All practice questions

Question 4601Question →

When carbon(IV) oxide gas is passed over red-hot coke in a industrial furnace, it undergoes reduction to form a poisonous gas that serves as a vital reducing agent in blast furnace operations. What is the IUPAC name of the gas produced?

Show answer & explanation

Answer: Carbon(II) oxide; carbon(II) oxide; Carbon (II) oxide; carbon (II) oxide; Carbon monoxide; carbon monoxide

Answer

Carbon(II) oxide

Passing carbon(IV) oxide gas over red-hot carbon (coke) reduces

\text{CO}_2

to

\text{CO}

. The IUPAC name for

\text{CO}

is carbon(II) oxide, which is a major reducing agent in industrial metal extraction.

Step-by-Step Solution

1

Identify the chemical reaction between carbon(IV) oxide and carbon (coke).

The balanced chemical equation is

\text{CO}_2(g) + \text{C}(s) \rightarrow 2\text{CO}(g)

.

Red-hot coke acts as a reducing agent, reducing carbon(IV) oxide to carbon(II) oxide.

2

Determine the IUPAC name of the gaseous product

\text{CO}

.

The IUPAC systematic name for

\text{CO}

is carbon(II) oxide.

The Roman numeral (II) represents the +2 oxidation state of carbon in the oxide.

Key Concept

Reduction of Carbon(IV) Oxide to Carbon(II) Oxide by hot carbon

Question 4602Question →

Which of the following statements correctly differentiates the circulatory pathway of a bony fish from that of a mammal?

In a bony fish, blood flows through the heart only once during a complete circuit, whereas in a mammal, blood flows through the heart twice.

In a bony fish, oxygenated blood returning from the respiratory organ enters the heart before being pumped to systemic tissues.

A bony fish possesses a three-chambered heart that prevents the mixing of oxygenated and deoxygenated blood.

In a mammal, deoxygenated blood returning from systemic tissues enters the left atrium first.

Show answer & explanation

Answer: In a bony fish, blood flows through the heart only once during a complete circuit, whereas in a mammal, blood flows through the heart twice.

Answer

In a bony fish, blood flows through the heart only once during a complete circuit, whereas in a mammal, blood flows through the heart twice.

Bony fish exhibit single circulation with a two-chambered heart, meaning blood passes through the heart only once per complete journey around the body. Mammals exhibit double circulation with a four-chambered heart, meaning blood passes through the heart twice per complete circuit.

Step-by-Step Solution

1

Analyze the circulatory pattern of Pisces (bony fish).

Fish have a single-circuit circulation powered by a two-chambered heart (one atrium and one ventricle). Deoxygenated blood is pumped from the ventricle to the gills, becomes oxygenated, and continues straight to body tissues.

Understanding single circulation clarifies how blood moves through fish heart chambers.

2

Analyze the circulatory pattern of Mammalia (mammals).

Mammals have a double-circuit circulation powered by a four-chambered heart. Blood passes through the right side of the heart to the lungs (pulmonary circuit), returns to the left side of the heart, and is pumped to body tissues (systemic circuit).

Comparing single versus double circulation highlights the fundamental physiological difference.

3

Match the comparative analysis to the correct option.

The statement highlighting single flow through the heart in fish versus double flow in mammals accurately reflects vertebrate circulatory evolution.

Identifies the correct physiological distinction.

Key Concept

Single versus Double Circulation in Vertebrates

Question 4603Question →

In the industrial refining of copper by electrolysis, a block of impure copper is used as the anode in an aqueous copper(II) tetraoxosulfate(VI) electrolyte. Which of the following half-equations correctly represents the primary reaction taking place at the anode?

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

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

4OH^-_{(aq)} \rightarrow 2H_2O_{(l)} + O_{2(g)} + 4e^-

2H^+_{(aq)} + 2e^- \rightarrow H_{2(g)}

Show answer & explanation

Answer: $Cu_{(s)} \rightarrow Cu^{2+}_{(aq)} + 2e^-$

Answer

The primary reaction occurring at the anode is the oxidation of copper metal into copper(II) ions:

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

.

During the industrial electrorefining of copper, an active copper anode is oxidized by losing two electrons per atom, dissolving into the electrolyte as copper(II) ions (

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

). This maintains the concentration of copper(II) ions in solution while transferring pure copper onto the cathode.

Step-by-Step Solution

1

Identify the general electrode process at the positive terminal (anode).

Oxidation (loss of electrons) always occurs at the anode.

Electrons are removed from species at the anode during electrolysis.

2

Determine whether the anode electrode material is inert or active (reactive).

The anode is made of impure copper metal, which is an active electrode.

Active metal anodes dissolve into solution more readily than anions in the solution can discharge.

3

Write the oxidation half-reaction for the dissolution of the copper anode.

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

Neutral copper atoms in the impure block lose two electrons each to enter the solution as soluble copper(II) cations.

Key Concept

Electrorefining of copper and active anode oxidation

Question 4604Question →

Match each group of spermatophytes on the left with its defining structural or reproductive feature on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Gymnosperms pair with producing naked seeds borne exposed on cones; Angiosperms pair with producing flowers and seeds enclosed within a fruit; Monocotyledons pair with seeds containing one cotyledon and leaves exhibiting parallel venation; Dicotyledons pair with seeds containing two cotyledons and leaves exhibiting reticulate venation.

Gymnosperms are distinguished by naked seeds exposed on cones. Angiosperms produce flowers and seeds enclosed inside fruits. Monocotyledons possess seeds with a single seed leaf (cotyledon) and parallel leaf venation. Dicotyledons possess seeds with two seed leaves (cotyledons) and reticulate (net-like) leaf venation.

Step-by-Step Solution

1

Identify the primary seed enclosure distinction between seed plant divisions

Gymnosperms produce naked seeds exposed on reproductive structures like cones, whereas Angiosperms produce seeds housed within fruits.

This establishes the fundamental division within Spermatophyta.

2

Distinguish between the two main classes of flowering plants (Angiosperms)

Monocots have one cotyledon and parallel leaf veins; Dicots have two cotyledons and reticulate leaf veins.

Cotyledon count and leaf venation pattern are key anatomical markers for classifying angiosperms.

Key Concept

Classification and diagnostic features of Spermatophytes (Gymnosperms vs Angiosperms, Monocots vs Dicots)

Question 4605Question →

Which of the following plant tissues provides flexible mechanical support to young growing stems through cell walls with localized cellulose and pectin thickenings at the corners?

Collenchyma tissue

Sclerenchyma tissue

Xylem vessel tissue

Parenchyma tissue

Show answer & explanation

Answer: Collenchyma tissue

Answer

Collenchyma tissue provides flexible mechanical support to young growing shoots due to uneven cellulose and pectin thickenings at cell wall corners.

Collenchyma cells remain living at maturity and possess unevenly thickened cell walls containing high amounts of cellulose and pectin at the corners. This structural arrangement allows young plant parts, such as growing stems and leaf petioles, to bend without breaking.

Step-by-Step Solution

1

Identify the primary supporting tissues in plants and their structural features.

Collenchyma tissue consists of living cells with localized wall thickenings composed of cellulose and pectin.

These specific corner thickenings allow the tissue to resist bending stress while permitting continued growth and flexibility.

Key Concept

Plant supporting tissues and cell wall thickenings

Question 4606Question →

During a withdrawal reflex when a person touches a hot object, electrical impulses travel through a specific sequence of neural structures to produce an immediate somatic motor response. Which of the following correctly traces the pathway of the nerve impulse from stimulus detection to the execution of the response?

Sensory receptor → Afferent neuron → Relay neuron → Efferent neuron → Skeletal muscle effector

Sensory receptor → Efferent neuron → Relay neuron → Afferent neuron → Skeletal muscle effector

Skeletal muscle effector → Efferent neuron → Relay neuron → Afferent neuron → Sensory receptor

Afferent neuron → Sensory receptor → Efferent neuron → Relay neuron → Skeletal muscle effector

Show answer & explanation

Answer: Sensory receptor → Afferent neuron → Relay neuron → Efferent neuron → Skeletal muscle effector

Answer

Sensory receptor → Afferent neuron → Relay neuron → Efferent neuron → Skeletal muscle effector

The correct response traces impulse flow in strict physiological sequence: starting at the sensory receptor (stimulus transduction), moving through the afferent neuron (sensory input to CNS), across the relay neuron in the spinal cord (integration), along the efferent neuron (motor output from CNS), and terminating at the skeletal muscle effector (response execution).

Step-by-Step Solution

1

Identify the initial site of stimulus reception

Sensory receptors in the skin detect thermal stimulus and generate action potentials.

Receptors transduce physical energy from stimuli into electrical nerve impulses.

2

Trace inward conduction toward the central nervous system

Impulses travel along the afferent (sensory) neuron into the dorsal horn of the spinal cord.

Afferent neurons transmit sensory information unidirectionally toward the spinal cord.

3

Identify synaptic integration in the spinal cord

The impulse synaptically transfers to a relay (interneuron) within the spinal cord grey matter.

Relay neurons connect sensory and motor pathways within the central nervous system for rapid reflex processing.

4

Trace outward conduction to the target organ

The relay neuron excites an efferent (motor) neuron, which conducts impulses to the skeletal muscle effector to cause contraction.

Efferent neurons conduct motor signals from the spinal cord to muscle effectors to carry out the physical response.

Key Concept

Unidirectional impulse flow in a spinal reflex arc

Question 4607Question →

In a $1.0\text{ dm}^3$ rigid reaction vessel, $4.0\text{ moles}$ of gas $A$ and $3.0\text{ moles}$ of gas $B$ are mixed and allowed to reach equilibrium at a constant temperature according to the equation:

2A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)}

If analysis shows that $1.0\text{ mole}$ of gas $C$ is present at equilibrium, what is the value of the equilibrium constant, $K_c$ , for this reaction?

0.125\text{ dm}^6\text{mol}^{-2}

0.250\text{ dm}^6\text{mol}^{-2}

0.056\text{ dm}^6\text{mol}^{-2}

8.000\text{ dm}^6\text{mol}^{-2}

Show answer & explanation

Answer: $0.125\text{ dm}^6\text{mol}^{-2}$

Answer

The equilibrium constant

K_c

is

0.125\text{ dm}^6\text{mol}^{-2}

.

To find

K_c

, set up an equilibrium concentration table. Producing

1.0\text{ mole}

of product

C

consumes

2.0\text{ moles}

of

A

and

1.0\text{ mole}

of

B

per

\text{dm}^3

. The equilibrium concentrations are

[A] = 2.0\text{ mol dm}^{-3}

,

[B] = 2.0\text{ mol dm}^{-3}

, and

[C] = 1.0\text{ mol dm}^{-3}

. Substituting these values into

K_c = \frac{[C]}{[A]^2[B]}

yields

\frac{1.0}{2.0^2 \times 2.0} = \frac{1.0}{8.0} = 0.125\text{ dm}^6\text{mol}^{-2}

.

Step-by-Step Solution

1

Determine initial molar concentrations

Since container volume is

1.0\text{ dm}^3

, initial concentrations are

[A]_0 = 4.0\text{ mol dm}^{-3}

,

[B]_0 = 3.0\text{ mol dm}^{-3}

, and

[C]_0 = 0\text{ mol dm}^{-3}

.

Concentration is moles divided by volume in

\text{dm}^3

.

2

Calculate equilibrium concentrations using stoichiometry (ICE Table)

At equilibrium,

[C]_{eq} = 1.0\text{ mol dm}^{-3}

.
According to

2A + B \rightleftharpoons C

:
Moles of

A

consumed =

2 \times 1.0 = 2.0\text{ mol dm}^{-3} \implies [A]_{eq} = 4.0 - 2.0 = 2.0\text{ mol dm}^{-3}

.
Moles of

B

consumed =

1 \times 1.0 = 1.0\text{ mol dm}^{-3} \implies [B]_{eq} = 3.0 - 1.0 = 2.0\text{ mol dm}^{-3}

.

Stoichiometric coefficients govern the mole ratio of reactants consumed to products formed.

3

Write the

K_c

expression and substitute equilibrium concentrations

K_c = \frac{[C]}{[A]^2[B]} = \frac{1.0}{(2.0)^2 \times 2.0} = \frac{1.0}{4.0 \times 2.0} = \frac{1.0}{8.0} = 0.125\text{ dm}^6\text{mol}^{-2}

The equilibrium constant expression raises each concentration to the power of its stoichiometric coefficient.

Key Concept

Equilibrium constant

K_c

expression and stoichiometric calculations

Estimated Time:2m 0s

Question 4608Question →

Match each transition metal complex species on the left with its corresponding IUPAC name and oxidation state of the central metal ion on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches are:

[Fe(CN)_6]^{3-}

matches Hexacyanoferrate(III) ion (

Fe

in

+3

state);

[Cu(NH_3)_4]^{2+}

matches Tetraamminecopper(II) ion (

Cu

in

+2

state);

[Co(H_2O)_6]^{2+}

matches Hexaaquacobalt(II) ion (

Co

in

+2

state); and

[Ni(CO)_4]

matches Tetracarbonylnickel(0) (

Ni

in

0

state).

Each complex is correctly paired based on the number and charge of its ligands, the resulting oxidation state of the central transition metal, and IUPAC nomenclature rules for cationic, neutral, and anionic complexes.

Step-by-Step Solution

1

Determine the charge of each ligand present in the transition metal complex.

Cyano (

CN^-

) carries a charge of

-1

, while ammine (

NH_3

), aqua (

H_2O

), and carbonyl (

CO

) are neutral (

0

charge).

Knowing ligand charges is required to set up the algebraic equation for the metal's oxidation state.

2

Calculate the oxidation state of the central metal ion in each complex.

For

[Fe(CN)_6]^{3-}

:

x + 6(-1) = -3 \Rightarrow x = +3

. For

[Cu(NH_3)_4]^{2+}

:

x + 4(0) = +2 \Rightarrow x = +2

. For

[Co(H_2O)_6]^{2+}

:

x + 6(0) = +2 \Rightarrow x = +2

. For

[Ni(CO)_4]

:

x + 4(0) = 0 \Rightarrow x = 0

.

The sum of oxidation numbers of the central atom and ligands equals the total net charge of the complex species.

3

Apply standard IUPAC naming rules for complex species.

Name ligands with multiplicative prefixes (tetra-, hexa-), follow with the metal name (using '-ate' suffix for anionic complexes like ferrate), and indicate the oxidation state in Roman numerals.

Anionic complexes modify the metal name root, whereas cationic and neutral complexes retain the standard metal element name.

Key Concept

IUPAC Nomenclature and Oxidation State Calculations for Transition Metal Complex Ions

Question 4609Question →

During binary fission in Paramecium caudatum, the cell undergoes nuclear division prior to transverse cytoplasmic division. Which statement accurately describes the division process of its nuclei?

The macronucleus divides amitotically by constriction, while the micronucleus divides mitotically.

Both the macronucleus and micronucleus divide by standard mitotic division.

The micronucleus disintegrates while the macronucleus undergoes meiotic division.

The macronucleus undergoes meiosis to produce haploid pronuclei for daughter cells.

Show answer & explanation

Answer: The macronucleus divides amitotically by constriction, while the micronucleus divides mitotically.

Answer

The macronucleus divides amitotically by constriction, while the micronucleus divides mitotically.

Paramecium displays nuclear dualism. During asexual binary fission, the small diploid micronucleus undergoes mitosis to distribute genetic material equally, whereas the large polyploid macronucleus elongates and divides amitotically by simple constriction.

Step-by-Step Solution

1

Identify the nuclear structure of Paramecium.

Paramecium exhibits nuclear dualism, possessing a small diploid micronucleus (germline/reproductive function) and a large polyploid macronucleus (somatic/metabolic function).

The different ploidy levels and roles determine their division mechanisms during asexual reproduction.

2

Determine the mode of division for each nucleus during binary fission.

The micronucleus duplicates its chromosomes and undergoes mitosis. The macronucleus elongates and splits amitotically by constriction without chromosome condensation or spindle assembly.

Mitotic distribution ensures equal genetic division of the micronucleus, while the somatic macronucleus divides its mass amitotically.

Key Concept

Nuclear Dualism and Binary Fission in Paramecium

Question 4610Question →

An ecology student investigated the population of *Tridax procumbens* in a cassava farmland in Ogun State using a $0.5\text{ m} \times 0.5\text{ m}$ quadrat frame. Across 10 random quadrat throws, the total number of *Tridax procumbens* plants counted was 80. What is the estimated population density of *Tridax procumbens* per square metre ( $\text{m}^{-2}$ ) in the farmland?

8\text{ plants m}^{-2}

32\text{ plants m}^{-2}

160\text{ plants m}^{-2}

320\text{ plants m}^{-2}

Show answer & explanation

Answer: $32\text{ plants m}^{-2}$

Answer

The population density of *Tridax procumbens* is

32\text{ plants m}^{-2}

.

The area of one quadrat frame is

0.5\text{ m} \times 0.5\text{ m} = 0.25\text{ m}^2

. Across 10 throws, the total area sampled is

10 \times 0.25\text{ m}^2 = 2.5\text{ m}^2

. Dividing the total number of plants counted (80) by the total sampled area (

2.5\text{ m}^2

) yields an accurate population density of

32\text{ plants m}^{-2}

.

Step-by-Step Solution

1

Calculate the area of a single quadrat frame.

\text{Area of 1 quadrat} = 0.5\text{ m} \times 0.5\text{ m} = 0.25\text{ m}^2

Determines the spatial coverage of one quadrat throw.

2

Calculate the total area sampled across all quadrat throws.

\text{Total sampled area} = 10 \times 0.25\text{ m}^2 = 2.5\text{ m}^2

Accounting for all 10 sampling units used during field sampling.

3

Divide the total count of organisms by the total sampled area.

\text{Population Density} = \frac{80\text{ plants}}{2.5\text{ m}^2} = 32\text{ plants m}^{-2}

Population density is defined as the number of individuals of a species per unit area.

Key Concept

Quadrat Sampling and Population Density Calculation

Question 4611Question →

In the mammalian kidney, ultrafiltration is driven by net filtration pressure across the glomerular capillaries into the Bowman's capsule. Which combination of vascular changes would produce the highest increase in the rate of glomerular filtration?

Vasodilation of the afferent arteriole combined with vasoconstriction of the efferent arteriole

Vasoconstriction of the afferent arteriole combined with vasodilation of the efferent arteriole

Vasoconstriction of both afferent and efferent arterioles to equal degrees

Vasodilation of the efferent arteriole combined with an increase in plasma protein concentration

Show answer & explanation

Answer: Vasodilation of the afferent arteriole combined with vasoconstriction of the efferent arteriole

Answer

Vasodilation of the afferent arteriole combined with vasoconstriction of the efferent arteriole

Dilation of the wide afferent arteriole increases blood flow into the glomerulus, while constriction of the narrow efferent arteriole impedes fluid exit. This raises the hydrostatic pressure within the glomerular capillaries, which is the main driving force for ultrafiltration across the basement membrane into the capsular space.

Step-by-Step Solution

1

Analyze the forces governing glomerular ultrafiltration in the nephron.

Net filtration pressure depends directly on glomerular capillary hydrostatic pressure pushing fluid out into Bowman's capsule, opposed by capsular hydrostatic pressure and plasma colloid osmotic pressure.

Glomerular hydrostatic pressure is the primary variable driving ultrafiltration.

2

Evaluate the hemodynamic impact of altering arteriolar resistance.

Opening (dilating) the inlet vessel (afferent arteriole) brings more blood under pressure into the glomerulus, while narrowing (constricting) the outlet vessel (efferent arteriole) creates backpressure.

This dual action traps blood in the glomerular capillaries, driving capillary hydrostatic pressure to its maximum level.

3

Select the option that maximizes net filtration pressure.

Afferent vasodilation paired with efferent vasoconstriction produces the greatest elevation in glomerular hydrostatic pressure, maximizing filtration rate.

It optimizes the pressure gradient required for ultrafiltration.

Key Concept

Renal Hemodynamics and Glomerular Ultrafiltration Mechanics

Question 4612Question →

When copper metal turnings are heated with concentrated tetraoxosulfate(VI) acid ( $H_2SO_4$ ), a dense gas is evolved alongside the formation of copper(II) tetraoxosulfate(VI) ( $CuSO_4$ ) and water ( $H_2O$ ). Which of the following correctly identifies the gaseous product evolved and the specific role of concentrated $H_2SO_4$ in this chemical reaction?

Sulfur(IV) oxide (

SO_2

); concentrated

H_2SO_4

acts as an oxidizing agent.

Hydrogen gas (

H_2

); concentrated

H_2SO_4

acts as a typical acid reacting with a metal.

Sulfur(VI) oxide (

SO_3

); concentrated

H_2SO_4

acts as a dehydrating agent.

Hydrogen sulfide (

H_2S

); concentrated

H_2SO_4

acts as a reducing agent.

Show answer & explanation

Answer: Sulfur(IV) oxide ( $SO_2$ ); concentrated $H_2SO_4$ acts as an oxidizing agent.

Answer

Sulfur(IV) oxide (

SO_2

); concentrated

H_2SO_4

acts as an oxidizing agent.

Copper is an unreactive transition metal located below hydrogen in the electrochemical series. Consequently, it does not react with dilute acids to liberate hydrogen gas. However, when heated with concentrated tetraoxosulfate(VI) acid, the acid behaves as a strong oxidizing agent rather than a typical acid. Copper is oxidized to copper(II) tetraoxosulfate(VI) (

CuSO_4

), while the sulfur in

H_2SO_4

is reduced from oxidation state

+6

to

+4

, liberating sulfur(IV) oxide gas (

SO_2

) along with water.

Step-by-Step Solution

1

Analyze the position of copper in the electrochemical activity series.

Copper is less electropositive than hydrogen (

E^\circ_{Cu^{2+}/Cu} = +0.34\text{ V}

), meaning it cannot displace

H_2

gas from dilute or non-oxidizing acids.

Direct single replacement of hydrogen by copper is thermodynamically unfavorable.

2

Determine the chemical action of hot concentrated tetraoxosulfate(VI) acid on copper.

Hot concentrated

H_2SO_4

is a strong oxidizing agent. It oxidizes copper from oxidation state

0

to

+2

(

Cu^{2+}

).

Oxidizing acids react with unreactive metals via redox pathways rather than acid-base displacement.

3

Formulate the balanced thermochemical equation.

Cu_{(s)} + 2H_2SO_{4(aq)} \rightarrow CuSO_{4(aq)} + 2H_2O_{(l)} + SO_{2(g)}

Sulfur in

H_2SO_4

(oxidation state

+6

) undergoes reduction to form sulfur(IV) oxide (

SO_2

, oxidation state

+4

).

Key Concept

Chemical properties of copper and oxidizing behavior of concentrated tetraoxosulfate(VI) acid

Estimated Time:2m 0s

Question 4613Question →

What volume of carbon(IV) oxide gas, in $\text{dm}^3$ , measured at STP, is produced by the complete thermal decomposition of $20.0\text{ g}$ of pure calcium trioxocarbonate(IV), $\text{CaCO}_3$ ? [ $\text{Ca} = 40$ , $\text{C} = 12$ , $\text{O} = 16$ ; Molar volume of gas at STP $= 22.4\text{ dm}^3\text{mol}^{-1}$ ]

Show answer & explanation

Answer: 4.48

Answer

4.48 dm³

Complete thermal decomposition of 20.0 g of CaCO₃ (molar mass 100 g/mol) generates 0.20 mol of CO₂ gas according to the equation CaCO₃(s) -> CaO(s) + CO₂(g). Since 1 mol of gas at STP occupies 22.4 dm³, 0.20 mol occupies 4.48 dm³.

Step-by-Step Solution

1

Determine the molar mass and number of moles of calcium trioxocarbonate(IV).

Molar mass of CaCO₃ = 100 g/mol; Moles of CaCO₃ = 20.0 g / 100 g/mol = 0.20 mol

Converting given mass to moles is required to apply stoichiometric ratios.

2

Apply the balanced reaction mole ratio to find moles of carbon(IV) oxide produced.

Moles of CO₂ = 0.20 mol

The equation CaCO₃(s) -> CaO(s) + CO₂(g) shows a 1:1 molar ratio between CaCO₃ and CO₂.

3

Multiply moles of CO₂ by molar gas volume at STP.

Volume of CO₂ = 0.20 mol × 22.4 dm³/mol = 4.48 dm³

At STP, 1 mole of any ideal gas occupies 22.4 dm³.

Key Concept

Thermal decomposition of trioxocarbonate(IV) salts and gas volume calculations at STP

Estimated Time:45s

Question 4614Question →

Match each organic functional group class on the left with its characteristic IUPAC naming suffix on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Alkanol (Alcohol) matches with '-ol', Alkanal (Aldehyde) matches with '-al', Alkanone (Ketone) matches with '-one', and Alkanoic acid (Carboxylic acid) matches with '-oic acid'.

Each organic functional group family corresponds to a specific IUPAC naming suffix: Alkanols take '-ol', Alkanals take '-al', Alkanones take '-one', and Alkanoic acids take '-oic acid'.

Step-by-Step Solution

1

Identify the primary functional group present in each class of organic compound.

Alkanols possess hydroxyl groups, alkanals possess aldehyde groups, alkanones possess ketone carbonyl groups, and alkanoic acids possess carboxyl groups.

The characteristic functional group determines both the homologous series and the IUPAC suffix rules.

2

Assign the standard IUPAC nomenclature suffix corresponding to each functional group.

Alkanol

\rightarrow

'-ol', Alkanal

\rightarrow

'-al', Alkanone

\rightarrow

'-one', Alkanoic acid

\rightarrow

'-oic acid'.

Standard IUPAC rules modify the parent alkane name by replacing the terminal '-e' with the designated functional group suffix.

Key Concept

IUPAC Suffixes for Functional Groups

Question 4615Question →

In a freshwater pond ecosystem, energy flows sequentially through distinct trophic levels following the principle of ecological energy transfer. Consider the following aquatic organisms: Freshwater pike, Water fleas (Daphnia), Microscopic green algae (Chlorella), and Minnows.

Arrange these organisms in order from the HIGHEST available energy to the LOWEST available energy.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from highest available energy to lowest available energy is: Microscopic green algae (Chlorella) → Water fleas (Daphnia) → Minnows → Freshwater pike.

In any ecosystem, primary producers (green algae) fix solar energy into biochemical energy and possess the greatest amount of available energy. As energy transfers to primary consumers (water fleas), secondary consumers (minnows), and tertiary consumers (freshwater pike), roughly 90% of the energy at each level is dissipated as metabolic heat. Consequently, available energy decreases continuously from producers up to apex carnivores.

Step-by-Step Solution

1

Identify the trophic role of each organism in the pond ecosystem.

Microscopic green algae are primary producers (TL1), Water fleas are primary consumers (TL2), Minnows are secondary consumers (TL3), and Freshwater pike are tertiary consumers (TL4).

Determining trophic positions is necessary to trace the direction of energy transfer along the food chain.

2

Apply the 10% law of energy transfer (Lindeman's efficiency principle).

Energy decreases by approximately 90% at each successive trophic level due to metabolic respiration, movement, excretion, and heat dissipation.

The second law of thermodynamics requires that energy available to subsequent trophic levels decreases steadily from producers to top carnivores.

3

Arrange the organisms from maximum available energy (Trophic Level 1) to minimum available energy (Trophic Level 4).

Microscopic green algae (Chlorella) [TL1] → Water fleas (Daphnia) [TL2] → Minnows [TL3] → Freshwater pike [TL4].

Energy pyramids are strictly upright, meaning energy content is highest at the base and lowest at the apex.

Key Concept

Trophic Energy Transfer and the 10% Law

Estimated Time:1m 15s

Question 4616Question →

An anatomical comparison across higher invertebrate phyla reveals distinct adaptations in excretory structures, appendage arrangement, and scientific nomenclature. Which of the following combinations correctly matches the organism with its physiological feature, structural organization, and properly formatted scientific name?

Earthworm (Lumbricus terrestris): Excretes via metanephridia and exhibits bilateral symmetry with metameric body segmentation

Centipede (Lithobius forficatus): Excretes via Malpighian tubules and exhibits bilateral symmetry with two pairs of jointed legs per body segment

Sea star (Asterias rubens): Excretes via Malpighian tubules and exhibits pentamerous radial symmetry in the adult stage

Garden snail (helix Pomatia): Excretes via metanephridia and exhibits bilateral symmetry modified by body torsion

Show answer & explanation

Answer: Earthworm (Lumbricus terrestris): Excretes via metanephridia and exhibits bilateral symmetry with metameric body segmentation

Answer

Earthworm (Lumbricus terrestris): Excretes via metanephridia and exhibits bilateral symmetry with metameric body segmentation

The correct answer accurately links the earthworm (*Lumbricus terrestris*) to its metanephridial excretory system, metameric segmentation, bilateral symmetry, and properly formatted scientific binomial nomenclature.

Step-by-Step Solution

1

Evaluate the excretory organ and structural traits of Annelida.

Earthworms (Lumbricus terrestris) utilize metanephridia for excretion, exhibit metameric segmentation, and are bilaterally symmetrical.

Annelids are coelomate invertebrates characterized by segmented bodies and paired nephridial tubules.

2

Analyze the appendage count per segment in Arthropoda classes (Chilopoda vs. Diplopoda).

Centipedes (Chilopoda) have one pair of legs per trunk segment, not two.

Diplopods (millipedes) have double segments (diplosomites) bearing two leg pairs, whereas chilopods (centipedes) have single segments with one leg pair.

3

Examine the excretory mechanism of Echinodermata.

Sea stars rely on cell diffusion via papulae and tube feet rather than Malpighian tubules.

Malpighian tubules are restricted to insects, arachnids, and myriapods.

4

Check binomial nomenclature formatting rules.

The binomial form 'helix Pomatia' is incorrectly capitalized.

Genus names must be capitalized and species names lowercase (Helix pomatia).

Key Concept

Diagnostic anatomical features and nomenclature across Higher Invertebrate Phyla (Annelida, Mollusca, Arthropoda, Echinodermata)

Estimated Time:1m 30s

Question 4617Question →

Match each blood component or circulatory vessel in Column I with its primary physiological function or characteristic in Column II.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Erythrocytes pair with oxygen transport via hemoglobin; Thrombocytes pair with initiating blood clotting via thromboplastin release; Lymphatic Vessels pair with draining excess interstitial tissue fluid; Pulmonary Vein pairs with carrying oxygenated blood from lungs to the left atrium.

Each transport component matches its exact anatomical and physiological role: erythrocytes transport oxygen via hemoglobin, thrombocytes initiate coagulation by releasing clotting factors, lymphatic vessels return extracellular interstitial fluid to systemic circulation, and pulmonary veins return oxygen-rich blood from the lungs into the heart's left atrium.

Step-by-Step Solution

1

Identify the biological role of red blood cells (erythrocytes).

Erythrocytes specialize in gas transport using hemoglobin and are enucleated in mature mammals.

Maximizing internal volume allows efficient binding and transport of oxygen gas.

2

Identify the function of blood platelets (thrombocytes).

Thrombocytes initiate blood coagulation at wound sites.

Disruption of platelets releases thromboplastin, converting prothrombin to thrombin.

3

Determine the role of the lymphatic system vessels.

Lymphatic capillaries absorb tissue fluid filtered out of blood capillaries and return lymph to venous blood.

This maintains fluid balance between blood tissue and extracellular space.

4

Trace the vessel returning oxygenated blood from pulmonary circuit.

Pulmonary veins carry oxygenated blood directly from alveoli capillaries back into the left atrium.

Although most veins carry deoxygenated blood, pulmonary veins carry oxygenated blood post-gas exchange.

Key Concept

Blood Cellular Composition and Vessel Functions in Mammalian Transport Systems

Estimated Time:1m 0s

Question 4618Question →

Given the following thermochemical equations:

\text{I. } 2\text{Fe}(s) + \frac{3}{2}\text{O}_2(g) \rightarrow \text{Fe}_2\text{O}_3(s) \quad \Delta H^\circ = -824.2\text{ kJ mol}^{-1}

\text{II. } \text{CO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H^\circ = -283.0\text{ kJ mol}^{-1}

What is the standard enthalpy change,

\Delta H^\circ

, for the reduction of iron(III) oxide by carbon monoxide according to the following reaction?

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

-24.8\text{ kJ mol}^{-1}

-1673.2\text{ kJ mol}^{-1}

+1673.2\text{ kJ mol}^{-1}

+541.2\text{ kJ mol}^{-1}

Show answer & explanation

Answer: $-24.8\text{ kJ mol}^{-1}$

Answer

-24.8\text{ kJ mol}^{-1}

According to Hess's Law, the total enthalpy change for a reaction is the sum of the enthalpy changes for individual intermediate steps. Reversing the formation equation of

\text{Fe}_2\text{O}_3(s)

flips its enthalpy sign from

-824.2\text{ kJ mol}^{-1}

to

+824.2\text{ kJ mol}^{-1}

. Multiplying the oxidation reaction of

\text{CO}(g)

by 3 scales its enthalpy from

-283.0\text{ kJ mol}^{-1}

to

-849.0\text{ kJ mol}^{-1}

. Adding both modified values yields

+824.2\text{ kJ mol}^{-1} - 849.0\text{ kJ mol}^{-1} = -24.8\text{ kJ mol}^{-1}

.

Step-by-Step Solution

1

Reverse Equation I so that

\text{Fe}_2\text{O}_3(s)

becomes a reactant.

\text{Fe}_2\text{O}_3(s) \rightarrow 2\text{Fe}(s) + \frac{3}{2}\text{O}_2(g) \quad \Delta H_1^\circ = +824.2\text{ kJ mol}^{-1}

Reversing a thermochemical equation changes the sign of its standard enthalpy change,

\Delta H^\circ

.

2

Multiply Equation II by 3 so that the stoichiometric coefficient of

\text{CO}(g)

matches the target reaction.

3\text{CO}(g) + \frac{3}{2}\text{O}_2(g) \rightarrow 3\text{CO}_2(g) \quad \Delta H_2^\circ = 3 \times (-283.0\text{ kJ mol}^{-1}) = -849.0\text{ kJ mol}^{-1}

Enthalpy change is an extensive property, so multiplying reaction coefficients requires multiplying

\Delta H^\circ

by the same factor.

3

Sum the modified thermochemical equations according to Hess's Law.

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

\Delta H^\circ = \Delta H_1^\circ + \Delta H_2^\circ = +824.2\text{ kJ mol}^{-1} + (-849.0\text{ kJ mol}^{-1}) = -24.8\text{ kJ mol}^{-1}

Hess's Law states that the overall enthalpy change for a chemical reaction is independent of the pathway taken.

Key Concept

Hess's Law of Constant Heat Summation

Estimated Time:2m 0s

Question 4619Question →

Which of the following characteristics distinguishes gymnosperms from angiosperms?

Possession of exposed seeds that are not enclosed within an ovary wall

Production of seeds enclosed entirely within a true fruit structure

Complete absence of specialized vascular tissues for transport

Conduction of organic food materials exclusively through xylem vessels

Show answer & explanation

Answer: Possession of exposed seeds that are not enclosed within an ovary wall

Answer

Possession of exposed seeds that are not enclosed within an ovary wall

Gymnosperms produce exposed or 'naked' seeds borne on cone scales because they lack carpels and ovaries to enclose their seeds, whereas angiosperms enclose their seeds within ovaries that develop into fruits.

Step-by-Step Solution

1

Identify the two main sub-groups of seed-bearing plants (spermatophytes).

Spermatophytes are divided into gymnosperms and angiosperms.

Both groups produce seeds, but differ fundamentally in how their seeds develop and are protected.

2

Compare the seed structures of gymnosperms and angiosperms.

Gymnosperms bear 'naked' seeds on megasporophylls (cones), whereas angiosperms bear seeds enclosed within carpels (ovaries) that mature into fruits.

The absence of an enclosed ovary wall in gymnosperms is the primary diagnostic feature separating the two groups.

Key Concept

Seed enclosure distinction between Gymnosperms and Angiosperms

Question 4620Question →

Match each specialized structure or cell modification of organisms in Kingdom Monera on the left with its corresponding biological function or characterization on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Heterocyst matches with atmospheric nitrogen fixation under anaerobic conditions; Akinete matches with dormant resting cell adapted to survive unfavorable conditions; Mesosome matches with invagination of plasma membrane involved in respiration and cell division; Plasmid matches with small extrachromosomal circular DNA molecule.

Each specialized cell or structure in Kingdom Monera is correctly matched with its biological role: heterocysts fix atmospheric nitrogen, akinetes serve as dormant survival cells, mesosomes aid respiration and cell division, and plasmids represent extrachromosomal DNA.

Step-by-Step Solution

1

Identify the primary role of heterocysts in filamentous cyanobacteria.

Recognize that heterocysts provide an oxygen-free site for nitrogenase activity during nitrogen fixation.

Cyanobacteria require specialized cells to protect nitrogenase from oxygen produced during photosynthesis.

2

Determine the protective function of akinetes.

Match akinetes with thick-walled resting spores resistant to environmental stress.

Akinetes accumulate food reserves and develop thick walls to survive desiccation or cold temperatures.

3

Analyze the structural nature of mesosomes in bacteria.

Associate mesosomes with plasma membrane folds that aid in respiration and septum creation.

Prokaryotes lack membrane-bound mitochondria, utilizing plasma membrane invaginations for enzymatic metabolic processes.

4

Define plasmids in prokaryotic genetics.

Match plasmids with extrachromosomal circular DNA pieces capable of independent replication.

Plasmids exist separately from the bacterial nucleoid and replicate autonomously.

Key Concept

Cellular structures and specialized functional adaptations in Kingdom Monera (Bacteria and Cyanobacteria)

Estimated Time:1m 0s

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