All practice questions

Question 8421Question →

Carbon(IV) oxide ( $CO_2$ ) is the major chemical agent responsible for the depletion of the stratospheric ozone layer because it reacts directly with ozone ( $O_3$ ) molecules upon absorbing ultraviolet radiation.

True

False

Show answer & explanation

Answer: False

Answer

False. Carbon(IV) oxide (

CO_2

) is a primary greenhouse gas causing global warming in the troposphere by trapping outgoing thermal infrared radiation, whereas stratospheric ozone depletion is catalyzed by chlorine free radicals released from chlorofluorocarbons (CFCs) under UV radiation.

The statement incorrectly attributes stratospheric ozone depletion to carbon(IV) oxide.

CO_2

causes global warming by absorbing thermal infrared radiation in the lower atmosphere (troposphere), while ozone layer depletion occurs in the stratosphere via catalytic cycles involving halogen radicals (such as

Cl^\bullet

from CFC photolysis).

Step-by-Step Solution

1

Identify the role of Carbon(IV) oxide (

CO_2

) in atmospheric chemistry.

CO_2

is a greenhouse gas present in the troposphere that absorbs re-radiated infrared (heat) energy from the Earth's surface.

Understanding the distinct mechanism of the greenhouse effect prevents confusing heat absorption with ozone layer breakdown.

2

Identify the chemical species responsible for stratospheric ozone layer depletion.

Chlorofluorocarbons (CFCs) undergo photolysis by solar UV light in the stratosphere to yield chlorine free radicals (

Cl^\bullet

), which catalytically destroy

O_3

molecules.

Pinpointing the specific catalyst for ozone destruction proves that

CO_2

is not the agent destroying stratospheric

O_3

.

Key Concept

Distinction between the Greenhouse Effect/Global Warming (

CO_2

,

CH_4

, IR absorption in troposphere) and Ozone Layer Depletion (CFCs,

Cl^\bullet

radicals, UV radiation in stratosphere).

Question 8422Question →

In a municipal water treatment system, the effluent water following alum coagulation and filtration is found to be slightly acidic and retains a persistent earthy odor. Which pair of chemical substances should be added to neutralize the acidity and eliminate the odor, respectively?

Slaked lime and activated carbon

Calcium tetraoxosulfate(VI) and chlorine gas

Washing soda and potash alum

Concentrated tetraoxosulfate(VI) acid and bleaching powder

Show answer & explanation

Answer: Slaked lime and activated carbon

Answer

Slaked lime (

\text{Ca(OH)}_2

) or soda ash is added to neutralize acidity (raising pH), while activated carbon (charcoal) is used to remove unpleasant odors and tastes via adsorption.

Slaked lime (

\text{Ca(OH)}_2

) acts as a base to neutralize hydrogen ions, correcting acidity, while activated carbon effectively adsorbs volatile organic compounds that cause unpleasant tastes and odors in water.

Step-by-Step Solution

1

Identify the cause of acidity and the substance required to raise pH.

Coagulation with alum leaves acidic residues due to hydrolysis. An alkali such as slaked lime,

\text{Ca(OH)}_2

, neutralizes acidity.

Acids are neutralized by adding basic compounds to adjust the water pH to a safe drinking level.

2

Identify the process and agent required for odor removal.

Activated carbon (charcoal) possesses a high surface area that adsorbs dissolved organic volatile impurities responsible for odor.

Physical adsorption on activated carbon removes taste and odor without introducing additional chemical pollutants.

Key Concept

Chemical roles in water purification: pH adjustment using slaked lime/soda ash and taste/odor removal using activated carbon.

Estimated Time:1m 0s

Question 8423Question →

A organic mixture containing naphthalene, chlorobenzene, acetophenone, and benzoic acid is separated using silica gel column chromatography with hexane (a non-polar solvent) as the mobile phase. Arrange these four compounds in order of their elution from the column, starting with the compound that elutes first and ending with the compound that elutes last.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order of elution from first to last is: Naphthalene, Chlorobenzene, Acetophenone, Benzoic acid.

In silica gel column chromatography using a non-polar solvent like hexane, separation is governed by adsorption efficiency. Silica gel contains polar silanol groups (

-Si-OH

). Non-polar compounds (naphthalene) do not interact significantly with the polar stationary phase and dissolve readily in the non-polar eluent, causing them to move rapidly down the column and elute first. As solute polarity increases from weakly polar (chlorobenzene) to moderately polar ketone (acetophenone) to strongly polar hydrogen-bonding acid (benzoic acid), the strength of adsorption increases, delaying elution accordingly.

Step-by-Step Solution

1

Identify the nature of the stationary and mobile phases.

The stationary phase (silica gel) is polar, while the mobile phase (hexane) is non-polar.

Normal-phase column chromatography relies on adsorption differences on a polar stationary matrix.

2

Determine the polarities of the four organic solutes based on their functional groups.

Naphthalene (non-polar hydrocarbon) < Chlorobenzene (weakly polar haloarene) < Acetophenone (moderately polar ketone) < Benzoic acid (strongly polar carboxylic acid).

Carboxylic acids can hydrogen-bond, ketones have strong dipoles, haloarenes have weak dipoles, and aromatic hydrocarbons are non-polar.

3

Correlate solute polarity with retention time and elution order.

Non-polar solutes adsorb weakly to silica gel and elute first; highly polar solutes adsorb strongly and elute last.

The non-polar mobile phase preferentially dissolves and transports less polar molecules down the column more quickly.

4

Arrange the compounds in sequence from least polar (first eluted) to most polar (last eluted).

Naphthalene

\rightarrow

Chlorobenzene

\rightarrow

Acetophenone

\rightarrow

Benzoic acid.

This represents the exact elution sequence in normal-phase column chromatography.

Key Concept

Adsorption Column Chromatography Elution Order based on Solute Polarity

Estimated Time:2m 0s

Question 8424Question →

During the industrial production of water gas, steam is blown over red-hot coke at elevated temperatures ( $1000^\circ\text{C}$ ). Why must the steam supply be periodically interrupted to blow air over the coke bed instead?

The reaction between steam and coke is endothermic, which continuously cools the coke bed below the required reaction temperature.

Steam acts as a homogeneous catalyst that shifts the position of dynamic equilibrium back toward reactants.

Air is required to convert amorphous carbon into graphite isomers before steam can react.

The gaseous water gas product expands beyond the molar volume of 22.4 dm³ per mole, stopping the flow of steam.

Show answer & explanation

Answer: The reaction between steam and coke is endothermic, which continuously cools the coke bed below the required reaction temperature.

Answer

The steam supply must be interrupted because the reaction between steam and red-hot coke (

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

) is strongly endothermic. This lowers the temperature of the coke bed below

1000^\circ\text{C}

, requiring air to be blown in so that carbon burns exothermically in oxygen (

\text{C} + \text{O}_2 \rightarrow \text{CO}_2

) to reheat the bed.

The reaction

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

is endothermic, absorbing heat from the red-hot coke. To maintain the high temperatures necessary for continuous reaction, the steam blow is periodically stopped and an air blow is introduced to exothermically burn some coke and reheat the bed.

Step-by-Step Solution

1

Analyze the thermochemical nature of the water gas reaction.

The formation of water gas (

\text{CO} + \text{H}_2

) from coke and steam absorbs heat (

\Delta H > 0

).

Endothermic reactions remove heat from the reaction vessel, causing the solid coke bed temperature to drop.

2

Determine the industrial necessity of alternating steam and air runs.

Blowing air causes exothermic combustion of carbon, restoring the red-hot temperature necessary for water gas synthesis.

High temperatures (

1000^\circ\text{C}

) are required to maintain high reaction rates and favor product yield.

Key Concept

Industrial production and energetics of water gas synthesis from coal/coke

Estimated Time:1m 0s

Question 8425Question →

Match each environmental contaminant or waste management process in Column A with its corresponding chemical characteristic or primary effect in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Lead (

\text{Pb}

) matches with 'Toxic heavy metal from battery manufacturing and old paints causing neurological harm'. DDT (Organochlorine) matches with 'Persistent synthetic pesticide that bioaccumulates up the agricultural food chain'. Incineration matches with 'High-temperature thermal waste disposal process that converts combustible refuse into ash'.

Each item correctly matches its specific chemical, environmental, or waste management role: Lead is a heavy metal associated with battery waste and nervous system toxicity; DDT is a non-biodegradable pesticide causing bioaccumulation; Incineration is thermal combustion reducing waste to ash.

Step-by-Step Solution

1

Identify the primary environmental impact of Lead (

\text{Pb}

).

Lead is a non-biodegradable heavy metal commonly sourced from lead-acid batteries and paints, causing neurological dysfunction.

Heavy metal contamination in soil primarily stems from industrial effluents and battery production.

2

Identify the characteristic behavior of DDT in soil and ecosystems.

DDT is an organochlorine pesticide notorious for environmental persistence and fat-solubility, causing bioaccumulation in higher trophic levels.

Organochlorine pesticides resist rapid chemical and biological decomposition.

3

Identify the operational principle of waste Incineration.

Incineration thermal treatment burns solid waste at high temperatures, turning organic matter into gaseous products and non-combustible ash.

It is a thermal waste volume reduction method commonly employed in municipal waste management.

Key Concept

Soil Pollution, Heavy Metal Toxicity, Pesticide Persistence, and Waste Disposal Methods

Question 8426Question →

What is the pH of an aqueous solution prepared by dissolving $0.365\text{ g}$ of pure hydrogen chloride gas ( $\text{HCl}$ ) in distilled water to make a total solution volume of $10.0\text{ dm}^3$ at $25^\circ\text{C}$ ? [Molar mass: $\text{H} = 1.0\text{ g mol}^{-1}, \text{Cl} = 35.5\text{ g mol}^{-1}$ ]

1.0

2.0

3.0

11.0

Show answer & explanation

Answer: 3.0

Answer

The pH of the resulting solution is 3.0.

Dissolving

0.365\text{ g}

of

\text{HCl}

(molar mass

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

) yields

0.01\text{ mol}

of

\text{HCl}

. In a

10.0\text{ dm}^3

solution, the hydrogen ion concentration is

\frac{0.01\text{ mol}}{10.0\text{ dm}^3} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

. Taking the negative logarithm gives a pH of

3.0

.

Step-by-Step Solution

1

Calculate the molar mass of

\text{HCl}

and find the number of moles dissolved.

\text{Molar mass of HCl} = 1.0 + 35.5 = 36.5\text{ g mol}^{-1}

. Moles of

\text{HCl} = \frac{0.365\text{ g}}{36.5\text{ g mol}^{-1}} = 0.01\text{ mol} = 1.0 \times 10^{-2}\text{ mol}

.

pH depends on the molar concentration of hydrogen ions, which requires knowing the total moles of solute.

2

Determine the molarity (

[\text{H}^+]

) of the solution.

[\text{H}^+] = \frac{\text{Moles of solute}}{\text{Volume in dm}^3} = \frac{0.01\text{ mol}}{10.0\text{ dm}^3} = 0.001\text{ mol dm}^{-3} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}

.

\text{HCl}

is a strong monoprotic acid and ionizes completely in water to yield equal moles of

\text{H}^+

ions.

3

Calculate the pH using the pH definition formula.

\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(1.0 \times 10^{-3}) = 3.0

.

The negative logarithm of the hydrogen ion concentration gives the pH value of the solution.

Key Concept

pH Calculation of a Strong Acid from Mass and Volume

Estimated Time:1m 30s

Question 8427Question →

Match each chemical system modification on the left with the primary kinetic mechanism on the right that accounts for the observed increase in reaction rate.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Ultraviolet irradiation corresponds to supplying photon energy for bond cleavage and free-radical generation; pulverizing calcium carbonate corresponds to maximizing reactant contact area and collision frequency; raising temperature by 10 K corresponds to elevating kinetic energy to expand the fraction of collisions exceeding activation energy; introducing a catalyst corresponds to establishing a lower activation energy pathway for both forward and reverse processes.

Each factor influences the rate through a distinct physical or chemical mechanism: light supplies photochemical activation energy; surface area controls collision site availability; temperature dictates the population of molecules with

E \ge E_a

; and catalysts lower the energy barrier for both reaction directions.

Step-by-Step Solution

1

Analyze the photochemical process (UV light application)

Light acts as an energy source for specific photochemical reactions, breaking bonds to yield radicals.

Light intensity and wavelength directly influence rates of photochemical reactions like halogenation of alkanes.

2

Analyze the effect of particle size (pulverization of solid)

Breaking solid chunks into powder exposes more surface atoms to acid molecules.

Greater exposed surface area increases the total frequency of particle collisions per second.

3

Analyze thermal energy and kinetic distribution

A temperature rise increases average kinetic energy and Maxwell-Boltzmann tail population.

The rate increases exponentially with temperature because a much larger percentage of collisions meet the activation energy threshold.

4

Analyze catalytic action in reversible systems

Catalysts lower activation energy (

E_a

) for both forward and reverse reactions without altering equilibrium position.

Catalysts alter the reaction mechanism to provide an alternative pathway with lower energy requirements.

Key Concept

Collision theory principles underlying reaction rate factors (light intensity, surface area, thermal kinetic energy distribution, and catalytic pathways)

Question 8428Question →

A metallurgical research laboratory analyzed three different metallic alloys ( $P$ , $Q$ , and $R$ ) to evaluate their internal lattice structures and physical property modifications relative to their primary base metals:

- Sample $P$ : Consists of copper with $30\%$ zinc solute atoms.
- Sample $Q$ : Consists of iron with $1.0\%$ carbon solute atoms.
- Sample $R$ : Consists of aluminium alloyed with small amounts of copper, magnesium, and manganese.

Which of the following statements correctly classifies the lattice alloy types of $P$ , $Q$ , and $R$ and accurately describes their electrical conductivity relative to their pure base metals?

Sample

P

and Sample

R

are substitutional alloys, while Sample

Q

is an interstitial alloy; all three alloys exhibit lower electrical conductivity than their pure base metals due to lattice distortion.

Sample

P

and Sample

Q

are interstitial alloys, while Sample

R

is a substitutional alloy; all three alloys exhibit higher electrical conductivity than their pure base metals due to enhanced metallic bonding.

Sample

Q

is a substitutional alloy, while Sample

P

and Sample

R

are interstitial alloys; Sample

Q

exhibits lower electrical conductivity, whereas Sample

P

and Sample

R

exhibit unchanged conductivity.

Sample

P

, Sample

Q

, and Sample

R

are all interstitial alloys; their electrical conductivity remains higher than that of pure copper, iron, and aluminium respectively because solute atoms donate free electrons to the delocalized sea.

Show answer & explanation

Answer: Sample $P$ and Sample $R$ are substitutional alloys, while Sample $Q$ is an interstitial alloy; all three alloys exhibit lower electrical conductivity than their pure base metals due to lattice distortion.

Answer

Sample P and Sample R are substitutional alloys, while Sample Q is an interstitial alloy; all three alloys exhibit lower electrical conductivity than their pure base metals due to lattice distortion.

The correct option accurately identifies that brass (Sample P) and duralumin (Sample R) are substitutional alloys because their solute atoms have comparable atomic radii to their host atoms, whereas carbon steel (Sample Q) is an interstitial alloy because carbon atoms are small enough to enter the voids of the iron lattice. Furthermore, it correctly states that all alloys suffer a reduction in electrical conductivity relative to pure host metals due to electron scattering caused by lattice distortion.

Step-by-Step Solution

1

Determine the alloy lattice type for Sample P (brass) and Sample R (duralumin).

Zinc atoms (in brass) and copper/magnesium/manganese atoms (in duralumin) have atomic radii within

15\%

of the host copper and aluminium radii, so they replace host atoms directly in the crystal lattice, forming substitutional alloys.

When solute and solvent atomic radii are comparable, solute atoms substitute for host metal atoms at lattice sites.

2

Determine the alloy lattice type for Sample Q (carbon steel).

Carbon has a significantly smaller atomic radius than iron, allowing carbon atoms to fit into the interstitial voids between iron atoms, forming an interstitial alloy.

Solute atoms with radii much smaller than the host metal fit into spaces (interstices) between lattice atoms.

3

Evaluate the effect of alloying on electrical conductivity.

The presence of foreign solute atoms (whether substitutional or interstitial) causes lattice distortion, which disrupts the uniform periodic potential of the metal lattice and scatters conduction electrons, resulting in lower electrical conductivity compared to pure base metals.

Electron mobility is reduced by lattice irregularities and strain fields created by solute atoms.

Key Concept

Classification of substitutional vs interstitial alloys and the effect of lattice distortion on physical properties like electrical conductivity.

Estimated Time:2m 0s

Question 8429Question →

A sample of hydrated copper(II) tetraoxosulfate(VI), $\text{CuSO}_4\cdot 5\text{H}_2\text{O}$ , with a mass of $12.5\text{ g}$ is dissolved completely in distilled water to make $250\text{ cm}^3$ of solution. What is the concentration of the copper(II) tetraoxosulfate(VI) solution in $\text{mol dm}^{-3}$ ?

( $\text{Relative atomic masses: Cu} = 64,\text{ S} = 32,\text{ O} = 16,\text{ H} = 1$ )

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

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

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

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

Show answer & explanation

Answer: $0.20\text{ mol dm}^{-3}$

Answer

The molar concentration of the copper(II) tetraoxosulfate(VI) solution is

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

.

The correct concentration is

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

. The molar mass of hydrated copper(II) tetraoxosulfate(VI) (

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

) is

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

, so

12.5\text{ g}

corresponds to

0.05\text{ mol}

. Dividing

0.05\text{ mol}

by the volume of

0.25\text{ dm}^3

gives

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

.

Step-by-Step Solution

1

Calculate the molar mass of hydrated copper(II) tetraoxosulfate(VI),

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

.

\text{Molar mass} = 64 + 32 + (4 \times 16) + 5 \times (2 \times 1 + 16) = 250\text{ g mol}^{-1}

.

The total molar mass must include the 5 molecules of water of crystallization present in the solid salt.

2

Determine the number of moles of

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

dissolved.

\text{Moles} = \frac{12.5\text{ g}}{250\text{ g mol}^{-1}} = 0.05\text{ mol}

.

Number of moles is equal to the given mass divided by the molar mass.

3

Convert the volume of the solution from

\text{cm}^3

to

\text{dm}^3

.

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

.

Molar concentration requires the volume of solution to be expressed in cubic decimetres.

4

Calculate the molar concentration in

\text{mol dm}^{-3}

.

\text{Concentration} = \frac{0.05\text{ mol}}{0.25\text{ dm}^3} = 0.20\text{ mol dm}^{-3}

.

Molarity is defined as moles of solute per cubic decimetre of solution.

Key Concept

Molar Concentration of Hydrated Copper Compounds

Estimated Time:1m 30s

Question 8430Question →

A steady electric current is passed through an aqueous solution of zinc tetraoxosulfate(VI) for $4825\text{ s}$ . If $3.25\text{ g}$ of zinc is deposited at the cathode, what is the magnitude of the electric current, in Amperes, used?

$[\text{Zn} = 65,\text{ 1 F} = 96500\text{ C mol}^{-1}]$

Show answer & explanation

Answer: 2

Answer

The magnitude of the electric current required is

2.0\text{ A}

.

Depositing

3.25\text{ g}

of

\text{Zn}

(atomic mass

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

) requires

0.05\text{ mol}

of zinc metal. Since each

\text{Zn}^{2+}

ion requires

2

electrons to be reduced,

0.10\text{ mol}

of electrons (

9650\text{ C}

) must pass through the electrolyte. Dividing this charge by time (

4825\text{ s}

) gives

2.0\text{ A}

.

Step-by-Step Solution

1

Calculate the moles of zinc deposited at the cathode.

Moles of

\text{Zn} = \frac{3.25\text{ g}}{65\text{ g mol}^{-1}} = 0.05\text{ mol}

.

Dividing the mass of metal deposited by its relative atomic mass gives the number of moles deposited.

2

Determine the quantity of electricity in Coulombs needed for the deposition.

Reduction half-reaction:

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

. Moles of

e^- = 2 \times 0.05\text{ mol} = 0.10\text{ mol}

. Quantity of electricity

Q = 0.10\text{ mol} \times 96500\text{ C mol}^{-1} = 9650\text{ C}

.

Faraday's second law relates the mole ratio of electrons to metal ion charge.

3

Calculate the steady electric current in Amperes.

I = \frac{Q}{t} = \frac{9650\text{ C}}{4825\text{ s}} = 2.0\text{ A}

.

Electric current is defined as the rate of charge flow over time (

I = \frac{Q}{t}

).

Key Concept

Faraday's Laws of Electrolysis and Quantitative Calculations

Question 8431Question →

Match each chemical process or reaction involving alkanes and petroleum refining in Column A with its corresponding chemical description or primary purpose in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Catalytic Cracking matches with the thermal breakdown of long-chain hydrocarbons into shorter alkanes and alkenes. Reforming matches with converting straight-chain alkanes into branched or aromatic hydrocarbons to boost octane rating. Complete Combustion matches with reacting alkanes in excess oxygen to produce

CO_2

and

H_2O

. Free-Radical Substitution matches with replacing hydrogen atoms with halogens under UV light.

Catalytic Cracking breaks larger hydrocarbon molecules into smaller, more useful molecules (alkanes and alkenes). Reforming increases fuel quality (octane rating) by isomerizing straight chains to branched chains or aromatics. Complete Combustion converts alkanes in excess oxygen to carbon dioxide and water. Free-Radical Substitution halogenates alkanes in the presence of UV light.

Step-by-Step Solution

1

Identify the primary function of Catalytic Cracking.

Cracking involves breaking heavy petroleum fractions into smaller alkanes and alkenes.

Heavy oils have low demand, whereas lighter fractions like petrol and gases have high industrial demand.

2

Identify the structural transformation involved in Reforming.

Reforming converts straight-chain alkanes into branched-chain alkanes and aromatic compounds.

Straight-chain alkanes cause engine knocking; branched and aromatic structures improve fuel efficiency by increasing the octane rating.

3

Determine the products of Complete Combustion of alkanes.

Alkanes react completely with excess oxygen to yield

CO_2(g)

and

H_2O(g)

.

Hydrocarbon oxidation in excess

O_2

yields fully oxidized carbon dioxide and water.

4

Determine the mechanism for alkane halogenation.

Halogenation of alkanes requires ultraviolet light to generate free radicals for substitution.

Alkanes are unreactive saturated hydrocarbons (paraffins) and require UV light to initiate homeolytic fission of chlorine or bromine molecules.

Key Concept

Chemical reactions of alkanes and industrial petroleum refining processes

Question 8432Question →

Match each of the following chemical species with the predominant intermolecular force operating between its molecules in the liquid or solid state.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Hydrogen fluoride matches with Hydrogen bonding; Trichloromethane matches with Permanent dipole-dipole interactions; Solid iodine matches with London dispersion forces (in a non-polar crystalline lattice); Methane matches with Weak London dispersion forces (in a small non-polar molecule).

Hydrogen fluoride forms hydrogen bonds due to the extreme electronegativity difference between H and F. Trichloromethane exhibits permanent dipole-dipole attractions because of its permanent net molecular dipole. Solid iodine is non-polar but has a large polarizable electron cloud leading to substantial London dispersion forces in its solid crystal. Methane is non-polar and small, possessing only weak London dispersion forces.

Step-by-Step Solution

1

Analyze the polarity and chemical structure of each given substance.

HF

is highly polar with

H-F

bonds;

CHCl_3

is a polar asymmetrical molecule;

I_2

is a non-polar diatomic solid;

CH_4

is a non-polar tetrahedral gas.

Intermolecular forces depend strictly on molecular polarity, presence of

N-H

,

O-H

, or

F-H

bonds, and molecular size/polarizability.

2

Identify specific conditions for hydrogen bonding.

HF

satisfies the requirement of hydrogen attached to highly electronegative fluorine, giving rise to intermolecular hydrogen bonds.

Hydrogen bonding requires a hydrogen atom covalently bonded to

N

,

O

, or

F

interacting with a lone pair on a neighbouring electronegative atom.

3

Differentiate dipole-dipole forces from dispersion forces in neutral covalent compounds.

CHCl_3

possesses a permanent dipole moment giving dipole-dipole forces, while

I_2

and

CH_4

are non-polar and rely on London dispersion forces, with

I_2

having larger dispersion forces due to greater electron cloud polarizability.

Dispersion forces scale with molecular size and electron count, while dipole-dipole forces require permanent polar bonds in asymmetrical shapes.

Key Concept

Classification and Origin of Intermolecular Forces

Question 8433Question →

Which physical method is most suitable for separating tiny solid precipitate particles or fat globules suspended in a liquid medium when ordinary filtration is ineffective because the particles clog or pass through filter paper pores?

Centrifugation

Sublimation

Magnetization

Fractional distillation

Show answer & explanation

Answer: Centrifugation

Answer

Centrifugation is the correct method because it relies on spinning at high speeds to force denser suspended particles or lighter liquid droplets to separate according to their density.

Centrifugation spins the suspension at high velocities, subjecting the particles to centrifugal force that sediment them according to their density, making it ideal for fine suspensions or emulsions where filtration fails.

Step-by-Step Solution

1

Analyze the physical state and nature of the mixture

The mixture consists of fine solid particles or fat globules suspended in a liquid medium that cannot be trapped by filter paper.

Filtration relies on pore size relative to particle size, which fails when particles are colloidal or microscopic.

2

Evaluate the mechanism of centrifugation

High-speed spinning creates centrifugal acceleration, pulling denser particles outward/downward and forcing lighter components upward.

Centrifugation accelerates gravity-based sedimentation for fine suspensions.

Key Concept

Centrifugation as a physical separation technique based on density and particle sedimentation under centrifugal force

Question 8434Question →

What is the total number of nitrogen atoms contained in a $5.60\text{ dm}^3$ sample of dinitrogen monoxide gas ( $\text{N}_2\text{O}$ ) measured at standard temperature and pressure ( $\text{STP}$ )? $[N_A = 6.02 \times 10^{23}\text{ mol}^{-1}\text{; Molar volume of gas at STP } = 22.4\text{ dm}^3\text{mol}^{-1}]$

3.01 \times 10^{23}

2.81 \times 10^{23}

1.51 \times 10^{23}

4.52 \times 10^{23}

Show answer & explanation

Answer: $3.01 \times 10^{23}$

Answer

3.01 \times 10^{23}

nitrogen atoms

The correct answer of

3.01 \times 10^{23}

is calculated by determining the amount in moles of dinitrogen monoxide gas (

5.60\text{ dm}^3 / 22.4\text{ dm}^3\text{mol}^{-1} = 0.25\text{ mol}

), multiplying by 2 because there are two nitrogen atoms per

\text{N}_2\text{O}

molecule (

0.50\text{ mol}

of N atoms), and multiplying by Avogadro's constant (

0.50 \times 6.02 \times 10^{23} = 3.01 \times 10^{23}

).

Step-by-Step Solution

1

Calculate the moles of dinitrogen monoxide gas at STP

\text{Moles of N}_2\text{O} = \frac{5.60\text{ dm}^3}{22.4\text{ dm}^3\text{mol}^{-1}} = 0.25\text{ mol}

At STP, one mole of any ideal gas occupies a molar volume of 22.4 dm³.

2

Determine the number of moles of nitrogen atoms

\text{Moles of N atoms} = 0.25\text{ mol} \times 2 = 0.50\text{ mol}

Each formula unit of dinitrogen monoxide (N₂O) contains 2 atoms of nitrogen.

3

Convert moles of nitrogen atoms to the total number of atoms using Avogadro's constant

\text{Number of N atoms} = 0.50\text{ mol} \times 6.02 \times 10^{23}\text{ atoms/mol} = 3.01 \times 10^{23}\text{ atoms}

Avogadro's constant provides the number of particles per mole of substance.

Key Concept

Mole Concept and Avogadro's Constant applied to Gas Molar Volume

Question 8435Question →

For the endothermic steam-reforming process represented by the thermochemical equation below:

CH_4(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g) \quad \Delta H = +206\text{ kJ mol}^{-1}

Match each applied change (stress) on the system with its corresponding effect on the equilibrium position or system performance.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Increasing temperature shifts the equilibrium to the right; increasing pressure shifts the equilibrium to the left; adding a nickel catalyst increases reaction rates without changing the equilibrium position; continuous removal of carbon monoxide shifts the equilibrium to the right.

According to Le Chatelier's principle: heating an endothermic reaction shifts equilibrium toward products; increasing total pressure shifts equilibrium toward the side with fewer gas molecules (reactants, 2 moles vs 4 moles); adding a catalyst speeds up reaching equilibrium without shifting its position; and removing a product shifts equilibrium toward products to compensate for the decrease in concentration.

Step-by-Step Solution

1

Analyze the temperature change using enthalpy sign

The positive enthalpy change (

\Delta H = +206\text{ kJ mol}^{-1}

) indicates an endothermic forward reaction. Raising the temperature shifts the equilibrium to the right.

Le Chatelier's principle dictates that adding thermal energy favors the heat-absorbing (endothermic) direction.

2

Count gaseous moles on both sides to determine the pressure effect

Reactant gas moles = 2 (

CH_4 + H_2O

), Product gas moles = 4 (

CO + 3H_2

). Increasing pressure shifts the system to the left.

An increase in pressure shifts the equilibrium position toward the side with fewer moles of gas to reduce pressure.

3

Determine the role of a catalyst

The nickel catalyst accelerates both forward and backward reaction rates equally.

Catalysts do not alter the position of dynamic equilibrium or change product yield.

4

Determine the concentration change effect

Removing

CO(g)

causes a rightward shift in equilibrium.

Removing a component drives the system to shift in the direction that replaces it.

Key Concept

Le Chatelier's Principle

Question 8436Question →

Match each chemical term or phenomenon on the left with its correct defining characteristic on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Deliquescence matches the spontaneous absorption of atmospheric moisture until dissolving into a solution; Efflorescence matches the loss of water vapor to dry air forming a powdery residue; Hygroscopy matches the absorption of water vapor without dissolving; Water of crystallization matches the definite ratio of water molecules chemically bound in a crystal lattice.

Deliquescence describes solids absorbing moisture until they dissolve into a solution. Efflorescence describes hydrated crystals spontaneously losing water of crystallization into dry air to become powdery. Hygroscopy describes substances absorbing water vapor without turning into a solution. Water of crystallization is the fixed stoichiometric quantity of water built into the salt crystal structure.

Step-by-Step Solution

1

Define deliquescence and match it with its defining process.

Deliquescence is matched with spontaneous absorption of moisture until the solid completely dissolves into a liquid solution.

Deliquescent solids absorb so much water from moist air that they form a liquid solution.

2

Define efflorescence and match it with its defining process.

Efflorescence is matched with the process in which a crystalline salt loses water vapor to dry air, forming a powdery residue.

This occurs because the vapor pressure of the hydrated crystal exceeds the ambient atmospheric water vapor pressure.

3

Define hygroscopy and distinguish it from deliquescence.

Hygroscopy is matched with the absorption of water vapor from the surrounding atmosphere without dissolving or forming a liquid solution.

Hygroscopic materials absorb water but remain in their original state without liquefying into a solution.

4

Define water of crystallization.

Water of crystallization is matched with the definite ratio of water molecules stoichiometrically locked inside a salt's crystal framework.

It represents the chemically bound water necessary for maintaining the specific crystal structure of hydrated salts.

Key Concept

Atmospheric Behavior and Hydration of Salts

Estimated Time:1m 15s

Question 8437Question →

Chromium is a first-row transition element with an atomic number of 24. During a chemical reaction, a neutral chromium atom loses three electrons to form the chromium(III) cation, $Cr^{3+}$ . Which electronic structure correctly represents this $Cr^{3+}$ ion in its ground state?

[Ar] 3d^3

[Ar] 3d^1 4s^2

[Ar] 3d^2 4s^1

[Ar] 3d^4

Show answer & explanation

Answer: $[Ar] 3d^3$

Answer

The ground-state electronic configuration of the chromium(III) ion,

Cr^{3+}

, is

[Ar] 3d^3

.

Neutral chromium has an electronic configuration of

[Ar] 3d^5 4s^1

due to half-filled subshell stability. When forming a

Cr^{3+}

cation, the atom loses three electrons: the single valence electron in the

4s

orbital is lost first, followed by two electrons from the

3d

subshell, leaving a final ground-state configuration of

[Ar] 3d^3

.

Step-by-Step Solution

1

Determine the ground-state electronic configuration of neutral chromium (

Cr

,

Z = 24

).

Neutral chromium has the anomalous configuration

[Ar] 3d^5 4s^1

.

Chromium exhibits an exception to the standard Aufbau principle because a half-filled

d

-subshell (

3d^5

) provides extra thermodynamic stability.

2

Apply the rule for cation formation in transition metals.

Electrons in the outermost principal quantum shell (

4s

) are removed prior to removing electrons from the inner

(n-1)d

subshell (

3d

).

Once filled, the

3d

orbitals experience greater nuclear attraction and drop lower in energy than the

4s

orbital.

3

Deduct three electrons to account for the

+3

charge of

Cr^{3+}

.

Remove the single electron from

4s

(

[Ar] 3d^5 4s^0

), then remove two electrons from

3d

to obtain

[Ar] 3d^3

.

Removing 3 electrons total converts neutral

Cr

into

Cr^{3+}

.

Key Concept

Electronic Configuration of Transition Metal Cations

Estimated Time:2m 0s

Question 8438Question →

Which of the following species features a central atom with $sp^2$ hybridization and a trigonal planar molecular geometry?

CO_3^{2-}

H_3O^+

PCl_3

SO_3^{2-}

Show answer & explanation

Answer: $CO_3^{2-}$

Answer

The trioxocarbonate(IV) ion (

CO_3^{2-}

) possesses a central carbon atom with

sp^2

hybridization and a trigonal planar molecular shape.

In the trioxocarbonate(IV) ion (

CO_3^{2-}

), the central carbon atom forms three sigma bonds with three oxygen atoms and contains zero non-bonding lone pairs. According to VSEPR theory, three electron charge clouds position themselves as far apart as possible (

120^\circ

bond angles), adopting a trigonal planar geometry which requires

sp^2

hybridization of the central carbon atom.

Step-by-Step Solution

1

Determine the valence electron count and steric number for the central carbon in

CO_3^{2-}

Carbon brings 4 valence electrons, plus 2 electrons from the net charge (-2), total 6 valence electrons available for bonding. Carbon forms 3 sigma bonds with three oxygen atoms and has 0 lone pairs. Steric number = 3.

VSEPR theory uses the number of electron domains (steric number) around the central atom to determine orbital hybridization and electron geometry.

2

Determine the hybridization and molecular geometry based on VSEPR theory

Steric number 3 with 0 lone pairs corresponds to

sp^2

hybridization and a trigonal planar molecular shape.

Three electron domains orient themselves at

120^\circ

bond angles to minimize electron pair repulsion.

3

Compare with the distractor species (

H_3O^+

,

PCl_3

, and

SO_3^{2-}

)

Each of

H_3O^+

,

PCl_3

, and

SO_3^{2-}

has 3 bonding pairs and 1 non-bonding lone pair on its central atom (steric number 4).

A steric number of 4 corresponds to

sp^3

hybridization, and 3 bonding pairs with 1 lone pair produces a trigonal pyramidal geometry rather than trigonal planar.

Key Concept

VSEPR Theory and Hybridization of Species with Three Attached Atoms

Question 8439Question →

The sulfite ion ( $SO_3^{2-}$ ) is formed when sulfur dioxide dissolves in aqueous basic media. Based on Valence Shell Electron Pair Repulsion (VSEPR) theory, which of the following correctly describes the hybridization of the central sulfur atom and the molecular geometry of the $SO_3^{2-}$ ion?

sp^3

hybridization with trigonal pyramidal molecular geometry

sp^3

hybridization with tetrahedral molecular geometry

sp^2

hybridization with trigonal planar molecular geometry

sp^2

hybridization with bent molecular geometry

Show answer & explanation

Answer: $sp^3$ hybridization with trigonal pyramidal molecular geometry

Answer

The central sulfur atom in the sulfite ion (

SO_3^{2-}

) exhibits

sp^3

hybridization and has a trigonal pyramidal molecular geometry.

The central sulfur atom in

SO_3^{2-}

possesses 4 electron domains (3 bonding pairs and 1 non-bonding lone pair), yielding

sp^3

hybridization. According to VSEPR theory, an

AX_3E

electron arrangement produces a trigonal pyramidal molecular shape.

Step-by-Step Solution

1

Determine total valence electrons for

SO_3^{2-}

Sulfur provides 6, each of the three oxygens provides 6, plus 2 electrons from the

2-

charge:

6 + (3 \times 6) + 2 = 26

valence electrons (13 pairs).

Accurate valence electron counting is necessary to determine bonding and non-bonding electron distribution.

2

Determine electron domain count and steric number on the central atom

Sulfur forms 3 single sigma bonds to oxygen atoms and retains 1 lone pair of non-bonding electrons (

3 \text{ bonding pairs} + 1 \text{ lone pair} = 4 \text{ total electron domains}

).

Steric number determines the orbital hybridization of the central atom.

3

Determine hybridization and molecular geometry using VSEPR theory

A steric number of 4 corresponds to

sp^3

hybridization. With 3 bonding pairs and 1 lone pair (an

AX_3E

system), the electron geometry is tetrahedral, but the actual molecular geometry is trigonal pyramidal.

Molecular geometry is named based only on the spatial arrangement of the bonded atoms, not the non-bonding lone pair.

Key Concept

VSEPR theory and hybridization determination for polyatomic ions with lone pairs

Estimated Time:1m 30s

Question 8440Question →

An aqueous solution of chromium(III) tetraoxosulfate(VI) is electrolyzed using inert platinum electrodes. A steady current of $5.00\text{ A}$ is passed through the electrolyte for $96.5\text{ minutes}$ . If the cathodic current efficiency for chromium deposition is $75.0\%$ , calculate the mass, in grams, of chromium metal deposited at the cathode. [Molar mass of $\text{Cr} = 52.0\text{ g/mol}$ , $1\text{ F} = 96500\text{ C/mol}$ ]

Show answer & explanation

Answer: 3.9

Answer

3.90\text{ g}

To find the mass of chromium deposited, calculate total charge (

Q = I \times t = 5.00 \times 5790 = 28950\text{ C}

), adjust for

75.0\%

current efficiency (

Q_{\text{eff}} = 21712.5\text{ C}

), convert to Faradays (

0.225\text{ F}

), divide by the valency of 3 for

\text{Cr}^{3+}

to find moles of chromium (

0.075\text{ mol}

), and multiply by molar mass (

52.0\text{ g/mol}

) to yield

3.90\text{ g}

.

Step-by-Step Solution

1

Convert the electrolysis time into seconds

t = 96.5 \times 60 = 5790\text{ s}

Standard SI unit of time (seconds) is required for charge calculation (

Q = I \times t

).

2

Calculate the total charge transferred

Q = 5.00\text{ A} \times 5790\text{ s} = 28950\text{ C}

Determines total quantity of electricity delivered by the current source.

3

Apply the current efficiency percentage

Q_{\text{eff}} = 28950\text{ C} \times 0.750 = 21712.5\text{ C}

Only 75% of the total current is utilized specifically for reducing chromium ions.

4

Convert effective charge into moles of electrons

n_e = \frac{21712.5\text{ C}}{96500\text{ C/mol}} = 0.225\text{ mol } e^-

Faraday's constant gives the charge carried per mole of electrons.

5

Relate moles of electrons to moles of chromium deposited

n_{\text{Cr}} = \frac{0.225}{3} = 0.075\text{ mol}

Reduction of one mole of

\text{Cr}^{3+}

requires three moles of electrons (3 Faradays).

6

Calculate the mass of chromium deposited

m = 0.075\text{ mol} \times 52.0\text{ g/mol} = 3.90\text{ g}

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

Key Concept

Faraday's Laws of Electrolysis and Current Efficiency

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