All practice questions

Question 8601Question →

Match each nitrogen oxide or nitrogen cycle component in Column I with its correct physical property or biological role in Column II.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Dinitrogen monoxide (

N_2O

) matches with 'A sweet-smelling, neutral gas that relights a glowing splint'. Nitrogen dioxide (

NO_2

) matches with 'A reddish-brown, acidic gas that dissolves in water to form a mixture of two acids'. Nitrosomonas bacteria matches with 'Converts soil ammonium ions (

NH_4^+

) into trioxonitrate(III) ions (

NO_2^-

)'. Nitrobacter bacteria matches with 'Converts soil trioxonitrate(III) ions (

NO_2^-

) into trioxonitrate(V) ions (

NO_3^-

)'.

Each nitrogen oxide and nitrifying bacterium is correctly associated with its characteristic chemical behavior or distinct biochemical pathway in the nitrogen cycle.

Step-by-Step Solution

1

Differentiate between the physical and chemical properties of the nitrogen oxides.

N_2O

is neutral and sweet-smelling while supporting combustion.

NO_2

is acidic, reddish-brown, and forms

HNO_2

and

HNO_3

upon reaction with water.

Oxides of nitrogen vary in color, acidity, and combustion-supporting capabilities depending on the oxidation state of nitrogen.

2

Differentiate the roles of nitrifying bacteria in the nitrogen cycle.

Nitrosomonas oxidizes ammonium to nitrite (

NO_2^-

), whereas Nitrobacter oxidizes nitrite (

NO_2^-

) to nitrate (

NO_3^-

).

Nitrification proceeds in two distinct enzymatic steps mediated by specialized microbial species.

Key Concept

Properties of Nitrogen Oxides and Biological Nitrification Stages

Question 8602Question →

In a monohybrid cross between two heterozygous green-pod pea plants ( $Gg$ ), Mendel's Law of Segregation predicts that $75\%$ of the $F_2$ offspring will possess the heterozygous ( $Gg$ ) genotype.

True

False

Show answer & explanation

Answer: False

Answer

False

The statement is false because a monohybrid cross of two heterozygous individuals (

Gg \times Gg

) yields a genotypic ratio of

1\ GG : 2\ Gg : 1\ gg

. Exactly

50\%

(

2

out of

4

) of the

F_2

offspring inherit the heterozygous genotype (

Gg

). The claimed

75\%

proportion applies to the dominant phenotype, improperly interchanging genotypic composition with phenotypic expression.

Step-by-Step Solution

1

Identify parental genotypes and apply Mendel's First Law

Both parents are heterozygous (

Gg \times Gg

). During gamete formation, alleles segregate so each gamete receives either

G

or

g

with equal probability (

50\%

).

Mendel's Law of Segregation states that allele pairs separate during gametogenesis.

2

Determine expected offspring genotypic proportions

Random combination of gametes yields

1/4\ GG

,

2/4\ Gg

, and

1/4\ gg

. The genotypic ratio is

1 : 2 : 1

.

A standard Punnett square for a monohybrid cross combines

G

and

g

alleles from both parents.

3

Evaluate the statement's genotypic percentage claim

The heterozygous (

Gg

) fraction is

2/4

, which equals

50\%

. The

75\%

value corresponds to the dominant phenotype (

GG + Gg = 25\% + 50\% = 75\%

).

The statement incorrectly equates the dominant phenotypic percentage (

75\%

) with the heterozygous genotypic percentage (

50\%

).

Key Concept

Mendel's Law of Segregation and Monohybrid Genotypic vs Phenotypic Ratios

Estimated Time:1m 30s

Question 8603Question →

A biology student investigating edaphic factors collected $100\text{ cm}^3$ of an undisturbed garden soil sample in a graduated cylinder and added $100\text{ cm}^3$ of water. After stirring thoroughly and allowing all trapped air bubbles to escape, the final settled volume of the soil-water mixture measured $160\text{ cm}^3$ . What is the percentage porosity of this soil sample?

40\%

60\%

25\%

16\%

Show answer & explanation

Answer: $40\%$

Answer

The percentage porosity of the garden soil sample is

40\%

.

Soil porosity represents the percentage of pore space (voids filled with air or water) within a given volume of soil. When

100\text{ cm}^3

of water is added to

100\text{ cm}^3

of dry soil, the expected combined volume without pores would be

200\text{ cm}^3

. The actual mixture volume of

160\text{ cm}^3

indicates that

40\text{ cm}^3

of air was displaced by water. Expressed as a percentage of the original

100\text{ cm}^3

soil volume, the porosity is

\frac{40\text{ cm}^3}{100\text{ cm}^3} \times 100\% = 40\%

.

Step-by-Step Solution

1

Calculate the theoretical total volume if no air spaces were present.

Theoretical volume =

\text{Volume of soil} + \text{Volume of water} = 100\text{ cm}^3 + 100\text{ cm}^3 = 200\text{ cm}^3

.

Water fills the pore spaces previously occupied by air in the soil sample.

2

Determine the volume of air spaces (pores) in the soil sample.

Volume of air spaces =

\text{Theoretical volume} - \text{Actual mixture volume} = 200\text{ cm}^3 - 160\text{ cm}^3 = 40\text{ cm}^3

.

The reduction in expected total volume corresponds directly to the volume of displaced air.

3

Calculate the percentage porosity using the ratio of pore space volume to initial soil volume.

\text{Percentage Porosity} = \left(\frac{\text{Volume of air spaces}}{\text{Initial volume of soil}}\right) \times 100\% = \left(\frac{40\text{ cm}^3}{100\text{ cm}^3}\right) \times 100\% = 40\%

.

Porosity measures the percentage of total soil volume occupied by pore spaces.

Key Concept

Soil Porosity Measurement (Edaphic Factor Analysis)

Question 8604Question →

A marine biologist analyzed the ecological structure of a coastal reef ecosystem and categorized its functional components into four groups:

Component	Description
P	Unicellular phytoplankton performing carbon fixation in the photic zone
Q	Zooplankton and small herbivorous fish feeding on phytoplankton
R	Predatory fish consuming herbivorous aquatic organisms
S	Benthic bacteria and fungi decomposing organic fallout

During a seasonal survey, the standing crop biomass of component Q was recorded to be higher than that of component P. Which statement correctly evaluates the structural dynamics and thermodynamic principles of this ecosystem?

The ecosystem remains energetically stable because phytoplankton have a rapid turnover rate and high productivity, ensuring that the pyramid of energy remains upright despite an inverted biomass snapshot.

Energy flows bidirectionally between component P and component Q, causing the overall pyramid of energy to become permanently inverted in marine environments.

Component S operates as primary producers by ingesting solid detritus waste holozoically to synthesize new organic matter for component P.

The inverted biomass ratio between primary producers and primary consumers forces the ecosystem to restart via primary succession on bare rock substrate.

Show answer & explanation

Answer: The ecosystem remains energetically stable because phytoplankton have a rapid turnover rate and high productivity, ensuring that the pyramid of energy remains upright despite an inverted biomass snapshot.

Answer

The ecosystem remains energetically stable because phytoplankton have a rapid turnover rate and high productivity, ensuring that the pyramid of energy remains upright despite an inverted biomass snapshot.

In aquatic ecosystems, primary producers (phytoplankton) have a very high rate of reproduction and turnover. Even though their instantaneous standing crop biomass is lower than that of primary consumers (zooplankton), their cumulative rate of primary production supplies ample energy to sustain the consumer population. Thus, the pyramid of biomass appears inverted, while the pyramid of energy remains strictly upright and unidirectional.

Step-by-Step Solution

1

Analyze the functional roles of ecosystem components P, Q, R, and S.

P is the primary producer, Q is the primary consumer, R is the secondary consumer, and S represents the decomposers.

Establishing trophic roles is required to evaluate energy flow and biomass relationships.

2

Evaluate the relationship between standing crop biomass and energy flow in marine ecosystems.

Phytoplankton (P) reproduce and are consumed extremely rapidly. Although their standing crop biomass at any instant is small relative to zooplankton (Q), their total energy production over time is much larger.

Pyramids of standing crop biomass in aquatic systems can be inverted, but pyramids of energy are always upright due to the Second Law of Thermodynamics.

3

Identify the correct physical law governing energy pyramids.

Energy transfer is strictly unidirectional and decreases at successive trophic levels due to metabolic heat loss.

An inverted energy pyramid would violate fundamental laws of thermodynamics.

Key Concept

Standing crop biomass vs. energy flow in ecosystem structure

Question 8605Question →

The table below shows the solubility of potassium trioxonitrate(V), $\text{KNO}_3$ , in water at two different temperatures:

Temperature ( $^\circ\text{C}$ )	Solubility ( $\text{g / } 100\text{ g } \text{H}_2\text{O}$ )
60	110.0
20	32.0

What mass of $\text{KNO}_3$ will deposit when $420.0\text{ g}$ of a saturated solution of $\text{KNO}_3$ at $60^\circ\text{C}$ is cooled to $20^\circ\text{C}$ ?

78.0\text{ g}

156.0\text{ g}

220.0\text{ g}

327.6\text{ g}

Show answer & explanation

Answer: $156.0\text{ g}$

Answer

156.0\text{ g}

At

60^\circ\text{C}

, every

210.0\text{ g}

of saturated solution contains

100.0\text{ g}

of water and

110.0\text{ g}

of solute. Thus,

420.0\text{ g}

of solution contains

200.0\text{ g}

of water and

220.0\text{ g}

of dissolved

\text{KNO}_3

. Upon cooling to

20^\circ\text{C}

, the solubility drops to

32.0\text{ g}

per

100\text{ g}

of water, so

200.0\text{ g}

of water can retain only

64.0\text{ g}

of solute. The excess solute that crystallizes out is

220.0\text{ g} - 64.0\text{ g} = 156.0\text{ g}

.

Step-by-Step Solution

1

Determine the composition of a saturated solution at

60^\circ\text{C}

At

60^\circ\text{C}

,

100.0\text{ g}

of water dissolves

110.0\text{ g}

of

\text{KNO}_3

, yielding

100.0 + 110.0 = 210.0\text{ g}

of saturated solution.

Solubility is expressed per

100\text{ g}

of water, so total solution mass is the sum of solvent and solute masses.

2

Calculate the mass of water and solute in

420.0\text{ g}

of saturated solution at

60^\circ\text{C}

Mass of water =

420.0 \times \frac{100.0}{210.0} = 200.0\text{ g}

. Mass of

\text{KNO}_3

=

420.0 \times \frac{110.0}{210.0} = 220.0\text{ g}

.

Scaling the ratio of components to the given solution mass.

3

Calculate the mass of solute remaining dissolved at

20^\circ\text{C}

At

20^\circ\text{C}

,

100.0\text{ g}

of water holds

32.0\text{ g}

of

\text{KNO}_3

. Therefore,

200.0\text{ g}

of water holds

2 \times 32.0 = 64.0\text{ g}

of

\text{KNO}_3

.

Solubility at the lower temperature determines the maximum solute that stays in solution.

4

Calculate the mass of salt crystallized (deposited)

Mass deposited =

220.0\text{ g} - 64.0\text{ g} = 156.0\text{ g}

.

Subtracting the solute remaining in solution from the initial mass of dissolved solute.

Key Concept

Solubility curves and fractional crystallization calculations

Estimated Time:1m 30s

Question 8606Question →

An organic compound $X$ with the molecular formula $C_3H_9N$ reacts with cold nitrous acid ( $HNO_2$ ) at $0-5^\circ\text{C}$ to produce a yellow, oily liquid without the evolution of nitrogen gas. Which of the following is the IUPAC name of compound $X$ ?

Propan-1-amine

N

-methylethanamine

Propan-2-amine

N,N

-dimethylmethanamine

Show answer & explanation

Answer: $N$ -methylethanamine

Answer

The correct compound is

N

-methylethanamine because secondary aliphatic amines react with nitrous acid (

HNO_2

) at low temperatures to produce insoluble, yellow oily

N

-nitrosamines without liberating nitrogen gas.

Secondary aliphatic amines such as

N

-methylethanamine react with cold nitrous acid (

HNO_2

) to form

N

-nitrosamines. These compounds are insoluble in water and appear as yellow oily liquids. Because no aliphatic diazonium intermediate breaks down to release gas, no nitrogen gas effervescence is observed.

Step-by-Step Solution

1

Classify the given structural isomers of

C_3H_9N

by amine degree.

Propan-1-amine and propan-2-amine are primary (

1^\circ

) amines;

N

-methylethanamine is a secondary (

2^\circ

) amine;

N,N

-dimethylmethanamine is a tertiary (

3^\circ

) amine.

Amine classification determines the distinct reaction pathway and observable products with nitrous acid.

2

Analyze the reaction behavior of each amine class with nitrous acid (

HNO_2

) at

0-5^\circ\text{C}

.

Primary amines evolve

N_2

gas and form alcohols; secondary amines form yellow oily

N

-nitrosamines without gas evolution; tertiary amines form soluble nitrite salts.

Nitrous acid is used as a qualitative reagent to distinguish between

1^\circ

,

2^\circ

, and

3^\circ

amines.

3

Match the observation (yellow oily liquid, no nitrogen gas) to the correct compound.

The observation corresponds uniquely to a secondary amine, which is

N

-methylethanamine.

Only secondary amines undergo nitrosation at the nitrogen atom to form neutral, oily

N

-nitrosamine layers.

Key Concept

Distinction tests for primary, secondary, and tertiary amines using nitrous acid (

HNO_2

)

Question 8607Question →

Excess solid carbon,

\text{C}(s)

, is allowed to react with

1.0\text{ mol}

of carbon dioxide gas,

\text{CO}_2(g)

, in a sealed

2.0\text{ dm}^3

reaction vessel at a constant high temperature according to the equation:

\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g)

If

1.0\text{ mol}

of carbon monoxide gas,

\text{CO}(g)

, is present at equilibrium, calculate the numerical value of the equilibrium constant,

K_c

, in

\text{mol dm}^{-3}

.

Show answer & explanation

Answer: 1

Answer

1.0

To find

K_c

, convert initial and equilibrium amounts to molar concentrations by dividing by the volume (

2.0\text{ dm}^3

). Initial concentration of

\text{CO}_2

is

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

. At equilibrium,

[\text{CO}] = \frac{1.0}{2.0} = 0.50\text{ mol dm}^{-3}

. From the

1:2

stoichiometry,

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

of

\text{CO}_2

reacted, leaving

[\text{CO}_2] = 0.25\text{ mol dm}^{-3}

. For a heterogeneous system, solid carbon is excluded from the equilibrium constant expression. Thus,

K_c = \frac{[\text{CO}]^2}{[\text{CO}_2]} = \frac{(0.50)^2}{0.25} = 1.0\text{ mol dm}^{-3}

.

Step-by-Step Solution

1

Calculate initial concentrations from the given moles and volume

Initial

[\text{CO}_2] = 0.50\text{ mol dm}^{-3}

, initial

[\text{CO}] = 0.0\text{ mol dm}^{-3}

Concentration is calculated using

C = \frac{n}{V}

where

V = 2.0\text{ dm}^3

.

2

Determine equilibrium concentrations using stoichiometric mole ratios

Equilibrium

[\text{CO}] = 0.50\text{ mol dm}^{-3}

and

[\text{CO}_2] = 0.25\text{ mol dm}^{-3}

Producing

1.0\text{ mol}

of

\text{CO}

consumes

0.50\text{ mol}

of

\text{CO}_2

. Remaining moles of

\text{CO}_2 = 1.0 - 0.50 = 0.50\text{ mol}

.

3

Write the equilibrium constant expression for the heterogeneous system

K_c = \frac{[\text{CO}]^2}{[\text{CO}_2]}

Pure solids like

\text{C}(s)

have constant activity and are omitted from the

K_c

expression.

4

Substitute concentrations into the

K_c

expression and solve

K_c = \frac{(0.50)^2}{0.25} = 1.0\text{ mol dm}^{-3}

Calculates the numerical value of the equilibrium constant.

Key Concept

Heterogeneous equilibrium constant expression and ICE calculations

Question 8608Question →

When propanal ( $CH_3CH_2CHO$ ) is warmed with Fehling's solution, a brick-red precipitate is formed. What is the chemical formula of this precipitate, and what organic product is formed from the reaction?

Cu_2O

and propanoic acid

CuO

and propan-1-ol

Cu_2O

and propanone

Cu(OH)_2

and propanoic acid

Show answer & explanation

Answer: $Cu_2O$ and propanoic acid

Answer

The precipitate formed is copper(I) oxide (

Cu_2O

) and the organic oxidation product is propanoic acid.

Fehling's solution contains alkaline solution of copper(II) sulfate complexed with tartrate ions. When heated with an alkanal like propanal, the aldehyde group is oxidized to a carboxylic acid (propanoic acid), while

Cu^{2+}

is reduced to insoluble brick-red copper(I) oxide (

Cu_2O

).

Step-by-Step Solution

1

Identify the functional group and reaction type

Propanal is an alkanal containing the terminal carbonyl group (

-CHO

). Warming with Fehling's solution causes a redox distinction test.

Alkanals act as reducing agents and readily undergo oxidation, whereas Fehling's solution contains complexed

Cu^{2+}

ions.

2

Determine the inorganic reduction product (precipitate)

The deep blue

Cu^{2+}

ions are reduced to copper(I) oxide (

Cu_2O

), which precipitates as a insoluble brick-red solid.

The reduction half-reaction is

2Cu^{2+} + 2OH^- + 2e^- → Cu_2O + H_2O

.

3

Determine the organic oxidation product

Propanal (

CH_3CH_2CHO

) is oxidized to propanoic acid (

CH_3CH_2COOH

).

Oxidation of an alkanal inserts an oxygen atom into the

C-H

bond of the aldehyde group without altering the carbon backbone length.

Key Concept

Fehling's distinction test for alkanals and their oxidation products

Question 8609Question →

Aluminium resists atmospheric corrosion better than iron because it rapidly forms a tough, non-porous coating of aluminium oxide ( $Al_2O_3$ ) that prevents oxygen and moisture from reaching the underlying metal.

True

False

Show answer & explanation

Answer: True

Answer

The statement is TRUE. Aluminium forms an adherent, non-porous protective oxide film (

Al_2O_3

) upon exposure to air, which renders the metal passive and prevents further atmospheric corrosion.

The statement is true because aluminium undergoes passivation in atmospheric air, forming an impermeable, self-healing oxide film (

Al_2O_3

) that insulates the underlying metal from environmental oxygen and moisture.

Step-by-Step Solution

1

Analyze the surface reactivity of aluminium exposed to atmospheric air.

Aluminium rapidly reacts with oxygen to form a thin surface layer of aluminium oxide (

Al_2O_3

).

Aluminium has a high affinity for oxygen due to its electropositive nature.

2

Compare the physical properties of aluminium oxide with those of iron rust.

Unlike iron rust (

Fe_2O_3 \cdot xH_2O

), which is porous, brittle, and flakes off, the

Al_2O_3

coating is continuous, tough, and impermeable.

The non-porous nature of

Al_2O_3

prevents further diffusion of oxygen and moisture to the unreacted metal underneath.

Key Concept

Metal Passivation via Protective Oxide Layer

Estimated Time:45s

Question 8610Question →

In an industrial chlor-alkali membrane cell, concentrated sodium chloride solution (brine) is electrolyzed using a constant current of $19.3\text{ A}$ for $50\text{ minutes}$ . What is the mass, in grams, of sodium hydroxide ( $\text{NaOH}$ ) produced in the solution? [Molar mass of $\text{NaOH} = 40.0\text{ g mol}^{-1}$ , Faraday constant $F = 96,500\text{ C mol}^{-1}$ ]

Show answer & explanation

Answer: 24

Answer

The mass of sodium hydroxide produced is

24.0\text{ g}

.

Converting the electrolysis time to seconds (

3000\text{ s}

) yields a total charge of

Q = 19.3\text{ A} \times 3000\text{ s} = 57,900\text{ C}

. Dividing by Faraday's constant (

96,500\text{ C mol}^{-1}

) gives

0.6\text{ mol}

of electrons. In the chlor-alkali process, reduction of water produces

1\text{ mol}

of

\text{OH}^-

ions per mole of electrons, producing

0.6\text{ mol}

of

\text{NaOH}

. Multiplying by the molar mass (

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

) gives a mass of

24.0\text{ g}

.

Step-by-Step Solution

1

Convert electrolysis time into seconds and calculate total charge.

Q = 19.3\text{ A} \times (50 \times 60\text{ s}) = 57,900\text{ C}

.

Faraday's equations require time in seconds (

Q = I \times t

).

2

Calculate the moles of electrons transferred using the Faraday constant.

n(e^-) = \frac{57,900\text{ C}}{96,500\text{ C mol}^{-1}} = 0.6\text{ mol}

.

One Faraday (

96,500\text{ C}

) corresponds to one mole of electrons.

3

Determine the stoichiometry of the cathode reaction.

Cathode reaction:

2\text{H}_2\text{O}_{(l)} + 2e^- \rightarrow \text{H}_{2(g)} + 2\text{OH}^-_{(aq)}

. Thus,

1\text{ mol } e^-

forms

1\text{ mol } \text{OH}^-

, giving

0.6\text{ mol}

of

\text{NaOH}

.

During brine electrolysis, water is preferentially reduced at the cathode, generating hydroxide ions that pair with sodium ions.

4

Calculate the mass of sodium hydroxide produced.

\text{Mass} = 0.6\text{ mol} \times 40.0\text{ g mol}^{-1} = 24.0\text{ g}

.

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

Key Concept

Industrial Electrolysis Stoichiometry (Chlor-Alkali Process) and Faraday's First Law

Estimated Time:1m 30s

Question 8611Question →

A

2.50\text{ g}

sample of impure iron(III) oxide,

\text{Fe}_2\text{O}_3

, was completely reduced by excess carbon(II) oxide gas to yield

1.40\text{ g}

of pure iron metal according to the equation:

\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g)

What is the percentage purity of the iron(III) oxide sample? [

\text{Fe} = 56, \text{O} = 16, \text{C} = 12

]

80.0\%

70.0\%

56.0\%

40.0\%

Show answer & explanation

Answer: $80.0\%$

Answer

The percentage purity of the iron(III) oxide sample is 80.0%.

According to the balanced chemical equation,

1\text{ mole}

of

\text{Fe}_2\text{O}_3

(

160\text{ g}

) yields

2\text{ moles}

of

\text{Fe}

(

112\text{ g}

). To obtain

1.40\text{ g}

of pure iron metal, the mass of pure

\text{Fe}_2\text{O}_3

required is

1.40 \times \frac{160}{112} = 2.00\text{ g}

. The percentage purity is calculated by taking the mass of pure

\text{Fe}_2\text{O}_3

divided by the total sample mass (

2.50\text{ g}

) multiplied by

100

, yielding

80.0\%

.

Step-by-Step Solution

1

Calculate the molar mass of iron(III) oxide (

\text{Fe}_2\text{O}_3

) and total mass of iron produced per mole.

Molar mass of

\text{Fe}_2\text{O}_3 = 2(56) + 3(16) = 112 + 48 = 160\text{ g/mol}

. Mass of

2\text{ moles}

of

\text{Fe} = 2 \times 56 = 112\text{ g}

.

Stoichiometry shows

1\text{ mole}

of

\text{Fe}_2\text{O}_3

(

160\text{ g}

) produces

2\text{ moles}

of

\text{Fe}

(

112\text{ g}

).

2

Determine the mass of pure

\text{Fe}_2\text{O}_3

in the sample required to produce

1.40\text{ g}

of

\text{Fe}

.

\text{Mass of pure Fe}_2\text{O}_3 = 1.40\text{ g Fe} \times \frac{160\text{ g Fe}_2\text{O}_3}{112\text{ g Fe}} = 2.00\text{ g}

.

Mass proportions allow determination of the mass of reacting pure compound.

3

Calculate the percentage purity of the sample.

\text{Percentage purity} = \left(\frac{2.00\text{ g}}{2.50\text{ g}}\right) \times 100 = 80.0\%

.

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

Key Concept

Determining percentage purity using stoichiometric mass calculations

Estimated Time:1m 30s

Question 8612Question →

Pair each basic genetic term listed on the left with its correct biological description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Heterozygous matches with possessing two non-identical alleles; Genotype matches with the specific allele composition; Dominant allele matches with an allele that masks its contrasting form; Phenotype matches with the observable physical traits.

Each genetic term matches its corresponding definition precisely: Heterozygous indicates possessing two non-identical alleles at a locus; Genotype represents the organism's specific allele composition; Dominant allele describes an allele that masks the expression of its contrasting alternative form; and Phenotype refers to the observable physical or physiological traits.

Step-by-Step Solution

1

Define Heterozygous

Heterozygous refers to having different alleles at a given gene locus.

The prefix 'hetero-' means different, indicating non-identical alleles.

2

Define Genotype

Genotype represents the internal genetic code or allele constitution.

Genotype specifies the genetic information rather than the outward appearance.

3

Define Dominant allele

A dominant allele masks the phenotypic effect of its alternative allele.

Dominance means the trait is expressed whenever at least one copy of the allele is present.

4

Define Phenotype

Phenotype represents the observable attributes and physical characteristics.

Phenotype is the external expression resulting from genotypic and environmental factors.

Key Concept

Basic Genetics Terminology and Concepts

Estimated Time:1m 0s

Question 8613Question →

Match each organism or ecological role on the left with its corresponding trophic level or function in energy flow on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Green plants and phytoplankton match with Primary producers (Trophic Level 1); Herbivores such as grasshoppers match with Primary consumers (Trophic Level 2); Carnivores such as frogs match with Secondary consumers (Trophic Level 3); Saprophytic fungi and bacteria match with Decomposers recycling organic matter.

Organisms are categorized into trophic levels based on their source of energy: photosynthetic autotrophs are primary producers (Level 1), plant-eaters are primary consumers (Level 2), animal-eaters preying on herbivores are secondary consumers (Level 3), and decay organisms recycle organic matter as decomposers.

Step-by-Step Solution

1

Identify the energy-capturing organisms in an ecosystem.

Green plants and phytoplankton capture solar energy to synthesize food, placing them at Trophic Level 1 as primary producers.

Primary producers form the foundational trophic level in all food chains.

2

Identify organisms that directly feed on producers.

Herbivores such as grasshoppers consume plant matter directly, placing them at Trophic Level 2 as primary consumers.

Direct consumers of autotrophs occupy the second trophic position.

3

Identify organisms that prey on primary consumers.

Carnivores such as frogs feed on primary consumers (herbivores), placing them at Trophic Level 3 as secondary consumers.

Predators of herbivores occupy the third trophic position in energy transfer.

4

Identify organisms responsible for breaking down dead organic waste.

Saprophytic fungi and bacteria digest non-living organic matter, functioning as decomposers.

Decomposers facilitate nutrient recycling back into abiotic ecosystem pools.

Key Concept

Trophic level classification and functional roles in food chains

Question 8614Question →

Two synthetic polymers, Polymer X (a polyamide formed by condensation polymerization) and Polymer Y (a polyalkene formed by addition polymerization), are buried in soil at a waste disposal site. Over time, Polymer X undergoes gradual microbial breakdown, whereas Polymer Y remains chemically intact for decades. What is the primary structural reason for this marked difference in biodegradability?

Polymer X contains hydrolyzable polar linkages vulnerable to enzymatic cleavage, whereas Polymer Y possesses a non-polar carbon-carbon backbone resistant to microbial attack.

Polymer Y has a significantly higher average molecular mass than Polymer X, preventing soil bacteria from absorbing its polymer chains.

Polymer Y undergoes spontaneous cross-linking in damp soil, forming a rigid network that repels microbial enzymes.

Polymer X is synthesized from naturally occurring monomers, while Polymer Y is made exclusively from toxic petrochemically derived alkene monomers.

Show answer & explanation

Answer: Polymer X contains hydrolyzable polar linkages vulnerable to enzymatic cleavage, whereas Polymer Y possesses a non-polar carbon-carbon backbone resistant to microbial attack.

Answer

Polymer X contains hydrolyzable polar linkages vulnerable to enzymatic cleavage, whereas Polymer Y possesses a non-polar carbon-carbon backbone resistant to microbial attack.

Biodegradability depends on the presence of polar functional linkages (such as amide –CO–NH– or ester –CO–O– bonds) within the main chain that soil micro-organisms can cleave via enzymatic hydrolysis. Polymer X (a polyamide) contains these hydrolyzable bonds. In contrast, Polymer Y (an addition polyalkene) has a saturated, non-polar carbon-carbon backbone that micro-organisms lack the specific enzymes to decompose, causing it to persist indefinitely.

Step-by-Step Solution

1

Analyze the backbone chemical structure of Polymer X (condensation polyamide) versus Polymer Y (addition polyalkene).

Polymer X features repeating amide linkages (–CO–NH–), while Polymer Y consists of a continuous chain of single carbon-carbon bonds (–C–C–).

Chemical degradation in nature relies on microbial enzymes recognizing specific functional groups.

2

Evaluate the mechanism of microbial breakdown for these functional groups.

Microbial enzymes secrete hydrolases that break polar amide and ester bonds via hydrolysis, whereas non-polar C-C backbones lack reactive sites for enzymatic cleavage.

Enzymatic hydrolysis requires polar or reactive functional groups embedded within the polymer chain.

3

Select the option that correctly attributes biodegradability to structural functional group susceptibility.

The option stating that Polymer X contains hydrolyzable polar linkages while Polymer Y has a non-polar carbon-carbon backbone is correct.

This represents the fundamental chemical basis of polymer degradation in the environment.

Key Concept

Structural Basis of Polymer Biodegradability

Estimated Time:1m 30s

Question 8615Question →

Match each salt in aqueous solution to the correct chemical description of its hydrolysis behavior and resulting pH at $25^\circ\text{C}$ .

Click a left item, then click its matching right item

Show answer & explanation

Answer

Ammonium sulfate pairs with the acidic cation-hydrolyzing description; Sodium propanoate pairs with the alkaline anion-hydrolyzing description; Potassium nitrate pairs with the neutral non-hydrolyzing description; Ammonium cyanide pairs with the alkaline dual-hydrolyzing description where

K_b > K_a

.

The solution pH resulting from salt hydrolysis depends directly on the relative strengths of the parent acid and base. Salts derived from weak bases and strong acids yield acidic solutions via cation hydrolysis. Salts derived from strong bases and weak acids yield alkaline solutions via anion hydrolysis. Salts of strong acids and strong bases do not undergo net hydrolysis, remaining neutral. For salts derived from both a weak acid and a weak base, both ions undergo hydrolysis, and the solution acidity or alkalinity is determined by comparing the

K_b

of the weak base to the

K_a

of the weak acid.

Step-by-Step Solution

1

Identify the parent acid and parent base for each salt.

(NH_4)_2SO_4

comes from

NH_3

(weak base) and

H_2SO_4

(strong acid).

CH_3CH_2COONa

comes from

NaOH

(strong base) and

CH_3CH_2COOH

(weak acid).

KNO_3

comes from

KOH

(strong base) and

HNO_3

(strong acid).

NH_4CN

comes from

NH_3

(weak base) and

HCN

(weak acid).

The strength of parent acids and bases determines which ions undergo hydrolysis in water.

2

Determine which ion hydrolyzes for single-weak component salts.

In

(NH_4)_2SO_4

,

NH_4^+

hydrolyzes to produce

H_3O^+

(acidic,

pH < 7

). In

CH_3CH_2COONa

,

CH_3CH_2COO^-

hydrolyzes to produce

OH^-

(alkaline,

pH > 7

). In

KNO_3

, neither ion hydrolyzes (neutral,

pH = 7

).

Conjugate ions of weak species react with water, whereas conjugate ions of strong species do not hydrolyze.

3

Compare ionization constants for the weak acid-weak base salt.

For

NH_4CN

, compare

K_b(NH_3) = 1.8 \times 10^{-5}

with

K_a(HCN) = 6.2 \times 10^{-10}

. Since

K_b > K_a

,

CN^-

anion hydrolysis produces more

OH^-

than

NH_4^+

cation hydrolysis produces

H_3O^+

, resulting in an alkaline solution (

pH > 7

).

When both ions hydrolyze, the relative magnitudes of

K_a

and

K_b

govern whether

OH^-

or

H_3O^+

is in excess.

Key Concept

Salt Hydrolysis and Solution Acidity/Alkalinity

Estimated Time:1m 30s

Question 8616Question →

Excess solid carbon,

\text{C}(s)

, and

4.0\text{ mol}

of carbon dioxide gas,

\text{CO}_2(g)

, are introduced into an evacuated

2.0\text{ dm}^3

rigid reaction vessel at a constant temperature. The system reaches equilibrium according to the reaction equation:

\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g)

If the equilibrium concentration of carbon monoxide gas,

\text{CO}(g)

, is determined to be

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

, calculate the numerical value of the equilibrium constant,

K_c

, in

\text{mol dm}^{-3}

.

Show answer & explanation

Answer: 0.4

Answer

The numerical value of the equilibrium constant,

K_c

, is

0.4

(or

0.40

).

To find

K_c

, first convert moles of

\text{CO}_2

to initial concentration:

[\text{CO}_2]_0 = \frac{4.0\text{ mol}}{2.0\text{ dm}^3} = 2.00\text{ mol dm}^{-3}

. From stoichiometry (

\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g)

), creating

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

of

\text{CO}

consumes

\frac{0.80}{2} = 0.40\text{ mol dm}^{-3}

of

\text{CO}_2

. At equilibrium,

[\text{CO}_2]_{eq} = 2.00 - 0.40 = 1.60\text{ mol dm}^{-3}

. Omitting the solid carbon

\text{C}(s)

from the expression gives

K_c = \frac{[\text{CO}]^2}{[\text{CO}_2]} = \frac{(0.80)^2}{1.60} = 0.40\text{ mol dm}^{-3}

.

Step-by-Step Solution

1

Calculate initial concentration of reactant gas

[\text{CO}_2]_0 = 2.00\text{ mol dm}^{-3}

Concentration must be calculated in

\text{mol dm}^{-3}

by dividing moles by vessel volume (

2.0\text{ dm}^3

).

2

Determine equilibrium concentrations using stoichiometric ratios

[\text{CO}_2]_{eq} = 1.60\text{ mol dm}^{-3}

and

[\text{CO}]_{eq} = 0.80\text{ mol dm}^{-3}

The stoichiometric mole ratio of

\text{CO}_2

to

\text{CO}

is

1:2

. Thus, consuming

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

of

\text{CO}_2

yields

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

of

\text{CO}

.

3

Formulate the equilibrium constant expression for the heterogeneous reaction

K_c = \frac{[\text{CO}]^2}{[\text{CO}_2]}

Pure solids such as

\text{C}(s)

have constant concentration and are omitted from the equilibrium constant expression.

4

Calculate the value of

K_c

K_c = \frac{(0.80)^2}{1.60} = 0.40

Substitute equilibrium concentrations into the

K_c

expression and solve.

Key Concept

Equilibrium constant expression for heterogeneous equilibria and ICE table calculations.

Question 8617Question →

A standard solution is prepared by dissolving $4.0\text{ g}$ of pure solid sodium hydroxide ( $\text{NaOH}$ ) in distilled water to make $1.0\text{ dm}^3$ of solution. What is the molar concentration of this solution? $[\text{Na} = 23,\text{ O} = 16,\text{ H} = 1]$

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

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

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

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

Show answer & explanation

Answer: $0.10\text{ mol dm}^{-3}$

Answer

The molar concentration of the sodium hydroxide solution is

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

.

The solution containing

4.0\text{ g}

of

\text{NaOH}

per

\text{dm}^3

is converted to molar concentration by dividing by the molar mass of

\text{NaOH}

(

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

), yielding

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

.

Step-by-Step Solution

1

Calculate the molar mass of sodium hydroxide (

\text{NaOH}

).

Molar mass

= 23 + 16 + 1 = 40\text{ g mol}^{-1}

.

Molar mass is required to convert mass to moles.

2

Calculate the concentration in

\text{mol dm}^{-3}

by dividing mass concentration by molar mass.

\text{Molar concentration} = \frac{4.0\text{ g dm}^{-3}}{40\text{ g mol}^{-1}} = 0.10\text{ mol dm}^{-3}.

Molar concentration is defined as moles of solute per unit volume of solution in

\text{dm}^3

.

Key Concept

Conversion from mass concentration (g/dm³) to molar concentration (mol/dm³)

Estimated Time:45s

Question 8618Question →

The chemical breakdown of dietary proteins requires sequential enzymatic activity across different regions of the mammalian alimentary canal. What is the correct sequence of these processes from the start of chemical digestion to nutrient absorption?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of protein digestion and absorption is: acidic cleavage into polypeptides in the stomach, alkaline breakdown into oligopeptides in the duodenum, brush-border cleavage into free amino acids in the ileum, and active absorption of amino acids into intestinal capillaries.

Protein digestion follows a strict anatomical and biochemical progression. First, stomach hydrochloric acid and pepsin denature and break complex proteins into polypeptides under acidic conditions. Next, pancreatic trypsin and chymotrypsin in the alkaline environment of the duodenum break polypeptides into short oligopeptides. Then, brush-border peptidases in the ileum convert oligopeptides into absorbable free amino acids. Finally, these free amino acids are actively absorbed across microvilli into blood capillaries.

Step-by-Step Solution

1

Identify the initial organ of chemical protein digestion.

Chemical digestion of protein begins in the stomach where gastric juice containing hydrochloric acid activates pepsinogen to pepsin, hydrolyzing native proteins into polypeptides.

Salivary amylase in the mouth acts only on carbohydrates; no protein-digesting enzymes are secreted in the buccal cavity.

2

Determine the subsequent region of the digestive tract and its corresponding enzymatic activity.

Chyme enters the duodenum where pancreatic juice neutralizes acid. Pancreatic proteases (trypsin and chymotrypsin) hydrolyze polypeptides into short oligopeptides.

Pancreatic endopeptidases require an alkaline pH optimal for their catalytic activity in the duodenum.

3

Identify the terminal stage of enzymatic hydrolysis.

Intestinal peptidases (erepsin) on the brush border of the ileum hydrolyze oligopeptides into individual amino acids.

Final breakdown to amino acid monomers must occur before absorption across cell membranes can take place.

4

Identify the final nutrient absorption process.

Free amino acids are actively transported across villi epithelial membranes into blood capillaries of the hepatic portal system.

Absorbed amino acids travel via the hepatic portal vein directly to the liver for metabolic processing.

Key Concept

Sequential Protein Digestion and Absorption Pathway

Question 8619Question →

A $16.1\text{ g}$ sample of hydrated sodium tetraoxosulfate(VI), $\text{Na}_2\text{SO}_4 \cdot x\text{H}_2\text{O}$ , is heated in a crucible until all the water of crystallization is driven off. The mass of the remaining anhydrous salt is $7.1\text{ g}$ . Calculate the value of $x$ . [Relative atomic masses: $\text{Na} = 23$ , $\text{S} = 32$ , $\text{O} = 16$ , $\text{H} = 1$ ]

Show answer & explanation

Answer: 10

Answer

The integer value of

x

is 10.

Heating

16.1\text{ g}

of hydrated sodium tetraoxosulfate(VI) yields

7.1\text{ g}

of anhydrous

\text{Na}_2\text{SO}_4

(

0.05\text{ mol}

) and releases

9.0\text{ g}

of water (

0.5\text{ mol}

). The mole ratio of

\text{H}_2\text{O}

to

\text{Na}_2\text{SO}_4

is

0.5 / 0.05 = 10

, giving

x = 10

.

Step-by-Step Solution

1

Calculate the mass of water lost upon heating

Mass of

\text{H}_2\text{O} = 16.1\text{ g} - 7.1\text{ g} = 9.0\text{ g}

The difference between the initial hydrated mass and final anhydrous mass represents the driven-off water of crystallization.

2

Calculate molar masses of anhydrous salt and water

Molar mass of

\text{Na}_2\text{SO}_4 = 142\text{ g/mol}

, Molar mass of

\text{H}_2\text{O} = 18\text{ g/mol}

Required to convert sample masses to mole quantities.

3

Calculate moles of anhydrous salt and water, and find the mole ratio

Moles of

\text{Na}_2\text{SO}_4 = 0.05\text{ mol}

, Moles of

\text{H}_2\text{O} = 0.5\text{ mol}

,

x = \frac{0.5}{0.05} = 10

The subscript

x

gives the ratio of moles of water to moles of anhydrous salt per mole of compound.

Key Concept

Stoichiometric determination of water of crystallization in hydrated salts

Question 8620Question →

Organophosphate pesticides persist in agricultural soils for significantly longer periods than organochlorine insecticides because organophosphates resist chemical hydrolysis and microbial degradation.

True

False

Show answer & explanation

Answer: False

Answer

False

The correct evaluation is False. Organochlorines contain non-polar carbon-chlorine bonds that resist chemical hydrolysis and biological breakdown, causing them to persist in agricultural soils and bioaccumulate up the food chain. Organophosphates possess ester linkages that hydrolyze relatively quickly in moist soil, resulting in much shorter environmental persistence.

Step-by-Step Solution

1

Analyze the chemical bonding and functional groups present in organochlorines vs organophosphates.

Organochlorines feature stable C-Cl bonds, while organophosphates contain reactive phosphate ester bonds.

Bond strength and polarity determine susceptibility to chemical and biological degradation pathways in soil.

2

Evaluate degradation mechanisms and environmental persistence in soil.

Organophosphates undergo rapid hydrolysis and microbial enzymatic cleavage in moist soil, whereas organochlorines resist weathering and remain in soil for long periods.

Ester linkages in organophosphates are easily cleaved by soil moisture and enzymes, unlike non-polar chlorinated hydrocarbons.

Key Concept

Chemical stability, environmental persistence, and bioaccumulation pathways of organochlorine vs organophosphate pesticides in soil

Estimated Time:1m 0s

Answer

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