When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of the MnMn reacts, raising the temperature of the solution from 23.1 ∘C∘C to 28.9 ∘C∘C. Find ΔHrxnΔHrxn for the reaction as written. (Assume that the specific heat capacity of the solution is 4.18 J/g∘CJ/g∘C and the density is 1.00 g/mLg/mL.)

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat =

= final temperature =

= initial temperature =

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

Now we have to calculate the enthalpy change during the reaction.

where,

= enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose =

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

i'm assuming that it's c.) red bank..

the increase of protons   hope this