Where do I even start here? **Text for Educational Website:**---**Problem Statement:**Assume that 0.321 g of diborane is combusted in a calorimeter whose heat capacity (\(C_{\text{calorimeter}}\)) is 7.854 kJ/°C at 16.31°C. What is the final temperature of the calorimeter?\[\text{Final Temperature:} \ \_\_\_\_ \ °C\]**Instructions for Solution Attempt:**1. **Calculate the Heat Released:** - Use the molecular weight of diborane and its combustion enthalpy to determine the heat released upon combustion.2. **Apply Calorimetry Principle:** - Use the formula \( q = C \times \Delta T \), where \( q \) is the heat absorbed by the calorimeter, \( C \) is the calorimeter's heat capacity, and \( \Delta T \) is the temperature change.3. **Determine the Final Temperature:** - Rearrange the equation to solve for the final temperature.**First Attempt:**- Complete the calculation using the steps above and compare with experimental data if available.---**Graph or Diagram:**- *There is no graph or diagram provided in the image.*This structured approach will help students understand how to approach calorimetry problems and apply thermodynamic principles in a systematic manner. A bomb calorimeter can be used to measure the enthalpy of reaction for combustions, such as the combustion of diborane represented by the following balanced equation:\[ \text{B}_2\text{H}_6 (g) + 3\text{O}_2 (g) \rightarrow \text{B}_2\text{O}_3 (s) + 3\text{H}_2\text{O} (g) \]\(\Delta H_{\text{combustion}}\) for this process is \(-1958 \, \text{kJ/mol}\).

Q = C x (T2 - T1) Where, Q = heat evolved/absorbed C = heat capacity of calorie meter T2 = final temperature T1 = initial temperature Balanced equation for the combustion of diborane: B2H6 (g) + 3O2 (g) ----------> B2O3 (s) + 3H2O (g) ∆H = -1958 kJ/mol Mass of diborane = 0.321 g Molar mass of diborane = 27.66 g/mol Number of moles of diborane = Mass/molar mass = 0.321/27.66 = 0.01161 Heat evolved = 0.01161 x 1958 = 22.73238 kJ = 22732.38 J Heat given = heat evolved Q = 22732.38 J C = 7.854 kJ = 7854 J/C T1 = 16.31 deg CQ = C x (T2 - T1) 22732.38 = 7854 (T2 - 16.31) T2 - 16.31 = 22732.38/7854 T2 - 16.31 = 2.89 T2 = 16.32 + 2.89 T2 = 19.21 deg C