All practice questions

Question 12801Question →

Complete the following comparison of industrial chemical manufacturing methods by filling in the missing operational process type.

Fill in the blanks below

In industrial chemistry, heavy chemicals such as tetraoxosulfate(VI) acid are synthesized in high tonnage using continuous operations, whereas fine chemicals such as pharmaceuticals and analytical reagents are produced in small, high-purity quantities using processes.

Show answer & explanation

Answer

The correct term to complete the statement is 'batch'.

Fine chemicals (such as drugs, dyes, and specialized reagents) require strict purity standards and are synthesized in relatively small quantities using batch processes. In contrast, heavy chemicals (such as sulfuric acid, sodium hydroxide, and ammonia) are produced in bulk via continuous processes.

Step-by-Step Solution

1

Analyze the operational differences between heavy chemical and fine chemical manufacturing.

Heavy chemicals are produced continuously on a large scale due to high global demand, whereas fine chemicals are high-value substances produced in limited quantities.

Fine chemical production requires precise purity controls and flexibility in manufacturing.

2

Identify the process mode characteristic of fine chemical production.

Manufacturing in discrete, specific production runs or lots is defined as a batch process.

Batch processing allows for thorough quality control, specialized reaction conditions, and equipment cleaning between synthesis cycles.

Key Concept

Operational process distinction between continuous processing (heavy chemicals) and batch processing (fine chemicals)

Estimated Time:1m 0s

Question 12802Question →

A Uranium-235 nucleus undergoes nuclear fission after absorbing a thermal neutron according to the reaction equation:

{^{235}_{92}\text{U}} + {^{1}_{0}\text{n}} \rightarrow {^{140}_{54}\text{Xe}} + {^{94}_{38}\text{Sr}} + x\ {^{1}_{0}\text{n}}

Given the rest masses:
- Mass of

{^{235}_{92}\text{U}} = 235.0439\text{ u}

- Mass of

{^{140}_{54}\text{Xe}} = 139.9216\text{ u}

- Mass of

{^{94}_{38}\text{Sr}} = 93.9154\text{ u}

- Mass of

{^{1}_{0}\text{n}} = 1.0087\text{ u}

-

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

Calculate the total energy released during this fission process in MeV.

Show answer & explanation

Answer: 184.62

Answer

184.62 MeV

The total energy released is calculated by first balancing the nuclear equation to determine that 2 neutrons are emitted (

x = 2

). The total mass of reactants is

236.0526\text{ u}

and the products is

235.8544\text{ u}

. The mass defect of

0.1982\text{ u}

multiplied by

931.5\text{ MeV/u}

yields

184.62\text{ MeV}

.

Step-by-Step Solution

1

Balance the mass numbers in the nuclear equation to find the number of emitted neutrons (

x

)

x = 2

neutrons

Conservation of mass number requires

235 + 1 = 140 + 94 + x

, giving

236 = 234 + x

, so

x = 2

.

2

Calculate the total mass of the reactants before fission

236.0526\text{ u}

Summing the mass of U-235 (

235.0439\text{ u}

) and the incident neutron (

1.0087\text{ u}

).

3

Calculate the total mass of the products after fission

235.8544\text{ u}

Summing masses of Xe-140 (

139.9216\text{ u}

), Sr-94 (

93.9154\text{ u}

), and 2 neutrons (

2 \times 1.0087\text{ u} = 2.0174\text{ u}

).

4

Determine the mass defect (mass difference)

\Delta m = 0.1982\text{ u}

Mass defect

\Delta m = m_{\text{reactants}} - m_{\text{products}} = 236.0526 - 235.8544 = 0.1982\text{ u}

.

5

Convert mass defect to energy using

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

184.62\text{ MeV}

Multiplying mass defect

0.1982\text{ u}

by

931.5\text{ MeV/u}

gives

184.6233\text{ MeV}

, which rounds to

184.62\text{ MeV}

.

Key Concept

Mass defect and mass-energy equivalence in nuclear fission reactions

Estimated Time:2m 30s

Question 12803Question →

A bar magnet is pulled away from a stationary circular coil such that its South pole moves directly away from the front face of the coil. Based on Lenz's law, which polarity is induced on the front face of the coil?

North pole, to attract the receding South pole and oppose its motion

South pole, to repel the receding South pole and oppose its motion

South pole, to attract the receding South pole and assist its motion

North pole, to repel the receding South pole and assist its motion

Show answer & explanation

Answer: North pole, to attract the receding South pole and oppose its motion

Answer

A North pole is induced on the front face of the coil to attract the receding South pole and oppose its motion.

According to Lenz's law, the direction of induced current creates a magnetic field that opposes the change causing it. As the South pole moves away, the decreasing magnetic flux is opposed by an attractive force pulling the magnet back. Therefore, an opposite pole (North pole) is induced on the coil's front face.

Step-by-Step Solution

1

Identify the change causing electromagnetic induction

The South pole of the magnet is moving away from the coil, causing a decrease in magnetic flux through the coil.

Induction is driven by a change in magnetic flux according to Faraday's law.

2

Apply Lenz's law to determine the direction of induced magnetic effect

The induced magnetic field must attempt to pull the magnet back to oppose its withdrawal.

Lenz's law dictates that the direction of an induced current always opposes the motion or change causing it.

3

Determine the required magnetic polarity on the front face

To attract the departing South pole, an opposite magnetic pole (North pole) must be induced on the front face.

Unlike magnetic poles attract each other.

Key Concept

Lenz's Law and Direction of Induced Current

Estimated Time:45s

Question 12804Question →

A metal block of mass $4.0\text{ kg}$ absorbs $12\text{ kJ}$ of thermal energy, causing its temperature to rise by $15\text{ K}$ . What is the specific heat capacity of the metal block in $\text{J kg}^{-1}\text{K}^{-1}$ ?

Show answer & explanation

Answer: 200

Answer

The specific heat capacity of the metal block is

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

.

Specific heat capacity

c

is determined using

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

. Substituting

Q = 12,000\text{ J}

,

m = 4.0\text{ kg}

, and

\Delta T = 15\text{ K}

yields

c = \frac{12000}{4.0 \times 15} = 200\text{ J kg}^{-1}\text{K}^{-1}

.

Step-by-Step Solution

1

Convert given energy to standard SI units

Q = 12\text{ kJ} = 12,000\text{ J}

Calculation of specific heat capacity requires heat energy in Joules.

2

Relate heat energy, mass, specific heat capacity, and temperature change

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

Specific heat capacity

c

represents heat energy per unit mass per Kelvin temperature change.

3

Substitute the values and compute the result

c = \frac{12000}{4.0 \times 15} = 200\text{ J kg}^{-1}\text{K}^{-1}

Evaluating the expression gives

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

.

Key Concept

Specific heat capacity is the quantity of heat required to raise the temperature of

1\text{ kg}

of a substance by

1\text{ K}

, expressed as

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

.

Question 12805Question →

An $\alpha$ -particle (charge $+2e$ , mass $m_\alpha$ ) and a $\beta^-$ -particle (charge $-e$ , mass $m_\beta$ ) emitted during natural radioactive decay enter a region containing uniform, mutually perpendicular electric ( $E$ ) and magnetic ( $B$ ) fields. Both particles move along paths perpendicular to both fields and traverse the region without undergoing any deflection. What is the ratio of the kinetic energy of the $\alpha$ -particle to that of the $\beta^-$ -particle, $\frac{E_{k,\alpha}}{E_{k,\beta}}$ ?

\frac{m_\alpha}{m_\beta}

1

2

\frac{m_\alpha}{2m_\beta}

Show answer & explanation

Answer: $\frac{m_\alpha}{m_\beta}$

Answer

The ratio of their kinetic energies is equal to the ratio of their masses,

\frac{m_\alpha}{m_\beta}

.

In crossed uniform electric and magnetic fields acting as a velocity selector, a charged particle passes undeflected when the electric force

qE

balances the magnetic Lorentz force

qvB

. Equating these forces yields

v = \frac{E}{B}

, which depends only on field strengths and is independent of mass and charge. Because both the

\alpha

-particle and

\beta^-

-particle traverse undeflected, both possess the same speed

v

. Substituting equal speeds into the kinetic energy formula

E_k = \frac{1}{2}mv^2

leaves the ratio of their kinetic energies equal to the ratio of their rest masses,

\frac{m_\alpha}{m_\beta}

.

Step-by-Step Solution

1

Apply the force balance condition for particles moving undeflected in crossed electric and magnetic fields.

Electric force magnitude

F_E = qE

must equal magnetic force magnitude

F_B = qvB

, giving

qE = qvB \implies v = \frac{E}{B}

.

For zero net deflection, the electrostatic force and magnetic Lorentz force must be equal in magnitude and opposite in direction.

2

Determine the velocities of the

\alpha

-particle and

\beta^-

-particle.

v_\alpha = \frac{E}{B}

and

v_\beta = \frac{E}{B}

, so

v_\alpha = v_\beta = v

.

The velocity selection equation

v = \frac{E}{B}

is completely independent of particle mass

m

and charge

q

.

3

Express the ratio of the kinetic energies of the two emissions using

E_k = \frac{1}{2}mv^2

.

\frac{E_{k,\alpha}}{E_{k,\beta}} = \frac{\frac{1}{2} m_\alpha v^2}{\frac{1}{2} m_\beta v^2} = \frac{m_\alpha}{m_\beta}

.

Since the speed

v

is identical for both particles, the

\frac{1}{2}v^2

terms cancel out completely.

Key Concept

Velocity selector behavior and kinetic energy dependence of radiation emissions in electromagnetic fields

Question 12806Question →

A concave shaving mirror has a radius of curvature of $60\text{ cm}$ . A person places their face in front of the mirror such that an upright image magnified $3$ times is formed. What is the distance of the face from the mirror, in centimeters?

Show answer & explanation

Answer: 20

Answer

The distance of the person's face from the mirror is

20\text{ cm}

.

For a concave mirror with a radius of curvature of

60\text{ cm}

, the focal length is

f = +30\text{ cm}

. An upright image is virtual, corresponding to a positive magnification

m = +3

. Since

m = -\frac{v}{u}

, the image distance is

v = -3u

. Substituting these into the mirror formula

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

gives

\frac{1}{30} = \frac{1}{u} - \frac{1}{3u} = \frac{2}{3u}

, solving to

u = 20\text{ cm}

.

Step-by-Step Solution

1

Determine the focal length of the concave mirror.

f = 30\text{ cm}

The focal length is half the radius of curvature (

f = \frac{r}{2} = \frac{60\text{ cm}}{2} = 30\text{ cm}

).

2

Express the image distance

v

in terms of the object distance

u

using the magnification relationship.

v = -3u

An upright image produced by a spherical mirror is virtual, so linear magnification

m = +3

. Using

m = -\frac{v}{u} = +3

, we obtain

v = -3u

.

3

Substitute

f

and

v

into the mirror equation to solve for

u

.

u = 20\text{ cm}

Applying the mirror formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \implies \frac{1}{30} = \frac{1}{u} - \frac{1}{3u} = \frac{2}{3u} \implies 3u = 60 \implies u = 20\text{ cm}

.

Key Concept

Mirror equation and sign conventions for virtual images formed by concave mirrors

Question 12807Question →

Two long, straight parallel wires separated by a distance of $5.0\text{ cm}$ in air carry equal currents in opposite directions. If the repulsive force per unit length between the wires is $1.6 \times 10^{-3}\text{ N/m}$ , determine the magnitude of the current flowing through each wire in amperes. (Take $\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}$ )

Show answer & explanation

Answer: 20

Answer

The magnitude of the current flowing through each wire is

20\text{ A}

.

Using the parallel conductor force formula

\frac{F}{L} = \frac{\mu_0 I^2}{2\pi d}

, we substitute

\frac{F}{L} = 1.6 \times 10^{-3}\text{ N/m}

,

d = 0.05\text{ m}

, and

\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}

. Simplifying yields

1.6 \times 10^{-3} = 4 \times 10^{-6} I^2

, giving

I^2 = 400

and

I = 20\text{ A}

.

Step-by-Step Solution

1

Recall the expression for force per unit length between two current-carrying parallel wires

\frac{F}{L} = \frac{\mu_0 I^2}{2\pi d}

The magnetic field generated by one wire exerts a magnetic force on the current in the adjacent wire.

2

Convert distance to meters and substitute all given values into the formula

d = 0.05\text{ m}

, leading to

1.6 \times 10^{-3} = \frac{(4\pi \times 10^{-7}) I^2}{2\pi (0.05)}

Standard SI unit for distance is meters, necessary for dimensional consistency.

3

Simplify the equation and compute the current magnitude

1.6 \times 10^{-3} = 4 \times 10^{-6} I^2 \implies I^2 = 400 \implies I = 20\text{ A}

Solving the quadratic term gives the scalar current magnitude in amperes.

Key Concept

Force per unit length between parallel current-carrying conductors

Question 12808Question →

A sealed room with a volume of $50\text{ m}^3$ at a temperature of $20^\circ\text{C}$ contains $0.40\text{ kg}$ of water vapour. If the mass of water vapour required to saturate $1\text{ m}^3$ of air at $20^\circ\text{C}$ is $0.016\text{ kg}$ , what is the relative humidity of the air in the room?

25\%

40\%

50\%

80\%

Show answer & explanation

Answer: $50\%$

Answer

50\%

Relative humidity is defined as the ratio of the actual mass of water vapour present in a given volume of air to the mass of water vapour required to saturate the same volume at the same temperature. For

50\text{ m}^3

of air, the mass needed for saturation is

50\text{ m}^3 \times 0.016\text{ kg/m}^3 = 0.80\text{ kg}

. Dividing the actual mass (

0.40\text{ kg}

) by

0.80\text{ kg}

and multiplying by

100\%

gives

50\%

.

Step-by-Step Solution

1

Calculate the total mass of water vapour required to saturate the entire room volume.

\text{Total saturation mass} = 50\text{ m}^3 \times 0.016\text{ kg/m}^3 = 0.80\text{ kg}

The saturation density gives the maximum water vapour

1\text{ m}^3

can hold, so it must be scaled by the room volume.

2

Apply the formula for relative humidity.

\text{Relative Humidity} = \left(\frac{\text{Actual mass of water vapour}}{\text{Saturation mass of water vapour}}\right) \times 100\%

Relative humidity measures the degree of saturation of an air sample at a specific temperature.

3

Substitute the known values to find the relative humidity.

\text{Relative Humidity} = \left(\frac{0.40\text{ kg}}{0.80\text{ kg}}\right) \times 100\% = 50\%

Simplifying the fraction

\frac{0.40}{0.80} = 0.5

, which equals

50\%

.

Key Concept

Relative Humidity

Estimated Time:1m 15s

Question 12809Question →

At a certain location, the horizontal component of the Earth's magnetic field is $3.0 \times 10^{-5}\text{ T}$ . If an additional uniform horizontal magnetic field of $4.0 \times 10^{-5}\text{ T}$ is applied perpendicular to the magnetic meridian, what is the magnitude of the resultant horizontal magnetic flux density experienced by a compass needle at this location?

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

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

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

1.2 \times 10^{-4}\text{ T}

Show answer & explanation

Answer: $5.0 \times 10^{-5}\text{ T}$

Answer

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

Because the Earth's horizontal field component acts along the magnetic meridian and the external field is applied perpendicular to it, the two fields form a right-angled triangle. Applying Pythagoras' theorem gives

\sqrt{(3.0 \times 10^{-5})^2 + (4.0 \times 10^{-5})^2} = 5.0 \times 10^{-5}\text{ T}

, which represents the true resultant field.

Step-by-Step Solution

1

Identify the vector orientation of the two magnetic fields.

The Earth's horizontal component

B_H

acts along the magnetic meridian (North-South), while the applied field

B_{\text{ext}}

acts perpendicular to it (East-West) at an angle of

\theta = 90^\circ

.

Magnetic flux density is a vector quantity, so direction matters when combining fields.

2

Apply the perpendicular vector addition formula.

B_R = \sqrt{B_H^2 + B_{\text{ext}}^2}

For two vectors acting at right angles (

90^\circ

), the resultant magnitude is given by the Pythagorean theorem.

3

Substitute the given numerical values into the equation.

B_R = \sqrt{(3.0 \times 10^{-5})^2 + (4.0 \times 10^{-5})^2} = \sqrt{(9.0 + 16.0) \times 10^{-10}} = \sqrt{25.0 \times 10^{-10}} = 5.0 \times 10^{-5}\text{ T}

Simplifying the square root yields the exact magnitude of the total horizontal field.

Key Concept

Vector Superposition of Magnetic Fields

Estimated Time:2m 0s

Question 12810Question →

During the industrial smelting stage of copper extraction from chalcopyrite ( $\text{CuFeS}_2$ ), silicon(IV) oxide ( $\text{SiO}_2$ ) is added to the furnace charge. What is the primary chemical role of silicon(IV) oxide in this process?

To act as an acidic flux that combines with iron(II) oxide impurity to form a fusible slag of iron(II) trioxosilicate(IV)

To serve as the primary reducing agent that converts copper(I) sulfide directly to metallic copper

To lower the melting point of the copper electrolyte during subsequent electro-refining

To react with acidic sulfur impurities and neutralize them into non-volatile salts

Show answer & explanation

Answer: To act as an acidic flux that combines with iron(II) oxide impurity to form a fusible slag of iron(II) trioxosilicate(IV)

Answer

Silicon(IV) oxide acts as an acidic flux that reacts with basic iron(II) oxide to form molten iron(II) trioxosilicate(IV) slag (

\text{FeSiO}_3

).

In the extraction of copper from sulfide ores, the ore contains significant iron impurities. Smelting oxidized iron into basic iron(II) oxide (

\text{FeO}

). Adding sand or silicon(IV) oxide (

\text{SiO}_2

), which acts as an acidic flux, causes a chemical reaction forming molten iron(II) trioxosilicate(IV) (

\text{FeSiO}_3

). This slag is less dense than the copper matte, allowing easy separation by skimming or tapping off.

Step-by-Step Solution

1

Identify the nature of impurities in copper ore roasting/smelting

Partial roasting of copper pyrites (

\text{CuFeS}_2

) produces iron(II) oxide (

\text{FeO}

), which is a basic oxide impurity.

Iron impurities must be separated from copper matte (

\text{Cu}_2\text{S} \cdot \text{FeS}

) prior to copper reduction.

2

Determine the role of the added flux

An acidic flux, silicon(IV) oxide (

\text{SiO}_2

), reacts with basic

\text{FeO}

according to the equation:

\text{FeO}(s) + \text{SiO}_2(s) \rightarrow \text{FeSiO}_3(l)

.

An acid-base reaction between flux and gangue forms a light, molten silicate layer known as slag.

3

Conclude the function of slag in pyrometallurgy

The molten slag (

\text{FeSiO}_3

) floats on top of the heavier copper matte layer and is tapped off.

Slag formation allows continuous mechanical removal of iron impurities.

Key Concept

Role of fluxes and slag formation in metallurgy

Estimated Time:1m 0s

Question 12811Question →

A car of mass $1200\text{ kg}$ ascends a straight road inclined at an angle $\theta$ to the horizontal, where $\sin\theta = 0.1$ , at a steady speed of $15\text{ m s}^{-1}$ . If the total frictional resistance to motion is $400\text{ N}$ , what is the useful mechanical power output of the engine in kilowatts? (Take $g = 10\text{ m s}^{-2}$ )

Show answer & explanation

Answer: 24

Answer

24 kW

To maintain a constant ascending speed, the engine must supply a force equal to the sum of the component of weight parallel to the incline (

mg\sin\theta = 1200\text{ N}

) and the frictional force (

400\text{ N}

), resulting in a total force of

1600\text{ N}

. Multiplying this total force by the constant speed of

15\text{ m s}^{-1}

yields a power output of

24000\text{ W}

, which corresponds to

24\text{ kW}

.

Step-by-Step Solution

1

Calculate the gravitational force component acting down the slope

1200 N

The component of the car's weight parallel to the incline opposes upward motion:

F_g = m g \sin\theta = 1200 \text{ kg} \times 10 \text{ m s}^{-2} \times 0.1 = 1200 \text{ N}

.

2

Determine the total tractive force required from the car engine

1600 N

Because the velocity is constant, the net force is zero; hence, the engine force must balance both the gravitational slope component and the frictional resistance:

F_{\text{engine}} = F_g + F_{\text{friction}} = 1200 \text{ N} + 400 \text{ N} = 1600 \text{ N}

.

3

Calculate the mechanical power delivered by the engine

24 kW

Power is the product of tractive force and constant speed:

P = F_{\text{engine}} \times v = 1600 \text{ N} \times 15 \text{ m s}^{-1} = 24000 \text{ W} = 24 \text{ kW}

.

Key Concept

Power required to maintain motion against opposing forces on an inclined plane

Question 12812Question →

Which of the following types of radiation emitted during natural radioactive decay has the greatest ionizing power?

\alpha

-particles

\beta

-particles

\gamma

-rays

X-rays

Show answer & explanation

Answer: $\alpha$ -particles

Answer

Alpha particles (

\alpha

-particles) have the highest ionizing power among natural radioactive emissions.

Alpha particles have the largest charge (

+2e

) and mass among natural nuclear emissions, allowing them to exert strong electrostatic forces on atoms in their path and strip away electrons easily, giving them the highest ionizing capability.

Step-by-Step Solution

1

Identify the nature, mass, and charge of natural radiation emissions

\alpha

-particles consist of helium nuclei (

^4_2\text{He}

) with mass

4\text{ u}

and charge

+2e

;

\beta

-particles are high-speed electrons with negligible mass and charge

-e

;

\gamma

-rays are neutral photons.

Ionization depends directly on charge magnitude and kinetic energy transfer capacity during interactions with matter.

2

Compare the ionizing power of the emissions

Because

\alpha

-particles carry a large

+2e

charge and move relatively slowly due to their larger mass, they interact strongly with orbital electrons, knocking them off atoms easily along a short path.

Relative ionizing power follows the ratio of roughly

10,000 : 100 : 1

for

\alpha : \beta : \gamma

emissions.

Key Concept

Ionizing power vs. Penetrating power of radioactive emissions

Estimated Time:45s

Question 12813Question →

During the laboratory preparation of oxygen gas, a sample of

17.0\text{ g}

of hydrogen peroxide (

\text{H}_2\text{O}_2

) decomposes completely in the presence of a manganese(IV) oxide catalyst according to the equation:

2\text{H}_2\text{O}_2(\text{aq}) \xrightarrow{\text{MnO}_2} 2\text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g})

What volume of oxygen gas, measured at standard temperature and pressure (STP), is released in this process?
[Molar mass of

\text{H}_2\text{O}_2 = 34.0\text{ g mol}^{-1}

; Molar volume of gas at STP

= 22.4\text{ dm}^3\text{ mol}^{-1}

]

2.80\text{ dm}^3

5.60\text{ dm}^3

6.00\text{ dm}^3

11.2\text{ dm}^3

Show answer & explanation

Answer: $5.60\text{ dm}^3$

Answer

5.60\text{ dm}^3

The decomposition of

17.0\text{ g}

of

\text{H}_2\text{O}_2

yields

0.50\text{ mol}

of reactant. According to the balanced chemical equation,

2\text{ moles}

of

\text{H}_2\text{O}_2

yield

1\text{ mole}

of

\text{O}_2

, producing

0.25\text{ mol}

of oxygen gas. At standard temperature and pressure (STP),

0.25\text{ mol}

occupies

0.25 \times 22.4\text{ dm}^3\text{ mol}^{-1} = 5.60\text{ dm}^3

.

Step-by-Step Solution

1

Calculate the amount in moles of hydrogen peroxide (

\text{H}_2\text{O}_2

) reactant.

n(\text{H}_2\text{O}_2) = \frac{17.0\text{ g}}{34.0\text{ g mol}^{-1}} = 0.50\text{ mol}

Converting mass to moles using the molar mass provides the quantity of reactant available.

2

Determine the moles of oxygen gas (

\text{O}_2

) formed using equation stoichiometry.

Since

2\text{ mol } \text{H}_2\text{O}_2 \rightarrow 1\text{ mol } \text{O}_2

,

n(\text{O}_2) = \frac{0.50\text{ mol}}{2} = 0.25\text{ mol}

The balanced chemical equation shows a 2:1 molar ratio between reactant and gaseous product.

3

Calculate the volume of oxygen gas produced at STP.

V(\text{O}_2) = 0.25\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 5.60\text{ dm}^3

Multiplying the calculated moles of gas by the standard molar gas volume gives the volume at STP.

Key Concept

Stoichiometric calculations and gas molar volume at STP for oxygen preparation.

Estimated Time:1m 30s

Question 12814Question →

Under the same conditions of temperature and pressure, how many times faster does protium gas ( $\text{H}_2$ ) diffuse through a porous plug compared to tritium gas ( $\text{T}_2$ )? [Relative atomic masses: protium, $^{1}\text{H} = 1.0$ ; tritium, $^{3}\text{H} = 3.0$ ]

1.73

3.00

0.58

9.00

Show answer & explanation

Answer: $1.73$

Answer

Protium gas diffuses approximately

1.73

times faster than tritium gas.

The relative rate of diffusion of protium gas (

\text{H}_2

) relative to tritium gas (

\text{T}_2

) is governed by Graham's Law, which states that

R_1 / R_2 = \sqrt{M_2 / M_1}

. Given

M(\text{H}_2) = 2\text{ g mol}^{-1}

and

M(\text{T}_2) = 6\text{ g mol}^{-1}

, the ratio is

\sqrt{6 / 2} = \sqrt{3} \approx 1.73

. Thus, protium gas diffuses

1.73

times faster.

Step-by-Step Solution

1

Calculate the molar mass of diatomic protium gas (

\text{H}_2

) and tritium gas (

\text{T}_2

).

M(\text{H}_2) = 2 \times 1.0 = 2.0\text{ g mol}^{-1}

and

M(\text{T}_2) = 2 \times 3.0 = 6.0\text{ g mol}^{-1}

.

Graham's law of diffusion requires the molar masses of the gaseous species.

2

Apply Graham's Law of diffusion:

\frac{R_{\text{H}_2}}{R_{\text{T}_2}} = \sqrt{\frac{M_{\text{T}_2}}{M_{\text{H}_2}}}

.

\frac{R_{\text{H}_2}}{R_{\text{T}_2}} = \sqrt{\frac{6.0}{2.0}} = \sqrt{3.0}

.

The rate of effusion or diffusion of a gas is inversely proportional to the square root of its molar mass.

3

Evaluate the square root to determine the ratio.

\sqrt{3.0} \approx 1.73

.

This yields the relative diffusion rate of protium gas compared to tritium gas.

Key Concept

Graham's Law of Diffusion applied to Hydrogen Isotopes

Estimated Time:1m 15s

Question 12815Question →

A small object lies at the bottom of a transparent vessel containing two immiscible liquid layers, $A$ and $B$ . Layer $A$ (top) has a real thickness of $7.0\text{ cm}$ and a refractive index of $1.40$ . Layer $B$ (bottom) has a real thickness of $8.0\text{ cm}$ and a refractive index of $1.60$ . Calculate the apparent displacement of the object, in centimeters, when viewed vertically from directly above.

Show answer & explanation

Answer: 5

Answer

The apparent displacement of the object is

5.0\text{ cm}

.

For multiple refractive layers viewed normally, the total apparent depth is the sum of individual layer apparent depths:

\frac{7.0}{1.40} + \frac{8.0}{1.60} = 5.0 + 5.0 = 10.0\text{ cm}

. Subtracting this total apparent depth from the total real depth of

15.0\text{ cm}

gives an apparent vertical shift (displacement) of

5.0\text{ cm}

.

Step-by-Step Solution

1

Calculate the apparent depth of the top liquid layer (Layer A)

Apparent depth of Layer A =

5.0\text{ cm}

Apparent depth for a single medium is obtained by dividing real depth by its refractive index:

\frac{7.0}{1.40} = 5.0\text{ cm}

.

2

Calculate the apparent depth of the bottom liquid layer (Layer B)

Apparent depth of Layer B =

5.0\text{ cm}

Apparent depth for Layer B is

\frac{8.0}{1.60} = 5.0\text{ cm}

.

3

Calculate total real depth and total apparent depth

Total real depth =

15.0\text{ cm}

; Total apparent depth =

10.0\text{ cm}

Depths in composite media are additive.

4

Calculate vertical apparent displacement

Apparent displacement =

5.0\text{ cm}

Displacement is the difference between total real depth and total apparent depth:

15.0\text{ cm} - 10.0\text{ cm} = 5.0\text{ cm}

.

Key Concept

Refraction through composite media and vertical apparent displacement

Question 12816Question →

An electric water heater operating at a voltage of $240\,\text{V}$ has a heating element of resistance $48\,\Omega$ . It is used to heat $1.5\,\text{kg}$ of water from $20^\circ\text{C}$ to $100^\circ\text{C}$ . If the thermal efficiency of the heating process is $80\%$ , calculate the total electrical energy consumed by the heater in kilojoules ( $\text{kJ}$ ). [Take specific heat capacity of water = $4200\,\text{J}\cdot\text{kg}^{-1}\cdot\text{K}^{-1}$ ]

Show answer & explanation

Answer: 630

Answer

The total electrical energy consumed by the heater is

630\,\text{kJ}

.

The thermal energy required to raise the temperature of

1.5\,\text{kg}

of water by

80^\circ\text{C}

is

Q = 1.5 \times 4200 \times 80 = 504,000\,\text{J} = 504\,\text{kJ}

. Taking into account the

80\%

thermal efficiency, the total electrical energy consumed is

E_{\text{elec}} = \frac{504\,\text{kJ}}{0.80} = 630\,\text{kJ}

.

Step-by-Step Solution

1

Calculate the useful heat energy needed to heat the water.

Q = m c (T_2 - T_1) = 1.5 \times 4200 \times (100 - 20) = 504,000\,\text{J} = 504\,\text{kJ}

.

The thermal energy transferred to the water depends on its mass, specific heat capacity, and temperature increase.

2

Account for the efficiency of the heating element to find total electrical energy input.

E_{\text{elec}} = \frac{Q}{\text{Efficiency}} = \frac{504\,\text{kJ}}{0.80} = 630\,\text{kJ}

.

Since only

80\%

of the electrical energy is converted into useful heat energy for the water, the input electrical energy must be greater than the output heat energy.

Key Concept

Conversion of electrical energy to thermal energy and application of thermal efficiency.

Question 12817Question →

A hypermetropic (far-sighted) person has a near point located at a distance of $100\text{ cm}$ from the eye. What power of spectacle lens, in dioptres, is required to enable this person to read print comfortably held at the standard near point of $25\text{ cm}$ ?

+3.0\text{ D}

-3.0\text{ D}

+5.0\text{ D}

+0.75\text{ D}

Show answer & explanation

Answer: $+3.0\text{ D}$

Answer

+3.0\text{ D}

To correct hypermetropia, a converging (convex) lens is required to bend incoming rays so that an object placed at the standard near point of

25\text{ cm}

(

+0.25\text{ m}

) forms a virtual image at the defective eye's near point of

100\text{ cm}

(

-1.0\text{ m}

). Substituting

u = +0.25\text{ m}

and

v = -1.0\text{ m}

into the power formula

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

yields

P = +4.0\text{ D} - 1.0\text{ D} = +3.0\text{ D}

.

Step-by-Step Solution

1

Identify object distance (

u

) and required virtual image distance (

v

)

u = +25\text{ cm} = +0.25\text{ m}

and

v = -100\text{ cm} = -1.0\text{ m}

The lens must create a virtual image (on the same side as the object) at the person's actual near point when an object is placed at the standard reading distance.

2

Apply the thin lens formula to determine lens power

P

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

Power in dioptres is the reciprocal of the focal length in metres.

3

Calculate the numerical value of lens power

P = \frac{1}{0.25\text{ m}} + \frac{1}{-1.0\text{ m}} = +4.0\text{ D} - 1.0\text{ D} = +3.0\text{ D}

Adding the reciprocal quantities yields a positive focal power of

+3.0\text{ D}

.

Key Concept

Correction of Hypermetropia using Converging Lenses

Estimated Time:1m 30s

Question 12818Question →

The mass defect of a nitrogen nucleus $^{14}_{7}\text{N}$ is calculated to be $0.112\text{ u}$ . Taking $1\text{ u} = 931.5\text{ MeV}$ , what is the total binding energy of the nucleus in $\text{MeV}$ ?

Show answer & explanation

Answer: 104.328

Answer

The total binding energy of the

^{14}_{7}\text{N}

nucleus is

104.328\text{ MeV}

.

The total binding energy of a nucleus is equal to the mass defect multiplied by the energy equivalent of one atomic mass unit (

931.5\text{ MeV/u}

). Calculating

0.112\text{ u} \times 931.5\text{ MeV/u}

gives

104.328\text{ MeV}

.

Step-by-Step Solution

1

Apply the binding energy formula

E_b = \Delta m \times 931.5\text{ MeV/u}

.

E_b = 0.112\text{ u} \times 931.5\text{ MeV/u} = 104.328\text{ MeV}

.

The binding energy is obtained by converting the mass defect directly into its energy equivalent.

Key Concept

Mass Defect and Binding Energy Conversion

Question 12819Question →

Match each optical instrument listed on the left with its corresponding lens configuration and image characteristics on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Astronomical Telescope pairs with the configuration having

f_o > f_e

forming an image at infinity; Simple Microscope pairs with a single converging lens forming an erect virtual image; Compound Microscope pairs with two converging lenses having

f_o < f_e

; and Projection Lantern pairs with a converging lens forming a real, inverted image on a screen.

Each instrument matches its distinct optical construction: telescopes use

f_o > f_e

for distant viewing at infinity, simple microscopes use a single convex lens for virtual magnifying, compound microscopes use

f_o < f_e

for double magnification of tiny objects, and projectors use a single convex lens to cast real images onto a distant surface.

Step-by-Step Solution

1

Determine the lens setup and final image position of an astronomical telescope in normal adjustment.

The objective has a larger focal length than the eyepiece (

f_o > f_e

), and the final image is formed at infinity.

Telescopes gather light from distant objects, requiring a larger objective focal length for high angular magnification and comfortable viewing at infinity.

2

Determine the configuration of a simple microscope.

It consists of a single convex lens producing an erect, virtual, and magnified image.

When an object is placed within the focal length of a single convex lens, it acts as a magnifying glass.

3

Determine the focal length relationship of a compound microscope.

It uses two convex lenses where the objective focal length is shorter than the eyepiece focal length (

f_o < f_e

).

A very short objective focal length maximizes linear magnification of small, near objects before the eyepiece further magnifies the intermediate image.

4

Determine the type of image produced by a projection lantern (slide projector).

It forms a real, inverted, and magnified image on a screen.

Projecting images onto a screen requires a real image formed by a converging lens.

Key Concept

Optical Instrument Lens Configurations and Image Properties

Question 12820Question →

An object is placed at a distance $u$ in front of a concave mirror of focal length $12\text{ cm}$ . A plane mirror is placed perpendicular to the principal axis at a distance of $32\text{ cm}$ in front of the concave mirror, between the object and the concave mirror. If the real image formed by the concave mirror coincides in space with the virtual image formed by the plane mirror, what is the value of $u$ in centimeters?

Show answer & explanation

Answer: 48

Answer

The correct object distance

u

is

48\text{ cm}

.

The object is located at distance

u

from the concave mirror. With the plane mirror at

32\text{ cm}

from the concave mirror, the object distance from the plane mirror is

u - 32

. The plane mirror forms an image at distance

u - 32

behind itself, which corresponds to

32 - (u - 32) = 64 - u

from the concave mirror. Setting

v = 64 - u

in the mirror formula

\frac{1}{12} = \frac{1}{u} + \frac{1}{64 - u}

gives

u^2 - 64u + 768 = 0

. Factoring yields

u = 48\text{ cm}

or

u = 16\text{ cm}

. Because the plane mirror is between the object and the concave mirror,

u > 32\text{ cm}

, so

u = 48\text{ cm}

.

Step-by-Step Solution

1

Find the position of the image formed by the plane mirror in terms of

u

.

The object is at a distance

(u - 32)\text{ cm}

in front of the plane mirror. Its virtual image is formed

(u - 32)\text{ cm}

behind the plane mirror, which places it at

32 - (u - 32) = (64 - u)\text{ cm}

in front of the concave mirror.

A plane mirror forms an image behind it at a distance equal to the object distance in front of it.

2

Equate the image distance of the concave mirror

v

to the position of the plane mirror image.

v = 64 - u

The question states that the image formed by the concave mirror coincides in position with the image formed by the plane mirror.

3

Substitute

f = 12\text{ cm}

and

v = 64 - u

into the mirror equation.

\frac{1}{12} = \frac{1}{u} + \frac{1}{64 - u}

The standard mirror formula relates focal length, object distance, and image distance.

4

Solve the algebraic equation for

u

.

\frac{1}{12} = \frac{(64 - u) + u}{u(64 - u)} \implies 64u - u^2 = 768 \implies u^2 - 64u + 768 = 0

Combining fractions and multiplying across gives a quadratic equation in standard form.

5

Factor the quadratic equation and select the physical root.

(u - 48)(u - 16) = 0 \implies u = 48\text{ cm}

or

u = 16\text{ cm}

. Since

u > 32\text{ cm}

,

u = 48\text{ cm}

.

The plane mirror is situated between the object and the concave mirror at

32\text{ cm}

, so the object distance

u

must be greater than

32\text{ cm}

.

Key Concept

Image coincidence in combined plane and curved optical systems

Estimated Time:3m 0s

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