. Eathalpy Heat) of Neutralization for an Acid-Base Reaction HNO, +NaOH Trial 1
ID: 551823 • Letter: #
Question
. Eathalpy Heat) of Neutralization for an Acid-Base Reaction HNO, +NaOH Trial 1 Trial I Trial 2 Trial 2 1. Volume of acid (mL) 23 R.7 50 mL 2. Tempernture of scid CC) 3. Volume of NaOH (mL) 4. Temperature of NaOH Cc 5. Exact molar concentration of NaOH (molL) 6. Maximam temperature from graph (c) 7. Instructor's approval of graph Calculations for Enthalpy (Heat) of Neutralization for an Acid-Base Reaction 1. Average initial temperature of acid and NaOHCC) 2. Temperature change. ATCC) 3. Volume of final mixture (mL) 4. Mass of final mixture () (Assume the density of the solution is 1.0 g/mL.) S. Specific heat of mixture 6. Heat evolved J) 4.183/rc 4.18 Wg'C 8. Moles of H,O formed (mol) 9, ahl,OLIinol 11,0), equation 25.8 10. Average AH, lmol HO Show cakulations for Trial I using the correct number of significant figures Comment on your two values of 14 302 CalorimetryExplanation / Answer
Energy Change = Mass of water Specific heat of water Temperature change of water
= 16.692*1.00 * (100-21.2) Cal = 1315.3296 Cal. 1160.094
Since the energy gained by the water equals the energy released by the metal, the specific heat of the metal is calculated
Specific heat of the metal = Energy released by the metal/(Mass of metalTemperature change of metal)
=1315.3296/(14.0978*69.5) = 1315.3296/979.7971 = 1.3424 cal/gm degreecelcius.
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