All practice questions

Question 8241Question →

Which of the following reagents forms a characteristic precipitate when reacted with propyne, but produces no precipitate when reacted with propene?

Ammoniacal silver nitrate solution

Bromine water

Acidified potassium tetraoxomanganate(VII) solution

Concentrated tetraoxosulfate(VI) acid

Show answer & explanation

Answer: Ammoniacal silver nitrate solution

Answer

Ammoniacal silver nitrate solution

Ammoniacal silver nitrate solution contains complexed silver ions

[Ag(NH_3)_2]^+

which react selectively with the acidic hydrogen attached to the triply bonded carbon in terminal alkynes like propyne (

CH_3C\equiv CH

), forming a insoluble white precipitate of silver propynide. Propene lacks this acidic acetylenic hydrogen and gives no precipitate.

Step-by-Step Solution

1

Identify the structural difference between propyne and propene

Propyne (

CH_3C\equiv CH

) is a terminal alkyne containing a weakly acidic hydrogen attached to a triply-bonded carbon atom, whereas propene (

CH_3CH=CH_2

) is an alkene.

Terminal alkynes possess acidic acetylenic hydrogens (

R-C\equiv C-H

), unlike alkenes.

2

Evaluate the chemical reagents for specific reactivity with terminal acetylenic hydrogen

Ammoniacal silver nitrate solution (

[Ag(NH_3)_2]^+

) reacts with terminal alkynes to precipitate silver dicarbide/alkynide (a white precipitate). Alkenes do not undergo this substitution reaction.

Reagents like bromine water and acidified

KMnO_4

test for general double/triple bond unsaturation and react with both compounds.

Key Concept

Distinction between terminal alkynes and alkenes using ammoniacal silver nitrate solution

Estimated Time:45s

Question 8242Question →

A solution is formed by mixing $300\text{ cm}^3$ of $0.10\text{ mol dm}^{-3}$ tetraoxosulfate(VI) acid ( $\text{H}_2\text{SO}_4$ ) solution with $700\text{ cm}^3$ of $0.10\text{ mol dm}^{-3}$ sodium hydroxide ( $\text{NaOH}$ ) solution at $25^\circ\text{C}$ . Assuming complete dissociation of both electrolytes, what is the pH of the resulting mixture?

Show answer & explanation

Answer: 12

Answer

The pH of the resulting mixture is 12.0.

The mixture contains excess hydroxide ions (0.010 mol in 1.0 dm³ solution), resulting in a pOH of 2.0. Subtracting this from 14.0 gives a pH of 12.0.

Step-by-Step Solution

1

Calculate the moles of hydrogen ions (H⁺) contributed by the acid solution.

n(\text{H}^+) = 0.060\text{ mol}

Tetraoxosulfate(VI) acid is diprotic (dibasic), releasing 2 moles of

\text{H}^+

per mole of acid:

0.300\text{ dm}^3 \times 0.10\text{ mol dm}^{-3} \times 2 = 0.060\text{ mol}

.

2

Calculate the moles of hydroxide ions (OH⁻) contributed by the base solution.

n(\text{OH}^-) = 0.070\text{ mol}

Sodium hydroxide is a monobasic base:

0.700\text{ dm}^3 \times 0.10\text{ mol dm}^{-3} = 0.070\text{ mol}

.

3

Determine the unneutralized excess ions and calculate their molar concentration in the total volume.

[\text{OH}^-] = 0.010\text{ mol dm}^{-3}

The neutralization reaction is

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

. Excess

\text{OH}^- = 0.070 - 0.060 = 0.010\text{ mol}

. Divided by the total mixture volume of

1.0\text{ dm}^3

,

[\text{OH}^-] = 0.010\text{ mol dm}^{-3}

.

4

Calculate pOH and convert it to pH using the water autoionization relation.

\text{pH} = 12.0

\text{pOH} = -\log_{10}(0.010) = 2.0

. Since

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

at

25^\circ\text{C}

,

\text{pH} = 14.0 - 2.0 = 12.0

.

Key Concept

Neutralization stoichiometry of diprotic acids and strong bases followed by pH determination from excess hydroxide concentration.

Question 8243Question →

During the formation of the hydronium ion ( $H_3O^+$ ), a water molecule reacts with a proton ( $H^+$ ). Which condition must be satisfied by the oxygen atom in water to enable this coordinate (dative) bond formation?

It must possess at least one unshared lone pair of electrons available for donation.

It must completely transfer one of its valence electrons to the proton.

It must receive one electron from the proton to complete a shared electron pair.

It must expand its valence shell beyond eight electrons to accept the proton.

Show answer & explanation

Answer: It must possess at least one unshared lone pair of electrons available for donation.

Answer

It must possess at least one unshared lone pair of electrons available for donation.

A coordinate (dative) covalent bond is formed when a single donor atom provides both electrons of a shared pair to an acceptor species with an empty orbital. In the reaction of

H_2O

with

H^+

, the oxygen atom acts as the donor because it possesses unshared lone pairs of valence electrons.

Step-by-Step Solution

1

Analyze the electronic configuration of the reacting species.

The water molecule (

H_2O

) has two single covalent bonds and two unshared lone pairs on the oxygen atom, while the hydrogen ion (

H^+

) has an empty valence shell with no electrons.

Identifying the electron distribution helps determine how the bond is formed between the two species.

2

Apply the definition of coordinate (dative) covalent bonding.

Because

H^+

carries no electrons, the shared pair forming the new

O-H

bond must come entirely from one of the lone pairs on the oxygen atom.

A coordinate bond is formed when one atom provides both electrons of the shared bonding pair to an electron-deficient species.

Key Concept

Coordinate (Dative) Covalent Bonding

Question 8244Question →

In the industrial synthesis of ammonia via the Haber process, finely divided iron is employed as the catalyst. Which substance is added to the system as a promoter to enhance the catalytic activity of iron?

Aluminium oxide (

\text{Al}_2\text{O}_3

) combined with potassium oxide (

\text{K}_2\text{O}

)

Vanadium(V) oxide (

\text{V}_2\text{O}_5

)

Platinum-rhodium gauze

Finely divided nickel

Show answer & explanation

Answer: Aluminium oxide ( $\text{Al}_2\text{O}_3$ ) combined with potassium oxide ( $\text{K}_2\text{O}$ )

Answer

Aluminium oxide (

\text{Al}_2\text{O}_3

) combined with potassium oxide (

\text{K}_2\text{O}

)

In the Haber process, finely divided iron acts as the primary catalyst. Its activity and thermal durability are enhanced by adding small quantities of metal oxide promoters such as aluminium oxide (

\text{Al}_2\text{O}_3

) and potassium oxide (

\text{K}_2\text{O}

). Aluminium oxide prevents the iron particles from sintering (clumping together) at high operating temperatures, maintaining a high catalytic surface area.

Step-by-Step Solution

1

Identify the industrial chemical process described

The process described is the Haber process for the industrial manufacture of ammonia (

\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3

).

Understanding the specific process allows identification of the correct catalyst and promoter.

2

Distinguish between the main catalyst and the catalyst promoter

Finely divided iron (

\text{Fe}

) is the primary catalyst, while aluminium oxide (

\text{Al}_2\text{O}_3

) and potassium oxide (

\text{K}_2\text{O}

) are added as promoters.

Promoters are substances that increase the activity, thermal resistance, and operational lifespan of a catalyst without themselves acting as catalysts.

Key Concept

Industrial Catalyst Promoters in the Haber Process

Estimated Time:1m 0s

Question 8245Question →

A solid mixture contains insoluble barium sulfate ( $\text{BaSO}_4$ ) and soluble potassium trioxonitrate(V) ( $\text{KNO}_3$ ). Which of the following procedures should be used to recover pure crystals of $\text{KNO}_3$ from the mixture?

Add water to dissolve the mixture, filter off the residue, heat the filtrate until a saturated solution is formed, and allow it to cool slowly.

Add water to dissolve the mixture, filter off the residue, and heat the filtrate strongly until all water evaporates completely to dryness.

Add water to dissolve the mixture, collect the residue on the filter paper as potassium trioxonitrate(V), and discard the filtrate.

Heat the solid mixture directly to evaporate barium sulfate and leave pure potassium trioxonitrate(V) in the container.

Show answer & explanation

Answer: Add water to dissolve the mixture, filter off the residue, heat the filtrate until a saturated solution is formed, and allow it to cool slowly.

Answer

Add water to dissolve the mixture, filter off the residue, heat the filtrate until a saturated solution is formed, and allow it to cool slowly.

Dissolving the mixture in water solubilizes potassium trioxonitrate(V) while leaving barium sulfate insoluble. Filtering removes the solid barium sulfate as a residue. Heating the filtrate to saturation and cooling it slowly allows pure crystals of potassium trioxonitrate(V) to precipitate out.

Step-by-Step Solution

1

Separate insoluble solid from soluble solid using water.

Barium sulfate (

\text{BaSO}_4

) remains insoluble while potassium trioxonitrate(V) (

\text{KNO}_3

) dissolves completely.

Difference in solubility allows physical separation.

2

Filter the mixture.

Barium sulfate is collected on the filter paper as the residue, and the aqueous solution of potassium trioxonitrate(V) passes through as the filtrate.

Filtration separates insoluble solids from liquids.

3

Concentrate the filtrate to saturation and cool slowly.

Pure crystals of potassium trioxonitrate(V) precipitate out of solution.

Crystallization yields pure, well-formed crystals without thermal decomposition or loss of hydration structure.

Key Concept

Separation of soluble and insoluble solids by dissolution, filtration, and crystallization.

Estimated Time:1m 30s

Question 8246Question →

Arrange the following crude oil (petroleum) fractions in order of increasing boiling point range, starting from the fraction with the lowest boiling point to the one with the highest boiling point.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from lowest to highest boiling point range is Refinery gas, followed by Petrol (Gasoline), Kerosene (Paraffin), and Diesel oil (Gas oil).

In the fractional distillation of petroleum, fractions condense and exit the fractionating column at different levels depending on their boiling point ranges. Smaller alkane molecules have lower molecular masses, weaker intermolecular forces, and lower boiling points. Thus, Refinery gas (

C_1-C_4

) has the lowest boiling point, followed by Petrol (

C_5-C_{10}

), Kerosene (

C_{10}-C_{16}

), and Diesel oil (

C_{14}-C_{20}

) with the highest boiling point among the listed options.

Step-by-Step Solution

1

Determine the relationship between carbon chain length and boiling point in alkane fractions.

Smaller hydrocarbon molecules have weaker intermolecular van der Waals forces and therefore lower boiling points.

Boiling point increases as the number of carbon atoms per molecule increases.

2

Identify the carbon chain lengths for each given fraction.

Refinery gas (

C_1-C_4

), Petrol (

C_5-C_{10}

), Kerosene (

C_{10}-C_{16}

), and Diesel oil (

C_{14}-C_{20}

).

Fractional distillation separates crude oil based on boiling point ranges governed by molecular sizes.

3

Arrange the fractions from smallest carbon number to largest carbon number.

Refinery gas

\rightarrow

Petrol

\rightarrow

Kerosene

\rightarrow

Diesel oil.

This sequence directly corresponds to increasing boiling point range.

Key Concept

Fractional Distillation and Boiling Point Trends of Petroleum Fractions

Estimated Time:45s

Question 8247Question →

Ammonium trioxonitrate(V), $NH_4NO_3$ , decomposes on heating to produce dinitrogen(I) oxide and water vapor according to the balanced chemical equation:

NH_4NO_3(s) \rightarrow N_2O(g) + 2H_2O(g)

What volume of dinitrogen(I) oxide gas, measured at STP, is produced from the complete decomposition of $8.0\text{ g}$ of pure ammonium trioxonitrate(V)?
$[\text{Molar mass of } NH_4NO_3 = 80\text{ g mol}^{-1},\text{ Molar volume of gas at STP } = 22.4\text{ dm}^3\text{ mol}^{-1}]$

2.24\text{ dm}^3

2.40\text{ dm}^3

4.48\text{ dm}^3

6.72\text{ dm}^3

Show answer & explanation

Answer: $2.24\text{ dm}^3$

Answer

The volume of dinitrogen(I) oxide gas produced at STP is

2.24\text{ dm}^3

.

The correct response calculates the mole value of ammonium trioxonitrate(V) as

0.10\text{ mol}

(

8.0\text{ g} / 80\text{ g mol}^{-1}

). Using the

1:1

stoichiometric ratio from the balanced decomposition reaction,

0.10\text{ mol}

of dinitrogen(I) oxide is produced. At standard temperature and pressure (STP), one mole of any ideal gas occupies

22.4\text{ dm}^3

, giving

0.10 \times 22.4 = 2.24\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the number of moles of

NH_4NO_3

reacted

\text{Moles of } NH_4NO_3 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{8.0\text{ g}}{80\text{ g mol}^{-1}} = 0.10\text{ mol}

Converting given mass into moles is necessary to apply stoichiometric relationships.

2

Determine the mole ratio between

NH_4NO_3

and

N_2O

From the balanced equation,

1\text{ mol of } NH_4NO_3

produces

1\text{ mol of } N_2O

. Thus,

0.10\text{ mol of } NH_4NO_3

yields

0.10\text{ mol of } N_2O

.

Stoichiometry dictates the theoretical mole yield of the target product.

3

Calculate the volume of

N_2O

gas at STP

\text{Volume at STP} = 0.10\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 2.24\text{ dm}^3

Multiplying the mole amount of gas by the molar volume at STP (

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

) gives the volume.

Key Concept

Stoichiometric gas volume calculation at standard temperature and pressure (STP) for thermal decomposition reactions of nitrogen compounds.

Estimated Time:1m 30s

Question 8248Question →

In atmospheric chemistry, certain active species initiate catalytic cycles that destroy stratospheric ozone. Which of the following species acts as a direct catalyst in the destruction of stratospheric ozone?

Chlorine free radicals (

\text{Cl}^\bullet

)

Carbon(IV) oxide (

\text{CO}_2

)

Methane (

\text{CH}_4

)

Water vapour (

\text{H}_2\text{O}

)

Show answer & explanation

Answer: Chlorine free radicals ( $\text{Cl}^\bullet$ )

Answer

Chlorine free radicals (

\text{Cl}^\bullet

)

Chlorine free radicals participate directly in a catalytic cycle in the stratosphere, converting ozone molecules into oxygen molecules without being permanently consumed in the overall process.

Step-by-Step Solution

1

Identify the cause of catalytic ozone layer depletion in the stratosphere.

Ultraviolet light breaks down chlorofluorocarbon (CFC) molecules, releasing reactive chlorine free radicals (

\text{Cl}^\bullet

).

Photolysis produces reactive atomic chlorine species in the upper atmospheric layers.

2

Examine the catalytic reaction step.

Chlorine radicals react with ozone:

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

, depleting the ozone layer.

The chlorine radical is regenerated during subsequent reactions, allowing a single radical to break down thousands of ozone molecules.

Key Concept

Catalytic Role of Chlorine Free Radicals in Stratospheric Ozone Depletion

Estimated Time:45s

Question 8249Question →

A river water sample collected for a township supply system is found to contain fine suspended clay particles, objectionable taste caused by decomposing organic matter, and pathogenic bacteria, but has negligible dissolved calcium and magnesium salts. Which combination of chemical reagents must be utilized during the treatment process to render this water clean, odorless, and biologically safe?

Potash alum, activated charcoal, and chlorine

Slaked lime, washing soda, and chlorine

Potash alum, calcium chloride, and slaked lime

Sodium carbonate, activated charcoal, and boiling water

Show answer & explanation

Answer: Potash alum, activated charcoal, and chlorine

Answer

Potash alum, activated charcoal, and chlorine form the correct combination of reagents for purifying soft, turbid, contaminated surface water for public distribution.

The correct answer combines potash alum, activated charcoal, and chlorine because each chemical directly addresses one of the identified raw water issues: potash alum coagulates colloidal clay particles into larger flocs during sedimentation; activated charcoal adsorbs dissolved organic matter causing unpleasant tastes and smells; and chlorine kills disease-causing bacteria during final disinfection.

Step-by-Step Solution

1

Analyze the specific impurities present in the raw water sample

Impurities identified: fine suspended colloidal clay (requires coagulation), organic taste/odor compounds (requires adsorption), and pathogenic microbes (requires disinfection/sterilization). Hardness minerals are absent.

Different water treatment chemicals address distinct types of physical, chemical, and biological contamination.

2

Select the appropriate chemical reagent for colloidal suspension removal

Potash alum (

KAl(SO_4)_2 \cdot 12H_2O

) provides

Al^{3+}

ions to neutralize negative charges on clay particles, causing coagulation into settleable flocs.

Physical filtration alone cannot remove extremely fine colloidal clay without prior chemical coagulation.

3

Select the reagent for taste and odor removal

Activated charcoal (carbon) adsorbs volatile organic compounds responsible for objectionable smells and tastes.

Porous activated carbon has a very large surface area optimized for organic molecule adsorption.

4

Select the chemical required for biological safety

Chlorine (or chlorine compounds) acts as a powerful oxidizing disinfectant to destroy disease-causing bacteria.

Water for public distribution must undergo chemical sterilization to eliminate waterborne pathogens.

Key Concept

Municipal water treatment requires distinct functional reagents: coagulants (potash alum) for suspended particles, adsorbents (activated carbon) for taste/odor, and disinfectants (chlorine) for sterilization. Softening reagents (slaked lime/washing soda) are only needed when hardness ions (

Ca^{2+}

,

Mg^{2+}

) are present.

Question 8250Question →

A $0.88\text{ g}$ sample of a purified gas in a laboratory experiment occupies a volume of $400\text{ cm}^3$ at $27^\circ\text{C}$ and a pressure of $1.23\text{ atm}$ . Given that the gas constant $R = 0.082\text{ dm}^3\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$ , what is the molar mass of the gas?

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

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

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

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

Show answer & explanation

Answer: $44\text{ g mol}^{-1}$

Answer

The molar mass of the gas is

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

.

Converting all parameters to standard SI/gas units yields

V = 0.40\text{ dm}^3

and

T = 300\text{ K}

. Substituting these into the ideal gas equation

n = \frac{PV}{RT}

gives

n = 0.020\text{ mol}

. Dividing the sample mass (

0.88\text{ g}

) by the calculated number of moles gives the correct molar mass of

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

.

Step-by-Step Solution

1

Convert given values into appropriate units matching the gas constant

R

V = \frac{400}{1000} = 0.40\text{ dm}^3

,

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

,

P = 1.23\text{ atm}

The ideal gas law requires temperature in Kelvin and volume in

\text{dm}^3

to match the units of

R

.

2

Calculate the number of moles (

n

) using the ideal gas equation

PV = nRT

n = \frac{PV}{RT} = \frac{1.23 \times 0.40}{0.082 \times 300} = \frac{0.492}{24.6} = 0.020\text{ mol}

Rearranging

PV = nRT

allows direct computation of gas moles.

3

Compute the molar mass (

M

) using

M = \frac{\text{mass}}{n}

M = \frac{0.88\text{ g}}{0.020\text{ mol}} = 44\text{ g mol}^{-1}

Molar mass is the ratio of given mass in grams to the amount of substance in moles.

Key Concept

Ideal Gas Equation and Molar Mass Determination

Question 8251Question →

Consider the exothermic reaction $2\text{NO}_2\text{(g)} \rightleftharpoons \text{N}_2\text{O}_4\text{(g)}$ , where the potential energy of the reactants is $180\text{ kJ mol}^{-1}$ and that of the products is $70\text{ kJ mol}^{-1}$ . If the uncatalyzed forward activation energy is $90\text{ kJ mol}^{-1}$ and a catalyst lowers the activation energy of the forward reaction by $35\text{ kJ mol}^{-1}$ , what is the activation energy for the catalyzed reverse reaction?

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

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

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

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

Show answer & explanation

Answer: $165\text{ kJ mol}^{-1}$

Answer

The activation energy for the catalyzed reverse reaction is

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

.

A catalyst lowers the energy barrier peak for both forward and reverse paths by the same magnitude (

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

). Starting with reactants at

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

and a catalyzed forward activation energy of

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

, the transition state peak is located at

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

. Since the products lie at

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

, the energy required for the reverse reaction to reach the transition state peak is

235 - 70 = 165\text{ kJ mol}^{-1}

.

Step-by-Step Solution

1

Calculate the uncatalyzed transition state peak energy

E_{\text{peak, uncatalyzed}} = 180\text{ kJ mol}^{-1} + 90\text{ kJ mol}^{-1} = 270\text{ kJ mol}^{-1}

The peak energy equals the reactant energy plus the forward activation energy.

2

Calculate the catalyzed transition state peak energy

E_{\text{peak, catalyzed}} = 270\text{ kJ mol}^{-1} - 35\text{ kJ mol}^{-1} = 235\text{ kJ mol}^{-1}

A catalyst lowers the transition state energy barrier by

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

.

3

Determine the catalyzed reverse activation energy

E_{a, \text{reverse, catalyzed}} = 235\text{ kJ mol}^{-1} - 70\text{ kJ mol}^{-1} = 165\text{ kJ mol}^{-1}

The reverse activation energy is the difference between the catalyzed peak energy and the energy level of the products.

Key Concept

Activation Energy and Catalyzed Energy Profiles

Estimated Time:2m 0s

Question 8252Question →

An aqueous solution at $25^\circ\text{C}$ has a hydrogen ion concentration, $[\text{H}^+]$ , of $1.0 \times 10^{-5}\text{ mol dm}^{-3}$ . What is the pOH of this solution?

5.0

9.0

14.0

1.0 \times 10^{-9}\text{ mol dm}^{-3}

Show answer & explanation

Answer: $9.0$

Answer

The pOH of the solution is

9.0

.

First, find the pH using

\text{pH} = -\log_{10}([\text{H}^+]) = -\log_{10}(1.0 \times 10^{-5}) = 5.0

. Then, apply the relationship

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

at

25^\circ\text{C}

to calculate

\text{pOH} = 14.0 - 5.0 = 9.0

. Thus, the option specifying

9.0

is correct.

Step-by-Step Solution

1

Calculate the pH of the solution from the given hydrogen ion concentration

\text{pH} = -\log_{10}([\text{H}^+]) = -\log_{10}(1.0 \times 10^{-5}) = 5.0

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

2

Use the relationship between pH and pOH at

25^\circ\text{C}

to find pOH

\text{pOH} = 14.0 - \text{pH} = 14.0 - 5.0 = 9.0

For any aqueous solution at 25°C,

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

.

Key Concept

Relationship between pH, pOH, and ion concentrations in aqueous solutions at 25°C

Estimated Time:45s

Question 8253Question →

Consider four main group elements $P$ , $Q$ , $R$ , and $S$ with atomic numbers $11$ , $12$ , $16$ , and $17$ respectively. Which combination of these elements forms an electrovalent compound with the highest melting point?

The combination of

Q

and

S

, because the

+2

and

-2

ionic charges maximize the electrostatic lattice attraction.

The combination of

P

and

R

, because single electron transfer minimizes interionic repulsion within the crystal lattice.

The combination of

Q

and

R

, because a

1:2

stoichiometric ratio produces the highest crystal packing density.

The combination of

P

and

S

, because two

P^+

cations double the total electrostatic pull on a single

S^{2-}

anion.

Show answer & explanation

Answer: The combination of $Q$ and $S$ , because the $+2$ and $-2$ ionic charges maximize the electrostatic lattice attraction.

Answer

The combination of element Q (atomic number 12) and element S (atomic number 16) forms QS, which has the highest melting point due to the +2 and -2 ionic charges maximizing electrostatic lattice attraction.

Element Q (atomic number 12) has electronic configuration 2,8,2 and loses two electrons to form Q2+. Element S (atomic number 16) has electronic configuration 2,8,6 and gains two electrons to form S2-. The compound formed between Q and S (QS) consists of divalent ions. By Coulomb's Law, lattice energy is proportional to the product of ionic charges (|q1 * q2|). The charge product for QS is 4, which is double that of QR2 or P2S (charge product 2) and four times that of PR (charge product 1). Consequently, QS possesses the highest lattice energy and highest melting point.

Step-by-Step Solution

1

Determine the identity and valency of each element from its atomic number

P

(

Z=11

, Sodium) forms

P^+

cations;

Q

(

Z=12

, Magnesium) forms

Q^{2+}

cations;

S

(

Z=16

, Sulfur) forms

S^{2-}

anions;

R

(

Z=17

, Chlorine) forms

R^-

anions.

Electronic configurations determine the number of valence electrons lost or gained to achieve stable octet structures.

2

Write the chemical formulas for the electrovalent compounds formed by valid metal-nonmetal pairs

Possible ionic compounds are

PR

(

P^+ R^-

),

P_2S

(

(P^+)_2 S^{2-}

),

QR_2

(

Q^{2+} (R^-)_2

), and

QS

(

Q^{2+} S^{2-}

).

Electrovalent compounds form when metals transfer electrons to non-metals to achieve electrical neutrality.

3

Compare the electrostatic lattice energies of the resulting crystal lattices

Lattice energy is directly proportional to the product of ionic charges (

E_{\text{lattice}} \propto |q_1 q_2|

). For

QS

,

|q_1 q_2| = |(+2)(-2)| = 4

. For

QR_2

and

P_2S

,

|q_1 q_2| = 2

. For

PR

,

|q_1 q_2| = 1

.

According to Coulomb's Law, higher ionic charges create substantially stronger electrostatic forces of attraction between ions in the solid lattice.

4

Relate lattice energy to the physical property of melting point

Higher lattice energy requires significantly more thermal energy to break the ionic bonds, making

QS

the compound with the highest melting point.

The melting point of an electrovalent compound increases as the strength of the lattice attraction increases.

Key Concept

Lattice energy dependence on ionic charge magnitude (Coulomb's Law in ionic crystals)

Question 8254Question →

Match each feature of a chemical reaction energy profile diagram on the left with its correct description or definition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Activation Energy (

E_a

) matches the minimum energy difference between the reactants and the peak of the energy barrier; Activated Complex matches the unstable, high-energy arrangement at the maximum point of the energy curve; Enthalpy Change (

\Delta H

) matches the net energy difference between products and reactants; Effect of a Catalyst matches lowering the potential energy barrier peak.

Each feature of an energy profile diagram describes a specific thermodynamic or kinetic aspect: Activation Energy measures the hurdle from reactants to the transition state; Activated Complex is the transient species at the barrier apex; Enthalpy Change is the net heat change from reactants to products; and a Catalyst reduces the activation barrier height.

Step-by-Step Solution

1

Identify the definition of Activation Energy (

E_a

)

It corresponds to the energy gap between reactant potential energy and the peak energy.

Reactants must absorb this energy threshold to initiate bond transformation.

2

Identify the Activated Complex (Transition State)

It is located at the peak of the curve where potential energy is at its maximum.

This state is highly unstable and short-lived as bonds are actively breaking and forming.

3

Identify the Enthalpy Change (

\Delta H

)

It represents the overall potential energy gap between products and reactants.

\Delta H = E_{\text{products}} - E_{\text{reactants}}

shows whether a reaction is exothermic or endothermic.

4

Identify the role of a Catalyst in the profile diagram

It decreases the height of the transition state peak.

Catalysts lower activation energy without shifting initial reactant or final product energy levels.

Key Concept

Key parameters of reaction energy profile diagrams

Estimated Time:1m 0s

Question 8255Question →

An aerosol spray can contains a fixed mass of gas at a pressure of $100\text{ kPa}$ and a temperature of $17^\circ\text{C}$ . If the volume of the container remains constant, what is the pressure inside the can when its temperature increases to $307^\circ\text{C}$ ?

200\text{ kPa}

106\text{ kPa}

1806\text{ kPa}

50\text{ kPa}

Show answer & explanation

Answer: $200\text{ kPa}$

Answer

The pressure inside the can when heated to

307^\circ\text{C}

is

200\text{ kPa}

.

The correct answer is

200\text{ kPa}

. According to Gay-Lussac's Pressure Law, for a gas at constant volume, pressure is directly proportional to absolute temperature in Kelvin. Converting both temperatures to Kelvin gives

T_1 = 290\text{ K}

and

T_2 = 580\text{ K}

. Since absolute temperature doubles, the pressure also doubles from

100\text{ kPa}

to

200\text{ kPa}

.

Step-by-Step Solution

1

Convert given temperatures from Celsius to Kelvin.

T_1 = 17^\circ\text{C} + 273 = 290\text{ K}

and

T_2 = 307^\circ\text{C} + 273 = 580\text{ K}

.

Gas law equations require absolute temperatures in Kelvin.

2

Apply Gay-Lussac's Pressure Law formula at constant volume.

\frac{P_1}{T_1} = \frac{P_2}{T_2} \implies P_2 = P_1 \times \frac{T_2}{T_1}

.

Pressure of a fixed mass of gas is directly proportional to its absolute temperature.

3

Substitute the values and calculate the final pressure

P_2

.

P_2 = 100\text{ kPa} \times \frac{580\text{ K}}{290\text{ K}} = 100\text{ kPa} \times 2 = 200\text{ kPa}

.

Multiplying the initial pressure by the ratio of absolute temperatures gives the final pressure.

Key Concept

Pressure Law (Gay-Lussac's Law of Temperature-Pressure)

Question 8256Question →

A sample of nitrogen gas is collected over water at a total pressure of $760\text{ mmHg}$ . If the saturated vapor pressure of water at the collection temperature is $24\text{ mmHg}$ , what is the partial pressure of the dry nitrogen gas?

736\text{ mmHg}

784\text{ mmHg}

760\text{ mmHg}

24\text{ mmHg}

Show answer & explanation

Answer: $736\text{ mmHg}$

Answer

The partial pressure of the dry nitrogen gas is

736\text{ mmHg}

.

According to Dalton's Law of Partial Pressures, the total pressure of a gas collected over water is the sum of the partial pressure of the dry gas and the vapor pressure of water (

P_{\text{total}} = P_{\text{dry gas}} + P_{\text{water}}

). To find the pressure of the dry gas alone, subtract the saturated water vapor pressure from the total pressure:

760\text{ mmHg} - 24\text{ mmHg} = 736\text{ mmHg}

.

Step-by-Step Solution

1

State Dalton's Law of Partial Pressures for a gas collected over water.

P_{\text{total}} = P_{\text{dry gas}} + P_{\text{H}_2\text{O}}

When a gas is collected over water, the measured total pressure is the sum of the partial pressure of the dry gas and the saturated vapor pressure of water.

2

Rearrange the equation to solve for the partial pressure of the dry nitrogen gas.

P_{\text{dry gas}} = P_{\text{total}} - P_{\text{H}_2\text{O}}

To isolate the pressure exerted by the dry gas, the water vapor pressure (aqueous tension) must be subtracted from the total pressure.

3

Substitute the given values into the formula and calculate.

P_{\text{dry gas}} = 760\text{ mmHg} - 24\text{ mmHg} = 736\text{ mmHg}

Performing simple subtraction gives the final partial pressure of dry nitrogen.

Key Concept

Dalton's Law of Partial Pressures and Collection of Gas over Water

Question 8257Question →

Match each physical or chemical process on the left with its corresponding entropy change ( $\Delta S$ ) characteristic on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Each process is matched based on the change in molecular disorder: dissolving solid NaCl produces mobile hydrated ions (positive entropy change), condensing water vapor transitions gas to liquid (negative entropy change), subliming dry ice transitions solid to gas (positive entropy change), and synthesizing ammonia decreases gaseous moles from 4 to 2 (negative entropy change).

Entropy (

\Delta S

) is a measure of randomness or molecular disorder. Phase transitions from solid to aqueous or solid to gas increase disorder (

\Delta S > 0

). Phase transitions from gas to liquid or reactions that reduce the number of gas moles decrease disorder (

\Delta S < 0

).

Step-by-Step Solution

1

Analyze state changes and particle freedom for physical processes.

Solid to aqueous ions increases disorder (

\Delta S > 0

). Gas to liquid decreases disorder (

\Delta S < 0

). Solid to gas increases disorder (

\Delta S > 0

).

Gases have the highest entropy, followed by aqueous solutions, liquids, and solids.

2

Analyze gas mole stoichiometry for chemical reactions involving gases.

For

\text{N}_{2(g)} + 3\text{H}_{2(g)} \rightarrow 2\text{NH}_{3(g)}

, reactant gas moles =

1 + 3 = 4

, product gas moles =

2

. Since gas moles decrease,

\Delta S < 0

.

A decrease in the number of gaseous particles decreases the available microstates of the system.

Key Concept

Predicting the sign of entropy change (

\Delta S

) from physical state transitions and changes in total moles of gas

Question 8258Question →

Match each common calcium compound on the left with its correct chemical name, formula, and primary application on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Quicklime matches Calcium oxide (

\text{CaO}

); Slaked lime matches Calcium hydroxide (

\text{Ca(OH)}_2

); Gypsum matches Calcium tetraoxosulfate(VI) dihydrate (

\text{CaSO}_4 \cdot 2\text{H}_2\text{O}

); Plaster of Paris matches Calcium tetraoxosulfate(VI) hemihydrate (

\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}

).

Each calcium compound is accurately paired with its chemical formula and practical use: Quicklime (

\text{CaO}

) as a drying agent for basic gases, Slaked lime (

\text{Ca(OH)}_2

) for soil liming, Gypsum (

\text{CaSO}_4 \cdot 2\text{H}_2\text{O}

) for retarding cement setting, and Plaster of Paris (

\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}

) for surgical casts.

Step-by-Step Solution

1

Identify the chemical composition of Quicklime and Slaked lime

Quicklime is

\text{CaO}

(calcium oxide) and Slaked lime is

\text{Ca(OH)}_2

(calcium hydroxide).

Thermal decomposition of limestone (

\text{CaCO}_3

) yields

\text{CaO}

, which reacts with water to form

\text{Ca(OH)}_2

.

2

Distinguish between Gypsum and Plaster of Paris based on hydration state

Gypsum contains two water molecules per sulfate unit (

\text{CaSO}_4 \cdot 2\text{H}_2\text{O}

), whereas Plaster of Paris contains half a water molecule per sulfate unit (

\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}

).

Heating gypsum to around

120^\circ\text{C}

removes three-quarters of its crystallization water to form Plaster of Paris.

3

Correlate each compound with its characteristic industrial application

Quicklime dries ammonia gas, Slaked lime neutralizes acidic soil, Gypsum regulates cement setting time, and Plaster of Paris forms orthopedic casts.

Chemical properties direct specific industrial uses as specified in standard JAMB UTME chemistry syllabus guidelines.

Key Concept

Nomenclature, formulas, and practical applications of major calcium compounds.

Estimated Time:45s

Question 8259Question →

During the fractional distillation of a miscible mixture of volatile liquids, a thermal gradient is established along the fractionating column. As vapor ascends from the bottom to the top of the column, which of the following correctly describes the changes in temperature and vapor composition?

The temperature decreases, and the vapor becomes increasingly enriched in the component with the lower boiling point.

The temperature decreases, and the vapor becomes increasingly enriched in the component with the higher boiling point.

The temperature increases, and the vapor becomes increasingly enriched in the component with the lower boiling point.

The temperature increases, and the vapor becomes increasingly enriched in the component with the higher boiling point.

Show answer & explanation

Answer: The temperature decreases, and the vapor becomes increasingly enriched in the component with the lower boiling point.

Answer

The temperature decreases from bottom to top along the column, while the ascending vapor becomes increasingly enriched in the component with the lower boiling point (more volatile liquid).

In a fractionating column, the base nearest the boiling liquid is hottest and the top near the thermometer and condenser outlet is coolest. As mixed vapors ascend, continuous condensation and re-evaporation occur along the packing material (glass beads). The component with the higher boiling point condenses preferentially on the cooler surfaces and returns down the column as reflux, whereas the component with the lower boiling point remains in the vapor state and concentrates at the top to exit into the condenser.

Step-by-Step Solution

1

Analyze the temperature profile along the fractionating column.

The heat source is at the bottom (distillation flask), so the temperature is highest at the base and decreases continuously toward the top of the column.

Heat dissipates as vapor moves further away from the boiling liquid flask.

2

Determine the condensation and evaporation behavior of the mixture components.

The component with the higher boiling point has a lower vapor pressure (is less volatile) and condenses easily on the cool packing surfaces, dripping back down into the flask as liquid reflux.

Higher boiling point substances require higher thermal energy to stay in the gaseous phase.

3

Evaluate the composition of the vapor reaching the top of the column.

Repeated condensation-evaporation cycles along the glass beads enrich the ascending vapor in the component with the lower boiling point (more volatile component), which distills over first into the condenser.

The more volatile component evaporates more readily at lower temperatures established at the upper region of the column.

Key Concept

Temperature Gradient and Vapor Enrichment in Fractional Distillation

Estimated Time:1m 30s

Question 8260Question →

A laboratory technician needs to separate a dry solid mixture containing iron filings, ammonium chloride ( $NH_4Cl$ ), and fine sand. In what correct sequential order should the steps below be carried out from first to last?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of steps is: first, pass a strong magnet to remove the iron filings; second, heat the remaining mixture to sublime ammonium chloride (

NH_4Cl

); third, collect the non-volatile sand residue.

The process begins with magnetization to extract ferromagnetic iron filings. Next, heating the remaining mixture causes ammonium chloride (

NH_4Cl

) to sublime directly into vapor, leaving fine sand behind as the non-volatile residue.

Step-by-Step Solution

1

Perform magnetization on the dry mixture.

Iron filings adhere to the magnet and are completely removed from the container.

Iron is ferromagnetic and can be physically isolated without affecting the other non-magnetic solids.

2

Apply heat to the remaining mixture using an inverted funnel collector.

Ammonium chloride (

NH_4Cl

) sublimes into gas and re-crystallizes as solid deposit on the cool funnel surface.

Ammonium chloride readily sublimes upon heating, separating it cleanly from sand.

3

Isolate the remaining solid in the dish.

Pure sand remains in the evaporating dish.

Sand does not sublime under normal heating conditions and is non-magnetic, making it the final isolated component.

Key Concept

Sequential separation of solid mixtures by exploiting differences in magnetic property (magnetization) and thermal volatility (sublimation).

Answer

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