0.042 moles of Hydrogen evolved
<h3>Further explanation</h3>
Given
I = 1.5 A
t = 1.5 hr = 5400 s
Required
Number of Hydrogen evolved
Solution
Electrolysis of water ⇒ decomposition reaction of water into Oxygen and Hydrogen gas.
Cathode(reduction-negative pole) : 2H₂O(l)+2e⁻ ⇒ H₂(g)+2OH⁻(aq)
Anode(oxidation-positive pole) : 2H₂O(l)⇒O₂(g)+4H⁻(aq)+4e⁻
Total reaction : 2H₂O(l)⇒2H₂(g)+O₂(g)
So at the cathode H₂ gas is produced
Faraday : 1 mole of electrons (e⁻) contains a charge of 96,500 C
Q = i.t
Q = 1.5 x 5400
Q = 8100 C
mol e⁻ = 8100 : 96500 = 0.084
From equation at cathode , mol ratio e⁻ : H₂ = 2 : 1, so mol H₂ = 0.042
Answer:
1. pH = 1.23.
2. 
Explanation:
Hello!
1. In this case, for the ionization of H2C2O4, we can write:
It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:
![pH=pKa+log(\frac{[base]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
Whereas the pKa is:
The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:
2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:
It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:
Which is also shown in net ionic notation.
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Answer:
C. sorry if I'm wrong ...................
Mass CoCl2 = 10.27 g
moles CoCl2 = 10.27 g/ 129.839 g/mol=0.07910
mass water = 17.40 - 10.27=7.13 g
moles water = 7.13 / 18.02 g/mol=0.396
0.396/ 0.07910=5
CoCl2 * 5 H2O
moles CaF2 = 85.8 g/ 78.0748 g/mol=1.10
moles Ca = 1.10
mass Ca = 1.10 x 40.078 g/mol=44.1 g
V = 44.1 / 1.55 =28.5 mL
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