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Iridium(IV) fluoride react with ammonium fluoride

IrF4 + 2NH4F → (NH4)2[IrF6]

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Iridium(IV) fluoride react with ammonium fluoride to produce ammonium hexafluoroiridate(IV). This reaction takes place at a temperature near 200°C.

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Сoding to search: IrF4 + 2 NH4F cnd [ temp ] = NH42IrF6

Add / Edited: 01.06.2015 /

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