Respuesta :
Answer:
The molar heat of combustion of hydrazine is -663.82 kJ/mole.
Explanation:
First we have to calculate the heat gained by the calorimeter.
[tex]q=c\times (T_{final}-T_{initial})[/tex]
where,
q = heat gained = ?
c = specific heat = [tex]5860 J/^oC[/tex]
[tex]T_{final}[/tex] = final temperature = [tex]28.16^oC[/tex]
[tex]T_{initial}[/tex] = initial temperature = [tex]24.62^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]q=5860 J/^oC\times (28.16-24.62)^oC[/tex]
[tex]q=20,744.4 J=20.7444 kJ[/tex]
Now we have to calculate the enthalpy change during the reaction.
[tex]\Delta H=-\frac{q}{n}[/tex]
where,
[tex]\Delta H[/tex] = enthalpy change = ?
q = heat gained = 20.7444 kJ
n = number of moles fructose = [tex]\frac{\text{Mass of hydrazine}}{\text{Molar mass of hydrazine}}=\frac{1.000 g}{32 g/mol}=0.03125 mole[/tex]
[tex]\Delta H=-\frac{20.7444 kJ}{0.03125 mole}=-663.82 kJ/mole[/tex]
The molar heat of combustion of hydrazine is -663.82 kJ/mole.
Answer: -20.7
Explanation:
qrxn=−(qwater+qbomb)qrxn=−(cwatermΔT+ccalorimeterΔT)qrxn=−[(4.184Jg∘C)(1200. g)(3.54∘C)+(840. J∘C)(3.54∘C)]
qrxn=−20,700 J=−20.7 kJ
