What is the volume of 0.100 g of C2H2F4
vapor at 1.09 atm and 43.9◦C?
Answer in units of L
5 answers
YES(:
V=nRT/P
change 0.100g of C2H2F4 to moles n
change temp to kelvins
pressure units ok.
R=22.4L*1atm/1mole*273K=0.082 L*atm/mole*K
change 0.100g of C2H2F4 to moles n
change temp to kelvins
pressure units ok.
R=22.4L*1atm/1mole*273K=0.082 L*atm/mole*K
so then would it be (22.4)(.0821)(316.0) divided by 1.09? which is ultimately 534.671
it said that it was wrong
T is wrong. What's the 22.4? It doesn't belong there. Where is n? It belongs there. No surprise the data base gave you a thumbs down.
1 answer