I don't know who it was named for but the law is
P1/T1 = P2/T2
Make sure the pressure units are the same (atmospheres or kPa usually) and that the temperature is in Degrees Kelvin which is derived from Celsius degrees.
Try Charles' Law for the name.
Answer:
![\Delta _{comb}H=-2,093\frac{kJ}{mol}](https://tex.z-dn.net/?f=%5CDelta%20_%7Bcomb%7DH%3D-2%2C093%5Cfrac%7BkJ%7D%7Bmol%7D)
Explanation:
Hello!
In this case, since these calorimetry problems are characterized by the fact that the calorimeter absorbs the heat released by the combustion of the substance, we can write:
![Q_{rxn}+Q_{cal}=0](https://tex.z-dn.net/?f=Q_%7Brxn%7D%2BQ_%7Bcal%7D%3D0)
Thus, given the temperature change and the total heat capacity, we obtain the following total heat of reaction:
![Q_{rxn}=-14.01kJ/K*3.69K\\\\Q_{rxn}=-51.70kJ](https://tex.z-dn.net/?f=Q_%7Brxn%7D%3D-14.01kJ%2FK%2A3.69K%5C%5C%5C%5CQ_%7Brxn%7D%3D-51.70kJ)
Now, by dividing by the moles in 1.04 g of cyclopropane (42.09 g/mol) we obtain the enthalpy of combustion of this fuel:
![n=\frac{1.04g}{42.09g/mol}=0.0247mol\\\\\Delta _{comb}H=\frac{Q_{rxn}}{n}\\\\ \Delta _{comb}H=-2,093\frac{kJ}{mol}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B1.04g%7D%7B42.09g%2Fmol%7D%3D0.0247mol%5C%5C%5C%5C%5CDelta%20_%7Bcomb%7DH%3D%5Cfrac%7BQ_%7Brxn%7D%7D%7Bn%7D%5C%5C%5C%5C%20%20%5CDelta%20_%7Bcomb%7DH%3D-2%2C093%5Cfrac%7BkJ%7D%7Bmol%7D)
Best regards!
Answer:
lower mantle
Explanation:
because the core and the crust are solid.