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
Answer:
(1) the surface area of the solute,
(2) the temperature of the solvent,
(3) the amount of agitation that occurs when the solute and the solvent are mixed.
Explanation:
Answer:
Explanation:
1=i
4=d
5=m
6=c
8=o
9=j
10=a
13=b
14=h
15=l
Theses are the ones i knew
Answer:
Li2S> Na2S> K2S> CsS
Explanation:
The lattice energy of ionic species depends on the relative sizes of ions in the ionic compounds. As the size of ions increases, the lattice energy decreases and vice versa.
When the size of the anions are the same, the lattice energy now depends on the relative sizes of the cations. Therefore, since all the compounds are sulphides and the order of magnitude of ionic sizes is: Li^+ < Na^+ < K^+ < Cs^+.
Therefore, the order of decrease in lattice energy is; Li2S> Na2S> K2S> CsS
