A 0.708 mole copper piece at -15.4 degree C absorbed 613 J of heat and the final

Question

A 0.708 mole copper piece at -15.4 degree C absorbed 613 J of heat and the final temperature is measured at 19.6 degree C, what is the specific heat of copper? a. 0.389 J/mol middot degree C b. 17.5 J/mol middot degree C c. 12.4 J/mol middot degree C d. 206 J/mol middot degree C e. 24.7 J/mol middot degree C When 0.0129 mol of KI dissolved in 11.57 degree C water and made 20.0 g of solution, the temperature of the solution decrease 10.00 degree C. Assuming the container did not absorb/release any heat, what is the Delta H degree of the following reaction? (solution Cp = 4.184 J/g middot degree C) Kl (s) rightarrow K^+ (aq) + l^- (aq) Delta H degree = ? a. -20.4 kJ b. +10.2 kJ c. +20.4 kJ d.+5.10 kJ e -10.2 kJ

Explanation / Answer

Q10

mol of KI = 0.0129

T = 11.57°C

m = 20 g of solution,

Tmix = 10°C

find dHsolution

given

HRxn = Qsolution/mol

Qsolutin = -Qwater

Qwater = m*C*(Tf-Ti)

Qwater = 20*4.184*(10-11.57) = -131.3776 J

so

HRxn = Qsolution/mol

HRxn = 131.3776 /0.0129

HRxn = 10184.31 J/mol = 10.184 kJ/mol

must be possitive, since it is ENDOTHERMIC

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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