Given the reaction P,O10 (s) + H2O → H3 PO4 (aq) AH = - 96, %3D .2 kJ what the AH for P4010 (s) + 6 H20 → 4 H3 PO4 (aq) be? 192.4 kJ -192.4 kJ -384.8 kJ 384.8 kJ > 79°F Clou
Given the reaction P,O10 (s) + H2O → H3 PO4 (aq) AH = - 96, %3D .2 kJ what the AH for P4010 (s) + 6 H20 → 4 H3 PO4 (aq) be? 192.4 kJ -192.4 kJ -384.8 kJ 384.8 kJ > 79°F Clou
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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![Given the reaction:
\[
\frac{1}{4} \text{P}_4 \text{O}_{10} \, (s) + \frac{3}{2} \text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 \, (aq) \quad \Delta H = -96.2 \, \text{kJ}
\]
What is the \(\Delta H\) for the reaction:
\[
\text{P}_4 \text{O}_{10} \, (s) + 6 \text{H}_2\text{O} \rightarrow 4 \text{H}_3\text{PO}_4 \, (aq)
\]
Options:
- \(192.4 \, \text{kJ}\)
- \(-192.4 \, \text{kJ}\)
- \(-384.8 \, \text{kJ}\)
- \(384.8 \, \text{kJ}\)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7f7d658b-0bac-4f55-9528-936034acb7b3%2F63ba6927-96a4-4594-8923-29048c52202f%2Fx3ykc1t_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Given the reaction:
\[
\frac{1}{4} \text{P}_4 \text{O}_{10} \, (s) + \frac{3}{2} \text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 \, (aq) \quad \Delta H = -96.2 \, \text{kJ}
\]
What is the \(\Delta H\) for the reaction:
\[
\text{P}_4 \text{O}_{10} \, (s) + 6 \text{H}_2\text{O} \rightarrow 4 \text{H}_3\text{PO}_4 \, (aq)
\]
Options:
- \(192.4 \, \text{kJ}\)
- \(-192.4 \, \text{kJ}\)
- \(-384.8 \, \text{kJ}\)
- \(384.8 \, \text{kJ}\)
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