50.00 mL of 1.0 M HCl and 50.00 mL of 1.0 M NaOH are mixed. The temperature is observed to increase by 10.0 degrees C. How much energy (in J) was absorbed by the calorimeter? Assume a density of the reaction mixture of 1 g/mL. Specific heat of the solution is 5.00 J/g/degree C. The calorimeter constant is 10.0 J/degree C.

Please help, thanks a lot!

To calculate the energy absorbed by the calorimeter, we need to use the equation:

q = m * c * ΔT

where:
- q is the energy absorbed (in Joules),
- m is the mass of the reaction mixture (in grams),
- c is the specific heat of the solution (in J/g/°C),
- ΔT is the change in temperature (in °C).

Let's break down the problem step by step:

1. Calculate the mass of the reaction mixture:
- The density of the reaction mixture is given as 1 g/mL.
- The total volume of the reaction mixture is 50.00 mL + 50.00 mL = 100.00 mL.
- Since 1 mL of the reaction mixture weighs 1 gram (as given by the density), the mass of the reaction mixture is 100.00 grams.

2. Calculate the change in temperature:
- The temperature increased by 10.0 degrees C (as given in the question).

3. Plug the values into the equation q = m * c * ΔT:
- q = 100.00 g * 5.00 J/g/°C * 10.0 °C
- q = 5000 J

4. Lastly, consider the calorimeter constant:
- The calorimeter constant is given as 10.0 J/°C.
- Since the temperature increased by 10.0 degrees C, the energy absorbed by the calorimeter is:
- Energy absorbed by the calorimeter = 10.0 J/°C * 10.0 °C
- Energy absorbed by the calorimeter = 100.0 J

Therefore, the total energy absorbed by the calorimeter is the sum of the energy absorbed by the reaction mixture and the energy absorbed by the calorimeter:
- Total energy absorbed = 5000 J + 100.0 J
- Total energy absorbed = 5100 J

Hence, the energy absorbed by the calorimeter is 5100 Joules.