previous | 2 of 3 | next Experiment 11 - Post-lab-Question 3 Part A A 50.0-mL sa

Question

previous | 2 of 3 | next Experiment 11 - Post-lab-Question 3 Part A A 50.0-mL sample of a 1.00 M solution of is mixed with 50.0 mlL of 2.00 M KOH in a 26.1d after mixing. The heat capacity of the calonmeter is 121 JK From these data calculate AH (in kJ mol) for the process. calorimeter. The temperature of both solutions was 20 2 C before mixing and CuSO4 (1 M) + 2KOH(2M)- Cu(OH)2(s) + K2SO. (0.5M) Assume the specific heat and density of the solution after mixing are the same as those of pure water alue Units Submit My Answers Give Up Continue

Explanation / Answer

total volume = 50+50 = 100 mL

mass = D*V = 100*1 = 100 g

Qwater = m*C*(Tf-Ti)

Qcal = CCal*(Tf-Ti)

Qtotal = m*C*(Tf-Ti)+CCal*(Tf-Ti)= 100*4.184*(26.1-20.2) + 12.1(26.1-20.2) = 2539.95 J

mol of CuSO4= MV = 1*50*10^-3 = 0.05

HRXN = -Qtotal /n = 2539.95 /-0.05

HRXN = -50799 J/mol

HRXN = -50.8 kJ/mol

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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