Question

weight of CO2 in a 10L cylinder at 5atm and 27degree​

Answer 1

Answer:

$PV=nRT=MmRT</p><p>P=5 atm,V=10 L,T=300 K,</p><p>R=0.0821 L atm K−1mol−1</p><p>m=RTPVM=0.0821×3005×10×44=89.3 g.</p><p></p><p>$

Explanation:

hope it helped you friend if it helps you mark it as brainliest