All practice questions

Question 8041Question →

Which type of electromagnetic radiation is primarily detected using a thermopile or a bolometer due to its heating effect?

Infrared radiation

Ultraviolet radiation

Gamma radiation

Radio waves

Show answer & explanation

Answer: Infrared radiation

Answer

Infrared radiation is the component of the electromagnetic spectrum primarily detected by a thermopile or bolometer.

Infrared radiation is absorbed by matter primarily as heat, raising the temperature of the absorber. Thermopiles measure this temperature rise by generating a small voltage across thermocouple junctions, making infrared radiation the primary band detected by thermopiles.

Step-by-Step Solution

1

Identify the characteristic mechanism of detection for thermopiles

Thermopiles convert thermal energy (heat) into an electrical voltage via the thermoelectric effect.

Understanding the working principle of the detector points to the radiation band with prominent heating properties.

2

Match the detector to the corresponding electromagnetic band

Infrared radiation causes significant thermal excitation and temperature rise upon absorption, making thermopiles ideal detectors.

Infrared rays are also known as heat waves because they transfer thermal energy efficiently.

Key Concept

Detection mechanisms of electromagnetic waves

Estimated Time:45s

Question 8042Question →

Metals positioned near the top of the electrochemical reactivity series, such as sodium and aluminium, are extracted from their purified ores primarily through which method?

Electrolysis of their molten compounds

Reduction of their oxides using carbon monoxide

Roasting their sulfide ores directly in excess air

Thermal decomposition of their trioxocarbonate(IV) salts

Show answer & explanation

Answer: Electrolysis of their molten compounds

Answer

Electrolysis of their molten compounds

Metals positioned high in the electrochemical series (such as sodium, potassium, calcium, and aluminium) have a strong affinity for oxygen and chlorine. Chemical reducing agents like carbon or carbon monoxide are weaker reducing agents than these metals and cannot reduce their oxides. Consequently, these metals are extracted by electrolysis of their molten (fused) ores or salts.

Step-by-Step Solution

1

Assess metal reactivity based on position in the electrochemical series

Sodium and aluminium are highly electropositive and hold onto oxygen/halogens strongly.

Metals at the top of the reactivity series form compounds with very high thermal and chemical stability.

2

Select the appropriate reduction method

Standard chemical reducing agents (such as carbon or hydrogen) are ineffective.

Only powerful electrical energy supplied during electrolysis of fused compounds can force electron gain at the cathode to yield free metal.

Key Concept

Selection of metal extraction method based on reactivity series

Question 8043Question →

During daytime in coastal regions, land heats up faster than the sea. The air above the land expands, becomes less dense, and rises, allowing cooler air from the ocean to move inland to form a sea breeze. Which mode of heat transfer is primarily responsible for this bulk movement of fluid, and what physical property change drives it?

Convection, driven by a decrease in air density due to thermal expansion

Conduction, driven by energy transfer through stationary molecular collisions at the land-air boundary

Radiation, driven by direct absorption of solar electromagnetic waves by atmospheric air

Convection, driven by an increase in air density due to thermal expansion

Show answer & explanation

Answer: Convection, driven by a decrease in air density due to thermal expansion

Answer

Convection, driven by a decrease in air density due to thermal expansion

Convection is the mode of heat transfer in liquids and gases where heat is carried from one place to another by the actual bulk movement of the heated fluid. As air over the land absorbs heat, thermal expansion causes its volume to increase and its density to decrease. The lighter, warm air rises and is replaced by cooler, denser air from over the water, creating a sea breeze.

Step-by-Step Solution

1

Identify the mode of heat transfer involved in fluids moving in bulk currents

Heat transfer in fluids involving actual physical displacement/movement of matter is convection.

Conduction occurs without net motion of the medium, and radiation occurs via electromagnetic waves.

2

Analyze the physical driver of convective currents

When air above the land is heated, it expands (

ΔV > 0

), reducing its density (

ρ = m/V

).

Lower density air experiences a net upward buoyant force, creating a low-pressure area near the ground into which cooler, denser air flows.

Key Concept

Convection in fluids and buoyancy driven by thermal expansion

Question 8044Question →

A solid block floats in water of density $1000\text{ kg/m}^3$ with $60\%$ of its total volume submerged. When the same block is placed in an unknown liquid $X$ , $80\%$ of its total volume is submerged. What is the density of liquid $X$ ?

750\text{ kg/m}^3

1333\text{ kg/m}^3

480\text{ kg/m}^3

800\text{ kg/m}^3

Show answer & explanation

Answer: $750\text{ kg/m}^3$

Answer

The density of liquid

X

is

750\text{ kg/m}^3

.

For any floating body, its weight equals the upthrust exerted by the liquid. The upthrust is given by

U = \rho_{\text{fluid}} \cdot V_{\text{submerged}} \cdot g

. Since the weight of the block is unchanged,

\rho_{\text{water}} \cdot V_{\text{sub, water}} = \rho_X \cdot V_{\text{sub, X}}

. Substituting the given values:

1000 \times 0.60V = \rho_X \times 0.80V

, yielding

\rho_X = 750\text{ kg/m}^3

.

Step-by-Step Solution

1

Apply the Law of Flotation to the block in water

Weight of block

W = \rho_w \cdot V_{\text{sub, water}} \cdot g = 1000 \cdot 0.60V \cdot g = 600 V g

A floating object displaces its own weight of fluid.

2

Apply the Law of Flotation to the block in liquid X

Weight of block

W = \rho_X \cdot V_{\text{sub, X}} \cdot g = \rho_X \cdot 0.80V \cdot g

The weight of the block remains constant regardless of the fluid.

3

Equate the two expressions for the weight of the block and solve for

\rho_X

\rho_X \cdot 0.80V \cdot g = 600 V g \implies \rho_X = \frac{600}{0.80} = 750\text{ kg/m}^3

Since both buoyant forces equal the block's weight, set them equal to each other.

Key Concept

Law of Flotation and Archimedes' Principle

Question 8045Question →

A sample of a radioactive nuclide with a half-life of $6\text{ hours}$ initially has a mass of $200\text{ mg}$ . What mass of the nuclide has decayed after an elapsed time of $24\text{ hours}$ ?

187.5\text{ mg}

12.5\text{ mg}

50.0\text{ mg}

150.0\text{ mg}

Show answer & explanation

Answer: $187.5\text{ mg}$

Answer

The mass of the nuclide that has decayed after

24\text{ hours}

is

187.5\text{ mg}

.

After

24\text{ hours}

, exactly

4

half-lives (

24 / 6 = 4

) have passed. The remaining radioactive mass is

200\text{ mg} / 2^4 = 12.5\text{ mg}

. Subtracting the remaining mass from the initial mass (

200\text{ mg} - 12.5\text{ mg}

) yields

187.5\text{ mg}

as the decayed mass.

Step-by-Step Solution

1

Determine the number of half-lives (

n

) that have elapsed.

n = \frac{t}{T_{1/2}} = \frac{24\text{ hours}}{6\text{ hours}} = 4\text{ half-lives}

The total elapsed time divided by the half-life duration gives the total number of decay cycles.

2

Calculate the remaining mass (

N

) after

4

half-lives.

N = N_0 \left(\frac{1}{2}\right)^n = 200\text{ mg} \times \left(\frac{1}{2}\right)^4 = \frac{200\text{ mg}}{16} = 12.5\text{ mg}

The radioactive decay law states that after

n

half-lives, the initial quantity reduces by a factor of

2^n

.

3

Calculate the mass of the nuclide that has decayed (

N_{\text{decayed}}

).

N_{\text{decayed}} = N_0 - N = 200\text{ mg} - 12.5\text{ mg} = 187.5\text{ mg}

The amount decayed is the initial mass minus the mass remaining.

Key Concept

Radioactive Decay Law and Half-life Calculation

Question 8046Question →

A fixed mass of helium gas is sealed inside a rigid container at a pressure of $2.0\text{ atm}$ and a temperature of $27^\circ\text{C}$ . If the gas is heated until its pressure reaches $3.5\text{ atm}$ while maintaining a constant volume, what is its final absolute temperature in Kelvin?

Fill in the blanks below

The final absolute temperature of the gas is K.

Show answer & explanation

Answer

The final absolute temperature of the gas is 525 K.

According to Gay-Lussac's Pressure Law, pressure and absolute temperature are directly proportional at constant volume (

P_1/T_1 = P_2/T_2

). First, convert

27^\circ\text{C}

to absolute temperature:

27 + 273 = 300\text{ K}

. Then substitute the pressures and initial temperature into the equation to find

T_2

:

T_2 = (3.5\text{ atm} \times 300\text{ K}) / 2.0\text{ atm} = 525\text{ K}

.

Step-by-Step Solution

1

Convert the initial temperature from degrees Celsius to the Kelvin scale.

T_1 = 27^\circ\text{C} + 273 = 300\text{ K}

All thermodynamic gas law calculations require absolute temperature in Kelvin.

2

Apply the Pressure Law (Gay-Lussac's Law) equation for constant volume.

\frac{P_1}{T_1} = \frac{P_2}{T_2}

The pressure of a fixed mass of gas at constant volume is directly proportional to its absolute temperature.

3

Substitute the given values into the formula and solve for the final temperature

T_2

.

T_2 = \frac{P_2 \times T_1}{P_1} = \frac{3.5 \times 300}{2.0} = 525\text{ K}

Multiplying the new pressure by the initial absolute temperature and dividing by the initial pressure yields the final absolute temperature.

Key Concept

Pressure Law (Gay-Lussac's Law of Temperature-Pressure)

Question 8047Question →

A solid object has a mass of $0.50\text{ kg}$ and a volume of $2.0 \times 10^{-4}\text{ m}^3$ . If it is completely immersed in a liquid of density $800\text{ kg/m}^3$ , what is the magnitude of the upthrust exerted on the object? [Take $g = 10\text{ m/s}^2$ ]

1.6\text{ N}

5.0\text{ N}

3.4\text{ N}

0.16\text{ N}

Show answer & explanation

Answer: $1.6\text{ N}$

Answer

The magnitude of the upthrust exerted on the object is

1.6\text{ N}

.

By Archimedes' Principle, upthrust is equal to the weight of the liquid displaced:

U = \rho_{\text{liquid}} V g

. Substituting

\rho = 800\text{ kg/m}^3

,

V = 2.0 \times 10^{-4}\text{ m}^3

, and

g = 10\text{ m/s}^2

yields

1.6\text{ N}

.

Step-by-Step Solution

1

Identify the given values and state Archimedes' Principle

Volume of displaced liquid

V = 2.0 \times 10^{-4}\text{ m}^3

, density of liquid

\rho = 800\text{ kg/m}^3

, acceleration due to gravity

g = 10\text{ m/s}^2

. Upthrust formula is

U = \rho V g

.

According to Archimedes' Principle, the upward buoyant force (upthrust) equals the weight of the liquid displaced by the submerged object.

2

Calculate the upthrust

U = 800\text{ kg/m}^3 \times (2.0 \times 10^{-4}\text{ m}^3) \times 10\text{ m/s}^2 = 1.6\text{ N}

.

Multiplying fluid density by submerged volume and gravitational acceleration gives the force in newtons.

Key Concept

Archimedes' Principle and Upthrust

Question 8048Question →

During the industrial extraction of zinc from its principal ore, zinc blende ( $ZnS$ ), the concentrated sulfide ore undergoes thermal conversion in excess air before metal recovery. Which set of balanced chemical equations correctly represents the roasting step and the subsequent reduction step using carbon (coke)?

Roasting:

2ZnS(s) + 3O_2(g) \rightarrow 2ZnO(s) + 2SO_2(g)

; Reduction:

ZnO(s) + C(s) \rightarrow Zn(g) + CO(g)

Roasting:

ZnS(s) + 2O_2(g) \rightarrow ZnSO_4(s)

; Reduction:

ZnSO_4(s) + C(s) \rightarrow Zn(s) + SO_2(g) + CO(g)

Roasting:

2ZnS(s) + O_2(g) \rightarrow 2Zn(s) + 2SO_2(g)

; Reduction:

Zn(s) + C(s) \rightarrow ZnC(s)

Roasting:

ZnS(s) + H_2O(g) \rightarrow ZnO(s) + H_2S(g)

; Reduction:

ZnO(s) + CO_2(g) \rightarrow Zn(s) + CO(g) + O_2(g)

Show answer & explanation

Answer: Roasting: $2ZnS(s) + 3O_2(g) \rightarrow 2ZnO(s) + 2SO_2(g)$ ; Reduction: $ZnO(s) + C(s) \rightarrow Zn(g) + CO(g)$

Answer

The correct sequence of reactions involves roasting zinc blende in excess oxygen to produce zinc oxide and sulfur(IV) oxide (

2ZnS + 3O_2 \rightarrow 2ZnO + 2SO_2

), followed by reduction of zinc oxide with coke at high temperature to produce zinc gas and carbon(II) oxide (

ZnO + C \rightarrow Zn + CO

).

Roasting zinc blende (

ZnS

) in excess oxygen gas converts the sulfide ore into zinc oxide (

ZnO

) while liberating sulfur(IV) oxide (

SO_2

). In the reduction furnace, carbon (coke) reduces zinc oxide to zinc metal vapor (

Zn

) and carbon(II) oxide gas (

CO

) because the reaction takes place above the boiling point of zinc.

Step-by-Step Solution

1

Identify the chemical nature of the roasting process in metallurgy.

Roasting involves heating concentrated sulfide ores (

ZnS

) in an abundant supply of atmospheric oxygen (

O_2

).

Sulfide ores are difficult to reduce directly to metals; converting them to oxides makes subsequent reduction thermodynamically feasible.

2

Write and balance the roasting reaction equation.

2ZnS(s) + 3O_2(g) \rightarrow 2ZnO(s) + 2SO_2(g)

Zinc sulfide reacts with oxygen gas to yield solid zinc oxide and sulfur(IV) oxide gas byproduct.

3

Identify the reduction process of zinc oxide using coke.

ZnO(s) + C(s) \rightarrow Zn(g) + CO(g)

Carbon (coke) acts as a reducing agent at elevated temperatures (~1400 °C), stripping oxygen from zinc oxide. Because the temperature exceeds zinc's boiling point (907 °C), zinc is collected as a gas (vapor) and condensed.

Key Concept

Metallurgical stages of sulfide ore extraction: Roasting (conversion to oxide) followed by pyrometallurgical carbon reduction.

Estimated Time:2m 0s

Question 8049Question →

When smoke particles suspended in air are observed under a microscope, they display a continuous, random zigzag movement known as Brownian motion. According to the kinetic theory of matter, what causes this phenomenon?

Unbalanced, rapid collisions between invisible air molecules and the larger smoke particles

Thermal convection currents created within the container carrying smoke particles along fixed streamlines

Electrostatic repulsion occurring constantly between identical electric charges on adjacent smoke particles

Gravitational attraction continuously pulling smoke particles downward against upward buoyant forces

Show answer & explanation

Answer: Unbalanced, rapid collisions between invisible air molecules and the larger smoke particles

Answer

Unbalanced, rapid collisions between invisible air molecules and the larger smoke particles

According to the kinetic theory of matter, gas particles are in continuous, rapid, and random motion. Smoke particles, though microscopic, are significantly larger than individual air molecules. At any given moment, air molecules strike a smoke particle unevenly from various directions, creating a net unbalanced force that causes the smoke particle to change direction abruptly and produce Brownian motion.

Step-by-Step Solution

1

Recall the fundamental postulate of kinetic theory regarding particle motion in gases

Gas molecules (such as air) are in constant, rapid, and random straight-line motion.

Gas particles possess thermal kinetic energy proportional to the absolute temperature.

2

Analyze the impact of moving gas molecules on suspended microscopic particles

Because air molecules hit a smoke particle unequally from different sides at any given instant, the net unbalanced impulse shifts the particle's direction constantly.

This continuous molecular bombardment directly demonstrates the existence of perpetual molecular motion postulated by kinetic theory.

Key Concept

Brownian motion as empirical evidence for continuous random molecular motion

Question 8050Question →

A spherical ball bearing of radius $3.0\text{ mm}$ and density $5400\text{ kg/m}^3$ falls vertically through a viscous oil of density $900\text{ kg/m}^3$ and dynamic viscosity coefficient $0.10\text{ Pa}\cdot\text{s}$ . Assuming the motion obeys Stokes' law and taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the terminal velocity of the sphere?

0.90\text{ m/s}

1.08\text{ m/s}

0.68\text{ m/s}

0.18\text{ m/s}

Show answer & explanation

Answer: $0.90\text{ m/s}$

Answer

The terminal velocity of the sphere is

0.90\text{ m/s}

.

At terminal velocity, the downward force of gravity (weight of the sphere) is balanced by the sum of two upward forces: the buoyant force (upthrust) and the viscous drag force given by Stokes' law (

F_v = 6\pi \eta r v_T

). Using the formula

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

with

r = 3.0 \times 10^{-3}\text{ m}

,

\rho_s - \rho_f = 4500\text{ kg/m}^3

,

\eta = 0.10\text{ Pa}\cdot\text{s}

, and

g = 10\text{ m/s}^2

yields

0.90\text{ m/s}

.

Step-by-Step Solution

1

Convert given parameters to standard SI units

Radius

r = 3.0\text{ mm} = 3.0 \times 10^{-3}\text{ m}

, density of sphere

\rho_s = 5400\text{ kg/m}^3

, density of liquid

\rho_f = 900\text{ kg/m}^3

, viscosity

\eta = 0.10\text{ Pa}\cdot\text{s}

,

g = 10\text{ m/s}^2

.

Ensures dimensional consistency across all terms in the physical equations.

2

Apply the equilibrium condition at terminal velocity

At terminal velocity

v_T

, downward weight equals upward forces:

W = U + F_v

, where

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

,

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

, and

F_v = 6\pi \eta r v_T

.

Terminal velocity is reached when net acceleration is zero.

3

Rearrange Stokes' law formula for terminal velocity

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

Isolates the target unknown variable

v_T

.

4

Substitute the physical values and solve

v_T = \frac{2 \times (3.0 \times 10^{-3})^2 \times (5400 - 900) \times 10}{9 \times 0.10} = \frac{2 \times (9.0 \times 10^{-6}) \times 4500 \times 10}{0.90} = \frac{0.81}{0.90} = 0.90\text{ m/s}

.

Calculates the final quantitative answer.

Key Concept

Viscosity and Stokes' Law for terminal velocity of a sphere in a viscous fluid

Estimated Time:2m 0s

Question 8051Question →

A pharmaceutical analyst evaluated an organic solid compound $Z$ suspected of containing a soluble non-volatile impurity. When heated slowly in a melting point apparatus, the sample began to melt at $141.2^\circ\text{C}$ and completely liquefied at $147.8^\circ\text{C}$ (literature melting point of pure $Z = 151.0^\circ\text{C}$ ). A thin-layer chromatography (TLC) plate developed with a $12.0\text{ cm}$ solvent front produced a major spot at a distance of $4.8\text{ cm}$ from the baseline and a minor impurity spot at $9.6\text{ cm}$ . Which of the following correctly identifies the main component's $R_f$ value and describes the melting characteristics of pure compound $Z$ relative to this impure sample?

The main component has an

R_f

value of

0.40

, and pure compound

Z

melts sharply at

151.0^\circ\text{C}

over a very narrow temperature range.

The main component has an

R_f

value of

2.50

, and pure compound

Z

melts at a lower temperature over the same broad range.

The main component has an

R_f

value of

0.40

, and the presence of the soluble impurity elevates the melting point while narrowing the melting range.

The main component has an

R_f

value of

0.50

obtained by comparing the two spot distances, and pure compound

Z

melts over a broad temperature range.

Show answer & explanation

Answer: The main component has an $R_f$ value of $0.40$ , and pure compound $Z$ melts sharply at $151.0^\circ\text{C}$ over a very narrow temperature range.

Answer

The main component has an

R_f

value of

0.40

, and pure compound

Z

melts sharply at

151.0^\circ\text{C}

over a very narrow temperature range.

Calculating the retention factor gives

R_f = 4.8 / 12.0 = 0.40

. Pure chemical substances exhibit sharp, well-defined physical constants. For a solid compound, chemical purity is confirmed when the substance melts sharply at its exact literature value (

151.0^\circ\text{C}

) over a narrow range of less than

1^\circ\text{C}

. The crude sample's lowered melting point (

141.2^\circ\text{C}

) and wide range (

6.6^\circ\text{C}

) are classic indicators of impurity.

Step-by-Step Solution

1

Calculate the retention factor (

R_f

) of the main component spot from the TLC data.

R_f = \frac{\text{Distance moved by main spot}}{\text{Distance moved by solvent front}} = \frac{4.8\text{ cm}}{12.0\text{ cm}} = 0.40

R_f

is defined as the ratio of solute migration distance to solvent front migration distance.

2

Analyze the impact of a soluble non-volatile impurity on the melting point of a solid compound.

The presence of impurities depresses the melting point (sample melts below

151.0^\circ\text{C}

) and broadens the melting range (

141.2^\circ\text{C} - 147.8^\circ\text{C}

, a span of

6.6^\circ\text{C}

).

Foreign molecules disrupt the crystal lattice of the solid, requiring less kinetic energy to collapse the structure over a variable range of local impurity concentrations.

3

Determine the melting criteria for a completely pure sample of compound

Z

.

A pure sample melts sharply at its characteristic literature temperature (

151.0^\circ\text{C}

) within a range of

0.5^\circ\text{C}

to

1.0^\circ\text{C}

.

Chemical purity is characterized by fixed physical constants, including a sharp melting point and single chromatographic spot.

Key Concept

Criteria of purity for solids: sharp melting point at literature value and single spot on TLC with correct Rf.

Question 8052Question →

A simple pendulum of length $L$ and bob mass $m$ has a period of oscillation $T$ on the surface of the Earth. The length of the pendulum is increased by $44\%$ , its bob mass is doubled to $2m$ , and the entire apparatus is transported to a planet where the acceleration due to gravity is $36\%$ less than that on Earth. What is the new period of oscillation of the pendulum?

3.0T

1.5T

1.2T

0.67T

Show answer & explanation

Answer: $1.5T$

Answer

1.5T

The period of a simple pendulum is determined by

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

. It is entirely independent of the mass of the bob. With a length increase of

44\%

(

L' = 1.44L

) and a gravity reduction of

36\%

(

g' = 0.64g

), the new period becomes

\sqrt{\frac{1.44}{0.64}}T = \frac{1.2}{0.8}T = 1.5T

.

Step-by-Step Solution

1

State the equation for the period of a simple pendulum

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

The period depends solely on length

L

and acceleration due to gravity

g

, and is independent of mass

m

.

2

Determine the new length and new acceleration due to gravity

L' = 1.44L

and

g' = 0.64g

A

44\%

increase in length yields

1 + 0.44 = 1.44

, while a

36\%

decrease in gravity yields

1 - 0.36 = 0.64

.

3

Calculate the factor of change in the period

\frac{T'}{T} = \sqrt{\frac{L'}{L} \times \frac{g}{g'}} = \sqrt{\frac{1.44}{0.64}} = \sqrt{\frac{144}{64}} = \frac{12}{8} = 1.5

Substituting the relative changes into the pendulum period ratio yields the scaling factor.

4

Express the new period in terms of

T

T' = 1.5T

Multiplying the initial period by the calculated scaling factor gives the final period.

Key Concept

Mass independence and parametric scaling of simple pendulum period in Simple Harmonic Motion

Question 8053Question →

A water treatment facility needs to soften $100\text{ dm}^3$ of well water containing $0.005\text{ mol dm}^{-3}$ of dissolved calcium hydrogentrioxocarbonate(IV), $\text{Ca(HCO}_3)_2$ , using Clark's process. What is the minimum mass of calcium hydroxide, $\text{Ca(OH)}_2$ , required to completely precipitate the calcium ions responsible for this temporary hardness? $[\text{Ca} = 40,\text{ O} = 16,\text{ H} = 1]$

37.0\text{ g}

74.0\text{ g}

18.5\text{ g}

81.0\text{ g}

Show answer & explanation

Answer: $37.0\text{ g}$

Answer

The minimum mass of calcium hydroxide required is

37.0\text{ g}

.

The correct answer of

37.0\text{ g}

is obtained by finding the moles of dissolved

\text{Ca(HCO}_3)_2

(

0.005\text{ mol dm}^{-3} \times 100\text{ dm}^3 = 0.5\text{ mol}

) and applying the

1:1

stoichiometric ratio from the reaction equation

\text{Ca(HCO}_3)_2 + \text{Ca(OH)}_2 \rightarrow 2\text{CaCO}_3 + 2\text{H}_2\text{O}

. Multiplying

0.5\text{ mol}

by the molar mass of

\text{Ca(OH)}_2

(

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

) yields

37.0\text{ g}

.

Step-by-Step Solution

1

Calculate the amount in moles of dissolved calcium hydrogentrioxocarbonate(IV) in the water sample.

\text{Moles of Ca(HCO}_3)_2 = \text{Concentration} \times \text{Volume} = 0.005\text{ mol dm}^{-3} \times 100\text{ dm}^3 = 0.5\text{ mol}

.

Determining the exact molar quantity of solute is the first step in stoichiometric calculations.

2

Write the balanced chemical equation for Clark's process (slaked lime softening).

\text{Ca(HCO}_3)_2\text{(aq)} + \text{Ca(OH)}_2\text{(aq)} \rightarrow 2\text{CaCO}_3\text{(s)} + 2\text{H}_2\text{O(l)}

. The mole ratio of

\text{Ca(HCO}_3)_2

to

\text{Ca(OH)}_2

is

1:1

.

Clark's process uses calculated amounts of calcium hydroxide to convert soluble hydrogentrioxocarbonates into insoluble trioxocarbonate(IV) precipitates.

3

Calculate the molar mass of calcium hydroxide,

\text{Ca(OH)}_2

.

\text{Molar mass} = 40 + 2(16 + 1) = 74\text{ g mol}^{-1}

.

Molar mass is required to convert moles of reagent into mass in grams.

4

Determine the required mass of

\text{Ca(OH)}_2

.

\text{Mass} = \text{Moles} \times \text{Molar mass} = 0.5\text{ mol} \times 74\text{ g mol}^{-1} = 37.0\text{ g}

.

Multiplying the required moles by molar mass gives the required mass of slaked lime.

Key Concept

Removal of temporary water hardness using Clark's process (addition of calculated lime).

Estimated Time:2m 0s

Question 8054Question →

A solid sphere of mass $0.60\text{ kg}$ and volume $2.0 \times 10^{-4}\text{ m}^3$ is released from rest in a tall vessel filled with a viscous liquid of density $1000\text{ kg/m}^3$ . As the sphere falls, it eventually reaches a constant terminal velocity of $4.0\text{ m/s}$ . Assuming that the viscous drag force is directly proportional to the speed of the sphere, what is the magnitude of the viscous drag force acting on the sphere when its speed is $1.5\text{ m/s}$ ? (Take $g = 10\text{ m/s}^2$ )

0.75\text{ N}

1.50\text{ N}

2.25\text{ N}

2.50\text{ N}

Show answer & explanation

Answer: $1.50\text{ N}$

Answer

1.50\text{ N}

At terminal velocity, the sphere is in translational equilibrium under three forces: downward weight (

6.0\text{ N}

), upward upthrust (

2.0\text{ N}

), and upward viscous drag (

4.0\text{ N}

). Since viscous drag is directly proportional to velocity (

F_v = k v

), the proportionality constant

k

is

1.0\text{ N}\cdot\text{s/m}

. Therefore, at

1.5\text{ m/s}

, the viscous force is

1.0 \times 1.5 = 1.50\text{ N}

.

Step-by-Step Solution

1

Calculate the downward gravitational force (weight) acting on the sphere.

W = m \times g = 0.60\text{ kg} \times 10\text{ m/s}^2 = 6.0\text{ N}

Weight is the force pulling the sphere downward.

2

Calculate the buoyant force (upthrust) exerted by the liquid using Archimedes' principle.

U = \rho_l \times V \times g = 1000\text{ kg/m}^3 \times (2.0 \times 10^{-4}\text{ m}^3) \times 10\text{ m/s}^2 = 2.0\text{ N}

Upthrust equals the weight of the fluid displaced by the submerged sphere.

3

Determine the viscous drag force at terminal velocity (

v_t = 4.0\text{ m/s}

) using equilibrium of forces.

F_v(v_t) = W - U = 6.0\text{ N} - 2.0\text{ N} = 4.0\text{ N}

At terminal velocity, the net acceleration is zero, so downward weight is balanced by upward upthrust and viscous drag.

4

Find the constant of proportionality

k

for viscous drag (

F_v = k v

).

k = \frac{F_v(v_t)}{v_t} = \frac{4.0\text{ N}}{4.0\text{ m/s}} = 1.0\text{ N}\cdot\text{s/m}

Viscous force is given as directly proportional to speed.

5

Calculate the viscous drag force at a speed of

1.5\text{ m/s}

.

F_v(1.5) = k \times 1.5\text{ m/s} = 1.0\text{ N}\cdot\text{s/m} \times 1.5\text{ m/s} = 1.50\text{ N}

Applying the constant

k

to the specified speed.

Key Concept

Terminal velocity in viscous fluids and Archimedes' Principle

Estimated Time:2m 30s

Question 8055Question →

Consider the nuclear fusion reaction below:

\text{^{2}_{1}H} + \text{^{3}_{1}H} \rightarrow \text{^{4}_{2}He} + \text{^{1}_{0}n} + Q

Given the rest masses:
Mass of

\text{^{2}_{1}H} = 2.0141\text{ u}

Mass of

\text{^{3}_{1}H} = 3.0160\text{ u}

Mass of

\text{^{4}_{2}He} = 4.0015\text{ u}

Mass of

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

If

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

, what is the total energy released (

Q

) in this fusion reaction?

18.53\text{ MeV}

957.69\text{ MeV}

2.96 \times 10^{-12}\text{ MeV}

18.5\text{ MeV}

Show answer & explanation

Answer: $18.53\text{ MeV}$

Answer

18.53\text{ MeV}

The total initial mass of the reactants is

2.0141\text{ u} + 3.0160\text{ u} = 5.0301\text{ u}

. The total final mass of the products is

4.0015\text{ u} + 1.0087\text{ u} = 5.0102\text{ u}

. Subtracting the product mass from the reactant mass gives a mass defect of

\Delta m = 0.0199\text{ u}

. Multiplying this mass defect by the conversion factor

931\text{ MeV/u}

yields an energy release of

18.53\text{ MeV}

.

Step-by-Step Solution

1

Calculate the total mass of the reactants

\text{Mass of reactants} = 2.0141\text{ u} + 3.0160\text{ u} = 5.0301\text{ u}

The total initial mass must be determined from the sum of the masses of deuterium and tritium.

2

Calculate the total mass of the products

\text{Mass of products} = 4.0015\text{ u} + 1.0087\text{ u} = 5.0102\text{ u}

The total final mass includes both the helium nucleus and the released neutron.

3

Determine the mass defect (

\Delta m

)

\Delta m = 5.0301\text{ u} - 5.0102\text{ u} = 0.0199\text{ u}

Mass defect is the difference between the total initial mass and the total final mass.

4

Convert the mass defect into energy released (

Q

)

Q = 0.0199 \times 931\text{ MeV} = 18.5269\text{ MeV} \approx 18.53\text{ MeV}

Using the mass-energy equivalence factor

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

yields the energy in

\text{MeV}

.

Key Concept

Calculation of energy released in nuclear fusion reactions using mass defect

Estimated Time:2m 0s

Question 8056Question →

A chemical manufacturing firm in Nigeria plans to establish a large-scale heavy chemical plant producing tetraoxosulfate(VI) acid ( $H_2SO_4$ ) and superphosphate fertilizers using imported elemental sulfur and phosphate rock. Considering raw material transportation, energy requirements, environmental trade-offs, and product distribution logistics, which of the following is the primary economic factor dictating that this plant be sited near a coastal seaport rather than inland near agricultural consumer regions?

Minimizing the high freight costs of transporting bulky, low-value imported raw materials inland

Utilizing untreated seawater as a direct solvent in diluting concentrated acid to commercial grade

Discharging acidic gaseous effluents directly into oceanic trade winds to circumvent atmospheric emissions standards

Exploiting coastal water currents to catalyze the exothermic conversion of sulfur(IV) oxide to sulfur(VI) oxide

Show answer & explanation

Answer: Minimizing the high freight costs of transporting bulky, low-value imported raw materials inland

Answer

Minimizing the high freight costs of transporting bulky, low-value imported raw materials inland

Heavy chemical manufacturing involving bulky imported raw materials (such as sulfur and rock phosphate) is primarily raw-material-oriented. Transporting heavy solid raw materials inland from ports incurs massive transport costs, so siting the facility near a coastal port minimizes overall supply chain expenditure.

Step-by-Step Solution

1

Classify the industry type and identify the primary inputs.

Production of tetraoxosulfate(VI) acid and fertilizers is a heavy chemical industry requiring large volumes of heavy imported solid raw materials (sulfur and phosphate rock).

Heavy chemical industries are material-oriented or market-oriented depending on the weight-loss or bulkiness of inputs versus outputs.

2

Analyze transport economics for imported bulky raw materials versus finished product.

Transporting thousands of tonnes of imported solid minerals inland over long distances incurs enormous freight costs, whereas transporting concentrated acid or packaged fertilizer is more manageable.

Proximity to raw material entry points (seaports) minimizes overland haulage costs for heavy imported inputs.

3

Evaluate potential chemical and operational misattributions.

Seawater cannot be used directly in chemical formulations due to salt contamination, emissions regulations apply anywhere, and physical water flow does not catalyze reactions.

Eliminates incorrect technical and regulatory distractors.

Key Concept

Siting Factors for Heavy Chemical Industries (Raw Material Proximity vs. Market Proximity)

Estimated Time:2m 0s

Question 8057Question →

An RLC series circuit connected across a $100\text{ V}$ (RMS) AC voltage source operates at resonance, dissipating an average power of $400\text{ W}$ . If the inductive reactance at resonance is $25\ \Omega$ , what is the total impedance of the circuit when the capacitance is adjusted such that the capacitive reactance increases by $60\ \Omega$ ?

65\ \Omega

85\ \Omega

60\ \Omega

110\ \Omega

Show answer & explanation

Answer: $65\ \Omega$

Answer

The total impedance of the circuit after adjusting the capacitance is

65\ \Omega

.

At resonance, the net reactance is zero and the circuit behaves purely resistively. Using

P = \frac{V_{\text{rms}}^2}{R}

, the resistance is

R = \frac{100^2}{400} = 25\ \Omega

. Since

X_L = X_C = 25\ \Omega

at resonance, increasing capacitive reactance by

60\ \Omega

gives a new capacitive reactance of

85\ \Omega

. The net reactance magnitude is

|25\ \Omega - 85\ \Omega| = 60\ \Omega

. Combining resistance and net reactance in quadrature yields an impedance of

Z = \sqrt{25^2 + 60^2} = 65\ \Omega

.

Step-by-Step Solution

1

Determine the resistance of the circuit at resonance using the power dissipation formula.

At resonance, impedance equals resistance (

Z = R

) and phase angle is zero, so

P = \frac{V_{\text{rms}}^2}{R} \implies R = \frac{(100)^2}{400} = 25\ \Omega

.

At resonance, inductive reactance and capacitive reactance cancel each other out completely.

2

Identify the initial and modified reactances.

At resonance,

X_L = X_C = 25\ \Omega

. After adjustment, the new capacitive reactance is

X_C' = 25\ \Omega + 60\ \Omega = 85\ \Omega

.

Capacitive reactance was increased by

60\ \Omega

from its resonant value.

3

Calculate the net reactance of the modified circuit.

X_{\text{net}} = |X_L - X_C'| = |25\ \Omega - 85\ \Omega| = 60\ \Omega

.

Net reactance is the magnitude of the difference between inductive and capacitive reactances.

4

Calculate the new total impedance using phasor addition.

Z = \sqrt{R^2 + (X_L - X_C')^2} = \sqrt{25^2 + 60^2} = \sqrt{625 + 3600} = \sqrt{4225} = 65\ \Omega

.

Resistance and net reactance are

90^\circ

out of phase, requiring vector summation (Pythagorean theorem).

Key Concept

Resonance and Impedance in AC Circuits

Question 8058Question →

Two concentric circular conducting loops, $P$ and $Q$ , lie flat in the same horizontal plane, with loop $Q$ situated inside loop $P$ . Loop $P$ is connected to a DC power source through a variable resistor and initially carries a steady clockwise current. If the resistance of the variable resistor is suddenly decreased, which of the following correctly describes the direction of the induced current in loop $Q$ and the nature of the magnetic force exerted on loop $Q$ ?

The induced current in

Q

is counter-clockwise, and the magnetic force on

Q

is repulsive.

The induced current in

Q

is clockwise, and the magnetic force on

Q

is attractive.

The induced current in

Q

is counter-clockwise, and the magnetic force on

Q

is attractive.

The induced current in

Q

is clockwise, and the magnetic force on

Q

is repulsive.

Show answer & explanation

Answer: The induced current in $Q$ is counter-clockwise, and the magnetic force on $Q$ is repulsive.

Answer

The induced current in loop

Q

flows in a counter-clockwise direction, and the magnetic force between loop

P

and loop

Q

is repulsive.

Decreasing the variable resistance increases the clockwise current in loop

P

, which increases the downward magnetic flux through loop

Q

. By Lenz's law, loop

Q

generates an opposing upward magnetic flux, which corresponds to a counter-clockwise induced current. Because the two loops carry currents flowing in opposite directions, they exert a repulsive magnetic force on each other.

Step-by-Step Solution

1

Determine the initial magnetic field direction created by loop

P

Using the right-hand grip rule, a clockwise current in outer loop

P

produces a magnetic field directed perpendicularly downward (into the plane of the page) inside the loop.

Current in a circular loop generates an axial magnetic field whose direction is given by the right-hand rule.

2

Analyze the change in magnetic flux through inner loop

Q

Decreasing the resistance increases the current in loop

P

, thereby increasing the downward magnetic flux passing through loop

Q

.

Ohm's law (

I = V/R

) indicates that reducing resistance increases current, which proportionally strengthens the magnetic field (

B \propto I

).

3

Apply Faraday's and Lenz's laws to find the induced current direction in

Q

To oppose the increasing downward flux, the induced current in loop

Q

must produce a magnetic field directed upward (out of the page). By the right-hand grip rule, an upward field requires a counter-clockwise current in loop

Q

.

Lenz's law states that an induced current always flows in such a direction that its magnetic effect opposes the change producing it.

4

Determine the nature of the magnetic force between the two loops

Loop

P

carries a clockwise current while loop

Q

carries a counter-clockwise current. Concentric circular conductors carrying currents in opposite directions repel each other.

Opposing electric currents produce magnetic fields that result in mutual magnetic repulsion.

Key Concept

Lenz's Law and Electromagnetic Induction in Coaxial Loops

Question 8059Question →

Under identical conditions of temperature and pressure, a given volume of deuterium gas ( $\text{D}_2$ ) requires $40\text{ seconds}$ to diffuse through a porous membrane. How long will it take for an equal volume of protium gas ( $\text{H}_2$ ) to diffuse through the same membrane? (Relative atomic masses: $\text{H} = 1.0$ , $\text{D} = 2.0$ )

28.3\text{ s}

20.0\text{ s}

56.6\text{ s}

80.0\text{ s}

Show answer & explanation

Answer: $28.3\text{ s}$

Answer

The time required for an equal volume of protium gas to diffuse is

28.3\text{ s}

.

According to Graham's Law of diffusion, the time required for a fixed volume of gas to diffuse is directly proportional to the square root of its molar mass (

t \propto \sqrt{M}

). Since protium gas (

\text{H}_2

,

M = 2.0\text{ g mol}^{-1}

) is lighter than deuterium gas (

\text{D}_2

,

M = 4.0\text{ g mol}^{-1}

), it diffuses faster, taking

t = 40 \times \sqrt{2/4} = 28.3\text{ s}

.

Step-by-Step Solution

1

Calculate the relative molecular masses of deuterium gas (

\text{D}_2

) and protium gas (

\text{H}_2

).

M(\text{D}_2) = 2 \times 2.0 = 4.0\text{ g mol}^{-1}

and

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

.

Molecular masses are needed to apply Graham's Law of diffusion.

2

State Graham's Law relating diffusion time to molar mass for a constant volume.

\frac{t_{\text{H}_2}}{t_{\text{D}_2}} = \sqrt{\frac{M(\text{H}_2)}{M(\text{D}_2)}}

.

The time required for diffusion of a fixed volume is directly proportional to the square root of the molar mass of the gas.

3

Substitute the given values into the formula and solve for

t_{\text{H}_2}

.

t_{\text{H}_2} = 40 \times \sqrt{\frac{2.0}{4.0}} = 40 \times \sqrt{0.5} \approx 40 \times 0.7071 = 28.3\text{ s}

.

Performing the calculation yields the exact diffusion time for protium gas.

Key Concept

Graham's Law of Diffusion applied to hydrogen isotopes

Question 8060Question →

Match each state of matter or gaseous behavior with the corresponding kinetic molecular postulate that explains its microscopic thermodynamic properties.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Solid phase rigidity pairs with strong cohesive forces restricting motion to fixed lattice vibrations; Liquid phase fluidity pairs with translational kinetic energy allowing particle sliding under cohesive contact; Real gas liquefaction pairs with non-negligible intermolecular attractions at low thermal energy and high density; Ideal gas energy distribution pairs with average kinetic energy being directly proportional to absolute temperature without attractive forces.

The Kinetic Molecular Theory correlates macroscopic bulk properties of matter (solids, liquids, gases, and real gas deviations) to microscopic balances between thermal kinetic energy and intermolecular forces. Solids are dominated by strong attractive forces restricting particles to fixed vibration points. Liquids possess comparable kinetic energy and cohesive forces allowing fluid translational motion while preserving volume. Real gases condense because intermolecular forces become significant when thermal motion slows at low temperatures and high pressures. Ideal gases assume zero intermolecular forces, where absolute temperature directly dictates average translational kinetic energy.

Step-by-Step Solution

1

Analyze the solid phase property of fixed shape and volume.

In solids, cohesive intermolecular forces significantly exceed thermal kinetic energy, restricting motion to vibration about fixed equilibrium positions.

Explains why solids maintain rigid geometric structures.

2

Analyze the liquid phase balance between kinetic energy and cohesive forces.

Liquid particles have sufficient energy to execute translational motion over short distances, giving liquids fluidity while cohesion maintains a constant volume.

Distinguishes liquid dynamic structure from rigid solids.

3

Evaluate real gas behavior under condensation conditions.

At low temperatures (low kinetic energy) and high pressures (small intermolecular distances), gas molecules interact noticeably, invalidating ideal gas assumptions and leading to liquefaction.

Identifies the kinetic origin of deviations from ideal gas postulates.

4

Evaluate the fundamental thermodynamic postulate for ideal gas particles.

The average kinetic energy of gas molecules is defined entirely by absolute temperature (

E_k \propto T

), with no potential energy component from intermolecular attractions.

Establishes the quantitative relation governing ideal gas thermal motion.

Key Concept

Kinetic Molecular Theory Postulates across States of Matter

Answer

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