39122.4.33 molar mass 23.4 O. 12=mplar mass of cl₂ = 40.90° d. If 464mL of an ideal gas is at 27°C and 1.05atm, what will be new volume at STP? Gas Law: Solve for the Unknown:
39122.4.33 molar mass 23.4 O. 12=mplar mass of cl₂ = 40.90° d. If 464mL of an ideal gas is at 27°C and 1.05atm, what will be new volume at STP? Gas Law: Solve for the Unknown:
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Please help with 2(d)
![## Educational Transcription of Gas Law Calculations
### 1. Conversion Problems for Pressure
**c. Conversion to kPa**
- Convert atm to kPa:
\[
0.9899 \, \text{atm} = 0.9899 \, \text{atm} \times \frac{101.325 \, \text{kPa}}{1 \, \text{atm}} = 100.3 \, \text{kPa}
\]
---
### 2. Gas Law Problems
**With the given information:**
1. **State the gas law that can be used and:**
2. **Solve for the unknown:**
**a. If a 425 mL sample of He gas is cooled from 76°C to 22°C, what is the new volume of the gas?**
- **Gas Law:** Charles's Law
\[
\frac{V_1}{T_1} = \frac{V_2}{T_2}
\]
- **Solve for the Unknown:**
- Initial and final temperatures:
\[
T_1 = 76^\circ C = 76 + 273 = 349 \, \text{K}
\]
\[
T_2 = 22^\circ C = 22 + 273 = 295 \, \text{K}
\]
- Volume at initial temperature: \( V_1 = 425 \, \text{mL} \)
- \[
V_2 = \frac{425 \, \text{mL} \times 295 \, \text{K}}{349 \, \text{K}} = 359.24 \, \text{mL}
\]
**b. The new pressure of a gas when it expands to 1.25 L when originally 235 mL and 5.34 atm.**
- **Gas Law:** Boyle's Law
\[
P_1V_1 = P_2V_2
\]
- **Solve for the Unknown:**
- Initial pressure and volume: \( 235 \, \text{mL} \times 5.34 \, \text{atm} \)
- Final volume: \( P_2 = \frac{1250}{1250} \times](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa7e1480f-493b-4334-b48e-1a7032a37693%2Fc6220bed-1b40-4846-bef7-76d894a13e36%2Fta95xdj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Educational Transcription of Gas Law Calculations
### 1. Conversion Problems for Pressure
**c. Conversion to kPa**
- Convert atm to kPa:
\[
0.9899 \, \text{atm} = 0.9899 \, \text{atm} \times \frac{101.325 \, \text{kPa}}{1 \, \text{atm}} = 100.3 \, \text{kPa}
\]
---
### 2. Gas Law Problems
**With the given information:**
1. **State the gas law that can be used and:**
2. **Solve for the unknown:**
**a. If a 425 mL sample of He gas is cooled from 76°C to 22°C, what is the new volume of the gas?**
- **Gas Law:** Charles's Law
\[
\frac{V_1}{T_1} = \frac{V_2}{T_2}
\]
- **Solve for the Unknown:**
- Initial and final temperatures:
\[
T_1 = 76^\circ C = 76 + 273 = 349 \, \text{K}
\]
\[
T_2 = 22^\circ C = 22 + 273 = 295 \, \text{K}
\]
- Volume at initial temperature: \( V_1 = 425 \, \text{mL} \)
- \[
V_2 = \frac{425 \, \text{mL} \times 295 \, \text{K}}{349 \, \text{K}} = 359.24 \, \text{mL}
\]
**b. The new pressure of a gas when it expands to 1.25 L when originally 235 mL and 5.34 atm.**
- **Gas Law:** Boyle's Law
\[
P_1V_1 = P_2V_2
\]
- **Solve for the Unknown:**
- Initial pressure and volume: \( 235 \, \text{mL} \times 5.34 \, \text{atm} \)
- Final volume: \( P_2 = \frac{1250}{1250} \times
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