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Li(s) →Li(g) enthalpy of sublimation of Li(s) = 166 kJ/molHBr(g) → H(g) + Br(g) bond energy of HBr = 363 kJ/molLi(g) → Li+(g) + e- ionization energy of Li(g) = 520. kJ/molBr(g) + e- → Br-(g) electron affinity of Br(g) = -325 kJ/molLi+(g) + Br-(g) → LiBr(s) lattice energy of LiBr(s) = -809 kJ/molH2(g) → 2H(g) bond energy of H2 = 432 kJ/mol
Calculate the change in enthalpy for: 2Li(s) + 2HBr(g) → H2(g) + 2LiBr(s)
What are those byproducts? to remove the byproducts from the organic phase
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when 92.0 grams of nonionic compound were dissolved in 985 grams of water the freezing point of the solution was found
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