Enthalpy change for the forward reaction of 1 mole of (CH3)2NNH2 is -1694 kJ.
When the direction of reaction is reversed then sign of ∆H is also reversed.
Hence for this reaction sign of ∆H will be positive.
Since reaction is multiplied by 2 , enthalpy will also be multiplied by 2.
∆H = 1694 × 2
= 3388 kJ
Answer ---- ∆H = +3388 kJ
! The combustion reaction of dimethylhydrazine is used to fuel rockets (CH3)2NNH, (1) +40,(8) —— N,(8)+44,0...
1st attempt Feedback The combustion reaction of dimethylhydrazine is used to fuel rockets. (CH3)2NNH, (1) +402(8) —— N2(g) + 4H2O(g) +2002(g) AH x = -1694 kJ Calculate the enthalpy change for the following reaction: 2N,(g) + 8H,O(g) + 4CO2(g) → 2(CH3)2NNH, (1) +802 (8) AH - 3388 6 OF 11 QUESTIONS COMPLETED < 05/11 > Bite Balance for olb070 9 Smartwork5 - Goog.
The combustion reaction of dimethylhydrazine is used to fuel
rockets.
(CH3)2NNH2(l)+4O2(g)N2(g)+4H2O(g)+2CO2(g)
ΔHrxn= –1694 kJ
Calculate the enthalpy change for the following reaction:
2N2(g)+8H2O(g)+4CO2(g)2(CH3)2NNH2(l)+8O2(g)
ΔHrxn= kJ
Given the following balanced equation for the combustion reaction of an unknown fuel: 4 Fuel (1) + 902(9) -> 8C02(g) + 10H 20(1) + AHørxn = -3857 kJ/mol 2N2(g) Given that AH°f{CO2(g)) = -393.5 kJ/mol and AH°f[H2001)) = -285.8 kJ/mol, calculate the enthalpy of formation of this fuel. Write answer to four significant figures. Numeric Response
HQ11.47 Homework. Answered The combustion reaction of nitromethane fuel occurs as follows: 2 CH3 NO2 (1) + 3/2O2(g) → 2 CO2(g) + 3 H2O(l) + N2(g) The standard enthalpy of this reaction is -1418 kJ. What is the standard enthalpy of formation of nitromethane in kJ/mol?
11 Question (3 points) See page 243 Rocket Fuels The payload of a rocket Includes fuel and oxygen for combustion of the fuel, Reactions 1 and 2 describe the combustion of dimethylhydrazine and hydrogen, respectively. 1st attempt Part 1 (1 point) See Periodic Table See Hint Calculate the energy change for the below reaction in k per pound. (CH3)2NNH, (1) + 4H2O(g) - N,(s) + 4H2O(g) +200,() AH --1694k) КЛЬ We were unable to transcribe this image
can someone explain all of this to me?
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Consider the two reactions. 2 NH,(g) + 3 N, O(g) - 4N, (g) + 3H,O(1) 4NH,(g) + 30,(g) — 2N, (g) + 6H, 0(1) AH = -1010 KJ AH = 1531 kJ Using these two reactions, calculate and enter the enthalpy change for the reaction below. N,(g) + 40,(8) NO(g) AH- AHL
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