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:
The upper and lower limits for the room-temperature thermal conductivity of a magnesium oxide material having a volume fraction of 0.10 of pores that are filled with still air are
Ku = 38.252 W/mK
K lower = 0.199 W/mK
Explanation:
As we know
Ku = Vp * Kair + Vmagnesium * K metal
Ku = 0.10 *0.02 + (1-0.25) * 51
Ku = 38.252 W/mK
The lower limit
K lower = Kmetal* Kair/( Vp * Kmetal + Vmetal * K air)
K lower = (0.02*51)/(0.10*51 + 0.90 * 0.02)
K lower = 0.199 W/mK
Answer:
10.6 grams is approximately 0.10 moles. So you would need about 0.10 moles of sulfuric acid. That converts to about 9.80 grams.
Explanation:
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Answer:
The new partial pressures after equilibrium is reestablished:



Explanation:

At equilibrium before adding chlorine gas:
Partial pressure of the 
Partial pressure of the 
Partial pressure of the 
The expression of an equilibrium constant is given by :


At equilibrium after adding chlorine gas:
Partial pressure of the 
Partial pressure of the 
Partial pressure of the 
Total pressure of the system = P = 263.0 Torr




At initail
(13.2) Torr (32.8) Torr (13.2) Torr
At equilbriumm
(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr


Solving for x;
x = 6.402 Torr
The new partial pressures after equilibrium is reestablished:


