Answer:
[tex][A]=0.012M[/tex]
[tex][B]=0.008M[/tex]
Explanation:
Hello,
In this case, given the concentration and volume of each reactant, we first compute their moles in the solution:
[tex]n_A=0.003L*0.04mol/L=1.2x10^{-4}mol\\\\n_B=0.004L*0.02mol/L=8x10^{-5}mol[/tex]
Thus, since the total volume is computed given the volume of each reactant as well as the extra water, we have:
[tex]V=0.003L+0.004L+0.003L=0.010L[/tex]
Thus, the concentrations of A and B are:
[tex][A]=\frac{1.2x10^{-4}mol}{0.010L}=0.012M[/tex]
[tex][B]=\frac{8x10^{-5}mol}{0.010L}=0.008M[/tex]
Best regards.
