Chemistry, 18.07.2019 18:30 lilytimpsonx

The rate of effusion of oxygen to an unknown gas is 1.49. what is the other gas?

Answers: 1

Show answers

Answers

Answer from: rainyshea72

The unknown gas is $\boxed{{\text{chlorine}}}$ .

Further Explanation:

Effusion is defined as the process by which molecules of gas travel through a small hole from high pressure to the low pressure. According to Graham’s law, effusion rate of a gas is inversely proportional to the square root of the molar mass of gas.

${\text{Rate of effusion}}\left({\text{R}}\right)\propto\frac{1}{{\sqrt {{\mu}}}}$ ......(1)

Where $\mu$ is the molar mass of gas.

The rate of effusion of oxygen is expressed as

${{\text{R}}_{{\text{oxygen}}}}\propto \frac{1}{{\sqrt {{{{\mu }}__{{\text{oxygen}}}}}}}$ ......(2)

And, the rate of effusion of other gas is expressed as

${{\text{R}}_{{\text{gas}}}}\propto\frac{1}{{\sqrt{{{{\mu }}__{{\text{gas}}}}}}}$ ......(3)

Since rate of effusion of oxygen is 1.49 times of rate of effusion of the unknown gas. Therefore,

${{\text{R}}_{{\text{oxygen}}}}=1.49{{\text{R}}_{{\text{gas}}}}$ ......(4)

Substitute the value of ${{\text{R}}_{{\text{white phosphorus}}}}$ and ${{\text{R}}_{{\text{Neon}}}}$ from equation (2) and equation (3) respectively in equation (4).

$\frac{1}{{\sqrt{{{{\mu }}_{_{{\text{oxygen}}}}}}}}= 1.49\frac{1}{{\sqrt{{{{\mu }}__{{\text{gas}}}}}}}$ ......(5)

Simplify equation (5) as follows:

$\begin{aligned}\frac{1}{{\sqrt{{{{\mu }}__{{\text{oxygen}}}}}}}&=1.49\frac{1}{{\sqrt{{{{\mu }}__{{\text{gas}}}}}}}\hfill\\\sqrt{{{{\mu }}__{{\text{gas}}}}}&=1.49\left({\sqrt {{{{\mu }}__{{\text{oxygen}}}}}}\right) \hfill\\{{{\mu }}__{{\text{gas}}}}&={\left({1.49\left({\sqrt {{{{\mu }}__{{\text{oxygen}}}}}} \right)}\right)^2}\hfill\\\end{gathered}$

Therefore the molar mass of white phosphorus is,

${{{\mu }}__{{\text{gas}}}}={\left({1.49\left({\sqrt{{{{\mu }}__{{\text{oxygen}}}}}}\right)} \right)^2}$ ......(6)

The molar mass of oxygen is 32 g/mol.

Substitute the value of molar mass of oxygen in the equation (6).

$\begin{aligned}{{{\mu }}__{{\text{gas}}}}&= {\left( {1.49\left({\sqrt {32\;{\text{g/mol}}} } \right)} \right)^2}\\&=71.0432\;{\text{g/mol}}\\\end{gathered}$

Therefore, the calculated molar mass of the unknown gas is 71.0432 g/mol and we know the gas with the molar mass of 71.0432 g/mol is chlorine. So the unknown gas is chlorine.

Learn more:

1. Which is the oxidation-reduction reaction:

2. Calculation of volume of gas:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ideal gas equation

Keywords: Effusion, rate of effusion, oxygen, unknown gas, chlorine, 1.49, molar mass, atomic mass, rate of effusion of oxygen, rate of effusion of gas, molar mass of oxygen, Graham’s law of effusion.

Answer from: Quest

the solar system used to have much dust particles option no d

Which of the following best illustrates the role that gravity played in the formation of the solar s

Answer from: Quest

the answer is minerals.

Another question on Chemistry

Chemistry, 22.06.2019 09:00

What term is missing from the central region that describes hypotheses, theories, and laws? popular predictable mathematical falsifiable

Answers: 2

Answer

Chemistry, 22.06.2019 12:00

Which statement best explains the relationship between an area is geography and the temperature of its surface water

Answers: 1

Answer

Chemistry, 22.06.2019 16:00

What rule is used to determine how many covalent bonds an element can form? a. the number of covalent bonds is equal to six c the number of covalent bonds is equal to five minus the group number plus the group number b. the number of covalent bonds is equal to eight d. none of the above minus the group number select the best answer from the choices provided

Answers: 2

Answer

Chemistry, 22.06.2019 22:00