Question 2 of 8 Submit When 0.0801 mol of an unknown hydrocarbon is burned in a bomb calorimeter, the calorimeter increases in temperature by 2.19°C. If the heat capacity of the bomb calorimeter is 1.229 kJ/°C, what is the heat of combustion for the unknown hydrocarbon, in kJ/mol? kJ/mol 1 2 3 4 6. C 7 8 9. +/- x 100 II <

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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### Calorimetry Problem: Heat of Combustion Calculation

**Scenario:**
When 0.0801 mol of an unknown hydrocarbon is burned in a bomb calorimeter, the calorimeter's temperature increases by 2.19°C. Given that the heat capacity of the bomb calorimeter is 1.229 kJ/°C, calculate the heat of combustion for the unknown hydrocarbon in kJ/mol.

**Solution Steps:**

1. **Determine Heat Absorbed by Calorimeter:**
   - Formula: \( q = C \times \Delta T \)
   - Where \( q \) is the heat absorbed, \( C \) is the heat capacity, and \( \Delta T \) is the change in temperature.
   - Substitute the values: \( q = 1.229 \, \text{kJ/°C} \times 2.19 \, \text{°C} \)
   - Calculate \( q \).

2. **Calculate Heat of Combustion:**
   - Since the reaction involved 0.0801 mol of the hydrocarbon, find the heat of combustion (in kJ/mol) using the formula:
   - \( \text{Heat of Combustion} = \frac{q}{\text{moles of hydrocarbon}} \)

3. **Final Calculation:**
   - Use the values derived to find the heat of combustion per mole.

This setup involves a numerical keypad for input, allowing direct calculation of results, enhancing the interactive learning experience for students.

**Note:** Exact calculations are omitted as it requires step-by-step computational work.
Transcribed Image Text:### Calorimetry Problem: Heat of Combustion Calculation **Scenario:** When 0.0801 mol of an unknown hydrocarbon is burned in a bomb calorimeter, the calorimeter's temperature increases by 2.19°C. Given that the heat capacity of the bomb calorimeter is 1.229 kJ/°C, calculate the heat of combustion for the unknown hydrocarbon in kJ/mol. **Solution Steps:** 1. **Determine Heat Absorbed by Calorimeter:** - Formula: \( q = C \times \Delta T \) - Where \( q \) is the heat absorbed, \( C \) is the heat capacity, and \( \Delta T \) is the change in temperature. - Substitute the values: \( q = 1.229 \, \text{kJ/°C} \times 2.19 \, \text{°C} \) - Calculate \( q \). 2. **Calculate Heat of Combustion:** - Since the reaction involved 0.0801 mol of the hydrocarbon, find the heat of combustion (in kJ/mol) using the formula: - \( \text{Heat of Combustion} = \frac{q}{\text{moles of hydrocarbon}} \) 3. **Final Calculation:** - Use the values derived to find the heat of combustion per mole. This setup involves a numerical keypad for input, allowing direct calculation of results, enhancing the interactive learning experience for students. **Note:** Exact calculations are omitted as it requires step-by-step computational work.
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