whats the correct answer please = 06/21/16 are osidized or reduced, in anry )-2H\"(aq ) +2NO? (aq) + NO(g) 2nd attempt Part 1 (0.9 point) 3 Feedback l See Periodic Table 0 See Hnt N O o O 044> Solution 1) Oxidation number of O = -2 lets the oxidation number of N be x use: 2* oxidation number (O) + 1* oxidation number (N) = net charge 2*(-2)+1* x = 0 -4 + 1 * x = 0 x = 4 So oxidation number of N = +4 Oxidation number of each element in NO2 is O=-2 N=+4 2) Oxidation number of H = +1 lets the oxidation number of O be x use: 2* oxidation number (H) + 1* oxidation number (O) = net charge 2*(+1)+1* x = 0 2 + 1 * x = 0 x = -2 So oxidation number of O = -2 Oxidation number of each element in H2O is H=+1 O=-2 3) Oxidation number of each element in H+1 is H=+1 4) Oxidation number of O = -2 lets the oxidation number of N be x use: 3* oxidation number (O) + 1* oxidation number (N) = net charge 3*(-2)+1* x = -1 -6 + 1 * x = -1 x = 5 So oxidation number of N = +5 Oxidation number of each element in NO3-1 is O=-2 N=+5 5) Oxidation number of O = -2 lets the oxidation number of N be x use: 1* oxidation number (O) + 1* oxidation number (N) = net charge 1*(-2)+1* x = 0 -2 + 1 * x = 0 x = 2 So oxidation number of N = +2 Oxidation number of each element in NO is O=-2 N=+2 .

1. whats the correct answer please
= 06/21/16 are osidized or reduced, in anry )-2H"(aq ) +2NO? (aq) + NO(g) 2nd attempt Part 1
(0.9 point) 3 Feedback l See Periodic Table 0 See Hnt N O o O 044>
Solution
1)
Oxidation number of O = -2
lets the oxidation number of N be x
use:
2* oxidation number (O) + 1* oxidation number (N) = net charge
2*(-2)+1* x = 0
-4 + 1 * x = 0
x = 4
So oxidation number of N = +4
Oxidation number of each element in NO2 is
O=-2
N=+4
2)
Oxidation number of H = +1
lets the oxidation number of O be x

2. use:
2* oxidation number (H) + 1* oxidation number (O) = net charge
2*(+1)+1* x = 0
2 + 1 * x = 0
x = -2
So oxidation number of O = -2
Oxidation number of each element in H2O is
H=+1
O=-2
3)
Oxidation number of each element in H+1 is
H=+1
4)
Oxidation number of O = -2
lets the oxidation number of N be x
use:
3* oxidation number (O) + 1* oxidation number (N) = net charge
3*(-2)+1* x = -1
-6 + 1 * x = -1
x = 5
So oxidation number of N = +5
Oxidation number of each element in NO3-1 is
O=-2

3. N=+5
5)
Oxidation number of O = -2
lets the oxidation number of N be x
use:
1* oxidation number (O) + 1* oxidation number (N) = net charge
1*(-2)+1* x = 0
-2 + 1 * x = 0
x = 2
So oxidation number of N = +2
Oxidation number of each element in NO is
O=-2
N=+2