The Enthalpy of Vaporization of 1-chlorobutane, C4H9Cl, is 30.4 kJ/mol at a boiling point of 78.6 °C.  How much energy must be added to a 121.35g sample of 1-chlorobutane at its boiling point in order to vaporize the entire sample?

The complete combustion of salicylic acid releases 21.90 kJ of energy per gram of salicylic acid.In a particular bomb calorimeter (initially at room temperature), the combustion of 0.2745 g of salicylic acid, in the presence of excess oxygen, causes the temperature of the calorimeter to rise by 1.85 °C.When a 0.2999-g sample of an unknown organic substance is similarly burned in the same calorimeter, the temperature rises by 3.59 °C.What is the energy of combustion per unit mass of the unknown substance?

A bomb calorimeter has a heat capacity of 4.66 kJ/°C. When 1.772 g of glucose, C6H₁2O6(s), is combusted in the bomb calorimeter, the temperature of the calorimeter and its contents increased from 22.32 °C to 28.24 °C. Calculate the change in internal energy, AU, for the combustion of 1.772 g of glucose in the bomb calorimeter. AU = -716.36 kJ

An electrical heater adds 25.3 kJ of heat to a constant-volume calorimeter. The temperature of the calorimeter rises by 4.55 °C. When 1.75 g of methanol is burned in the same calorimeter, the temperature increases by 7.14 °C. Calculate the molar energy of combustion of methanol. kJ/mol

A 1.22 g sample of acetic acid (HC2H3O2) was burned in excess oxygen in a bomb calorimeter. The calorimeter, which alone had a heat capacity of 2.67 kJ/oC, contained 7 g of water. The temperature of the calorimeter and its contents increased from 23.47oC to 27.95oC. What is ΔH (in kJ) for the combustion per 1.00 mol of acetic acid? MW of acetic acid is 60 g/mol. (Round off the final answer to ONE decimal place. Do not include the unit.)

If we measure the heat change (q) of everything in a system except the calorimeter, we can use the first law of thermodynamics (∆Euniverse = 0) to calculate the heat change of the calorimeter (qcal).  In this reaction, we combine 5.85 g of 63.7 ºC water with 6.63 g of 18.9 ºC water in a calorimeter.  Since some of the heat is absorbed by the calorimeter, the final temperature of the water is 28.8ºC.  Since the hot and cold water are combined, the final temp of the water is the final temp for the hot water and the cold water.  Calculate the heat change of the calorimeter in J. Remember that all of the heat changes must add up to zero so you can calculate the heat change of the hot water (qhotwater) and the heat change of the cold water (qcoldwater) using for each where Cs = 4.186 J/gºC.  The first law of thermodynamics tells us that all of these changes sum up to zero so   Enter your answer to one decimal place (tenths)