Sodium metal forms at the cathode
Answer:
sp3d
Explanation:
The ground state electronic configuration of tin is written as; [Kr] 5s²4d¹⁰5p². Hybridization is a concept used to explain the combination of orbitals of appropriate energy to produce suitable orbitals that could be used for bonding.
In forming the compound Snf5^ -1, we have to hybridize the following orbitals on tin; 5p, 5d and 6s orbitals. This gives us a set of sp3d hybrid hence the answer.
Answer:
V = 2.32 Liters
Explanation:
PV = nRT => V = nRT/P
n = 25.8g/122g/mole = 0.21 mole
R = 0.08206 L·atm/mol·K
T = 25.44°C + 273 = 298.44K
P = 2.22 atm (given in problem)
V = (0.21mol)(0.08206 L·atm/mol·K)(298.44K)/(2.22atm) = 2.32 Liters at 25.44°C & 2.22atm
Answer:
The answer is 465.6 mg of MgI₂ to be added.
Explanation:
We find the mole of ion I⁻ in the final solution
C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol
But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.
So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.
Hence, the weight of MgI₂ must be added is
Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg
Answer:
613 mg
Explanation:
Number of fargday's 
Here, I = 9.20 A
t = 10.5 min
= 10.5 x 60 seconds
So, 
= 0.0208 F
Here, 2e, 2F
2F = 1 mol of Ni
1 mol = 59 gm of Ni
0.0104 mol = 59 x0.0104 gm Ni
= 0.613 gm Ni
= (0.613 x 1000 ) mg of Ni
= 613 mg of Ni
