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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![**Chapter 4 - Stoichiometry: Quantitative Information About Chemical Reactions**
**Problem:**
If you dilute 35.0 mL of 2.25 M hydrochloric acid to 550. mL, what is the molar concentration of the dilute acid?
Molar concentration = _______ M
---
**Explanation:**
To solve this dilution problem, use the dilution formula:
\[ C_1V_1 = C_2V_2 \]
Where:
- \( C_1 \) is the initial concentration (2.25 M),
- \( V_1 \) is the initial volume (35.0 mL),
- \( C_2 \) is the final concentration (unknown),
- \( V_2 \) is the final volume (550. mL).
Rearrange the equation to solve for \( C_2 \):
\[ C_2 = \frac{C_1 \times V_1}{V_2} \]
This equation allows calculating the final molar concentration after dilution. Plug the values into the equation to find the answer.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff7ba9c54-af2b-48c5-8dd4-4f214c676725%2Fc4a83974-36c0-449f-812e-78af52bf3329%2Fcbn01u_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Chapter 4 - Stoichiometry: Quantitative Information About Chemical Reactions**
**Problem:**
If you dilute 35.0 mL of 2.25 M hydrochloric acid to 550. mL, what is the molar concentration of the dilute acid?
Molar concentration = _______ M
---
**Explanation:**
To solve this dilution problem, use the dilution formula:
\[ C_1V_1 = C_2V_2 \]
Where:
- \( C_1 \) is the initial concentration (2.25 M),
- \( V_1 \) is the initial volume (35.0 mL),
- \( C_2 \) is the final concentration (unknown),
- \( V_2 \) is the final volume (550. mL).
Rearrange the equation to solve for \( C_2 \):
\[ C_2 = \frac{C_1 \times V_1}{V_2} \]
This equation allows calculating the final molar concentration after dilution. Plug the values into the equation to find the answer.
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