A 8.55-L container hold a mixture of two gases at 45 degrees C the partial pressures of gas A and gas B, respectively, are .150 atm and .751 atm. If .230 mol a third gas is added with no change in volume or temperature, what will the total pressure become?

Question

anonymous · Answer

PV=nRT
P = Pressure
V = Volume
n = Number of Moles
R = Gas Constant 
T = Temperature in Kelvin
[(.15+.751)*8.55]=n[0.08206*(45+273.15)]
n=[(.15+.751)*8.55]/[0.08206*(45+273.15)]
We've solved for n now but we've got to add .23 because you're adding that many moles to the mix.
n={[(.15+.751)*8.55]/[0.08206*(45+273.15)]+.23}
P*8.55 = {[(.15+.751)*8.55]/[0.08206*(45+273.15)]+.23}*[0.08206*(45+273.15)]
P = {{[(.15+.751)*8.55]/[0.08206*(45+273.15)]+.23}*[0.08206*(45+273.15)]}/8.55
P is about 1.603304032 atm

abb0t · Answer

Watch your sig figs.

anonymous · Answer

calculate the morality of a solution made by adding 36.0mL of concentrated ammonia (28.0%) by mass, density 0.880 g?mL) to some water in a volumetric flask, then adding water to the mark to make exactly 250 mL of solution. (it is important to add concentrated acid or base to water, rather than the other way, to minimize splashing and maximize safety.) 
[NH3] = blank M