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: Option (c) is the correct answer.
Explanation:
The bottom of pencil is placed at the starting point of scale. Whereas the tip of pencil depicts the end point of its length.
The bottom of pencil is at 0 mm and tip of pencil is at 18.73 mm. The appropriate amount of significant figures is 18.73 mm.
Therefore, we can conclude that out of the given options, pencil is 18.73 mm long.
Answer:
a) ΔGrxn = 6.7 kJ/mol
b) K = 0.066
c) PO2 = 0.16 atm
Explanation:
a) The reaction is:
M₂O₃ = 2M + 3/2O₂
The expression for Gibbs energy is:
ΔGrxn = ∑Gproducts - ∑Greactants
Where
M₂O₃ = -6.7 kJ/mol
M = 0
O₂ = 0
b) To calculate the constant we have the following expression:
Where
ΔGrxn = 6.7 kJ/mol = 6700 J/mol
T = 298 K
R = 8.314 J/mol K
c) The equilibrium pressure of O₂ over M is:
According to the law of conservation of mass, the mass of reactants will be equal to the mass of the products. The mass of products and reactants will only differ during a nuckear reaction
Changing of the physical state of water is not a nuclear reaction. So becoz of that the mass will remain constant without any change.
