Answer:
[tex]Q=-76.7kJ[/tex]
Explanation:
Hello,
In this case, for such formation of sulfur hexafluoride, the standard enthalpy of formation is -1220.47 kJ/mol (data extracted from NIST database). Next, we compute the moles in 10.0 grams of sulfur hexafluoride as shown below:
[tex]n_{SF_6}=10.0gSF_6*\frac{1molSF_6}{146.06 gSF_6}=0.0685mol[/tex]
Next, for the given energy, we compute the total heat that is liberated:
[tex]Q=-1220.47\frac{kJ}{mol}*0.0685 mol\\\\Q=-76.7kJ[/tex]
Finally, we conclude such symbol has sense since negative heat is related with liberated heat.
Best regards.
