Question #af8d7

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Question #af8d7

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Answer 1 · 2017-05-11T13:44:35

$\frac{16}{3}$ $16/3$ .

Explanation:

The formula for tribromooctaoxide:
$B r_{3} O_{8}$ $Br_(3)O_(8)$
Oxygen has an oxidation state of $- 2$ $−2$ when it is present in most compounds. ([https://en.wikipedia.org/wiki/Oxidation_state](https://en.wikipedia.org/wiki/Oxidation_state))
Since the sum of oxidation state number of neutral compound is $0$ $0$ , sum of the oxidation number of bromine should equal to the opposite of oxygen's.
$- (- 2) \cdot \frac{8}{3} = \frac{16}{3}$ $-(-2)*8/3=16/3$

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