Vraag 22

Een voorbeeld van een juist antwoord is:

$K=\frac{[\text { Actine-Dys-A }]}{[\text { Actine }][\text { Dys-A }]}

\text{\lbrack Actine-Dys-A \rbrack}=\frac{28}{10^{2}}\times3{,}0\cdot10^{-6}=8{,}4\cdot10^{-7}(\mathrm{~mol~L}^{-1})\left\lbrack\text{\lbrack Actine-Dys-A \rbrack}=\frac{28}{10^{2}}\times3{,}0\cdot10^{-6}=8{,}4\cdot10^{-7}(\mathrm{~mol~L}^{-1})\right\rbrack\left\lbrack\text{A\lbrack Actine-Dys-A \rbrack}=\frac{28}{10^{2}}\times3{,}0\cdot10^{-6}=8{,}4\cdot10^{-7}(\mathrm{~mol~L}^{-1})\right\rbrack\left\lbrack\text{A\lbrack ctine-Dys-A \rbrack}=\frac{28}{10^{2}}\times3{,}0\cdot10^{-6}=8{,}4\cdot10^{-7}(\mathrm{~mol~L}^{-1})\right\rbrack\left\lbrack\text{Actine-Dys-A \rbrack}=\frac{28}{10^{2}}\times3{,}0\cdot10^{-6}=8{,}4\cdot10^{-7}(\mathrm{~mol~L}^{-1})\right\rbrack\left\lbrack\text{Actine-Dys-A }=\frac{28}{10^{2}}\times3{,}0\cdot10^{-6}=8{,}4\cdot10^{-7}(\mathrm{~mol~L}^{-1})\right\rbrack\left\lbrack\text{Actine-Dys-A }]=\frac{28}{10^{2}}\times3{,}0\cdot10^{-6}=8{,}4\cdot10^{-7}(\mathrm{~mol~L}^{-1})\right\rbrack\left\lbrack\text{Actine-Dys-A \rbrack}]=\frac{28}{10^{2}}\times3{,}0\cdot10^{-6}=8{,}4\cdot10^{-7}(\mathrm{~mol~L}^{-1})\right\rbrack\left\lbrack\text{Actine-Dys-A }]=\frac{28}{10^{2}}\times3{,}0\cdot10^{-6}=8{,}4\cdot10^{-7}(\mathrm{~mol~L}^{-1})\right\rbrack\text{Actine-Dys-A }]=\frac{28}{10^{2}}\times3{,}0\cdot10^{-6}=8{,}4\cdot10^{-7}(\mathrm{~mol~L}^{-1})\left.\text{Actine-Dys-A }]=\frac{28}{10^{2}}\times3{,}0\cdot10^{-6}=8{,}4\cdot10^{-7}(\mathrm{~mol~L}^{-1})\right\rbrack\left\lbrack\text{Actine-Dys-A }]=\frac{28}{10^{2}}\times3{,}0\cdot10^{-6}=8{,}4\cdot10^{-7}(\mathrm{~mol~L}^{-1})\right\rbrack\left.\text{Actine-Dys-A }]=\frac{28}{10^{2}}\times3{,}0\cdot10^{-6}=8{,}4\cdot10^{-7}(\mathrm{~mol~L}^{-1})\right\rbrack\left\lbrack\text{Actine-Dys-A }]=\frac{28}{10^{2}}\times3{,}0\cdot10^{-6}=8{,}4\cdot10^{-7}(\mathrm{~mol~L}^{-1})\right\rbrack\left\lbrack\left\lbrack\text{Actine-Dys-A }]=\frac{28}{10^{2}}\times3{,}0\cdot10^{-6}=8{,}4\cdot10^{-7}(\mathrm{~mol~L}^{-1})\right\rbrack\right\rbrack\left\lbrack\text{Actine-Dys-A }]=\frac{28}{10^{2}}\times3{,}0\cdot10^{-6}=8{,}4\cdot10^{-7}(\mathrm{~mol~L}^{-1})\right\rbrack\left\lbrack=\frac{28}{10^{2}}\times3{,}0\cdot10^{-6}=8{,}4\cdot10^{-7}(\mathrm{~mol~L}^{-1})\right\rbrack=\frac{28}{10^{2}}\times3{,}0\cdot10^{-6}=8{,}4\cdot10^{-7}(\mathrm{~mol~L}^{-1})$]=\frac{28}{10^{2}} \times 3{,}0 \cdot 10^{-6}=8{,}4 \cdot 10^{-7}(\mathrm{~mol} \mathrm{~L}^{-1})$K=\frac{8{,}4 \cdot 10^{-7}}{(3{,}0 \cdot 10^{-6}-8{,}4 \cdot 10^{-7}) \times(6{,}0 \cdot 10^{-6}-8{,}4 \cdot 10^{-7})}=7{,}5 \cdot 10^{4}