All practice questions

Question 8181Question →

An industrial fuel gas mixture derived from coal processing contains $40.0\%$ carbon(II) oxide ( $\text{CO}$ ), $40.0\%$ hydrogen gas ( $\text{H}_2$ ), and $20.0\%$ carbon(IV) oxide ( $\text{CO}_2$ ) by volume. Assuming air contains $20.0\%$ oxygen ( $\text{O}_2$ ) by volume, what is the total volume of air required at s.t.p. for the complete combustion of $100.0\text{ dm}^3$ of this fuel gas mixture?

200.0\text{ dm}^3

40.0\text{ dm}^3

400.0\text{ dm}^3

500.0\text{ dm}^3

Show answer & explanation

Answer: $200.0\text{ dm}^3$

Answer

The total volume of air required at s.t.p. for complete combustion is

200.0\text{ dm}^3

.

In the

100.0\text{ dm}^3

sample of fuel gas, the combustible gases are

40.0\text{ dm}^3

of

\text{CO}

and

40.0\text{ dm}^3

of

\text{H}_2

, while

\text{CO}_2

is already fully oxidized and non-combustible. According to reaction stoichiometry,

2\text{CO} + \text{O}_2 \rightarrow 2\text{CO}_2

and

2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}

,

40.0\text{ dm}^3

of

\text{CO}

requires

20.0\text{ dm}^3

of

\text{O}_2

and

40.0\text{ dm}^3

of

\text{H}_2

requires

20.0\text{ dm}^3

of

\text{O}_2

. The total pure oxygen needed is

40.0\text{ dm}^3

. Since air is

20.0\%

oxygen by volume, the required volume of air is

40.0\text{ dm}^3 \div 0.20 = 200.0\text{ dm}^3

.

Step-by-Step Solution

1

Determine the volumes of combustible and non-combustible components in the

100.0\text{ dm}^3

fuel gas mixture.

Volume of

\text{CO} = 40.0\text{ dm}^3

, volume of

\text{H}_2 = 40.0\text{ dm}^3

, and volume of

\text{CO}_2 = 20.0\text{ dm}^3

(non-combustible).

By Gay-Lussac's Law of Combining Volumes, volume percentage corresponds directly to gas volume at s.t.p.

2

Write the balanced chemical equations and determine oxygen requirements for each combustible component.

2\text{CO}(g) + \text{O}_2(g) \rightarrow 2\text{CO}_2(g)

requires

20.0\text{ dm}^3

of

\text{O}_2

;

2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l)

requires

20.0\text{ dm}^3

of

\text{O}_2

. Total pure

\text{O}_2 = 40.0\text{ dm}^3

.

Each combustible gas (

\text{CO}

and

\text{H}_2

) reacts with oxygen in a

2:1

mole (and volume) ratio.

3

Calculate the total volume of air needed to supply the required

40.0\text{ dm}^3

of pure oxygen.

\text{Volume of air} = \frac{40.0\text{ dm}^3}{0.20} = 200.0\text{ dm}^3

.

Air contains only

20.0\%

oxygen by volume, so five times the volume of oxygen is needed in total air.

Key Concept

Industrial Fuel Gas Stoichiometry and Gas Volumetric Calculations

Question 8182Question →

An industrial effluent contains suspended colloidal particles of clay carrying negative surface charges, with particle diameters ranging from $1\text{ nm}$ to $100\text{ nm}$ . To treat the effluent, equal volumes of four different $0.01\text{ mol dm}^{-3}$ electrolyte solutions— $Al_2(SO_4)_3$ , $CaCl_2$ , $NaCl$ , and $Na_3PO_4$ —are evaluated for their ability to induce coagulation. Which electrolyte has the highest precipitating power (lowest coagulation value) for this clay colloid, and what is the physical mechanism causing coagulation?

Al_2(SO_4)_3

, because the trivalent cation (

Al^{3+}

) carries the highest positive charge to neutralize the negative charges on the colloidal particles, causing them to aggregate into particles larger than

100\text{ nm}

.

Na_3PO_4

, because the trivalent phosphate anion (

PO_4^{3-}

) provides the highest charge density to neutralize the colloidal dispersion.

NaCl

, because its low formula mass enables rapid diffusion into the colloidal particles, converting the colloid into a true solution of particle size under

1\text{ nm}

.

All four electrolytes possess equal precipitating power because coagulation depends strictly on the total molar concentration of solute particles rather than ionic charge.

Show answer & explanation

Answer: $Al_2(SO_4)_3$ , because the trivalent cation ( $Al^{3+}$ ) carries the highest positive charge to neutralize the negative charges on the colloidal particles, causing them to aggregate into particles larger than $100\text{ nm}$ .

Answer

The electrolyte with the highest precipitating power is

Al_2(SO_4)_3

because the trivalent cation (

Al^{3+}

) effectively neutralizes the negative surface charge on the clay particles according to the Hardy-Schulze rule.

The clay colloidal particles are negatively charged with diameters between

1\text{ nm}

and

100\text{ nm}

. According to the Hardy-Schulze rule, the coagulating power of an electrolyte is determined by the ion bearing a charge opposite to that of the colloidal particles, and it increases rapidly with the valency of the active ion. For a negative sol, cations are active. The trivalent

Al^{3+}

ion from aluminium tetraoxosulfate(VI) has a significantly higher precipitating power than divalent

Ca^{2+}

or monovalent

Na^{+}

, neutralizing the charge and aggregating particles beyond

100\text{ nm}

.

Step-by-Step Solution

1

Identify the charge carried by the colloidal clay particles.

The clay particles are negatively charged lyophobic colloidal particles in the size range

1\text{ nm}

to

100\text{ nm}

.

Coagulation requires neutralizing the electrostatic repulsion between colloidal particles using an oppositely charged ion.

2

Apply the Hardy-Schulze rule to determine which ion causes precipitation.

Positively charged cations (

Al^{3+}

,

Ca^{2+}

,

Na^{+}

) are responsible for coagulating the negative colloid.

Ions possessing a charge opposite to that of the colloidal sol are effective for coagulation.

3

Compare the valencies of the effective cations present in the electrolytes.

Al^{3+}

has a valency of

+3

,

Ca^{2+}

has

+2

, and

Na^{+}

has

+1

.

Precipitating power increases exponentially with increasing valency of the active ion (

Al^{3+} > Ca^{2+} > Na^{+}

).

4

Relate charge neutralization to particle size change.

Neutralized particles coalesce into larger aggregates (

>100\text{ nm}

) that precipitate out.

Removal of surface charge reduces electrostatic repulsion, allowing van der Waals forces to aggregate colloidal particles into suspension-sized precipitates.

Key Concept

Hardy-Schulze Rule and Coagulation of Colloidal Systems

Question 8183Question →

A sample of sulfur(IV) oxide gas ( $\text{SO}_2$ ) occupies a volume of $1.20\text{ dm}^3$ at room temperature and pressure ( $\text{RTP}$ ). What is the total number of oxygen atoms present in this sample? $[\text{Molar volume of gas at RTP} = 24.0\text{ dm}^3\text{ mol}^{-1}, \text{Avogadro's constant } N_A = 6.02 \times 10^{23}\text{ mol}^{-1}]$

6.02 \times 10^{22}

6.45 \times 10^{22}

3.01 \times 10^{22}

3.23 \times 10^{22}

Show answer & explanation

Answer: $6.02 \times 10^{22}$

Answer

6.02 \times 10^{22}

oxygen atoms

The correct answer is

6.02 \times 10^{22}

. Dividing the gas volume (

1.20\text{ dm}^3

) by the molar volume at RTP (

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

) gives

0.05\text{ mol}

of

\text{SO}_2

gas. Since each molecule of

\text{SO}_2

contains 2 atoms of oxygen, the sample contains

0.10\text{ mol}

of oxygen atoms. Multiplying

0.10\text{ mol}

by Avogadro's constant (

6.02 \times 10^{23}\text{ mol}^{-1}

) yields

6.02 \times 10^{22}

oxygen atoms.

Step-by-Step Solution

1

Calculate the number of moles of

\text{SO}_2

gas at RTP

n(\text{SO}_2) = \frac{1.20\text{ dm}^3}{24.0\text{ dm}^3\text{ mol}^{-1}} = 0.050\text{ mol}

Molar volume at room temperature and pressure is

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

.

2

Determine the number of moles of oxygen atoms present

n(\text{O}) = 2 \times 0.050\text{ mol} = 0.100\text{ mol}

Each molecule of

\text{SO}_2

contains 2 oxygen atoms.

3

Calculate the total number of oxygen atoms using Avogadro's constant

N(\text{O}) = 0.100\text{ mol} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 6.02 \times 10^{22}\text{ atoms}

Multiplying moles of atoms by Avogadro's constant gives the exact atom count.

Key Concept

Molar volume of gases at RTP and stoichometric atom counting within polyatomic molecules.

Question 8184Question →

Match each microscopic structural feature of metallic bonding on the left with the macroscopic physical property of metals on the right that directly results from it.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The mobile sea of delocalized electrons accounts for high electrical and thermal conductivity. Non-directional bonding between cation layers accounts for malleability and ductility. Strong multi-directional electrostatic attraction accounts for high melting and boiling points.

Each physical property of metals stems directly from the electron-sea structural model. Electrical and thermal conductivity rely on mobile delocalized valence electrons. Malleability and ductility arise from non-directional bonding that allows cation layers to slide smoothly past each other under force. High melting and boiling points are due to the strong electrostatic forces holding the giant metallic lattice together.

Step-by-Step Solution

1

Link electrical/thermal transport to microscopic charge carriers

Free-moving (delocalized) electrons drift when an electric field or thermal gradient is applied, explaining high conductivity.

Electrical conduction requires mobile charged particles, which in metals are delocalized electrons.

2

Analyze the mechanical behavior of metal cation layers during deformation

When hammered or drawn into wires, cation layers slide without repulsive disruption because the electron sea flexibly adjusts.

Non-directional bonding prevents catastrophic cleavage, providing malleability and ductility.

3

Connect lattice stability to thermal energy requirements for melting

Substantial thermal energy is necessary to overcome strong electrostatic attraction between positive ions and negative electrons.

High bond energy across the giant metallic structure leads directly to elevated melting and boiling points.

Key Concept

Relationship between metallic bonding structure (electron sea model) and physical properties of metals

Question 8185Question →

Fill in the missing numerical values in the statement below regarding the pH and pOH scale at $25^\circ\text{C}$ .

Fill in the blanks below

For an aqueous solution at

25^\circ\text{C}

, a hydrogen ion concentration of

1.0 \times 10^{-4}\text{ mol dm}^{-3}

corresponds to a pH of and a pOH of .

Show answer & explanation

Answer

The pH of the solution is 4 (or 4.0) and the pOH is 10 (or 10.0).

Taking the negative logarithm of

[H^+] = 1.0 \times 10^{-4}\text{ mol dm}^{-3}

gives a pH of 4. Since

\text{pH} + \text{pOH} = 14

at

25^\circ\text{C}

, subtracting 4 from 14 gives a pOH of 10.

Step-by-Step Solution

1

Calculate the pH from the given hydrogen ion concentration

[H^+]

.

\text{pH} = -\log_{10}(1.0 \times 10^{-4}) = 4

pH is defined as the negative logarithm to base 10 of the hydrogen ion concentration.

2

Calculate the pOH using the relationship

\text{pH} + \text{pOH} = 14

.

\text{pOH} = 14 - 4 = 10

At

25^\circ\text{C}

, the sum of pH and pOH for an aqueous solution equals 14.

Key Concept

Logarithmic pH and pOH calculations from hydrogen ion concentration

Question 8186Question →

Complete the statement below regarding chemical bonding.

Fill in the blanks below

A chemical bond formed when a single atom donates both electrons of a shared pair is known as a coordinate covalent bond or a bond.

Show answer & explanation

Answer

dative

In standard chemistry terminology, a coordinate covalent bond and a dative bond are synonymous. Both refer to a covalent linkage where one atom contributes both electrons to the shared pair.

Step-by-Step Solution

1

Recall the definition and alternative chemical terminology for coordinate covalent bonding.

A coordinate bond is also termed a dative covalent bond.

Both terms describe a special type of covalent bond where the shared pair of electrons is provided exclusively by a single donor atom possessing a lone pair.

Key Concept

Definition and terminology of coordinate (dative) covalent bonding

Question 8187Question →

Which of the following best explains why solid sodium chloride ( $NaCl$ ) does not conduct electricity, whereas molten sodium chloride conducts electricity readily?

In the solid state, the ions are held in fixed positions within the crystal lattice and cannot move freely.

Solid sodium chloride consists of neutral molecules, whereas molten sodium chloride breaks into free electrons.

Electrons are transferred from sodium to chlorine only when the compound is heated into the molten state.

The ionic bonds in sodium chloride solid are covalent, but they turn electrovalent when melted.

Show answer & explanation

Answer: In the solid state, the ions are held in fixed positions within the crystal lattice and cannot move freely.

Answer

In the solid state, the ions are held in fixed positions within the crystal lattice and cannot move freely.

Ionic (electrovalent) compounds conduct electricity only when charge carriers (ions) are free to move. In the solid state, ions are locked tightly into fixed positions within the giant ionic lattice, preventing electrical conduction. When melted, the high thermal energy breaks the rigid lattice structure, enabling the positive (

Na^+

) and negative (

Cl^-

) ions to move freely and carry electrical current.

Step-by-Step Solution

1

Identify the nature of charge carriers in ionic compounds.

Electrical conductivity in electrovalent (ionic) compounds relies on mobile ions (

Na^+

and

Cl^-

), not free electrons.

Ionic compounds do not contain free delocalized electrons like metals.

2

Analyze the structural state of solid sodium chloride.

In solid

NaCl

, electrostatic forces lock ions in fixed positions in a giant 3D crystal lattice structure.

Because ions cannot move from place to place, solid

NaCl

acts as an electrical insulator.

3

Compare the solid state with the molten state.

When melted (molten state), thermal energy overcomes the rigid lattice forces, allowing

Na^+

and

Cl^-

ions to move freely toward electrodes.

Mobile ions enable the flow of electric current in the liquid/molten state.

Key Concept

Electrical Conductivity of Electrovalent Compounds

Question 8188Question →

In a municipal water treatment plant, raw water undergoes several physical and chemical processing steps. What is the primary purpose of introducing activated charcoal during the purification of water for town supply?

To remove unpleasant odors, objectionable tastes, and colored organic compounds

To destroy disease-causing bacteria and pathogenic microorganisms

To cause fine suspended colloidal clay particles to aggregate into large flocs

To precipitate calcium and magnesium ions that cause water hardness

Show answer & explanation

Answer: To remove unpleasant odors, objectionable tastes, and colored organic compounds

Answer

The primary purpose of activated charcoal in municipal water treatment is to remove unpleasant odors, objectionable tastes, and colored organic compounds.

Activated charcoal (carbon) has a micro-porous structure with an exceptionally high surface area. It physically adsorbs dissolved organic pollutants, pigments, and volatile substances that cause objectionable odors and tastes, thereby clarifying and improving the aesthetic quality of drinking water.

Step-by-Step Solution

1

Identify the chemical reagent or material specified in the question.

The specified material is activated charcoal (activated carbon).

Activated charcoal is an adsorbent material used in filtration beds during water treatment.

2

Recall the specific functional role of activated charcoal in water purification.

Activated charcoal physically adsorbs dissolved organic impurities, volatile organic compounds, pigments, and substances responsible for foul tastes and odors.

Its extremely porous structure provides a vast surface area suitable for physical adsorption.

3

Distinguish activated charcoal's role from other water treatment chemicals.

Chlorine sterilizes water, potash alum coagulates suspended particles, and slaked lime adjusts pH or precipitates hardness.

Each chemical reagent added during municipal water treatment has a distinct, specialized role.

Key Concept

Role of Activated Charcoal in Water Treatment

Question 8189Question →

An agricultural economist recorded the prices per bag of fertilizer (in thousands of Naira) across five regional markets as $₦12$ , $₦15$ , $₦18$ , $₦21$ , and $₦24$ . What is the mean deviation of the fertilizer prices (in thousands of Naira)?

Show answer & explanation

Answer: 3.6

Answer

The mean deviation of the fertilizer prices is

3.6

thousand Naira.

The mean deviation is

3.6

because the arithmetic mean of the prices is

18

. The absolute differences of the data values from

18

are

6

,

3

,

0

,

3

, and

6

. The sum of these absolute deviations is

18

, and dividing by

5

observations yields

3.6

.

Step-by-Step Solution

1

Calculate the arithmetic mean (

\bar{x}

) of the data set

\bar{x} = \frac{12 + 15 + 18 + 21 + 24}{5} = \frac{90}{5} = 18

The arithmetic mean provides the central reference point required to evaluate individual deviations.

2

Determine the absolute deviation

|x - \bar{x}|

for each value

|12 - 18| = 6

,

|15 - 18| = 3

,

|18 - 18| = 0

,

|21 - 18| = 3

,

|24 - 18| = 6

Mean deviation measures dispersion using absolute differences to prevent positive and negative deviations from canceling out.

3

Compute the average of the absolute deviations

\text{Mean Deviation} = \frac{6 + 3 + 0 + 3 + 6}{5} = \frac{18}{5} = 3.6

Dividing the sum of absolute deviations by the total number of observations gives the mean deviation.

Key Concept

Mean Deviation

Question 8190Question →

In a bimolecular gas-phase reaction between nitrogen monoxide ( $\text{NO}$ ) and ozone ( $\text{O}_3$ ), collision theory states that molecules must collide with appropriate orientation and sufficient energy. Which of the following best explains why increasing the partial pressure of $\text{NO}$ increases the overall rate of product formation at constant temperature?

It increases the average kinetic energy of the reacting gas molecules.

It lowers the activation energy barrier required for successful chemical transformation.

It increases the collision frequency per unit volume, leading to a greater number of effective collisions per unit time.

It increases the proportion of colliding molecules that possess energy greater than or equal to the activation energy.

Show answer & explanation

Answer: It increases the collision frequency per unit volume, leading to a greater number of effective collisions per unit time.

Answer

Increasing the partial pressure increases the collision frequency per unit volume, leading to a greater number of effective collisions per unit time.

Increasing the partial pressure of a gaseous reactant increases its concentration (number of molecules per unit volume). According to collision theory, a higher concentration increases the total collision frequency. Because the fraction of fruitful collisions (determined by kinetic energy and orientation) stays constant at fixed temperature, a greater total number of collisions produces a higher rate of effective collisions per second.

Step-by-Step Solution

1

Analyze the effect of increasing partial pressure on molecular density

Higher partial pressure means more reactant molecules per unit volume (increased concentration).

According to the ideal gas law, pressure is directly proportional to concentration at constant temperature.

2

Relate molecular density to collision frequency

More molecules per unit volume lead to more total collisions occurring per unit time.

Collision frequency is proportional to the concentration of reacting species.

3

Evaluate the rate of effective collisions and distinguish from temperature effects

While the fraction of effective collisions remains constant (since kinetic energy and temperature do not change), the absolute count of effective collisions per unit time increases.

Rate of reaction = (Total collision frequency) × (Fraction of effective collisions).

Key Concept

Collision Theory and Reactant Concentration/Pressure Effects

Estimated Time:1m 15s

Question 8191Question →

How many acyclic structural isomers exist for the haloalkane with the molecular formula $C_4H_9Cl$ ?

2

3

4

5

Show answer & explanation

Answer: 4

Answer

There are 4 acyclic structural isomers for

C_4H_9Cl

.

The molecular formula

C_4H_9Cl

allows for both chain isomerism (butane vs. methylpropane carbon backbones) and positional isomerism (placement of the chlorine atom). This results in four distinct IUPAC structures: 1-chlorobutane, 2-chlorobutane, 1-chloro-2-methylpropane, and 2-chloro-2-methylpropane.

Step-by-Step Solution

1

Identify structural isomers based on the straight-chain butyl skeleton (

CH_3-CH_2-CH_2-CH_3

).

Two positional isomers are obtained: 1-chlorobutane (

CH_3CH_2CH_2CH_2Cl

) and 2-chlorobutane (

CH_3CH_2CH(Cl)CH_3

).

Varying the position of the chlorine atom on a four-carbon unbranched chain yields two distinct structures.

2

Identify structural isomers based on the branched methylpropane skeleton (

(CH_3)_2CH-CH_3

).

Two chain/positional isomers are obtained: 1-chloro-2-methylpropane (

(CH_3)_2CHCH_2Cl

) and 2-chloro-2-methylpropane (

(CH_3)_3CCl

).

Attaching the chlorine atom to a primary carbon versus the tertiary carbon of the methylpropane backbone yields two additional unique isomers.

3

Sum the distinct structural isomers found.

Total number of isomers =

2 + 2 = 4

.

All possible constitutional connectivities for

C_4H_9Cl

have been evaluated without double counting.

Key Concept

Structural Isomerism in Haloalkanes

Estimated Time:1m 15s

Question 8192Question →

What is the IUPAC name of the major organic product formed when one mole of hydrogen bromide ( $HBr$ ) reacts with prop-1-ene in the absence of peroxides?

Show answer & explanation

Answer: 2-bromopropane; 2-Bromopropane

Answer

2-bromopropane

Electrophilic addition of hydrogen bromide to prop-1-ene without peroxides follows Markovnikov's rule. The hydrogen atom attaches to the terminal carbon (

CH_2

), leading to the formation of a secondary carbocation intermediate. Subsequent nucleophilic attack by the bromide ion yields 2-bromopropane as the major product.

Step-by-Step Solution

1

Identify the reactants and the reaction type

The reaction takes place between an unsymmetrical alkene, prop-1-ene (

CH_3CH=CH_2

), and hydrogen bromide (

HBr

). This is an electrophilic addition reaction.

Alkenes readily undergo electrophilic addition across the unsaturated carbon-carbon double bond.

2

Apply Markovnikov's rule to determine carbocation stability

In the absence of peroxides, the reaction follows Markovnikov's rule. The proton (

H^+

) adds to the double-bonded carbon holding more hydrogen atoms (

C1

,

CH_2

), generating a more stable secondary carbocation (

CH_3CH^+CH_3

).

A secondary carbocation is more stable than a primary carbocation due to alkyl group electron-donating inductive effects.

3

Attach the bromide ion and name the resulting haloalkane

The bromide ion (

Br^-

) attacks the secondary carbocation to form

CH_3CH(Br)CH_3

. The systematic IUPAC name for this compound is 2-bromopropane.

Numbering the three-carbon parent chain gives the bromine substituent locant position 2.

Key Concept

Markovnikov's rule in electrophilic addition reactions of alkenes

Question 8193Question →

Match each thermodynamic sign combination of enthalpy change ( $\Delta H$ ) and entropy change ( $\Delta S$ ) with its corresponding condition for reaction spontaneity.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The thermodynamic combinations match as follows: negative enthalpy change and positive entropy change match spontaneous at all temperatures; positive enthalpy change and negative entropy change match non-spontaneous at all temperatures; negative enthalpy change and negative entropy change match spontaneous only at low temperatures; positive enthalpy change and positive entropy change match spontaneous only at high temperatures.

A reaction is spontaneous when

\Delta G < 0

, according to the Gibbs free energy relationship

\Delta G = \Delta H - T\Delta S

. Combining a negative enthalpy change with a positive entropy change ensures

\Delta G

is negative at all temperatures. Combining a positive enthalpy change with a negative entropy change ensures

\Delta G

is positive at all temperatures. When enthalpy and entropy changes share the same sign, temperature dictates spontaneity: negative signs require low temperatures, while positive signs require high temperatures.

Step-by-Step Solution

1

Recall the Gibbs Free Energy equation

\Delta G = \Delta H - T\Delta S

, where

T

is temperature in Kelvin (

T > 0

).

Reaction spontaneity depends on the sign of

\Delta G

; a process is spontaneous if

\Delta G < 0

.

2

Analyze each sign combination in the equation

1. Negative

\Delta H

and positive

\Delta S

gives

\Delta G = (-) - T(+) = (-)

, always spontaneous.
2. Positive

\Delta H

and negative

\Delta S

gives

\Delta G = (+) - T(-) = (+)

, always non-spontaneous.
3. Negative

\Delta H

and negative

\Delta S

gives

\Delta G = (-) + T(+)

, which is negative only at low

T

.
4. Positive

\Delta H

and positive

\Delta S

gives

\Delta G = (+) - T(+)

, which is negative only at high

T

.

Evaluating the magnitude of

T\Delta S

relative to

\Delta H

determines the temperature dependency of spontaneity.

Key Concept

Gibbs Free Energy and Reaction Spontaneity

Question 8194Question →

Match each mixture separation scenario on the left with its corresponding application of simple or fractional distillation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Ethanol-water mixture matches laboratory fractional distillation of miscible liquids with close boiling points; recovery of pure water from salt solution matches simple distillation of volatile solvent from non-volatile solute; refining crude oil matches industrial fractional distillation of petroleum; liquid air separation matches cryogenic industrial fractional distillation.

Simple distillation is used when separating a volatile solvent from a non-volatile solute (e.g., pure water from sea water/salt solution). Fractional distillation is required when separating miscible liquids with close boiling points (e.g., ethanol and water, petroleum crude oil, or liquefied air components).

Step-by-Step Solution

1

Identify the nature of the solute and solvent in each mixture.

Determine whether components are volatile miscible liquids or a liquid containing non-volatile dissolved solids.

Simple distillation is suitable when separating a liquid solvent from a non-volatile solid, whereas fractional distillation is required for two or more miscible volatile liquids.

2

Compare boiling points for miscible liquid pairs.

Ethanol (

78^\circ\text{C}

) and water (

100^\circ\text{C}

) have close boiling points requiring repeated condensation-vaporization cycles provided by a fractionating column.

A fractionating column packed with glass beads provides surface area for multiple simple distillations to occur simultaneously.

3

Match industrial processes to their scaled applications.

Crude oil refining uses tall fractionating towers, while air liquefaction separates sub-zero boiling gases like nitrogen and oxygen.

Industrial separation scales fractional distillation principles to handle petroleum fractions and liquefied gas mixtures.

Key Concept

Distinguishing between simple distillation (volatile solvent from non-volatile solute) and fractional distillation (miscible liquids with close boiling points).

Estimated Time:1m 0s

Question 8195Question →

Magnesium oxide ( $MgO$ ) has a significantly higher melting point ( $2,852^\circ\text{C}$ ) than sodium fluoride ( $NaF$ ) ( $996^\circ\text{C}$ ) primarily because the product of the ionic charges in $MgO$ is four times greater than in $NaF$ , resulting in stronger electrostatic forces within the crystal lattice. Is this statement true or false?

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because the strength of electrovalent bonding and lattice energy increases proportionally with the product of the ionic charges.

The statement is correct because electrovalent bond strength is governed by Coulomb's Law of electrostatic attraction. Doubly charged cations and anions (

Mg^{2+}

and

O^{2-}

) experience an electrostatic force four times stronger than singly charged ions (

Na^+

and

F^-

) of comparable size, yielding a substantially higher lattice energy and melting point.

Step-by-Step Solution

1

Identify the constituent ions and their respective charges for both ionic compounds.

In

MgO

, the ions are

Mg^{2+}

and

O^{2-}

. In

NaF

, the ions are

Na^+

and

F^-

.

Electrostatic attraction depends fundamentally on the magnitude of nuclear charges transferred during ionic bond formation.

2

Calculate the product of the ionic charges for each compound.

For

MgO

:

|(+2) \times (-2)| = 4

. For

NaF

:

|(+1) \times (-1)| = 1

.

Coulomb's Law states that electrostatic force is proportional to the product of the charges (

q_1 q_2

).

3

Relate the charge product to lattice energy and physical properties such as melting point.

A higher lattice energy requires more thermal energy to break the rigid giant ionic lattice, leading to a much higher melting point for

MgO

than

NaF

.

Melting point directly reflects the magnitude of the electrostatic attraction holding the ions in their solid lattice positions.

Key Concept

Factors Affecting Ionic Lattice Energy and Melting Points

Question 8196Question →

A water sample contains both temporary hardness due to dissolved $\text{Ca(HCO}_3)_2$ and permanent hardness due to dissolved $\text{MgSO}_4$ . Which sequence correctly arranges the operational steps required to completely soften this water sample using slaked lime followed by washing soda?

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Answer

The correct operational sequence is: first, adding slaked lime to precipitate calcium carbonate from temporary hardness; second, filtering off the solid precipitate; third, adding washing soda to precipitate magnesium carbonate from permanent hardness; and fourth, performing a final filtration to obtain clear, soft water.

The removal process follows a logically structured chemical workflow. Temporary hardness caused by

\text{Ca(HCO}_3)_2

is treated first with a calculated amount of slaked lime (Clark's process), producing insoluble

\text{CaCO}_3

, which is filtered out. The remaining filtrate contains permanent hardness from

\text{MgSO}_4

, which is treated with washing soda (sodium trioxocarbonate(IV)) to precipitate

\text{MgCO}_3

. A second filtration removes this precipitate, leaving pure soft water.

Step-by-Step Solution

1

Identify the chemical reaction for temporary hardness removal.

Adding

\text{Ca(OH)}_2

precipitates temporary hardness:

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

.

Slaked lime specifically removes hydrogentrioxocarbonate(IV) salts causing temporary hardness.

2

Separate the solid calcium carbonate formed.

The solid

\text{CaCO}_3

is filtered out of the solution.

Precipitates must be removed so they do not redissolve or contaminate subsequent steps.

3

Identify the chemical reaction for permanent hardness removal.

Adding

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

precipitates permanent hardness:

\text{MgSO}_4 + \text{Na}_2\text{CO}_3 \rightarrow \text{MgCO}_3\downarrow + \text{Na}_2\text{SO}_4

.

Soluble carbonate ions from washing soda precipitate divalent magnesium cations as insoluble trioxocarbonate(IV) salts.

4

Conduct final liquid-solid separation.

Filtering removes insoluble

\text{MgCO}_3

, producing soft water.

Soluble sodium tetraoxosulfate(VI) remaining in solution does not react with soap to form scum.

Key Concept

Sequential chemical removal of temporary and permanent hardness using Clark's process followed by precipitation with washing soda.

Question 8197Question →

Match each condition or graphical representation of a gas sample obeying Charles's law with its corresponding physical or mathematical outcome.

Click a left item, then click its matching right item

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Answer

The correct pairings match heating from 27 °C to 54 °C with a volume increase factor of 1.09, heating from 300 K to 600 K with volume doubling, extrapolation to zero volume with intersecting the temperature axis at absolute zero (-273 °C), and the V versus T (K) plot with a straight line through the origin.

Each item correctly links a concept of Charles's law to its outcome: heating from

27^\circ\text{C}

(

300\text{ K}

) to

54^\circ\text{C}

(

327\text{ K}

) increases volume by a factor of

1.09

; heating from

300\text{ K}

to

600\text{ K}

doubles absolute temperature and volume; extrapolating a

V

-

T

(

^\circ\text{C}

) graph to zero volume yields absolute zero (

-273^\circ\text{C}

); and plotting

V

against absolute temperature (

T

in

\text{K}

) produces a straight line passing through the origin.

Step-by-Step Solution

1

Convert Celsius temperatures to Kelvin for the first scenario

T_1 = 27 + 273 = 300\text{ K}

and

T_2 = 54 + 273 = 327\text{ K}

.

Charles's law requires temperatures to be expressed in Kelvin.

2

Determine the volume ratio for heating from

27^\circ\text{C}

to

54^\circ\text{C}

\frac{V_2}{V_1} = \frac{T_2}{T_1} = \frac{327\text{ K}}{300\text{ K}} = 1.09.

Doubling Celsius temperature does not double absolute temperature, so volume increases by a factor of 1.09.

3

Apply direct proportionality for the Kelvin heating scenario

Heating from

300\text{ K}

to

600\text{ K}

doubles the absolute temperature, so

V_2 = 2V_1

.

Volume is directly proportional to temperature on the Kelvin scale.

4

Analyze graphical characteristics of Charles's law

A

V

vs

T

(K) plot is a straight line through the origin

(0,0)

, and extrapolating

V

vs

T

(

^\circ\text{C}

) to zero volume yields absolute zero (

-273^\circ\text{C}

).

Absolute zero is the theoretical temperature at which gas volume extrapolates to zero.

Key Concept

Charles's Law and Absolute Temperature Scale

Estimated Time:1m 30s

Question 8198Question →

Match each municipal water purification stage or chemical additive on the left with its precise operational mechanism on the right.

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Answer

Aeration matches with expelling volatile gases and oxidizing soluble iron(II); Addition of Potash Alum matches with neutralizing negative charges on clay colloids for flocculation; Sand Filtration matches with straining out fine suspended particles remaining after sedimentation; Addition of Slaked Lime matches with adjusting pH of acidic water post-coagulation.

Aeration oxidizes soluble iron and removes dissolved gases. Potash alum supplies trivalent cations to neutralize negative charges on colloidal clay. Sand filtration physically retains fine suspended particles. Slaked lime neutralizes acidity induced by alum coagulation.

Step-by-Step Solution

1

Analyze the primary chemical and physical effects of Aeration.

Spraying raw water into air maximizes gas exchange, driving off dissolved volatile species (

H_2S

) and converting soluble

Fe^{2+}

ions to insoluble

Fe(OH)_3

precipitate.

Aeration targets volatile odors/tastes and dissolved metals rather than solid particulate filtration.

2

Analyze the coagulating action of Potash Alum (

KAl(SO_4)_2 \cdot 12H_2O

).

Trivalent

Al^{3+}

ions neutralize negative charges on suspended clay colloids, causing them to coalesce into larger settleable flocs.

Fine clay particles do not settle spontaneously due to mutual electrostatic repulsion.

3

Determine the function of Sand Filtration.

Water percolating through sand layer beds leaves behind non-settled micro-particles.

Filtration serves as a final physical straining stage after bulk sedimentation.

4

Evaluate the requirement for Slaked Lime (

Ca(OH)_2

).

Neutralizes acidity produced during alum hydrolysis, raising the pH to safe alkaline levels.

Acidic water damages metal distribution pipes through corrosive action.

Key Concept

Operational mechanisms of chemical and physical processes in town water supply treatment

Question 8199Question →

A molecule possessing two identical chiral carbon atoms, such as $2,3$ -dichlorobutane, yields a total of four optically active stereoisomers.

True

False

Show answer & explanation

Answer: False

Answer

The statement is false. A molecule with two identical chiral carbon atoms generates three stereoisomers in total: one pair of optically active enantiomers and one optically inactive meso compound.

The statement is false because

2,3

-dichlorobutane contains two identical chiral carbon atoms. Its internal plane of symmetry in the meso configuration causes internal compensation of optical rotation, yielding only two optically active enantiomers and one optically inactive meso form (three stereoisomers in total).

Step-by-Step Solution

1

Identify the chiral carbon atoms and their bonded groups in

2,3

-dichlorobutane.

Carbon-2 and Carbon-3 are both chiral centers, each attached to four groups:

-H

,

-Cl

,

-CH_3

, and

-CH(Cl)CH_3

. Because both carbons have identical sets of substituents, the chiral centers are identical.

Recognizing identical chiral centers is essential for identifying internal planes of symmetry.

2

Evaluate theoretical stereoisomer combinations using the

2^n

rule.

The theoretical maximum without symmetry considerations is

2^2 = 4

configurations:

(2R, 3R)

,

(2S, 3S)

,

(2R, 3S)

, and

(2S, 3R)

.

The formula

2^n

determines the upper limit of stereoisomers for

n

chiral centers.

3

Examine internal symmetry and optical activity of the configurations.

The

(2R, 3R)

and

(2S, 3S)

forms are non-superimposable mirror images (enantiomers) and are optically active. However, the

(2R, 3S)

and

(2S, 3R)

structures are identical to each other due to an internal plane of symmetry. This single achiral meso compound is optically inactive via internal compensation.

Internal symmetry cancels optical rotation, leaving only two optically active stereoisomers and one optically inactive meso stereoisomer (3 total).

Key Concept

Meso compounds and internal optical compensation in molecules with identical chiral centers.

Question 8200Question →

Match each physical property of metallic elements listed on the left with the atomic-scale mechanism on the right that best accounts for it.

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Answer

High thermal conductivity matches rapid transfer of kinetic energy by mobile delocalized electrons; Malleability and ductility matches non-directional electrostatic attractions permitting cation layers to slide; High melting point matches strong multi-directional electrostatic attraction between cations and the electron sea; Metallic luster matches oscillation and re-emission of incident photons by surface delocalized electrons.

Each macro-level property corresponds directly to specific behaviors of the delocalized electron sea and cation lattice: thermal conduction relies on mobile electron kinetic transport; malleability depends on non-directional bonding allowing cation layers to slip; high melting points result from strong multi-directional electrostatic attractions; and luster is caused by surface electron excitation and photon re-emission.

Step-by-Step Solution

1

Analyze the microscopic origin of thermal transport in metals.

Identify mobile delocalized valence electrons as the primary carriers of thermal kinetic energy.

Delocalized electrons move rapidly through the lattice when a temperature gradient is applied, transferring kinetic energy much faster than localized atomic vibrations.

2

Examine the mechanism of mechanical deformation under applied stress.

Identify non-directional electrostatic attraction enabling cation layers to slide without fracture.

Unlike ionic crystals where sliding brings like charges into repelling contact, metallic electron clouds shield shifting cations, preserving lattice cohesion.

3

Evaluate the structural requirements for melting and high mechanical strength.

Identify strong multi-directional electrostatic binding throughout the 3D lattice.

Overcoming the structural stability requires substantial energy to disrupt the strong net electrostatic pull between positive metal ions and delocalized electrons.

4

Determine the interaction of metal surfaces with electromagnetic radiation.

Connect surface delocalized electrons to photon absorption and rapid re-emission.

Unbound surface valence electrons absorb incoming light energy and immediately vibrate and re-emit the photons, producing specular reflection.

Key Concept

Connecting macroscopic physical properties of metals to the delocalized electron sea model and non-directional metallic bonding.

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