Answer : The correct option is, +91 kJ/mole
Solution :
The balanced cell reaction will be,

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.
First we have to calculate the standard electrode potential of the cell.
![E^0_{[Pb^{2+}/Pb]}=-0.13V](https://tex.z-dn.net/?f=E%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D%3D-0.13V)
![E^0_{[Cu^{2+}/Cu]}=+0.34V](https://tex.z-dn.net/?f=E%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D%3D%2B0.34V)

![E^0_{cell}=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cu^{2+}/Cu]}](https://tex.z-dn.net/?f=E%5E0_%7Bcell%7D%3DE%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D-E%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D)

Now we have to calculate the standard Gibbs free energy.
Formula used :

where,
= standard Gibbs free energy = ?
n = number of electrons = 2
F = Faraday constant = 96500 C/mole
= standard e.m.f of cell = -0.47 V
Now put all the given values in this formula, we get the Gibbs free energy.

Therefore, the standard Gibbs free energy is +91 kJ/mole
Through hemoglobin transport in blood, material exchange is carried out on the alveolar membrane.
I think the answer is 7mm but I'm not sure.
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All of them are soluble salt.
First one dissociates into two ions.
The second one dissociates into 3 ions.
The third dissociate into 4 ions. therefore, Al(NO3)3
Answer:
Igneous=B
Sedimentary=C
Metamorphic=A
Weathering is when=B
If sedimentary rock.....=C