The oxidation numbers of nitrogen in NH3, HNO3, and NO2 are, respectively: -3, -5, +4 +3, +5, +4 -3, +5, -4 -3, +5, +4
Evgesh-ka [11]
In NH3 , let oxidation number of N be x
x + (+1)3 = 0
x = -3
In HNO3 , let oxidation number of N be x
1 + x + (-2)3 = 0
x = +5
In NO2 , let oxidation number of N be x
x + (-2)2 = 0
x = +4
Answer:
The molarity of the acid HX is 6.0 M.
Explanation:
We determine the amount of moles of KOH used to neutralize the acid:
=0.12 moles KOH
Then, we calculate the amount of moles of acid:
0.12 moles KOH×
=0.12 moles HX
The molarity of HX is:
=6.0 M
Br2 == 2Br
24% dissociated => n total moles, 0.24 mol*n of Br, and 0.76*n mol of Br2
=> partial pressure of Br, P Br = 0.24 bar, and
partical pressure of Br2, P Br2 = 0.76 bar
kp = (P Br)^2 / P Br2 = (0.24)^2 / 0.76 = 0.0758
