This is the second reaction:

HCl(aq)+NaOH(aq)-->NaCl(aq)+H2O(l)
(Heat of neutralization)

This reaction involves mixing two solutions: 1.00 mol/L NaOH and 1.00 mol/L HCl.

Trial 1: 48.0 mL of the NaOH solution is mixed with 47.5 mL of the HCl. The temperature rises from 22.00°C to 27.90°C.

HINT:
Remember that you are heating water. Use the formula Q = C x M x DT. The volume of water used equals the volume of whatever solution you are using, be it NaOH or HCl. The volume must be converted to mass (1 mL of water has a mass of 1 g). The mass must be reported in kilograms.
In the second reaction you are using a solution of NaOH. To calculate the number of moles you must use the equation c= n/V. Also the final mass of water is the sum of the volumes of the NaOH solution and the HCl solution.

Questions:
1) Calculate the heat involved in the trial(Q)
2) Calculate the moles of NaOH used in the trial(mol)
3)Calculate how much heat would be evolved if 1 mole of NaOH was used (Q/mol)

Just follow the instructions given in questions 1, 2, and 3 along with the hint which tells you what to do.

Post your work if you have trouble.

I don't understand. in the formula Q=C*M*deltaT, do you use 4.200 kj/mol as C. Is the answer to question one, calculate the heat involved in the trial, 1.19 KJ?

The heat capacity of water is 4.18 J/g*K which is 4.18 kJ/kg*K (which I assume is why the question asks for mass in kg.

You have 48.0 mL + 47.5 mL = 95.5 mL and since the density of water is 1 g/mL, the solution has a mass of 95.5 g or 0.0955 kg.

To answer these questions, we need to use the formula Q = C x M x ΔT, where Q represents the heat involved, C represents the specific heat capacity of water, M represents the mass of the water, and ΔT represents the change in temperature.

1) Calculate the heat involved in the trial (Q):
First, we need to convert the volumes of NaOH and HCl solutions used into masses of water. Given that 1 mL of water has a mass of 1 g, the mass of water used equals the volume of the NaOH solution plus the volume of the HCl solution.

Mass of water = Volume of NaOH solution + Volume of HCl solution
= 48.0 mL + 47.5 mL
= 95.5 g

Next, convert the mass of water to kilograms:
Mass of water = 95.5 g = 0.0955 kg

Since the specific heat capacity of water is approximately 4.18 J/g°C, we can calculate the heat involved using the formula:

Q = C x M x ΔT
= 4.18 J/g°C x 0.0955 kg x (27.90°C - 22.00°C)
= 20.02 J

Therefore, the heat involved in the trial is 20.02 J.

2) Calculate the moles of NaOH used in the trial (mol):
To calculate the moles of NaOH used, we can use the equation c = n/V, where c represents the concentration, n represents the number of moles, and V represents the volume.

Given that the concentration of the NaOH solution is 1.00 mol/L and the volume used is 48.0 mL, we can plug these values into the equation:

c = n/V
1.00 mol/L = n / 0.0480 L

Solving for n, we get:
n = 1.00 mol/L x 0.0480 L
= 0.0480 mol

Therefore, the moles of NaOH used in the trial is 0.0480 mol.

3) Calculate how much heat would be evolved if 1 mole of NaOH was used (Q/mol):
Since we already know the heat involved in the trial (Q = 20.02 J) and the moles of NaOH used (mol = 0.0480 mol), we can use the formula:

Q/mol = Q / mol
= 20.02 J / 0.0480 mol
= 417.92 J/mol

Therefore, if 1 mole of NaOH was used, the heat evolved would be 417.92 J/mol.