Explanation:
The given data is as follows.
Current (I) = 3.50 amp, Mass deposited = 100.0 g
Molar mass of Cr = 52 g
It is known that 1 faraday of electricity will deposit 1 mole of chromium. As 1 faraday means 96500 C and 1 mole of Cr means 52 g.
Therefore, 100 g of Cr will be deposited by "z" grams of electricity.
z = 
= 185576.9 C
As we know that, Q = I × t
Hence, putting the given values into the above equation as follows.
Q = I × t
185576.9 C = 
t = 53021.9 sec
Thus, we can conclude that 100 g of Cr will be deposited in 53021.9 sec.
Answer:
16
Explanation:
The proton number is always the number in the corner of the box on a periodic table
Answer: 
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
The balanced chemical equation for reaction of potassium superoxide with carbon dioxide to produce oxygen and potassium carbonate will be:
Answer : The chemical symbol for the element that results from this process is, (Ar) for argon.
Explanation :
Electron capture : In this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. Simply, a proton and an electron combines together to form a neutron. Mass number does not change in this process.
The equation for the given reaction is,
Thus, the chemical symbol for the element that results from this process is, argon (Ar).
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
