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
What quantity in moles is BaI2 present in ?
![**Chemistry Problem: Reaction of BaI₂ and Na₃PO₄**
- **Question:** Consider the reaction of 30.0 mL of 0.244 M BaI₂ with 20.0 mL of 0.315 M Na₃PO₄. What quantity in moles of BaI₂ are present in the solution?
- **Calculation Area:**
- Input pad with numbers 1-9, 0, decimal point, and clear options.
- The calculated result is displayed as "391 mol."
**Explanation:**
To find the moles of BaI₂, use the formula:
\[ \text{Moles of BaI₂} = \text{Volume (L)} \times \text{Molarity (M)} \]
Convert 30.0 mL to liters:
\[ 30.0 \, \text{mL} = 0.0300 \, \text{L} \]
Then calculate:
\[ \text{Moles of BaI₂} = 0.0300 \, \text{L} \times 0.244 \, \text{M} = 0.00732 \, \text{mol} \]
(Note: The displayed result of "391 mol" seems incorrect based on standard calculations.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F163dc8ba-851f-4764-b6aa-51728fdb757c%2Fb0bf2bfc-a515-4eea-8aa7-cb7e818cd11a%2F00l6v5e_processed.png&w=3840&q=75)
Transcribed Image Text:**Chemistry Problem: Reaction of BaI₂ and Na₃PO₄**
- **Question:** Consider the reaction of 30.0 mL of 0.244 M BaI₂ with 20.0 mL of 0.315 M Na₃PO₄. What quantity in moles of BaI₂ are present in the solution?
- **Calculation Area:**
- Input pad with numbers 1-9, 0, decimal point, and clear options.
- The calculated result is displayed as "391 mol."
**Explanation:**
To find the moles of BaI₂, use the formula:
\[ \text{Moles of BaI₂} = \text{Volume (L)} \times \text{Molarity (M)} \]
Convert 30.0 mL to liters:
\[ 30.0 \, \text{mL} = 0.0300 \, \text{L} \]
Then calculate:
\[ \text{Moles of BaI₂} = 0.0300 \, \text{L} \times 0.244 \, \text{M} = 0.00732 \, \text{mol} \]
(Note: The displayed result of "391 mol" seems incorrect based on standard calculations.)
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