HNO₃ + H₂S → S + NO + H₂<span>O
Assign Oxidation Number:
L.H.S R.H.S
N in HNO</span>₃ = +5 +2 = N in NO
S in H₂S = -2 0 = S in S
Write Half cell Reactions:
Reduction Reaction:
3e⁻ + HNO₃ → NO -------(1)
Oxidation Reaction:
H₂S → S + 2e⁻ -------(2)
Multiply eq. 1 with 2 and eq. 2 with 3 to balance electrons.
6e⁻ + 2 HNO₃ → 2 NO
3 H₂S → 3 S + 6e⁻
Cancel e⁻s,
______________________________
2 HNO₃ + 3 H₂S → 2 NO + 3 S + H₂O
Balance Oxygen Atoms by multiplying H₂O with 4, Hydrogen atoms will automatically get balance.
2 HNO₃ + 3 H₂S → 2 NO + 3 S + 4H₂O
<span> The Reactants are Zinc (Zn) and Sulfur (S).
The Product is Zinc Sulfide (ZnS).
All of them are solids.
The combined masses of the reactants must be 14 grams, too. Later in
Chemistry you'll learn that's not really true, but it is for now.
Hope This Helps:)
</span>
Answer:
Groups 14, 15, and 16 have 2,3, and 4 electrons in the p sublevel (p sublevel has 3 "spaces" AKA orbitals), because Hunds says one in each orbital before doubling up if you had 2 electrons, group 14, they would both be in the first orbital, with 3 electrons, group 15, two in the first orbital one in the 2nd none in the 3rd. With 4 electrons, group 16, then you would have 2 in the first 2 orbitals and NONE in the 3rd.
Explanation:
If you are in group 13 you only have 1 electron so it can only be in one orbital. with group 17, you have 5 electrons, so 2 in the first 2 in the second and 1 in the 3rd, correct for Hunds rule anyway. Noble gasses, group 18, have 6 elecctrons, so every orbital is full any way you look at it.
