Chemistry, 18.11.2020 16:40 rose200215
If the concentration of the stock (provided) Cu(NH3)42 was 0.041 M, what concentration will the Cu2 be in beaker?
Answers: 3
![[Cu^{2+}]=0.041 M](/tpl/images/0908/5461/0056f.png)
![[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}](/tpl/images/0908/5461/e82ed.png)
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If the concentration of the stock (provided) Cu(NH3)42 was 0.041 M, what concentration will the Cu2...
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