A nuclear explosion may release tremendous amounts of energy in the form of noise, heat, visible light, radiation and an atmosph
2 answers:
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Answer:
1 m
Explanation:
L = 100 m
A = 1 mm^2 = 1 x 10^-6 m^2
Y = 1 x 10^11 N/m^2
F = 1000 N
Let the cable stretch be ΔL.
By the formula of Young's modulus
ΔL = 1 m
Thus, the cable stretches by 1 m.
Missing data in the text of the exercise: The molar concentration of Zinc is 10 times the molar concentration of copper.
Solution:
1) First of all, let's calculate the standard electrode potential difference at standard temperature. This is given by:
where
is the standard potential at the cathode, while 
is the standard potential at the anode. For a Daniel Cell, at the cathode we have copper: 
, while at the anode we have zinc: 
. Therefore, at standard temperature the electrode potential difference of the Daniel Cell is
2) To calculate
at any temperature T, we should use Nerst equation:
![E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]}](https://tex.z-dn.net/?f=E%5E0%28T%29%3DE%5E0-%20%5Cfrac%7BR%20T%7D%7Bz%20F%7D%20%5Cln%20%20%5Cfrac%7B%5BZn%5D%7D%7B%5BCu%5D%7D%20%20)
where
is the temperature in our problem
is the number of electrons transferred in the cell's reaction
is the Faraday's constant
![[Zn]](https://tex.z-dn.net/?f=%5BZn%5D)
and ![[Cu]](https://tex.z-dn.net/?f=%5BCu%5D)
are the molar concentrations of zinc and in copper, and in our problem we have ![[Zn]=10[Cu]](https://tex.z-dn.net/?f=%5BZn%5D%3D10%5BCu%5D)
.
Using all these data inside the equation, and using
, in the end we find:
![E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]}=+1.053 V](https://tex.z-dn.net/?f=E%5E0%28T%29%3DE%5E0-%20%5Cfrac%7BR%20T%7D%7Bz%20F%7D%20%5Cln%20%5Cfrac%7B%5BZn%5D%7D%7B%5BCu%5D%7D%3D%2B1.053%20V)
Solution:
1) First of all, let's calculate the standard electrode potential difference at standard temperature. This is given by:
where
2) To calculate
where
Using all these data inside the equation, and using