It 1
cause yea it is im doing it in class and the answers is 1
Answer:
During the initial cell operation, each reaction is thermodynamically favorable, but the larger operating potential of the lithium-iodine cell indicates that its cell reaction is more thermodynamically favorable. ( B )
During the initial cell operation, the oxidation of iodine is thermodynamically favorable but the oxidation of mercury is not. ( C )
Explanation:
<u>The major Differences between The Zinc mercury cell and Lithium-iodine cell are :</u>
During the initial cell operation, each reaction is thermodynamically favorable, but the larger operating potential of the lithium-iodine cell indicates that its cell reaction is more thermodynamically favorable. and
During the initial cell operation, the oxidation of iodine is thermodynamically favorable but the oxidation of mercury is not.
Given the relationship below,
Δ G = -nFE
E = emf of cell , G = free energy.
This relationship shows that if E is positive the reaction will be thermodynamically favorable also if E is large it will increase the negativity of free energy also From the question we can see that with the reduction of mercury the value of E is more positive and this shows that Mercury is thermodynamically unfavorable
I believe your answer should be C because the heat would move from the soup to the spoon.
<span>CH3-N=O, [with one more O attached to N] <=====> CH2=N-O- + H+ ; this negative charge on O can be with the other O too, thus it gets stabilises itself.This is called the conjugate acid, as it has lost a proton. This is called Nitro- Acinitroso tautomerism .
I hope my answers has come to your help. God bless and have a nice day ahead!</span>
<span>
1. Remember (sum of products) - (sum of reactants)
So ΔHrxn = 2 ΔHf [H2(g)] + ΔHf [Ca(OH)2(s)] - 2 ΔHf [H2O(l)] - ΔHf [Ca(s)]
= 2*0 + -986.09 kJ/mol - 2*(-285.8 kJ/mol) - 0
Do the math and you'll have the answer. BTW the ΔHf [H2(g)] and ΔHf
[Ca(s)] were 0 because these are elements in their standard states.
</span>HOPE THIS HELPS ;)