A sample of gas at 57oC and 1.02atm occupies a volume of 2.20L. What volume would this gas occupy at 107oC and 0.789atm?

(p1v1/t1) = (p2v2/t2)

Remember T must be in kelvin.

1.06atm x 1.25L/250.0K

1.325/250.0K
m=.0053

I don't understand anything you wrote. Are you working this problem or another one? Plug values into that equation I gave you and solve. What you have posted is gibberish.

To solve this problem, we can use the combined gas law, which states that the ratio of the initial pressure, initial volume, and initial temperature is equal to the ratio of the final pressure, final volume, and final temperature.

The formula for the combined gas law is:

(P1 * V1) / (T1) = (P2 * V2) / (T2)

where P1 and P2 are the initial and final pressures, V1 and V2 are the initial and final volumes, and T1 and T2 are the initial and final temperatures.

Let's plug in the given values:

P1 = 1.02 atm
V1 = 2.20 L
T1 = 57°C (which needs to be converted to Kelvin by adding 273.15)
P2 = 0.789 atm
V2 = ? (what we need to find)
T2 = 107°C (which needs to be converted to Kelvin by adding 273.15)

Using the formula, we can rearrange it to find V2:

V2 = (P1 * V1 * T2) / (P2 * T1)

Now let's plug in the values and calculate:

V2 = (1.02 atm * 2.20 L * (107°C + 273.15)) / (0.789 atm * (57°C + 273.15))

V2 = (1.02 atm * 2.20 L * 380.15 K) / (0.789 atm * 330.15 K)

V2 = (898.32 L atm K) / (261.12 atm K)

V2 ≈ 3.440 L

Therefore, the gas would occupy approximately 3.440L at 107°C and 0.789 atm.