Thank you so much Steve
I just have one more on this module. I think this question is really easy, but I just can't get it!
How many grams of helium are required to fill a 2.3 L balloon to a pressure of 1.10 atm at 25C?
I did PV=nRT, so n= RT / PV
So I went, (0.08206 x 298.15) / (1.1 x 2.3) and I got 9.67
and then 9.67 x 4.003 = 38.68g
But the answer is 0.414g |
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