4 p estion 2 In a second experiment, 16.00 g of sodium acetate trihydrate (136.0

Question

4 p estion 2 In a second experiment, 16.00 g of sodium acetate trihydrate (136.08 g/mol) was dissolved in 100.00 g of water. The initial temperature was 25.00 °C, and the final temperature was 20.06 °C The heat capacity of the reaction mixture was 4.045 J/g K C (2 dec places) (hint: magnitude of the sign T for this experiment was matters). IJoules (4 sig figs) (hint: magnitude of the sign The qsys for the reaction is matters). mol (4 sig The number of moles of sodium acetate trihydrate dissolved is figs) kJ/mol (4 sig The H of dissolution for sodium acetate trihydrate is figs) (hint: magnitude of the sign matters)

Explanation / Answer

from the given data

1) DT = 25-20.06 = 4.94 C

2) Qsys = C*DT

C = heat capacity of the reaction mixture = 4.045 j/g.k

DT = 4.94

Qsys = 4.045*4.94 = +19.98 j

      = +0.01998 kj

3) NO of mole of NaCH3COO*3H2O = 16/136.08 = 0.1176 mol

4) DHdis = +q/n = 0.01998/0.1176 = 0.1699 kj/mol

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.