If I have 4.4 moles of nitrous oxide (laughing gas) that is kept at 32.9 °C in a container under 1.6 atm, what is the volume of the container? (R = 0.0821 (L*atm)/(mol*K)) Round your answer to 1 decimal place.

Explanation:

Using the ideal gas equation as follows:

PV = nRT

Where:

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

Based on the information provided in this question;

P = 1.6atm

n = 4.4 moles

R = 0.0821 L*atm/mol*K

T = 32.9°C = 32.9 + 273 = 305.9K

V = ?

answer; i believe that the correct answer is (a) potassium(k)