Tüm alıştırma soruları

Soru 441Soru →

A heavy-duty truck tire contains a fixed mass of air at an initial absolute pressure of $2.00 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . After traveling a long distance, friction causes the temperature of the air inside the tire to increase to $57^\circ\text{C}$ while its volume remains constant. What is the new absolute pressure of the air inside the tire in pascals ( $\text{Pa}$ )?

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Cevap: 220000

Cevap

The new absolute pressure of the air inside the tire is

220,000\text{ Pa}

(or

2.20 \times 10^5\text{ Pa}

).

According to Gay-Lussac's Law, at constant volume, pressure is directly proportional to absolute temperature (

P \propto T

). Converting temperatures to Kelvin gives

T_1 = 300\text{ K}

and

T_2 = 330\text{ K}

. Calculating

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

gives

2.00 \times 10^5\text{ Pa} \times \frac{330}{300} = 220,000\text{ Pa}

.

Adım Adım Çözüm

1

Convert the initial and final temperatures from Celsius to the absolute Kelvin scale.

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

and

T_2 = 57^\circ\text{C} + 273 = 330\text{ K}

.

All gas law equations require absolute temperatures in Kelvin.

2

Apply the Pressure Law (Gay-Lussac's Law) for a fixed volume of gas.

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

When volume is constant, gas pressure is directly proportional to absolute temperature.

3

Substitute the known values into the equation to calculate the final pressure

P_2

.

P_2 = 2.00 \times 10^5\text{ Pa} \times \frac{330\text{ K}}{300\text{ K}} = 2.20 \times 10^5\text{ Pa} = 220,000\text{ Pa}

.

Multiplying the initial pressure by the ratio of absolute temperatures gives the new pressure.

Anahtar Kavram

Pressure Law (Gay-Lussac's Law)

Soru 442Soru →

An electron in an atom transitions from an excited state with energy $-1.7 \times 10^{-19}\text{ J}$ to a lower state with energy $-5.0 \times 10^{-19}\text{ J}$ . What is the wavelength of the emitted photon in nanometers ( $\text{nm}$ )? (Take Planck's constant $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$ and the speed of light $c = 3.0 \times 10^8\text{ m/s}$ )

Cevabı ve açıklamayı göster

Cevap: 600

Cevap

The wavelength of the emitted photon is 600 nm.

The energy released during the atomic transition is

\Delta E = (-1.7 \times 10^{-19}\text{ J}) - (-5.0 \times 10^{-19}\text{ J}) = 3.3 \times 10^{-19}\text{ J}

. Substituting this into

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

yields

\lambda = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{3.3 \times 10^{-19}} = 6.0 \times 10^{-7}\text{ m}

. Converting to nanometers (

1\text{ m} = 10^9\text{ nm}

) gives

600\text{ nm}

.

Adım Adım Çözüm

1

Calculate energy of the emitted photon

\Delta E = 3.3 \times 10^{-19}\text{ J}

The energy of the photon equals the difference between the initial higher energy level and the final lower energy level.

2

Calculate wavelength in meters using Planck's relation

\lambda = 6.0 \times 10^{-7}\text{ m}

Rearranging

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

gives

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

.

3

Convert wavelength to nanometers

600\text{ nm}

Multiply by

10^9\text{ nm/m}

to obtain the final value in nanometers.

Anahtar Kavram

Energy level transitions and photon emission wavelength

Soru 443Soru →

Find the positive constant $p$ such that the value of the definite integral $\int_{0}^{p} (3x^2 + 2x) \, dx$ is equal to $12$ .

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Cevap: 2

Cevap

The positive constant

p

is equal to

2

.

Evaluating the definite integral

\int_{0}^{p} (3x^2 + 2x) \, dx

gives

p^3 + p^2

. Setting this expression equal to

12

forms the cubic equation

p^3 + p^2 - 12 = 0

. Solving for the positive real solution yields

p = 2

, since

2^3 + 2^2 = 8 + 4 = 12

.

Adım Adım Çözüm

1

Find the antiderivative of the integrand

\int (3x^2 + 2x) \, dx = x^3 + x^2 + C

Applying the power rule for integration

\int x^n dx = \frac{x^{n+1}}{n+1}

term by term gives

\frac{3x^3}{3} + \frac{2x^2}{2} = x^3 + x^2

.

2

Evaluate the antiderivative at the limits

0

and

p

\left[ x^3 + x^2 \right]_0^p = (p^3 + p^2) - (0^3 + 0^2) = p^3 + p^2

Using the Fundamental Theorem of Calculus: substitute the upper limit

p

and subtract the value at the lower limit

0

.

3

Set up the polynomial equation using the given integral value

p^3 + p^2 = 12 \implies p^3 + p^2 - 12 = 0

The problem states that the definite integral equals 12.

4

Solve for the positive real value of

p

p = 2

Testing

p=2

:

2^3 + 2^2 = 8 + 4 = 12

. Factoring gives

(p-2)(p^2 + 3p + 6) = 0

, where the quadratic factor has a negative discriminant

3^2 - 4(1)(6) = -15 < 0

, making

p=2

the only real solution.

Anahtar Kavram

Definite integration with an unknown limit of integration

Soru 444Soru →

A merchant's business records show capital employed of $\text{₦}2,200,000$ , fixed assets of $\text{₦}1,500,000$ , a long-term loan of $\text{₦}400,000$ , current liabilities of $\text{₦}250,000$ , trade debtors of $\text{₦}300,000$ , and cash at bank of $\text{₦}150,000$ . What is the value of the merchant's closing inventory (stock) in Naira?

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Cevap: 500000

Cevap

The value of the merchant's closing inventory is ₦500,000.

Working Capital is calculated as

\text{Capital Employed} - \text{Fixed Assets} = \text{₦}2,200,000 - \text{₦}1,500,000 = \text{₦}700,000

. Total Current Assets is then found by adding Current Liabilities:

\text{₦}700,000 + \text{₦}250,000 = \text{₦}950,000

. Subtracting the known current assets (trade debtors of

\text{₦}300,000

and cash at bank of

\text{₦}150,000

) leaves closing inventory equal to

\text{₦}500,000

. Note that the long-term loan of

\text{₦}400,000

is a non-current liability and does not affect the working capital calculation.

Adım Adım Çözüm

1

Calculate Working Capital from Capital Employed and Fixed Assets

\text{Working Capital} = \text{₦}700,000

Capital Employed equals Fixed Assets plus Working Capital. Rearranging gives Working Capital = Capital Employed - Fixed Assets.

2

Calculate Total Current Assets

\text{Current Assets} = \text{₦}950,000

Working Capital equals Current Assets minus Current Liabilities. Rearranging gives Current Assets = Working Capital + Current Liabilities.

3

Deduct trade debtors and cash at bank from total Current Assets to isolate closing inventory

\text{Inventory} = \text{₦}500,000

Current Assets consists of inventory, trade debtors, and cash at bank. Subtracting the known items leaves the inventory value.

Anahtar Kavram

Backwards Calculation of Capital Structure and Working Capital Components

Tahmini Süre:2m 30s

Soru 445Soru →

A trading firm recorded current assets of ₦450,000, current liabilities of ₦180,000, fixed assets of ₦850,000, and long-term liabilities of ₦200,000. What is the working capital of the firm in Naira?

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Cevap: 270000

Cevap

The working capital of the firm is ₦270,000.

Working capital is calculated as Current Assets minus Current Liabilities. Subtracting ₦180,000 from ₦450,000 yields ₦270,000.

Adım Adım Çözüm

1

Identify the components required for working capital.

Current Assets = ₦450,000 and Current Liabilities = ₦180,000.

Working capital measures short-term liquidity using short-term assets and short-term obligations.

2

Calculate working capital by subtracting current liabilities from current assets.

₦450,000 - ₦180,000 = ₦270,000.

The formula for working capital is Current Assets minus Current Liabilities.

Anahtar Kavram

Working Capital

Soru 446Soru →

A particle of mass $6.63 \times 10^{-27}\text{ kg}$ has a de Broglie wavelength of $2.0 \times 10^{-13}\text{ m}$ . What is the speed of the particle in $\text{m/s}$ ? (Take Planck's constant $h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$ )

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Cevap: 500000

Cevap

The speed of the particle is

500,000\text{ m/s}

(or

5.0 \times 10^5\text{ m/s}

).

According to de Broglie's hypothesis, the matter wavelength

\lambda

of a particle is related to its momentum

p = m v

by

\lambda = \frac{h}{m v}

. Rearranging for speed yields

v = \frac{h}{m \lambda}

. Substituting

h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}

,

m = 6.63 \times 10^{-27}\text{ kg}

, and

\lambda = 2.0 \times 10^{-13}\text{ m}

gives

v = \frac{6.63 \times 10^{-34}}{6.63 \times 10^{-27} \times 2.0 \times 10^{-13}} = 500,000\text{ m/s}

.

Adım Adım Çözüm

1

Identify the given physical quantities and formula

m = 6.63 \times 10^{-27}\text{ kg}

,

\lambda = 2.0 \times 10^{-13}\text{ m}

,

h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}

. The de Broglie equation is

\lambda = \frac{h}{m v}

.

The de Broglie wavelength formula relates matter wave properties to particle momentum.

2

Isolate the target variable (speed

v

)

v = \frac{h}{m \lambda}

Algebraic manipulation is needed to solve directly for speed.

3

Substitute parameters and compute the result

v = \frac{6.63 \times 10^{-34}}{(6.63 \times 10^{-27})(2.0 \times 10^{-13})} = \frac{6.63 \times 10^{-34}}{1.326 \times 10^{-39}} = 500,000\text{ m/s}

Performing scientific notation division yields the numerical particle speed.

Anahtar Kavram

de Broglie Wavelength and Particle Speed

Soru 447Soru →

Find the value of the definite integral $\int_{1}^{2} \left(x + \frac{1}{x^2}\right) dx$ .

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Cevap: 2

Cevap

The value of the definite integral is 2.

Integrating

x + x^{-2}

gives

\frac{x^2}{2} - \frac{1}{x}

. Evaluating from

x=1

to

x=2

yields

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

.

Adım Adım Çözüm

1

Find the antiderivative of

f(x) = x + x^{-2}

F(x) = \frac{x^2}{2} - \frac{1}{x} + C

Apply the power rule of integration

\int x^n dx = \frac{x^{n+1}}{n+1}

for

n \neq -1

.

2

Evaluate

F(x)

at the upper boundary

x = 2

F(2) = \frac{2^2}{2} - \frac{1}{2} = 2 - \frac{1}{2} = \frac{3}{2}

Substitute

x = 2

into the antiderivative.

3

Evaluate

F(x)

at the lower boundary

x = 1

F(1) = \frac{1^2}{2} - \frac{1}{1} = \frac{1}{2} - 1 = -\frac{1}{2}

Substitute

x = 1

into the antiderivative.

4

Calculate

F(2) - F(1)

\frac{3}{2} - \left(-\frac{1}{2}\right) = \frac{3}{2} + \frac{1}{2} = 2

Apply the Fundamental Theorem of Calculus:

\int_{a}^{b} f(x) dx = F(b) - F(a)

.

Anahtar Kavram

Definite Integration using the Power Rule

Soru 448Soru →

A business owner acquires a piece of industrial equipment under a hire purchase agreement. The cash price of the equipment is $₦2,500,000$ . The contract requires an initial deposit of $25\%$ of the cash price, with the remaining balance subject to simple interest of $12\%$ per annum calculated over a term of $2$ years. If the principal balance and total interest are repaid in equal monthly installments over the $2$ -year period, what is the amount of each monthly installment in Naira?

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Cevap: 96875

Cevap

The amount of each monthly installment is ₦96,875.

To determine the monthly installment in a hire purchase transaction, deducting the

25\%

deposit (

₦625,000

) from the cash price (

₦2,500,000

) yields an unpaid balance of

₦1,875,000

. Applying

12\%

per annum interest for

2

years yields total interest of

₦450,000

. Combining the balance and total interest gives

₦2,325,000

. Dividing this total by

24

months yields

₦96,875

per month.

Adım Adım Çözüm

1

Calculate the initial cash deposit required

0.25 \times ₦2,500,000 = ₦625,000

The buyer must pay 25% of the total cash price upfront.

2

Find the remaining unpaid principal balance

₦2,500,000 - ₦625,000 = ₦1,875,000

Interest is charged only on the balance remaining after paying the deposit.

3

Calculate the total simple interest accrued over 2 years

₦1,875,000 \times 0.12 \times 2 = ₦450,000

Simple interest formula is

I = P \times R \times T

where

P = ₦1,875,000

,

R = 12\%

, and

T = 2

years.

4

Calculate the total sum to be spread across monthly installments

₦1,875,000 + ₦450,000 = ₦2,325,000

The total installment debt consists of the outstanding principal balance plus the accrued interest.

5

Determine the value of each monthly installment

\frac{₦2,325,000}{24} = ₦96,875

A 2-year period comprises 24 equal monthly payments.

Anahtar Kavram

Calculation of Hire Purchase Interest and Monthly Installments

Soru 449Soru →

A sledge of mass $8.0\text{ kg}$ sliding on ice with an initial velocity of $15\text{ m s}^{-1}$ enters a rough patch that exerts a constant retarding force of $24\text{ N}$ . Calculate the time, in seconds, required for the sledge to come to a complete stop.

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The time required for the sledge to come to a complete stop is

5.0\text{ s}

.

By Newton's second law in terms of momentum, the rate of change of momentum is equal to the applied net force (

F = \frac{\Delta p}{\Delta t}

). Rearranging gives

\Delta t = \frac{m(v - u)}{F}

. Substituting

m = 8.0\text{ kg}

,

u = 15\text{ m s}^{-1}

,

v = 0\text{ m s}^{-1}

, and retarding force

F = -24\text{ N}

yields

\Delta t = \frac{8.0 \times (0 - 15)}{-24} = 5.0\text{ s}

.

Adım Adım Çözüm

1

Determine the change in linear momentum of the sledge.

The change in linear momentum is

\Delta p = m(v - u) = 8.0\text{ kg} \times (0\text{ m s}^{-1} - 15\text{ m s}^{-1}) = -120\text{ kg m s}^{-1}

.

Linear momentum is defined as the product of mass and velocity.

2

Apply the impulse-momentum theorem to determine the time duration.

\Delta t = \frac{\Delta p}{F} = \frac{-120\text{ kg m s}^{-1}}{-24\text{ N}} = 5.0\text{ s}

.

Impulse delivered by a net force over a time interval equals the change in linear momentum.

Anahtar Kavram

Newton's Second Law and Impulse-Momentum Theorem

Soru 450Soru →

An electric heater rated at $100\text{ W}$ is used to heat a metal block of mass $2.5\text{ kg}$ for $5\text{ minutes}$ . If the temperature of the block increases from $25^\circ\text{C}$ to $45^\circ\text{C}$ and no heat energy is lost to the surroundings, what is the specific heat capacity of the metal?

Cevabı ve açıklamayı göster

Cevap: 600

Cevap

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

The energy transferred by the

100\text{ W}

heater over

300\text{ seconds}

is

Q = 30,000\text{ J}

. Dividing this by the product of mass (

2.5\text{ kg}

) and temperature rise (

20\text{ K}

) yields the specific heat capacity

c = 600\text{ J kg}^{-1}\text{ K}^{-1}

.

Adım Adım Çözüm

1

Calculate total electrical heat energy supplied to the metal block

Q = P \times t = 100\text{ W} \times (5 \times 60\text{ s}) = 30,000\text{ J}

Heat energy supplied by an electric heater is given by power multiplied by heating time in seconds.

2

Calculate the change in temperature

\Delta T = 45^\circ\text{C} - 25^\circ\text{C} = 20\text{ K}

Temperature difference is calculated by subtracting initial temperature from final temperature.

3

Solve for the specific heat capacity

c = \frac{Q}{m \Delta T} = \frac{30,000\text{ J}}{2.5\text{ kg} \times 20\text{ K}} = 600\text{ J kg}^{-1}\text{ K}^{-1}

Specific heat capacity is the amount of heat energy required to raise the temperature of unit mass by one kelvin.

Anahtar Kavram

Specific Heat Capacity and Electrical Energy

Soru 451Soru →

A point charge of $+5.0 \times 10^{-8}\text{ C}$ is situated in a vacuum. Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}$ , what is the magnitude of the electric field intensity, in $\text{N C}^{-1}$ , at a distance of $0.3\text{ m}$ from the charge?

Cevabı ve açıklamayı göster

Cevap: 5000

Cevap

The magnitude of the electric field intensity at a distance of

0.3\text{ m}

is

5000\text{ N C}^{-1}

.

The electric field intensity

E

at a distance

r

from a point charge

q

in free space is given by

E = \frac{k |q|}{r^2}

. Substituting

k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}

,

q = 5.0 \times 10^{-8}\text{ C}

, and

r = 0.3\text{ m}

into the expression yields

E = \frac{9.0 \times 10^9 \times 5.0 \times 10^{-8}}{(0.3)^2} = \frac{450}{0.09} = 5000\text{ N C}^{-1}

.

Adım Adım Çözüm

1

Identify the given values and formula

q = 5.0 \times 10^{-8}\text{ C}

,

r = 0.3\text{ m}

,

k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}

, and

E = \frac{k |q|}{r^2}

Electric field intensity surrounding a point charge depends on the charge magnitude and inversely on the square of the distance.

2

Calculate the square of the distance

r^2 = (0.3)^2 = 0.09\text{ m}^2

The inverse-square law requires using

r^2

in the denominator.

3

Substitute values and solve for field intensity

E = \frac{9.0 \times 10^9 \times 5.0 \times 10^{-8}}{0.09} = \frac{450}{0.09} = 5000\text{ N C}^{-1}

Multiplying the terms in the numerator yields

450\text{ N}\cdot\text{m}^2\text{C}^{-1}

, and dividing by

0.09\text{ m}^2

gives

5000\text{ N C}^{-1}

.

Anahtar Kavram

Electric field intensity due to a isolated point charge

Soru 452Soru →

Calculate the percentage by mass of calcium in a pure sample of calcium carbonate ( $\text{CaCO}_3$ ). $[\text{Ca} = 40, \text{C} = 12, \text{O} = 16]$

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Cevap: 40

Cevap

The percentage by mass of calcium in calcium carbonate is 40%.

The molar mass of calcium carbonate (

\text{CaCO}_3

) is

100\text{ g/mol}

. Since one formula unit contains

40\text{ g}

of calcium, the percentage composition by mass of calcium is

\frac{40}{100} \times 100\% = 40\%

.

Adım Adım Çözüm

1

Calculate the molar mass of calcium carbonate (

\text{CaCO}_3

).

Molar mass =

40 + 12 + (3 \times 16) = 100\text{ g/mol}

.

Sum the relative atomic masses of all constituent atoms in one formula unit of

\text{CaCO}_3

.

2

Determine the mass contribution of calcium in one mole of

\text{CaCO}_3

.

Mass of Ca =

40\text{ g}

.

Each formula unit of

\text{CaCO}_3

contains 1 atom of calcium.

3

Calculate the percentage composition of calcium.

Percentage of Ca =

\frac{40}{100} \times 100\% = 40\%

.

Divide the mass of calcium by the total molar mass of the compound and multiply by 100.

Anahtar Kavram

Percentage composition of an element in a compound

Soru 453Soru →

An electric motor with an efficiency of $80\%$ is used to pull a $100\text{ kg}$ object up a smooth incline inclined at $30^\circ$ to the horizontal at a constant speed of $2\text{ m s}^{-1}$ . Taking acceleration due to gravity $g = 10\text{ m s}^{-2}$ , what is the electrical power input to the motor, in watts?

Cevabı ve açıklamayı göster

Cevap: 1250

Cevap

1250 W

The force needed to move the mass up the smooth incline at constant speed is the parallel component of weight,

F = mg \sin(30^\circ) = 500\text{ N}

. The useful power output is

P_{\text{out}} = Fv = 500 \times 2 = 1000\text{ W}

. Dividing by the efficiency of

0.80

yields the electrical power input

P_{\text{in}} = 1250\text{ W}

.

Adım Adım Çözüm

1

Determine the force required along the inclined plane.

F = mg \sin(30^\circ) = 100\text{ kg} \times 10\text{ m s}^{-2} \times 0.5 = 500\text{ N}

At constant velocity, the applied force balances the component of weight parallel to the incline.

2

Calculate the useful power output delivered by the motor.

P_{\text{out}} = F \times v = 500\text{ N} \times 2\text{ m s}^{-1} = 1000\text{ W}

Mechanical power output is the product of pulling force and constant speed.

3

Calculate the total electrical power input required.

P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}} = \frac{1000\text{ W}}{0.80} = 1250\text{ W}

Efficiency is defined as the ratio of useful power output to total power input.

Anahtar Kavram

Mechanical power on inclined planes and system efficiency

Soru 454Soru →

A point charge of $+6.0 \times 10^{-6}\text{ C}$ experiences an attractive electrostatic force of $0.54\text{ N}$ when placed at a distance of $1.0\text{ m}$ from a second point charge in a vacuum. Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}$ , what is the magnitude of the second charge in microcoulombs ( $\mu\text{C}$ )?

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

The magnitude of the second charge is 10 µC.

Using Coulomb's law

F = \frac{k |q_1 q_2|}{r^2}

, substituting

F = 0.54\text{ N}

,

q_1 = 6.0 \times 10^{-6}\text{ C}

,

r = 1.0\text{ m}

, and

k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}

yields

|q_2| = 1.0 \times 10^{-5}\text{ C}

, which equals

10\text{ }\mu\text{C}

.

Adım Adım Çözüm

1

State Coulomb's Law formula

F = \frac{k |q_1 q_2|}{r^2}

Coulomb's Law describes the electrostatic force between two point charges.

2

Substitute given parameters into the equation

0.54 = \frac{9.0 \times 10^9 \times 6.0 \times 10^{-6} \times |q_2|}{1.0^2}

Knowns:

F = 0.54\text{ N}

,

q_1 = 6.0 \times 10^{-6}\text{ C}

,

r = 1.0\text{ m}

,

k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}

.

3

Solve for the unknown charge magnitude

q_2

|q_2| = 1.0 \times 10^{-5}\text{ C}

Rearranging yields

|q_2| = \frac{0.54}{5.4 \times 10^4} = 1.0 \times 10^{-5}\text{ C}

.

4

Convert the value from Coulombs to microcoulombs

10\text{ }\mu\text{C}

1\text{ }\mu\text{C} = 10^{-6}\text{ C}

, so

1.0 \times 10^{-5}\text{ C} = 10\text{ }\mu\text{C}

.

Anahtar Kavram

Coulomb's Law

Soru 455Soru →

A radioactive detector records an initial disintegration rate of $6400\text{ counts per minute}$ from a freshly prepared isotope. If the half-life of the isotope is $5\text{ hours}$ , determine the total time, in hours, required for the count rate to decrease to $400\text{ counts per minute}$ .

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

The total time required for the disintegration rate to decrease to

400\text{ counts per minute}

is

20\text{ hours}

.

The fraction of activity remaining is

\frac{400}{6400} = \frac{1}{16}

. Expressing this as a power of one-half,

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

, shows that

4

half-lives have elapsed. Multiplying

4

half-lives by

5\text{ hours}

per half-life yields a total duration of

20\text{ hours}

.

Adım Adım Çözüm

1

Calculate the ratio of remaining activity to initial activity

\frac{N}{N_0} = \frac{400}{6400} = \frac{1}{16}

To find the fraction of the original radioactive substance that remains undecayed.

2

Determine the number of elapsed half-lives

n = 4

Since \left(\frac{1}{2}\right)^n = \frac{1}{16} = \left(\frac{1}{2}\right)^4, four complete half-lives have passed.

3

Compute the total elapsed time

t = 4 \times 5 = 20\text{ hours}

Total time equals the number of half-lives multiplied by the duration of one half-life.

Anahtar Kavram

Radioactive Decay Law and Half-life

Tahmini Süre:1m 30s

Soru 456Soru →

A $100\text{ cm}^3$ sample of dry air is passed slowly over excess heated copper turnings in a combustion tube until no further contraction in volume occurs. Assuming oxygen accounts for $21\%$ of air by volume, what is the volume of the remaining unreacted gas in $\text{cm}^3$ at room temperature and pressure?

Cevabı ve açıklamayı göster

Cevap: 79

Cevap

The volume of the remaining unreacted gas is

79\text{ cm}^3

.

Dry atmospheric air is made up of approximately

21\%

oxygen gas and

79\%

non-reactive gases (primarily nitrogen along with argon and carbon dioxide). When passed over heated copper, only oxygen reacts to form solid copper(II) oxide. The volume of oxygen removed is

21\text{ cm}^3

, leaving

100\text{ cm}^3 - 21\text{ cm}^3 = 79\text{ cm}^3

of unreacted gaseous mixture.

Adım Adım Çözüm

1

Calculate the volume of oxygen absorbed by the heated copper

Volume of

\text{O}_2 = \frac{21}{100} \times 100\text{ cm}^3 = 21\text{ cm}^3

Air contains approximately

21\%

oxygen by volume, which reacts quantitatively with heated copper turnings.

2

Determine the remaining unreacted gas volume

Volume remaining =

100\text{ cm}^3 - 21\text{ cm}^3 = 79\text{ cm}^3

The unreacted component consists mainly of nitrogen (approx.

78\%

) and noble gases (approx.

1\%

) which do not react with heated copper.

Anahtar Kavram

Composition of dry air and quantitative removal of oxygen gas

Tahmini Süre:45s

Soru 457Soru →

A $6.60\text{ g}$ sample of impure ammonium tetraoxosulfate(VI), $(\text{NH}_4)_2\text{SO}_4$ , is heated with excess sodium hydroxide solution. The evolved ammonia gas, $\text{NH}_3$ , is absorbed completely in $100.0\text{ cm}^3$ of $0.50\text{ mol dm}^{-3}$ tetraoxosulfate(VI) acid solution, $\text{H}_2\text{SO}_4$ . The unreacted acid requires $40.0\text{ cm}^3$ of $0.50\text{ mol dm}^{-3}$ sodium hydroxide solution for complete neutralization. What is the percentage purity of the ammonium tetraoxosulfate(VI) sample? $[\text{N} = 14, \text{H} = 1, \text{S} = 32, \text{O} = 16]$

Cevabı ve açıklamayı göster

Cevap: 80

Cevap

80%

The correct answer is 80.0%. Through back-titration analysis, 0.020 mol of NaOH neutralizes 0.010 mol of unreacted excess H₂SO₄ out of the initial 0.050 mol, leaving 0.040 mol of H₂SO₄ to react with 0.080 mol of evolved NH₃ gas. Since 1 mole of pure ammonium tetraoxosulfate(VI) produces 2 moles of NH₃ gas, the sample contained 0.040 mol of pure (NH₄)₂SO₄. Multiplying by its molar mass (132 g/mol) yields 5.28 g of pure compound. Dividing 5.28 g by the total sample mass of 6.60 g and multiplying by 100 gives exactly 80.0%.

Adım Adım Çözüm

1

Calculate the initial moles of H₂SO₄ acid solution used for absorbing ammonia.

0.050 mol H₂SO₄

Total acid available = Volume (in dm³) × Concentration (in mol dm⁻³).

2

Calculate the unreacted moles of H₂SO₄ from the titration with NaOH.

0.010 mol excess H₂SO₄

1 mole of H₂SO₄ reacts with 2 moles of NaOH, so excess H₂SO₄ = 0.5 × moles of NaOH used.

3

Calculate moles of H₂SO₄ neutralized by evolved NH₃ gas.

0.040 mol H₂SO₄ reacted

Reacted acid = Initial total acid - Excess unreacted acid.

4

Calculate the moles of NH₃ evolved from the sample.

0.080 mol NH₃

2 moles of NH₃ react with 1 mole of H₂SO₄.

5

Determine the mass of pure (NH₄)₂SO₄ present in the original sample.

5.28 g of pure (NH₄)₂SO₄

1 mole of (NH₄)₂SO₄ yields 2 moles of NH₃. Mass = moles (0.040 mol) × molar mass (132 g/mol).

6

Compute the percentage purity of the sample.

80%

Percentage Purity = (Mass of pure substance / Total mass of impure sample) × 100%.

Anahtar Kavram

Back-titration quantitative analysis for determining percentage purity

Tahmini Süre:3m 0s

Soru 458Soru →

A sample of pure methane ( $\text{CH}_4$ ) contains $3.00\text{ g}$ of carbon. When this sample undergoes complete combustion in excess oxygen gas, all the hydrogen present is converted into water vapor ( $\text{H}_2\text{O}$ ). Based on the Law of Definite Proportions, what is the total mass (in grams) of water vapor produced? [Atomic masses: $\text{H} = 1$ , $\text{C} = 12$ , $\text{O} = 16$ ]

Cevabı ve açıklamayı göster

Cevap: 9

Cevap

The total mass of water vapor produced is

9.00\text{ g}

.

According to the Law of Definite Proportions, a chemical compound always contains its component elements in a fixed ratio by mass. In methane (

\text{CH}_4

), the ratio of mass of carbon to hydrogen is

12 : 4

(

3 : 1

). Therefore,

3.00\text{ g}

of carbon is combined with

1.00\text{ g}

of hydrogen. When methane undergoes complete combustion, all

1.00\text{ g}

of hydrogen is converted into water (

\text{H}_2\text{O}

). Since hydrogen makes up

\frac{2}{18}

of the mass of water,

1.00\text{ g}

of hydrogen yields

1.00 \times \frac{18}{2} = 9.00\text{ g}

of water.

Adım Adım Çözüm

1

Calculate the mass of hydrogen present in the methane sample using the Law of Definite Proportions.

The mass of hydrogen in the sample is

1.00\text{ g}

.

In

\text{CH}_4

, the mass ratio of carbon to hydrogen is

12 : 4 = 3 : 1

. Given

3.00\text{ g}

of carbon, the mass of hydrogen is

\frac{3.00\text{ g}}{3} = 1.00\text{ g}

.

2

Determine the mass fraction of hydrogen in water (

\text{H}_2\text{O}

).

Hydrogen accounts for

\frac{2}{18}

of the total mass of water.

The molar mass of

\text{H}_2\text{O}

is

2(1) + 16 = 18\text{ g/mol}

, of which

2\text{ g}

is hydrogen.

3

Calculate the total mass of water vapor formed from the hydrogen.

The total mass of water produced is

9.00\text{ g}

.

All

1.00\text{ g}

of hydrogen from methane is converted into water. Mass of

\text{H}_2\text{O} = 1.00\text{ g} \times \frac{18}{2} = 9.00\text{ g}

.

Anahtar Kavram

Law of Definite Proportions (Constant Composition)

Soru 459Soru →

A sample of hydrated copper(II) sulfate ( $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$ ) has a mass of $24.95\text{ g}$ . What is the total number of moles of oxygen atoms contained in this sample? $[\text{Relative atomic masses: } \text{Cu} = 63.5, \text{S} = 32.0, \text{O} = 16.0, \text{H} = 1.0]$

Cevabı ve açıklamayı göster

Cevap: 0.9

Cevap

The total number of moles of oxygen atoms contained in the sample is

0.90\text{ mol}

.

The molar mass of hydrated copper(II) sulfate (

\text{CuSO}_4 \cdot 5\text{H}_2\text{O}

) is

249.5\text{ g/mol}

. Dividing

24.95\text{ g}

by

249.5\text{ g/mol}

gives

0.10\text{ mol}

of the compound. Since each mole of

\text{CuSO}_4 \cdot 5\text{H}_2\text{O}

contains

9\text{ moles}

of oxygen atoms (

4

from

\text{CuSO}_4

and

5

from

5\text{H}_2\text{O}

), the total quantity of oxygen atoms is

0.10 \times 9 = 0.90\text{ mol}

.

Adım Adım Çözüm

1

Calculate the molar mass of

\text{CuSO}_4 \cdot 5\text{H}_2\text{O}

Molar mass =

249.5\text{ g/mol}

Sum the relative atomic masses of all atoms present in one formula unit of the hydrated compound.

2

Calculate the moles of the hydrated salt

Moles of compound =

0.10\text{ mol}

Divide the mass of the sample (

24.95\text{ g}

) by its molar mass (

249.5\text{ g/mol}

).

3

Determine the stoichiometric multiplier for oxygen atoms

9 moles of oxygen atoms per mole of compound

Each formula unit contains 4 oxygen atoms in the sulfate group and 5 oxygen atoms in the water of crystallization.

4

Calculate total moles of oxygen atoms

Moles of oxygen atoms =

0.90\text{ mol}

Multiply the moles of compound (

0.10\text{ mol}

) by the 9 moles of oxygen atoms per mole of compound.

Anahtar Kavram

Stoichiometric relationship of constituent atoms in a hydrated compound

Soru 460Soru →

A constant horizontal force acts on a body of mass $10\text{ kg}$ , accelerating it from rest to a speed of $12\text{ m s}^{-1}$ in a time of $4\text{ s}$ along a smooth horizontal surface. What is the average power delivered by the force during this time interval?

Cevabı ve açıklamayı göster

Cevap: 180

Cevap

The average power delivered by the force during the 4-second interval is

180\text{ W}

.

By the work-energy theorem, the total work done by the constant force equals the gain in kinetic energy:

W = \frac{1}{2} m v^2 = \frac{1}{2} \times 10 \times 144 = 720\text{ J}

. The average power is the rate at which work is performed over time:

P = \frac{W}{t} = \frac{720\text{ J}}{4\text{ s}} = 180\text{ W}

.

Adım Adım Çözüm

1

Calculate the final kinetic energy acquired by the body.

E_k = \frac{1}{2} m v^2 = \frac{1}{2} (10\text{ kg})(12\text{ m s}^{-1})^2 = 720\text{ J}

.

Since the body starts from rest on a smooth surface, all work done by the net force goes into increasing its kinetic energy.

2

Divide the total work done by the elapsed time to find the average power.

P_{\text{avg}} = \frac{W}{t} = \frac{720\text{ J}}{4\text{ s}} = 180\text{ W}

.

Average power is defined as the rate of doing work over a given time interval.

Anahtar Kavram

Work-Energy Theorem and Average Power

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