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Answer :
We start by determining the number of moles of aniline in the 6.55 g sample. Using the molar mass of aniline, which is 93.13 g/mol, the number of moles is calculated as
[tex]$$
n = \frac{\text{mass}}{\text{molar mass}} = \frac{6.55\, \text{g}}{93.13\, \text{g/mol}} \approx 0.07033\, \text{mol}.
$$[/tex]
Next, we determine the total energy released by the combustion of this sample. The energy released per mole of aniline is given as [tex]$3.20\times10^3$[/tex] kJ/mol (we use the absolute value since the energy is released). Therefore, the energy released by the sample is
[tex]$$
q = n \times 3.20\times10^3\, \text{kJ/mol} \approx 0.07033 \times 3200\, \text{kJ} \approx 225.06\, \text{kJ}.
$$[/tex]
Finally, the heat capacity of the calorimeter ([tex]$C_{\text{cal}}$[/tex]) can be found using the relation
[tex]$$
q = C_{\text{cal}}\, \Delta T.
$$[/tex]
Rearranging to solve for [tex]$C_{\text{cal}}$[/tex], we have
[tex]$$
C_{\text{cal}} = \frac{q}{\Delta T} = \frac{225.06\, \text{kJ}}{32.9^\circ\text{C}} \approx 6.84\, \text{kJ/}^\circ\text{C}.
$$[/tex]
Thus, the heat capacity of the calorimeter is approximately
[tex]$$
\boxed{6.84\, \text{kJ/}^\circ\text{C}}.
$$[/tex]
[tex]$$
n = \frac{\text{mass}}{\text{molar mass}} = \frac{6.55\, \text{g}}{93.13\, \text{g/mol}} \approx 0.07033\, \text{mol}.
$$[/tex]
Next, we determine the total energy released by the combustion of this sample. The energy released per mole of aniline is given as [tex]$3.20\times10^3$[/tex] kJ/mol (we use the absolute value since the energy is released). Therefore, the energy released by the sample is
[tex]$$
q = n \times 3.20\times10^3\, \text{kJ/mol} \approx 0.07033 \times 3200\, \text{kJ} \approx 225.06\, \text{kJ}.
$$[/tex]
Finally, the heat capacity of the calorimeter ([tex]$C_{\text{cal}}$[/tex]) can be found using the relation
[tex]$$
q = C_{\text{cal}}\, \Delta T.
$$[/tex]
Rearranging to solve for [tex]$C_{\text{cal}}$[/tex], we have
[tex]$$
C_{\text{cal}} = \frac{q}{\Delta T} = \frac{225.06\, \text{kJ}}{32.9^\circ\text{C}} \approx 6.84\, \text{kJ/}^\circ\text{C}.
$$[/tex]
Thus, the heat capacity of the calorimeter is approximately
[tex]$$
\boxed{6.84\, \text{kJ/}^\circ\text{C}}.
$$[/tex]
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