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Question 4561Question

Match each chemical reaction or process involving alkanoic acids, esters, fats, or oils on the left with its corresponding chemical outcome or product description on the right.

Click a left item, then click its matching right item

Items

Acid-catalyzed esterification of propane-1,2,3-triol with hexadecanoic acid
Alkaline hydrolysis of glyceryl tristearate using excess aqueous sodium hydroxide
High-pressure catalytic hydrogenation of glyceryl trioleate in the presence of nickel
Acid-catalyzed reflux of ethyl ethanoate with an excess of water

Matches

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Answer

The correct pairings match: (1) Acid-catalyzed esterification of propane-1,2,3-triol with hexadecanoic acid to producing tripalmitin fat and water; (2) Alkaline hydrolysis of glyceryl tristearate with aqueous NaOH to yielding propane-1,2,3-triol and sodium octadecanoate soap; (3) Catalytic hydrogenation of glyceryl trioleate to converting an unsaturated liquid triacylglycerol into a saturated solid fat; and (4) Acid-catalyzed reflux of ethyl ethanoate with water to reversibly producing ethanoic acid and ethanol.
The correct pairings logically connect each organic process to its definitive product or reaction characteristic: esterification of hexadecanoic acid with glycerol yields tripalmitin; saponification of glyceryl tristearate using NaOH irreversibly yields glycerol and sodium octadecanoate soap; catalytic hydrogenation saturates double bonds in glyceryl trioleate turning liquid oil into solid fat; and acid hydrolysis of ethyl ethanoate is a reversible equilibrium yielding ethanoic acid and ethanol.

Step-by-Step Solution

1
Analyze the reaction of propane-1,2,3-triol (glycerol) with hexadecanoic acid (palmitic acid).
Identified as esterification forming tripalmitin and water.
Hexadecanoic acid (C15H31COOHC_{15}H_{31}COOH) esterifies with glycerol to form glyceryl tripalmitin (C51H98O6C_{51}H_{98}O_6), a saturated fat.
2
Analyze the alkaline hydrolysis of glyceryl tristearate using aqueous NaOHNaOH.
Identified as saponification yielding glycerol and sodium octadecanoate.
Base hydrolysis of fats irreversibly converts ester groups into glycerol and carboxylate salts (soap).
3
Analyze the catalytic hydrogenation of glyceryl trioleate.
Identified as hardening of oils.
Addition of H2H_2 across the C=CC=C double bonds of unsaturated oleic acid residues converts liquid oil into saturated solid fat.
4
Analyze the acid-catalyzed reaction of ethyl ethanoate with water.
Identified as reversible ester hydrolysis.
Acid hydrolysis of esters is reversible, establishing an equilibrium mixture of the parent alkanoic acid (ethanoic acid) and alkanol (ethanol).

Key Concept

Reactivity and Interconversion of Alkanoic Acids, Esters, Fats, and Oils
Question 4562Question

An ecologist conducting a field study along a coastal cliff needs to measure both the angle of slope of the terrain and the atmospheric pressure at different elevations. Which pair of ecological instruments should the ecologist select for these measurements?

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Answer: Clinometer and barometer

Answer

The ecologist should select a clinometer to measure the angle of slope and a barometer to measure atmospheric pressure.
The correct answer correctly pairs the clinometer, which measures slope gradient and terrain inclination, with the barometer, which quantifies atmospheric pressure.

Step-by-Step Solution

1
Identify the ecological factors specified in the study
The parameters are terrain slope angle (a topographic factor) and atmospheric pressure (a climatic factor).
Matching abiotic parameters to their appropriate measurement instruments requires categorizing each parameter correctly.
2
Determine the correct instrument for measuring slope angle
A clinometer (or abney level) is specifically designed to measure gradient or angle of inclination.
Topographic measurements of slope steepness require an instrument that quantifies angles relative to the horizontal.
3
Determine the correct instrument for measuring atmospheric pressure
A barometer measures atmospheric pressure in units such as mmHg or hPa.
Climatic variations with altitude involve changes in air pressure, which are measured using a barometer.

Key Concept

Measurement of topographic and climatic abiotic factors using ecological instruments
Question 4563Question

During humid nights when atmospheric humidity is high and transpiration is minimal, liquid water droplets are observed exuding from leaf margins of herbaceous plants. Which transport mechanism and plant structure account for this liquid exudation?

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Answer: Root pressure forcing water out through hydathodes

Answer

Root pressure forcing water out through hydathodes
The phenomenon described is guttation, which occurs when low transpiration rates combined with high soil moisture cause positive root pressure to push water up the stem and out through specialized pores called hydathodes at leaf margins.

Step-by-Step Solution

1
Analyze the environmental conditions described in the stem.
High atmospheric humidity and nighttime conditions reduce transpiration to near zero, preventing tension-driven transpiration pull.
Stomata close or atmospheric humidity limits the water potential gradient required for evaporation.
2
Identify the primary force driving xylem sap upward under low transpiration.
Active accumulation of mineral ions in root xylem causes osmotic water influx, generating positive hydrostatic root pressure.
Root pressure pushes water upward through the xylem when upward pulling forces are absent.
3
Determine the exit point for liquid water exudation (guttation).
Excess liquid sap is exuded through specialized pores called hydathodes located at leaf tips and margins.
Hydathodes are permanently open vascular terminal structures designed for liquid water discharge.

Key Concept

Guttation and Root Pressure
Question 4564Question

A microscopic unicellular organism isolated from a freshwater pond exhibits mixotrophic nutrition: it photosynthesizes in sunlight using chloroplasts, but absorbs organic nutrients heterotrophically when kept in the dark. It moves by means of a single flagellum and lacks a cellulose cell wall, possessing a flexible proteinaceous pellicle beneath its plasma membrane instead. Which of the following protists is described, and what is its primary carbohydrate storage product?

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Answer: Euglena, which stores carbohydrate as paramylon

Answer

Euglena, which stores carbohydrate as paramylon
The correct answer identifies Euglena, which uniquely combines animal-like features (locomotion by a single flagellum, absence of a cellulose cell wall, presence of a flexible pellicle) and plant-like features (chloroplasts for autotrophic nutrition). When photosynthetic product is accumulated, it is stored as paramylon rather than typical plant starch.

Step-by-Step Solution

1
Analyze the structural features described in the stem
Unicellular, single flagellum, lacks cellulose cell wall, possesses a flexible proteinaceous pellicle.
Euglena lacks a rigid cell wall made of cellulose; instead, it is bounded by a proteinaceous layer called the pellicle which allows shape flexibility.
2
Analyze the nutritional mode described in the stem
Mixotrophic (autotrophic via chloroplasts in light, heterotrophic in dark).
Euglena contains chlorophyll for photosynthesis, but can survive saprophytically/heterotrophically in darkness.
3
Identify the unique storage carbohydrate of this organism
Paramylon (a β-1,3-glucan polymer).
Unlike green algae (which store true starch) or animals/fungi (which store glycogen), Euglena stores reserve food as paramylon granules.

Key Concept

Distinctive structural, nutritional, and biochemical features of Euglena
Estimated Time:1m 30s
Question 4565Question

Match each organism in Column I with its corresponding structure used for gaseous exchange in Column II.

Click a left item, then click its matching right item

Items

Amoeba
Tilapia (Fish)
Grasshopper (Insect)
Flowering plant

Matches

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Answer

Amoeba pairs with Cell membrane, Tilapia pairs with Filamentous gills, Grasshopper pairs with Tracheoles, and Flowering plant pairs with Stomata.
Each organism is correctly paired with its evolutionary adaptation for gas exchange: Amoeba uses its cell membrane, Tilapia uses filamentous gills, Grasshoppers use tracheoles, and flowering plants use stomata.

Step-by-Step Solution

1
Determine the gaseous exchange surface for single-celled protozoans such as Amoeba.
Amoeba matches with Cell membrane.
Due to its high surface area-to-volume ratio, Amoeba does not require specialized respiratory organs and relies on direct diffusion across the cell membrane.
2
Determine the organ adapted for aquatic respiration in bony fish like Tilapia.
Tilapia matches with Filamentous gills.
Water passes over the gill filaments where oxygen diffuses into blood capillaries while carbon dioxide diffuses out.
3
Identify the respiratory structures in terrestrial insects like the Grasshopper.
Grasshopper matches with Tracheoles.
Insects transport gases directly to tissue cells via a system of chitinous tubes that divide into fine tracheoles.
4
Identify the primary structure for gas exchange in angiosperm leaves.
Flowering plant matches with Stomata.
Stomata are specialized leaf pores bounded by guard cells that open and close to facilitate carbon dioxide uptake and oxygen release.

Key Concept

Adaptive diversity of specialized respiratory surfaces across unicellular, multicellular animal, and plant groups.
Question 4566Question

When purified and isolated outside a host organism, viral particles can form crystals and exhibit no measurable respiration or energy production. Which of the following features explains why viruses are metabolically inert outside a host cell?

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Answer: The complete absence of cytoplasm, cell organelles, and metabolic machinery

Answer

The complete absence of cytoplasm, cell organelles, and metabolic machinery accounts for the metabolic inactivity of extracellular viruses.
The correct option states that viruses lack cytoplasm, cellular organelles, and metabolic machinery. Because viruses are acellular nucleoprotein particles without cytosol, enzymes for cellular respiration, or organelle systems, they cannot synthesize ATP or perform metabolic work when isolated from host cells.

Step-by-Step Solution

1
Analyze the structural nature of viruses
Viruses are classified as acellular (non-cellular) nucleoprotein particles.
Understanding whether viruses possess cellular organelles determines their metabolic capabilities.
2
Evaluate viral components required for metabolism
Viruses possess only a nucleic acid core (DNA or RNA) surrounded by a protein coat (capsid), lacking cytoplasm, enzymes, and ATP-generating machinery.
Metabolism requires cytosol, cellular organelles, and functional metabolic pathways.
3
Deduce extracellular behavior from structural composition
Without cellular machinery, viruses cannot generate energy or perform metabolic functions outside a living host cell, functioning as obligate intracellular parasites.
Connects viral acellular structure to extracellular metabolic inertness.

Key Concept

Acellular Nature and Obligate Intracellular Parasitism of Viruses
Question 4567Question

An organic compound XX with the molecular formula C4H10O\text{C}_4\text{H}_{10}\text{O} resists oxidation when treated with acidified potassium dichromate(VI) (K2Cr2O7/H+\text{K}_2\text{Cr}_2\text{O}_7/\text{H}^+). When compound XX is heated with concentrated tetraoxosulfate(VI) acid (H2SO4\text{H}_2\text{SO}_4) at 170C170^\circ\text{C}, it undergoes dehydration to produce a major organic product YY. What is the IUPAC name of compound YY?

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Answer: 2-methylpropene

Answer

2-methylpropene
The compound resisting oxidation must be a tertiary alkanol because the hydroxyl-bearing carbon lacks an alpha-hydrogen atom. The only tertiary alkanol with formula C4H10O\text{C}_4\text{H}_{10}\text{O} is 2-methylpropan-2-ol. Subjecting 2-methylpropan-2-ol to intra-molecular dehydration using concentrated tetraoxosulfate(VI) acid at 170C170^\circ\text{C} removes water to form 2-methylpropene as the major alkene product.

Step-by-Step Solution

1
Determine the structural class of compound X from its resistance to oxidation.
Compound X is a tertiary alkanol.
Primary and secondary alkanols are readily oxidized by acidified potassium dichromate(VI), whereas tertiary alkanols resist mild oxidation because the carbon atom bonded to the hydroxyl group (OH-OH) carries no hydrogen atoms.
2
Identify the specific isomer of formula C4H10O\text{C}_4\text{H}_{10}\text{O} corresponding to a tertiary alkanol.
Compound X is 2-methylpropan-2-ol, (CH3)3C-OH(\text{CH}_3)_3\text{C-OH}.
Among the four structural isomers of C4H10O\text{C}_4\text{H}_{10}\text{O} alkanols, only 2-methylpropan-2-ol is tertiary.
3
Determine the elimination product when 2-methylpropan-2-ol undergoes acid-catalyzed dehydration.
Dehydration yields 2-methylpropene, (CH3)2C=CH2(\text{CH}_3)_2\text{C=CH}_2.
Heating with concentrated H2SO4\text{H}_2\text{SO}_4 at 170C170^\circ\text{C} removes a molecule of water (the OH-OH group and a hydrogen atom from an adjacent methyl group), forming an alkene.

Key Concept

Classification of alkanols based on oxidation behavior and acid-catalyzed dehydration to alkenes
Estimated Time:2m 0s
Question 4568Question

An investigation into the excretory mechanisms of four diverse invertebrate species revealed distinct structures for nitrogenous waste elimination:
- Structure P: Uses ciliated flame cells to propel fluid through a network of branching tubules.
- Structure Q: Relies on ciliated funnels (nephrostomes) opening into the coelom for fluid filtration and tubular reabsorption.
- Structure R: Consists of blind-ending tubules lying free in the hemolymph that actively transport nitrogenous wastes as uric acid into the hindgut.
- Structure S: Relies entirely on simple diffusion across the general body wall without specialized excretory organs.

Which of the following correctly matches excretory structures P, Q, R, and S with their respective animal representatives?

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Answer: P: Planaria, Q: Earthworm, R: Cockroach, S: Hydra

Answer

Structure P is found in Planaria, Structure Q in Earthworm, Structure R in Cockroach, and Structure S in Hydra.
The correct alignment pairs Planaria with flame cells (protonephridia), Earthworm with ciliated nephridial funnels (metanephridia), Cockroach with Malpighian tubules eliminating uric acid via hemolymph, and Hydra with unspecialized body surface diffusion.

Step-by-Step Solution

1
Identify Structure P
Flame cells (protonephridia) with beating cilia are specialized excretory units characteristic of Platyhelminthes (e.g., Planaria).
Flame cells maintain osmoregulation and excretion by driving fluid through excretory canals.
2
Identify Structure Q
Nephridia (metanephridia) with ciliated funnels collecting coelomic fluid belong to Annelida (e.g., Earthworm).
Coelomic fluid enters the nephrostome and undergoes selective reabsorption along the nephridial tubule.
3
Identify Structure R
Malpighian tubules floating in hemolymph and excreting solid uric acid into the digestive tract are unique to terrestrial Arthropoda, specifically Insecta (e.g., Cockroach).
This adaptation conserves water by converting waste into insoluble uric acid.
4
Identify Structure S
Absence of specialized excretory organs, relying on direct diffusion of ammonia across body layers, is characteristic of simple diploblastic organisms like Cnidaria (e.g., Hydra).
High surface-area-to-volume ratio in thin body walls allows direct diffusion into the aquatic environment.

Key Concept

Invertebrate Excretory Structures and Evolutionary Adaptations
Estimated Time:2m 0s
Question 4569Question

During the complete aerobic oxidation of one molecule of glucose in a cell, a total of 10 NADHNADH and 2 FADH2FADH_2 reduced coenzymes are generated across glycolysis, the link reaction, and the Krebs cycle. Assuming that each NADHNADH yields 3 ATP molecules and each FADH2FADH_2 yields 2 ATP molecules during electron transport, what is the net number of ATP molecules produced specifically through oxidative phosphorylation?

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Answer: 34 ATP34\text{ ATP}

Answer

The net number of ATP molecules synthesized specifically through oxidative phosphorylation is 34 ATP.
Oxidative phosphorylation generates ATP via electron transport and chemiosmosis using reduced coenzymes (NADHNADH and FADH2FADH_2). Multiplying 10 NADHNADH by 3 ATP gives 30 ATP, and multiplying 2 FADH2FADH_2 by 2 ATP gives 4 ATP. Summing these yields exactly 34 ATP.

Step-by-Step Solution

1
Calculate ATP yield from NADH coenzymes via the electron transport chain.
10 NADH×3 ATP/NADH=30 ATP10\text{ NADH} \times 3\text{ ATP/NADH} = 30\text{ ATP}
Each NADH molecule donates electrons to the respiratory chain to pump sufficient protons for generating 3 ATP molecules.
2
Calculate ATP yield from FADH2 coenzymes via the electron transport chain.
2 FADH2×2 ATP/FADH2=4 ATP2\text{ FADH}_2 \times 2\text{ ATP/FADH}_2 = 4\text{ ATP}
FADH2 enters the electron transport chain at a lower energy level (Complex II), yielding 2 ATP molecules per FADH2.
3
Sum the ATP produced by both coenzymes to find total oxidative phosphorylation yield.
30 ATP+4 ATP=34 ATP30\text{ ATP} + 4\text{ ATP} = 34\text{ ATP}
Oxidative phosphorylation refers exclusively to ATP synthesized via chemiosmosis powered by electron transport, separate from substrate-level phosphorylation.

Key Concept

Differentiation between substrate-level phosphorylation and oxidative phosphorylation ATP yields in cellular respiration
Estimated Time:2m 0s
Question 4570Question

A farmer notices a rapid loss of soil nitrogen in a waterlogged maize field that had been treated with ammonium-based fertilizer. Soil analysis confirms that ammonium ions (NH4+\text{NH}_4^+) were first oxidized to nitrites (NO2\text{NO}_2^-), then further oxidized to nitrates (NO3\text{NO}_3^-), which were subsequently reduced to gaseous nitrogen (N2\text{N}_2) under anaerobic conditions. Which sequence of bacteria is sequentially responsible for these three specific biochemical transformations?

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Answer: Nitrosomonas, Nitrobacter, and Pseudomonas

Answer

The sequence Nitrosomonas, Nitrobacter, and Pseudomonas correctly identifies the organisms performing ammonium oxidation, nitrite oxidation, and denitrification respectively.
The first step of nitrification (oxidation of ammonium NH4+\text{NH}_4^+ to nitrite NO2\text{NO}_2^-) is performed by Nitrosomonas. The second step (oxidation of nitrite NO2\text{NO}_2^- to nitrate NO3\text{NO}_3^-) is performed by Nitrobacter. Under waterlogged, oxygen-depleted soil conditions, anaerobic denitrifying bacteria such as Pseudomonas convert soil nitrates back into gaseous elemental nitrogen (N2\text{N}_2), causing a loss of soil fertility.

Step-by-Step Solution

1
Identify the bacterium responsible for converting ammonium ions (NH4+\text{NH}_4^+) to nitrite ions (NO2\text{NO}_2^-).
Nitrosomonas is the nitrifying bacterium that carries out the first step of nitrification.
Ammonium oxidation is an aerobic process mediated by chemoautotrophic bacteria like Nitrosomonas.
2
Identify the bacterium responsible for converting nitrite ions (NO2\text{NO}_2^-) to nitrate ions (NO3\text{NO}_3^-).
Nitrobacter completes nitrification by oxidizing nitrite to nitrate.
Nitrate is the primary form of nitrogen absorbed by plants, generated through nitrite oxidation by Nitrobacter.
3
Identify the bacterium responsible for converting nitrates (NO3\text{NO}_3^-) into gaseous nitrogen (N2\text{N}_2) in waterlogged, anaerobic soil.
Pseudomonas (or Thiobacillus denitrificans) conducts denitrification, returning nitrogen gas to the atmosphere.
Waterlogged soils lack molecular oxygen, forcing facultative anaerobes like Pseudomonas to use nitrate as a terminal electron acceptor.

Key Concept

Nitrification and Denitrification Pathways in the Nitrogen Cycle
Estimated Time:2m 0s
Question 4571Question

Which of the following excretory structures is characteristic of flatworms such as *Planaria*?

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Answer: Flame cells

Answer

Flame cells are the characteristic excretory structures found in flatworms (Platyhelminthes).
Flame cells are cilia-bearing cells involved in excretion and osmoregulation specifically in flatworms like *Planaria*.

Step-by-Step Solution

1
Identify the taxonomic group of the target organism.
*Planaria* is a free-living flatworm belonging to the phylum Platyhelminthes.
Excretory structures vary across invertebrate phyla.
2
Match the phylum to its specialized excretory organ.
Platyhelminthes use flame cells (protonephridia) to remove metabolic wastes and regulate water balance.
Flame cells feature beating cilia that draw fluid into excretory tubules for waste elimination.

Key Concept

Excretory structures in invertebrate phyla
Question 4572Question

In a savanna ecosystem, primary producers have a gross primary productivity (GPP) of 20000 kcal m2 yr120{}000\text{ kcal m}^{-2}\text{ yr}^{-1}, but consume 50%50\% of this energy through autotrophic cellular respiration (RAR_A). Assuming a constant ecological efficiency of 10%10\% for energy transfer between consecutive trophic levels, what is the total amount of energy per square meter per year available to tertiary consumers?

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Answer: 10 kcal m2 yr110\text{ kcal m}^{-2}\text{ yr}^{-1}

Answer

The total energy available to tertiary consumers is 10 kcal m2 yr110\text{ kcal m}^{-2}\text{ yr}^{-1}.
To find the energy available to tertiary consumers, first calculate Net Primary Productivity: NPP=GPPRA=2000010000=10000 kcal m2 yr1\text{NPP} = \text{GPP} - R_A = 20{}000 - 10{}000 = 10{}000\text{ kcal m}^{-2}\text{ yr}^{-1}. Then apply the 10%10\% transfer efficiency across three consumer steps: primary consumers receive 1000 kcal m2 yr11{}000\text{ kcal m}^{-2}\text{ yr}^{-1}, secondary consumers receive 100 kcal m2 yr1100\text{ kcal m}^{-2}\text{ yr}^{-1}, and tertiary consumers receive 10 kcal m2 yr110\text{ kcal m}^{-2}\text{ yr}^{-1}.

Step-by-Step Solution

1
Calculate Net Primary Productivity (NPP) of the primary producers
NPP=GPPRA=20000(0.50×20000)=10000 kcal m2 yr1\text{NPP} = \text{GPP} - R_A = 20{}000 - (0.50 \times 20{}000) = 10{}000\text{ kcal m}^{-2}\text{ yr}^{-1}
Only energy stored as organic biomass after metabolic respiration is available to herbivores.
2
Calculate energy transferred to primary consumers (trophic level 2)
Energy2=10% of 10000=1000 kcal m2 yr1\text{Energy}_2 = 10\% \text{ of } 10{}000 = 1{}000\text{ kcal m}^{-2}\text{ yr}^{-1}
According to the 10%10\% law of energy transfer, only one-tenth of available energy passes to primary consumers.
3
Calculate energy transferred to secondary consumers (trophic level 3)
Energy3=10% of 1000=100 kcal m2 yr1\text{Energy}_3 = 10\% \text{ of } 1{}000 = 100\text{ kcal m}^{-2}\text{ yr}^{-1}
Apply the 10%10\% transfer efficiency from primary consumers to secondary consumers.
4
Calculate energy transferred to tertiary consumers (trophic level 4)
Energy4=10% of 100=10 kcal m2 yr1\text{Energy}_4 = 10\% \text{ of } 100 = 10\text{ kcal m}^{-2}\text{ yr}^{-1}
Apply the 10%10\% transfer efficiency from secondary consumers to tertiary consumers.

Key Concept

Calculation of net primary productivity and progressive thermodynamic energy attenuation across trophic levels.
Question 4573Question

Calculate the quantity of electricity, in Coulombs, required to deposit 0.108 g0.108\text{ g} of silver at the cathode during the electrolysis of silver trioxonitrate(V) solution. (Ag=108 g/mol\text{Ag} = 108\text{ g/mol}, 1 F=96,500 C/mol1\text{ F} = 96,500\text{ C/mol})

Show answer & explanation

Answer: 96.5

Answer

The quantity of electricity required is 96.5 C96.5\text{ C}.
Depositing 0.108 g0.108\text{ g} of silver (molar mass 108 g/mol108\text{ g/mol}) requires 0.001 moles0.001\text{ moles} of silver atoms. According to the cathodic reaction Ag++eAg\text{Ag}^+ + \text{e}^- \rightarrow \text{Ag}, 1 mole1\text{ mole} of electrons (96,500 C96,500\text{ C}) is required to deposit 1 mole1\text{ mole} of Ag\text{Ag}. Therefore, the total charge required is 0.001×96,500 C=96.5 C0.001 \times 96,500\text{ C} = 96.5\text{ C}.

Step-by-Step Solution

1
Calculate the number of moles of silver deposited
Moles of Ag=0.108 g108 g/mol=0.001 mol\text{Moles of Ag} = \frac{0.108\text{ g}}{108\text{ g/mol}} = 0.001\text{ mol}
Number of moles is calculated by dividing mass by molar mass.
2
Determine the quantity of electricity (charge) required
Q=0.001 mol×96,500 C/mol=96.5 CQ = 0.001\text{ mol} \times 96,500\text{ C/mol} = 96.5\text{ C}
The reduction reaction Ag++eAg\text{Ag}^+ + \text{e}^- \rightarrow \text{Ag} shows 1 mole1\text{ mole} of electrons (1 F=96,500 C1\text{ F} = 96,500\text{ C}) deposits 1 mole1\text{ mole} of silver.

Key Concept

Faraday's First Law of Electrolysis and Quantitative Mass-Charge Relationship
Question 4574Question
In the industrial Contact process for the manufacture of tetraoxosulfate(VI) acid, the conversion of sulfur(IV) oxide to sulfur(VI) oxide proceeds according to the thermochemical equation:
2SO2(g)+O2(g)2SO3(g)ΔH=197 kJ mol12SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \quad \Delta H = -197 \text{ kJ mol}^{-1}
Although Le Chatelier's principle predicts a higher equilibrium yield of SO3(g)SO_3(g) at lower temperatures, the process is industrially operated at a compromise temperature of approximately 450C450^\circ\text{C} using a vanadium(V) oxide (V2O5V_2O_5) catalyst. Which of the following best explains why operating at a significantly lower temperature is commercially unviable?
Show answer & explanation

Answer: Lowering the temperature significantly reduces the kinetic energy of the reactant molecules and renders the V2O5V_2O_5 catalyst inactive, resulting in an unacceptably slow reaction rate despite the favorable equilibrium position.

Answer

Operating at a lower temperature is commercially unviable because lowering the temperature significantly reduces the kinetic energy of reactant molecules and renders the vanadium(V) oxide catalyst inactive, making the rate of reaching equilibrium extremely slow despite a favorable equilibrium yield.
The correct explanation emphasizes the crucial distinction between chemical equilibrium (yield) and reaction rate (kinetics) in industrial chemistry. For exothermic reactions like the oxidation of SO2SO_2, decreasing temperature shifts equilibrium to the right, yielding more product. However, at lower temperatures, the rate of reaction drops exponentially, and the V2O5V_2O_5 catalyst loses its catalytic activity (which requires temperatures 400C\ge 400^\circ\text{C}). Thus, 450C450^\circ\text{C} is chosen as an optimum compromise temperature.

Step-by-Step Solution

1
Analyze the thermodynamic profile of the reaction
The forward reaction 2SO2(g)+O2(g)2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) is exothermic (ΔH=197 kJ mol1\Delta H = -197 \text{ kJ mol}^{-1}). According to Le Chatelier's principle, lowering the temperature favors the exothermic direction, increasing the equilibrium concentration of SO3SO_3.
Understanding thermodynamic equilibrium factors governing yield.
2
Analyze kinetic factor and catalyst requirements
Reaction rate depends on activation energy and temperature. At lower temperatures, fewer molecules possess energy EEaE \ge E_a. Furthermore, vanadium(V) oxide (V2O5V_2O_5) catalyst is only active above 400C\sim 400^\circ\text{C}.
Commercial viability requires a balance between yield (thermodynamics) and rate (kinetics).
3
Evaluate the commercial compromise
Operating at 450C450^\circ\text{C} provides an optimum compromise: a satisfactory rate of conversion (98%\sim 98\% yield) in a reasonable timeframe.
Industrial chemical processes must maximize production rate alongside conversion efficiency.

Key Concept

Compromise Conditions in Industrial Chemical Synthesis (Kinetics vs. Equilibrium Yield)
Estimated Time:2m 0s
Question 4575Question

Arrange the following stages of sexual reproduction (zygospore formation) in the bread mould *Rhizopus* in the correct chronological sequence from start to finish:

Drag items to arrange them in the correct order

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Answer

The correct chronological sequence of zygospore formation in Rhizopus begins with hyphae of opposite mating strains making contact, followed by the development of progametangia, then the formation of septa to isolate gametangia, and culminates in the dissolution of intervening walls to allow nuclear fusion into a thick-walled zygospore.
Zygospore formation in *Rhizopus* follows a strict sequence: contact between compatible (+ and -) hyphae occurs first, inducing progametangia outgrowth, followed by septum formation to isolate multinucleate gametangia, and ending with wall dissolution and nuclear fusion to produce the mature zygospore.

Step-by-Step Solution

1
Identify the initial stimulus for sexual reproduction.
Hyphae of opposite mating strains (+ and -) grow towards each other and make physical contact.
Physical interaction between compatible mating strains initiates the sexual pathway in mucoralean fungi.
2
Determine the early morphological response to hyphal contact.
Progametangia form as lateral, swollen outgrowths.
Hormonal signaling causes the hyphal tips at the contact zone to swell.
3
Identify the cellular isolation step.
Cross-walls (septa) form behind the tips, delimiting gametangia.
Septa isolate the apical multinucleate gametangia from the rest of the hyphae (suspensors).
4
Determine the final cell fusion and spore maturation event.
The touching walls dissolve, cytoplasm and nuclei fuse, and a thick, dark wall surrounds the resulting zygospore.
Plasmogamy and karyogamy yield a diploid zygospore capable of enduring harsh environmental conditions.

Key Concept

Sexual Reproduction and Zygospore Formation in Rhizopus
Question 4576Question

During the electrolysis of dilute tetraoxosulfate(VI) acid, a steady current of 2.50 A2.50\text{ A} is passed through the electrolyte for 1930 seconds1930\text{ seconds}. What volume of hydrogen gas, measured at STP, is liberated at the cathode?

[1 Faraday=96500 C mol1,Molar volume of gas at STP=22.4 dm3mol1][1\text{ Faraday} = 96500\text{ C mol}^{-1}, \text{Molar volume of gas at STP} = 22.4\text{ dm}^3\text{mol}^{-1}]

Show answer & explanation

Answer: 0.56 dm30.56\text{ dm}^3

Answer

The volume of hydrogen gas liberated at STP is 0.56 dm30.56\text{ dm}^3.
The correct answer is derived by calculating the electric charge passed (Q=2.50×1930=4825 CQ = 2.50 \times 1930 = 4825\text{ C}), converting to moles of electrons (0.050 mol0.050\text{ mol}), applying the electrode reduction stoichiometry (2 mol e2\text{ mol } e^- per 1 mol H21\text{ mol } H_2), and converting the resulting 0.025 mol H20.025\text{ mol } H_2 to volume at STP using 22.4 dm3mol122.4\text{ dm}^3\text{mol}^{-1}, which yields 0.56 dm30.56\text{ dm}^3.

Step-by-Step Solution

1
Calculate total quantity of electricity (QQ) passed.
Q=I×t=2.50 A×1930 s=4825 CQ = I \times t = 2.50\text{ A} \times 1930\text{ s} = 4825\text{ C}
Faraday's first law states quantity of charge is the product of current and time in seconds.
2
Calculate the amount of electrons in moles transferred.
n(e)=QF=4825 C96500 C mol1=0.050 moln(e^-) = \frac{Q}{F} = \frac{4825\text{ C}}{96500\text{ C mol}^{-1}} = 0.050\text{ mol}
One Faraday (96500 C96500\text{ C}) represents one mole of electrons.
3
Determine moles of H2H_2 gas produced using the cathode half-reaction stoichiometry.
Cathode reaction: 2H++2eH2(g)2H^+ + 2e^- \rightarrow H_2(g). Moles of H2=0.050 mol e2=0.025 mol H2H_2 = \frac{0.050\text{ mol } e^-}{2} = 0.025\text{ mol } H_2
Two moles of electrons are required to reduce hydrogen ions to produce one mole of hydrogen gas.
4
Calculate the volume of hydrogen gas at STP.
V=0.025 mol×22.4 dm3mol1=0.56 dm3V = 0.025\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 0.56\text{ dm}^3
One mole of any ideal gas occupies 22.4 dm322.4\text{ dm}^3 at STP.

Key Concept

Quantitative application of Faraday's Laws of Electrolysis for gas evolution at electrodes.
Estimated Time:1m 30s
Question 4577Question

An aquatic unicellular protist possessing a rigid cellulose cell wall, two equal anterior flagella, a cup-shaped chloroplast with a pyrenoid, and a prominent eyespot is transferred from freshwater to a hypertonic saline medium under continuous illumination. Which of the following responses describes the physiological and subcellular changes occurring in this organism?

Show answer & explanation

Answer: Contractile vacuole discharge rate decreases while water leaves the cytoplasm by osmosis.

Answer

Contractile vacuole discharge rate decreases while water leaves the cytoplasm by osmosis.
Freshwater unicellular protists rely on contractile vacuoles to pump out excess water that enters by endosmosis. When transferred to a hypertonic saline medium, exosmosis occurs, drawing water out of the cell and eliminating the osmotic influx, which causes the contractile vacuole pulsation rate to decline sharply.

Step-by-Step Solution

1
Identify the unicellular protist from diagnostic structural traits
The organism with a cellulose cell wall, two equal anterior flagella, cup-shaped chloroplast, pyrenoid, and eyespot is Chlamydomonas.
These combined morphological traits uniquely define Chlamydomonas within unicellular chlorophyte algae.
2
Evaluate the osmotic movement of water in a hypertonic environment
Water moves passively out of the cell across the semi-permeable membrane down a water potential gradient.
A hypertonic medium has a lower water potential compared to the internal cytoplasm of a freshwater organism.
3
Determine the functional response of the contractile vacuole
The rate of contractile vacuole pulsation decreases or stops.
Contractile vacuoles actively eliminate excess water gained by freshwater organisms; when water loss occurs osmotic gain stops, rendering rapid vacuolar pumping unnecessary.

Key Concept

Osmoregulation and diagnostic cell structure in Chlamydomonas (Kingdom Protista)
Question 4578Question

A mixture of 50 cm350\text{ cm}^3 of carbon(II) oxide and 30 cm330\text{ cm}^3 of oxygen was ignited in an eudiometer tube to complete reaction. After cooling to room temperature and pressure, the resulting gas mixture was passed through concentrated potassium hydroxide solution. What is the volume of the residual gas remaining?

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Answer: 5 cm35\text{ cm}^3

Answer

The volume of residual gas remaining is 5 cm35\text{ cm}^3.
According to Gay-Lussac's Law of Combining Volumes, 2 cm32\text{ cm}^3 of CO\text{CO} combines with 1 cm31\text{ cm}^3 of O2\text{O}_2 to produce 2 cm32\text{ cm}^3 of CO2\text{CO}_2. For 50 cm350\text{ cm}^3 of CO\text{CO}, exactly 25 cm325\text{ cm}^3 of O2\text{O}_2 is required, leaving 5 cm35\text{ cm}^3 of O2\text{O}_2 unreacted. The reaction generates 50 cm350\text{ cm}^3 of CO2\text{CO}_2. Passing the mixture through concentrated potassium hydroxide (KOH\text{KOH}) removes all 50 cm350\text{ cm}^3 of CO2\text{CO}_2 via trioxocarbonate(IV) salt formation, leaving behind only the 5 cm35\text{ cm}^3 of unreacted oxygen gas.

Step-by-Step Solution

1
Write the balanced chemical equation for the combustion of carbon(II) oxide.
2CO(g)+O2(g)2CO2(g)2\text{CO}_{(g)} + \text{O}_{2(g)} \rightarrow 2\text{CO}_{2(g)}
Establishing the combining volume ratio according to Gay-Lussac's Law.
2
Determine the reacting volumes and the limiting reactant.
2 volumes of CO\text{CO} react with 1 volume of O2\text{O}_2. Therefore, 50 cm350\text{ cm}^3 of CO\text{CO} reacts with 502=25 cm3\frac{50}{2} = 25\text{ cm}^3 of O2\text{O}_2. Oxygen is in excess.
To find how much oxygen is consumed and how much remains unreacted.
3
Calculate the volume of products formed and unreacted gas remaining after combustion.
Volume of CO2\text{CO}_2 produced = 50 cm350\text{ cm}^3. Volume of excess O2\text{O}_2 remaining = 30 cm325 cm3=5 cm330\text{ cm}^3 - 25\text{ cm}^3 = 5\text{ cm}^3.
Stoichiometric yield of CO2\text{CO}_2 equals the initial volume of CO\text{CO} burned.
4
Account for the absorption of carbon oxides by concentrated KOH.
CO2\text{CO}_2 is an acidic oxide and reacts with potassium hydroxide: CO2(g)+2KOH(aq)K2CO3(aq)+H2O(l)\text{CO}_{2(g)} + 2\text{KOH}_{(aq)} \rightarrow \text{K}_2\text{CO}_{3(aq)} + \text{H}_2\text{O}_{(l)}. All 50 cm350\text{ cm}^3 of CO2\text{CO}_2 is absorbed. Residual gas = 5 cm35\text{ cm}^3 of O2\text{O}_2.
Potassium hydroxide selectively absorbs carbon(IV) oxide gas.

Key Concept

Gay-Lussac's Law of Combining Volumes and chemical absorption properties of carbon oxides
Question 4579Question

Cyanobacteria perform oxygenic photosynthesis within membrane-bound chloroplasts, distinguishing them structurally from bacterial groups in Kingdom Monera.

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Answer: False

Answer

The statement is False. Cyanobacteria are prokaryotes belonging to Kingdom Monera and lack membrane-bound organelles such as chloroplasts; their photosynthetic pigments are distributed on thylakoids in the cytoplasm.
The statement is false because cyanobacteria are prokaryotes belonging to Kingdom Monera. They lack membrane-bound organelles such as chloroplasts, carrying out photosynthesis on thylakoid membranes located in the cytoplasm.

Step-by-Step Solution

1
Analyze the cellular domain and kingdom classification of cyanobacteria.
Cyanobacteria are prokaryotic organisms classified under Kingdom Monera.
All members of Kingdom Monera possess a prokaryotic cellular organization lacking membrane-bound organelles.
2
Evaluate the subcellular location of photosynthetic pigments in cyanobacteria.
Photosynthetic pigments like chlorophyll a and phycobilins are located on thylakoid membranes free in the cytoplasm, not inside chloroplasts.
Chloroplasts are membrane-bound organelles restricted to eukaryotic cells (Kingdom Protista and Plantae).

Key Concept

Prokaryotic structural characteristics of cyanobacteria in Kingdom Monera
Question 4580Question

Match each ecological concept or trophic entity on the left with its corresponding characteristic regarding energy flow and pyramid structure on the right.

Click a left item, then click its matching right item

Items

Primary Producers
Primary Consumers
Pyramid of Energy
Pyramid of Numbers (Parasitic)

Matches

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Answer

Primary Producers match with converting light energy into chemical energy at the foundation level; Primary Consumers match with occupying the second trophic level; Pyramid of Energy matches with being strictly upright due to thermodynamic heat loss; Pyramid of Numbers (Parasitic) matches with an inverted shape where one host supports many parasites.
Each item correctly aligns with fundamental ecological principles: autotrophs fix solar energy into biomass; herbivores occupy the second trophic level; energy pyramids remain exclusively upright due to thermodynamic dissipation at each level; and parasitic pyramids of numbers invert because many organisms feed on a single larger host.

Step-by-Step Solution

1
Identify the biological role of Primary Producers
They fix solar energy into organic matter at the base trophic level.
Autotrophs are the entry point of energy into ecosystems.
2
Identify the position and role of Primary Consumers
They occupy the second trophic level (herbivores).
They obtain energy by consuming primary producers.
3
Analyze the thermodynamic constraint on the Pyramid of Energy
It must always be upright.
Energy transfer between trophic levels is never 100% efficient due to metabolic heat loss.
4
Analyze structural exceptions in Pyramids of Numbers
Parasitic chains yield inverted pyramids.
A single tree or animal host supports many smaller parasites.

Key Concept

Energy Flow, Food Chains, and Ecological Pyramids
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