All practice questions

Question 8081Question →

In the froth flotation process used for concentrating low-grade copper sulfide ores such as chalcopyrite ( $CuFeS_2$ ), pine oil and sodium ethyl xanthate are added to the aqueous suspension of the pulverized ore. What specific roles do pine oil and sodium ethyl xanthate play in this metallurgical separation process?

Pine oil acts as a frothing agent to produce stable foam, while sodium ethyl xanthate acts as a collector to render sulfide mineral particles hydrophobic.

Pine oil acts as a chemical reducing agent to convert copper sulfide into elemental copper, while sodium ethyl xanthate acts as a flux to convert gangue into slag.

Pine oil preferentially dissolves silicate gangue impurities, while sodium ethyl xanthate causes copper ions to precipitate via selective electrolytic discharge.

Pine oil neutralizes amphoteric metal oxides in the matrix, while sodium ethyl xanthate oxidizes the sulfide ions to elemental sulfur.

Show answer & explanation

Answer: Pine oil acts as a frothing agent to produce stable foam, while sodium ethyl xanthate acts as a collector to render sulfide mineral particles hydrophobic.

Answer

Pine oil serves as the frothing agent to generate stable air bubbles, while sodium ethyl xanthate functions as a collector that selectively coats sulfide ore particles to make them hydrophobic.

In froth flotation concentration of sulfide ores, pine oil acts as a frothing agent that creates a stable foam matrix when air is agitated through the ore slurry. Sodium ethyl xanthate serves as a collector reagent; its molecules selectively adsorb onto the surface of sulfide mineral particles, turning them hydrophobic. The water-repellent sulfide particles attach to the rising froth bubbles and are skimmed off at the top, leaving water-wetted silicate gangue behind.

Step-by-Step Solution

1

Identify the purpose of the froth flotation method in metallurgy.

Froth flotation is a physical concentration process used specifically for sulfide ores to separate valuable mineral particles from earthy gangue based on surface wettability.

Sulfide ore particles are naturally or artificially hydrophobic (water-repellent), whereas silicate gangue is hydrophilic (water-attracting).

2

Analyze the function of pine oil in the process.

Pine oil acts as a frother.

It lowers the surface tension of water, enabling the formation of stable oil-coated air bubbles when air is blown into the suspension.

3

Analyze the function of sodium ethyl xanthate in the process.

Sodium ethyl xanthate acts as a collector (polar-nonpolar surfactant).

Its polar group attaches to the sulfide mineral surface while its nonpolar hydrocarbon tail points outward, rendering the mineral particle strongly hydrophobic so it attaches to rising air bubbles.

Key Concept

Froth Flotation Principle and Additive Roles in Ore Concentration

Question 8082Question →

A converging lens of focal length $20\text{ cm}$ is placed in thin coaxial contact with a diverging lens of focal length $50\text{ cm}$ . An object is placed $30\text{ cm}$ in front of this lens combination. What is the position and nature of the final image formed?

300\text{ cm}

in front of the combination (virtual image)

60\text{ cm}

behind the combination (real image)

27.3\text{ cm}

behind the combination (real image)

15.8\text{ cm}

behind the combination (real image)

Show answer & explanation

Answer: $300\text{ cm}$ in front of the combination (virtual image)

Answer

The image is virtual and formed

300\text{ cm}

in front of the lens combination.

Combining a converging lens (

f = +20\text{ cm}

) and a diverging lens (

f = -50\text{ cm}

) yields an effective focal length of

F = +\frac{100}{3}\text{ cm}

. Using the lens formula

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

with object distance

u = 30\text{ cm}

gives

\frac{1}{v} = \frac{3}{100} - \frac{1}{30} = -\frac{1}{300}\text{ cm}^{-1}

, resulting in

v = -300\text{ cm}

. The negative sign confirms the image is virtual and located

300\text{ cm}

in front of the combination.

Step-by-Step Solution

1

Calculate the effective focal length (

F

) of the two lenses in contact.

\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{20\text{ cm}} + \frac{1}{-50\text{ cm}} = \frac{5 - 2}{100\text{ cm}} = \frac{3}{100}\text{ cm}^{-1}

, so

F = +\frac{100}{3}\text{ cm}

.

Thin lenses in contact combine algebraically according to their optical powers, taking signs into account (positive for converging, negative for diverging).

2

Apply the lens formula to find the image distance (

v

).

\frac{1}{F} = \frac{1}{u} + \frac{1}{v} \implies \frac{3}{100} = \frac{1}{30} + \frac{1}{v} \implies \frac{1}{v} = \frac{3}{100} - \frac{1}{30} = \frac{9 - 10}{300} = -\frac{1}{300}\text{ cm}^{-1}

.

Rearranging the thin lens equation allows us to solve for the image distance

v

given object distance

u = 30\text{ cm}

.

3

Interpret the sign and magnitude of

v

.

v = -300\text{ cm}

, which signifies a virtual image located

300\text{ cm}

in front of the lens combination (on the object side).

A negative image distance in the standard real-is-positive convention denotes a virtual image.

Key Concept

Combination of thin lenses in contact and lens sign conventions

Estimated Time:2m 0s

Question 8083Question →

Heavy water is water in which the hydrogen atoms are replaced by the hydrogen isotope deuterium ( $^{2}_{1}\text{H}$ or $\text{D}$ ). Given that the atomic mass of deuterium is $2\text{ g mol}^{-1}$ and oxygen ( $^{16}_{8}\text{O}$ ) is $16\text{ g mol}^{-1}$ , what is the molar mass of heavy water ( $\text{D}_2\text{O}$ ) in $\text{g mol}^{-1}$ ?

Show answer & explanation

Answer: 20

Answer

The molar mass of heavy water (

\text{D}_2\text{O}

) is

20\text{ g mol}^{-1}

.

Heavy water (

\text{D}_2\text{O}

) contains two atoms of deuterium (

^2\text{H}

, atomic mass =

2

) and one atom of oxygen (

^{16}\text{O}

, atomic mass =

16

). The molar mass is calculated as

(2 \times 2) + 16 = 20\text{ g mol}^{-1}

.

Step-by-Step Solution

1

Determine the molecular composition of heavy water

Heavy water (

\text{D}_2\text{O}

) contains 2 deuterium atoms and 1 oxygen atom.

Deuterium is an isotope of hydrogen containing one proton and one neutron, giving it a mass number of 2.

2

Sum the relative atomic masses of all constituent atoms

(2 \times 2) + 16 = 20\text{ g mol}^{-1}

Multiplying the mass of deuterium by two and adding the mass of one oxygen atom yields the molar mass of the compound.

Key Concept

Heavy Water and Deuterium Molar Mass

Question 8084Question →

A solid wooden block of density $600\text{ kg/m}^3$ and volume $4.0 \times 10^{-3}\text{ m}^3$ floats in water of density $1000\text{ kg/m}^3$ . A metal block is placed on top of the wooden block so that the wooden block is just completely submerged while the metal block remains entirely above the water surface. What is the mass of the metal block? (Take $g = 10\text{ m/s}^2$ )

1.6\text{ kg}

2.4\text{ kg}

4.0\text{ kg}

6.4\text{ kg}

Show answer & explanation

Answer: $1.6\text{ kg}$

Answer

1.6\text{ kg}

When the wooden block is completely submerged, it displaces

4.0 \times 10^{-3}\text{ m}^3

of water, creating an upthrust of

40\text{ N}

. The weight of the wooden block is

24\text{ N}

. For equilibrium, the total downward weight must equal the upthrust (

24\text{ N} + W_{\text{metal}} = 40\text{ N}

), which gives

W_{\text{metal}} = 16\text{ N}

and a corresponding mass of

1.6\text{ kg}

.

Step-by-Step Solution

1

Calculate the mass and weight of the wooden block.

m_{\text{wood}} = \rho_{\text{wood}} \times V_{\text{wood}} = 600\text{ kg/m}^3 \times 4.0 \times 10^{-3}\text{ m}^3 = 2.4\text{ kg}

, so

W_{\text{wood}} = 2.4 \times 10 = 24\text{ N}

.

The weight of the wood contributes to the total downward force of the floating system.

2

Calculate the total upthrust exerted by the water when the wooden block is completely submerged.

U = \rho_{\text{water}} \times V_{\text{wood}} \times g = 1000\text{ kg/m}^3 \times 4.0 \times 10^{-3}\text{ m}^3 \times 10\text{ m/s}^2 = 40\text{ N}

.

By Archimedes' principle, upthrust equals the weight of the displaced liquid.

3

Apply the law of flotation to solve for the weight and mass of the metal block.

W_{\text{wood}} + W_{\text{metal}} = U \implies 24\text{ N} + W_{\text{metal}} = 40\text{ N} \implies W_{\text{metal}} = 16\text{ N}

. Thus,

m_{\text{metal}} = \frac{16\text{ N}}{10\text{ m/s}^2} = 1.6\text{ kg}

.

For the system to float in equilibrium just submerged, total downward weight must equal total upward buoyant force.

Key Concept

Archimedes' Principle and Law of Flotation

Estimated Time:2m 0s

Question 8085Question →

A vertical light spring stretches by $0.10\text{ m}$ when a block is suspended from it in equilibrium. The block is then pulled down an additional $0.05\text{ m}$ from its equilibrium position and released from rest to undergo simple harmonic motion. Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the speed of the block when it is at a displacement of $0.03\text{ m}$ from its equilibrium position?

0.40\text{ m/s}

0.50\text{ m/s}

0.30\text{ m/s}

0.04\text{ m/s}

Show answer & explanation

Answer: $0.40\text{ m/s}$

Answer

0.40\text{ m/s}

The system's angular frequency

\omega

is determined by the equilibrium extension

e

using

\omega = \sqrt{\frac{g}{e}} = 10\text{ rad/s}

. Combining this with the amplitude

A = 0.05\text{ m}

in the SHM speed relation

v = \omega \sqrt{A^2 - y^2}

at

y = 0.03\text{ m}

yields

v = 10 \sqrt{(0.05)^2 - (0.03)^2} = 0.40\text{ m/s}

.

Step-by-Step Solution

1

Calculate the angular frequency of the mass-spring system using static equilibrium conditions.

\omega = 10\text{ rad/s}

At equilibrium,

mg = ke \implies \frac{k}{m} = \frac{g}{e}

. Therefore,

\omega = \sqrt{\frac{g}{e}} = \sqrt{\frac{10}{0.10}} = 10\text{ rad/s}

.

2

Identify the amplitude of simple harmonic motion.

A = 0.05\text{ m}

The initial displacement from the equilibrium position when released from rest defines the amplitude of oscillation.

3

Apply the SHM velocity-displacement formula

v = \omega \sqrt{A^2 - y^2}

at

y = 0.03\text{ m}

.

v = 0.40\text{ m/s}

v = 10 \times \sqrt{(0.05)^2 - (0.03)^2} = 10 \times \sqrt{0.0025 - 0.0009} = 10 \times 0.04 = 0.40\text{ m/s}

.

Key Concept

Relating static extension to angular frequency and calculating instantaneous speed in Simple Harmonic Motion

Question 8086Question →

Consider the cumulated diene compound penta-1,2-diene, represented by the condensed structure $\text{CH}_2=\text{C}=\text{CH}-\text{CH}_2-\text{CH}_3$ . What are the hybridization states of the central carbon atom (C2) and the methylene carbon atom (C4), respectively?

sp

and

sp^3

sp^2

and

sp^3

sp

and

sp^2

sp^2

and

sp^2

Show answer & explanation

Answer: $sp$ and $sp^3$

Answer

The central carbon atom (C2) is

sp

hybridized and the methylene carbon atom (C4) is

sp^3

hybridized.

In penta-1,2-diene, the central allene carbon atom at position 2 (C2) forms two double bonds (

2\ \sigma

bonds and

2\ \pi

bonds), which necessitates

sp

hybridization. The methylene carbon atom at position 4 (C4) forms four single

\sigma

bonds with two hydrogen atoms and two carbon atoms, which corresponds to

sp^3

tetrahedral hybridization.

Step-by-Step Solution

1

Determine the number of

\sigma

and

\pi

bonds on carbon-2 (C2).

C2 forms two double bonds, which consists of

2\ \sigma

bonds and

2\ \pi

bonds.

Carbon atoms involved in two double bonds (cumulated dienes/allenes) use two

sp

hybrid orbitals to form

\sigma

bonds at an angle of

180^\circ

.

2

Determine the hybridization state of C2.

C2 is

sp

hybridized with linear geometry.

Two

\sigma

bonding domains correlate to

sp

hybridization.

3

Determine the bonding domains and hybridization of carbon-4 (C4).

C4 is bonded to two hydrogen atoms, C3, and C5 via single covalent bonds, giving

4\ \sigma

bonds.

Four single

\sigma

bonding domains require

sp^3

hybridization with tetrahedral geometry.

Key Concept

Hybridization in Cumulated Dienes and Saturated Carbons

Estimated Time:1m 30s

Question 8087Question →

In an industrial plant, dry air is cooled to $-200^\circ\text{C}$ to form liquid air after removing carbon dioxide and water vapor. When this liquid air undergoes fractional distillation in a fractionating column as its temperature is slowly raised, which statement correctly identifies the constituent that boils off first and the rationale behind its separation?

Nitrogen boils off first at

-196^\circ\text{C}

because it has a lower boiling point than argon and oxygen.

Oxygen boils off first at

-183^\circ\text{C}

because

-183^\circ\text{C}

represents a higher numerical boiling point than

-196^\circ\text{C}

.

Carbon dioxide boils off first as a gas because it is the lightest component remaining in the liquid mixture.

Argon boils off first at

-186^\circ\text{C}

because unreactive noble gases always distill before diatomic gases.

Show answer & explanation

Answer: Nitrogen boils off first at $-196^\circ\text{C}$ because it has a lower boiling point than argon and oxygen.

Answer

Nitrogen boils off first at

-196^\circ\text{C}

because it has the lowest boiling point among the main liquefied atmospheric constituents.

During the fractional distillation of liquid air, components separate according to their boiling points. As the column temperature warms up from

-200^\circ\text{C}

, nitrogen (boiling point

-196^\circ\text{C}

) reaches its boiling point first, vaporizes, and is collected at the top of the fractionating column.

Step-by-Step Solution

1

Identify the main liquefied components of air after pre-purification

Purified liquid air consists primarily of liquid nitrogen, argon, and liquid oxygen. Water vapor and carbon dioxide are removed prior to liquefaction to prevent clogging the equipment.

Water freezes at 0°C and carbon dioxide sublimes at -78.5°C, making them solids well above the liquefaction temperature of air (-200°C).

2

Compare the boiling points of the liquefied components

Nitrogen boils at

-196^\circ\text{C}

(77 K), Argon boils at

-186^\circ\text{C}

(87 K), and Oxygen boils at

-183^\circ\text{C}

(90 K).

Lower boiling point values on the Celsius scale correspond to colder temperatures (e.g., -196°C is colder than -183°C).

3

Determine the distillation sequence as temperature rises from

-200^\circ\text{C}

Nitrogen reaches its boiling point first at

-196^\circ\text{C}

and turns into gas, leaving argon and oxygen behind until further warming.

In fractional distillation, the liquid with the lowest boiling point boils off first at the top of the column.

Key Concept

Fractional Distillation of Liquid Air

Estimated Time:1m 30s

Question 8088Question →

A spherical black body of radius $r$ at an initial temperature of $27^\circ\text{C}$ emits thermal radiation at a rate of $W$ . If the radius of the sphere is doubled and its temperature is increased to $327^\circ\text{C}$ , what is the new rate of thermal radiation emitted by the sphere in terms of $W$ ?

64W

32W

16W

8W

Show answer & explanation

Answer: $64W$

Answer

The new rate of thermal radiation emitted by the sphere is

64W

.

According to the Stefan-Boltzmann law, the rate of energy radiation from a black body is given by

P = \sigma A T^4

. Doubling the radius of a sphere increases its surface area by a factor of

2^2 = 4

. Converting temperatures from Celsius to Kelvin gives

T_1 = 27 + 273 = 300\text{ K}

and

T_2 = 327 + 273 = 600\text{ K}

, showing that the absolute temperature doubles (

T_2 / T_1 = 2

). Raising this temperature ratio to the fourth power yields

2^4 = 16

. Combining the area factor of

4

and the temperature factor of

16

results in a total radiation rate increase of

4 \times 16 = 64

times the original rate

W

.

Step-by-Step Solution

1

Express initial radiation rate using Stefan-Boltzmann law and sphere surface area formula

The total power radiated by a black body is given by

P = \sigma A T^4

. For a sphere of radius

r

,

A_1 = 4\pi r^2

. Absolute temperature

T_1 = 27 + 273 = 300\text{ K}

. Thus,

W = \sigma (4\pi r^2) (300)^4

.

Stefan's law requires absolute temperature in Kelvin and total surface area of the radiator.

2

Determine the scaled surface area and absolute temperature for the final state

New radius

r_2 = 2r \implies A_2 = 4\pi (2r)^2 = 4 A_1

. New temperature

T_2 = 327 + 273 = 600\text{ K} = 2 T_1

.

Surface area of a sphere scales quadratically with radius, and temperatures must be converted to Kelvin.

3

Calculate the ratio of the new radiation rate to the initial radiation rate

\frac{P_2}{P_1} = \frac{A_2}{A_1} \times \left(\frac{T_2}{T_1}\right)^4 = 4 \times (2)^4 = 4 \times 16 = 64

.

Radiated power is directly proportional to surface area and to the fourth power of absolute temperature.

4

State the new power in terms of

W

P_2 = 64 W

.

Multiplying the initial rate

W

by the overall scaling factor of 64 gives the final answer.

Key Concept

Stefan-Boltzmann Law of Radiation (

P = \epsilon \sigma A T^4

)

Question 8089Question →

Match each chemical species or system on the left with its correct theoretical acid-base role or behavior on the right according to the Arrhenius, Brønsted-Lowry, or Lewis concepts.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are:

BF_3

matches Lewis acid (electron-pair acceptor);

HSO_4^-

acting to produce

H_3O^+

matches Brønsted-Lowry acid (proton donor);

HCO_3^-

matches amphoteric/amphiprotic species; and

NH_2^-

from liquid ammonia matches conjugate base in a non-aqueous solvent system.

Each species is correctly categorized based on fundamental acid-base definitions: Lewis theory accounts for electron-pair transfer (

BF_3

), Brønsted-Lowry theory accounts for proton donor/acceptor roles (

HSO_4^-

and

HCO_3^-

), and solvent self-ionization describes non-aqueous acid-base equilibria (

NH_2^-

in liquid

NH_3

).

Step-by-Step Solution

1

Analyze

BF_3

in

BF_3 + :NH_3 \rightarrow F_3B:NH_3

Boron has six valence electrons and accepts an electron pair from nitrogen.

According to Lewis theory, an electron-pair acceptor is defined as a Lewis acid.

2

Analyze

HSO_4^-

converting to

SO_4^{2-}

in water

HSO_4^-

transfers a proton (

H^+

) to

H_2O

to form

H_3O^+

.

According to Brønsted-Lowry theory, a proton donor is an acid.

3

Analyze

HCO_3^-

double-behavior in water

HCO_3^-

can accept

H^+

to become

H_2CO_3

or donate

H^+

to become

CO_3^{2-}

.

Species that can act as either a proton donor or proton acceptor are termed amphiprotic/amphoteric.

4

Analyze

NH_2^-

in liquid ammonia auto-ionization

Ammonia undergoes auto-protolysis:

2NH_3 \rightleftharpoons NH_4^+ + NH_2^-

.

The amide ion (

NH_2^-

) is formed when

NH_3

loses a proton, acting as the characteristic conjugate base of the liquid ammonia solvent system.

Key Concept

Distinction and application of Arrhenius, Brønsted-Lowry, Lewis, and solvent-system theories of acids and bases.

Question 8090Question →

A particle of mass $0.20\text{ kg}$ executes simple harmonic motion along a straight line. When its displacement from the equilibrium position is $0.03\text{ m}$ , its speed is $0.16\text{ m/s}$ . When its displacement is $0.04\text{ m}$ , its speed is $0.12\text{ m/s}$ . What is the total mechanical energy of the particle in millijoules ( $\text{mJ}$ )?

Show answer & explanation

Answer: 4

Answer

The total mechanical energy of the particle is

4\text{ mJ}

.

Using the relation

v^2 = \omega^2(A^2 - x^2)

for the two given state points

(0.03\text{ m}, 0.16\text{ m/s})

and

(0.04\text{ m}, 0.12\text{ m/s})

forms a set of simultaneous equations. Subtracting them yields

\omega^2 = 16\text{ rad}^2/\text{s}^2

, leading to

A^2 = 0.0025\text{ m}^2

. Substituting these values into

E = \frac{1}{2}m\omega^2 A^2

gives

E = 0.004\text{ J}

, which converts to

4\text{ mJ}

.

Step-by-Step Solution

1

Set up kinematic equations for both displacement points using

v^2 = \omega^2(A^2 - x^2)

.

0.0256 = \omega^2(A^2 - 0.0009)

and

0.0144 = \omega^2(A^2 - 0.0016)

.

The equation relates linear speed, angular frequency, amplitude, and instantaneous displacement in SHM.

2

Subtract the two simultaneous equations to eliminate

A^2

and find

\omega^2

.

0.0112 = 0.0007\omega^2 \implies \omega^2 = 16\text{ rad}^2/\text{s}^2

.

Eliminating amplitude isolates the angular frequency squared.

3

Determine

A^2

by substituting

\omega^2 = 16

back into one of the state equations.

A^2 = 0.0025\text{ m}^2 \implies A = 0.05\text{ m}

.

Amplitude is required to calculate the maximum potential or total mechanical energy.

4

Calculate total mechanical energy

E = \frac{1}{2}m\omega^2 A^2

and convert to millijoules.

E = \frac{1}{2} \times 0.20 \times 16 \times 0.0025 = 0.004\text{ J} = 4\text{ mJ}

.

Total energy in SHM is constant and proportional to mass, square of angular frequency, and square of amplitude.

Key Concept

Conservation of energy and phase-space relationship between velocity and displacement in simple harmonic motion.

Question 8091Question →

A small spherical lead shot of radius $3.0 \times 10^{-3}\text{ m}$ and density $8000\text{ kg/m}^3$ falls vertically through a viscous oil of density $800\text{ kg/m}^3$ and dynamic viscosity $0.18\text{ Pa}\cdot\text{s}$ . What is the magnitude of the terminal velocity of the sphere in $\text{m/s}$ ? (Take $g = 10\text{ m/s}^2$ )

Show answer & explanation

Answer: 0.8

Answer

The terminal velocity of the lead shot is

0.8\text{ m/s}

.

At terminal velocity, acceleration is zero because the upward forces (viscous drag

6\pi \eta r v_t

plus buoyant upthrust

\frac{4}{3}\pi r^3 \rho_f g

) completely balance the downward weight of the sphere (

\frac{4}{3}\pi r^3 \rho_s g

). Solving for velocity gives

v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta} = 0.8\text{ m/s}

.

Step-by-Step Solution

1

Formulate the force balance equation at terminal velocity.

Weight (

W

) = Upthrust (

U

) + Viscous drag (

F_v

), which simplifies to

v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}

.

When terminal velocity is attained, the net acceleration of the sphere is zero, so upward forces balance downward force.

2

Substitute the physical parameters into the terminal velocity formula.

v_t = \frac{2 \times (3.0 \times 10^{-3})^2 \times (8000 - 800) \times 10}{9 \times 0.18} = 0.8\text{ m/s}

.

Direct calculation using Stokes' law and Archimedes' principle.

Key Concept

Terminal Velocity and Stokes' Law in a Viscous Medium

Question 8092Question →

The boiling point of a liquid decreases when the external atmospheric pressure acting on its surface is reduced.

True

False

Show answer & explanation

Answer: True

Answer

The statement is True.

Boiling occurs when the saturated vapour pressure of a liquid equals the external atmospheric pressure. Decreasing the external atmospheric pressure reduces the saturated vapour pressure threshold required for boiling, allowing the liquid to boil at a lower temperature.

Step-by-Step Solution

1

Identify the condition required for boiling to occur.

Boiling takes place when the saturated vapour pressure of the liquid equals the external atmospheric pressure.

This is the fundamental physical definition of boiling.

2

Analyze the effect of reducing external atmospheric pressure.

Lower external pressure means the required saturated vapour pressure is reached at a lower temperature.

Saturated vapour pressure increases with temperature, so a lower target pressure corresponds to a lower boiling temperature.

Key Concept

Dependence of Boiling Point on External Atmospheric Pressure

Question 8093Question →

A uniform horizontal rod $AB$ of length $2.0\text{ m}$ and mass $6.0\text{ kg}$ is suspended horizontally by two light vertical strings attached at end $A$ and at a point $C$ located $0.5\text{ m}$ from end $B$ . Taking $g = 10\text{ m/s}^2$ , what is the tension in the string at point $C$ in Newtons?

Show answer & explanation

Answer: 40

Answer

The tension in the string at point C is 40 N.

The weight of the rod (

60\text{ N}

) acts at its midpoint (

1.0\text{ m}

from end

A

). Point

C

is located

1.5\text{ m}

from end

A

. Taking moments about end

A

gives

60\text{ N} \times 1.0\text{ m} = T_C \times 1.5\text{ m}

, which evaluates to

T_C = 40\text{ N}

.

Step-by-Step Solution

1

Calculate total weight and identify the center of gravity position

Weight

W = 60\text{ N}

acting at

1.0\text{ m}

from end

A

For a uniform rod, the weight acts vertically downwards at its geometric center.

2

Set up the moment equilibrium equation taking end A as pivot

60\text{ N} \times 1.0\text{ m} = T_C \times 1.5\text{ m}

Taking moments about point

A

eliminates the force at

A

and equates clockwise moment from weight to counter-clockwise moment from tension at

C

.

3

Solve for the tension force at point C

T_C = 40\text{ N}

Dividing the total moment of

60\text{ N}\cdot\text{m}

by the moment arm of

1.5\text{ m}

yields

40\text{ N}

.

Key Concept

Principle of Moments and Rotational Equilibrium

Estimated Time:1m 30s

Question 8094Question →

A sample of a radioactive isotope has an initial activity of $8000\text{ Bq}$ . After an elapsed time of $18\text{ minutes}$ , its activity reduces to $1000\text{ Bq}$ . Calculate the decay constant $\lambda$ of the isotope in $\text{min}^{-1}$ .

Show answer & explanation

Answer: 0.1155

Answer

The decay constant of the radioactive isotope is

0.1155\text{ min}^{-1}

.

The activity decreases from

8000\text{ Bq}

to

1000\text{ Bq}

, which is a reduction to

\frac{1}{8}

of its initial value. Since

\left(\frac{1}{2}\right)^3 = \frac{1}{8}

, exactly

3

half-lives have elapsed in

18\text{ minutes}

, meaning

T_{1/2} = 6\text{ minutes}

. Using the relationship

\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.69315}{6\text{ min}}

, we obtain

\lambda \approx 0.1155\text{ min}^{-1}

.

Step-by-Step Solution

1

Determine the number of elapsed half-lives from the activity reduction ratio.

The fraction of remaining activity is

\frac{A}{A_0} = \frac{1000}{8000} = \frac{1}{8} = \left(\frac{1}{2}\right)^3

, giving

n = 3

half-lives.

Radioactive decay follows the relation

A = A_0 (1/2)^n

.

2

Determine the half-life

T_{1/2}

of the isotope.

T_{1/2} = \frac{t}{n} = \frac{18\text{ min}}{3} = 6\text{ minutes}

.

Total elapsed time is equal to the number of half-lives multiplied by the duration of one half-life.

3

Compute the decay constant

\lambda

in

\text{min}^{-1}

.

\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.69315}{6\text{ min}} = 0.1155\text{ min}^{-1}

.

The decay constant is fundamental to decay rate and related to half-life via

\lambda = \frac{\ln 2}{T_{1/2}}

.

Key Concept

Radioactive Decay Law and Decay Constant

Estimated Time:2m 0s

Question 8095Question →

Match each chemical reaction or behavior in Column A with the acid-base theory in Column B that uniquely or best explains it.

Click a left item, then click its matching right item

Show answer & explanation

Answer

1. Dissociation yielding H+ as the only positive ion matches Arrhenius Theory. 2. Proton transfer forming conjugate base SO42- matches Brønsted-Lowry Theory. 3. Electron-pair donation forming a dative bond with AlCl3 matches Lewis Theory.

Each statement matches its respective historical acid-base theory based on foundational criteria: Arrhenius requires aqueous H+ generation, Brønsted-Lowry centers on proton transfer and conjugate species, and Lewis broadens the scope to electron-pair donation and coordinate bond formation.

Step-by-Step Solution

1

Analyze the first item regarding HCl yielding H+ as the only positive ion in water.

Identified as Arrhenius Theory.

Arrhenius strictly defined acids by their ability to ionize in water to yield hydrogen ions as sole positive ions.

2

Analyze the second item involving HSO4- donating a proton to H2O to generate SO42-.

Identified as Brønsted-Lowry Theory.

Brønsted-Lowry focuses on proton transfer, where the donor is the acid and the resulting species is its conjugate base.

3

Analyze the third item involving NH3 donating a lone pair of electrons to AlCl3 in a non-protic coordinate covalent bond context.

Identified as Lewis Theory.

Lewis theory encompasses electron pair transfer, defining electron pair donors as bases and acceptors as acids.

Key Concept

Definitions and Theories of Acids and Bases (Arrhenius, Brønsted-Lowry, and Lewis)

Estimated Time:1m 0s

Question 8096Question →

Consider the following thermochemical reactions:

\text{S}(s) + \text{O}_2(g) \rightarrow \text{SO}_2(g) \quad \Delta H = -297\text{ kJ mol}^{-1}

2\text{SO}_2(g) + \text{O}_2(g) \rightarrow 2\text{SO}_3(g) \quad \Delta H = -198\text{ kJ mol}^{-1}

What is the standard enthalpy change for the reaction represented by the equation below?

2\text{S}(s) + 3\text{O}_2(g) \rightarrow 2\text{SO}_3(g)

-792\text{ kJ mol}^{-1}

-495\text{ kJ mol}^{-1}

-396\text{ kJ mol}^{-1}

+792\text{ kJ mol}^{-1}

Show answer & explanation

Answer: $-792\text{ kJ mol}^{-1}$

Answer

-792\text{ kJ mol}^{-1}

According to Hess's law, the total enthalpy change is the sum of the enthalpy changes for each step. Multiplying the first equation by 2 yields an enthalpy change of

-594\text{ kJ mol}^{-1}

. Adding the second equation directly (

\Delta H = -198\text{ kJ mol}^{-1}

) gives a combined enthalpy change of

-594 + (-198) = -792\text{ kJ mol}^{-1}

.

Step-by-Step Solution

1

Multiply the first thermochemical equation by 2 to match the 2 moles of sulfur in the target equation.

2\text{S}(s) + 2\text{O}_2(g) \rightarrow 2\text{SO}_2(g) \quad \Delta H_1 = 2 \times (-297\text{ kJ mol}^{-1}) = -594\text{ kJ mol}^{-1}

The stoichiometric coefficient of solid sulfur in the target reaction is 2.

2

Add the modified first equation to the second equation.

2\text{S}(s) + 3\text{O}_2(g) \rightarrow 2\text{SO}_3(g)

Intermediate sulfur dioxide (

2\text{SO}_2

) on opposite sides of the combined reaction cancels out.

3

Sum the enthalpy changes for both steps according to Hess's law.

\Delta H = -594\text{ kJ mol}^{-1} + (-198\text{ kJ mol}^{-1}) = -792\text{ kJ mol}^{-1}

Hess's law states that the net enthalpy change of a reaction is the sum of the enthalpy changes of its constituent steps.

Key Concept

Hess's Law of Constant Heat Summation

Question 8097Question →

Match each chemical industry operation in Nigeria with its primary siting factor and chemical raw material requirement.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings align the petrochemical urea plant with natural gas fields (feedstock and fuel), the glass factory with silica sand deposits (bulk solid reduction), the chlor-alkali facility with brine deposits and power supply (electrolysis requirements), and the mini-steel mill with urban scrap markets (market-oriented secondary metallurgy).

The matches correctly link each industry to its primary chemical input and siting logic: urea synthesis requires natural gas feedstock, glass manufacturing requires heavy silica sand deposits, chlor-alkali requires brine and power for electrolysis, and secondary mini-steel mills require scrap metal availability and market access.

Step-by-Step Solution

1

Analyze the chemical synthesis pathways for urea fertilizer manufacture.

Identify natural gas (

CH_4

) as the hydrogen source for steam reforming to make ammonia (

NH_3

), which is combined with carbon dioxide (

CO_2

) to yield urea.

Natural gas serves a dual role as both chemical feedstock and thermal energy source, necessitating plant siting near gas fields.

2

Examine the material transport economics for commercial glass manufacturing.

Determine that high-purity silica sand (

SiO_2

) forms the bulk of the raw material mass.

Transporting heavy, low-unit-cost raw silica across long distances is economically inefficient compared to locating near riverine/coastal sand deposits.

3

Evaluate the inputs for industrial chlor-alkali production.

Connect concentrated aqueous sodium chloride (

NaCl(aq)

) and electric power to the production of sodium hydroxide (

NaOH

) and chlorine gas (

Cl_2

).

Electrochemical chlor-alkali cells rely on high power input and continuous brine supply.

4

Differentiate primary steelmaking from secondary scrap-based steel mills.

Associate secondary mini-mills with urban scrap centers and market proximity.

Unlike blast furnace primary steel production (which locates near iron ore and metallurgical coal), mini-mills re-melt local metal scrap in electric arc furnaces, making them market-oriented.

Key Concept

Raw material dependencies and industrial siting economics in chemical and metallurgical industries

Question 8098Question →

If the mass defect of a nucleus is $0.030\text{ u}$ , what is its binding energy? (Take $1\text{ u} = 931\text{ MeV}$ )

27.93\text{ MeV}

27.93\text{ J}

0.030\text{ MeV}

31.033\text{ MeV}

Show answer & explanation

Answer: $27.93\text{ MeV}$

Answer

27.93\text{ MeV}

The binding energy is calculated directly by multiplying the mass defect

\Delta m

by the conversion factor

931\text{ MeV/u}

. Thus,

E_b = 0.030 \times 931 = 27.93\text{ MeV}

.

Step-by-Step Solution

1

Identify the given values

Mass defect

\Delta m = 0.030\text{ u}

and mass-energy equivalent

1\text{ u} = 931\text{ MeV}

.

To calculate the binding energy, the mass defect in atomic mass units must be converted to energy.

2

Apply mass-energy equivalence conversion

E_b = \Delta m \times 931\text{ MeV/u} = 0.030 \times 931 = 27.93\text{ MeV}

.

Binding energy is equal to mass defect multiplied by the energy equivalent per atomic mass unit.

Key Concept

Mass Defect and Binding Energy

Question 8099Question →

A composite cylindrical bar consists of two uniform sections of equal length joined end-to-end. Section A has a radius of $2.0\text{ cm}$ and a thermal conductivity of $300\text{ W m}^{-1}\text{K}^{-1}$ . Section B has a radius of $4.0\text{ cm}$ and a thermal conductivity of $150\text{ W m}^{-1}\text{K}^{-1}$ . The outer end of Section A is maintained at a constant temperature of $120^\circ\text{C}$ , while the outer end of Section B is held at $0^\circ\text{C}$ . Assuming the curved surfaces of both sections are perfectly insulated and heat flow is steady, what is the temperature at the junction between the two sections in $^\circ\text{C}$ ?

Show answer & explanation

Answer: 40

Answer

The steady-state temperature at the junction between Section A and Section B is 40.0°C.

At steady state, the rate of heat conduction through Section A equals that through Section B. Because Section B has double the radius of Section A, its cross-sectional area is four times as large. Equating the heat flow rates gives 300 * A_A * (120 - T_J) = 150 * (4 A_A) * T_J, which simplifies directly to 120 - T_J = 2 T_J, yielding a junction temperature of 40.0°C.

Step-by-Step Solution

1

Determine the relationship between the cross-sectional areas of Section A and Section B.

The area ratio A_B / A_A = (r_B / r_A)² = (4.0 cm / 2.0 cm)² = 4.

The cross-sectional area of a cylinder is proportional to the square of its radius.

2

Write the steady-state heat flow equation for each section.

H_A = (k_A * A_A * (120 - T_J)) / L and H_B = (k_B * A_B * (T_J - 0)) / L.

According to Fourier's law of thermal conduction, the heat transfer rate through a uniform layer is proportional to thermal conductivity, cross-sectional area, and temperature difference, and inversely proportional to length.

3

Equate the heat transfer rates H_A and H_B and solve for the junction temperature T_J.

300 * A_A * (120 - T_J) = 150 * (4 A_A) * T_J => 300 * (120 - T_J) = 600 * T_J => 120 - T_J = 2 T_J => 3 T_J = 120 => T_J = 40.0°C.

At steady state with insulated sides, heat does not accumulate or escape, so the rate of heat conduction through both sections must be identical.

Key Concept

Steady-state thermal conduction through composite conductors with differing cross-sectional areas and thermal conductivities

Question 8100Question →

In liquid ammonia as a solvent, ammonium ions react with amide ions according to the following ionic equilibrium equation:

NH_4^+ + NH_2^- \rightleftharpoons 2NH_3

Based on the Brønsted-Lowry theory of acids and bases, which of the following statements correctly identifies the acid and its corresponding conjugate base in the forward reaction?

NH_4^+

is the acid, and

NH_3

is its conjugate base.

NH_4^+

is the acid, and

NH_2^-

is its conjugate base.

NH_2^-

is the acid, and

NH_3

is its conjugate base.

NH_3

is the acid, and

NH_4^+

is its conjugate base.

Show answer & explanation

Answer: $NH_4^+$ is the acid, and $NH_3$ is its conjugate base.

Answer

The species

NH_4^+

acts as the acid, and

NH_3

is its corresponding conjugate base.

According to the Brønsted-Lowry concept, an acid is a proton (

H^+

) donor, and a conjugate base is the species formed when an acid loses a proton. In the given reaction,

NH_4^+

donates a proton to

NH_2^-

, producing

NH_3

. Therefore,

NH_4^+

is the acid and

NH_3

is its conjugate base.

Step-by-Step Solution

1

Analyze the forward reaction and track proton transfer

In

NH_4^+ + NH_2^- \rightleftharpoons 2NH_3

,

NH_4^+

loses a proton (

H^+

) to form one molecule of

NH_3

.

The Brønsted-Lowry definition specifies that an acid is a proton (

H^+

) donor.

2

Identify the Brønsted-Lowry acid

NH_4^+

is the proton donor, so it is the Brønsted-Lowry acid.

It donates

H^+

to the amide ion (

NH_2^-

).

3

Determine the conjugate base formed from the acid

When

NH_4^+

loses a proton, it becomes

NH_3

.

A conjugate base is the species remaining after a Brønsted-Lowry acid donates a proton.

Key Concept

Brønsted-Lowry Acid-Base Theory and Conjugate Pairs

Estimated Time:1m 30s

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