b) The heat absorbed by the solution.
c) The heat absorbed by the calorimeter
d) The heat released by the reaction
e) The enthalpy change of the reaction
3. 25.0 mL of a 0.50 M solution of acid HA was combined with 25.0 mL of a 0.50 solution of base MOH
in a calorimeter with a calorimeter constant of 13.5 J/ C. The initial temperature of the solution was 23.3
C and the maximum temperature was 34.7 C. The resulting solution had a specific heat capacity of 3.92
J/g· C and a density of 1.04 g/mL.
Calculate each of the following
a) The mass of the resulting solution.
5 answers
How much do you know how to do? What is your hang up? That is, what do you not understand.
I honestly don't know where to begin on this problem.
(a). You have 25.0 mL added to 25.0 mL or 50 mL total. Then you know density; use that to convert 50.0 mL to grams.
(b) heat absorbed by solution = q.
q = mass soln x specific heat soln x (Tfinal-Tinitial)
(c)heat absorbed by calorimeter = Ccal x (Tfinal-Tinitial)
(d) (b) + (c)
(e)delta H = q in this case. Usually these are reported as dH/mol. You had 0.5M x 0.025L = 0.0125 mol so
q/0.0125 = ? J/mol and change that to kJ/mol
(b) heat absorbed by solution = q.
q = mass soln x specific heat soln x (Tfinal-Tinitial)
(c)heat absorbed by calorimeter = Ccal x (Tfinal-Tinitial)
(d) (b) + (c)
(e)delta H = q in this case. Usually these are reported as dH/mol. You had 0.5M x 0.025L = 0.0125 mol so
q/0.0125 = ? J/mol and change that to kJ/mol
A) 50.0mlx 1.0g/ml= 50.0g
B) 50.0g(4.814)(34.7c-23.3c)= 2743.98
C) 13.5j/c(34.8-23.3)= 153.9
D)2743.98-153.9=2590.08
E)0.0125(1000/1)= 12.5 J?
Is this correct? By the way thank you for your help.
B) 50.0g(4.814)(34.7c-23.3c)= 2743.98
C) 13.5j/c(34.8-23.3)= 153.9
D)2743.98-153.9=2590.08
E)0.0125(1000/1)= 12.5 J?
Is this correct? By the way thank you for your help.
1 answer