All practice questions

Question 8301Question →

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

Click a left item, then click its matching right item

Show answer & explanation

Answer

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

s

-electron delocalization and

d

-orbital overlap.

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

s

-electron delocalization and

d

-orbital bonding.

Step-by-Step Solution

1

Analyze the mechanism for heat conduction in metals.

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

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

2

Analyze how temperature affects electrical resistance/conductivity in metals.

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

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

3

Analyze the optical reflection property of metals.

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

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

4

Compare cohesive energy differences between alkali metals and transition metals.

Transition elements utilize both outer

s

valence electrons and partially filled inner

d

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

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

Key Concept

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

Question 8302Question →

For a particular gasification process, a chemical reaction has a standard enthalpy change ( $\Delta H^\circ$ ) of $+136.5\text{ kJ mol}^{-1}$ and a standard entropy change ( $\Delta S^\circ$ ) of $+325.0\text{ J K}^{-1}\text{ mol}^{-1}$ . What is the minimum temperature, in degrees Celsius ( $^\circ\text{C}$ ), above which the reaction becomes thermodynamically spontaneous?

Show answer & explanation

Answer: 147

Answer

The minimum temperature above which the reaction becomes spontaneous is 147 °C.

At the boundary of spontaneity,

\Delta G^\circ = 0

. Substituting this into

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

yields

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

. Converting

\Delta H^\circ

to Joules yields

+136,500\text{ J mol}^{-1}

. Dividing by

+325.0\text{ J K}^{-1}\text{ mol}^{-1}

gives

T = 420\text{ K}

. Converting to Celsius gives

420 - 273 = 147^\circ\text{C}

.

Step-by-Step Solution

1

Convert enthalpy change from kilojoules per mole to joules per mole

\Delta H^\circ = +136.5\text{ kJ mol}^{-1} = +136,500\text{ J mol}^{-1}

\Delta H^\circ

must be expressed in Joules to match the unit of

\Delta S^\circ

(

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

).

2

Determine the threshold condition for reaction spontaneity using Gibbs free energy equation

\Delta G^\circ = 0 \implies T = \frac{\Delta H^\circ}{\Delta S^\circ}

A reaction is spontaneous when

\Delta G^\circ < 0

. The minimum temperature where spontaneity begins occurs when

\Delta G^\circ = 0

.

3

Calculate the absolute temperature in Kelvin

T = \frac{136,500\text{ J mol}^{-1}}{325.0\text{ J K}^{-1}\text{ mol}^{-1}} = 420\text{ K}

Dividing the enthalpy change in Joules by the entropy change in Joules per Kelvin yields the temperature in Kelvin.

4

Convert temperature from Kelvin to degrees Celsius

T(^\circ\text{C}) = 420 - 273 = 147^\circ\text{C}

The question explicitly requests the answer in degrees Celsius (

T(^\circ\text{C}) = T(\text{K}) - 273

).

Key Concept

Gibbs Free Energy and Temperature Dependence of Spontaneity

Estimated Time:2m 0s

Question 8303Question →

An organic compound $X$ with the molecular formula $C_4H_6$ rapidly decolorizes bromine water in tetrachloromethane, but produces no precipitate when treated with ammoniacal silver nitrate solution. What is the IUPAC name of compound $X$ ?

But-2-yne

But-1-yne

But-1-ene

But-2-ene

Show answer & explanation

Answer: But-2-yne

Answer

But-2-yne

The compound But-2-yne (

CH_3-C\equiv C-CH_3

) has the molecular formula

C_4H_6

and contains an internal carbon-carbon triple bond. As an unsaturated hydrocarbon, it readily decolorizes bromine water. However, because its triple bond is located between carbon-2 and carbon-3, it lacks a terminal acetylenic hydrogen atom (

-C\equiv C-H

). Therefore, it cannot react with ammoniacal silver nitrate to form a precipitate, matching all given experimental observations.

Step-by-Step Solution

1

Determine the structural class using the given molecular formula

C_4H_6

.

The formula

C_4H_6

fits the general formula

C_nH_{2n-2}

(alkynes or alkadienes), indicating two degrees of unsaturation.

Alkynes with 4 carbon atoms have the formula

C_4H_6

.

2

Analyze the response to bromine water in tetrachloromethane.

Decolorization confirms the presence of carbon-carbon multiple bonds (unsaturation).

Electrophilic addition of bromine occurs across the triple bond.

3

Analyze the reaction with ammoniacal silver nitrate solution.

The absence of a precipitate indicates that the alkyne is non-terminal (internal).

Only terminal alkynes containing an acidic acetylenic hydrogen atom (

-C\equiv C-H

) react with Tollens' reagent (

[Ag(NH_3)_2]^+

) to yield a insoluble silver alkynide precipitate.

4

Identify the correct IUPAC name among the

C_4H_6

isomers.

But-2-yne (

CH_3-C\equiv C-CH_3

) is an internal alkyne, whereas But-1-yne (

CH_3-CH_2-C\equiv CH

) is a terminal alkyne.

Since compound

X

does not form a precipitate, it must be the internal alkyne, But-2-yne.

Key Concept

Distinction between terminal and internal alkynes using Tollens' reagent (ammoniacal silver nitrate)

Estimated Time:1m 30s

Question 8304Question →

An element $M$ has the ground-state electronic configuration $1s^2 2s^2 2p^6 3s^2$ , while element $X$ has the configuration $1s^2 2s^2 2p^5$ . What is the chemical formula of the compound formed between $M$ and $X$ , and what type of bonding holds the compound together?

MX_2

; electrovalent (ionic) bonding

MX

; electrovalent (ionic) bonding

M_2X

; covalent bonding

MX_2

; covalent bonding

Show answer & explanation

Answer: $MX_2$ ; electrovalent (ionic) bonding

Answer

The correct compound formula is

MX_2

and the bonding is electrovalent (ionic).

The correct answer states that the compound formula is

MX_2

formed by electrovalent (ionic) bonding. Element

M

has 2 valence electrons (

3s^2

) which it transfers to two atoms of element

X

(each having 7 valence electrons,

2s^2 2p^5

), forming

M^{2+}

and two

X^-

ions. Bonding via complete electron transfer between metal and non-metal is electrovalent.

Step-by-Step Solution

1

Analyze the electronic configuration of element

M

to determine its valency and ion charge.

Element

M

(

1s^2 2s^2 2p^6 3s^2

) has 2 valence electrons in its outermost shell (

3s

). It donates 2 electrons to achieve a stable octet, forming the cation

M^{2+}

.

Elements with 1 or 2 outer electrons readily lose them to achieve stable noble gas electron configurations.

2

Analyze the electronic configuration of element

X

to determine its valency and ion charge.

Element

X

(

1s^2 2s^2 2p^5

) has 7 valence electrons in its outermost shell (

2s^2 2p^5

). It accepts 1 electron to complete its octet, forming the anion

X^-

.

Non-metals with 7 valence electrons require 1 additional electron for octet stability.

3

Balance the ionic charges to establish the formula unit and identify the bond type.

To maintain electrical neutrality, one

M^{2+}

ion combines with two

X^-

ions, resulting in the chemical formula

MX_2

. The complete transfer of electrons from a metal to a non-metal yields electrostatic attraction, which constitutes electrovalent (ionic) bonding.

Ionic compounds must be electrically neutral, and bonding formed by complete electron transfer is electrovalent.

Key Concept

Ionic (Electrovalent) Bond Formation and Formula Determination

Question 8305Question →

An organic compound containing carbon, hydrogen, and nitrogen was analyzed and found to contain $61.02\%$ carbon and $15.25\%$ hydrogen by mass, with the remainder being nitrogen. Given that the relative molecular mass of the compound is $118\text{ g/mol}$ , what is the value of the integer multiplier $n$ that relates its empirical formula to its molecular formula? $[C = 12, H = 1, N = 14]$

Show answer & explanation

Answer: 2

Answer

The value of the integer multiplier n is 2.

Subtracting the carbon (

61.02\%

) and hydrogen (

15.25\%

) percentages from

100\%

gives a nitrogen content of

23.73\%

. Converting these mass percentages to mole ratios by dividing by relative atomic masses (

C=12, H=1, N=14

) gives

5.085\text{ mol}

of C,

15.25\text{ mol}

of H, and

1.695\text{ mol}

of N. Dividing by the smallest value (

1.695

) produces the mole ratio

3:9:1

, establishing the empirical formula as

C_3H_9N

. The empirical mass is

(3 \times 12) + (9 \times 1) + 14 = 59\text{ g/mol}

. Dividing the relative molecular mass (

118\text{ g/mol}

) by the empirical mass (

59\text{ g/mol}

) yields

n = 2

.

Step-by-Step Solution

1

Calculate percentage composition of nitrogen

23.73%

The sum of percentages of all constituent elements in a compound must equal 100%.

2

Calculate relative number of moles for each element

C = 5.085 mol, H = 15.25 mol, N = 1.695 mol

Dividing the mass percentage of each element by its relative atomic mass gives its relative mole quantity.

3

Determine simplest whole number atomic ratio

C : H : N = 3 : 9 : 1

Dividing all mole amounts by the smallest value (1.695) gives the empirical mole ratio.

4

Determine the empirical formula mass

59 g/mol

Summing the atomic masses of elements in C3H9N yields (3 x 12) + (9 x 1) + 14 = 59 g/mol.

5

Calculate the integer multiplier n

2

Dividing relative molecular mass (118 g/mol) by empirical formula mass (59 g/mol) gives n = 2.

Key Concept

Calculation of Empirical Formula and Molecular Formula Multiplier

Question 8306Question →

Equal volumes of $0.10\text{ mol dm}^{-3}$ solutions of methanoic acid ( $\text{HCOOH}$ ) and hydrochloric acid ( $\text{HCl}$ ) are prepared at $25^\circ\text{C}$ . Which of the following statements correctly accounts for the lower electrical conductivity observed in the methanoic acid solution?

Methanoic acid has a lower molar concentration of dissolved acid molecules in solution than hydrochloric acid.

Methanoic acid undergoes incomplete ionization in water, resulting in a lower concentration of mobile hydronium ions.

Methanoic acid requires an indicator to undergo complete ionization in an aqueous environment.

Methanoic acid molecules neutralize hydronium ions rapidly to reform un-ionized acid molecules, reducing ionic mobility.

Show answer & explanation

Answer: Methanoic acid undergoes incomplete ionization in water, resulting in a lower concentration of mobile hydronium ions.

Answer

Methanoic acid undergoes incomplete ionization in water, resulting in a lower concentration of mobile hydronium ions.

Electrical conductivity in aqueous solutions depends on the concentration of free, mobile ions. Hydrochloric acid is a strong monobasic acid that dissociates completely in water to yield a high concentration of

\text{H}_3\text{O}^+

and

\text{Cl}^-

ions. In contrast, methanoic acid is a weak monobasic acid that ionizes only partially in aqueous solution, leaving most acid molecules un-ionized. This lower ion concentration directly accounts for the reduced electrical conductivity.

Step-by-Step Solution

1

Analyze acid strength and ionization behavior of both acids

Hydrochloric acid (

\text{HCl}

) is a strong acid that ionizes completely:

\text{HCl}_{(aq)} + \text{H}_2\text{O}_{(l)} \rightarrow \text{H}_3\text{O}^+_{(aq)} + \text{Cl}^-_{(aq)}

. Methanoic acid (

\text{HCOOH}

) is a weak acid that ionizes partially:

\text{HCOOH}_{(aq)} + \text{H}_2\text{O}_{(l)} \rightleftharpoons \text{HCOO}^-_{(aq)} + \text{H}_3\text{O}^+_{(aq)}

.

Electrical conductivity in aqueous solutions depends on the concentration of mobile ions present.

2

Compare ion concentrations at equal molar concentration

For

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

\text{HCl}

,

[\text{H}_3\text{O}^+] = 0.10\text{ mol dm}^{-3}

. For

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

\text{HCOOH}

,

[\text{H}_3\text{O}^+] \ll 0.10\text{ mol dm}^{-3}

due to partial ionization.

Fewer ions per unit volume results in lower electrical current transport through the methanoic acid solution.

Key Concept

Relative strength of acids depends on the extent of ionization in water, where weak acids establish an equilibrium yielding fewer mobile ions than strong acids of equal concentration.

Question 8307Question →

During the industrial separation of liquefied air to obtain pure nitrogen gas, nitrogen vaporizes at $-196^\circ\text{C}$ while oxygen remains in the liquid state at $-183^\circ\text{C}$ . Which of the following statements correctly explains why nitrogen distills over first?

Nitrogen has a lower boiling point than oxygen, so it vaporizes at a lower temperature.

Nitrogen has a higher boiling point than oxygen, causing it to evaporate before liquid oxygen.

Nitrogen diffuses faster than oxygen because it is less dense, forcing it out of the fractionating column first.

Nitrogen's boiling point appears lower only because negative Celsius temperatures convert inversely to the Kelvin scale.

Show answer & explanation

Answer: Nitrogen has a lower boiling point than oxygen, so it vaporizes at a lower temperature.

Answer

Nitrogen has a lower boiling point than oxygen, so it vaporizes at a lower temperature.

Nitrogen has a lower boiling point (

-196^\circ\text{C}

or

77\text{ K}

) than oxygen (

-183^\circ\text{C}

or

90\text{ K}

). In fractional distillation, liquid mixtures are separated based on differences in boiling points, and the component with the lower boiling point boils off and distills over first.

Step-by-Step Solution

1

Compare the boiling point values on the Celsius and Kelvin temperature scales.

Nitrogen boils at

-196^\circ\text{C}

(

77\text{ K}

) while oxygen boils at

-183^\circ\text{C}

(

90\text{ K}

).

On the Celsius scale,

-196

is smaller (lower) than

-183

.

2

Apply the physical principle of fractional distillation.

As liquefied air warms, nitrogen reaches its boiling point first and converts into gas.

The component with the lower boiling point always distills over first during fractional distillation.

Key Concept

Fractional Distillation of Liquid Air

Question 8308Question →

Increasing the temperature of a gaseous reaction mixture increases the rate of reaction primarily because it lowers the activation energy of the reaction.

True

False

Show answer & explanation

Answer: False

Answer

False. Increasing the temperature increases the kinetic energy of particles and the frequency of effective collisions, but it does not alter the activation energy of the reaction.

The statement is false because temperature affects molecular kinetic energy and the proportion of effective collisions, but it does not alter the activation energy of the reaction. Lowering the activation energy is achieved exclusively by adding a catalyst.

Step-by-Step Solution

1

Analyze the role of temperature in collision theory.

Higher temperature increases the average kinetic energy of the reactant particles.

Temperature is directly proportional to the average kinetic energy of molecules.

2

Evaluate the effect of temperature on activation energy (

E_a

).

The activation energy barrier remains unchanged.

Activation energy is a fixed energy threshold for a specific reaction mechanism; temperature change does not alter this threshold.

3

Determine why the rate of reaction increases at higher temperatures.

A significantly larger fraction of colliding particles possess energy

\geq E_a

, increasing the frequency of effective collisions.

The Maxwell-Boltzmann distribution shifts to higher kinetic energy values without shifting the position of

E_a

.

Key Concept

Temperature effect on molecular energy versus catalyst effect on activation energy

Question 8309Question →

Agricultural runoff containing excess synthetic fertilizers frequently drains into freshwater bodies. Which of the following chemical species present in such runoff is primarily responsible for triggering rapid algal growth and subsequent dissolved oxygen depletion (eutrophication)?

Trioxophosphates(V)

Chlorofluorocarbons

Calcium tetraoxosulfate(VI)

Concentrated ethanoic acid

Show answer & explanation

Answer: Trioxophosphates(V)

Answer

Trioxophosphates(V) (phosphates) are nutrient compounds present in agricultural runoff and detergents that stimulate algal growth, leading to oxygen depletion in water bodies.

Trioxophosphates(V) derived from fertilizers and synthetic detergents act as major plant nutrients. When introduced into rivers and lakes, they cause excessive growth of algae. Upon dying, aerobic bacteria decompose the algae, consuming large amounts of dissolved oxygen and causing fish kill and water degradation.

Step-by-Step Solution

1

Identify the chemical pollutant associated with agricultural fertilizer runoff.

Agricultural fertilizers predominantly contain trioxonitrates(V) and trioxophosphates(V).

Phosphorus compounds act as limiting nutrients in aquatic ecosystems.

2

Determine the environmental mechanism triggered by nutrient enrichment.

Excess trioxophosphates(V) cause rapid accumulation of algae (algal bloom), which cuts off light and consumes dissolved oxygen during decomposition.

This process is known as eutrophication.

Key Concept

Eutrophication caused by industrial and agricultural nutrient pollution

Question 8310Question →

Which of the following observations is made when aqueous ammonia is added gradually until in excess to a solution containing aluminium ions, $Al^{3+}(aq)$ ?

A white gelatinous precipitate forms which dissolves in excess aqueous ammonia to yield a clear, colorless solution.

A white gelatinous precipitate forms which remains insoluble in excess aqueous ammonia.

No precipitate is formed because aluminium hydroxide is completely soluble in weak ammonium solutions.

A reddish-brown precipitate forms which turns white upon adding excess aqueous ammonia.

Show answer & explanation

Answer: A white gelatinous precipitate forms which remains insoluble in excess aqueous ammonia.

Answer

A white gelatinous precipitate of aluminium hydroxide forms, which remains insoluble in excess aqueous ammonia.

When aqueous ammonia is added to a solution containing

Al^{3+}

ions, hydroxide ions precipitate aluminium hydroxide,

Al(OH)_3

, as a white gelatinous solid. Because aqueous ammonia is a weak base, it cannot provide the high concentration of

OH^-

ions required to react with amphoteric

Al(OH)_3

to form a soluble complex ion. Consequently, the white precipitate remains insoluble in excess aqueous ammonia.

Step-by-Step Solution

1

Identify the initial precipitation reaction when aqueous ammonia is added to

Al^{3+}(aq)

.

Aqueous ammonia provides hydroxide ions (

OH^-

), reacting with

Al^{3+}

to form a white gelatinous precipitate of aluminium hydroxide,

Al(OH)_3(s)

.

The precipitation ionic equation is

Al^{3+}(aq) + 3OH^-(aq) \rightarrow Al(OH)_3(s)

.

2

Evaluate the effect of adding excess aqueous ammonia.

Aluminium hydroxide is amphoteric and dissolves in strong alkalis (like excess

NaOH

) to form aluminate complex ions

[Al(OH)_4]^-

, but it does NOT dissolve in weak alkalis like aqueous ammonia (

NH_3(aq)

).

Aqueous ammonia does not supply a high enough hydroxide concentration to form the complex ion, nor does aluminium form a soluble ammine complex.

Key Concept

Distinct qualitative analysis reactions of

Al^{3+}

ions with aqueous ammonia versus sodium hydroxide

Estimated Time:1m 0s

Question 8311Question →

A sample of $0.62\text{ g}$ of sodium oxide ( $\text{Na}_2\text{O}$ ) is reacted completely with distilled water to prepare $200\text{ cm}^3$ of stock solution. If $50\text{ cm}^3$ of this stock solution is diluted with distilled water to a final volume of $500\text{ cm}^3$ at $25^\circ\text{C}$ , what is the pH of the resulting diluted solution? $[\text{Na} = 23, \text{O} = 16, \text{H} = 1]$

Show answer & explanation

Answer: 12

Answer

The pH of the diluted solution is 12.0.

Dissolving

0.62\text{ g}

(

0.01\text{ mol}

) of

\text{Na}_2\text{O}

produces

0.02\text{ mol}

of

\text{OH}^-

ions in

200\text{ cm}^3

, giving a stock concentration of

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

. Diluting

50\text{ cm}^3

of this stock solution to

500\text{ cm}^3

reduces the

[\text{OH}^-]

tenfold to

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

. Taking the negative logarithm gives

\text{pOH} = 2.0

, which corresponds to a

\text{pH}

of

14.0 - 2.0 = 12.0

.

Step-by-Step Solution

1

Calculate the molar mass of sodium oxide (

\text{Na}_2\text{O}

) and determine the amount of moles dissolved.

\text{Molar mass of Na}_2\text{O} = (2 \times 23) + 16 = 62\text{ g mol}^{-1}

. Moles of

\text{Na}_2\text{O} = \frac{0.62\text{ g}}{62\text{ g mol}^{-1}} = 0.01\text{ mol}

.

Converting mass to moles is required to apply chemical stoichiometry.

2

Determine the moles of hydroxide ions (

\text{OH}^-

) formed upon complete reaction with water.

The balanced equation is

\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH} \rightarrow 2\text{Na}^+ + 2\text{OH}^-

. Therefore,

0.01\text{ mol}

of

\text{Na}_2\text{O}

yields

0.02\text{ mol}

of

\text{OH}^-

.

Sodium oxide is a basic oxide that reacts with water in a 1:2 mole ratio to yield hydroxide ions.

3

Calculate the hydroxide ion concentration in the

200\text{ cm}^3

(

0.2\text{ dm}^3

) stock solution.

[\text{OH}^-]_{\text{stock}} = \frac{0.02\text{ mol}}{0.2\text{ dm}^3} = 0.1\text{ mol dm}^{-3}

.

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

4

Apply the dilution formula

C_1 V_1 = C_2 V_2

to find the hydroxide ion concentration after dilution.

[\text{OH}^-]_{\text{diluted}} = \frac{0.1\text{ mol dm}^{-3} \times 50\text{ cm}^3}{500\text{ cm}^3} = 0.01\text{ mol dm}^{-3} = 1.0 \times 10^{-2}\text{ mol dm}^{-3}

.

Diluting

50\text{ cm}^3

to

500\text{ cm}^3

decreases the concentration by a factor of 10.

5

Calculate the pOH and subsequently the pH of the diluted solution.

\text{pOH} = -\log_{10}(1.0 \times 10^{-2}) = 2.0

. Using

\text{pH} + \text{pOH} = 14.0

,

\text{pH} = 14.0 - 2.0 = 12.0

.

The logarithmic scale defines

\text{pOH} = -\log_{10}[\text{OH}^-]

and at

25^\circ\text{C}

,

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

.

Key Concept

Stoichiometric reaction of basic oxides with water combined with dilution calculations to determine solution pH and pOH.

Question 8312Question →

In a paper chromatography experiment, a mixture of plant pigments was resolved using an organic solvent mixture. The solvent front travelled a total distance of $15.0\text{ cm}$ from the baseline. Spot $A$ was found to have a retardation factor ( $R_f$ ) value of $0.60$ . If spot $B$ migrated $3.0\text{ cm}$ further than spot $A$ along the chromatogram, what is the $R_f$ value of spot $B$ ?

0.20

0.40

0.80

1.25

Show answer & explanation

Answer: 0.80

Answer

0.80

The retardation factor (

R_f

) is defined as

R_f = \frac{\text{distance moved by solute}}{\text{distance moved by solvent front}}

. First, calculate the distance moved by spot

A

:

0.60 \times 15.0\text{ cm} = 9.0\text{ cm}

. Spot

B

moved

3.0\text{ cm}

further, so its total distance from the baseline is

9.0\text{ cm} + 3.0\text{ cm} = 12.0\text{ cm}

. Thus,

R_f(B) = \frac{12.0\text{ cm}}{15.0\text{ cm}} = 0.80

.

Step-by-Step Solution

1

Calculate the distance travelled by spot A from the baseline.

\text{Distance}_A = R_f(A) \times \text{Solvent Front Distance} = 0.60 \times 15.0\text{ cm} = 9.0\text{ cm}

.

The

R_f

value is defined as the ratio of solute distance to solvent front distance.

2

Determine the total distance travelled by spot B from the baseline.

\text{Distance}_B = \text{Distance}_A + 3.0\text{ cm} = 9.0\text{ cm} + 3.0\text{ cm} = 12.0\text{ cm}

.

Spot B migrated

3.0\text{ cm}

further than spot A along the paper.

3

Calculate the

R_f

value of spot B.

R_f(B) = \frac{\text{Distance}_B}{\text{Solvent Front Distance}} = \frac{12.0\text{ cm}}{15.0\text{ cm}} = 0.80

.

Dividing the distance moved by spot B by the total solvent front distance yields its

R_f

value.

Key Concept

Calculation of retardation factor (

R_f

) in paper chromatography.

Question 8313Question →

A sample of pure hydrated aluminum nitrate, $\text{Al(NO}_3)_3 \cdot 9\text{H}_2\text{O}$ , has a mass of $75.0\text{ g}$ . What is the mass, in grams, of oxygen contained in this sample? $[\text{Al} = 27, \text{N} = 14, \text{O} = 16, \text{H} = 1]$

Show answer & explanation

Answer: 57.6

Answer

57.6

The molar mass of

\text{Al(NO}_3)_3 \cdot 9\text{H}_2\text{O}

is

375\text{ g/mol}

. A

75.0\text{ g}

sample equals

0.20\text{ mol}

of the hydrated salt. Because each formula unit contains

18

oxygen atoms (

9

from the nitrate groups and

9

from the water molecules),

0.20\text{ mol}

of the salt contains

3.60\text{ mol}

of oxygen atoms. Multiplying

3.60\text{ mol}

by the atomic mass of oxygen (

16\text{ g/mol}

) yields

57.6\text{ g}

.

Step-by-Step Solution

1

Calculate the molar mass of hydrated aluminum nitrate,

\text{Al(NO}_3)_3 \cdot 9\text{H}_2\text{O}

.

375\text{ g/mol}

Sum the atomic masses of all constituent atoms:

\text{Al} = 27

,

3 \times \text{NO}_3 = 3 \times 62 = 186

,

9 \times \text{H}_2\text{O} = 9 \times 18 = 162

. Total

= 27 + 186 + 162 = 375\text{ g/mol}

.

2

Determine the amount (in moles) of the compound in the

75.0\text{ g}

sample.

0.20\text{ mol}

Moles of compound

= \frac{\text{mass}}{\text{molar mass}} = \frac{75.0\text{ g}}{375\text{ g/mol}} = 0.20\text{ mol}

.

3

Determine the total moles of oxygen atoms per mole of the hydrated compound.

18\text{ mol of O}

Each formula unit contains

9

oxygen atoms in the nitrate groups (

3 \times 3

) and

9

oxygen atoms in the water molecules (

9 \times 1

), giving a total of

18

oxygen atoms.

4

Calculate the total mass of oxygen atoms in the sample.

57.6\text{ g}

Mass of oxygen

= \text{moles of compound} \times 18 \times \text{molar mass of O} = 0.20 \times 18 \times 16 = 3.60 \times 16 = 57.6\text{ g}

.

Key Concept

Mole Concept and Stoichiometric Mass Composition in Hydrated Compounds

Question 8314Question →

An aqueous solution of sodium chloride ( $NaCl$ ) undergoes salt hydrolysis, causing it to turn blue litmus paper red.

True

False

Show answer & explanation

Answer: False

Answer

The statement is false. Sodium chloride (

NaCl

) is a neutral salt of a strong acid (

HCl

) and a strong base (

NaOH

); its ions do not undergo hydrolysis in aqueous solution.

The statement is false because sodium chloride (

NaCl

) originates from a strong acid (

HCl

) and a strong base (

NaOH

). Since neither cation (

Na^+

) nor anion (

Cl^-

) reacts hydrolytically with water, the aqueous solution maintains a neutral

\text{pH}

of 7 and has no effect on blue litmus paper.

Step-by-Step Solution

1

Identify the parent acid and base that form sodium chloride (

NaCl

).

The parent acid is hydrochloric acid (

HCl

, a strong acid) and the parent base is sodium hydroxide (

NaOH

, a strong base).

Knowing the relative strengths of the parent acid and base determines whether hydrolysis occurs.

2

Determine the hydrolysis tendency of the constituent ions (

Na^+

and

Cl^-

).

Neither

Na^+

nor

Cl^-

undergoes hydrolysis in aqueous solution.

Conjugate ions of strong acids and strong bases are extremely weak and do not react with water molecules.

3

Deduce the solution acidity/alkalinity and its effect on litmus paper.

The concentrations of

H^+

and

OH^-

remain equal, giving a neutral solution (

\text{pH} = 7

) that does not turn blue litmus paper red.

Without hydrolysis generating excess

H^+

ions, the solution cannot display acidic behavior.

Key Concept

Salts derived from strong acids and strong bases yield neutral aqueous solutions because their ions do not hydrolyze.

Estimated Time:45s

Question 8315Question →

A $2.016\text{ g}$ sample of a hydrated dicarboxylic acid, $(\text{COOH})_2 \cdot n\text{H}_2\text{O}$ , was dissolved in distilled water and made up to $250.0\text{ cm}^3$ in a volumetric flask. A $25.0\text{ cm}^3$ portion of this solution required exactly $20.0\text{ cm}^3$ of $0.160\text{ mol dm}^{-3}$ sodium hydroxide ( $\text{NaOH}$ ) solution for complete neutralization. What is the value of the integer $n$ in the formula of the hydrated acid? $[\text{H} = 1, \text{C} = 12, \text{O} = 16]$

Show answer & explanation

Answer: 2

Answer

The value of the integer n is 2.

Titration of 20.0 cm³ of 0.160 mol dm⁻³ NaOH consumes 0.0032 mol of NaOH. Because the dicarboxylic acid is diprotic ((COOH)₂), it reacts in a 1:2 ratio with NaOH, giving 0.0016 mol of acid in the 25.0 cm³ aliquot. Scaling to the full 250.0 cm³ flask yields 0.0160 mol of hydrated acid. The molar mass of the hydrated acid is 2.016 g / 0.0160 mol = 126 g/mol. Since the anhydrous formula mass of (COOH)₂ is 90 g/mol, the water of crystallization contributes 126 - 90 = 36 g/mol. Dividing 36 by 18 (the molar mass of H₂O) gives n = 2.

Step-by-Step Solution

1

Calculate the amount in moles of NaOH used in the titration.

Moles of NaOH = 0.0032 mol

Moles = concentration × volume = 0.160 mol dm⁻³ × (20.0 / 1000) dm³ = 0.0032 mol.

2

Determine the moles of acid present in the 25.0 cm³ titration sample using the stoichiometric mole ratio.

Moles of acid in 25.0 cm³ = 0.0016 mol

Ethanoic/oxalic acid is diprotic ((COOH)₂ + 2NaOH → (COONa)₂ + 2H₂O), requiring 2 moles of NaOH per mole of acid. Moles of acid = 0.0032 / 2 = 0.0016 mol.

3

Scale up to find the total moles of acid in the original 250.0 cm³ solution.

Total moles of acid = 0.0160 mol

Total moles = 0.0016 mol × (250.0 cm³ / 25.0 cm³) = 0.0160 mol.

4

Calculate the molar mass of the hydrated dicarboxylic acid.

Molar mass = 126 g mol⁻¹

Molar mass M = sample mass / total moles = 2.016 g / 0.0160 mol = 126 g mol⁻¹.

5

Calculate the value of integer n by comparing the molar mass to the anhydrous acid mass.

n = 2

Formula mass of anhydrous (COOH)₂ = 2(12) + 2(1) + 4(16) = 90 g mol⁻¹. The water component mass is 18n = 126 - 90 = 36 g mol⁻¹, giving n = 36 / 18 = 2.

Key Concept

Empirical and Molecular Formula Calculations with Water of Crystallization and Volumetric Stoichiometry

Question 8316Question →

The standard reduction potentials for tin and silver half-cells are given below:

\text{Sn}^{2+}(aq) + 2e^- \rightarrow \text{Sn}(s) \quad E^\circ = -0.14\text{ V}

\text{Ag}^+(aq) + e^- \rightarrow \text{Ag}(s) \quad E^\circ = +0.80\text{ V}

What is the standard cell potential (

E^\circ_{\text{cell}}

) for the overall redox reaction:

\text{Sn}(s) + 2\text{Ag}^+(aq) \rightarrow \text{Sn}^{2+}(aq) + 2\text{Ag}(s)

+0.94\text{ V}

+0.66\text{ V}

-0.94\text{ V}

+1.74\text{ V}

Show answer & explanation

Answer: $+0.94\text{ V}$

Answer

+0.94\text{ V}

In the given cell reaction, silver ions undergo reduction at the cathode (

E^\circ = +0.80\text{ V}

) while tin metal undergoes oxidation at the anode (

E^\circ = -0.14\text{ V}

). The standard cell potential is calculated using

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = +0.80\text{ V} - (-0.14\text{ V}) = +0.94\text{ V}

. Because

E^\circ

values are intensive properties, the stoichiometric multiplier for the silver half-reaction does not scale its potential value.

Step-by-Step Solution

1

Identify the reduction and oxidation half-reactions from the given reaction equation.

Tin metal (

\text{Sn}

) is oxidized to

\text{Sn}^{2+}

at the anode, while silver ions (

\text{Ag}^+

) are reduced to silver metal at the cathode.

Oxidation occurs at the anode (loss of electrons) and reduction occurs at the cathode (gain of electrons).

2

Apply the formula for standard cell potential

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

.

E^\circ_{\text{cell}} = (+0.80\text{ V}) - (-0.14\text{ V}) = +0.80\text{ V} + 0.14\text{ V} = +0.94\text{ V}

.

Standard electrode potentials are intensive properties and are not multiplied by stoichiometric coefficients.

Key Concept

Calculation of Standard Cell Potential and Reaction Spontaneity

Question 8317Question →

A

3.25\text{ g}

sample of impure zinc granules reacts completely with excess dilute tetraoxosulfate(VI) acid according to the equation:

\text{Zn(s)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{H}_2\text{(g)}

If

896\text{ cm}^3

of hydrogen gas is collected at STP, calculate the percentage purity of the zinc sample.

[\text{Zn} = 65\text{; Molar volume of gas at STP} = 22.4\text{ dm}^3\text{mol}^{-1}]

Show answer & explanation

Answer: 80

Answer

80%

The volume of hydrogen gas produced at STP (

896\text{ cm}^3 = 0.896\text{ dm}^3

) corresponds to

0.04\text{ mol}

of

\text{H}_2

. Based on the

1:1

stoichiometric reaction between zinc and tetraoxosulfate(VI) acid,

0.04\text{ mol}

of pure zinc was present in the sample. The mass of pure zinc is

0.04\text{ mol} \times 65\text{ g/mol} = 2.60\text{ g}

. Dividing this pure mass by the total sample mass of

3.25\text{ g}

and multiplying by

100\%

yields a percentage purity of

80\%

.

Step-by-Step Solution

1

Convert the collected volume of hydrogen gas from cm³ to dm³

\text{Volume of H}_2 = \frac{896\text{ cm}^3}{1000\text{ cm}^3/\text{dm}^3} = 0.896\text{ dm}^3

Molar gas volume is given in dm³/mol, so gas volume must be expressed in dm³.

2

Calculate the amount in moles of hydrogen gas produced at STP

\text{Moles of H}_2 = \frac{0.896\text{ dm}^3}{22.4\text{ dm}^3\text{mol}^{-1}} = 0.04\text{ mol}

Number of moles of gas at STP equals volume divided by molar volume.

3

Determine the moles and mass of pure zinc that reacted

\text{Moles of pure Zn} = 0.04\text{ mol}

;

\text{Mass of pure Zn} = 0.04\text{ mol} \times 65\text{ g/mol} = 2.60\text{ g}

The reaction stoichiometry shows a 1:1 mole ratio between Zn and H₂.

4

Calculate the percentage purity of the zinc sample

\text{Percentage Purity} = \left( \frac{2.60\text{ g}}{3.25\text{ g}} \right) \times 100\% = 80\%

Percentage purity is the ratio of pure substance mass to total sample mass expressed as a percentage.

Key Concept

Percentage purity determination via gas volume stoichiometry at STP

Question 8318Question →

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

Click a left item, then click its matching right item

Show answer & explanation

Answer

Heavy metal ions (

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

) pair with precipitation using

Ca(OH)_2

or

Na_2S

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

H_2SO_4

) pairs with lime/limestone neutralization.

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

Step-by-Step Solution

1

Analyze heavy metal waste remediation chemistry

Heavy metal cations (

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

) react with

OH^-

or

S^{2-}

to form insoluble salts (

PbS

,

Cd(OH)_2

) precipitating out of solution.

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

2

Evaluate textile dye removal processes

Refractory organic dye molecules adsorb onto activated carbon porous structures.

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

3

Identify phosphate pollution control

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

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

4

Determine acidic wastewater treatment

H_2SO_4

reacts with

CaCO_3

or

Ca(OH)_2

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

Adjusting effluent

\text{pH}

to near neutral (

6.5-8.5

) is mandatory prior to discharge to safeguard aquatic life.

Key Concept

Industrial Effluents and Specific Water Remediation Techniques

Question 8319Question →

When transparent crystals of washing soda, $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$ , are left exposed to dry laboratory air, they gradually lose water of crystallization to become a white powdery monohydrate, $\text{Na}_2\text{CO}_3 \cdot \text{H}_2\text{O}$ . Which phenomenon is demonstrated by this observation?

Deliquescence

Efflorescence

Hygroscopy

Hydrolysis

Show answer & explanation

Answer: Efflorescence

Answer

Efflorescence is demonstrated when washing soda crystals lose water of crystallization upon exposure to dry air.

Efflorescence is the phenomenon where a hydrated compound loses its water of crystallization spontaneously when exposed to air because its hydration vapor pressure exceeds atmospheric moisture vapor pressure.

Step-by-Step Solution

1

Analyze the observed physical change.

Hydrated washing soda crystals (decahydrate) lose 9 molecules of water of crystallization to form a monohydrate powder in dry air.

The vapor pressure of water in the hydrated crystal is greater than the partial pressure of water vapor in the surrounding air.

2

Match the observed behavior with the appropriate chemical term.

The spontaneous loss of water of crystallization to the atmosphere is defined as efflorescence.

Efflorescent salts lose moisture to dry air until an equilibrium is reached.

Key Concept

Efflorescence in hydrated salts

Question 8320Question →

Complete the following statement regarding salt classification and laboratory preparation methods by filling in the missing terms.

Fill in the blanks below

Potash alum,

KAl(SO_4)_2 \cdot 12H_2O

, is classified as a salt because it completely dissociates in water into all its constituent simple ions, whereas an insoluble salt like lead(II) sulfate (

PbSO_4

) is prepared in the laboratory using .

Show answer & explanation

Answer

The first blank is 'double' (or 'double salt') and the second blank is 'double decomposition' (or 'precipitation').

Potash alum dissociates into

K^+

,

Al^{3+}

, and

SO_4^{2-}

ions in solution, making it a double salt. Lead(II) sulfate (

PbSO_4

) is insoluble in water and must be prepared by double decomposition (precipitation) by combining two aqueous solutions containing the requisite ions.

Step-by-Step Solution

1

Identify the classification of potash alum

Potash alum,

KAl(SO_4)_2 \cdot 12H_2O

, contains two different cations (

K^+

and

Al^{3+}

) and ionizes completely into simple ions in aqueous solution, which defines a double salt.

Double salts retain the chemical properties of their constituent individual salts when dissolved in water, unlike complex salts which form complex ions.

2

Determine the appropriate laboratory preparation method for lead(II) sulfate (

PbSO_4

)

Since

PbSO_4

is an insoluble salt, it is prepared by mixing solutions of two soluble salts containing

Pb^{2+}

and

SO_4^{2-}

ions respectively.

Insoluble salts are synthesized via precipitation (double decomposition) reactions.

Key Concept

Classification of salts (double salts vs complex salts) and preparation of insoluble salts via precipitation/double decomposition.

Estimated Time:1m 0s

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