Part a: three gases (8.00 g of methane, ch_4, 18.0g of ethane, c_2h_6, and an unknown amount of propane, c_3h_8) were added to the same 10.0- l container. at 23.0 degrees c, the total pressure in the container is 5.40 atm. calculate the partial pressure of each gas in the container. part b: a gaseous mixture of o_2 and n_2 contains 37.8 % nitrogen by mass. what is the partial pressure of oxygen in the mixture if the total pressure is 405 mmhg?

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

Moles of methane gas =

Moles of ethane gas  =

Moles of propane gas =

Mole fraction of methane =

Similarly, mole fraction of ethane and propane :

Partial pressure of each gas can be calculated by the help of Dalton's' law:

Partial pressure of methane gas:

Partial pressure of ethane gas:

Partial pressure of propane gas:

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas =

Moles of oxygen gas  =

Mole fraction of nitrogen=

Similarly, mole fraction of oxygen

Partial pressure of each gas can be calculated by the help of Dalton's' law:

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

Partial pressure of oxygen gas:

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.