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Check all that apply.

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

-4 for C in CH4

+4 for C in CO2

-2 for O in all substances

+1 for H in both CH4 and H2O

+4 for O in H2O

1 Answer

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Answer:

4 for C in CH4

+4 for C in CO2

-2 for O in all substances

+1 for H in both CH4 and H2O

option 1,2,3 and 4 are correct. Option 5 is not correct

Step-by-step explanation:

Step 1: Data given

Oxidation number of H = +1

Oxidation number of O = -2

Step 2: The balanced equation

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Step 3: The oxidation numbers

-4 for C in CH4

⇒ Oxidation number of H = +1

⇒ 4x +1 = +4

C has an oxidation number of -4

This is correct.

+4 for C in CO2

⇒ Oxidation number of O = -2

⇒ 2x -2 = -4

C has an oxidation number of +4

This is correct.

-2 for O in all substances

⇒ this is correct, the oxidation number of O is always -2 (except in H2O2 and Na2O2)

This is correct.

+1 for H in both CH4 and H2O

⇒ this is correct, the oxidation number of H is always +1 (except in metal hydrides).

This is correct.

+4 for O in H2O

⇒ Oxidation number of H = +1

⇒ 2x +1 = +2

The oxidation number of O is -2

This is not correct

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