The reaction between the reactants would be:
CH₃NH₂ + HCl ↔ CH₃NH₃⁺ + Cl⁻
Let the conjugate acid undergo hydrolysis. Then, apply the ICE approach.
CH₃NH₃⁺ + H₂O → H₃O⁺ + CH₃NH₂
I 0.11 0 0
C -x +x +x
E 0.11 - x x x
Ka = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
Since the given information is Kb, let's find Ka in terms of Kb.
Ka = Kw/Kb, where Kw = 10⁻¹⁴
So,
Ka = 10⁻¹⁴/5×10⁻⁴ = 2×10⁻¹¹ = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
2×10⁻¹¹ = [x][x]/[0.11-x]
Solving for x,
x = 1.483×10⁻⁶ = [H₃O⁺]
Since pH = -log[H₃O⁺],
pH = -log(1.483×10⁻⁶)
<em>pH = 5.83</em>
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>
The pressure is 19.3 N/cm².
<em>p</em> = (28.0 lb/1 in²) × (4.45 N/1 lb) × (1 in²/6.45 cm²) = 19.3 N/cm²
Answer:
48
Explanation:
because you add 6 and 6 and 12 to get it
Answer:
% composition O = 19.9%
% composition Cu = 80.1%
Explanation:
Given data:
Total mass of compound = 3.12 g
Mass of copper = 2.50 g
Mass of oxygen = 3.12 - 2.50 = 0.62 g
% composition = ?
Solution:
Formula:
<em>% composition = ( mass of element/ total mass)×100</em>
% composition Cu = (2.50 g / 3.12 g)×100
% composition Cu = 0.80 ×100
% composition Cu = 80.1%
For oxygen:
<em>% composition = ( mass of element/ total mass)×100</em>
% composition O = (0.62 g / 3.12 g)×100
% composition O = 0.199 ×100
% composition O = 19.9%
