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How many millilitres of 0.5 M `H_(2)SO_(4)` are needed to dissolve 0.5 g of copper (II) carbonate ?

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`N_(1)V_(1)=N_(2)V_(2)`
`N_(1)= "Normality of " H_(2)SO_(4)=0.5xx2=1N`
`V_(1)= " Vol of " H_(2)SO_(4)`
`N_(2)`= Normality of copper (II) carbonate `=(0.5xx2)/(123.5)N`
`V_(2)`=Volume of copper (II) carbonate =1000 mL
Thus, `1.0xxV_(1)=(0.5xx2)/(123.5)xx1000`
or `V_(1)=8.09 mL`

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