# Chemistry Problem: Gas Pressure Conversion ## Given Data - Mass of Mg: **25. mg** (not directly needed for this pressure problem) - Volume of \( \mathrm{H_2(g)} \): **26. mL** - Atmospheric pressure: **30.68 inches of Hg** - Find: **Pressure of \( \mathrm{H_2O} \) (in mm) above the water in the beaker** --- ## Step 1: Convert Atmospheric Pressure to mm Hg 1 inch = 25.4 mm \[ 30.68 \text{ in Hg} \times 25.4 \frac{\text{mm}}{\text{in}} = 779.27 \text{ mm Hg} \] --- ## Step 2: Convert mm Hg to mm H\(_2\)O 1 mm Hg = 13.6 mm H\(_2\)O \[ 779.27 \text{ mm Hg} \times 13.6 \frac{\text{mm H}_2\text{O}}{\text{mm Hg}} = 10,593.07 \text{ mm H}_2\text{O} \] --- ## Final Answer **The pressure of the atmosphere is equivalent to approximately _10,600 mm H₂O_ (rounded to three significant figures) above the water in the beaker.** --- ### Notes - If you want the _difference_ in pressure between the inside and outside of the beaker, you’d need more info (e.g., vapor pressure of water, height of water column, etc.) - This answer gives the atmospheric pressure converted into mm H\(_2\)O, which is commonly how height above water is measured in such experiments. --- **Summary table** | Quantity | Value | |---------------------------|------------| | Atmospheric pressure | 30.68 in Hg| | In mm Hg | 779 mm Hg | | In mm H\(_2\)O | 10,600 mm H\(_2\)O |