3. Charles Law: What is the volume of the air in a Helium balloon that has 135 mL at 67.6 degrees Fahrenheit if the temperature is raised to 100. degrees Fahrenheit?

199.7mL

Explanation:

Using Charles law equation as follows:

V1/T1 = V2/T2

Where;

V1 = initial volume (mL)

T1 = initial temperature (°F)

V2 = final volume (mL)

T2 = final temperature (°F)

According to the information provided in this question, V1 = 135mL, V2 = ?, T1 = 67.6°F, T2 = 100°F

Using V1/T1 = V2/T2

135/67.6 = V2/100

Cross multiply

135 × 100 = 67.6 × V2

13500 = 67.6V2

V2 = 13500/67.6

V2 = 199.704

The new volume in the air in the helium ballon is 199.7mL.

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.