Answer:
it is possible to remove 99.99% Cu2 by converting it to Cu(s)
Explanation:
So, from the question/problem above we are given the following ionic or REDOX equations of reactions;
Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V
Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V
In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:
Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.
Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.
Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.
Thus, ΔG° = - 92640.
This is less than zero[0]. Therefore, it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.
First, make sure to balance your equation.
3H2(g) + N2(g) ⇄ 2NH3(g)
Now, you can write your Kc expression. Remember that Kc is products over reactants, and the exponent for each product or reactant is based on its coefficient.
Kc = [NH3]^2 / [H2]^3[N2]
The mass of carbon needed is 56g to get it to stop
to be honest I'm trying to give an educated guess hope I've helped
D. I, II, and IV. Basically everything but the temperature.
