Answer:
The standard potential, E cell, for this galvanic cell is 0.5670V
Explanation:
Ni²⁺(aq) + 2e⁻ → Ni(s) E red = - 0.23V ANODE
Cu²⁺(aq) + 2e- → Cu(s) E red = + 0.337V CATHODE
ΔE° = E cathode - E anode
ΔE° = 0.337V - (0.23V) = 0.5670 V
Answer:
90.3 L
Explanation:
Given data:
Volume of water produced = 77.4 L
Volume of oxygen required = ?
Solution:
Chemical equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
It is known that,
1 mole = 22.414 L
There are 7 moles of oxygen = 7×22.414 = 156.9 L
There are 6 moles of water = 6×22.414 = 134.5 L
Now we will compare:
H₂O : O₂
134.5 : 156.9
77.4 : 156.9/134.5×77.4 =90.3 L
So for the production of 77.4 L water 90.3 L oxygen is required.
The balanced equation would be 
<h3>Electrochemical equations</h3>
Zn reacts with Cu solution according to the following equation:
In the reaction, 
is reduced according to the following: 
While Zn is oxidized according to the following: 
Thus, giving the overall equation of;
More oxidation-reduction equations can be found here: brainly.com/question/13699873
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Since we are only asked for the number of moles, we don't need the information of density. The concentration is expressed in terms of 0.135 M AgCl or 0.135 moles of AgCl per liter solution. The solution is as follows:
Moles AgCl = Molarity * Volume
Moles AgCl = 0.135 mol/L * 244 mL * 1 L/1000 mL
<em>Moles AgCl = 0.03294 mol </em>
They contain Carbon, Nitrogen, Hydrogen, Oxygen, and Sulfur
