InterviewSolution
Saved Bookmarks
| 1. |
8. From the following molar conductivities at infinite dilution, ∧^° for Al_2(SO_4)_3=858 S cm^2 mol^{-1} ∧^° for NH_4OH = 238.3 S cm^2 mol^{-1} ∧^° for (NH_4)_2SO_4 = 238.4 S cm^2 mol^{-1} Calculate ∧^° for Al(OH)_3 |
| Answer» 8. From the following molar conductivities at infinite dilution, ∧^° for Al_2(SO_4)_3=858 S cm^2 mol^{-1} ∧^° for NH_4OH = 238.3 S cm^2 mol^{-1} ∧^° for (NH_4)_2SO_4 = 238.4 S cm^2 mol^{-1} Calculate ∧^° for Al(OH)_3 | |