find the oxidation numbers of the following show work

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The P atom in H2PO3- 

The N atom in NH3 

The N atom NO2

 The N atom in N2

The P atom in P2O5 

The S atom in CuSO4 

The K atom in KClO4 
 The Cl atom in KClO2 
The N atom in HNO2 
The C atom in CO2 

The Cl atom in KClO2

The S atom in SF6

The S atom in HSO4- 

May 12th, 2015

The P atom in H2PO3-

Total charge is (-1), Note that always H is +1, and almost in all cases Oxygen is (-2)

then we have: 

H: 2*(+1)=+2

O: 3*(-2)=-6

Total Charge = (-1) = +2 -6 +P; Solving for P

P= + 3

The N atom in NH3

the oxidation numbers of the 1 nitrogen and 3 hydrogen atoms of the neutral NH 3 ammonia molecule sum to 0:

 equation

The N atom NO2

Total charge =0

Oxygen = -2

2*(-2)=-4

then N=+4

The N atom in N2O 

N=+1

The P atom in P2O5

2(P ox #) +5(O ox #) = 0
2(P ox #) + 5(-2) = 0
2(P ox #) -10 = 0
2(P ox #) = +10
P ox # = +5

The S atom in CuSO4

S=+6

The K atom in KClO4

K=+1

 The Cl atom in KClO2

Cl=+3

The N atom in HNO2

N=+3

The C atom in CO2

C=+4 since Oxygen is -2--> 2*(-2)=-4 then C must be +4 because is a neutral compound

The Cl atom in KClO2

Cl=+3

The S atom in SF6

The oxidation number of fluorine in its compounds is –1. Since there are 6 fluorine atoms in SF6, the total contribution of fluorine is –6. Therefore, the oxidation number of S in SF6 is +6, and the oxidation number of F in SF6 is –1.

The S atom in HSO4-

S=+6

May 12th, 2015

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May 12th, 2015
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