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 $N_2$?

Md Ariful Islam

0 like 0 dislike
1 views
Share with your friends

1 Answers

Md Ariful Islam

Call

$chi_i=(n_i)/(n_"total")$.

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 $"temperature"$ was included to distract you.

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
The temperature at a camp was -4 deg C on Saturday. Sunday morning, the temperature was 3 deg C lower than what it was on Saturday. By Sunday evening, the temperature was 1 deg C lower than what it was that morning. What was the final temperature?

Add the temperatures in the order they were measured. $overbrace(-4^@"C")^color(purple)"Sat"+ overbrace((-3^@"C"))^color(purple)"Sun morn"+overbrace((-1^@"C"))^color(purple)"Sun eve"=-8^@"C"$

1 Answers 1 views
A physicist must make sure that the temperature of a metal at 0 deg C gets no colder than -80 deg C. The physicist changes the metal's temperature at a steady rate of -4 deg C per hour. For how long can the physicist change the temperature?

First let us take the metal as $ x $ If the rate of decreasing temperature is -4C/1hr and the physician cannot decrease the temperature beyond -80C. The...

1 Answers 1 views
$Mg$ and $O_2$ react in a 2.1 molar ratio. 2 moles $M_g$ = 1 mole $O_2$. If a reaction used 32.5 g of $O_2$, how many g of $Mg$ reacted?

Step 1. Calculate the moles of $"O"_2$. $32.5 cancel("g O"_2) × ("1 mol O"2)/(32.00 cancel("g O"2)) = "1.016 mol O"_2$ Step 2. Use the molar ratio to calculate the...

1 Answers 1 views
What is the mole fraction of $NO$ in a 55.0 L gas cylinder at 30.0°C which comes from a mixture of $N_2$ and $NO$ if you have 3.238 mol of $N_2$ and the gas cylinder has a total pressure of 2.14 atm?

Your 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 views
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...

1 Answers 1 views
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...

1 Answers 1 views
What are the mole fractions of each component in a mixture of 15.08 g of $O_2$, 8.17 g of $N_2$, and 2.64 g of $H_2$? What is the partial pressure in atm of each component of this mix ture if it is held in a 15.50-L vessel at 15°C?

In 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 views
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...

1 Answers 1 views
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...

1 Answers 1 views
When 2 moles of $H_2(g)$ and 1 mole of $O_2(g)$ react to give liquid water, 572 kJ of heat evolve. $2H_2(g)$ + $O_2(g)$ $rarr$ $2H_2O(l)$ ; $DeltaH$ = -572 kJ a) What is the themochemical eq. for 1 mole of liquid water?

You know that when $2$ moles of hydrogen gas react with $1$ mole of oxygen gas, you get $2$ moles of water and $"572 kJ"$ of heat are evolved....

1 Answers 1 views
X