All practice questions

Question 12861Question →

In a biological survey of invertebrate excretory mechanisms, four organisms (I, II, III, and IV) were observed to possess the following primary excretory organs:

- Organism I: Flame cells (protonephridia)
- Organism II: Nephridia
- Organism III: Malpighian tubules
- Organism IV: Antennal (green) glands

Which of the following correctly identifies organisms I, II, III, and IV respectively?

I: Planarian, II: Earthworm, III: Cockroach, IV: Prawn

I: Earthworm, II: Planarian, III: Prawn, IV: Cockroach

I: Planarian, II: Cockroach, III: Earthworm, IV: Prawn

I: Cockroach, II: Earthworm, III: Planarian, IV: Prawn

Show answer & explanation

Answer: I: Planarian, II: Earthworm, III: Cockroach, IV: Prawn

Answer

Organism I is a planarian, Organism II is an earthworm, Organism III is a cockroach, and Organism IV is a prawn.

Flame cells function in flatworms (Planarian), nephridia function in annelids (Earthworm), Malpighian tubules function in insects (Cockroach), and green glands function in crustaceans (Prawn). Therefore, the correct combination lists these four organisms in that specific sequence.

Step-by-Step Solution

1

Identify the organism group corresponding to Organism I

Flame cells (protonephridia) are characteristic of flatworms (Platyhelminthes, such as planarians).

Flame cells serve as simple excretory structures that propel waste fluids through cilia.

2

Identify the organism group corresponding to Organism II

Nephridia are characteristic excretory structures of annelids (such as earthworms).

Nephridia filter coelomic fluid and reabsorb essential nutrients in segmented worms.

3

Identify the organism groups corresponding to Organisms III and IV

Malpighian tubules are found in terrestrial arthropods like insects (cockroaches), whereas antennal (green) glands are found in crustaceans (prawns).

Different arthropod subphyla feature specialized structures tailored to terrestrial water conservation or aquatic ion balance.

Key Concept

Comparative anatomy of invertebrate excretory structures across major phyla.

Question 12862Question →

A rectangular coil consisting of $50$ turns of wire has dimensions of $0.10\text{ m}$ by $0.05\text{ m}$ and carries a steady electric current of $2.0\text{ A}$ . The coil is suspended in a uniform magnetic field of flux density $0.40\text{ T}$ . If the normal to the plane of the coil makes an angle of $30^\circ$ with the direction of the magnetic field, what is the magnitude of the torque exerted on the coil?

0.10\text{ N}\cdot\text{m}

0.17\text{ N}\cdot\text{m}

0.20\text{ N}\cdot\text{m}

3.00\text{ N}\cdot\text{m}

Show answer & explanation

Answer: $0.10\text{ N}\cdot\text{m}$

Answer

The magnitude of the torque exerted on the coil is

0.10\text{ N}\cdot\text{m}

.

The magnitude of torque on a current-carrying coil situated in a uniform magnetic field is given by

\tau = N I A B \sin\theta

, where

\theta

is the angle between the normal vector of the coil and the magnetic field vector. Substituting

N = 50

,

I = 2.0\text{ A}

,

A = 0.005\text{ m}^2

,

B = 0.40\text{ T}

, and

\theta = 30^\circ

yields

\tau = 50 \times 2.0 \times 0.005 \times 0.40 \times 0.5 = 0.10\text{ N}\cdot\text{m}

.

Step-by-Step Solution

1

Calculate the area

A

of the rectangular coil.

A = \text{length} \times \text{width} = 0.10\text{ m} \times 0.05\text{ m} = 0.005\text{ m}^2

The torque on a coil depends on its total surface area.

2

Identify the given values and formula for magnetic torque.

Formula:

\tau = N I A B \sin\theta

, where

N = 50

,

I = 2.0\text{ A}

,

A = 0.005\text{ m}^2

,

B = 0.40\text{ T}

, and

\theta = 30^\circ

.

When

\theta

is measured relative to the normal of the coil's plane, the sine component determines the effective perpendicular force arm.

3

Substitute the values into the formula and solve for torque

\tau

.

\tau = 50 \times 2.0\text{ A} \times 0.005\text{ m}^2 \times 0.40\text{ T} \times \sin 30^\circ = 100 \times 0.002 \times 0.5 = 0.10\text{ N}\cdot\text{m}

Direct evaluation yields the final magnetic torque.

Key Concept

Torque on a Current-Carrying Loop in a Uniform Magnetic Field

Estimated Time:1m 30s

Question 12863Question →

A mixture of $15\text{ cm}^3$ of ethene ( $C_2H_4$ ) and $50\text{ cm}^3$ of oxygen gas was exploded at constant temperature and pressure. If the resulting mixture was cooled to room temperature, what is the total volume of the residual gas in $\text{cm}^3$ ?

Show answer & explanation

Answer: 35

Answer

35 cm³

According to Gay-Lussac's Law,

15\text{ cm}^3

of ethene reacts with

45\text{ cm}^3

of oxygen to produce

30\text{ cm}^3

of carbon(IV) oxide gas. Since

50\text{ cm}^3

of oxygen was initially present,

5\text{ cm}^3

of unreacted oxygen gas remains. Upon cooling to room temperature, water condenses to liquid, leaving a total residual gaseous volume of

5\text{ cm}^3 + 30\text{ cm}^3 = 35\text{ cm}^3

.

Step-by-Step Solution

1

Write the balanced equation for the complete combustion of ethene gas.

C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(l)

By Gay-Lussac's Law of Combining Volumes, gases react in simple whole-number volume ratios equal to their stoichiometric coefficients at constant temperature and pressure.

2

Determine the volume of oxygen required and the unreacted excess volume.

Volume of

O_2

consumed =

3 \times 15\text{ cm}^3 = 45\text{ cm}^3

. Remaining unreacted

O_2 = 50\text{ cm}^3 - 45\text{ cm}^3 = 5\text{ cm}^3

.

Since

15\text{ cm}^3

of

C_2H_4

requires only

45\text{ cm}^3

of

O_2

, oxygen is in excess and ethene is the limiting reactant.

3

Calculate the volume of carbon(IV) oxide gas produced.

Volume of

CO_2

produced =

2 \times 15\text{ cm}^3 = 30\text{ cm}^3

.

1 volume of

C_2H_4

produces 2 volumes of

CO_2

gas.

4

Calculate the total volume of the residual gaseous mixture after cooling to room temperature.

Total residual gaseous volume =

5\text{ cm}^3 \text{ (excess } O_2\text{)} + 30\text{ cm}^3 \text{ (produced } CO_2\text{)} = 35\text{ cm}^3

.

Water formed is in the liquid state at room temperature and contributes negligible volume to the gaseous mixture.

Key Concept

Gay-Lussac's Law of Combining Volumes and Residual Gas Volume Calculations

Estimated Time:1m 30s

Question 12864Question →

Match each physical analytical observation of a chemical sample with its corresponding interpretation regarding purity.

Click a left item, then click its matching right item

Show answer & explanation

Answer

1. Solid with a sharp melting point at

156.5^\circ\text{C}

matches 'Confirms the sample is a pure crystalline compound'.
2. Solid melting over

142.0^\circ\text{C}

to

150.5^\circ\text{C}

matches 'Indicates the presence of impurities that depress and broaden the melting point'.
3. Liquid boiling at a steady temperature higher than literature value matches 'Indicates the presence of a non-volatile dissolved impurity in the liquid'.
4. Single spot on paper chromatography matches 'Demonstrates chemical homogeneity and single-component composition'.

Pure compounds exhibit constant physical properties such as sharp melting points, constant boiling points, and single chromatographic spots. Impurities alter lattice stability and vapor pressure, leading to melting point depression, boiling point elevation, and broad temperature transition ranges.

Step-by-Step Solution

1

Analyze solid sample melting characteristics

A sharp melting point indicates high purity, while a broad melting range below the expected temperature indicates contamination.

Impurities disrupt the orderly crystal lattice, requiring less thermal energy to melt and causing melting across a range of temperatures.

2

Analyze liquid boiling point behavior

Boiling point elevation at constant pressure signifies a non-volatile solute dissolved in the liquid.

Solute particles reduce the escaping tendency of liquid molecules into the vapor phase, requiring a higher temperature to match atmospheric pressure.

3

Evaluate chromatographic criteria

A pure substance contains only one chemical species and produces a single chromatogram spot.

If multiple components were present, differing partition coefficients would separate them into multiple spots.

Key Concept

Criteria of Purity for Substances

Question 12865Question →

A trapped gas sample has a volume of $4.0\text{ dm}^3$ at a pressure of $570\text{ mmHg}$ under isothermal conditions. If the pressure is altered to $1.5\text{ atm}$ , what is the final volume occupied by the gas? ( $1\text{ atm} = 760\text{ mmHg}$ )

2.0\text{ dm}^3

8.0\text{ dm}^3

1520.0\text{ dm}^3

1.52\text{ dm}^3

Show answer & explanation

Answer: $2.0\text{ dm}^3$

Answer

The final volume occupied by the gas is

2.0\text{ dm}^3

.

According to Boyle's Law, the pressure and volume of a fixed mass of gas at constant temperature are inversely proportional (

P_1 V_1 = P_2 V_2

). First, convert

570\text{ mmHg}

into atmospheres:

\frac{570}{760} = 0.75\text{ atm}

. Substituting the values gives

0.75\text{ atm} \times 4.0\text{ dm}^3 = 1.5\text{ atm} \times V_2

, which yields

V_2 = 2.0\text{ dm}^3

.

Step-by-Step Solution

1

Convert the initial pressure

P_1

from

\text{mmHg}

to

\text{atm}

to match the unit of

P_2

.

P_1 = \frac{570\text{ mmHg}}{760\text{ mmHg/atm}} = 0.75\text{ atm}

Units of pressure must be consistent before substituting into gas law equations.

2

Apply Boyle's Law formula for an isothermal process (

P_1 V_1 = P_2 V_2

).

0.75\text{ atm} \times 4.0\text{ dm}^3 = 1.5\text{ atm} \times V_2

Boyle's Law states that for a fixed mass of gas at constant temperature, pressure and volume are inversely proportional.

3

Solve the algebraic equation for the final volume

V_2

.

V_2 = \frac{0.75 \times 4.0}{1.5} = 2.0\text{ dm}^3

Dividing both sides by

1.5\text{ atm}

isolates the unknown final volume.

Key Concept

Boyle's Law (

P_1 V_1 = P_2 V_2

at constant temperature)

Estimated Time:2m 0s

Question 12866Question →

Comparative biochemical analysis shows that organisms with a more recent common evolutionary ancestor possess a higher degree of similarity in the amino acid sequences of conserved proteins like Cytochrome c, irrespective of their modern geographic locations.

True

False

Show answer & explanation

Answer: True

Answer

The statement is TRUE.

The statement accurately reflects the principles of comparative biochemistry: the degree of amino acid sequence similarity in universally conserved proteins directly indicates how recently two species shared a common ancestor, regardless of where those species are located today.

Step-by-Step Solution

1

Identify the biological role of conserved proteins in comparative biochemistry.

Proteins such as Cytochrome c perform essential respiratory functions across diverse taxa, making them reliable indicators of phylogenetic ancestry.

Because these proteins exist across virtually all eukaryotic life, their sequences can be directly compared to gauge evolutionary relationships.

2

Relate sequence divergence to time since common ancestry.

Fewer amino acid differences correspond to a shorter time elapsed since divergence from a shared ancestor.

Genetic mutations alter protein sequences incrementally over time, acting as a molecular clock.

3

Assess the effect of geography on biochemical sequence similarity.

Molecular similarity reflects evolutionary lineage rather than geographic proximity or ecological habitat.

While geography influences speciation and external adaptation, primary biochemical sequences strictly track ancestral genetic inheritance.

Key Concept

Comparative Biochemistry and Molecular Homology

Estimated Time:1m 0s

Question 12867Question →

Calculate the thermal energy, in joules, required to completely melt $0.50\text{ kg}$ of ice at its melting point of $0^\circ\text{C}$ . (Take the specific latent heat of fusion of ice as $3.36 \times 10^5\text{ J kg}^{-1}$ ).

Show answer & explanation

Answer: 168000

Answer

The total heat required to completely melt the ice is

168,000\text{ J}

.

During a state change from solid to liquid at constant temperature, the thermal energy absorbed is given directly by

Q = m L_f

. Multiplying mass (

0.50\text{ kg}

) by specific latent heat (

3.36 \times 10^5\text{ J kg}^{-1}

) gives

168,000\text{ J}

.

Step-by-Step Solution

1

Identify the given values and formula

Mass

m = 0.50\text{ kg}

, Specific latent heat of fusion

L_f = 3.36 \times 10^5\text{ J kg}^{-1}

, Formula:

Q = m L_f

Since the phase change occurs at constant temperature (

0^\circ\text{C}

), sensible heat calculation (

mc\Delta T

) is zero and only latent heat is needed.

2

Substitute values into the formula and calculate

Q = 0.50\text{ kg} \times 3.36 \times 10^5\text{ J kg}^{-1} = 168,000\text{ J}

Direct multiplication of mass and specific latent heat yields the energy required in Joules.

Key Concept

Specific Latent Heat of Fusion

Question 12868Question →

A series alternating current (AC) circuit consists of a resistor of resistance $R = 40\ \Omega$ , an inductor with inductive reactance $X_L = 70\ \Omega$ , and a capacitor with capacitive reactance $X_C = 40\ \Omega$ , connected to an AC voltage supply of $100\ \text{V}$ . What is the power factor of the circuit?

Show answer & explanation

Answer: 0.8

Answer

The power factor of the circuit is 0.8.

The total impedance

Z

of the series circuit is found using

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + (70 - 40)^2} = 50\ \Omega

. The power factor is the ratio of resistance to impedance:

\cos\phi = \frac{R}{Z} = \frac{40}{50} = 0.8

.

Step-by-Step Solution

1

Calculate the net reactance of the circuit.

X = X_L - X_C = 70\ \Omega - 40\ \Omega = 30\ \Omega

In a series RLC circuit, the net reactance is the arithmetic difference between the inductive reactance and the capacitive reactance.

2

Calculate the total impedance of the circuit.

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + 30^2} = \sqrt{2500} = 50\ \Omega

Impedance represents the combined opposition to current flow from resistance and net reactance in quadrature.

3

Determine the power factor of the circuit.

\cos\phi = \frac{R}{Z} = \frac{40}{50} = 0.8

The power factor is equal to the cosine of the phase angle, which is defined as the ratio of resistance to total impedance.

Key Concept

Power Factor in AC Circuits

Estimated Time:1m 30s

Question 12869Question →

Match each carbon species or bond descriptor on the left with its corresponding hybridization state, geometric configuration, or orbital overlap mode on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The central carbon in methane matches

sp^3

hybridization with tetrahedral geometry (

109.5^\circ

); the

\pi

bond component matches sideways overlap of unhybridized

p

orbitals; the central carbon in carbon dioxide matches

sp

hybridization with linear geometry (

180^\circ

); and the carbon in ethene matches

sp^2

hybridization with trigonal planar geometry (

120^\circ

).

Each carbon atom's hybridization and spatial arrangement depend directly on its steric number (number of

\sigma

bonds). Methane features four

\sigma

bonds (

sp^3

,

109.5^\circ

tetrahedral). Ethene features three

\sigma

bonds per carbon (

sp^2

,

120^\circ

trigonal planar). Carbon dioxide features two

\sigma

bonds (

sp

,

180^\circ

linear).

\Pi

bonds are characterized by the sideways overlap of unhybridized

2p

atomic orbitals.

Step-by-Step Solution

1

Determine the steric number and geometry of the carbon in methane (

\text{CH}_4

).

Four single

\sigma

bonds give a steric number of 4, which dictates

sp^3

hybridization and a tetrahedral angle of

109.5^\circ

.

Mixing one

s

and three

p

orbitals forms four equivalent

sp^3

hybrid orbitals pointing to tetrahedral corners.

2

Identify how a carbon-carbon

\pi

bond is formed.

It forms via lateral/sideways overlap of parallel, unhybridized

p

orbitals above and below the internuclear axis.

Head-on overlap forms

\sigma

bonds, whereas parallel side-by-side overlap creates

\pi

electron clouds.

3

Analyze the steric environment around the carbon in carbon dioxide (

\text{CO}_2

).

The carbon forms two

\sigma

bonds (one to each oxygen atom) and two

\pi

bonds, yielding a steric number of 2, corresponding to

sp

hybridization and

180^\circ

linear geometry.

Two hybrid orbitals position themselves as far apart as possible at

180^\circ

.

4

Determine the hybridization and bond angles of carbon in ethene (

\text{C}_2\text{H}_4

).

Each carbon atom forms three

\sigma

bonds (steric number 3), requiring

sp^2

hybridization with a trigonal planar shape and

120^\circ

bond angles.

Three hybrid orbitals lie in a single plane separated by

120^\circ

.

Key Concept

Carbon Hybridization, Geometry, and Orbital Overlap

Question 12870Question →

An electric generator supplies power to a workshop through transmission lines having a total resistance of $2\,\Omega$ . The workshop operates electrical equipment drawing a power of $4\,\text{kW}$ at a terminal voltage of $200\,\text{V}$ . What is the total power generated by the generator?

4.8\,\text{kW}

4.0\,\text{kW}

4.04\,\text{kW}

24.0\,\text{kW}

Show answer & explanation

Answer: $4.8\,\text{kW}$

Answer

The total power generated by the generator is

4.8\,\text{kW}

.

The electrical current flowing through the system is

20\,\text{A}

based on the load power of

4000\,\text{W}

at

200\,\text{V}

. The power lost in transmission lines is

P = I^2 R = 20^2 \times 2 = 800\,\text{W}

(

0.8\,\text{kW}

). Therefore, the total electrical power generated by the source must be the sum of the load power and line losses, giving

4.0\,\text{kW} + 0.8\,\text{kW} = 4.8\,\text{kW}

.

Step-by-Step Solution

1

Calculate the current flowing through the circuit using the power and voltage at the workshop.

I = \frac{P_{\text{workshop}}}{V} = \frac{4000\,\text{W}}{200\,\text{V}} = 20\,\text{A}

Current is uniform across the series transmission line.

2

Calculate the power loss dissipated as heat in the transmission lines.

P_{\text{loss}} = I^2 R = (20\,\text{A})^2 \times 2\,\Omega = 400 \times 2 = 800\,\text{W} = 0.8\,\text{kW}

By Joule's law of heating, power dissipated in a resistor carrying current

I

is

I^2 R

.

3

Sum the useful power delivered to the workshop and the transmission power loss to obtain total generated power.

P_{\text{total}} = P_{\text{workshop}} + P_{\text{loss}} = 4.0\,\text{kW} + 0.8\,\text{kW} = 4.8\,\text{kW}

Total energy generated per unit time equals energy consumed by load plus energy lost.

Key Concept

Power Loss in Transmission Lines and Total Source Power

Question 12871Question →

A proton of mass $1.67 \times 10^{-27}\text{ kg}$ and charge $1.60 \times 10^{-19}\text{ C}$ enters a uniform magnetic field of flux density $0.50\text{ T}$ at a speed of $4.00 \times 10^6\text{ m s}^{-1}$ . If the path of the proton makes an angle of $30^\circ$ with the magnetic field lines, what is the magnitude of the initial acceleration of the proton?

9.58 \times 10^{13}\text{ m s}^{-2}

1.92 \times 10^{14}\text{ m s}^{-2}

1.66 \times 10^{14}\text{ m s}^{-2}

9.58 \times 10^{12}\text{ m s}^{-2}

Show answer & explanation

Answer: $9.58 \times 10^{13}\text{ m s}^{-2}$

Answer

The magnitude of the initial acceleration of the proton is

9.58 \times 10^{13}\text{ m s}^{-2}

.

The magnetic force on a moving charged particle is given by

F = qvB\sin\theta

. Substituting

q = 1.60 \times 10^{-19}\text{ C}

,

v = 4.00 \times 10^6\text{ m s}^{-1}

,

B = 0.50\text{ T}

, and

\theta = 30^\circ

gives

F = 1.60 \times 10^{-13}\text{ N}

. By Newton's second law, acceleration

a = \frac{F}{m} = \frac{1.60 \times 10^{-13}\text{ N}}{1.67 \times 10^{-27}\text{ kg}} = 9.58 \times 10^{13}\text{ m s}^{-2}

.

Step-by-Step Solution

1

Calculate the magnetic force acting on the proton using

F = qvB\sin\theta

.

F = (1.60 \times 10^{-19}\text{ C}) \times (4.00 \times 10^6\text{ m s}^{-1}) \times (0.50\text{ T}) \times \sin 30^\circ = 1.60 \times 10^{-13}\text{ N}

.

The magnetic force on a moving charge in a magnetic field depends on the component of velocity perpendicular to the field.

2

Apply Newton's second law of motion (

F = ma

) to find acceleration.

a = \frac{F}{m} = \frac{1.60 \times 10^{-13}\text{ N}}{1.67 \times 10^{-27}\text{ kg}} \approx 9.58 \times 10^{13}\text{ m s}^{-2}

.

Dividing the magnetic force by the mass yields the magnitude of the resulting acceleration.

Key Concept

Magnetic force on a moving charged particle and Newton's second law

Question 12872Question →

A body of mass $0.50\text{ kg}$ connected to a light helical spring of force constant $32\text{ N/m}$ executes simple harmonic motion on a smooth horizontal surface. If the total mechanical energy of the oscillating system is $0.16\text{ J}$ , what is the maximum speed of the body in $\text{m/s}$ ?

Show answer & explanation

Answer: 0.8

Answer

The maximum speed of the body is

0.8\text{ m/s}

.

The total mechanical energy in simple harmonic motion is equal to the maximum kinetic energy at the equilibrium position:

E = \frac{1}{2} m v_{\text{max}}^2

. Substituting

E = 0.16\text{ J}

and

m = 0.50\text{ kg}

yields

0.16 = 0.25 v_{\text{max}}^2

, so

v_{\text{max}}^2 = 0.64

and

v_{\text{max}} = 0.8\text{ m/s}

.

Step-by-Step Solution

1

Relate total energy to maximum kinetic energy

E = \frac{1}{2} m v_{\text{max}}^2

At the equilibrium position, potential energy is zero and total mechanical energy is entirely kinetic.

2

Substitute given values into the equation

0.16 = \frac{1}{2} (0.50) v_{\text{max}}^2

Mass

m = 0.50\text{ kg}

and total energy

E = 0.16\text{ J}

are provided.

3

Solve for the maximum speed

v_{\text{max}} = \sqrt{\frac{2 \times 0.16}{0.50}} = \sqrt{0.64} = 0.8\text{ m/s}

Solving for

v_{\text{max}}

gives

0.8\text{ m/s}

.

Key Concept

Conservation of Energy in Simple Harmonic Motion

Question 12873Question →

An object is placed $24\text{ cm}$ in front of a concave mirror with a focal length of $8\text{ cm}$ . What is the distance of the image from the mirror in $\text{cm}$ ?

Show answer & explanation

Answer: 12

Answer

The distance of the image from the mirror is

12\text{ cm}

.

Applying the spherical mirror formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

with focal length

f = 8\text{ cm}

and object distance

u = 24\text{ cm}

yields

\frac{1}{v} = \frac{1}{8} - \frac{1}{24} = \frac{1}{12}

, giving an image distance of

12\text{ cm}

.

Step-by-Step Solution

1

Identify the given parameters and select the appropriate relation.

Focal length

f = 8\text{ cm}

, object distance

u = 24\text{ cm}

, using the mirror equation

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

.

The mirror formula relates focal length, object distance, and image distance for spherical mirrors.

2

Substitute the given values into the formula and solve for

\frac{1}{v}

.

\frac{1}{8} = \frac{1}{24} + \frac{1}{v} \implies \frac{1}{v} = \frac{1}{8} - \frac{1}{24} = \frac{3-1}{24} = \frac{2}{24} = \frac{1}{12}

Subtracting

\frac{1}{24}

from

\frac{1}{8}

isolates the reciprocal of the image distance.

3

Invert the result to determine the image distance

v

.

v = 12\text{ cm}

Taking the reciprocal of

\frac{1}{12}

gives the image distance in centimeters.

Key Concept

Mirror Formula for Concave Mirrors

Estimated Time:45s

Question 12874Question →

When dilute tetraoxosulfate(VI) acid ( $H_2SO_4$ ) reacts with zinc metal, hydrogen gas ( $H_2$ ) is liberated. However, when concentrated trioxonitrate(V) acid ( $HNO_3$ ) reacts with zinc metal, hydrogen gas is not evolved. Which property of concentrated trioxonitrate(V) acid accounts for this difference in behavior?

Its strong oxidizing property, which oxidizes the evolved hydrogen to water

Its dehydrating property, which removes water molecules from the metal surface

Its weak acidic strength when present in concentrated solution

Its high volatility, which causes the acid to evaporate before reacting with the metal

Show answer & explanation

Answer: Its strong oxidizing property, which oxidizes the evolved hydrogen to water

Answer

Concentrated trioxonitrate(V) acid does not liberate hydrogen gas with zinc because of its strong oxidizing property, which oxidizes hydrogen to water.

Unlike typical mineral acids that liberate hydrogen gas when reacting with electropositive metals, concentrated trioxonitrate(V) acid (

HNO_3

) is a powerful oxidizing agent. It oxidizes hydrogen to water (

H_2O

) as soon as it is produced, while the acid itself undergoes reduction to nitrogen dioxide (

NO_2

).

Step-by-Step Solution

1

Analyze the typical reaction of metals with dilute acids

Dilute acids such as

H_2SO_4

or

HCl

act as typical acids where reactive metals displace hydrogen:

Zn_{(s)} + H_2SO_{4(aq)} \rightarrow ZnSO_{4(aq)} + H_{2(g)}

.

This is a standard displacement reaction driven by hydrogen ion reduction.

2

Examine the behavior of concentrated trioxonitrate(V) acid (

HNO_3

)

Concentrated

HNO_3

is a powerful oxidizing acid. Instead of releasing

H_2

gas, any formed hydrogen is immediately oxidized to water (

H_2O

), while

HNO_3

is reduced to brown nitrogen dioxide gas (

NO_2

).

The equation is:

Zn_{(s)} + 4HNO_{3(aq)} \rightarrow Zn(NO_3)_{2(aq)} + 2NO_{2(g)} + 2H_2O_{(l)}

.

3

Identify the chemical property responsible

The absence of hydrogen gas is due to the strong oxidizing nature of concentrated

HNO_3

.

The acid acts primarily as an oxidizing agent rather than a simple proton donor.

Key Concept

Oxidizing property of trioxonitrate(V) acid

Question 12875Question →

During a normal mammalian cardiac cycle, electrical excitation coordinates the synchronized contraction of the heart chambers. What is the correct sequence of electrical impulse propagation through the cardiac tissues starting from the initiation of the heartbeat?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of cardiac conduction begins with depolarization at the sinoatrial (SA) node, followed by the spread of excitation across the atrial myocardium, a brief delay at the atrioventricular (AV) node, rapid signal conduction through the Bundle of His and Purkinje fibers, and finally contraction of the ventricular myocardium.

In the mammalian heart, the electrical signal originates spontaneously at the sinoatrial (SA) node in the right atrium. The impulse spreads across the atrial myocardium, causing atrial systole. It then reaches the atrioventricular (AV) node, where conduction is delayed to allow complete ventricular filling. From the AV node, the impulse travels rapidly through the Bundle of His and Purkinje fibers, causing coordinated contraction of the ventricular myocardium from the apex upward.

Step-by-Step Solution

1

Identify the primary pacemaker site where the cardiac action potential originates.

The heartbeat starts at the sinoatrial (SA) node located in the right atrium wall.

The SA node possesses the intrinsic capability for spontaneous rhythm generation.

2

Trace the path of electrical excitation across the upper chambers.

The signal spreads through the muscle cells of both atria, causing atrial contraction.

Atrial muscle fibers are electrically connected via intercalated discs.

3

Determine the role and location of the secondary conducting node.

The signal reaches the atrioventricular (AV) node, where its transmission is momentarily delayed.

This delay ensures ventricular filling occurs prior to ventricular systole.

4

Follow the specialized rapid conducting pathway down the ventricles.

The signal travels down the Bundle of His and through Purkinje fibers in the ventricular walls.

Standard muscle conduction would be too slow to produce coordinated ventricular pumping.

5

Identify the final mechanical response of the lower chambers.

The ventricular myocardium depolarizes and contracts from the apex toward the base of the heart.

This upward squeezing motion efficiently ejects blood into the major arterial trunks.

Key Concept

Mammalian Cardiac Conduction Pathway

Question 12876Question →

An electric heater rated at $500\text{ W}$ is used to melt $0.20\text{ kg}$ of ice initially at $0^\circ\text{C}$ and heat the resulting water to $20^\circ\text{C}$ . Assuming no heat losses to the surroundings, what is the time required in seconds? (Specific latent heat of fusion of ice = $3.36 \times 10^5\text{ J kg}^{-1}$ , specific heat capacity of water = $4.2 \times 10^3\text{ J kg}^{-1}\text{ K}^{-1}$ )

Show answer & explanation

Answer: 168

Answer

168 s

The total energy required is the sum of the heat required for melting (

mL_f = 67,200\text{ J}

) and the heat required for warming the liquid (

mc\Delta T = 16,800\text{ J}

), giving

84,000\text{ J}

. Dividing this by the heater power rating of

500\text{ W}

gives

168\text{ s}

.

Step-by-Step Solution

1

Calculate thermal energy needed to change phase from ice to water at

0^\circ\text{C}

Q_1 = m L_f = 0.20\text{ kg} \times 3.36 \times 10^5\text{ J kg}^{-1} = 67,200\text{ J}

Latent heat of fusion accounts for phase change at constant temperature.

2

Calculate thermal energy needed to raise water temperature from

0^\circ\text{C}

to

20^\circ\text{C}

Q_2 = m c \Delta T = 0.20\text{ kg} \times 4.2 \times 10^3\text{ J kg}^{-1}\text{K}^{-1} \times 20\text{ K} = 16,800\text{ J}

Sensible heat increases kinetic energy of liquid water molecules.

3

Sum energy required for both phase change and temperature change

Q_{\text{total}} = Q_1 + Q_2 = 67,200\text{ J} + 16,800\text{ J} = 84,000\text{ J}

Total energy is the sum of sensible heat and latent heat.

4

Calculate heating time using heater power rating

t = \frac{Q_{\text{total}}}{P} = \frac{84,000\text{ J}}{500\text{ W}} = 168\text{ s}

Power is defined as energy transferred per unit time (

P = \frac{Q}{t}

).

Key Concept

Energy balance combining latent heat and sensible heat

Estimated Time:2m 0s

Question 12877Question →

A mixture of

10\text{ cm}^3

of hydrogen sulfide (

H_2S

) gas and

40\text{ cm}^3

of oxygen gas was sparked to react completely at constant temperature and pressure according to the equation:

2H_2S(g) + 3O_2(g) \rightarrow 2SO_2(g) + 2H_2O(l)

Assuming the water formed condenses into liquid, what is the total volume of the residual gas mixture?

35\text{ cm}^3

25\text{ cm}^3

10\text{ cm}^3

45\text{ cm}^3

Show answer & explanation

Answer: $35\text{ cm}^3$

Answer

The total volume of the residual gas mixture is

35\text{ cm}^3

.

According to Gay-Lussac's law,

2\text{ cm}^3

of

H_2S

reacts with

3\text{ cm}^3

of

O_2

to form

2\text{ cm}^3

of

SO_2

gas. Thus,

10\text{ cm}^3

of

H_2S

consumes

15\text{ cm}^3

of

O_2

and yields

10\text{ cm}^3

of

SO_2

. The remaining excess

O_2

is

40\text{ cm}^3 - 15\text{ cm}^3 = 25\text{ cm}^3

. Combining the unreacted oxygen (

25\text{ cm}^3

) and produced sulfur(IV) oxide (

10\text{ cm}^3

) gives a total residual gas volume of

35\text{ cm}^3

.

Step-by-Step Solution

1

Determine the stoichiometric volume ratios from the balanced chemical equation

2 volumes of

H_2S(g)

react with 3 volumes of

O_2(g)

to produce 2 volumes of

SO_2(g)

.

According to Gay-Lussac's Law of Combining Volumes, gases react in simple whole-number volume ratios at constant temperature and pressure.

2

Calculate the volume of oxygen required to react with

10\text{ cm}^3

of

H_2S

Volume of

O_2

used =

10\text{ cm}^3 \times \frac{3}{2} = 15\text{ cm}^3

.

Since

10\text{ cm}^3

of

H_2S

is available, it is the limiting reactant and requires

15\text{ cm}^3

of

O_2

.

3

Find the volume of unreacted excess oxygen

Unreacted

O_2 = 40\text{ cm}^3 - 15\text{ cm}^3 = 25\text{ cm}^3

.

Subtracting the reacted oxygen volume from the initial volume gives the excess.

4

Calculate the volume of gaseous

SO_2

produced

Volume of

SO_2 = 10\text{ cm}^3 \times \frac{2}{2} = 10\text{ cm}^3

.

The mole ratio of

H_2S

to

SO_2

is

2:2

, so

10\text{ cm}^3

of

H_2S

produces

10\text{ cm}^3

of

SO_2

gas. Water is liquid so its volume is neglected.

5

Calculate total residual gas volume

Total residual volume =

25\text{ cm}^3 \text{ (excess } O_2) + 10\text{ cm}^3 \text{ (produced } SO_2) = 35\text{ cm}^3

.

The residual gas consists of both unreacted excess reactant and gaseous product.

Key Concept

Gay-Lussac's Law of Combining Volumes

Question 12878Question →

A student records the mass of a chemical sample using a digital balance as $0.02050\text{ kg}$ . How many significant figures are contained in this measurement?

3

4

5

2

Show answer & explanation

Answer: 4

Answer

4 significant figures

In the measurement

0.02050\text{ kg}

, the leading zeros (

0.0

) are placeholders and do not count as significant. The non-zero digits (

2

and

5

), the captive zero between them, and the trailing zero following the decimal point are all significant. This gives a total of four significant figures.

Step-by-Step Solution

1

Identify and evaluate leading zeros in

0.02050\text{ kg}

.

The zeros before the digit '2' (

0.0

) are leading zeros.

Leading zeros serve only as decimal placeholders and are never significant.

2

Identify non-zero digits and zeros between non-zero digits.

The digits '2' and '5' are non-zero, and the zero between them is a captive zero.

All non-zero digits and zeros trapped between non-zero digits are always significant.

3

Evaluate the trailing zero.

The zero after '5' is a trailing zero in a decimal number.

Trailing zeros to the right of a decimal point indicate measurement precision and are significant.

4

Sum the total number of significant figures.

Digits '2', '0', '5', and '0' give a total of 4 significant figures.

Combining two non-zero digits, one captive zero, and one trailing decimal zero gives four significant figures.

Key Concept

Rules for counting significant figures in physical measurements

Estimated Time:45s

Question 12879Question →

A simple pendulum on Earth ( $g = 10\text{ m/s}^2$ ) completes $50$ full oscillations in $40\text{ s}$ . The pendulum is then transferred to a lunar station where the acceleration due to gravity is $1.6\text{ m/s}^2$ , and its length is reduced by $64\%$ . What is the time taken, in seconds, for this modified pendulum to complete $30$ oscillations on the lunar station?

Show answer & explanation

Answer: 36

Answer

The time taken for the modified pendulum to complete 30 oscillations on the lunar station is 36 s.

The initial period on Earth is

T_1 = \frac{40}{50} = 0.8\text{ s}

. The formula for the period of a simple pendulum is

T = 2\pi\sqrt{\frac{L}{g}}

. When length decreases by

64\%

, the remaining length ratio is

\frac{L_2}{L_1} = 0.36

. The ratio of gravity is

\frac{g_1}{g_2} = \frac{10}{1.6} = 6.25

. Taking the ratio gives

\frac{T_2}{T_1} = \sqrt{0.36 \times 6.25} = \sqrt{2.25} = 1.5

. Thus, the new period is

T_2 = 1.5 \times 0.8\text{ s} = 1.2\text{ s}

. For

30

oscillations, the total time is

t = 30 \times 1.2\text{ s} = 36\text{ s}

.

Step-by-Step Solution

1

Calculate the initial period of oscillation on Earth

T_1 = 0.8\text{ s}

Period

T_1

is total time divided by the number of oscillations:

T_1 = \frac{40\text{ s}}{50} = 0.8\text{ s}

.

2

Set up the ratio for period under altered length and gravitational field

\frac{T_2}{T_1} = 1.5

Using

T = 2\pi \sqrt{\frac{L}{g}}

, we have

\frac{T_2}{T_1} = \sqrt{\frac{L_2}{L_1} \cdot \frac{g_1}{g_2}} = \sqrt{(1 - 0.64) \cdot \frac{10}{1.6}} = \sqrt{0.36 \cdot 6.25} = \sqrt{2.25} = 1.5

.

3

Determine the new period of oscillation

T_2 = 1.2\text{ s}

T_2 = 1.5 \times T_1 = 1.5 \times 0.8\text{ s} = 1.2\text{ s}

.

4

Calculate the total time required for 30 oscillations

t_2 = 36\text{ s}

Total time

t_2 = N_2 \times T_2 = 30 \times 1.2\text{ s} = 36\text{ s}

.

Key Concept

Period of a simple pendulum and its dependence on length and gravitational acceleration

Question 12880Question →

In a nuclear fusion reaction, a lithium-6 nucleus ( ${^{6}_{3}\text{Li}}$ ) fuses with a deuterium nucleus ( ${^{2}_{1}\text{H}}$ ) to form two alpha particles ( ${^{4}_{2}\text{He}}$ ). Given the atomic masses of ${^{6}_{3}\text{Li}} = 6.0151\text{ u}$ , ${^{2}_{1}\text{H}} = 2.0141\text{ u}$ , and ${^{4}_{2}\text{He}} = 4.0026\text{ u}$ , and taking $1\text{ u} = 931.5\text{ MeV}$ , what is the total energy released in this reaction in $\text{MeV}$ ?

Show answer & explanation

Answer: 22.356

Answer

The total energy released in the fusion reaction is 22.356 MeV.

The total mass before fusion (

{^{6}_{3}\text{Li}} + {^{2}_{1}\text{H}}

) is

8.0292\text{ u}

, while the total mass after fusion (

2 \times {^{4}_{2}\text{He}}

) is

8.0052\text{ u}

. The resulting mass defect is

\Delta m = 0.0240\text{ u}

. Multiplying this mass loss by

931.5\text{ MeV/u}

gives an energy release of

22.356\text{ MeV}

.

Step-by-Step Solution

1

Calculate the total initial mass of the reactants

m_{\text{initial}} = 6.0151\text{ u} + 2.0141\text{ u} = 8.0292\text{ u}

Add the rest masses of the lithium-6 nucleus and the deuterium nucleus together.

2

Calculate the total final mass of the products

m_{\text{final}} = 2 \times 4.0026\text{ u} = 8.0052\text{ u}

The reaction produces two helium-4 nuclei (alpha particles), so multiply the mass of one alpha particle by 2.

3

Determine the mass defect

\Delta m = m_{\text{initial}} - m_{\text{final}} = 8.0292\text{ u} - 8.0052\text{ u} = 0.0240\text{ u}

Subtract the final mass of products from the initial mass of reactants to find the mass lost.

4

Convert the mass defect into energy released

E = \Delta m \times 931.5\text{ MeV/u} = 0.0240 \times 931.5 = 22.356\text{ MeV}

Apply Einstein's mass-energy equivalence principle using the standard conversion factor

1\text{ u} = 931.5\text{ MeV}

.

Key Concept

Mass Defect and Energy Release in Nuclear Fusion Reactions

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept