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Boron(III) bromide react with sodium hydroxide

4BBr3 + 14NaOH → Na2B4O7 + 12NaBr + 7H2O




Boron(III) bromide react with sodium hydroxide to produce sodium tetraborate, sodium bromide and water. Sodium hydroxide - diluted solution.


Picture of reaction: img1 BB html

Сoding to search: 4 BBr3 + 14 NaOH = Na2B4O7 + 12 NaBr + 7 H2O

Add / Edited: 25.01.2015 /

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


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