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Boron(III) iodide react with sodium hydroxide

BI3 + 4NaOH → Na[B(OH)4] + 3NaI




Boron(III) iodide react with sodium hydroxide to produce sodium tetrahydroxoborate(III) and sodium iodide. Sodium hydroxide - concentrated solution.


Picture of reaction: img1 BB html

Сoding to search: BI3 + 4 NaOH = NaBOH4 + 3 NaI

Add / Edited: 02.02.2015 /

Evaluation of information: 5.0 out of 5 / number of votes: 1


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