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9 December, 14:42

If you start with 10ml of 0.75 m cu (no3) 2 how much cu (s) in grams should be recovered in step #7

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  1. 9 December, 16:13
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    0.48 grams. Not a well worded question since it's assuming I know the reactions. But I'll assume that since there's just 1 atom of copper per molecule of Cu (NO3) 2, that the reaction will result in 1 atom of copper per molecule of Cu (NO3) 2 used. With that in mind, we will have 0.010 l * 0.75 mol/l = 0.0075 moles of copper produced. To convert the amount in moles, multiply by the atomic weight of copper, which is 63.546 g/mol. So 0.0075 mol * 63.546 g/mol = 0.476595 g. Round the results to 2 significant figures, giving 0.48 grams.
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