if you did not heat hydrated copper (ii) sulfate long enough and it was not completely converted to the anhydrous product, how would that affect the percent of water that you calculate in the hydrate? would your value be too large, too small, or not affected? explain your answer
1
Expert's answer
2020-02-25T04:50:40-0500
The percentage of water in the hydrate is lowered because some water is still present in the assumed anhydrous copper (ii) sulfate, hence the value will be too small.
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