please help with these chemistry questions! 1. A piece of (specifie capacity -04

Question

please help with these chemistry questions!

1. A piece of (specifie capacity -0452 J g i C i) with a mass of 54.3218 g is laeated water bath until it reaches a perature of 98.2 C. The iron is then transferred inlo a calorimeter (nested styrofoam cups) that contains 68 53 water initially 22.1 C. The final of the entire c. specific hel capacity o Based on this information, determine the value of the calo 2. When 60.0 mL of a 0.400 M solution of HNO3 is combined with 60.0 mL of a 0.400 M solution of NaOH in the calorimeter described in question #1, the final temperature of the solution is measured to be 26.6 The initial temperature of the solutions is 24.0 c Assuming the specific heat capacity of the final solution is 3.90 J g Cland the density of the final solution is 1.04 g/ml calculate gns. Hint: stat with Waut quan qeal 0. HNO (aq) NaOH (aq

Explanation / Answer

1) energy lost by the hot water,
q1 = mCpT = 68.5314 gms * 4.184 J/gm oC * (98.2-27.5)oC = 20272.2J
energy gained by Iron,
   q2 = mCpT = 54.3218 gms * 0.452 J/gm oC * (27.5 - 22.1)o C = 132.588 J
The remaining heat is absorbed by calorimeter,
qcal = 20272.2 - 132.588 = 20139.6 J
Calorimeter constant, Ccal = 20139.6 J / (98.2 - 22.1)oC = 264.65 J/oC

2) qrxn + qsoln + qcal = 0
   calculating qsoln ,
amount of energy absorbed = (120 mL*1.04gm/mL)* 3.90J/gm oC *(26.6-24)oC = 1265.472 J
from #1 qcal = 20139.6 J
qrxn + 1265.472 J + 20139.6 J = 0
qrxn = -21405 J (energy released)

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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