` [log "k=-(2000)/(T)+6.0]` `E_(a)=?`, 1) 9.212(kcal)/(mole) ) ,2.)2000(kcal)/(mole) 3.)4.606(kcal)/(mole) 4.)2.500(kcal)/(mole)

1kcal =4.18 joules.

Calculate the electronegativity of fluorine from following data : E_(H - H) = 104.2 kcal mol^(-1) E_(F - F) = 36.6 kcal mol^(-1) E_(H -F) = 134.6 kcal mol^(-1) Electronegativity of H is 2.05.

E_(n)=-313.6//n^(2) kcal/mol. If the value of E=-34.84 kcal/mol, to which value does 'n' correspond?

Calculate the electronegativity of fluorine from the following data: E_(H-H) = 104.4 kcal mol^(-1), E_(F-F) = 36.6 kcal mol^(-1) E_(H-F) = 134.3 kcal mol^(-1), chi_(H) = 2.1