ dm}^3

.

3

Use the mole ratio to determine moles of copper oxidized.

\text{Moles of Cu} = 0.05\text{ mol NO} \times \left(\frac{3\text{ mol Cu}}{2\text{ mol NO}}\right) = 0.075\text{ mol Cu}

The stoichiometric ratio between

\text{Cu}

and

\text{NO}

is 3:2.

4

Convert moles of copper to mass in grams.

\text{Mass of Cu} = 0.075\text{ mol} \times 64\text{ g mol}^{-1} = 4.8\text{ g}

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

Key Concept

Stoichiometry of redox reactions involving transition metals and oxidizing acids

Question 8300Question →

Match each atmospheric chemical species or pollutant listed on the left with its corresponding atmospheric role or environmental impact listed on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Chlorofluorocarbons pair with solar UV photolysis releasing chlorine free radicals; Carbon(IV) oxide pairs with absorbing terrestrial infrared radiation in the lower troposphere; Stratospheric ozone pairs with filtering solar ultraviolet radiation; Methane pairs with being a potent greenhouse gas from anaerobic decomposition.

Each chemical species is accurately matched to its distinct atmospheric layer and chemical property: Chlorofluorocarbons photolyze into free radicals that destroy ozone, Carbon(IV) oxide absorbs Earth's infrared heat, Stratospheric ozone shields the surface from UV radiation, and Methane acts as a greenhouse gas from biological anaerobic processes.

Step-by-Step Solution

1

Identify the primary environmental mechanism associated with Chlorofluorocarbons (CFCs).

CFCs are photolyzed by UV light in the stratosphere, generating atomic chlorine radicals that deplete ozone.

CFC molecules are unreactive in the troposphere but breakdown under high-energy UV light in the stratosphere.

2

Determine the atmospheric function of Carbon(IV) oxide (

\text{CO}_2

).

\text{CO}_2

absorbs re-radiated heat (infrared radiation) emitted by Earth, causing tropospheric warming.

\text{CO}_2

molecules possess vibrational modes that absorb thermal infrared wavelengths.

3

Determine the protective role of Stratospheric Ozone (

\text{O}_3

).

Stratospheric ozone absorbs biological harmful UV-B radiation.

The Chapman mechanism demonstrates how photolysis and reformation of ozone absorb solar UV light.

4

Identify the primary source and impact of Methane (

\text{CH}_4

).

Methane is a strong greenhouse gas emitted from anaerobic habitats like wetlands and livestock digestion.

Methanogenic bacteria produce

\text{CH}_4

under anaerobic conditions.

Key Concept

Distinguishing the chemical roles and environmental impacts of atmospheric pollutants causing global warming versus those causing ozone layer depletion.

Estimated Time:1m 30s

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