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Question 1281Question →

Match each thermodynamic sign combination of enthalpy change (ΔH\Delta H) and entropy change (ΔS\Delta S) with its corresponding condition for reaction spontaneity.

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Items

ΔH<0\Delta H < 0 and ΔS>0\Delta S > 0
ΔH>0\Delta H > 0 and ΔS<0\Delta S < 0
ΔH<0\Delta H < 0 and ΔS<0\Delta S < 0
ΔH>0\Delta H > 0 and ΔS>0\Delta S > 0

Matches

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Answer

The thermodynamic combinations match as follows: negative enthalpy change and positive entropy change match spontaneous at all temperatures; positive enthalpy change and negative entropy change match non-spontaneous at all temperatures; negative enthalpy change and negative entropy change match spontaneous only at low temperatures; positive enthalpy change and positive entropy change match spontaneous only at high temperatures.
A reaction is spontaneous when ΔG<0\Delta G < 0, according to the Gibbs free energy relationship ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S. Combining a negative enthalpy change with a positive entropy change ensures ΔG\Delta G is negative at all temperatures. Combining a positive enthalpy change with a negative entropy change ensures ΔG\Delta G is positive at all temperatures. When enthalpy and entropy changes share the same sign, temperature dictates spontaneity: negative signs require low temperatures, while positive signs require high temperatures.

Step-by-Step Solution

1
Recall the Gibbs Free Energy equation
ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S, where TT is temperature in Kelvin (T>0T > 0).
Reaction spontaneity depends on the sign of ΔG\Delta G; a process is spontaneous if ΔG<0\Delta G < 0.
2
Analyze each sign combination in the equation
1. Negative ΔH\Delta H and positive ΔS\Delta S gives ΔG=(−)−T(+)=(−)\Delta G = (-) - T(+) = (-), always spontaneous.
2. Positive ΔH\Delta H and negative ΔS\Delta S gives ΔG=(+)−T(−)=(+)\Delta G = (+) - T(-) = (+), always non-spontaneous.
3. Negative ΔH\Delta H and negative ΔS\Delta S gives ΔG=(−)+T(+)\Delta G = (-) + T(+), which is negative only at low TT.
4. Positive ΔH\Delta H and positive ΔS\Delta S gives ΔG=(+)−T(+)\Delta G = (+) - T(+), which is negative only at high TT.
Evaluating the magnitude of TΔST\Delta S relative to ΔH\Delta H determines the temperature dependency of spontaneity.

Key Concept

Gibbs Free Energy and Reaction Spontaneity
Question 1282Question →

Match each mixture separation scenario on the left with its corresponding application of simple or fractional distillation on the right.

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Items

Separation of ethanol (boiling point 78∘C78^\circ\text{C}) and water (boiling point 100∘C100^\circ\text{C})
Recovery of pure water from a sodium chloride salt solution
Refining crude oil into petrol, kerosene, and gas oil
Separation of liquid nitrogen (boiling point −196∘C-196^\circ\text{C}) and liquid oxygen (boiling point −183∘C-183^\circ\text{C})

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Answer

Ethanol-water mixture matches laboratory fractional distillation of miscible liquids with close boiling points; recovery of pure water from salt solution matches simple distillation of volatile solvent from non-volatile solute; refining crude oil matches industrial fractional distillation of petroleum; liquid air separation matches cryogenic industrial fractional distillation.
Simple distillation is used when separating a volatile solvent from a non-volatile solute (e.g., pure water from sea water/salt solution). Fractional distillation is required when separating miscible liquids with close boiling points (e.g., ethanol and water, petroleum crude oil, or liquefied air components).

Step-by-Step Solution

1
Identify the nature of the solute and solvent in each mixture.
Determine whether components are volatile miscible liquids or a liquid containing non-volatile dissolved solids.
Simple distillation is suitable when separating a liquid solvent from a non-volatile solid, whereas fractional distillation is required for two or more miscible volatile liquids.
2
Compare boiling points for miscible liquid pairs.
Ethanol (78∘C78^\circ\text{C}) and water (100∘C100^\circ\text{C}) have close boiling points requiring repeated condensation-vaporization cycles provided by a fractionating column.
A fractionating column packed with glass beads provides surface area for multiple simple distillations to occur simultaneously.
3
Match industrial processes to their scaled applications.
Crude oil refining uses tall fractionating towers, while air liquefaction separates sub-zero boiling gases like nitrogen and oxygen.
Industrial separation scales fractional distillation principles to handle petroleum fractions and liquefied gas mixtures.

Key Concept

Distinguishing between simple distillation (volatile solvent from non-volatile solute) and fractional distillation (miscible liquids with close boiling points).
Estimated Time:1m 0s
Question 1283Question →

Match each condition or graphical representation of a gas sample obeying Charles's law with its corresponding physical or mathematical outcome.

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Items

A gas sample is heated from 27∘C27^\circ\text{C} to 54∘C54^\circ\text{C} at constant pressure
A gas sample is heated from 300 K300\text{ K} to 600 K600\text{ K} at constant pressure
Extrapolation of a volume versus temperature (∘C^\circ\text{C}) plot to zero volume
Plot of volume (VV) versus absolute temperature (TT in K\text{K}) at constant pressure

Matches

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Answer

The correct pairings match heating from 27 °C to 54 °C with a volume increase factor of 1.09, heating from 300 K to 600 K with volume doubling, extrapolation to zero volume with intersecting the temperature axis at absolute zero (-273 °C), and the V versus T (K) plot with a straight line through the origin.
Each item correctly links a concept of Charles's law to its outcome: heating from 27∘C27^\circ\text{C} (300 K300\text{ K}) to 54∘C54^\circ\text{C} (327 K327\text{ K}) increases volume by a factor of 1.091.09; heating from 300 K300\text{ K} to 600 K600\text{ K} doubles absolute temperature and volume; extrapolating a VV-TT (∘C^\circ\text{C}) graph to zero volume yields absolute zero (−273∘C-273^\circ\text{C}); and plotting VV against absolute temperature (TT in K\text{K}) produces a straight line passing through the origin.

Step-by-Step Solution

1
Convert Celsius temperatures to Kelvin for the first scenario
T1=27+273=300 KT_1 = 27 + 273 = 300\text{ K} and T2=54+273=327 KT_2 = 54 + 273 = 327\text{ K}.
Charles's law requires temperatures to be expressed in Kelvin.
2
Determine the volume ratio for heating from 27∘C27^\circ\text{C} to 54∘C54^\circ\text{C}
\frac{V_2}{V_1} = \frac{T_2}{T_1} = \frac{327\text{ K}}{300\text{ K}} = 1.09.
Doubling Celsius temperature does not double absolute temperature, so volume increases by a factor of 1.09.
3
Apply direct proportionality for the Kelvin heating scenario
Heating from 300 K300\text{ K} to 600 K600\text{ K} doubles the absolute temperature, so V2=2V1V_2 = 2V_1.
Volume is directly proportional to temperature on the Kelvin scale.
4
Analyze graphical characteristics of Charles's law
A VV vs TT (K) plot is a straight line through the origin (0,0)(0,0), and extrapolating VV vs TT (∘C^\circ\text{C}) to zero volume yields absolute zero (−273∘C-273^\circ\text{C}).
Absolute zero is the theoretical temperature at which gas volume extrapolates to zero.

Key Concept

Charles's Law and Absolute Temperature Scale
Estimated Time:1m 30s
Question 1284Question →

Match each municipal water purification stage or chemical additive on the left with its precise operational mechanism on the right.

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Items

Aeration
Addition of Potash Alum
Sand Filtration
Addition of Slaked Lime

Matches

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Answer

Aeration matches with expelling volatile gases and oxidizing soluble iron(II); Addition of Potash Alum matches with neutralizing negative charges on clay colloids for flocculation; Sand Filtration matches with straining out fine suspended particles remaining after sedimentation; Addition of Slaked Lime matches with adjusting pH of acidic water post-coagulation.
Aeration oxidizes soluble iron and removes dissolved gases. Potash alum supplies trivalent cations to neutralize negative charges on colloidal clay. Sand filtration physically retains fine suspended particles. Slaked lime neutralizes acidity induced by alum coagulation.

Step-by-Step Solution

1
Analyze the primary chemical and physical effects of Aeration.
Spraying raw water into air maximizes gas exchange, driving off dissolved volatile species (H2SH_2S) and converting soluble Fe2+Fe^{2+} ions to insoluble Fe(OH)3Fe(OH)_3 precipitate.
Aeration targets volatile odors/tastes and dissolved metals rather than solid particulate filtration.
2
Analyze the coagulating action of Potash Alum (KAl(SO4)2⋅12H2OKAl(SO_4)_2 \cdot 12H_2O).
Trivalent Al3+Al^{3+} ions neutralize negative charges on suspended clay colloids, causing them to coalesce into larger settleable flocs.
Fine clay particles do not settle spontaneously due to mutual electrostatic repulsion.
3
Determine the function of Sand Filtration.
Water percolating through sand layer beds leaves behind non-settled micro-particles.
Filtration serves as a final physical straining stage after bulk sedimentation.
4
Evaluate the requirement for Slaked Lime (Ca(OH)2Ca(OH)_2).
Neutralizes acidity produced during alum hydrolysis, raising the pH to safe alkaline levels.
Acidic water damages metal distribution pipes through corrosive action.

Key Concept

Operational mechanisms of chemical and physical processes in town water supply treatment
Question 1285Question →

Match each physical property of metallic elements listed on the left with the atomic-scale mechanism on the right that best accounts for it.

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Items

High thermal conductivity
Malleability and ductility
High melting point and tensile strength
Metallic luster and opacity

Matches

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Answer

High thermal conductivity matches rapid transfer of kinetic energy by mobile delocalized electrons; Malleability and ductility matches non-directional electrostatic attractions permitting cation layers to slide; High melting point matches strong multi-directional electrostatic attraction between cations and the electron sea; Metallic luster matches oscillation and re-emission of incident photons by surface delocalized electrons.
Each macro-level property corresponds directly to specific behaviors of the delocalized electron sea and cation lattice: thermal conduction relies on mobile electron kinetic transport; malleability depends on non-directional bonding allowing cation layers to slip; high melting points result from strong multi-directional electrostatic attractions; and luster is caused by surface electron excitation and photon re-emission.

Step-by-Step Solution

1
Analyze the microscopic origin of thermal transport in metals.
Identify mobile delocalized valence electrons as the primary carriers of thermal kinetic energy.
Delocalized electrons move rapidly through the lattice when a temperature gradient is applied, transferring kinetic energy much faster than localized atomic vibrations.
2
Examine the mechanism of mechanical deformation under applied stress.
Identify non-directional electrostatic attraction enabling cation layers to slide without fracture.
Unlike ionic crystals where sliding brings like charges into repelling contact, metallic electron clouds shield shifting cations, preserving lattice cohesion.
3
Evaluate the structural requirements for melting and high mechanical strength.
Identify strong multi-directional electrostatic binding throughout the 3D lattice.
Overcoming the structural stability requires substantial energy to disrupt the strong net electrostatic pull between positive metal ions and delocalized electrons.
4
Determine the interaction of metal surfaces with electromagnetic radiation.
Connect surface delocalized electrons to photon absorption and rapid re-emission.
Unbound surface valence electrons absorb incoming light energy and immediately vibrate and re-emit the photons, producing specular reflection.

Key Concept

Connecting macroscopic physical properties of metals to the delocalized electron sea model and non-directional metallic bonding.
Question 1286Question →

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

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Items

Addition of calcium fluoride (CaF2\text{CaF}_2) during the electrolytic extraction of calcium metal
Exothermic hydration of quicklime (CaO\text{CaO}) to yield slaked lime
Reaction of dry slaked lime (Ca(OH)2\text{Ca(OH)}_2) with chlorine gas at room temperature
Rehydration and setting mechanism of Plaster of Paris (CaSO4⋅12H2O\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O})

Matches

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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 CaF2\text{CaF}_2 as a flux that lowers the melting point of fused CaCl2\text{CaCl}_2 from 800∘C800^\circ\text{C} to 600∘C600^\circ\text{C}.
Electrolysis of pure CaCl2\text{CaCl}_2 requires high temperature where molten calcium would dissolve in the electrolyte, so CaF2\text{CaF}_2 is added as a flux.
2
Evaluate the chemical properties of calcium oxide and its slaked product
Slaking CaO\text{CaO} gives Ca(OH)2\text{Ca(OH)}_2, which precipitates Ca(HCO3)2\text{Ca(HCO}_3)_2 in Clark's method.
Calcium hydroxide reacts with hydrogen carbonate ions to precipitate insoluble CaCO3\text{CaCO}_3, removing temporary hardness.
3
Determine the industrial reaction between slaked lime and chlorine
Chlorination of dry Ca(OH)2\text{Ca(OH)}_2 yields bleaching powder, CaOCl2⋅H2O\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 1287Question →

Match each set of thermodynamic conditions for a chemical reaction on the left with its corresponding temperature-dependent spontaneity behavior on the right, based on the Gibbs free energy relationship ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S.

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Items

Endothermic process (ΔH>0\Delta H > 0) accompanied by an increase in entropy (ΔS>0\Delta S > 0)
Exothermic process (ΔH<0\Delta H < 0) accompanied by a decrease in entropy (ΔS<0\Delta S < 0)
Endothermic process (ΔH>0\Delta H > 0) accompanied by a decrease in entropy (ΔS<0\Delta S < 0)
Exothermic process (ΔH<0\Delta H < 0) accompanied by an increase in entropy (ΔS>0\Delta S > 0)

Matches

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Answer

1. Endothermic with ΔS>0\Delta S > 0 matches Spontaneous only at high temperatures (T>ΔHΔST > \frac{\Delta H}{\Delta S}).
2. Exothermic with ΔS<0\Delta S < 0 matches Spontaneous only at low temperatures (T<ΔHΔST < \frac{\Delta H}{\Delta S}).
3. Endothermic with ΔS<0\Delta S < 0 matches Non-spontaneous at all temperatures.
4. Exothermic with ΔS>0\Delta S > 0 matches Spontaneous at all temperatures.
Each pair is matched by evaluating the sign of ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S. An endothermic reaction with positive entropy change requires high temperature to make TΔS>ΔHT\Delta S > \Delta H. An exothermic reaction with negative entropy change requires low temperature to keep ∣ΔH∣>T∣ΔS∣|\Delta H| > T|\Delta S|. An endothermic reaction with negative entropy change yields positive ΔG\Delta G at all temperatures. An exothermic reaction with positive entropy change yields negative ΔG\Delta G at all temperatures.

Step-by-Step Solution

1
Recall the fundamental thermodynamic criterion for spontaneity
A reaction is spontaneous when Gibbs free energy change is negative (ΔG<0\Delta G < 0), given by ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S.
Temperature TT in Kelvin is always positive, so the sign of ΔG\Delta G depends on the combination of signs of ΔH\Delta H and ΔS\Delta S.
2
Analyze Case 1: ΔH>0\Delta H > 0 and ΔS>0\Delta S > 0
ΔG=(+value)−T(+value)\Delta G = (+\text{value}) - T(+\text{value}). For ΔG<0\Delta G < 0, TΔS>ΔH  ⟹  T>ΔHΔST\Delta S > \Delta H \implies T > \frac{\Delta H}{\Delta S}.
The reaction is driven by entropy increase at elevated temperatures.
3
Analyze Case 2: ΔH<0\Delta H < 0 and ΔS<0\Delta S < 0
ΔG=(−value)−T(−value)=−∣ΔH∣+T∣ΔS∣\Delta G = (-\text{value}) - T(-\text{value}) = -|\Delta H| + T|\Delta S|. For ΔG<0\Delta G < 0, ∣ΔH∣>T∣ΔS∣  ⟹  T<ΔHΔS|\Delta H| > T|\Delta S| \implies T < \frac{\Delta H}{\Delta S}.
The reaction is enthalpy-driven and requires low temperatures so that the positive −TΔS-T\Delta S term does not outweigh ΔH\Delta H.
4
Analyze Case 3: ΔH>0\Delta H > 0 and ΔS<0\Delta S < 0
ΔG=(+value)−T(−value)=(+value)+T(+value)>0\Delta G = (+\text{value}) - T(-\text{value}) = (+\text{value}) + T(+\text{value}) > 0 always.
Both enthalpy and entropy factors oppose spontaneity.
5
Analyze Case 4: ΔH<0\Delta H < 0 and ΔS>0\Delta S > 0
ΔG=(−value)−T(+value)<0\Delta G = (-\text{value}) - T(+\text{value}) < 0 always.
Both enthalpy release and entropy increase favor spontaneity under all conditions.

Key Concept

Dependence of Gibbs Free Energy and Reaction Spontaneity on the Signs of Enthalpy and Entropy Changes
Question 1288Question →

Match each industrial chemical process or biotechnology application listed on the left with its corresponding catalyst or biological agent on the right.

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Items

Haber process for ammonia synthesis
Contact process for tetraoxosulfate(VI) acid synthesis
Fermentation of glucose to ethanol
Hydrogenation of vegetable oils to margarine

Matches

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Answer

Haber process matches Finely divided iron (FeFe); Contact process matches Vanadium(V) oxide (V2O5V_2O_5); Fermentation of glucose matches Zymase enzyme; Hydrogenation of vegetable oils matches Finely divided nickel (NiNi).
Each process relies on a specific catalyst or biological agent: the Haber process utilizes finely divided iron, the Contact process uses vanadium(V) oxide, glucose fermentation relies on the biological enzyme zymase, and vegetable oil hydrogenation uses finely divided nickel.

Step-by-Step Solution

1
Identify the catalyst used in ammonia production via the Haber process.
Finely divided iron accelerates the reversible reaction between N2N_2 and H2H_2.
Iron lowers the activation energy required to break the strong triple bond in nitrogen.
2
Identify the catalyst in the Contact process stage converting SO2SO_2 to SO3SO_3.
Vanadium(V) oxide (V2O5V_2O_5) is the modern industrial catalyst employed.
V2O5V_2O_5 provides high efficiency and is resistant to catalytic poisoning compared to platinum.
3
Identify the biocatalyst in ethanol fermentation.
Zymase, an enzyme complex produced by yeast cells, catalyzes glucose breakdown.
Fermentation is an anaerobic biotechnological process relying on enzymatic activity.
4
Identify the catalyst used in margarine synthesis.
Finely divided nickel is used during unsaturated oil hydrogenation.
Nickel adsorbs hydrogen gas and unsaturated hydrocarbon chains onto its surface to facilitate addition.

Key Concept

Industrial catalysts and biological enzymes in commercial chemical processes
Question 1289Question →

Match each type of organic isomerism on the left with its corresponding example pair of compounds on the right.

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Items

Chain isomerism
Positional isomerism
Functional group isomerism
Geometric isomerism

Matches

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Answer

Chain isomerism pairs with butane and 22-methylpropane; Positional isomerism pairs with drop-in pair butan-11-ol and butan-22-ol; Functional group isomerism pairs with ethanoic acid and methyl methanoate; Geometric isomerism pairs with ciscis-but-22-ene and transtrans-but-22-ene.
Each type of isomerism is matched to its definitive structural characteristic: chain isomerism involves skeleton branching changes (butane and 22-methylpropane), positional isomerism involves relocations of the same functional group (butan-11-ol and butan-22-ol), functional group isomerism involves different organic families sharing a formula (ethanoic acid and methyl methanoate), and geometric isomerism involves spatial orientations across a double bond (ciscis-but-22-ene and transtrans-but-22-ene).

Step-by-Step Solution

1
Analyze the carbon skeletons of butane and 22-methylpropane.
Both have formula C4H10C_4H_{10}, but one is unbranched while the other is branched, defining chain isomerism.
Chain isomers possess identical molecular formulas but different carbon chain arrangements.
2
Examine the functional group positions in butan-11-ol and butan-22-ol.
The −OH-\text{OH} group is on carbon-11 in butan-11-ol and carbon-22 in butan-22-ol, defining positional isomerism.
Positional isomers have the same functional group located on different carbon atoms along the same parent chain.
3
Compare the functional groups of ethanoic acid and methyl methanoate.
Ethanoic acid is a carboxylic acid (CH3COOHCH_3COOH) while methyl methanoate is an ester (HCOOCH3HCOOCH_3). Both have the formula C2H4O2C_2H_4O_2, defining functional group isomerism.
Functional group isomers share a molecular formula but belong to different organic homologous families.
4
Evaluate the spatial arrangement in ciscis-but-22-ene and transtrans-but-22-ene.
The double bond restricts rotation, placing methyl groups on the same side (ciscis) or opposite sides (transtrans), defining geometric isomerism.
Geometric isomerism is a stereoisomerism type arising from restricted double-bond rotation.

Key Concept

Classification of Structural Isomerism and Stereoisomerism
Question 1290Question →

Match each redox reaction mixture (left) with its corresponding characteristic laboratory observation (right).

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Items

Acidified KMnO4(aq)KMnO_4(aq) mixed with sulfur(IV) oxide gas (SO2SO_2)
Acidified K2Cr2O7(aq)K_2Cr_2O_7(aq) treated with iron(II) tetraoxosulfate(VI) solution (FeSO4FeSO_4)
Chlorine gas (Cl2Cl_2) bubbled through potassium iodide solution (KIKI)
Hydrogen sulfide gas (H2SH_2S) passed into iron(III) chloride solution (FeCl3FeCl_3)

Matches

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Answer

Acidified KMnO4KMnO_4 with SO2SO_2 changes from purple to colorless (Mn2+Mn^{2+}); Acidified K2Cr2O7K_2Cr_2O_7 with FeSO4FeSO_4 changes from orange to green (Cr3+Cr^{3+}); Chlorine gas with KIKI forms brown iodine (I2I_2); Hydrogen sulfide gas with FeCl3FeCl_3 reduces Fe3+Fe^{3+} to Fe2+Fe^{2+} (pale green) with a yellow sulfur precipitate.
Each test pair matches an oxidizing or reducing reagent with its definitive qualitative laboratory test observation: permanganate decolorizes upon reduction, dichromate turns green upon reduction, iodide oxidizes to brown iodine, and iron(III) reduces to pale green iron(II) alongside yellow sulfur precipitation.

Step-by-Step Solution

1
Determine the redox behavior of SO2SO_2 with acidified KMnO4KMnO_4
SO2SO_2 is oxidized while reducing purple MnO4−MnO_4^- to colorless Mn2+Mn^{2+}.
Potassium permanganate test for reducing agents involves reduction of manganese from oxidation state +7 to +2.
2
Determine the redox behavior of FeSO4FeSO_4 with acidified K2Cr2O7K_2Cr_2O_7
Fe2+Fe^{2+} ions oxidize to Fe3+Fe^{3+} while Cr2O72−Cr_2O_7^{2-} (orange) is reduced to Cr3+Cr^{3+} (green).
Dichromate(VI) is a standard oxidizing agent whose reduced form contains green chromium(III) ions.
3
Determine the reaction of Cl2Cl_2 with KIKI
Cl2Cl_2 oxidizes colorless I−I^- ions to elemental iodine (I2I_2), turning the solution brown.
Chlorine is a stronger oxidizing agent than iodine and displaces iodide from solution.
4
Determine the reaction of H2SH_2S with FeCl3FeCl_3
H2SH_2S reduces yellow-brown Fe3+Fe^{3+} to pale green Fe2+Fe^{2+} while forming a insoluble yellow deposit of sulfur.
Hydrogen sulfide acts as a reducing agent and deposits elemental sulfur upon oxidation.

Key Concept

Laboratory tests and characteristic color changes for oxidizing and reducing agents.
Question 1291Question →

Match each redox reaction scenario involving an oxidizing or reducing agent on the left with its corresponding characteristic laboratory observation on the right.

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Items

Bubbling sulfur(IV) oxide (SO2SO_2) gas into acidified potassium tetraoxomanganate(VII) (KMnO4KMnO_4) solution
Passing chlorine (Cl2Cl_2) gas into aqueous potassium iodide (KIKI) solution
Adding concentrated trioxonitrate(V) acid (HNO3HNO_3) to freshly prepared iron(II) tetraoxosulfate(VI) (FeSO4FeSO_4) solution
Bubbling hydrogen sulfide (H2SH_2S) gas through iron(III) chloride (FeCl3FeCl_3) solution

Matches

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Answer

The correct pairings match each redox reagent with its specific electron-transfer observation: SO2SO_2 with acidified KMnO4KMnO_4 produces a purple to colorless change; Cl2Cl_2 with aqueous KIKI turns colorless solution brown; conc. HNO3HNO_3 with FeSO4FeSO_4 converts pale green solution to brown; and H2SH_2S with FeCl3FeCl_3 converts yellow/brown solution to pale green with yellow sulfur deposit.
Each pair correctly links the specific chemical species undergoing oxidation or reduction to its empirical qualitative test result. Sulfur(IV) oxide decolorizes acidified potassium tetraoxomanganate(VII); chlorine oxidizes iodide ions to brown iodine; concentrated trioxonitrate(V) acid converts green iron(II) to brown iron(III); and hydrogen sulfide reduces brown iron(III) to green iron(II) with yellow sulfur precipitation.

Step-by-Step Solution

1
Analyze the redox roles of the reagents in each left item.
SO2SO_2 and H2SH_2S act as reducing agents; Cl2Cl_2 and conc. HNO3HNO_3 act as oxidizing agents.
Identifying whether a species donates or accepts electrons determines the expected chemical transformation of the target solution.
2
Determine the oxidation state change and color change for SO2SO_2 + acidified KMnO4KMnO_4.
MnO4−MnO_4^- (oxidation state +7, purple) is reduced to Mn2+Mn^{2+} (oxidation state +2, colorless).
Manganate(VII) reduction is the standard test for reducing agents like SO2SO_2.
3
Determine the oxidation state change and color change for Cl2Cl_2 + aqueous KIKI.
I−I^- (oxidation state -1, colorless) is oxidized to I2I_2 (oxidation state 0, brown).
Halogen displacement shows chlorine's higher electronegativity and oxidizing strength compared to iodine.
4
Determine the oxidation state change for conc. HNO3HNO_3 + FeSO4FeSO_4 and H2SH_2S + FeCl3FeCl_3.
Conc. HNO3HNO_3 oxidizes pale green Fe2+Fe^{2+} to brown Fe3+Fe^{3+}. H2SH_2S reduces yellow/brown Fe3+Fe^{3+} to pale green Fe2+Fe^{2+} with precipitate of sulfur.
Iron transitions between +2 (pale green) and +3 (yellow/brown) depending on whether an oxidant or reductant is introduced.

Key Concept

Laboratory identification of oxidizing and reducing agents via characteristic color changes and oxidation state transitions
Estimated Time:2m 0s
Question 1292Question →

Match each observed physical property or phenomenon of ionic (electrovalent) compounds on the left with its correct underlying thermodynamic or structural explanation on the right.

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Items

Magnesium oxide (MgOMgO) exhibits an exceptionally high melting point (2852∘C2852^\circ\text{C}) compared to sodium chloride (NaClNaCl, 801∘C801^\circ\text{C}).
Anhydrous aluminium iodide (AlI3AlI_3) exhibits marked covalent character and a low melting point (191∘C191^\circ\text{C}) despite forming between a metal and a non-metal.
Sodium hydroxide (NaOHNaOH) dissolves exothermically in water despite requiring energy to break its crystal lattice.
Solid calcium fluoride (CaF2CaF_2) is an electrical insulator, but conducts electricity readily when melted.

Matches

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Answer

1 matches with the explanation of charge product dependence on lattice energy; 2 matches with the explanation of Fajans' rules of polarization; 3 matches with the explanation of hydration enthalpy exceeding lattice enthalpy; 4 matches with the explanation of ion mobility in molten versus solid states.
Each physical property directly corresponds to its underlying quantum mechanical or thermodynamic principle: lattice energy scales with charge product (MgOMgO vs NaClNaCl), polarization of anion electron clouds by small high-charge cations creates covalent character (AlI3AlI_3), exothermic dissolution occurs when hydration energy exceeds lattice energy (NaOHNaOH), and electrical conduction requires mobile ions that are locked in solids but liberated upon melting (CaF2CaF_2).

Step-by-Step Solution

1
Analyze the high melting point of MgOMgO versus NaClNaCl
Lattice energy is governed by Coulomb's law: E∝∣q1q2∣rE \propto \frac{|q_1 q_2|}{r}. MgOMgO consists of Mg2+Mg^{2+} and O2−O^{2-} (product = 4), while NaClNaCl consists of Na+Na^+ and Cl−Cl^- (product = 1). Higher charge product leads to stronger lattice attraction and a higher melting point.
Identify the primary thermodynamic factor controlling lattice strength in ionic crystals.
2
Analyze the anomalous covalent behavior of AlI3AlI_3
Apply Fajans' rules: Covalency increases with high cation charge density and large anion size. Al3+Al^{3+} has high charge density and I−I^- is large and easily polarized, leading to electron cloud sharing (covalent character).
Explain deviations from purely electrovalent behavior using polarization principles.
3
Analyze the thermochemistry of dissolution of NaOHNaOH
Dissolution enthalpy ΔHsoln=ΔHlat+ΔHhyd\Delta H_{soln} = \Delta H_{lat} + \Delta H_{hyd}. If hydration enthalpy released is greater in magnitude than the lattice enthalpy required to separate ions, the net process is exothermic.
Relate lattice energy and hydration energy to dissolution energetics.
4
Analyze electrical conductivity in solid versus molten CaF2CaF_2
Solid ionic compounds contain ions held rigidly in a lattice structure. When melted, thermal energy breaks the lattice, producing free-moving ions capable of carrying electrical current.
Distinguish between mobile charge carriers (molten state) and immobile lattice positions (solid state).

Key Concept

Thermodynamic and structural factors governing ionic lattice stability, polarization (Fajans' rules), solution energetics, and state-dependent conductivity.
Question 1293Question →

Match each organic reagent setup or chemical process in Column A with its corresponding chemical reaction product or characteristic diagnostic observation in Column B.

Click a left item, then click its matching right item

Items

Bubbling but−1−ynebut-1-yne gas into ammoniacal silver nitrate solution, [Ag(NH3)2]NO3[Ag(NH_3)_2]NO_3
Passing propenepropene gas into cold, dilute alkaline potassium tetraoxomanganate(VII) solution, KMnO4KMnO_4
Heating excess ethanol with concentrated tetraoxosulfate(VI) acid, H2SO4H_2SO_4, at 170∘C170^\circ\text{C}
Controlled addition of cold water to calcium dicarbide, CaC2CaC_2

Matches

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Answer

Bubbling but−1−ynebut-1-yne into ammoniacal silver nitrate forms a white precipitate of silver but-1-ynide due to acidic acetylenic hydrogen replacement; passing propenepropene into cold dilute alkaline KMnO4KMnO_4 decolourizes purple MnO4−MnO_4^- forming a diol and brown MnO2MnO_2; heating ethanol with excess concentrated H2SO4H_2SO_4 at 170∘C170^\circ\text{C} produces ethene via intramolecular dehydration; adding water to CaC2CaC_2 yields ethyne gas via hydrolysis.
The matches correctly pair each chemical reaction with its distinctive mechanism or diagnostic test outcome: terminal alkyne acidity forming silver salts, alkene hydroxylation via Baeyer's reagent, alcohol elimination yielding ethene at elevated temperature, and carbide hydrolysis generating ethyne.

Step-by-Step Solution

1
Analyze the reaction of terminal alkynes with ammoniacal silver nitrate
but−1−ynebut-1-yne is a 1-alkyne containing a hydrogen atom bonded to an spsp-hybridized carbon (CH3CH2C≡CH \text{CH}_3\text{CH}_2\text{C}\equiv\text{CH}). This hydrogen is slightly acidic and is readily displaced by Ag+Ag^+ to form a insoluble white precipitate of silver but-1-ynide.
To distinguish terminal alkynes from internal alkynes and alkenes.
2
Analyze the reaction of alkenes with Baeyer's reagent
propenepropene (CH3CH=CH2 \text{CH}_3\text{CH}=\text{CH}_2) undergoes syn-hydroxylation with cold, dilute alkaline KMnO4KMnO_4 to form propane−1,2−diolpropane-1,2-diol. The purple trioxomanganate(VII) is reduced to brown manganese(IV) oxide (MnO2MnO_2).
To identify mild oxidation of double bonds in unsaturation testing.
3
Analyze the acid-catalyzed dehydration condition of ethanol
Heating ethanol with concentrated H2SO4H_2SO_4 at a high temperature (170∘C170^\circ\text{C}) favours intramolecular elimination of water yielding ethene (C2H4C_2H_4).
Lower temperatures (140∘C140^\circ\text{C}) yield ethoxyethane instead, so temperature controls product selectivity.
4
Analyze the laboratory preparation of ethyne
Calcium dicarbide (CaC2CaC_2) reacts directly with water to yield ethyne gas (C2H2C_2H_2) and calcium hydroxide (Ca(OH)2Ca(OH)_2).
This is the primary laboratory synthesis method for ethyne.

Key Concept

Chemical reactions, laboratory preparation methods, and diagnostic unsaturation tests for alkenes and alkynes.
Question 1294Question →

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

Items

Thermal decomposition of siderite ore (FeCO3\text{FeCO}_3)
Reaction of iron metal with hot concentrated tetraoxosulfate(VI) acid (H2SO4\text{H}_2\text{SO}_4)
Galvanization of iron sheets
Fluxing action of calcium oxide (CaO\text{CaO}) in the blast furnace

Matches

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Answer

The correct matches pair thermal decomposition of siderite with production of FeO\text{FeO} and CO2\text{CO}_2; reaction of iron with hot concentrated H2SO4\text{H}_2\text{SO}_4 with formation of iron(III) sulfate, water, and SO2\text{SO}_2; galvanization with zinc acting as a sacrificial anode; and fluxing action of CaO\text{CaO} with molten slag (CaSiO3\text{CaSiO}_3) formation.
Each process matches its corresponding chemical principle: siderite decomposes into FeO\text{FeO} and CO2\text{CO}_2; hot concentrated H2SO4\text{H}_2\text{SO}_4 oxidizes iron to iron(III) sulfate releasing SO2\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 (FeCO3\text{FeCO}_3) decomposes upon heating to yield iron(II) oxide (FeO\text{FeO}) and carbon(IV) oxide (CO2\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 H2SO4\text{H}_2\text{SO}_4 oxidizes iron metal to the iron(III) oxidation state, producing Fe2(SO4)3\text{Fe}_2(\text{SO}_4)_3, SO2\text{SO}_2 gas, and H2O\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 (CaO\text{CaO}) reacts with acidic sandy impurities (SiO2\text{SiO}_2) to form molten slag (CaSiO3\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 1295Question →

Match each redox reagent mixture or reaction system with its corresponding characteristic diagnostic observation.

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Items

Acidified potassium heptaoxodichromate(VI) solution (K2Cr2O7/H+K_2Cr_2O_7 / H^+) treated with sulfur(IV) oxide gas (SO2SO_2)
Freshly prepared iron(II) tetraoxosulfate(VI) solution (FeSO4FeSO_4) treated with concentrated trioxonitrate(V) acid (HNO3HNO_3)
Aqueous potassium iodide solution (KIKI) treated with chlorine gas (Cl2Cl_2) in the presence of starch indicator
Acidified potassium tetraoxomanganate(VII) solution (KMnO4/H+KMnO_4 / H^+) treated with hydrogen peroxide (H2O2H_2O_2)

Matches

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Answer

Acidified potassium heptaoxodichromate(VI) with sulfur(IV) oxide matches the orange to green color change. Freshly prepared iron(II) tetraoxosulfate(VI) with concentrated trioxonitrate(V) acid matches the pale green to reddish-brown change. Aqueous potassium iodide with chlorine gas and starch matches the blue-black complex formation. Acidified potassium tetraoxomanganate(VII) with hydrogen peroxide matches decolorization from purple to colorless with oxygen gas effervescence.
Each test reagent matches its specific chemical observation based on electron transfer and oxidation state changes: K2Cr2O7K_2Cr_2O_7 turns green when reduced by SO2SO_2; Fe2+Fe^{2+} turns brown when oxidized by conc. HNO3HNO_3; KIKI yields free I2I_2 which gives a blue-black color with starch when oxidized by Cl2Cl_2; and acidified KMnO4KMnO_4 is decolorized with O2O_2 evolution when reduced by H2O2H_2O_2.

Step-by-Step Solution

1
Analyze the reduction of acidified K2Cr2O7K_2Cr_2O_7 by SO2SO_2
Chromium decreases in oxidation number from +6+6 in Cr2O72−Cr_2O_7^{2-} to +3+3 in Cr3+Cr^{3+}, causing a distinct color shift from orange to green.
SO2SO_2 acts as a reductant and K2Cr2O7K_2Cr_2O_7 as an oxidant.
2
Analyze the oxidation of FeSO4FeSO_4 by concentrated HNO3HNO_3
Iron increases in oxidation state from +2+2 (Fe2+Fe^{2+}, pale green) to +3+3 (Fe3+Fe^{3+}, brown/yellowish-brown).
Concentrated HNO3HNO_3 is a strong oxidizing agent.
3
Analyze halogen displacement of KIKI by Cl2Cl_2
Chlorine oxidizes I−I^- to I2I_2. Liberated I2I_2 forms a blue-black starch-iodine inclusion complex.
Chlorine has a higher standard reduction potential than iodine.
4
Analyze the redox reaction between acidified KMnO4KMnO_4 and H2O2H_2O_2
Manganese is reduced from +7+7 (MnO4−MnO_4^-, purple) to +2+2 (Mn2+Mn^{2+}, colorless), while peroxide oxygen is oxidized from −1-1 to 00 (O2O_2 gas bubbles).
In the presence of a stronger oxidant (KMnO4KMnO_4), H2O2H_2O_2 behaves as a reducing agent.

Key Concept

Oxidizing and Reducing Agents and Diagnostic Chemical Tests
Estimated Time:1m 30s
Question 1296Question →

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

Click a left item, then click its matching right item

Items

Activation Energy (EaE_a)
Activated Complex (Transition State)
Enthalpy Change (ΔH\Delta H)
Effect of a Catalyst

Matches

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Answer

Activation Energy (EaE_a) matches the minimum energy difference between the reactants and the peak of the energy barrier; Activated Complex matches the unstable, high-energy arrangement at the maximum point of the energy curve; Enthalpy Change (ΔH\Delta H) matches the net energy difference between products and reactants; Effect of a Catalyst matches lowering the potential energy barrier peak.
Each feature of an energy profile diagram describes a specific thermodynamic or kinetic aspect: Activation Energy measures the hurdle from reactants to the transition state; Activated Complex is the transient species at the barrier apex; Enthalpy Change is the net heat change from reactants to products; and a Catalyst reduces the activation barrier height.

Step-by-Step Solution

1
Identify the definition of Activation Energy (EaE_a)
It corresponds to the energy gap between reactant potential energy and the peak energy.
Reactants must absorb this energy threshold to initiate bond transformation.
2
Identify the Activated Complex (Transition State)
It is located at the peak of the curve where potential energy is at its maximum.
This state is highly unstable and short-lived as bonds are actively breaking and forming.
3
Identify the Enthalpy Change (ΔH\Delta H)
It represents the overall potential energy gap between products and reactants.
ΔH=Eproducts−Ereactants\Delta H = E_{\text{products}} - E_{\text{reactants}} shows whether a reaction is exothermic or endothermic.
4
Identify the role of a Catalyst in the profile diagram
It decreases the height of the transition state peak.
Catalysts lower activation energy without shifting initial reactant or final product energy levels.

Key Concept

Key parameters of reaction energy profile diagrams
Estimated Time:1m 0s
Question 1297Question →

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

Click a left item, then click its matching right item

Items

Dissolution of solid sodium chloride in water: NaCl(s)→Na(aq)++Cl(aq)−\text{NaCl}_{(s)} \rightarrow \text{Na}^+_{(aq)} + \text{Cl}^-_{(aq)}
Condensation of water vapor into liquid: H2O(g)→H2O(l)\text{H}_2\text{O}_{(g)} \rightarrow \text{H}_2\text{O}_{(l)}
Sublimation of solid carbon dioxide: \text{CO}_{2(s)} \rightarrow \text{CO}_{2(g)}$
Industrial synthesis of ammonia gas: N2(g)+3H2(g)→2NH3(g)\text{N}_{2(g)} + 3\text{H}_{2(g)} \rightarrow 2\text{NH}_{3(g)}

Matches

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Answer

Each process is matched based on the change in molecular disorder: dissolving solid NaCl produces mobile hydrated ions (positive entropy change), condensing water vapor transitions gas to liquid (negative entropy change), subliming dry ice transitions solid to gas (positive entropy change), and synthesizing ammonia decreases gaseous moles from 4 to 2 (negative entropy change).
Entropy (ΔS\Delta S) is a measure of randomness or molecular disorder. Phase transitions from solid to aqueous or solid to gas increase disorder (ΔS>0\Delta S > 0). Phase transitions from gas to liquid or reactions that reduce the number of gas moles decrease disorder (ΔS<0\Delta S < 0).

Step-by-Step Solution

1
Analyze state changes and particle freedom for physical processes.
Solid to aqueous ions increases disorder (ΔS>0\Delta S > 0). Gas to liquid decreases disorder (ΔS<0\Delta S < 0). Solid to gas increases disorder (ΔS>0\Delta S > 0).
Gases have the highest entropy, followed by aqueous solutions, liquids, and solids.
2
Analyze gas mole stoichiometry for chemical reactions involving gases.
For N2(g)+3H2(g)→2NH3(g)\text{N}_{2(g)} + 3\text{H}_{2(g)} \rightarrow 2\text{NH}_{3(g)}, reactant gas moles = 1+3=41 + 3 = 4, product gas moles = 22. Since gas moles decrease, ΔS<0\Delta S < 0.
A decrease in the number of gaseous particles decreases the available microstates of the system.

Key Concept

Predicting the sign of entropy change (ΔS\Delta S) from physical state transitions and changes in total moles of gas
Question 1298Question →

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

Items

Quicklime
Slaked lime
Gypsum
Plaster of Paris

Matches

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Answer

Quicklime matches Calcium oxide (CaO\text{CaO}); Slaked lime matches Calcium hydroxide (Ca(OH)2\text{Ca(OH)}_2); Gypsum matches Calcium tetraoxosulfate(VI) dihydrate (CaSO4⋅2H2O\text{CaSO}_4 \cdot 2\text{H}_2\text{O}); Plaster of Paris matches Calcium tetraoxosulfate(VI) hemihydrate (CaSO4⋅12H2O\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 (CaO\text{CaO}) as a drying agent for basic gases, Slaked lime (Ca(OH)2\text{Ca(OH)}_2) for soil liming, Gypsum (CaSO4⋅2H2O\text{CaSO}_4 \cdot 2\text{H}_2\text{O}) for retarding cement setting, and Plaster of Paris (CaSO4⋅12H2O\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 CaO\text{CaO} (calcium oxide) and Slaked lime is Ca(OH)2\text{Ca(OH)}_2 (calcium hydroxide).
Thermal decomposition of limestone (CaCO3\text{CaCO}_3) yields CaO\text{CaO}, which reacts with water to form Ca(OH)2\text{Ca(OH)}_2.
2
Distinguish between Gypsum and Plaster of Paris based on hydration state
Gypsum contains two water molecules per sulfate unit (CaSO4⋅2H2O\text{CaSO}_4 \cdot 2\text{H}_2\text{O}), whereas Plaster of Paris contains half a water molecule per sulfate unit (CaSO4⋅12H2O\text{CaSO}_4 \cdot \frac{1}{2}\text{H}_2\text{O}).
Heating gypsum to around 120∘C120^\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 1299Question →

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

Items

Calcium fluoride (CaF2\text{CaF}_2)
Calcium oxide (CaO\text{CaO})
Calcium sulfate hemihydrate (CaSO4⋅12H2O\text{CaSO}_4\cdot\frac{1}{2}\text{H}_2\text{O})
Calcium hydroxide (Ca(OH)2\text{Ca(OH)}_2)

Matches

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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 CaCl2\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 CaCO3\text{CaCO}_3 yields basic CaO\text{CaO}, which does not react with basic gases like NH3\text{NH}_3.
Acidic drying agents such as concentrated H2SO4\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 (CaSO4⋅12H2O\text{CaSO}_4\cdot\frac{1}{2}\text{H}_2\text{O}).
Heating at 120∘C120^\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 CaO\text{CaO} produces Ca(OH)2\text{Ca(OH)}_2, whose aqueous solution forms an insoluble milky CaCO3\text{CaCO}_3 precipitate with CO2\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 1300Question →

In water purification and industrial wastewater management, specific chemical and physical processes are used to eliminate target pollutants. Match each water treatment process on the left with its corresponding chemical function on the right.

Click a left item, then click its matching right item

Items

Coagulation
Aeration
Chlorination
Activated Carbon Filtration

Matches

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Answer

Coagulation matches with clumping fine suspended solids using coagulants such as potash alum; Aeration matches with expelling dissolved volatile gases and oxidising soluble iron compounds; Chlorination matches with destroying pathogenic microorganisms to sanitize water; Activated Carbon Filtration matches with adsorbing dissolved organic impurities, dyes, and unpleasant odours.
Each process corresponds directly to its functional role in water purification: Coagulation uses coagulants like alum to clump fine suspended matter; Aeration strips unpleasant volatile gases and oxidises soluble metals; Chlorination kills disease-causing microorganisms; Activated Carbon Filtration adsorbs dissolved organic impurities and odours.

Step-by-Step Solution

1
Identify the primary chemical mechanism of Coagulation
Coagulation uses salts like alum to neutralize particle charges, resulting in the aggregation of fine suspended matter into larger settled masses.
Alum provides trivalent cations (Al3+Al^{3+}) to destabilize colloidal suspensions.
2
Identify the primary chemical mechanism of Aeration
Aeration increases dissolved oxygen to oxidize dissolved ferrous iron (Fe2+Fe^{2+}) to ferric iron (Fe3+Fe^{3+}) and strips out foul-smelling gases like H2SH_2S.
Physical gas exchange and oxidation improve water taste and clarity.
3
Identify the primary biological mechanism of Chlorination
Chlorination serves as the final disinfection stage to eliminate harmful biological pathogens.
Chlorine generates active oxidizing species (HOCl/OCl−HOCl / OCl^-) that rupture bacterial cell walls.
4
Identify the primary physical mechanism of Activated Carbon Filtration
Activated carbon adsorbs non-polar organic contaminants, synthetic detergents, and residual pigments due to its extremely porous structure.
High internal surface area promotes strong Van der Waals forces to trap organic pollutants.

Key Concept

Municipal and Industrial Water Treatment Processes
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