The answer is FeCI3. I took this on the exam and got it right. :)
Answer:
a beam of electrons emitted from the cathode of a high-vacuum tube.
Explanation:
Answer:
The heat capacity of the calorimeter is 5.11 J/g°C
Explanation:
Step 1: Given data
50.0 mL of water with temperature of 80.0 °C
Specific heat capacity of water = 4.184 J/g°C
Consider the density of water = 1g/mL
50.0 mL of water in a calorimeter at 20.0 °C
Final temperature = 47.0 °C
Step 2: Calculate specific heat capacity of the water in calorimeter
Q = Q(cal) + Q(water)
Q(cal) = mass * C(cal) * ΔT
Qwater = mass * Cwater * ΔT
Qcal = -Qwater
mass(cal) * C(cal) * ΔT(cal) = mass(water) * C(water) * ΔT(water)
50 grams * C(cal) * (47.0 - 20.0) =- 50grams * 4.184 J/g°C * (47-80)
1350 * C(cal) = 6903.6
C(cal) = 5.11 J/g°C
The heat capacity of the calorimeter is 5.11 J/g°C
