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Datlon's law of Partial Pressures states that in a gaseous mixture, the exerted by a gaseous component is the same as the pressure it would exert if it ALONE occupied the container. The total pressure is the SUM of the partial pressures.
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In a mixture of oxygen and nitrogen gas, $80.0%$ of the total gas pressure is exerted by the nitrogen. If the total pressure is $"2.0 atm"$, what pressure does the oxygen exert?
$80%$of pressure is exerted by nitrogen gas. So, $20%$of pressure will be exerted by oxygen gas. $"p"_("O"_2) = 20% "of 2 atm" = 20/100 × "2 atm" = "0.4 atm"$
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A mixture of neon and oxygen gases, in a 9.77 L flask at 65 °C, contains 2.84 grams of neon and 7.67 grams of oxygen. The partial pressure of oxygen in the flask is ? atm and the total pressure in the flask is atm?
We can use the to calculate the partial pressures. $color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "$ or $p = (nRT)/V$ The number of moles $n$...
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A mixture of oxygen, hydrogen, and nitrogen exerts a total pressure of 378 kPa. If the partial pressures of oxygen and hydrogen are 212 kPa and 101 kPa respectively, what is the partial pressure exerted by nitrogen?
Old Dalton's law of partial pressures states that in a gaseous mixture, the exerted by a component gas is the same as the pressure it would exert IF IT ALONE...
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A 1.2-liter tank at a temperature of 27°C contains a mixture of 1.5 moles of oxygen, 1.2 moles of nitrogen, and 1.6 moles of hydrogen. What is the total pressure of the mixture inside the tank?
Dalton's law of partial pressures states that in a gaseous mixture, the pressure exerted by a component is the same as it would exert if it alone occupied the container....
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A mixture of gases with a pressure of 800.0 Hg contains 60% nitrogen and 40% oxygen by volume. What is the partial pressure of oxygen in this mixture?
I am unhappy in quoting a pressure measurement of , as it is something you would try to avoid in the lab....... We have $P_"Total"=1.052*atm$. But by definition, $P_"Total"=P_"dioxygen"+P_"dinitrogen"=1.052*atmxx40%+1.052*atmxx60%=1.052*atm$ ..............and...
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Mixture containing .538mol $He$ gas, .315mol $Ne$ gas .103 $Ar$ gas, is confined in 7.00-L vessel at $25^o$Celsius. How do you calculate partial pressure of each gas in mixture? How would you calculate the total pressure of the mixture?
$"Dalton's law of partial pressures states that in a gaseous"$ $"partial pressure exerted by a component gas is the same as"$ $"the pressure it would exert if it alone occupied...
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A container of gas with a pressure of 450 kPa contains three different gasses -hydrogen, oxygen, and nitrogen. If the partial pressure of the hydrogen is 210 kPa and the partial pressure of the oxygen is 125 kPa, what is the partial pressure of nitrogen?
According to Dalton's law of partial pressures, the total pressure is equivalent to the sum of the pressures of the individual gases (if they were present alone). Therefore, the...
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A heliox deep-sea diving mixture contains 2.0 g of oxygen to every 98.0 g of helium. What is the partial pressure of oxygen when this mixture is delivered at a total pressure of 8.2 atm?
And so we calculate the individual mole fractions.... $chi_(O_2)=((2.0*g*mol^-1)/(32.0*g*mol^-1))/{(2.0*g)/(32.0*g*mol^-1)+(98.0*g)/(4.0*g*mol^-1)}=2.54xx10^-3$ $chi_(He)=((98.0*g*mol^-1)/(4.0*g*mol^-1))/{(2.0*g)/(32.0*g*mol^-1)+(98.0*g)/(4.0*g*mol^-1)}=0.998$ And so $p_(O_2)=2.54xx10^-3xx8.2*atm=0.0208*atm$ The diver is at a serious depth....and had he or she been inhaling dinitrogen, i.e. from a...
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A mixture of $"N"_2$, $"O"_2$. and $"CO"_2$ has a total pressure of $"740 torr"$ In this mixture, the partial pressure of $"N"_2$ is $"120 torr"$ and the partial pressure of $"O"_2$ is $"400 torr"$. What is the partial pressure of $"CO"_2$ ?
From 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...
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At some temperature, a 100-L reaction vessel contains a mixture that is initially 1.0 mol Co and 2.0 mol H2. The vessel also contains a catalyst so that the following equilibrium is attained:
CO(g) + 2H2(g) <--> CH3OH(g)
At equilibrium, the mixture contains 0.100 mol CH3OH. In a later experiment in the same vessel, you start with 1.0 mol CH3OH. How much methanol is there at equilibrium?
Methanol's concentration at equilibrium is $0.001M$. Before solving for the concentration, try and intuitively imagine what will happen. Since the second experiment is done under the same conditions as the...
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