The density of a substance is given by the formula ρ = m/v, where m is the mass of the substance, and v is the volume.
ρ = m/v
ρ = 3.1 g / 4.1 mL
ρ = 31/41 g/mL = 0.756 g/mL = 0.756 g/cm^3
When: (CH3)3COOH(CH3)3⇒ C2H6 + 2CH3CoCH3
A B 2C
SO first for A:
When feed rate to reactor = FA(o), and Effluent rate from reactor= FA=Fa(o)(1-x)
Second for B:
When feed rate to reactor = 0, and Effluent rate from reactor FB= FA(o)X
Third for C:
When feed rate tp reactor = 0, and the Effluent rate from reactor FC= 2FA(o)X
So, the total feed rate to reactor FT(o) = FA(o)
and the total Effluent rate from reactor FT = FA(o)(1+2X)
For the CSTR:
V= FA(o) X / γa where -γ = KCA
We can get the total concentration at any point CT and the entrance CT(o) from this equation:
CT= FT/Q= P/ZRT
and CT(o) = FT(o)/ Q(o) = P(o)/ Z(o)RT(o)
We can neglect the changes in Z (the compressibility factor) So:
Q = Q(o) FT/FT(o) P(o)/P T/T(o)
Because there is no pressure drop and for the isothermal system we can assume P & T are constants:
∴ Q = Q(o) FT/ FT(o)
∴ CA = FA / Q = (FA(o)(1-X)) / (Q(o)(1+2X))
= CA(o)((1-X)/(1+2X)) -----> eq 1
When K (the reaction rate constant) at 127C° can be evaluated with R (the gas constant) 8.314 J/mole.K° = 0.08205 L.atm/mole.°K
When, K2=K1(exp)[E/R ( (1/T1) - (1/T2)]
when T1 by kelvin = 50+273= 323 K° and T2= 127+273 = 400 K°
and activition energy = 85 KJ/mol = 85000 J/mol
∴ K2 =10^-4 (exp) [85000/8314 ((1/323) - (1/400)] = 0.0443 min^-1
We can get the concentration of (A) at 10 atm and 127C° (400K°) by:
CA(o)= PA(o) /RT = 10 / ( 0.08205 * 400 ) = 0.305 mol / L
∴ The reactor volume of the CSTR is:
V = FA(o)X / KCA---> eq 2
by substitute eq 1 in eq 2
∴ V=[ FA(o)x * (1+2x)] / [ KCA(o) * (1-X) ]
= [ 2.5 * 0.9 * (1+1.8)] / [0.0443 * 0.305* 0.1]
= 4662.7 L
- The equilibrium conversion:
according to the reaction:
A⇔B + 2C
and we have:
-γ = KfCA - Kb CB( C(C)^2) = 0 and KC = Kf / Kb = (CB* CC^2)/ CA= 0.025(as it is agiven)
∴[ ( XCA(o) ) ( 2XCA(o)^2 ) ] / [ (1+2X)^2 * CA(o) * (1-X) ] = 0.025
when we have CA(o) (the initial concentration)= 0.305 mol/L, So we have this eq:
[(XCA(o)) (2XCA(o))^2] / [(1+2X)^2 CA(o) (1-X) ] = 0.025
By the matlab function solve
∴ X= X eq = 0.512 for 90% of the equilibrium conversion
∴ X= 0.9 X eq = 0.4608
A & C.
much debated nuclear power plants uses nuclear fission power stations, with uranium-235 as the source if fission. It is "non-renewable" according to the Energy Information Administration.
burning wood is also a non-renewable energy source
Answer:
Heated water begins circulating in a fish tank.
Explanation:
