The reaction of hydrogen(g) with oxygen(g) to form water(g) proceeds as follows: 2H2(g) + O2(g)-2H20(g) When 0.637 g H2(g) reacts with sufficient O2(g), 76.5 kJ is evolved. Calculate the value of A,H for the chemical equation given. kJ/mol
The reaction of hydrogen(g) with oxygen(g) to form water(g) proceeds as follows: 2H2(g) + O2(g)-2H20(g) When 0.637 g H2(g) reacts with sufficient O2(g), 76.5 kJ is evolved. Calculate the value of A,H for the chemical equation given. kJ/mol
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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![### Enthalpy Changes I
The reaction of hydrogen (\( \text{H}_2 \)(g)) with oxygen (\( \text{O}_2 \)(g)) to form water (\( \text{H}_2\text{O} \)(g)) proceeds as follows:
\[ 2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(g) \]
When 0.637 g of \( \text{H}_2 \)(g) reacts with sufficient \( \text{O}_2 \)(g), 76.5 kJ is evolved.
Calculate the value of \( \Delta H \) for the chemical equation given.
\[ \boxed{} \, \text{kJ/mol} \]
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Transcribed Image Text:### Enthalpy Changes I
The reaction of hydrogen (\( \text{H}_2 \)(g)) with oxygen (\( \text{O}_2 \)(g)) to form water (\( \text{H}_2\text{O} \)(g)) proceeds as follows:
\[ 2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(g) \]
When 0.637 g of \( \text{H}_2 \)(g) reacts with sufficient \( \text{O}_2 \)(g), 76.5 kJ is evolved.
Calculate the value of \( \Delta H \) for the chemical equation given.
\[ \boxed{} \, \text{kJ/mol} \]
**Instructions:**
1. Fill in the text box with your calculated value of \( \Delta H \).
2. Click on "Submit" to check your answer.
3. If needed, you can view the approach or detailed steps by selecting "Show Approach" or "Show Tutor Steps".
**Note:**
You have 4 item attempts remaining.
**Options:**
- Submit Answer
- Try Another Version
---
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