class 12 Missing JEE Advanced

For the following reaction, equilibrium constant $K_{c}$ at 298 K is $1.6×10_{17}$

$Fe_{(aq)}+S_{(aq)}⇔FeS(s)$

When equal volume of 0.06 $MFe_{2+}$ and 0.2 $Ms_{−2}$ solution are mixed, then equilibrium concentration of $Fe_{2+}$ is found to be $Y×10_{−17}M$. Y is

$Fe_{(aq)}+S_{(aq)}⇔FeS(s)$

When equal volume of 0.06 $MFe_{2+}$ and 0.2 $Ms_{−2}$ solution are mixed, then equilibrium concentration of $Fe_{2+}$ is found to be $Y×10_{−17}M$. Y is

Similar topics

jee 2019

jee mains

jee 2019

jee mains

Related Questions

Related Questions