化学例

$I B r$

ステップ 1

$I B r$ is soluble because most bromides are soluble except $A g B r$ , $H g B r_{2}$ , $H g_{2} B r_{2}$ , and $P b B r_{2}$ .

$I B r$ は可溶性です

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$