Volume of acid (mL) Temperature of acid (degree C) Volume of BNaO (mL) Temperatu
ID: 1064153 • Letter: V
Question
Volume of acid (mL) Temperature of acid (degree C) Volume of BNaO (mL) Temperature of NaOH (degree C) Exact molar concentration of NaOH (mol/L) Maximum temperature from graph (degree C) Instructor's approval of graph Calculations for Enthalpy (Heat) of Neutralization for an Acid-Base Reaction Average initial temperature of acid and NaOH (degree C) Temperature change, Delta T(degree C) Volume of final mixture (mL) Mass of final mixture (g) (Assume the density of the solution is 1.0 g/mL.) Specific heat of mixture Heat evolved (J) Moles of OH reacted, the limiting reactant (mol) Moles of H_2O formed (mol) Delta H_n (kJ/mol H_2 O). equation 25.8 Average Delta H_n (kJ/mol H_2O)Explanation / Answer
1.
Temperature of NaOH =
22.0 C
2 Temperature change = 27.9 -22.0= 5.9 C
3 total volume of solution = 12.5 ml+12.5 ml
= 25.0 ml
4
Given that density = mass / volume
Mass = density * volume
= 1..0 g/ ml* 25.0 ml
= 25 g
5.
Specific heat of mixture = 4.18 J/ g-C
6 Amount of heat evolved ; Q = mcdT
= 25.0 g * 4.18 J / g-C* 5.9 C
= 616.55 J
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