A 255mL gas sample contains argon and nitrogen at a temperature of 65 degrees celsius. The total pressure of the sample is 725mmHg and the partial pressure of argon is 231 mmHg. what mass of nitrogen is present in the sample?

Ptot= pN2 + pAR; (725mmHg-231mmHg=494mmHg)

494mmHg*(1atm/760mmHg)= 0.65atm
65celsius+273 = 338K
V=0.255L
R= 0.0821 Latm/mol K

Ideal Gas Law: PV=nRT
n N2 = (PV/RT) plug in above values to equation
n= (0.65atm * 0.255L) / (0.0821 Latm/molK * 338K)
moles N2 (n) = 0.00597 mols

mass N2= (0.00597 mols)/ (28.02 g/mol)= 0.167 g N2 present in sample

Ptotal = pN2 + pAr

You know Ptotal and pAr. Solve for pN2.
Then PV = nRT, substitute and solve for n N2 then use mols N2 = n = grams/molar mass and solve for grams.

Well, it seems like we have a gas party going on inside that sample! So, let's get cracking on solving this riddle.

First, we need to find the partial pressure of nitrogen. To do that, we subtract the partial pressure of argon from the total pressure of the sample:

Partial pressure of nitrogen = Total pressure - Partial pressure of argon
Partial pressure of nitrogen = 725 mmHg - 231 mmHg
Partial pressure of nitrogen = 494 mmHg

Now that we have the partial pressure of nitrogen, we can use the ideal gas law to find the number of moles of nitrogen present in the sample. Remember, the ideal gas law goes like this: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, let's convert the temperature from Celsius to Kelvin:
65 degrees Celsius + 273 = 338 Kelvin

Now, let's rearrange the ideal gas law to solve for n (number of moles):
n = PV / RT

Substituting the values we have:
n = (494 mmHg * 255 mL) / (0.0821 L*atm/(mol*K) * 338 K)

Calculating it out, we find that the number of moles of nitrogen is approximately 3.75 moles (rounded to two decimal places).

But, we're not done yet! We need to convert moles of nitrogen to grams, using the molar mass of nitrogen (which is around 28 grams/mol). So, we multiply the number of moles by the molar mass:

Mass of nitrogen = 3.75 moles * 28 g/mol
Mass of nitrogen = 105 grams

Voila! The mass of nitrogen present in the sample is approximately 105 grams.

I hope my mathematical circus act brought a smile to your face!

To find the mass of nitrogen present in the sample, we need to first determine the partial pressure of nitrogen in the gas sample.

Since the total pressure of the sample is 725 mmHg and the partial pressure of argon is 231 mmHg, we can subtract the partial pressure of argon from the total pressure to find the partial pressure of nitrogen:

Partial pressure of nitrogen = Total pressure - Partial pressure of argon
Partial pressure of nitrogen = 725 mmHg - 231 mmHg
Partial pressure of nitrogen = 494 mmHg

Now, we can use the ideal gas law equation to find the number of moles of nitrogen in the sample. The ideal gas law equation is given as:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Let's solve for the number of moles of nitrogen:

n = (PV) / (RT)

First, let's convert the volume of the sample from mL to L:

Volume = 255 mL * (1 L / 1000 mL)
Volume = 0.255 L

Next, let's convert the temperature from degrees Celsius to Kelvin:

Temperature (K) = 65 degrees Celsius + 273.15
Temperature (K) = 338.15 K

Now, we can calculate the number of moles of nitrogen:

n = (494 mmHg * 0.255 L) / (0.0821 L·atm / K·mol * 338.15 K)
n = 0.0511 mol

Since the ratio of the number of moles of nitrogen to its mass is 1:28.02 (using the molar mass of nitrogen), we can find the mass of nitrogen:

Mass of nitrogen = 0.0511 mol * 28.02 g/mol
Mass of nitrogen = 1.43 grams

Therefore, the mass of nitrogen present in the sample is approximately 1.43 grams.

To find the mass of nitrogen present in the gas sample, we need to use the ideal gas law and the molar mass of nitrogen.

The ideal gas law equation is:

PV = nRT

Where:
P = pressure of the gas sample
V = volume of the gas sample
n = number of moles of gas
R = ideal gas constant
T = temperature of the gas sample

First, let's convert the given volume from milliliters (mL) to liters (L):
255 mL = 255/1000 L = 0.255 L

Next, let's convert the given temperature from degrees Celsius to Kelvin (K):
T(K) = T(°C) + 273.15
T = 65 °C + 273.15 = 338.15 K

Now we can rearrange the ideal gas law equation to solve for the number of moles of gas (n):

n = (PV)/(RT)

Substituting the given values:
n = (725 mmHg * 0.255 L) / (0.0821 L*atm/mol*K * 338.15 K)
(Note: The ideal gas constant we are using is 0.0821 L*atm/mol*K)

Now we need to calculate the number of moles of nitrogen present. Since the partial pressure of argon is known (231 mmHg), we can subtract this from the total pressure to find the partial pressure of nitrogen:

Partial pressure of nitrogen = Total pressure - Partial pressure of argon
Partial pressure of nitrogen = 725 mmHg - 231 mmHg = 494 mmHg

Finally, using the ideal gas law equation again, we can solve for the mass of nitrogen:

n(N2) = (Partial pressure of nitrogen * V) / (R * T)

Substituting the values:
n(N2) = (494 mmHg * 0.255 L) / (0.0821 L*atm/mol*K * 338.15 K)

To convert moles of nitrogen to grams, we multiply by the molar mass of nitrogen, which is approximately 28 g/mol.

Mass of nitrogen = n(N2) * molar mass of nitrogen

Substituting the molar mass:
Mass of nitrogen = n(N2) * 28 g/mol

Using these calculations, we can find the mass of nitrogen present in the gas sample.