Question

Partial pressure:

Answer 1

A mixture of argon and krypton gases at a total pressure of 973 mm Hg contains argon at a partial pressure of 474 mm Hg. If the gas mixture contains 6.62 grams of argon, how many grams of krypton are present?

Answer 2

First, we already know the partial pressure of Krypton by subtracting the partial pres. of argon from the total. Next, we'll have to make use of the ideal gas formula PV=nRT. Given the mass of argon, we can find the moles of argon by dividing the mass by the molar mass (which is n in the gas equation).

Answer 3

DEEP BREATH

Answer 4

I guess we're assuming both gases are in the same container or whatever, so we can set up this comparison: P1/n1 = P2/n2, where the first side is argon #s and second side is krypton #s.

Answer 5

oh nice! thats as far as i got was the partial pressure of Kr and the moles or argon!

Answer 6

973mmHg/# of moles of argon = 474mmHg/? ----- Solve for the ?, which is the number of moles of krypton. Now we just multiply this by the molar mass and boom. Ooo kill em. Sounds like alot when writing it out. D:

Answer 7

oh wow! thats no so bad after all!

Answer 8

Whoa scratch those #s. 474mmHg/# moles of argon = P of Kr/moles of Kr. Solve for moles of Kr.

Answer 9

Comparing the #'s of argon to #'s of Krypton, assuming same volume, temperature in same container.

Answer 10

ok let me try it out:)

Answer 11

I don't have an answer cuz I'm too lazy to use a calc and find a table.

Answer 12

haha its ok! you've done more than enough!
:)

Answer 13

would i need to convert mmHg to atm? @iPwnBunnies

Answer 14

Doesn't matter in this problem b/c the units are staying the same throughout.

Answer 15

Correct! :D
Nice explanation!

Answer 16

Thanks!
Finally I reached my personal goal of 75! :D