All practice questions

Question 8681Question →

A $1.50\text{ g}$ sample of an impure hydrated dibasic acid, $\text{H}_2\text{X}\cdot 2\text{H}_2\text{O}$ (molar mass of anhydrous $\text{H}_2\text{X} = 90.0\text{ g mol}^{-1}$ ), was dissolved in distilled water and made up to $250.0\text{ cm}^3$ of solution in a volumetric flask. A $25.0\text{ cm}^3$ portion of this acid solution required $20.0\text{ cm}^3$ of $0.100\text{ mol dm}^{-3}$ sodium hydroxide ( $\text{NaOH}$ ) solution for complete neutralization. What is the percentage purity of the hydrated acid sample?

60.0\%

42.0\%

84.0\%

8.4\%

Show answer & explanation

Answer: $84.0\%$

Answer

The percentage purity of the hydrated acid sample is

84.0\%

.

The correct answer is

84.0\%

. Each mole of dibasic acid reacts with 2 moles of

\text{NaOH}

. The

20.0\text{ cm}^3

of

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

\text{NaOH}

contains

0.0020\text{ mol}

of base, neutralizing

0.0010\text{ mol}

of acid in

25.0\text{ cm}^3

. Scaling to the total

250.0\text{ cm}^3

gives

0.010\text{ mol}

of pure acid in the flask. Multiplying by the hydrated molar mass of

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

(

90.0 + 36.0

) yields

1.26\text{ g}

of pure acid, which corresponds to

(1.26 / 1.50) \times 100\% = 84.0\%

purity.

Step-by-Step Solution

1

Calculate molar mass of the hydrated acid and write balanced neutralization equation

Molar mass of

\text{H}_2\text{X}\cdot 2\text{H}_2\text{O} = 90.0 + 2(18.0) = 126.0\text{ g mol}^{-1}

. Reaction equation:

\text{H}_2\text{X} + 2\text{NaOH} \rightarrow \text{Na}_2\text{X} + 2\text{H}_2\text{O}

, so mole ratio

n_a : n_b = 1 : 2

.

The acid is dibasic, meaning each mole of acid reacts with two moles of sodium hydroxide, and the molar mass must include the water of crystallization.

2

Calculate the moles of base reacted and corresponding moles of acid in the

25.0\text{ cm}^3

aliquot

\text{Moles of NaOH} = \frac{20.0\text{ cm}^3}{1000} \times 0.100\text{ mol dm}^{-3} = 0.0020\text{ mol}

. Moles of acid in

25.0\text{ cm}^3 = \frac{0.0020}{2} = 0.0010\text{ mol}

.

Applying the stoichiometric ratio

n_a/n_b = 1/2

converts the moles of base used to moles of dibasic acid neutralized.

3

Scale the moles of pure acid to the total

250.0\text{ cm}^3

solution volume and determine pure mass

\text{Moles of pure acid in } 250.0\text{ cm}^3 = 0.0010\text{ mol} \times \left(\frac{250.0}{25.0}\right) = 0.010\text{ mol}

.

\text{Mass of pure hydrated acid} = 0.010\text{ mol} \times 126.0\text{ g mol}^{-1} = 1.26\text{ g}

.

The entire sample was dissolved to make 250 cm³, so multiplying by the dilution factor (10) gives the total moles of pure acid in the sample.

4

Calculate percentage purity of the sample

\text{Percentage purity} = \left(\frac{1.26\text{ g}}{1.50\text{ g}}\right) \times 100\% = 84.0\%

.

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

Key Concept

Volumetric Analysis and Percentage Purity Calculation of a Hydrated Dibasic Acid

Question 8682Question →

Arrange the following microbial and biochemical transformations of nitrogen in sequential order, beginning with the fixation of atmospheric dinitrogen ( $N_2$ ) gas by symbiotic root nodule bacteria and ending with the release of gaseous dinitrogen back into the atmosphere.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of transformations is: (1) Reduction of atmospheric dinitrogen gas (

N_2

) by symbiotic *Rhizobium* inside root nodules, followed by (2) Decomposition of organic nitrogen wastes into ammonium ions (

NH_4^+

) by ammonifying saprophytes, then (3) Oxidation of ammonium ions (

NH_4^+

) to nitrite (

NO_2^-

) by *Nitrosomonas*, followed by (4) Oxidation of nitrite (

NO_2^-

) to nitrate (

NO_3^-

) by *Nitrobacter*, and finally (5) Reduction of soil nitrates (

NO_3^-

) to dinitrogen gas (

N_2

) by *Pseudomonas* under anaerobic conditions.

The biological nitrogen cycle begins with nitrogen fixation by *Rhizobium*, which converts inert atmospheric dinitrogen (

N_2

) into organic amino acids and proteins in legumes. Upon plant death or excretion, ammonifying decomposers convert organic nitrogen into ammonium ions (

NH_4^+

). Two-step nitrification follows: first, *Nitrosomonas* oxidizes ammonium to nitrite (

NO_2^-

), and second, *Nitrobacter* oxidizes nitrite to nitrate (

NO_3^-

). Finally, anaerobic *Pseudomonas* carries out denitrification, reducing nitrates back to atmospheric dinitrogen gas (

N_2

), completing the cycle.

Step-by-Step Solution

1

Identify the initial process fixing elemental nitrogen gas (

N_2

) into biological systems.

Symbiotic fixation by *Rhizobium* in root nodules converts gaseous

N_2

into organic nitrogen compounds.

Atmospheric nitrogen cannot be directly utilized by plants without biological fixation by specialized prokaryotes.

2

Trace the movement of organic nitrogen through biological consumption and excretion to ammonification.

Saprophytic bacteria and fungi break down organic nitrogen compounds into inorganic ammonium ions (

NH_4^+

).

Ammonification is necessary to release bound organic nitrogen from dead tissues and excretions back into soil ionic forms.

3

Determine the first step of nitrification.

Chemoautotrophic *Nitrosomonas* bacteria oxidize ammonium ions (

NH_4^+

) to nitrite ions (

NO_2^-

).

Nitrification proceeds in two distinct obligate stages, starting with ammonium oxidation.

4

Determine the second step of nitrification.

*Nitrobacter* bacteria oxidize toxic nitrite ions (

NO_2^-

) into bioavailable nitrate ions (

NO_3^-

).

Nitrate is the chief chemical form of nitrogen absorbed and assimilated by terrestrial plants.

5

Identify the closing pathway of the cycle returning nitrogen to the gaseous state.

Anaerobic denitrifying bacteria such as *Pseudomonas* reduce nitrates (

NO_3^-

) back into atmospheric dinitrogen gas (

N_2

).

Denitrification prevents complete accumulation of soil nitrates and restores atmospheric dinitrogen balance.

Key Concept

Biogeochemical Nitrogen Cycle Transformation Pathway

Estimated Time:1m 30s

Question 8683Question →

Match each core concept of Charles Darwin's theory of natural selection on the left with its corresponding description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Overproduction matches with organisms producing more offspring than the environment can support; Struggle for existence matches with individuals competing for limited environmental resources; Survival of the fittest matches with organisms with advantageous traits being more likely to survive and reproduce; Inheritance of variation matches with favorable inherited traits being transmitted to subsequent generations.

Each core concept of natural selection directly aligns with a key evolutionary process: overproduction creates resource pressure; scarce resources lead to competition (struggle for existence); advantageous traits enable differential reproductive success (survival of the fittest); and passing those favorable traits to offspring alters population characteristics over time.

Step-by-Step Solution

1

Define overproduction within Darwin's theoretical framework.

Species have a natural tendency to produce more offspring than their habitat can sustain.

High reproductive potential creates population pressure against finite natural resources.

2

Identify the primary consequence of overproduction and limited resources.

Organisms must engage in a struggle for existence.

Scarcity of essential resources like food, light, and shelter leads to competition.

3

Relate individual differences to survival outcomes.

Organisms with advantageous variations undergo differential survival (survival of the fittest).

Traits that better adapt an individual to its environment increase its probability of surviving and reproducing.

4

Explain how natural selection affects future generations.

Advantageous variations are inherited by offspring.

Heritable favorable traits become increasingly common in subsequent generations.

Key Concept

Core Concepts and Principles of Natural Selection

Question 8684Question →

A farmer repeatedly sprays a synthetic insecticide on a vegetable farm over several growing seasons. Initially, the chemical kills nearly all target pests, but after several years, the insecticide loses its effectiveness and pest populations surge. According to Darwin's theory of natural selection, which of the following statements best explains how resistance arose in this insect population?

Insects exposed to non-lethal doses of the insecticide gradually built up immunity during their lifespan and passed this acquired tolerance to their offspring.

A small fraction of insects possessed pre-existing genetic variations for resistance, enabling them to survive the chemical, reproduce, and transmit the resistant genes to subsequent generations.

The chemical spray induced targeted mutations in individual insects to allow them to adapt directly to the toxic environment.

Susceptible insects modified their behavioral habits to avoid contact with the chemical, resulting in physiological immunity over time.

Show answer & explanation

Answer: A small fraction of insects possessed pre-existing genetic variations for resistance, enabling them to survive the chemical, reproduce, and transmit the resistant genes to subsequent generations.

Answer

Pre-existing genetic variation allowed resistant individuals to survive, reproduce, and pass advantageous alleles to their offspring.

According to Charles Darwin's theory of natural selection, evolution operates on pre-existing genetic variations within a population. When an environmental selective pressure (such as an insecticide) is introduced, individuals already possessing alleles for resistance are more likely to survive and reproduce. Over successive generations, these advantageous alleles become more common, increasing overall resistance in the population.

Step-by-Step Solution

1

Identify the key mechanism of Darwinian natural selection

Natural selection relies on natural, pre-existing genetic variation present within a population prior to environmental change.

Environmental factors act as selective pressures rather than mutational triggers or causes of physiological adaptation during an individual's lifespan.

2

Apply selective pressure to the population

The insecticide acts as a selective agent, destroying susceptible individuals while resistant individuals survive (differential survival).

Survival is determined by inherited traits that confer a reproductive advantage under specific environmental conditions.

3

Trace the frequency of the advantageous trait across generations

Surviving resistant individuals reproduce, increasing the frequency of resistance alleles in future generations.

Over time, the population shifts towards a higher proportion of resistant individuals, resulting in microevolution.

Key Concept

Darwinian Natural Selection and Antibiotic/Pesticide Resistance

Question 8685Question →

During the electrolysis of a very dilute solution of sodium chloride using inert platinum electrodes, which gas is evolved at the anode and what primary factor determines its discharge?

Oxygen gas, because the position of

OH^-

in the electrochemical series is lower than that of

Cl^-

Chlorine gas, because chloride ions are present in higher concentration than hydroxide ions

Chlorine gas, because chloride ions are lower than hydroxide ions in the electrochemical series

Hydrogen gas, because the nature of the platinum electrode promotes cation oxidation

Show answer & explanation

Answer: Oxygen gas, because the position of $OH^-$ in the electrochemical series is lower than that of $Cl^-$

Answer

Oxygen gas is evolved at the anode because the position of hydroxide ions (

OH^-

) in the electrochemical series is lower than that of chloride ions (

Cl^-

).

In a very dilute solution of sodium chloride, the position of ions in the electrochemical series is the primary factor determining preferential discharge. Because hydroxide ions (

OH^-

) lie below chloride ions (

Cl^-

) in the electrochemical series,

OH^-

ions are preferentially discharged at the anode to produce oxygen gas.

Step-by-Step Solution

1

Identify all anions migrating to the positive electrode (anode).

The anions present in aqueous

NaCl

are

Cl^-

from sodium chloride and

OH^-

from the auto-ionization of water.

Negatively charged ions migrate to the anode during electrolysis.

2

Determine which factor governs preferential discharge in this setup.

Because the solution is explicitly stated as very dilute, position in the electrochemical series predominates over concentration.

Ion concentration only overrides standard discharge potential when the concentration of a competing ion is significantly high.

3

Compare the electrochemical series positions of the anions.

OH^-

lies below

Cl^-

in the electrochemical series, so

OH^-

is oxidized preferentially to produce oxygen gas:

4OH^- \rightarrow 2H_2O + O_2 + 4e^-

.

Anions positioned lower in the series require less energy to give up electrons.

Key Concept

Factors affecting preferential discharge of ions in electrolysis: position in the electrochemical series vs. ion concentration.

Question 8686Question →

Stainless steel exhibits high resistance to atmospheric corrosion primarily because incorporated chromium reacts with oxygen to form a micro-thin, passive oxide surface layer that prevents further oxidation of the underlying iron.

True

False

Show answer & explanation

Answer: True

Answer

The statement is TRUE. The addition of chromium to iron in stainless steel leads to the formation of an adherent, passive chromium(III) oxide (

Cr_2O_3

) film on the surface, protecting the alloy from ongoing oxidation.

Chromium in stainless steel reacts with ambient oxygen to generate a thin, self-repairing surface film of

Cr_2O_3

. This passive layer serves as a barrier preventing moisture and oxygen from coming into direct contact with the iron matrix, stopping corrosion.

Step-by-Step Solution

1

Identify the primary alloying constituent responsible for corrosion resistance in stainless steel.

Stainless steel is an alloy composed of iron (

Fe

), carbon (

C

), chromium (

Cr

), and often nickel (

Ni

). Chromium is the element specifically added to impart corrosion resistance.

Understanding the chemical roles of individual constituent metals in an alloy is necessary to evaluate physical and chemical property modifications.

2

Analyze the surface chemical reaction between chromium in the alloy and atmospheric oxygen.

Chromium oxidizes preferentially over iron to form a continuous, insoluble layer of chromium oxide (

Cr_2O_3

).

This process, known as passivation, prevents reactive agents like water and oxygen from penetrating to the core iron atoms.

3

Evaluate the correctness of the statement based on passivation principles.

The statement accurately describes the passivation process of stainless steel.

Because chromium passivation is the accepted scientific mechanism for stainless steel's rust prevention, the statement is true.

Key Concept

Passivation and Corrosion Resistance of Chromium in Steel Alloys

Question 8687Question →

Match each ecological pyramid structural characteristic or anomaly on the left with its correct ecological or thermodynamic explanation on the right.

Click a left item, then click its matching right item

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Answer

Inverted marine biomass pyramid pairs with rapid turnover of phytoplankton; strictly upright energy pyramid pairs with second law of thermodynamics energy dissipation; inverted forest tree numbers pyramid pairs with a single large producer supporting many smaller organisms; and upright grassland biomass pyramid pairs with high standing crop biomass.

Each ecological pyramid structural phenomenon directly reflects how standing crop measurements, turnover rates, or thermodynamic energy dissipation shape the trophic structure of ecosystems.

Step-by-Step Solution

1

Analyze the inverted biomass pyramid in marine environments

Phytoplankton reproduce and are consumed rapidly, resulting in low standing crop biomass at any instant but high productivity, creating an inverted biomass pyramid.

Measures of standing crop biomass at a single moment differ from total energy production over time.

2

Analyze why energy pyramids are strictly upright

Energy transfer between trophic levels is inefficient (typically around 10%), as heat energy is lost via respiration (Second Law of Thermodynamics).

Energy cannot be recycled or inverted because total usable energy strictly decreases at each successive trophic level.

3

Analyze inverted numbers pyramid in a tree habitat

Physical size of individual organisms dictates the count; one massive oak tree supports thousands of caterpillars or birds.

Pyramids of numbers count individual organisms rather than biomass or energy content.

4

Analyze upright terrestrial biomass pyramid

Grasses and plants accumulate substantial structural plant matter, yielding a high standing crop biomass compared to herbivores.

Terrestrial producers have longer lifespans and lower turnover rates compared to aquatic phytoplankton.

Key Concept

Thermodynamics and Trophic Structure of Ecological Pyramids

Question 8688Question →

Arrange the following stages of the fern (pteridophyte) reproductive cycle in their correct chronological sequence, starting from spore germination:

Drag items to arrange them in the correct order

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Answer

The correct chronological sequence is: Germination of a haploid spore into a green cell filament → Development of a heart-shaped photosynthetic prothallus bearing sex organs → Transfer of flagellated sperm through water to fertilize the egg inside an archegonium → Emergence and growth of a diploid leafy sporophyte from the prothallus.

In pteridophytes such as ferns, the life cycle exhibits alternation of generations where the haploid spore germinates first into a small, photosynthetic, heart-shaped prothallus (gametophyte). The prothallus produces gametes in sex organs (antheridia and archegonia). Swimming flagellated sperm require water to reach the egg cell inside the archegonium for fertilization. Once fertilized, the diploid zygote grows into the familiar leafy vascular plant, which is the dominant sporophyte generation.

Step-by-Step Solution

1

Identify the initial reproductive unit.

Haploid spores dislodged from sori germinate in moist soil to form an initial filament.

Spores represent the start of the gametophyte generation.

2

Identify the mature gametophyte structure.

The filament grows into a photosynthetic, heart-shaped prothallus.

In pteridophytes, the prothallus is the free-living gametophyte stage.

3

Determine the fertilization requirement and process.

Flagellated sperm swim through environmental water to reach the egg inside the archegonium.

Fertilization unites haploid gametes into a diploid zygote.

4

Trace the growth of the new generation.

The diploid zygote divides and develops into the mature vascular fern plant (sporophyte).

The sporophyte eventually becomes independent as the prothallus degenerates.

Key Concept

Fern Life Cycle and Alternation of Generations in Pteridophytes

Estimated Time:1m 0s

Question 8689Question →

Match each prokaryotic cellular structure or morphological arrangement of Kingdom Monera listed on the left with its corresponding biological description or function on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Lophotrichous arrangement matches with presence of a tuft of flagella localized at a single pole; Streptococcal morphology matches with spherical prokaryotic cells adhering together in linear unbranched chains; Akinete matches with enlarged, thick-walled resting cell in cyanobacteria for survival during adverse conditions; Carboxysome matches with protein-bound microcompartment containing RuBisCO for inorganic carbon fixation.

Each Moneran cellular structure or arrangement correctly matches its definition: Lophotrichous arrangement corresponds to a cluster of flagella at one pole; Streptococcal morphology corresponds to spherical cells forming chains; Akinetes are cyanobacterial thick-walled resting spores; and Carboxysomes are protein compartments housing RuBisCO for carbon fixation.

Step-by-Step Solution

1

Analyze bacterial flagellar patterns

Lophotrichous flagellation refers specifically to a tuft of flagella located at one end (pole) of the bacterial cell.

Differentiating flagellar arrangements (monotrichous, amphitrichous, lophotrichous, peritrichous) is a key morphological classification feature in Kingdom Monera.

2

Analyze bacterial cellular groupings

Cocci that divide along one axis and remain attached in chain-like filaments are termed streptococci.

Distinguishing chain arrangements (streptococci) from cluster arrangements (staphylococci) is essential for bacterial identification.

3

Evaluate specialized cyanobacterial survival structures

Akinetes are enlarged, thick-walled, food-storing resting cells that allow filamentous cyanobacteria to endure freezing or desiccation.

Differentiating akinetes (resting survival cells) from heterocysts (nitrogen-fixing cells) is vital in cyanobacterial biology.

4

Identify prokaryotic carbon-fixation microcompartments

Carboxysomes are protein inclusions packed with RuBisCO that concentrate CO2 near the enzyme within autotrophic cyanobacteria.

Understanding sub-cellular compartmentation in prokaryotic autotrophy highlights metabolic adaptations in Monera.

Key Concept

Structural Diversity and Microcompartments in Kingdom Monera

Question 8690Question →

Match each ecosystem component or ecological factor on the left with its correct description on the right.

Click a left item, then click its matching right item

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Answer

Biotic Factor matches living organism that affects an ecosystem; Abiotic Factor matches non-living physical or chemical element; Primary Producer matches autotrophic organism capable of manufacturing organic food; Decomposer matches saprophytic organism that breaks down dead organic matter.

Each ecological concept is matched directly to its definition: Biotic Factor describes living components; Abiotic Factor describes non-living environmental elements; Primary Producer describes autotrophic organisms synthesizing organic food; and Decomposer describes saprophytic organisms recycling organic waste.

Step-by-Step Solution

1

Identify the definition of living vs non-living ecosystem factors.

Biotic refers to living components (plants, animals, microbes), while Abiotic refers to non-living physical components (light, temperature, soil pH).

Ecosystem structure is broadly split into biotic (living) and abiotic (non-living) parts.

2

Determine the role of food-producing organisms.

Primary producers generate organic nutrients via autotrophic processes like photosynthesis.

Producers convert solar energy into chemical energy stored in organic molecules.

3

Determine the role of organisms responsible for nutrient recycling.

Decomposers break down dead tissue and return simple nutrients back into the ecosystem.

Saprophytic action prevents organic matter build-up and closes the biogeochemical cycle.

Key Concept

Components and Structural Organization of an Ecosystem

Question 8691Question →

What volume of nitrogen(IV) oxide gas, measured at standard temperature and pressure (STP), is evolved when $12.7\text{ g}$ of copper completely reacts with excess concentrated trioxonitrate(V) acid according to the equation below?

Cu_{(s)} + 4HNO_{3(aq)} \rightarrow Cu(NO_3)_{2(aq)} + 2NO_{2(g)} + 2H_2O_{(l)}

(Relative atomic mass: $Cu = 63.5$ ; Molar volume of gas at STP = $22.4\text{ dm}^3\text{/mol}$ )

8.96\text{ dm}^3

9.60\text{ dm}^3

4.48\text{ dm}^3

17.92\text{ dm}^3

Show answer & explanation

Answer: $8.96\text{ dm}^3$

Answer

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

8.96\text{ dm}^3

.

According to the balanced chemical equation,

1\text{ mol}

of copper reacts with excess concentrated trioxonitrate(V) acid to produce

2\text{ mol}

of nitrogen(IV) oxide gas. Since

12.7\text{ g}

of copper corresponds to

0.2\text{ mol}

, the reaction generates

0.4\text{ mol}

of

NO_2

. At STP,

0.4\text{ mol}

occupies

0.4 \times 22.4 = 8.96\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the amount of copper in moles

\text{Moles of } Cu = \frac{12.7\text{ g}}{63.5\text{ g/mol}} = 0.2\text{ mol}

Dividing given mass by relative atomic mass yields moles of reactant.

2

Determine moles of

NO_2

gas produced using stoichiometric coefficients

\text{Moles of } NO_2 = 0.2\text{ mol } Cu \times \left(\frac{2\text{ mol } NO_2}{1\text{ mol } Cu}\right) = 0.4\text{ mol}

The balanced chemical equation shows that

1\text{ mol}

of

Cu

yields

2\text{ mol}

of

NO_2

gas.

3

Calculate the volume of

NO_2

gas evolved at STP

\text{Volume of } NO_2 = 0.4\text{ mol} \times 22.4\text{ dm}^3\text{/mol} = 8.96\text{ dm}^3

One mole of any gas occupies

22.4\text{ dm}^3

at standard temperature and pressure.

Key Concept

Redox stoichiometric calculation of gas volumes produced by trioxonitrate(V) acid reactions

Question 8692Question →

In an industrial electroplating plant, a steel component is coated with silver in an electrolytic bath. If a constant electric current of $2.0\text{ A}$ is passed through the bath for $48.25\text{ minutes}$ , what is the mass of silver, in grams, deposited on the cathode? [Molar mass of $\text{Ag} = 108\text{ g mol}^{-1}$ , $1\text{ F} = 96500\text{ C mol}^{-1}$ ]

Show answer & explanation

Answer: 6.48

Answer

The mass of silver deposited on the cathode during electroplating is 6.48 g.

According to Faraday's first law of electrolysis, charge

Q = 2.0\text{ A} \times (48.25 \times 60\text{ s}) = 5790\text{ C}

. The number of moles of electrons passed is

5790 / 96500 = 0.06\text{ mol}

. Since silver reduction (

\text{Ag}^+ + e^- \rightarrow \text{Ag}

) requires

1\text{ mole}

of electrons per mole of silver,

0.06\text{ mol}

of

\text{Ag}

is formed. The mass of silver deposited is

0.06\text{ mol} \times 108\text{ g mol}^{-1} = 6.48\text{ g}

.

Step-by-Step Solution

1

Convert time to seconds

t = 48.25 \text{ min} \times 60 \text{ s/min} = 2895 \text{ s}

Time must be expressed in seconds to calculate electric charge in coulombs.

2

Calculate total quantity of electricity

Q

Q = I \times t = 2.0 \text{ A} \times 2895 \text{ s} = 5790 \text{ C}

Electric charge is the product of current and duration of electrolysis.

3

Determine moles of electrons transferred

n(e^-) = \frac{5790 \text{ C}}{96500 \text{ C mol}^{-1}} = 0.06 \text{ mol}

Faraday's constant indicates that

96500\text{ C}

corresponds to

1\text{ mole}

of electrons.

4

Relate electron flow to silver discharge at cathode

\text{Ag}^+ + e^- \rightarrow \text{Ag}

, so

0.06 \text{ mol of } e^- \text{ yields } 0.06 \text{ mol of Ag}

Silver ion discharge requires one electron per silver atom deposited.

5

Calculate mass of silver deposited

\text{Mass} = 0.06 \text{ mol} \times 108 \text{ g mol}^{-1} = 6.48 \text{ g}

Multiplying the molar quantity by relative atomic mass yields the total mass deposited.

Key Concept

Quantitative application of Faraday's laws of electrolysis in industrial electroplating.

Question 8693Question →

Match each chemical species to its correct molecular geometry and central atom hybridization state.

Click a left item, then click its matching right item

Show answer & explanation

Answer

BeCl_2

matches Linear shape,

sp

hybridization;

BF_3

matches Trigonal planar shape,

sp^2

hybridization;

CH_4

matches Tetrahedral shape,

sp^3

hybridization;

SF_6

matches Octahedral shape,

sp^3d^2

hybridization.

Each chemical species is matched to its corresponding molecular geometry and central atom hybridization based on the number of sigma bonds and lone pairs present on the central atom.

Step-by-Step Solution

1

Determine steric number for

BeCl_2

Be

forms 2 single bonds with 0 lone pairs, giving a steric number of 2 (

sp

hybridization, linear shape).

Two electron domains arrange at 180° to minimize electron pair repulsion.

2

Determine steric number for

BF_3

B

forms 3 single bonds with 0 lone pairs, giving a steric number of 3 (

sp^2

hybridization, trigonal planar shape).

Three electron domains arrange at 120° in a single plane.

3

Determine steric number for

CH_4

C

forms 4 single bonds with 0 lone pairs, giving a steric number of 4 (

sp^3

hybridization, tetrahedral shape).

Four electron domains arrange symmetrically in three-dimensional space at 109.5°.

4

Determine steric number for

SF_6

S

forms 6 single bonds with 0 lone pairs, giving a steric number of 6 (

sp^3d^2

hybridization, octahedral shape).

Six electron domains arrange symmetrically at 90° axial/equatorial positions.

Key Concept

Valence Shell Electron Pair Repulsion (VSEPR) Theory and Orbital Hybridization

Estimated Time:1m 0s

Question 8694Question →

Match each synthetic polymer or biomolecule listed in Column A with its corresponding chemical linkage and structural classification in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Nylon-6,6 pairs with polyamide linkage formed from hexanedioic acid and hexane-1,6-diamine; Terylene pairs with polyester linkage formed from benzene-1,4-dicarboxylic acid and ethane-1,2-diol; Starch pairs with glycosidic linkage formed from

\alpha

-D-glucose monomers; Protein pairs with peptide linkage formed from

\alpha

-amino acid monomers.

Each polymer is correctly matched to its functional linkage and monomer constituents: Nylon-6,6 is a polyamide formed from hexanedioic acid and hexane-1,6-diamine; Terylene is a polyester formed from benzene-1,4-dicarboxylic acid and ethane-1,2-diol; Starch is a polysaccharide held together by glycosidic linkages between glucose units; Protein is a natural polymer made of amino acids linked by peptide bonds.

Step-by-Step Solution

1

Identify the monomer composition and functional groups of Nylon-6,6

Nylon-6,6 contains amide linkages formed between carboxylic acid (

-\text{COOH}

) groups of hexanedioic acid and amino (

-\text{NH}_2

) groups of hexane-1,6-diamine.

Synthetic polyamides require a di-acid and a di-amine reactant.

2

Identify the monomer composition and functional groups of Terylene

Terylene contains ester linkages (

-\text{COO}-

) formed between benzene-1,4-dicarboxylic acid and ethane-1,2-diol.

Polyesters are produced by reacting a dicarboxylic acid with a dihydric alcohol (diol).

3

Determine the structural linkages present in Starch

Starch is a polysaccharide composed of

\alpha

-D-glucose monomers linked via condensation through glycosidic bonds.

Carbohydrates form ether-like condensation links known as glycosidic linkages.

4

Determine the structural linkages present in Proteins

Proteins are natural polymers made of

\alpha

-amino acids joined together by peptide linkages (

-\text{CO}-\text{NH}-

).

The reaction between the carboxyl group of one amino acid and the amino group of another forms a peptide bond.

Key Concept

Structural linkages and monomeric constituents of synthetic condensation polymers and natural biomolecules

Estimated Time:1m 30s

Question 8695Question →

Cobalt is a first-row transition element with an atomic number of 27. What is the ground-state electronic configuration of the cobalt(II) ion, $Co^{2+}$ ?

[Ar] 3d^7

[Ar] 3d^5 4s^2

[Ar] 3d^6 4s^1

[Ar] 3d^9

Show answer & explanation

Answer: $[Ar] 3d^7$

Answer

The electronic configuration of the cobalt(II) ion,

Co^{2+}

, is

[Ar] 3d^7

.

Neutral cobalt (

Z = 27

) has the electronic configuration

[Ar] 3d^7 4s^2

. When transition metals ionize, electrons are lost first from the outermost

4s

orbital because

n=4

electrons experience lower electrostatic attraction from the nucleus than

n=3

electrons. Removing two electrons yields

[Ar] 3d^7

.

Step-by-Step Solution

1

Determine the ground-state electronic configuration of neutral cobalt (

Co

,

Z = 27

).

The neutral cobalt atom has 27 electrons, giving an electronic configuration of

[Ar] 3d^7 4s^2

.

According to the Aufbau principle, orbitals are filled in increasing order of energy, placing two electrons in the

4s

orbital and seven in the

3d

orbitals.

2

Apply cation formation rules to remove two electrons for the

Co^{2+}

ion.

Two electrons are removed from the outermost principal energy level (

n = 4

), which is the

4s

orbital.

Although

4s

is filled before

3d

, electrons in the principal shell with the highest principal quantum number (

n = 4

) are held least tightly and are lost first upon ionization.

3

Write the resulting electronic configuration for

Co^{2+}

.

[Ar] 3d^7

Removing the two

4s

electrons leaves seven electrons in the

3d

subshell.

Key Concept

Electronic Configuration of Transition Metal Cations

Question 8696Question →

Which of the following carbonyl compounds yields a secondary alcohol upon reduction with lithium tetrahydridoaluminate(III) ( $LiAlH_4$ )?

Propanone

Ethanal

Butanal

Methanal

Show answer & explanation

Answer: Propanone

Answer

Propanone yields a secondary alcohol (propan-2-ol) when reduced by lithium tetrahydridoaluminate(III).

Propanone is a ketone (alkanone). When reduced with lithium tetrahydridoaluminate(III) (

LiAlH_4

), the carbonyl carbon (

C=O

) is converted to a secondary alcohol group (

-CH(OH)-

). Specifically, propanone (

CH_3COCH_3

) is reduced to propan-2-ol (

CH_3CH(OH)CH_3

), which contains a carbon atom bonded to two other carbon atoms and the hydroxyl group.

Step-by-Step Solution

1

Identify the functional groups of the given carbonyl compounds.

Propanone (

CH_3COCH_3

) is a ketone (alkanone), whereas ethanal (

CH_3CHO

), butanal (

CH_3CH_2CH_2CHO

), and methanal (

HCHO

) are aldehydes (alkanals).

Alkanals and alkanones exhibit distinct reduction behaviors based on the position of the carbonyl group (

C=O

).

2

Apply the general reduction reaction rules for alkanals and alkanones using

LiAlH_4

.

Reduction of an alkanal (

R-CHO

) produces a primary alcohol (

R-CH_2OH

). Reduction of an alkanone (

R-CO-R'

) produces a secondary alcohol (

R-CH(OH)-R'

).

The hydride ion (

H^-

) adds to the carbonyl carbon atom, converting the ketone group into a secondary hydroxyl group.

3

Determine which compound forms a secondary alcohol.

Propanone (

CH_3COCH_3

) is reduced to propan-2-ol (

CH_3CH(OH)CH_3

), which is a secondary alcohol.

Propan-2-ol has the hydroxyl-bearing carbon attached to two other carbon atoms, fitting the definition of a secondary alcohol.

Key Concept

Reduction of Alkanals and Alkanones

Question 8697Question →

A sample of human salivary amylase was boiled at $100^\circ\text{C}$ for ten minutes and then cooled to $37^\circ\text{C}$ before being added to a starch solution at optimal $\text{pH}$ . After incubating for an hour, iodine testing revealed that starch was still present. Which of the following statements explains why starch digestion did not occur?

The high temperature permanently altered the 3D structure of the enzyme active site, causing denaturation.

The enzyme became temporarily inactive due to high temperature and required freezing to restore its catalytic activity.

Boiling changed the mode of digestion from intracellular enzyme action to extracellular saprophytic nutrition.

The amylase molecules were converted into starch molecules by the extreme heat treatment.

Show answer & explanation

Answer: The high temperature permanently altered the 3D structure of the enzyme active site, causing denaturation.

Answer

The high temperature permanently altered the 3D structure of the enzyme active site, causing denaturation.

Enzymes are protein molecules with precise three-dimensional active sites. Exposing salivary amylase to

100^\circ\text{C}

causes heat denaturation, breaking bonds that maintain its tertiary structure. This structural collapse permanently destroys the active site, preventing starch binding even after the temperature returns to

37^\circ\text{C}

.

Step-by-Step Solution

1

Identify the nature of salivary amylase and the impact of extreme heat.

Salivary amylase is a protein enzyme designed to function optimally around body temperature (

37^\circ\text{C}

).

Enzyme activity depends heavily on its specific three-dimensional tertiary structure and active site shape.

2

Analyze the biological effect of boiling (

100^\circ\text{C}

) on protein enzymes.

High temperatures break chemical bonds holding the tertiary protein structure together, leading to irreversible denaturation.

Once denatured, the active site can no longer bind its substrate (starch), so catalysis cannot take place even after cooling back to

37^\circ\text{C}

.

Key Concept

Effect of Temperature on Digestive Enzyme Structure and Function

Question 8698Question →

A man with blood group AB ( $I^A I^B$ ) marries a woman with blood group A who is heterozygous ( $I^A i$ ). What is the probability that their first child will have blood group A?

50\%

75\%

25\%

100\%

Show answer & explanation

Answer: $50\%$

Answer

The probability that their child will have blood group A is

50\%

(or

1/2

).

The option stating 50% is correct because crossing parental genotypes

I^A I^B

and

I^A i

yields four equally likely offspring genotypes:

I^A I^A

(blood group A),

I^A i

(blood group A),

I^A I^B

(blood group AB), and

I^B i

(blood group B). Combining the two genotypes that result in blood group A (

25\% + 25\%

) gives a total probability of

50\%

.

Step-by-Step Solution

1

Identify the parental genotypes and gametes produced.

Father's genotype is

I^A I^B

(gametes:

I^A

,

I^B

). Mother's genotype is

I^A i

(gametes:

I^A

,

i

).

Heterozygous blood group A carries the recessive allele

i

, while blood group AB expresses both

I^A

and

I^B

codominantly.

2

Construct a Punnett square for the cross

I^A I^B \times I^A i

.

The four possible offspring genotypes are

I^A I^A

(

25\%

),

I^A i

(

25\%

),

I^A I^B

(

25\%

), and

I^B i

(

25\%

).

Combining male and female gametes gives all expected genetic ratios.

3

Determine the phenotypic expression for each genotype.

I^A I^A

and

I^A i

both express blood group A (

25\% + 25\% = 50\%

).

I^A I^B

expresses blood group AB (

25\%

).

I^B i

expresses blood group B (

25\%

).

Alleles

I^A

and

I^B

are codominant with each other, and both are completely dominant over the recessive allele

i

.

Key Concept

Codominance and Multiple Alleles in Human ABO Blood Groups

Question 8699Question →

In sweet pea plants (*Lathyrus odoratus*), purple flower color ( $P$ ) is dominant over red flower color ( $p$ ), and long pollen grain ( $L$ ) is dominant over round pollen grain ( $l$ ). If a heterozygous dihybrid plant with the genotype $PpLl$ is self-pollinated and yields a total of $160$ F2 seeds, how many of these seeds are expected to produce plants with the double recessive phenotype of red flowers and round pollen grains?

Show answer & explanation

Answer: 10

Answer

The expected number of seeds producing plants with red flowers and round pollen grains is 10.

In a dihybrid cross involving two heterozygous parents (

PpLl \times PpLl

), independent assortment results in a 9:3:3:1 phenotypic ratio in the F2 generation. The double recessive phenotype (red flowers and round pollen grains, genotype

ppll

) accounts for 1 out of 16 total offspring. Multiplying this fraction (1/16) by the total yield of 160 seeds produces an expected value of 10 seeds.

Step-by-Step Solution

1

Determine the dihybrid F2 phenotypic ratio

The phenotypic ratio for a cross between two heterozygous dihybrid parents (

PpLl \times PpLl

) is 9:3:3:1.

According to Mendel's Law of Independent Assortment, the alleles for flower color and pollen shape segregate independently during gamete formation.

2

Calculate the proportion of double recessive offspring

The fraction of offspring displaying both recessive traits (red flowers and round pollen grains, genotype

ppll

) is 1/16.

Out of 16 equal Punnett square combinations, exactly 1 combination represents the homozygous double recessive phenotype.

3

Compute the expected number of double recessive seeds

(1 / 16) * 160 = 10 seeds.

Multiplying the phenotypic probability (1/16) by the total seed population (160) yields the expected quantity.

Key Concept

Mendel's Law of Independent Assortment and Dihybrid Phenotypic Ratios

Question 8700Question →

During non-cyclic photophosphorylation in plant photosynthesis, light energy drives a sequential flow of electrons across the thylakoid membrane. What is the correct chronological sequence of physiological events occurring during this light-dependent stage from initial photon absorption to final electron reduction?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct chronological sequence of non-cyclic photophosphorylation is: 1) Excitation of P680 in Photosystem II → 2) Photolysis of water to replace electrons → 3) Electron transport through cytochrome

b_6f

complex creating a proton gradient → 4) Re-excitation at P700 in Photosystem I → 5) Reduction of

NADP^+

to

NADPH

.

The non-cyclic light reaction (Z-scheme) begins with photon absorption at Photosystem II (P680). The loss of electrons from P680 triggers the enzymatic photolysis of water to replace those electrons. The released electrons move down an electron transport chain featuring the cytochrome

b_6f

complex (generating a proton gradient), after which they reach Photosystem I (P700) where photon absorption re-excites them. Finally, ferredoxin passes the electrons to

NADP^+

reductase to reduce

NADP^+

to

NADPH

.

Step-by-Step Solution

1

Identify the initiating trigger of non-cyclic photophosphorylation.

Photon absorption by P680 (Photosystem II) excites electrons to a primary electron acceptor.

Light absorption at PS II initiates the entire Z-scheme electron transport sequence.

2

Determine how electron deficiency in P680 is resolved.

Photolysis of water splits

H_2O

into electrons,

H^+

ions, and

O_2

, supplying replacement electrons to P680.

Oxidized P680 is a strong oxidizing agent that forces water splitting at the manganese-containing complex.

3

Trace the path of energized electrons from Photosystem II.

Electrons pass down the plastoquinone-cytochrome

b_6f

-plastocyanin chain into Photosystem I.

This electron transport generates the proton motive force required for ATP synthesis via chemiosmosis.

4

Follow the fate of electrons upon reaching Photosystem I.

Electrons are re-excited by light absorption at P700 (Photosystem I) and transferred to ferredoxin.

PS I absorbs light energy to boost electrons to a redox potential high enough to reduce

NADP^+

.

5

Identify the final electron acceptor step.

NADP^+

reductase transfers electrons from ferredoxin and stromal protons to form

NADPH

.

Terminal reduction of

NADP^+

stores chemical reducing power for subsequent use in the Calvin cycle.

Key Concept

Non-cyclic Photophosphorylation and Z-scheme Electron Transport

Answer

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