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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 density of air $20$ $km$ above Earths surface is $92$ $g$$/$$m^3.$ The pressure of the atmosphere is $42$ $mm$ $Hg$, and the temperature is $-63$ $°C$. What is the average molar mass of the atmosphere at this altitude?
The idea here is that you need to start from the standard form of the equation $color(blue)(PV = nRT) " " " "color(purple)((1))$ and try to manipulate it...
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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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If 5.0 moles of $O_2$ and 3.0 moles of $N_2$ are placed in a 30.0 L tank at a temperature of 25°C, what will the pressure of the resulting mixture of gases be?
In order to find the pressure of the mixture, you need to know the total number of moles present in the mixture. The problem provides you with the volume...
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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 tank contains $N_2$ at 1.0 atm and $O_2$ at 2.0 atm. Helium is added to this tank until the total pressure is 6.0 atm. What is the partial pressure of the helium?
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....
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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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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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