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.
Answer: The IUPAC name of compound is
6-Ethyl-2-Octene.
Explanation:First of all draw a straight chain of 8 carbon atoms.
Make a double bond between carbon number 6 and 7 numbering from left.
Add ethyl group at position 3 starting from left.
The structure sketched is attached below,
According to rules the longest chain containing a double bond is selected. Numbering is started from the end to which double bond is nearer. So, in our case the double bond starts at carbon 2, hence parent name of compound is 2-Octene. Then the substituent at position 6 is named.
I believe the answer is <span>0.035 N. Hope this helps. (ノ◕ヮ◕)ノ*:・゚✧ </span>
An indicator shows where the endpoint of a titration is. Different indicators will change colors at different equivalence points. Phenolphthalein will change colors at around pH7.5 or so, and others like Methylene Blue will change around a pH of 6. Different titrations will have different pH's at different equivalence points.