Question

HI i am having trouble with my post lab report. It involved Delta H, specific heat and other things i am having trouble calculating. I have uploaded the images onto a google document https://docs.google.com/document/d/1VQi5tjHzD-mmVQqBvdczQdkBRWSPpV7N5vY2xlL_Gmw/edit?usp=sharing I AM ONLY ASKING YOU TO complete the questions on PAGE 4 of the google doc.

I WOULD HIGHLY APPRECIATE IF IF YOU WOULD SHOW STEP BY STEP, AND EXACTLY WHAT NUMBERS REPRESENT WHICH VARIABLES. Please help me out, I am really struggling in my chemistry course since we are not supposed to post multiple questions together, that is why i am asking for this question to only answer page 6 on the google docs, the information regarding ENTHALPY OF NEUTRALIZATIOn, which is data sheet 2 in the google doc. On page 7 there are resources that you may need to calculate the answers

Answer 1

Now the highest temperature that trial 1 went to is 37.3 - 23.4 (This is the average of the two initial temperature since volumes are equal)

13.9 x 100 x 4.184 = 5815 J

Trial 2 - 35.8 - 23.2 = 12.6

12.6 x 100 x 4.184 = 5271 J

Average value is 5516 J will be for avergae temperature difference which is 13.25 ^oC

You limiting reagent is the acid 1.92 M

0.05L x 1.92 M = 0.096 moles

5515 J for 0.096 moles

The value of enthalpy change ofneutralisation of sodium hydroxidesolution with HCl = 5515/0.096 = 57447.9 = 57.48 kJ/mol