Given the following data: N2H4(1) + CH4O(1) CH2O(g) + N2(g) + 3H2(g) Part A) N2(g) + 3H2(g) → 2NH3(g) AH = -37 kJ AH -91.8 kJ A,H=-65 kJ CH4O(1) CH2O(g) + H2(g) -> Using Hess's Law, the enthalpy of reaction for the following equation: N2H4(1) + H2(g) → 2NH3(g) is calculated to be kJ. Given the following data: Part B) 2CIF(g) + O2(g) → Cl₂O(g) + F2O(g) AH +167.4 kJ 2CIF3(g) + 202(g) - Cl₂O(g) + 3F2O(g) AH +341.4 kJ A,H = -43.4 kJ 2F2(g) + O2(g) 2F₂O(g) Using Hess's Law, the enthalpy of reaction for the following equation: CIF(g) + F2(g) → CIF3(g) is calculated to be kJ. Record your 4-digit answer. Include sign; do not include units.

Introduction to General, Organic and Biochemistry
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Hi there! This is one question I have attached here. I know this may seem long but they really are in parts since the second part is related to part A) given I promise. From my side I had to take separate pictures of the parts because the parts are long as it won't let me take a full picture. Can you please please help answer the part A) and part B) for me please since I've been struggling to answer this question for days. Thank you so much. :)
Given the following data:
N2H4(1) + CH4O(1)CH2O(g) + N2(g) + 3H2(g)
N2(g) + 3H2(g) → 2NH3(g)
T
CH4O(1) CH2O(g) + H2(g)
AH = -37 kJ
A,H=-91.8 kJ
A,H=-65 kJ
Using Hess's Law, the enthalpy of reaction for the following equation: N₂H4(1) + H2(g) 2NH3(g) is calculated to be
kJ.
Part A)
Given the following data:
Part B)
T
2CIF(g) + O2(g) → Cl₂O(g) + F2O(g)
A,H+167.4 kJ
A,H+341.4 kJ
A,H = -43.4 kJ
2CIF3(g) + 202(g) → Cl₂O(g) + 3F₂O(g)
2F2(g) + O2(g) - 2F₂O(g)
Using Hess's Law, the enthalpy of reaction for the following equation: CIF(g) + F2(g) → CIF3(g) is calculated to be
KJ.
Record your 4-digit answer. Include sign; do not include units.
Transcribed Image Text:Given the following data: N2H4(1) + CH4O(1)CH2O(g) + N2(g) + 3H2(g) N2(g) + 3H2(g) → 2NH3(g) T CH4O(1) CH2O(g) + H2(g) AH = -37 kJ A,H=-91.8 kJ A,H=-65 kJ Using Hess's Law, the enthalpy of reaction for the following equation: N₂H4(1) + H2(g) 2NH3(g) is calculated to be kJ. Part A) Given the following data: Part B) T 2CIF(g) + O2(g) → Cl₂O(g) + F2O(g) A,H+167.4 kJ A,H+341.4 kJ A,H = -43.4 kJ 2CIF3(g) + 202(g) → Cl₂O(g) + 3F₂O(g) 2F2(g) + O2(g) - 2F₂O(g) Using Hess's Law, the enthalpy of reaction for the following equation: CIF(g) + F2(g) → CIF3(g) is calculated to be KJ. Record your 4-digit answer. Include sign; do not include units.
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