Chemistry help plz?

a mixture of 0.250 mol of oxygen and 0.750 mol of nitrogen is prepared in a container such as that the total pressure is 800 torr. calculate the partial pressure of oxygen in the mixture.

Answer:
We use the formula :
p(a) / p(t) = n(a) / n (t)

n(t) = 0.250 + 0.750 = 1
p (O2) / 800 = 0.250 / 1
p( O2) = 200 torr
p( N2) / 800 = 0.750 / 1
p (N2) = 600 torr
partial pressure = mole fraction x total pressure

mole fraction = moles of gas/total moles of gas

partial pressure of oxygen = 0.250moles /1.00 total moles x 800 torr

partial pressure of nitrogen = 0.750 moles/1.00 total moles x 800 torr

The answers post by the user, for information only, FunQA.com does not guarantee the right.



More Questions and Answers:
  • At constant pressure the temperature of a 4.00 liter sample of gas is decreased from 400 K to 300 K. ...?
  • Zinc nitrate + ammonia?
  • What is the advantage of using chromic acid test to alcohols over Lucas test?
  • Stoichiometry?
  • Please, tell me if this is correct?
  • Why would it be unwise for each country to choose its own names and symbols for elements?
  • Calculate the density of a gas if the mass is 2.94 grams, the volume is 756 mL at 20°C...?
  • Is the dissolution of sucrose in water due to miscibility or ionic dissociation?
  • Explain what happens when an antaacid (Maalox), which contains Al(OH)3, is consumed to relieve heartburn?