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 2 years ago
Can you find what is being oxidized and reduced (including their oxidation numbers), the oxidizing agent, and reducing agent for CH4 + 2O2 CO2 + 2H2O.
 2 years ago
Can you find what is being oxidized and reduced (including their oxidation numbers), the oxidizing agent, and reducing agent for CH4 + 2O2 CO2 + 2H2O.

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Theo177
 2 years ago
Best ResponseYou've already chosen the best response.0The equation has the correct numbers, you dont need to change anyghing. CH4 is oxidized. O2 is the oxidised agent CO2 is the reducing agent I think.

forever_confused
 2 years ago
Best ResponseYou've already chosen the best response.0So nothing has oxidation numbers?

forever_confused
 2 years ago
Best ResponseYou've already chosen the best response.0CH4 + 2O2 > CO2 + 2H2O sorry if me forgetting the yield sign confused you

Theo177
 2 years ago
Best ResponseYou've already chosen the best response.0Oh wait, I though you were talking about the equation numbers. I am not sure about the axidation numbers.

forever_confused
 2 years ago
Best ResponseYou've already chosen the best response.0Hm, no the oxidation numbers :/ I'm not sure. Like, would I solve the oxidation numbers for each one.. Like CH4 and then 2O2 and so on?

Theo177
 2 years ago
Best ResponseYou've already chosen the best response.0I m not sure about the oxidation numbers either, sry... gtg, I have some math exams right now:p. I think I m gonna fail so hard :P

forever_confused
 2 years ago
Best ResponseYou've already chosen the best response.0its fine. and thanks a lot. Good luck! You seem really smart (: You'll do fine

meera_yadav
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1341216983942:dw

meera_yadav
 2 years ago
Best ResponseYou've already chosen the best response.0hii, if u need any clarifications then let me know !!

meera_yadav
 2 years ago
Best ResponseYou've already chosen the best response.0here , C has oxidation no. = 4 in CH4 (since H has +1 oxidation state), while it's +4 in CO2 ( since O has 2 oxidation state) while O is in zero oxidation state in O2 (as oxidation states of elements in free state is zero) and 2 in H2O
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