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Boron(III) fluoride react with water

4BF3 + 3H2O → 3H[BF4] + H3BO3




Boron(III) fluoride react with water to produce tetrafluoroborate(III) hydrogen and orthoboric acid. This reaction takes place at a temperature of 20-80°C.


Picture of reaction: img1 BB html

Сoding to search: 4 BF3 + 3 H2O cnd [ temp ] = 3 HBF4 + H3BO3

Add / Edited: 26.01.2015 /

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


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