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Chemistry, 12.08.2020 06:01, munziruddin204
Consider the following: Li(s) + ½ I₂(g) --> LiI(s) ΔH = –292 kJ. LiI(s) has a lattice energy of –753 kJ/mol. The ionization energy of Li(g) is 520 kJ/mol, the bond energy of I₂(g) is 151 kJ/mol, and the electron affinity of I(g) is –295 kJ/mol. Use these data to determine the heat of sublimation of Li(s).
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Consider the following: Li(s) + ½ I₂(g) --> LiI(s) ΔH = –292 kJ. LiI(s) has a lattice energy of –...
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