All practice questions

Question 1301Question →

Match each mixture separation scenario on the left with its corresponding distillation requirement or thermal principle on the right.

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Answer

The correct matches pair liquid mixtures with small boiling point differences to fractional distillation involving fractionating columns, non-volatile solute solutions to simple distillation, industrial liquefied gases to cryogenic fractional distillation, and widely separated boiling point liquids to simple distillation.

Each mixture scenario is accurately paired according to volatility differences, non-volatility of solid solutes, temperature threshold limits (

25^\circ\text{C}

difference rule), and specialized industrial gas liquefaction conditions.

Step-by-Step Solution

1

Analyze the thermal criteria for distillation methods based on boiling point differences.

Miscible liquids with boiling point differences less than

25^\circ\text{C}

require a fractionating column (fractional distillation), whereas differences exceeding

50^\circ\text{C}

allow single-stage simple distillation.

Fractionating columns create a temperature gradient allowing multiple vaporizations and condensations.

2

Evaluate the volatility of solutes in liquid solutions.

Dissolved solid salts are non-volatile and remain in the boiling flask while solvent vapor turns into pure distillate.

Non-volatile compounds do not contribute to the vapor phase pressure at distillation temperatures.

3

Examine industrial gas separation requirements.

Gaseous air components must first be liquefied cryogenically before undergoing fractional distillation to separate components with close boiling points like argon and nitrogen.

Distillation requires liquid-vapor equilibrium, which only exists for air at cryogenic temperatures.

Key Concept

Simple vs Fractional Distillation Operational Criteria

Question 1302Question →

Match each redox testing reagent or indicator on the left with its characteristic diagnostic observation on the right when reacting with an oxidizing or reducing agent.

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Answer

Acidified potassium tetraoxomanganate(VII) turns from purple to colorless with reducing agents; starch-iodide paper turns blue-black with oxidizing agents; acidified potassium heptaoxodichromate(VI) turns from orange to green with reducing agents; and freshly prepared iron(II) sulfate changes from pale green to reddish-brown with oxidizing agents.

Each testing reagent displays a distinct diagnostic color change depending on whether it reacts with an oxidizing or reducing agent. Acidified

\text{KMnO}_4

changes from purple to colorless in the presence of a reducing agent due to reduction of

\text{MnO}_4^-

to

\text{Mn}^{2+}

. Starch-iodide paper turns blue-black in the presence of an oxidizing agent as iodide is oxidized to iodine. Acidified

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

changes from orange to green in the presence of a reducing agent as

\text{Cr}_2\text{O}_7^{2-}

is reduced to

\text{Cr}^{3+}

. Freshly prepared

\text{FeSO}_4

changes from pale green to reddish-brown when an oxidizing agent oxidizes

\text{Fe}^{2+}

to

\text{Fe}^{3+}

.

Step-by-Step Solution

1

Identify the role of acidified

\text{KMnO}_4

solution in redox testing.

Acidified

\text{KMnO}_4

contains manganese in the

+7

oxidation state (purple). When it oxidizes a reducing agent, manganese is reduced to

\text{Mn}^{2+}

(colorless).

This is the classic quantitative and qualitative test for reducing agents.

2

Determine the response of moist starch-iodide paper to oxidizing gases.

Oxidizing agents liberate free iodine (

I_2

) from iodide ions (

I^-

). Free iodine reacts with starch to yield a distinctive blue-black color.

This tests specifically for oxidizing agents such as chlorine or ozone.

3

Identify the color change associated with acidified

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

.

Dichromate ions (orange, chromium oxidation state

+6

) are reduced to

\text{Cr}^{3+}

ions (green, oxidation state

+3

) by reducing agents.

The reduction of dichromate(VI) to chromium(III) causes the orange-to-green color change.

4

Determine the oxidation behavior of iron(II) sulfate solution.

Iron(II) ions (pale green) act as a reducing agent and are oxidized to iron(III) ions (reddish-brown/yellow) by oxidizing agents.

The oxidation of

\text{Fe}^{2+}

to

\text{Fe}^{3+}

shifts the solution color from green to brown/yellow.

Key Concept

Laboratory Diagnostic Tests for Oxidizing and Reducing Agents

Question 1303Question →

Match each chemical mixture recovery requirement on the left with the most appropriate physical separation technique on the right.

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Answer

Insoluble chalk suspended in water is separated by filtration. Hydrated copper(II) tetraoxosulfate(VI) crystals require evaporation to saturation followed by cooling to avoid thermal decomposition. Dry sodium chloride is recovered by evaporation to complete dryness. Potassium trioxonitrate(V) is purified from sodium chloride impurity using fractional crystallization.

Each technique is uniquely suited to the physical properties of the mixture: filtration removes insoluble suspensions; evaporation to saturation followed by cooling yields hydrated crystals without thermal decomposition; direct evaporation to dryness efficiently recovers anhydrous thermally stable salts; fractional crystallization separates multiple soluble solutes using solubility-temperature gradients.

Step-by-Step Solution

1

Analyze the solubility and thermal stability of each solute/mixture component.

Chalk is insoluble; hydrated copper sulfate is heat-sensitive; sodium chloride is heat-stable; potassium nitrate and sodium chloride mixture contains two soluble salts with different solubility-temperature slopes.

Physical properties determine which technique yields pure product without thermal breakdown.

2

Match solid-liquid heterogeneous suspensions to filtration.

Chalk powder suspended in water matches filtration.

Filtration separates undissolved solid particles from liquid filtrates.

3

Distinguish between thermal evaporation to dryness and crystallization for soluble salts.

Thermally stable salt (NaCl) matches evaporation to complete dryness, while hydrated salt (CuSO4·5H2O) matches evaporation to saturation followed by cooling.

Evaporating hydrated salts to dryness destroys water of crystallization, yielding anhydrous powder rather than crystals.

4

Match multi-solute solution separation based on temperature-dependent solubility to fractional crystallization.

Separating KNO3 from NaCl impurity matches fractional crystallization.

Cooling a hot saturated solution of both salts causes KNO3 (steep solubility rise with temperature) to crystallize out first.

Key Concept

Selection of separation techniques (filtration, evaporation, crystallization, fractional crystallization) based on solubility and thermal stability.

Question 1304Question →

Match each structural feature of the electron sea model on the left with the macroscopic metal property it directly accounts for on the right.

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Answer

1. Movement of delocalized electrons toward a positive terminal matches High electrical conductivity. 2. Layers of cations sliding past each other matches Malleability and ductility. 3. Absorption and re-emission of light by free electrons matches Lustrous (shiny) appearance. 4. Strong non-directional electrostatic attraction matches High melting and boiling points.

Each structural feature in the electron sea model directly dictates a specific macroscopic behavior: delocalized electron motion provides electrical conductivity, cation layer flexibility allows deformation (malleability/ductility), light oscillation by surface electrons causes shiny luster, and extensive electrostatic forces produce high thermal melting thresholds.

Step-by-Step Solution

1

Relate electric charge transport to metallic conduction

Free electrons moving toward a positive potential corresponds to high electrical conductivity.

Electric current in solid metals consists of a net flow of delocalized valence electrons.

2

Analyze deformation behavior of metallic lattices under pressure

Sliding layers of cations buffered by the electron sea corresponds to malleability and ductility.

Metals deform without shattering because metallic bonding is non-directional.

3

Connect light interaction with free electron oscillations

Free surface electrons absorbing and re-emitting light photons corresponds to luster.

Unbound electrons respond dynamically to electromagnetic waves, reflecting light.

4

Examine thermal stability of the lattice bonding

Strong omnidirectional electrostatic attraction corresponds to high melting and boiling points.

Separating metallic particles requires inputting significant thermal energy to overcome electrostatic bonds.

Key Concept

Electron Sea Model of Metallic Bonding

Question 1305Question →

Match each atmospheric chemical species or pollutant listed on the left with its corresponding atmospheric role or environmental impact listed on the right.

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Answer

Chlorofluorocarbons pair with solar UV photolysis releasing chlorine free radicals; Carbon(IV) oxide pairs with absorbing terrestrial infrared radiation in the lower troposphere; Stratospheric ozone pairs with filtering solar ultraviolet radiation; Methane pairs with being a potent greenhouse gas from anaerobic decomposition.

Each chemical species is accurately matched to its distinct atmospheric layer and chemical property: Chlorofluorocarbons photolyze into free radicals that destroy ozone, Carbon(IV) oxide absorbs Earth's infrared heat, Stratospheric ozone shields the surface from UV radiation, and Methane acts as a greenhouse gas from biological anaerobic processes.

Step-by-Step Solution

1

Identify the primary environmental mechanism associated with Chlorofluorocarbons (CFCs).

CFCs are photolyzed by UV light in the stratosphere, generating atomic chlorine radicals that deplete ozone.

CFC molecules are unreactive in the troposphere but breakdown under high-energy UV light in the stratosphere.

2

Determine the atmospheric function of Carbon(IV) oxide (

\text{CO}_2

).

\text{CO}_2

absorbs re-radiated heat (infrared radiation) emitted by Earth, causing tropospheric warming.

\text{CO}_2

molecules possess vibrational modes that absorb thermal infrared wavelengths.

3

Determine the protective role of Stratospheric Ozone (

\text{O}_3

).

Stratospheric ozone absorbs biological harmful UV-B radiation.

The Chapman mechanism demonstrates how photolysis and reformation of ozone absorb solar UV light.

4

Identify the primary source and impact of Methane (

\text{CH}_4

).

Methane is a strong greenhouse gas emitted from anaerobic habitats like wetlands and livestock digestion.

Methanogenic bacteria produce

\text{CH}_4

under anaerobic conditions.

Key Concept

Distinguishing the chemical roles and environmental impacts of atmospheric pollutants causing global warming versus those causing ozone layer depletion.

Estimated Time:1m 30s

Question 1306Question →

Match each physical or chemical behavior of metallic substances on the left with its corresponding atomic-scale mechanism on the right.

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Answer

High thermal conductivity corresponds to rapid kinetic energy transfer by delocalized electrons; the decrease in electrical conductivity at higher temperatures corresponds to increased scattering from vibrating metal cations; metallic lustre corresponds to photon absorption and re-emission by surface delocalized electrons; and the higher melting points of transition metals correspond to combined

s

-electron delocalization and

d

-orbital overlap.

Each property is accurately matched with its fundamental physical cause: thermal conduction is driven by kinetic energy transfer by mobile electrons; thermal reduction of electrical conductivity stems from enhanced cation scattering; lustre arises from rapid light re-emission by surface electrons; and high transition metal melting points are due to combined

s

-electron delocalization and

d

-orbital bonding.

Step-by-Step Solution

1

Analyze the mechanism for heat conduction in metals.

Thermal conduction occurs because delocalized valence electrons move freely and quickly pass kinetic energy down the temperature gradient.

Free electrons carry kinetic energy much faster than localized lattice atom collisions alone.

2

Analyze how temperature affects electrical resistance/conductivity in metals.

Heating increases the vibrational amplitude of positive cations in the lattice, creating greater resistance (scattering) for moving electron streams.

Impeding the mean free path of drift electrons reduces electrical conductivity.

3

Analyze the optical reflection property of metals.

Incident light causes surface delocalized electrons to oscillate and instantly re-radiate light photons across continuous energy levels.

The sea of mobile electrons acts as a reflective barrier to light waves.

4

Compare cohesive energy differences between alkali metals and transition metals.

Transition elements utilize both outer

s

valence electrons and partially filled inner

d

subshells to form additional covalent bonds, significantly increasing lattice strength and melting point.

Greater electrostatic attraction and inter-atomic orbital overlap increase the energy required to break the lattice.

Key Concept

Metallic Bonding mechanisms relating atomic-scale electron sea and lattice structures to macroscopic physical properties

Question 1307Question →

Match each specific industrial effluent contaminant listed on the left with its standard chemical treatment or remediation technique on the right.

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Answer

Heavy metal ions (

Pb^{2+}, Cd^{2+}

) pair with precipitation using

Ca(OH)_2

or

Na_2S

; non-biodegradable synthetic dyes pair with activated carbon adsorption; phosphate surfactants pair with precipitation and biological nutrient removal; and acidic effluent (

H_2SO_4

) pairs with lime/limestone neutralization.

Each industrial pollutant is matched to its chemically specific treatment based on functional chemistry: heavy metal ions precipitate as sulfides/hydroxides, non-biodegradable organic dyes adsorb onto activated carbon, phosphates precipitate to control eutrophication, and acidic effluents are neutralized with basic compounds.

Step-by-Step Solution

1

Analyze heavy metal waste remediation chemistry

Heavy metal cations (

Pb^{2+}, Cd^{2+}

) react with

OH^-

or

S^{2-}

to form insoluble salts (

PbS

,

Cd(OH)_2

) precipitating out of solution.

Chemical precipitation renders toxic soluble metals insoluble and separable by filtration.

2

Evaluate textile dye removal processes

Refractory organic dye molecules adsorb onto activated carbon porous structures.

Synthetic organic dyes are resistant to standard oxidation/biodegradation but adsorb readily onto carbon surfaces.

3

Identify phosphate pollution control

Phosphates precipitate as calcium phosphate or aluminum phosphate during tertiary wastewater treatment.

Phosphate removal is critical to prevent algal blooms and eutrophication in receiving aquatic ecosystems.

4

Determine acidic wastewater treatment

H_2SO_4

reacts with

CaCO_3

or

Ca(OH)_2

in an acid-base neutralization producing neutral sulfate salts and water.

Adjusting effluent

\text{pH}

to near neutral (

6.5-8.5

) is mandatory prior to discharge to safeguard aquatic life.

Key Concept

Industrial Effluents and Specific Water Remediation Techniques

Question 1308Question →

Match each distillation term or phenomenon on the left with its correct chemical definition or description on the right.

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Answer

Azeotropic mixture matches with a constant-boiling liquid mixture; Distillate matches with the purified liquid collected after vapor condensation; Refluxing matches with the continuous return of condensed vapor back down the column; Theoretical plate matches with a hypothetical zone where liquid and vapor reach equilibrium.

Each distillation concept is correctly matched based on foundational principles: an azeotropic mixture maintains constant composition at its boiling point; distillate refers to the condensed liquid product; refluxing is the downward flow of condensate inside the column; and a theoretical plate quantifies column separation performance.

Step-by-Step Solution

1

Define an azeotropic mixture.

Identify that azeotropes boil at a fixed temperature with identical liquid and vapor compositions.

Azeotropes act like pure substances during boiling and cannot be separated further by simple or fractional distillation.

2

Define distillate.

Identify distillate as the final condensed liquid collected in the receiving flask.

Vapors leaving the still head are condensed into liquid distillate.

3

Define refluxing in fractional distillation.

Identify refluxing as returning condensed liquid back down the fractionating column.

This establishes a temperature gradient and continuous equilibrium steps along the column.

4

Define a theoretical plate.

Identify a theoretical plate as a stage representing one complete vaporization-condensation cycle.

More theoretical plates in a column correspond to higher separation efficiency for liquids with close boiling points.

Key Concept

Distillation Terminology and Theoretical Principles

Question 1309Question →

Pair each of the given chemical molecules with its corresponding molecular geometry as predicted by Valence Shell Electron Pair Repulsion (VSEPR) theory.

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Answer

BF_3

matches Trigonal planar,

CH_4

matches Tetrahedral,

BeCl_2

matches Linear, and

H_2O

matches Bent (V-shaped).

According to VSEPR theory, molecular shape depends on the total number of bonding pairs and lone pairs surrounding the central atom.

BeCl_2

has 2 bonding pairs with no lone pairs, yielding a linear geometry.

BF_3

has 3 bonding pairs with no lone pairs, producing a trigonal planar geometry.

CH_4

has 4 bonding pairs with no lone pairs, resulting in a tetrahedral geometry.

H_2O

has 2 bonding pairs and 2 non-bonding lone pairs, creating a bent (V-shaped) geometry.

Step-by-Step Solution

1

Determine the number of valence electron pairs (bonding pairs and lone pairs) surrounding the central atom for each chemical species.

BF_3

has 3 bonding pairs and 0 lone pairs;

CH_4

has 4 bonding pairs and 0 lone pairs;

BeCl_2

has 2 bonding pairs and 0 lone pairs;

H_2O

has 2 bonding pairs and 2 lone pairs.

VSEPR theory states that electron pairs around a central atom arrange themselves to minimize electrostatic repulsion.

2

Deduce the resulting molecular geometry for each molecule based on the arrangement of bonding and lone pairs.

3 bond pairs (0 lone pairs) = Trigonal planar; 4 bond pairs (0 lone pairs) = Tetrahedral; 2 bond pairs (0 lone pairs) = Linear; 2 bond pairs + 2 lone pairs = Bent.

Lone pairs exert greater repulsive force than bonding pairs, bending the molecular framework accordingly.

Key Concept

VSEPR Theory and Molecular Geometries

Question 1310Question →

Match each change in reaction conditions to its corresponding effect on the kinetic model of reaction rates.

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Answer

1. Increasing reactant concentration matches with increasing collision frequency by raising the number of solute particles per unit volume; 2. Adding a positive catalyst matches with providing an alternative pathway with lower activation energy; 3. Grinding solid reactant into fine powder matches with exposing more surface area for collisions; 4. Increasing temperature matches with raising average kinetic energy and the proportion of collisions with energy equal to or exceeding activation energy.

Each factor uniquely modifies a component of collision theory: concentration alters particle density and collision frequency; catalysts lower the activation energy threshold; surface area determines exposed reactive sites; temperature enhances molecular kinetic energy and the proportion of successful high-energy collisions.

Step-by-Step Solution

1

Evaluate the effect of concentration on particle distribution

More solute particles exist in a given volume when concentration increases.

Higher particle density directly increases collision frequency.

2

Evaluate the catalytic mechanism

A positive catalyst changes the reaction pathway.

This alternative pathway has a lower activation energy barrier (

E_a

).

3

Evaluate particle size reduction

Grinding a solid exposes interior atoms/molecules to the surrounding fluid.

Greater exposed surface area yields a higher number of effective collisions per unit time.

4

Evaluate kinetic energy distribution with temperature change

Higher temperature shifts the Maxwell-Boltzmann distribution curve to higher kinetic energies.

A significantly larger fraction of colliding species possess kinetic energy greater than or equal to

E_a

.

Key Concept

Collision theory explanations for factors influencing reaction rates

Question 1311Question →

Match each of the following chemical species or systems with the predominant type of intermolecular force or interaction present between its units in the liquid or solid state.

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Answer

Water matches with Hydrogen bonding; Trichloromethane matches with Permanent dipole-dipole interaction; Argon matches with London dispersion forces; Hydrated sodium ion matches with Ion-dipole interaction.

Water undergoes hydrogen bonding due to the presence of highly polar

\text{O-H}

bonds. Trichloromethane exhibits permanent dipole-dipole attractions because it is a polar molecule without hydrogen attached to

\text{N, O, F}

. Argon is a nonpolar noble gas that relies solely on temporary induced dipoles (London dispersion forces). The hydrated sodium ion experiences electrostatic ion-dipole interactions between the positive ion and polar water molecules.

Step-by-Step Solution

1

Analyze the chemical composition and molecular polarity of each species.

Water (

\text{H}_2\text{O}

) has

\text{O-H}

bonds; Trichloromethane (

\text{CHCl}_3

) is a polar molecule lacking

\text{N-H}

,

\text{O-H}

, or

\text{F-H}

bonds; Argon (

\text{Ar}

) is a nonpolar noble gas; Hydrated sodium ion (

\text{Na}^+_{(aq)}

) features an ionic species surrounded by polar solvent molecules.

Determining polarity, presence of electronegative elements bonded to hydrogen, and ionic charge is necessary to classify intermolecular forces.

2

Assign the predominant intermolecular force to each species.

\text{H}_2\text{O}

forms hydrogen bonds;

\text{CHCl}_3

experiences permanent dipole-dipole attractions;

\text{Ar}

relies on London dispersion forces;

\text{Na}^+_{(aq)}

displays ion-dipole attractions.

Hydrogen bonding requires

\text{H}

attached directly to

\text{F, O, N}

; dipole-dipole forces operate between permanent dipoles; dispersion forces exist in all species but predominate in nonpolar units; ion-dipole interactions occur between ions and polar molecules.

Key Concept

Intermolecular Forces and Hydrogen Bonding

Estimated Time:1m 30s

Question 1312Question →

Match each change in reaction conditions on the left with its corresponding microscopic mechanism under collision theory on the right.

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Answer

Concentration increase matches with higher particle density and collision frequency per unit volume; temperature increase matches with higher average kinetic energy and greater fraction of particles exceeding activation energy; adding a catalyst matches with providing an alternative pathway with lower activation energy; crushing a solid matches with increasing exposed surface area for collisions.

Each rate factor operates via a specific collision theory principle: concentration governs particle crowding and collision frequency per unit volume; temperature primarily governs the proportion of successful high-energy collisions; catalysts reduce the energy barrier via an alternate pathway; and particle size determines the surface contact area available for collisions.

Step-by-Step Solution

1

Analyze the effect of aqueous reactant concentration.

More particles occupy the same volume, causing more frequent collisions.

Collision frequency depends directly on the number of reactant particles present per unit volume.

2

Analyze the effect of increasing temperature.

Particles move faster and possess more thermal kinetic energy.

Temperature directly dictates the average kinetic energy distribution of particles and the fraction exceeding activation energy (

E_a

).

3

Analyze the mechanism of a positive catalyst.

Reaction proceeds via an alternative mechanism with a reduced activation energy threshold.

Catalysts alter the reaction pathway to lower

E_a

without affecting overall chemical equilibrium position.

4

Analyze the physical effect of reducing particle size.

Greater surface contact area is created for the solid reactant.

A higher ratio of surface area to mass allows more fluid particles to collide with the solid reactant simultaneously.

Key Concept

Collision Theory Mechanisms of Reaction Rate Factors

Question 1313Question →

Match each modification of reaction conditions on the left with its corresponding microscopic mechanism according to collision theory on the right.

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Answer

Increasing pressure matches increasing particle concentration per unit volume without altering kinetic energy distribution; Adding a catalyst matches lowering activation energy to increase successful collisions; Grinding solid into powder matches increasing exposed surface area and contact sites; Increasing temperature matches shifting molecular energy distribution to increase the fraction of particles possessing

E \ge E_a

.

Each factor alters reaction rate through a distinct collision theory mechanism: pressure increases gas concentration and collision frequency; catalysts lower the activation energy barrier; surface area increases exposed contact points; and temperature increases average kinetic energy and the fraction of energetic collisions exceeding activation energy.

Step-by-Step Solution

1

Analyze the microscopic effect of increasing pressure on gases.

Gaseous molecules are forced into a smaller volume, increasing concentration and therefore the total frequency of collisions, while temperature-dependent kinetic energy stays constant.

Pressure directly alters volumetric density of gas particles.

2

Analyze the action of a catalyst.

A catalyst lowers the activation energy (

E_a

) barrier by offering an alternative route, allowing a larger percentage of existing collisions to be successful.

Catalysts change reaction energetics without altering particle kinetic energy.

3

Determine the physical result of reducing solid particle size.

Finely dividing a solid exposes interior atoms/molecules to the surrounding fluid phase, increasing collision contact opportunities per second.

Heterogeneous rates depend on exposed interfacial area.

4

Evaluate the kinetic effect of a temperature rise.

Higher temperatures broaden and flatten the Maxwell-Boltzmann distribution curve, exponentially increasing the population of molecules with energy

E \ge E_a

.

Temperature measures average molecular kinetic energy.

Key Concept

Collision Theory Mechanisms for Reaction Rate Factors

Question 1314Question →

Match each atmospheric component listed on the left with its corresponding environmental function or impact listed on the right.

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Answer

Chlorofluorocarbons (CFCs) match with decomposing under UV light to produce chlorine radicals that destroy ozone; Carbon(IV) oxide (

\text{CO}_2

) matches with absorbing terrestrial thermal infrared radiation to cause global warming; Stratospheric ozone (

\text{O}_3

) matches with filtering high-energy solar ultraviolet radiation.

Chlorofluorocarbons generate free chlorine radicals in the upper atmosphere that deplete stratospheric ozone; Carbon(IV) oxide absorbs Earth's re-radiated infrared radiation to drive global warming; and stratospheric ozone absorbs incoming solar ultraviolet radiation to protect life.

Step-by-Step Solution

1

Determine the atmospheric mechanism of Chlorofluorocarbons (CFCs)

CFCs photolyze under UV light to release free chlorine radicals, driving stratospheric ozone depletion.

Chlorine radicals act as catalysts in the cyclic destruction of

\text{O}_3

molecules.

2

Determine the atmospheric mechanism of Carbon(IV) oxide (

\text{CO}_2

)

\text{CO}_2

absorbs thermal infrared energy radiated from Earth's surface.

This trapped heat enhances the greenhouse effect, raising average global temperatures.

3

Determine the protective role of Stratospheric ozone (

\text{O}_3

)

Stratospheric ozone absorbs biological harmful UV-B and UV-C rays.

Ozone photolysis and regeneration absorb high-energy solar radiation before it hits the surface.

Key Concept

Distinction between greenhouse warming mechanisms (infrared absorption) and ozone layer depletion mechanisms (radical-catalyzed photolysis)

Question 1315Question →

In solid-state chemistry, the distinct physical behaviors of metals arise directly from the structural characteristics of metallic bonds. Match each observable metallic property or behavior on the left with its corresponding atomic-scale explanation on the right.

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Answer

High electrical conductivity corresponds to unconfined valence electrons drifting directionally under an applied potential difference. High malleability and ductility correspond to non-directional electrostatic attractions allowing cation layers to slide past each other while maintaining cohesive forces. Higher melting point of iron compared to sodium corresponds to the contribution of delocalized d-orbital electrons alongside s-electrons. Lustrous reflective appearance corresponds to the oscillation of free valence electrons absorbing and rapidly re-emitting incident photons.

High electrical conductivity is explained by the movement of unconfined valence electrons drifting directionally when a potential difference is applied. High malleability and ductility stem from non-directional electrostatic forces allowing metal cation planes to slide over each other without breaking cohesive bonds. The higher melting point of transition metals like iron compared to alkali metals like sodium is caused by extra binding strength provided by delocalized d-orbital electrons in addition to s-electrons. Metallic luster is caused by mobile valence electrons absorbing incident light energy and immediately re-emitting it.

Step-by-Step Solution

1

Analyze the microscopic origin of electrical conduction in metallic crystals.

Electrical conduction requires mobile charge carriers. In metals, delocalized valence electrons move freely across the lattice under an electric potential.

Relates macroscopic electric current to electron mobility.

2

Analyze how mechanical force affects metal cation layers.

Deformation causes layers of cations to slip over each other. Because metallic bonds are non-directional, the electron sea adjusts instantly to keep the lattice bound without brittle cleavage.

Explains malleability and ductility via non-directional bonding.

3

Compare the bonding strength of alkali metals versus transition metals.

Sodium donates only one s-electron per atom into the sea, whereas iron donates both s and unpaired inner d-electrons, greatly increasing the electrostatic cohesive energy and melting point.

Explains variation in thermal resistance and hardness across different metals.

4

Analyze the interaction between light waves and delocalized electron clouds.

Mobile surface electrons readily absorb light energy and oscillate, promptly re-radiating light photons to generate a high spectral reflectance (luster).

Connects optical reflectivity to electron sea excitation.

Key Concept

Electron Sea Model and Metal Property Mechanisms

Question 1316Question →

Match each chemical species or substance involved in water chemistry with its primary role or behavior regarding water hardness.

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Answer

Magnesium hydrogencarbonate causes temporary hardness that decomposes on boiling; magnesium tetraoxosulfate(VI) causes permanent hardness unaffected by boiling; sodium trioxocarbonate(IV) removes all hardness via chemical precipitation; sodium permutit removes all hardness via ion exchange.

Magnesium hydrogencarbonate is a soluble hydrogencarbonate salt causing temporary hardness removed by boiling. Magnesium tetraoxosulfate(VI) is a soluble sulfate salt causing permanent hardness that cannot be removed by boiling. Sodium trioxocarbonate(IV) removes all hardness types by precipitating calcium/magnesium ions as insoluble carbonates. Permutit softened water by exchanging hardness cations for soluble sodium ions.

Step-by-Step Solution

1

Identify the cause of temporary hardness.

Magnesium hydrogencarbonate,

\text{Mg(HCO}_3)_2

, dissolves in water to cause temporary hardness. Upon boiling, hydrogencarbonate ions decompose to insoluble carbonate precipitates:

\text{Mg(HCO}_3)_2(aq) \rightarrow \text{MgCO}_3(s) + \text{H}_2\text{O}(l) + \text{CO}_2(g)

.

Temporary hardness is specifically caused by hydrogen carbonate salts of calcium and magnesium, which are thermally unstable.

2

Identify the cause of permanent hardness.

Magnesium tetraoxosulfate(VI),

\text{MgSO}_4

, causes permanent hardness because sulfate salts do not undergo thermal decomposition upon boiling.

Permanent hardness is caused by soluble tetraoxosulfate(VI) or chloride salts of calcium and magnesium.

3

Analyze the action of washing soda (

\text{Na}_2\text{CO}_3

).

Adding washing soda introduces

\text{CO}_3^{2-}

ions, which combine with dissolved

\text{Mg}^{2+}

or

\text{Ca}^{2+}

ions to precipitate them as

\text{MgCO}_3

or

\text{CaCO}_3

.

Precipitation of divalent metallic cations as insoluble carbonates removes both temporary and permanent hardness.

4

Analyze the action of sodium permutit (zeolite).

Permutit acts as an ion exchanger:

\text{Na}_2\text{Z}(s) + \text{Mg}^{2+}(aq) \rightarrow \text{MgZ}(s) + 2\text{Na}^+(aq)

.

Hardness-causing divalent cations are bound to the zeolite matrix while harmless sodium ions are released into solution.

Key Concept

Classification of water hardness causes (hydrogencarbonate vs sulfate salts) and chemical removal mechanisms (boiling, precipitation, ion exchange).

Question 1317Question →

Match each chemical species with its corresponding central atom hybridization state and molecular geometry as predicted by Valence Shell Electron Pair Repulsion (VSEPR) theory.

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Answer

Chlorate ion (

ClO_3^-

) matches

sp^3

hybridization with a trigonal pyramidal shape; Xenon difluoride (

XeF_2

) matches

sp^3d

hybridization with a linear shape; Tetrachloroiodate ion (

ICl_4^-

) matches

sp^3d^2

hybridization with a square planar shape; Sulfur dioxide (

SO_2

) matches

sp^2

hybridization with a bent shape.

Each chemical species is accurately paired by identifying the total number of electron domains around its central atom:

ClO_3^-

has 4 domains (

sp^3

, trigonal pyramidal shape),

XeF_2

has 5 domains (

sp^3d

, linear shape),

ICl_4^-

has 6 domains (

sp^3d^2

, square planar shape), and

SO_2

has 3 domains (

sp^2

, bent shape).

Step-by-Step Solution

1

Determine the valence electron count and steric number (bonding domains + lone pairs) for the central atom of each species.

Chlorate ion (

ClO_3^-

): 3 bonds + 1 lone pair = steric number 4; Xenon difluoride (

XeF_2

): 2 bonds + 3 lone pairs = steric number 5; Tetrachloroiodate ion (

ICl_4^-

): 4 bonds + 2 lone pairs = steric number 6; Sulfur dioxide (

SO_2

): 2 double-bond domains + 1 lone pair = steric number 3.

The total number of electron domains determines both the hybridization of atomic orbitals and the underlying electron pair geometry.

2

Assign the hybridization corresponding to each steric number.

Steric number 4 corresponds to

sp^3

; steric number 5 corresponds to

sp^3d

; steric number 6 corresponds to

sp^3d^2

; steric number 3 corresponds to

sp^2

.

Hybrid orbital set size equals the number of electron domains.

3

Apply VSEPR theory rules to deduce the molecular geometry (accounting only for atomic positions).

ClO_3^-

is trigonal pyramidal (

sp^3

);

XeF_2

is linear (

sp^3d

);

ICl_4^-

is square planar (

sp^3d^2

);

SO_2

is bent (

sp^2

).

Non-bonding lone pairs cause greater repulsion and define the non-spherical molecular geometry relative to electron geometry.

Key Concept

VSEPR Theory and Hybridization of Molecular Species and Polyatomic Ions

Question 1318Question →

Match each atmospheric phenomenon or chemical process on the left with its precise molecular mechanism or cause on the right.

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Answer

Tropospheric thermal radiation trapping matches vibrational absorption of terrestrial infrared radiation; Stratospheric ozone catalytic destruction matches the chlorine radical propagation reaction cycle; Chlorofluorocarbon (CFC) photolysis matches homolytic cleavage of carbon-chlorine bonds by ultraviolet light; Natural stratospheric ozone layer formation matches solar UV decomposition of

\text{O}_2

into atomic oxygen followed by reaction with

\text{O}_2

.

Each atmospheric process is accurately paired with its chemical mechanism: tropospheric heat trapping is driven by infrared absorption by greenhouse gases; stratospheric ozone catalytic breakdown occurs through free-radical propagation steps involving atomic chlorine; CFC photolysis involves solar UV cleavage of C-Cl bonds; and natural ozone formation requires UV dissociation of diatomic oxygen into atomic oxygen.

Step-by-Step Solution

1

Analyze the process of tropospheric thermal radiation trapping

Identify that global warming / greenhouse effect involves absorption and re-emission of terrestrial infrared (heat) radiation by gases like

\text{CO}_2

and

\text{CH}_4

.

Greenhouse gases selectively interact with outgoing longwave thermal radiation in the troposphere.

2

Analyze stratospheric ozone depletion and CFC reactions

Distinguish between CFC photolysis (UV cleavage of C-Cl bonds releasing Cl radicals) and the catalytic propagation cycle (

\text{Cl} + \text{O}_3 \rightarrow \text{ClO} + \text{O}_2

).

Photolysis produces reactive radicals, whereas propagation steps directly decompose ozone repeatedly.

3

Analyze the formation mechanism of the ozone layer

Identify that natural stratospheric ozone is created when UV light dissociates

\text{O}_2

into oxygen atoms, which combine with

\text{O}_2

to form

\text{O}_3

.

This dynamic photochemical cycle maintains the ozone shield in the stratosphere.

Key Concept

Greenhouse Effect, Global Warming, and Ozone Layer Depletion

Question 1319Question →

Match each class of alkanol listed on the left with its characteristic oxidation behavior on the right when reacted with acidified potassium heptaoxodichromate(VI) solution.

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Answer

Primary alkanols pair with oxidation to an alkanal and then an alkanoic acid; secondary alkanols pair with oxidation to an alkanone; tertiary alkanols pair with resistance to oxidation under mild conditions.

Primary alkanols possess two alpha-hydrogens and oxidize in two steps to form alkanals and then alkanoic acids. Secondary alkanols possess one alpha-hydrogen and oxidize to form alkanones. Tertiary alkanols lack alpha-hydrogens entirely, rendering them resistant to oxidation under mild conditions.

Step-by-Step Solution

1

Examine the structural environment of primary alkanols (

R-CH_2OH

)

Primary alkanols have two

\alpha

-hydrogen atoms attached to the carbon holding the

-OH

group, permitting two sequential oxidation steps.

Oxidation requires the removal of hydrogen from the hydroxyl-bearing carbon atom.

2

Examine the structural environment of secondary alkanols (

R_2CHOH

)

Secondary alkanols have only one

\alpha

-hydrogen atom, yielding an alkanone (

R_2C=O

).

Alkanones resist further oxidation under mild conditions because no additional

\alpha

-hydrogens are available.

3

Examine the structural environment of tertiary alkanols (

R_3COH

)

Tertiary alkanols possess zero

\alpha

-hydrogen atoms on the hydroxyl-bearing carbon atom.

Without an

\alpha

-hydrogen, oxidation cannot proceed without breaking carbon-carbon bonds.

Key Concept

Classification and oxidation products of alkanols

Question 1320Question →

Match each type of solution state with its corresponding characteristic physical state and behavior at a given temperature.

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Answer

An unsaturated solution matches with holding less solute than the maximum limit and being able to dissolve more solute. A saturated solution matches with holding the maximum solute in dynamic equilibrium with undissolved solute. A supersaturated solution matches with holding excess dissolved solute beyond normal solubility and rapidly crystallizing upon seeding.

An unsaturated solution can dissolve more solute; a saturated solution maintains dynamic equilibrium with maximum dissolved solute; a supersaturated solution holds solute beyond normal solubility limits and crystallizes when seeded.

Step-by-Step Solution

1

Identify the characteristic of an unsaturated solution.

It has not reached maximum capacity, so adding solute leads to further dissolution.

Solvent molecules are available to interact with and dissolve additional solute particles at that temperature.

2

Identify the characteristic of a saturated solution.

It exists in dynamic equilibrium with undissolved solid.

At saturation, the solvent holds the maximum possible concentration of solute at that specific temperature.

3

Identify the characteristic of a supersaturated solution.

It contains dissolved solute in excess of the saturation concentration and is metastable/unstable.

Disturbing the solution with a seed crystal initiates rapid crystallization to relieve the unstable excess concentration.

Key Concept

Saturation States of Solutions

Answer

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