"n(Cu(NO3)2)=\\dfrac{6.68g}{187.57g\/mol}=0.0356moles"
"V=\\dfrac{n}{c}=\\dfrac{0.0356moles}{0.52mol\/L}=0.06846L=68.46mL"
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment