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Module 3 Exam 2022
MODULE 3 chem103
MODULE 3 chem103
MODULE 3 EXAM 2022 Question 1 - [22.8 g x 4.184 J/g oC x (Tmix - 74.6oC)] = [14.3 g x 4.184 J/g oC x (Tmix - 24.3oC)] - [95.3952 J/oC x (Tmix - 74.6oC)] = [59.8312 J/oC x (Tmix - 24.3oC)] Tmix = 55.2oC 2. ql↔s = m x ∆Hfusion = 54.3 g x 0.334 kJ/g = 18.14 kJ (since heat is removed) = - 18.14 kJ
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