Saved Bookmarks
| 1. |
The equilibrium constant `K_(c)` for the following reaction at `842^(@)`C is `7.90xx10^(-3)`. What is `K_(p)`at same temperature ?` (1)/(2)F_(2(g)) hArr F_(g)`A. `8.64xx10^(5)`B. `8.26xx10^(-4)`C. `7.90xx10^(-2)`D. `7.56xx10^(-2)` |
|
Answer» Correct Answer - D `K_(P)=(7.9xx10^(-3))xx(0.0821xx115)^(0.5)` `K_(P)=7.56xx10^(-2)` |
|