Calculate the Enthalpy of change in Kj/mol and % difference in the neutralization reaction below: HNO2 + NH3 = N2 + 2 H20 Initial Temp= 20 Deg celsius Final Temp= 22.3 Deg celsius Conc= 0.4 M [HNO3] and 0.4 M [NH3] Vol= 55 mL [HNO2] and 60 mL [NH3]= Mass is 115 g Specific Heat= 4.184 J/g For the other value in % difference use: -348 Kj/mol

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Calculate the Enthalpy of change in Kj/mol and
% difference in the neutralization reaction
below:
HNO2 + NH3 = N2 + 2 H20
Initial Temp= 20 Deg celsius
Final Temp= 22.3 Deg celsius
Conc= 0.4 M [HNO3] and 0.4 M [NH3]
Vol= 55 mL [HNO2] and 60 mL [NH3]= Mass is
115 g
Specific Heat= 4.184 J/g
For the other value in % difference use: -348
Kj/mol
Transcribed Image Text:Calculate the Enthalpy of change in Kj/mol and % difference in the neutralization reaction below: HNO2 + NH3 = N2 + 2 H20 Initial Temp= 20 Deg celsius Final Temp= 22.3 Deg celsius Conc= 0.4 M [HNO3] and 0.4 M [NH3] Vol= 55 mL [HNO2] and 60 mL [NH3]= Mass is 115 g Specific Heat= 4.184 J/g For the other value in % difference use: -348 Kj/mol
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