Answer:
Keq = 5.33*10²⁶
Explanation:
Based on the standard reduction potential table:
E°(Fe2+/Fe) = -0.45 V
E°(Cu2+/Cu) = +0.34 V
Since the reduction potential of copper is greater than iron, the former acts as the cathode and the latter as anode.
The half reactions are:
Cathode (Reduction): 
Anode (Oxidation):
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Overall reaction: 
The Gibbs free energy change at 25 C is related to the standard emf (E°) of the cell as well as the equilibrium constant K as:
here:
R = 8.314 J/mol-K
T = 25 C = 25+273 = 298 K
n = number of electrons involved = 2
F = 96500 Coulomb/mol e-
Keq = 5.33*10²⁶
Answer:
( About ) 110 kilograms
Explanation:
Take a look at the attachment below,
The answer is no because producers need the energy from the sun in order to produce. Hope this helped!
Answer:
the correct answer would be D Iron (III) oxide
Explanation:
hope this helps
Answer:
3.93 mol
Explanation:
The balanced equation is given as;
2 Fe₂S₃+ 9 O₂ --> 2 Fe₂O₃ + 6 SO₂
From the reaction;
2 mol of Fe₂S₃ reacts with 9 mol of O₂ to form 6 mol of 6 SO₂
This means;
1 mol of Fe₂S₃ requires 9/2 mol of O₂
Also,
1 mol of O₂ requires 1/9 mol of Fe₂S₃
In the question, we have;
1.31 moles of Fe2S3 and 22.8 moles of O2
The limiting reactant which determine how much of the product formed is; Fe₂S₃ because O₂ is in excess.
The relationship between Fe₂S₃ and SO₂ is;
2 mol Fe₂S₃ produces 6 mol of SO₂
1.31 mol of Fe₂S₃ would produce x mol ?
2 = 6
1.31 = x
x = 6 * 1.31 / 2 = 3.93 mol
