A gas mixture contains 1.52 atm of Ne, 766 mmHg of He and Ar. What is the partial pressure, in atmospheres, of At if the gas mixture has a total pressure of 3.27atm

0.74 atm.

Explanation:

From the question given above, the following data were obtained:

Pressure of Ne (Pₙₑ) = 1.52 atm

Pressure of He (Pₕₑ) = 766 mmHg

Total pressure (Pₜ) = 3.27 atm

Pressure of Ar (Pₐᵣ) =?

Next, we shall convert the pressure of He from mmHg to atm. This can be obtained as follow:

760 mmHg = 1 atm

Therefore,

766 mmHg = 766 mmHg × 1 atm / 760 mmHg

766 mmHg = 1.01 atm

Finally, we shall determine the partial pressure of Ar. This can be obtained as follow:

Pressure of Ne (Pₙₑ) = 1.52 atm

Pressure of He (Pₕₑ) = 1.01 atm

Total pressure (Pₜ) = 3.27 atm

Pressure of Ar (Pₐᵣ) =?

Pₜ = Pₙₑ + Pₕₑ + Pₐᵣ

3.27 = 1.52 + 1.01 + Pₐᵣ

3.27 = 2.53 + Pₐᵣ

Collect like terms

3.27 – 2.53 = Pₐᵣ

Pₐᵣ = 0.74 atm

Thus the partial pressure of Ar is 0.74 atm.

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b