All practice questions

Question 8141Question →

The electromagnetic spectrum consists of waves with varying frequencies, wavelengths, and photon energies, each associated with distinct physical detection mechanisms and applications. Consider the following types of electromagnetic radiation:

I. Radiation emitted by warm bodies, primarily detected using a thermopile.
II. Radiation utilized in radar systems and satellite communications.
III. Radiation emitted during nuclear decay processes, detected using a Geiger-Müller counter.
IV. Radiation responsible for sun tanning and detected by its ability to induce fluorescence on zinc sulfide screens.

Arrange these four types of electromagnetic radiation in order of increasing photon energy (from lowest photon energy to highest photon energy).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence in order of increasing photon energy is: Radiation utilized in radar systems (Microwaves) < Radiation emitted by warm bodies (Infrared) < Radiation causing sun tanning (Ultraviolet) < Radiation emitted during nuclear decay (Gamma rays).

Microwaves possess the lowest frequency among the four types, followed by infrared radiation, then ultraviolet radiation, and finally gamma rays which possess the highest frequency and photon energy.

Step-by-Step Solution

1

Identify the region of the electromagnetic spectrum corresponding to each property and detector described.

Item I corresponds to Infrared radiation; Item II corresponds to Microwaves; Item III corresponds to Gamma rays; Item IV corresponds to Ultraviolet radiation.

Thermopiles detect thermal radiation (IR); radar uses microwaves; Geiger-Müller counters detect nuclear ionizing radiation (Gamma rays); fluorescence on ZnS is caused by UV light.

2

Relate photon energy

E

to frequency

f

and wavelength

\lambda

using Planck's relation

E = hf = \frac{hc}{\lambda}

.

Photon energy is directly proportional to frequency (

E \propto f

) and inversely proportional to wavelength (

E \propto \frac{1}{\lambda}

).

Higher frequency radiation consists of more energetic individual photons.

3

Sequence the identified electromagnetic waves from lowest frequency to highest frequency.

Microwaves (

f \approx 10^9 - 10^{11}\text{ Hz}

) < Infrared (

f \approx 10^{11} - 4 \times 10^{14}\text{ Hz}

) < Ultraviolet (

f \approx 7.5 \times 10^{14} - 3 \times 10^{16}\text{ Hz}

) < Gamma rays (

f > 10^{19}\text{ Hz}

).

This sequence reflects the fundamental order of increasing photon energy across the spectrum.

Key Concept

Electromagnetic Spectrum Spectral Regions, Detection Devices, and Photon Energy Ordering

Question 8142Question →

A quality control analyst evaluated a synthesized sample of benzoic acid using thin-layer chromatography. The solvent front advanced $10.0\text{ cm}$ from the origin line, while the compound produced a single distinct spot that moved $7.5\text{ cm}$ . What is the retardation factor ( $R_f$ ) of the compound, and what does the result reveal about the purity of the sample?

R_f = 0.75

, indicating that the sample is pure.

R_f = 1.33

, indicating that the sample is pure.

R_f = 0.75

, indicating that the sample is impure because soluble impurities lower the boiling point.

R_f = 1.33

, indicating that the sample is an impure mixture.

Show answer & explanation

Answer: $R_f = 0.75$ , indicating that the sample is pure.

Answer

The retardation factor is

R_f = 0.75

, and the single chromatogram spot confirms that the sample is pure.

Calculating the retardation factor gives

R_f = \frac{7.5\text{ cm}}{10.0\text{ cm}} = 0.75

. In paper or thin-layer chromatography, pure substances yield a single spot under specified solvent conditions, confirming chemical purity.

Step-by-Step Solution

1

Calculate the retardation factor (

R_f

) using the chromatography ratio formula.

R_f = \frac{\text{Distance moved by solute}}{\text{Distance moved by solvent front}} = \frac{7.5\text{ cm}}{10.0\text{ cm}} = 0.75

The

R_f

value is the ratio of solute migration distance to solvent migration distance.

2

Interpret the number of spots on the chromatogram to assess purity.

A single distinct spot indicates only one component is present.

Impure samples produce multiple spots corresponding to different components in the mixture.

Key Concept

Criteria of Purity and Retardation Factor (

R_f

) Calculation

Estimated Time:1m 0s

Question 8143Question →

Complete the following statement regarding manganese redox species by filling in the correct numerical oxidation states.

Fill in the blanks below

In potassium manganate(VI),

K_2MnO_4

, the oxidation state of the central manganese atom is , whereas in potassium tetraoxomanganate(VII),

KMnO_4

, the oxidation state of manganese is .

Show answer & explanation

Answer

The oxidation state of manganese in

K_2MnO_4

is +6, and in

KMnO_4

it is +7.

In potassium manganate(VI),

K_2MnO_4

, two potassium ions yield a

+2

total charge and four oxide ions yield

-8

, requiring manganese to be

+6

to balance to zero. In potassium tetraoxomanganate(VII),

KMnO_4

, one potassium ion (

+1

) and four oxide ions (

-8

) leave manganese at

+7

.

Step-by-Step Solution

1

Calculate the oxidation state of manganese in

K_2MnO_4

.

Assign standard oxidation numbers:

K = +1

and

O = -2

. For the neutral molecule

K_2MnO_4

,

2(+1) + \text{Mn} + 4(-2) = 0 \implies +2 + \text{Mn} - 8 = 0 \implies \text{Mn} = +6

.

The sum of oxidation states in a neutral chemical compound must equal zero.

2

Calculate the oxidation state of manganese in

KMnO_4

.

Assign standard oxidation numbers:

K = +1

and

O = -2

. For the neutral molecule

KMnO_4

,

1(+1) + \text{Mn} + 4(-2) = 0 \implies +1 + \text{Mn} - 8 = 0 \implies \text{Mn} = +7

.

The sum of oxidation states in a neutral chemical compound must equal zero.

Key Concept

Determining oxidation states of central transition metal atoms in oxoanions and neutral salts.

Question 8144Question →

In the preparation of potassium ferrate, $K_2FeO_4$ , elemental iron ( $Fe$ ) is oxidized in an alkaline medium. What is the change in the oxidation number of iron per atom during this reaction, and what is the systematic IUPAC name of $K_2FeO_4$ ?

Increase of 6; potassium tetraoxoferrate(VI)

Increase of 4; potassium tetraoxoferrate(VI)

Increase of 4; potassium tetraoxoferrate(IV)

Increase of 8; potassium tetraoxoferrate(VIII)

Show answer & explanation

Answer: Increase of 6; potassium tetraoxoferrate(VI)

Answer

The oxidation number of iron increases by 6, and the systematic IUPAC name of

K_2FeO_4

is potassium tetraoxoferrate(VI).

Free uncombined iron (

Fe

) has an oxidation number of

0

. In

K_2FeO_4

, potassium is

+1

and oxygen is

-2

. Solving

2(+1) + Fe + 4(-2) = 0

yields

Fe = +6

. The change in oxidation number is

+6 - 0 = +6

. The polyatomic anion

FeO_4^{2-}

contains four oxo ligands attached to iron in the

+6

oxidation state, making its systematic IUPAC name potassium tetraoxoferrate(VI).

Step-by-Step Solution

1

Determine the oxidation state of uncombined elemental iron (

Fe

).

Oxidation state of

Fe = 0

.

By IUPAC rules, elements in their free uncombined elemental state always have an oxidation number of zero.

2

Calculate the oxidation state of iron in

K_2FeO_4

.

Oxidation state of

Fe = +6

.

Potassium has an oxidation state of

+1

and oxygen has

-2

. Setting up the charge balance equation:

2(+1) + Fe + 4(-2) = 0 \implies +2 + Fe - 8 = 0 \implies Fe = +6

.

3

Calculate the change in oxidation number.

Change

= +6 - 0 = +6

(an increase of 6).

The change in oxidation state is the final state minus the initial state.

4

Formulate the systematic IUPAC name for

K_2FeO_4

.

Potassium tetraoxoferrate(VI).

The salt consists of potassium cations and the polyatomic anion

FeO_4^{2-}

. With four oxygen atoms surrounding iron in oxidation state

+6

, the IUPAC name is potassium tetraoxoferrate(VI).

Key Concept

Determination of oxidation state changes from elemental forms and IUPAC Stock nomenclature of oxoanions

Question 8145Question →

A student investigates two unlabelled water samples, $X$ and $Y$ . Boiling sample $X$ eliminates its soap-consuming property, allowing it to lather readily, whereas sample $Y$ continues to form scum with soap after boiling. Which solute present in sample $Y$ accounts for its persistent hardness?

Calcium tetraoxosulfate(VI)

Calcium hydrogentrioxocarbonate(IV)

Magnesium hydrogentrioxocarbonate(IV)

Sodium chloride

Show answer & explanation

Answer: Calcium tetraoxosulfate(VI)

Answer

Calcium tetraoxosulfate(VI) accounts for the persistent (permanent) water hardness in sample Y.

Permanent water hardness is caused by dissolved calcium tetraoxosulfate(VI) or magnesium chloride. These salts do not decompose upon heating, so boiling does not precipitate the metal ions responsible for consuming soap.

Step-by-Step Solution

1

Classify the type of hardness based on response to boiling.

Sample X exhibits temporary hardness because boiling removes its soap-consuming capability. Sample Y exhibits permanent hardness because boiling fails to remove its scum-forming property.

Temporary hardness is caused by hydrogen trioxocarbonates which decompose on boiling, whereas permanent hardness is caused by soluble sulfate or chloride salts of calcium or magnesium.

2

Identify the specific chemical species responsible for permanent hardness in sample Y.

Calcium tetraoxosulfate(VI) (

\text{CaSO}_4

) remains dissolved after boiling.

Tetraoxosulfate(VI) ions do not undergo thermal decomposition at boiling temperatures to form insoluble precipitates.

Key Concept

Temporary vs. Permanent Water Hardness

Estimated Time:1m 0s

Question 8146Question →

A liquid sample collected from a reaction vessel is tested in a laboratory and observed to boil continuously over a temperature range of $102^\circ\text{C}$ to $106^\circ\text{C}$ at standard atmospheric pressure. Which classification of matter best describes this sample?

A mixture

A chemical compound

A pure element

A homogeneous pure substance

Show answer & explanation

Answer: A mixture

Answer

The correct answer is a mixture, because impure substances and mixtures boil over a range of temperatures rather than at a distinct, sharp boiling point.

A liquid that boils over a temperature range rather than at a single, fixed temperature is impure and constitutes a mixture. Pure substances (both elements and compounds) have definite compositions and sharp boiling points.

Step-by-Step Solution

1

Analyze the physical behavior described in the question.

The liquid boils across a temperature range of

102^\circ\text{C}

to

106^\circ\text{C}

.

Determining whether a substance has a sharp or broad boiling point is a key test of chemical purity.

2

Apply the criteria of purity for elements, compounds, and mixtures.

Pure elements and compounds have fixed compositions and fixed, sharp boiling points. Mixtures contain varying amounts of components and therefore boil over a temperature range.

The presence of multiple substances in variable proportions alters the vapor pressure continuously during boiling.

Key Concept

Distinction between pure substances (elements and compounds) which have sharp boiling points, and mixtures which boil over a range of temperatures.

Question 8147Question →

A sample of a radioactive nuclide has a half-life of $4\text{ hours}$ . What is the ratio of the number of nuclei that have decayed to the number of undecayed nuclei remaining after a total elapsed time of $16\text{ hours}$ ?

15 : 1

1 : 15

15 : 16

7 : 1

Show answer & explanation

Answer: $15 : 1$

Answer

The ratio of decayed nuclei to remaining undecayed nuclei is

15 : 1

.

In an elapsed time of

16\text{ hours}

(which is

4

half-lives), the remaining undecayed portion of the sample is

(1/2)^4 = 1/16

of the original amount. Consequently,

1 - 1/16 = 15/16

of the nuclei have decayed. The ratio of decayed nuclei to remaining nuclei is

(15/16) : (1/16) = 15 : 1

.

Step-by-Step Solution

1

Determine the number of half-lives (

n

) that have elapsed.

n = \frac{t}{T_{1/2}} = \frac{16\text{ hours}}{4\text{ hours}} = 4\text{ half-lives}

.

The number of half-lives is the total time divided by the half-life duration.

2

Calculate the fraction of original nuclei remaining undecayed (

N/N_0

).

\frac{N}{N_0} = \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^4 = \frac{1}{16}

.

After

n

half-lives, the remaining fraction is

(1/2)^n

.

3

Find the fraction of original nuclei that have decayed (

N_{\text{decayed}}/N_0

).

\frac{N_{\text{decayed}}}{N_0} = 1 - \frac{N}{N_0} = 1 - \frac{1}{16} = \frac{15}{16}.

Decayed fraction equals total initial fraction (1) minus the undecayed fraction.

4

Compute the ratio of decayed nuclei to remaining undecayed nuclei.

\text{Ratio} = \frac{N_{\text{decayed}}}{N} = \frac{15/16}{1/16} = 15 : 1.

The question asks for the ratio of decayed nuclei to remaining undecayed nuclei.

Key Concept

Radioactive Decay Law and Half-life calculations involving ratios of decayed versus remaining quantities.

Question 8148Question →

An alternating voltage source of root-mean-square (RMS) voltage $200\ \text{V}$ is connected across a series combination of a resistor with resistance $60\ \Omega$ and a capacitor with capacitive reactance $80\ \Omega$ . Calculate the root-mean-square current in the circuit.

Show answer & explanation

Answer: 2

Answer

The root-mean-square current flowing through the circuit is 2 A.

The total impedance of the series RC circuit is obtained via quadrature sum

Z = \sqrt{R^2 + X_C^2} = \sqrt{60^2 + 80^2} = 100\ \Omega

. Dividing the RMS supply voltage (

200\ \text{V}

) by this impedance gives an RMS current of

2\ \text{A}

.

Step-by-Step Solution

1

Calculate the total impedance of the series RC circuit.

Z = \sqrt{R^2 + X_C^2} = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = 100\ \Omega

In an AC circuit with resistance and capacitive reactance in series, the total Opposition (impedance) is found by phasor addition.

2

Apply Ohm's law for alternating current circuits to find RMS current.

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{200\ \text{V}}{100\ \Omega} = 2\ \text{A}

The RMS current is equal to the RMS voltage divided by the total circuit impedance.

Key Concept

Impedance and RMS Current in AC Series Circuits

Question 8149Question →

Complete the following statement regarding industrial chemical manufacturing processes by identifying the correct categories of chemicals.

Fill in the blanks below

Chemicals manufactured on a large scale in continuous processes with relatively lower purity for widespread utility are classified as chemicals, whereas specialized compounds like pharmaceuticals and analytical reagents produced in small batches with extremely high purity are known as chemicals.

Show answer & explanation

Answer

The first blank is heavy and the second blank is fine.

Heavy chemicals are manufactured in massive bulk quantities through continuous processes for general industrial applications, while fine chemicals are produced in small batch quantities with high purity standards for specialized uses such as medicine and research.

Step-by-Step Solution

1

Analyze the production scale, process type, and purity level of the first class of chemicals.

Chemicals produced in high tonnage, continuously, and with lower purity (e.g.,

H_2SO_4

,

NaOH

) are heavy chemicals.

Large-scale utility and industrial demand define heavy (bulk) chemicals.

2

Analyze the production scale, process type, and purity level of the second class of chemicals.

Chemicals produced in smaller quantities, via batch processes, with high degree of purity (e.g., drugs, dyes) are fine chemicals.

High purity and specific end-use application define fine chemicals.

Key Concept

Classification and properties of heavy vs fine chemicals

Question 8150Question →

When $10.0\text{ g}$ of pure calcium carbonate ( $\text{CaCO}_3$ ) is strongly heated, it completely decomposes into solid calcium oxide ( $\text{CaO}$ ) and carbon(IV) oxide gas ( $\text{CO}_2$ ). According to the Law of Conservation of Mass, if $5.6\text{ g}$ of calcium oxide remains in the container, what is the mass of carbon(IV) oxide gas released in grams?

Show answer & explanation

Answer: 4.4

Answer

The mass of carbon(IV) oxide gas released is

4.4\text{ g}

.

According to the Law of Conservation of Mass, the total mass of reactants must equal the total mass of products in a chemical change. For the reaction

\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)

, the initial mass of

10.0\text{ g}

of

\text{CaCO}_3

must equal the combined mass of

\text{CaO}

(

5.6\text{ g}

) and

\text{CO}_2

. Subtracting

5.6\text{ g}

from

10.0\text{ g}

yields

4.4\text{ g}

for the gas produced.

Step-by-Step Solution

1

Apply the Law of Conservation of Mass

Total mass of reactants (

10.0\text{ g}

) = Total mass of products (solid residue + gas)

Mass cannot be created or destroyed in a chemical reaction.

2

Calculate the missing mass of carbon(IV) oxide gas

\text{Mass of CO}_2 = 10.0\text{ g} - 5.6\text{ g} = 4.4\text{ g}

Subtracting the mass of the solid product from the initial mass of reactant gives the mass of the gaseous product evolved.

Key Concept

Law of Conservation of Mass

Estimated Time:45s

Question 8151Question →

A light rigid lever of length $2.5\text{ m}$ is pivoted horizontally at one end $O$ . A downward vertical weight of $60\text{ N}$ is hung from the lever at a distance of $1.5\text{ m}$ from $O$ . An upward force $F$ inclined at an angle of $30^\circ$ to the lever is applied at the free end. What is the magnitude of the force $F$ required to maintain horizontal equilibrium?

36\text{ N}

72\text{ N}

104\text{ N}

180\text{ N}

Show answer & explanation

Answer: $72\text{ N}$

Answer

The magnitude of the force

F

required to maintain equilibrium is

72\text{ N}

.

According to the Principle of Moments, for a body in rotational equilibrium, the total clockwise moment about a pivot equals the total counterclockwise moment. The load creates a clockwise moment of

60\text{ N} \times 1.5\text{ m} = 90\text{ N}\cdot\text{m}

. The force

F

applied at

30^\circ

has a perpendicular component of

F \sin 30^\circ = 0.5F

. The counterclockwise moment is

0.5F \times 2.5\text{ m} = 1.25F

. Setting

1.25F = 90\text{ N}\cdot\text{m}

gives

F = 72\text{ N}

.

Step-by-Step Solution

1

Calculate the clockwise moment about the pivot

O

caused by the hanging load.

\text{Clockwise Moment} = 60\text{ N} \times 1.5\text{ m} = 90\text{ N}\cdot\text{m}

The force of

60\text{ N}

acts perpendicularly at a distance of

1.5\text{ m}

from the pivot.

2

Express the counterclockwise moment about the pivot

O

exerted by the force

F

.

\text{Counterclockwise Moment} = F \sin(30^\circ) \times 2.5\text{ m} = 1.25 F\text{ N}\cdot\text{m}

Only the perpendicular component of the force,

F \sin(30^\circ)

, contributes to the moment about pivot

O

.

3

Equate the clockwise and counterclockwise moments according to the Principle of Moments.

1.25 F = 90 \implies F = \frac{90}{1.25} = 72\text{ N}

For rotational equilibrium, the sum of clockwise moments must equal the sum of counterclockwise moments.

Key Concept

Principle of Moments and Perpendicular Force Components

Question 8152Question →

Planet $P$ has a mean uniform density that is twice that of Planet $Q$ , and a radius that is half that of Planet $Q$ . What is the ratio of the escape velocity at the surface of Planet $P$ to the escape velocity at the surface of Planet $Q$ ?

\frac{1}{\sqrt{2}}

\frac{1}{2}

1

\sqrt{2}

Show answer & explanation

Answer: $\frac{1}{\sqrt{2}}$

Answer

\frac{1}{\sqrt{2}}

The correct answer is derived by substituting the planet's mass in terms of radius and density (

M = \frac{4}{3}\pi R^3 \rho

) into the escape velocity equation

v_e = \sqrt{\frac{2GM}{R}}

. Simplifying gives

v_e = R\sqrt{\frac{8}{3}\pi G \rho}

, which means escape velocity is proportional to

R\sqrt{\rho}

. Halving the radius and doubling the density yields a factor of

\frac{1}{2} \times \sqrt{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}

.

Step-by-Step Solution

1

Express the mass of a uniform spherical planet in terms of its radius

R

and density

\rho

.

M = \frac{4}{3}\pi R^3 \rho

Mass equals volume multiplied by mean density.

2

Substitute mass into the escape velocity formula

v_e = \sqrt{\frac{2GM}{R}}

.

v_e = \sqrt{\frac{2G \left(\frac{4}{3}\pi R^3 \rho\right)}{R}} = R\sqrt{\frac{8}{3}\pi G \rho}

This establishes the proportionality

v_e \propto R\sqrt{\rho}

.

3

Set up the ratio of escape velocity for Planet

P

to Planet

Q

using

R_P = \frac{1}{2}R_Q

and

\rho_P = 2\rho_Q

.

\frac{v_{e,P}}{v_{e,Q}} = \frac{R_P\sqrt{\rho_P}}{R_Q\sqrt{\rho_Q}} = \frac{\left(\frac{1}{2}R_Q\right)\sqrt{2\rho_Q}}{R_Q\sqrt{\rho_Q}} = \frac{1}{2}\sqrt{2} = \frac{1}{\sqrt{2}}

Evaluating the ratio of proportional quantities yields the final scale factor.

Key Concept

Dependence of Escape Velocity on Planet Density and Radius

Estimated Time:2m 0s

Question 8153Question →

What is the oxidation state of chlorine in uncombined chlorine gas, $Cl_2$ ?

0

-1

+1

-2

Show answer & explanation

Answer: $0$

Answer

The oxidation state of chlorine in uncombined chlorine gas (

Cl_2

) is

0

.

According to standard rules for assigning oxidation states, any element in its free or uncombined elemental state has an oxidation number of zero. Because

Cl_2

consists of identical chlorine atoms sharing bonding electrons equally, its oxidation state is

0

.

Step-by-Step Solution

1

Identify the chemical state of the given species.

Chlorine is present as free, uncombined diatomic chlorine gas (

Cl_2

).

Redox rules distinguish uncombined elemental states from elements bound in chemical compounds.

2

Apply the fundamental oxidation number rule for uncombined elements.

The oxidation number assigned to any atom in an uncombined element is

0

.

In a homonuclear diatomic molecule like

Cl_2

, electrons are shared equally between identical atoms, resulting in zero net charge allocation.

Key Concept

Oxidation state of uncombined elements

Question 8154Question →

Ammonia gas (

\text{NH}_3

) reacts with oxygen gas (

\text{O}_2

) over a platinum-rhodium catalyst to form nitrogen(II) oxide (

\text{NO}

) and water vapor (

\text{H}_2\text{O}

) according to the equation:

4\text{NH}_{3(g)} + 5\text{O}_{2(g)} \rightarrow 4\text{NO}_{(g)} + 6\text{H}_2\text{O}_{(g)}

Which of the following statements correctly interprets the oxidation process occurring in this reaction from both classical and modern redox perspectives?

Ammonia is oxidized because nitrogen loses hydrogen classically, which corresponds to an increase in the oxidation state of nitrogen from

-3

to

+2

.

Oxygen gas is reduced because elemental

\text{O}_2

starts with an oxidation state of

-2

and gains hydrogen to form

\text{H}_2\text{O}

.

Ammonia acts as an oxidizing agent because nitrogen loses electrons, resulting in a decrease in its oxidation state.

Oxygen gas acts as a reducing agent because its oxidation number increases from

-2

to

0

as it forms water.

Show answer & explanation

Answer: Ammonia is oxidized because nitrogen loses hydrogen classically, which corresponds to an increase in the oxidation state of nitrogen from $-3$ to $+2$ .

Answer

Ammonia is oxidized because nitrogen loses hydrogen classically, which corresponds to an increase in the oxidation state of nitrogen from

-3

to

+2

.

In classical terminology, oxidation involves the removal of hydrogen or addition of oxygen. Ammonia (

\text{NH}_3

) loses hydrogen to form nitrogen(II) oxide (

\text{NO}

), fulfilling classical oxidation. In modern electronic terms, the oxidation number of nitrogen increases from

-3

in

\text{NH}_3

to

+2

in

\text{NO}

, representing a loss of 5 electrons per nitrogen atom, which directly aligns with the modern definition of oxidation.

Step-by-Step Solution

1

Analyze classical redox definitions for ammonia

In

\text{NH}_3

, nitrogen is combined with hydrogen. In forming

\text{NO}

, hydrogen is removed and oxygen is added. Classically, loss of hydrogen and gain of oxygen both define oxidation.

Classical redox concepts define oxidation as the gain of oxygen or loss of hydrogen.

2

Determine oxidation states for nitrogen in reactants and products

In

\text{NH}_3

, hydrogen is

+1

, so

N + 3(+1) = 0 \Rightarrow N = -3

. In

\text{NO}

, oxygen is

-2

, so

N + (-2) = 0 \Rightarrow N = +2

.

Sum of oxidation numbers in a neutral molecule equals zero.

3

Correlate modern redox definition with classical observation

The change in oxidation state of nitrogen from

-3

to

+2

is an increase of

5

, which corresponds to loss of electrons (oxidation).

Modern redox concepts define oxidation as an increase in oxidation state due to loss of electrons.

Key Concept

Integration of Classical (Hydrogen/Oxygen Transfer) and Modern (Oxidation Number/Electron Transfer) Redox Concepts

Estimated Time:2m 0s

Question 8155Question →

A sample of neon gas occupies a volume of $300\text{ cm}^3$ at a pressure of $100\text{ kPa}$ under isothermal conditions. If the pressure on the gas is increased to $150\text{ kPa}$ , what is the final volume of the gas?

200\text{ cm}^3

450\text{ cm}^3

500\text{ cm}^3

150\text{ cm}^3

Show answer & explanation

Answer: $200\text{ cm}^3$

Answer

The final volume of the gas is

200\text{ cm}^3

.

According to Boyle's law, pressure and volume are inversely proportional at constant temperature (

P_1 V_1 = P_2 V_2

). Multiplying the initial pressure (

100\text{ kPa}

) by the initial volume (

300\text{ cm}^3

) gives

30000\text{ kPa}\cdot\text{cm}^3

. Dividing this product by the final pressure (

150\text{ kPa}

) yields the correct volume of

200\text{ cm}^3

.

Step-by-Step Solution

1

Identify the given parameters and constant conditions

P_1 = 100\text{ kPa}

,

V_1 = 300\text{ cm}^3

,

P_2 = 150\text{ kPa}

, temperature is constant.

Boyle's law applies when mass and temperature are kept constant.

2

State Boyle's Law formula

P_1 V_1 = P_2 V_2

Pressure and volume are inversely proportional for a fixed mass of gas at constant temperature.

3

Substitute the values into the formula and solve for the unknown volume (

V_2

)

V_2 = \frac{P_1 V_1}{P_2} = \frac{100 \times 300}{150} = 200\text{ cm}^3

Rearranging the equation yields the new volume following compression.

Key Concept

Boyle's Law states that for a fixed mass of gas at constant temperature, the volume of the gas is inversely proportional to its pressure (

P_1 V_1 = P_2 V_2

).

Question 8156Question →

A Plutonium-239 nucleus (

{^{239}_{94}\text{Pu}}

) captures a thermal neutron (

{^{1}_{0}\text{n}}

) and undergoes nuclear fission according to the reaction equation:

{^{239}_{94}\text{Pu}} + {^{1}_{0}\text{n}} \rightarrow {^{144}_{56}\text{Ba}} + {^{93}_{38}\text{Sr}} + x\,{^{1}_{0}\text{n}} + Q

Given the rest masses:
- Mass of ${^{239}_{94}\text{Pu}} = 239.0522\text{ u}$
- Mass of ${^{1}_{0}\text{n}} = 1.0087\text{ u}$
- Mass of ${^{144}_{56}\text{Ba}} = 143.9229\text{ u}$
- Mass of ${^{93}_{38}\text{Sr}} = 92.9154\text{ u}$
- $1\text{ u} = 931.5\text{ MeV}$

What is the number of emitted neutrons $x$ and the total energy $Q$ released in this fission process?

x = 3

neutrons and

Q = 183.04\text{ MeV}

x = 3

neutrons and

Q = 756.56\text{ MeV}

x = 2

neutrons and

Q = 183.04\text{ MeV}

x = 3

neutrons and

Q = 2.11 \times 10^{-4}\text{ MeV}

Show answer & explanation

Answer: $x = 3$ neutrons and $Q = 183.04\text{ MeV}$

Answer

The reaction releases

x = 3

neutrons and an energy

Q = 183.04\text{ MeV}

.

The conservation of nucleon number requires

239 + 1 = 144 + 93 + x

, giving

x = 3

neutrons. The mass defect is calculated by subtracting the mass of products (

143.9229 + 92.9154 + 3 \times 1.0087 = 239.8644\text{ u}

) from the mass of reactants (

239.0522 + 1.0087 = 240.0609\text{ u}

), yielding

\Delta m = 0.1965\text{ u}

. Converting this to energy gives

Q = 0.1965 \times 931.5\text{ MeV} = 183.04\text{ MeV}

.

Step-by-Step Solution

1

Balance mass number (

A

) to find the number of neutrons

x

239 + 1 = 144 + 93 + x \implies 240 = 237 + x \implies x = 3

Total mass number must be conserved in a nuclear reaction.

2

Calculate the total mass of the reactants (

m_{\text{reactants}}

)

m_{\text{reactants}} = 239.0522\text{ u} + 1.0087\text{ u} = 240.0609\text{ u}

The reactants consist of one Plutonium-239 nucleus and one thermal neutron.

3

Calculate the total mass of the products (

m_{\text{products}}

)

m_{\text{products}} = 143.9229\text{ u} + 92.9154\text{ u} + 3(1.0087\text{ u}) = 239.8644\text{ u}

The products consist of one Barium-144 nucleus, one Strontium-93 nucleus, and three neutrons.

4

Calculate the mass defect (

\Delta m

)

\Delta m = 240.0609\text{ u} - 239.8644\text{ u} = 0.1965\text{ u}

Mass defect is the difference between total mass of reactants and total mass of products.

5

Calculate the liberated energy (

Q

)

Q = 0.1965\text{ u} \times 931.5\text{ MeV/u} = 183.03975\text{ MeV} \approx 183.04\text{ MeV}

Using Einstein's mass-energy equivalence conversion factor

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

.

Key Concept

Nuclear fission conservation laws (mass/atomic number conservation) and mass defect energy equivalence (

Q = \Delta m \cdot 931.5\text{ MeV/u}

).

Estimated Time:2m 0s

Question 8157Question →

In a saturated acyclic alkane such as propane ( $\text{C}_3\text{H}_8$ ), each carbon atom exhibits tetrahedral geometry through $sp^3$ hybridization. What is the percentage of $s$ -orbital character present in each of these hybrid orbitals?

25\%

33.3\%

50\%

75\%

Show answer & explanation

Answer: $25\%$

Answer

The percentage of

s

-orbital character in each

sp^3

hybrid orbital of a tetrahedral carbon atom is

25\%

.

In tetrahedral carbon compounds,

sp^3

hybridization involves mixing one

s

orbital and three

p

orbitals to produce four degenerate orbitals. Therefore, the proportion of

s

-character in each hybrid orbital is

\frac{1}{4}

, which equals

25\%

.

Step-by-Step Solution

1

Determine the composition of orbitals in

sp^3

hybridization.

An

sp^3

hybrid orbital is produced by combining one

s

atomic orbital and three

p

atomic orbitals, yielding a total of 4 equivalent hybrid orbitals.

Hybridization mixes pure atomic orbitals to form equivalent hybrid orbitals around a central tetrahedral carbon atom.

2

Calculate the fractional contribution of the

s

orbital.

The fraction of

s

-character is

\frac{1}{1 + 3} = \frac{1}{4} = 0.25

.

One out of the four constituent atomic orbitals is an

s

orbital.

3

Convert the fraction into a percentage.

0.25 \times 100\% = 25\%

.

Multiplying the fractional contribution by 100 gives the percentage of

s

-character.

Key Concept

Orbital Hybridization and s/p Character in Tetrahedral Carbon

Question 8158Question →

A satellite revolves around a planet in a circular orbit of radius $1.0 \times 10^4 \text{ km}$ with an orbital period of $12 \text{ hours}$ . Calculate the orbital period, in hours, of a second satellite orbiting the same planet in a circular path of radius $4.0 \times 10^4 \text{ km}$ .

Show answer & explanation

Answer: 96

Answer

96 hours

According to Kepler's Third Law (

T^2 \propto r^3

), the orbital period

T

scales with radius

r

as

T \propto r^{3/2}

. Increasing the orbital radius by a factor of 4 increases the period by a factor of

4^{3/2} = 8

. Multiplying the original period of 12 hours by 8 yields 96 hours.

Step-by-Step Solution

1

Set up Kepler's Third Law equation relating orbital period and orbital radius.

\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3}

Kepler's Third Law states that the square of the orbital period of a body in circular orbit is directly proportional to the cube of the radius of its orbit.

2

Substitute the given orbital radii and evaluate the scaling factor.

\frac{r_2}{r_1} = \frac{4.0 \times 10^4 \text{ km}}{1.0 \times 10^4 \text{ km}} = 4

Simplifying the ratio of the two orbital radii gives a factor of 4 increase in radius.

3

Calculate the period multiplier by taking the ratio to the power of 3/2.

4^{3/2} = (\sqrt{4})^3 = 2^3 = 8

Taking

T_2 = T_1 \times \left(\frac{r_2}{r_1}\right)^{3/2}

shows the period scales by a factor of 8.

4

Multiply the initial orbital period by the scaling factor to find the final answer.

T_2 = 12 \text{ hours} \times 8 = 96 \text{ hours}

Multiplying the baseline period of 12 hours by 8 yields the new orbital period.

Key Concept

Kepler's Third Law of Planetary Motion

Estimated Time:1m 30s

Question 8159Question →

A concentrated solution of ethanoic acid ( $CH_3COOH$ ) exhibits higher electrical conductivity than a dilute solution of hydrochloric acid ( $HCl$ ) because the higher concentration provides a greater total number of free mobile ions.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Dilute hydrochloric acid conducts electricity more effectively than concentrated ethanoic acid because hydrochloric acid is a strong acid that ionizes completely to yield a higher concentration of free mobile ions.

The statement is false because the electrical conductivity of an electrolyte solution depends strictly on the concentration of free mobile ions available to carry electric current. Hydrochloric acid (

HCl

) is a strong acid that ionizes completely in water, yielding a high density of

H^+

and

Cl^-

ions. Ethanoic acid (

CH_3COOH

) is a weak acid that ionizes only slightly in water, leaving most molecules unionized. Consequently, even a dilute solution of

HCl

has a much higher concentration of mobile ions and a higher electrical conductivity than a concentrated solution of ethanoic acid.

Step-by-Step Solution

1

Analyze the acid strength and degree of ionization of both compounds.

Hydrochloric acid (

HCl

) is a strong acid that undergoes complete ionization (

HCl(aq) \rightarrow H^+(aq) + Cl^-(aq)

), whereas ethanoic acid (

CH_3COOH

) is a weak acid that undergoes partial ionization (

CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)

).

Electrical conductivity in aqueous solutions depends directly on the concentration of free, mobile charge-carrying ions.

2

Compare the concentration of mobile ions in concentrated weak acid vs. dilute strong acid.

Despite having a high molecular concentration, concentrated ethanoic acid contains mostly neutral, unionized molecules and very few free ions. Dilute

HCl

, even at lower molarity, yields a significantly higher concentration of free ions due to complete dissociation.

Only ionized species conduct electricity in solution; neutral molecules do not carry current.

3

Determine the relative electrical conductivity and evaluate the statement.

Dilute

HCl

is a far better conductor of electricity than concentrated

CH_3COOH

. Thus, the claim in the statement is false.

High acid concentration does not guarantee high conductivity if the acid is weak.

Key Concept

Electrical Conductivity and Ionization of Strong vs. Weak Acids

Question 8160Question →

Match each structural feature or carbon center of 2-methylbut-1-en-3-yne ( $HC\equiv C-C(CH_3)=CH_2$ ) on the left with its corresponding hybridization state, geometric descriptor, or orbital overlap description on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: (1) The

C_2-C_3

single bond matches with the

\sigma

-bond formed by head-on

sp-sp^2

hybrid orbital overlap. (2) The methyl carbon center matches with

sp^3

hybridization, tetrahedral geometry, and

\approx 109.5^\circ

bond angles. (3) The alkene double bond matches with one

\sigma

-bond (

sp^2-sp^2

) and one

\pi

-bond (

2p-2p

). (4) The terminal acetylenic carbon matches with

sp

hybridization, linear geometry, and

180^\circ

bond angle.

Each carbon atom in 2-methylbut-1-en-3-yne adopts a hybridization state determined by its steric number (number of attached atoms and lone pairs).

C_1

and

C_2

are

sp

-hybridized (linear,

180^\circ

),

C_3

and

C_4

are

sp^2

-hybridized (trigonal planar,

120^\circ

), and the methyl group carbon is

sp^3

-hybridized (tetrahedral,

109.5^\circ

). Consequently, single bonds between differently hybridized carbons utilize hybrid orbitals corresponding to each carbon (

sp-sp^2

for

C_2-C_3

), and double bonds consist of one

\sigma

bond (

sp^2-sp^2

) plus one

\pi

bond (

2p-2p

).

Step-by-Step Solution

1

Analyze the expanded structural formula of 2-methylbut-1-en-3-yne

H-C(1)\equiv C(2)-C(3)(CH_3)=C(4)H_2

Determining the bonding domains around each carbon atom establishes its hybridization state and structural role.

2

Determine hybridization state for each carbon center

C_1

(

sp

),

C_2

(

sp

),

C_3

(

sp^2

),

C_4

(

sp^2

), and methyl carbon (

sp^3

)

Carbons with 2 electron domains are

sp

(linear), 3 domains are

sp^2

(trigonal planar), and 4 domains are

sp^3

(tetrahedral).

3

Map orbital overlap types to specific bonds

C_2-C_3

is an

sp-sp^2

\sigma

-bond; double bond

C_3=C_4

comprises an

sp^2-sp^2

\sigma

-bond and a

2p-2p

\pi

-bond.

Single bonds are formed by head-on overlap of hybrid orbitals, while double bonds consist of one coaxial

\sigma

bond and one collateral

\pi

bond.

4

Match left items with their corresponding right item descriptions based on hybridization and geometry principles

All 4 items are accurately matched to their structural characteristics.

Ensures complete alignment between structural features and underlying orbital hybridization properties.

Key Concept

Orbital Hybridization and Overlap Types in Hydrocarbon Frameworks

Estimated Time:2m 0s

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