Make sure have same amounts of species on both sides
Cu (s) + 2 AgNO3 (aq) -> Cu(NO3)2 (aq) + 2 Ag (s)
Answer:
5 g / ml
Explanation:
Convert the values given to g and ml
12.5 kg = 12500 g
2.5 L = 2500 ml
12500 g / 2500 ml = 5 g/ml
Answer:
see explaination
Explanation:
We are given the (R)-3-bromo-2,3-dimethylpentane and asking to draw the curved arrow which is the showing the mechanism for first-order substitution and first-order elimination reactions. We know the formation of carbocation is the rate determining step in the first-order substitution and first-order elimination reactions.
So in the (R)-3-bromo-2,3-dimethylpentane there is –Br gets removed and formed the tertiary carbocation which is more stable, so the curved arrows in Box 1 to depict the flow of electrons and intermediate in Box 2.
Check attachment
To find the neutrons you add the mass number and the atomic number for neutrons
Hope this helps
-Zayn Malik
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
