All practice questions

Question 8061Question →

The presence of a soluble non-volatile impurity in a liquid sample lowers its boiling point below that of the pure liquid.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Soluble non-volatile impurities elevate the boiling point of a pure liquid rather than lowering it.

A soluble non-volatile solute reduces the vapor pressure of a liquid solvent. As a result, extra thermal energy (a higher temperature) is required for the vapor pressure to equal atmospheric pressure, causing boiling point elevation.

Step-by-Step Solution

1

Identify the physical effect of a dissolved non-volatile impurity on liquid vapor pressure.

Dissolved non-volatile solute particles occupy surface area and reduce the vapor pressure of the solvent liquid at a given temperature.

Fewer solvent molecules can escape into the gas phase due to solute-solvent interactions and physical surface blockage.

2

Determine the impact of lower vapor pressure on the boiling point.

The liquid must be heated to a higher temperature for its vapor pressure to equal external atmospheric pressure.

Boiling occurs when vapor pressure equals atmospheric pressure, leading to boiling point elevation.

Key Concept

Effect of Impurities on Boiling Point

Question 8062Question →

A binary solid mixture present at its exact eutectic composition melts sharply at a single fixed temperature, demonstrating that a narrow melting point range alone does not serve as absolute proof of chemical purity.

True

False

Show answer & explanation

Answer: True

Answer

True

The statement is true because a binary mixture at its exact eutectic composition melts completely at a single characteristic temperature (

T_{\text{eutectic}}

) without displaying the broad melting range typical of non-eutectic mixtures. Therefore, while a sharp melting point is a strong indicator of purity, it is not an absolute guarantee without further verification such as mixed melting point determination.

Step-by-Step Solution

1

Analyze standard physical criteria used to evaluate solid substance purity.

Pure solid compounds typically exhibit sharp melting points over a narrow range (

\le 0.5^\circ\text{C}

to

1.0^\circ\text{C}

), whereas ordinary impure mixtures melt over a broad temperature range below the pure substance melting point.

Establishing the general baseline behavior of pure and impure samples.

2

Examine the solid-liquid phase equilibria of binary mixtures at the eutectic point.

At the specific eutectic ratio, both solid components freeze and melt simultaneously at a constant minimum temperature (

T_{\text{eutectic}}

), behaving thermodynamically like a single pure component.

Identifying theoretical exceptions to the rule that mixtures always melt over broad ranges.

3

Evaluate the adequacy of a sharp melting range as an absolute criterion of purity.

Because eutectic mixtures also possess sharp melting points, a narrow melting range alone is an insufficient guarantee of chemical purity unless supported by mixed melting point determination or chromatography.

Confirming the statement is true.

Key Concept

Eutectic Mixtures and Criteria of Purity

Question 8063Question →

A $250\text{ dm}^3$ sample of hard water contains $0.012\text{ mol dm}^{-3}$ of dissolved magnesium tetraoxosulfate(VI), $\text{MgSO}_4$ . What mass, in grams, of anhydrous sodium trioxocarbonate(IV), $\text{Na}_2\text{CO}_3$ , is required to completely precipitate all the magnesium ions as magnesium trioxocarbonate(IV) and soften the water? $[\text{Molar mass of Na}_2\text{CO}_3 = 106\text{ g mol}^{-1}]$

Show answer & explanation

Answer: 318

Answer

318 g of anhydrous sodium trioxocarbonate(IV) is required.

Permanent water hardness caused by soluble magnesium salts like magnesium tetraoxosulfate(VI) (

\text{MgSO}_4

) is removed by reaction with sodium trioxocarbonate(IV) (

\text{Na}_2\text{CO}_3

). The balanced reaction

\text{MgSO}_4\text{(aq)} + \text{Na}_2\text{CO}_3\text{(aq)} \rightarrow \text{MgCO}_3\text{(s)} + \text{Na}_2\text{SO}_4\text{(aq)}

shows a 1:1 molar ratio. A

250\text{ dm}^3

volume at

0.012\text{ mol dm}^{-3}

contains

3.0\text{ moles}

of

\text{MgSO}_4

, which requires

3.0\text{ moles}

of

\text{Na}_2\text{CO}_3

. Multiplying by its molar mass (

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

) gives

318\text{ g}

.

Step-by-Step Solution

1

Calculate the total number of moles of magnesium tetraoxosulfate(VI) in the water sample.

n(\text{MgSO}_4) = 250\text{ dm}^3 \times 0.012\text{ mol dm}^{-3} = 3.0\text{ mol}

Molar amount is calculated by multiplying the solution volume by its molar concentration.

2

Write the balanced chemical equation for softening permanent hardness with sodium trioxocarbonate(IV).

\text{MgSO}_4(\text{aq}) + \text{Na}_2\text{CO}_3(\text{aq}) \rightarrow \text{MgCO}_3(\text{s}) + \text{Na}_2\text{SO}_4(\text{aq})

Soluble magnesium ions causing permanent hardness are removed by precipitation as insoluble magnesium trioxocarbonate(IV).

3

Calculate the mass of anhydrous sodium trioxocarbonate(IV) needed.

\text{Mass} = 3.0\text{ mol} \times 106\text{ g mol}^{-1} = 318\text{ g}

From the 1:1 stoichiometric ratio, 3.0 moles of sodium trioxocarbonate(IV) is required.

Key Concept

Quantitative removal of permanent water hardness using sodium trioxocarbonate(IV) (washing soda)

Question 8064Question →

A non-uniform wooden pole of length $6.0\text{ m}$ and weight $150\text{ N}$ is balanced horizontally on a pivot placed $2.4\text{ m}$ from its heavy end $P$ . The system achieves rotational equilibrium when a load of $50\text{ N}$ is hung directly from end $P$ . What is the distance of the center of gravity of the pole from end $P$ ?

Show answer & explanation

Answer: 3.2

Answer

The distance of the center of gravity of the pole from end

P

is

3.2\text{ m}

.

Taking moments about the pivot at

2.4\text{ m}

from end

P

, the counter-clockwise moment created by the

50\text{ N}

load (

50\text{ N} \times 2.4\text{ m} = 120\text{ N}\cdot\text{m}

) must balance the clockwise moment created by the

150\text{ N}

weight of the pole acting at its center of gravity (

150\text{ N} \times (d - 2.4\text{ m})

). Equating these gives

120 = 150(d - 2.4)

, leading to

d - 2.4 = 0.8\text{ m}

, so

d = 3.2\text{ m}

.

Step-by-Step Solution

1

Identify force positions relative to the pivot

The

50\text{ N}

load is

2.4\text{ m}

to the left of the pivot. The

150\text{ N}

weight acts at the center of gravity, which is

(d - 2.4\text{ m})

to the right of the pivot.

Moments are evaluated relative to the fulcrum to eliminate the unknown normal reaction force at the pivot.

2

Apply the Principle of Moments

Anti-clockwise moment =

50 \times 2.4 = 120\text{ N}\cdot\text{m}

. Clockwise moment =

150 \times (d - 2.4)

. Setting them equal:

120 = 150(d - 2.4)

.

For a body in rotational equilibrium, the total clockwise moment about any pivot equals the total anti-clockwise moment.

3

Solve the equation for distance

d

d - 2.4 = 0.8 \implies d = 3.2\text{ m}

.

Adding the displacement from the pivot (

0.8\text{ m}

) to the pivot position from end

P

(

2.4\text{ m}

) yields the position of the center of gravity from end

P

.

Key Concept

Rotational equilibrium and Principle of Moments for non-uniform rigid bodies

Estimated Time:1m 30s

Question 8065Question →

Match each chemical transformation on the left with the correct classical or modern redox definition that specifically describes it on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Conversion of methane to methanol matches classical oxidation via direct addition of oxygen; conversion of mercury(II) chloride to mercury(I) chloride matches classical reduction via removal of an electronegative element; transformation of magnesium to magnesium fluoride matches modern oxidation via electron loss (

0

to

+2

); conversion of dichromate to chromium(III) ion matches modern reduction via decrease in oxidation number (

+6

to

+3

).

Each chemical transformation is matched to its corresponding classical or modern definition based on fundamental redox principles: oxygen addition represents classical oxidation; removal of an electronegative element represents classical reduction; electron loss and oxidation number increase represent modern oxidation; and oxidation number decrease represents modern reduction.

Step-by-Step Solution

1

Analyze the conversion of

\text{CH}_4

to

\text{CH}_3\text{OH}

using classical definitions.

An oxygen atom is directly added to methane without removing hydrogen. Under classical rules, addition of oxygen is defined as oxidation.

Classical redox concepts classify oxygen gain as oxidation.

2

Analyze

2\text{HgCl}_2 + \text{SnCl}_2 \rightarrow \text{Hg}_2\text{Cl}_2 + \text{SnCl}_4

for

\text{HgCl}_2

.

In

\text{HgCl}_2

, mercury is bound to two chlorines per atom. In

\text{Hg}_2\text{Cl}_2

, mercury is bound to one chlorine per atom. The removal/loss of an electronegative element (

\text{Cl}

) is classical reduction.

Classical definitions extend reduction to include the removal of electronegative elements or addition of electropositive elements.

3

Examine

\text{Mg}_{(s)} \rightarrow \text{MgF}_{2(s)}

under modern concepts.

Elemental magnesium (

\text{Mg}^0

) forms

\text{Mg}^{2+}

ions by losing

2

electrons. Loss of electrons and an increase in oxidation state from

0

to

+2

represents modern oxidation.

Modern redox theory defines oxidation as loss of electrons (OIL) or increase in oxidation number.

4

Examine

\text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+}

using oxidation states.

In

\text{Cr}_2\text{O}_7^{2-}

,

2x + 7(-2) = -2 \Rightarrow x = +6

. In

\text{Cr}^{3+}

, the oxidation state is

+3

. The decrease in oxidation number from

+6

to

+3

represents modern reduction.

A decrease in oxidation state is the defining feature of reduction under modern IUPAC guidelines.

Key Concept

Definitions and Classical vs Modern Concepts of Redox

Question 8066Question →

A $4.00\text{ g}$ sample of a copper oxide is completely reduced by dry hydrogen gas to yield $3.20\text{ g}$ of metallic copper. According to the Law of Definite Proportions, what mass of this same copper oxide, in grams, will be produced when $5.00\text{ g}$ of pure copper is completely oxidized?

Show answer & explanation

Answer: 6.25

Answer

6.25 g

According to the Law of Definite Proportions (or Constant Composition), a chemical compound always contains its component elements in fixed mass ratios. In the first sample, copper makes up

3.20\text{ g} / 4.00\text{ g} = 0.80

or

80\%

of the total mass. Therefore, in any sample of this oxide,

5.00\text{ g}

of copper represents

80\%

of the total mass. Dividing

5.00\text{ g}

by

0.80

gives

6.25\text{ g}

of copper oxide.

Step-by-Step Solution

1

Determine the mass percentage (or mass fraction) of copper in the compound from the first experiment

Mass fraction of Cu = 3.20 / 4.00 = 0.80 (80%)

The first experiment provides quantitative data regarding the mass of copper contained in a known mass of oxide.

2

Apply the Law of Definite Proportions to calculate the required mass of copper oxide for 5.00 g of copper

Mass of Copper Oxide = 5.00 / 0.80 = 6.25 g

The Law of Definite Proportions dictates that the mass composition ratio remains constant regardless of the sample source or size.

Key Concept

Law of Definite Proportions

Question 8067Question →

Match each heat transfer scenario on the left with its dominant microscopic mechanism or physical pathway on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Heat transfer through a copper rod matches energy transport dominated by free electron movement; heat transfer across an evacuated space matches energy transport via electromagnetic waves; heat transfer throughout water heated from the bottom matches energy transport by bulk fluid movement driven by density changes; heat transfer through a porcelain ceramic plate matches energy transport restricted strictly to lattice vibrational waves.

Each heat transfer scenario correctly pairs with its governing physical mechanism: copper conducts heat via free electrons and lattice vibrations, an evacuated space allows thermal energy propagation only through electromagnetic radiation, heated water circulates via density-driven convection currents, and porcelain conducts heat slowly and exclusively via lattice vibrational waves.

Step-by-Step Solution

1

Analyze heat conduction pathways in metals versus non-metallic solids

Metals possess free electrons that diffuse rapidly to transfer kinetic energy along with lattice vibrations. Non-metallic insulators lack mobile free electrons, so thermal conduction occurs at a much slower rate exclusively via lattice vibrations.

Understanding the atomic-level distinction between metallic conductors and non-metallic insulators.

2

Evaluate heat transfer in a medium-free region (vacuum)

Conduction and convection both depend on molecular collisions or particle transport, whereas thermal radiation is an electromagnetic wave phenomenon requiring no material medium.

Identifying radiation as the sole mode capable of propagating across a vacuum.

3

Analyze thermal behavior in fluids heated from below

Thermal expansion reduces the fluid density at the bottom. Gravitational buoyancy forces push the less dense fluid upward while denser, cooler fluid sinks, forming convection currents.

Establishing buoyancy and density differentials as the driving forces of convection.

Key Concept

Microscopic mechanisms of heat conduction, convection, and radiation

Question 8068Question →

A particle of mass $0.50\text{ kg}$ executes simple harmonic motion along a straight line. When its displacement from the equilibrium position is $0.06\text{ m}$ , its speed is $0.80\text{ m/s}$ and its potential energy is $0.09\text{ J}$ . What is the magnitude of the maximum acceleration of the particle in $\text{m/s}^2$ ?

Show answer & explanation

Answer: 10

Answer

The magnitude of the maximum acceleration of the particle is

10\text{ m/s}^2

.

Using potential energy

E_p = \frac{1}{2}m\omega^2 x^2

, the angular frequency

\omega

is

10\text{ rad/s}

. Using

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

, the amplitude

A

is

0.10\text{ m}

. Substituting these values into

a_{\max} = \omega^2 A

yields

10\text{ m/s}^2

.

Step-by-Step Solution

1

Find angular frequency

\omega

from potential energy.

\omega = 10\text{ rad/s}

Using potential energy

E_p = \frac{1}{2}m\omega^2 x^2

:

0.09 = \frac{1}{2}(0.50)\omega^2 (0.06)^2 \implies 0.09 = 0.0009 \omega^2 \implies \omega^2 = 100 \implies \omega = 10\text{ rad/s}

.

2

Find amplitude

A

from speed.

A = 0.10\text{ m}

Using speed

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

:

0.80 = 10\sqrt{A^2 - 0.06^2} \implies 0.08 = \sqrt{A^2 - 0.0036} \implies 0.0064 = A^2 - 0.0036 \implies A^2 = 0.0100 \implies A = 0.10\text{ m}

.

3

Calculate maximum acceleration.

a_{\max} = 10\text{ m/s}^2

Using maximum acceleration formula

a_{\max} = \omega^2 A

:

a_{\max} = 100 \times 0.10 = 10\text{ m/s}^2

.

Key Concept

Simple Harmonic Motion Energy and Kinematic Relations

Question 8069Question →

Match each physical heat transfer scenario on the left with its underlying physical mechanism or governing property on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Heat propagation along a solid copper bar matches free electron diffusion and lattice vibrations. Vertical circulation of water in a vessel matches temperature-dependent density variations causing buoyant fluid motion. Thermal energy transport from the Sun to the Earth matches propagation of electromagnetic waves requiring no material medium. Minimization of heat transport by silvered glass walls matches reflection of infrared radiation by low-emissivity surfaces.

Each physical scenario strictly corresponds to its defining heat transfer process: conduction in metals operates via free electron diffusion and lattice vibration; convection in heated liquids is driven by density changes under gravity; radiation from the Sun traverses space via electromagnetic waves without a physical medium; and silvered thermos coatings prevent radiative transfer by reflecting infrared radiation due to low emissivity.

Step-by-Step Solution

1

Identify the primary mechanism of heat conduction in metals

Conduction in metals relies on both atomic lattice vibrations and the motion of free conduction electrons.

Solids maintain fixed positions, preventing bulk mass displacement, so heat transfers microscopically.

2

Analyze fluid movement under thermal expansion

Heating fluid decreases its local density, causing warm regions to experience upward buoyant forces.

This setup establishes free thermal convection, which requires both a fluid medium and a gravitational field.

3

Evaluate energy transfer through a vacuum

Energy moving through empty space propagates as thermal electromagnetic waves.

Radiation is the unique mode of heat transfer that functions without a physical medium.

4

Examine radiative reflection by low-emissivity coatings

Polished silver coating acts as a mirror to infrared rays, reflecting radiant energy.

Low emissivity directly reduces the rate of radiant heat emission and absorption.

Key Concept

Distinct mechanisms of conduction, convection, and thermal radiation

Estimated Time:1m 30s

Question 8070Question →

According to Gay-Lussac's Law of Combining Volumes, what volume of ammonia gas ( $NH_3$ ) is produced when $20\text{ cm}^3$ of nitrogen gas ( $N_2$ ) reacts completely with excess hydrogen gas ( $H_2$ ) at constant temperature and pressure?

10\text{ cm}^3

20\text{ cm}^3

40\text{ cm}^3

60\text{ cm}^3

Show answer & explanation

Answer: $40\text{ cm}^3$

Answer

40\text{ cm}^3

of ammonia gas (

NH_3

) is produced.

According to the balanced chemical equation

N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)

, 1 mole (or volume) of

N_2

reacts to form 2 moles (or volumes) of

NH_3

. Therefore,

20\text{ cm}^3

of

N_2

produces

2 \times 20\text{ cm}^3 = 40\text{ cm}^3

of

NH_3

gas.

Step-by-Step Solution

1

Write the balanced chemical equation for the synthesis of ammonia

N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)

Stoichiometric coefficients determine the combining volume ratios of reacting and product gases under constant temperature and pressure.

2

Determine the volume ratio between

N_2

and

NH_3

1\text{ volume of } N_2 : 2\text{ volumes of } NH_3

By Gay-Lussac's Law of Combining Volumes, gases react in simple whole-number ratios by volume.

3

Calculate the volume of

NH_3

formed from

20\text{ cm}^3

of

N_2

\text{Volume of } NH_3 = 20\text{ cm}^3 \times 2 = 40\text{ cm}^3

Directly scaling the volume of

N_2

by the coefficient ratio (2/1).

Key Concept

Gay-Lussac's Law of Combining Volumes states that when gases react, they do so in volumes which bear a simple whole-number ratio to one another and to the volume of the product if gaseous, provided temperature and pressure remain constant.

Estimated Time:45s

Question 8071Question →

Two samples of pure sodium chloride obtained from different sources were analyzed quantitatively. The first sample contained $4.60\text{ g}$ of sodium combined with $7.10\text{ g}$ of chlorine. If the second sample contains $14.20\text{ g}$ of chlorine, what mass of sodium is present in the second sample to satisfy the Law of Definite Proportions?

4.60\text{ g}

9.20\text{ g}

11.70\text{ g}

18.80\text{ g}

Show answer & explanation

Answer: $9.20\text{ g}$

Answer

9.20\text{ g}

According to the Law of Definite Proportions (or Constant Composition), a pure chemical compound always contains its elements combined in a fixed ratio by mass, regardless of its source. Since the chlorine mass in the second sample (

14.20\text{ g}

) is twice that of the first sample (

7.10\text{ g}

), the mass of sodium must also be twice as large, giving

2 \times 4.60\text{ g} = 9.20\text{ g}

.

Step-by-Step Solution

1

Determine the mass ratio of sodium to chlorine in the first sample.

The ratio of mass of sodium to mass of chlorine is

\frac{4.60\text{ g}}{7.10\text{ g}} = \frac{46}{71}

.

The Law of Definite Proportions states that a chemical compound always contains its constituent elements in a fixed ratio by mass.

2

Set up the proportion for the second sample with

14.20\text{ g}

of chlorine.

\frac{m_{\text{Na}}}{14.20\text{ g}} = \frac{4.60\text{ g}}{7.10\text{ g}}

.

The mass ratio must remain constant across all samples of the same compound.

3

Solve for the unknown mass of sodium (

m_{\text{Na}}

).

m_{\text{Na}} = 4.60\text{ g} \times \left(\frac{14.20\text{ g}}{7.10\text{ g}}\right) = 4.60\text{ g} \times 2 = 9.20\text{ g}

.

Since

14.20\text{ g}

is exactly twice

7.10\text{ g}

, the mass of sodium required is twice

4.60\text{ g}

.

Key Concept

Law of Definite Proportions (Law of Constant Composition)

Estimated Time:1m 0s

Question 8072Question →

A closed vessel contains a sample of oxygen gas at an initial temperature of $127^\circ\text{C}$ . According to the kinetic theory of gases, if heat is added until the average kinetic energy of the gas molecules is exactly doubled, what is the final temperature of the gas in degrees Celsius?

527^\circ\text{C}

254^\circ\text{C}

800^\circ\text{C}

1073^\circ\text{C}

Show answer & explanation

Answer: $527^\circ\text{C}$

Answer

The final temperature of the gas is

527^\circ\text{C}

.

According to the kinetic molecular theory, the average kinetic energy of gas particles is directly proportional to the absolute temperature in Kelvin (

E_k \propto T

). Converting the initial temperature to Kelvin gives

127 + 273 = 400\text{ K}

. Doubling the average kinetic energy doubles the absolute temperature to

800\text{ K}

. Converting

800\text{ K}

back to degrees Celsius yields

800 - 273 = 527^\circ\text{C}

.

Step-by-Step Solution

1

Convert the initial temperature from degrees Celsius to Kelvin

T_1 = 127^\circ\text{C} + 273 = 400\text{ K}

According to the kinetic theory of gases, average kinetic energy is directly proportional to absolute temperature in Kelvin (

E_k \propto T

).

2

Calculate the final absolute temperature after doubling the average kinetic energy

T_2 = 2 \times 400\text{ K} = 800\text{ K}

Since average kinetic energy is doubled, the absolute temperature must also double.

3

Convert the final temperature back to degrees Celsius

T_{2,^\circ\text{C}} = 800\text{ K} - 273 = 527^\circ\text{C}

The question asks for the final temperature in degrees Celsius.

Key Concept

Postulates of Kinetic Theory and States of Matter

Estimated Time:1m 30s

Question 8073Question →

Consider the gas-phase combustion of methane represented by the balanced equation:

\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)

Given the following average bond dissociation energies:
- $\text{C--H} = 414\text{ kJ mol}^{-1}$
- $\text{O=O} = 498\text{ kJ mol}^{-1}$
- $\text{C=O} = 803\text{ kJ mol}^{-1}$
- $\text{O--H} = 464\text{ kJ mol}^{-1}$

What is the overall enthalpy change ( $\Delta H$ ) for this reaction, and how is the process classified thermodynamically?

-810\text{ kJ mol}^{-1}

, and the reaction is exothermic

+810\text{ kJ mol}^{-1}

, and the reaction is endothermic

-810\text{ kJ mol}^{-1}

, but the heat released is doubled upon adding a catalyst

-1308\text{ kJ mol}^{-1}

, and the reaction is exothermic

Show answer & explanation

Answer: $-810\text{ kJ mol}^{-1}$ , and the reaction is exothermic

Answer

-810\text{ kJ mol}^{-1}

, and the reaction is exothermic

The correct answer is obtained by summing the energies needed to break reactant bonds (

4 \times \text{C--H} + 2 \times \text{O=O} = 2652\text{ kJ}

) and subtracting the energy released from forming product bonds (

2 \times \text{C=O} + 4 \times \text{O--H} = 3462\text{ kJ}

). The result

\Delta H = -810\text{ kJ mol}^{-1}

shows that heat is liberated to the surroundings, defining an exothermic reaction.

Step-by-Step Solution

1

Calculate the total energy required to break all reactant bonds.

Energy absorbed

= 4(\text{C--H}) + 2(\text{O=O}) = 4(414) + 2(498) = 1656 + 996 = 2652\text{ kJ mol}^{-1}

.

Bond breaking is an endothermic process requiring energy input.

2

Calculate the total energy released when forming all product bonds.

Energy released

= 2(\text{C=O}) + 4(\text{O--H}) = 2(803) + 4(464) = 1606 + 1856 = 3462\text{ kJ mol}^{-1}

.

Bond formation is an exothermic process that releases energy.

3

Determine the net enthalpy change (

\Delta H

) and thermodynamic classification.

\Delta H = \text{Energy absorbed} - \text{Energy released} = 2652 - 3462 = -810\text{ kJ mol}^{-1}

. Since

\Delta H < 0

, the reaction is exothermic.

A negative enthalpy change indicates that more energy is released in bond formation than absorbed during bond breaking.

Key Concept

Bond Energy and Enthalpy Change of Reaction

Question 8074Question →

An electromagnetic wave propagating in a vacuum has a frequency of $6.0 \times 10^{14}\text{ Hz}$ . It passes from the vacuum into a dense glass block with a refractive index of $1.50$ . Given that the speed of light in vacuum is $c = 3.0 \times 10^8\text{ m/s}$ , what is the wavelength of the wave inside the glass block, and to which region of the electromagnetic spectrum does the wave belong based on its vacuum properties?

3.33 \times 10^{-7}\text{ m}

, Visible light

5.00 \times 10^{-7}\text{ m}

, Visible light

7.50 \times 10^{-7}\text{ m}

, Infrared

2.25 \times 10^{-7}\text{ m}

, Ultraviolet

Show answer & explanation

Answer: $3.33 \times 10^{-7}\text{ m}$ , Visible light

Answer

The wavelength of the wave inside the glass block is

3.33 \times 10^{-7}\text{ m}

, and the wave belongs to the Visible light region.

The vacuum wavelength of the wave is

\lambda_0 = c/f = 5.0 \times 10^{-7}\text{ m}

, placing it in the visible light spectrum. Upon entering the glass medium (

n = 1.50

), the wave frequency remains unchanged while its wavelength is compressed by the factor

n

, giving

\lambda = (5.0 \times 10^{-7})/1.50 = 3.33 \times 10^{-7}\text{ m}

.

Step-by-Step Solution

1

Calculate the vacuum wavelength of the electromagnetic wave

\lambda_0 = \frac{c}{f} = \frac{3.0 \times 10^8\text{ m/s}}{6.0 \times 10^{14}\text{ Hz}} = 5.0 \times 10^{-7}\text{ m}

The wave equation in vacuum relates wave speed, frequency, and wavelength by

c = f \lambda_0

.

2

Classify the electromagnetic spectral region

Visible light region

A vacuum wavelength of

5.0 \times 10^{-7}\text{ m}

(

500\text{ nm}

) falls within the visible light band (

400\text{ nm} - 700\text{ nm}

).

3

Calculate the wavelength inside the glass block

\lambda = \frac{\lambda_0}{n} = \frac{5.0 \times 10^{-7}\text{ m}}{1.50} = 3.33 \times 10^{-7}\text{ m}

When passing into a medium with refractive index

n

, frequency remains constant while the wave speed and wavelength are reduced by a factor of

n

.

Key Concept

Electromagnetic Wave Refraction and Spectrum Classification

Question 8075Question →

A solid alloy specimen weighs $240\text{ g}$ in air, $180\text{ g}$ when completely immersed in water, and $195\text{ g}$ when completely immersed in an unknown liquid $X$ . What is the relative density of liquid $X$ ?

0.75

0.81

1.33

0.25

Show answer & explanation

Answer: $0.75$

Answer

The relative density of liquid X is

0.75

.

According to Archimedes' principle, relative density of a liquid is the ratio of the upthrust experienced by a solid in that liquid to the upthrust experienced by the same solid in water. Upthrust in water is

240\text{ g} - 180\text{ g} = 60\text{ g}

, and upthrust in liquid X is

240\text{ g} - 195\text{ g} = 45\text{ g}

. Dividing

45

by

60

gives

0.75

.

Step-by-Step Solution

1

Calculate the upthrust in water

240\text{ g} - 180\text{ g} = 60\text{ g}

By Archimedes' principle, upthrust in water equals the mass loss when immersed in water.

2

Calculate the upthrust in liquid X

240\text{ g} - 195\text{ g} = 45\text{ g}

Upthrust in liquid X equals the mass loss when immersed in liquid X.

3

Determine the relative density of liquid X

\text{Relative Density} = \frac{\text{Upthrust in liquid X}}{\text{Upthrust in water}} = \frac{45\text{ g}}{60\text{ g}} = 0.75

Relative density of a liquid is defined as the ratio of the weight of a given volume of the liquid to the weight of an equal volume of water.

Key Concept

Archimedes' Principle and Relative Density of Liquids

Estimated Time:1m 30s

Question 8076Question →

A $0.40\text{ kg}$ mass attached to a light helical spring undergoes simple harmonic motion on a smooth horizontal surface. If the force constant of the spring is $160\text{ N/m}$ and the amplitude of oscillation is $0.05\text{ m}$ , what is the maximum speed of the mass?

1.0\text{ m/s}

20.0\text{ m/s}

0.40\text{ m/s}

4.00\text{ m/s}

Show answer & explanation

Answer: $1.0\text{ m/s}$

Answer

The maximum speed of the mass is

1.0\text{ m/s}

.

The angular frequency of the mass-spring system is calculated using $\omega = \sqrt{k/m} = \sqrt{160/0.40} = 20\text{ rad/s}$. Multiplying this by the amplitude $A = 0.05\text{ m}$ yields a maximum speed of $v_{\text{max}} = 1.0\text{ m/s}$.

Step-by-Step Solution

1

Calculate the angular frequency ($\omega$) of the mass-spring system.

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{160\text{ N/m}}{0.40\text{ kg}}} = \sqrt{400} = 20\text{ rad/s}$

The angular frequency of a spring-mass oscillator depends on the spring constant and the mass.

2

Determine the maximum speed ($v_{\text{max}}$) using the amplitude.

$v_{\text{max}} = \omega A = 20\text{ rad/s} \times 0.05\text{ m} = 1.0\text{ m/s}$

In simple harmonic motion, maximum speed occurs at the equilibrium position and equals the product of angular frequency and amplitude.

Key Concept

Maximum velocity in simple harmonic motion for a mass-spring system

Question 8077Question →

When a $2.0\text{ g}$ sample of a solid solute is completely dissolved in $100.0\text{ g}$ of distilled water initially at $25.0^\circ\text{C}$ , an exothermic process occurs and the temperature of the water rises to $30.0^\circ\text{C}$ . Assuming the specific heat capacity of water is $4.2\text{ J g}^{-1}\text{ K}^{-1}$ and neglecting the heat capacity of the calorimeter container, what is the amount of heat energy, in Joules, absorbed by the water?

Show answer & explanation

Answer: 2100

Answer

The amount of heat energy absorbed by the water is

2100\text{ J}

.

In an exothermic process, heat is transferred to the surroundings (water). Using the calorimetric relation

Q = m c \Delta T

with

m = 100.0\text{ g}

,

c = 4.2\text{ J g}^{-1}\text{ K}^{-1}

, and

\Delta T = (30.0 - 25.0)\text{ K} = 5.0\text{ K}

, the heat gained by the water is

Q = 100.0 \times 4.2 \times 5.0 = 2100\text{ J}

.

Step-by-Step Solution

1

Determine the change in temperature (

\Delta T

) of the water.

\Delta T = T_{\text{final}} - T_{\text{initial}} = 30.0^\circ\text{C} - 25.0^\circ\text{C} = 5.0^\circ\text{C} = 5.0\text{ K}

The heat calculation requires the temperature difference resulting from the exothermic dissolution.

2

Apply the heat formula

Q = m c \Delta T

.

Q = 100.0\text{ g} \times 4.2\text{ J g}^{-1}\text{ K}^{-1} \times 5.0\text{ K}

Heat absorbed depends directly on the mass of water, its specific heat capacity, and the temperature rise.

3

Calculate the total heat energy absorbed.

Q = 2100\text{ J}

Multiplying the mass, heat capacity, and temperature change yields

2100\text{ J}

.

Key Concept

Calorimetric Heat Calculation (

Q = m c \Delta T

)

Question 8078Question →

When dilute tetraoxosulfate(VI) acid ( $\text{H}_2\text{SO}_4$ ) is added to sodium peroxide ( $\text{Na}_2\text{O}_2$ ), hydrogen peroxide ( $\text{H}_2\text{O}_2$ ) is liberated. In contrast, treating lead(IV) oxide ( $\text{PbO}_2$ ) with dilute acids does not produce hydrogen peroxide. Which of the following best accounts for this chemical distinction?

Sodium peroxide contains the peroxide ion (

\text{O}_2^{2-}

), whereas lead(IV) oxide is a true dioxide containing discrete oxide ions (

\text{O}^{2-}

).

Lead(IV) oxide is an amphoteric oxide, whereas sodium peroxide is a neutral oxide.

Sodium peroxide acts purely as a dehydrating agent in acid solution, while lead(IV) oxide is a reducing agent.

Oxygen in sodium peroxide has an oxidation state of

-2

, whereas in lead(IV) oxide oxygen has an oxidation state of

-1

.

Show answer & explanation

Answer: Sodium peroxide contains the peroxide ion ( $\text{O}_2^{2-}$ ), whereas lead(IV) oxide is a true dioxide containing discrete oxide ions ( $\text{O}^{2-}$ ).

Answer

Sodium peroxide contains the peroxide ion (

\text{O}_2^{2-}

), whereas lead(IV) oxide is a true dioxide containing discrete oxide ions (

\text{O}^{2-}

).

Peroxides such as

\text{Na}_2\text{O}_2

contain the peroxide linkage (

-\text{O}-\text{O}-

or

\text{O}_2^{2-}

ion) where oxygen has an oxidation state of

-1

. When treated with dilute acids, the peroxide ion combines with hydrogen ions to yield hydrogen peroxide (

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

). Dioxides such as

\text{PbO}_2

contain standard oxide ions (

\text{O}^{2-}

) paired with a quadrivalent metal ion (

\text{Pb}^{4+}

); they do not contain the peroxide link and therefore cannot produce hydrogen peroxide upon reaction with dilute acids.

Step-by-Step Solution

1

Analyze the structural composition and oxidation states of oxygen in sodium peroxide (

\text{Na}_2\text{O}_2

).

In

\text{Na}_2\text{O}_2

, sodium is

+1

, so oxygen exists as the peroxide ion

\text{O}_2^{2-}

with an oxidation number of

-1

.

Peroxides are characterized by the oxygen-oxygen single bond (

-\text{O}-\text{O}-

) present in the

\text{O}_2^{2-}

group.

2

Analyze the structural composition and oxidation states of oxygen in lead(IV) oxide (

\text{PbO}_2

).

In

\text{PbO}_2

, lead is in the

+4

oxidation state, bonded to two separate oxide ions (

\text{O}^{2-}

), where oxygen has an oxidation number of

-2

.

Dioxides contain metal atoms in high oxidation states paired with simple oxide ions (

\text{O}^{2-}

), rather than peroxide ions.

3

Evaluate the chemical behavior of both compounds with dilute acids.

\text{Na}_2\text{O}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O}_2

(liberates

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

).

\text{PbO}_2

does not yield

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

because it lacks the

\text{O}_2^{2-}

link.

Only compounds containing the structural peroxide ion can form hydrogen peroxide upon acidification.

Key Concept

Distinction between peroxides (

\text{O}_2^{2-}

) and dioxides (

\text{O}^{2-}

)

Estimated Time:1m 15s

Question 8079Question →

A sulfide ion is represented by the symbol $^{32}_{16}\text{S}^{2-}$ . How many neutrons, protons, and electrons are present in this ion?

16 neutrons, 16 protons, and 18 electrons

16 neutrons, 16 protons, and 14 electrons

32 neutrons, 16 protons, and 16 electrons

16 neutrons, 16 protons, and 16 electrons

Show answer & explanation

Answer: 16 neutrons, 16 protons, and 18 electrons

Answer

16 neutrons, 16 protons, and 18 electrons

For the nuclide symbol

^{32}_{16}\text{S}^{2-}

, the atomic number is 16, which dictates that the species has 16 protons. The mass number is 32, so the number of neutrons is

32 - 16 = 16

. Because the species is a divalent anion (

\text{S}^{2-}

), it has gained 2 electrons beyond its neutral count of 16, giving a total of 18 electrons.

Step-by-Step Solution

1

Determine the number of protons and neutrons from the nuclide symbol

Protons = 16, Neutrons = 16

The lower subscript represents the atomic number (

Z = 16

), which equals the number of protons. The upper superscript represents the mass number (

A = 32

). The number of neutrons is calculated as

A - Z = 32 - 16 = 16

.

2

Determine the number of electrons for the charged species

Electrons = 18

The ion carries a charge of

-2

, meaning it has gained 2 extra electrons compared to its neutral atomic state (

16 + 2 = 18

).

Key Concept

Subatomic Particle Calculations in Ions

Question 8080Question →

A solid block of wood has a mass of $0.60\text{ kg}$ . When placed in a vessel of water, it floats freely on the surface. What is the magnitude of the upthrust exerted by the water on the block? (Take $g = 10\text{ m/s}^2$ )

6.0\text{ N}

0.60\text{ N}

0.0\text{ N}

60\text{ N}

Show answer & explanation

Answer: $6.0\text{ N}$

Answer

The upthrust exerted by the water on the floating block is

6.0\text{ N}

.

According to the Law of Flotation, a body floating freely in a fluid displaces a weight of fluid equal to its own weight. Therefore, the upward force (upthrust) exerted by the fluid is equal to the weight of the block:

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

.

Step-by-Step Solution

1

Determine the weight of the floating block in air.

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

Weight is the force of gravity acting on the block's mass.

2

Apply the Law of Flotation to find the upthrust.

Upthrust

U = W = 6.0\text{ N}

A freely floating body displaces a volume of fluid whose weight is equal to the total weight of the body.

Key Concept

Law of Flotation and Archimedes' Principle

Estimated Time:45s

Answer

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