All practice questions

Question 8561Question →

During primary ecological succession on a newly emerged volcanic island, which of the following structural and functional shifts characterizes the transition from the pioneer stage to a climax community?

Accumulation of soil organic matter and increased species diversity, with gross primary productivity eventually balancing total community respiration

Rapid initial colonization by deep-rooted herbaceous species that draw nutrients from pre-existing topsoil reserves beneath the volcanic substrate

Pioneer bryophytes absorbing water and mineral salts rapidly through specialized internal xylem and phloem conductive tissues

Immediate dominance by large canopy trees that suppress soil formation by exhausting organic humus deposited by pioneer colonizers

Show answer & explanation

Answer: Accumulation of soil organic matter and increased species diversity, with gross primary productivity eventually balancing total community respiration

Answer

Accumulation of soil organic matter and increased species diversity, with gross primary productivity eventually balancing total community respiration

Primary succession on bare volcanic rock requires pioneer organisms to weather substrate and contribute organic humus, forming soil over time. This enables higher plant species to invade, leading to higher species richness, structural complexity, and an energy balance where gross primary production equals community respiration at climax equilibrium.

Step-by-Step Solution

1

Identify the type of succession described in the stem.

The scenario describes primary succession on a newly emerged volcanic island, which starts on bare rock substrate without pre-existing soil.

Determining whether succession is primary or secondary establishes the baseline environmental conditions (e.g., presence or absence of soil).

2

Trace ecological changes from pioneer species to the climax stage.

Pioneer colonizers (lichens/microorganisms) secrete acids that weather bare rock. As they die and decompose, organic matter mixes with weathered minerals to form thin soil, allowing mosses, herbs, shrubs, and eventually trees to establish sequentially.

Succession involves progressive soil formation, increase in species diversity, food web complexity, and total biomass over time.

3

Analyze energy dynamics and ecosystem maturity at the climax stage.

In a mature climax community, total primary productivity (gross primary productivity) stabilizes and matches the total respiratory demands of all organisms in the community (

P/R \approx 1

).

Climax ecosystems reach a homeostatic equilibrium between gross production and total community respiration.

Key Concept

Primary Ecological Succession and Climax Community Dynamics

Question 8562Question →

A $24.4\text{ g}$ sample of hydrated barium chloride, $\text{BaCl}_2 \cdot x\text{H}_2\text{O}$ , was heated strongly in a crucible to constant mass. The residue of anhydrous barium chloride obtained weighed $20.8\text{ g}$ . What is the value of $x$ ? $[\text{Ba} = 137, \text{Cl} = 35.5, \text{H} = 1, \text{O} = 16]$

2

1

4

5

Show answer & explanation

Answer: 2

Answer

The value of

x

is 2, giving the formula

\text{BaCl}_2 \cdot 2\text{H}_2\text{O}

.

Heating the hydrated salt removes all water of crystallization, leaving

20.8\text{ g}

of anhydrous

\text{BaCl}_2

(

0.10\text{ mol}

) and releasing

3.6\text{ g}

of water vapor (

0.20\text{ mol}

). The ratio of moles of water to moles of anhydrous salt is

\frac{0.20}{0.10} = 2

, so

x = 2

.

Step-by-Step Solution

1

Calculate the mass of water of crystallization lost during heating.

\text{Mass of H}_2\text{O} = 24.4\text{ g} - 20.8\text{ g} = 3.6\text{ g}

Heating drives off all water of crystallization, leaving only anhydrous salt.

2

Determine the molar masses of anhydrous

\text{BaCl}_2

and

\text{H}_2\text{O}

.

\text{Molar mass of BaCl}_2 = 137 + 2(35.5) = 208\text{ g/mol}

;

\text{Molar mass of H}_2\text{O} = 2(1) + 16 = 18\text{ g/mol}

Molar masses are needed to convert mass quantities into mole quantities.

3

Calculate the number of moles of anhydrous

\text{BaCl}_2

and water.

\text{Moles of BaCl}_2 = \frac{20.8\text{ g}}{208\text{ g/mol}} = 0.10\text{ mol}

;

\text{Moles of H}_2\text{O} = \frac{3.6\text{ g}}{18\text{ g/mol}} = 0.20\text{ mol}

The stoichiometric coefficient

x

represents the mole ratio of water molecules per mole of salt.

4

Find the mole ratio

x = \frac{\text{Moles of H}_2\text{O}}{\text{Moles of BaCl}_2}

.

x = \frac{0.20\text{ mol}}{0.10\text{ mol}} = 2

Simplifying the mole ratio determines the integer coefficient

x

in the hydrated salt formula.

Key Concept

Determination of Water of Crystallization by Stoichiometric Gravimetric Heating

Question 8563Question →

For the reversible reaction $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$ , the equilibrium concentrations of $\text{PCl}_5$ , $\text{PCl}_3$ , and $\text{Cl}_2$ in a $1.0\text{ dm}^3$ vessel are $0.20\text{ mol dm}^{-3}$ , $0.60\text{ mol dm}^{-3}$ , and $0.30\text{ mol dm}^{-3}$ respectively. What is the numerical value of the equilibrium constant, $K_c$ ?

Show answer & explanation

Answer: 0.9

Answer

The equilibrium constant

K_c

is

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

.

The equilibrium constant

K_c

is calculated by substituting the equilibrium concentrations into

K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \frac{0.60 \times 0.30}{0.20} = 0.90\text{ mol dm}^{-3}

.

Step-by-Step Solution

1

Formulate the equilibrium constant expression

K_c

for the reaction.

K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}

The equilibrium constant expression is defined as the product of the concentrations of the reaction products divided by the product of the concentrations of the reactants, each raised to the power of their stoichiometric coefficient.

2

Substitute the equilibrium concentrations into the expression.

K_c = \frac{0.60 \times 0.30}{0.20}

Given equilibrium values:

[\text{PCl}_5] = 0.20\text{ mol dm}^{-3}

,

[\text{PCl}_3] = 0.60\text{ mol dm}^{-3}

, and

[\text{Cl}_2] = 0.30\text{ mol dm}^{-3}

.

3

Calculate the arithmetic result.

K_c = \frac{0.18}{0.20} = 0.90\text{ mol dm}^{-3}

Multiplying

0.60 \times 0.30

gives

0.18

, and dividing by

0.20

yields

0.90

.

Key Concept

Direct calculation of equilibrium constant

K_c

from equilibrium concentrations

Question 8564Question →

Four organic substances—starch, nylon-6,6, polyethene, and natural rubber—were analyzed to compare their chemical structures, reaction behaviors, and environmental properties. Which of the following statements provides a correct scientific comparison among these materials?

Starch consists of monomeric glucose units joined by glycosidic linkages that undergo acid hydrolysis, whereas nylon-6,6 contains recurring amide linkages formed via condensation polymerization.

Polyethene is a synthetic addition polymer that undergoes rapid microbial degradation in soil in a manner similar to starch.

The charring of starch when exposed to concentrated

\text{H}_2\text{SO}_4

occurs via a reversible acid-base neutralization reaction rather than chemical dehydration.

Natural rubber is an addition polymer whose double bonds react with ammoniacal silver nitrate solution to form a characteristic silver precipitate.

Show answer & explanation

Answer: Starch consists of monomeric glucose units joined by glycosidic linkages that undergo acid hydrolysis, whereas nylon-6,6 contains recurring amide linkages formed via condensation polymerization.

Answer

Starch consists of monomeric glucose units joined by glycosidic linkages that undergo acid hydrolysis, whereas nylon-6,6 contains recurring amide linkages formed via condensation polymerization.

Starch is a carbohydrate (polysaccharide) composed of glucose units connected by glycosidic bonds that can be broken down by acid hydrolysis. Nylon-6,6 is a synthetic condensation polymer (polyamide) made from a dicarboxylic acid and a diamine, containing amide linkages.

Step-by-Step Solution

1

Analyze the structural classification and linkages of starch and nylon-6,6.

Starch is a naturally occurring polysaccharide composed of glucose monomer units linked via

\alpha

-glycosidic bonds. Nylon-6,6 is a synthetic polyamide composed of hexanedioic acid and hexane-1,6-diamine monomers joined by amide (

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

) linkages with the elimination of water.

Identifying monomer units and linkage types distinguishes condensation polymers and carbohydrates.

2

Evaluate the biodegradability of synthetic addition polymers like polyethene versus natural polymers like starch.

Starch is readily degraded by enzymatic action and micro-organisms. Polyethene is a synthetic addition polymer containing strong

\text{C}-\text{C}

single bonds that resist microbial enzymatic attack, rendering it non-biodegradable.

Synthetic addition polymers lack functional linkages susceptible to hydrolysis.

3

Assess the chemical action of concentrated

\text{H}_2\text{SO}_4

on carbohydrates.

Concentrated

\text{H}_2\text{SO}_4

acts as a dehydrating agent, removing water elements from starch according to

(\text{C}_6\text{H}_{10}\text{O}_5)_n \xrightarrow{\text{conc. H}_2\text{SO}_4} 6n\text{C} + 5n\text{H}_2\text{O}

, turning the sample black.

This is a dehydration reaction, not an acid-base neutralization.

4

Analyze the chemical test reagents for unsaturation in rubber versus terminal alkynes.

Bromine water or acidified

\text{KMnO}_4

is used to test for unsaturation in alkenes and dienes like natural rubber. Ammoniacal silver nitrate selectively reacts with terminal alkynes to give a silver acetylide precipitate.

Different unsaturated functional groups require specific chemical reagents for qualitative identification.

Key Concept

Classification of polymers and biomolecules by linkage type (glycosidic vs amide), reaction mode (addition vs condensation), and chemical reactivity.

Estimated Time:2m 0s

Question 8565Question →

During the laboratory preparation and isolation of nitrogen gas from atmospheric air, a student must pass atmospheric air through several reagents in a specific sequence to remove impurities. What is the correct order of steps to isolate nitrogen gas from atmospheric air, starting from the initial removal of carbon(IV) oxide to the final collection of the gas?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence to isolate nitrogen gas from atmospheric air is: 1. Pass air through concentrated caustic alkali solution (

KOH

/

NaOH

) to remove

CO_2

. 2. Pass the gas through concentrated

H_2SO_4

to remove water vapor. 3. Pass dry gas over red-hot copper turnings to remove

O_2

. 4. Collect the remaining nitrogen gas over water.

Atmospheric air consists primarily of

N_2

(78%),

O_2

(21%),

CO_2

(0.03%), water vapor, and noble gases. To isolate nitrogen, impurities are removed according to chemical reactivity:

CO_2

is removed first by neutralisation with an alkali (

KOH

), moisture is absorbed by a dehydrating agent (concentrated

H_2SO_4

), oxygen is removed by reduction of red-hot copper turnings to

CuO

, and the remaining nitrogen gas (mixed with trace noble gases like argon) is collected over water.

Step-by-Step Solution

1

Identify the acidic gas impurity in air and select its removal agent.

Carbon(IV) oxide (

CO_2

) is acidic and must be scrubbed first using concentrated

KOH

or

NaOH

solution:

2KOH_{(aq)} + CO_{2(g)} \rightarrow K_2CO_{3(aq)} + H_2O_{(l)}

.

Removing

CO_2

first prevents it from contaminating subsequent drying and heating apparatus.

2

Dry the remaining gas mixture.

Passing the remaining gases (

O_2

,

N_2

, water vapor, noble gases) through concentrated

H_2SO_4

absorbs moisture.

Gas must be thoroughly dried before passing over hot copper turnings to prevent thermal shock and unwanted reactions.

3

Remove oxygen gas chemically.

Passing dry air over red-hot copper turnings removes

O_2

:

2Cu_{(s)} + O_{2(g)} \rightarrow 2CuO_{(s)}

.

Hot copper chemically binds oxygen, leaving only unreactive nitrogen and traces of noble gases.

4

Collect the purified nitrogen gas.

Nitrogen gas is collected over water.

Nitrogen has very low solubility in water, making water displacement ideal for gas collection.

Key Concept

Laboratory Isolation of Nitrogen from Atmospheric Air

Estimated Time:1m 30s

Question 8566Question →

A mixture containing $0.40\text{ mol}$ of neon and $0.10\text{ mol}$ of argon is collected over water at $27^\circ\text{C}$ . If the partial pressure of neon in the dry gas mixture is $196.0\text{ kPa}$ and the saturated vapor pressure of water at $27^\circ\text{C}$ is $4.0\text{ kPa}$ , what is the total pressure of the wet gas mixture?

249.0\text{ kPa}

245.0\text{ kPa}

241.0\text{ kPa}

200.0\text{ kPa}

Show answer & explanation

Answer: $249.0\text{ kPa}$

Answer

The total pressure of the wet gas mixture is

249.0\text{ kPa}

.

The total dry gas pressure is calculated from the partial pressure of neon and its mole fraction (

0.80

), yielding

245.0\text{ kPa}

. Adding the saturated vapor pressure of water (

4.0\text{ kPa}

) gives the total pressure of the wet gas as

249.0\text{ kPa}

.

Step-by-Step Solution

1

Calculate total moles of dry gas and the mole fraction of neon.

Total dry moles

= 0.40\text{ mol} + 0.10\text{ mol} = 0.50\text{ mol}

. Mole fraction of neon

\chi_{\text{Ne}} = 0.40 / 0.50 = 0.80

.

Dalton's Law states that partial pressure is proportional to mole fraction in a dry gas mixture.

2

Calculate the total pressure of the dry gas mixture.

P_{\text{dry total}} = P_{\text{Ne}} / \chi_{\text{Ne}} = 196.0\text{ kPa} / 0.80 = 245.0\text{ kPa}

.

Rearranging

P_{\text{Ne}} = \chi_{\text{Ne}} \times P_{\text{dry total}}

yields the total dry pressure.

3

Calculate the total pressure of the wet gas mixture collected over water.

P_{\text{total}} = P_{\text{dry total}} + P_{\text{H}_2\text{O}} = 245.0\text{ kPa} + 4.0\text{ kPa} = 249.0\text{ kPa}

.

Gas collected over water is saturated with water vapor, so total pressure is the sum of dry gas pressure and saturated vapor pressure of water.

Key Concept

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

Estimated Time:1m 30s

Question 8567Question →

Match each chlorine oxoacid listed on the left with its corresponding chlorine oxidation state and defining chemical characteristics on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Hypochlorous acid (HClO) matches with oxidation state +1 and bleaching/germicidal properties; Chlorous acid (HClO2) matches with oxidation state +3 and dioxochlorate(III) salt formation; Chloric acid (HClO3) matches with oxidation state +5 and trioxochlorate(V) salt formation; Perchloric acid (HClO4) matches with oxidation state +7 and being the strongest oxoacid.

Each chlorine oxoacid is correctly paired based on the oxidation state of chlorine (ranging from +1 in hypochlorous acid to +7 in perchloric acid) and its associated chemical behavior, where acid strength and oxidizing power in concentrated form increase with increasing oxygen content.

Step-by-Step Solution

1

Calculate the oxidation number of chlorine in each oxoacid using standard oxidation states (H = +1, O = -2).

HClO: 1 + Cl + (-2) = 0 → Cl = +1. HClO2: 1 + Cl + 2(-2) = 0 → Cl = +3. HClO3: 1 + Cl + 3(-2) = 0 → Cl = +5. HClO4: 1 + Cl + 4(-2) = 0 → Cl = +7.

Determining oxidation states is the first step in differentiating chlorine oxoacids.

2

Correlate the oxidation states with acid strength trends in halogen oxoacids.

Acid strength increases as the number of oxygen atoms increases (HClO < HClO2 < HClO3 < HClO4). Thus, HClO4 is the strongest oxoacid.

Additional oxygen atoms pull electron density away from the O-H bond, weakening it and stabilizing the resulting oxoanion.

3

Pair each acid with its systematic IUPAC nomenclature and chemical properties.

HClO (+1) is hypochlorous acid (oxochlorate(I)), HClO2 (+3) is chlorous acid (dioxochlorate(III)), HClO3 (+5) is chloric acid (trioxochlorate(V)), and HClO4 (+7) is perchloric acid (tetraoxochlorate(VII)).

This establishes the exact matching pairs between left and right items.

Key Concept

Oxidation States and Acid Strength Trends of Chlorine Oxoacids

Estimated Time:2m 0s

Question 8568Question →

A mixture containing

10.8\text{ g}

of aluminium powder is reacted with

16.0\text{ g}

of oxygen gas according to the balanced chemical equation:

4\text{Al}_{(s)} + 3\text{O}_{2(g)} \rightarrow 2\text{Al}_2\text{O}_{3(s)}

What is the mass in grams of the excess reactant remaining unreacted at the end of the reaction?

[\text{Al} = 27, \text{O} = 16]

Show answer & explanation

Answer: 6.4

Answer

The mass of the excess reactant (oxygen gas) remaining unreacted is 6.4 g.

To determine the unreacted mass of the excess reactant, first convert given masses to moles: 10.8 g of Al corresponds to 0.40 mol and 16.0 g of O₂ corresponds to 0.50 mol. Using the mole ratio from the balanced equation (4 moles Al : 3 moles O₂), 0.40 mol of Al reacts completely with 0.30 mol of O₂. Thus, Al is the limiting reactant and O₂ is in excess. The unreacted amount of O₂ is 0.50 mol - 0.30 mol = 0.20 mol. Converting 0.20 mol of O₂ back to mass using its molar mass of 32 g/mol yields 6.4 g.

Step-by-Step Solution

1

Calculate the mole amounts of reactants provided

n(Al) = 0.40 mol, n(O₂) = 0.50 mol

Converting masses to moles using molar masses (Al = 27 g/mol, O₂ = 32 g/mol) is necessary for stoichiometric comparison.

2

Determine the theoretical moles of oxygen needed to react with all aluminium

n(O₂) required = 0.30 mol

From the balanced equation, 4 moles of Al require 3 moles of O₂, so 0.40 mol Al requires 0.40 × (3/4) = 0.30 mol O₂.

3

Identify the excess reactant and compute remaining moles

O₂ is in excess by 0.20 mol

Available O₂ (0.50 mol) exceeds required O₂ (0.30 mol), leaving 0.50 - 0.30 = 0.20 mol of O₂ unreacted.

4

Convert remaining moles of excess reactant back to mass

Mass of excess O₂ = 6.4 g

Multiplying 0.20 mol by the molar mass of O₂ (32 g/mol) gives the unreacted mass of oxygen.

Key Concept

Limiting and excess reactant calculations based on stoichiometric coefficients and mole conversions

Estimated Time:2m 0s

Question 8569Question →

Jean-Baptiste Lamarck proposed that physical characteristics acquired by an adult organism through the frequent use or disuse of its organs can be directly transmitted to the next generation.

True

False

Show answer & explanation

Answer: True

Answer

The statement is True. Lamarck's theory asserts that somatic modifications acquired during an individual's lifetime due to environmental adaptation and organ use are inherited by offspring.

The statement is True because Jean-Baptiste Lamarck hypothesized that traits acquired by an organism during its lifetime through environmental influence and organ usage are passed down to subsequent generations.

Step-by-Step Solution

1

Identify the core evolutionary mechanism described in the stem.

The statement describes the passage of phenotypic modifications acquired during an organism's life to its descendants.

Understanding the specific mechanism is required to evaluate historical evolutionary hypotheses.

2

Evaluate whether this concept matches Lamarck's postulates.

Lamarckism is specifically defined by the Law of Use and Disuse and the Law of Inheritance of Acquired Characteristics.

Since the stem accurately restates Lamarck's second postulate, the statement is True.

Key Concept

Inheritance of Acquired Characteristics in Lamarckism

Question 8570Question →

In a diploid eukaryotic organism, alternative forms of a single gene (alleles) controlling a specific trait occupy different loci on non-homologous chromosomes.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Alleles of a given gene occupy identical loci on homologous chromosomes, rather than different loci on non-homologous chromosomes.

The statement is false because homologous chromosomes carry alleles controlling the exact same character at identical loci. Non-homologous chromosomes contain non-allelic genes governing unrelated traits.

Step-by-Step Solution

1

Define the terms allele, locus, and homologous chromosomes.

Alleles are variant forms of a gene. A locus is the specific physical location of a gene on a chromosome. Homologous chromosomes are matching pairs that carry the same genes in the same sequence.

Clear definitions are needed to determine chromosome and locus arrangements.

2

Determine where alleles of the same gene reside in diploid cells.

In diploid cells, one allele is inherited from each parent, placing them at the exact same relative locus on each member of a homologous chromosome pair.

Non-homologous chromosomes carry completely different sets of genes, not alternative alleles of the same gene.

3

Evaluate the validity of the statement.

Since the statement asserts that alleles occupy different loci on non-homologous chromosomes, it contradicts fundamental principles of chromosome structure.

Concluding that the statement is false.

Key Concept

Gene Locus and Homologous Chromosomes

Question 8571Question →

Match each chlorine oxoacid or oxoanion listed in Column I with its corresponding oxidation state, IUPAC designation, or chemical property in Column II.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Hypochlorous acid (HClO) matches with the weak, unstable acid (+1 oxidation state) decomposing in sunlight to release O2 gas; Chloric(V) acid (HClO3) matches with the strong oxidizing acid (+5 oxidation state) prepared from barium chlorate and dilute H2SO4; Perchloric acid (HClO4) matches with the strongest oxoacid (+7 oxidation state); Oxochlorate(I) anion (ClO-) matches with the active bleaching conjugate base formed in cold aqueous NaOH.

Each chlorine species is accurately paired according to oxidation state calculations, resonance stability of conjugate bases, and established laboratory synthesis routes.

Step-by-Step Solution

1

Determine the oxidation state of chlorine in each specified oxoacid and oxoanion species

In

\text{HClO}

, chlorine is

+1

. In

\text{HClO}_3

, chlorine is

+5

. In

\text{HClO}_4

, chlorine is

+7

. In

\text{ClO}^-

, chlorine is

+1

.

Oxidation numbers dictate IUPAC nomenclature and help categorize chemical reactivity.

2

Analyze acid strength trends among chlorine oxoacids

Acid strength increases with increasing number of terminal oxygen atoms:

\text{HClO} < \text{HClO}_2 < \text{HClO}_3 < \text{HClO}_4

. Thus,

\text{HClO}_4

is the strongest oxoacid.

Electronegative terminal oxygen atoms withdraw electron density from the

\text{O-H}

bond, stabilizing the conjugate base via resonance.

3

Correlate specific preparation methods and stability characteristics to their respective species

\text{HClO}

decomposes into

\text{HCl}

and

\text{O}_2

.

\text{HClO}_3

is synthesized via

\text{Ba(ClO}_3)_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4\downarrow + 2\text{HClO}_3

.

\text{ClO}^-

is generated in cold alkaline chlorination:

\text{Cl}_2 + 2\text{OH}^- \rightarrow \text{ClO}^- + \text{Cl}^- + \text{H}_2\text{O}

.

Matching unique reaction mechanisms and industrial/laboratory preparation routes identifies each chlorine compound.

Key Concept

Oxoacids of chlorine, oxidation states, relative acid strengths, and chemical preparation methods.

Question 8572Question →

The solubility product ( $K_{sp}$ ) of calcium tetraoxosulfate(VI), $\text{CaSO}_4$ , is $2.5 \times 10^{-5}\text{ mol}^2\text{ dm}^{-6}$ at $25^\circ\text{C}$ . What is the molar solubility of $\text{CaSO}_4$ in water at this temperature?

( $\text{Molar mass of CaSO}_4 = 136\text{ g mol}^{-1}$ )

5.0 \times 10^{-3}\text{ mol dm}^{-3}

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

1.25 \times 10^{-5}\text{ mol dm}^{-3}

2.5 \times 10^{-5}\text{ mol dm}^{-3}

Show answer & explanation

Answer: $5.0 \times 10^{-3}\text{ mol dm}^{-3}$

Answer

The molar solubility of calcium tetraoxosulfate(VI) is

5.0 \times 10^{-3}\text{ mol dm}^{-3}

.

For a 1:1 sparingly soluble salt like

\text{CaSO}_4

, dissociation yields equal molar concentrations of

\text{Ca}^{2+}

and

\text{SO}_4^{2-}

ions (

s

). The solubility product expression is

K_{sp} = s^2

. Taking the square root of

2.5 \times 10^{-5}\text{ mol}^2\text{ dm}^{-6}

gives

5.0 \times 10^{-3}\text{ mol dm}^{-3}

.

Step-by-Step Solution

1

Write the solubility equilibrium equation and

K_{sp}

expression for

\text{CaSO}_4

.

\text{CaSO}_4(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{SO}_4^{2-}(aq)

, so

K_{sp} = [\text{Ca}^{2+}][\text{SO}_4^{2-}]

.

Establishing the stoichiometric relationship between dissolved ions and molar solubility.

2

Substitute molar solubility

s

into the

K_{sp}

expression.

If

[\text{Ca}^{2+}] = s

and

[\text{SO}_4^{2-}] = s

, then

K_{sp} = s \times s = s^2

.

Expressing

K_{sp}

in terms of a single variable

s

.

3

Calculate the value of

s

by taking the square root of

K_{sp}

.

s = \sqrt{2.5 \times 10^{-5}} = \sqrt{25 \times 10^{-6}} = 5.0 \times 10^{-3}\text{ mol dm}^{-3}

.

Solving the algebraic equation for molar solubility.

Key Concept

Solubility product constant (

K_{sp}

) and molar solubility relation for AB-type binary ionic salts.

Question 8573Question →

Fill in the missing terms to complete the statement describing the electrochemical roles of different regions during the rusting of iron.

Fill in the blanks below

In the electrochemical mechanism of rusting, the site on the iron surface where metallic iron is oxidized to iron(II) ions functions as the , while the region exposed to moisture and oxygen where reduction occurs acts as the .

Show answer & explanation

Answer

The site where metallic iron is oxidized functions as the anode, and the region where reduction occurs acts as the cathode.

During the rusting of iron, a micro-galvanic cell is established on the surface of the metal. The region where iron loses electrons to become

Fe^{2+}

ions undergoes oxidation and therefore functions as the anode. The oxygen-rich droplet region where electrons are consumed to reduce oxygen to

OH^-

ions functions as the cathode.

Step-by-Step Solution

1

Identify the region of oxidation during rusting.

Iron atoms lose electrons (

Fe(s) \rightarrow Fe^{2+}(aq) + 2e^-

), which corresponds to oxidation.

By electrochemical definition, oxidation always occurs at the anode.

2

Identify the region of reduction during rusting.

Dissolved oxygen accepts electrons in the presence of water (

O_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq)

), which corresponds to reduction.

By electrochemical definition, reduction always takes place at the cathode.

Key Concept

Electrochemical mechanism of metallic corrosion

Question 8574Question →

Complete the following statement regarding the electronic structure, hybridization, and molecular geometry of phosphorus trichloride ( $PCl_3$ ).

Fill in the blanks below

In a molecule of phosphorus trichloride (

PCl_3

), the central phosphorus atom undergoes hybridization and exhibits a molecular geometry.

Show answer & explanation

Answer

The central phosphorus atom in

PCl_3

undergoes

sp^3

hybridization and adopts a trigonal pyramidal molecular geometry.

The central phosphorus atom in

PCl_3

has 5 valence electrons. It uses 3 valence electrons to form single covalent bonds with three chlorine atoms, leaving 2 non-bonding valence electrons as a single lone pair. The total steric number is 4 (3 bonding pairs + 1 lone pair), which corresponds to

sp^3

hybridization. According to VSEPR theory, four electron pairs arrange tetrahedrally, but the presence of three bonded atoms and one lone pair produces a trigonal pyramidal molecular shape.

Step-by-Step Solution

1

Determine the valence electron count and steric number for the central atom.

Phosphorus (Group 15) has 5 valence electrons. It forms 3 single covalent bonds with chlorine atoms and retains 1 lone pair of electrons. Steric number = 3 bonding pairs + 1 lone pair = 4.

The steric number dictates the set of hybridized orbitals used by the central atom.

2

Determine the hybridization of the central atom.

A steric number of 4 corresponds to

sp^3

hybridization.

Four electron domains require four degenerate hybridized orbitals formed from one s orbital and three p orbitals.

3

Differentiate between electron pair geometry and molecular geometry.

The electron pair geometry is tetrahedral, but because one position is occupied by a non-bonding lone pair, the molecular geometry is trigonal pyramidal.

VSEPR theory specifies that molecular geometry considers only the spatial arrangement of the atomic nuclei.

Key Concept

VSEPR Theory, Steric Number, and Hybridization

Estimated Time:1m 0s

Question 8575Question →

In sorghum plants, tall stem height ( $T$ ) is completely dominant over dwarf stem height ( $t$ ). In a monohybrid cross between two heterozygous tall plants ( $Tt \times Tt$ ), the conditional probability that a plant displaying the tall phenotype is homozygous dominant ( $TT$ ) is $1/3$ .

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because restricting the sample space to offspring displaying the dominant tall phenotype (

TT

and

Tt

) yields 3 out of 4 total genotypic combinations, of which exactly 1 is homozygous dominant (

TT

), giving a conditional probability of

1/3

.

The statement is correct because out of the 3 possible genotype combinations that produce the dominant tall phenotype (

1\ TT

and

2\ Tt

), exactly 1 is homozygous dominant (

TT

), making the conditional probability

1/3

.

Step-by-Step Solution

1

Determine the genotypic ratio resulting from a monohybrid cross of two heterozygotes (

Tt \times Tt

).

The genotypic ratio is

1\ TT : 2\ Tt : 1\ tt

.

According to Mendel's Law of Segregation, alleles segregate independently during gamete formation, resulting in a

1:2:1

genotypic ratio.

2

Identify the genotypes that express the dominant tall phenotype.

Tall plants possess either the homozygous dominant genotype (

TT

) or the heterozygous genotype (

Tt

), totaling

3

out of

4

genotypic units (

1\ TT + 2\ Tt

).

Dominance dictates that both

TT

and

Tt

produce the tall phenotype.

3

Calculate the conditional probability that a tall plant is homozygous dominant (

TT

).

The probability is

\frac{\text{Number of } TT \text{ combinations}}{\text{Total number of tall combinations}} = \frac{1}{1 + 2} = \frac{1}{3}

.

The question restricts the sample space to only tall offspring.

Key Concept

Conditional probability in monohybrid crosses and Mendel's Law of Segregation

Question 8576Question →

An ecologist monitored a freshwater pond and recorded the following observations:

1. A group of *Oreochromis niloticus* (Tilapia) feeding on floating algae near the surface.
2. Water lilies absorbing dissolved mineral nutrients and solar radiation.
3. Decomposing bacteria breaking down organic detritus on the muddy bottom.
4. Variations in water temperature, pH, and dissolved oxygen concentration throughout the day.

Which ecological concept is best described by the structural and functional interaction of all these living organisms together with their non-living physical environment?

An ecosystem

An ecological niche

A biological community

A population

Show answer & explanation

Answer: An ecosystem

Answer

An ecosystem is the correct term representing the interactive system of biotic organisms and abiotic environmental factors.

An ecosystem is defined as a self-sustaining structural and functional unit formed by the interaction of a biological community (plants, animals, microorganisms) with its abiotic physical environment (water, light, temperature, chemical factors).

Step-by-Step Solution

1

Identify the components present in the pond observation.

The observations list biotic components (tilapia, water lilies, bacteria) and abiotic factors (temperature, pH, dissolved oxygen).

Determining whether both living organisms and physical environment are included helps define the level of ecological organization.

2

Distinguish between population, community, niche, and ecosystem.

A population includes one species; a community includes multiple interacting species; an ecological niche is a species' functional role; an ecosystem incorporates both the community and abiotic factors.

Comparing definitions isolates the unique requirement of abiotic-biotic integration.

3

Select the term that encompasses both biotic and abiotic interactions.

An ecosystem correctly describes this complete structural and functional unit.

The stem explicitly asks for the concept describing organisms interacting together with their non-living environment.

Key Concept

Levels of Ecological Organization and Ecosystem Components

Question 8577Question →

A high-pressure diving cylinder contains $5.00\text{ dm}^3$ of compressed air at a pressure of $3.00\text{ atm}$ and a temperature of $27^\circ\text{C}$ . If the air is expanded into a flexible chamber where the pressure drops to $1.50\text{ atm}$ and the temperature rises to $54^\circ\text{C}$ , what is the final volume of the gas?

20.00\text{ dm}^3

10.90\text{ dm}^3

9.17\text{ dm}^3

2.73\text{ dm}^3

Show answer & explanation

Answer: $10.90\text{ dm}^3$

Answer

The final volume of the gas is

10.90\text{ dm}^3

.

Applying the General Gas Law formula

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

with temperatures converted to Kelvin (

T_1 = 300\text{ K}

,

T_2 = 327\text{ K}

) correctly yields

V_2 = \frac{3.00 \times 5.00 \times 327}{1.50 \times 300} = 10.90\text{ dm}^3

.

Step-by-Step Solution

1

Convert initial and final temperatures from degrees Celsius to Kelvin.

T_1 = 27^\circ\text{C} + 273 = 300\text{ K}

and

T_2 = 54^\circ\text{C} + 273 = 327\text{ K}

.

Gas laws strictly require absolute temperature in Kelvin for proportional relationships to hold true.

2

State the General Gas Law equation and rearrange for the final volume (

V_2

).

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \implies V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}

.

The equation relates changes in pressure, volume, and absolute temperature for a fixed mass of gas.

3

Substitute the given values into the rearranged equation and compute the result.

V_2 = \frac{3.00\text{ atm} \times 5.00\text{ dm}^3 \times 327\text{ K}}{1.50\text{ atm} \times 300\text{ K}} = \frac{4905}{450} = 10.90\text{ dm}^3

.

Simplifying the numerical expression yields the final expanded volume.

Key Concept

General Gas Law (Combined Gas Law) and Absolute Temperature Scale

Estimated Time:1m 30s

Question 8578Question →

A researcher observed that waterbirds developed broader webbed feet over their individual lifetimes through constant stretching while swimming, and asserted that their offspring inherited these enlarged webs directly. Which evaluation best critiques this assertion according to modern biological principles?

The assertion relies on the inheritance of acquired characteristics, which is invalid because somatic modifications caused by environmental use do not alter germline genetics.

The assertion correctly describes natural selection, where environment-induced somatic changes immediately mutate reproductive genes.

The assertion accurately applies Lamarck's law of use and disuse, proving that somatic hypertrophy directly alters chromosomal allele frequencies.

The assertion is flawed because use and disuse can only cause the complete atrophy and loss of vestigial structures, not morphological enlargement.

Show answer & explanation

Answer: The assertion relies on the inheritance of acquired characteristics, which is invalid because somatic modifications caused by environmental use do not alter germline genetics.

Answer

The assertion relies on the inheritance of acquired characteristics, which is invalid because somatic modifications caused by environmental use do not alter germline genetics.

The correct answer highlights the fundamental flaw in Lamarckian evolution. Physical changes occurring in body cells (somatic tissue) due to use or environmental strain are not encoded in gametes (germline tissue), meaning they cannot be inherited by offspring.

Step-by-Step Solution

1

Analyze the researcher's claim

The claim states that physical stretching of webbed feet during an individual's lifetime (a somatic modification) is directly passed to offspring.

This proposal reflects Jean-Baptiste Lamarck's hypothesis of the inheritance of acquired characteristics.

2

Evaluate against modern evolutionary biology principles

Modern genetics establishes that evolutionary inheritance occurs via germline DNA in gametes, not somatic changes in body tissues resulting from use or disuse.

Somatic adaptations acquired during an organism's lifetime do not alter gametic genotype, rendering Lamarck's mechanism scientifically invalid.

Key Concept

Lamarck's Theory of Inheritance of Acquired Characteristics vs. Modern Genetics

Question 8579Question →

A

4.20\text{ g}

sample of impure sodium hydrogentrioxocarbonate(IV),

\text{NaHCO}_3

, was thermally decomposed according to the equation:

2\text{NaHCO}_3\text{(s)} \rightarrow \text{Na}_2\text{CO}_3\text{(s)} + \text{H}_2\text{O(g)} + \text{CO}_2\text{(g)}

If

0.448\text{ dm}^3

of carbon(IV) oxide gas was collected at s.t.p., what is the percentage purity of the

\text{NaHCO}_3

sample?

[\text{Na} = 23, \text{H} = 1, \text{C} = 12, \text{O} = 16, \text{molar volume of gas at s.t.p.} = 22.4\text{ dm}^3\text{ mol}^{-1}]

Show answer & explanation

Answer: 80

Answer

The percentage purity of the sodium hydrogentrioxocarbonate(IV) sample is 80.0%.

The volume of CO₂ gas collected (

0.448\text{ dm}^3

) corresponds to

0.020\text{ mol}

at s.t.p. According to the balanced equation,

2\text{ mol}

of

\text{NaHCO}_3

decompose to form

1\text{ mol}

of

\text{CO}_2

, meaning

0.040\text{ mol}

of pure

\text{NaHCO}_3

reacted. Multiplying by the molar mass of

\text{NaHCO}_3

(

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

) gives

3.36\text{ g}

of pure compound, which represents

80.0\%

of the original

4.20\text{ g}

sample.

Step-by-Step Solution

1

Calculate the moles of carbon(IV) oxide gas evolved at s.t.p.

0.020 mol of CO₂

Dividing the volume of gas collected by the molar volume of a gas at s.t.p. gives the chemical amount in moles.

2

Determine the moles of pure NaHCO₃ using the mole ratio from the balanced chemical equation.

0.040 mol of NaHCO₃

The reaction stoichiometry shows a 2:1 mole ratio between NaHCO₃ and CO₂.

3

Calculate the mass of pure NaHCO₃ by multiplying its moles by its molar mass (84 g/mol).

3.36 g of pure NaHCO₃

Mass is obtained by converting moles to grams using molar mass.

4

Divide the mass of pure NaHCO₃ by the initial mass of the impure sample (4.20 g) and multiply by 100%.

80.0%

Percentage purity expresses the proportion of active compound relative to the total mass of the sample.

Key Concept

Percentage purity calculation based on gas stoichiometry

Estimated Time:1m 30s

Question 8580Question →

A botanist cataloged four organisms from a conservation area along with their proposed scientific names:
1. *Periplaneta americana*
2. *Manihot esculenta* Crantz
3. *Saccharomyces cerevisiae*
4. *rhizobium leguminosarum*

Based on the fundamental principles of biological classification and Linnaean binomial nomenclature, which of the following conclusions is correct?

The scientific designation of the nitrogen-fixing bacterium is incorrectly written because the generic name begins with a lowercase letter.

The designation *Manihot esculenta* Crantz is improper because author citations following binomial names must always be italicized.

*Saccharomyces cerevisiae* is grouped in Kingdom Monera due to its unicellular structure and extracellular saprophytic digestion.

The cellular boundary of *Manihot esculenta* contains peptidoglycan, which serves as the primary structural polymer across all plant cell walls.

Show answer & explanation

Answer: The scientific designation of the nitrogen-fixing bacterium is incorrectly written because the generic name begins with a lowercase letter.

Answer

The scientific designation of the nitrogen-fixing bacterium is incorrectly written because the generic name begins with a lowercase letter.

The correct conclusion identifies that 'rhizobium leguminosarum' violates Linnaean binomial rules because the genus name must begin with a capital letter ('Rhizobium').

Step-by-Step Solution

1

Analyze the binomial nomenclature rules for genus and species names.

Binomial nomenclature requires the genus name to start with a capital letter and the species epithet to start with a lowercase letter. Both names must be italicized or underlined.

This establishes a standardized international naming convention across all biological disciplines.

2

Evaluate the four listed scientific names against standard formatting and classification rules.

The name 'rhizobium leguminosarum' starts with an uncapitalized genus name ('rhizobium'), violating Linnaean formatting rules.

Correct formatting demands 'Rhizobium leguminosarum'.

3

Verify author citation rules and kingdom-level cellular features in the remaining choices.

Author names (Crantz) are non-italicized additions; yeast is a eukaryotic fungus (not Monera); cassava cell walls contain cellulose (not peptidoglycan).

Confirming these principles validates the correct choice and refutes all distractors.

Key Concept

Rules of Linnaean Binomial Nomenclature and Kingdom Classification

Estimated Time:2m 0s

Answer

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