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SOVA2 [1]
3 years ago
14

A mixture of helium and methane gases, at a total pressure of 821 mm Hg, contains 0.723 grams of helium and 3.43 grams of methan

e. What is the partial pressure of each gas in the mixture?
PHe = mm Hg
PCH4 = mm Hg

2.) A mixture of nitrogen and carbon dioxide gases contains nitrogen at a partial pressure of 363 mm Hg and carbon dioxide at a partial pressure of 564 mm Hg. What is the mole fraction of each gas in the mixture?

XN2 =
XCO2 =
Chemistry
1 answer:
Nookie1986 [14]3 years ago
6 0

<u>Answer:</u>

<u>For 1:</u> The partial pressure of helium is 376 mmHg and that of methane gas is 445 mmHg

<u>For 2:</u> The mole fraction of nitrogen gas is 0.392 and that of carbon dioxide gas is 0.608

<u>Explanation:</u>

<u>For 1:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For helium:</u>

Given mass of helium = 0.723 g

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

\text{Moles of helium}=\frac{0.723g}{4g/mol}=0.181mol

  • <u>For methane gas:</u>

Given mass of methane gas = 3.43 g

Molar mass of methane gas = 16 g/mol

Putting values in equation 1, we get:

\text{Moles of methane gas}=\frac{3.43g}{16g/mol}=0.214mol

To calculate the mole fraction , we use the equation:

\chi_A=\frac{n_A}{n_A+n_B}     .......(2)

To calculate the partial pressure of gas, we use the equation given by Raoult's law, which is:

p_{A}=p_T\times \chi_{A}       ......(3)

  • <u>For Helium gas:</u>

We are given:

n_{He}=0.181mol\\n_{CH_4}=0.214mol

Putting values in equation 2, we get:

\chi_{He}=\frac{0.181}{0.181+0.214}=0.458

Calculating the partial pressure by using equation 3, we get:

p_T=821mmHg\\\\\chi_{He}=0.458

Putting values in equation 3, we get:

p_{He}=0.458\times 821mmHg=376mmHg

  • <u>For Methane gas:</u>

We are given:

n_{He}=0.181mol\\n_{CH_4}=0.214mol

Putting values in equation 2, we get:

\chi_{CH_4}=\frac{0.214}{0.181+0.214}=0.542

Calculating the partial pressure by using equation 3, we get:

p_T=821mmHg\\\\\chi_{CH_4}=0.542

Putting values in equation 3, we get:

p_{CH_4}=0.542\times 821mmHg=445mmHg

Hence, the partial pressure of helium is 376 mmHg and that of methane gas is 445 mmHg

  • <u>For 2:</u>

We are given:

Partial pressure of nitrogen gas = 363 mmHg

Partial pressure of carbon dioxide gas = 564 mmHg

Total pressure = (363 + 564) mmHg = 927 mmHg

Calculating the mole fraction of the gases by using equation 3:

<u>For nitrogen gas:</u>

363=\chi_{N_2}\times 927\\\\\chi_{N_2}=\frac{363}{927}=0.392

<u>For carbon dioxide gas:</u>

564=\chi_{CO_2}\times 927\\\\\chi_{CO_2}=\frac{564}{927}=0.608

Hence, the mole fraction of nitrogen gas is 0.392 and that of carbon dioxide gas is 0.608

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A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of
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Answer: The molecular of the compound is, C_2H_3O

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=3.095g

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We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 3.095g of carbon dioxide, \frac{12}{44}\times 3.095=0.844g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 1.902g of water, \frac{2}{18}\times 1.092=0.121g of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = (1.621)-[(0.844)+(0.121)]=0.656g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.041 moles.

For Carbon = \frac{0.0703}{0.041}=1.71\approx 2

For Hydrogen  = \frac{0.121}{0.041}=2.95\approx 3

For Oxygen  = \frac{0.041}{0.041}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is C_2H_3O_1=C_2H_3O

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}

n=\frac{46.06}{43}=1

Molecular formula = (C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O

Therefore, the molecular of the compound is, C_2H_3O

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Question is incomplete, complete question is;

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If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H_2SO_4(aq)? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of H_2SO_4(aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_1,M_2\text{ and }V_2  are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL

Putting values in above equation, we get:

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3 years ago
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Answer: The value of equilibrium constant for new reaction is 1.92\times 10^{-25}

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The equilibrium constant for the above equation is 5.77\times 10^{-9}

We need to calculate the equilibrium constant for the equation of 3 times of the above chemical equation, which is:

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The equilibrium constant for this reaction will be the cube of the initial reaction.

If the equation is multiplied by a factor of '3', the equilibrium constant of the new reaction will be the cube of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

K_{eq}'=(5.77\times 10^{-9})^3=1.92\times 10^{-25}

Hence, the value of equilibrium constant for new reaction is 1.92\times 10^{-25}

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