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Boron(III) fluoride react with water

BF3 + H2O → [B(H2O)F3]

Boron(III) fluoride react with water to produce tetrafluoroaquaboron(III). This reaction takes place in a temperature range 8-18°C.

Picture of reaction: img1 BB html

Сoding to search: BF3 + H2O = BH2OF3

Add / Edited: 26.01.2015 /

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

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