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...
Related questions
Question
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.
---](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ae764b7-d237-4053-a9e9-5b016ede3169%2F2af3eab1-866a-44c8-b631-9ebfaa70ff3d%2F8rr5m9u_processed.png&w=3840&q=75)
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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