All practice questions

Question 12781Question →

An atom of an element $X$ forms a stable monoatomic anion $X^{2-}$ containing 18 electrons and 18 neutrons. What is the mass number of element $X$ ?

16

34

36

38

Show answer & explanation

Answer: 34

Answer

34

The monoatomic anion

X^{2-}

carries a negative charge of 2, indicating that it has gained 2 electrons compared to its neutral atomic state. Since the ion has 18 electrons, the neutral atom of element

X

has

18 - 2 = 16

electrons, which corresponds to an atomic number of 16 (16 protons). The mass number (

A

) is defined as the total number of protons and neutrons in the nucleus. Thus, mass number

A = 16\text{ protons} + 18\text{ neutrons} = 34

.

Step-by-Step Solution

1

Determine the atomic number (number of protons) of element

X

from its anion

X^{2-}

.

Number of protons

Z = 18 - 2 = 16

.

An anion with a

2-

charge has gained 2 electrons. Therefore, the neutral atom has 2 fewer electrons than the ion, which equals its atomic number.

2

Calculate the mass number (

A

) of element

X

by summing the number of protons and neutrons.

Mass number

A = 16 + 18 = 34

.

Mass number is the total sum of protons and neutrons present in the nucleus of an atom (

A = Z + N

).

Key Concept

Relationship between ion charge, subatomic particles, atomic number, and mass number

Question 12782Question →

During the industrial extraction of sodium metal by the electrolysis of molten sodium chloride in a Downs cell, a steady current of $9.65\text{ A}$ is passed through the electrolytic cell for $50\text{ minutes}$ . What mass of pure sodium metal is collected at the cathode? [ $\text{Na} = 23$ , $1\text{ F} = 96,500\text{ C mol}^{-1}$ ]

6.9\text{ g}

3.45\text{ g}

13.8\text{ g}

6.72\text{ g}

Show answer & explanation

Answer: $6.9\text{ g}$

Answer

The mass of pure sodium metal collected at the cathode is

6.9\text{ g}

.

At the cathode of the Downs cell, sodium ions undergo single-electron reduction (

\text{Na}^+ + e^- \rightarrow \text{Na}

). Passing

28,950\text{ C}

of charge transfers

0.3\text{ mol}

of electrons. Multiplying

0.3\text{ mol}

by the atomic mass of sodium (

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

) yields exactly

6.9\text{ g}

.

Step-by-Step Solution

1

Calculate the total electric charge (

Q

) passed through the cell in seconds.

Q = I \times t = 9.65\text{ A} \times (50 \times 60\text{ s}) = 28,950\text{ C}

.

Electric charge is determined by multiplying current in amperes by time in seconds.

2

Calculate the amount of substance (in moles) of electrons transferred.

Moles of

e^- = \frac{Q}{F} = \frac{28,950\text{ C}}{96,500\text{ C mol}^{-1}} = 0.3\text{ mol}

.

One Faraday (

96,500\text{ C}

) corresponds to the electric charge of one mole of electrons.

3

Relate the moles of electrons to the moles of sodium metal deposited using the cathode reduction half-equation.

Cathode reaction:

\text{Na}^+ + e^- \rightarrow \text{Na}

. Thus,

1\text{ mol}

of

e^-

produces

1\text{ mol}

of

\text{Na}

, yielding

0.3\text{ mol}

of

\text{Na}

.

Sodium is a univalent alkali metal cation requiring 1 electron per discharged ion.

4

Calculate the mass of sodium metal deposited.

Mass of

\text{Na} = 0.3\text{ mol} \times 23\text{ g mol}^{-1} = 6.9\text{ g}

.

Mass is obtained by multiplying the number of moles by the relative atomic mass.

Key Concept

Quantitative Electrolysis of Molten Salts in Downs Cell Extraction

Estimated Time:2m 0s

Question 12783Question →

An object of unknown mass is suspended from a spring balance possessing a zero error of $+1.5\text{ N}$ inside a lift on an unexplored planet. When the lift accelerates vertically upwards at $2.0\text{ m s}^{-2}$ , the spring balance displays a reading of $33.5\text{ N}$ . Simultaneously, an equal-arm beam balance calibrated with standard masses measures the mass of the object inside the accelerating lift to be $4.0\text{ kg}$ . What is the local acceleration due to gravity on this planet and the true weight of the object when at rest on its surface?

6.0\text{ m s}^{-2}

and

24.0\text{ N}

6.4\text{ m s}^{-2}

and

25.5\text{ N}

10.0\text{ m s}^{-2}

and

40.0\text{ N}

8.0\text{ m s}^{-2}

and

32.0\text{ N}

Show answer & explanation

Answer: $6.0\text{ m s}^{-2}$ and $24.0\text{ N}$

Answer

The local acceleration due to gravity on the planet is

6.0\text{ m s}^{-2}

and the true weight of the object at rest is

24.0\text{ N}

.

The correct response identifies that an equal-arm beam balance measures invariant mass (

4.0\text{ kg}

) because acceleration affects both balance pans equally. Subtracting the

+1.5\text{ N}

zero error from the scale reading yields a true apparent weight of

32.0\text{ N}

. Applying Newton's second law in an upward accelerating lift gives

W_{\text{app}} = m(g + a)

, which yields

32.0 = 4.0(g + 2.0)

, resulting in

g = 6.0\text{ m s}^{-2}

. The true weight at rest is therefore

W = mg = 4.0 \times 6.0 = 24.0\text{ N}

.

Step-by-Step Solution

1

Determine the true mass of the object from the beam balance measurement.

m = 4.0\text{ kg}

An equal-arm beam balance operates by comparing gravitational moments on standard masses and the test object. Because the effective acceleration

(g + a)

acts equally on both pans, it cancels out, making the beam balance measure the true, invariant mass regardless of frame acceleration or location.

2

Correct the spring balance scale reading for zero error to find the true apparent weight.

W_{\text{app}} = 33.5\text{ N} - 1.5\text{ N} = 32.0\text{ N}

A positive zero error means the balance reads

+1.5\text{ N}

when unloaded, so the true force exerted on the spring is the scale reading minus the zero error.

3

Relate apparent weight to local gravity

g

in an upward accelerating lift.

g = 6.0\text{ m s}^{-2}

In an upward accelerating frame with acceleration

a = 2.0\text{ m s}^{-2}

, the normal force/apparent weight is

W_{\text{app}} = m(g + a)

. Substituting values gives

32.0 = 4.0(g + 2.0) \implies 8.0 = g + 2.0 \implies g = 6.0\text{ m s}^{-2}

.

4

Calculate the true weight of the object when at rest on the planet.

W_{\text{true}} = 24.0\text{ N}

True weight is the force of gravity acting on the mass at rest:

W_{\text{true}} = m \cdot g = 4.0\text{ kg} \times 6.0\text{ m s}^{-2} = 24.0\text{ N}

.

Key Concept

Distinction between mass (measured by beam balance, frame-invariant) and weight (measured by spring balance, dependent on frame acceleration and zero error).

Estimated Time:2m 0s

Question 12784Question →

A conveyor system pulls a $40\text{ kg}$ crate at a constant speed of $3\text{ m s}^{-1}$ up a rough inclined ramp. The ramp rises $3\text{ m}$ for every $5\text{ m}$ measured along its slope (giving $\sin\theta = 0.6$ and $\cos\theta = 0.8$ ). If the coefficient of kinetic friction between the crate and the ramp is $0.25$ and $g = 10\text{ m s}^{-2}$ , what is the power output of the conveyor system in watts?

Show answer & explanation

Answer: 960

Answer

The power output required by the conveyor system is

960\text{ W}

.

To pull the crate up the ramp at constant speed, the conveyor force must overcome both the parallel gravitational component (

240\text{ N}

) and friction (

80\text{ N}

), making the total force

320\text{ N}

. Multiplying this force by the constant speed of

3\text{ m s}^{-1}

gives a total power output of

960\text{ W}

.

Step-by-Step Solution

1

Calculate the component of weight parallel to the inclined plane

F_g = mg \sin\theta = 40 \times 10 \times 0.6 = 240\text{ N}

Gravity pulls the object back down along the slope with force

mg \sin\theta

.

2

Calculate the normal reaction force perpendicular to the plane

N = mg \cos\theta = 40 \times 10 \times 0.8 = 320\text{ N}

The normal force balances the perpendicular weight component.

3

Determine the magnitude of kinetic friction force

f_k = \mu N = 0.25 \times 320 = 80\text{ N}

Friction opposes motion up the slope and depends on the normal force.

4

Calculate the total pulling force needed for zero net acceleration

F = F_g + f_k = 240 + 80 = 320\text{ N}

At constant velocity, net force along the incline must equal zero, so

F = mg\sin\theta + f_k

.

5

Calculate the power output of the conveyor

P = F \times v = 320\text{ N} \times 3\text{ m s}^{-1} = 960\text{ W}

Power developed by a constant force moving an object at velocity

v

is given by

P = Fv

.

Key Concept

Work done against gravity and friction, and rate of doing work (Power

P = Fv

)

Question 12785Question →

In an experiment to determine the specific latent heat of vaporization of water at $100^\circ\text{C}$ using an electric immersion heater in an uninsulated vessel, if heat loss from the vessel to the cooler surrounding environment is neglected in the calculations, the experimentally determined value of the specific latent heat of vaporization will be higher than the true value.

True

False

Show answer & explanation

Answer: True

Answer

True

The statement is correct because the measured energy supplied by the heater includes both the energy required for the phase change and the thermal energy dissipated to the surroundings. Dividing this larger total energy by the mass of vaporized liquid results in an overestimation of the specific latent heat of vaporization.

Step-by-Step Solution

1

Formulate the thermal energy balance equation including heat loss.

Q_{\text{electrical}} = Q_{\text{vaporization}} + Q_{\text{loss}}

, where

Q_{\text{electrical}} = P \cdot t

and

Q_{\text{vaporization}} = m \cdot L_{\text{true}}

.

Energy conservation dictates that the electrical energy supplied by the heater equals the thermal energy used for vaporization plus the heat lost to the cooler ambient surroundings.

2

Express the calculated specific latent heat

L_{\text{calc}}

under the assumption of zero heat loss.

L_{\text{calc}} = \frac{Q_{\text{electrical}}}{m} = \frac{m \cdot L_{\text{true}} + Q_{\text{loss}}}{m}

.

The experimenter measures the total electrical power and time, assuming all of this energy converted liquid into vapor.

3

Compare the calculated specific latent heat

L_{\text{calc}}

with the true value

L_{\text{true}}

.

L_{\text{calc}} = L_{\text{true}} + \frac{Q_{\text{loss}}}{m} > L_{\text{true}}

.

Because

Q_{\text{loss}} > 0

and mass

m > 0

, the ratio

\frac{Q_{\text{loss}}}{m}

is positive, meaning

L_{\text{calc}}

is higher than

L_{\text{true}}

.

Key Concept

Experimental Heat Loss Error Propagation in Latent Heat Calculations

Question 12786Question →

Consider the following standard enthalpies of combustion at

298\text{ K}

:

\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H^\circ = -394\text{ kJ mol}^{-1}

\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H^\circ = -286\text{ kJ mol}^{-1}

\text{C}_2\text{H}_2(g) + \frac{5}{2}\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + \text{H}_2\text{O}(l) \quad \Delta H^\circ = -1300\text{ kJ mol}^{-1}

Using Hess's law, calculate the standard enthalpy of formation of ethyne gas, $\text{C}_2\text{H}_2(g)$ , in $\text{kJ mol}^{-1}$ .

Show answer & explanation

Answer: 226

Answer

The standard enthalpy of formation of ethyne gas is

+226\text{ kJ mol}^{-1}

(or

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

).

Applying Hess's law involves expressing the enthalpy of formation of ethyne as the sum of twice the enthalpy of combustion of carbon, once the enthalpy of combustion of hydrogen, minus the enthalpy of combustion of ethyne:

\Delta H_f^\circ = 2(-394) + (-286) - (-1300) = -788 - 286 + 1300 = +226\text{ kJ mol}^{-1}

.

Step-by-Step Solution

1

Write the standard formation equation for ethyne

2\text{C}(s) + \text{H}_2(g) \rightarrow \text{C}_2\text{H}_2(g)

The enthalpy of formation represents the formation of one mole of a compound from its constituent elements in their standard states.

2

Manipulate given thermochemical equations to match the target equation

Multiply equation 1 by 2 (

\Delta H = -788\text{ kJ}

), keep equation 2 unchanged (

\Delta H = -286\text{ kJ}

), and reverse equation 3 (

\Delta H = +1300\text{ kJ}

)

According to Hess's Law, changing stoichiometric coefficients multiplies

\Delta H

by the same factor, and reversing a reaction flips the sign of

\Delta H

.

3

Sum the enthalpy values for the target reaction

\Delta H_f^\circ = -788 - 286 + 1300 = 226\text{ kJ mol}^{-1}

The overall enthalpy change of a reaction is equal to the sum of the enthalpy changes for each intermediate step.

Key Concept

Hess's Law of Constant Heat Summation

Question 12787Question →

In a photoelectric cell experiment, monochromatic light of frequency $f$ and intensity $I$ illuminates a potassium surface, emitting photoelectrons with a stopping potential of $1.5\text{ V}$ . When the light frequency is increased to $1.5f$ and the intensity is set to $2I$ , the stopping potential rises to $3.5\text{ V}$ . What is the stopping potential if the light intensity is further increased to $4I$ while maintaining the frequency constant at $1.5f$ ?

3.5\text{ V}

7.0\text{ V}

6.0\text{ V}

14.0\text{ V}

Show answer & explanation

Answer: $3.5\text{ V}$

Answer

3.5\text{ V}

According to Einstein's photoelectric theory, stopping potential

V_s

is directly proportional to maximum kinetic energy (

e V_s = hf - W_0

). It depends exclusively on the frequency of incident radiation and the target work function. Changing light intensity from

2I

to

4I

increases the number of emitted photoelectrons per second, but does not alter maximum kinetic energy. Therefore, at frequency

1.5f

, the stopping potential remains

3.5\text{ V}

.

Step-by-Step Solution

1

Identify the relationship between stopping potential, frequency, and light intensity in the photoelectric effect.

Einstein's photoelectric equation states that

e V_s = K_{\text{max}} = h f - W_0

, where

V_s

is stopping potential,

f

is frequency, and

W_0

is work function.

Stopping potential measures the maximum kinetic energy of emitted photoelectrons.

2

Analyze the impact of changing light intensity while holding frequency constant at

1.5f

.

Increasing intensity from

2I

to

4I

increases the number of incident photons per second (hence increasing emission current), but individual photon energy

E = hf

remains unchanged.

Photon energy depends solely on frequency

f

, not on radiation intensity.

3

Determine the new stopping potential value.

Since the frequency remains fixed at

1.5f

,

V_s

stays at

3.5\text{ V}

.

Maximum kinetic energy and stopping potential are independent of light intensity.

Key Concept

Independence of photoelectron kinetic energy and stopping potential from light intensity

Question 12788Question →

A solid metallic sphere of mass $0.4\text{ kg}$ and specific heat capacity $500\text{ J kg}^{-1}\text{ K}^{-1}$ is heated to $100^\circ\text{C}$ and then placed into a well-insulated calorimeter of heat capacity $100\text{ J K}^{-1}$ . The calorimeter contains $0.5\text{ kg}$ of a liquid initially at $20^\circ\text{C}$ . If the final equilibrium temperature of the system is $40^\circ\text{C}$ and heat loss to the surroundings is negligible, what is the specific heat capacity of the liquid?

1200\text{ J kg}^{-1}\text{ K}^{-1}

1000\text{ J kg}^{-1}\text{ K}^{-1}

1100\text{ J kg}^{-1}\text{ K}^{-1}

200\text{ J kg}^{-1}\text{ K}^{-1}

Show answer & explanation

Answer: $1000\text{ J kg}^{-1}\text{ K}^{-1}$

Answer

1000\text{ J kg}^{-1}\text{ K}^{-1}

The heat lost by the cooling metallic sphere is

Q = 0.4 \times 500 \times 60 = 12000\text{ J}

. This energy raises the temperature of both the calorimeter container and the liquid by

20^\circ\text{C}

. Setting

(100 + 0.5 c_{\text{liquid}}) \times 20 = 12000

yields

100 + 0.5 c_{\text{liquid}} = 600

, giving

c_{\text{liquid}} = 1000\text{ J kg}^{-1}\text{ K}^{-1}

.

Step-by-Step Solution

1

Calculate the total heat energy lost by the cooling metallic sphere.

Q_{\text{lost}} = m_1 c_1 (T_{\text{initial}} - T_{\text{final}}) = 0.4 \times 500 \times (100 - 40) = 12000\text{ J}

Heat lost depends on mass, specific heat capacity, and temperature decrease of the hot body.

2

Formulate the thermal energy absorption by the calorimeter and liquid.

Q_{\text{gained}} = (C_{\text{cal}} + m_2 c_2) (T_{\text{final}} - T_{\text{initial, liquid}}) = (100 + 0.5 c_2) \times (40 - 20)

Heat capacity of the container (

C_{\text{cal}}

) is an extensive property already accounting for container mass, whereas specific heat capacity (

c_2

) must be multiplied by liquid mass.

3

Equate heat lost to heat gained using conservation of thermal energy and solve for

c_2

.

12000 = (100 + 0.5 c_2) \times 20 \implies 600 = 100 + 0.5 c_2 \implies 0.5 c_2 = 500 \implies c_2 = 1000\text{ J kg}^{-1}\text{ K}^{-1}

Assuming no external losses, energy conservation dictates that thermal energy lost equals thermal energy absorbed.

Key Concept

Method of Mixtures and Distinction Between Heat Capacity and Specific Heat Capacity

Question 12789Question →

A wheel and axle system consists of a wheel with a radius of $25\text{ cm}$ attached to an axle with a radius of $5\text{ cm}$ . What is the velocity ratio of this machine?

Show answer & explanation

Answer: 5

Answer

The velocity ratio of the machine is 5.

The velocity ratio (VR) of a wheel and axle is the ratio of the radius of the wheel (

R

) to the radius of the axle (

r

). Calculating

\frac{25\text{ cm}}{5\text{ cm}}

gives a velocity ratio of

5

.

Step-by-Step Solution

1

Identify the formula for the velocity ratio of a wheel and axle system.

\text{VR} = \frac{R}{r}

Velocity ratio is defined as the distance moved by the effort (proportional to wheel radius) divided by the distance moved by the load (proportional to axle radius).

2

Substitute the given values into the formula.

\text{VR} = \frac{25\text{ cm}}{5\text{ cm}}

The radius of the wheel

R = 25\text{ cm}

and the radius of the axle

r = 5\text{ cm}

.

3

Calculate the final ratio.

\text{VR} = 5

Dividing 25 by 5 yields 5. The ratio is dimensionless because the units of centimeters cancel out.

Key Concept

Velocity Ratio of a Wheel and Axle

Question 12790Question →

An alternating current (AC) circuit consists of a resistor of resistance $R = 30\ \Omega$ connected in series with a pure inductor across an AC supply of root-mean-square (RMS) voltage $100\ \text{V}$ . If the average power dissipated in the circuit is $120\ \text{W}$ , what is the inductive reactance of the inductor?

20\ \Omega

40\ \Omega

50\ \Omega

70\ \Omega

Show answer & explanation

Answer: $40\ \Omega$

Answer

The inductive reactance of the inductor is

40\ \Omega

.

In an AC circuit containing a resistor and a pure inductor, power is dissipated solely by the resistance. Using

P = I_{\text{rms}}^2 R

, the current is

I_{\text{rms}} = \sqrt{120 / 30} = 2\ \text{A}

. The total impedance

Z

is

V_{\text{rms}} / I_{\text{rms}} = 100 / 2 = 50\ \Omega

. Applying the phasor formula for impedance

Z = \sqrt{R^2 + X_L^2}

, solving for

X_L

yields

X_L = \sqrt{50^2 - 30^2} = 40\ \Omega

.

Step-by-Step Solution

1

Calculate the RMS current in the circuit using the average power formula

I_{\text{rms}} = 2\ \text{A}

In an AC circuit with a resistor and a pure inductor, average power is dissipated only by the resistor:

P = I_{\text{rms}}^2 R \implies 120 = I_{\text{rms}}^2 \times 30 \implies I_{\text{rms}}^2 = 4 \implies I_{\text{rms}} = 2\ \text{A}

.

2

Determine the total impedance of the circuit

Z = 50\ \Omega

The total impedance is the ratio of RMS voltage to RMS current:

Z = \frac{V_{\text{rms}}}{I_{\text{rms}}} = \frac{100\ \text{V}}{2\ \text{A}} = 50\ \Omega

.

3

Calculate the inductive reactance using the impedance relationship for a series RL circuit

X_L = 40\ \Omega

Impedance in a series RL circuit is given by

Z = \sqrt{R^2 + X_L^2}

. Substituting the known values gives

50 = \sqrt{30^2 + X_L^2} \implies 2500 = 900 + X_L^2 \implies X_L^2 = 1600 \implies X_L = 40\ \Omega

.

Key Concept

Power Dissipation and Impedance in Series RL AC Circuits

Question 12791Question →

A uniform horizontal wooden rod $XY$ of length $3.0\text{ m}$ and weight $50\text{ N}$ rests horizontally on two smooth supports located at $X$ (the left end) and at a point $Z$ which is $0.6\text{ m}$ from end $Y$ . A block of weight $120\text{ N}$ is placed on the rod at a distance of $0.9\text{ m}$ from end $X$ . What is the magnitude of the upward reaction force, in newtons, at support $Z$ ?

Show answer & explanation

Answer: 76.25

Answer

The magnitude of the upward reaction force at support

Z

is

76.25\text{ N}

.

Taking moments about support

X

, the total clockwise moment is the sum of the moment due to the load (

120\text{ N} \times 0.9\text{ m} = 108\text{ N}\cdot\text{m}

) and the weight of the rod (

50\text{ N} \times 1.5\text{ m} = 75\text{ N}\cdot\text{m}

), giving

183\text{ N}\cdot\text{m}

. Equating this to the counterclockwise moment of the reaction force at

Z

(

R_Z \times 2.4\text{ m}

) yields

R_Z = \frac{183}{2.4} = 76.25\text{ N}

.

Step-by-Step Solution

1

Identify the perpendicular distance of each force and support from pivot point

X

.

Center of gravity position

x_{cg} = 1.5\text{ m}

, load position

x_{L} = 0.9\text{ m}

, and support

Z

position

x_{Z} = 3.0 - 0.6 = 2.4\text{ m}

.

Taking moments about

X

requires knowing the exact moment arm for each force from

X

.

2

Set up the equation for rotational equilibrium about point

X

.

Total clockwise moment =

(120 \times 0.9) + (50 \times 1.5) = 183\text{ N}\cdot\text{m}

; Total counterclockwise moment =

R_Z \times 2.4

.

Choosing pivot

X

eliminates the unknown reaction force

R_X

because its distance from

X

is zero.

3

Equate clockwise moments to counterclockwise moments and solve for

R_Z

.

R_Z = \frac{183}{2.4} = 76.25\text{ N}

.

For a body in rotational equilibrium, the algebraic sum of moments about any point must equal zero.

Key Concept

Principle of Moments and Rotational Equilibrium

Question 12792Question →

A wheel and axle machine has a wheel of radius $25\text{ cm}$ and an axle of radius $5\text{ cm}$ . If an effort force of $100\text{ N}$ is applied to lift a load of $400\text{ N}$ , what is the efficiency of the machine?

80\%

125\%

20\%

5\%

Show answer & explanation

Answer: $80\%$

Answer

The efficiency of the wheel and axle machine is

80\%

.

First, calculate the velocity ratio as the ratio of wheel radius to axle radius:

\text{VR} = \frac{25}{5} = 5

. Next, calculate mechanical advantage as load over effort:

\text{MA} = \frac{400}{100} = 4

. Finally, calculate efficiency by dividing mechanical advantage by velocity ratio:

\text{Efficiency} = \frac{4}{5} \times 100\% = 80\%

.

Step-by-Step Solution

1

Calculate the Velocity Ratio (VR) of the wheel and axle system

\text{VR} = \frac{\text{Radius of Wheel}}{\text{Radius of Axle}} = \frac{25\text{ cm}}{5\text{ cm}} = 5

For a wheel and axle, the velocity ratio is the ratio of the radius of the wheel to the radius of the axle.

2

Calculate the Mechanical Advantage (MA) of the machine

\text{MA} = \frac{\text{Load}}{\text{Effort}} = \frac{400\text{ N}}{100\text{ N}} = 4

Mechanical advantage measures the force magnification of a machine, defined as the ratio of load raised to effort applied.

3

Calculate the Efficiency of the machine

\text{Efficiency} = \frac{\text{MA}}{\text{VR}} \times 100\% = \frac{4}{5} \times 100\% = 80\%

Efficiency is defined as the ratio of Mechanical Advantage to Velocity Ratio expressed as a percentage.

Key Concept

Efficiency of Simple Machines

Question 12793Question →

The mass $m$ of an object is measured as $(2.0 \pm 0.1)\text{ kg}$ and its speed $v$ is measured as $(5.0 \pm 0.2)\text{ m s}^{-1}$ . What is the maximum percentage error in the calculated kinetic energy of the object?

13\%

9\%

8\%

21\%

Show answer & explanation

Answer: $13\%$

Answer

The maximum percentage error in the calculated kinetic energy is

13\%

.

For a calculated quantity

E = \frac{1}{2}mv^2

, the constant factor

\frac{1}{2}

has no error. The fractional error propagation formula gives

\frac{\Delta E}{E} = \frac{\Delta m}{m} + 2\frac{\Delta v}{v}

. Substituting

\frac{\Delta m}{m} = 0.05

(

5\%

) and

\frac{\Delta v}{v} = 0.04

(

4\%

) yields

5\% + 2(4\%) = 13\%

. Thus, the option stating

13\%

is correct.

Step-by-Step Solution

1

Calculate the fractional error and percentage error in mass

m

\frac{\Delta m}{m} = \frac{0.1}{2.0} = 0.05 = 5\%

Percentage error in mass is the absolute error divided by the measured value multiplied by 100.

2

Calculate the fractional error and percentage error in speed

v

\frac{\Delta v}{v} = \frac{0.2}{5.0} = 0.04 = 4\%

Percentage error in speed is the absolute error divided by the measured value multiplied by 100.

3

Apply the error propagation formula for kinetic energy

E = \frac{1}{2}m v^2

\frac{\Delta E}{E} = \frac{\Delta m}{m} + 2\left(\frac{\Delta v}{v}\right)

When physical quantities are raised to a power and multiplied, their fractional errors are multiplied by the respective power index and added.

4

Compute the maximum percentage error in kinetic energy

\%\text{ error in } E = 5\% + 2(4\%) = 5\% + 8\% = 13\%

Combining the weighted percentage errors yields the total maximum percentage error.

Key Concept

Error Propagation in Power and Product Relationships

Estimated Time:2m 0s

Question 12794Question →

Match each physical observation or analytical test result with its corresponding purity condition or chemical interpretation.

Click a left item, then click its matching right item

Show answer & explanation

Answer

A solid melting sharply at

122.5^\circ\text{C}

corresponds to the criterion for a pure solid compound. Distilling across

74^\circ\text{C}-81^\circ\text{C}

indicates an impure liquid mixture. Producing a single chromatogram spot confirms a single chemical component. A boiling point elevated above literature value indicates the presence of a non-volatile dissolved impurity.

Each physical criterion uniquely identifies pure vs. impure substances: sharp melting point defines pure solids; broad boiling ranges define liquid mixtures; single chromatographic spots confirm a single substance; and elevated boiling points indicate non-volatile solutes in liquids.

Step-by-Step Solution

1

Analyze the solid melting point behavior

A sharp melting point indicates high purity for a solid sample.

Pure crystalline solids have uniform intermolecular forces that break simultaneously at a specific temperature.

2

Evaluate the liquid boiling range

A broad boiling range indicates a mixture or an impure liquid.

Different components or impurities alter the vapor pressure progressively as temperature changes.

3

Interpret the chromatographic result

A single spot on chromatography confirms homogeneity/purity.

Multiple components would separate into distinct spots based on differing solubilities and affinities.

4

Determine the cause of boiling point elevation

A boiling point higher than the literature value points to a non-volatile dissolved impurity.

Solute particles reduce the solvent's vapor pressure, requiring higher kinetic energy (temperature) to match atmospheric pressure.

Key Concept

Criteria of purity (fixed sharp melting point, fixed boiling point, single chromatographic spot, density, refractive index).

Question 12795Question →

A closed rigid container holds a mixture of dry air and water vapour at a temperature of $27^\circ\text{C}$ under a total pressure of $740\text{ mmHg}$ . The relative humidity of the air inside the container is $80\%$ , and the saturated vapour pressure of water at $27^\circ\text{C}$ is $25\text{ mmHg}$ . If the container is heated at constant volume to $127^\circ\text{C}$ , what is the partial pressure of the dry air in $\text{mmHg}$ at this higher temperature?

Show answer & explanation

Answer: 960

Answer

The partial pressure of dry air inside the container at

127^\circ\text{C}

is

960\text{ mmHg}

.

First, the partial pressure of water vapour at

27^\circ\text{C}

is determined by multiplying relative humidity (

80\%

) by the saturated vapour pressure (

25\text{ mmHg}

), yielding

20\text{ mmHg}

. Next, subtracting this vapour pressure from the total pressure of

740\text{ mmHg}

gives the partial pressure of dry air alone as

720\text{ mmHg}

at

27^\circ\text{C}

(

300\text{ K}

). Finally, applying the Pressure Law (

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

) for the dry air between

300\text{ K}

and

400\text{ K}

(

127^\circ\text{C}

) yields

P_2 = 720 \times \frac{400}{300} = 960\text{ mmHg}

.

Step-by-Step Solution

1

Calculate the partial pressure of water vapour at

27^\circ\text{C}

P_{\text{vapour}, 1} = 0.80 \times 25\text{ mmHg} = 20\text{ mmHg}

Relative humidity is the ratio of actual partial vapour pressure to the saturated vapour pressure at that temperature.

2

Determine the initial partial pressure of the dry air at

27^\circ\text{C}

using Dalton's Law

P_{\text{dry air}, 1} = 740\text{ mmHg} - 20\text{ mmHg} = 720\text{ mmHg}

Total pressure of a gas mixture is the sum of the partial pressures of its individual components.

3

Convert temperatures to Kelvin and apply Gay-Lussac's Pressure Law for dry air at constant volume

T_1 = 27 + 273 = 300\text{ K}

,

T_2 = 127 + 273 = 400\text{ K}

.

P_{\text{dry air}, 2} = 720\text{ mmHg} \times \left(\frac{400\text{ K}}{300\text{ K}}\right) = 960\text{ mmHg}

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

Key Concept

Dalton's Law of Partial Pressures and Gay-Lussac's Pressure Law applied to gas-vapour mixtures

Question 12796Question →

A sample of hydrated sodium trioxocarbonate(IV), $\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}$ , is heated strongly in a crucible until a constant mass is reached. If the salt loses $62.94\%$ of its initial mass as water vapor during heating, what is the integer value of $x$ ?

[Relative atomic masses: $\text{Na} = 23$ , $\text{C} = 12$ , $\text{O} = 16$ , $\text{H} = 1$ ]

Show answer & explanation

Answer: 10

Answer

The integer value of

x

is 10.

Heating hydrated sodium trioxocarbonate(IV) drives off all water of crystallization as steam. Since

62.94\%

of the total mass is lost, water accounts for

62.94\%

of the molar mass of

\text{Na}_2\text{CO}_3 \cdot x\text{H}_2\text{O}

. Solving

\frac{18x}{106 + 18x} = 0.6294

yields

x = 10

, representing decahydrate crystals,

\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}

.

Step-by-Step Solution

1

Calculate the formula masses of the anhydrous salt

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

and water

\text{H}_2\text{O}

.

Molar mass of

\text{Na}_2\text{CO}_3 = 106\text{ g/mol}

and molar mass of

\text{H}_2\text{O} = 18\text{ g/mol}

.

Establishing the molar component masses is required to determine the percentage ratio of water of crystallization to the total mass of the hydrated salt.

2

Set up an algebraic ratio relating the mass of water lost to the total hydrated mass using the given percentage.

\frac{18x}{106 + 18x} = 0.6294

.

Heating to constant mass removes all water of crystallization, meaning the mass lost corresponds to the

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

component of the hydrated crystal.

3

Solve the algebraic equation for

x

.

x = 10

.

Isolating

x

yields the exact stoichiometric coefficient of water molecules per mole of hydrated salt.

Key Concept

Quantitative determination of water of crystallization using percentage mass loss.

Question 12797Question →

A hydraulic press has a small piston with a diameter of $4\text{ cm}$ and a large piston with a diameter of $20\text{ cm}$ . If an effort force of $80\text{ N}$ applied to the small piston raises a load of $1500\text{ N}$ placed on the large piston, what is the efficiency of the machine?

133.3\%

75.0\%

375.0\%

26.7\%

Show answer & explanation

Answer: $75.0\%$

Answer

The efficiency of the hydraulic press is

75.0\%

.

The mechanical advantage is

MA = \frac{1500}{80} = 18.75

. The velocity ratio for pistons of diameters

20\text{ cm}

and

4\text{ cm}

is

VR = \left(\frac{20}{4}\right)^2 = 25

. Efficiency is

\frac{MA}{VR} \times 100\% = \frac{18.75}{25} \times 100\% = 75.0\%

.

Step-by-Step Solution

1

Calculate Mechanical Advantage (MA)

MA = \frac{\text{Load}}{\text{Effort}} = \frac{1500\text{ N}}{80\text{ N}} = 18.75

Mechanical advantage is defined as the ratio of load force to effort force.

2

Calculate Velocity Ratio (VR) for the hydraulic press

VR = \frac{A_2}{A_1} = \left(\frac{d_2}{d_1}\right)^2 = \left(\frac{20\text{ cm}}{4\text{ cm}}\right)^2 = 5^2 = 25

The velocity ratio of a hydraulic press equals the ratio of the cross-sectional areas of the pistons, which simplifies to the square of the ratio of their diameters.

3

Calculate Efficiency (\eta)

\eta = \frac{MA}{VR} \times 100\% = \frac{18.75}{25} \times 100\% = 75.0\%

Efficiency is the ratio of mechanical advantage to velocity ratio expressed as a percentage.

Key Concept

Hydraulic Press Efficiency and Velocity Ratio

Question 12798Question →

Match each chemical formula of the inorganic redox species on the left with its corresponding IUPAC name on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are:

HClO_2

matches with Dioxochloric(III) acid;

H_2S_2O_7

matches with Heptaoxodisulfuric(VI) acid;

NaNO_2

matches with Sodium dioxonitrate(III); and

KIO_4

matches with Potassium tetraoxoiodate(VII).

Each chemical species is systematically named by calculating the oxidation state of its central non-metal atom and prefixing the number of oxygen atoms present (dioxo-, tetraoxo-, heptaoxo-).

HClO_2

has chlorine in

+3

state (Dioxochloric(III) acid),

H_2S_2O_7

has sulfur in

+6

state across two atoms (Heptaoxodisulfuric(VI) acid),

NaNO_2

has nitrogen in

+3

state (Sodium dioxonitrate(III)), and

KIO_4

has iodine in

+7

state (Potassium tetraoxoiodate(VII)).

Step-by-Step Solution

1

Determine the oxidation state of chlorine in

HClO_2

.

Assign

+1

to

H

and

-2

to

O

:

(+1) + Cl + 2(-2) = 0 ⇒ Cl = +3

. Combined with two oxo groups, the IUPAC name is Dioxochloric(III) acid.

Oxoacids are named by specifying the number of oxygen atoms with oxo prefixes followed by the central element and its Roman numeral oxidation state.

2

Calculate the oxidation state of sulfur in

H_2S_2O_7

.

Assign

+1

to

H

and

-2

to

O

:

2(+1) + 2(S) + 7(-2) = 0 ⇒ 2S = +12 ⇒ S = +6

. With seven oxygen atoms and two sulfur atoms, the IUPAC name is Heptaoxodisulfuric(VI) acid.

The prefix 'heptaoxo-' accounts for seven oxygens and 'disulfuric' indicates two sulfur atoms.

3

Calculate the oxidation state of nitrogen in

NaNO_2

.

Assign

+1

to

Na

and

-2

to

O

:

(+1) + N + 2(-2) = 0 ⇒ N = +3

. The anion is dioxonitrate(III), making the salt Sodium dioxonitrate(III).

Salts of oxoanions state the cation name first followed by the IUPAC name of the oxoanion.

4

Calculate the oxidation state of iodine in

KIO_4

.

Assign

+1

to

K

and

-2

to

O

:

(+1) + I + 4(-2) = 0 ⇒ I = +7

. The anion is tetraoxoiodate(VII), making the salt Potassium tetraoxoiodate(VII).

Four oxygen atoms dictate the prefix 'tetraoxo-' and the iodine state

+7

gives Roman numeral (VII).

Key Concept

IUPAC Nomenclature and Oxidation State Calculation of Oxoacids and Oxosalts

Question 12799Question →

The radius $r$ of a solid cylinder is measured as $(2.0 \pm 0.1)\text{ cm}$ and its height $h$ is measured as $(5.0 \pm 0.1)\text{ cm}$ . What is the maximum percentage error in the calculated volume of the cylinder?

7.0\%

8.0\%

10.0\%

12.0\%

Show answer & explanation

Answer: $12.0\%$

Answer

12.0\%

The formula for the volume of a cylinder is

V = \pi r^2 h

. According to the principles of error propagation, the fractional error in

V

is

\frac{\Delta V}{V} = 2\frac{\Delta r}{r} + \frac{\Delta h}{h}

. Substituting the values:

\frac{\Delta r}{r} = \frac{0.1}{2.0} = 0.05

(

5.0\%

) and

\frac{\Delta h}{h} = \frac{0.1}{5.0} = 0.02

(

2.0\%

). Thus, the maximum percentage error is

2(5.0\%) + 2.0\% = 12.0\%

.

Step-by-Step Solution

1

Calculate the percentage error in the radius measurement

\text{Percentage error in } r = \left(\frac{0.1}{2.0}\right) \times 100\% = 5.0\%

Percentage error is given by the ratio of absolute uncertainty to the measured value multiplied by 100.

2

Calculate the percentage error in the height measurement

\text{Percentage error in } h = \left(\frac{0.1}{5.0}\right) \times 100\% = 2.0\%

Percentage error of a single linear measurement.

3

Apply the error propagation formula for the volume of a cylinder

\frac{\Delta V}{V} \times 100\% = 2\left(\frac{\Delta r}{r} \times 100\%\right) + \left(\frac{\Delta h}{h} \times 100\%\right)

Since

V = \pi r^2 h

, fractional errors add up, with the power of any variable acting as a multiplier for its fractional error.

4

Compute the total maximum percentage error in volume

\text{Percentage error in } V = 2(5.0\%) + 2.0\% = 10.0\% + 2.0\% = 12.0\%

Adding the individual fractional error contributions gives the maximum percentage error.

Key Concept

Error propagation in derived quantities with exponent powers

Question 12800Question →

A light, rigid horizontal bar $AB$ of length $1.5\text{ m}$ is smoothly pivoted at end $A$ . A vertical downward load of $40\text{ N}$ is hung from end $B$ . The bar is kept in horizontal equilibrium by a light string attached at point $C$ , located $1.0\text{ m}$ from $A$ . The string exerts a tension force $T$ pulling upwards at an angle of $30^\circ$ relative to the horizontal bar. What is the magnitude of the tension $T$ in newtons?

Show answer & explanation

Answer: 120

Answer

The magnitude of the tension

T

in the string is

120\text{ N}

.

For the bar to maintain rotational equilibrium, the clockwise moment about pivot

A

must equal the counterclockwise moment about

A

. The

40\text{ N}

load exerts a clockwise moment of

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

. The string tension

T

exerts a counterclockwise moment given by its vertical component multiplied by the distance from the pivot:

(T \sin 30^\circ) \times 1.0\text{ m} = 0.5T \text{ N}\cdot\text{m}

. Equating the two moments gives

0.5T = 60\text{ N}\cdot\text{m}

, yielding

T = 120\text{ N}

.

Step-by-Step Solution

1

Calculate the clockwise moment about the pivot at end

A

τ_{\text{clockwise}} = 40\text{ N} \times 1.5\text{ m} = 60\text{ N}\cdot\text{m}

The weight at

B

acts vertically downward at a perpendicular distance of

1.5\text{ m}

from pivot

A

.

2

Determine the perpendicular component of tension

T

relative to the bar

F_{\perp} = T \sin 30^\circ = 0.5T

Only the component of force perpendicular to the bar produces a moment about the pivot.

3

Set up the counterclockwise moment about pivot

A

τ_{\text{counterclockwise}} = (0.5T) \times 1.0\text{ m} = 0.5T \text{ N}\cdot\text{m}

The string is attached at point

C

, which is

1.0\text{ m}

away from pivot

A

.

4

Apply the principle of moments for rotational equilibrium and solve for

T

0.5T = 60 \implies T = 120\text{ N}

For rotational equilibrium, total clockwise moments must equal total counterclockwise moments about any pivot.

Key Concept

Principle of moments and rotational equilibrium for forces acting at non-perpendicular angles.

Estimated Time:1m 30s

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