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A sample of CuSO4*5H2O was heated to 110 degrees C, where it lost water and gave another hydrate of copper (II) ion that
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This is as far as I have got in the past few hours: CuSo4 x 5H2O --> CuSO4 x ? H2O CuSO4 x ? H2O + Ba (NO2)2 --> BaSO4 116.66 mg BaSO4 98.77 mg CuSO4 x ? H2O At some point once I can get the % of the CuSO4 x ? H2O I can turn it into an empirical formula
