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 :
Here ya go if you need the link we’re I found this answer key https://studylib.net/doc/8708211/exam-2---chemistry
Answer:
The empirical formula is SF6 (option E)
Explanation:
Step 1: Data given
Mass of sulfur = 3.21 grams
Mass of fluorine = 11.4 grams
Molar mass sulfur = 32.065 g/mol
Molar mass fluorine = 19.00 g/mol
Step 2: Calculate moles
Moles = mass /molar mass
Moles sulfur = 3.21 grams / 32.065 g/mol
Moles sulfur = 0.100 moles
Moles fluorine = 11.4 grams / 19.00 g/mol
Moles fluorine = 0.600 moles
Step 3: Calculate mol ratio
We divide by the smallest amount of moles
S: 0.100 / 0.100 = 1
F : 0.600 / 0.100 = 6
The empirical formula is SF6 (option E)
It is an example of a molecule
