169mol=0.45m0) of CHy . Latm mol-k 3Kןגו 0)- 76ommNg | atm why is whiy =45Y:6 mmhg

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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 check my calculations plz..

**Educational Website Transcription**

---

**Question Part 1A:**

Given Information:
- Mass (\( n \)) = 7.2 g
- Molar Mass = 16 g/mol (for CH₄)
- Number of moles (\( n \)) = 0.45 mol of CH₄

**Equation Used:**
\[ P = \frac{nRT}{V} \]

Calculations:
\[
P = \frac{(0.45 \, \text{mol}) \times (0.082 \, \text{L atm} / \text{mol K}) \times (10 + 273 \, \text{K})}{17.2 \, \text{L}} \times \frac{760 \, \text{mmHg}}{1 \, \text{atm}}
\]

Result:
\[ P = 45.6 \, \text{mmHg} \]

Note: The result is labeled as **"why is wrong"** and **highlighted**.

---

**Question Part 1B:**

**Two methods to reduce the pressure:**
1. Lower the temperature.
2. Decrease the amount of CH₄.

**Question:**
“How can I add more detail atom scale explanation?”

---

**Part 2:**

**Equation Used:**
\[ M = \frac{mRT}{P} \]

Calculations:
\[
M = \frac{7.2 \, \text{g} \times 0.082 \, \text{L atm} / \text{mol K} \times (79.1 + 273 \, \text{K})}{1.15 \, \text{atm}}
\]

Result:
\[ M = 143.5 \, \text{g/mol} \]

Note: This result is accompanied by a note indicating **"too many sfics"** referring to significant figures.

--- 

**Notes for Understanding:**
- Conversions between units and settings for significant figures are important to avoid errors.
- Correct any discrepancies by ensuring consistent units and proper application of gas laws.

---
Transcribed Image Text:**Educational Website Transcription** --- **Question Part 1A:** Given Information: - Mass (\( n \)) = 7.2 g - Molar Mass = 16 g/mol (for CH₄) - Number of moles (\( n \)) = 0.45 mol of CH₄ **Equation Used:** \[ P = \frac{nRT}{V} \] Calculations: \[ P = \frac{(0.45 \, \text{mol}) \times (0.082 \, \text{L atm} / \text{mol K}) \times (10 + 273 \, \text{K})}{17.2 \, \text{L}} \times \frac{760 \, \text{mmHg}}{1 \, \text{atm}} \] Result: \[ P = 45.6 \, \text{mmHg} \] Note: The result is labeled as **"why is wrong"** and **highlighted**. --- **Question Part 1B:** **Two methods to reduce the pressure:** 1. Lower the temperature. 2. Decrease the amount of CH₄. **Question:** “How can I add more detail atom scale explanation?” --- **Part 2:** **Equation Used:** \[ M = \frac{mRT}{P} \] Calculations: \[ M = \frac{7.2 \, \text{g} \times 0.082 \, \text{L atm} / \text{mol K} \times (79.1 + 273 \, \text{K})}{1.15 \, \text{atm}} \] Result: \[ M = 143.5 \, \text{g/mol} \] Note: This result is accompanied by a note indicating **"too many sfics"** referring to significant figures. --- **Notes for Understanding:** - Conversions between units and settings for significant figures are important to avoid errors. - Correct any discrepancies by ensuring consistent units and proper application of gas laws. ---
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