Metals and Their Compounds Practice Questions — JAMB UTME — Page 2

Question 21Question →

Arrange the following sequential steps involved in the industrial production of sodium trioxocarbonate(IV) ( $\text{Na}_2\text{CO}_3$ ) via the Solvay process in the correct chronological order from first to last.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence starts with saturating brine with ammonia, bubbling carbon dioxide to precipitate sodium hydrogen trioxocarbonate(IV), filtering the precipitate, and finally heating (calcinating) the sodium hydrogen trioxocarbonate(IV) to produce sodium trioxocarbonate(IV).

The Solvay process begins by dissolving ammonia gas into brine (concentrated

\text{NaCl}

) to form ammoniacal brine. Next, carbon dioxide gas is bubbled through this mixture to precipitate sodium hydrogen trioxocarbonate(IV) (

\text{NaHCO}_3

). The solid precipitate is then separated by filtration. Finally, the dried

\text{NaHCO}_3

is subjected to thermal decomposition (calcination) to produce sodium trioxocarbonate(IV) (

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

).

Step-by-Step Solution

1

Identify the primary raw preparation step

Ammonia gas is dissolved in concentrated sodium chloride solution (brine) to create ammoniacal brine.

Ammonia acts as a base and reacts with carbon dioxide in the subsequent step to form hydrogen trioxocarbonate ions.

2

Identify the carbonation and precipitation step

Carbon dioxide gas is passed through the ammoniacal brine in a carbonating tower.

The reaction produces ammonium chloride (

\text{NH}_4\text{Cl}

) and sodium hydrogen trioxocarbonate (

\text{NaHCO}_3

), which precipitates because of its low solubility in cold brine solution.

3

Identify the solid-liquid separation step

The mixture is filtered to isolate the solid

\text{NaHCO}_3

crystals from the solution containing

\text{NH}_4\text{Cl}

.

Filtration separates the insoluble sodium intermediate prior to thermal conversion.

4

Identify the final chemical transformation step

The filtered

\text{NaHCO}_3

is heated (calcinated) at around

300^\circ\text{C}

to form

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

,

\text{H}_2\text{O}

, and

\text{CO}_2

.

Thermal decomposition yields the target sodium trioxocarbonate(IV) product while recycling

\text{CO}_2

back into the carbonating tower.

Key Concept

The Solvay Process for Sodium Trioxocarbonate(IV) Production

Question 22Question →

Sodium metal reacts with excess oxygen gas at elevated temperatures to produce sodium peroxide ( $\text{Na}_2\text{O}_2$ ). What is the oxidation state of oxygen in sodium peroxide?

-1

-2

0

+1

Show answer & explanation

Answer: $-1$

Answer

The oxidation state of oxygen in sodium peroxide (

\text{Na}_2\text{O}_2

) is

-1

.

In sodium peroxide (

\text{Na}_2\text{O}_2

), the two sodium ions contribute a total charge of

+2

. To maintain electrical neutrality, the two oxygen atoms balance this with a total charge of

-2

, giving each oxygen atom an oxidation state of

-1

.

Step-by-Step Solution

1

Assign the known oxidation state for sodium in its compounds.

Sodium (

\text{Na}

) is an alkali metal (Group 1) and always has an oxidation state of

+1

in compounds.

Alkali metals lose one valence electron to achieve a stable noble gas electron configuration.

2

Set up the oxidation state equation for the neutral molecule

\text{Na}_2\text{O}_2

.

2(+1) + 2(x) = 0

, where

x

represents the oxidation state of oxygen.

The algebraic sum of oxidation numbers in a neutral compound must equal zero.

3

Solve for

x

.

+2 + 2x = 0 \implies 2x = -2 \implies x = -1

.

Algebraic evaluation yields the oxidation number for each oxygen atom.

Key Concept

Oxidation Numbers of Oxygen in Alkali Metal Peroxides

Estimated Time:1m 0s

Question 23Question →

Match each calcium compound or reagent listed on the left with its corresponding industrial preparation, chemical property, or role on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Quicklime (

\text{CaO}

) matches with thermal decomposition of limestone; Slaked lime (

\text{Ca(OH)}_2

) matches with adding water to quicklime; Plaster of Paris (

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

) matches with partial dehydration of gypsum; Calcium fluoride (

\text{CaF}_2

) matches with electrolyte flux in calcium extraction.

Each calcium compound is accurately paired with its definitive reaction or industrial application in standard JAMB UTME chemistry syllabus specifications.

Step-by-Step Solution

1

Identify the industrial reaction for Quicklime (

\text{CaO}

).

Quicklime is formed via

\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})

.

Thermal decomposition of limestone yields calcium oxide.

2

Identify the process of slaking lime to form Slaked lime (

\text{Ca(OH)}_2

).

Slaking quicklime

\text{CaO}(\text{s}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{Ca(OH)}_2(\text{s})

produces calcium hydroxide.

Calcium hydroxide solution (limewater) is used to detect

\text{CO}_2

gas.

3

Determine the formula and production method of Plaster of Paris.

Heating gypsum (

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

) yields hemihydrate

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

.

Controlled dehydration removes

1.5

molecules of crystallization water per mole of calcium sulfate.

4

Determine the function of Calcium fluoride (

\text{CaF}_2

) in metallurgy.

Acts as a flux in fused

\text{CaCl}_2

electrolysis.

It lowers the melting point of anhydrous

\text{CaCl}_2

from about

772^\circ\text{C}

to around

650^\circ\text{C}

.

Key Concept

Chemical transformations, properties, and metallurgical roles of calcium and its principal compounds.

Question 24Question →

During the smelting stage of copper extraction, copper pyrites (

CuFeS_2

) is roasted in air according to the following balanced equation:

2CuFeS_{2(s)} + 4O_{2(g)} \rightarrow Cu_2S_{(s)} + 2FeO_{(s)} + 3SO_{2(g)}

If

0.50\text{ mole}

of

CuFeS_2

is reacted with

0.80\text{ mole}

of

O_2

, which reagent is limiting, what volume of

SO_2

gas is produced at STP, and what is the oxidation state of iron in the resulting

FeO

?
[Molar volume of gas at STP =

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

]

O_2

is limiting;

13.44\text{ dm}^3

of

SO_2

; iron oxidation state is

+2

CuFeS_2

is limiting;

16.80\text{ dm}^3

of

SO_2

; iron oxidation state is

+2

O_2

is limiting;

14.40\text{ dm}^3

of

SO_2

; iron oxidation state is

+2

O_2

is limiting;

13.44\text{ dm}^3

of

SO_2

; iron oxidation state is

0

Show answer & explanation

Answer: $O_2$ is limiting; $13.44\text{ dm}^3$ of $SO_2$ ; iron oxidation state is $+2$

Answer

Oxygen gas (

O_2

) is the limiting reactant, yielding

13.44\text{ dm}^3

of

SO_2

gas at STP, and iron in

FeO

has an oxidation state of

+2

.

The correct answer identifies oxygen gas (

O_2

) as the limiting reactant because

0.50\text{ mole}

of

CuFeS_2

requires

1.00\text{ mole}

of

O_2

for complete reaction, but only

0.80\text{ mole}

is provided. Utilizing

O_2

to calculate product yield gives

0.60\text{ mole}

of

SO_2

, which translates to

13.44\text{ dm}^3

at STP (

0.60 \times 22.4

). Iron in

FeO

carries an oxidation state of

+2

.

Step-by-Step Solution

1

Determine the limiting reactant by comparing the mole ratio of available reactants to stoichiometric coefficients

From the balanced equation,

2\text{ moles}

of

CuFeS_2

require

4\text{ moles}

of

O_2

(ratio

1:2

).

0.50\text{ mole}

of

CuFeS_2

requires

1.00\text{ mole}

of

O_2

. Since only

0.80\text{ mole}

of

O_2

is available,

O_2

is the limiting reactant.

The reaction extent is governed entirely by the reactant that is fully consumed first.

2

Calculate the moles of

SO_2

produced using the limiting reactant

According to the stoichiometric ratio,

4\text{ moles}

of

O_2

produce

3\text{ moles}

of

SO_2

. Therefore,

0.80\text{ mole}

of

O_2

produces

0.80 \times \frac{3}{4} = 0.60\text{ mole}

of

SO_2

.

The yield of product depends directly on the moles of the limiting reactant.

3

Convert moles of

SO_2

gas to volume at STP

Volume of

SO_2 = 0.60\text{ mole} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 13.44\text{ dm}^3

.

At STP, one mole of any ideal gas occupies

22.4\text{ dm}^3

.

4

Determine the oxidation state of iron in

FeO

Oxygen has an oxidation state of

-2

. For neutral

FeO

,

\text{Fe} + (-2) = 0 \Rightarrow \text{Fe} = +2

.

The sum of oxidation states in a neutral compound equals zero.

Key Concept

Copper extraction roasting reaction stoichiometry and limiting reactant calculations

Estimated Time:2m 30s

Question 25Question →

Metals positioned near the top of the electrochemical reactivity series, such as sodium and aluminium, are extracted from their purified ores primarily through which method?

Electrolysis of their molten compounds

Reduction of their oxides using carbon monoxide

Roasting their sulfide ores directly in excess air

Thermal decomposition of their trioxocarbonate(IV) salts

Show answer & explanation

Answer: Electrolysis of their molten compounds

Answer

Electrolysis of their molten compounds

Metals positioned high in the electrochemical series (such as sodium, potassium, calcium, and aluminium) have a strong affinity for oxygen and chlorine. Chemical reducing agents like carbon or carbon monoxide are weaker reducing agents than these metals and cannot reduce their oxides. Consequently, these metals are extracted by electrolysis of their molten (fused) ores or salts.

Step-by-Step Solution

1

Assess metal reactivity based on position in the electrochemical series

Sodium and aluminium are highly electropositive and hold onto oxygen/halogens strongly.

Metals at the top of the reactivity series form compounds with very high thermal and chemical stability.

2

Select the appropriate reduction method

Standard chemical reducing agents (such as carbon or hydrogen) are ineffective.

Only powerful electrical energy supplied during electrolysis of fused compounds can force electron gain at the cathode to yield free metal.

Key Concept

Selection of metal extraction method based on reactivity series

Question 26Question →

During the industrial extraction of zinc from its principal ore, zinc blende ( $ZnS$ ), the concentrated sulfide ore undergoes thermal conversion in excess air before metal recovery. Which set of balanced chemical equations correctly represents the roasting step and the subsequent reduction step using carbon (coke)?

Roasting:

2ZnS(s) + 3O_2(g) \rightarrow 2ZnO(s) + 2SO_2(g)

; Reduction:

ZnO(s) + C(s) \rightarrow Zn(g) + CO(g)

Roasting:

ZnS(s) + 2O_2(g) \rightarrow ZnSO_4(s)

; Reduction:

ZnSO_4(s) + C(s) \rightarrow Zn(s) + SO_2(g) + CO(g)

Roasting:

2ZnS(s) + O_2(g) \rightarrow 2Zn(s) + 2SO_2(g)

; Reduction:

Zn(s) + C(s) \rightarrow ZnC(s)

Roasting:

ZnS(s) + H_2O(g) \rightarrow ZnO(s) + H_2S(g)

; Reduction:

ZnO(s) + CO_2(g) \rightarrow Zn(s) + CO(g) + O_2(g)

Show answer & explanation

Answer: Roasting: $2ZnS(s) + 3O_2(g) \rightarrow 2ZnO(s) + 2SO_2(g)$ ; Reduction: $ZnO(s) + C(s) \rightarrow Zn(g) + CO(g)$

Answer

The correct sequence of reactions involves roasting zinc blende in excess oxygen to produce zinc oxide and sulfur(IV) oxide (

2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2

), followed by reduction of zinc oxide with coke at high temperature to produce zinc gas and carbon(II) oxide (

ZnO + C \rightarrow Zn + CO

).

Roasting zinc blende (

ZnS

) in excess oxygen gas converts the sulfide ore into zinc oxide (

ZnO

) while liberating sulfur(IV) oxide (

SO_2

). In the reduction furnace, carbon (coke) reduces zinc oxide to zinc metal vapor (

Zn

) and carbon(II) oxide gas (

CO

) because the reaction takes place above the boiling point of zinc.

Step-by-Step Solution

1

Identify the chemical nature of the roasting process in metallurgy.

Roasting involves heating concentrated sulfide ores (

ZnS

) in an abundant supply of atmospheric oxygen (

O_2

).

Sulfide ores are difficult to reduce directly to metals; converting them to oxides makes subsequent reduction thermodynamically feasible.

2

Write and balance the roasting reaction equation.

2ZnS(s) + 3O_2(g) \rightarrow 2ZnO(s) + 2SO_2(g)

Zinc sulfide reacts with oxygen gas to yield solid zinc oxide and sulfur(IV) oxide gas byproduct.

3

Identify the reduction process of zinc oxide using coke.

ZnO(s) + C(s) \rightarrow Zn(g) + CO(g)

Carbon (coke) acts as a reducing agent at elevated temperatures (~1400 °C), stripping oxygen from zinc oxide. Because the temperature exceeds zinc's boiling point (907 °C), zinc is collected as a gas (vapor) and condensed.

Key Concept

Metallurgical stages of sulfide ore extraction: Roasting (conversion to oxide) followed by pyrometallurgical carbon reduction.

Estimated Time:2m 0s

Question 27Question →

In the froth flotation process used for concentrating low-grade copper sulfide ores such as chalcopyrite ( $CuFeS_2$ ), pine oil and sodium ethyl xanthate are added to the aqueous suspension of the pulverized ore. What specific roles do pine oil and sodium ethyl xanthate play in this metallurgical separation process?

Pine oil acts as a frothing agent to produce stable foam, while sodium ethyl xanthate acts as a collector to render sulfide mineral particles hydrophobic.

Pine oil acts as a chemical reducing agent to convert copper sulfide into elemental copper, while sodium ethyl xanthate acts as a flux to convert gangue into slag.

Pine oil preferentially dissolves silicate gangue impurities, while sodium ethyl xanthate causes copper ions to precipitate via selective electrolytic discharge.

Pine oil neutralizes amphoteric metal oxides in the matrix, while sodium ethyl xanthate oxidizes the sulfide ions to elemental sulfur.

Show answer & explanation

Answer: Pine oil acts as a frothing agent to produce stable foam, while sodium ethyl xanthate acts as a collector to render sulfide mineral particles hydrophobic.

Answer

Pine oil serves as the frothing agent to generate stable air bubbles, while sodium ethyl xanthate functions as a collector that selectively coats sulfide ore particles to make them hydrophobic.

In froth flotation concentration of sulfide ores, pine oil acts as a frothing agent that creates a stable foam matrix when air is agitated through the ore slurry. Sodium ethyl xanthate serves as a collector reagent; its molecules selectively adsorb onto the surface of sulfide mineral particles, turning them hydrophobic. The water-repellent sulfide particles attach to the rising froth bubbles and are skimmed off at the top, leaving water-wetted silicate gangue behind.

Step-by-Step Solution

1

Identify the purpose of the froth flotation method in metallurgy.

Froth flotation is a physical concentration process used specifically for sulfide ores to separate valuable mineral particles from earthy gangue based on surface wettability.

Sulfide ore particles are naturally or artificially hydrophobic (water-repellent), whereas silicate gangue is hydrophilic (water-attracting).

2

Analyze the function of pine oil in the process.

Pine oil acts as a frother.

It lowers the surface tension of water, enabling the formation of stable oil-coated air bubbles when air is blown into the suspension.

3

Analyze the function of sodium ethyl xanthate in the process.

Sodium ethyl xanthate acts as a collector (polar-nonpolar surfactant).

Its polar group attaches to the sulfide mineral surface while its nonpolar hydrocarbon tail points outward, rendering the mineral particle strongly hydrophobic so it attaches to rising air bubbles.

Key Concept

Froth Flotation Principle and Additive Roles in Ore Concentration

Question 28Question →

Concentrated trioxonitrate(V) acid can be safely stored and transported in containers constructed from aluminium metal. Which of the following chemical phenomena accounts for this behavior?

Formation of an impervious, protective surface oxide film

High standard reduction potential preventing acid attack

Precipitation of an insoluble trioxonitrate(V) salt lining the container

Neutralization of the acid by the basic nature of aluminium metal

Show answer & explanation

Answer: Formation of an impervious, protective surface oxide film

Answer

Formation of an impervious, protective surface oxide film

Concentrated trioxonitrate(V) acid is a powerful oxidizing agent that renders aluminium passive by forming a thin, dense, and non-porous oxide coating over its surface. This continuous protective barrier prevents the acid from contacting the bulk metal underneath.

Step-by-Step Solution

1

Examine the chemical action of concentrated trioxonitrate(V) acid on aluminium metal.

Concentrated trioxonitrate(V) acid acts as a powerful oxidizing agent.

Upon contact with aluminium, it immediately oxidizes the metal surface.

2

Determine the physical consequence of the surface oxidation.

A thin, tough, and impermeable oxide layer covers the surface of the metal.

This process, termed passivity, shields the underlying bulk aluminium from undergoing further chemical attack by the acid.

Key Concept

Aluminium oxide passivity

Question 29Question →

A

25.0\text{ g}

sample of limestone containing

80.0\%

calcium trioxocarbonate(IV) by mass is strongly heated until decomposition is complete according to the equation:

CaCO_3(s) \xrightarrow{\Delta} CaO(s) + CO_2(g)

What volume of carbon(IV) oxide gas, measured at room temperature and pressure (RTP), is liberated in this reaction?
(

M(CaCO_3) = 100\text{ g mol}^{-1}

; Molar volume of gas at RTP

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

)

4.80\text{ dm}^3

4.48\text{ dm}^3

6.00\text{ dm}^3

5.60\text{ dm}^3

Show answer & explanation

Answer: $4.80\text{ dm}^3$

Answer

The volume of carbon(IV) oxide gas liberated at RTP is

4.80\text{ dm}^3

.

The mass of active

CaCO_3

is

80.0\%

of

25.0\text{ g}

, which equals

20.0\text{ g}

. Dividing by the molar mass (

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

) gives

0.20\text{ mol}

of

CaCO_3

. By stoichiometry,

0.20\text{ mol}

of

CO_2

is evolved. Multiplying by the molar volume at RTP (

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

) yields

4.80\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the mass of pure calcium trioxocarbonate(IV) (

CaCO_3

) present in the limestone sample.

Mass of pure

CaCO_3 = \frac{80.0}{100} \times 25.0\text{ g} = 20.0\text{ g}

.

Impurities in the limestone do not produce

CO_2

gas upon heating.

2

Determine the amount in moles of pure

CaCO_3

decomposed.

Moles of

CaCO_3 = \frac{20.0\text{ g}}{100\text{ g mol}^{-1}} = 0.20\text{ mol}

.

Converting mass to amount in moles allows stoichiometric evaluation using balanced chemical equations.

3

Use the stoichiometric ratio from the balanced chemical equation to find the moles of

CO_2

produced.

Since

1\text{ mol } CaCO_3 \rightarrow 1\text{ mol } CO_2

, moles of

CO_2 = 0.20\text{ mol}

.

The thermal decomposition ratio between

CaCO_3

and

CO_2

is

1:1

.

4

Calculate the volume of

CO_2

gas at room temperature and pressure (RTP).

Volume of

CO_2 = 0.20\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 4.80\text{ dm}^3

.

Molar gas volume at RTP is

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

.

Key Concept

Thermal decomposition of calcium carbonate and percentage purity stoichiometry at non-STP conditions

Question 30Question →

Match each calcium-based compound or reagent listed on the left with its corresponding industrial process, chemical behavior, or metallurgical function on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Calcium fluoride addition matches lowering the electrolyte melting point and enhancing conductivity; Calcium oxide hydration matches forming slaked lime used in water softening; Calcium hydroxide reaction with chlorine matches forming bleaching powder; Plaster of Paris rehydration matches forming dihydrate gypsum with volume expansion.

Each calcium compound or reagent is paired strictly according to its industrial function or chemical reaction behavior as specified in the JAMB UTME syllabus.

Step-by-Step Solution

1

Analyze the metallurgical role of calcium fluoride in calcium extraction

Identified

\text{CaF}_2

as a flux that lowers the melting point of fused

\text{CaCl}_2

from

800^\circ\text{C}

to

600^\circ\text{C}

.

Electrolysis of pure

\text{CaCl}_2

requires high temperature where molten calcium would dissolve in the electrolyte, so

\text{CaF}_2

is added as a flux.

2

Evaluate the chemical properties of calcium oxide and its slaked product

Slaking

\text{CaO}

gives

\text{Ca(OH)}_2

, which precipitates

\text{Ca(HCO}_3)_2

in Clark's method.

Calcium hydroxide reacts with hydrogen carbonate ions to precipitate insoluble

\text{CaCO}_3

, removing temporary hardness.

3

Determine the industrial reaction between slaked lime and chlorine

Chlorination of dry

\text{Ca(OH)}_2

yields bleaching powder,

\text{CaOCl}_2 \cdot \text{H}_2\text{O}

.

This specific gas-solid reaction forms active bleaching agents used in water treatment and industrial oxidation.

4

Examine the hydration chemistry of calcium sulfate hemihydrate

Plaster of Paris absorbs water to form gypsum with a slight increase in solid volume.

The crystallization process of gypsum yields interlocking monoclinic crystals that expand slightly to fill mold details perfectly.

Key Concept

Extraction, reactions, and industrial applications of alkaline earth metal (calcium) compounds.

Question 31Question →

During the industrial extraction of iron from hematite in the blast furnace, distinct chemical reactions occur across different temperature zones. Arrange the following key reaction stages in sequential order from the top of the furnace (coolest zone, approx. $250^\circ\text{C}-400^\circ\text{C}$ ) down to the hearth/tuyere region at the bottom (hottest zone, approx. $1500^\circ\text{C}-1900^\circ\text{C}$ ):

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequential order from top (coolest zone) to bottom (hottest zone) is: (1) Reduction of hematite to triiron tetroxide, (2) Reduction of triiron tetroxide to iron(II) oxide, (3) Reduction of iron(II) oxide to metallic iron, (4) Combination of calcium oxide with silica to form slag, and (5) Exothermic combustion of coke with preheated air.

The blast furnace operates with a temperature gradient rising from top to bottom. Near the top (

200^\circ\text{C}-400^\circ\text{C}

), hematite (

\text{Fe}_2\text{O}_3

) is reduced to

\text{Fe}_3\text{O}_4

. As the mixture descends to warmer regions (

500^\circ\text{C}-700^\circ\text{C}

),

\text{Fe}_3\text{O}_4

is reduced to

\text{FeO}

. Deeper down (

800^\circ\text{C}-1000^\circ\text{C}

),

\text{FeO}

is reduced to spongy iron (

\text{Fe}

). At around

1000^\circ\text{C}-1200^\circ\text{C}

, limestone-derived

\text{CaO}

reacts with

\text{SiO}_2

to form molten slag (

\text{CaSiO}_3

). Finally, at the base near the tuyeres (

1500^\circ\text{C}-1900^\circ\text{C}

), coke reacts exothermically with oxygen to fuel the entire process.

Step-by-Step Solution

1

Analyze the thermal gradient inside the blast furnace

Temperatures increase from top (

200^\circ\text{C}

) to bottom hearth region (

>1500^\circ\text{C}

).

Cold raw materials enter from the top while hot air blasts enter from the tuyeres at the bottom.

2

Identify top-zone reactions (

200^\circ\text{C}-400^\circ\text{C}

)

3\text{Fe}_2\text{O}_3 + \text{CO} \rightarrow 2\text{Fe}_3\text{O}_4 + \text{CO}_2

Hematite is initially converted to magnetite at relatively low temperatures by upflowing carbon(II) oxide.

3

Identify upper-middle zone reactions (

500^\circ\text{C}-700^\circ\text{C}

)

\text{Fe}_3\text{O}_4 + \text{CO} \rightarrow 3\text{FeO} + \text{CO}_2

Further reduction converts magnetite into iron(II) oxide as the charge descends.

4

Identify lower-middle zone reactions (

800^\circ\text{C}-1000^\circ\text{C}

)

\text{FeO} + \text{CO} \rightarrow \text{Fe} + \text{CO}_2

Complete reduction of iron(II) oxide to spongy metallic iron takes place here.

5

Identify slag formation zone (

1000^\circ\text{C}-1200^\circ\text{C}

)

\text{CaO} + \text{SiO}_2 \rightarrow \text{CaSiO}_3

Limestone decomposes into calcium oxide, which reacts with acidic silicon dioxide impurities to form molten calcium trioxosilicate(IV) slag.

6

Identify bottom tuyere region reactions (

1500^\circ\text{C}-1900^\circ\text{C}

)

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

Preheated air blown in through tuyeres reacts violently with coke to generate heat and carbon dioxide, which is subsequently reduced by hot coke to carbon monoxide.

Key Concept

Blast furnace temperature zones and sequential chemical reduction stages of iron ore

Question 32Question →

Underground steel pipelines are often connected to sacrificial blocks of magnesium to prevent rusting. Which statement best explains the electrochemical principle behind this method of corrosion protection?

Carbon(II) oxide in the surrounding soil reduces any oxidized iron ions back to metallic iron, preventing corrosion.

Magnesium is more electropositive than iron, so it loses electrons preferentially and serves as a sacrificial anode.

Magnesium acts as the cathode where oxidation takes place, while the iron pipe functions as the protective anode.

The rate of iron oxidation is halted because magnesium acts as the limiting reactant based purely on its initial mass.

Show answer & explanation

Answer: Magnesium is more electropositive than iron, so it loses electrons preferentially and serves as a sacrificial anode.

Answer

Magnesium is more electropositive than iron, so it loses electrons preferentially and serves as a sacrificial anode.

Magnesium is higher than iron in the electrochemical series. When connected electrically in a moist environment, magnesium oxidizes preferentially by donating electrons to the iron structure, causing magnesium to dissolve as sacrificial anode while keeping iron uncorroded as the cathode.

Step-by-Step Solution

1

Compare the positions of magnesium and iron in the electrochemical series.

Magnesium (

Mg

) has a higher standard oxidation potential than iron (

Fe

).

Metals positioned higher in the activity series lose valence electrons more readily.

2

Determine which metal undergoes oxidation in the galvanic cell formed by moisture.

Magnesium undergoes oxidation at the sacrificial anode (

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

).

The more electropositive metal preferentially oxidizes, suppressing the oxidation of iron (

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

).

3

Identify the overall protection mechanism.

Cathodic protection (sacrificial protection) keeps the iron pipeline safe as long as magnesium is present to corrode sacrificial anodes.

Iron is forced to remain the cathode, maintaining its elemental state.

Key Concept

Cathodic (Sacrificial) Protection of Iron

Estimated Time:1m 0s

Question 33Question →

In the reduction zone of a blast furnace, hematite (

\text{Fe}_2\text{O}_3

) reacts with gaseous carbon(II) oxide (

\text{CO}

) according to the balanced equation:

\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(l) + 3\text{CO}_2(g)

If a furnace charge containing $160.0\text{ kg}$ of pure $\text{Fe}_2\text{O}_3$ is reacted with $40.32\text{ m}^3$ of $\text{CO}$ gas measured at STP, which of the following correctly identifies the limiting reactant and the maximum mass of iron produced?
[ $\text{Fe} = 56$ , $\text{O} = 16$ , $\text{C} = 12$ ; Molar volume of gas at STP $= 22.4\text{ dm}^3\text{mol}^{-1}$ ]

\text{CO}

is the limiting reactant, yielding

67.2\text{ kg}

of iron.

\text{Fe}_2\text{O}_3

is the limiting reactant, yielding

112.0\text{ kg}

of iron.

Carbon (

\text{C}

) acts as the direct reducing agent in this zone, yielding

112.0\text{ kg}

of iron.

\text{CO}

is the limiting reactant, yielding

100.8\text{ kg}

of iron.

Show answer & explanation

Answer: $\text{CO}$ is the limiting reactant, yielding $67.2\text{ kg}$ of iron.

Answer

Carbon(II) oxide (

\text{CO}

) is the limiting reactant, yielding

67.2\text{ kg}

of iron.

Carbon(II) oxide (

\text{CO}

) is the limiting reactant because

1800\text{ mol}

of

\text{CO}

can only reduce

600\text{ mol}

of

\text{Fe}_2\text{O}_3

out of the available

1000\text{ mol}

. Based on the stoichiometric ratio of

3\text{ mol CO} : 2\text{ mol Fe}

,

1800\text{ mol}

of

\text{CO}

yields

1200\text{ mol}

of iron metal, which corresponds to

67.2\text{ kg}

.

Step-by-Step Solution

1

Calculate the moles of reactants provided

Moles of

\text{Fe}_2\text{O}_3 = \frac{160,000\text{ g}}{160\text{ g/mol}} = 1000\text{ mol}

. Moles of

\text{CO} = \frac{40,320\text{ dm}^3}{22.4\text{ dm}^3/\text{mol}} = 1800\text{ mol}

.

Converting quantities to moles allows stoichiometric comparison.

2

Determine the limiting reactant using the balanced equation stoichiometry

According to

\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2

,

1000\text{ mol}

of

\text{Fe}_2\text{O}_3

requires

3000\text{ mol}

of

\text{CO}

. Since only

1800\text{ mol}

of

\text{CO}

is available,

\text{CO}

is the limiting reactant.

The reactant that produces the lesser amount of product limits the reaction.

3

Calculate the theoretical yield of iron in kilograms

Moles of

\text{Fe}

produced

= 1800\text{ mol CO} \times \frac{2\text{ mol Fe}}{3\text{ mol CO}} = 1200\text{ mol Fe}

. Mass of

\text{Fe} = 1200\text{ mol} \times 56\text{ g/mol} = 67,200\text{ g} = 67.2\text{ kg}

.

Multiply moles of limiting reactant by the mole ratio and molar mass of iron.

Key Concept

Stoichiometric limiting reactant analysis in the blast furnace reduction of iron ore

Question 34Question →

Arrange the following sequential electrochemical and physical steps occurring during the reduction of alumina in the Hall-Héroult cell to extract molten aluminium metal, from initial electrolyte preparation to final anode gas emission.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of steps in the Hall-Héroult cell is: (1) Dissolving alumina in molten cryolite, (2) Dissociation of alumina into

Al^{3+}

and

O^{2-}

ions, (3) Migration of

Al^{3+}

ions to the cathode, (4) Reduction of

Al^{3+}

to form molten aluminium, and (5) Oxidation of

O^{2-}

at the graphite anodes yielding carbon dioxide gas.

The Hall-Héroult process operates sequentially by first dissolving alumina (

Al_2O_3

) in molten cryolite (

Na_3AlF_6

) at around

950^\circ\text{C}

to create a conducting medium. Upon melting, alumina dissociates into mobile

Al^{3+}

and

O^{2-}

ions. Applying an electric current causes

Al^{3+}

cations to migrate to the carbon cathode at the bottom, where they are reduced to liquid aluminium metal (

Al^{3+} + 3e^- \rightarrow Al_{(l)}

). Concurrently,

O^{2-}

anions migrate to the top graphite anodes and undergo oxidation to oxygen gas (

2O^{2-} \rightarrow O_{2(g)} + 4e^-

), which reacts with the hot carbon anodes to form carbon dioxide gas (

C + O_2 \rightarrow CO_2

).

Step-by-Step Solution

1

Identify the electrolyte preparation step in the Hall-Héroult cell.

Alumina is dissolved in molten cryolite at about

950^\circ\text{C}

.

Cryolite acts as a solvent and flux to lower the high melting point of pure alumina and enhance conductivity.

2

Determine the ionization behavior of the dissolved alumina.

Alumina dissociates into mobile

Al^{3+}

cations and

O^{2-}

anions.

Liquid state ionic dissociation is necessary for current transport through the electrolyte.

3

Trace the movement of cations under the applied electric field.

Al^{3+}

cations migrate to the negatively charged carbon cathode lining at the cell floor.

Electrostatic attraction draws positive ions toward the negative electrode.

4

Determine the chemical reaction occurring at the cathode.

Al^{3+}

ions gain electrons to form molten aluminium metal (

Al^{3+} + 3e^- \rightarrow Al_{(l)}

).

Cation gain of electrons at the cathode represents the reduction process that isolates elemental aluminium.

5

Determine the chemical reaction occurring at the anode and the fate of the anode material.

O^{2-}

ions lose electrons to produce oxygen gas, which reacts with graphite anodes to produce

CO_2

gas.

Anode oxidation releases oxygen gas at high temperature, causing carbon anodes to burn away continuously.

Key Concept

Electrolytic reduction of alumina in the Hall-Héroult process

Estimated Time:1m 30s

Question 35Question →

Match each chemical process or application involving iron and its compounds on the left with its corresponding chemical reaction characteristic or product on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches pair thermal decomposition of siderite with production of

\text{FeO}

and

\text{CO}_2

; reaction of iron with hot concentrated

\text{H}_2\text{SO}_4

with formation of iron(III) sulfate, water, and

\text{SO}_2

; galvanization with zinc acting as a sacrificial anode; and fluxing action of

\text{CaO}

with molten slag (

\text{CaSiO}_3

) formation.

Each process matches its corresponding chemical principle: siderite decomposes into

\text{FeO}

and

\text{CO}_2

; hot concentrated

\text{H}_2\text{SO}_4

oxidizes iron to iron(III) sulfate releasing

\text{SO}_2

; galvanization uses zinc as a sacrificial anode; and calcium oxide combines with silica to form slag.

Step-by-Step Solution

1

Analyze the thermal decomposition of siderite

Siderite (

\text{FeCO}_3

) decomposes upon heating to yield iron(II) oxide (

\text{FeO}

) and carbon(IV) oxide (

\text{CO}_2

).

Transition metal carbonates decompose thermally into the corresponding metal oxide and carbon dioxide.

2

Examine the reaction of iron with hot concentrated oxidizing acid

Hot concentrated

\text{H}_2\text{SO}_4

oxidizes iron metal to the iron(III) oxidation state, producing

\text{Fe}_2(\text{SO}_4)_3

,

\text{SO}_2

gas, and

\text{H}_2\text{O}

.

Hot concentrated acid acts as a strong oxidizing agent rather than liberating hydrogen gas as dilute acid would.

3

Evaluate galvanization for rusting prevention

Galvanization coats iron with zinc metal, which undergoes preferential cathodic protection as a sacrificial anode.

Zinc is higher in the electrochemical series than iron, so it corrodes preferentially even if scratched.

4

Identify the fluxing action of basic oxides in the blast furnace

Calcium oxide (

\text{CaO}

) reacts with acidic sandy impurities (

\text{SiO}_2

) to form molten slag (

\text{CaSiO}_3

).

Slag floats on top of molten iron, preventing re-oxidation of the extracted metal while removing silicon impurities.

Key Concept

Extraction, Reaction Properties, and Corrosion Prevention of Iron

Question 36Question →

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 37Question →

Arrange the following chemical and electrochemical stages of the rusting of iron in chronological sequence, from initial anodic oxidation to the final formation of rust.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct chronological sequence is: Oxidation of iron metal to

\text{Fe}^{2+}

ions

\rightarrow

Reduction of dissolved oxygen to

\text{OH}^-

ions

\rightarrow

Precipitation of

\text{Fe(OH)}_2

\rightarrow

Oxidation of

\text{Fe(OH)}_2

to hydrated

\text{Fe}_2\text{O}_3\cdot x\text{H}_2\text{O}

.

Rusting is an electrochemical process. Iron metal initially oxidizes at anodic sites to produce

\text{Fe}^{2+}

ions and electrons. The released electrons migrate through the iron to cathodic regions, where dissolved atmospheric oxygen is reduced to

\text{OH}^-

ions. These ions react in the electrolyte solution to form insoluble green

\text{Fe(OH)}_2

, which is subsequently oxidized by dissolved oxygen to form reddish-brown hydrated iron(III) oxide (rust,

\text{Fe}_2\text{O}_3\cdot x\text{H}_2\text{O}

).

Step-by-Step Solution

1

Identify the initial anodic oxidation process

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

Rusting begins as an electrochemical reaction where iron metal acts as the anode and undergoes oxidation.

2

Identify the cathodic reduction process

\text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4\text{e}^- \rightarrow 4\text{OH}^-\text{(aq)}

Electrons released during anodic oxidation flow to cathodic sites where atmospheric oxygen dissolved in water is reduced.

3

Determine the ionic precipitation reaction

\text{Fe}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} \rightarrow \text{Fe(OH)}_2\text{(s)}

The formed cations and anions diffuse towards each other in the aqueous electrolyte layer, forming an insoluble precipitate.

4

Determine the final oxidation step to rust

4\text{Fe(OH)}_2\text{(s)} + \text{O}_2\text{(g)} + 2x\text{H}_2\text{O(l)} \rightarrow 2(\text{Fe}_2\text{O}_3\cdot x\text{H}_2\text{O})\text{(s)}

Dissolved oxygen further oxidizes iron(II) hydroxide to hydrated iron(III) oxide, commonly known as rust.

Key Concept

Electrochemical mechanism of iron rusting

Question 38Question →

Match each calcium-containing substance on the left with its correct chemical function or industrial preparation description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Calcium fluoride matches with serving as a flux in calcium extraction; Calcium oxide matches with calcination product of limestone used to dry ammonia; Calcium sulfate hemihydrate matches with partial dehydration product of gypsum; Calcium hydroxide matches with slaked lime used to detect carbon(IV) oxide.

Each calcium compound is correctly paired according to standard industrial practices and chemical properties: calcium fluoride lowers the electrolytic bath melting point; calcium oxide is a basic desiccant produced from limestone; calcium sulfate hemihydrate is formed by partially dehydrating gypsum; and slaked lime solution forms a precipitate with carbon(IV) oxide.

Step-by-Step Solution

1

Identify the industrial metallurgical role of calcium fluoride in calcium metal extraction.

Calcium fluoride acts as a flux in fused

\text{CaCl}_2

electrolysis to decrease the operating temperature.

Lowering the melting point improves electrical conductivity and reduces thermal energy consumption.

2

Analyze the industrial preparation and chemical nature of calcium oxide.

Thermal decomposition of

\text{CaCO}_3

yields basic

\text{CaO}

, which does not react with basic gases like

\text{NH}_3

.

Acidic drying agents such as concentrated

\text{H}_2\text{SO}_4

would react with ammonia, making basic quicklime the required choice.

3

Determine the formula and thermal origin of Plaster of Paris.

Controlled heating of gypsum yields calcium sulfate hemihydrate (

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

).

Heating at

120^\circ\text{C}

drives off part of the water of crystallization without causing complete dehydration to anhydrous anhydrite.

4

Relate calcium hydroxide to its slaking reaction and analytical application.

Slaking

\text{CaO}

produces

\text{Ca(OH)}_2

, whose aqueous solution forms an insoluble milky

\text{CaCO}_3

precipitate with

\text{CO}_2

.

The reaction of dissolved calcium hydroxide with carbon(IV) oxide produces insoluble calcium trioxocarbonate(IV).

Key Concept

Chemical properties, industrial preparations, and extraction roles of calcium and its major compounds.

Question 39Question →

When excess copper turnings are reacted with dilute trioxonitrate(V) acid ( $\text{HNO}_3$ ), $1.12\text{ dm}^3$ of nitrogen(II) oxide ( $\text{NO}$ ) gas, measured at STP, is evolved. What mass of copper metal was oxidized during this reaction?

$(\text{Molar mass of Cu} = 64\text{ g mol}^{-1},\text{ molar volume of gas at STP} = 22.4\text{ dm}^3\text{ mol}^{-1})$

4.8\text{ g}

3.2\text{ g}

1.6\text{ g}

2.1\text{ g}

Show answer & explanation

Answer: $4.8\text{ g}$

Answer

4.8\text{ g}

of copper metal was oxidized.

The option specifying

4.8\text{ g}

is correct because the balanced chemical reaction of copper turnings with dilute trioxonitrate(V) acid is

3\text{Cu} + 8\text{HNO}_3 \rightarrow 3\text{Cu(NO}_3)_2 + 2\text{NO} + 4\text{H}_2\text{O}

. Converting

1.12\text{ dm}^3

of

\text{NO}

at STP gives

0.05\text{ mol}

. Multiplying by the stoichiometric ratio

\frac{3}{2}

gives

0.075\text{ mol}

of

\text{Cu}

, which equals

0.075 \times 64 = 4.8\text{ g}

.

Step-by-Step Solution

1

Write the balanced chemical equation for the reaction of copper with dilute trioxonitrate(V) acid.

3\text{Cu}(s) + 8\text{HNO}_3(aq) \rightarrow 3\text{Cu(NO}_3)_2(aq) + 2\text{NO}(g) + 4\text{H}_2\text{O}(l)

Dilute

\text{HNO}_3

acts as an oxidizing agent, reducing to nitrogen(II) oxide (

\text{NO}

) gas rather than hydrogen gas.

2

Calculate the moles of nitrogen(II) oxide (

\text{NO}

) gas produced at STP.

\text{Moles of NO} = \frac{1.12\text{ dm}^3}{22.4\text{ dm}^3\text{ mol}^{-1}} = 0.05\text{ mol}

At STP, one mole of any ideal gas occupies

22.4\text{ dm}^3

.

3

Use the mole ratio to determine moles of copper oxidized.

\text{Moles of Cu} = 0.05\text{ mol NO} \times \left(\frac{3\text{ mol Cu}}{2\text{ mol NO}}\right) = 0.075\text{ mol Cu}

The stoichiometric ratio between

\text{Cu}

and

\text{NO}

is 3:2.

4

Convert moles of copper to mass in grams.

\text{Mass of Cu} = 0.075\text{ mol} \times 64\text{ g mol}^{-1} = 4.8\text{ g}

Mass is calculated by multiplying the number of moles by the molar mass.

Key Concept

Stoichiometry of redox reactions involving transition metals and oxidizing acids

Question 40Question →

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

Metals and Their Compounds

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept