where, E^{o} (Ag+/Ag) = std. reduction potential of Ag+ = 0.7994 v
and Sn2+/Sn = std. reduction potential of Sn2+ = -0.14 v
Thus, E^{o}cell = 0.7994v - (-0.14v) = 0.9394 v
Now, ΔG^{o} = -nF
,
where, n = number of electrons = 2
F = Faraday's constant = 96500 C
∴ΔG^{o} = 2 X 96500 X 0.9394 = -1.18 X
Now, using Nernst's Equation we have,
![[tex]E_{cell} = 0.9394 - \frac{2.303X298}{2X96500}log \frac{0.0115}{ 3.5^{2} }](https://tex.z-dn.net/?f=%20%5Btex%5DE_%7Bcell%7D%20%3D%200.9394%20-%20%5Cfrac%7B2.303X298%7D%7B2X96500%7Dlog%20%5Cfrac%7B0.0115%7D%7B%203.5%5E%7B2%7D%20%7D%20)
E_{cell} = 0.9765 v
Finally, ΔG = -nFE = -2 X 96500 X 0.9765 = -1.88 X
Answer: The amount of heat needed is = 4.3kJ
Explanation:
Amount of heat H = M × C × ΔT
M= mass of benzene = 64.7g
C= specific heat capacity = 1.74J/gK
ΔT = T2-T1
Where T1 is initai temperature = 41.9C
T2 is the final temperature( boiling point of benzene) = 80.1C
H= 64.7×1.74×80.7
H= 4300J
H=4.3kJ
Therefore, the amount of heat needed is 4.3kJ
Answer:
See the answer below
Explanation:
<em>Since the experiment is set out to determine the melting point of the white solid, after missing the melting point due to distraction, there are two possible solutions and both involves a repeat of the experiment.</em>
1. The first one is to allow the molten substance to solidify again and then repeat the experiment. This time around, a critical attention should be paid to be able to notice the melting point temperature once the temperature gets to 132 C.
2. The second solution would be discard the molten substance and repeat the experiment with the a new solid one. Similarly, critical attention should be paid once the temperature gets to 132 C since it is sure that the melting point lies within 132 and 138 C.
<span>c. Passing electric charge through the reactants Is the answer to you're question.
</span>
The <span>Anaphase stage
Happy studying!</span>
