All practice questions

Question 8121Question →

Element $X$ exists naturally as three isotopes with mass numbers $24$ , $25$ , and $26$ . The relative atomic mass of element $X$ is $24.32$ . If the natural abundance of the isotope $^{25}X$ is $10\%$ , what is the percentage abundance of the isotope $^{26}X$ ?

Show answer & explanation

Answer: 11

Answer

The percentage abundance of

^{26}X

is

11\%

.

By setting the percentage abundance of

^{24}X

as

x

and

^{26}X

as

z

, given

^{25}X = 10\%

, we have

x + z = 90\%

. Using the relative atomic mass formula

24x + 25(10) + 26z = 2432

and substituting

x = 90 - z

yields

2410 + 2z = 2432

, solving to

z = 11\%

.

Step-by-Step Solution

1

Formulate the abundance relation for the three isotopes.

The sum of abundances is

x + 10 + z = 100\%

, which gives

x = 90 - z

.

The sum of percentage abundances of all naturally occurring isotopes of an element must equal 100%.

2

Set up the weighted relative atomic mass equation.

24x + 25(10) + 26z = 2432

.

Relative atomic mass is the weighted average of isotopic masses based on fractional abundance.

3

Substitute

x = 90 - z

and solve for

z

.

24(90 - z) + 250 + 26z = 2432 \implies 2410 + 2z = 2432 \implies z = 11\%

.

Substituting

x

reduces the equation to a single variable

z

, representing the percentage abundance of

^{26}X

.

Key Concept

Calculation of Isotopic Abundances from Relative Atomic Mass

Question 8122Question →

Match each chemical species listed on the left with its correct systematic IUPAC name on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are:

MnO_4^{2-}

matches Tetraoxomanganate(VI) ion;

ClO_4^-

matches Tetraoxochlorate(VII) ion;

N_2O

matches Dinitrogen(I) oxide; and

Fe_2O_3

matches Iron(III) oxide.

Each chemical formula correctly corresponds to its IUPAC name based on the calculated oxidation number of the electropositive element and standard IUPAC nomenclature rules for oxoanions and oxides.

Step-by-Step Solution

1

Determine the oxidation number of manganese in

MnO_4^{2-}

Manganese has an oxidation state of

+6

.

Applying the algebraic rule for polyatomic ions:

x + 4(-2) = -2 \implies x = +6

. The four oxygen atoms prefix as 'tetraoxo-', giving tetraoxomanganate(VI) ion.

2

Determine the oxidation number of chlorine in

ClO_4^-

Chlorine has an oxidation state of

+7

.

Applying the algebraic rule:

x + 4(-2) = -1 \implies x = +7

. The species is named tetraoxochlorate(VII) ion.

3

Determine the oxidation number of nitrogen in

N_2O

Nitrogen has an oxidation state of

+1

.

Applying the neutrality rule:

2x + (-2) = 0 \implies 2x = +2 \implies x = +1

. The IUPAC name is dinitrogen(I) oxide.

4

Determine the oxidation number of iron in

Fe_2O_3

Iron has an oxidation state of

+3

.

Applying the neutrality rule:

2x + 3(-2) = 0 \implies 2x = +6 \implies x = +3

. The IUPAC name is iron(III) oxide.

Key Concept

Calculation of oxidation numbers for central atoms in oxoanions and binary oxides to deduce standard IUPAC names.

Question 8123Question →

Element $M$ has a relative atomic mass of $24.32$ . It exists naturally as three isotopes: $^{24}M$ with a relative abundance of $79\%$ , $^{25}M$ with a relative abundance of $10\%$ , and an isotope $^{x}M$ with a relative abundance of $11\%$ . What is the mass number ( $x$ ) of the third isotope?

Show answer & explanation

Answer: 26

Answer

The mass number of the third isotope is 26.

The relative atomic mass is the weighted average of isotopic mass numbers:

24.32 = \frac{(24 \times 79) + (25 \times 10) + (x \times 11)}{100}

. Simplifying gives

2432 = 1896 + 250 + 11x

, which simplifies to

11x = 286

, yielding

x = 26

.

Step-by-Step Solution

1

Set up the relative atomic mass equation based on percentage abundances.

\text{RAM} = (24 \times 0.79) + (25 \times 0.10) + (x \times 0.11)

The relative atomic mass of an element is the weighted average of the mass numbers of its naturally occurring isotopes.

2

Calculate the mass contributions of the first two isotopes.

24 \times 0.79 = 18.96

and

25 \times 0.10 = 2.50

, giving a combined sum of

21.46

Evaluating the contribution of known isotopes isolates the variable term.

3

Subtract the combined contribution from the given relative atomic mass to find the contribution of the third isotope.

0.11x = 24.32 - 21.46 = 2.86

The remaining mass contribution must come entirely from the third isotope.

4

Divide by the fractional abundance of the third isotope to solve for

x

.

x = \frac{2.86}{0.11} = 26

Dividing the mass contribution by fractional abundance yields the integer mass number.

Key Concept

Relative Atomic Mass Calculation from Isotopic Abundances

Estimated Time:1m 15s

Question 8124Question →

During a nuclear reaction, a mass defect of $0.02\text{ u}$ is observed. Given that $1\text{ u} = 931\text{ MeV}$ , what is the total energy released in this reaction?

18.62\text{ MeV}

46\,550\text{ MeV}

931.02\text{ MeV}

18.62\text{ J}

Show answer & explanation

Answer: $18.62\text{ MeV}$

Answer

18.62\text{ MeV}

The energy released in a nuclear reaction is determined by multiplying the mass defect

\Delta m

by the energy equivalence factor (

931\text{ MeV}

per atomic mass unit). Multiplying

0.02\text{ u}

by

931\text{ MeV/u}

yields

18.62\text{ MeV}

.

Step-by-Step Solution

1

Identify the mass defect and the energy equivalence factor.

Mass defect

\Delta m = 0.02\text{ u}

and

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

.

The energy released in a nuclear reaction is directly proportional to its mass defect.

2

Calculate the total energy released by multiplying mass defect by energy equivalent per unit mass.

E = 0.02\text{ u} \times 931\text{ MeV/u} = 18.62\text{ MeV}

.

Applying the conversion factor yields the energy released in

\text{MeV}

.

Key Concept

Mass defect and energy conversion in nuclear reactions

Question 8125Question →

What volume of dry oxygen gas measured at STP is produced when $12.25\text{ g}$ of potassium trioxochlorate(V), $\text{KClO}_3$ , is completely decomposed by heating in the presence of manganese(IV) oxide catalyst?

$[\text{K} = 39.0, \text{Cl} = 35.5, \text{O} = 16.0, \text{Molar volume of gas at STP} = 22.4\text{ dm}^3\text{mol}^{-1}]$

2.24\text{ dm}^3

3.36\text{ dm}^3

3.60\text{ dm}^3

6.72\text{ dm}^3

Show answer & explanation

Answer: $3.36\text{ dm}^3$

Answer

3.36\text{ dm}^3

Thermal decomposition of potassium trioxochlorate(V) follows

2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2

. Since

12.25\text{ g}

of

\text{KClO}_3

corresponds to

0.10\text{ mol}

, the reaction yields

0.15\text{ mol}

of

\text{O}_2

. At STP,

0.15\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 3.36\text{ dm}^3

of oxygen gas.

Step-by-Step Solution

1

Write the balanced chemical equation for the thermal decomposition of potassium trioxochlorate(V).

2\text{KClO}_3\text{(s)} \rightarrow 2\text{KCl(s)} + 3\text{O}_2\text{(g)}

Establishing the correct stoichiometric mole ratio between reactant and gaseous product is essential for calculations.

2

Calculate the molar mass of

\text{KClO}_3

and determine the number of moles reacted.

Molar mass of

\text{KClO}_3 = 39.0 + 35.5 + (3 \times 16.0) = 122.5\text{ g mol}^{-1}

. Moles of

\text{KClO}_3 = \frac{12.25\text{ g}}{122.5\text{ g mol}^{-1}} = 0.10\text{ mol}

.

Mass must be converted to moles to utilize equation stoichiometry.

3

Determine the moles of

\text{O}_2

produced using the mole ratio.

Moles of

\text{O}_2 = 0.10\text{ mol KClO}_3 \times \frac{3\text{ mol O}_2}{2\text{ mol KClO}_3} = 0.15\text{ mol O}_2

.

Two moles of

\text{KClO}_3

yield three moles of

\text{O}_2

.

4

Calculate the volume of

\text{O}_2

produced at STP.

\text{Volume of } \text{O}_2 = 0.15\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 3.36\text{ dm}^3$.

Multiplying the amount of gas in moles by the molar gas volume at STP gives the volume.

Key Concept

Laboratory preparation of oxygen gas via catalytic thermal decomposition of potassium trioxochlorate(V) and stoichiometric volume calculations at STP.

Question 8126Question →

Match each quantum rule or principle with its correct statement regarding electronic configuration.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Hund's Rule matches single occupancy of degenerate orbitals before pairing; Pauli Exclusion Principle matches the restriction that no two electrons share four identical quantum numbers; Aufbau Principle matches filling lowest energy orbitals first.

Hund's rule describes filling degenerate orbitals singly first with parallel spins. The Pauli exclusion principle mandates that no two electrons in an atom possess identical sets of four quantum numbers. The Aufbau principle specifies filling orbitals starting from the lowest energy level.

Step-by-Step Solution

1

Identify the definition of Hund's Rule of Maximum Multiplicity.

Hund's rule specifies that degenerate orbitals are occupied singly first to minimize electron-electron repulsion.

Electrons in different orbitals with parallel spins lower electrostatic energy.

2

Identify the definition of the Pauli Exclusion Principle.

Pauli's principle states that an orbital can hold at most two electrons of opposite spin.

This guarantees that every electron in an atom has a unique set of four quantum numbers (

n, l, m_l, m_s

).

3

Identify the definition of the Aufbau Principle.

The Aufbau principle dictates that subshells fill in order of increasing energy according to the

(n+l)

rule.

Ground states require minimum total electronic potential energy.

Key Concept

Fundamental Quantum Rules Governing Ground-State Electronic Configuration

Question 8127Question →

A Geiger-Müller counter records a total count rate of $340\text{ counts per minute}$ near a radioactive source. The background radiation in the laboratory produces a steady count rate of $20\text{ counts per minute}$ . If the total count rate recorded by the counter drops to $60\text{ counts per minute}$ after an elapsed time of $15\text{ minutes}$ , what is the half-life of the radioactive source in minutes?

Show answer & explanation

Answer: 5

Answer

The half-life of the radioactive source is

5\text{ minutes}

.

To find the true activity of the radioactive source, the constant background radiation of

20\text{ cpm}

must be subtracted from all detector readings. The initial source activity is

340 - 20 = 320\text{ cpm}

and the activity after

15\text{ minutes}

is

60 - 20 = 40\text{ cpm}

. The fraction of source activity remaining is

40 / 320 = 1/8 = (1/2)^3

, which means

3

half-lives have elapsed in

15\text{ minutes}

. Dividing the total time by the number of half-lives (

15 / 3

) gives a half-life of

5\text{ minutes}

.

Step-by-Step Solution

1

Calculate the initial activity of the radioactive source by subtracting the background count rate.

A_0 = 340\text{ cpm} - 20\text{ cpm} = 320\text{ cpm}

Background radiation contributes to the detector reading and must be isolated from the source activity.

2

Calculate the activity of the source after 15 minutes by subtracting the background count rate.

A(t) = 60\text{ cpm} - 20\text{ cpm} = 40\text{ cpm}

The background count remains constant at

20\text{ cpm}

, so the actual count due to the source is

40\text{ cpm}

.

3

Determine the remaining fraction of the radioactive source.

\frac{A(t)}{A_0} = \frac{40}{320} = \frac{1}{8}

Radioactive decay follows an exponential decay law based on the fraction of initial undecayed nuclei.

4

Calculate the number of elapsed half-lives

n

.

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

The remaining fraction equals

(1/2)^n

where

n

is the number of half-lives.

5

Compute the half-life

T_{1/2}

.

T_{1/2} = \frac{t}{n} = \frac{15\text{ minutes}}{3} = 5\text{ minutes}

The total elapsed time is the product of the number of half-lives and the duration of one half-life.

Key Concept

Radioactive Decay Law and Half-life with Background Radiation Correction

Question 8128Question →

A sample of a radioactive substance has a decay constant of $0.0154\text{ h}^{-1}$ . What is the elapsed time, in hours, required for $87.5\%$ of the original sample to decay? (Take $\ln 2 = 0.693$ )

Show answer & explanation

Answer: 135

Answer

135 hours

First, calculate the half-life of the substance using the relation

T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{0.0154\text{ h}^{-1}} = 45\text{ hours}

. Since

87.5\%

of the sample has decayed, the remaining fraction of the sample is

100\% - 87.5\% = 12.5\% = \frac{1}{8}

. Expressing

\frac{1}{8}

as a power of

\frac{1}{2}

gives

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

, indicating that

3

half-lives have passed. The total elapsed time is therefore

3 \times 45\text{ hours} = 135\text{ hours}

.

Step-by-Step Solution

1

Calculate the half-life from the given decay constant

Half-life

T_{1/2} = 45\text{ hours}

The decay constant

\lambda

and half-life

T_{1/2}

are related by the formula

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

2

Find the remaining percentage and fraction of the sample

Remaining fraction is

12.5\%

or

\frac{1}{8}

Radioactive decay equations use the undecayed remaining amount, which is

100\% - 87.5\% = 12.5\%

3

Calculate the number of half-lives elapsed

Number of half-lives

n = 3

Since

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

, solving for

n

gives

n = 3

4

Compute the total elapsed time

Total elapsed time

t = 135\text{ hours}

Total elapsed time is the product of the number of half-lives and the half-life duration (

3 \times 45\text{ hours}

)

Key Concept

Radioactive Decay Law and Half-life Relationship

Question 8129Question →

A chemical engineer is selecting a location for a new heavy chemical manufacturing plant producing fertilizer in Nigeria. How should the following siting factors be arranged in decreasing order of priority (from most critical to least critical)?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order of priority from most critical to least critical is: Proximity to bulk raw materials, Availability of continuous industrial power and water supply, Access to heavy transport infrastructure, and Proximity to immediate urban consumer retail markets.

For heavy chemical manufacturing plants, proximity to raw materials is paramount because raw inputs are bulky and costly to transport. Availability of continuous power and water ranks second to ensure uninterrupted continuous-flow operations. Heavy transport infrastructure (rail and sea links) ranks third to facilitate bulk freight movements. Proximity to local urban consumer markets is the lowest priority because heavy chemicals are intermediate goods sold to other industries, not end-user retail products.

Step-by-Step Solution

1

Analyze the primary cost driver for heavy chemical manufacturing.

Heavy chemical industries process huge volumes of bulky, low-value-per-unit raw inputs.

Locating the plant close to raw material deposits minimizes the massive freight expenses associated with transporting crude inputs.

2

Evaluate key processing utility requirements.

Industrial power grids and high-capacity water sources represent the second highest priority.

Heavy chemical reactions operate continuously and demand vast amounts of water for cooling and steam generation along with reliable electric power.

3

Compare transport logistics against market location.

Access to heavy transport networks (railways/ports) takes priority over urban retail market proximity.

Heavy chemicals serve as raw materials for secondary factories across wide geographic regions rather than being sold in local urban retail markets.

Key Concept

Priority of siting factors in heavy vs light chemical industries

Estimated Time:1m 30s

Question 8130Question →

A uniform beam $PQ$ of length $4.0\text{ m}$ and mass $20\text{ kg}$ is supported horizontally on a pivot at end $P$ and by a vertical wire attached at end $Q$ . A load of mass $30\text{ kg}$ is placed on the beam at a distance of $1.0\text{ m}$ from $P$ . Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the tension in the vertical wire attached at $Q$ in newtons?

Show answer & explanation

Answer: 175

Answer

The tension in the vertical wire attached at end

Q

is

175\text{ N}

.

By applying the principle of moments about the pivot at

P

, the sum of downward clockwise moments produced by the

30\text{ kg}

load (

300\text{ N} \times 1.0\text{ m} = 300\text{ N}\cdot\text{m}

) and the beam's center of gravity (

200\text{ N} \times 2.0\text{ m} = 400\text{ N}\cdot\text{m}

) equals

700\text{ N}\cdot\text{m}

. Equating this to the counterclockwise moment of the tension force (

T \times 4.0\text{ m}

) gives

T = 175\text{ N}

.

Step-by-Step Solution

1

Determine forces and their perpendicular distances from the pivot at

P

.

The load exerts a downward force of

300\text{ N}

at

1.0\text{ m}

from

P

. The uniform beam's weight of

200\text{ N}

acts at its midpoint (

2.0\text{ m}

from

P

). The vertical tension

T

acts upward at

Q

(

4.0\text{ m}

from

P

).

Before applying the principle of moments, all force magnitudes and their distance arms relative to the pivot point must be identified.

2

Equate total clockwise moments to total counterclockwise moments about

P

.

(300\text{ N} \times 1.0\text{ m}) + (200\text{ N} \times 2.0\text{ m}) = T \times 4.0\text{ m}

For rotational equilibrium, the sum of clockwise moments about any pivot must equal the sum of counterclockwise moments about that same pivot.

3

Calculate the value of the tension force

T

.

T = \frac{700}{4.0} = 175\text{ N}

Simplifying the moment equation gives the magnitude of the upward supporting force.

Key Concept

Principle of Moments and Rotational Equilibrium

Question 8131Question →

Two solid cylindrical copper rods, $P$ and $Q$ , are maintained under identical temperature differences across their ends. Rod $P$ has twice the radius and half the length of Rod $Q$ . What is the ratio of the rate of heat conduction through Rod $P$ to that through Rod $Q$ ?

8 : 1

4 : 1

2 : 1

1 : 8

Show answer & explanation

Answer: $8 : 1$

Answer

The ratio of the rate of heat conduction through Rod

P

to that through Rod

Q

is

8 : 1

.

The rate of heat conduction is given by

\frac{Q}{t} = \frac{k \pi r^2 \Delta T}{L}

. For Rod

P

, substituting

r_P = 2 r_Q

and

L_P = 0.5 L_Q

gives

\frac{(2)^2}{0.5} = 8

times the rate of Rod

Q

. Thus, the ratio of heat conduction rate is

8 : 1

.

Step-by-Step Solution

1

Write the fundamental equation for the rate of thermal conduction through a uniform conductor.

\frac{Q}{t} = \frac{k A \Delta T}{L}

Heat conduction rate is proportional to thermal conductivity

k

, cross-sectional area

A

, temperature difference

\Delta T

, and inversely proportional to length

L

.

2

Express the cross-sectional area

A

in terms of radius

r

for a cylindrical rod.

A = \pi r^2 \implies \frac{Q}{t} = \frac{k \pi r^2 \Delta T}{L}

The cross-section of a cylindrical rod is a circle.

3

Set up the ratio of heat conduction rates for Rod

P

and Rod

Q

given

r_P = 2 r_Q

and

L_P = 0.5 L_Q

.

\frac{(Q/t)_P}{(Q/t)_Q} = \frac{r_P^2 / L_P}{r_Q^2 / L_Q} = \frac{(2 r_Q)^2 / (0.5 L_Q)}{r_Q^2 / L_Q} = \frac{4}{0.5} = 8

Material (

k

) and temperature difference (

\Delta T

) are identical for both rods and cancel out.

Key Concept

Thermal Conduction Rate Formula

Question 8132Question →

An analyst examined a sample of benzoic acid synthesized in a laboratory. During paper chromatography, the solvent front migrated $12.0\text{ cm}$ from the origin line, while the primary spot migrated $4.8\text{ cm}$ and a faint secondary spot appeared at $9.6\text{ cm}$ . Subsequent thermal analysis showed that the sample melted over a range of $115.0^\circ\text{C}$ to $120.5^\circ\text{C}$ , whereas pure benzoic acid has a sharp literature melting point of $122.3^\circ\text{C}$ .

What is the retention factor ( $R_f$ ) of the primary component, and what does the data reveal regarding the sample's purity?

R_f = 0.40

; the sample is impure because it exhibits a broad melting range below the literature value and forms multiple chromatographic spots.

R_f = 2.50

; the sample is impure because the presence of impurities inverts the solvent-to-solute ratio during retention factor calculations.

R_f = 0.40

; the sample is pure because non-volatile impurities elevate boiling points while leaving solid melting ranges unaffected.

R_f = 2.50

; the sample is pure because the secondary spot represents a solvent artifact and the primary spot moved faster than the solvent front.

Show answer & explanation

Answer: $R_f = 0.40$ ; the sample is impure because it exhibits a broad melting range below the literature value and forms multiple chromatographic spots.

Answer

The retention factor (

R_f

) of the primary component is

0.40

, and the sample is impure because it exhibits a depressed, broad melting point range and multiple spots on the chromatogram.

The retention factor is given by

R_f = \frac{4.8\text{ cm}}{12.0\text{ cm}} = 0.40

. The criteria of purity state that a pure solid melts sharply at a single characteristic temperature. A melting range (

115.0^\circ\text{C} - 120.5^\circ\text{C}

) lower and wider than the literature value (

122.3^\circ\text{C}

), combined with multiple chromatographic spots, conclusively proves the sample is impure.

Step-by-Step Solution

1

Calculate the retention factor (

R_f

) of the primary component.

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

By definition,

R_f

is the ratio of solute distance to solvent front distance from the origin line.

2

Analyze the chromatography spot evidence for chemical purity.

The presence of two distinct spots (at

4.8\text{ cm}

and

9.6\text{ cm}

) indicates at least two components in the mixture.

A pure chemical compound produces exactly one spot on a developed chromatogram under uniform conditions.

3

Evaluate the melting point behavior against purity criteria.

The sample melts gradually over a

5.5^\circ\text{C}

temperature span (

115.0^\circ\text{C} - 120.5^\circ\text{C}

) below the pure literature value (

122.3^\circ\text{C}

).

Impure solids melt over a broad temperature range and at temperatures lower than the pure compound's sharp melting point.

Key Concept

Criteria of Purity: Sharp melting point / boiling point, fixed physical constants, and single-spot chromatogram.

Estimated Time:1m 30s

Question 8133Question →

A $250\text{ cm}^3$ sample of dry air is passed over excess heated phosphorus in a closed tube to remove all the oxygen gas present. Assuming oxygen constitutes $21\%$ by volume of dry air, what is the volume of the remaining gas mixture in $\text{cm}^3$ ?

Show answer & explanation

Answer: 197.5

Answer

The volume of the remaining gas mixture is

197.5\text{ cm}^3

.

Because oxygen makes up

21\%

by volume of dry air, a

250\text{ cm}^3

sample contains

0.21 \times 250\text{ cm}^3 = 52.5\text{ cm}^3

of oxygen. Heated phosphorus reacts with all the oxygen to form solid phosphorus oxide, leaving behind

250\text{ cm}^3 - 52.5\text{ cm}^3 = 197.5\text{ cm}^3

of unreacted gases.

Step-by-Step Solution

1

Calculate the volume of oxygen gas in the initial sample

52.5\text{ cm}^3

Oxygen makes up

21\%

by volume of dry air, so

0.21 \times 250\text{ cm}^3 = 52.5\text{ cm}^3

.

2

Determine the remaining gas volume after complete removal of oxygen

197.5\text{ cm}^3

Phosphorus reacts completely with oxygen, leaving the unreacted components of air:

250\text{ cm}^3 - 52.5\text{ cm}^3 = 197.5\text{ cm}^3

.

Key Concept

Percentage composition of air by volume

Question 8134Question →

An electromagnetic microwave signal used in telecommunication has a wavelength of $0.02\text{ m}$ in a vacuum. Given that the speed of light in a vacuum is $3.0 \times 10^8\text{ m/s}$ , calculate the frequency of the signal in gigahertz ( $\text{GHz}$ ).

Show answer & explanation

Answer: 15

Answer

The frequency of the microwave signal is

15\text{ GHz}

.

Using the electromagnetic wave equation

c = f \lambda

, the frequency in Hz is calculated as

f = \frac{c}{\lambda} = \frac{3.0 \times 10^8\text{ m/s}}{0.02\text{ m}} = 1.5 \times 10^{10}\text{ Hz}

. Dividing by

10^9

to convert into gigahertz gives

15\text{ GHz}

.

Step-by-Step Solution

1

Identify the given parameters and formula.

Speed of light

c = 3.0 \times 10^8\text{ m/s}

, wavelength

\lambda = 0.02\text{ m}

, and wave equation

c = f \lambda

.

Electromagnetic waves propagate at speed

c

in a vacuum, relating frequency and wavelength.

2

Calculate the frequency in Hertz (Hz).

f = \frac{3.0 \times 10^8}{0.02} = 1.5 \times 10^{10}\text{ Hz}

.

Dividing the wave speed by the wavelength yields the frequency.

3

Convert the unit from Hz to GHz.

\frac{1.5 \times 10^{10}\text{ Hz}}{10^9\text{ Hz/GHz}} = 15\text{ GHz}

.

One gigahertz (

1\text{ GHz}

) equals

10^9\text{ Hz}

.

Key Concept

Relationship between speed of light, frequency, and wavelength (

c = f \lambda

) for electromagnetic radiation.

Question 8135Question →

A thin converging lens forms a real image of an object on a screen placed $60\text{ cm}$ from the lens. If the object is located $30\text{ cm}$ in front of the lens, what is the focal length of the lens in centimeters ( $\text{cm}$ )?

Show answer & explanation

Answer: 20

Answer

The focal length of the converging lens is

20\text{ cm}

.

Using the thin lens formula

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

with an object distance

u = 30\text{ cm}

and a real image distance

v = 60\text{ cm}

yields

\frac{1}{f} = \frac{1}{30} + \frac{1}{60} = \frac{1}{20}

, giving

f = 20\text{ cm}

.

Step-by-Step Solution

1

Identify given parameters and apply correct sign conventions

u = +30\text{ cm}

(real object) and

v = +60\text{ cm}

(real image on screen)

In thin lens calculations for real objects and images formed on screens, both distances are positive.

2

Substitute parameters into the thin lens formula

\frac{1}{f} = \frac{1}{30} + \frac{1}{60}

The thin lens equation relates focal length

f

, object distance

u

, and image distance

v

.

3

Perform fraction addition and solve for focal length

\frac{1}{f} = \frac{3}{60} = \frac{1}{20} \implies f = 20\text{ cm}

Taking the common denominator gives

\frac{1}{20}\text{ cm}^{-1}

, which yields

f = 20\text{ cm}

.

Key Concept

Thin Lens Formula for Real Image Formation

Question 8136Question →

Two miscible liquids, $P$ (boiling point $78^\circ\text{C}$ ) and $Q$ (boiling point $100^\circ\text{C}$ ), are thoroughly mixed to form a uniform, single-phase solution. Which of the following observations correctly describes the thermal behavior and nature of this mixture during distillation?

The liquid mixture boils at a single, fixed temperature of

89^\circ\text{C}

.

The mixture distills over a temperature range between

78^\circ\text{C}

and

100^\circ\text{C}

.

Liquid

Q

with the higher boiling point distills over first before liquid

P

.

The components can only be separated back into

P

and

Q

through chemical decomposition.

Show answer & explanation

Answer: The mixture distills over a temperature range between $78^\circ\text{C}$ and $100^\circ\text{C}$ .

Answer

The mixture distills over a temperature range between

78^\circ\text{C}

and

100^\circ\text{C}

.

A mixture of miscible liquids retains the distinct physical properties of its constituents and lacks a fixed chemical composition. Consequently, it vaporizes and distills over a temperature range starting from the boiling point of the more volatile liquid (

78^\circ\text{C}

) up to that of the less volatile liquid (

100^\circ\text{C}

).

Step-by-Step Solution

1

Identify the type of matter described in the scenario

A uniform blend of two miscible liquids without chemical reaction forms a liquid-liquid homogeneous mixture.

No chemical bonds are formed or broken when mixing two miscible liquids.

2

Recall the fundamental distinctions in physical properties between pure compounds and mixtures

Pure compounds have sharp, fixed boiling points, while mixtures boil over a range of temperatures.

The vapor pressure and composition of a liquid mixture change continuously as the lower boiling component boils off.

3

Determine the expected distillation order and boiling behavior

Liquid

P

(lower boiling point of

78^\circ\text{C}

) vaporizes faster and distills first, causing distillation to occur over the temperature interval

78^\circ\text{C}

to

100^\circ\text{C}

.

Fractional distillation separates components based on differences in volatility and boiling point.

Key Concept

Distinction Between Compounds and Mixtures Based on Physical Constants

Question 8137Question →

Water hardness is classified as either temporary or permanent depending on the dissolved salts present. Which of the following chemical compounds causes temporary hardness in water that can be removed simply by boiling?

Calcium hydrogentrioxocarbonate(IV)

Calcium tetraoxosulfate(VI)

Magnesium chloride

Calcium chloride

Show answer & explanation

Answer: Calcium hydrogentrioxocarbonate(IV)

Answer

Calcium hydrogentrioxocarbonate(IV)

The correct compound is Calcium hydrogentrioxocarbonate(IV). Temporary water hardness is uniquely caused by dissolved hydrogentrioxocarbonate(IV) salts of calcium or magnesium. Heating or boiling the water decomposes this soluble compound into insoluble calcium trioxocarbonate(IV) precipitate, water, and carbon(IV) oxide gas, thereby removing the calcium ions responsible for hardness.

Step-by-Step Solution

1

Identify the cause of temporary water hardness.

Temporary hardness is caused by dissolved hydrogentrioxocarbonate(IV) salts of calcium and magnesium, such as

\text{Ca(HCO}_3)_2

.

Hydrogentrioxocarbonate(IV) ions decompose thermally when heated.

2

Analyze the chemical effect of boiling on calcium hydrogentrioxocarbonate(IV).

\text{Ca(HCO}_3)_2\text{(aq)} \xrightarrow{\Delta} \text{CaCO}_3\text{(s)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}

Boiling converts the soluble hydrogentrioxocarbonate(IV) into insoluble calcium trioxocarbonate(IV), which precipitates out as fur or scale.

Key Concept

Temporary Water Hardness and Removal by Boiling

Estimated Time:45s

Question 8138Question →

Match each oxoanion on the left with the correct oxidation number of its central element on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches pair

PO_4^{3-}

with

+5

,

NO_2^-

with

+3

,

SO_4^{2-}

with

+6

, and

CO_3^{2-}

with

+4

.

Each central atom's oxidation state is determined by setting the sum of the oxidation states equal to the net ionic charge, using

-2

for each oxygen atom.

Step-by-Step Solution

1

Assign the standard oxidation number of

-2

to oxygen in oxoanions.

Each oxygen atom contributes an oxidation state of

-2

.

Oxygen is more electronegative than phosphorus, nitrogen, sulfur, and carbon.

2

Set up an algebraic sum where the total oxidation numbers equal the ion's net charge.

For

PO_4^{3-}

:

P + 4(-2) = -3

; for

NO_2^-

:

N + 2(-2) = -1

; for

SO_4^{2-}

:

S + 4(-2) = -2

; for

CO_3^{2-}

:

C + 3(-2) = -2

.

The sum of oxidation states in a polyatomic species equals the charge on the species.

3

Solve each linear equation for the oxidation number of the central atom.

P = +5

,

N = +3

,

S = +6

, and

C = +4

.

Algebraic isolation of the unknown variable determines the oxidation state.

Key Concept

Assigning Oxidation Numbers in Polyatomic Oxoanions

Question 8139Question →

In chemical analysis, oxidation-reduction reactions are interpreted using either classical transfer principles (oxygen/hydrogen) or modern electronic and oxidation-state theories. Match each chemical transformation on the left with the specific classical or modern redox definition concept on the right that governs it.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The conversion of ammonia to nitrogen gas matches classical oxidation via hydrogen removal. The oxidation of magnesium metal to magnesium ions matches modern electronic oxidation via electron loss. The change of permanganate ion to manganese(II) ion matches modern reduction via a decrease in oxidation state from +7 to +2. The conversion of ethene to ethane matches classical reduction via hydrogen addition.

Each pair correctly matches a specific chemical transformation with its governing classical or modern redox rule. The removal of hydrogen from ammonia is classical oxidation. The loss of electrons from magnesium metal is modern electronic oxidation. The reduction in oxidation number of manganese from +7 to +2 is modern reduction. The addition of hydrogen to ethene is classical reduction.

Step-by-Step Solution

1

Analyze the conversion of

2\text{NH}_3

to

\text{N}_2

Ammonia loses hydrogen atoms during the reaction.

According to classical redox concepts, the removal of hydrogen from a compound constitutes oxidation.

2

Analyze the half-reaction

\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-

Magnesium loses two electrons.

Under modern electronic theory, oxidation is defined as electron loss (OIL - Oxidation Is Loss).

3

Analyze the reduction of

\text{MnO}_4^-

to

\text{Mn}^{2+}

Manganese's oxidation state drops from

+7

to

+2

.

According to modern oxidation number conventions, a drop/decrease in oxidation number represents reduction.

4

Analyze the hydrogenation of ethene

\text{C}_2\text{H}_4 + \text{H}_2 \rightarrow \text{C}_2\text{H}_6

Hydrogen is added across the carbon-carbon double bond.

Classical redox principles define reduction as the gain or addition of hydrogen.

Key Concept

Classical vs Modern Concepts of Redox

Estimated Time:2m 0s

Question 8140Question →

Match each heat transfer process or physical phenomenon on the left with its underlying governing mechanism or quantitative relationship on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The steady-state rate of heat conduction through a uniform slab corresponds to being directly proportional to the temperature gradient. The radiant energy emitted per unit area by an ideal blackbody corresponds to being directly proportional to the fourth power of absolute temperature. Natural heat transport in fluids under gravity corresponds to being driven by buoyant forces resulting from density variations. The microscopic conduction mechanism in electrical insulators corresponds to propagation via quantized lattice vibrations (phonons).

Each heat transfer mechanism matches its fundamental law and microscopic process: conduction across a plane wall is governed by Fourier's law and proportional to the temperature gradient; thermal radiation from a blackbody obeys Stefan's law and scales with the fourth power of absolute temperature; natural convection in fluids requires gravity to drive density-based buoyant circulation; and thermal conduction in non-metallic insulators relies on atomic lattice vibrations (phonons) due to the absence of free electrons.

Step-by-Step Solution

1

Analyze conduction governing equation (Fourier's Law)

Heat current

\frac{Q}{t} = kA \frac{\Delta T}{d}

, showing that heat flow per unit area depends directly on the temperature gradient

\frac{\Delta T}{\Delta x}

.

Identify the quantitative relationship governing thermal conduction in solid materials.

2

Analyze radiation power equation (Stefan-Boltzmann Law)

Total power per unit area

P/A = \sigma T^4

, establishing fourth-power dependence on thermodynamic temperature

T

.

Identify the law governing thermal radiation emissions.

3

Examine natural convection mechanics

Thermal expansion leads to density differences

\Delta \rho

, causing buoyant forces under gravity to set up fluid circulation currents.

Identify the physical drive behind natural convection in fluids.

4

Examine microscopic heat transfer mechanisms in insulators

Insulators lack mobile valence electrons, leaving atomic lattice vibrations (phonons) as the sole mechanism for thermal energy transport.

Distinguish between electronic conduction in metals and lattice/phonon conduction in non-metals.

Key Concept

Physical Principles and Microscopic Mechanisms of Conduction, Convection, and Radiation

Answer

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