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The reaction between palladium chloride(II) and sulfur with the formation of sulfide palladium(IV) and detialed

PdCl2 + 4S → PdS2 + S2Cl2


The reaction between palladium chloride(II) and sulfur with the formation of sulfide palladium(IV) and detialed.


Picture of reaction: img1 BB html

Сoding to search: PdCl2 + 4 S = PdS2 + S2Cl2

Add / Edited: 11.12.2019 /

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