All practice questions

Question 12841Question →

Match each characteristic functional group structure listed on the left with its corresponding organic class name on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are:

-\text{NH}_2

matches with Primary amine,

-\text{CONH}_2

matches with Amide,

-\text{O}-

matches with Ether, and

-\text{COOR}

matches with Ester.

Each organic compound class is identified by its specific functional group arrangement:

-\text{NH}_2

specifies primary amines,

-\text{CONH}_2

specifies primary amides, an oxygen atom bridging two carbon groups (

-\text{O}-

) specifies ethers, and

-\text{COOR}

specifies esters.

Step-by-Step Solution

1

Identify the amino functional group

-\text{NH}_2

is recognized as an amino group attached to a carbon chain.

Compounds containing the

-\text{NH}_2

group attached directly to an alkyl carbon belong to the class of primary amines.

2

Identify the carboxamide functional group

-\text{CONH}_2

contains a carbonyl bonded to an amino group.

This structural unit defines the primary amide family.

3

Identify the ether linkage

-\text{O}-

represents a single oxygen atom bridging two carbon atoms.

An oxygen atom connected to two alkyl or aryl groups (

\text{R-O-R'}

) characterizes an ether.

4

Identify the ester group

-\text{COOR}

contains a carbonyl carbon attached to an alkoxy group.

This group is derived from an alkanoic acid and an alkanol, forming an ester.

Key Concept

Identification of characteristic functional groups in organic compounds

Question 12842Question →

A neutral copper atom undergoes oxidation to form a copper(II) ion, $Cu^{2+}$ . Given that the atomic number of copper ( $Cu$ ) is 29, which of the following represents the correct ground-state electronic configuration of $Cu^{2+}$ ?

1s^2 2s^2 2p^6 3s^2 3p^6 3d^9

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7

1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^8

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^9

Show answer & explanation

Answer: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^9$

Answer

The correct ground-state electronic configuration of the copper(II) ion (

Cu^{2+}

) is

1s^2 2s^2 2p^6 3s^2 3p^6 3d^9

.

The correct option correctly applies the rules of electronic configuration for transition metals. Neutral copper (

Z=29

) has an ground state of

1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1

. When ionized to

Cu^{2+}

, two electrons are removed: the outermost

4s^1

electron is lost first, followed by one electron from the

3d^{10}

subshell, leaving

1s^2 2s^2 2p^6 3s^2 3p^6 3d^9

.

Step-by-Step Solution

1

Determine the total number of electrons in neutral copper (

Cu

) and write its ground-state configuration.

Atomic number

Z = 29

, so neutral

Cu

has 29 electrons. Due to the extra stability of a completely filled

d

-subshell, the configuration is

1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1

.

Full subshell stability causes one electron from the

4s

orbital to promote to the

3d

subshell in neutral copper.

2

Calculate the number of electrons lost to form the

Cu^{2+}

ion.

Cu^{2+}

loses 2 electrons, leaving

29 - 2 = 27

electrons.

A

+2

charge indicates the loss of two valence electrons.

3

Remove two electrons following the rule for transition metal cation formation.

Remove the 1 electron from the outermost

4s

orbital first, then 1 electron from the

3d

orbital, leaving

3d^9

.

Electrons in the highest principal energy level (

n=4

) are lost before inner

(n-1)d

electrons (

n=3

).

Key Concept

Electronic configuration of d-block cations

Estimated Time:1m 0s

Question 12843Question →

In the industrial synthesis of sodium trioxocarbonate(IV) via the Solvay process, ammonia gas is recovered and continuously recycled within the manufacturing loop. Which two naturally occurring raw materials are consumed as the primary starting feedstocks, making proximity to their natural deposits the key siting factor for the plant?

Brine (concentrated sodium chloride) and limestone (calcium carbonate)

Ammonia gas and concentrated tetraoxosulfate(VI) acid

Natural gas (methane) and bauxite mineral ore

Calcium chloride and carbon(IV) oxide gas

Show answer & explanation

Answer: Brine (concentrated sodium chloride) and limestone (calcium carbonate)

Answer

Brine (concentrated sodium chloride) and limestone (calcium carbonate) are the primary naturally occurring raw materials consumed in the Solvay process.

The Solvay process for manufacturing sodium trioxocarbonate(IV) (soda ash) requires concentrated sodium chloride solution (brine) and calcium carbonate (limestone) as primary, naturally occurring starting raw materials. Because these materials are heavy and consumed in bulk, factories are sited near salt deposits/sea coast and limestone quarries to minimize freight costs.

Step-by-Step Solution

1

Identify the chemical reactions and overall stoichiometry of the Solvay process.

The overall chemical reaction is

2\text{NaCl} + \text{CaCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{CaCl}_2

.

Understanding the net balanced equation shows which substances are consumed as starting reactants.

2

Distinguish between primary naturally occurring raw materials, intermediates/reagents, and by-products.

Brine (

\text{NaCl}

) and limestone (

\text{CaCO}_3

) are directly extracted natural resources consumed in massive quantities. Ammonia (

\text{NH}_3

) is recycled internally.

Heavy chemical industries are sited near bulky primary raw material deposits to minimize raw material transportation costs.

Key Concept

Solvay Process Raw Materials and Siting Factors

Estimated Time:1m 0s

Question 12844Question →

In a nuclear fusion reaction, a deuterium nucleus (

{^{2}_{1}\text{H}}

) fuses with a tritium nucleus (

{^{3}_{1}\text{H}}

) according to the equation:

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

Given the atomic masses:
- Mass of ${^{2}_{1}\text{H}} = 2.0141\text{ u}$
- Mass of ${^{3}_{1}\text{H}} = 3.0160\text{ u}$
- Mass of ${^{4}_{2}\text{He}} = 4.0026\text{ u}$
- Mass of ${^{1}_{0}\text{n}} = 1.0087\text{ u}$
- $1\text{ u} = 931.5\text{ MeV}$

What is the total energy $Q$ released in this reaction?

17.51\text{ MeV}

957.12\text{ MeV}

1.75\text{ MeV}

4686.38\text{ MeV}

Show answer & explanation

Answer: $17.51\text{ MeV}$

Answer

The total energy released in the reaction is

17.51\text{ MeV}

.

The correct answer of

17.51\text{ MeV}

is obtained by subtracting the combined mass of the helium-4 nucleus and neutron (

5.0113\text{ u}

) from the combined mass of the deuterium and tritium nuclei (

5.0301\text{ u}

) to yield a mass defect of

0.0188\text{ u}

, which equates to

17.51\text{ MeV}

when multiplied by

931.5\text{ MeV/u}

.

Step-by-Step Solution

1

Calculate the total mass of the reactants before the fusion reaction.

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

To find the mass defect, we must first sum the masses of the initial nuclei.

2

Calculate the total mass of the reaction products.

\text{Mass of products} = 4.0026\text{ u} + 1.0087\text{ u} = 5.0113\text{ u}

The products include both the helium-4 nucleus and the emitted neutron.

3

Determine the mass defect (

\Delta m

).

\Delta m = 5.0301\text{ u} - 5.0113\text{ u} = 0.0188\text{ u}

The mass defect represents the mass converted into energy during fusion.

4

Convert the mass defect into released energy

Q

in

\text{MeV}

.

Q = 0.0188\text{ u} \times 931.5\text{ MeV/u} = 17.5122\text{ MeV} \approx 17.51\text{ MeV}

Using Einstein's mass-energy equivalence with

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

.

Key Concept

Mass defect and energy release in nuclear fusion reactions

Estimated Time:1m 30s

Question 12845Question →

A light ray traveling inside a transparent glass prism of refractive index $1.50$ strikes the boundary with air. What is the sine of the critical angle ( $\sin C$ ) for total internal reflection to occur at this boundary?

0.67

1.50

0.50

0.87

Show answer & explanation

Answer: $0.67$

Answer

The sine of the critical angle (

\sin C

) for total internal reflection at the glass-air boundary is

0.67

.

For light moving from a medium with refractive index

n

into air, total internal reflection occurs when the angle of incidence exceeds the critical angle

C

. The critical angle satisfies

\sin C = \frac{1}{n}

. Substituting

n = 1.50

gives

\sin C = \frac{1}{1.50} = 0.67

.

Step-by-Step Solution

1

Identify the relationship between the refractive index (

n

) and the critical angle (

C

) when light travels from a medium to air.

\sin C = \frac{1}{n}

By Snell's law, at the critical angle of incidence, the angle of refraction in air is

90^\circ

(so

\sin 90^\circ = 1

).

2

Substitute the given refractive index (

n = 1.50

) into the formula.

\sin C = \frac{1}{1.50} = \frac{2}{3} \approx 0.67

Dividing

1

by

1.50

yields

0.666...

, which rounds to

0.67

.

Key Concept

Critical Angle and Total Internal Reflection

Estimated Time:45s

Question 12846Question →

According to the kinetic theory of gases, the average kinetic energy of gas molecules is directly proportional to the absolute temperature. If a sample of nitrogen gas at an initial temperature of $27^\circ\text{C}$ is heated until the average kinetic energy of its molecules is tripled, what is the final temperature of the gas in $^\circ\text{C}$ ?

627^\circ\text{C}

81^\circ\text{C}

900^\circ\text{C}

1173^\circ\text{C}

Show answer & explanation

Answer: $627^\circ\text{C}$

Answer

The final temperature of the gas is

627^\circ\text{C}

.

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

E_k \propto T

). Converting the initial temperature

27^\circ\text{C}

to Kelvin yields

27 + 273 = 300\text{ K}

. Tripling the average kinetic energy means tripling the absolute temperature to

3 \times 300\text{ K} = 900\text{ K}

. Converting this back to degrees Celsius gives

900 - 273 = 627^\circ\text{C}

.

Step-by-Step Solution

1

Convert the initial temperature from Celsius to Kelvin.

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

Kinetic molecular theory states that average kinetic energy is directly proportional to absolute temperature (

E_k \propto T

in Kelvin).

2

Calculate the new absolute temperature after tripling the kinetic energy.

T_2 = 3 \times T_1 = 3 \times 300\text{ K} = 900\text{ K}

Tripling the average kinetic energy directly triples the absolute temperature in Kelvin.

3

Convert the final absolute temperature back to degrees Celsius.

t_2 = 900\text{ K} - 273 = 627^\circ\text{C}

The question explicitly requests the final temperature in degrees Celsius (

t = T - 273

).

Key Concept

Relationship between average kinetic energy and absolute temperature in the kinetic molecular theory

Question 12847Question →

A uniform rigid beam $MN$ of length $2.0\text{ m}$ and weight $80\text{ N}$ is smoothly pivoted at end $M$ . A vertical load of $120\text{ N}$ is hung at a distance of $1.5\text{ m}$ from $M$ . The beam is held horizontally in equilibrium by a cable attached at end $N$ pulling upward at an angle of $30^\circ$ to the horizontal beam. What is the magnitude of the tension in the cable?

100\text{ N}

130\text{ N}

260\text{ N}

340\text{ N}

Show answer & explanation

Answer: $260\text{ N}$

Answer

The magnitude of the tension in the cable is

260\text{ N}

.

For rotational equilibrium about pivot

M

, the sum of clockwise moments must equal the counterclockwise moment. Clockwise moment from the beam's weight and load is

(80\text{ N} \times 1.0\text{ m}) + (120\text{ N} \times 1.5\text{ m}) = 260\text{ N}\cdot\text{m}

. Counterclockwise moment from the cable tension is

T \times 2.0\text{ m} \times \sin(30^\circ) = 1.0 T\text{ N}\cdot\text{m}

. Equating the two gives

T = 260\text{ N}

.

Step-by-Step Solution

1

Identify the positions and lines of action of all forces relative to pivot

M

.

The weight of the uniform beam (

80\text{ N}

) acts at its center of gravity (

1.0\text{ m}

from

M

). The suspended load (

120\text{ N}

) acts at

1.5\text{ m}

from

M

. Cable tension

T

acts at

2.0\text{ m}

from

M

at an angle of

30^\circ

to the beam.

Rotational equilibrium requires evaluating moments created by all forces about the pivot point.

2

Calculate the sum of clockwise moments about pivot

M

.

\sum \tau_{\text{clockwise}} = (80\text{ N} \times 1.0\text{ m}) + (120\text{ N} \times 1.5\text{ m}) = 80\text{ N}\cdot\text{m} + 180\text{ N}\cdot\text{m} = 260\text{ N}\cdot\text{m}

.

Both downward forces exert clockwise turning effects about pivot

M

.

3

Determine the counterclockwise moment exerted by the inclined cable tension

T

.

\tau_{\text{counterclockwise}} = T \times d \sin\theta = T \times 2.0\text{ m} \times \sin(30^\circ) = 1.0 T\text{ N}\cdot\text{m}.

Only the component of tension perpendicular to the beam (

T \sin 30^\circ

) produces a moment about the pivot.

4

Apply the Principle of Moments to calculate tension

T

.

1.0 T = 260 \implies T = 260\text{ N}.

For the beam to remain horizontally balanced, total clockwise moment must equal total counterclockwise moment.

Key Concept

Principle of Moments and Rotational Equilibrium with Inclined Forces

Question 12848Question →

An object of mass $4\text{ kg}$ is released from rest at a height of $5\text{ m}$ above the ground. Neglecting air resistance and taking $g = 10\text{ m s}^{-2}$ , what is its kinetic energy just before striking the ground?

200\text{ J}

100\text{ J}

40\text{ J}

20\text{ J}

Show answer & explanation

Answer: $200\text{ J}$

Answer

The kinetic energy of the object just before striking the ground is

200\text{ J}

.

The total mechanical energy is conserved during free fall. The initial gravitational potential energy

E_p = mgh = 4 \times 10 \times 5 = 200\text{ J}

is completely converted into kinetic energy

E_k

just before impact, making

200\text{ J}

the correct answer.

Step-by-Step Solution

1

Calculate the initial potential energy at maximum height

E_p = mgh = 4\text{ kg} \times 10\text{ m s}^{-2} \times 5\text{ m} = 200\text{ J}

At the top of the fall, all mechanical energy is stored as gravitational potential energy.

2

Apply the principle of conservation of mechanical energy

E_k = E_p = 200\text{ J}

In the absence of resistive forces such as air drag, potential energy lost converts completely into kinetic energy gained.

Key Concept

Conservation of Mechanical Energy

Estimated Time:45s

Question 12849Question →

Match each organic functional group class on the left with its corresponding characteristic IUPAC suffix on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Carboxylic acid matches '-oic acid', Acid amide matches '-amide', Acyl chloride matches '-oyl chloride', and Ester matches '-oate'.

Each organic class is correctly matched to its IUPAC principal suffix: carboxylic acids use '-oic acid', amides use '-amide', acyl chlorides use '-oyl chloride', and esters use '-oate'.

Step-by-Step Solution

1

Identify the characteristic functional group structure for each carbonyl derivative class.

Carboxylic acids have

-\text{COOH}

, amides have

-\text{CONH}_2

, acyl chlorides have

-\text{COCl}

, and esters have

-\text{COOR}

.

Functional groups define the chemical family and dictate IUPAC naming rules.

2

Pair each class with its designated IUPAC principal suffix.

-\text{COOH} \rightarrow \text{-oic acid}

,

-\text{CONH}_2 \rightarrow \text{-amide}

,

-\text{COCl} \rightarrow \text{-oyl chloride}

, and

-\text{COOR} \rightarrow \text{-oate}

.

Standard IUPAC rules assign specific characteristic suffixes to represent the main functional group in systematic names.

Key Concept

IUPAC Nomenclature Suffixes for Carbonyl Derivatives

Question 12850Question →

A thin converging lens of focal length $15\text{ cm}$ forms an erect image that is magnified three times. What is the distance of the object from the lens?

10\text{ cm}

20\text{ cm}

5\text{ cm}

45\text{ cm}

Show answer & explanation

Answer: $10\text{ cm}$

Answer

10\text{ cm}

An erect image produced by a thin converging lens is virtual, which means the linear magnification is positive (

m = +3

) and the image distance is negative relative to the real object (

v = -3u

). Substituting

f = +15\text{ cm}

and

v = -3u

into the thin lens formula

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

gives

\frac{1}{15} = \frac{1}{u} - \frac{1}{3u} = \frac{2}{3u}

. Solving for

u

yields

u = 10\text{ cm}

.

Step-by-Step Solution

1

Determine the nature of the image and establish the relationship between image distance and object distance

Since the image formed by a converging lens is erect, it must be virtual. Thus, linear magnification

m = +3 = -\frac{v}{u}

, giving

v = -3u

.

A single convex lens produces an erect image only when the image is virtual, requiring a negative image distance under standard optical sign conventions.

2

Substitute focal length and image distance into the thin lens formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \implies \frac{1}{15} = \frac{1}{u} + \frac{1}{-3u}

The thin lens equation relates focal length, object distance, and image distance.

3

Simplify the algebraic expression and solve for object distance u

\frac{1}{15} = \frac{3 - 1}{3u} = \frac{2}{3u} \implies 3u = 30 \implies u = 10\text{ cm}

Finding a common denominator allows direct solution for the object distance

u

.

Key Concept

Thin lens formula and sign conventions for virtual images

Question 12851Question →

A particle moves along a circular track of radius $r$ with a constant speed $v$ . If the radius of the track is measured with a percentage error of $1.2\%$ , and the speed is measured as $(10.0 \pm 0.3)\text{ m s}^{-1}$ , what is the percentage error in the calculated centripetal acceleration of the particle?

Show answer & explanation

Answer: 7.2

Answer

The percentage error in the calculated centripetal acceleration is 7.2%.

The centripetal acceleration is

a_c = \frac{v^2}{r}

. The percentage error in

v

is

\frac{0.3}{10.0} \times 100\% = 3.0\%

. According to error propagation rules, the percentage error in

a_c

is

2 \times (\text{percentage error in } v) + (\text{percentage error in } r) = 2(3.0\%) + 1.2\% = 7.2\%

.

Step-by-Step Solution

1

Calculate the percentage error in the speed measurement

v

.

\text{Percentage error in } v = \frac{0.3}{10.0} \times 100\% = 3.0\%

Percentage error is given by dividing the absolute error by the measured value and multiplying by 100%.

2

Write the relationship for centripetal acceleration

a_c

in terms of

v

and

r

.

a_c = \frac{v^2}{r}

Centripetal acceleration is directly proportional to the square of speed and inversely proportional to radius.

3

Apply the fractional error propagation law for a quantity involving powers and quotients.

\frac{\Delta a_c}{a_c} = 2\left(\frac{\Delta v}{v}\right) + \frac{\Delta r}{r}

When a measured quantity is raised to a power

n

, its fractional error contribution is multiplied by

n

. Errors accumulate additively.

4

Substitute the individual percentage errors to find the total percentage error in

a_c

.

\text{Percentage error in } a_c = 2(3.0\%) + 1.2\% = 7.2\%

Multiplying the speed's percentage error by 2 and adding the radius percentage error yields the total relative error.

Key Concept

Propagation of percentage errors in physical formulas containing powers and ratios.

Question 12852Question →

For a fixed mass of an ideal gas held under isothermal conditions, a graph of pressure ( $P$ ) plotted against inverse volume ( $\frac{1}{V}$ ) forms a straight line passing through the origin, whose slope increases proportionally with an increase in the system's absolute temperature.

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because the mathematical formulation of Boyle's Law extended by the ideal gas relation,

P = (nRT)\left(\frac{1}{V}\right)

, matches the linear equation

y = mx

where the slope

m = nRT

is directly proportional to the absolute temperature

T

.

The statement accurately reflects the mathematical structure of Boyle's Law under ideal gas behavior. Expressing pressure as

P = (nRT)\left(\frac{1}{V}\right)

shows that a plot of

P

versus

\frac{1}{V}

yields a straight line with y-intercept at zero and slope equal to

nRT

. Because the slope is directly proportional to absolute temperature

T

, raising the temperature increases the gradient of the resulting line.

Step-by-Step Solution

1

Express Boyle's Law incorporating absolute temperature

PV = nRT \implies P = (nRT)\left(\frac{1}{V}\right)

Relate pressure, volume, and absolute temperature mathematically for a fixed mass of gas.

2

Compare the equation to the standard straight-line equation

y = mx + c

y = P

,

x = \frac{1}{V}

, slope

m = nRT

, and y-intercept

c = 0

Determine the graphical relationship when

P

is plotted against

\frac{1}{V}

.

3

Analyze the dependence of the slope on absolute temperature

Since

n

(moles of gas) and

R

(molar gas constant) are fixed,

m \propto T

Evaluate how increasing absolute temperature affects the gradient of the isotherm.

Key Concept

Boyle's Law Graphical Representation and Slope-Temperature Relationship

Question 12853Question →

An electron of mass $9.0 \times 10^{-31}\text{ kg}$ and charge $1.6 \times 10^{-19}\text{ C}$ is accelerated from rest through an electric potential difference $V$ . If the associated de Broglie wavelength of the electron is $1.1 \times 10^{-10}\text{ m}$ and Planck's constant is $6.6 \times 10^{-34}\text{ J s}$ , what is the value of the potential difference $V$ ?

125\text{ V}

250\text{ V}

11.2\text{ V}

62.5\text{ V}

Show answer & explanation

Answer: $125\text{ V}$

Answer

The accelerating potential difference required is

125\text{ V}

.

The de Broglie wavelength of an electron accelerated from rest through a potential difference

V

is given by

\lambda = \frac{h}{\sqrt{2meV}}

. Rearranging for

V

gives

V = \frac{h^2}{2me\lambda^2}

. Substituting

h = 6.6 \times 10^{-34}\text{ J s}

,

m = 9.0 \times 10^{-31}\text{ kg}

,

e = 1.6 \times 10^{-19}\text{ C}

, and

\lambda = 1.1 \times 10^{-10}\text{ m}

gives

V = 125\text{ V}

.

Step-by-Step Solution

1

Relate kinetic energy to momentum and accelerating potential.

E_k = eV

and

p = \sqrt{2m E_k} = \sqrt{2meV}

An electron accelerated through potential

V

gains kinetic energy equal to

eV

.

2

Substitute momentum into the de Broglie wavelength equation.

\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}

De Broglie wavelength is defined as Planck's constant divided by momentum.

3

Rearrange the equation to solve for potential difference

V

.

V = \frac{h^2}{2me\lambda^2}

Squaring both sides allows isolating

V

.

4

Substitute the given numerical values and calculate

V

.

$V = \frac{(6.6 \times 10^{-34})^2}{2(9.0 \times 10^{-31})(1.6 \times 10^{-19})(1.1 \times 10^{-10})^2} = 125\text{ V}$

Performing the algebraic substitution yields the final potential.

Key Concept

de Broglie wavelength of an electron accelerated through a potential difference

Estimated Time:1m 30s

Question 12854Question →

An alternating voltage source described by $V(t) = 210\sqrt{2}\sin(100\pi t)\ \text{V}$ is connected in series with a $40\ \Omega$ resistor, an inductor of inductive reactance $100\ \Omega$ , and a capacitor of capacitive reactance $70\ \Omega$ . What is the root-mean-square (RMS) current flowing through the circuit?

4.2\ \text{A}

1.0\ \text{A}

3.0\ \text{A}

7.0\ \text{A}

Show answer & explanation

Answer: $4.2\ \text{A}$

Answer

The root-mean-square (RMS) current flowing through the circuit is

4.2\ \text{A}

.

The peak voltage is

V_0 = 210\sqrt{2}\ \text{V}

, giving an RMS voltage of

V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = 210\ \text{V}

. The impedance of the series RLC circuit is calculated by

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + (100 - 70)^2} = 50\ \Omega

. Therefore, the RMS current is

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{210}{50} = 4.2\ \text{A}

.

Step-by-Step Solution

1

Determine the RMS voltage from the voltage equation

V_{\text{rms}} = 210\ \text{V}

The standard equation for AC voltage is

V(t) = V_0 \sin(\omega t)

, where

V_0 = 210\sqrt{2}\ \text{V}

. The RMS voltage is

V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = 210\ \text{V}

.

2

Calculate the total impedance of the series RLC circuit

Z = 50\ \Omega

Impedance is determined using phasor addition:

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + (100 - 70)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50\ \Omega

.

3

Calculate the RMS current using Ohm's law for AC circuits

I_{\text{rms}} = 4.2\ \text{A}

I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{210\ \text{V}}{50\ \Omega} = 4.2\ \text{A}

.

Key Concept

Impedance and RMS current calculation in series RLC alternating current circuits

Estimated Time:2m 0s

Question 12855Question →

A nation's economic records for a given fiscal year provide the following national income figures:

- Gross Domestic Product ( $\text{GDP}$ ): $\text{N}5,400\text{ million}$
- Income earned by domestic citizens working abroad: $\text{N}450\text{ million}$
- Income earned by foreign nationals operating domestically: $\text{N}600\text{ million}$
- Capital Consumption Allowance ( $\text{CCA}$ ): $\text{N}380\text{ million}$

Calculate the Net National Product ( $\text{NNP}$ ) of the country in millions of Naira.

Show answer & explanation

Answer: 4870

Answer

The Net National Product (NNP) of the country is N4,870 million.

Net National Product (NNP) is obtained by adding Net Factor Income from Abroad (NFIA) to Gross Domestic Product (GDP) to get Gross National Product (GNP), and then subtracting Capital Consumption Allowance (CCA). Here, NFIA = N450 million - N600 million = -N150 million. Thus, GNP = N5,400 million - N150 million = N5,250 million. Finally, NNP = N5,250 million - N380 million = N4,870 million.

Step-by-Step Solution

1

Calculate Net Factor Income from Abroad (NFIA)

NFIA = N450 million - N600 million = -N150 million

NFIA measures the net flow of factor payments between domestic citizens abroad and foreign residents domestically.

2

Calculate Gross National Product (GNP)

GNP = N5,400 million + (-N150 million) = N5,250 million

GNP adjusts GDP for net factor receipts from abroad.

3

Calculate Net National Product (NNP)

NNP = N5,250 million - N380 million = N4,870 million

NNP reflects the net output available to an economy after accounting for capital depreciation.

Key Concept

Basic National Income Aggregates (GDP, GNP, NNP, NDP)

Question 12856Question →

Match each specific electronic structure scenario or electron assignment with the corresponding quantum principle or thermodynamic factor that governs it.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings match copper's anomalous configuration to full-subshell exchange energy stability, identical quantum set restriction to the Pauli exclusion principle, unpaired degenerate orbital filling in nitrogen to Hund's rule, and

4s

filling prior to

3d

to the Aufbau

(n+l)

principle.

Each scenario directly aligns with its foundational quantum principle: copper's ground-state configuration (

[Ar]3d^{10}4s^1

) is stabilized by high exchange energy of the full

d

-subshell; identical quantum numbers are forbidden by the Pauli exclusion principle; single filling of degenerate

2p

orbitals in nitrogen obeys Hund's rule of maximum multiplicity; and subshell filling order (

4s

before

3d

) is governed by the Aufbau

(n+l)

rule.

Step-by-Step Solution

1

Analyze the copper configuration anomaly (

[Ar]3d^{10}4s^1

).

Identified as a deviation due to exchange energy stabilization of a completely filled

d

subshell.

Completely filled subshells provide enhanced stability via maximum exchange interactions and spherical symmetry.

2

Examine the prohibition of identical four-quantum-number sets

(n, l, m_l, m_s)

.

Identified as a direct statement of the Pauli exclusion principle.

Two electrons in an orbital must have opposite spin quantum numbers (

+\frac{1}{2}

and

-\frac{1}{2}

).

3

Evaluate nitrogen's

2p^3

configuration (

2p_x^1 2p_y^1 2p_z^1

).

Identified as an application of Hund's rule of maximum multiplicity.

Electrons occupy degenerate subshell orbitals singly with parallel spins to minimize inter-electronic coulomb repulsion.

4

Assess the sequence of filling

4s

prior to

3d

.

Identified as following the Aufbau principle via the

(n+l)

rule.

For

4s

,

n+l = 4+0 = 4

; for

3d

,

n+l = 3+2 = 5

. Lower

(n+l)

subshells fill first.

Key Concept

Quantum Rules, Subshell Energies, and Electronic Configuration Anomalies

Question 12857Question →

Duchenne muscular dystrophy is an X-linked recessive genetic condition in humans. If a female carrier ( $X^D X^d$ ) marries a male affected by the condition ( $X^d Y$ ), what is the probability that their first child will be an affected female?

25\%

50\%

75\%

0\%

Show answer & explanation

Answer: $25\%$

Answer

The probability that their first child will be an affected female is 25% (or 1/4).

The correct answer is 25%. A cross between a carrier mother (

X^D X^d

) and an affected father (

X^d Y

) yields four equally likely offspring genotypes:

X^D X^d

(carrier female),

X^d X^d

(affected female),

X^D Y

(unaffected male), and

X^d Y

(affected male). Since only 1 of the 4 possible outcomes represents an affected female (

X^d X^d

), the probability is 1 out of 4, or 25%.

Step-by-Step Solution

1

Determine the parental genotypes

Mother =

X^D X^d

(carrier female); Father =

X^d Y

(affected male).

Duchenne muscular dystrophy is an X-linked recessive disorder, so the father must carry the recessive allele on his single X chromosome.

2

Construct a Punnett square for the cross

Gametes: Mother produces

X^D

and

X^d

; Father produces

X^d

and

Y

.
Offspring Genotypes:

X^D X^d

(carrier female),

X^d X^d

(affected female),

X^D Y

(unaffected male),

X^d Y

(affected male).

Each of the 4 combinations is equally likely (25% chance each).

3

Identify the target phenotype among all possible offspring

The target genotype for an affected female is

X^d X^d

, which accounts for 1 out of 4 possible total outcomes.

Probability =

\frac{1}{4} = 25\%

.

Key Concept

X-linked recessive inheritance patterns and probability calculations across total offspring versus sex-specific subsets.

Question 12858Question →

An $\alpha$ -particle has a mass approximately $4$ times that of a proton and carries $2$ times the elementary charge of a proton. If both particles are accelerated from rest through the same potential difference $V$ , what is the ratio of the de Broglie wavelength of the $\alpha$ -particle to that of the proton?

1 : 2\sqrt{2}

1 : 8

2\sqrt{2} : 1

1 : 2

Show answer & explanation

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

Answer

The ratio of the de Broglie wavelength of the

\alpha

-particle to that of the proton is

1 : 2\sqrt{2}

.

The de Broglie wavelength of a particle accelerated by a potential difference

V

is given by

\lambda = \frac{h}{\sqrt{2mqV}}

. Taking the ratio of wavelengths yields

\frac{\lambda_\alpha}{\lambda_p} = \sqrt{\frac{m_p q_p}{m_\alpha q_\alpha}} = \sqrt{\frac{m_p e}{(4m_p)(2e)}} = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}}

.

Step-by-Step Solution

1

Relate de Broglie wavelength to accelerating potential

\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE_k}} = \frac{h}{\sqrt{2mqV}}

The kinetic energy gained by a charged particle accelerated through a potential difference

V

is

E_k = qV

.

2

Express wavelengths for both particles

\lambda_p = \frac{h}{\sqrt{2 m_p e V}}

and

\lambda_\alpha = \frac{h}{\sqrt{2 m_\alpha q_\alpha V}}

Setting up the parametric expressions for proton (

m_p, e

) and

\alpha

-particle (

m_\alpha, q_\alpha

).

3

Substitute relative values

m_\alpha = 4m_p

and

q_\alpha = 2e

\lambda_\alpha = \frac{h}{\sqrt{2(4m_p)(2e)V}} = \frac{h}{\sqrt{16 m_p e V}} = \frac{h}{4\sqrt{m_p e V}}

Using the given relationships between mass and charge of the two particles.

4

Calculate the ratio

\frac{\lambda_\alpha}{\lambda_p}

\frac{\lambda_\alpha}{\lambda_p} = \frac{\frac{h}{4\sqrt{m_p e V}}}{\frac{h}{\sqrt{2 m_p e V}}} = \frac{\sqrt{2}}{4} = \frac{1}{2\sqrt{2}}

Simplifying the fractional ratio of wavelengths.

Key Concept

de Broglie Wavelength of Charged Particles Accelerated in Electric Fields

Question 12859Question →

A sample of a radioactive isotope used in medical imaging has an initial activity of $320\text{ MBq}$ . If its activity decreases to $20\text{ MBq}$ after an elapsed time of $15\text{ hours}$ , what is the half-life of the isotope in hours?

Show answer & explanation

Answer: 3.75

Answer

The half-life of the radioactive isotope is

3.75\text{ hours}

.

The initial activity of

320\text{ MBq}

drops to

20\text{ MBq}

, which is a reduction to

\frac{20}{320} = \frac{1}{16}

of its original value. Since

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

, exactly 4 half-lives have elapsed over the period of

15\text{ hours}

. Therefore, one half-life is

\frac{15\text{ hours}}{4} = 3.75\text{ hours}

.

Step-by-Step Solution

1

Calculate the fraction of initial activity remaining

Fraction remaining =

\frac{20\text{ MBq}}{320\text{ MBq}} = \frac{1}{16}

The ratio of current activity to initial activity determines the fraction of un-decayed nuclei remaining.

2

Determine the number of half-lives (

n

) that have elapsed

n = 4

half-lives, since

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

Each half-life reduces the remaining sample activity by half.

3

Calculate the half-life duration (

T_{1/2}

)

T_{1/2} = \frac{t}{n} = \frac{15\text{ hours}}{4} = 3.75\text{ hours}

Dividing total elapsed time by the number of half-lives gives the duration of one half-life.

Key Concept

Radioactive Decay Law and relationship between remaining activity fraction, number of half-lives, and total elapsed time.

Question 12860Question →

Match each physical electromagnetic configuration listed under Column I with its corresponding net magnetic force or motion characteristic listed under Column II.

Click a left item, then click its matching right item

Show answer & explanation

Answer

1. Charged particle moving parallel to magnetic field lines

\rightarrow

Experiences zero magnetic force (

F = 0

) and maintains its initial linear trajectory.
2. Parallel conductors carrying opposite currents

\rightarrow

Experience a mutually repulsive force per unit length given by

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

.
3. Current-carrying rectangular coil parallel to magnetic field

\rightarrow

Experiences maximum net magnetic torque (

\tau = N I A B

) while net translational force is zero.
4. Charged particle injected perpendicularly into magnetic field

\rightarrow

Follows a uniform circular path due to a constant magnetic centripetal force (

F = qvB

).

Each electromagnetic system is correctly paired based on the vector cross-product rules governing magnetic force (

F = q(\mathbf{v} \times \mathbf{B})

and

F = I(\mathbf{L} \times \mathbf{B})

). Parallel motion produces zero force due to zero angle; opposite currents repel per unit length according to Ampère's law; a parallel loop experiences maximal couple/torque without linear displacement; and perpendicular particle velocity results in a constant centripetal deflection into circular motion.

Step-by-Step Solution

1

Analyze the angle

\theta

between velocity and magnetic field for a parallel moving charge

Since velocity is parallel to the field,

\theta = 0^\circ

. The magnetic Lorentz force formula

F = qvB\sin\theta

yields

F = 0

, meaning no deflection occurs.

Magnetic forces require a non-zero perpendicular component of motion relative to the magnetic field direction.

2

Determine the interaction force between parallel conductors with opposite currents

By applying the magnetic field rule for long straight conductors and Fleming's left-hand rule for the resulting force, currents flowing in opposite directions repel each other with force per unit length

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

.

Opposite current directions generate magnetic field lines between the wires that reinforce each other, creating a high-density magnetic field region that pushes the conductors apart.

3

Evaluate forces and torque on a rectangular coil aligned parallel to magnetic field lines

The forces on opposite arms are equal in magnitude and opposite in direction, canceling out net linear translation (

F_{\text{net}} = 0

). However, they act along different lines of action, producing a maximum torque

\tau = N I A B \cos 0^\circ = N I A B

.

Maximum torque occurs when the plane of the coil is parallel to the field because the lever arm for the magnetic forces on the side conductors is at its maximum length.

4

Evaluate the trajectory of a charged particle entering perpendicularly into a magnetic field

At

\theta = 90^\circ

,

\sin 90^\circ = 1

, giving a constant magnetic force

F = qvB

. Because vector force is perpendicular to velocity at every instant, it changes only the direction of velocity, driving the particle into a circular orbit of radius

r = \frac{mv}{qB}

.

A constant magnitude force acting perpendicular to the instantaneous velocity vectors fulfills the exact condition for centripetal acceleration.

Key Concept

Magnetic forces on moving charges and current-carrying conductors under specific geometric alignments and current configurations

Estimated Time:3m 0s

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept