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Question 12901Question →

Match each chemical reaction or treatment involving an acid or base on the left with its corresponding characteristic observation or underlying chemical property on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Warming

NH_4Cl

with

NaOH

liberates

NH_3

gas which turns moist red litmus blue; heating

NaCl

with concentrated

H_2SO_4

liberates

HCl

gas which forms white fumes with ammonia vapor; adding excess

NaOH

to

Zn(OH)_2

dissolves the precipitate due to its amphoteric nature; reacting copper with concentrated

HNO_3

yields brown

NO_2

gas due to the acid's strong oxidizing power.

Each pair correctly matches the specific acid/base transformation to its corresponding property: liberation of alkaline

NH_3

gas from ammonium salt displacement, liberation of volatile

HCl

gas by non-volatile

H_2SO_4

, dissolution of amphoteric

Zn(OH)_2

in excess base, and evolution of

NO_2

gas due to the oxidizing nature of concentrated

HNO_3

.

Step-by-Step Solution

1

Examine the reaction between ammonium chloride (

NH_4Cl

) and sodium hydroxide (

NaOH

).

Heating ammonium salts with soluble alkalis yields sodium chloride, water, and ammonia gas (

NH_3

), which is alkaline and turns red litmus blue.

This demonstrates the general chemical property of alkalis reacting with ammonium salts to displace volatile ammonia.

2

Examine the reaction between solid sodium chloride (

NaCl

) and concentrated tetraoxosulfate(VI) acid (

H_2SO_4

).

Concentrated

H_2SO_4

displaces hydrogen chloride (

HCl

) gas because

H_2SO_4

is significantly less volatile (higher boiling point) than

HCl

.

This illustrates the principle that non-volatile acids displace volatile acids from their salts.

3

Examine the effect of adding excess sodium hydroxide (

NaOH

) to zinc hydroxide (

Zn(OH)_2

).

The insoluble precipitate dissolves to form a clear solution containing the soluble complex anion

[Zn(OH)_4]^{2-}

.

This exhibits the amphoteric property of zinc hydroxide, allowing it to react with strong bases.

4

Examine the reaction of copper metal (

Cu

) with concentrated trioxonitrate(V) acid (

HNO_3

).

Copper is oxidized to

Cu^{2+}

while

HNO_3

is reduced to brown nitrogen(IV) oxide gas (

NO_2

), rather than evolving hydrogen gas.

This highlights that

HNO_3

functions primarily as an oxidizing agent when reacting with metals below hydrogen in the reactivity series.

Key Concept

Distinct physical and chemical properties of acids and bases, including acid/base displacement volatility, amphoterism, and oxidizing acid behavior.

Question 12902Question →

Arrange the following hydrocarbons in order of increasing boiling point, starting with the compound that has the lowest boiling point:

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from lowest to highest boiling point is 2,2-dimethylpropane, 2-methylbutane, pentane, and hexane.

Boiling points of alkanes depend on molecular mass and surface area contact. Hexane has the highest molecular mass and largest surface area, yielding the highest boiling point. Among the structural isomers of pentane, increased branching produces a more spherical shape with smaller surface contact area, reducing intermolecular van der Waals forces. Consequently, 2,2-dimethylpropane (most branched) has the lowest boiling point, followed by 2-methylbutane, straight-chain pentane, and finally hexane.

Step-by-Step Solution

1

Compare molecular mass and carbon chain length among the given hydrocarbons.

hexane (

C_6H_{14}

) has a larger molecular mass and longer chain than the pentane isomers (

C_5H_{12}

), giving it the strongest London dispersion forces and highest boiling point.

Boiling point increases with molar mass in a homologous series due to increased electron cloud polarizability.

2

Compare the degree of branching among the structural isomers of pentane (

C_5H_{12}

).

2,2-dimethylpropane is highly branched (compact spherical shape), 2-methylbutane is moderately branched, and pentane is a straight chain.

Increased branching makes the molecule more spherical, reducing the surface area available for intermolecular van der Waals contact.

3

Rank all four compounds in order of increasing boiling point.

2,2-dimethylpropane < 2-methylbutane < pentane < hexane.

More branched isomers have lower boiling points than less branched isomers of the same molecular formula, and higher molar mass alkanes boil at higher temperatures than lower mass alkanes.

Key Concept

Effect of molecular mass and chain branching on alkane boiling points

Estimated Time:1m 0s

Question 12903Question →

Match each electromagnetic rule or concept in Column I with its corresponding physical description or application in Column II.

Click a left item, then click its matching right item

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Answer

Fleming's Left-Hand Rule matches with determining the direction of magnetic force on a current-carrying conductor; Right-Hand Grip Rule matches with determining the direction of the magnetic field around a straight wire;

F = qvB\sin\theta

matches with calculating the magnitude of magnetic force on a moving electric charge.

Fleming's Left-Hand Rule establishes the perpendicular direction of magnetic force relative to current and magnetic field vectors. The Right-Hand Grip Rule relates linear current direction to circular magnetic field lines. The equation

F = qvB\sin\theta

directly calculates the magnitude of force on a free charge moving through a uniform magnetic field.

Step-by-Step Solution

1

Analyze Fleming's Left-Hand Rule.

The thumb, forefinger, and middle finger represent Force, Magnetic Field, and Current respectively.

It predicts the direction of mechanical motion/force produced in motors and conductors.

2

Analyze the Right-Hand Grip Rule.

Pointing the right thumb in the current's direction makes the curled fingers show the magnetic field line direction.

It provides a simple geometric method to determine field orientation around conductors.

3

Analyze the force relation

F = qvB\sin\theta

.

It quantifies the Lorentz force acting on an isolated moving electric charge.

The magnetic force magnitude depends on charge, velocity, magnetic field strength, and the angle between velocity and field.

Key Concept

Fundamental rules and formulas of electromagnetism

Question 12904Question →

At a geographical research station, a bar magnet placed along the magnetic meridian neutralizes the horizontal component of the Earth's magnetic field at its neutral points. If the measured horizontal component of the Earth's magnetic field at this site is $1.8 \times 10^{-5}\text{ T}$ and the angle of dip (inclination) is $60^\circ$ , what is the magnitude of the Earth's total magnetic field intensity?

3.6 \times 10^{-5}\text{ T}

0.9 \times 10^{-5}\text{ T}

1.56 \times 10^{-5}\text{ T}

3.12 \times 10^{-5}\text{ T}

Show answer & explanation

Answer: $3.6 \times 10^{-5}\text{ T}$

Answer

The magnitude of the Earth's total magnetic field intensity is

3.6 \times 10^{-5}\text{ T}

.

The horizontal component

B_h

of the Earth's magnetic field is related to the total magnetic intensity

B

and the angle of dip

\theta

by

B_h = B \cos\theta

. Dividing

1.8 \times 10^{-5}\text{ T}

by

\cos 60^\circ = 0.5

gives

3.6 \times 10^{-5}\text{ T}

.

Step-by-Step Solution

1

Identify given parameters and formula relating field components.

Horizontal component

B_h = 1.8 \times 10^{-5}\text{ T}

, Angle of dip

\theta = 60^\circ

. Formula:

B_h = B \cos\theta

.

The horizontal component of the Earth's magnetic field is the projection of the total field onto the horizontal plane.

2

Rearrange formula to solve for the total magnetic field intensity

B

.

B = \frac{B_h}{\cos\theta}

.

To isolate total magnetic field intensity

B

from the given horizontal component.

3

Substitute numerical values and evaluate.

B = \frac{1.8 \times 10^{-5}\text{ T}}{\cos 60^\circ} = \frac{1.8 \times 10^{-5}}{0.5} = 3.6 \times 10^{-5}\text{ T}

.

\cos 60^\circ = 0.5

, yielding a simple exact calculation.

Key Concept

Resolution of Earth's Magnetic Field Components

Question 12905Question →

Unlike most other molluscs, members of the class Cephalopoda (such as squids and octopuses) exhibit which combination of physiological and anatomical adaptations?

A closed circulatory system, a well-developed nervous system with image-forming eyes, and a foot modified into tentacles

An open circulatory system, excretion via Malpighian tubules, and a heavy external calcareous shell

Metameric body segmentation, jointed chitinous appendages, and a hydrostatic skeleton

Secondary pentaradial body symmetry, a water vascular system, and tube feet for movement

Show answer & explanation

Answer: A closed circulatory system, a well-developed nervous system with image-forming eyes, and a foot modified into tentacles

Answer

Cephalopods are characterized by a closed circulatory system, a highly developed nervous system with complex image-forming eyes, and a foot modified into arms or tentacles.

Members of the class Cephalopoda differ from other molluscs by having a closed circulatory system with capillaries to deliver oxygen rapidly during active locomotion. They also possess large, image-forming eyes similar in efficiency to vertebrate eyes, and their muscular foot has evolved into tentacles surrounding the head for prey capture.

Step-by-Step Solution

1

Analyze the general characteristics of the phylum Mollusca.

Molluscs are typically unsegmented, soft-bodied organisms with a muscular foot, visceral mass, mantle, and usually an open circulatory system.

Establishing baseline molluscan features highlights the specialized evolutionary divergences in Cephalopoda.

2

Identify the unique adaptations of the class Cephalopoda.

Cephalopods (e.g., squids, octopuses) adapted to an active predatory lifestyle by developing a closed circulatory system with systemic and branchial hearts, advanced sensory organs including image-forming eyes, and tentacles derived from the foot.

High metabolic rates required for active swimming demand efficient closed blood circulation, unlike slow-moving gastropods or bivalves.

3

Match these adaptations with the correct response option.

The option specifying a closed circulatory system, image-forming eyes, and tentacles correctly describes Cephalopoda.

This option accurately captures both physiological and morphological specializations distinguishing cephalopods from other molluscan classes.

Key Concept

Cephalopod Physiological and Anatomical Adaptations

Question 12906Question →

Element $E$ has a relative atomic mass of $12.011$ and exists as two naturally occurring isotopes: $^{12}\text{E}$ with a natural abundance of $98.9\%$ and $^{A}\text{E}$ with a natural abundance of $1.1\%$ . What is the mass number ( $A$ ) of the second isotope?

Show answer & explanation

Answer: 13

Answer

The mass number of the second isotope is 13.

Relative atomic mass is calculated as the weighted average of the mass numbers of all naturally occurring isotopes. Setting up the equation

12.011 = \frac{(98.9 \times 12) + (1.1 \times A)}{100}

yields

1201.1 = 1186.8 + 1.1A

, which simplifies to

1.1A = 14.3

and gives

A = 13

.

Step-by-Step Solution

1

State the relationship for relative atomic mass based on isotopic abundance.

\text{RAM} = \frac{(\text{Abundance}_1 \times \text{Mass}_1) + (\text{Abundance}_2 \times \text{Mass}_2)}{100}

Relative atomic mass is the weighted average of the atomic masses of naturally occurring isotopes of an element.

2

Substitute the given numerical values into the formula.

12.011 = \frac{(98.9 \times 12) + (1.1 \times A)}{100}

Inserting the known abundances (

98.9\%

and

1.1\%

) and mass number (

12

) sets up an algebraic equation for the unknown mass number

A

.

3

Solve the algebraic equation for

A

.

1201.1 = 1186.8 + 1.1A \implies 1.1A = 14.3 \implies A = 13

Subtracting the contribution of the first isotope and dividing by the abundance of the second isotope gives the integer mass number

13

.

Key Concept

Calculating isotopic mass number from relative atomic mass and fractional abundances

Question 12907Question →

Match each optical instrument component or device on the left with its correct focal length requirement and image formation condition on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Compound Microscope Objective Lens pairs with very short focal length producing a real, inverted, and magnified intermediate image. Astronomical Telescope Objective Lens pairs with long focal length forming a real, inverted, and diminished image of a distant body. Simple Magnifying Glass pairs with single convex lens with object placed within focal length producing an erect, virtual, and magnified image. Slide / Film Projector Lens pairs with convex lens with object positioned between f and 2f producing a real, inverted, and enlarged image on a screen.

The matching correct pairs align each optical device with its precise optical parameter: microscope objectives utilize short focal lengths for high magnification of near objects, telescope objectives utilize long focal lengths for distant objects, simple magnifiers place objects closer than the focal point to form virtual erect images, and projectors place objects between one and two focal lengths to cast real enlarged images on a screen.

Step-by-Step Solution

1

Analyze the objective lens of a compound microscope

It requires a very short focal length to achieve high linear magnification of near objects.

The specimen is placed just outside the focal point (

f < u < 2f

), producing a real, inverted, magnified image inside the tube.

2

Analyze the objective lens of an astronomical telescope

It requires a long focal length and large aperture.

Distant celestial bodies subtend tiny angles at the eye; a long focal length objective creates a larger real intermediate image for the eyepiece to magnify.

3

Analyze the ray path condition of a simple magnifying glass

The object is held within the principal focus (

u < f

).

Light rays emerging from the lens diverge, forming an enlarged, virtual, and erect image on the same side of the lens as the object.

4

Analyze the image projection setup in a slide projector

The slide is placed between

f

and

2f

of a converging lens.

According to the thin lens formula, placing an object between

f

and

2f

yields a real, inverted, magnified image beyond

2f

.

Key Concept

Operating principles and ray placement parameters of optical instruments

Question 12908Question →

The table below presents the mass, quantity of heat absorbed, and resulting temperature change for three solid metal blocks, $P$ , $Q$ , and $R$ :

Block	Mass ( $\text{kg}$ )	Heat Absorbed ( $\text{J}$ )	Temperature Change ( $\text{K}$ )
$P$	$0.20$	$800$	$10$
$Q$	$0.50$	$1500$	$15$
$R$	$0.40$	$1200$	$20$

Which of the following statements correctly compares the thermal properties of the blocks?

Block

P

has the highest specific heat capacity, while block

Q

has the highest total heat capacity.

Block

Q

has both the highest specific heat capacity and the highest total heat capacity.

Block

R

has the highest specific heat capacity, while block

P

has the lowest total heat capacity.

Block

P

has the lowest specific heat capacity, while block

R

has the highest total heat capacity.

Show answer & explanation

Answer: Block $P$ has the highest specific heat capacity, while block $Q$ has the highest total heat capacity.

Answer

Block

P

has the highest specific heat capacity, while block

Q

has the highest total heat capacity.

The total heat capacity

C

is found using

C = \frac{Q}{\Delta T}

, giving

80\text{ J K}^{-1}

for

P

,

100\text{ J K}^{-1}

for

Q

, and

60\text{ J K}^{-1}

for

R

. Dividing heat capacity by mass yields the specific heat capacity

c = \frac{C}{m}

, resulting in

400\text{ J kg}^{-1}\text{ K}^{-1}

for

P

,

200\text{ J kg}^{-1}\text{ K}^{-1}

for

Q

, and

150\text{ J kg}^{-1}\text{ K}^{-1}

for

R

. Therefore, block

P

has the highest specific heat capacity and block

Q

has the highest total heat capacity.

Step-by-Step Solution

1

Calculate the total heat capacity (

C = \frac{Q}{\Delta T}

) for each block

C_P = \frac{800}{10} = 80\text{ J K}^{-1}

,

C_Q = \frac{1500}{15} = 100\text{ J K}^{-1}

,

C_R = \frac{1200}{20} = 60\text{ J K}^{-1}

Heat capacity measures the total thermal energy required to raise the entire object's temperature by

1\text{ K}

.

2

Calculate the specific heat capacity (

c = \frac{C}{m} = \frac{Q}{m \Delta T}

) for each block

c_P = \frac{80}{0.20} = 400\text{ J kg}^{-1}\text{ K}^{-1}

,

c_Q = \frac{100}{0.50} = 200\text{ J kg}^{-1}\text{ K}^{-1}

,

c_R = \frac{60}{0.40} = 150\text{ J kg}^{-1}\text{ K}^{-1}

Specific heat capacity measures the heat capacity per unit mass of the material.

3

Compare the computed values

Highest specific heat capacity: Block

P

(

400\text{ J kg}^{-1}\text{ K}^{-1}

); Highest total heat capacity: Block

Q

(

100\text{ J K}^{-1}

).

Block

P

requires the most heat per kilogram per Kelvin, while Block

Q

requires the most heat overall to raise its temperature by

1\text{ K}

due to its larger mass.

Key Concept

Distinction between Heat Capacity (

C = \frac{Q}{\Delta T}

) and Specific Heat Capacity (

c = \frac{Q}{m \Delta T}

)

Estimated Time:1m 30s

Question 12909Question →

A student records the time taken for a simple pendulum to complete $20$ complete oscillations as $30\text{ s}$ . What is the period of oscillation of the pendulum, in seconds?

Show answer & explanation

Answer: 1.5

Answer

The period of oscillation of the pendulum is

1.5\text{ s}

.

The period of oscillation (

T

) is the time required for one complete cycle. Dividing the total time (

30\text{ s}

) by the number of oscillations (

20

) gives

1.5\text{ s}

.

Step-by-Step Solution

1

Identify the given values for total time and total number of oscillations.

Total time

t = 30\text{ s}

and number of oscillations

N = 20

.

The period is defined as the time taken for a single complete oscillation.

2

Divide the total time by the number of oscillations to calculate the period.

T = \frac{30\text{ s}}{20} = 1.5\text{ s}

.

Applying the period formula

T = \frac{t}{N}

yields the duration of one period.

Key Concept

Period of Oscillation

Estimated Time:45s

Question 12910Question →

The threshold wavelength for photoelectric emission from a metallic emitter is $\lambda_0$ . When light of wavelength $\frac{\lambda_0}{3}$ illuminates the emitter, photoelectrons are ejected with a maximum kinetic energy of $K_1$ . If the incident light is changed to a wavelength of $\frac{\lambda_0}{4}$ , what is the new maximum kinetic energy $K_2$ of the ejected photoelectrons in terms of $K_1$ ?

\frac{3}{2} K_1

\frac{4}{3} K_1

2 K_1

\frac{2}{3} K_1

Show answer & explanation

Answer: $\frac{3}{2} K_1$

Answer

The new maximum kinetic energy

K_2

is equal to

\frac{3}{2} K_1

.

According to Einstein's photoelectric equation, maximum kinetic energy is

K = E - W_0

. With threshold wavelength

\lambda_0

, the work function is

W_0 = \frac{hc}{\lambda_0}

. For light of wavelength

\frac{\lambda_0}{3}

, the photon energy is

3W_0

, yielding

K_1 = 3W_0 - W_0 = 2W_0

. For light of wavelength

\frac{\lambda_0}{4}

, the photon energy is

4W_0

, yielding

K_2 = 4W_0 - W_0 = 3W_0

. Comparing

K_2

and

K_1

gives

K_2 = \frac{3}{2} K_1

.

Step-by-Step Solution

1

Express the work function

W_0

in terms of threshold wavelength

\lambda_0

.

W_0 = \frac{hc}{\lambda_0}

The threshold wavelength defines the minimum photon energy required to liberate an electron from the metal surface.

2

Apply Einstein's photoelectric equation for the first incident wavelength

\lambda_1 = \frac{\lambda_0}{3}

.

K_1 = \frac{hc}{\lambda_0/3} - W_0 = 3\left(\frac{hc}{\lambda_0}\right) - W_0 = 3W_0 - W_0 = 2W_0

Einstein's photoelectric equation states that maximum kinetic energy equals photon energy minus work function.

3

Apply Einstein's photoelectric equation for the second incident wavelength

\lambda_2 = \frac{\lambda_0}{4}

.

K_2 = \frac{hc}{\lambda_0/4} - W_0 = 4\left(\frac{hc}{\lambda_0}\right) - W_0 = 4W_0 - W_0 = 3W_0

Calculate maximum kinetic energy for the shorter incident wavelength.

4

Determine the ratio of

K_2

to

K_1

.

\frac{K_2}{K_1} = \frac{3W_0}{2W_0} = \frac{3}{2} \implies K_2 = \frac{3}{2} K_1

Express the new kinetic energy in terms of the initial kinetic energy.

Key Concept

Einstein's Photoelectric Equation and Work Function Threshold

Question 12911Question →

Match each phenotypic trait or population distribution profile on the left with its corresponding characteristic pattern of variation or graphical feature on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Human height and skin color matches with Continuous variation influenced by polygenes and environmental factors. ABO blood group and tongue rolling ability matches with Discontinuous variation controlled by monogenic inheritance with minimal environmental effect. Bell-shaped frequency curve matches with Graphical representation showing a continuous spectrum of intermediate phenotypes. Discrete bar chart with clear gaps matches with Graphical representation showing non-overlapping, distinct phenotypic categories.

Human height and skin color exhibit continuous variation caused by polygenic inheritance and environmental influences. ABO blood group and tongue rolling ability show discontinuous variation controlled by single genes with minimal environmental impact. Population data for continuous variation forms a smooth bell-shaped curve showing intermediate phenotypes, whereas discontinuous variation forms a discrete bar chart with distinct non-overlapping categories.

Step-by-Step Solution

1

Identify the biological nature and genetic control of phenotypic traits

Human height and skin color exhibit gradual transitions (polygenic continuous variation), while ABO blood group and tongue rolling show clear, separate phenotypic classes (monogenic discontinuous variation).

Continuous traits show a continuum of phenotypes due to multiple genes, whereas discontinuous traits show distinct classes due to single-gene control.

2

Analyze the graphical representations associated with population data for both variation types

Continuous data produces a smooth, bell-shaped normal distribution curve, while discontinuous data forms separate bars representing distinct non-overlapping categories.

Quantitative measurements across a spectrum form a continuous curve, while qualitative discrete categories plot as separate bars.

3

Pair each left item with its corresponding genetic mechanism or graphical characteristic

All four pairs are correctly matched based on variation patterns and graphical profiles.

Ensures complete concept alignment across examples, mechanisms, and graphical representations.

Key Concept

Distinction between continuous and discontinuous variation in inheritance patterns, underlying genetic mechanisms, and graphical representations.

Question 12912Question →

The electrical resistance $R$ of a uniform wire of length $L$ and diameter $d$ is given by $R = \frac{4\rho L}{\pi d^2}$ , where $\rho$ is the constant resistivity of the material. If the length $L$ is measured with a percentage error of $2.0\%$ and the diameter $d$ is measured with a percentage error of $1.5\%$ , what is the maximum percentage error in the calculated value of $R$ ?

3.5\%

5.0\%

0.5\%

4.25\%

Show answer & explanation

Answer: $5.0\%$

Answer

The maximum percentage error in the calculated resistance

R

is

5.0\%

.

For a physical quantity defined by

R = \frac{4\rho L}{\pi d^2}

, the maximum percentage error is determined by adding the percentage error of

L

to twice the percentage error of

d

. Calculating

2.0\% + 2(1.5\%) = 5.0\%

yields the correct maximum percentage error.

Step-by-Step Solution

1

Identify the relation for maximum percentage error propagation in a physical formula.

For

R = \frac{4\rho L}{\pi d^2}

, the fractional error equation is

\frac{\Delta R}{R} = \frac{\Delta L}{L} + 2\left(\frac{\Delta d}{d}\right)

.

When quantities are multiplied or divided, their relative errors add, and any exponent becomes a multiplier for that quantity's relative error.

2

Express the relation in terms of percentage errors.

\% \text{ Error in } R = (\% \text{ Error in } L) + 2 \times (\% \text{ Error in } d)

Multiplying the fractional error expression by

100\%

converts all relative errors into percentage errors.

3

Substitute the given values into the formula.

\% \text{ Error in } R = 2.0\% + 2 \times (1.5\%) = 2.0\% + 3.0\% = 5.0\%

Substituting

2.0\%

for length error and

1.5\%

for diameter error yields the combined maximum percentage error.

Key Concept

Propagation of errors in physical quantities involving exponents

Question 12913Question →

Polyploidy is a primary mechanism of rapid sympatric speciation in flowering plants. Arrange the following steps of allopolyploid speciation in the correct chronological sequence from initial interbreeding to the formation of a fertile new species.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of allopolyploid speciation begins with interspecific hybridization between two parental species, followed by hybrid sterility due to unpaired non-homologous chromosomes, subsequent spontaneous chromosome doubling (polyploidization), and finally the restoration of fertility establishing a novel, reproductively isolated polyploid species.

Allopolyploidy represents a mode of sympatric speciation driven by hybridization followed by genome duplication. The process initiates when two related but distinct species interbreed to produce an interspecific hybrid. This hybrid is initially sterile because its chromosomes are non-homologous and cannot pair during meiosis. When spontaneous chromosome doubling occurs, every chromosome acquires an identical partner, which restores balanced meiotic division and fertility. The resulting polyploid population is fertile among itself but reproductively isolated from both parent species, constituting a new species.

Step-by-Step Solution

1

Identify the initiating event of allopolyploid speciation.

Two distinct species cross-pollinate to generate an interspecific hybrid offspring.

Allopolyploidy requires genomes originating from two distinct ancestral species.

2

Determine the initial reproductive status of the interspecific hybrid.

The hybrid is sterile because non-homologous chromosomes cannot synapse properly during prophase I of meiosis.

Lack of matching homologous chromosomes prevents balanced chromosome segregation into viable gametes.

3

Identify the cellular modification that overcomes hybrid sterility.

A chromosome doubling event (somatic non-disjunction or unreduced gamete fusion) takes place.

Doubling every chromosome provides an identical sister homologue for pairing during meiosis.

4

Conclude the final evolutionary outcome of the polyploid lineage.

Meiotic stability and fertility are achieved, establishing a novel, reproductively isolated species.

The newly formed polyploid cannot successfully backcross with either diploid parent, completing sympatric speciation.

Key Concept

Allopolyploid Speciation

Estimated Time:1m 30s

Question 12914Question →

A copper calorimeter of mass $0.15\text{ kg}$ contains $0.25\text{ kg}$ of water at $20.0^\circ\text{C}$ . Dry steam at $100.0^\circ\text{C}$ is passed into the water until the final temperature of the mixture reaches $40.0^\circ\text{C}$ . Assuming no heat is lost to the surroundings, what is the mass of steam condensed, in grams?

(Take specific heat capacity of water $= 4200\text{ J kg}^{-1}\text{ K}^{-1}$ , specific heat capacity of copper $= 400\text{ J kg}^{-1}\text{ K}^{-1}$ , and specific latent heat of vaporization of water $= 2.26 \times 10^6\text{ J kg}^{-1}$ )

Show answer & explanation

Answer: 8.84

Answer

8.84 g

By energy conservation, the heat gained by the cold water and copper vessel must equal the total heat released by the condensing steam and the cooling of that condensed water. The heat gained is (0.25 kg × 4200 J/kg·K + 0.15 kg × 400 J/kg·K) × 20.0 K = 22,200 J. The heat lost per kilogram of steam is 2,260,000 J/kg + 4200 J/kg·K × 60.0 K = 2,512,000 J/kg. Dividing 22,200 J by 2,512,000 J/kg gives 0.0088376 kg, which corresponds to 8.84 g.

Step-by-Step Solution

1

Calculate the thermal energy absorbed by the cold water and copper calorimeter

Q_gained = 22,200 J

Both the water and copper calorimeter increase in temperature from 20.0°C to 40.0°C (ΔT = 20.0 K).

2

Express the heat released by mass m_s of steam as it condenses and cools to 40.0°C

Q_lost = m_s * 2,512,000 J

Steam releases latent heat when condensing at 100.0°C (m_s * L_v) and sensible heat when the condensed water cools from 100.0°C to 40.0°C (m_s * c_w * 60.0).

3

Apply the principle of conservation of thermal energy and solve for mass m_s in grams

m_s = 8.84 g

Setting Q_gained equal to Q_lost yields m_s = 22,200 / 2,512,000 = 0.0088376 kg, which equals 8.84 g.

Key Concept

Thermal energy balance involving phase change (latent heat of vaporization) and sensible heat exchange

Question 12915Question →

In domestic cats, coat color is an X-linked trait where the allele for black fur ( $X^B$ ) and the allele for orange fur ( $X^O$ ) are codominant, with heterozygous females ( $X^B X^O$ ) exhibiting a tortoiseshell phenotype. If an orange male cat ( $X^O Y$ ) is mated with a tortoiseshell female cat ( $X^B X^O$ ), what percentage of their male offspring is expected to have orange fur?

Show answer & explanation

Answer: 50

Answer

50%

The female parent (

X^B X^O

) transmits her

X^B

allele or her

X^O

allele to male offspring with equal probability (50% each). Because male offspring inherit the Y chromosome from their father, half of the male offspring will inherit

X^O

and have orange fur (

X^O Y

), giving an expected percentage of 50%.

Step-by-Step Solution

1

Determine the gametes produced by each parent

The male parent (

X^O Y

) produces

X^O

and

Y

gametic types in equal proportion (1:1). The female parent (

X^B X^O

) produces

X^B

and

X^O

gametic types in equal proportion (1:1).

Segregation of chromosomes during meiosis separates alleles into gametes.

2

Determine the genotypes specifically for male offspring

Male offspring inherit the

Y

chromosome from the male parent and one

X

chromosome from the female parent. The possible male genotypes are

X^B Y

(black fur) and

X^O Y

(orange fur).

Sex determination follows the XX-XY system where male sex is determined by inheriting the paternal Y chromosome.

3

Calculate the percentage of orange males among total male offspring

Out of 2 possible male genotypes (

X^B Y

and

X^O Y

), 1 is orange (

X^O Y

). The ratio is 1 out of 2, which equals 50%.

The question asks for the proportion strictly within male offspring.

Key Concept

Sex-linked codominant inheritance and sex determination in XX-XY systems

Question 12916Question →

A ticker-tape timer connected to a power supply operates at a frequency of $50\text{ Hz}$ . A continuous strip of paper tape pulled through the timer records a sequence of dots. Calculate the total time interval, in seconds, between the $1\text{st}$ dot and the $21\text{st}$ dot on the tape.

Show answer & explanation

Answer: 0.4

Answer

0.4 s

On a ticker tape, the time interval between consecutive dots represents one period

T = \frac{1}{f} = \frac{1}{50\text{ Hz}} = 0.02\text{ s}

. The total elapsed time for

N

dots corresponds to

(N - 1)

spaces. For 21 dots, there are 20 spaces, giving a total time of

20 \times 0.02\text{ s} = 0.40\text{ s}

.

Step-by-Step Solution

1

Find the number of spaces between dots

20 spaces

Between

N

dots on a ticker-tape, there are

(N - 1)

time intervals.

2

Calculate the period per space

0.02 s

The period

T

is the reciprocal of the operating frequency

f = 50\text{ Hz}

.

3

Multiply the number of spaces by the period per space

0.4 s

Total time duration is the product of the total number of intervals and the time for one interval.

Key Concept

Calculation of time interval using ticker-tape timer frequency and dot count

Question 12917Question →

An industrial chemist evaluates four chemical substances ( $W$ , $X$ , $Y$ , and $Z$ ) based on their production scale, manufacturing process, purity standards, and economic value:

- **Substance $W$ **: Synthesized via continuous plant operation in high tonnage for fertilizer manufacturing; requires industrial-grade purity.
- **Substance $X$ **: Formulated in small-batch reactors under stringent quality controls for pharmaceutical use; commands a high price per gram.
- **Substance $Y$ **: Manufactured continuously in bulk quantities as a primary alkaline agent for soap making.
- **Substance $Z$ **: Produced in multi-step batch synthesis to yield high-purity analytical indicators.

Which of the following options correctly groups these substances into heavy chemicals and fine chemicals?

Heavy chemicals:

W

and

Y

; Fine chemicals:

X

and

Z

Heavy chemicals:

X

and

Z

; Fine chemicals:

W

and

Y

Heavy chemicals:

W

and

X

; Fine chemicals:

Y

and

Z

Heavy chemicals:

Y

and

Z

; Fine chemicals:

W

and

X

Show answer & explanation

Answer: Heavy chemicals: $W$ and $Y$ ; Fine chemicals: $X$ and $Z$

Answer

Heavy chemicals:

W

and

Y

; Fine chemicals:

X

and

Z

Heavy chemicals are manufactured continuously in vast quantities (bulk scale) for use as primary industrial feedstocks (such as fertilizers and soap reagents), which describes substances

W

and

Y

. Fine chemicals are produced in small batch operations with strict purity standards for specialized high-value uses (such as pharmaceuticals and analytical reagents), which describes substances

X

and

Z

.

Step-by-Step Solution

1

Analyze the operational parameters of Heavy Chemicals

Heavy chemicals are characterized by continuous production processes, massive tonnage/bulk scale, relatively lower unit cost, and application as basic raw materials (e.g., fertilizers, saponification agents).

Substances

W

(fertilizer raw material via continuous high tonnage) and

Y

(bulk alkaline agent for soap making via continuous synthesis) fit all criteria for heavy chemicals.

2

Analyze the operational parameters of Fine Chemicals

Fine chemicals are characterized by batch manufacturing processes, high purity standards, multi-stage synthesis, specialized applications, and high unit cost.

Substances

X

(pharmaceutical formulation via small-batch reactors with high price per gram) and

Z

(analytical indicators via multi-step high-purity batch synthesis) fit all criteria for fine chemicals.

3

Match the identified categories to the options

Substances

W

and

Y

are heavy chemicals, while substances

X

and

Z

are fine chemicals.

This corresponds to the correct classification option.

Key Concept

Distinction between Heavy Chemicals (bulk scale, continuous process, raw material applications) and Fine Chemicals (small batch, high purity, high unit value).

Question 12918Question →

A straight wire of length $0.50\text{ m}$ carrying a current of $4.0\text{ A}$ is placed in a uniform magnetic field of flux density $0.20\text{ T}$ . If the wire experiences a magnetic force of $0.20\text{ N}$ , what is the angle between the wire and the direction of the magnetic field?

30^\circ

45^\circ

60^\circ

90^\circ

Show answer & explanation

Answer: $30^\circ$

Answer

The angle between the wire and the direction of the magnetic field is

30^\circ

.

The magnetic force on a straight current-carrying wire in a uniform magnetic field is given by

F = BIL\sin\theta

. Substituting

F = 0.20\text{ N}

,

B = 0.20\text{ T}

,

I = 4.0\text{ A}

, and

L = 0.50\text{ m}

gives

0.20 = 0.40 \sin\theta

, which simplifies to

\sin\theta = 0.50

. Therefore,

\theta = 30^\circ

.

Step-by-Step Solution

1

Identify the formula for the magnetic force on a current-carrying conductor in a magnetic field.

F = B I L \sin\theta

The magnetic force depends on magnetic flux density

B

, current

I

, length

L

, and the angle

\theta

between the conductor and the magnetic field.

2

Substitute the given values into the magnetic force formula.

0.20 = 0.20 \times 4.0 \times 0.50 \times \sin\theta

Given values are

F = 0.20\text{ N}

,

B = 0.20\text{ T}

,

I = 4.0\text{ A}

, and

L = 0.50\text{ m}

.

3

Solve for

\sin\theta

and calculate

\theta

.

\sin\theta = \frac{0.20}{0.40} = 0.50 \implies \theta = \arcsin(0.50) = 30^\circ

Taking the inverse sine of

0.50

yields an angle of

30^\circ

.

Key Concept

Magnetic force on a current-carrying conductor (

F = BIL \sin\theta

)

Question 12919Question →

Consider the following three representative poikilothermic vertebrate species: *Agama agama* (rainbow lizard), *Tilapia zillii* (redbelly tilapia), and *Bufo regularis* (African toad). What is the correct sequence of these organisms when arranged in order of increasing structural complexity of their heart and circulatory system?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct evolutionary order of cardiac structural complexity is *Tilapia zillii* (Pisces), followed by *Bufo regularis* (Amphibia), and finally *Agama agama* (Reptilia).

The correct sequence follows the evolutionary advancement of the circulatory system across poikilothermic classes: Class Pisces (*Tilapia zillii*) features a simple two-chambered heart; Class Amphibia (*Bufo regularis*) possesses a three-chambered heart consisting of two atria and one ventricle; and Class Reptilia (*Agama agama*) displays an advanced three-chambered heart with a partial septum dividing the ventricle, reducing cardiac mixing.

Step-by-Step Solution

1

Identify the vertebrate class and heart structure for each organism

*Tilapia zillii* belongs to Class Pisces (bony fish) and has a 2-chambered heart. *Bufo regularis* belongs to Class Amphibia and has a 3-chambered heart. *Agama agama* belongs to Class Reptilia and has a 3-chambered heart with a partially partitioned ventricle.

Evolutionary progression in poikilothermic vertebrates advances from 2-chambered single circulation in fish to 3-chambered double circulation in amphibians and partially divided 4-chambered systems in reptiles.

2

Order the organisms based on cardiac chamber separation and efficiency

Fish (*Tilapia zillii*) < Amphibian (*Bufo regularis*) < Reptile (*Agama agama*).

Pisces has no separation of pulmonary and systemic circuits in the heart, Amphibia has separated atria but an undivided ventricle, and Reptilia introduces partial ventricular septation for improved oxygen transport efficiency.

Key Concept

Evolutionary trends in cardiac anatomy among poikilothermic vertebrate classes (Pisces, Amphibia, Reptilia)

Question 12920Question →

Consider the following thermochemical equations at $298\text{ K}$ :

1. $\text{CH}_3\text{OH}(l) + \frac{3}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta H^\circ = -726\text{ kJ mol}^{-1}$
2. $\text{CO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H^\circ = -283\text{ kJ mol}^{-1}$
3. $\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H^\circ = -286\text{ kJ mol}^{-1}$

What is the standard enthalpy change for the synthesis of liquid methanol from carbon monoxide and hydrogen gas according to the equation:

\text{CO}(g) + 2\text{H}_2(g) \rightarrow \text{CH}_3\text{OH}(l)

-1581\text{ kJ mol}^{-1}

-129\text{ kJ mol}^{-1}

+129\text{ kJ mol}^{-1}

+157\text{ kJ mol}^{-1}

Show answer & explanation

Answer: $-129\text{ kJ mol}^{-1}$

Answer

-129\text{ kJ mol}^{-1}

Applying Hess's Law requires manipulating the given equations to yield the target reaction

\text{CO}(g) + 2\text{H}_2(g) \rightarrow \text{CH}_3\text{OH}(l)

. Reversing equation 1 gives

\Delta H = +726\text{ kJ mol}^{-1}

. Keeping equation 2 gives

\Delta H = -283\text{ kJ mol}^{-1}

. Doubling equation 3 gives

\Delta H = -572\text{ kJ mol}^{-1}

. Summing these values gives

+726 - 283 - 572 = -129\text{ kJ mol}^{-1}

.

Step-by-Step Solution

1

Reverse Equation (1) so that

\text{CH}_3\text{OH}(l)

is on the product side.

\text{CO}_2(g) + 2\text{H}_2\text{O}(l) \rightarrow \text{CH}_3\text{OH}(l) + \frac{3}{2}\text{O}_2(g) \quad \Delta H_1 = +726\text{ kJ mol}^{-1}

Reversing a reaction changes the sign of its enthalpy change according to Hess's Law.

2

Use Equation (2) as written to supply

\text{CO}(g)

on the reactant side.

\text{CO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_2 = -283\text{ kJ mol}^{-1}

\text{CO}(g)

is required as a reactant with a coefficient of 1.

3

Multiply Equation (3) by 2 to supply

2\text{H}_2(g)

on the reactant side.

2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l) \quad \Delta H_3 = 2 \times (-286\text{ kJ mol}^{-1}) = -572\text{ kJ mol}^{-1}

The target reaction requires 2 moles of

\text{H}_2(g)

, so the enthalpy change must be multiplied by 2.

4

Sum the modified equations and their respective

\Delta H

values.

\Delta H^\circ = +726 + (-283) + (-572) = -129\text{ kJ mol}^{-1}

Hess's Law states that the overall enthalpy change is the sum of the enthalpy changes of individual steps.

Key Concept

Hess's Law of Constant Heat Summation

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