How many moles of O2(g) are present in a 93 mL vessel at a pressure of 2 atm and at 325 °C?  Provide you answer to 4 decimal places.R = 0.08205 L·atm·mol-1·K-1

To solve this problem, we will use the ideal gas law equation, which is:

PV = nRT

Where: P = pressure (in atm) V = volume (in L) n = number of moles R = ideal gas constant (0.08205 L·atm·mol-1·K-1) T = temperature (in K)

Step 1: Convert the volume from mL to L. 1 L = 1000 mL So, 93 mL = 93/1000 = 0.093 L

Step 2: Convert the temperature from °C to K. K = °C + 273.15 So, 325 °C = 325 + 273.15 = 598.15 K

Step 3: Substitute the values into the ideal gas law equation and solve for n (number of moles). 2 atm * 0.093 L = n * 0.08205 L·atm·mol-1·K-1 * 598.15 K

Solving for n gives: n = (2 atm * 0.093 L) / (0.08205 L·atm·mol-1·K-1 * 598.15 K) = 0.0044 mol

So, there are 0.0044 moles of O2(g) present in a 93 mL vessel at a pressure of 2 atm and at 325 °C.