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How many grams of `Cu(NO_(3))_(2)` would you need to take to get `1.00 g` of copper? `Cu = 63.5, N = 14, O = 16`.

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How many grams of `Cu(NO_(3))_(2)` would you need to take to get `1.00 g` of copper? `Cu = 63.5, N = 14, O = 16`.
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`Mw` of `Cu(NO_(3))_(2) = 63.5 + 2(14 + 16 xx 3) = 187.5 g`
`Cu(NO_(3))_(2) rarr Ci`
`1 mol` `1 mol`
`= 187.5 g` `= 63.5 g`
`63.5 g of Cu` is obtained from `187.5 g of Cu(NO_(3))_(2)`
`1 g of Cu` is obtained from `= (187.5)/(63.5) = 2.9528 g`
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