Answer:
[tex]q^{combustion}_{acetylene}=-1100.2kJ/mol[/tex]
Explanation:
Hello.
In this case, for this calorimetry problem, since the acetylene is burned into the calorimeter causing a temperature increase, we can infer that the combustion of acetylene releases heat which is absorbed by the calorimeter, therefore, the following equation can be written:
[tex]Q_{acetylene}=-Q_{calorimeter}[/tex]
Which can be written in terms of the energy of combustion and moles of the acetylene and the heat capacity and temperature change of the calorimeter:
[tex]n_{acetylene}q^{combustion}_{acetylene}=-C_{calorimeter}\Delta T_{calorimeter}[/tex]
Thus, since the molar mass of acetylene is 26.04 g/mol, the resulting energy of combustion of the acetylene turns out:
[tex]q^{combustion}_{acetylene}=\frac{-C_{calorimeter}\Delta T_{calorimeter}}{n_{acetylene}} =\frac{-31.5kJ/\°C*16.9\°C}{12.6g*\frac{1mol}{26.04g} } \\\\q^{combustion}_{acetylene}=-1100.2kJ/mol[/tex]
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