All practice questions

Question 8641Question →

In nitrogen-fixing filamentous cyanobacteria, specialized cells called heterocysts lack Photosystem II activity, thereby preventing oxygen production that would otherwise inhibit the enzyme nitrogenase.

True

False

Show answer & explanation

Answer: True

Answer

The statement is TRUE.

Heterocysts are specialized cells in certain cyanobacteria (such as Anabaena) adapted for nitrogen fixation. Because nitrogenase is oxygen-sensitive, heterocysts modify their photosynthetic apparatus to eliminate Photosystem II, preventing internal oxygen generation while retaining Photosystem I for ATP production.

Step-by-Step Solution

1

Analyze the oxygen sensitivity of nitrogenase during prokaryotic nitrogen fixation.

The nitrogenase enzyme complex reduces atmospheric nitrogen (

N_2

) to ammonia (

NH_3

), but it is highly sensitive to molecular oxygen (

O_2

) and rapidly denatured by it.

Understanding enzyme inhibition by oxygen is required to analyze prokaryotic adaptation strategies.

2

Examine structural and photosynthetic modifications in cyanobacterial heterocysts.

Heterocysts undergo specialized cellular differentiation wherein Photosystem II is inactivated (stopping

O_2

generation), while Photosystem I remains active to supply ATP via cyclic photophosphorylation.

This spatial separation isolates oxygen-producing photosynthesis in vegetative cells from nitrogen fixation in heterocysts.

3

Determine the truth value of the given statement.

Because heterocysts specifically lack Photosystem II to protect nitrogenase from oxygen inhibition, the statement is accurate.

The physiological mechanism stated aligns directly with cyanobacterial biochemistry.

Key Concept

Cellular specialization for nitrogen fixation in Cyanobacteria (Heterocysts)

Question 8642Question →

In a terrestrial grassland ecosystem, the total energy fixed by green plants (producers) is measured at $20{}000\text{ kJ m}^{-2}\text{ yr}^{-1}$ . According to Lindeman's $10\%$ law of ecological efficiency, what amount of energy will be available to the secondary consumers in this ecosystem?

2{}000\text{ kJ m}^{-2}\text{ yr}^{-1}

200\text{ kJ m}^{-2}\text{ yr}^{-1}

20\text{ kJ m}^{-2}\text{ yr}^{-1}

20{}000\text{ kJ m}^{-2}\text{ yr}^{-1}

Show answer & explanation

Answer: $200\text{ kJ m}^{-2}\text{ yr}^{-1}$

Answer

The amount of energy available to secondary consumers is

200\text{ kJ m}^{-2}\text{ yr}^{-1}

.

Secondary consumers occupy the third trophic level. Beginning with

20{}000\text{ kJ m}^{-2}\text{ yr}^{-1}

at the producer level, primary consumers (level 2) obtain

10\%

of this energy (

2{}000\text{ kJ m}^{-2}\text{ yr}^{-1}

). Secondary consumers (level 3) retain

10\%

of the energy from primary consumers, which equals

200\text{ kJ m}^{-2}\text{ yr}^{-1}

.

Step-by-Step Solution

1

Identify the trophic level of secondary consumers.

Producers belong to Trophic Level 1, Primary Consumers to Trophic Level 2, and Secondary Consumers to Trophic Level 3.

Energy flows sequentially from producers to consumers across discrete trophic steps.

2

Calculate energy available at Trophic Level 2 (Primary Consumers).

20{}000\text{ kJ m}^{-2}\text{ yr}^{-1} \times 0.10 = 2{}000\text{ kJ m}^{-2}\text{ yr}^{-1}

.

Approximately

10\%

of net primary production is assimilated by herbivores, while

90\%

is lost as metabolic heat and unconsumed waste.

3

Calculate energy available at Trophic Level 3 (Secondary Consumers).

2{}000\text{ kJ m}^{-2}\text{ yr}^{-1} \times 0.10 = 200\text{ kJ m}^{-2}\text{ yr}^{-1}

.

Applying the

10\%

transfer efficiency once more determines the energy incorporated at the carnivore/secondary consumer level.

Key Concept

10% Law of Energy Transfer across Ecological Trophic Levels

Estimated Time:1m 15s

Question 8643Question →

In a dihybrid organism with the genotype $AaBb$ , the two gene pairs assort independently during meiosis according to Mendel's Second Law. What is the expected phenotypic or genotypic ratio of the gametes produced by this individual?

1\ AB : 1\ Ab : 1\ aB : 1\ ab

9\ AB : 3\ Ab : 3\ aB : 1\ ab

3\ AB : 1\ ab

1\ Aa : 1\ Bb

Show answer & explanation

Answer: $1\ AB : 1\ Ab : 1\ aB : 1\ ab$

Answer

The expected ratio of gametes produced is 1 AB : 1 Ab : 1 aB : 1 ab.

The correct answer states a 1:1:1:1 ratio of

AB

,

Ab

,

aB

, and

ab

. A dihybrid organism heterozygous for two unlinked genes (

AaBb

) undergoes independent assortment during meiosis. Allele

A

or

a

has a 50% chance of combining with allele

B

or

b

, generating four distinct haploid gametes in equal proportions (25% each).

Step-by-Step Solution

1

Determine the possible allele combinations for gamete formation from genotype

AaBb

.

Gene

A

produces alleles

A

and

a

; Gene

B

produces alleles

B

and

b

.

Meiosis separates homologous chromosomes so each gamete receives one allele per gene.

2

Apply Mendel's Law of Independent Assortment to combine alleles from both genes.

The four possible gamete combinations are

AB

,

Ab

,

aB

, and

ab

.

Alleles of unlinked genes segregate independently into gametes.

3

Calculate the probability of each gamete combination.

Each combination has a probability of

\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

, yielding a ratio of

1 : 1 : 1 : 1

.

Since segregation of each allele pair is equally likely, all four gamete types occur in equal frequencies.

Key Concept

Mendel's Law of Independent Assortment states that alleles of different genes segregate independently of one another during gamete formation, resulting in equal proportions of all possible haploid allele combinations.

Estimated Time:45s

Question 8644Question →

Concentrated tetraoxosulfate(VI) acid reacts with copper metal as an oxidizing agent according to the balanced chemical equation:

Cu(s) + 2H_2SO_4(aq) \rightarrow CuSO_4(aq) + 2H_2O(l) + SO_2(g)

If

31.75\text{ g}

of copper turnings react completely with an excess of concentrated tetraoxosulfate(VI) acid, what volume of sulfur(IV) oxide (

SO_2

) gas, in

\text{dm}^3

, is evolved at standard temperature and pressure (STP)? [Molar mass of

Cu = 63.5\text{ g mol}^{-1}

; Molar volume of gas at STP =

22.4\text{ dm}^3\text{ mol}^{-1}

]

Show answer & explanation

Answer: 11.2

Answer

The volume of sulfur(IV) oxide gas evolved at STP is

11.2\text{ dm}^3

.

Concentrated tetraoxosulfate(VI) acid acts as an oxidizing agent when heated with copper metal, being reduced to sulfur(IV) oxide gas. From the balanced reaction equation,

1\text{ mole}

(

63.5\text{ g}

) of copper yields

1\text{ mole}

(

22.4\text{ dm}^3

at STP) of

SO_2

gas. Therefore,

31.75\text{ g}

(

0.5\text{ moles}

) of copper yields

0.5 \times 22.4\text{ dm}^3 = 11.2\text{ dm}^3

of

SO_2

gas.

Step-by-Step Solution

1

Calculate the moles of copper metal reacted

n(Cu) = \frac{31.75\text{ g}}{63.5\text{ g mol}^{-1}} = 0.5\text{ mol}

Converting the given mass of copper to moles allows stoichiometric comparison with the reaction products.

2

Determine the moles of sulfur(IV) oxide (

SO_2

) gas produced

n(SO_2) = 0.5\text{ mol}

From the balanced chemical equation,

1\text{ mole}

of

Cu

reacts to produce

1\text{ mole}

of

SO_2

gas.

3

Calculate the volume of

SO_2

gas evolved at STP

V(SO_2) = 0.5\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 11.2\text{ dm}^3

At STP,

1\text{ mole}

of any ideal gas occupies a standard molar volume of

22.4\text{ dm}^3

.

Key Concept

Oxidizing action of concentrated tetraoxosulfate(VI) acid on metals and gas volume stoichiometry at STP

Question 8645Question →

Arrange the following sequential stages of carbon assimilation in the Calvin cycle of photosynthesis in the correct order from start to finish.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence from first to last is: Fixation of carbon dioxide by RuBP, Reduction of 3-phosphoglycerate to triose phosphate, and Regeneration of RuBP.

Carbon assimilation during the light-independent stage occurs in three distinct steps: initial carbon fixation where carbon dioxide is accepted by RuBP, reduction of 3-phosphoglycerate to sugar intermediates (triose phosphate) utilizing ATP and NADPH, and regeneration of RuBP to allow continuous carbon dioxide capture.

Step-by-Step Solution

1

Identify the primary carbon uptake event.

Atmospheric carbon dioxide is combined with RuBP to form 3-phosphoglycerate.

Enzymatic carbon fixation must occur first to introduce inorganic carbon into the biological system.

2

Trace energy input and chemical reduction.

3-phosphoglycerate is converted to triose phosphate.

ATP and reduced NADP (NADPH) generated from light-dependent reactions drive chemical reduction.

3

Identify the reset mechanism for the cycle.

Triose phosphate molecules are rearranged into RuBP.

Regenerating the primary acceptor RuBP allows the Calvin cycle to continue assimilating carbon dioxide.

Key Concept

Stages of Light-Independent Stage (Calvin Cycle)

Estimated Time:45s

Question 8646Question →

In biological classification, every species is assigned a universal two-part scientific name according to the principles of binomial nomenclature. Which of the following represents the correctly written scientific name for the African lion?

*Panthera leo*

*Panthera Leo*

*panthera leo*

*panthera Leo*

Show answer & explanation

Answer: *Panthera leo*

Answer

The scientific name formatted as *Panthera leo* is correct because the genus name (*Panthera*) is capitalized, the specific epithet (*leo*) is lowercase, and both words are italicized.

Under Linnaean rules of binomial nomenclature, a scientific name consists of two parts: the genus name (which must always begin with a capital letter) and the specific epithet (which must be written entirely in lowercase). Both terms are italicized when printed. Therefore, the formatting *Panthera leo* is correct.

Step-by-Step Solution

1

Identify the two components of the scientific name

The first term represents the Genus and the second term represents the Specific Epithet (species).

Binomial nomenclature uses a formal two-part naming convention established by Carl Linnaeus.

2

Apply Linnaean capitalization rules

The Genus name *Panthera* starts with a capital letter, while the species epithet *leo* begins with a lowercase letter.

Taxonomic convention strictly requires Genus names to be capitalized and species epithets to remain lowercase.

3

Apply typographical formatting rules

The full binomial name is written in italics.

Scientific names are Latinized terms that must be distinguished visually from surrounding text.

Key Concept

Binomial Nomenclature Capitalization and Formatting Rules

Question 8647Question →

Match each structural component of a virus on the left with its corresponding chemical composition or biological function on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Capsid matches with the protein coat enclosing the nucleic acid; Genetic core matches with DNA or RNA carrying hereditary instructions; Envelope matches with lipid membrane derived from host cell; Tail fibers match with protein appendages for host receptor attachment.

Each viral component serves a specific structural or biochemical role: Capsid protects genetic material via a protein coat; Genetic core contains DNA or RNA genome; Envelope is host-derived lipid bilayer; Tail fibers facilitate target cell attachment.

Step-by-Step Solution

1

Identify the primary protective layer of a virus

The capsid is composed of capsomeres (proteins) and surrounds the nucleic acid core.

Proteins form the outer structural coat of all non-enveloped and enveloped viruses.

2

Identify the nucleic acid component

The genetic core consists of either single-stranded or double-stranded DNA or RNA.

Viruses possess a single type of nucleic acid carrying their genetic code.

3

Identify the lipid-containing structure

The viral envelope consists of lipids acquired from host cellular membranes.

Enveloped viruses exit host cells by budding, picking up host membrane lipids.

4

Identify the host attachment apparatus

Tail fibers function in host cell recognition and anchoring.

Bacteriophage tail fibers bind specifically to bacterial cell wall receptors.

Key Concept

Structural organization and chemical composition of viral components

Question 8648Question →

Which group of plants possesses true vascular tissues (xylem and phloem) for internal transport but reproduces via spores rather than seeds?

Pteridophytes

Bryophytes

Thallophytes

Fungi

Show answer & explanation

Answer: Pteridophytes

Answer

Pteridophytes are the plant group that has well-developed vascular tissues (xylem and phloem) while relying on spores instead of seeds for reproduction.

Pteridophytes (such as ferns) represent the earliest group of vascular plants. They possess true vascular tissues consisting of xylem and phloem for transport, yet they remain seedless, producing spores for reproduction.

Step-by-Step Solution

1

Identify the key structural requirement given in the stem.

The plant must possess true conducting tissues, namely xylem and phloem.

This separates non-vascular plants (thallophytes and bryophytes) from vascular plants.

2

Identify the reproductive requirement given in the stem.

The plant reproduces by spore formation and does not produce seeds.

This distinguishes seedless vascular plants (pteridophytes) from seed-bearing vascular plants (spermatophytes).

Key Concept

Classification of seedless vascular plants (Pteridophytes)

Question 8649Question →

Zinc metal reacts with hydrochloric acid according to the balanced chemical equation:

\text{Zn}_{(s)} + 2\text{HCl}_{(aq)} \rightarrow \text{ZnCl}_{2(aq)} + \text{H}_{2(g)}

If

13.0\text{ g}

of zinc is added to a solution containing

7.3\text{ g}

of hydrochloric acid, what mass of zinc remains unreacted after the reaction goes to completion?
(

\text{Zn} = 65.0,\text{ H} = 1.0,\text{ Cl} = 35.5

)

6.5\text{ g}

5.7\text{ g}

0.0\text{ g}

3.25\text{ g}

Show answer & explanation

Answer: $6.5\text{ g}$

Answer

The mass of zinc remaining unreacted is

6.5\text{ g}

.

Converting the given masses into moles shows that 0.20 mol of zinc and 0.20 mol of hydrochloric acid are present. Since 1 mole of zinc requires 2 moles of hydrochloric acid, 0.20 mol of hydrochloric acid reacts with only 0.10 mol of zinc. Hydrochloric acid is completely consumed, leaving 0.10 mol (6.5 g) of zinc unreacted.

Step-by-Step Solution

1

Calculate the molar masses of the reactants

Molar mass of

\text{Zn} = 65.0\text{ g/mol}

; Molar mass of

\text{HCl} = 1.0 + 35.5 = 36.5\text{ g/mol}

.

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

2

Determine the initial mole quantities of each reactant

Moles of

\text{Zn} = \frac{13.0\text{ g}}{65.0\text{ g/mol}} = 0.20\text{ mol}

; Moles of

\text{HCl} = \frac{7.3\text{ g}}{36.5\text{ g/mol}} = 0.20\text{ mol}

.

Chemical reactions occur according to mole ratios, not mass ratios.

3

Identify the limiting reactant and calculate the moles of zinc consumed

From the equation,

1\text{ mol of Zn}

reacts with

2\text{ mol of HCl}

. Thus,

0.20\text{ mol of HCl}

requires

\frac{0.20}{2} = 0.10\text{ mol of Zn}

.

\text{HCl}

is the limiting reactant.

The limiting reactant determines the extent of the reaction.

4

Calculate the unreacted moles and mass of zinc

Unreacted moles of

\text{Zn} = 0.20\text{ mol} - 0.10\text{ mol} = 0.10\text{ mol}

. Unreacted mass of

\text{Zn} = 0.10\text{ mol} \times 65.0\text{ g/mol} = 6.5\text{ g}

.

Subtracting consumed moles from initial moles gives the remaining amount.

Key Concept

Limiting and Excess Reactants

Question 8650Question →

Match each industrial electrolytic process on the left with its corresponding chemical characteristic or operating condition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Hall-Héroult Process matches with 'Electrolysis of molten alumina dissolved in molten cryolite using carbon electrodes'. Electrorefining of Copper matches with 'Impure metal serves as the dissolving anode while pure metal deposits at the cathode'. Electroplating of Iron with Silver matches with 'The article to be coated serves as the cathode in an electrolyte containing silver ions'. Chlor-Alkali Process matches with 'Electrolysis of concentrated brine yielding chlorine gas, hydrogen gas, and sodium hydroxide'.

Each industrial process relies on specific redox reactions and cell configurations: Hall-Héroult process uses molten cryolite as a solvent for alumina; copper refining uses impure copper as the dissolving anode; electroplating places the article to be coated at the cathode; chlor-alkali electrolyzes brine to yield chlorine, hydrogen, and sodium hydroxide.

Step-by-Step Solution

1

Identify the key components and conditions of aluminium extraction.

Aluminium is extracted by electrolyzing molten alumina (

Al_2O_3

) dissolved in cryolite (

Na_3AlF_6

) to lower its melting point.

This corresponds to the Hall-Héroult process operating condition.

2

Identify the electrode roles in metal purification.

Impure metal dissolves at the anode and deposits as pure metal at the cathode.

This defines electrorefining of copper.

3

Identify the arrangement for electroplating.

The target object to be coated forms the cathode where metal cations are reduced.

This describes the electroplating of iron with silver.

4

Identify the reactants and products of brine electrolysis.

Concentrated

NaCl

solution electrolyzed gives

Cl_2

,

H_2

, and

NaOH

.

This defines the chlor-alkali industrial process.

Key Concept

Industrial applications of electrolysis including metal extraction, refining, electroplating, and chlor-alkali synthesis

Question 8651Question →

Match each Lamarckian evolutionary postulate or related historical concept on the left with its corresponding descriptive statement or experimental critique on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Principle of Use and Disuse matches the statement regarding organ hypertrophy through exertion and atrophy through disuse. Inheritance of Acquired Characteristics matches the statement regarding somatic modifications being passed to progeny. Environmental Need (Besoin) matches the statement about environmental shifts creating new demands that drive behavioral and structural adaptations. Germplasm Theory Critique matches the experimental proof that somatic changes do not affect germ cells.

Each concept correctly aligns with its historical definition or scientific critique. The principle of use and disuse concerns organ development through exertion or atrophy through neglect. Inheritance of acquired characteristics describes the transfer of somatic traits to progeny. Environmental need explains how changing surroundings prompt adaptive responses. Germplasm theory refutes Lamarckism by demonstrating the separation of somatic and reproductive cells.

Step-by-Step Solution

1

Analyze the core postulates of Lamarckism and the primary historical counter-evidence.

Identify the definitions of use and disuse, acquired traits, environmental needs, and Weismann's germplasm theory.

Clear differentiation between Lamarck's mechanisms and their experimental refutation is necessary to pair each concept accurately.

2

Match 'Principle of Use and Disuse' to its mechanism.

Pairs with the description detailing organ enlargement from frequent use and degeneration from disuse.

Lamarck proposed that physical exertion directly alters organ structure within an individual's lifetime.

3

Match 'Inheritance of Acquired Characteristics' to its definition.

Pairs with the statement that lifespan modifications are transmitted to subsequent generations.

This postulate erroneously assumed somatic changes could be inherited.

4

Match 'Environmental Need (Besoin)' and 'Germplasm Theory Critique' to their remaining descriptions.

Environmental Need pairs with environmental shifts creating new demands, and Germplasm Theory Critique pairs with the distinction between somatic and germ cells.

Lamarck emphasized environmental stimulus for adaptation, while Weismann proved germ cells are isolated from somatic changes.

Key Concept

Lamarckian Evolutionary Principles and Germplasm Counter-Evidence

Question 8652Question →

In an agricultural soil plot rich in decaying organic matter, ammonium ions are rapidly produced by decomposers. Soil biochemical analysis reveals that while ammonium is converted into nitrites ( $NO_2^-$ ), the subsequent transformation of nitrites into nitrates ( $NO_3^-$ ) is completely blocked, leading to a toxic buildup of nitrites. Which soil bacterium is deficient or inactive in this ecosystem?

Nitrobacter

Rhizobium

Clostridium

Pseudomonas

Show answer & explanation

Answer: Nitrobacter

Answer

Nitrobacter is the bacterium responsible for the second stage of nitrification, transforming nitrites (

NO_2^-

) into nitrates (

NO_3^-

).

Nitrobacter is the obligate aerobic chemoautotroph responsible for the second oxidation step in nitrification, converting nitrites (

NO_2^-

) into nitrates (

NO_3^-

). If Nitrobacter is inactive, nitrites accumulate because the pathway cannot proceed to completion.

Step-by-Step Solution

1

Identify the specific biochemical step that is blocked in the soil.

The reaction converting nitrites (

NO_2^-

) to nitrates (

NO_3^-

) is inhibited.

Ammonium has already been oxidized to nitrite, so the second stage of nitrification is where the block occurs.

2

Recall the micro-organism responsible for nitrite oxidation.

Nitrobacter chemoautotrophically oxidizes

NO_2^-

to

NO_3^-

.

Different nitrifying bacteria specialize in distinct steps of the nitrification pathway.

3

Match the missing biological function to the candidate bacteria.

Inactivity or absence of Nitrobacter leads directly to the accumulation of toxic nitrites.

Without Nitrobacter, nitrites cannot be converted to nitrates, which plants absorb for amino acid synthesis.

Key Concept

Nitrification Pathway and Microbial Roles

Estimated Time:1m 30s

Question 8653Question →

In garden pea plants (*Pisum sativum*), tall stem height ( $T$ ) is dominant over dwarf stem height ( $t$ ), and yellow seed color ( $Y$ ) is dominant over green seed color ( $y$ ). According to Mendel's Law of Independent Assortment, how many genetically distinct types of gametes can be produced by an $F_1$ plant with the genotype $TtYy$ ?

2 types

4 types

8 types

16 types

Show answer & explanation

Answer: 4 types

Answer

4 types of gametes (

TY

,

Ty

,

tY

, and

ty

)

According to Mendel's Second Law (Law of Independent Assortment), alleles for different traits segregate independently during meiosis. For a dihybrid individual with genotype

TtYy

, the two alleles of the stem height gene (

T

and

t

) combine randomly with the two alleles of the seed color gene (

Y

and

y

). Using the formula

2^n

(where

n

is the number of heterozygous gene pairs, here

n = 2

), the number of distinct gamete types produced is

2^2 = 4

, which are

TY

,

Ty

,

tY

, and

ty

.

Step-by-Step Solution

1

Identify the number of heterozygous gene pairs in the given genotype.

The genotype

TtYy

has two heterozygous gene pairs (

n = 2

).

Mendel's Law of Independent Assortment states that allele pairs separate independently during gamete formation.

2

Calculate the number of possible unique gamete types using the formula

2^n

.

2^2 = 4

distinct gamete types.

Where

n

represents the number of heterozygous loci,

2^2

yields the total combinations.

3

List the distinct allele combinations to verify.

The four possible gametes are

TY

,

Ty

,

tY

, and

ty

.

Each gamete receives one allele from each gene pair independently.

Key Concept

Mendel's Law of Independent Assortment and Gamete Formation in Dihybrids

Question 8654Question →

Match each organic reaction involving an amine or amide on the left with its corresponding principal product on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Reduction of propanamide with

LiAlH_4

pairs with propylamine; Reaction of propanamide with

Br_2/KOH

pairs with ethylamine; Alkaline hydrolysis of propanamide pairs with sodium propanoate and ammonia; Acylation of methylamine with ethanoyl chloride pairs with

N

-methylethanamide.

Each reaction pair is determined by its specific mechanistic pathway:

LiAlH_4

reduces the carbonyl group to methylene (retaining carbon count to form propylamine);

Br_2/KOH

undergoes Hofmann degradation to lose the carbonyl carbon (forming ethylamine); basic hydrolysis cleaves the

C-N

bond (yielding sodium propanoate and ammonia); and acylation of methylamine with ethanoyl chloride produces the substituted amide (

N

-methylethanamide).

Step-by-Step Solution

1

Analyze the reduction reaction of primary amides.

Reducing

CH_3CH_2CONH_2

with

LiAlH_4

reduces the

C=O

bond to a

-CH_2-

group without altering the total carbon count, yielding

CH_3CH_2CH_2NH_2

(propylamine).

Amide reduction retains the full carbon skeleton.

2

Identify the reaction of primary amides with

Br_2

and

KOH

.

This is Hofmann degradation, which removes the carbonyl carbon (

C=O

) as carbonate, reducing the carbon length by 1. Propanamide (3 carbons) yields ethylamine (2 carbons).

Hofmann degradation shortens the carbon chain by one atom.

3

Examine the basic hydrolysis of amides.

Nucleophilic attack of

OH^-

on the carbonyl carbon of propanamide cleaves the amide bond to generate propanoate anion (forming sodium propanoate with

Na^+

) and ammonia gas.

Base hydrolysis of amides yields a carboxylate salt and ammonia.

4

Examine the nucleophilic substitution between methylamine and ethanoyl chloride.

The nitrogen lone pair of methylamine attacks ethanoyl chloride, releasing

HCl

to form a secondary amide,

N

-methylethanamide (

CH_3CONHCH_3

).

Primary amines undergo acylation to form secondary amides.

Key Concept

Chemical Reactions and Interconversions of Amines and Amides

Estimated Time:1m 30s

Question 8655Question →

Consider the gas-phase reversible reaction involved in the steam reforming of methane:

\text{CH}_4(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}(g) + 3\text{H}_2(g)

What are the units of the equilibrium constant

K_c

when concentrations are expressed in

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

?

Show answer & explanation

Answer: mol^2 dm^-6; mol^2dm^-6; mol2 dm-6; mol2dm-6; mol^2/dm^6; mol2/dm6; (mol dm^-3)^2; (mol/dm^3)^2

Answer

The units of

K_c

for this reaction are

\text{mol}^2\text{dm}^{-6}

(or

(\text{mol}\cdot\text{dm}^{-3})^2

).

The reaction produces 4 moles of gaseous products (1 mol

\text{CO}

+ 3 mol

\text{H}_2

) from 2 moles of gaseous reactants (1 mol

\text{CH}_4

+ 1 mol

\text{H}_2\text{O}

). Substituting concentration units into

K_c = \frac{[\text{CO}][\text{H}_2]^3}{[\text{CH}_4][\text{H}_2\text{O}]}

yields

\frac{(\text{mol}\cdot\text{dm}^{-3})^4}{(\text{mol}\cdot\text{dm}^{-3})^2} = (\text{mol}\cdot\text{dm}^{-3})^2 = \text{mol}^2\text{dm}^{-6}

.

Step-by-Step Solution

1

Write the equilibrium constant expression for the reaction.

K_c = \frac{[\text{CO}][\text{H}_2]^3}{[\text{CH}_4][\text{H}_2\text{O}]}

The equilibrium constant expression places product concentrations in the numerator and reactant concentrations in the denominator, each raised to the power of their stoichiometric coefficients.

2

Substitute the concentration unit (

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

) for each species into the

K_c

expression.

\text{Units of } K_c = \frac{(\text{mol}\cdot\text{dm}^{-3}) \times (\text{mol}\cdot\text{dm}^{-3})^3}{(\text{mol}\cdot\text{dm}^{-3}) \times (\text{mol}\cdot\text{dm}^{-3})} = \frac{(\text{mol}\cdot\text{dm}^{-3})^4}{(\text{mol}\cdot\text{dm}^{-3})^2}

Replacing the concentration terms with the standard concentration units isolates the dimensional units of

K_c

.

3

Simplify the powers of the concentration units.

(\text{mol}\cdot\text{dm}^{-3})^{4 - 2} = (\text{mol}\cdot\text{dm}^{-3})^2 = \text{mol}^2\text{dm}^{-6}

Dividing exponential terms with the same base subtracts the powers, yielding the final net units.

Key Concept

Determining the units of equilibrium constants (

K_c

) from reaction stoichiometry.

Question 8656Question →

A gas mixture containing $8.0\text{ g}$ of oxygen ( $\text{O}_2$ ) and $7.0\text{ g}$ of nitrogen ( $\text{N}_2$ ) is collected over water at a total pressure of $750\text{ mmHg}$ . If the saturated vapor pressure of water at the collection temperature is $30\text{ mmHg}$ , what is the partial pressure exerted by oxygen in the dry gas mixture? (Molar masses: $\text{O}_2 = 32\text{ g/mol}$ , $\text{N}_2 = 28\text{ g/mol}$ )

360\text{ mmHg}

375\text{ mmHg}

390\text{ mmHg}

400\text{ mmHg}

Show answer & explanation

Answer: $360\text{ mmHg}$

Answer

360\text{ mmHg}

The total pressure exerted by a gas collected over water is the sum of the partial pressures of the dry gases and the saturated vapor pressure of water. Subtracting the vapor pressure of

30\text{ mmHg}

from the total pressure of

750\text{ mmHg}

yields a dry gas pressure of

720\text{ mmHg}

. Since

8.0\text{ g}

of

\text{O}_2

(

0.25\text{ mol}

) and

7.0\text{ g}

of

\text{N}_2

(

0.25\text{ mol}

) give equal mole fractions of

0.50

each, the partial pressure of oxygen is half of the dry total pressure, which equals

360\text{ mmHg}

.

Step-by-Step Solution

1

Calculate the dry gas mixture total pressure by subtracting aqueous tension from total pressure.

P_{\text{dry}} = P_{\text{total}} - P_{\text{water}} = 750\text{ mmHg} - 30\text{ mmHg} = 720\text{ mmHg}

Gas collected over water contains water vapor, so total pressure is the sum of dry gas pressure and vapor pressure (Dalton's Law).

2

Calculate the number of moles of each gas in the mixture.

n(\text{O}_2) = \frac{8.0}{32} = 0.25\text{ mol}

,

n(\text{N}_2) = \frac{7.0}{28} = 0.25\text{ mol}

Moles are required to determine the mole fraction of oxygen.

3

Determine the mole fraction of oxygen in the dry mixture.

\chi(\text{O}_2) = \frac{0.25\text{ mol}}{0.25\text{ mol} + 0.25\text{ mol}} = 0.50

Mole fraction represents the ratio of moles of a specific gas component to total dry moles.

4

Calculate the partial pressure of oxygen.

P(\text{O}_2) = \chi(\text{O}_2) \times P_{\text{dry}} = 0.50 \times 720\text{ mmHg} = 360\text{ mmHg}

By Dalton's Law of Partial Pressures, the partial pressure of a gas is equal to its mole fraction multiplied by the dry total pressure.

Key Concept

Dalton's Law of Partial Pressures and Gas Collection Over Water

Estimated Time:1m 30s

Question 8657Question →

In watermelon plants (*Citrullus lanatus*), solid green rind color ( $G$ ) is dominant over striped rind color ( $g$ ), and short fruit shape ( $R$ ) is dominant over long fruit shape ( $r$ ). If a watermelon plant heterozygous for both traits ( $GgRr$ ) is self-pollinated and yields $800$ offspring, how many of these offspring are expected to possess solid green rind and long fruit shape?

450

150

50

600

Show answer & explanation

Answer: $150$

Answer

150 offspring exhibit the solid green rind and long fruit shape phenotype.

In a dihybrid cross between two heterozygous parents (

GgRr \times GgRr

), the genes assort independently, yielding an

F_2

phenotypic ratio of

9 : 3 : 3 : 1

(9 solid green short : 3 solid green long : 3 striped short : 1 striped long). The proportion of offspring showing solid green rind (

G\_

) and long fruit (

rr

) is

\frac{3}{16}

. Among

800

offspring, the expected number is

\frac{3}{16} \times 800 = 150

.

Step-by-Step Solution

1

Determine the parental genotypes and set up the cross.

Parental cross is

GgRr \times GgRr

.

Self-pollination of a dihybrid plant involves crossing two organisms heterozygous for both genes.

2

Calculate the expected phenotypic ratio for the

F_2

generation according to Mendel's Second Law.

The standard dihybrid phenotypic ratio is

9 : 3 : 3 : 1

.

Genes for rind color and fruit shape assort independently during gamete formation.

3

Identify the probability of the target phenotype (solid green rind, long fruit).

Target phenotype genotype is

G\_rr

, which corresponds to a probability of

\frac{3}{16}

.

Dominant for rind color (

G\_

) has probability

\frac{3}{4}

, and recessive for fruit shape (

rr

) has probability

\frac{1}{4}

. Thus,

\frac{3}{4} \times \frac{1}{4} = \frac{3}{16}

.

4

Multiply the phenotype probability by the total offspring count.

\frac{3}{16} \times 800 = 150

.

Multiplying total progeny count by the expected probability yields the expected phenotype frequency.

Key Concept

Mendel's Law of Independent Assortment and Dihybrid Phenotypic Calculations

Question 8658Question →

A student needs to extract a non-polar organic solute from an aqueous mixture using diethyl ether in a separating funnel. Arrange the following procedural steps in the correct logical sequence from first to last.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of steps is: First, pour the liquids into the funnel, stopper, shake, and periodically vent pressure; second, clamp the funnel and allow the mixture to settle into two clear layers; third, unstopper the funnel and drain the denser lower aqueous layer through the stopcock; fourth, collect the lighter organic layer containing the solute from the funnel.

In solvent extraction using a separating funnel, the mixture and extracting solvent must first be thoroughly shaken together with periodic pressure venting to maximize solute partition into the solvent. The funnel is then supported upright until the immiscible liquids separate into two distinct layers. The top stopper is removed to equalize atmospheric pressure before opening the stopcock to run off the lower (denser) aqueous layer. Finally, the upper (less dense) organic layer containing the solute is collected separately.

Step-by-Step Solution

1

Identify the initial extraction step.

Mixing the aqueous mixture with diethyl ether in the stoppered funnel while venting vapor pressure.

Solute extraction requires intimate contact between the two immiscible liquid phases under safe conditions.

2

Determine the phase separation step.

Allowing the mixture to settle vertically in a stand.

Gravity causes the two immiscible liquids to separate into discrete layers based on density differences.

3

Identify the first draining operation.

Removing the stopper and draining the bottom aqueous layer via the stopcock.

Unstoppering prevents a vacuum, allowing the denser bottom layer to flow out smoothly.

4

Determine the final collection step.

Collecting the remaining upper organic layer in a clean flask.

The organic solvent containing the extracted solute remains in the funnel after the lower layer is removed.

Key Concept

Procedural execution of liquid-liquid solvent extraction using a separating funnel

Estimated Time:1m 30s

Question 8659Question →

Arrange the following ecological spatial units in order of increasing geographic scale and structural complexity, starting from the most localized micro-environment to the broadest regional ecological zone.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence from smallest to largest ecological scale is Microhabitat, Habitat, Ecosystem, and Biome.

The correct sequence begins with the microhabitat, which is the most localized physical space (such as the space under a rotting log). This exists within a habitat, which describes the general physical environment where organisms live. Combining living communities in a habitat with non-living abiotic factors creates a functional ecosystem. Finally, a biome represents the broadest regional ecological zone defined by macro-climate and dominant vegetation.

Step-by-Step Solution

1

Identify the smallest and most localized physical environmental unit.

The Microhabitat is a specialized, small-scale physical location providing immediate specific microclimatic conditions.

Microhabitats exist as sub-units within larger habitat environments.

2

Identify the broader physical environment encompassing multiple micro-environments.

The Habitat is the natural address or localized area where populations of organisms live.

A habitat provides the overall living space for organisms across a community.

3

Integrate living biological communities with their non-living physical surroundings.

The Ecosystem incorporates all biotic factors interacting with abiotic factors (light, soil, water, temperature).

An ecosystem extends beyond physical space to include functional energy and nutrient interactions.

4

Determine the broad regional ecological classification.

The Biome encompasses major geographical zones sharing characteristic climate patterns and vegetation types.

Biomes are regional aggregations of similar ecosystems worldwide.

Key Concept

Hierarchy of Ecological Spatial Scale and Ecosystem Structure

Question 8660Question →

Arrange the following steps in the correct chronological sequence to illustrate how modern evolutionary theory (Neo-Darwinism) explains a change in allele frequency within a population.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct chronological sequence begins with a random germline mutation introducing a new allele, followed by phenotypic expression of a beneficial trait, differential reproductive success of favored individuals, and finally an increase in allele frequency in the population's gene pool.

Modern evolutionary theory specifies that random gene mutations generate novel alleles, which express phenotypic traits. Environmental selective pressures favor individuals bearing advantageous phenotypes, leading to higher reproductive success. Consequently, the frequency of those advantageous alleles increases in the population's gene pool over time.

Step-by-Step Solution

1

Identify the origin of new genetic variation in population genetics.

Random gene mutation in germline cells creates a new allele.

According to the modern synthesis, mutation is the ultimate primary source of novel genetic variation.

2

Determine how the new allele manifests in the organism.

The novel allele is expressed phenotypically and provides an adaptive advantage.

Natural selection operates on phenotypic variations produced by underlying genotypes.

3

Analyze the impact of the adaptive trait on reproduction.

Organisms with the trait survive better and leave more offspring (differential reproduction).

Advantageous traits improve fitness, enabling higher reproductive output.

4

Trace the population-level genetic outcome over generations.

The beneficial allele's frequency rises in the gene pool across generations.

Evolution at the microevolutionary scale is defined as a change in population allele frequencies over time.

Key Concept

Neo-Darwinian Mechanism of Evolution

Answer

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