Calculate the lattice enthalpy of AgF (s) using the following thermodynamic data. Note that the data given has been perturbed, so looking up the answer is probably not a good idea. FF(g) Enthalpy of dissociation =138kJ/mol Ag(g) Enthalpy of formation =265kJ/mol F(g) Electron attachment enthalpy =348kJ/mol Ag (g) Enthalpy of ionization =711kJ/mol AgF (s) Enthalpy of formation =225kJ/mol kJ/mol Do you expect the magnitude (absolute value) of this to be larger or smaller than the lattice enthalpy of AgBr (s)? At a pressure of 1atm, the temperature 14K is called the of H2. The normal boiling point for H2 is at K The triple point pressure for H2 is atm The critical pressure for H2 is 12.8atm. At temperatures above 33.2K and pressures above 12.8atm,H2 is a H2 is a at 7.10102atm and 27.0K. H2 is a at 1.00atm and 16.6K. H2 is a at 10.8atm and 7.23K