Tüm alıştırma soruları

Soru 241Soru →

Monochromatic light with a photon energy of $4.5\text{ eV}$ strikes the surface of a sodium plate inside a vacuum tube. If the work function of sodium is $2.1\text{ eV}$ , what is the maximum kinetic energy of the emitted photoelectrons in electron-volts ( $\text{eV}$ )?

Cevabı ve açıklamayı göster

Cevap: 2.4

Cevap

The maximum kinetic energy of the emitted photoelectrons is 2.4 eV.

According to Einstein's photoelectric theory, energy is conserved such that the energy of an incident photon (

E

) is partly used to liberate an electron from the metal surface (work function

W_0

) and the remaining energy appears as the maximum kinetic energy (

K_{\text{max}}

) of the photoelectron. Subtracting

2.1\text{ eV}

from

4.5\text{ eV}

gives

2.4\text{ eV}

.

Adım Adım Çözüm

1

Identify the given values from the problem statement

Photon energy

E = 4.5\text{ eV}

, Work function

W_0 = 2.1\text{ eV}

Establishing known variables helps determine the correct photoelectric relation to apply.

2

Apply Einstein's photoelectric equation

K_{\text{max}} = E - W_0

The maximum kinetic energy of photoelectrons equals the excess energy of incident photons after overcoming the work function.

3

Calculate the value

K_{\text{max}} = 4.5\text{ eV} - 2.1\text{ eV} = 2.4\text{ eV}

Subtracting the work function from the photon energy yields the final numerical answer.

Anahtar Kavram

Einstein's Photoelectric Equation and Energy Conservation

Soru 242Soru →

A uniform string of length $0.50\text{ m}$ and mass $2.0\text{ g}$ is fixed at both ends under a tension of $90\text{ N}$ . When vibrating, the second harmonic of this string resonates with the first overtone of an air column in a pipe closed at one end. Assuming the speed of sound in air is $340\text{ m/s}$ , calculate the length of the pipe in meters.

Cevabı ve açıklamayı göster

Cevap: 0.85

Cevap

The length of the pipe is

0.85\text{ m}

.

The wave speed on the string is computed from tension and linear mass density as $150\text{ m/s}$, yielding a second harmonic frequency of $300\text{ Hz}$. Equating this to the first overtone (third harmonic) frequency formula of a closed air pipe, $f = \frac{3v_{air}}{4L_p}$, yields an exact pipe length of $0.85\text{ m}$.

Adım Adım Çözüm

1

Calculate the linear mass density ($\mu$) of the string

$\mu = \frac{m}{L_s} = \frac{0.0020\text{ kg}}{0.50\text{ m}} = 4.0 \times 10^{-3}\text{ kg/m}$

Mass must be converted to kilograms before determining mass per unit length.

2

Determine the wave speed ($v_s$) along the stretched string

$v_s = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{90\text{ N}}{4.0 \times 10^{-3}\text{ kg/m}}} = \sqrt{22500} = 150\text{ m/s}$

The velocity of a transverse wave on a string depends on tension and linear mass density.

3

Calculate the second harmonic frequency of the string ($f_{2,s}$)

$f_{2,s} = \frac{v_s}{L_s} = \frac{150\text{ m/s}}{0.50\text{ m}} = 300\text{ Hz}$

The fundamental frequency is $f_{1,s} = \frac{v_s}{2L_s} = 150\text{ Hz}$, so the second harmonic is twice the fundamental frequency.

4

Set up the resonance equation for the first overtone of a closed pipe

$f_{3,p} = \frac{3 v_{air}}{4 L_p} = 300\text{ Hz}$

A pipe closed at one end produces only odd harmonics, so the first overtone is the 3rd harmonic.

5

Solve for the length of the closed pipe ($L_p$)

$L_p = \frac{3 \times 340\text{ m/s}}{4 \times 300\text{ Hz}} = \frac{1020}{1200} = 0.85\text{ m}$

Rearranging the frequency formula gives the required air column length.

Anahtar Kavram

Coupled resonance between standing waves on strings and air columns in closed pipes

Soru 243Soru →

If $y = (x^2 + 1)(2x - 3)^3$ , find the value of $\frac{dy}{dx}$ at $x = 2$ .

Cevabı ve açıklamayı göster

Cevap: 34

Cevap

The value of

\frac{dy}{dx}

at

x = 2

is

34

.

Applying the product rule

\frac{dy}{dx} = u'v + uv'

alongside the chain rule gives

\frac{dy}{dx} = 2x(2x - 3)^3 + 6(x^2 + 1)(2x - 3)^2

. Evaluating at

x = 2

yields

4(1) + 30(1) = 34

.

Adım Adım Çözüm

1

Identify component functions for the product rule

Let

u(x) = x^2 + 1

and

v(x) = (2x - 3)^3

.

The function

y

is a product of two differentiable functions.

2

Differentiate each component function

u'(x) = 2x

and

v'(x) = 3(2x - 3)^2 \cdot 2 = 6(2x - 3)^2

.

The power rule gives

u'(x)

and the chain rule gives

v'(x)

by multiplying by the derivative of the inner function

(2x - 3)

.

3

Apply the product rule formula

\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)

\frac{dy}{dx} = 2x(2x - 3)^3 + 6(x^2 + 1)(2x - 3)^2

.

To find the general derivative of a product of functions.

4

Evaluate the derivative at

x = 2

\frac{dy}{dx}\Big|_{x=2} = 2(2)(2(2) - 3)^3 + 6(2^2 + 1)(2(2) - 3)^2 = 4(1) + 30(1) = 34

.

To calculate the specific numerical value of the derivative at

x = 2

.

Anahtar Kavram

Product and Chain Rules of Differentiation

Tahmini Süre:1m 30s

Soru 244Soru →

A $3\text{ }\mu\text{F}$ capacitor and a $6\text{ }\mu\text{F}$ capacitor are connected in series across a direct current voltage source. What is the total equivalent capacitance of the combination, in microfarads ( $\mu\text{F}$ )?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The total equivalent capacitance of the combination is

2\text{ }\mu\text{F}

.

For capacitors connected in series, the reciprocal of the total equivalent capacitance is equal to the sum of the reciprocals of the individual capacitances. Substituting

3\text{ }\mu\text{F}

and

6\text{ }\mu\text{F}

gives

\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}\text{ }\mu\text{F}^{-1}

, which yields an equivalent capacitance of

2\text{ }\mu\text{F}

.

Adım Adım Çözüm

1

State the formula for equivalent capacitance of two capacitors in series

\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}

Capacitors connected in series combine reciprocally, unlike resistors connected in series.

2

Substitute the values of

C_1

and

C_2

\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}\text{ }\mu\text{F}^{-1}

Find a common denominator and add the fractions.

3

Calculate the reciprocal to determine

C_{eq}

C_{eq} = 2\text{ }\mu\text{F}

Inverting

\frac{1}{2}

yields the total equivalent capacitance.

Anahtar Kavram

Equivalent Capacitance in Series

Soru 245Soru →

A body of mass $2\text{ kg}$ is projected vertically upward from ground level with an initial speed of $30\text{ m s}^{-1}$ . During its entire flight, it experiences a constant resistive force due to air resistance of $5\text{ N}$ . Taking $g = 10\text{ m s}^{-2}$ , calculate the kinetic energy of the body in Joules when it returns to the ground level.

Cevabı ve açıklamayı göster

Cevap: 540

Cevap

The kinetic energy of the body when it returns to ground level is

540\text{ J}

.

At launch, the body possesses an initial kinetic energy of

E_i = \frac{1}{2}(2)(30)^2 = 900\text{ J}

. During ascent, both gravity (

20\text{ N}

) and air resistance (

5\text{ N}

) retard the motion, giving a net downward force of

25\text{ N}

and a deceleration of

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

. The maximum height reached is

h = \frac{30^2}{2(12.5)} = 36\text{ m}

. Since the body travels up and down, total distance covered is

72\text{ m}

. Non-conservative work done against air resistance is

W = 5\text{ N} \times 72\text{ m} = 360\text{ J}

. Therefore, the remaining kinetic energy upon returning to the ground is

900\text{ J} - 360\text{ J} = 540\text{ J}

.

Adım Adım Çözüm

1

Calculate initial kinetic energy of launch

E_i = 900\text{ J}

Kinetic energy is given by

\frac{1}{2}m u^2 = \frac{1}{2}(2\text{ kg})(30\text{ m s}^{-1})^2 = 900\text{ J}

.

2

Determine maximum height reached during ascent

h = 36\text{ m}

Net upward retarding force

F = mg + F_{\text{air}} = 20 + 5 = 25\text{ N}

, yielding a deceleration

a = 12.5\text{ m s}^{-2}

. Using

0 = u^2 - 2ah

,

h = \frac{900}{25} = 36\text{ m}

.

3

Calculate energy lost to air resistance over total trajectory

W_{\text{air}} = 360\text{ J}

Air resistance acts continuously over both ascent and descent (total distance

2h = 72\text{ m}

). Work dissipated

= F_{\text{air}} \times 2h = 5 \times 72 = 360\text{ J}

.

4

Subtract non-conservative work loss from initial mechanical energy

E_f = 540\text{ J}

By mechanical energy balance, final kinetic energy

E_f = E_i - W_{\text{air}} = 900 - 360 = 540\text{ J}

.

Anahtar Kavram

Work-Energy Theorem and Mechanical Energy Dissipation by Non-Conservative Forces

Soru 246Soru →

The sum of the interior angles of a regular polygon is $1440^\circ$ . What is the measure, in degrees, of one exterior angle of this polygon?

Cevabı ve açıklamayı göster

Cevap: 36

Cevap

The measure of one exterior angle of the polygon is

36^\circ

.

The sum of the interior angles of an

n

-sided polygon is given by

(n-2) \times 180^\circ

. Setting

(n-2) \times 180^\circ = 1440^\circ

gives

n-2 = 8

, so the polygon has

n = 10

sides (a decagon). The measure of each exterior angle of a regular polygon is

\frac{360^\circ}{n} = \frac{360^\circ}{10} = 36^\circ

.

Adım Adım Çözüm

1

Set up the equation for the sum of interior angles of an

n

-sided polygon.

(n - 2) \times 180^\circ = 1440^\circ

The sum of interior angles of any convex

n

-sided polygon is

(n - 2) \times 180^\circ

.

2

Solve for

n

, the number of sides.

n - 2 = \frac{1440}{180} = 8 \implies n = 10

Dividing the interior angle sum by

180^\circ

gives

n - 2

.

3

Calculate the measure of one exterior angle.

Exterior angle

= \frac{360^\circ}{10} = 36^\circ

The sum of exterior angles of any convex polygon is

360^\circ

, so each exterior angle of a regular polygon with

n

sides is

\frac{360^\circ}{n}

.

Anahtar Kavram

Relationship between interior angle sum, number of sides, and exterior angles of a regular polygon.

Soru 247Soru →

In a telecommunications network, two independent relay switches, $R_1$ and $R_2$ , operate during a data transmission. The probability that switch $R_1$ functions successfully is $0.80$ , and the probability that switch $R_2$ functions successfully is $0.75$ . What is the probability that at least one of the two switches functions successfully during the transmission?

Cevabı ve açıklamayı göster

Cevap: 0.95

Cevap

The probability that at least one switch functions successfully is 0.95.

Since the two switches operate independently, the probability of both failing is the product of their individual failure probabilities:

(1 - 0.80) \times (1 - 0.75) = 0.20 \times 0.25 = 0.05

. Therefore, the probability that at least one switch functions successfully is

1 - 0.05 = 0.95

. Alternatively, using the addition law for independent events:

P(R_1 \cup R_2) = P(R_1) + P(R_2) - P(R_1 \cap R_2) = 0.80 + 0.75 - (0.80 \times 0.75) = 1.55 - 0.60 = 0.95

.

Adım Adım Çözüm

1

Determine the probabilities of individual switch failure.

P(R_1') = 0.20 and P(R_2') = 0.25

The probability of an event failing is the complement of its success probability: P(E') = 1 - P(E).

2

Compute the probability that both switches fail simultaneously.

P(R_1' ∩ R_2') = 0.20 × 0.25 = 0.05

Since the switches operate independently, the multiplication law for independent events applies: P(A ∩ B) = P(A) × P(B).

3

Calculate the probability that at least one switch functions successfully.

P(at least one) = 1 - 0.05 = 0.95

The complement of 'neither switch functioning' is 'at least one switch functioning'.

Anahtar Kavram

Probability laws for independent compound events and the complement rule

Tahmini Süre:1m 30s

Soru 248Soru →

A point charge $q_1 = +9.0 \times 10^{-9}\text{ C}$ is fixed at the origin ( $x = 0\text{ m}$ ), and a second point charge $q_2 = -4.0 \times 10^{-9}\text{ C}$ is fixed on the x-axis at $x = 0.5\text{ m}$ . At what position $x$ (in meters) along the x-axis is the net electric field intensity equal to zero?

Cevabı ve açıklamayı göster

Cevap: 1.5

Cevap

The net electric field intensity is zero at

x = 1.5\text{ m}

.

The correct position is

x = 1.5\text{ m}

. At this point, the electric field from

+q_1

points in the

+x

direction with magnitude

E_1 = \frac{9.0 \times 10^9 \times 9.0 \times 10^{-9}}{1.5^2} = 36\text{ N C}^{-1}

, and the electric field from

-q_2

points in the

-x

direction with magnitude

E_2 = \frac{9.0 \times 10^9 \times 4.0 \times 10^{-9}}{1.0^2} = 36\text{ N C}^{-1}

. The two vectors are equal in magnitude and opposite in direction, yielding a net electric field of zero.

Adım Adım Çözüm

1

Determine the physical region where electric fields can cancel

The point of zero field lies to the right of

q_2

, i.e.,

x > 0.5\text{ m}

.

Between the charges, the fields due to

+q_1

and

-q_2

point in the same direction (+x). To the left of

q_1

,

q_1

is both larger in magnitude and closer, so

E_1 > E_2

everywhere. Hence, balance can only occur to the right of the smaller magnitude charge

q_2

.

2

Set up the condition for equal electric field magnitudes

\frac{k |q_1|}{x^2} = \frac{k |q_2|}{(x - 0.5)^2}

For the net field to be zero, the vector sum of

E_1

and

E_2

must equal zero, meaning their magnitudes must be equal.

3

Substitute values and simplify the algebraic equation

\frac{9.0 \times 10^{-9}}{x^2} = \frac{4.0 \times 10^{-9}}{(x - 0.5)^2} \implies \frac{9}{x^2} = \frac{4}{(x - 0.5)^2}

Coulomb's constant

k

and the power factor

10^{-9}

cancel from both sides.

4

Take square root on both sides to solve for x

\frac{3}{x} = \frac{2}{x - 0.5} \implies 3(x - 0.5) = 2x \implies x = 1.5\text{ m}

Taking the principal square root reduces the quadratic relation to a simple linear equation.

Anahtar Kavram

Electric Field Superposition and Zero Field Condition for Point Charges

Soru 249Soru →

Given the function $y = \frac{2x - 3}{(x^2 + 1)^2}$ , find the numerical value of $\frac{dy}{dx}$ at $x = 1$ .

Cevabı ve açıklamayı göster

Cevap: 1

Cevap

The numerical value of

\frac{dy}{dx}

at

x = 1

is 1.

Applying the quotient rule

\frac{dy}{dx} = \frac{u'v - uv'}{v^2}

alongside the chain rule for the denominator yields

\frac{dy}{dx} = \frac{2(x^2 + 1)^2 - (2x - 3) \cdot 4x(x^2 + 1)}{(x^2 + 1)^4}

. Substituting

x = 1

evaluates to

\frac{2(4) - (-1)(8)}{16} = \frac{16}{16} = 1

.

Adım Adım Çözüm

1

Identify the numerator and denominator functions

Let

u(x) = 2x - 3

and

v(x) = (x^2 + 1)^2

.

The given function is structured as a quotient

y = \frac{u}{v}

, requiring the quotient rule.

2

Find the derivatives

u'(x)

and

v'(x)

u'(x) = 2

and

v'(x) = 2(x^2 + 1) \cdot 2x = 4x(x^2 + 1)

.

Differentiating

u(x)

follows standard polynomial rules;

v(x)

requires the chain rule.

3

Evaluate

u(1), u'(1), v(1),

and

v'(1)

at

x = 1

u(1) = -1

,

u'(1) = 2

,

v(1) = 4

, and

v'(1) = 8

.

Evaluating components before substitution simplifies the arithmetic.

4

Apply the quotient rule formula

\frac{dy}{dx} = \frac{u'v - uv'}{v^2}

at

x = 1

\frac{dy}{dx} = \frac{(2)(4) - (-1)(8)}{4^2} = \frac{8 + 8}{16} = 1.

Substitute the calculated component values into the quotient rule formula.

Anahtar Kavram

Quotient Rule and Chain Rule of Differentiation

Tahmini Süre:1m 30s

Soru 250Soru →

A uniform conductor of length $50\,\text{m}$ and cross-sectional area $2.0 \times 10^{-6}\,\text{m}^2$ is made of a material with a resistivity of $4.0 \times 10^{-7}\,\Omega\cdot\text{m}$ . What is the electrical resistance of the conductor in ohms?

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

The resistance of the conductor is

10\,\Omega

.

The resistance of a uniform conductor is given by

R = \frac{\rho L}{A}

. Substituting the values

L = 50\,\text{m}

,

A = 2.0 \times 10^{-6}\,\text{m}^2

, and

\rho = 4.0 \times 10^{-7}\,\Omega\cdot\text{m}

yields

R = \frac{(4.0 \times 10^{-7})(50)}{2.0 \times 10^{-6}} = 10\,\Omega

.

Adım Adım Çözüm

1

Identify given physical quantities

L = 50\,\text{m}

,

A = 2.0 \times 10^{-6}\,\text{m}^2

,

\rho = 4.0 \times 10^{-7}\,\Omega\cdot\text{m}

Extracting given parameter values clearly sets up the mathematical relationship.

2

Apply the resistivity formula for resistance

R = \frac{\rho L}{A}

Resistance varies directly with length and resistivity, and inversely with cross-sectional area.

3

Perform the calculation

R = \frac{(4.0 \times 10^{-7}\,\Omega\cdot\text{m})(50\,\text{m})}{2.0 \times 10^{-6}\,\text{m}^2} = 10\,\Omega

Multiplying the numerator gives

2.0 \times 10^{-5}\,\Omega\cdot\text{m}^2

; dividing by

2.0 \times 10^{-6}\,\text{m}^2

yields

10\,\Omega

.

Anahtar Kavram

Direct computation of electrical resistance using resistivity, length, and cross-sectional area

Tahmini Süre:45s

Soru 251Soru →

A progressive wave traveling along a stretched string is represented by the equation $y = 0.02 \sin(50\pi t - 4\pi x)$ , where $x$ and $y$ are in meters and $t$ is in seconds. What is the speed of the wave in meters per second ( $\text{m/s}$ )?

Cevabı ve açıklamayı göster

Cevap: 12.5

Cevap

The speed of the wave is

12.5\text{ m/s}

.

Comparing the wave equation

y = 0.02 \sin(50\pi t - 4\pi x)

with the standard form

y = A \sin(\omega t - kx)

shows that the angular frequency is

\omega = 50\pi\text{ rad/s}

and the wave number is

k = 4\pi\text{ rad/m}

. Using the relation

v = \frac{\omega}{k}

, the speed of the wave is calculated as

v = \frac{50\pi}{4\pi} = 12.5\text{ m/s}

.

Adım Adım Çözüm

1

Extract angular frequency and wave number from the equation

Comparing

y = 0.02 \sin(50\pi t - 4\pi x)

with

y = A \sin(\omega t - kx)

gives

\omega = 50\pi\text{ rad/s}

and

k = 4\pi\text{ rad/m}

.

Matching corresponding terms yields the wave parameters.

2

Calculate the wave propagation speed

v = \frac{\omega}{k} = \frac{50\pi}{4\pi} = 12.5\text{ m/s}

.

Wave speed equals the ratio of angular frequency to wave number.

Anahtar Kavram

Calculating wave speed from a mathematical wave equation

Soru 252Soru →

A car initially traveling at a constant speed of $15\text{ m/s}$ accelerates uniformly at $2\text{ m/s}^2$ for $5\text{ s}$ . It then maintains the acquired maximum speed for $10\text{ s}$ before coming to rest under uniform retardation in $4\text{ s}$ . What is the total distance covered by the car during the entire motion?

Cevabı ve açıklamayı göster

Cevap: 400

Cevap

The total distance covered by the car during the entire motion is

400\text{ m}

.

The total distance is calculated by summing the distances covered in the three distinct phases of motion: acceleration (

100\text{ m}

), uniform velocity (

250\text{ m}

), and uniform retardation (

50\text{ m}

), yielding a total distance of

400\text{ m}

.

Adım Adım Çözüm

1

Calculate final speed and distance for the acceleration phase.

Final speed

v = 25\text{ m/s}

and distance

s_1 = 100\text{ m}

.

Using kinematic equations

v = u + a t_1 = 15 + (2)(5) = 25\text{ m/s}

and

s_1 = u t_1 + \frac{1}{2}a t_1^2 = 15(5) + \frac{1}{2}(2)(5^2) = 75 + 25 = 100\text{ m}

.

2

Calculate the distance covered during the constant speed phase.

Distance

s_2 = 250\text{ m}

.

The car maintains the acquired speed of

25\text{ m/s}

for

10\text{ s}

, giving

s_2 = v \cdot t_2 = 25 \times 10 = 250\text{ m}

.

3

Calculate the distance covered during the retardation phase.

Distance

s_3 = 50\text{ m}

.

Using average velocity for uniform retardation to rest:

s_3 = \frac{v + 0}{2} t_3 = \frac{25}{2} \times 4 = 50\text{ m}

.

4

Sum the distances from all three stages to determine total distance.

Total distance

S = 400\text{ m}

.

S = s_1 + s_2 + s_3 = 100 + 250 + 50 = 400\text{ m}

.

Anahtar Kavram

Multi-stage linear motion and equations of uniform acceleration

Soru 253Soru →

A radioactive isotope has a half-life of $4\text{ hours}$ . If a sample initially contains $80\text{ g}$ of the isotope, what mass of the isotope, in grams, will remain undecayed after $12\text{ hours}$ ?

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

The mass of the radioactive isotope remaining undecayed after

12\text{ hours}

is

10\text{ g}

.

After

3

half-lives (

12\text{ hours}

total elapsed time with a half-life of

4\text{ hours}

), the fraction of the initial sample remaining is

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

. Multiplying this fraction by the initial mass of

80\text{ g}

gives

10\text{ g}

.

Adım Adım Çözüm

1

Determine the number of elapsed half-lives (

n

)

n = \frac{12\text{ hours}}{4\text{ hours}} = 3

half-lives

Dividing the total time elapsed by the half-life period gives the number of decay cycles.

2

Calculate the mass remaining after

3

half-lives

N = 80 \times \left(\frac{1}{2}\right)^3 = 80 \times \frac{1}{8} = 10\text{ g}

The remaining mass halves during each half-life interval according to the exponential decay rule

N = N_0 (1/2)^n

.

Anahtar Kavram

Radioactive Decay Law and Half-life

Soru 254Soru →

A $5.00\text{ g}$ sample of impure limestone ( $\text{CaCO}_3$ ) is strongly heated until decomposition is complete. If the loss in mass due to the escape of carbon dioxide ( $\text{CO}_2$ ) gas is $1.76\text{ g}$ , what is the percentage purity of the limestone sample? $[\text{Relative atomic masses: } \text{Ca} = 40, \text{C} = 12, \text{O} = 16]$

Cevabı ve açıklamayı göster

Cevap: 80

Cevap

The percentage purity of the limestone sample is 80%.

Thermal decomposition of calcium carbonate yields calcium oxide and carbon dioxide. The mass loss of 1.76 g corresponds to the evolved CO2. From the molar masses (CaCO3 = 100 g/mol, CO2 = 44 g/mol), 44 g of CO2 is released by 100 g of pure CaCO3. Thus, 1.76 g of CO2 is released by 4.00 g of pure CaCO3. Dividing the pure mass (4.00 g) by the original sample mass (5.00 g) and multiplying by 100 yields a percentage purity of 80%.

Adım Adım Çözüm

1

Write the balanced equation for the decomposition reaction.

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

The decrease in mass is entirely due to the evolved carbon dioxide gas.

2

Calculate the relative formula mass of calcium carbonate and carbon dioxide.

\text{Molar mass of } \text{CaCO}_3 = 100\text{ g/mol}, \quad \text{Molar mass of } \text{CO}_2 = 44\text{ g/mol}

Required to relate the mass of evolved gas to the mass of reacting calcium carbonate.

3

Calculate the mass of pure calcium carbonate in the sample.

\text{Mass of pure } \text{CaCO}_3 = \left(\frac{100}{44}\right) \times 1.76\text{ g} = 4.00\text{ g}

Direct stoichiometric ratio derived from 1 mol CaCO3 producing 1 mol CO2.

4

Calculate percentage purity.

\text{Percentage purity} = \left(\frac{4.00\text{ g}}{5.00\text{ g}}\right) \times 100 = 80\%

Ratio of pure reactant mass to total sample mass expressed as a percentage.

Anahtar Kavram

Calculating percentage purity using stoichiometry and gravimetric decomposition data.

Soru 255Soru →

In a regular polygon, the measure of each interior angle is $132^\circ$ greater than the measure of each exterior angle. How many sides does this polygon have?

Cevabı ve açıklamayı göster

Cevap: 15

Cevap

The polygon has 15 sides.

Since the interior angle

I

and exterior angle

E

of a regular polygon sum to

180^\circ

(

I + E = 180^\circ

) and their given difference is

I - E = 132^\circ

, subtracting the difference equation from the sum equation gives

2E = 48^\circ

, which simplifies to

E = 24^\circ

. The number of sides is

n = \frac{360^\circ}{E} = \frac{360^\circ}{24^\circ} = 15

.

Adım Adım Çözüm

1

Set up the linear pair equation for interior and exterior angles

I + E = 180^\circ

An interior angle and its adjacent exterior angle at any vertex of a polygon lie on a straight line and sum to

180^\circ

.

2

Set up the given condition equation

I - E = 132^\circ

The question states that each interior angle is

132^\circ

greater than each exterior angle.

3

Solve for the exterior angle

E

E = 24^\circ

Subtracting

I - E = 132^\circ

from

I + E = 180^\circ

yields

2E = 48^\circ

, giving

E = 24^\circ

.

4

Calculate the number of sides

n

n = 15

The sum of all exterior angles of any convex polygon is

360^\circ

, so

n = \frac{360^\circ}{E} = \frac{360^\circ}{24^\circ} = 15

.

Anahtar Kavram

Interior and Exterior Angle Properties of Regular Polygons

Soru 256Soru →

A ray of light passes symmetrically through an equilateral glass prism of refractive index $\sqrt{2}$ . What is the angle of minimum deviation of the ray, in degrees?

Cevabı ve açıklamayı göster

Cevap: 30

Cevap

The angle of minimum deviation of the ray is

30^\circ

.

For an equilateral triangular prism, the apex angle

A

is

60^\circ

. At minimum deviation, light travels symmetrically through the prism and obeys the exact relation

n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}

. Substituting

n = \sqrt{2}

and

A = 60^\circ

gives

\sin\left(\frac{60^\circ + D_m}{2}\right) = \frac{\sqrt{2}}{2} = \sin(45^\circ)

. Equating arguments yields

\frac{60^\circ + D_m}{2} = 45^\circ

, leading directly to

D_m = 30^\circ

.

Adım Adım Çözüm

1

Determine the refracting angle of the prism

A = 60^\circ

An equilateral prism has interior angles of

60^\circ

each, so the apex angle

A = 60^\circ

.

2

Set up the prism formula for minimum deviation

n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}

When light passes symmetrically through a prism, the deviation is at its minimum value

D_m

.

3

Substitute known values into the equation

\sqrt{2} = \frac{\sin\left(\frac{60^\circ + D_m}{2}\right)}{\sin(30^\circ)}

Given

n = \sqrt{2}

and

A = 60^\circ

, with

\sin(30^\circ) = 0.5

.

4

Calculate the sine of the half-angle

\sin\left(\frac{60^\circ + D_m}{2}\right) = \sqrt{2} \times 0.5 = \frac{\sqrt{2}}{2}

Multiplying both sides by

\sin(30^\circ) = 0.5

.

5

Solve for the minimum deviation angle

D_m

D_m = 30^\circ

Since

\arcsin\left(\frac{\sqrt{2}}{2}\right) = 45^\circ

, we set

\frac{60^\circ + D_m}{2} = 45^\circ \implies 60^\circ + D_m = 90^\circ \implies D_m = 30^\circ

.

Anahtar Kavram

Refraction of light through a prism at the angle of minimum deviation

Soru 257Soru →

A series alternating current (AC) circuit contains an inductor of inductance $L = \frac{1}{\pi^2}\ \text{H}$ , a capacitor of capacitance $C = 25\ \mu\text{F}$ , and a resistor of resistance $R = 50\ \Omega$ . What is the resonant frequency of the circuit in hertz ( $\text{Hz}$ )?

Cevabı ve açıklamayı göster

Cevap: 100

Cevap

The resonant frequency of the circuit is

100\ \text{Hz}

.

At resonance, the inductive reactance

X_L = 2\pi f L

equals the capacitive reactance

X_C = \frac{1}{2\pi f C}

. Equating both yields

f_0 = \frac{1}{2\pi\sqrt{LC}}

. Substituting

L = \frac{1}{\pi^2}\ \text{H}

and

C = 25 \times 10^{-6}\ \text{F}

gives

\sqrt{LC} = \frac{5 \times 10^{-3}}{\pi}\ \text{s}

, leading to

f_0 = \frac{1}{2\pi \left(\frac{5 \times 10^{-3}}{\pi}\right)} = 100\ \text{Hz}

.

Adım Adım Çözüm

1

Write down the formula for the resonant frequency of a series RLC circuit.

f_0 = \frac{1}{2\pi\sqrt{LC}}

Resonance occurs when the inductive reactance equals the capacitive reactance (

X_L = X_C

).

2

Substitute the values of inductance

L = \frac{1}{\pi^2}\ \text{H}

and capacitance

C = 25 \times 10^{-6}\ \text{F}

into

\sqrt{LC}

.

\sqrt{LC} = \sqrt{\frac{1}{\pi^2} \times 25 \times 10^{-6}} = \frac{5 \times 10^{-3}}{\pi}\ \text{s}

Simplifying the square root removes the fraction containing

\pi

.

3

Calculate the resonant frequency

f_0

.

f_0 = \frac{1}{2\pi \times \frac{5 \times 10^{-3}}{\pi}} = \frac{1}{10^{-2}} = 100\ \text{Hz}

The factor of

\pi

cancels out in the denominator, resulting in a whole number value.

Anahtar Kavram

Resonant Frequency in AC Series Circuits

Tahmini Süre:1m 30s

Soru 258Soru →

A $45.0\text{ g}$ sample of impure glucose containing $80.0\%$ pure glucose ( $C_6H_{12}O_6$ ) by mass undergoes complete fermentation in the presence of zymase enzyme at suitable conditions. What is the volume of carbon dioxide gas, in $\text{dm}^3$ , released at standard temperature and pressure (STP)?

(Relative atomic masses: $C = 12.0$ , $H = 1.0$ , $O = 16.0$ ; Molar volume of gas at STP $= 22.4\text{ dm}^3\text{mol}^{-1}$ )

Cevabı ve açıklamayı göster

Cevap: 8.96

Cevap

The volume of carbon dioxide gas released at STP is

8.96\text{ dm}^3

.

The complete fermentation of glucose is represented by the equation

C_6H_{12}O_6 \xrightarrow{\text{zymase}} 2C_2H_5OH + 2CO_2

. Taking into account the

80.0\%

purity, the mass of active glucose is

0.800 \times 45.0\text{ g} = 36.0\text{ g}

, which corresponds to

\frac{36.0}{180.0} = 0.200\text{ mol}

. Because

1\text{ mol}

of glucose yields

2\text{ mol}

of

CO_2

,

0.400\text{ mol}

of

CO_2

is produced. At STP,

0.400\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 8.96\text{ dm}^3

.

Adım Adım Çözüm

1

Determine the mass of pure glucose in the impure sample

36.0 g of pure glucose

Only the active pure glucose undergoes fermentation.

2

Calculate the molar mass of glucose (

C_6H_{12}O_6

)

180.0 g/mol

Needed to convert mass of reactant into molar amount.

3

Calculate the number of moles of glucose fermented

0.200 mol of glucose

Moles = Mass / Molar mass.

4

Determine moles of CO2 evolved using reaction stoichiometry

0.400 mol of CO2

Fermentation of 1 mole of hexose sugar produces 2 moles of ethanol and 2 moles of carbon dioxide.

5

Calculate the volume of CO2 gas at STP

8.96 dm^3

Volume = Moles × Molar volume at STP.

Anahtar Kavram

Fermentation Stoichiometry and Molar Yield of Alkanols

Soru 259Soru →

An electric current of $5.00\text{ A}$ is passed through an aqueous solution of a metal chloride using inert electrodes for $3860\text{ seconds}$ . Calculate the volume of chlorine gas, in $\text{dm}^3$ , liberated at standard temperature and pressure (STP). (Take $1\text{ Faraday} = 96,500\text{ C mol}^{-1}$ , Molar volume of gas at $\text{STP} = 22.4\text{ dm}^3\text{ mol}^{-1}$ )

Cevabı ve açıklamayı göster

Cevap: 2.24

Cevap

The volume of chlorine gas liberated at STP is

2.24\text{ dm}^3

.

Passing

5.00\text{ A}

for

3860\text{ s}

transfers

19,300\text{ C}

of charge, corresponding to

0.200\text{ mol}

of electrons. Because oxidation of chloride ions (

2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-

) requires

2\text{ moles}

of electrons per mole of diatomic chlorine gas,

0.100\text{ mol}

of

\text{Cl}_2

gas is produced. At STP, this occupies

0.100\text{ mol} \times 22.4\text{ dm}^3\text{ mol}^{-1} = 2.24\text{ dm}^3

.

Adım Adım Çözüm

1

Calculate the total electric charge passed during electrolysis.

Q = 19,300\text{ C}

Using

Q = I \times t

, where current

I = 5.00\text{ A}

and time

t = 3860\text{ s}

.

2

Determine the amount of substance of electrons transferred in moles.

n(e^-) = 0.200\text{ mol}

Dividing total charge by Faraday's constant (

96,500\text{ C mol}^{-1}

).

3

Apply stoichiometric ratio from the anode half-reaction to find moles of chlorine gas.

n(\text{Cl}_2) = 0.100\text{ mol}

The reaction

2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-

indicates that

2\text{ moles}

of electrons liberate

1\text{ mole}

of

\text{Cl}_2

.

4

Calculate the volume of chlorine gas produced at STP.

V = 2.24\text{ dm}^3

Multiplying the moles of

\text{Cl}_2

by the molar volume of gas at STP (

22.4\text{ dm}^3\text{ mol}^{-1}

).

Anahtar Kavram

Faraday's laws of electrolysis applied to gas volume calculations at STP

Soru 260Soru →

Calculate the volume of oxygen gas, in $\text{cm}^3$ measured at STP, liberated at the anode during the electrolysis of dilute tetraoxosulfate(VI) acid when a steady current of $1.93\text{ A}$ is passed through the electrolyte for $50\text{ minutes}$ .

[Take Faraday's constant $F = 96500\text{ C mol}^{-1}$ , Molar volume of gas at STP $= 22.4\text{ dm}^3\text{ mol}^{-1}$ ]

Cevabı ve açıklamayı göster

Cevap: 336

Cevap

The volume of oxygen gas liberated at STP is

336\text{ cm}^3

.

Passing a current of

1.93\text{ A}

for

3000\text{ s}

delivers

5790\text{ C}

of charge, equivalent to

0.06\text{ moles}

of electrons. Because the anodic discharge of hydroxide ions (

4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4e^-

) requires

4\text{ moles}

of electrons per mole of

\text{O}_2

,

0.015\text{ moles}

of

\text{O}_2

are generated. Multiplying by the molar volume at STP (

22400\text{ cm}^3\text{mol}^{-1}

) gives

336\text{ cm}^3

.

Adım Adım Çözüm

1

Convert the duration of electrolysis from minutes to seconds

t = 50\text{ min} \times 60\text{ s/min} = 3000\text{ s}

Current calculations require time in SI units (seconds).

2

Calculate the total electric charge passed through the electrolyte

Q = I \times t = 1.93\text{ A} \times 3000\text{ s} = 5790\text{ C}

Charge passed is the product of electric current and time.

3

Calculate the quantity of electrons passed in moles

n(e^-) = \frac{5790\text{ C}}{96500\text{ C mol}^{-1}} = 0.06\text{ mol}

One mole of electrons corresponds to 1 Faraday (

96500\text{ C}

).

4

Use the anodic half-reaction equation to determine the molar ratio of electrons to oxygen gas

4\text{OH}^-(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g) + 4e^-

;

n(\text{O}_2) = \frac{0.06\text{ mol}}{4} = 0.015\text{ mol}

The discharge of hydroxide ions requires 4 moles of electrons per mole of oxygen gas evolved.

5

Calculate the volume of liberated oxygen gas at STP in

\text{cm}^3

V = 0.015\text{ mol} \times 22400\text{ cm}^3\text{ mol}^{-1} = 336\text{ cm}^3

One mole of gas occupies

22.4\text{ dm}^3 = 22400\text{ cm}^3

at standard temperature and pressure.

Anahtar Kavram

Quantitative electrochemistry using Faraday's laws and stoichiometric electron-to-gas relationships at electrodes.

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