Answer:
C. H₂SO₄
Explanation:
Given expression:
H₂O + SO₃ → H₂SO₄
In the chemical reaction, the product and the reactant have the same species.
Chemical reactions obeys the law of conservation of matter.
This way, atoms combine with one another to give a desired product. The products will contain a mix of the reactants.
- Fluorine is not part of the reactants
- Carbon is not part of the reactants
- Carbon is not part of the reactants.
Answer:
You would get 19.969 moles.
Explanation:
You would get 19.969 moles.
Answer:
17.04%
Explanation:
Actual Value = 173.1
Measured Value = 143.6
Percent error is obtained using the equation;
Percent error = (Measured - Actual) / Actual ]* 100
Percent error = [ (143.6 - 173.1) / 173.1 ] * 100
The absolute value of (Measured - Actual) is taken,
Percent Error = [29.5 / 173.1 ] * 100
Percent Error = 0.1704 * 100 = 17.04%
Explanation:
Bernoulli equation for the flow between bottom of the tank and pipe exit point is as follows.
= 
![\frac{(100 \times 144)}{62.43} + 0 + h[tex] = [tex]\frac{(50 \times 144)}{(62.43)} + \frac{(70)^{2}}{2(32.2)} + 0 + 40 + 60](https://tex.z-dn.net/?f=%5Cfrac%7B%28100%20%5Ctimes%20144%29%7D%7B62.43%7D%20%2B%200%20%2B%20h%5Btex%5D%20%3D%20%5Btex%5D%5Cfrac%7B%2850%20%5Ctimes%20144%29%7D%7B%2862.43%29%7D%20%2B%20%5Cfrac%7B%2870%29%5E%7B2%7D%7D%7B2%2832.2%29%7D%20%2B%200%20%2B%2040%20%2B%2060)
h = 
= 60.76 ft
Hence, formula to calculate theoretical power produced by the turbine is as follows.
P = mgh
= 
= 6076 lb.ft/s
= 11.047 hp
Efficiency of the turbine will be as follows.
= 
× 100%
= 
= 52.684%
Thus, we can conclude that the efficiency of the turbine is 52.684%.
Answer:
the water concentration at equilibrium is
⇒ [ H2O(g) ] = 0.0510 mol/L
Explanation:
- CH4(g) + H2O(g) ↔ CO(g) + 3H2(g)
∴ Kc = ( [ CO(g) ] * [ H2 ]³ ) / ( [ CH4(g) ] * [ H2O(g) ] ) = 0,30
⇒ [ CO(g) ] = 0.206 mol / 0.778 L = 0.2648 mol/L
⇒ [ H2(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L
⇒ [ CH4(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L
replacing in Kc:
⇒ ((0.2648) * (0.2404)³) / ([ H2O(g) ] * 0.2404 ) = 0.30
⇒ 0.0721 [ H2O(g) ] = 3.679 E-3
⇒ [ H2O(g) ] = 0.0510 mol/L
