What is the pressure due to just dry oxygen gas at \( 21.0^{\circ} \mathrm{C} \) in a wet gas sample that has total pressure of 751.8 mmHg ? Use Table 1. Show all work and circle your correctly rounded answer.

To find the pressure due to just dry oxygen gas in a wet gas sample, we need to use Dalton’s Law of Partial Pressures. The total pressure is the sum of the partial pressures of the gases present. 1. Look up the vapor pressure of water at \( 21.0^{\circ} \mathrm{C} \) from Table 1; let’s say it’s approximately 18.6 mmHg (check actual data). 2. Subtract the vapor pressure of water from the total pressure: \[ P_{\text{dry O}_2} = P_{\text{total}} - P_{\text{water}} \] \[ P_{\text{dry O}_2} = 751.8 \, \text{mmHg} - 18.6 \, \text{mmHg} \] \[ P_{\text{dry O}_2} = 733.2 \, \text{mmHg} \] Therefore, the pressure due to just dry oxygen gas is **733.2 mmHg**. (Make sure to circle this final answer.) --- Wet gas pressure calculations have been around for centuries, often utilized in laboratories and industrial settings. In the 19th century, scientists like John Dalton laid down foundational theories about gas behavior, paving the way for our understanding of gases in mixtures. The clever interplay of various gas pressures and vapor pressures was a key part of this scientific evolution. To avoid common pitfalls when working with gas calculations, always ensure you have the correct vapor pressure for the specific temperature you’re working with! It’s easy to mix up values or forget to subtract the vapor pressure from the total pressure. Always double-check your references and calculations to keep those pesky mistakes at bay.