All practice questions

Question 8481Question →

In the stratosphere, ozone ( $O_3$ ) serves as a protective layer by absorbing high-energy ultraviolet radiation, whereas in the troposphere, ozone functions as an air pollutant and a greenhouse gas.

True

False

Show answer & explanation

Answer: True

Answer

The statement is True.

The statement is correct because stratospheric ozone acts as a vital protective shield against biologically damaging solar ultraviolet radiation, while tropospheric ozone traps outgoing terrestrial heat radiation, thereby functioning as a greenhouse gas and pollutant.

Step-by-Step Solution

1

Examine the function of ozone (

O_3

) in the stratosphere.

Stratospheric ozone filters out harmful solar ultraviolet (

UV-B

and

UV-C

) radiation through photolytic dissociation reactions.

UV radiation breaks the bonds in ozone molecules, dissipating the radiation energy as heat before it reaches the surface.

2

Examine the role of ozone (

O_3

) in the troposphere.

Near ground level, tropospheric ozone absorbs thermal infrared (

IR

) radiation emitted by the Earth, contributing to radiative forcing and global warming.

Tropospheric ozone possesses vibrational modes that absorb outgoing terrestrial heat energy, acting as a greenhouse gas.

3

Compare the dual impacts of atmospheric ozone.

Ozone is beneficial in the stratosphere ('good ozone') but toxic and warming in the troposphere ('bad ozone').

The environmental effect of ozone depends strictly on its atmospheric region.

Key Concept

Dual Environmental Roles of Stratospheric and Tropospheric Ozone

Question 8482Question →

Consider the catalytic oxidation of ammonia gas represented by the balanced chemical equation below:

4NH_3(g) + 5O_2(g) \rightarrow 4NO(g) + 6H_2O(g)

If

6.8\text{ g}

of ammonia gas reacts completely with excess oxygen gas at STP, calculate the volume of nitrogen(II) oxide gas and the volume of steam produced.

[H = 1, N = 14, O = 16, \text{Molar volume of gas at STP} = 22.4\text{ dm}^3\text{ mol}^{-1}]

Fill in the missing values in the statement below.

Fill in the blanks below

The volume of nitrogen(II) oxide gas (

NO

) produced at STP is dm³, and the volume of steam (

H_2O

) produced at STP is dm³.

Show answer & explanation

Answer

The volume of nitrogen(II) oxide produced at STP is 8.96 dm³ and the volume of steam produced at STP is 13.44 dm³.

Molar mass of ammonia is 17 g/mol, meaning 6.8 g equals 0.4 mol of ammonia. According to the balanced equation, 4 moles of ammonia produce 4 moles of nitrogen(II) oxide gas and 6 moles of steam. Thus, 0.4 mol of ammonia yields 0.4 mol of nitrogen(II) oxide gas (0.4 * 22.4 = 8.96 dm³) and 0.6 mol of steam (0.6 * 22.4 = 13.44 dm³).

Step-by-Step Solution

1

Calculate the molar mass of ammonia (

NH_3

)

Molar mass of

NH_3 = 14 + (3 \times 1) = 17\text{ g mol}^{-1}

Required to convert the given mass of reactant into moles.

2

Calculate the number of moles of

NH_3

reacted

\text{Moles of } NH_3 = \frac{6.8\text{ g}}{17\text{ g mol}^{-1}} = 0.4\text{ mol}

Stoichiometric relations in chemical equations are expressed in mole ratios.

3

Determine the moles of

NO(g)

and

H_2O(g)

produced using stoichiometric coefficients

From the balanced equation,

4\text{ mol } NH_3 \rightarrow 4\text{ mol } NO

, so

0.4\text{ mol } NH_3 \rightarrow 0.4\text{ mol } NO

.
Also,

4\text{ mol } NH_3 \rightarrow 6\text{ mol } H_2O(g)

, so

\text{Moles of } H_2O = \frac{6}{4} \times 0.4 = 0.6\text{ mol}

.

The coefficients in the balanced equation define the molar ratio between reactants and products.

4

Convert the moles of each gaseous product to volume at STP using molar gas volume

\text{Volume of } NO = 0.4\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 8.96\text{ dm}^3

\text{Volume of } H_2O = 0.6\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 13.44\text{ dm}^3

At STP,

1\text{ mole}

of any ideal gas occupies

22.4\text{ dm}^3

.

Key Concept

Mass-Volume Stoichiometry at STP

Estimated Time:2m 0s

Question 8483Question →

What mass of copper is deposited at the cathode when a steady current of

0.50\text{ A}

is passed through an aqueous solution of copper(II) tetraoxosulfate(VI) for

1930\text{ s}

? [

\text{Molar mass of Cu} = 64\text{ g mol}^{-1}, 1\text{ F} = 96,500\text{ C mol}^{-1}

]

0.16\text{ g}

0.32\text{ g}

0.64\text{ g}

1.28\text{ g}

Show answer & explanation

Answer: $0.32\text{ g}$

Answer

The correct mass of copper deposited is

0.32\text{ g}

.

Passing

0.50\text{ A}

for

1930\text{ s}

transfers

965\text{ C}

of electricity, which equals

0.01\text{ mol}

of electrons. Because reduction of

Cu^{2+}

requires

2

moles of electrons per mole of copper metal deposited (

Cu^{2+} + 2e^- \rightarrow Cu

),

0.005\text{ mol}

of copper is produced. Multiplying

0.005\text{ mol}

by the molar mass of copper (

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

) gives

0.32\text{ g}

.

Step-by-Step Solution

1

Calculate the total quantity of electricity (

Q

) transferred in Coulombs.

Q = I \times t = 0.50\text{ A} \times 1930\text{ s} = 965\text{ C}

Electric charge is the product of current in amperes and time in seconds.

2

Determine the number of moles of electrons transferred.

\text{Moles of } e^- = \frac{965\text{ C}}{96,500\text{ C mol}^{-1}} = 0.01\text{ mol}

One Faraday (

96,500\text{ C}

) corresponds to one mole of electrons.

3

Use the cathode half-reaction stoichiometry to find the moles of copper deposited.

Cu^{2+} + 2e^- \rightarrow Cu(s) \implies \text{Moles of Cu} = \frac{0.01\text{ mol } e^-}{2} = 0.005\text{ mol}

Copper(II) ions require 2 moles of electrons per mole of copper metal deposited.

4

Calculate the mass of copper deposited.

\text{Mass} = \text{moles} \times \text{molar mass} = 0.005\text{ mol} \times 64\text{ g mol}^{-1} = 0.32\text{ g}

Mass is found by multiplying the quantity in moles by the molar mass.

Key Concept

Faraday's First Law of Electrolysis and Quantitative Stoichiometry of Electrode Reactions

Question 8484Question →

Match each thermodynamic state or free energy parameter on the left with its correct spontaneity or equilibrium interpretation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches pair

\Delta G = 0

with dynamic equilibrium,

\Delta G < 0

with spontaneous forward reaction,

\Delta G > 0

with non-spontaneous forward reaction, and

\Delta G^\circ < 0

with product favorability at standard equilibrium (

K_{eq} > 1

).

The correct matches pair each thermodynamic free energy symbol with its fundamental definition:

\Delta G = 0

represents dynamic equilibrium,

\Delta G < 0

represents a spontaneous forward process,

\Delta G > 0

represents a non-spontaneous forward process, and

\Delta G^\circ < 0

represents standard product favorability (

K_{eq} > 1

).

Step-by-Step Solution

1

Analyze the physical meaning of

\Delta G = 0

.

\Delta G = 0

corresponds to a state of dynamic equilibrium where the rate of the forward reaction equals the rate of the reverse reaction.

At equilibrium, there is no net change in free energy.

2

Analyze the condition for spontaneity (

\Delta G < 0

).

\Delta G < 0

indicates an exergonic, spontaneous forward reaction under the specified conditions.

Thermodynamics dictates that spontaneous processes decrease free energy to reach stability.

3

Analyze non-spontaneous conditions (

\Delta G > 0

).

\Delta G > 0

means energy input is required for the forward reaction, making the reverse reaction spontaneous.

A positive free energy change signifies an endergonic process.

4

Evaluate standard free energy

\Delta G^\circ < 0

in relation to the equilibrium constant.

Using

\Delta G^\circ = -RT \ln K_{eq}

, a negative

\Delta G^\circ

implies

\ln K_{eq} > 0

, so

K_{eq} > 1

.

When

K_{eq} > 1

, products are favored over reactants at standard equilibrium.

Key Concept

Gibbs Free Energy and Reaction Spontaneity Criteria

Question 8485Question →

In a laboratory setup, a student attempts to isolate an organic solute from an aqueous solution by solvent extraction using a separating funnel. If the student incorrectly selects ethanol instead of ethoxyethane as the extracting solvent, why will the separation fail?

Ethanol is completely miscible with water, preventing the formation of two distinct liquid layers.

Ethanol has a lower boiling point than water, causing the solvent to vaporize before extraction can occur.

Ethanol is less dense than water, which forces the organic solute to remain permanently in the lower layer.

Ethanol reacts chemically with aqueous solutes, converting them into insoluble trioxocarbonate salts.

Show answer & explanation

Answer: Ethanol is completely miscible with water, preventing the formation of two distinct liquid layers.

Answer

Ethanol is completely miscible with water, preventing the formation of two distinct liquid layers.

The successful use of a separating funnel for solvent extraction requires two immiscible liquid phases. Ethanol is completely miscible with water due to extensive hydrogen bonding, forming a single homogeneous mixture rather than two separate layers.

Step-by-Step Solution

1

Identify the key requirement for using a separating funnel in solvent extraction.

The extracting solvent must be immiscible with the solvent containing the solute (usually water) so that two separate physical phases form.

Solvent extraction relies on the partitioning of a solute between two immiscible liquid layers.

2

Analyze the solubility behavior of ethanol when added to water.

Ethanol forms strong hydrogen bonds with water molecules, making them completely miscible in all proportions.

Because ethanol and water mix homogeneously into a single phase, no interface or separate layers can form in a separating funnel.

3

Conclude why ethoxyethane (diethyl ether) works while ethanol fails.

Ethoxyethane is immiscible with water and forms a separate upper layer, whereas ethanol mixes completely and cannot separate.

Immiscibility is mandatory for phase separation.

Key Concept

Liquid-liquid immiscibility requirement in solvent extraction

Estimated Time:1m 0s

Question 8486Question →

Solve the following solubility and crystallization problem. What is the mass of the salt that crystallizes out of the solution upon cooling?

Fill in the blanks below

A saturated solution of a divalent metal salt

\text{MX}_2

(molar mass =

120\text{ g/mol}

) has a total mass of

120.0\text{ g}

at

60^\circ\text{C}

. The solubility of

\text{MX}_2

is

5.0\text{ mol/dm}^3

at

60^\circ\text{C}

and

2.0\text{ mol/dm}^3

at

25^\circ\text{C}

. Assuming the density of water is

1.00\text{ g/cm}^3

, the mass of

\text{MX}_2

deposited when the solution is cooled from

60^\circ\text{C}

to

25^\circ\text{C}

is g.

Show answer & explanation

Answer

The mass of

\text{MX}_2

deposited on cooling is

27.0\text{ g}

.

At

60^\circ\text{C}

, a solubility of

5.0\text{ mol/dm}^3

corresponds to

5.0 \times 120 = 600\text{ g}

of

\text{MX}_2

per

1000\text{ g}

of water. Thus,

1600\text{ g}

of saturated solution contains

600\text{ g}

solute and

1000\text{ g}

water. Proportionally,

120.0\text{ g}

of saturated solution contains

45.0\text{ g}

of

\text{MX}_2

dissolved in

75.0\text{ g}

of water. At

25^\circ\text{C}

, the solubility decreases to

2.0\text{ mol/dm}^3

, which equals

2.0 \times 120 = 240\text{ g}

of

\text{MX}_2

per

1000\text{ g}

of water. In

75.0\text{ g}

of water, the maximum mass of solute that remains dissolved is

75.0 \times (240 / 1000) = 18.0\text{ g}

. The mass of solid

\text{MX}_2

that crystallizes out is

45.0\text{ g} - 18.0\text{ g} = 27.0\text{ g}

.

Step-by-Step Solution

1

Calculate the mass of solute per

1000\text{ g}

of water at

60^\circ\text{C}

.

Mass of

\text{MX}_2

per

1000\text{ g}

water =

5.0\text{ mol/dm}^3 \times 120\text{ g/mol} = 600.0\text{ g}

.

Converting solubility from

\text{mol/dm}^3

to

\text{g/dm}^3

(or grams per

1000\text{ g}

of water, given water density is

1.00\text{ g/cm}^3

).

2

Determine the composition of the

120.0\text{ g}

saturated solution at

60^\circ\text{C}

.

Total mass of saturated solution containing

1000\text{ g}

water =

1000\text{ g} + 600\text{ g} = 1600.0\text{ g}

. Mass of water in

120.0\text{ g}

solution =

120.0\text{ g} \times \frac{1000}{1600} = 75.0\text{ g}

. Mass of

\text{MX}_2

dissolved =

120.0\text{ g} - 75.0\text{ g} = 45.0\text{ g}

.

To find how much solute and solvent are actually present in the given portion of solution.

3

Calculate the mass of solute that remains dissolved in

75.0\text{ g}

of water at

25^\circ\text{C}

.

At

25^\circ\text{C}

, mass of solute per

1000\text{ g}

water =

2.0\text{ mol/dm}^3 \times 120\text{ g/mol} = 240.0\text{ g}

. Mass dissolved in

75.0\text{ g}

water =

75.0\text{ g} \times \frac{240.0}{1000} = 18.0\text{ g}

.

Determining the maximum amount of solute that

75.0\text{ g}

of water can hold at the lower temperature.

4

Calculate the mass of

\text{MX}_2

deposited upon cooling.

Mass deposited =

45.0\text{ g} - 18.0\text{ g} = 27.0\text{ g}

.

The difference between the initial mass dissolved at

60^\circ\text{C}

and the remaining mass dissolved at

25^\circ\text{C}

represents the crystallized solid.

Key Concept

Quantitative solubility calculations involving crystallization from saturated solutions

Question 8487Question →

An industrial synthesis reaction carried out at $27^\circ\text{C}$ has a standard enthalpy change ( $\Delta H^\circ$ ) of $-45.0\text{ kJ mol}^{-1}$ and a standard Gibbs free energy change ( $\Delta G^\circ$ ) of $-15.0\text{ kJ mol}^{-1}$ . What is the standard entropy change ( $\Delta S^\circ$ ) for this reaction?

+100.0\text{ J K}^{-1}\text{ mol}^{-1}

-100.0\text{ J K}^{-1}\text{ mol}^{-1}

-1111.1\text{ J K}^{-1}\text{ mol}^{-1}

-0.10\text{ J K}^{-1}\text{ mol}^{-1}

Show answer & explanation

Answer: $-100.0\text{ J K}^{-1}\text{ mol}^{-1}$

Answer

-100.0\text{ J K}^{-1}\text{ mol}^{-1}

Using the standard thermodynamic relation

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

, convert

27^\circ\text{C}

to

300\text{ K}

. Solving for

\Delta S^\circ

yields

\frac{-45.0 - (-15.0)}{300} = -0.100\text{ kJ K}^{-1}\text{ mol}^{-1}

, which equals

-100.0\text{ J K}^{-1}\text{ mol}^{-1}

.

Step-by-Step Solution

1

Convert temperature from Celsius to Kelvin

T = 27 + 273.15 = 300\text{ K}

Thermodynamic calculations involving temperature require absolute temperature in Kelvin.

2

Rearrange the Gibbs free energy equation

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

to solve for

\Delta S^\circ

\Delta S^\circ = \frac{\Delta H^\circ - \Delta G^\circ}{T}

To isolate the standard entropy change variable.

3

Substitute given values and calculate

\Delta S^\circ

in

\text{kJ K}^{-1}\text{ mol}^{-1}

\Delta S^\circ = \frac{-45.0\text{ kJ mol}^{-1} - (-15.0\text{ kJ mol}^{-1})}{300\text{ K}} = \frac{-30.0}{300} = -0.100\text{ kJ K}^{-1}\text{ mol}^{-1}

Evaluate standard free energy and enthalpy difference divided by temperature.

4

Convert

\Delta S^\circ

to standard units of

\text{J K}^{-1}\text{ mol}^{-1}

\Delta S^\circ = -0.100 \times 1000 = -100.0\text{ J K}^{-1}\text{ mol}^{-1}

Entropy changes are conventionally reported in joules per kelvin per mole.

Key Concept

The relationship between Gibbs free energy, enthalpy, absolute temperature, and entropy is governed by

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

.

Estimated Time:1m 30s

Question 8488Question →

Potassium trioxonitrate(V) decomposes upon heating according to the balanced chemical equation:

2KNO_3(s) \rightarrow 2KNO_2(s) + O_2(g)

What volume of oxygen gas measured at STP is produced by the complete thermal decomposition of

50.5\text{ g}

of

KNO_3

?

[K = 39, N = 14, O = 16\text{; Molar gas volume at STP } = 22.4\text{ dm}^3\text{ mol}^{-1}]

2.8\text{ dm}^3

5.6\text{ dm}^3

6.0\text{ dm}^3

11.2\text{ dm}^3

Show answer & explanation

Answer: $5.6\text{ dm}^3$

Answer

The correct volume of oxygen gas produced at STP is

5.6\text{ dm}^3

.

The complete decomposition of

50.5\text{ g}

(

0.5\text{ mol}

) of

KNO_3

yields

0.25\text{ mol}

of

O_2

gas according to the

2:1

stoichiometric mole ratio. Multiplying

0.25\text{ mol}

by the standard molar gas volume at STP (

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

) gives

5.6\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the molar mass of

KNO_3

Molar mass of

KNO_3 = 39 + 14 + (3 \times 16) = 101\text{ g mol}^{-1}

Molar mass is needed to convert the given mass of reactant into moles.

2

Calculate the number of moles of

KNO_3

reacted

\text{Moles of } KNO_3 = \frac{50.5\text{ g}}{101\text{ g mol}^{-1}} = 0.5\text{ mol}

Determines the exact mole amount of reactant supplied.

3

Use the mole ratio from the balanced equation to find moles of

O_2

produced

\text{Moles of } O_2 = \frac{1}{2} \times 0.5\text{ mol} = 0.25\text{ mol}

The equation shows that

2\text{ moles}

of

KNO_3

yield

1\text{ mole}

of

O_2

.

4

Convert moles of

O_2

to volume at STP

\text{Volume of } O_2 = 0.25\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 5.6\text{ dm}^3

Molar gas volume at standard temperature and pressure (STP) is

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

.

Key Concept

Mass-Volume Stoichiometric Calculations at STP

Estimated Time:1m 30s

Question 8489Question →

Arrange the following sequential steps of a paper chromatography experiment in the correct procedural order from start to finish.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct procedural order is: (1) Draw a light baseline using a lead pencil about

2\text{ cm}

from the bottom edge of the filter paper strip, (2) Apply a small, concentrated spot of the sample mixture onto the baseline and allow it to dry, (3) Suspend the paper strip in a sealed chamber, ensuring the solvent level remains strictly below the baseline, and (4) Remove the strip when the mobile phase nears the top and immediately pencil-mark the solvent front.

The correct sequence follows standard analytical chromatography protocol: first, a graphite pencil baseline is drawn at the base of the paper strip. Next, the mixture is spotted onto this origin baseline. The paper is then placed into the developing tank with the mobile phase level kept below the baseline to allow upward capillary migration. Finally, the paper is removed before the solvent reaches the top edge, and the solvent front is marked immediately for accurate distance measurements.

Step-by-Step Solution

1

Determine paper preparation step

Draw a pencil baseline near the bottom edge of the paper.

Graphite from a lead pencil is insoluble in the mobile phase and will not contaminate the chromatogram.

2

Determine sample application step

Spot the sample mixture onto the pencil baseline and dry it.

Deposition onto the origin line must occur before the paper contacts the developing solvent.

3

Determine development step

Place the paper into the chamber with the solvent line positioned below the baseline.

Maintaining the solvent level below the baseline allows the solvent to ascend by capillary action, carrying solute components upward rather than dissolving them into the bulk solvent.

4

Determine experiment termination step

Remove the paper strip and mark the solvent front.

Immediate marking of the solvent front ensures accurate measurement of mobile phase migration distance for

R_f

calculations.

Key Concept

Laboratory procedural steps and operational principles of paper chromatography

Question 8490Question →

Phosphorus reacts with oxygen gas to produce phosphorus(V) oxide according to the balanced chemical equation:

4\text{P}_{(s)} + 5\text{O}_{2(g)} \rightarrow \text{P}_4\text{O}_{10(s)}

If a mixture containing

12.4\text{ g}

of phosphorus and

20.0\text{ g}

of oxygen gas is allowed to react to completion, what is the mass of the excess reactant remaining unreacted in grams?

[\text{P} = 31.0, \text{O} = 16.0]

Show answer & explanation

Answer: 4

Answer

The mass of unreacted excess oxygen gas remaining is

4.0\text{ g}

.

The correct calculation shows that

0.40\text{ mol}

of phosphorus requires

0.50\text{ mol}

of oxygen gas for complete reaction according to the

4:5

mole ratio in

4\text{P} + 5\text{O}_2 \rightarrow \text{P}_4\text{O}_{10}

. Subtracting the

0.50\text{ mol}

consumed from the initial

0.625\text{ mol}

leaves

0.125\text{ mol}

of unreacted oxygen gas, which equals

4.0\text{ g}

.

Step-by-Step Solution

1

Calculate the moles of phosphorus and oxygen gas present initially.

\text{Moles of P} = \frac{12.4\text{ g}}{31.0\text{ g/mol}} = 0.40\text{ mol}

;

\text{Moles of O}_2 = \frac{20.0\text{ g}}{32.0\text{ g/mol}} = 0.625\text{ mol}

.

Molar mass of

\text{P}

is

31.0\text{ g/mol}

and molar mass of

\text{O}_2

is

2 \times 16.0 = 32.0\text{ g/mol}

.

2

Determine the limiting reactant by comparing the required mole ratio to the available mole ratio.

\text{P}

is the limiting reactant, and

\text{O}_2

is the excess reactant.

From the balanced equation,

4\text{ moles of P}

require

5\text{ moles of O}_2

, so

1\text{ mole of P}

requires

1.25\text{ moles of O}_2

. Thus,

0.40\text{ mol}

of

\text{P}

requires

0.40 \times 1.25 = 0.50\text{ mol}

of

\text{O}_2

. Since

0.625\text{ mol}

of

\text{O}_2

is available,

\text{O}_2

is in excess.

3

Calculate the unreacted moles of oxygen gas remaining.

\text{Excess moles of O}_2 = 0.625\text{ mol} - 0.50\text{ mol} = 0.125\text{ mol}.

Subtracting the reacted moles from the initial moles gives the remaining amount.

4

Convert the unreacted moles of oxygen gas to mass in grams.

\text{Mass of remaining O}_2 = 0.125\text{ mol} \times 32.0\text{ g/mol} = 4.0\text{ g}.

Multiplying the excess moles by the molar mass of

\text{O}_2

(

32.0\text{ g/mol}

) gives the mass in grams.

Key Concept

Limiting and excess reactant stoichiometry

Question 8491Question →

A waste management facility categorizes organic polymers into synthetic addition polymers, such as polyethene, and natural or condensation polymers, such as starch and nylon-6,6. Upon microbial action, polyethene exhibits extreme resistance to decomposition. Which of the following statements correctly explains the chemical basis for the non-biodegradability of synthetic addition polymers compared to biodegradable natural polymers?

Synthetic addition polymers consist of long, non-polar carbon-carbon single bond chains lacking hydrolyzable functional groups necessary for microbial enzymatic cleavage.

Synthetic addition polymers readily undergo microbial oxidation because their backbone contains highly reactive ester linkages that release methane gas upon bacterial action.

Synthetic addition polymers resist biodegradation solely because they are thermosetting plastics with extensive cross-linking that prevents soil water absorption.

Synthetic addition polymers decompose into toxic chlorofluorocarbons during bacterial degradation, which inhibits further microbial activity in the soil.

Show answer & explanation

Answer: Synthetic addition polymers consist of long, non-polar carbon-carbon single bond chains lacking hydrolyzable functional groups necessary for microbial enzymatic cleavage.

Answer

Synthetic addition polymers consist of long, non-polar carbon-carbon single bond chains lacking hydrolyzable functional groups necessary for microbial enzymatic cleavage.

The correct answer highlights that microbial biodegradation depends on enzymatic hydrolysis of functional groups. Natural polymers (like starch) and synthetic condensation polymers (like nylon) possess ester, amide, or glycosidic linkages that enzymes recognize and break. Synthetic addition polymers like polyethene consist of long, non-polar carbon-carbon (

\text{C}-\text{C}

) chains without hydrolyzable functional groups, making them highly resistant to microbial decay.

Step-by-Step Solution

1

Identify the structural differences between addition polymers and natural/condensation polymers.

Addition polymers like polyethene feature a continuous backbone of non-polar carbon-carbon single bonds (

\text{C}-\text{C}

), whereas condensation polymers contain functional linkages such as esters, amides, or glycosidic bonds.

Enzymatic breakdown requires specific structural motifs for substrate-enzyme binding.

2

Analyze microbial enzymatic mechanisms of degradation.

Soil micro-organisms produce enzymes (e.g., esterases, peptidases, glucosidases) specialized in hydrolyzing polar functional groups containing heteroatoms (

\text{O}

,

\text{N}

).

Hydrolysis converts polymers into soluble monomers or oligomers that microbes can metabolize.

3

Evaluate why addition polymers resist enzymatic attack.

The high molecular weight, hydrophobic nature, and inert

\text{C}-\text{C}

single-bonded backbone of polyethene lack target sites for hydrolytic cleavage by microbial enzymes.

Without active hydrolyzable sites, microbial breakdown is extremely slow, leading to environmental accumulation.

Key Concept

Chemical basis of polymer biodegradability and structural enzymatic specificity

Estimated Time:1m 30s

Question 8492Question →

The solubility of a salt $X$ is $0.80\text{ mol dm}^{-3}$ at $60^\circ\text{C}$ and $0.30\text{ mol dm}^{-3}$ at $20^\circ\text{C}$ . Calculate the mass of salt $X$ (in grams) that will crystallize out of solution when $1.0\text{ dm}^3$ of its saturated solution is cooled from $60^\circ\text{C}$ to $20^\circ\text{C}$ . (Molar mass of salt $X = 100\text{ g mol}^{-1}$ )

Show answer & explanation

Answer: 50

Answer

The mass of salt

X

that crystallizes out of the solution is

50\text{ g}

.

At

60^\circ\text{C}

,

1.0\text{ dm}^3

of saturated solution contains

0.80\text{ mol}

of salt

X

. When cooled to

20^\circ\text{C}

, the solution can only hold

0.30\text{ mol}

. The excess amount that crystallizes out is

0.80\text{ mol} - 0.30\text{ mol} = 0.50\text{ mol}

. Converting this amount to mass yields

0.50\text{ mol} \times 100\text{ g mol}^{-1} = 50\text{ g}

.

Step-by-Step Solution

1

Calculate the difference in molar solubility between the two temperatures

0.80\text{ mol dm}^{-3} - 0.30\text{ mol dm}^{-3} = 0.50\text{ mol dm}^{-3}

This difference represents the amount of solute in moles per cubic decimeter that can no longer remain dissolved when cooled to

20^\circ\text{C}

.

2

Convert the precipitated moles into mass in grams for

1.0\text{ dm}^3

of solution

0.50\text{ mol} \times 100\text{ g mol}^{-1} = 50\text{ g}

Multiplying the precipitated amount in moles by the molar mass gives the total mass in grams that crystallizes out.

Key Concept

Solubility and crystallization calculations upon cooling

Question 8493Question →

A river downstream from a processing factory exhibits a high Biochemical Oxygen Demand (BOD) reading following the discharge of organic effluent. Which process directly accounts for the depletion of dissolved oxygen in this aquatic environment?

Aerobic micro-organisms consuming dissolved oxygen while breaking down the organic waste

Rapid microbial degradation of non-biodegradable synthetic addition polymers present in the water

Thermal decomposition of hydrogen trioxocarbonate(IV) during the removal of temporary hardness

Depletion of oxygen gas in water caused directly by chlorofluorocarbon destruction of the atmospheric ozone layer

Show answer & explanation

Answer: Aerobic micro-organisms consuming dissolved oxygen while breaking down the organic waste

Answer

Aerobic micro-organisms consuming dissolved oxygen while breaking down the organic waste

When organic effluent enters a water body, aerobic micro-organisms proliferate and decompose the organic matter via oxidation. This process consumes large amounts of dissolved oxygen, resulting in a high Biochemical Oxygen Demand (BOD) and reduced dissolved oxygen levels in the aquatic habitat.

Step-by-Step Solution

1

Identify the source of pollution

The factory effluent supplies large amounts of organic material to the water body.

Organic waste acts as a nutrient substrate for aerobic bacteria.

2

Analyze the biochemical process of organic waste breakdown

Aerobic bacteria oxidize the organic compounds into simpler inorganic molecules, using up dissolved oxygen in the process.

Biochemical Oxygen Demand (BOD) measures the quantity of oxygen needed by aerobic organisms to break down organic matter.

3

Relate BOD to dissolved oxygen levels

A high BOD value indicates intensive microbial respiration, leading directly to oxygen depletion in the river.

Oxygen consumption by micro-organisms exceeds the re-aeration rate of the water.

Key Concept

Biochemical Oxygen Demand (BOD) and Organic Water Pollution

Question 8494Question →

A laboratory technician prepares a mixture by thoroughly stirring gelatin in warm water. The resulting mixture passes completely through standard filter paper without leaving any solid residue, but when a concentrated beam of light passes through it in a darkened box, the path of the light beam becomes illuminated and visible. Which type of mixture is represented by this system?

A colloidal system

A true solution

A suspension

A coarse sediment

Show answer & explanation

Answer: A colloidal system

Answer

A colloidal system

Gelatin dispersed in warm water is a colloidal system. Colloidal particles range in diameter from

1\text{ nm}

to

100\text{ nm}

. This particle size is small enough to pass through standard filter paper pores, but large enough to scatter visible light rays (Tyndall effect).

Step-by-Step Solution

1

Analyze the filtration behavior described in the stem.

The mixture passes through standard filter paper without leaving a residue.

This eliminates suspensions, which have particle diameters exceeding

100\text{ nm}

and get trapped by filter paper pores.

2

Analyze the optical behavior (light scattering) described in the stem.

The mixture demonstrates the Tyndall effect by illuminating the light path.

True solutions have solute particles smaller than

1\text{ nm}

that cannot scatter visible light, whereas colloidal particles (

1\text{ nm}

to

100\text{ nm}

) effectively scatter light.

3

Synthesize the particle size characteristics to identify the mixture class.

The mixture is a colloidal system.

Exhibiting light scattering while passing through filter paper is the hallmark property distinguishing colloidal systems from true solutions and suspensions.

Key Concept

Distinguishing characteristics of true solutions, colloidal systems, and suspensions based on particle size and Tyndall effect.

Question 8495Question →

The solubility of sodium trioxonitrate(V), $\text{NaNO}_3$ , in water is $4.5\text{ mol dm}^{-3}$ at $60^\circ\text{C}$ and $2.0\text{ mol dm}^{-3}$ at $20^\circ\text{C}$ . If the molar mass of $\text{NaNO}_3$ is $85\text{ g mol}^{-1}$ , what mass of $\text{NaNO}_3$ in grams will crystallize out when $400\text{ cm}^3$ of a saturated solution of the salt at $60^\circ\text{C}$ is cooled to $20^\circ\text{C}$ ?

Show answer & explanation

Answer: 85

Answer

The mass of

\text{NaNO}_3

that will crystallize out is

85\text{ g}

.

The difference in solubility between

60^\circ\text{C}

and

20^\circ\text{C}

is

4.5 - 2.0 = 2.5\text{ mol dm}^{-3}

. In

400\text{ cm}^3

(

0.4\text{ dm}^3

) of solution, the amount of salt precipitated is

2.5\text{ mol dm}^{-3} \times 0.4\text{ dm}^3 = 1.0\text{ mol}

. Multiplying by the molar mass (

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

) gives

85\text{ g}

.

Step-by-Step Solution

1

Calculate the difference in solubility between

60^\circ\text{C}

and

20^\circ\text{C}

\Delta S = 4.5\text{ mol dm}^{-3} - 2.0\text{ mol dm}^{-3} = 2.5\text{ mol dm}^{-3}

Cooling causes the excess solute to precipitate out based on the difference in saturation concentration.

2

Convert solution volume to

\text{dm}^3

V = \frac{400\text{ cm}^3}{1000\text{ cm}^3\text{ dm}^{-3}} = 0.4\text{ dm}^3

Solubility is given per

\text{dm}^3

, so volume must be in

\text{dm}^3

.

3

Calculate the moles of solute precipitated

n = 2.5\text{ mol dm}^{-3} \times 0.4\text{ dm}^3 = 1.0\text{ mol}

Multiplying the concentration difference by the solution volume yields total precipitated moles.

4

Convert moles to mass in grams

m = 1.0\text{ mol} \times 85\text{ g mol}^{-1} = 85\text{ g}

Mass is found by multiplying moles by molar mass.

Key Concept

Crystallization and solubility change with temperature

Question 8496Question →

In a petroleum refinery's fractionating tower, crude oil is separated into different fractions based on boiling points. Considering four fractions—gasoline (boiling range $40^\circ\text{C} - 170^\circ\text{C}$ ), kerosene (boiling range $170^\circ\text{C} - 250^\circ\text{C}$ ), gas oil (boiling range $250^\circ\text{C} - 350^\circ\text{C}$ ), and lubricating oil (boiling range $> 350^\circ\text{C}$ )—which of the following correctly lists these fractions in order of their collection from the bottom to the top of the fractionating column?

Lubricating oil, gas oil, kerosene, gasoline

Gasoline, kerosene, gas oil, lubricating oil

Gasoline, gas oil, kerosene, lubricating oil

Lubricating oil, kerosene, gas oil, gasoline

Show answer & explanation

Answer: Lubricating oil, gas oil, kerosene, gasoline

Answer

Lubricating oil, gas oil, kerosene, gasoline

In industrial petroleum refining, the fractionating tower operates on a temperature gradient where the bottom is hottest and the top is coolest. Hydrocarbons with high boiling points (like lubricating oil) condense near the bottom where temperatures remain high. Hydrocarbons with progressively lower boiling points (gas oil, kerosene, gasoline) rise higher in the column before reaching their condensation temperatures. Thus, the order from bottom to top is lubricating oil, gas oil, kerosene, gasoline.

Step-by-Step Solution

1

Analyze the temperature gradient in an industrial fractionating column

The bottom of the column is kept at the highest temperature, while the top of the column is the coolest.

Vapors rise through the column and condense when the surrounding temperature drops to their respective boiling points.

2

Correlate fraction boiling points with condensation height

Higher boiling fractions condense lower in the column (near the bottom), whereas lower boiling fractions remain gaseous longer and condense higher up (near the top).

Substances with high boiling points require higher temperatures to remain in the vapor phase.

3

Arrange the given fractions from highest boiling point to lowest boiling point

Lubricating oil (

> 350^\circ\text{C}

)

\rightarrow

Gas oil (

250^\circ\text{C} - 350^\circ\text{C}

)

\rightarrow

Kerosene (

170^\circ\text{C} - 250^\circ\text{C}

)

\rightarrow

Gasoline (

40^\circ\text{C} - 170^\circ\text{C}

).

This sequence matches bottom-to-top collection in the fractionating tower.

Key Concept

Industrial Fractional Distillation of Petroleum

Question 8497Question →

What is the solubility in $\text{mol/dm}^3$ of potassium trioxonitrate(V), $\text{KNO}_3$ , at $40^\circ\text{C}$ if $50.5\text{ g}$ of the salt dissolves completely in $250\text{ cm}^3$ of water to form a saturated solution? $[\text{K} = 39, \text{N} = 14, \text{O} = 16]$

0.50\text{ mol/dm}^3

2.00\text{ mol/dm}^3

5.05\text{ mol/dm}^3

202.00\text{ mol/dm}^3

Show answer & explanation

Answer: $2.00\text{ mol/dm}^3$

Answer

The solubility of potassium trioxonitrate(V) at

40^\circ\text{C}

is

2.00\text{ mol/dm}^3

.

The correct answer is

2.00\text{ mol/dm}^3

. First, determine the molar mass of

\text{KNO}_3

, which is

39 + 14 + (3 \times 16) = 101\text{ g/mol}

. Converting

50.5\text{ g}

of

\text{KNO}_3

into moles gives

0.50\text{ mol}

. Converting

250\text{ cm}^3

of water into

\text{dm}^3

gives

0.25\text{ dm}^3

. Dividing

0.50\text{ mol}

by

0.25\text{ dm}^3

yields a molar solubility of

2.00\text{ mol/dm}^3

.

Step-by-Step Solution

1

Calculate the molar mass of

\text{KNO}_3

\text{Molar mass of KNO}_3 = 39 + 14 + 3(16) = 101\text{ g/mol}

Molar mass is required to convert the given mass of salt into number of moles.

2

Convert the mass of

\text{KNO}_3

to moles

\text{Moles of KNO}_3 = \frac{50.5\text{ g}}{101\text{ g/mol}} = 0.50\text{ mol}

Solubility in

\text{mol/dm}^3

requires solute quantity to be in moles.

3

Convert the volume of water from

\text{cm}^3

to

\text{dm}^3

\text{Volume} = \frac{250\text{ cm}^3}{1000} = 0.25\text{ dm}^3

Standard molarity and molar solubility use cubic decimeters (

\text{dm}^3

) as the volume unit.

4

Calculate molar solubility

\text{Solubility} = \frac{0.50\text{ mol}}{0.25\text{ dm}^3} = 2.00\text{ mol/dm}^3

Solubility in

\text{mol/dm}^3

is defined as moles of solute per cubic decimeter of saturated solution.

Key Concept

Solubility in

\text{mol/dm}^3

measures the maximum number of moles of solute that dissolve in

1\text{ dm}^3

of solvent at a specific temperature.

Question 8498Question →

In the industrial refining of crude oil, which physical property forms the basis for separating petroleum into its various fractions inside a fractionating column?

Differences in boiling points

Differences in densities

Differences in solubilities in water

Differences in melting points

Show answer & explanation

Answer: Differences in boiling points

Answer

Differences in boiling points

Industrial fractional distillation separates crude oil into distinct hydrocarbon fractions because each component liquid has a unique boiling point. Components with lower boiling points boil off first and are collected near the top of the column, while those with higher boiling points condense lower down.

Step-by-Step Solution

1

Identify the industrial separation process used for crude oil

The industrial process used to separate crude oil into petroleum fractions (gasoline, kerosene, diesel, etc.) is fractional distillation.

Crude oil consists of a complex mixture of miscible hydrocarbons.

2

Determine the governing physical property of fractional distillation

Fractional distillation relies on components vaporizing and condensing at different temperatures due to different boiling points.

Vapors of liquids with lower boiling points rise higher up the fractionating column before condensing, allowing distinct fraction collection.

Key Concept

Industrial Fractional Distillation

Question 8499Question →

Equal volumes of nitrogen ( $N_2$ ) and oxygen ( $O_2$ ) are allowed to effuse through a microscopic orifice under identical conditions of temperature and pressure. Which gas will effuse at a faster rate?

Show answer & explanation

Answer: Nitrogen; N2; nitrogen; N₂; Nitrogen gas; nitrogen gas

Answer

Nitrogen (

N_2

) will effuse at a faster rate because it has a lower molar mass (

28\text{ g/mol}

) compared to oxygen (

32\text{ g/mol}

).

According to Graham's Law, the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass (

r \propto \frac{1}{\sqrt{M}}

). Since nitrogen (

N_2

) has a molar mass of

28\text{ g/mol}

and oxygen (

O_2

) has a molar mass of

32\text{ g/mol}

, nitrogen is lighter and therefore effuses faster under identical conditions.

Step-by-Step Solution

1

Calculate the molar masses of both gases.

Molar mass of

N_2 = 2 \times 14 = 28\text{ g/mol}

. Molar mass of

O_2 = 2 \times 16 = 32\text{ g/mol}

.

Graham's Law relates the rate of effusion of a gas directly to its molar mass.

2

Apply Graham's Law of Effusion.

Effusion rate is inversely proportional to the square root of molar mass (

r \propto \frac{1}{\sqrt{M}}

).

Lighter gas molecules move faster at a given temperature, resulting in higher rates of effusion.

3

Compare rates of effusion.

Since

28\text{ g/mol} < 32\text{ g/mol}

, nitrogen (

N_2

) effuses faster than oxygen (

O_2

).

The gas with lower molar mass has the higher effusion rate.

Key Concept

Graham's Law of Diffusion and Effusion

Estimated Time:1m 0s

Question 8500Question →

What is the volume, in $\text{dm}^3$ , occupied by $6.80\text{ g}$ of hydrogen sulfide gas ( $\text{H}_2\text{S}$ ) measured at standard temperature and pressure (STP)? $[\text{H} = 1.0, \text{S} = 32.0, \text{Molar volume of gas at STP} = 22.4\text{ dm}^3\text{mol}^{-1}]$

4.48\text{ dm}^3

4.80\text{ dm}^3

4.76\text{ dm}^3

2.24\text{ dm}^3

Show answer & explanation

Answer: $4.48\text{ dm}^3$

Answer

The volume occupied by

6.80\text{ g}

of hydrogen sulfide gas at STP is

4.48\text{ dm}^3

.

To find the volume of gas at STP, first determine the molar mass of

\text{H}_2\text{S}

:

(2 \times 1.0) + 32.0 = 34.0\text{ g mol}^{-1}

. Next, convert the mass to moles:

\frac{6.80\text{ g}}{34.0\text{ g mol}^{-1}} = 0.20\text{ mol}

. Finally, multiply the moles by the molar volume at STP (

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

):

0.20 \times 22.4 = 4.48\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the molar mass of hydrogen sulfide (

\text{H}_2\text{S}

)

\text{Molar Mass} = (2 \times 1.0) + 32.0 = 34.0\text{ g mol}^{-1}

Molar mass is required to convert mass to moles.

2

Calculate the number of moles in

6.80\text{ g}

of

\text{H}_2\text{S}

\text{Moles} = \frac{6.80\text{ g}}{34.0\text{ g mol}^{-1}} = 0.20\text{ mol}

Dividing given mass by molar mass yields amount of substance in moles.

3

Calculate the volume occupied at STP

\text{Volume} = 0.20\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 4.48\text{ dm}^3

1 mole of any ideal gas occupies

22.4\text{ dm}^3

at STP.

Key Concept

Molar Volume of Gases at STP

Estimated Time:1m 30s

Answer

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