The atmosphere in a sealed diving bell contained oxygen and helium. if the gas mixture has 0.200 atm of oxygen and a total pressure of 3.00 atm, calculate the mass of helium in 10.0 l of the gas mixture at 20oc. [1] g

the total pressure in the system is 3.00 atm

the mixture contains only oxygen and helium. Pressure of gases in a mixture is additive. pressure exerted by individual gases are called partial pressures.

the addition of all the partial pressures of the gases making up the mixture is the total pressure of the system.

partial pressure of oxygen + partial pressure of helium = total pressure

partial pressure of oxygen = 0.200 atm

0.200 atm + partial pressure of helium = 3.00 atm

partial pressure of helium = 2.80 atm

pressure exerted by helium gas alone is 2.80 atm

we can use the ideal gas law equation to find the number of moles of helium in the mixture

PV = nRT

where P - 2.80 atm

V - volume - 10.0 L

n - number of moles

R - universal gas constant - 0.0821 L.atm/mol.K

T - temperature in kelvin - 273 + 20 °C = 293 K

substituting the values in the equation

2.80 atm x 10.0 L = n x 0.0821 L.atm/mol.K x 293 K

n = 1.16 mol

mass of He = number of moles x molar mass

mass = 1.16 mol x 4.00 g/mol = 4.64 g

mass of He present = 4.64 g

b) mutualism

hope this : )