All practice questions

Question 8581Question →

Match each of the following physical phenomena or chemical systems on the left with the predominant type of intermolecular force or interaction responsible for it on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Sodium chloride dissolution matches Ion-dipole interactions; Liquefaction of argon gas matches London dispersion forces; Lower density of ice matches Extensive three-dimensional hydrogen bonding; Higher boiling point of polar

\text{HCl}

over

\text{Ar}

matches Permanent dipole-dipole interactions.

Each system exhibits physical behaviors dictated by its specific intermolecular interaction: ion-dipole attractions enable ionic solvation; temporary induced dipoles (dispersion forces) allow nonpolar noble gases to condense; directional 3D hydrogen bonding creates an expanded lattice in ice; permanent dipole-dipole forces provide extra attraction in polar compounds like

\text{HCl}

.

Step-by-Step Solution

1

Analyze the nature of the chemical species involved in each phenomenon (ionic, polar, nonpolar, or hydrogen-bonded).

\text{NaCl}

in water involves ions and polar molecules; argon gas consists of isolated nonpolar atoms; ice involves water molecules forming a rigid lattice;

\text{HCl}

consists of polar molecules.

Identifying molecular polarity and ionic state determines which category of intermolecular interaction dominates.

2

Pair each phenomenon with the correct fundamental intermolecular force.

Ions + polar solvent

\rightarrow

Ion-dipole; Nonpolar atoms

\rightarrow

London dispersion forces; Water lattice expansion

\rightarrow

Extensive 3D hydrogen bonding; Permanent molecular dipoles

\rightarrow

Permanent dipole-dipole forces.

Connecting microscopic force definitions to observed physical properties yields the correct matches.

Key Concept

Intermolecular Forces and Hydrogen Bonding

Question 8582Question →

Match each nitrogen-containing organic compound on the left with the statement on the right that accurately accounts for its aqueous basicity and lone-pair electronic behavior.

Click a left item, then click its matching right item

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Answer

Dimethylamine matches the statement describing higher aqueous basicity than ammonia due to inductive donation and solvation; Phenylamine matches the statement describing reduced basicity from aromatic resonance delocalization; Ethanamide matches the statement describing neutrality caused by carbonyl resonance; Triethylamine matches the statement describing steric hindrance affecting conjugate acid solvation.

The correct matches reflect fundamental physical-organic chemistry principles governing nitrogen basicity: Dimethylamine combines inductive donation with high conjugate acid solvation stability; Phenylamine suffers basicity loss from aromatic resonance delocalization; Ethanamide lone-pair delocalization into the carbonyl group yields a neutral compound; Triethylamine basicity in water is moderated by steric crowding that interferes with hydration of the ammonium cation.

Step-by-Step Solution

1

Analyze the electronic structure of Dimethylamine (

(CH_3)_2NH

).

Two methyl groups supply electron density via

+I

inductive effects, enhancing nitrogen lone-pair availability, while the secondary cation remains readily solvated by water.

Secondary aliphatic amines are generally the strongest bases in aqueous media.

2

Analyze the resonance interactions in Phenylamine (

C_6H_5NH_2

).

The unshared electron pair on nitrogen participates in resonance with the benzene ring, lowering lone-pair availability.

Aromatic amines are significantly weaker bases than ammonia and aliphatic amines.

3

Examine the functional group characteristics of Ethanamide (

CH_3CONH_2

).

Resonance delocalization between nitrogen's lone pair and the adjacent

C=O

double bond (

O=C-N \leftrightarrow ^-O-C=N^+

) deprives nitrogen of basic character.

Amides behave as neutral organic compounds in aqueous solution.

4

Evaluate steric effects in Triethylamine (

(C_2H_5)_3N

).

Three ethyl groups create steric crowding around the nitrogen cation, hindering stabilization through hydration in water.

In aqueous solution, tertiary aliphatic amines are often weaker bases than secondary aliphatic amines due to solvation factors.

Key Concept

Relative basicity of aliphatic amines, aromatic amines, and amides governed by inductive, resonance, and solvation steric effects.

Estimated Time:2m 0s

Question 8583Question →

Match each organic nitrogen compound on the left with its corresponding relative basicity or acid-base structural property on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Phenylamine matches with being weaker than ammonia due to aromatic delocalization; Methylamine matches with being stronger than ammonia due to +I inductive effect; Ethanamide matches with being neutral due to carbonyl resonance; Dimethylamine matches with being stronger than methylamine in aqueous solution.

Phenylamine is less basic than ammonia because its lone pair is delocalized across the aromatic ring. Methylamine is more basic than ammonia due to inductive electron donation by the methyl group. Ethanamide is neutral because its lone pair participates in resonance with the carbonyl group. Dimethylamine is more basic than methylamine in water due to two electron-donating methyl groups.

Step-by-Step Solution

1

Analyze how structural features influence nitrogen lone pair availability.

Electron-donating alkyl groups (+I effect) enhance lone pair availability (increasing basicity), while electron-withdrawing groups or resonance delocalization decrease lone pair availability (decreasing basicity).

Lewis/Brønsted-Lowry basicity of nitrogen compounds depends directly on lone pair availability to accept a proton.

2

Evaluate phenylamine and ethanamide.

Phenylamine delocalizes its lone pair into the benzene ring, making it weaker than

NH_3

. Ethanamide delocalizes its lone pair into the

C=O

double bond, making it neutral in aqueous solution.

Resonance delocalization significantly stabilizes the unprotonated state and lowers basicity.

3

Compare methylamine and dimethylamine.

Methylamine has one alkyl group increasing basicity over

NH_3

. Dimethylamine has two alkyl groups supplying greater electron density, making it more basic than methylamine in aqueous solution.

Inductive electron donation by methyl groups stabilizes the positive conjugate ammonium ion.

Key Concept

Relative basicity of amines and amides based on inductive and resonance effects

Question 8584Question →

Alloying pure copper with zinc to produce brass enhances its mechanical hardness and tensile strength, but leads to a decrease in its electrical conductivity compared to pure copper.

True

False

Show answer & explanation

Answer: True

Answer

True. Alloying copper with zinc to form brass increases mechanical strength and hardness via lattice distortion, but decreases electrical conductivity due to increased electron scattering.

The statement accurately reflects physical metallurgy principles. Foreign zinc atoms create strain fields in the copper lattice that block dislocation movement (increasing hardness), while simultaneously disrupting the periodic potential required for efficient electron transport (decreasing electrical conductivity).

Step-by-Step Solution

1

Analyze the composition and structure of brass.

Brass is a substitutional alloy formed by dissolving zinc atoms into the copper metallic lattice.

Understanding the atomic mixture establishes how foreign atoms fit into the primary metal matrix.

2

Evaluate the effect of foreign zinc atoms on mechanical properties.

Zinc atoms differ in size from copper atoms, causing localized lattice distortion that impedes dislocation motion.

Restricting the slipping of crystal planes increases the alloy's hardness and tensile strength compared to pure copper.

3

Evaluate the effect of lattice distortion on electrical conductivity.

The loss of periodic lattice symmetry increases the scattering of free conduction electrons.

Increased electron scattering reduces the mean free path of charge carriers, lowering overall electrical conductivity relative to pure copper.

Key Concept

Influence of alloying on mechanical hardness and electrical conductivity

Question 8585Question →

At dynamic chemical equilibrium in a closed vessel, the concentration of the reactants must always be equal to the concentration of the products.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Dynamic equilibrium requires equality of reaction rates, not equality of concentrations.

The correct answer is False because dynamic chemical equilibrium is defined by equal rates of the forward and reverse reactions, which maintains constant (un-changing) concentrations of reactants and products over time. However, the numerical concentrations of reactants and products do not need to be equal.

Step-by-Step Solution

1

Analyze the condition for dynamic equilibrium.

Dynamic equilibrium is achieved when the rate of the forward reaction equals the rate of the reverse reaction (

\text{Rate}_{\text{forward}} = \text{Rate}_{\text{reverse}}

).

Equal rates prevent any net change in the concentrations of reactants and products.

2

Evaluate the relationship between rates and concentrations.

While rates are equal and concentrations remain constant, the absolute concentrations of reactants and products depend on the equilibrium position and the value of

K_c

.

If

K_c > 1

, product concentrations exceed reactant concentrations at equilibrium, whereas if

K_c < 1

, reactant concentrations predominate.

Key Concept

Equal Rates vs. Constant Concentrations in Dynamic Equilibrium

Question 8586Question →

Unlike aliphatic amines, amides such as ethanamide ( $CH_3CONH_2$ ) do not exhibit basic properties in aqueous solution. Which of the following best explains this neutral behavior?

The lone pair of electrons on the nitrogen atom is delocalized by resonance with the adjacent carbonyl group.

The nitrogen atom in ethanamide lacks an unshared pair of electrons.

Ethanamide dissociates completely in water to yield hydrogen ions (

H^+

).

The methyl group (

CH_3-

) withdraws electron density from the nitrogen atom through an inductive effect.

Show answer & explanation

Answer: The lone pair of electrons on the nitrogen atom is delocalized by resonance with the adjacent carbonyl group.

Answer

The lone pair of electrons on the nitrogen atom is delocalized by resonance with the adjacent carbonyl group.

The neutral character of amides is due to resonance stabilization. The non-bonding lone pair of electrons on the nitrogen atom overlaps with the

\pi

-system of the adjacent carbonyl group (

C=O

). This delocalization significantly decreases the electron density on nitrogen, preventing it from accepting a proton (

H^+

) and acting as a base.

Step-by-Step Solution

1

Identify the functional group and structural elements of ethanamide (

CH_3CONH_2

).

Ethanamide contains an amino group (

-NH_2

) directly attached to a carbonyl group (

-C=O

).

Understanding the adjacent arrangement of the carbonyl carbon and nitrogen is essential for evaluating electronic effects.

2

Analyze the availability of the nitrogen lone pair for protonation.

The nitrogen atom has a lone pair of electrons, but it interacts with the

\pi

-orbital of the carbonyl group, forming a resonance structure:

CH_3-C(O^-)=NH_2^+

.

Basicity depends on the availability of a lone pair to accept a proton (

H^+

). Delocalization drastically reduces lone pair availability.

3

Conclude the acid-base nature of the molecule.

Because the lone pair is delocalized, ethanamide cannot readily act as a proton acceptor, rendering it neutral in aqueous solution.

This explains the contrast between basic amines (localized lone pair) and neutral amides (delocalized lone pair).

Key Concept

Neutrality of amides due to resonance delocalization of the nitrogen lone pair into the adjacent carbonyl group

Estimated Time:45s

Question 8587Question →

Match each noble gas listed on the left with its corresponding primary industrial application or characteristic use on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Helium matches with weather balloons and diving gas mixtures; Neon matches with advertising signs; Argon matches with arc welding inert atmosphere; Radon matches with cancer radiotherapy.

Each noble gas is matched to its unique application based on its physical and chemical properties: Helium for low density and low blood solubility, Neon for glowing light discharge, Argon for unreactive shielding during welding, and Radon for cancer radiotherapy.

Step-by-Step Solution

1

Identify the primary application of Helium.

Helium's low density and minimal blood solubility pair it with weather balloons and deep-sea diving mixtures.

Prevents decompression sickness in divers and provides buoyancy in balloons.

2

Identify the primary application of Neon.

Neon emits a distinct reddish-orange light in electrical discharge tubes used for advertising signs.

Excited neon gas emits characteristic light when electrical discharge occurs.

3

Identify the primary application of Argon.

Argon serves as an inert shielding gas in electric arc welding.

Prevents atmospheric oxygen and nitrogen from reacting with hot metals being welded.

4

Identify the primary application of Radon.

Radon is radioactive and used in cancer radiotherapy.

Radiation emitted during radioactive decay destroys targeted cancer cells.

Key Concept

Industrial applications and chemical inertness of noble gases

Question 8588Question →

Match each industrial separation task on the left with the appropriate physical separation method used in the chemical industry on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Removal of iron impurities from cassiterite matches Magnetic separation; Separation of crude oil into petrol, kerosene, and diesel matches Fractional distillation; Recovery of vegetable oil using hexane matches Solvent extraction; Obtaining pure sugar crystals matches Fractional crystallization.

Each industrial process isolates target substances based on distinct physical properties: magnetic properties allow removal of magnetic iron from non-magnetic tin ore; boiling point differences allow separation of petroleum fractions via fractional distillation; organic solubility allows solvent extraction of vegetable oil; and solubility differences upon cooling allow recovery of pure sugar by fractional crystallization.

Step-by-Step Solution

1

Identify the physical properties exploited in each industrial scenario.

Magnetic susceptibility, differences in boiling points, selective solubility in liquid solvents, and temperature-dependent crystallization/solubility.

Industrial separation techniques rely directly on differences in physical properties between components of raw materials or mixtures.

2

Match each industrial scenario to its corresponding separation technique.

Iron impurities in tin ore are removed magnetically; crude oil fractions are separated by boiling point in a fractionating column; vegetable oil is dissolved out of seeds using solvent extraction; sugar is purified by crystallization from solution.

Connecting physical property differences to their correct industrial unit operations.

Key Concept

Industrial applications of physical separation methods based on unique physical properties.

Question 8589Question →

At $273\text{ K}$ and $1.0\text{ atm}$ , four separate $1.0\text{ mol}$ samples of sulfur dioxide ( $\text{SO}_2$ ), methane ( $\text{CH}_4$ ), nitrogen ( $\text{N}_2$ ), and helium ( $\text{He}$ ) are maintained under identical conditions. Which of these gases will exhibit the greatest deviation from ideal gas behavior?

Sulfur dioxide (

\text{SO}_2

)

Methane (

\text{CH}_4

)

Nitrogen (

\text{N}_2

)

Helium (

\text{He}

)

Show answer & explanation

Answer: Sulfur dioxide ( $\text{SO}_2$ )

Answer

Sulfur dioxide (

\text{SO}_2

) exhibits the greatest deviation from ideal gas behavior due to its strong polar intermolecular forces and relatively large molecular size.

Sulfur dioxide (

\text{SO}_2

) has a bent molecular shape and a permanent dipole moment, creating strong intermolecular dipole-dipole attractions. Additionally, its larger molecular mass and size give it a substantial molecular volume. These properties cause

\text{SO}_2

to deviate most significantly from ideal gas behavior compared to non-polar gases.

Step-by-Step Solution

1

Analyze the postulates of kinetic molecular theory for ideal gases.

Ideal gases assume zero intermolecular attractive forces and zero molecular volume.

Deviations occur when attractive forces become significant or molecular volume cannot be neglected.

2

Evaluate polarity and molecular structure for each given gas species.

\text{SO}_2

is polar (bent structure with dipole moment), while

\text{CH}_4

,

\text{N}_2

, and

\text{He}

are non-polar.

Polar molecules possess permanent dipole-dipole attractions, which are considerably stronger than London dispersion forces.

3

Compare molecular size and constant values (van der Waals

a

and

b

).

\text{SO}_2

has the largest molecular mass and size among the options, giving it the largest van der Waals attraction parameter (

a

).

Stronger attractive forces cause real gas pressure to be significantly less than predicted by

PV = nRT

.

Key Concept

Intermolecular Forces and Molecular Size in Real Gas Deviations

Question 8590Question →

An organic compound $X$ with the molecular formula $\text{C}_4\text{H}_8\text{O}_2$ reacts with aqueous sodium hydroxide upon heating to yield ethanol and a salt $Y$ . What is the IUPAC name of compound $Y$ ?

Sodium ethanoate

Ethanoic acid

Sodium propanoate

Sodium methanoate

Show answer & explanation

Answer: Sodium ethanoate

Answer

Sodium ethanoate

Alkaline hydrolysis of the four-carbon ester ethyl ethanoate (

\text{CH}_3\text{COOCH}_2\text{CH}_3

) with sodium hydroxide breaks the ester linkage to produce ethanol (

\text{C}_2\text{H}_5\text{OH}

) and the sodium salt of ethanoic acid, which is sodium ethanoate (

\text{CH}_3\text{COONa}

).

Step-by-Step Solution

1

Identify the structure of the ester from its molecular formula and hydrolysis alcohol product.

The ester has molecular formula

\text{C}_4\text{H}_8\text{O}_2

and produces ethanol (

\text{C}_2\text{H}_5\text{OH}

). The alkyl group attached to oxygen contains 2 carbons, leaving 2 carbons for the acyl group. Thus, the ester is ethyl ethanoate (

\text{CH}_3\text{COOCH}_2\text{CH}_3

).

Esters have the general formula

\text{RCOOR}'

, where

\text{R}'\text{OH}

is the alcohol component.

2

Determine the products of alkaline hydrolysis.

Heating ethyl ethanoate with aqueous sodium hydroxide (

\text{NaOH}

) cleaves the ester link to yield ethanol and sodium ethanoate (

\text{CH}_3\text{COONa}

).

Alkaline hydrolysis (saponification) yields the alcohol and the alkali metal salt of the alkanoic acid.

Key Concept

Alkaline Hydrolysis (Saponification) of Esters

Question 8591Question →

A

4.0\text{ mol}

sample of hydrogen iodide,

\text{HI}(g)

, is placed in a

2.0\text{ dm}^3

rigid vessel and allowed to dissociate according to the equation:

2\text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g)

If

20\%

of the

\text{HI}(g)

decomposes at equilibrium, what is the value of the equilibrium constant,

K_c

?

0.0156

0.0625

0.0250

0.2500

Show answer & explanation

Answer: $0.0156$

Answer

0.0156

To calculate

K_c

, first find the equilibrium amounts of all species. Starting with

4.0\text{ mol}

of

\text{HI}

, a

20\%

decomposition means

0.8\text{ mol}

decomposes, leaving

3.2\text{ mol}

of

\text{HI}

at equilibrium. Based on the balanced reaction stoichiometry

2\text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g)

, decomposing

0.8\text{ mol}

of

\text{HI}

forms

0.4\text{ mol}

of

\text{H}_2

and

0.4\text{ mol}

of

\text{I}_2

. Converting to concentrations in a

2.0\text{ dm}^3

vessel yields

[\text{HI}] = 1.6\text{ mol dm}^{-3}

,

[\text{H}_2] = 0.2\text{ mol dm}^{-3}

, and

[\text{I}_2] = 0.2\text{ mol dm}^{-3}

. Substituting into

K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2}

gives

\frac{(0.2)(0.2)}{(1.6)^2} = 0.0156

.

Step-by-Step Solution

1

Determine initial moles and amount decomposed at equilibrium

Initial

\text{HI} = 4.0\text{ mol}

. Amount decomposed

= 0.20 \times 4.0\text{ mol} = 0.8\text{ mol}

.

The reaction states that 20% of the initial HI dissociates at equilibrium.

2

Calculate equilibrium mole quantities

Equilibrium

\text{HI} = 4.0 - 0.8 = 3.2\text{ mol}

. According to

2\text{HI} \rightarrow \text{H}_2 + \text{I}_2

, equilibrium

\text{H}_2 = 0.4\text{ mol}

and

\text{I}_2 = 0.4\text{ mol}

.

Every 2 moles of HI decomposed produces 1 mole of H2 and 1 mole of I2.

3

Calculate equilibrium concentrations and solve for

K_c

[\text{HI}] = \frac{3.2}{2.0} = 1.6\text{ mol dm}^{-3}

,

[\text{H}_2] = \frac{0.4}{2.0} = 0.2\text{ mol dm}^{-3}

,

[\text{I}_2] = \frac{0.4}{2.0} = 0.2\text{ mol dm}^{-3}

. Thus,

K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} = \frac{(0.2)(0.2)}{(1.6)^2} = \frac{0.04}{2.56} = 0.015625 \approx 0.0156

.

Substitute equilibrium concentrations into the equilibrium constant expression.

Key Concept

Equilibrium Constant (

K_c

) Calculation from Percent Dissociation

Question 8592Question →

In pea plants, the allele for green pod color ( $G$ ) is completely dominant over the allele for yellow pod color ( $g$ ). What is the expected genotypic ratio of the offspring when two heterozygous green-podded plants ( $Gg$ ) are crossed?

1 : 2 : 1

3 : 1

1 : 1

9 : 3 : 3 : 1

Show answer & explanation

Answer: $1 : 2 : 1$

Answer

The expected genotypic ratio of the offspring is

1 : 2 : 1

(

1\ GG : 2\ Gg : 1\ gg

).

In a monohybrid cross between two heterozygous individuals (

Gg \times Gg

), the Law of Segregation dictates that gametes carry either allele

G

or

g

with equal probability. Combining these gametes produces offspring with genotypes

1/4\ GG

,

2/4\ Gg

, and

1/4\ gg

, resulting in a genotypic ratio of

1 : 2 : 1

.

Step-by-Step Solution

1

Identify parental genotypes

Both parent plants are heterozygous green-podded (

Gg \times Gg

).

The stem specifies crossing two heterozygous green-podded plants.

2

Determine gamete formation according to Mendel's Law of Segregation

Each parent produces two types of gametes:

G

and

g

in equal proportions (

50\%

G

,

50\%

g

).

Alleles separate during gamete formation so each gamete carries only one allele for the trait.

3

Perform the monohybrid cross using a Punnett square

Genotype combinations are

1/4\ GG

,

2/4\ Gg

, and

1/4\ gg

.

Random fertilization yields

1\ GG

,

2\ Gg

, and

1\ gg

among four equal outcomes.

4

State the genotypic ratio

The genotypic ratio is

1 : 2 : 1

.

The ratio compares homozygous dominant (

GG

), heterozygous (

Gg

), and homozygous recessive (

gg

) combinations.

Key Concept

Monohybrid Cross Genotypic Ratio

Estimated Time:45s

Question 8593Question →

According to the rules of binomial nomenclature established by Carl Linnaeus, which of the following represents the correct scientific designation for the African baobab tree when printed in a biology textbook?

*Adansonia digitata*

*Adansonia Digitata*

Adansonia digitata

*digitata Adansonia*

Show answer & explanation

Answer: *Adansonia digitata*

Answer

*Adansonia digitata*

The correct answer strictly adheres to all Linnaean formatting rules: the generic name is capitalized, the specific epithet is lowercase, the genus precedes the species, and both words are italicized in printed text.

Step-by-Step Solution

1

Identify the structural order of Linnaean binomial nomenclature.

The genus name comes first, followed by the specific epithet (species name).

Binomial nomenclature uses a two-part naming system where the genus identifies the broader group and the species identifies the specific organism.

2

Apply capitalization rules to the binomial components.

The genus name ('Adansonia') must start with an uppercase letter, while the specific epithet ('digitata') must be entirely lowercase.

Standard botanical and zoological codes of nomenclature require genus capitalization and lowercase species epithets.

3

Apply typographical styling rules for printed text.

Both the genus and species names must be italicized.

Scientific names are Latin or Latinized terms and must be distinguished from surrounding text through italics in print or underlining in manuscript.

Key Concept

Rules of Linnaean Binomial Nomenclature

Estimated Time:1m 0s

Question 8594Question →

Match each industrial separation requirement on the left with its corresponding separation technique applied in chemical manufacturing on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The extraction of heat-sensitive essential oils matches steam distillation; the concentration of low-grade copper sulfide ore matches froth flotation; the recovery of residual vegetable oil from oilseed cake matches solvent extraction; and the harvesting of sodium chloride from seawater brine matches solar evaporative crystallization.

Each industrial process matches a specific physical property of the mixture: steam distillation isolates thermolabile oils without heat damage; froth flotation separates sulfide ores from rock gangue based on differential surface wettability; solvent extraction uses non-polar liquids like hexane to dissolve residual oils from crushed seeds; and solar evaporative crystallization deposits sodium chloride salts as water evaporates naturally from brine.

Step-by-Step Solution

1

Identify the separation technique suitable for thermolabile volatile compounds.

Essential oils decompose when heated directly to high temperatures, making steam distillation the ideal process because steam lowers the effective boiling temperature of immiscible volatile liquids.

Reduces thermal degradation of delicate plant extracts.

2

Determine the ore beneficiation method based on surface wettability.

Copper sulfide minerals are hydrophobic and readily adhere to air bubbles generated in a froth flotation cell, while hydrophilic rock gangue sinks to the bottom.

Exploits differences in surface tension and hydrophobic properties.

3

Analyze how residual non-polar compounds are extracted from solid agricultural residue.

Solvent extraction using non-polar organic solvents (e.g., hexane) efficiently dissolves lipids from oilseed cakes where mechanical pressing is insufficient.

Utilizes selective solubility in organic solvents.

4

Select the industrial technique for salt recovery from saline water.

Solar evaporative crystallization uses solar radiation to continuously remove water from seawater until sodium chloride exceeds its solubility product and precipitates out.

Cost-effective, large-scale crystallization driven by solar evaporation.

Key Concept

Industrial applications of separation methods based on physical properties such as thermal stability, surface wettability, solubility, and volatility.

Question 8595Question →

Primary aliphatic amines react with nitrous acid ( $HNO_2$ ) at room temperature to yield an alcohol along with rapid effervescence. Which gas is evolved during this reaction?

Nitrogen gas (

N_2

)

Nitrogen(IV) oxide (

NO_2

)

Ammonia gas (

NH_3

)

Hydrogen gas (

H_2

)

Show answer & explanation

Answer: Nitrogen gas ( $N_2$ )

Answer

Nitrogen gas (

N_2

)

Primary aliphatic amines react with cold nitrous acid (

HNO_2

) to form an unstable diazonium compound that decomposes rapidly at room temperature. This decomposition produces an alcohol, water, and liberates nitrogen gas (

N_2

), which is observed as effervescence.

Step-by-Step Solution

1

Identify the functional group and reagent

The reaction involves a primary aliphatic amine (

R-NH_2

) and nitrous acid (

HNO_2

).

Nitrous acid is generated in situ from sodium trioxonitrate(III) (

NaNO_2

) and dilute hydrochloric acid (

HCl

).

2

Write the chemical reaction equation

R-NH_2 + HNO_2 \rightarrow R-OH + N_2 \uparrow + H_2O

The aliphatic diazonium salt formed as an intermediate is highly unstable and decomposes immediately at room temperature.

3

Determine the gas released

The effervescence observed is due to the release of nitrogen gas (

N_2

).

Deamination of aliphatic primary amines yields nitrogen gas as the characteristic gaseous product.

Key Concept

Reaction of primary aliphatic amines with nitrous acid to yield alcohols and nitrogen gas

Estimated Time:1m 0s

Question 8596Question →

In the Contact Process for the manufacture of tetraoxosulfate(VI) acid, $44.8\text{ dm}^3$ of sulfur(IV) oxide ( $SO_2$ ) gas measured at standard temperature and pressure (STP) is reacted with excess oxygen over a vanadium(V) oxide ( $V_2O_5$ ) catalyst. If the catalytic conversion efficiency of $SO_2$ to sulfur(VI) oxide ( $SO_3$ ) is $85\%$ , and all produced $SO_3$ is subsequently absorbed in concentrated $H_2SO_4$ and hydrated to form pure $H_2SO_4$ , what mass of pure $H_2SO_4$ in grams is produced? (Molar mass of $H_2SO_4 = 98\text{ g mol}^{-1}$ , Molar volume of gas at STP $= 22.4\text{ dm}^3\text{ mol}^{-1}$ )

Show answer & explanation

Answer: 166.6

Answer

The mass of pure H2SO4 produced is 166.6 g.

The correct answer of 166.6 g is derived by converting 44.8 dm³ of SO2 at STP to 2.0 moles, taking 85% of that value to find the actual 1.70 moles of SO3 produced, and multiplying by the molar mass of H2SO4 (98 g/mol).

Step-by-Step Solution

1

Calculate the moles of SO2 gas supplied at STP

Moles of SO2 = 2.0 mol

Dividing the gas volume at STP (44.8 dm³) by the molar gas volume (22.4 dm³/mol) gives the molar quantity.

2

Determine theoretical yield of SO3

Theoretical moles of SO3 = 2.0 mol

From the stoichiometric mole ratio in 2SO2 + O2 -> 2SO3, 2 moles of SO2 produce 2 moles of SO3.

3

Calculate actual moles of SO3 produced considering catalytic efficiency

Actual moles of SO3 = 1.70 mol

Multiplying the theoretical yield (2.0 mol) by the 85% conversion efficiency gives the actual yield of 1.70 mol.

4

Calculate mass of H2SO4 produced from the actual SO3 formed

Mass of H2SO4 = 166.6 g

Overall absorption and hydration converts SO3 to H2SO4 in a 1:1 mole ratio (SO3 + H2O -> H2SO4). Multiplying 1.70 mol by molar mass 98 g/mol yields 166.6 g.

Key Concept

Stoichiometry of the Contact Process involving molar volume at STP and percentage conversion efficiency

Question 8597Question →

The hydronium ion ( $H_3O^+$ ) is formed when a hydrogen ion bonds to a water molecule. Based on Valence Shell Electron Pair Repulsion (VSEPR) theory, what is the molecular geometry and the hybridization state of the central oxygen atom in $H_3O^+$ ?

Trigonal pyramidal geometry with

sp^3

hybridization

Tetrahedral geometry with

sp^3

hybridization

Trigonal planar geometry with

sp^2

hybridization

Bent geometry with

sp^2

hybridization

Show answer & explanation

Answer: Trigonal pyramidal geometry with $sp^3$ hybridization

Answer

The central oxygen atom in

H_3O^+

is

sp^3

hybridized, and the ion exhibits a trigonal pyramidal molecular geometry.

In

H_3O^+

, the central oxygen atom forms three single covalent bonds with hydrogen atoms and retains one lone pair. The total of four electron pairs creates a tetrahedral electron arrangement, which corresponds to

sp^3

hybridization. Because one of the four domains is a non-bonding lone pair, the positions of the atomic nuclei define a trigonal pyramidal molecular geometry.

Step-by-Step Solution

1

Calculate valence electron pairs around the central oxygen atom in

H_3O^+

Oxygen supplies 6 valence electrons, three hydrogens supply 3, and subtracting 1 for the positive charge leaves 8 valence electrons (4 electron pairs).

Determining steric number requires counting both bonding and non-bonding electron pairs.

2

Determine the electron-pair geometry and hybridization

Four electron pairs arrange tetrahedrally around oxygen, requiring

sp^3

hybridization.

Four steric domains around a central atom always correspond to an

sp^3

set of hybrid orbitals.

3

Deduce the molecular shape

With 3 bonding pairs and 1 non-bonding lone pair, the molecular shape is trigonal pyramidal.

Molecular geometry reflects only the arrangement of the atomic nuclei around the central atom.

Key Concept

VSEPR Theory and Hybridization of Polyatomic Ions

Question 8598Question →

Unlike pyramids of numbers or biomass, a pyramid of energy in a natural ecosystem can never be inverted. Which of the following best explains why a pyramid of energy always maintains an upright shape?

Energy is progressively lost as heat during metabolic processes at each successive trophic level

Primary producers always have lower total biomass than the primary consumers that feed on them

Energy accumulates and increases in total magnitude as it passes up the food chain

Decomposers continuously recycle dissipated heat energy back to autotrophs for photosynthesis

Show answer & explanation

Answer: Energy is progressively lost as heat during metabolic processes at each successive trophic level

Answer

A pyramid of energy is always upright because energy is progressively lost as heat through metabolic activities at each successive trophic level.

The correct answer correctly identifies that as energy flows from producers to higher trophic levels, a large percentage is lost as heat through respiration and metabolic activities. Consequently, less energy is available to support successive levels, ensuring the pyramid of energy is strictly upright in all ecosystems.

Step-by-Step Solution

1

Recall the second law of thermodynamics as applied to ecological energy flow.

Energy transformations are never 100% efficient, and a significant portion of usable energy is converted to unrecoverable heat at every transfer.

Organisms utilize energy for respiration, movement, and excretion, releasing heat energy into the surroundings.

2

Determine the direction of net energy availability across trophic levels.

Producers store the maximum total energy, while primary, secondary, and tertiary consumers receive exponentially smaller quantities of available energy.

Only about 10% of the energy stored in biomass at one level is converted to biomass at the next level.

3

Conclude why an energy pyramid cannot be inverted.

Since a higher trophic level can never contain more energy than the lower level supplying it, the energy pyramid must always remain upright.

An inverted energy pyramid would require energy creation from nothing, violating physical laws.

Key Concept

Thermodynamic energy loss and the non-inversion of pyramids of energy

Estimated Time:45s

Question 8599Question →

An organic compound $P$ with the molecular formula $\text{C}_4\text{H}_8\text{O}_2$ reacts with aqueous sodium hydroxide upon heating to produce a sodium salt $Q$ and an alkanol $R$ . Complete oxidation of alkanol $R$ with acidified potassium dichromate(VI) yields an alkanoic acid identical to the acid produced when salt $Q$ is acidified with dilute hydrochloric acid. What is the IUPAC name of compound $P$ ?

Methyl propanoate

Ethyl ethanoate

Propyl methanoate

Butanoic acid

Show answer & explanation

Answer: Ethyl ethanoate

Answer

Ethyl ethanoate

Ethyl ethanoate has the molecular formula

\text{C}_4\text{H}_8\text{O}_2

. Alkaline hydrolysis of ethyl ethanoate with sodium hydroxide yields sodium ethanoate (salt

Q

) and ethanol (alkanol

R

). Acidification of sodium ethanoate yields ethanoic acid (2 carbons). Complete oxidation of ethanol with acidified potassium dichromate(VI) also yields ethanoic acid (2 carbons). Because both pathways yield the exact same acid (ethanoic acid), ethyl ethanoate satisfies all conditions.

Step-by-Step Solution

1

Determine the functional group of compound P.

Compound P (

\text{C}_4\text{H}_8\text{O}_2

) reacts with

\text{NaOH}

to yield a salt and an alkanol, identifying

P

as an ester with general formula

\text{R}^1\text{COOR}^2

.

Alkaline hydrolysis (saponification) of esters produces a carboxylate salt and an alkanol.

2

Analyze the carbon distribution from the reaction products.

Acidifying salt

Q

(

\text{R}^1\text{COONa}

) gives alkanoic acid

\text{R}^1\text{COOH}

(containing

n_1 + 1

carbon atoms). Oxidation of primary alkanol

R

(

\text{R}^2\text{OH}

) gives alkanoic acid

\text{R}'\text{COOH}

(containing

n_2

carbon atoms).

Primary alkanols undergo complete oxidation with acidified

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

to produce alkanoic acids with the same number of carbon atoms as the alkanol.

3

Equate the carbon counts of the two acids formed.

Since both processes yield the identical acid, the acid must have 2 carbon atoms (ethanoic acid). Thus,

n_1 + 1 = 2 \implies n_1 = 1

(methyl group

\text{CH}_3-

) and

n_2 = 2

(ethyl group

-\text{C}_2\text{H}_5

).

The total number of carbon atoms in ester

P

is 4 (

1 + 1 + 2 = 4

). Dividing 4 total carbons equally between the acyl and alkoxy portions yields ethanoic acid derivative and ethanol derivative.

4

Deduce the structure and IUPAC name of ester P.

Ester

P

is

\text{CH}_3\text{COOCH}_2\text{CH}_3

, which has the IUPAC name ethyl ethanoate.

The IUPAC name of an ester consists of the alkyl group attached to the oxygen followed by the alkanoate chain.

Key Concept

Ester Saponification and Oxidation of Alkanols

Question 8600Question →

Despite hydrogen fluoride ( $HF$ ) possessing stronger individual hydrogen bonds than water ( $H_2O$ ), liquid water has a significantly higher boiling point ( $100^\circ\text{C}$ ) than liquid hydrogen fluoride ( $19.5^\circ\text{C}$ ). What is the primary structural reason for this higher boiling point in water?

Each water molecule can form an average of four hydrogen bonds in an extensive three-dimensional network, whereas each hydrogen fluoride molecule forms an average of only two hydrogen bonds.

The intramolecular covalent bond between hydrogen and oxygen in water is significantly stronger than the intramolecular covalent bond in hydrogen fluoride.

Oxygen has a higher electronegativity than fluorine, resulting in a larger permanent dipole moment in water molecules.

Hydrogen fluoride molecules in the liquid state are predominantly attracted by weak London dispersion forces rather than hydrogen bonds.

Show answer & explanation

Answer: Each water molecule can form an average of four hydrogen bonds in an extensive three-dimensional network, whereas each hydrogen fluoride molecule forms an average of only two hydrogen bonds.

Answer

Each water molecule can form an average of four hydrogen bonds in an extensive three-dimensional network, whereas each hydrogen fluoride molecule forms an average of only two hydrogen bonds.

The correct answer correctly identifies that the number of hydrogen bonds per molecule determines the bulk physical property. Water possesses 2 hydrogen atoms and 2 lone pairs on oxygen, creating an optimal ratio that allows 4 hydrogen bonds per molecule in a 3D network. Hydrogen fluoride has 3 lone pairs but only 1 hydrogen atom, creating a hydrogen deficit that limits the system to an average of 2 hydrogen bonds per molecule.

Step-by-Step Solution

1

Analyze the structural capacity for hydrogen bonding in hydrogen fluoride (

HF

).

An

HF

molecule has

1

hydrogen atom and

3

lone pairs on the fluorine atom. Because hydrogen atoms are the limiting factor, each molecule can only participate in an average of

2

hydrogen bonds (

1

donated,

1

accepted).

Hydrogen bonding requires both a hydrogen atom bonded to a highly electronegative atom and an available unshared electron pair.

2

Analyze the structural capacity for hydrogen bonding in water (

H_2O

).

An

H_2O

molecule has

2

hydrogen atoms and

2

lone pairs on the central oxygen atom. This 1:1 stoichiometry of hydrogens to lone pairs allows each water molecule to form an average of

4

hydrogen bonds in a 3D network.

The equal number of hydrogen atoms and lone pairs maximizes the overall density of the hydrogen-bonding network.

3

Compare the total intermolecular energy required to separate the molecules during boiling.

Even though a single

F-H\cdots F

bond is stronger than a single

O-H\cdots O

bond, twice as many hydrogen bonds must be broken per mole of water, requiring significantly more thermal energy to vaporize liquid water.

Boiling point depends on the total energy required to overcome all intermolecular attractions in the liquid bulk.

Key Concept

Hydrogen bonding capacity and network density

Estimated Time:1m 0s

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