All practice questions

Question 8521Question →

In chemical manufacturing and hydrometallurgy, specific separation methods are chosen according to the physicochemical properties of the mixtures involved. Match each industrial separation technique listed on the left with its corresponding operational principle and application on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Froth flotation matches the separation of hydrophobic mineral ores based on surface wettability. Vacuum distillation matches the separation of heavy petroleum residues at reduced pressure to prevent thermal cracking. Solvent extraction matches the selective recovery of metal ions into an immiscible organic phase. Precipitative crystallization matches the separation of solid

NaHCO_3

in the Solvay process by exploiting solubility limits.

Each technique is matched to its primary industrial application based on physical/chemical properties: Froth flotation relies on surface wettability differences; Vacuum distillation uses reduced pressure to lower boiling points of thermal-sensitive heavy oil; Solvent extraction uses differential solubility in immiscible liquid phases for hydrometallurgical metal purification; Precipitative crystallization relies on solubility thresholds to precipitate

NaHCO_3

in the Solvay process.

Step-by-Step Solution

1

Analyze Froth Flotation

Froth flotation utilizes surfactants and air bubbles to selectively attach to hydrophobic ore particles (e.g.,

ZnS

), floating them away from hydrophilic waste rock (gangue).

Difference in surface wettability is the core physical property utilized in flotation.

2

Analyze Vacuum Distillation

Heavy petroleum crude fractions decompose at high temperatures required for boiling at atmospheric pressure; reducing pressure lowers boiling points.

Lowering ambient pressure allows distillation below thermal decomposition (cracking) thresholds.

3

Analyze Solvent Extraction

In hydrometallurgy, metal ions in dilute aqueous solution are extracted into an immiscible organic liquid containing specific chelating ligands.

Different solubilities and partition coefficients drive transfer between immiscible liquid phases.

4

Analyze Precipitative Crystallization

In the Solvay process, passing carbon(IV) oxide into ammoniacal brine forms

NaHCO_3

, which precipitates due to low solubility relative to

NH_4Cl

.

Temperature-controlled crystallization isolates solid products from soluble reaction side-products.

Key Concept

Industrial Applications of Separation Methods

Question 8522Question →

Match each physical property or characteristic of ionic (electrovalent) bonding on the left with its correct microscopic or structural explanation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

High melting and boiling points match with the need for extensive thermal energy to disrupt the giant 3D lattice; Electrical conductivity in molten/aqueous state matches with lattice destruction freeing mobile charge carriers; Solubility in water matches with hydration energy overcoming lattice energy; Non-directional nature matches with the electrostatic field acting uniformly in all directions.

Each property of ionic compounds directly stems from its underlying electrostatic structure: High melting points are caused by the strong 3D electrostatic attractions requiring high thermal energy to break; Electrical conductivity in molten/dissolved states occurs because ions are set free as mobile charge carriers; Solubility in water occurs when hydration energy exceeds lattice energy; Non-directionality arises because an ion's electrostatic field attracts opposite charges equally in all spatial directions.

Step-by-Step Solution

1

Analyze the high melting/boiling points of ionic compounds.

Recognize that ions are held in a giant lattice by strong electrostatic forces in all dimensions, requiring high heat energy to overcome.

Relates macro property (melting point) to micro structure (lattice binding energy).

2

Examine the electrical conduction mechanism in ionic substances.

In solid state, ions are fixed in lattice positions. Melting or dissolving releases these ions as mobile charge carriers.

Conduction requires free charge carriers, which are absent in solid ionic crystals.

3

Evaluate the dissolution of ionic compounds in polar solvents.

Polar water molecules surround separated ions (solvation/hydration), releasing energy that overcomes the lattice energy holding the crystal together.

Solubility depends on the thermodynamic balance between hydration enthalpy and lattice enthalpy.

4

Assess the directional nature of ionic bonding.

Because electrostatic attraction operates spherically in space, ionic bonds have no preferred angle or directional vector.

Charges attract equally in all directions, unlike localized shared electron pairs in covalent bonds.

Key Concept

Physical Properties and Structural Basis of Electrovalent (Ionic) Bonding

Question 8523Question →

In the reaction between phosphine ( $PH_3$ ) and a hydrogen ion ( $H^+$ ) to form a phosphonium ion ( $PH_4^+$ ), a dative covalent bond is formed. Which of the following statements correctly describes the electronic transfer mechanism during this bond formation?

PH_3

donates an unshared electron pair to

H^+

, which provides an empty orbital.

H^+

donates an electron pair to

PH_3

to complete the phosphorus valence shell.

Both

PH_3

and

H^+

contribute one valence electron each to form the shared pair.

PH_3

completely transfers one electron to

H^+

, forming an electrovalent bond.

Show answer & explanation

Answer: $PH_3$ donates an unshared electron pair to $H^+$ , which provides an empty orbital.

Answer

Phosphine (

PH_3

) donates an unshared electron pair to the hydrogen ion (

H^+

), which provides an empty orbital.

The phosphorus atom in phosphine (

PH_3

) has a non-bonding lone pair of valence electrons. The hydrogen ion (

H^+

) possesses an empty 1s orbital. When

PH_3

reacts with

H^+

, the lone pair on phosphorus is donated into the empty orbital of

H^+

, forming a dative (coordinate) covalent bond where both shared electrons originate from phosphine.

Step-by-Step Solution

1

Identify the valence shell electron structure of the reactants

Phosphorus (Group 15) in

PH_3

forms three single covalent bonds with hydrogen atoms and retains one unshared lone pair of electrons. The hydrogen ion (

H^+

) has lost its electron, leaving an empty 1s orbital.

Determining available lone pairs and empty orbitals is necessary to identify electron donor and acceptor species.

2

Apply the concept of coordinate (dative) covalent bonding

A dative covalent bond occurs when one atom/molecule provides both electrons of the shared pair (the Lewis base/donor), and another atom/ion accepts the pair into an empty orbital (the Lewis acid/acceptor).

This distinguishes dative bonding from normal covalent bonding where each participating atom contributes one electron.

3

Deduce the specific roles of

PH_3

and

H^+

PH_3

serves as the lone pair donor and

H^+

serves as the electron pair acceptor, forming the

[PH_4]^+

ion.

Matching the electronic structures to donor/acceptor definitions confirms the correct mechanism.

Key Concept

Coordinate (Dative) Covalent Bonding and Electron Pair Donation

Estimated Time:1m 0s

Question 8524Question →

Which of the following statements correctly describes a chemical reaction system that has attained dynamic equilibrium in a closed vessel?

The rates of the forward and reverse reactions are equal, and the concentrations of reactants and products remain constant.

The addition of a positive catalyst increases the net equilibrium yield of the forward reaction products.

All chemical activity ceases completely once the concentrations of reactants and products become equal.

The total molar concentration of the products must equal the total molar concentration of the reactants.

Show answer & explanation

Answer: The rates of the forward and reverse reactions are equal, and the concentrations of reactants and products remain constant.

Answer

The correct statement is that the rates of the forward and reverse reactions are equal, while the concentrations of reactants and products remain constant.

Dynamic chemical equilibrium is defined by equal rates of the forward and reverse reactions in a closed system. Because reactants are converted to products at the exact same speed that products revert back to reactants, the macroscopic concentrations of all participating species remain constant over time.

Step-by-Step Solution

1

Define dynamic equilibrium in chemical systems.

Dynamic equilibrium occurs in a closed system when a reversible reaction proceeds in opposite directions at identical speeds.

Understanding the dynamic nature distinguishes chemical equilibrium from static equilibrium.

2

Analyze macroscopic versus microscopic behavior.

Microscopically, forward and reverse processes continuously take place; macroscopically, measurable parameters such as concentration, pressure, and color remain unvarying.

Equal rates mean the formation rate of any species equals its consumption rate.

3

Evaluate option statements against dynamic equilibrium criteria.

The statement asserting equal reaction rates and constant concentrations aligns perfectly with dynamic equilibrium principles.

This directly satisfies the fundamental definition.

Key Concept

Characteristics of Dynamic Chemical Equilibrium

Question 8525Question →

Match each noble gas on the left with its corresponding primary application or characteristic property on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Helium matches with filling weather balloons and deep-sea diving gas mixtures; Neon matches with advertising signs producing reddish-orange glow; Argon matches with inert shielding in arc welding and light bulbs; Radon matches with radioactive cancer treatment.

Each noble gas possesses specific physical characteristics: Helium's light weight and low solubility suit balloons and diving gas; Neon's electrical discharge color suits signage; Argon's abundance and chemical inertness suit welding and lighting; Radon's radioactivity suits cancer treatment.

Step-by-Step Solution

1

Identify the key physical and chemical properties of each Group 0 (noble gas) element.

Helium is the lightest non-flammable gas with minimal blood solubility; Neon exhibits characteristic light emission; Argon is an abundant inert gas; Radon is radioactive.

Matching noble gases requires aligning their unique electronic stability and physical properties with industrial and medical uses.

2

Pair each gas to its correct industrial or medical application.

Helium pairs with weather balloons/diving gas; Neon pairs with advertising glow lamps; Argon pairs with welding/bulbs; Radon pairs with radiotherapy.

Each application specifically relies on the unique physical state or reactivity profile of that element.

Key Concept

Noble gases are unreactive Group 8/0 elements with stable octet (or duplet) electron configurations whose distinct physical properties dictate specific industrial and medical uses.

Question 8526Question →

For the endothermic gaseous reaction $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$ at dynamic equilibrium in a closed vessel, decreasing the container volume at constant temperature increases the equilibrium constant $K_c$ because the system shifts toward the side with fewer moles of gas.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Changing the container volume or pressure alters the equilibrium position (partial pressures/concentrations of species), but the equilibrium constant

K_c

is solely dependent on temperature.

The assertion is false because temperature is the only parameter capable of changing the value of the equilibrium constant

K_c

. Adjusting container volume alters total system pressure, which causes a shift in the position of equilibrium to re-establish the same

K_c

value.

Step-by-Step Solution

1

Analyze the impact of volume reduction on pressure and equilibrium position

Decreasing container volume increases the concentration and partial pressure of all gaseous species, causing a shift toward the side with fewer moles of gas (left side, 1 mole of

\text{N}_2\text{O}_4

vs 2 moles of

\text{NO}_2

).

Le Chatelier's principle dictates that a system at equilibrium will counteract an applied stress (increased pressure).

2

Determine the effect of volume/pressure changes on the equilibrium constant

K_c

The equilibrium constant

K_c

remains unchanged.

The equilibrium constant

K_c

is a thermodynamic property defined by temperature alone. Changes in pressure or volume shift reactant/product ratios back to the exact ratio satisfying the same

K_c

value.

Key Concept

Independence of Equilibrium Constant (

K_c

) from Pressure and Volume Changes

Question 8527Question →

Match each industrial separation requirement on the left with the corresponding specialized separation technique employed in chemical industries on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings link sulfide ore concentration with froth flotation, stack gas particle removal with electrostatic precipitation, atmospheric gas separation with fractional distillation of liquid air, and heavy oil refining without thermal cracking with vacuum distillation.

Each industrial process is accurately paired with its governing separation method: mineral beneficiation uses froth flotation, particulate air pollution control uses electrostatic precipitation, bulk atmospheric gas isolation uses liquid air fractional distillation, and heat-sensitive hydrocarbon refining uses vacuum distillation.

Step-by-Step Solution

1

Analyze the requirement for ore concentration

Sulfide minerals preferentially adhere to oil-coated air bubbles while water wets the gangue, floating the ore to the surface via froth flotation.

Beneficiation of solid ore slurries relies on differences in surface tension and hydrophobic properties.

2

Analyze flue gas purification in industrial chimneys

Particulate matter gains electric charge in an electrostatic precipitator and migrates to oppositely charged electrode plates.

Removing fine particulate aerosols from high-velocity chimney gas requires non-mechanical electrical attraction.

3

Analyze atmospheric gas separation

Liquefied air undergoes fractional distillation where nitrogen vaporizes first at

-196^\circ\text{C}

, leaving liquid oxygen behind at

-183^\circ\text{C}

.

Gaseous components of liquefied air are isolated according to their distinct boiling points.

4

Analyze heavy petroleum residue refining

Vacuum distillation reduces boiling points by lowering vessel pressure, preventing thermal breakdown of long-chain hydrocarbons.

High-molecular-weight fractions decompose if heated to their normal boiling points at atmospheric pressure.

Key Concept

Industrial separation techniques exploit distinct physical properties such as surface wettability, particle electrical charge, boiling points, and vapor pressure tailored for large-scale operations.

Question 8528Question →

Match each specialized metallic alloy listed on the left with its correct elemental composition and primary functional application on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches pair German Silver with the copper-zinc-nickel silver-free alloy, Alnico with the aluminium-nickel-cobalt-iron magnetic alloy, Type Metal with the lead-tin-antimony alloy that expands upon solidification, and Wood's Metal with the low-melting bismuth-lead-tin-cadmium fusible alloy.

Each alloy aligns strictly with its chemical formula and specialized physical property: German Silver (

\text{Cu}-\text{Zn}-\text{Ni}

) is silver-free with high resistance; Alnico (

\text{Al}-\text{Ni}-\text{Co}-\text{Fe}

) forms permanent magnets; Type Metal (

\text{Pb}-\text{Sn}-\text{Sb}

) expands upon freezing; Wood's Metal (

\text{Bi}-\text{Pb}-\text{Sn}-\text{Cd}

) melts at

65\ ^\circ\text{C}

.

Step-by-Step Solution

1

Analyze the chemical composition and key characteristic of German Silver.

German Silver consists of

\text{Cu}

,

\text{Zn}

, and

\text{Ni}

without any silver content, valued for high electrical resistivity and silvery appearance.

Identifying naming misnomers prevents confusing German Silver with silver-bearing precious metal alloys.

2

Deconstruct the constituent components and physical property of Alnico.

Alnico combines

\text{Al}

,

\text{Ni}

,

\text{Co}

, and

\text{Fe}

to form hard ferromagnetic structures.

The alloy acronym highlights its elements (

\text{Al}-\text{Ni}-\text{Co}

), which deliver superior permanent magnetic strength.

3

Examine the solid-phase volume change characteristic of Type Metal.

Type Metal contains

\text{Pb}

,

\text{Sn}

, and

\text{Sb}

; the presence of antimony induces volumetric expansion upon cooling.

While most metals shrink when freezing, antimony forces Type Metal to expand into fine matrix details during printing press type manufacture.

4

Evaluate the thermal melting point and application of Wood's Metal.

Wood's Metal is a eutectic combination of

\text{Bi}

,

\text{Pb}

,

\text{Sn}

, and

\text{Cd}

melting at

65\ ^\circ\text{C}

.

Combining four metals in specific proportions disrupts individual crystal lattice stability, dropping the melting point below

100\ ^\circ\text{C}

for safety sprinkler valves.

Key Concept

Alloy Classifications, Compositions, Phase-Change Properties, and Industrial Uses

Estimated Time:3m 0s

Question 8529Question →

Calcium carbide reacts with water according to the balanced chemical equation:

\text{CaC}_{2(s)} + 2\text{H}_2\text{O}_{(l)} \rightarrow \text{Ca(OH)}_{2(aq)} + \text{C}_2\text{H}_{2(g)}

If

32.0\text{ g}

of

\text{CaC}_2

is reacted with

21.6\text{ g}

of

\text{H}_2\text{O}

, calculate the mass of the excess reactant remaining unreacted upon completion of the reaction.

[\text{Relative atomic masses: } \text{Ca} = 40, \text{C} = 12, \text{O} = 16, \text{H} = 1]

Show answer & explanation

Answer: 3.6

Answer

3.6 g

Converting initial masses to mole values yields 0.50 mol of CaC₂ and 1.20 mol of H₂O. Based on the 1:2 mole ratio in the balanced equation, 0.50 mol of CaC₂ requires 1.00 mol of H₂O to react completely. This leaves 0.20 mol of H₂O unreacted. Converting 0.20 mol of H₂O to mass gives 0.20 mol × 18 g/mol = 3.6 g of excess reactant remaining.

Step-by-Step Solution

1

Calculate molar masses of CaC₂ and H₂O

Molar mass of CaC₂ = 64 g/mol; Molar mass of H₂O = 18 g/mol

Molar mass is required to convert given mass values into mole quantities.

2

Convert initial masses to moles

Moles of CaC₂ = 0.50 mol; Moles of H₂O = 1.20 mol

Stoichiometric relationships depend strictly on mole ratios rather than direct mass ratios.

3

Determine limiting and excess reactants using mole ratios

CaC₂ is the limiting reactant; H₂O is in excess

According to the balanced equation coefficient ratio (1:2), 0.50 mol of CaC₂ requires 1.00 mol of H₂O. Because 1.20 mol of H₂O is present, H₂O is in excess.

4

Calculate remaining unreacted mass of excess reactant

3.6 g of excess H₂O remaining

Unreacted moles of H₂O = 1.20 - 1.00 = 0.20 mol. Mass = 0.20 mol × 18 g/mol = 3.6 g.

Key Concept

Determining limiting and excess reagents in chemical reactions and calculating unreacted leftover mass using mole ratios from balanced equations.

Estimated Time:2m 0s

Question 8530Question →

Magnesium is an essential mineral nutrient required for healthy plant growth and metabolic function. When crop plants are cultivated in magnesium-deficient soil, older leaves exhibit severe interveinal chlorosis, which leads to a steep decline in the rate of photosynthetic carbon dioxide assimilation. Which primary biochemical consequence of magnesium deficiency directly accounts for this reduced rate of carbon fixation during the light-independent reactions?

A severe reduction in chlorophyll synthesis within thylakoid membranes, which diminishes light absorption and impairs the production of ATP and NADPH required to power the Calvin cycle

The failure of water photolysis in the chloroplast stroma, preventing the release of molecular oxygen necessary to directly oxidize carbon dioxide during carbon fixation

An accumulation of mineral salts within xylem vessels, which forces organic sugars to translocate backward from root sinks into photosynthetic leaves

Immediate thermal denaturation of ribulose 1,5-bisphosphate carboxylase (RuBisCO) enzymes caused by low ambient temperatures in the chloroplast

Show answer & explanation

Answer: A severe reduction in chlorophyll synthesis within thylakoid membranes, which diminishes light absorption and impairs the production of ATP and NADPH required to power the Calvin cycle

Answer

Magnesium deficiency impairs chlorophyll synthesis in thylakoid membranes, reducing light absorption and the synthesis of ATP and NADPH needed for carbon fixation in the Calvin cycle.

Magnesium is the central metallic element in the chlorophyll porphyrin ring. A deficiency impairs chlorophyll synthesis (causing chlorosis), reducing photon absorption during the light-dependent stage. Consequently, fewer ATP and NADPH molecules are synthesized via photophosphorylation, starving the Calvin cycle of the energy needed for carbon dioxide fixation.

Step-by-Step Solution

1

Identify the structural role of magnesium in plant cells.

Magnesium (

Mg^{2+}

) is the central atom of the porphyrin ring of chlorophyll molecules and serves as an enzyme cofactor.

Understanding structural mineral roles determines which physiological process is primary impaired.

2

Trace the primary impact of magnesium deficiency on light-dependent reactions.

Deficiency causes interveinal chlorosis, drastically reducing chlorophyll content and decreasing photon absorption by photosystems I and II.

Fewer functional chlorophyll molecules result in lower light energy harvesting.

3

Connect light-dependent outputs to light-independent carbon fixation.

Reduced electron transport yields insufficient ATP and reduced NADPH, which are indispensable chemical energy sources required to reduce 3-phosphoglycerate (PGA) to triose phosphate in the Calvin cycle.

Carbon assimilation rate drops directly because the dark reaction depends strictly on ATP and NADPH produced during the light reaction.

Key Concept

Role of Magnesium in Chlorophyll Structure and Photosynthetic Light Energy Conversion

Estimated Time:1m 30s

Question 8531Question →

When aqueous ammonia is added dropwise until in excess to an aqueous solution containing copper(II) ions, the pale blue precipitate that initially forms dissolves to produce a characteristic deep blue solution. Which complex ion is responsible for this deep blue color?

[\text{Cu}(\text{NH}_3)_4]^{2+}

[\text{Cu}(\text{NH}_3)_2]^+

[\text{Cu}(\text{OH})_4]^{2-}

[\text{Cu}(\text{H}_2\text{O})_4]^{2+}

Show answer & explanation

Answer: $[\text{Cu}(\text{NH}_3)_4]^{2+}$

Answer

The complex ion responsible for the deep blue solution is the tetraamminecopper(II) ion,

[\text{Cu}(\text{NH}_3)_4]^{2+}

.

When excess aqueous ammonia is added to a solution containing

\text{Cu}^{2+}

ions, ammonia molecules act as ligands to coordinate with the

\text{Cu}^{2+}

ion, forming the deep blue tetraamminecopper(II) complex ion,

[\text{Cu}(\text{NH}_3)_4]^{2+}

.

Step-by-Step Solution

1

Identify the initial reaction of copper(II) ions with aqueous ammonia.

A pale blue precipitate of copper(II) hydroxide,

\text{Cu(OH)}_2

, is formed when small amounts of aqueous ammonia are added:

\text{Cu}^{2+}_{(\text{aq})} + 2\text{NH}_{3(\text{aq})} + 2\text{H}_2\text{O}_{(\text{l})} \rightarrow \text{Cu(OH)}_{2(\text{s})} + 2\text{NH}_{4(\text{aq})}^+

.

Aqueous ammonia acts as a weak base, generating hydroxide ions.

2

Determine the effect of adding excess aqueous ammonia.

The copper(II) hydroxide precipitate dissolves due to ligand substitution, forming the soluble complex ion

[\text{Cu}(\text{NH}_3)_4]^{2+}_{(\text{aq})}

.

Ammonia molecules act as unidentate neutral ligands that bind strongly to copper(II) ions, displacing hydroxide ions and forming the deep blue tetraamminecopper(II) ion.

Key Concept

Complex ion formation and qualitative test for copper(II) ions using aqueous ammonia.

Estimated Time:1m 0s

Question 8532Question →

A student needs to prepare a pure, dry sample of barium tetraoxosulfate(VI), $BaSO_4$ , in the laboratory. Which of the following methods provides the most suitable procedure for obtaining this salt?

Mixing aqueous barium nitrate with aqueous sodium tetraoxosulfate(VI), filtering the precipitate, washing it with distilled water, and drying it

Titrating aqueous barium hydroxide with dilute tetraoxosulfate(VI) acid, evaporating the resulting solution to dryness, and crystallizing

Reacting solid barium carbonate with dilute tetraoxosulfate(VI) acid, filtering off the unreacted solid, and evaporating the filtrate to dryness

Heating barium metal directly in a stream of sulfur dioxide gas and collecting the crystalline residue

Show answer & explanation

Answer: Mixing aqueous barium nitrate with aqueous sodium tetraoxosulfate(VI), filtering the precipitate, washing it with distilled water, and drying it

Answer

Mixing aqueous barium nitrate with aqueous sodium tetraoxosulfate(VI), filtering the precipitate, washing it with distilled water, and drying it

Barium tetraoxosulfate(VI) (

BaSO_4

) is insoluble in water. The standard laboratory procedure for preparing insoluble salts is double decomposition (precipitation), where two aqueous solutions containing the constituent ions (in this case,

Ba(NO_3)_2(aq)

and

Na_2SO_4(aq)

) are mixed. The insoluble salt precipitates immediately, is collected by filtration, washed with distilled water to eliminate spectator ions, and dried.

Step-by-Step Solution

1

Determine the solubility of the target salt

Barium tetraoxosulfate(VI),

BaSO_4

, is insoluble in water according to general solubility rules.

The method of salt preparation depends primarily on whether the target salt is soluble or insoluble.

2

Select the appropriate laboratory synthesis technique

Double decomposition (precipitation) between two soluble compounds containing the required cations and anions (

Ba^{2+}

and

SO_4^{2-}

) is required.

Insoluble salts are best prepared by combining solutions of two soluble salts to precipitate the insoluble product.

3

Identify the separation and purification steps

Filter the solid precipitate from the mixture, wash the residue with distilled water to remove soluble byproduct ions (

Na^+

and

NO_3^-

), and dry the solid.

Washing and drying ensure that a pure, dry sample of the insoluble salt is obtained.

Key Concept

Laboratory Preparation of Insoluble Salts by Double Decomposition (Precipitation)

Question 8533Question →

When aluminium metal is dissolved in a concentrated aqueous solution of sodium hydroxide, a gas is evolved and a soluble complex ion is formed in the solution. What are the chemical formulas of the gas evolved and the complex ion formed, respectively?

H_2

and

[Al(OH)_4]^-

O_2

and

[Al(OH)_4]^-

H_2

and

Al^{3+}

NH_3

and

[Al(OH)_4]^-

Show answer & explanation

Answer: $H_2$ and $[Al(OH)_4]^-$

Answer

The gas evolved is hydrogen gas (

H_2

) and the complex ion formed is the tetrahydroxoaluminate(III) ion (

[Al(OH)_4]^-

).

Aluminium metal exhibits amphoteric behavior, dissolving in strong alkaline solutions such as aqueous sodium hydroxide. The reaction oxidizes aluminium metal to form the soluble aluminate complex ion,

[Al(OH)_4]^-

, while reducing hydrogen species to yield hydrogen gas,

H_2

.

Step-by-Step Solution

1

Identify the amphoteric property of aluminium metal in basic solutions.

Aluminium reacts with both acids and strong bases (alkalis) such as sodium hydroxide.

Aluminium forms protective oxide layers and has amphoteric chemical behavior.

2

Write the balanced ionic reaction between aluminium metal, hydroxide ions, and water.

2Al(s) + 2OH^-(aq) + 6H_2O(l) \rightarrow 2[Al(OH)_4]^-(aq) + 3H_2(g)

Aluminium is oxidized to the soluble tetrahydroxoaluminate(III) complex ion, while water/hydrogen species are reduced to hydrogen gas.

3

Match the products to the question requirements.

Gas evolved:

H_2

; Soluble complex ion:

[Al(OH)_4]^-

The reaction produces hydrogen gas and tetrahydroxoaluminate(III).

Key Concept

Amphoteric Nature of Aluminium and Reaction with Strong Alkalis

Question 8534Question →

When solid lead(II) trioxonitrate(V), $Pb(NO_3)_2$ , is heated strongly in a dry test tube, it decomposes to yield a yellow solid residue, oxygen gas, and a reddish-brown gas. Which formula represents this reddish-brown gas, and what is the oxidation state of nitrogen in it?

NO_2

with an oxidation state of

+4

N_2O

with an oxidation state of

+1

NO

with an oxidation state of

+2

N_2O_5

with an oxidation state of

+5

Show answer & explanation

Answer: $NO_2$ with an oxidation state of $+4$

Answer

The reddish-brown gas is nitrogen(IV) oxide (

NO_2

), in which nitrogen has an oxidation state of

+4

.

Heating lead(II) trioxonitrate(V) decomposes it into lead(II) oxide (

PbO

), nitrogen(IV) oxide (

NO_2

), and oxygen (

O_2

). Nitrogen(IV) oxide is a characteristic reddish-brown gas, and assigned oxidation state calculations give

+4

for nitrogen in

NO_2

.

Step-by-Step Solution

1

Write the balanced chemical equation for the thermal decomposition of lead(II) trioxonitrate(V).

2Pb(NO_3)_2(s) \rightarrow 2PbO(s) + 4NO_2(g) + O_2(g)

Heavy metal nitrates decompose on heating to yield the metal oxide, nitrogen(IV) oxide gas, and oxygen gas.

2

Identify the physical property of the gaseous products.

PbO

is a yellow solid residue (when hot/cold depending on form),

O_2

is a colorless gas, and

NO_2

is a distinctive reddish-brown acidic gas.

Nitrogen(IV) oxide (

NO_2

) is the only brown oxide of nitrogen produced in this reaction.

3

Calculate the oxidation state of nitrogen in

NO_2

.

Let

x

be the oxidation state of N.

x + 2(-2) = 0 \implies x = +4

.

Oxygen has an oxidation number of

-2

in neutral covalent oxides.

Key Concept

Thermal Decomposition of Metal Nitrates and Oxides of Nitrogen

Question 8535Question →

During the industrial isolation of noble gases from liquid air, argon is collected in a fraction between nitrogen and oxygen. What physical property explains why argon distills over after nitrogen but before oxygen?

Argon has a higher boiling point than nitrogen but a lower boiling point than oxygen.

Argon has a lower boiling point than nitrogen but a higher boiling point than oxygen.

Argon has a higher vapor density than oxygen but a lower vapor density than nitrogen.

Argon has a lower molar mass than nitrogen but a higher molar mass than oxygen.

Show answer & explanation

Answer: Argon has a higher boiling point than nitrogen but a lower boiling point than oxygen.

Answer

Argon has a higher boiling point than nitrogen but a lower boiling point than oxygen.

Fractional distillation of liquefied air separates gases according to their boiling points. Nitrogen distills first at

-196\ ^{\circ}\text{C}

(

77\text{ K}

), argon distills next at

-186\ ^{\circ}\text{C}

(

87\text{ K}

), and oxygen distills last at

-183\ ^{\circ}\text{C}

(

90\text{ K}

). Therefore, argon distills over after nitrogen but before oxygen because its boiling point is higher than that of nitrogen and lower than that of oxygen.

Step-by-Step Solution

1

Identify the separation principle of liquid air distillation.

Components of liquid air are separated based on differences in their boiling points during fractional distillation.

Fractional distillation separates liquids with different boiling points as the liquid mixture is warmed.

2

Compare the boiling points of nitrogen, argon, and oxygen.

Nitrogen boils at

-196\ ^{\circ}\text{C}

, argon boils at

-186\ ^{\circ}\text{C}

, and oxygen boils at

-183\ ^{\circ}\text{C}

.

Lower boiling point components boil and distill off first as vapor.

3

Deduce the sequence of distillation for argon.

Since

-196\ ^{\circ}\text{C} < -186\ ^{\circ}\text{C} < -183\ ^{\circ}\text{C}

, nitrogen distills off first, followed by argon, while oxygen remains liquid longest.

Argon distills after nitrogen because its boiling point is higher than nitrogen's, and before oxygen because its boiling point is lower than oxygen's.

Key Concept

Isolation of noble gases from liquid air by fractional distillation

Estimated Time:1m 0s

Question 8536Question →

An aqueous solution of copper(II) sulfate ( $CuSO_4$ ) undergoes electrolysis using copper sheets as both the anode and cathode. Which statement correctly describes the reaction occurring at the anode and identifies the primary factor governing this process?

The copper anode dissolves into solution as

Cu^{2+}

ions, governed by the nature of the electrode material.

Hydroxide ions (

OH^-

) are discharged to liberate oxygen gas, governed strictly by their higher position in the electrochemical series relative to sulfate ions (

SO_4^{2-}

).

Sulfate ions (

SO_4^{2-}

) are preferentially discharged to produce sulfur dioxide gas, governed by the high concentration of sulfate ions relative to hydroxide ions (

OH^-

).

Hydrogen ions (

H^+

) are discharged to liberate hydrogen gas, governed by the low electrode potential of water molecules.

Show answer & explanation

Answer: The copper anode dissolves into solution as $Cu^{2+}$ ions, governed by the nature of the electrode material.

Answer

The copper anode dissolves into solution as

Cu^{2+}

ions, governed by the nature of the electrode material.

In the electrolysis of copper(II) sulfate solution using active copper electrodes, oxidation occurs at the anode. Because copper is an active metal electrode, dissolving copper metal into copper(II) ions (

Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-

) requires less energy than discharging hydroxide ions or sulfate ions. Thus, the nature of the electrode material is the governing factor.

Step-by-Step Solution

1

Identify the ions present in aqueous

CuSO_4

solution.

Cations:

Cu^{2+}

and

H^+

. Anions:

SO_4^{2-}

and

OH^-

.

Electrolysis of aqueous solutions involves ions from both the dissolved salt and water dissociation.

2

Analyze the nature of the electrode at the anode.

The anode is made of copper, which is an active (reactive) electrode, not an inert electrode like platinum or graphite.

Active electrodes participate directly in the electrochemical reaction.

3

Determine the preferred anodic oxidation reaction.

Copper atoms from the anode oxidize preferentially (

Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-

) instead of discharging

OH^-

or

SO_4^{2-}

anions.

The energy required to oxidize metallic copper atoms is lower than that needed to discharge anions from the solution, demonstrating that the nature of the electrode overrides electrochemical series position and ion concentration.

Key Concept

Influence of the Nature of Electrodes on Preferential Discharge

Estimated Time:2m 0s

Question 8537Question →

In an industrial plant for the fractional distillation of liquefied air, atmospheric air is compressed, cooled to $-200^\circ\text{C}$ , and fed into a fractionating column. Given the boiling points of nitrogen ( $-196^\circ\text{C}$ ), argon ( $-186^\circ\text{C}$ ), and oxygen ( $-183^\circ\text{C}$ ), which of the following statements correctly accounts for the sequence and physical state of the components as the liquid air mixture is gradually warmed?

Nitrogen vaporizes first at

-196^\circ\text{C}

because it has the lowest boiling point, followed by argon at

-186^\circ\text{C}

, leaving oxygen as a liquid residue until

-183^\circ\text{C}

.

Oxygen vaporizes first because its boiling point of

-183^\circ\text{C}

is higher than that of nitrogen (

-196^\circ\text{C}

), requiring less thermal energy to transition into gas.

Argon vaporizes before nitrogen because monoatomic noble gases possess higher kinetic energy at

-200^\circ\text{C}

than diatomic nitrogen molecules.

Nitrogen and oxygen vaporize simultaneously at

-183^\circ\text{C}

as a constant-boiling azeotropic mixture, leaving liquid argon at the base of the fractionating tower.

Show answer & explanation

Answer: Nitrogen vaporizes first at $-196^\circ\text{C}$ because it has the lowest boiling point, followed by argon at $-186^\circ\text{C}$ , leaving oxygen as a liquid residue until $-183^\circ\text{C}$ .

Answer

Nitrogen vaporizes first at

-196^\circ\text{C}

because it has the lowest boiling point, followed by argon at

-186^\circ\text{C}

, leaving oxygen as a liquid residue until

-183^\circ\text{C}

.

The correct answer is the statement noting that nitrogen vaporizes first at

-196^\circ\text{C}

, followed by argon at

-186^\circ\text{C}

, leaving oxygen as a liquid. In fractional distillation of liquefied air, warming the mixture from

-200^\circ\text{C}

causes the component with the lowest boiling point (nitrogen at

-196^\circ\text{C}

) to reach its boiling threshold first and turn into gas at the top of the fractionating column. Argon (

-186^\circ\text{C}

) boils next, while oxygen (

-183^\circ\text{C}

) remains liquid longest and collects near the bottom.

Step-by-Step Solution

1

Analyze the starting condition and temperature change

The mixture starts in the liquid state at

-200^\circ\text{C}

and temperature gradually increases (warming up).

Fractional distillation of liquid air operates by gradually supplying heat to a cryogenic liquid mixture.

2

Compare boiling points of components

Boiling points on the Celsius scale: Nitrogen (

-196^\circ\text{C}

) < Argon (

-186^\circ\text{C}

) < Oxygen (

-183^\circ\text{C}

).

-196^\circ\text{C}

is a lower temperature than

-186^\circ\text{C}

and

-183^\circ\text{C}

.

3

Determine the phase change order during temperature rise

As the temperature ascends from

-200^\circ\text{C}

, it first reaches

-196^\circ\text{C}

(nitrogen boils), then

-186^\circ\text{C}

(argon boils), and finally

-183^\circ\text{C}

(oxygen boils).

The component with the lowest boiling point (most volatile) boils off first at the top of the fractionating column.

Key Concept

Fractional distillation of liquid air depends on differences in boiling points, where the component with the lowest boiling point vaporizes first.

Question 8538Question →

Match each liquid mixture to its correct behavior or layer placement when processed in a separating funnel.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Water and tetrachloromethane match with forming a lower organic layer; Water and ethoxyethane match with forming an upper organic layer; Water and ethanol match with forming a single homogeneous solution that cannot be separated by a funnel.

Tetrachloromethane is immiscible with water and denser than water, forming the lower layer. Ethoxyethane is immiscible with water and less dense than water, forming the upper layer. Ethanol is completely miscible with water, so no distinct phases form, making separating funnel isolation impossible.

Step-by-Step Solution

1

Evaluate the miscibility of each liquid pair with water.

Tetrachloromethane and ethoxyethane are immiscible with water (forming two distinct layers). Ethanol forms strong hydrogen bonds with water and is miscible in all proportions.

Separating funnels require two immiscible liquid phases to function.

2

Determine the relative position of layers for the immiscible pairs using density values.

Tetrachloromethane (

1.59\text{ g/cm}^3

) is denser than water (

1.00\text{ g/cm}^3

), so it sits below water. Ethoxyethane (

0.71\text{ g/cm}^3

) is less dense than water, so it floats above water.

In a liquid-liquid mixture, the denser liquid forms the bottom layer.

Key Concept

Liquid-Liquid Immiscibility and Density Placement in Separating Funnels

Question 8539Question →

Consider the high-temperature reversible reaction occurring in a closed vessel, represented by the thermochemical equation below:

N_2(g) + O_2(g) \rightleftharpoons 2NO(g) \quad \Delta H = +180.5\text{ kJ mol}^{-1}

Which of the following conditions will shift the equilibrium position to favor the yield of nitrogen(II) oxide?

Increasing the temperature of the system

Increasing the total pressure on the reaction vessel

Adding a suitable platinum catalyst

Removing nitrogen gas from the mixture

Show answer & explanation

Answer: Increasing the temperature of the system

Answer

Increasing the temperature of the system shifts the equilibrium position to the right, favoring the formation of nitrogen(II) oxide.

The forward synthesis of nitrogen(II) oxide is endothermic, absorbing heat from the surroundings. According to Le Chatelier's principle, when temperature is increased, the system shifts in the direction that absorbs heat (the forward endothermic direction) to relieve the thermal stress, resulting in a higher yield of nitrogen(II) oxide.

Step-by-Step Solution

1

Analyze the enthalpy change of the forward reaction

The positive sign of $\Delta H = +180.5\text{ kJ mol}^{-1}$ indicates that the forward reaction is endothermic.

According to Le Chatelier's principle, supplying thermal energy by raising the temperature causes an endothermic reaction to shift forward to absorb the excess heat.

2

Analyze the volume/mole change across gaseous species

Reactant gas moles = $1 + 1 = 2$ moles; Product gas moles = $2$ moles.

Since the total number of gaseous moles is identical on both sides of the equation, pressure changes produce no net shift in equilibrium position.

3

Evaluate the effect of a catalyst and reactant removal

Adding a catalyst accelerates attainment of equilibrium equally in both directions; removing a reactant causes a leftward shift.

Neither catalyst addition nor removal of reactants increases product yield at equilibrium.

Key Concept

Effect of temperature, pressure, and catalyst on dynamic chemical equilibrium via Le Chatelier's Principle

Question 8540Question →

A chemist needs to recover solid solutes separately from two aqueous mixtures: Solution X containing sodium chloride ( $\text{NaCl}$ ) to obtain anhydrous salt, and Solution Y containing hydrated iron(II) tetraoxosulfate(VI) ( $\text{FeSO}_4 \cdot 7\text{H}_2\text{O}$ ) to obtain pure hydrated crystals. Which combination of techniques should be used to obtain the solutes from Solution X and Solution Y respectively?

Evaporation to dryness for Solution X, and crystallization for Solution Y

Crystallization for Solution X, and evaporation to dryness for Solution Y

Evaporation to dryness for both Solution X and Solution Y

Filtration for Solution X, and simple distillation for Solution Y

Show answer & explanation

Answer: Evaporation to dryness for Solution X, and crystallization for Solution Y

Answer

Evaporation to dryness for Solution X, and crystallization for Solution Y

The correct response identifies that sodium chloride solution can be safely evaporated to dryness because the salt is heat-stable and has no water of crystallization to preserve. Conversely, hydrated iron(II) tetraoxosulfate(VI) requires crystallization (partial evaporation to a saturated state followed by cooling) to retain its water of crystallization and avoid thermal decomposition.

Step-by-Step Solution

1

Analyze thermal stability and solute characteristics of Solution X

Sodium chloride (

\text{NaCl}

) is thermally stable at heating temperatures and does not rely on water of crystallization.

Evaporation to dryness completely removes the solvent water, leaving behind dry anhydrous sodium chloride.

2

Analyze thermal stability and solute characteristics of Solution Y

Hydrated iron(II) tetraoxosulfate(VI) (

\text{FeSO}_4 \cdot 7\text{H}_2\text{O}

) loses its water of crystallization and decomposes if heated to dryness.

Crystallization involves concentrating the solution partially and cooling it slowly so pure hydrated crystals precipitate without thermal decomposition.

Key Concept

Selecting evaporation to dryness versus crystallization based on thermal stability and water of crystallization

Answer

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