All practice questions

Question 8341Question →

During industrial water treatment, temporary hardness caused by dissolved calcium hydrogencarbonate is removed chemically via Clark's process by adding a controlled quantity of slaked lime, $\text{Ca(OH)}_2$ . Which balanced chemical equation correctly represents this water-softening reaction?

\text{Ca(HCO}_3)_2\text{(aq)} + \text{Ca(OH)}_2\text{(aq)} \rightarrow 2\text{CaCO}_3\text{(s)} + 2\text{H}_2\text{O(l)}

\text{Ca(HCO}_3)_2\text{(aq)} + \text{Ca(OH)}_2\text{(aq)} \rightarrow \text{CaCO}_3\text{(s)} + \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}

\text{CaSO}_4\text{(aq)} + \text{Ca(OH)}_2\text{(aq)} \rightarrow 2\text{CaCO}_3\text{(s)} + \text{H}_2\text{SO}_4\text{(aq)}

\text{Ca(HCO}_3)_2\text{(aq)} + \text{CaO(s)} \rightarrow \text{CaCO}_3\text{(s)} + \text{H}_2\text{O(l)}

Show answer & explanation

Answer: $\text{Ca(HCO}_3)_2\text{(aq)} + \text{Ca(OH)}_2\text{(aq)} \rightarrow 2\text{CaCO}_3\text{(s)} + 2\text{H}_2\text{O(l)}$

Answer

\text{Ca(HCO}_3)_2\text{(aq)} + \text{Ca(OH)}_2\text{(aq)} \rightarrow 2\text{CaCO}_3\text{(s)} + 2\text{H}_2\text{O(l)}

The equation

\text{Ca(HCO}_3)_2\text{(aq)} + \text{Ca(OH)}_2\text{(aq)} \rightarrow 2\text{CaCO}_3\text{(s)} + 2\text{H}_2\text{O(l)}

represents Clark's process. Slaked lime provides hydroxide ions that neutralize the hydrogencarbonate ions in temporary hard water, precipitating all calcium as insoluble calcium trioxocarbonate(IV), leaving soft water.

Step-by-Step Solution

1

Identify the chemical cause of temporary hardness in the water sample

Temporary hardness is caused by dissolved calcium hydrogencarbonate,

\text{Ca(HCO}_3)_2\text{(aq)}

.

Hydrogencarbonate salts of calcium and magnesium decompose or precipitate when treated with appropriate alkaline agents.

2

Apply the principles of Clark's process for temporary hardness removal

Clark's process involves adding a calculated amount of slaked lime,

\text{Ca(OH)}_2\text{(aq)}

, which supplies hydroxide ions to convert hydrogencarbonate ions into insoluble trioxocarbonate(IV) ions.

Adding excess lime would re-introduce calcium ions and cause artificial hardness, so a stoichiometric amount is used.

3

Balance the precipitation chemical equation

\text{Ca(HCO}_3)_2\text{(aq)} + \text{Ca(OH)}_2\text{(aq)} \rightarrow 2\text{CaCO}_3\text{(s)} + 2\text{H}_2\text{O(l)}

One mole of calcium hydrogencarbonate reacts with one mole of calcium hydroxide to produce two moles of insoluble calcium trioxocarbonate(IV) and two moles of water.

Key Concept

Clark's process (slaked lime precipitation) for removal of temporary water hardness

Question 8342Question →

A $5.00\text{ g}$ sample of an impure hydrated calcium tetraoxosulfate(VI) salt, $\text{CaSO}_4 \cdot x\text{H}_2\text{O}$ , containing $14\%$ non-volatile, inert impurities by mass, is heated strongly to constant mass to remove all water of crystallization. If the mass of the remaining dry residue is $4.10\text{ g}$ , what is the value of $x$ ? (Relative atomic masses: $\text{Ca} = 40$ , $\text{S} = 32$ , $\text{O} = 16$ , $\text{H} = 1$ )

1

2

3

5

Show answer & explanation

Answer: 2

Answer

The value of

x

is 2.

The original

5.00\text{ g}

sample contains

14\%

non-volatile impurity (

0.70\text{ g}

). Upon heating, only water evaporates, leaving

4.10\text{ g}

of residue. Subtracting the

0.70\text{ g}

of impurity gives

3.40\text{ g}

of pure anhydrous

\text{CaSO}_4

(

0.025\text{ mol}

). The mass of evaporated water is

5.00\text{ g} - 4.10\text{ g} = 0.90\text{ g}

(

0.050\text{ mol}

). Dividing moles of water by moles of anhydrous salt yields

x = 2

.

Step-by-Step Solution

1

Calculate the mass of the inert, non-volatile impurity in the original sample.

\text{Mass of impurity} = 14\% \times 5.00\text{ g} = 0.70\text{ g}

.

The sample consists of pure hydrated salt and

14\%

inert non-volatile impurity.

2

Determine the mass of pure anhydrous

\text{CaSO}_4

in the dry residue.

\text{Mass of pure } \text{CaSO}_4 = 4.10\text{ g} - 0.70\text{ g} = 3.40\text{ g}

.

Since the impurity is non-volatile, it remains in the

4.10\text{ g}

dry residue along with the anhydrous salt.

3

Calculate the mass of water of crystallization driven off.

\text{Mass of } \text{H}_2\text{O} = 5.00\text{ g} - 4.10\text{ g} = 0.90\text{ g}

.

Heating to constant mass removes only the volatile water of crystallization.

4

Find the molar amounts of pure

\text{CaSO}_4

and

\text{H}_2\text{O}

.

\text{Moles of } \text{CaSO}_4 = \frac{3.40\text{ g}}{136\text{ g/mol}} = 0.025\text{ mol}

;

\text{Moles of } \text{H}_2\text{O} = \frac{0.90\text{ g}}{18\text{ g/mol}} = 0.050\text{ mol}

.

Molar masses are

136\text{ g/mol}

for

\text{CaSO}_4

and

18\text{ g/mol}

for

\text{H}_2\text{O}

.

5

Determine the mole ratio

x = \frac{n(\text{H}_2\text{O})}{n(\text{CaSO}_4)}

.

x = \frac{0.050}{0.025} = 2

.

The formula ratio requires the relative mole ratio of water to anhydrous salt.

Key Concept

Accounting for non-volatile impurities in hydrated salt analysis to determine water of crystallization.

Question 8343Question →

In petroleum refining, catalytic cracking breaks down long-chain hydrocarbons into smaller, economically valuable fractions. If one mole of dodecane ( $C_{12}H_{26}$ ) undergoes thermal-catalytic cracking to yield one mole of hexane ( $C_6H_{14}$ ) and two moles of an alkene product $X$ , what is the molecular formula of $X$ ?

C_3H_6

C_3H_8

C_6H_{12}

C_6H_{14}

Show answer & explanation

Answer: $C_3H_6$

Answer

The molecular formula of alkene

X

is

C_3H_6

(propene).

In catalytic cracking, atomic mass is conserved. Subtracting the atoms of hexane (

C_6H_{14}

) from dodecane (

C_{12}H_{26}

) leaves 6 carbon atoms and 12 hydrogen atoms. Since 2 moles of product

X

are formed (

2X = C_6H_{12}

), dividing by 2 yields

C_3H_6

, which is propene, a valid alkene.

Step-by-Step Solution

1

Write the balanced stoichiometric equation for the cracking reaction.

C_{12}H_{26} \rightarrow C_6H_{14} + 2X

Cracking conserves the total number of carbon and hydrogen atoms between reactants and products.

2

Determine the remaining number of carbon and hydrogen atoms allocated to

2X

.

Carbons:

12 - 6 = 6

, Hydrogens:

26 - 14 = 12

. Total fragment formula is

C_6H_{12}

.

Subtract the atoms present in one mole of hexane from dodecane.

3

Divide the carbon and hydrogen count by 2 to find the formula of 1 mole of product

X

.

Carbon count for

X = \frac{6}{2} = 3

, Hydrogen count for

X = \frac{12}{2} = 6

. Formula of

X = C_3H_6

.

Since two moles of alkene

X

are produced, each mole contains half the remaining atoms, fitting the general formula for alkenes (

C_nH_{2n}

).

Key Concept

Catalytic Cracking Stoichiometry of Alkanes

Question 8344Question →

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

Click a left item, then click its matching right item

Show answer & explanation

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 8345Question →

An aqueous solution of ammonium ethanoate ( $CH_3COONH_4$ ) is approximately neutral with a pH close to 7 because neither the ammonium cation ( $NH_4^+$ ) nor the ethanoate anion ( $CH_3COO^-$ ) undergoes hydrolysis in water.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Both the cation (

NH_4^+

) and anion (

CH_3COO^-

) undergo hydrolysis in water, but because the acid dissociation constant of the parent weak acid (

K_a

) equals the base dissociation constant of the parent weak base (

K_b

), the concentrations of generated hydronium and hydroxide ions are equal, producing a neutral solution.

The statement incorrectly attributes the neutral pH of aqueous ammonium ethanoate to an absence of hydrolysis. In reality, both the conjugate acid (

NH_4^+

) and conjugate base (

CH_3COO^-

) hydrolyze simultaneously; neutrality arises solely because

K_a

of ethanoic acid equals

K_b

of ammonia.

Step-by-Step Solution

1

Identify the parent acid and parent base of ammonium ethanoate (

CH_3COONH_4

).

Ammonium ethanoate is derived from ethanoic acid (

CH_3COOH

), a weak acid, and aqueous ammonia (

NH_3

), a weak base.

Salts formed from weak acids and weak bases yield conjugate species that are strong enough to react with water.

2

Analyze the hydrolysis behavior of both ions in aqueous solution.

Cation hydrolysis:

NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+

. Anion hydrolysis:

CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-

.

Both

NH_4^+

(weak acid) and

CH_3COO^-

(weak base) hydrolyze simultaneously in water.

3

Determine the net effect of mutual hydrolysis on solution pH.

Since

K_a(CH_3COOH) = 1.8 \times 10^{-5}

and

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

,

[H_3O^+] = [OH^-]

, leading to

\text{pH} \approx 7.0

.

Equal ionization constants produce balanced amounts of hydronium and hydroxide ions, resulting in a neutral solution despite significant hydrolysis of both ions.

Key Concept

Hydrolysis of salts derived from a weak acid and a weak base

Question 8346Question →

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

Click a left item, then click its matching right item

Show answer & explanation

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 8347Question →

Sulfur tetrafluoride ( $SF_4$ ) and xenon tetrafluoride ( $XeF_4$ ) are both covalent fluorides containing four fluorine atoms bonded to a central atom. Based on VSEPR theory and hybridization models, which of the following correctly pairs the molecular geometry and central atom hybridization state for $SF_4$ and $XeF_4$ , respectively?

SF_4

: Seesaw geometry with

sp^3d

hybridization;

XeF_4

: Square planar geometry with

sp^3d^2

hybridization

SF_4

: Trigonal bipyramidal geometry with

sp^3d

hybridization;

XeF_4

: Octahedral geometry with

sp^3d^2

hybridization

SF_4

: Tetrahedral geometry with

sp^3

hybridization;

XeF_4

: Tetrahedral geometry with

sp^3

hybridization

SF_4

: Square planar geometry with

sp^3d^2

hybridization;

XeF_4

: Seesaw geometry with

sp^3d

hybridization

Show answer & explanation

Answer: $SF_4$ : Seesaw geometry with $sp^3d$ hybridization; $XeF_4$ : Square planar geometry with $sp^3d^2$ hybridization

Answer

SF_4

has a seesaw molecular geometry with

sp^3d

hybridization on the sulfur atom, while

XeF_4

has a square planar molecular geometry with

sp^3d^2

hybridization on the xenon atom.

For

SF_4

, sulfur has 6 valence electrons forming 4 single bonds and retaining 1 lone pair, giving 5 electron domains (

sp^3d

hybridization). The lone pair occupies an equatorial site of the trigonal bipyramid, yielding a seesaw shape. For

XeF_4

, xenon has 8 valence electrons forming 4 single bonds and retaining 2 lone pairs, giving 6 electron domains (

sp^3d^2

hybridization). The two lone pairs lie opposite each other along the axial axis, resulting in a square planar geometry.

Step-by-Step Solution

1

Determine valence electron count and electron domains for

SF_4

Sulfur has 6 valence electrons. Bonding 4 fluorine atoms consumes 4 electrons, leaving 2 non-bonding electrons (1 lone pair). Total steric number = 4 bonding pairs + 1 lone pair = 5 electron domains.

VSEPR theory requires calculating the total number of electron pairs around the central atom.

2

Assign hybridization and molecular geometry for

SF_4

A steric number of 5 corresponds to

sp^3d

hybridization and trigonal bipyramidal electron geometry. With 1 equatorial lone pair, the molecular shape is seesaw.

Lone pairs prefer equatorial positions in trigonal bipyramidal systems to minimize electron repulsion, leading to a seesaw geometry.

3

Determine valence electron count and electron domains for

XeF_4

Xenon has 8 valence electrons. Bonding 4 fluorine atoms consumes 4 electrons, leaving 4 non-bonding electrons (2 lone pairs). Total steric number = 4 bonding pairs + 2 lone pairs = 6 electron domains.

Steric number governs hybridization and electron arrangement for the central xenon atom.

4

Assign hybridization and molecular geometry for

XeF_4

A steric number of 6 corresponds to

sp^3d^2

hybridization and octahedral electron geometry. With 2 axial lone pairs opposing each other (180° apart), the molecular geometry is square planar.

Axial placement of two lone pairs in an octahedral arrangement maximizes separation and produces a planar square arrangement of bonded fluorine atoms.

Key Concept

Molecular Shapes, VSEPR Theory, and Hybridization

Estimated Time:2m 0s

Question 8348Question →

Arrange the following sequential transformations of nitrogen in the biological nitrogen cycle, beginning with atmospheric molecular nitrogen ( $N_2$ ) and concluding with its re-release into the atmosphere:

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of nitrogen cycle transformations is: Nitrogen fixation (

N_2 \rightarrow NH_4^+

), Nitrosification (

NH_4^+ \rightarrow NO_2^-

), Nitration (

NO_2^- \rightarrow NO_3^-

), Assimilation (

NO_3^- \rightarrow \text{organic plant proteins}

), and Denitrification (

NO_3^- \rightarrow N_2

).

The biological nitrogen cycle proceeds in a specific logical sequence. First, atmospheric nitrogen gas (

N_2

) must undergo nitrogen fixation to become ammonium ions (

NH_4^+

). Next, nitrification occurs in two steps: nitrosification converts ammonium (

NH_4^+

) to nitrite (

NO_2^-

) via *Nitrosomonas*, followed by nitration converting nitrite (

NO_2^-

) to nitrate (

NO_3^-

) via *Nitrobacter*. Plants then assimilate these nitrate ions (

NO_3^-

) into organic plant proteins. Finally, denitrifying bacteria reduce residual soil nitrates (

NO_3^-

) back into atmospheric nitrogen gas (

N_2

) under anaerobic conditions.

Step-by-Step Solution

1

Identify the initial entry point of atmospheric nitrogen into the biosphere.

Gaseous nitrogen (

N_2

) is fixed into ammonium ions (

NH_4^+

) by nitrogen-fixing bacteria (e.g., *Rhizobium*, *Azotobacter*).

Atmospheric

N_2

has a strong triple covalent bond (

N \equiv N

) and cannot be directly utilized by higher plants without preliminary biological fixation.

2

Determine the initial oxidation stage of nitrification (nitrosification).

Ammonium ions (

NH_4^+

) are oxidized to nitrite ions (

NO_2^-

) by *Nitrosomonas*.

Nitrification occurs in two distinct microbial steps, starting with ammonium conversion to nitrite.

3

Determine the second oxidation stage of nitrification (nitration).

Nitrite ions (

NO_2^-

) are oxidized to nitrate ions (

NO_3^-

) by *Nitrobacter*.

Nitrate (

NO_3^-

) is the most readily absorbed and utilized form of inorganic nitrogen for plants.

4

Trace the biological uptake of bioavailable soil nitrogen.

Plants absorb

NO_3^-

ions and assimilate them into plant proteins and nucleic acids.

Inorganic nitrate is reduced inside plant tissues and converted into organic amino acids.

5

Identify the pathway responsible for returning gaseous nitrogen to the atmosphere.

Denitrifying bacteria (e.g., *Pseudomonas denitrificans*) convert unabsorbed soil nitrates (

NO_3^-

) into atmospheric

N_2

gas under anaerobic conditions.

Denitrification closes the global nitrogen loop by replenishing free atmospheric

N_2

.

Key Concept

Nitrogen Cycle Transformations and Microorganisms

Estimated Time:2m 0s

Question 8349Question →

During the industrial extraction of aluminium by the Hall-Héroult electrolytic process, why must the graphite anode rods be replaced at regular intervals?

The oxygen gas discharged at the anode reacts with the graphite at high operating temperatures to form carbon dioxide

Molten aluminium is deposited directly onto the anode rods, causing severe corrosion and electrical shorting

Molten cryolite reacts chemically with graphite to produce gaseous carbon tetrafluoride

Graphite oxidizes directly by releasing carbon cations into the molten electrolyte solution

Show answer & explanation

Answer: The oxygen gas discharged at the anode reacts with the graphite at high operating temperatures to form carbon dioxide

Answer

The oxygen gas discharged at the anode reacts with the graphite at high operating temperatures to form carbon dioxide

During the Hall-Héroult process, oxide ions (

O^{2-}

) are discharged at the carbon (graphite) anode to produce oxygen gas. Because the cell operates at approximately 950°C, the freshly liberated oxygen gas reacts with the carbon anodes to form carbon dioxide (

CO_2

) and carbon monoxide (

CO

). As a result, the graphite anodes are burned away over time and must be routinely replaced.

Step-by-Step Solution

1

Identify the anode half-reaction during the electrolysis of molten alumina in cryolite

Oxide ions (

O^{2-}

) migrate to the positive graphite anode and undergo oxidation to form oxygen gas:

2O^{2-} \rightarrow O_2(g) + 4e^-

Anodes are the site of oxidation where negative ions discharge electrons.

2

Analyze the chemical interaction between liberated oxygen gas and the carbon anode at high cell operating temperatures

At temperatures around 950°C, the liberated oxygen gas reacts with the hot graphite anode to form carbon dioxide:

C(s) + O_2(g) \rightarrow CO_2(g)

Graphite carbon combusts in the presence of oxygen gas at elevated temperatures, leading to continuous depletion of the anodes.

Key Concept

Anode consumption due to oxygen reaction in the Hall-Héroult process

Estimated Time:1m 0s

Question 8350Question →

Complete the following passage regarding the redox behavior of hydrogen sulfide gas when reacted with an iron(III) compound by filling in the blanks with the correct chemical terms.

Fill in the blanks below

When hydrogen sulfide gas is bubbled through an acidified solution of iron(III) chloride, the characteristic yellow color of the solution changes to due to the reduction of iron(III) ions to iron(II) ions, while a fine precipitate of elemental sulfur is deposited.

Show answer & explanation

Answer

The solution turns pale green (or green) due to the formation of iron(II) ions (

Fe^{2+}

), accompanied by the deposition of a yellow precipitate of elemental sulfur (

S

).

Hydrogen sulfide gas (

H_2S

) acts as a strong reducing agent. When passed into an acidified solution of iron(III) chloride (

FeCl_3

), it reduces yellow

Fe^{3+}

ions to light green

Fe^{2+}

ions, while

H_2S

itself gets oxidized to form a yellow precipitate of elemental sulfur (

S

). The overall ionic equation for the reaction is:

2Fe^{3+}_{(aq)} + H_2S_{(g)} \rightarrow 2Fe^{2+}_{(aq)} + 2H^+_{(aq)} + S_{(s)}

.

Step-by-Step Solution

1

Identify the role of hydrogen sulfide (

H_2S

) in the reaction.

H_2S

acts as a reducing agent in acidic solution.

Sulfur in

H_2S

has an oxidation state of

-2

and undergoes oxidation to elemental sulfur (

0

oxidation state).

2

Determine the change in oxidation state of iron.

Iron(III) ions (

Fe^{3+}

) are reduced to iron(II) ions (

Fe^{2+}

).

Iron(III) chloride solution is yellow/brown due to

Fe^{3+}_{(aq)}

, whereas iron(II) ions (

Fe^{2+}_{(aq)}

) impart a light green/pale green color to the aqueous solution.

3

Identify the physical appearance of oxidized sulfur product.

Elemental sulfur precipitates out of the solution.

Insoluble sulfur particles form a yellow suspension or precipitate in the solution.

Key Concept

Reducing property of Hydrogen Sulfide (

H_2S

)

Question 8351Question →

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.

Click a left item, then click its matching right item

Show answer & explanation

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 8352Question →

According to the Valence Shell Electron Pair Repulsion (VSEPR) theory, what is the molecular shape and central atom hybridization of sulfur dioxide ( $SO_2$ )?

Bent (V-shaped) with

sp^2

hybridization

Trigonal planar with

sp^2

hybridization

Linear with

sp

hybridization

Bent (V-shaped) with

sp^3

hybridization

Show answer & explanation

Answer: Bent (V-shaped) with $sp^2$ hybridization

Answer

Bent (V-shaped) geometry with

sp^2

hybridization of the central sulfur atom.

The central sulfur atom in

SO_2

has three electron regions around it (two bonding regions to oxygen atoms and one lone pair). Three electron regions dictate a trigonal planar electron geometry and

sp^2

hybridization. Because one of these regions is a lone pair, the observable molecular geometry formed by the atoms is bent (V-shaped).

Step-by-Step Solution

1

Determine the valence electron count and electron domains around the central sulfur atom in

SO_2

.

Sulfur (Group 16) brings 6 valence electrons. It forms 2 double bonds with 2 oxygen atoms (2 bonding domains) and retains 1 lone pair of electrons. Total electron domains = 3.

VSEPR theory uses the total number of electron domains (bonding regions + lone pairs) to determine spatial arrangement.

2

Determine the hybridization of the central atom.

3 electron domains correspond to an

sp^2

hybridization state.

Mixing one s orbital and two p orbitals yields three equivalent

sp^2

hybrid orbitals directed towards the corners of an equilateral triangle.

3

Determine the molecular geometry based on bonding domains and lone pairs.

With 3 electron domains (2 bonding domains and 1 lone pair), the electron geometry is trigonal planar, but the molecular geometry (shape formed by atoms) is Bent (V-shaped) with a bond angle of slightly less than

120^\circ

.

Lone pair-bonding pair repulsion compresses the

O-S-O

bond angle slightly below the ideal

120^\circ

trigonal planar angle.

Key Concept

VSEPR Theory, Molecular Geometry, and Hybridization

Estimated Time:1m 0s

Question 8353Question →

When exposed to moist atmospheric air, certain solid compounds absorb sufficient water vapor to completely dissolve in it and form a liquid solution. Which of the following substances exhibits this phenomenon known as deliquescence?

Calcium chloride (

\text{CaCl}_2

)

Concentrated tetraoxosulfate(VI) acid (

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

)

Sodium trioxocarbonate(IV) decahydrate (

\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}

)

Copper(II) oxide (

\text{CuO}

)

Show answer & explanation

Answer: Calcium chloride ( $\text{CaCl}_2$ )

Answer

Calcium chloride (

\text{CaCl}_2

) is deliquescent because it absorbs atmospheric water vapor to form a liquid solution.

Calcium chloride (

\text{CaCl}_2

) is a deliquescent solid. When exposed to air with sufficient humidity, its surface vapor pressure is much lower than the partial pressure of water vapor in the atmosphere, causing it to absorb water continuously until it fully dissolves into an aqueous solution.

Step-by-Step Solution

1

Define deliquescence

Deliquescence is the process in which a solid substance absorbs moisture from the atmosphere until it completely dissolves in the absorbed water, forming a concentrated solution.

Understanding the precise chemical definition distinguishes deliquescent solids from hygroscopic liquids or efflorescent hydrated salts.

2

Evaluate the options based on their behavior in moist air

Solid calcium chloride (

\text{CaCl}_2

) absorbs water vapor and turns into a liquid solution, making it deliquescent.

Calcium chloride has a very low saturated solution vapor pressure compared to atmospheric water vapor pressure.

Key Concept

Atmospheric Phenomena of Salts: Deliquescence vs Hygroscopy vs Efflorescence

Question 8354Question →

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

Click a left item, then click its matching right item

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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 8355Question →

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 8356Question →

In a gas effusion experiment conducted at constant temperature and pressure, a sample of sulfur(IV) oxide ( $SO_2$ ) effuses through a micro-porous aperture at a rate of $0.625\text{ cm}^3\text{ s}^{-1}$ . A gaseous alkane, designated as hydrocarbon $X$ , effuses through the exact same aperture at a rate of $1.25\text{ cm}^3\text{ s}^{-1}$ . Given relative atomic masses ( $S = 32$ , $O = 16$ , $C = 12$ , $H = 1$ ), determine the molecular formula of hydrocarbon $X$ .

Show answer & explanation

Answer: CH4; CH_4; methane; CH4 (methane)

Answer

The molecular formula of hydrocarbon X is CH4 (methane).

According to Graham's Law of Effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. The ratio of the rate of effusion of hydrocarbon X to that of sulfur(IV) oxide (

SO_2

, molar mass

64\text{ g/mol}

) is

1.25 / 0.625 = 2

. Squaring this ratio gives

4 = 64 / M_X

, yielding a molar mass of

16\text{ g/mol}

for hydrocarbon X. For an alkane with general formula

C_n H_{2n+2}

, a molar mass of

16\text{ g/mol}

corresponds to

n = 1

, giving the molecular formula

CH_4

(methane).

Step-by-Step Solution

1

Calculate the molar mass of sulfur(IV) oxide (

SO_2

).

M(SO_2) = 32 + 2(16) = 64\text{ g/mol}

The molar mass of the reference gas is needed to apply Graham's law.

2

Apply Graham's Law of Diffusion/Effusion relating effusion rates to molar masses.

\frac{r_X}{r_{SO_2}} = \sqrt{\frac{M(SO_2)}{M_X}} \implies \frac{1.25}{0.625} = \sqrt{\frac{64}{M_X}}

Graham's law states that the rate of diffusion/effusion of a gas is inversely proportional to the square root of its molar mass.

3

Solve for the molar mass of hydrocarbon

X

(

M_X

).

2 = \sqrt{\frac{64}{M_X}} \implies 4 = \frac{64}{M_X} \implies M_X = 16\text{ g/mol}

Squaring both sides allows direct calculation of the unknown molar mass.

4

Determine the molecular formula of the alkane with molar mass

16\text{ g/mol}

.

Alkane general formula:

C_n H_{2n+2} \implies 12n + 1(2n+2) = 16 \implies 14n + 2 = 16 \implies 14n = 14 \implies n = 1

. Thus, formula is

CH_4

.

Comparing the calculated molar mass to the general molecular formula of alkanes identifies the specific hydrocarbon.

Key Concept

Graham's Law of Diffusion and Effusion combined with alkane stoichiometry

Question 8357Question →

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 8358Question →

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 8359Question →

Arrange the following sequential processing stages involved in the treatment of raw river water for municipal supply in the correct order from initial treatment to final distribution.

Drag items to arrange them in the correct order

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Answer

The correct sequence for municipal water purification is: Aeration → Coagulation (addition of alum) → Sedimentation → Sand Filtration → Chlorination (Disinfection).

Municipal water treatment follows a logical progression designed to clear large and dissolved impurities first before fine physical filtering and final chemical disinfection. Aeration removes volatile gases and oxidizes metals; coagulation aggregates colloidal particles using alum; sedimentation settles the heavy flocs; sand filtration removes micro-particles; and chlorination destroys disease-causing microorganisms as the final safeguard before piping to households.

Step-by-Step Solution

1

Identify the initial physical treatment step for raw water.

Aeration is performed first to remove volatile odors/gases and oxidize dissolved iron salts into insoluble forms.

Air exposure improves taste and prepares dissolved metals for precipitation.

2

Determine the step that precipitates fine colloidal particles.

Coagulation with potash alum follows aeration.

Alum causes fine, non-settling colloidal particles to aggregate into larger flocs.

3

Identify how the aggregated flocs are removed bulk-wise.

Sedimentation allows these heavy flocs to settle to the basin floor.

Gravity settling removes the bulk of suspended solids before filtration.

4

Identify the step that removes residual microscopic suspended solids.

Sand filtration passes clear water through sand and gravel beds.

Filtration traps any residual suspended particles that escaped sedimentation.

5

Identify the final biological safety step prior to distribution.

Chlorination is carried out as the final step.

Sterilization must occur after physical clarity is achieved so chlorine acts effectively on pathogenic bacteria.

Key Concept

Sequential Stages of Municipal Water Treatment

Question 8360Question →

During a quality control procedure in a chemical plant, a liquid sample containing microscopic colloidal solids suspended in a dense solvent is found to pass through filter paper without separating. Which technique utilizes high-speed rotation to exert an enhanced artificial gravitational force, forcing the suspended particles to settle rapidly at the bottom based on particle density?

Centrifugation

Sublimation

Magnetization

Fractional distillation

Show answer & explanation

Answer: Centrifugation

Answer

Centrifugation is the physical separation technique that uses rapid spinning to generate centrifugal force, causing dense suspended particles in a liquid medium to settle out rapidly.

Centrifugation is the appropriate technique because rapid rotation generates centrifugal force, accelerating the settling rate of fine, insoluble suspended particles based on density differences when gravity settling or filtration is ineffective.

Step-by-Step Solution

1

Analyze the physical state of the mixture described in the stem.

The mixture consists of microscopic colloidal solid particles suspended in a liquid medium that cannot be trapped by standard filter paper pores.

Identifying component sizes and physical properties determines the correct separation mechanism.

2

Evaluate the separation principle required for fine suspensions.

When gravity settling is too slow and filtration fails, high-speed rotational motion creates a strong outward centrifugal force that forces dense solid particles to sediment quickly at the tube bottom.

Centrifugation relies on density differences between suspended solids and the surrounding fluid medium under elevated pseudo-gravitational forces.

Key Concept

Centrifugation principles for colloidal suspensions and fine liquid-solid mixtures

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