58. What is the temperature if 30.1 g of N, in a 1.00-L container is at a pressure of 25.2 atm? DUE TONIGHT. LINKS DONT WORK SO PLS SHOW WORK. THANK YOU SO MUCH

14°C

Explanation:

In order to use the ideal gas law, you need the number of moles of N2 present.

30.1 g x (mol / 28.02 g) = 1.07 mol N2

Use the ideal gas law: PV = nRT

Now isolate for T(temperature) and plug in your other values:

T = PV / nR

T = (25.2 atm)(1.00 L) / (1.07 mol)(0.08206)

T = 287 K

T(°C) = 287 K - 273 K = 14°C

Therefore, the temperature is 14°C

im not exactly sure but i think its b!