Ask question

Ask question

All categories

3 years ago

5

A sample of gas isolated from unrefined petroleum contains 90.0% CH4, 8.9% C2H6, and 1.1% C3H8 at a total pressure of 3.34 atm.

What is the partial pressure of each component of this gas

1 answer:

Leona [35]3 years ago

7 0

Answer:

The partial pressure of each component of this gas is:

Partial pressure of methane= 3.006 atm

Partial pressure of ethane= 0.29726 atm

Partial pressure of propane= 0.03674 atm

Explanation:

The partial pressure of a gas in a mixture is the pressure that gas would exert if it were alone in the container.

On the other hand, the mole fraction is a dimensionless quantity that expresses the ratio of the number of moles of a component with the number of moles of all the components present.

Dalton's Law of Partial Pressures states that the partial pressure of each gas is equal to the total pressure of the gas mixture multiplied by the mole fraction of that gas.

The partial pressures then, can be calculated by multiplication of the percents by the total pressure.

So, the partial pressure of each component of this gas is: :

$Partial pressure of methane= \frac{90}{100} *3.34 atm=0.9*3.34 atm$

<u><em>Partial pressure of methane= 3.006 atm</em></u>

$Partial pressure of ethane= \frac{8.9}{100} *3.34 atm=0.089*3.34 atm$

<u><em>Partial pressure of ethane= 0.29726 atm</em></u>

$Partial pressure of propane= \frac{1.1}{100} *3.34 atm=0.011*3.34 atm$

<u><em>Partial pressure of propane= 0.03674 atm</em></u>

You might be interested in

Which statement accurately represents the arrangement of electrons in bohr’s atomic model?

Alexus [3.1K]

Electrons travel around the nucleus in fixed energy levels with energies that vary from level to level

4 0

3 years ago

compound consists of carbon, hydrogen and fluorine. In one experiment, combustion of 2.50 g of the compound produced 3.926 g of

arlik [135]

<u>Answer:</u> The empirical and molecular formula of the compound is $CH_2F$ and $C_{14}H_{28}F_{14}$ respectively

<u>Explanation:</u>

We are given:

Mass of $CO_2=3.926g$

<u>For calculating the mass of carbon:</u>

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

So, in 3.926 g of carbon dioxide, $\frac{12}{44}\times 3.926=1.071g$ of carbon will be contained.

To calculate the percentage composition of element in sample, we use the equation:

$\%\text{ composition of element}=\frac{\text{Mass of element}}{\text{Mass of sample}}\times 100$ ......(1)

<u>For Carbon:</u>

Mass of carbon = 1.071 g

Mass of sample = 2.50 g

Putting values in equation 1, we get:

$\%\text{ composition of carbon}=\frac{1.071g}{2.50g}\times 100=42.84\%$

<u>For Fluorine:</u>

Mass of fluorine = 2.54 g

Mass of sample = 5.00 g

Putting values in equation 1, we get:

$\%\text{ composition of fluorine}=\frac{2.54g}{5.00g}\times 100=50.8\%$

Percent composition of hydrogen = [100 - 42.84 - 50.8] % = 6.36 %

We are given:

Percentage of C = 42.84 %

Percentage of F = 50.8 %

Percentage of H = 6.36 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 42.84 g

Mass of F = 50.8 g

Mass of H = 6.36 g

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

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.84g}{12g/mole}=3.57moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{6.36g}{1g/mole}=6.36moles$

Moles of Fluorine = $\frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{50.8g}{19g/mole}=2.67moles$

<u>Step 2:</u> 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 2.67 moles.

For Carbon = $\frac{0.072}{2.67}=1.34\approx 1$

For Hydrogen = $\frac{6.36}{2.67}=2.38\approx 2$

For Fluorine = $\frac{2.67}{2.67}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : F = 1 : 2 : 1

The empirical formula for the given compound is $CH_2F$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 448.4 g/mol

Mass of empirical formula = $12+(2\times 1)+19]=33g/mol$

Putting values in above equation, we get:

$n=\frac{448.4g/mol}{33g/mol}=13.6\approx 14$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(14\times 1)}H_{(14\times 2)}F_{(14\times 1)}=C_{14}H_{28}F_{14}$

Hence, the empirical and molecular formula of the compound is $CH_2F$ and $C_{14}H_{28}F_{14}$ respectively

3 0

3 years ago

A base is a substance that can __________ a hydroxide ion.

ivanzaharov [21]

Answer:

A base is a substance that can donate a hydroxide ion.

Please Mark Brainliest If This Helped!

7 0

3 years ago

What is voltage? A. The pressure that pushes electrons to the anode; it is derived from the negative charge of electrons at the

Vlad [161]

Answer:

D

Explanation:

I think it's D but not 100% sure

4 0

3 years ago

Anyone noes this i need help please anybody

Sholpan [36]

ACID ARE SOUR IN TASTE

7 0

3 years ago