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
2.2 x 10^-2
0.055 / 250 = 0.00022 - This would be 2.2 x 10^-4, but the question is asking for percent, not proportion, so multiply by 100% to get the percentage.
0.00022 * 100% = 0.022% = 2.2 * 10^-2
We assume that we have Ka= 4.2x10^-13 (missing in the question)
and when we have this equation:
H2PO4 (-) → H+ + HPO4-
and form the Ka equation we can get [H+]:
Ka= [H+] [HPO4-] / [H2PO4] and we have Ka= 4.2x10^-13 & [H2PO4-] = 0.55m
by substitution:
4.2x10^-13 = (z)(z)/ 0.55
z^2 = 2.31x 10^-13
z= 4.81x10^-7
∴[H+] = 4.81x10^-7
when PH equation is:
PH= -㏒[H+]
= -㏒(4.81x10^-7) = 6.32
The number of Ml of a 0.40 %w/v solution of ,nalorphine that must be injected to obtain a dose of 1.5 mg is calculated as below
since M/v% is mass of solute in grams per 100 ml
convert Mg to g
1 g = 1000 mg what about 1.5 mg =? grams
= 1.5 /1000 = 0.0015 grams
volume is therefore = 100 ( mass/ M/v%)
= 100 x( 0.0015/ 0.4) = 0.375 ML
Answer:
Mass of one electron is 9.1 × 10⁻³¹ kg
Mass of one proton is 1.673 × 10⁻²⁷ Kg
Mass of one neutron is 1.675 × 10⁻²⁷ Kg
<u>-TheUnknownScientist</u><u> 72</u>
