The answer would be A because it’s a byeeeee
The answer is 15 bc you multiply the coefficient but the subscript
Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.
The formation reaction of
will be,

The intermediate balanced chemical reaction will be,
(1)

(2)

(3)

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equation, we get :
(1)

(2)

(3)

The expression for enthalpy of formation of
will be,



Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ
Answer:
ΔS = +541.3Jmol⁻¹K⁻¹
Explanation:
Given parameters:
Standard Entropy of Fe₂O₃ = 90Jmol⁻¹K⁻¹
Standard Entropy of C = 5.7Jmol⁻¹K⁻¹
Standard Entropy of Fe = 27.2Jmol⁻¹K⁻¹
Standard Entropy of CO = 198Jmol⁻¹K⁻¹
To find the entropy change of the reaction, we first write a balanced reaction equation:
Fe₂O₃ + 3C → 2Fe + 3CO
To calculate the entropy change of the reaction we simply use the equation below:
ΔS = ∑S
- ∑S
Therefore:
ΔS = [(2x27.2) + (3x198)] - [(90) + (3x5.7)] = 648.4 - 107.1
ΔS = +541.3Jmol⁻¹K⁻¹
Answer: pH = 2,897 , basic![[H+][OH-] = 10^{-14} ==> [H+] = \frac{10^{-14}}{7,89*10^{-12} } =\frac{1}{789} \\pH= -lg([H+]) = 2,897 \\pH basic](https://tex.z-dn.net/?f=%5BH%2B%5D%5BOH-%5D%20%3D%2010%5E%7B-14%7D%20%3D%3D%3E%20%5BH%2B%5D%20%3D%20%5Cfrac%7B10%5E%7B-14%7D%7D%7B7%2C89%2A10%5E%7B-12%7D%20%7D%20%3D%5Cfrac%7B1%7D%7B789%7D%20%5C%5CpH%3D%20-lg%28%5BH%2B%5D%29%20%3D%202%2C897%20%5C%5CpH%3C7%20%3D%3D%3E%20basic)
Explanation: