sample calculations

sample calculations - = 3022.64 J Moles of water produced...

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A. Heat of Neutralization for HCl and NaOH for Trial 1 T for reaction, o C = (T f -T 1 ) = 31.0 o C – 24.0 o C = 7 o C Heat gained by solution (J) = ( T x 100g x 4.18J/g o C) = 7 o C x 100g x 4.18J/g o C = 2926.0 J Heat absorbed by calorimeter (J) = ( T x Calorimeter Heat Capacity) = 7 o C x 13.52 J/ o C = 96.64 J Total heat released by reaction (J) = (heat gained by solution, J) + (Heat absorbed by calorimeter, J) = 2926.0 J + 96.64 J
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Unformatted text preview: = 3022.64 J Moles of water produced (moles) by using the HCl information = 0.05 L x . 1 0 moles of HClL x 1 mol of H2O1 mol of HCl =0.05 mol of H 2 O Molar heat of reaction, H rxn in kJ = (total heat released / moles of water produced) = . . 3022 64J0 05moles = 60452.8 J = 60.45 kJ Average value of H rxn = . + . + . 60 45 kJ 65 59 kJ 65 50 kJ3 = 63.88 kJ...
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This note was uploaded on 03/18/2010 for the course COS 101121501 taught by Professor Langner during the Fall '08 term at RIT.

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sample calculations - = 3022.64 J Moles of water produced...

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