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The reaction between iodide zirconium(IV) and ethyl alcohol with the formation of diiodide Zirconia, idtana and water

ZrI4 + 2C2H5OH → ZrOI2 + 2C2H5I + H2O


The reaction between iodide zirconium(IV) and ethyl alcohol with the formation of diiodide Zirconia, idtana and water.


Picture of reaction: img1 BB html

Сoding to search: ZrI4 + 2 C2H5OH = ZrOI2 + 2 C2H5I + H2O

Add / Edited: 12.12.2019 /

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