7.56 g of lithium iodide (4 H = -63.3 kJ/mol) is placed into 175 mL of...

Question

Question

7.56 g of lithium iodide (4 H = -63.3 kJ/mol) is placed into 175 mL of water at room temperature (25 °C). What is the final t

Answer 1

Answer #1

LII(s) = Li caq) +1°(aq) 4xH = -63.362/mol molar mass= 133.85g Imola Lithium iodide = Li I 7.56g Lil = 7.56g - 133185 g/mol =

Answer 2

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