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In 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.
We know that the total pressure is
Since we have been given the other partial pressures, calculation of
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A gas mixture contains 1.19 g $N_2$ and 0.79 g $O_2$ in a 1.60-L container at 25 deg C. How do you calculate the mole fraction of $N_2$?
$chi_i=(n_i)/(n_"total")$. 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 $"temperature"$...
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The partial pressures of the gases in a miture are 0.00461 atm $O_2$, 1.78 atm $N_2$, and 0.975 atm $Ar$. What is the total pressure of the mixture?
$P_"Total"=P_(O_2)+P_(N_2)+ P_(Ar)$ $=(4.61xx10^-2 + 1.78+0.975)*atm$ $~=2.76*atm$ This is a simple result of $"Dalton's Law of Partial Pressure"$, for which
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Five gases combined in a gas cylinder have the following partial pressures 3.00 atm ($N_2$), 1.80 atm ($O_2), 0.29 atm (Ar), 0.18 atm (He), and 0.10 atm (H). What is the total pressure that is exerted by the gases?
Old $"Dalton's Law of Partial Pressures"$ states that in a gaseous mixture, the exerted by a component gas is the same as it would have exerted if it alone occupied...
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The atmospheric pressure at the summit of Mt. Everest is 255 torr. The percentage of $O_2$ and $N_2$ is 20% and 79%, respectively. How many $N_2$ and $O_2$ molecules are there in a volume of 10.0L at 12°C?
Your 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...
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A steel tank with a volume of 9.583 L contains $N_2$ gas under a pressure of 4.972 atm at 31.8°C. How many moles of $N_2$ are in the tank?
For the old Ideal Gas equation, $n=(PV)/(RT)$... And so $n=(4.972*atmxx9.583*L)/(0.0821*(L*atm)/(K*mol)*305*K)$ $=1.90*mol$
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When a container is filled with 3.00 moles of $H_2$, 2.00 moles of $O_2$, and 1.00 mole of $N_2$, the pressure in thecontainer is 768 kPa. What is the partial pressure of $O_2$?
Partial 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...
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A sample containing 4.80 g of $O_2$ gas has a volume of 15.0 L at constant pressure and temperature. What is the new volume, in liters, after 0.500 mole of $O_2$ gas is added to the initial sample of 4.80 g of $O_2$?
Initially, you have $4.80color(red)(cancel(color(black)("g O"_2))) × "1 mol O"_2/(32.00 color(red)(cancel(color(black)("g O"_2)))) = "0.1500 mol O"_2$ Thus, you have $"0.1500 mol O"_2/"15.0 L" = "0.0100 mol O"_2/"1 L"$ If you...
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A tank contains a mixture of helium, neon, and argon gases. lf the total pressure in the tank is 490. mmHg and the partial pressures of helium and argon are 21.5 mmHg and 10.2 mmHg, respectively, what is the partial pressure of neon?
And so... $P_"Total"=P_"He"+P_"Ne"+P_"Ar"=490*mm*Hg$ But we are given that $P_"He"=21.5*mm*Hg$, and that $P_"Ar"=10.2*mm*Hg$.. And so...$P_"Ne"={490-21.5-10.2}*mm*Hg}=458.3*mm*Hg$..
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For the reaction: $N_2(g) + 2O_2(s) rightleftharpoons 2NO_2(g)$, $K_c$ - $8.3x10^-10$ at 25 °C. What is the concentration of $N_2$ gas at equilibrium when the concentration of $NO_2$ is five times the concentration of $O_2$ gas?
$Kc = [NO_2]^2/ ([N2][O2]^2) $ The Kc is measured at the equilibrum $8.3 * 10^-10 = x^2/([N_2][x]^2)$ Conc of $N_2 "when" NO_2 = O_2$ = 1204819277.108434mol $8.3 * 10^-10...
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What would happen if $O_2$ were added to $N_2(g) + O_2 (g) rightleftharpoons 2NO(g)$ at equilibrium?
For the following system at equilibrium: $N_2(g)+O_2(g)rightleftharpoons2NO(g)$ According to le Châtelier's principle , when a stress is applied to a system at equilibrium, the system will readjust to reach equilibrium...
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