When the equation zns+o zno+so isbalanced the coefficientof o is
1, 5, 3,or2

ZnS + O --> ZnO + SO

Okay so first you have to count up the number of elements on each side of the equation. Your objective is to have the same number of each element on both sides.

Left Side:
Zn - 1
S - 1
O - 1

Right Side:
Zn - 1
S - 1
O - 2

Since there are two oxygens on the right side, you have to add a coefficient of 2 to the oxygen on the left side. The coefficient tells us that that element or molecule is being multiplied by the value of coefficient. Since we're adding a coefficient of 2 to the oxygen on the left side, there are now 2 oxygens on that side. Because that is the same amount of oxygen as on the right, the equation is now balanced.

Your final equation should look like this: ZnS + 2O --> ZnO + SO

In conclusion, the answer is 2.

Answer is to avoid experimental error

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.