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This image appears to be part of a chemical calculation related to the mass of oxygen gas (\( \text{O}_2 \)). The equation shown seems to relate the molar mass and molar volume of oxygen gas to temperature and pressure, potentially using the Ideal Gas Law.

The formula being used appears to be:

\[
\text{Massa } \text{O}_2 = \frac{V_{\text{O}_2} + M_r\text{O}_2}{22.414} \times \frac{273 \, \text{K} + \text{rata-rata suhu}}{273 \, \text{K}}
\]

Breaking down the terms:
- \( V_{\text{O}_2} \) is the volume of oxygen gas (in mL or liters).
- \( M_r\text{O}_2 \) is the molar mass of oxygen gas (\( 32 \, \text{g/mol} \)).
- \( 22.414 \, \text{L/mol} \) is the molar volume of an ideal gas at standard temperature and pressure (STP), 273 K and 1 atm.
- The ratio \( \frac{273 \, \text{K} + \text{rata-rata suhu}}{273 \, \text{K}} \) accounts for temperature variation from standard conditions.

In the second step:
\[
\frac{1.7 \, \text{mL} + 32 \, \text{g/mol}}{22.414} \times \frac{273 \, \text{K} + 25^\circ C}{273 \, \text{K}}
\]

Here, the temperature difference is \( 25^\circ C \), meaning the final temperature is \( 273 + 25 = 298 \, \text{K} \).

The mass of \( \text{O}_2 \) will likely be computed after applying these values.

Would you like help in performing the full calculation or clarifying any specific part of the equation? 

Here are some related questions:
1. What does the molar volume of an ideal gas represent?
2. How does temperature affect gas volume and mass in ideal gas law calculations?
3. What is the significance of using \( 22.414 \, \text{L/mol} \) in these types of problems?
4. Why do we add \( 273 \, \text{K} \) to convert Celsius to Kelvin?
5. How would this calculation change at a different pressure?

**Tip**: Always convert Celsius to Kelvin in gas law calculations to avoid negative temperature values.

What is the mass of O2 based on the given chemical calculation involving molar volume and temperature conversion?

Learn how to calculate the mass of O2 gas using molar volume and the Ideal Gas Law. This problem involves temperature conversion from Celsius to Kelvin and applies principles from thermodynamics. Suitable for high school chemistry students.

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