You mix 200. mL of 0.400M HCl with 200. mL of 0.400M NaOH in a coffee cup calorimeter. The temperature of the solution goes from 25.10°C to 27.78° C. What is the molar enthalpy of neutralization of the acid in kJ/mol? Assume all densities are 1.00 g/mL and the specific heat capacities are 4.184 J/g*K.

Answer : The enthalpy of neutralization is, 56.012 kJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

The balanced chemical reaction will be,

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH

Thus, the number of neutralized moles = 0.08 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water =

Now we have to calculate the heat absorbed during the reaction.

where,

q = heat absorbed = ?

= specific heat of water =

m = mass of water = 400 g

= final temperature of water =

= initial temperature of metal =

Now put all the given values in the above formula, we get:

Thus, the heat released during the neutralization = -4480.96 J

Now we have to calculate the enthalpy of neutralization.

where,

= enthalpy of neutralization = ?

q = heat released = -4480.96 J

n = number of moles used in neutralization = 0.08 mole

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.012 kJ/mole

8 m/s because for him to get the same amount of energy as ray is putting out, bobby, who weighs half as much as ray, needs to run twice as fast.

the correct answer would be b) volume versus temperature, because they are also directly proportional.