Question

How many grams of N2 gas are present in 1.13 L of gas at 2.09 atm and 291 K?

Answer 1

Mass= 2.77g

Step-by-step explanation:

Applying

P=2.09atm, V= 1.13L, R= 0.082, T= 291K, Mm of N2= 28

PV=nRT

NB

Moles(n) = m/M

PV=m/M×RT

m= PVM/RT

Substitute and Simplify

m= (2.09×1.13×28)/(0.082×291)

m= 2.77g