First, we write the half equations for the reduction of the chemical species present:
Cu⁺² + 2e → Cu; E° = 0.34 V
Ni⁺² + 2e → Ni; E° = - 0.23 V
In order to determine the potential of the cell, we find the difference between the two values. For this:
E(cell) = 0.34 - (-0.23)
E(cell) = 0.57 V
The second option is correct. (The difference in values is due to different values in literature, and it is negligible)
<span> E = h v where v = frequency h = plancks constant 6.626 * 10 - 34
4.8 * 6.626 = 31.8 </span>
<span>they both had their conclusions based on solid evidence</span>
K5O2
convert grams to moles, divide both by the smallest mole mass, multiply that until hole.
30.5 g K ÷ 39.10 = .78 mol
6.24 g O ÷ 16 = .39 mol
.78 mol ÷ .39 mol = 2.5
.39 mol ÷ .39 mol = 1
2.5 x 2 = 5
1 x 2 = 2
K5O2
Answer:
Answer is Ca2+(aq)+S2-(aq)=>CaS(s)
Explanation:
I hope it's helpful!
