A dilute solution of hydrochloric acid with a mass of 614.86 g and containing 0.33004 mol of HCl was exactly neutralized in a calorimeter by the sodium hydroxide in 619.41 g of a comparably dilute solution. The temperature increased from 15.529 to 19.691°C. The specific heat of the HCl solution was 4.031 J·g-1·°C-1; that of the NaOH solution was 4.046 J·g-1·°C-1. The heat capacity of the calorimeter was 77.99 J·°C-1.

Question

A dilute solution of hydrochloric acid with a mass of 614.86 g and containing 0.33004 mol of HCl was exactly neutralized in a calorimeter by the sodium hydroxide in 619.41 g of a comparably dilute solution. The temperature increased from 15.529 to 19.691°C. The specific heat of the HCl solution was 4.031 J·g-1·°C-1; that of the NaOH solution was 4.046 J·g-1·°C-1. The heat capacity of the calorimeter was 77.99 J·°C-1.

Answer 1

I don't see a question here. I assume you want to calculate the heat of neutralization (in what units?)

Answer 2

To find the heat of neutralization, we can use the formula:

q = mcΔT

Where:
q = heat (in J)
m = mass (in g)
c = specific heat capacity (in J/g·°C)
ΔT = change in temperature (in °C)

First, let's calculate the heat of the hydrochloric acid (HCl):

qHCl = mHCl × cHCl × ΔTHCl

qHCl = 614.86g × 4.031 J·g^(-1)·°C^(-1) × (19.691 - 15.529)°C

Now, let's calculate the heat of the sodium hydroxide (NaOH):

qNaOH = mNaOH × cNaOH × ΔTNaOH

qNaOH = 619.41g × 4.046 J·g^(-1)·°C^(-1) × (19.691 - 15.529)°C

The overall heat change in the calorimeter is given by:

qcalorimeter = -qHCl = -qNaOH

Now, let's calculate qcalorimeter:

qcalorimeter = -(qHCl + qNaOH)

Finally, we can add up the heat changes to find the heat of neutralization:

qneutralization = qHCl + qNaOH

qneutralization = -qcalorimeter

This will give us the heat of neutralization in joules (J).