Assume 100 g:

$\mathbf{mol}\mathbf{}\mathbf{Br}\mathbf{=}\mathbf{47}\mathbf{.}\mathbf{85}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{Br}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{Br}}{\mathbf{79}\mathbf{.}\mathbf{904}\mathbf{}\overline{)\mathbf{g}\mathbf{}\mathbf{Br}}}\mathbf{=}$**0.599 mol Br**

An oxybromate compound, KBrO_{x}, where *x *is unknown, is analyzed and found to contain 47.85 % Br.

What is the value of x?

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