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Ann [662]
2 years ago
11

Suppose a gas mixture used for anesthesia contains 4.60 mol oxygen (O₂) and 6.00 mol nitrous oxide (N₂O). The total pressure of

the mixture is 2.00 atm. A mole fraction is defined as the moles of a specific component divided by the total number of moles present. What is the mole fraction of O₂ in this mixture?
Chemistry
1 answer:
julia-pushkina [17]2 years ago
3 0

Considering the definition of mole fraction, the mole fraction of O₂ in the mixture is 0.434.

<h3>Definition of mole fraction</h3>

The molar fraction is a way of measuring the concentration that expresses the proportion in which a substance is found with respect to the total moles of the solution.

In other words, the mole fraction expresses the concentration of solute in a solution as the ratio of moles of substance to total moles of solution:

mole fraction=\frac{moles of substance}{moles of solution}

<h3>Mole fraction of O₂ in this mixture</h3>

In this case, you know a gas mixture used for anesthesia contains 4.60 mol oxygen (O₂) and 6.00 mol nitrous oxide (N₂O).

So, the total moles of the solution can be calculated as:

Total moles = moles of oxygen (O₂) + moles of nitrous oxide (N₂O)

Then:

Total moles= 4.60 moles + 6 moles

Total moles= 10.60 moles

Finally, the more fraction of O₂ can be calculated as follow:

Mole fraction of O_{2} =\frac{moles of O_{2}}{total moles}

Mole fraction of O_{2} =\frac{4.60 moles}{10.6o moles}

Solving:

<u><em>Mole fraction O₂ = 0.434</em></u>

Finally, the mole fraction of O₂ in the mixture is 0.434.

Learn more about mole fraction:

brainly.com/question/14434096

brainly.com/question/10095502

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1) An aerosol can contains gases under a pressure of 4.50 atm at 20.0 degrees Celsius. If the can is left on a hot, sandy beach,
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Answer:

Explanation:

1) An aerosol can contains gases under a pressure of 4.50 atm at 20.0 degrees Celsius. If the can is left on a hot, sandy beach, the pressure of the gases increases to 4.78 atm. What is the Celsius temperature on the beach?

Given data:

Initial pressure = 4.50 atm

Initial temperature = 20.0°C (20 +273 = 293 K)

Final pressure = 4.78 atm

Final temperature = ?  (in °C)

Solution:

According to the Gay-Lussac law,

The temperature of given constant amount of a gas at constant volume is directly proportional to its absolute temperature.

Mathematical expression:

P₁/T₁ = P₂/T₂

P₁ = Initial pressure

T₁ = Initial temperature

P₂ = Final pressure

T₂ = Final temperature

Now we will put the values:

P₁/T₁ = P₂/T₂

4.50 atm / 293 k = 4.78 atm / T₂

T₂ = 4.78 atm. 293 k / 4.50 atm

T₂ = 1400.54  atm.K  / 4.50 atm

T₂ = = 311.23 k

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2) A sample of gas contains NO, NO2, and N2O. The pressure of the gas mixture is 4.68 atm. The pressure of NO is 501.6 mm Hg, whereas the pressure of NO2 is 2.54 atm. What is the pressure of N2O? HINT: All pressure units must be the same.

Given data:

Total pressure of gaseous mixture = 4.68 atm

Pressure of NO = 501.6 mmHg

Pressure of NO₂ = 2.54 atm

Pressure of N₂O = ?

Solution:

The given problem will be solve through the Dalton law of partial pressure.

According to this law,

" The total pressure of mixture of a gas is equal to the sum of partial pressure of all the component of gas"

Now we will convert the pressure of NO₂  in to atm.

Pressure of NO = 501.6/760 = 0.66 atm

Formula:

Total pressure = partial pressure of NO +  partial pressure of NO₂  +  partial pressure of N₂O

4.68 atm = 0.66 atm +  2.54 atm +  partial pressure of N₂O

4.68 atm = 3.2 atm +  partial pressure of N₂O

Partial pressure of N₂O = 4.68 atm - 3.2 atm

Partial pressure of N₂O = 1.48 atm

To confirm the answer:

Total pressure = partial pressure of NO +  partial pressure of NO₂  +  partial pressure of N₂O

4.68 atm =  0.66 atm +  2.54 atm +  1.48 atm

4.68 atm = 4.68 atm

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