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How many molecules of water of hydration are present in 39.2 mg of Mohr's salt? |
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Answer» ass = 392 G/mol(39.2 mg) X (1 g/1000 mg) x (1MOL / 392 g) x (6.022 x 1023 formula units / mole) x (6 molecules water / Formula UNIT) = 3.6132 x 1020 molecules.hope it HELPS u ❤️❤️ |
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