For the following reaction, equilibrium constant `K_(c)` at 298 K is `1.6xx10^(17)` `Fe_((aq))^(2+)+S_((aq))^(2) hArr FeS(s)` When equal volume of 0.06 `M Fe^(2+)` and 0.2 `Ms^(-2)` solution are mixed, then equilibrium concentration of `Fe^(2+)` is found to be `Yxx10^(-17) M`. Y is

For the following reaction, equilibrium constant `K_(c)` at 298 K is `

1.	For the following reaction, equilibrium constant `K_(c)` at 298 K is `1.6xx10^(17)` `Fe_((aq))^(2+)+S_((aq))^(2) hArr FeS(s)` When equal volume of 0.06 `M Fe^(2+)` and 0.2 `Ms^(-2)` solution are mixed, then equilibrium concentration of `Fe^(2+)` is found to be `Yxx10^(-17) M`. Y is
Answer» Correct Answer - 8.92 & 8.93 `{:(,Fe_((aq))^(2+),+,S_((aq))^(2-),hArr FeS(s),K_(C)=1.6xx10^(17)),(,0.06 M,,0.2 M,,),("After mixing",0.03 M,,0.1 M,,),(,?,,0.07M,,):}` `1.6xx10^(17) =1/([Fe^(2+)]xx0.07)` or `[Fe^(2+)]=(10^(-17))/(1.6xx0.07)=(10^(-15))/11.2=100/11.2xx10^(-17)=8.928xx10^(-17) =Yxx10^(-17)` Answer after rounding of is =8.93 Answer after truncation of is 8.92

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For the following reaction, equilibrium constant `K_(c)` at 298 K is `1.6xx10^(17)` `Fe_((aq))^(2+)+S_((aq))^(2) hArr FeS(s)` When equal volume of 0.06 `M Fe^(2+)` and 0.2 `Ms^(-2)` solution are mixed, then equilibrium concentration of `Fe^(2+)` is found to be `Yxx10^(-17) M`. Y is

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