Amixture of neon and methane gases, in a 8.53 l flask at 72 °c, contains 1.56 grams of neon and 4.85 grams of methane. the partial pressure of methane in the flask is atm and the total pressure in the flask is atm.

The partial pressure of methane in the flask is 1.00 atm and the total pressure in the flask is 1.26 atm.

Explanation:

You must apply the Ideal Gases Law to solve this;

Pressure . volume = n° moles . Gases constant (R) . T° in K

T° in K = T°C + 273

72 + 273 = 345K

Gases constant (R) = 0.082 L.atm /mol.K

n° moles = in mixture, the total moles are the sum of each gas.

Moles = mass / molar mass

Moles Ne = 1.56 g / 20.17 g/m = 0.077 moles

Moles CH₄ = 4.85 g/ 16.04 g/m = 0.302 moles

0.077 + 0.302 = 0.379 moles

Pressure . 8.53L = 0.379 moles . 0.082 L.atm /mol.K . 345K

Pressure = ( 0.379 moles . 0.082 L.atm /mol.K . 345K ) / 8.53L

Pressure = 1.26 atm (The total pressure in system)

We can use the molar fraction to know, the partial pressure of CH₄.

Moles CH₄ / Total moles = Partial pressure CH₄ / Total pressure

0.302 / 0.379 = Partial pressure CH₄ / 1.26 atm

(0.302 / 0.379 ) . 1.26 atm = Partial pressure CH₄

1.00 atm = Partial pressure CH₄

it had to transfer to a care and maintenance mode

the atomic mass is the bottom number on the element.

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