A mixture contains 5.0 g of He, 1.0 g of Ar, and 3.5 g of Ne. Calculate the partial pressure of each gas in the mixture at STP and total pressure of mixture.

Total pressure of the mixture = 1 atm

Partial pressure of He = 0.863 atm

Partial pressure of Ar  = 0.0173 atm

Partial pressure of Ne = 0.1197 atm

We confirm that the sum of partial pressure of each gas valu

Explanation:

STP conditions are:

1 atm of pressure and

273K of T°

We can also say that those value reffers to 1 mol.

We determine the moles of each:

5 g . 1mol / 4g = 1.25 moles of He

1g . 1mol / 39.9 g = 0.0250 moles of Ar

3.5g . 1mol / 20.18 g = 0.173 moles of Ne

Total pressure of a mixture can be obtained from the Ideal Gases Law.

P . V = n . R . T

Total moles: 1.25 + 0.025 + 0.173 = 1.448 moles

But now, we need the volume. We know that 1 mol is contained at 22.4L at STP.

1.448 mol . 22.4L / 1 mol = 32.4 L

That's the volume for our mixture. We replace:

32.4 L . P  = 1.448 mol . 0.082 . 273K

P = (1.448 mol . 0.082 . 273K) / 32.4L = 1 atm

Now we can obtained the partial pressure of each gas from mole fraction

Mol of gas / Total pressure = Partial pressure of gas / Total pressure

Partial pressure of He = (1.25 / 1.448) . 1 atm = 0.863 atm

Partial pressure of Ar = (0.025 / 1.448) . 1 atm = 0.0173 atm

We know that sum of partial pressure is 1 - (Mole fraction He + Mole fr. Ar)

Partial pressure of Ne = 1 - (0.863 + 0.0173) = 0.1197 atm

We confirm that the sum of partial pressure of each gas values the total pressure of the mixture: 0.863 + 0.0173 + 0.1197 = 1

clouds form

solid ice forms on leaves.

water droplets form on grass.

the answer is definitely c.