6K.1 7th edition
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Katie Frei 1L
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6K.1 7th edition
Could someone explain why for the balanced redox reaction in problem 6K.1) of the 7th edition of the textbook, there ends up being 3 C2H5OH(aq) and 3 C2H4O(aq) as opposed to just the coefficient of 1 for each of these compounds? Thank you!
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Kobe_Wright
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Re: 6K.1 7th edition
In order to balance the waters, and hydrogens, and electrons of both half reactions you'll need to multiply by those coefficients.
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