Asked by Anonymous
when 100mL of 1.00 mol/L sodium hydroxide is added to 100mL of 1.00mol/L hydrochloric acid in an insulated container, the temperature rises from 21.0 degress to 34.6 degrees. Calculate the heat of neutralisation for the reaction.
Answers
Answered by
DrBob222
Total q = mass H2O x specific heat H2O x delta T = 200g x 4.184 J/g x (34.6-21.0) = ?? (If you want this in calories substitute 1 cal/g for the 4.184 J/g.
This usually is expressed as J/mol, or cal/mol.
This usually is expressed as J/mol, or cal/mol.
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