---**Calculate \(\Delta H\) for the reaction:**\[ 2\text{NH}_3(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{N}_2\text{H}_4(l) + \text{H}_2\text{O}(l) \]**given the following data:**\[ 2\text{NH}_3(g) + 3\text{N}_2\text{O}(g) \rightarrow 4\text{N}_2(g) + 3\text{H}_2\text{O}(l) \quad \Delta H = -1010. \text{kJ} \]\[ \text{N}_2\text{O}(g) + 3\text{H}_2(g) \rightarrow \text{N}_2\text{H}_4(l) + \text{H}_2\text{O}(l) \quad \Delta H = -317 \text{kJ} \]\[ \text{N}_2\text{H}_4(l) + \text{O}_2(g) \rightarrow \text{N}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta H = -623 \text{kJ} \]\[ \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H = -286 \text{kJ} \]**\(\Delta H =\) \_\_\_\_ \text{kJ}**---*Buttons below:*- **Submit Answer**- **Try Another Version**(Note: 3 item attempts remaining)---

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2NH3 (g) + 02(9) → N2H4(1) + H20(1) given the following data: 2NH3 (g) + 3N20(g) → 4N2(9) + 3H2O(1) AH = -1010. kJ N20(g) + 3H2(g) → N½H4(1) + H2O(1) AH = -317 kJ N2H4 (1) + O2 (9) → N2(g) +2H2O(1) AH = -623 kJ H2 (9) + 02 (9) → H2O(1) AH = -286 kJ ΔΗ= kJ %D

2NH3 (g) + 02(9) → N2H4(1) + H20(1) given the following data: 2NH3 (g) + 3N20(g) → 4N2(9) + 3H2O(1) AH = -1010. kJ N20(g) + 3H2(g) → N½H4(1) + H2O(1) AH = -317 kJ N2H4 (1) + O2 (9) → N2(g) +2H2O(1) AH = -623 kJ H2 (9) + 02 (9) → H2O(1) AH = -286 kJ ΔΗ= kJ %D

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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