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7 April, 02:11

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

+3
Answers (1)
  1. 7 April, 04:24
    0
    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

    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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