q = [mass H2O x specific heat H2O x (Tfinal-Tinitial)
Solve for q = delta H in joules.
Then delta H/0.06 mol = delta H in J/mol. Most of these are given in kJ/mol; I recommend you convert to kJ/mol.
When 50.0mL of 1.20 M of HCl (aq)is combined with 50.0mL of 1.30 M of NaOH (aq) in a coffee-cup calorimeter, the temperature of the solution increases by 8.01 Degrees C. What is the change in enthalpy for this balanced reaction?
HCl(aq) + NaOH(aq) --yields-- NaCl(aq) + H20(l)
Assume that the solution density is 1.00 g/mL and the specific heat capacity of the solution is 4.18 J/ g x Degrees C
3 answers
since q= 100 mL x 4.18 x 8.01
=3348.18/.06= 55803. J
to kJ= 55.803 should be the final answer, right? Because the answer key on my quiz says it's -55.8 kJ. I'm so confused because I thought for sure it would positive, especially since in the equation it says increases by 8.01, meaning it would be endothermic (+)? Am I missing something here?
=3348.18/.06= 55803. J
to kJ= 55.803 should be the final answer, right? Because the answer key on my quiz says it's -55.8 kJ. I'm so confused because I thought for sure it would positive, especially since in the equation it says increases by 8.01, meaning it would be endothermic (+)? Am I missing something here?
it's because the reaction is exothermic reaction so the answer should be negative, because in exothermic reaction the heat is released to the surrounding