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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Question
Confused about #6
![**Educational Content - Heat of Neutralization Report Sheet**
---
**144. Report Sheet - Heat of Neutralization**
6. **Complete:**
\( \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \)
7. **The number of moles of HCl in 50 mL of 1.0 M HCl (show calculations):**
\[ = 0.05 \, \text{mol} \]
8. **The number of moles of H\(_2\)O produced in reaction of 50 mL 1.0 M HCl and 50 mL 1.0 M NaOH (show calculations):**
\[ \text{mol}(\text{H}_2\text{O}) = \text{mol}(\text{HCl}) \]
\[ = 0.05 \, \text{mol} \]
9. **Joules released per mole of water formed:**
\[
\text{total joules released (5)} = 2903.3 \, \text{J}
\]
\[
\frac{\text{2903.3 J}}{\text{0.05 mol}} = 57986.6 \, \text{J/mol} \approx 57999 \, \text{J/mol}
\]
(Note: Slight rounding differences might occur in calculations.)
---
This information is designed to guide students through the heat of neutralization process using a mixture of HCl and NaOH, illustrating how to calculate moles and energy changes in the reaction.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a86590d-c1f5-4894-93d9-46c435b15aa5%2Fc95a4c50-527d-4f43-9b8e-318a3bd6d5b9%2Fdbbo3jf.jpeg&w=3840&q=75)
Transcribed Image Text:**Educational Content - Heat of Neutralization Report Sheet**
---
**144. Report Sheet - Heat of Neutralization**
6. **Complete:**
\( \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \)
7. **The number of moles of HCl in 50 mL of 1.0 M HCl (show calculations):**
\[ = 0.05 \, \text{mol} \]
8. **The number of moles of H\(_2\)O produced in reaction of 50 mL 1.0 M HCl and 50 mL 1.0 M NaOH (show calculations):**
\[ \text{mol}(\text{H}_2\text{O}) = \text{mol}(\text{HCl}) \]
\[ = 0.05 \, \text{mol} \]
9. **Joules released per mole of water formed:**
\[
\text{total joules released (5)} = 2903.3 \, \text{J}
\]
\[
\frac{\text{2903.3 J}}{\text{0.05 mol}} = 57986.6 \, \text{J/mol} \approx 57999 \, \text{J/mol}
\]
(Note: Slight rounding differences might occur in calculations.)
---
This information is designed to guide students through the heat of neutralization process using a mixture of HCl and NaOH, illustrating how to calculate moles and energy changes in the reaction.
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