Given the data: Volume of 1.00M HCI (mL) 45.0 Volume of 1.00 M NAOH (mL) 50.0 Initial Temperature (C) 22.1 Final Temperature ("C) 31.6 Calculate the heat of the solution, qsolution, in units of J. NOTE: q= mc AT Specific heat for solution: c = 4.18 J/ g C %3D Density of the solution = 1.00 g/mL %3D 3776.06
Given the data: Volume of 1.00M HCI (mL) 45.0 Volume of 1.00 M NAOH (mL) 50.0 Initial Temperature (C) 22.1 Final Temperature ("C) 31.6 Calculate the heat of the solution, qsolution, in units of J. NOTE: q= mc AT Specific heat for solution: c = 4.18 J/ g C %3D Density of the solution = 1.00 g/mL %3D 3776.06
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
7) did i do this right?
![### Transcription and Explanation for Educational Website
#### Given Data:
- **Volume of 1.00 M HCl (mL):** 45.0
- **Volume of 1.00 M NaOH (mL):** 50.0
- **Initial Temperature (°C):** 22.1
- **Final Temperature (°C):** 31.6
#### Calculation:
Calculate the heat of the solution, \( q_{\text{solution}} \), in units of Joules (J).
**Formula:**
\[ q = m \cdot c \cdot \Delta T \]
**Where:**
- \( q \) = heat energy (Joules)
- \( m \) = mass of the solution (grams)
- \( c \) = specific heat capacity (J/g°C)
- \( \Delta T \) = change in temperature (°C)
**Specifications:**
- **Specific heat for the solution:** \( c = 4.18 \, \text{J/g°C} \)
- **Density of the solution:** \( 1.00 \, \text{g/mL} \)
**Calculated Result:**
The calculated heat of the solution is **3776.06 J**.
---
#### Explanation:
To solve for the heat of the solution, we use the formula:
1. **Calculate the mass of the solution:**
Since the density is 1.00 g/mL, the mass (m) is equal to the volume in mL. Therefore, the total volume is \( 45.0 \, \text{mL (HCl)} + 50.0 \, \text{mL (NaOH)} = 95.0 \, \text{mL} \).
Thus, \( m = 95.0 \, \text{g} \).
2. **Determine the temperature change (\( \Delta T \)):**
\[ \Delta T = \text{Final Temperature} - \text{Initial Temperature} \]
\[ \Delta T = 31.6^\circ \text{C} - 22.1^\circ \text{C} = 9.5^\circ \text{C} \]
3. **Substitute into the formula:**
\[ q = 95.0 \, \text{g} \times](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7f284677-4b29-4aaa-9fa9-5c6d32626ad8%2F5d6d6dd1-fe2b-498c-b59d-d41dc8d61969%2Fexjiynk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Transcription and Explanation for Educational Website
#### Given Data:
- **Volume of 1.00 M HCl (mL):** 45.0
- **Volume of 1.00 M NaOH (mL):** 50.0
- **Initial Temperature (°C):** 22.1
- **Final Temperature (°C):** 31.6
#### Calculation:
Calculate the heat of the solution, \( q_{\text{solution}} \), in units of Joules (J).
**Formula:**
\[ q = m \cdot c \cdot \Delta T \]
**Where:**
- \( q \) = heat energy (Joules)
- \( m \) = mass of the solution (grams)
- \( c \) = specific heat capacity (J/g°C)
- \( \Delta T \) = change in temperature (°C)
**Specifications:**
- **Specific heat for the solution:** \( c = 4.18 \, \text{J/g°C} \)
- **Density of the solution:** \( 1.00 \, \text{g/mL} \)
**Calculated Result:**
The calculated heat of the solution is **3776.06 J**.
---
#### Explanation:
To solve for the heat of the solution, we use the formula:
1. **Calculate the mass of the solution:**
Since the density is 1.00 g/mL, the mass (m) is equal to the volume in mL. Therefore, the total volume is \( 45.0 \, \text{mL (HCl)} + 50.0 \, \text{mL (NaOH)} = 95.0 \, \text{mL} \).
Thus, \( m = 95.0 \, \text{g} \).
2. **Determine the temperature change (\( \Delta T \)):**
\[ \Delta T = \text{Final Temperature} - \text{Initial Temperature} \]
\[ \Delta T = 31.6^\circ \text{C} - 22.1^\circ \text{C} = 9.5^\circ \text{C} \]
3. **Substitute into the formula:**
\[ q = 95.0 \, \text{g} \times
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Recommended textbooks for you
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY