The mixture flow rate in lbm/h = 117.65 lbm/h
<h3>Further explanation</h3>
Given
15.0 wt% methanol
The flow rate of the methyl acetate :100 lbm/h
Required
the mixture flow rate in lbm/h
Solution
mass of methanol(CH₃OH, Mw= 32 kg/kmol) in mixture :

mass of the methyl acetate(C₃H₆O₂,MW=74 kg/kmol,85% wt) in 200 kg :

Flow rate of the methyl acetate in the mixture is to be 100 lbm/h.
1 kg mixture = 0.85 .methyl acetate
So flow rate for mixture :

Answer: There are
molecules
gas are in 756.2 L.
Explanation:
It is known that 1 mole of any gas equals 22.4 L at STP. Hence, number of moles present in 756.2 L are calculated as follows.

According to mole concept, 1 mole of every substance contains
molecules.
Therefore, molecules of S present in 33.76 moles are calculated as follows.

Thus, we can conclude that there are
molecules
gas are in 756.2 L.
Answer:
letter A. i hope this is correct answer
The second one, vocab and definition doesn't add up
I don't understand what you mean??