Let's break down the question step by step: --- ### **Question Recap** - The atmospheric pressure at the summit of Mt. McKinley is **610. mmHg**. - You are asked to convert this pressure into: 1. **atm (atmospheres)** 2. **kPa (kilopascals)** --- ## **Step 1: Conversion to atm** ### **Conversion Factor** - \( 1 \text{ atm} = 760 \text{ mmHg} \) ### **Calculation** \[ \text{Pressure in atm} = \frac{610 \text{ mmHg}}{760 \text{ mmHg/atm}} \] \[ \text{Pressure in atm} = 0.803 \text{ atm} \] --- ## **Step 2: Conversion to kPa** ### **Conversion Factor** - \( 1 \text{ atm} = 101.325 \text{ kPa} \) - \( 1 \text{ mmHg} = 0.133322 \text{ kPa} \) (Alternatively, you can use this direct conversion.) ### **Calculation** \[ \text{Pressure in kPa} = 610 \text{ mmHg} \times 0.133322 \text{ kPa/mmHg} \] \[ \text{Pressure in kPa} = 81.327 \text{ kPa} \] - Rounded to three significant figures (since 610 has three): **81.3 kPa** --- ## **Summary Table** | Given | To Find | Conversion Factor | Answer | |--------------|---------|------------------------|---------------| | 610 mmHg | atm | 1 atm = 760 mmHg | 0.803 atm | | 610 mmHg | kPa | 1 mmHg = 0.133322 kPa | 81.3 kPa | --- ## **Conclusion** - **610 mmHg = 0.803 atm** - **610 mmHg = 81.3 kPa** These are the atmospheric pressures at the summit of Mt. McKinley in different units. Always use the correct number of significant digits based on the data provided. If you need a more detailed explanation for each step, let me know!