Calculate the The molar enthalpy of neutralization AHneut given the followinginformation.qíci soln = 3.9 x 10² J9NAOH Soln = 4.5 x 10² J9cal = 43.7 Jmoles of limiting reagent = 0.145 moles

qcal = qNaOH + qHCl + n × ΔHneut On substituting the values into it, we get ; 43.7 J = 450 J + 390 J…

Calculate the The molar enthalpy of neutralization AHneut given the following information. CHci soln = 3.9 x 10² J 9NAOH Soln = 4.5 x 10² J ¶cal = 43.7 J %3D moles of limiting reagent = 0.145 moles %3D

Calculate the The molar enthalpy of neutralization AHneut given the following information. CHci soln = 3.9 x 10² J 9NAOH Soln = 4.5 x 10² J ¶cal = 43.7 J %3D moles of limiting reagent = 0.145 moles %3D

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter21: Potentiometry

Section: Chapter Questions

Problem 21.22QAP

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Stoichiometry

Stoichiometry is a chemical division that evaluates the quantitative content of the chemical process with the help of the reacting molecules or products taking part in the reaction. The term “stoicheion” in “Stoichiometry”, is an old Greek term for "element" whereas “metron” means "measure".

Question

Calculate the The molar enthalpy of neutralization AHneut given the following
information.
qíci soln = 3.9 x 10² J
9NAOH Soln = 4.5 x 10² J
9cal = 43.7 J
moles of limiting reagent = 0.145 moles

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