The half-reaction are:
Cd ---> Cd(OH)₂
The oxidation number of Cd changed from 0 to +2. So, the number of mol electron transferred here is 2.
NiO(OH) --> Ni(OH)₂
The oxidation number of Cd changed from +3 to +2. So, the number of mol electron transferred here is 1.
Now, the greatest common factor would be 2. So, we use n=2 for the formula for ΔG°. F is Faraday's constant equal to 96,485 J/mol e.
ΔG° = nFE° = (2)(96,485)(1.5) =<em> 289,455 J</em>
Answer:
41.67 mol
Explanation:
1 Litre of water = 1000g
Mole = mass / molar mass
Mass of 1 L of water = 1000 g
Molar mass of water (H2O) :
(H = 1, O = 16)
H2O = (1 * 2) + 16 = (2 + 16) = 18g/mol
Amount of water consumed = (3/4) of 1 litre
= (3/4) * 1000g
= 750g
Therefore mass of water consumed = 750g
Mole = 750g / 18g/mol
Mole of water consumed = 41.6666
= 41.67 mol
Hey There!
At neutralisation moles of H⁺ from HCl = moles of OH⁻ from Ca(OH)2 so :
0.204 * 42.8 / 1000 => 0.0087312 moles
Moles of Ca(OH)2 :
2 HCl + Ca(OH)2 = CaCl2 + 2 H2O
0.0087312 / 2 => 0.0043656 moles ( since each Ca(OH)2 ives 2 OH⁻ ions )
Therefore:
Molar mass Ca(OH)2 = 74.1 g/mol
mass = moles of Ca(OH)2 * molar mass
mass = 0.0043656 * 74.1
mass = 0.32 g of Ca(OH)2
Hope that helps!
