A 1 L flask is filled with 1.40 g of argon at 25 degrees celcius. A sample of ethane vapor is added to the same flask until the total pressure is 1.00 atm.

Whatis the total pressure of Argon in the flask ?

What is the partial pressure of ethane in the flask?

ddd

Use PV = nRT to calculate the pressure of Ar in the flask at the beginning.

Then total pressure = partial pressure Ar + partial pressure ethane.

Please, can you give the answer???

i want to double check

the partial pressure of Argon is .888 atm

1. PV=nRT

Par(1L)=(0.035mol)(0.0821)(25+273)
=0.856 atm

2.Ptotal = Par + Pethane
Pethane= 1.00atm-0.856atm
= 0.144atm

Well, ain't that a gas-tly situation! To find the partial pressure of argon, we need to first find out its mole fraction in the mixture.

Using the ideal gas law, we can calculate the number of moles of argon:

n = PV / (RT)

where P is the pressure, V is the volume, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.

Since we're dealing with a 1 L flask, the volume is 1 L. The temperature needs to be converted to Kelvin, so 25 degrees Celsius becomes 298 K.

Now, let's find the number of moles of argon:

n_argon = (1 atm * 1 L) / (0.0821 L·atm/(mol·K) * 298 K) = 0.0406 moles

Since we have the number of moles of argon, we can now find its mole fraction:

X_argon = moles of argon / total moles = 0.0406 / (0.0406 + moles of ethane)

Now here's where things get funny. We need to find the partial pressure of ethane, but we don't know its moles. Talk about being a mystery guest at a gas party!

Wait, let me check my joke book. Ah, here's something: Why did the ethane go to therapy? Because it couldn't handle the pressure!

Anyway, back to the problem. We know the total pressure is 1 atm, so we can now find the partial pressure of ethane using Dalton's Law of Partial Pressure:

P_ethane = X_ethane * total pressure

But since we don't know X_ethane, we need to use a trickery. We can rearrange the equation for mole fraction to solve for X_ethane:

X_ethane = 1 - X_argon

Now we can finally find the partial pressure of ethane:

P_ethane = (1 - X_argon) * 1 atm

So, depending on the value of X_argon, the partial pressure of ethane can be calculated. Now, isn't that a gas-tly riddle?!

To calculate the total pressure of Argon in the flask, we need to know the partial pressure of Argon and the partial pressure of ethane.

First, let's calculate the partial pressure of Argon.
We know the total pressure of the flask is 1.00 atm and the partial pressure of any gas is the pressure that gas would exert if it occupied the entire volume alone. This means the partial pressure of Argon is the same as the total pressure since Argon is the only gas present initially.

Therefore, the total pressure of Argon in the flask is 1.00 atm.

Now, let's calculate the partial pressure of ethane.
We need to use the ideal gas law, which states:

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.

We'll use the ideal gas law to find the number of moles of ethane in the flask. Then, we can use the number of moles to calculate the partial pressure.

First, let's find the number of moles of ethane:
1 mole of any gas at STP (Standard Temperature and Pressure) occupies 22.4 L.
So, the number of moles of argon initially in the flask is:
1.00 mol / 22.4 L = 0.04464 mol

Now, let's find the number of moles of ethane:
To find the number of moles, we can use the equation:

n = PV / RT

Where:
P is the pressure (1.00 atm),
V is the volume (1 L),
R is the ideal gas constant (0.0821 atm·L/mol·K), and
T is the temperature (298 K).

n = (1.00 atm) * (1 L) / (0.0821 atm·L/mol·K) * (298 K) = 0.04064 mol

Since we're adding ethane to the initial Argon, the total number of moles of gas is the sum of the initial moles of Argon and the moles of added ethane:
Total moles of gas = moles of Argon + moles of ethane
Total moles of gas = 0.04464 mol + 0.04064 mol = 0.08528 mol

Finally, let's calculate the partial pressure of ethane using the ideal gas law again:
Pethane = nethane * R * T / V

Pethane = (0.04064 mol) * (0.0821 atm·L/mol·K) * (298 K) / (1 L) = 1.00 atm

Therefore, the partial pressure of ethane in the flask is 1.00 atm.