Answer:
SiO2 + 2C - - > Si + 2CO
Explanation:
Everything should balance out now. The same number of atoms of each kind on each side
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
The answer is "Location E, because Earth's axis is tilted towards the sun" :D
Answer : The mass of copper(I) oxide is, 1180.5 grams.
Explanation : Given,
Moles of copper(I) oxide = 8.250 mol
Molar mass of copper(I) oxide = 143.09 g/mol
Formula used:
Now put all the given values in this formula, we get:
Therefore, the mass of copper(I) oxide is, 1180.5 grams.
<span>Find
the elements sodium, oxygen, and phosphorus on the periodic table. Describe the
monoatomic ions each would form. Then, given what you know about chemical
bonding, describe and give the formulas for the types of ionic compounds they
would produce with each other. </span>
