Answer:
2Na(s) + 2HCl(aq) —> 2NaCl(aq) + H2(g)
Explanation:
Na(s) + HCl(aq) —> NaCl(aq) + H2(g)
Writing an ionic equation will actually help us to understand the equation and also to balance it. This is illustrated below:
Na + H+Cl-
Na is higher than H in the activity series and as such, it will displaces H from the solution and form NaCl with H2 liberated as shown below
Na + H+Cl- —> Na+Cl- + H2
Now, put 2 in front of Na, H+Cl- and Na+Cl- to balance the equation as shown below:
2Na + 2H+Cl- —> 2Na+Cl- + H2
Now we can write the elemental equation as follow:
2Na(s) + 2HCl(aq) —> 2NaCl(aq) + H2(g)
Answer:
d. at ground state
Explanation:
Electrons in the Bohr's model of the atom are at the lowest possible energy state when they are their ground state.
The lowest energy state available to an electron is the ground state.
- According to the the Bohr's model of the atom, the extranuclear part is made up of electrons in specific spherical orbits around the nucleus.
- In the excited state, the level is higher than the ground state.
- When electrons changes state they either absorb or gain energy.
Answer: 
Explanation:
The balanced chemical equation will be:

Here Ag undergoes oxidation by loss of electrons, thus act as anode. Nickel undergoes reduction by gain of electrons and thus act as cathode.

Where both
are standard reduction potentials.
![E^0_{[Ag^{+}/Mg]}=+0.80V](https://tex.z-dn.net/?f=E%5E0_%7B%5BAg%5E%7B%2B%7D%2FMg%5D%7D%3D%2B0.80V)
![E^0_{[Ni^{2+}/Ni]}=-0.25V](https://tex.z-dn.net/?f=E%5E0_%7B%5BNi%5E%7B2%2B%7D%2FNi%5D%7D%3D-0.25V)
![E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Ag^{+}/Ag]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BNi%5E%7B2%2B%7D%2FNi%5D%7D-%20E%5E0_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D)

The standard emf of a cell is related to Gibbs free energy by following relation:

= gibbs free energy
n= no of electrons gained or lost =?
F= faraday's constant
= standard emf

The Gibbs free energy is related to equilibrium constant by following relation:

R = gas constant = 8.314 J/Kmol
T = temperature in kelvin =
K = equilibrium constant



Thus the value of the equilibrium constant at
is 
The answer is -1.742 x 10^2 which the answer is C