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.

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) 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

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

To calculate the mass of the resulting solution, we need to know the densities of both the acid HA and the base MOH solutions. However, in the given information, we only have the density of the resulting solution, which is 1.04 g/mL. Therefore, we'll assume that the density of the resulting solution is the same as that of the acid HA or the base MOH solutions.

Let's first calculate the moles of acid and base present in the solution using the following equation:

moles = concentration (M) x volume (L)

Moles of acid HA = 0.50 M x 0.025 L = 0.0125 mol
Moles of base MOH = 0.50 M x 0.025 L = 0.0125 mol

Since the acid HA and base MOH react in a 1:1 ratio, the number of moles of acid and base are the same in the solution.

Next, we need to calculate the total heat transfer, which can be determined using the formula:

q = C x m x ΔT

where q is the heat transfer, C is the specific heat capacity, m is the mass of the solution, and ΔT is the change in temperature.

In this case, the heat transfer (q) is equal to zero because the acid and base are completely neutralized, and there is no change in enthalpy. Therefore, we can set up the equation as:

0 = C x m x ΔT

Solving for mass (m), we get:

m = 0 / (C x ΔT)

m = 0 / (3.92 J/g·C x (34.7 - 23.3) C)

m = 0 g (since we are dividing by zero)

However, since we have assumed the density of the resulting solution to be the same as the acid HA or base MOH, we can multiply the volume of the resulting solution by the density to find the mass.

Mass of resulting solution = volume x density

= (25.0 mL + 25.0 mL) x 1.04 g/mL

= 52.0 g

Therefore, the mass of the resulting solution is 52.0 grams.