The equation Eºcell = 0.0592/n logK must be used to find n and also Eºcell
2 Al(s) + 3 Mg2+(aq) → 2 Al3+(aq) + 3 Mg(s) Al3+ +3e- --> Al Eº = -1.66 V Mg2+ +2e- -->Mg Eº = -2.37V
To balance the equation, 6 moles of electrons must be transferred (2 Al and 3 Mg). This will be the value of n in the equation.
To find Eºcell, you need the reduction potentials which should be given in a table, and given above. Eºcell = -1.66 - (-2.37) = 0.71 V log K = Eºcell x n/0.0592 = 0.71 x 6/0.0592 log K = 71.95 K = 10^71.95 K = 1.1x10^72
Two radius of an atom is equal to the diameter. Adding up all the diameter of the atoms, it should be equal to 9.5 mm. Therefore, we simply convert the units to the same units then divide 1.35 A to 9.5 mm. We calculate as follows:
no. of atoms = 0.0095 m / 1.35x10^-9 m = 7037037 atoms
Hope this answers the question. Have a nice day.
C6H6(l) + 6 Cl2(g) = C6Cl6(s) + 6 HCl(g)
Number of moles : n₂ = 1.775 moles
<h3>Further explanation</h3>
Given
Moles = n₁ = 1.4
Volume = V₁=22.4 L
V₂=28.4 L
Required
Moles-n₂
Solution
Avogadro's hypothesis, at the same temperature and pressure, the ratio of gas volume will be equal to the ratio of gas moles
The ratio of gas volume will be equal to the ratio of gas moles
Input the values :
n₂ = (V₂ x n₁)/V₁
n₂ = (28.4 x 1.4)/22.4
n₂ = 1.775 moles
