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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![A student mixes sodium hydroxide solution with a monoprotic acid, HA. The collected data are listed in the following table:
| Property | Value |
|-----------------------------------------|----------------|
| [HA] (M) | 4.75 |
| Initial temperature of HA (°C) | 21.0 |
| Volume of HA (mL) | 56.2 |
| [NaOH] (M) | 2.45 |
| Initial temperature of NaOH (°C) | 21.0 |
| Volume of NaOH (mL) | 85.0 |
| Final solution temperature (°C) | 35.5 |
| Density of solution (g/mL) | 1.70 |
| Specific heat of solution (J/g·°C) | 4.004 |
| Calorimeter constant (J/°C) | 38.76 |
**Exercise:**
Part 1 of 4: Identify the limiting reagent for the reaction. [NaOH]
*Note: The part about identifying the limiting reagent likely involves a dropdown menu selection as part of an interactive exercise.*](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9ce1fa32-745b-446a-832b-1aa8581f8e40%2F66d12a7e-d9d6-48ce-9eef-bc890d9aa985%2F9niei3g_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A student mixes sodium hydroxide solution with a monoprotic acid, HA. The collected data are listed in the following table:
| Property | Value |
|-----------------------------------------|----------------|
| [HA] (M) | 4.75 |
| Initial temperature of HA (°C) | 21.0 |
| Volume of HA (mL) | 56.2 |
| [NaOH] (M) | 2.45 |
| Initial temperature of NaOH (°C) | 21.0 |
| Volume of NaOH (mL) | 85.0 |
| Final solution temperature (°C) | 35.5 |
| Density of solution (g/mL) | 1.70 |
| Specific heat of solution (J/g·°C) | 4.004 |
| Calorimeter constant (J/°C) | 38.76 |
**Exercise:**
Part 1 of 4: Identify the limiting reagent for the reaction. [NaOH]
*Note: The part about identifying the limiting reagent likely involves a dropdown menu selection as part of an interactive exercise.*
![**Transcription for Educational Website**
**Exercise: Calculating Enthalpy of Neutralization**
**Part 2 of 4**
*Progress Bar: Displayed as a horizontal blue progress indicator halfway filled.*
**Part 3 of 4**
**Task:**
Find the enthalpy of neutralization (in kJ/mol). Round your answer to 3 significant digits.
**Input Section:**
- \(\Delta H_{\text{neutralization}} =\) [Input Box] \(\frac{\text{kJ}}{\text{mol}}\)
**Additional Features:**
- A calculator tool is provided, which includes:
- An input box with an "x10" superscript option.
- Buttons for operations like clear (marked as 'X') and reset (indicated by a circular arrow).
**Navigation:**
- A "Next Part" button is located at the bottom to proceed further.
**Note:** Ensure accurate input with three significant digits for precise calculation results.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9ce1fa32-745b-446a-832b-1aa8581f8e40%2F66d12a7e-d9d6-48ce-9eef-bc890d9aa985%2F07ipuh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Transcription for Educational Website**
**Exercise: Calculating Enthalpy of Neutralization**
**Part 2 of 4**
*Progress Bar: Displayed as a horizontal blue progress indicator halfway filled.*
**Part 3 of 4**
**Task:**
Find the enthalpy of neutralization (in kJ/mol). Round your answer to 3 significant digits.
**Input Section:**
- \(\Delta H_{\text{neutralization}} =\) [Input Box] \(\frac{\text{kJ}}{\text{mol}}\)
**Additional Features:**
- A calculator tool is provided, which includes:
- An input box with an "x10" superscript option.
- Buttons for operations like clear (marked as 'X') and reset (indicated by a circular arrow).
**Navigation:**
- A "Next Part" button is located at the bottom to proceed further.
**Note:** Ensure accurate input with three significant digits for precise calculation results.
Expert Solution
Step 1
Heat Evolved when one gram equivalent of acid is completely neutralized by a base in a dilute solution is known as Heat of Neutralization.
Neutralization Process is a exothermic process and heat released by the reaction is absorbed by the solution and calorimeter.
Therefore,
Heat of Neutralization + Heat absorbed by Solution + Heat absorbed by Calorimeter = 0
We will use the above equation to solve the problem.
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