Answer:
0.7 M
Explanation:
We are given that
[tex]k=6.6\times 10^{-3}s^{-1}[/tex]
Initial concentration =[tex][A_0]=0.8 M[/tex]
Time=14 s
We have to fine the concentration after 14 s.
We know that
[tex]kt=ln\frac{[A_0]}{[A]}[/tex]
Using the formula
[tex]6.6\times 10^{-3}\times 14=ln\frac{0.8}{[A]}[/tex]
[tex]92.4\times 10^{-3}=ln(0.8)-ln[A][/tex]
By using identity ;[tex]ln\frac{m}{n}=ln m-ln n[/tex]
[tex]ln (0.8)-92.4\times 10^{-3}=ln[A][/tex]
[tex][A]=e^{ln0.8-92.4\times 10^{-3}}[/tex]
[tex]ln x=y\implies x=e^y[/tex]
[tex][A]=0.7[/tex] M
Hence, the concentration will be 0.7 M after 14 s
