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:
Height = 1.9493 cm
Width = 1.9493 cm
Depth = 1.9493 cm
Solution:
Data Given:
Mass = 20 g
Density = 2.7 g/mL
Step 1: Calculate the Volume,
As,
Density = Mass ÷ Volume
Or,
Volume = Mass ÷ Density
Putting values,
Volume = 20 g ÷ 2.7 g/mL
Volume = 7.407 mL or 7.407 cm³
Step 2: Calculate Dimensions of the Cube:
As we know,
Volume = length × width × depth
So, we will take the cube root of 7.407 cm³ which is 1.9493 cm.
Hence,
Volume = 1.9493 cm × 1.9493 cm × 1.9493 cm
Volume = 7.407 cm³