To find the average oxidation number of bromine in Br\(_3\)O\(_8\), we can assume that oxygen has an oxidation number of \(-2\) (since oxygen is typically \(-2\) in compounds). Let the oxidation number of bromine be \( x \). The compound is Br\(_3\)O\(_8\), so:
There are 3 bromine atoms, each with oxidation number \( x \).
There are 8 oxygen atoms, each with oxidation number \(-2\). The sum of oxidation numbers in the compound must be zero because the compound is neutral. Therefore, we can write the equation: \[ 3x + 8(-2) = 0 \] Simplifying: \[ 3x - 16 = 0 \quad \Rightarrow \quad 3x = 16 \quad \Rightarrow \quad x = \frac{16}{3} \]
The correct option is (A) : \(\frac{16}{3}\)
For the reaction:
\[ 2A + B \rightarrow 2C + D \]
The following kinetic data were obtained for three different experiments performed at the same temperature:
\[ \begin{array}{|c|c|c|c|} \hline \text{Experiment} & [A]_0 \, (\text{M}) & [B]_0 \, (\text{M}) & \text{Initial rate} \, (\text{M/s}) \\ \hline I & 0.10 & 0.10 & 0.10 \\ II & 0.20 & 0.10 & 0.40 \\ III & 0.20 & 0.20 & 0.40 \\ \hline \end{array} \]
The total order and order in [B] for the reaction are respectively: