Answer: Thus the cell potential of an electrochemical cell is +0.28 V
Explanation:
The calculation of cell potential is done by :

Where both
are standard reduction potentials.
![E^0_{[Fe^{2+}/Fe]}= -0.41V](https://tex.z-dn.net/?f=E%5E0_%7B%5BFe%5E%7B2%2B%7D%2FFe%5D%7D%3D%20-0.41V)
![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)
As Reduction takes place easily if the standard reduction potential is higher(positive) and oxidation takes place easily if the standard reduction potential is less(more negative). Thus iron acts as anode and lead acts as cathode.
![E^0=E^0_{[Pb^{2+}/Pb]}- E^0_{[Fe^{2+}/Fe]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D-%20E%5E0_%7B%5BFe%5E%7B2%2B%7D%2FFe%5D%7D)

Thus the cell potential of an electrochemical cell is +0.28 V
There are 6.022 × 10²³ atoms in 39.948 g of argon and 4.0026 g of helium.
Explanation:
39.945 g/mole is the molar mass of argon so 39.948 g of argon are equal to 1 mole of argon.
4.0026 g/mole is the molar mass of helium so 4.0026 g of helium are equal to 1 mole of helium.
We know that Avogadro's number tell us the number of particles in 1 mole of substance which is 6.022 × 10²³.
So in 39.948 g of argon and 4.0026 g of helium contains the same number of atoms, 6.022 × 10²³.
Learn more about:
Avogadro's number
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0.24529601530251 moles of neon