Tüm alıştırma soruları

Soru 981Soru →

Calculate the relative molecular mass of hydrated magnesium tetraoxosulfate(VI), $MgSO_4 \cdot 7H_2O$ . $[Mg = 24, S = 32, O = 16, H = 1]$

Cevabı ve açıklamayı göster

Cevap: 246

Cevap

246

The relative molecular mass of

MgSO_4 \cdot 7H_2O

is calculated by summing the atomic masses of all constituent atoms. The anhydrous part

MgSO_4

contributes

24 + 32 + (4 \times 16) = 120

. The water of crystallization

7H_2O

contributes

7 \times (2 + 16) = 126

. The total relative molecular mass is

120 + 126 = 246

.

Adım Adım Çözüm

1

Calculate the mass of the anhydrous salt component,

MgSO_4

24 + 32 + (4 \times 16) = 120

Sum the relative atomic masses of one magnesium atom, one sulfur atom, and four oxygen atoms.

2

Calculate the mass of the water of crystallization component,

7H_2O

7 \times (2 + 16) = 126

Multiply the molecular mass of one water molecule (18) by the coefficient 7.

3

Combine the mass of

MgSO_4

and

7H_2O

120 + 126 = 246

The total relative molecular mass of a hydrated salt is the sum of the anhydrous salt mass and the water of crystallization mass.

Anahtar Kavram

Relative Molecular Mass of Hydrated Salts

Soru 982Soru →

For a particular gasification process, a chemical reaction has a standard enthalpy change ( $\Delta H^\circ$ ) of $+136.5\text{ kJ mol}^{-1}$ and a standard entropy change ( $\Delta S^\circ$ ) of $+325.0\text{ J K}^{-1}\text{ mol}^{-1}$ . What is the minimum temperature, in degrees Celsius ( $^\circ\text{C}$ ), above which the reaction becomes thermodynamically spontaneous?

Cevabı ve açıklamayı göster

Cevap: 147

Cevap

The minimum temperature above which the reaction becomes spontaneous is 147 °C.

At the boundary of spontaneity,

\Delta G^\circ = 0

. Substituting this into

\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ

yields

T = \frac{\Delta H^\circ}{\Delta S^\circ}

. Converting

\Delta H^\circ

to Joules yields

+136,500\text{ J mol}^{-1}

. Dividing by

+325.0\text{ J K}^{-1}\text{ mol}^{-1}

gives

T = 420\text{ K}

. Converting to Celsius gives

420 - 273 = 147^\circ\text{C}

.

Adım Adım Çözüm

1

Convert enthalpy change from kilojoules per mole to joules per mole

\Delta H^\circ = +136.5\text{ kJ mol}^{-1} = +136,500\text{ J mol}^{-1}

\Delta H^\circ

must be expressed in Joules to match the unit of

\Delta S^\circ

(

325.0\text{ J K}^{-1}\text{ mol}^{-1}

).

2

Determine the threshold condition for reaction spontaneity using Gibbs free energy equation

\Delta G^\circ = 0 \implies T = \frac{\Delta H^\circ}{\Delta S^\circ}

A reaction is spontaneous when

\Delta G^\circ < 0

. The minimum temperature where spontaneity begins occurs when

\Delta G^\circ = 0

.

3

Calculate the absolute temperature in Kelvin

T = \frac{136,500\text{ J mol}^{-1}}{325.0\text{ J K}^{-1}\text{ mol}^{-1}} = 420\text{ K}

Dividing the enthalpy change in Joules by the entropy change in Joules per Kelvin yields the temperature in Kelvin.

4

Convert temperature from Kelvin to degrees Celsius

T(^\circ\text{C}) = 420 - 273 = 147^\circ\text{C}

The question explicitly requests the answer in degrees Celsius (

T(^\circ\text{C}) = T(\text{K}) - 273

).

Anahtar Kavram

Gibbs Free Energy and Temperature Dependence of Spontaneity

Tahmini Süre:2m 0s

Soru 983Soru →

An organic compound containing carbon, hydrogen, and nitrogen was analyzed and found to contain $61.02\%$ carbon and $15.25\%$ hydrogen by mass, with the remainder being nitrogen. Given that the relative molecular mass of the compound is $118\text{ g/mol}$ , what is the value of the integer multiplier $n$ that relates its empirical formula to its molecular formula? $[C = 12, H = 1, N = 14]$

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The value of the integer multiplier n is 2.

Subtracting the carbon (

61.02\%

) and hydrogen (

15.25\%

) percentages from

100\%

gives a nitrogen content of

23.73\%

. Converting these mass percentages to mole ratios by dividing by relative atomic masses (

C=12, H=1, N=14

) gives

5.085\text{ mol}

of C,

15.25\text{ mol}

of H, and

1.695\text{ mol}

of N. Dividing by the smallest value (

1.695

) produces the mole ratio

3:9:1

, establishing the empirical formula as

C_3H_9N

. The empirical mass is

(3 \times 12) + (9 \times 1) + 14 = 59\text{ g/mol}

. Dividing the relative molecular mass (

118\text{ g/mol}

) by the empirical mass (

59\text{ g/mol}

) yields

n = 2

.

Adım Adım Çözüm

1

Calculate percentage composition of nitrogen

23.73%

The sum of percentages of all constituent elements in a compound must equal 100%.

2

Calculate relative number of moles for each element

C = 5.085 mol, H = 15.25 mol, N = 1.695 mol

Dividing the mass percentage of each element by its relative atomic mass gives its relative mole quantity.

3

Determine simplest whole number atomic ratio

C : H : N = 3 : 9 : 1

Dividing all mole amounts by the smallest value (1.695) gives the empirical mole ratio.

4

Determine the empirical formula mass

59 g/mol

Summing the atomic masses of elements in C3H9N yields (3 x 12) + (9 x 1) + 14 = 59 g/mol.

5

Calculate the integer multiplier n

2

Dividing relative molecular mass (118 g/mol) by empirical formula mass (59 g/mol) gives n = 2.

Anahtar Kavram

Calculation of Empirical Formula and Molecular Formula Multiplier

Soru 984Soru →

A sample of $0.62\text{ g}$ of sodium oxide ( $\text{Na}_2\text{O}$ ) is reacted completely with distilled water to prepare $200\text{ cm}^3$ of stock solution. If $50\text{ cm}^3$ of this stock solution is diluted with distilled water to a final volume of $500\text{ cm}^3$ at $25^\circ\text{C}$ , what is the pH of the resulting diluted solution? $[\text{Na} = 23, \text{O} = 16, \text{H} = 1]$

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

The pH of the diluted solution is 12.0.

Dissolving

0.62\text{ g}

(

0.01\text{ mol}

) of

\text{Na}_2\text{O}

produces

0.02\text{ mol}

of

\text{OH}^-

ions in

200\text{ cm}^3

, giving a stock concentration of

0.1\text{ mol dm}^{-3}

. Diluting

50\text{ cm}^3

of this stock solution to

500\text{ cm}^3

reduces the

[\text{OH}^-]

tenfold to

0.01\text{ mol dm}^{-3}

. Taking the negative logarithm gives

\text{pOH} = 2.0

, which corresponds to a

\text{pH}

of

14.0 - 2.0 = 12.0

.

Adım Adım Çözüm

1

Calculate the molar mass of sodium oxide (

\text{Na}_2\text{O}

) and determine the amount of moles dissolved.

\text{Molar mass of Na}_2\text{O} = (2 \times 23) + 16 = 62\text{ g mol}^{-1}

. Moles of

\text{Na}_2\text{O} = \frac{0.62\text{ g}}{62\text{ g mol}^{-1}} = 0.01\text{ mol}

.

Converting mass to moles is required to apply chemical stoichiometry.

2

Determine the moles of hydroxide ions (

\text{OH}^-

) formed upon complete reaction with water.

The balanced equation is

\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH} \rightarrow 2\text{Na}^+ + 2\text{OH}^-

. Therefore,

0.01\text{ mol}

of

\text{Na}_2\text{O}

yields

0.02\text{ mol}

of

\text{OH}^-

.

Sodium oxide is a basic oxide that reacts with water in a 1:2 mole ratio to yield hydroxide ions.

3

Calculate the hydroxide ion concentration in the

200\text{ cm}^3

(

0.2\text{ dm}^3

) stock solution.

[\text{OH}^-]_{\text{stock}} = \frac{0.02\text{ mol}}{0.2\text{ dm}^3} = 0.1\text{ mol dm}^{-3}

.

Molarity is defined as moles of solute per cubic decimeter of solution.

4

Apply the dilution formula

C_1 V_1 = C_2 V_2

to find the hydroxide ion concentration after dilution.

[\text{OH}^-]_{\text{diluted}} = \frac{0.1\text{ mol dm}^{-3} \times 50\text{ cm}^3}{500\text{ cm}^3} = 0.01\text{ mol dm}^{-3} = 1.0 \times 10^{-2}\text{ mol dm}^{-3}

.

Diluting

50\text{ cm}^3

to

500\text{ cm}^3

decreases the concentration by a factor of 10.

5

Calculate the pOH and subsequently the pH of the diluted solution.

\text{pOH} = -\log_{10}(1.0 \times 10^{-2}) = 2.0

. Using

\text{pH} + \text{pOH} = 14.0

,

\text{pH} = 14.0 - 2.0 = 12.0

.

The logarithmic scale defines

\text{pOH} = -\log_{10}[\text{OH}^-]

and at

25^\circ\text{C}

,

\text{pH} + \text{pOH} = 14

.

Anahtar Kavram

Stoichiometric reaction of basic oxides with water combined with dilution calculations to determine solution pH and pOH.

Soru 985Soru →

A sample of pure hydrated aluminum nitrate, $\text{Al(NO}_3)_3 \cdot 9\text{H}_2\text{O}$ , has a mass of $75.0\text{ g}$ . What is the mass, in grams, of oxygen contained in this sample? $[\text{Al} = 27, \text{N} = 14, \text{O} = 16, \text{H} = 1]$

Cevabı ve açıklamayı göster

Cevap: 57.6

Cevap

57.6

The molar mass of

\text{Al(NO}_3)_3 \cdot 9\text{H}_2\text{O}

is

375\text{ g/mol}

. A

75.0\text{ g}

sample equals

0.20\text{ mol}

of the hydrated salt. Because each formula unit contains

18

oxygen atoms (

9

from the nitrate groups and

9

from the water molecules),

0.20\text{ mol}

of the salt contains

3.60\text{ mol}

of oxygen atoms. Multiplying

3.60\text{ mol}

by the atomic mass of oxygen (

16\text{ g/mol}

) yields

57.6\text{ g}

.

Adım Adım Çözüm

1

Calculate the molar mass of hydrated aluminum nitrate,

\text{Al(NO}_3)_3 \cdot 9\text{H}_2\text{O}

.

375\text{ g/mol}

Sum the atomic masses of all constituent atoms:

\text{Al} = 27

,

3 \times \text{NO}_3 = 3 \times 62 = 186

,

9 \times \text{H}_2\text{O} = 9 \times 18 = 162

. Total

= 27 + 186 + 162 = 375\text{ g/mol}

.

2

Determine the amount (in moles) of the compound in the

75.0\text{ g}

sample.

0.20\text{ mol}

Moles of compound

= \frac{\text{mass}}{\text{molar mass}} = \frac{75.0\text{ g}}{375\text{ g/mol}} = 0.20\text{ mol}

.

3

Determine the total moles of oxygen atoms per mole of the hydrated compound.

18\text{ mol of O}

Each formula unit contains

9

oxygen atoms in the nitrate groups (

3 \times 3

) and

9

oxygen atoms in the water molecules (

9 \times 1

), giving a total of

18

oxygen atoms.

4

Calculate the total mass of oxygen atoms in the sample.

57.6\text{ g}

Mass of oxygen

= \text{moles of compound} \times 18 \times \text{molar mass of O} = 0.20 \times 18 \times 16 = 3.60 \times 16 = 57.6\text{ g}

.

Anahtar Kavram

Mole Concept and Stoichiometric Mass Composition in Hydrated Compounds

Soru 986Soru →

A $2.016\text{ g}$ sample of a hydrated dicarboxylic acid, $(\text{COOH})_2 \cdot n\text{H}_2\text{O}$ , was dissolved in distilled water and made up to $250.0\text{ cm}^3$ in a volumetric flask. A $25.0\text{ cm}^3$ portion of this solution required exactly $20.0\text{ cm}^3$ of $0.160\text{ mol dm}^{-3}$ sodium hydroxide ( $\text{NaOH}$ ) solution for complete neutralization. What is the value of the integer $n$ in the formula of the hydrated acid? $[\text{H} = 1, \text{C} = 12, \text{O} = 16]$

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The value of the integer n is 2.

Titration of 20.0 cm³ of 0.160 mol dm⁻³ NaOH consumes 0.0032 mol of NaOH. Because the dicarboxylic acid is diprotic ((COOH)₂), it reacts in a 1:2 ratio with NaOH, giving 0.0016 mol of acid in the 25.0 cm³ aliquot. Scaling to the full 250.0 cm³ flask yields 0.0160 mol of hydrated acid. The molar mass of the hydrated acid is 2.016 g / 0.0160 mol = 126 g/mol. Since the anhydrous formula mass of (COOH)₂ is 90 g/mol, the water of crystallization contributes 126 - 90 = 36 g/mol. Dividing 36 by 18 (the molar mass of H₂O) gives n = 2.

Adım Adım Çözüm

1

Calculate the amount in moles of NaOH used in the titration.

Moles of NaOH = 0.0032 mol

Moles = concentration × volume = 0.160 mol dm⁻³ × (20.0 / 1000) dm³ = 0.0032 mol.

2

Determine the moles of acid present in the 25.0 cm³ titration sample using the stoichiometric mole ratio.

Moles of acid in 25.0 cm³ = 0.0016 mol

Ethanoic/oxalic acid is diprotic ((COOH)₂ + 2NaOH → (COONa)₂ + 2H₂O), requiring 2 moles of NaOH per mole of acid. Moles of acid = 0.0032 / 2 = 0.0016 mol.

3

Scale up to find the total moles of acid in the original 250.0 cm³ solution.

Total moles of acid = 0.0160 mol

Total moles = 0.0016 mol × (250.0 cm³ / 25.0 cm³) = 0.0160 mol.

4

Calculate the molar mass of the hydrated dicarboxylic acid.

Molar mass = 126 g mol⁻¹

Molar mass M = sample mass / total moles = 2.016 g / 0.0160 mol = 126 g mol⁻¹.

5

Calculate the value of integer n by comparing the molar mass to the anhydrous acid mass.

n = 2

Formula mass of anhydrous (COOH)₂ = 2(12) + 2(1) + 4(16) = 90 g mol⁻¹. The water component mass is 18n = 126 - 90 = 36 g mol⁻¹, giving n = 36 / 18 = 2.

Anahtar Kavram

Empirical and Molecular Formula Calculations with Water of Crystallization and Volumetric Stoichiometry

Soru 987Soru →

A

3.25\text{ g}

sample of impure zinc granules reacts completely with excess dilute tetraoxosulfate(VI) acid according to the equation:

\text{Zn(s)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{H}_2\text{(g)}

If

896\text{ cm}^3

of hydrogen gas is collected at STP, calculate the percentage purity of the zinc sample.

[\text{Zn} = 65\text{; Molar volume of gas at STP} = 22.4\text{ dm}^3\text{mol}^{-1}]

Cevabı ve açıklamayı göster

Cevap: 80

Cevap

80%

The volume of hydrogen gas produced at STP (

896\text{ cm}^3 = 0.896\text{ dm}^3

) corresponds to

0.04\text{ mol}

of

\text{H}_2

. Based on the

1:1

stoichiometric reaction between zinc and tetraoxosulfate(VI) acid,

0.04\text{ mol}

of pure zinc was present in the sample. The mass of pure zinc is

0.04\text{ mol} \times 65\text{ g/mol} = 2.60\text{ g}

. Dividing this pure mass by the total sample mass of

3.25\text{ g}

and multiplying by

100\%

yields a percentage purity of

80\%

.

Adım Adım Çözüm

1

Convert the collected volume of hydrogen gas from cm³ to dm³

\text{Volume of H}_2 = \frac{896\text{ cm}^3}{1000\text{ cm}^3/\text{dm}^3} = 0.896\text{ dm}^3

Molar gas volume is given in dm³/mol, so gas volume must be expressed in dm³.

2

Calculate the amount in moles of hydrogen gas produced at STP

\text{Moles of H}_2 = \frac{0.896\text{ dm}^3}{22.4\text{ dm}^3\text{mol}^{-1}} = 0.04\text{ mol}

Number of moles of gas at STP equals volume divided by molar volume.

3

Determine the moles and mass of pure zinc that reacted

\text{Moles of pure Zn} = 0.04\text{ mol}

;

\text{Mass of pure Zn} = 0.04\text{ mol} \times 65\text{ g/mol} = 2.60\text{ g}

The reaction stoichiometry shows a 1:1 mole ratio between Zn and H₂.

4

Calculate the percentage purity of the zinc sample

\text{Percentage Purity} = \left( \frac{2.60\text{ g}}{3.25\text{ g}} \right) \times 100\% = 80\%

Percentage purity is the ratio of pure substance mass to total sample mass expressed as a percentage.

Anahtar Kavram

Percentage purity determination via gas volume stoichiometry at STP

Soru 988Soru →

Copper(II) oxide ( $\text{CuO}$ ) is a black solid compound formed when copper metal is strongly heated in air. Given the relative atomic masses of copper ( $\text{Cu} = 64$ ) and oxygen ( $\text{O} = 16$ ), what is the percentage by mass of copper in pure copper(II) oxide?

Cevabı ve açıklamayı göster

Cevap: 80

Cevap

The percentage by mass of copper in pure copper(II) oxide (

\text{CuO}

) is

80\%

.

The molar mass of copper(II) oxide (

\text{CuO}

) is

64 + 16 = 80\text{ g/mol}

. The relative mass contributed by copper is

64\text{ g/mol}

. Dividing

64

by

80

and multiplying by

100

gives

80\%

.

Adım Adım Çözüm

1

Calculate the molar mass of copper(II) oxide (

\text{CuO}

).

Molar mass of

\text{CuO} = 64 + 16 = 80\text{ g/mol}

.

The molar mass of a binary compound is the sum of the relative atomic masses of its constituent elements.

2

Calculate the mass percentage of copper in the compound.

\text{Percentage of Cu} = \left(\frac{64}{80}\right) \times 100\% = 80\%

.

The mass percentage is found by dividing the mass contributed by copper by the total molar mass of the compound and multiplying by 100.

Anahtar Kavram

Percentage composition by mass of an element in a copper compound

Soru 989Soru →

A fixed mass of oxygen gas contained in a rigid metal vessel exerts a pressure of $1.20\text{ atm}$ at a temperature of $27^\circ\text{C}$ . If the volume of the vessel remains constant, at what absolute temperature in Kelvin (K) will the gas exert a pressure of $1.80\text{ atm}$ ?

Cevabı ve açıklamayı göster

Cevap: 450

Cevap

The final absolute temperature of the gas is

450\text{ K}

.

According to Gay-Lussac's Pressure Law, for a fixed mass of gas at constant volume,

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

. Converting the initial temperature

27^\circ\text{C}

to Kelvin gives

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

. Substituting

P_1 = 1.20\text{ atm}

,

P_2 = 1.80\text{ atm}

, and

T_1 = 300\text{ K}

gives

T_2 = \frac{1.80 \times 300}{1.20} = 450\text{ K}

.

Adım Adım Çözüm

1

Convert the given initial temperature from Celsius to Kelvin.

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

Gas laws require absolute temperature in Kelvin for proportional relationship calculations.

2

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

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

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

3

Substitute the given values into the Pressure Law equation.

\frac{1.20\text{ atm}}{300\text{ K}} = \frac{1.80\text{ atm}}{T_2}

Insert

P_1 = 1.20\text{ atm}

,

T_1 = 300\text{ K}

, and

P_2 = 1.80\text{ atm}

.

4

Rearrange and solve for

T_2

.

T_2 = \frac{1.80 \times 300}{1.20} = 450\text{ K}

Cross-multiplying yields the final absolute temperature.

Anahtar Kavram

Pressure Law (Gay-Lussac's Law)

Soru 990Soru →

A gas mixture containing $0.20\text{ mol}$ of oxygen ( $\text{O}_2$ ) and $0.30\text{ mol}$ of nitrogen ( $\text{N}_2$ ) is collected over water at $27^\circ\text{C}$ . The total pressure of the moist gas mixture is $775\text{ mmHg}$ and the saturated vapor pressure of water at $27^\circ\text{C}$ is $25\text{ mmHg}$ . If $0.50\text{ mol}$ of dry argon ( $\text{Ar}$ ) gas is subsequently added to the mixture at the same temperature while maintaining the total system pressure at $775\text{ mmHg}$ , what is the final partial pressure of nitrogen gas in the moist mixture in $\text{mmHg}$ ?

Cevabı ve açıklamayı göster

Cevap: 225

Cevap

The final partial pressure of nitrogen gas in the moist mixture is

225\text{ mmHg}

.

When a gas is collected over water, the total pressure measured includes the saturated vapor pressure of water (

P_{\text{H}_2\text{O}}

). Subtracting

P_{\text{H}_2\text{O}} = 25\text{ mmHg}

from

P_{\text{total}} = 775\text{ mmHg}

yields the total pressure of the dry gases (

P_{\text{dry}} = 750\text{ mmHg}

). Adding

0.50\text{ mol}

of argon increases the total dry gas quantity to

1.00\text{ mol}

(

0.20 + 0.30 + 0.50

). The mole fraction of nitrogen in this dry mixture is

\frac{0.30}{1.00} = 0.30

. Finally, multiplying this mole fraction by

P_{\text{dry}}

gives the partial pressure of nitrogen as

0.30 \times 750\text{ mmHg} = 225\text{ mmHg}

.

Adım Adım Çözüm

1

Calculate the total pressure exerted by the dry gas mixture.

P_{\text{dry}} = 750\text{ mmHg}

According to Dalton's Law of Partial Pressures, the total pressure of a gas collected over water is

P_{\text{total}} = P_{\text{dry}} + P_{\text{H}_2\text{O}}

. Therefore,

P_{\text{dry}} = 775\text{ mmHg} - 25\text{ mmHg} = 750\text{ mmHg}

.

2

Calculate total moles of dry gas after argon addition.

n_{\text{total, dry}} = 1.00\text{ mol}

Sum the moles of oxygen (

0.20\text{ mol}

), nitrogen (

0.30\text{ mol}

), and added argon (

0.50\text{ mol}

):

0.20 + 0.30 + 0.50 = 1.00\text{ mol}

.

3

Determine the mole fraction of nitrogen gas in the dry mixture.

\chi_{\text{N}_2} = 0.30

The mole fraction is the ratio of the moles of nitrogen to the total moles of dry gas:

\frac{0.30\text{ mol}}{1.00\text{ mol}} = 0.30

.

4

Calculate the partial pressure of nitrogen gas.

P_{\text{N}_2} = 225\text{ mmHg}

The partial pressure of an individual gas component in a dry mixture is given by

P_{\text{N}_2} = \chi_{\text{N}_2} \times P_{\text{dry}} = 0.30 \times 750\text{ mmHg} = 225\text{ mmHg}

.

Anahtar Kavram

Dalton's Law of Partial Pressures with Vapor Pressure Correction and Mole Fraction Calculation

Soru 991Soru →

A sample of argon gas occupies a volume of $400\text{ cm}^3$ at a temperature of $-23^\circ\text{C}$ . If the pressure remains constant and the gas expands to a volume of $600\text{ cm}^3$ , what is its final temperature in degrees Celsius ( $^\circ\text{C}$ )?

Cevabı ve açıklamayı göster

Cevap: 102

Cevap

102 °C

Converting the initial temperature to Kelvin gives

T_1 = -23 + 273 = 250\text{ K}

. Applying Charles's Law

\frac{V_1}{T_1} = \frac{V_2}{T_2}

gives

T_2 = \frac{600 \times 250}{400} = 375\text{ K}

. Subtracting

273

converts this back to degrees Celsius:

375 - 273 = 102^\circ\text{C}

.

Adım Adım Çözüm

1

Convert initial temperature from Celsius to Kelvin

T_1 = 250\text{ K}

Gas laws require temperature to be expressed on the absolute (Kelvin) scale.

2

Calculate final absolute temperature using Charles's Law

T_2 = \frac{600\text{ cm}^3 \times 250\text{ K}}{400\text{ cm}^3} = 375\text{ K}

At constant pressure, volume is directly proportional to absolute temperature (

V_1/T_1 = V_2/T_2

).

3

Convert final temperature back to degrees Celsius

t_2 = 375 - 273 = 102^\circ\text{C}

The question explicitly requests the final answer in degrees Celsius.

Anahtar Kavram

Charles's Law states that the volume of a fixed mass of gas is directly proportional to its absolute temperature at constant pressure (

V \propto T

). Temperatures must always be converted to Kelvin (

K = ^\circ\text{C} + 273

) prior to calculation.

Soru 992Soru →

The standard reduction potentials for two half-cell reactions at

25^\circ\text{C}

are given below:

\text{Mn}^{2+}(aq) + 2e^- \rightarrow \text{Mn}(s) \quad E^\circ = -1.18\text{ V}

\text{Pb}^{2+}(aq) + 2e^- \rightarrow \text{Pb}(s) \quad E^\circ = -0.13\text{ V}

Calculate the standard cell potential ( $E^\circ_{\text{cell}}$ ), in volts, for the spontaneous electrochemical reaction between these two half-cells.

Cevabı ve açıklamayı göster

Cevap: 1.05

Cevap

The standard cell potential for the spontaneous reaction is

+1.05\text{ V}

.

For a spontaneous redox reaction in a galvanic cell, the half-cell with the higher standard reduction potential acts as the cathode, and the one with the lower standard reduction potential acts as the anode. Lead(II) ions (

\text{Pb}^{2+}

,

E^\circ = -0.13\text{ V}

) have a higher reduction potential than manganese(II) ions (

\text{Mn}^{2+}

,

E^\circ = -1.18\text{ V}

). Substituting these values into

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

yields

E^\circ_{\text{cell}} = -0.13\text{ V} - (-1.18\text{ V}) = +1.05\text{ V}

.

Adım Adım Çözüm

1

Identify cathode and anode roles for a spontaneous galvanic cell.

Cathode (reduction):

\text{Pb}^{2+}(aq) + 2e^- \rightarrow \text{Pb}(s)

with

E^\circ = -0.13\text{ V}

. Anode (oxidation):

\text{Mn}(s) \rightarrow \text{Mn}^{2+}(aq) + 2e^-

with

E^\circ = -1.18\text{ V}

.

For a spontaneous process (

E^\circ_{\text{cell}} > 0

), the species with the more positive reduction potential acts as the oxidizing agent and undergoes reduction at the cathode.

2

Use the cell potential formula

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

.

E^\circ_{\text{cell}} = -0.13\text{ V} - (-1.18\text{ V})

Subtracting the standard reduction potential of the anode from that of the cathode gives the overall electromotive force of the cell.

3

Perform the subtraction to find the final numerical answer.

E^\circ_{\text{cell}} = +1.05\text{ V}

-0.13 + 1.18 = 1.05

, confirming a positive standard cell potential for the spontaneous reaction.

Anahtar Kavram

Standard Cell Potential and Reaction Spontaneity

Soru 993Soru →

Malachite is an important copper ore with the chemical formula

\text{CuCO}_3\cdot\text{Cu(OH)}_2

. Upon strong heating, it undergoes thermal decomposition according to the following balanced equation:

\text{CuCO}_3\cdot\text{Cu(OH)}_2(s) \rightarrow 2\text{CuO}(s) + \text{CO}_2(g) + \text{H}_2\text{O}(g)

What is the mass of copper(II) oxide (

\text{CuO}

), in grams, formed when

22.2\text{ g}

of malachite is completely decomposed? [Relative atomic masses:

\text{Cu} = 64

,

\text{C} = 12

,

\text{O} = 16

,

\text{H} = 1

]

Cevabı ve açıklamayı göster

Cevap: 16

Cevap

The mass of copper(II) oxide produced is 16.0 g.

Thermal decomposition of 1 mole of malachite (

\text{CuCO}_3\cdot\text{Cu(OH)}_2

, molar mass

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

) produces 2 moles of copper(II) oxide (

\text{CuO}

, molar mass

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

). Given

22.2\text{ g}

of malachite (

0.1\text{ mol}

), exactly

0.2\text{ mol}

of

\text{CuO}

is formed, giving a mass of

0.2\text{ mol} \times 80\text{ g mol}^{-1} = 16.0\text{ g}

.

Adım Adım Çözüm

1

Calculate the molar mass of malachite,

\text{CuCO}_3\cdot\text{Cu(OH)}_2

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

Summing the relative atomic masses:

2(64) + 12 + 5(16) + 2(1) = 222\text{ g mol}^{-1}

.

2

Calculate the number of moles of malachite in the sample

0.1\text{ mol}

Dividing given mass by molar mass:

\frac{22.2\text{ g}}{222\text{ g mol}^{-1}} = 0.1\text{ mol}

.

3

Determine the moles of

\text{CuO}

produced using stoichiometry

0.2\text{ mol}

of

\text{CuO}

The balanced chemical equation shows a

1:2

mole ratio between malachite and

\text{CuO}

.

4

Calculate the mass of

\text{CuO}

formed

16.0\text{ g}

Multiplying moles of

\text{CuO}

by its molar mass (

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

):

0.2 \times 80 = 16.0\text{ g}

.

Anahtar Kavram

Stoichiometry of copper compounds thermal decomposition

Soru 994Soru →

Under specific laboratory conditions of temperature and pressure, a $50\text{ cm}^3$ sample of sulfur(IV) oxide ( $SO_2$ ) gas diffuses through a tiny orifice in $24\text{ seconds}$ . What is the time, in seconds, required for an equal volume of helium ( $He$ ) gas to diffuse through the same orifice under identical conditions? [Relative atomic masses: $\text{He} = 4$ , $\text{O} = 16$ , $\text{S} = 32$ ]

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The time required for an equal volume of helium gas to diffuse under identical conditions is

6\text{ seconds}

.

According to Graham's Law of Diffusion, the time required for equal volumes of two gases to diffuse at constant temperature and pressure is directly proportional to the square root of their molar masses (

t_1 / t_2 = \sqrt{M_1 / M_2}

). Because helium (

4\text{ g/mol}

) has one-sixteenth the molar mass of sulfur(IV) oxide (

64\text{ g/mol}

), its diffusion time is

\sqrt{1/16} = 1/4

of the time taken by

SO_2

. Therefore,

t_{He} = 24 \text{ s} \times 0.25 = 6\text{ seconds}

.

Adım Adım Çözüm

1

Calculate the molar masses of both gases from their chemical formulas and relative atomic masses.

Molar mass of

SO_2 = 64\text{ g/mol}

; Molar mass of

He = 4\text{ g/mol}

.

Graham's Law requires the molar mass of each gas to determine relative diffusion rates and times.

2

Set up the Graham's Law expression relating diffusion time to molar mass for equal volumes.

\frac{t_{He}}{t_{SO_2}} = \sqrt{\frac{M_{He}}{M_{SO_2}}}

Diffusion time for a given volume is directly proportional to the square root of the molar mass.

3

Substitute given values and simplify the square root expression.

\frac{t_{He}}{24} = \sqrt{\frac{4}{64}} = \sqrt{\frac{1}{16}} = \frac{1}{4}

Simplifying the molar mass ratio gives a simple fraction.

4

Multiply to solve for the diffusion time of helium gas.

t_{He} = 24 \times \frac{1}{4} = 6\text{ seconds}

Helium is lighter and diffuses 4 times faster than

SO_2

, requiring 1/4th of the time.

Anahtar Kavram

Graham's Law of Diffusion and Effusion

Tahmini Süre:1m 30s

Soru 995Soru →

A $17.2\text{ g}$ sample of hydrated calcium tetraoxosulfate(VI), $\text{CaSO}_4 \cdot x\text{H}_2\text{O}$ , is heated at $120^\circ\text{C}$ until it partially dehydrates, losing $2.70\text{ g}$ of water vapor to form plaster of Paris, $\text{CaSO}_4 \cdot 0.5\text{H}_2\text{O}$ . What is the value of $x$ , the number of molecules of water of crystallization per formula unit in the original hydrated salt? $[\text{Ca} = 40, \text{S} = 32, \text{O} = 16, \text{H} = 1]$

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The value of

x

in the hydrated salt formula is 2.

Applying mass conservation, the mass of plaster of Paris (

\text{CaSO}_4 \cdot 0.5\text{H}_2\text{O}

) remaining is

17.20\text{ g} - 2.70\text{ g} = 14.50\text{ g}

, corresponding to

0.10\text{ mol}

. The mass of evolved water is

2.70\text{ g}

, which equals

0.15\text{ mol}

. The mole ratio of evolved water to salt formula units is

\frac{0.15}{0.10} = 1.5

. Since the reaction is

\text{CaSO}_4 \cdot x\text{H}_2\text{O} \rightarrow \text{CaSO}_4 \cdot 0.5\text{H}_2\text{O} + (x - 0.5)\text{H}_2\text{O}

,

x - 0.5 = 1.5 \implies x = 2

.

Adım Adım Çözüm

1

Calculate the molar masses of

\text{CaSO}_4

,

\text{H}_2\text{O}

, and the residue

\text{CaSO}_4 \cdot 0.5\text{H}_2\text{O}

.

Molar mass of

\text{CaSO}_4 = 40 + 32 + (4 \times 16) = 136\text{ g/mol}

;

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

;

\text{CaSO}_4 \cdot 0.5\text{H}_2\text{O} = 136 + (0.5 \times 18) = 145\text{ g/mol}

.

Molar masses are necessary to perform mole calculations from mass measurements.

2

Determine the mass and amount in moles of plaster of Paris formed.

Mass of residue

= 17.20\text{ g} - 2.70\text{ g} = 14.50\text{ g}

. Moles of

\text{CaSO}_4 \cdot 0.5\text{H}_2\text{O} = \frac{14.50\text{ g}}{145\text{ g/mol}} = 0.10\text{ mol}

.

Subtracting the mass of lost water gives the mass of solid product remaining.

3

Calculate the moles of water vapor driven off.

Moles of

\text{H}_2\text{O} = \frac{2.70\text{ g}}{18\text{ g/mol}} = 0.15\text{ mol}

.

Determining the quantity of lost water allows finding the mole ratio of lost water to salt units.

4

Relate the moles of lost water to the stoichiometry of partial dehydration to solve for

x

.

Moles of water lost per mole of salt

= \frac{0.15\text{ mol}}{0.10\text{ mol}} = 1.5\text{ mol}

. Since partial dehydration yields

(x - 0.5)

moles of lost water,

x - 0.5 = 1.5 \implies x = 2

.

Connecting empirical mole ratios to chemical formula coefficients yields the integer hydration number

x

.

Anahtar Kavram

Stoichiometric determination of water of crystallization from mass loss during partial dehydration.

Soru 996Soru →

A weather balloon is launched containing $3.20\text{ dm}^3$ of helium gas at an initial temperature of $47^\circ\text{C}$ and a pressure of $1.50\text{ atm}$ . As the balloon rises into the atmosphere, the temperature drops to $-23^\circ\text{C}$ and the pressure decreases to $0.60\text{ atm}$ . Calculate the final volume of the helium gas in $\text{dm}^3$ .

Cevabı ve açıklamayı göster

Cevap: 6.25

Cevap

The final volume of the helium gas is

6.25\text{ dm}^3

.

Converting the given temperatures to Kelvin yields

T_1 = 47 + 273 = 320\text{ K}

and

T_2 = -23 + 273 = 250\text{ K}

. Applying the General Gas Law

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

and rearranging for

V_2

gives

V_2 = \frac{1.50 \times 3.20 \times 250}{0.60 \times 320} = 6.25\text{ dm}^3

.

Adım Adım Çözüm

1

Convert both initial and final temperatures from Celsius to Kelvin.

T_1 = 47^\circ\text{C} + 273 = 320\text{ K}

and

T_2 = -23^\circ\text{C} + 273 = 250\text{ K}

.

Gas laws require absolute temperature units (Kelvin).

2

State the General Gas Law equation.

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

Relates initial and final values of pressure, volume, and temperature for a fixed mass of gas.

3

Rearrange the equation to solve for the final volume (

V_2

).

V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}

Isolates the target variable on one side.

4

Substitute the values into the equation and compute the result.

V_2 = \frac{1.50\text{ atm} \times 3.20\text{ dm}^3 \times 250\text{ K}}{0.60\text{ atm} \times 320\text{ K}} = 6.25\text{ dm}^3

Yields the exact final volume of the gas.

Anahtar Kavram

General Gas Law

Soru 997Soru →

In a gas counter-diffusion experiment using a horizontal glass tube of length $100\text{ cm}$ , gas $A$ with a molar mass of $36\text{ g/mol}$ is released from end $P$ , while an unknown gas $B$ is released simultaneously from end $Q$ under identical conditions of temperature and pressure. The two gases meet and form a visible reaction ring at a distance of $40\text{ cm}$ from end $P$ . What is the molar mass of gas $B$ in $\text{g/mol}$ ?

Cevabı ve açıklamayı göster

Cevap: 16

Cevap

16 g/mol

According to Graham's Law of Diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass (

r_A / r_B = \sqrt{M_B / M_A}

). Because both gases diffuse over the same duration, the distance ratio equals the rate ratio (

d_A / d_B = 40 / 60 = 2/3

). Substituting

M_A = 36\text{ g/mol}

into

2/3 = \sqrt{M_B / 36}

and squaring both sides gives

4/9 = M_B / 36

, which yields

M_B = 16\text{ g/mol}

.

Adım Adım Çözüm

1

Calculate the distance traveled by gas B

d_B = 100 cm - 40 cm = 60 cm

The total length of the diffusion tube is 100 cm, and gas A traveled 40 cm from end P before meeting gas B.

2

Determine the ratio of the rates of diffusion of gas A and gas B

r_A / r_B = 40 / 60 = 2/3

The rate of diffusion is directly proportional to the distance traveled in a given time period.

3

Apply Graham's Law of Diffusion relating diffusion rates to molar masses

2/3 = sqrt(M_B / 36)

Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

4

Solve for the unknown molar mass M_B

4/9 = M_B / 36 => M_B = 16 g/mol

Squaring both sides of the equation eliminates the square root, allowing straightforward algebraic solution.

Anahtar Kavram

Graham's Law of Diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass (

r_1 / r_2 = \sqrt{M_2 / M_1}

).

Soru 998Soru →

A steady electric current is passed through an aqueous solution of zinc tetraoxosulfate(VI) for $4825\text{ s}$ . If $3.25\text{ g}$ of zinc is deposited at the cathode, what is the magnitude of the electric current, in Amperes, used?

$[\text{Zn} = 65,\text{ 1 F} = 96500\text{ C mol}^{-1}]$

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The magnitude of the electric current required is

2.0\text{ A}

.

Depositing

3.25\text{ g}

of

\text{Zn}

(atomic mass

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

) requires

0.05\text{ mol}

of zinc metal. Since each

\text{Zn}^{2+}

ion requires

2

electrons to be reduced,

0.10\text{ mol}

of electrons (

9650\text{ C}

) must pass through the electrolyte. Dividing this charge by time (

4825\text{ s}

) gives

2.0\text{ A}

.

Adım Adım Çözüm

1

Calculate the moles of zinc deposited at the cathode.

Moles of

\text{Zn} = \frac{3.25\text{ g}}{65\text{ g mol}^{-1}} = 0.05\text{ mol}

.

Dividing the mass of metal deposited by its relative atomic mass gives the number of moles deposited.

2

Determine the quantity of electricity in Coulombs needed for the deposition.

Reduction half-reaction:

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

. Moles of

e^- = 2 \times 0.05\text{ mol} = 0.10\text{ mol}

. Quantity of electricity

Q = 0.10\text{ mol} \times 96500\text{ C mol}^{-1} = 9650\text{ C}

.

Faraday's second law relates the mole ratio of electrons to metal ion charge.

3

Calculate the steady electric current in Amperes.

I = \frac{Q}{t} = \frac{9650\text{ C}}{4825\text{ s}} = 2.0\text{ A}

.

Electric current is defined as the rate of charge flow over time (

I = \frac{Q}{t}

).

Anahtar Kavram

Faraday's Laws of Electrolysis and Quantitative Calculations

Soru 999Soru →

An aqueous solution of chromium(III) tetraoxosulfate(VI) is electrolyzed using inert platinum electrodes. A steady current of $5.00\text{ A}$ is passed through the electrolyte for $96.5\text{ minutes}$ . If the cathodic current efficiency for chromium deposition is $75.0\%$ , calculate the mass, in grams, of chromium metal deposited at the cathode. [Molar mass of $\text{Cr} = 52.0\text{ g/mol}$ , $1\text{ F} = 96500\text{ C/mol}$ ]

Cevabı ve açıklamayı göster

Cevap: 3.9

Cevap

3.90\text{ g}

To find the mass of chromium deposited, calculate total charge (

Q = I \times t = 5.00 \times 5790 = 28950\text{ C}

), adjust for

75.0\%

current efficiency (

Q_{\text{eff}} = 21712.5\text{ C}

), convert to Faradays (

0.225\text{ F}

), divide by the valency of 3 for

\text{Cr}^{3+}

to find moles of chromium (

0.075\text{ mol}

), and multiply by molar mass (

52.0\text{ g/mol}

) to yield

3.90\text{ g}

.

Adım Adım Çözüm

1

Convert the electrolysis time into seconds

t = 96.5 \times 60 = 5790\text{ s}

Standard SI unit of time (seconds) is required for charge calculation (

Q = I \times t

).

2

Calculate the total charge transferred

Q = 5.00\text{ A} \times 5790\text{ s} = 28950\text{ C}

Determines total quantity of electricity delivered by the current source.

3

Apply the current efficiency percentage

Q_{\text{eff}} = 28950\text{ C} \times 0.750 = 21712.5\text{ C}

Only 75% of the total current is utilized specifically for reducing chromium ions.

4

Convert effective charge into moles of electrons

n_e = \frac{21712.5\text{ C}}{96500\text{ C/mol}} = 0.225\text{ mol } e^-

Faraday's constant gives the charge carried per mole of electrons.

5

Relate moles of electrons to moles of chromium deposited

n_{\text{Cr}} = \frac{0.225}{3} = 0.075\text{ mol}

Reduction of one mole of

\text{Cr}^{3+}

requires three moles of electrons (3 Faradays).

6

Calculate the mass of chromium deposited

m = 0.075\text{ mol} \times 52.0\text{ g/mol} = 3.90\text{ g}

Mass is obtained by multiplying the number of moles by the molar mass.

Anahtar Kavram

Faraday's Laws of Electrolysis and Current Efficiency

Soru 1000Soru →

For a chemical reaction carried out at $27^\circ\text{C}$ , the standard Gibbs free energy change ( $\Delta G^\circ$ ) is $-54.0\text{ kJ mol}^{-1}$ . Given that the standard entropy change ( $\Delta S^\circ$ ) for the reaction is $-120.0\text{ J K}^{-1}\text{ mol}^{-1}$ , calculate the standard enthalpy change ( $\Delta H^\circ$ ) in $\text{kJ mol}^{-1}$ .

Cevabı ve açıklamayı göster

Cevap: -90

Cevap

The standard enthalpy change ($\Delta H^\circ$) for the reaction is $-90.0\text{ kJ mol}^{-1}$.

The standard enthalpy change is calculated using the rearranged Gibbs free energy equation

\Delta H^\circ = \Delta G^\circ + T\Delta S^\circ

. Converting

27^\circ\text{C}

to

300\text{ K}

and

-120.0\text{ J K}^{-1}\text{ mol}^{-1}

to

-0.120\text{ kJ K}^{-1}\text{ mol}^{-1}

yields

\Delta H^\circ = -54.0 + (300 \times -0.120) = -90.0\text{ kJ mol}^{-1}

.

Adım Adım Çözüm

1

Convert the temperature from degrees Celsius to Kelvin

T = 300 K

Thermodynamic calculations require absolute temperature in Kelvin: T = 27 + 273 = 300 K.

2

Convert the standard entropy change units from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹

ΔS° = -0.120 kJ K⁻¹ mol⁻¹

Since ΔG° is given in kJ mol⁻¹, ΔS° must be converted to kJ K⁻¹ mol⁻¹ by dividing by 1000.

3

Rearrange the Gibbs free energy equation to express ΔH° and substitute the given values

ΔH° = -90.0 kJ mol⁻¹

From ΔG° = ΔH° - TΔS°, rearranging gives ΔH° = ΔG° + TΔS° = -54.0 + (300 × -0.120) = -90.0 kJ mol⁻¹.

Anahtar Kavram

Gibbs Free Energy Equation and Unit Consistency

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