$chi_i=(n_i)/(n_"total")$. Now $n_"total"=(1.19*g)/(28.01*g*mol^-1)+(0.79*g)/(32.00*g*mol^-1)$ $=6.67xx10^-2*mol$ And $chi_"dinitrogen"$ $=$ $((1.19*g)/(28.01*g*mol^-1))/(6.67xx10^-2*mol)=0.64$. Given that it is a binary system, what is $chi_"dioxygen"?$ Note that the information given with respect to $"volume"$ and $"temperature"$...
1 Answers 1 viewsOld $"Dalton's Law of Partial Pressures"$ states that in a gaseous mixture, the exerted by a component gas is the same as it would have exerted if it alone occupied...
1 Answers 1 viewsYour strategy here will be to use the equation in order to find the total number of moles present in the mixture, then use the number of moles of nitrogen...
1 Answers 1 viewsYour strategy here will be to use the equation to determine the total number of moles of air you have in that $"10.0-L"$ sample under those conditions for pressure and...
1 Answers 1 viewsPartial Pressure of $O_2$ = 256 kPa. Dalton's law of partial pressures covers two concepts; Mole fraction Partial pressure Let's talk about Mole Fraction first. You...
1 Answers 1 viewsThe of a gas that's part of a gaseous mixture will depend on two things the mole fraction said gas has in the mixture the total pressure of...
1 Answers 1 viewsIn order to be able to calculate the mole fraction of each component in the mixture, you will have to convert the masses given to you to moles. To...
1 Answers 1 viewsIn a gaseous mixture, the exerted by a gas is the same if that gas alone had occupied the container. The total pressure is the sum of the partial pressure....
1 Answers 1 viewsAnd thus $P_"dinitrogen"$ $=$ $78.8%xx763*mm*Hg$. And $P_"dioxygen"$ $=$ $21.0%xx763*mm*Hg$.........etc. And the sum of the individual partial pressures is necessarily the overall pressure, $P=763*mm*Hg$
1 Answers 1 viewsFrom Dalton's Law of , the total pressure of a mixture of gases is the sum of the partial pressures of each gas. Therefore, the partial pressure of $"CO"_2$ is...
1 Answers 1 views