50.0mL of .500M NaOH is added to 25.0mL of .500M HCL in a calorimeter BOth are...

Free

60.1K

Verified Solution

Question

Chemistry

50.0mL of .500M NaOH is added to 25.0mL of .500M HCL in acalorimeter BOth are at an initial temperature of 25 C ans finaltemperature is 27.21 C. Assume the mass of the final solution is76.0 g and the Cs of the solution is 4.18 J/gC. what is the valueof q for water, what is the value of q for the reaction, what isthe change of H of the reaction in kj/mol?

Answer & Explanation Solved by verified expert
4.2 Ratings (819 Votes)

q = ms ΔT = = 76 x 4.18 x(27.21-25) = 702.07 joules

   NaOH    +     HCl -------->    H2O   + NaCl

t=0      25                     12.5                                       millimoles

t=t      12.5                   0                 12.5      12.5 millimoles

for   12.5 millimoles    --------->    energy   is 702.07 joules

for      1 millimoles     ---------->    702.07/12.5 Joules

for 1 mole= 1000millimoles ------->   1000x (702.07/12.5) =56165.6 joules

ΔH = 56.165.6KJ /moles


Get Answers to Unlimited Questions

Join us to gain access to millions of questions and expert answers. Enjoy exclusive benefits tailored just for you!

Membership Benefits:
  • Unlimited Question Access with detailed Answers
  • Zin AI - 3 Million Words
  • 10 Dall-E 3 Images
  • 20 Plot Generations
  • Conversation with Dialogue Memory
  • No Ads, Ever!
  • Access to Our Best AI Platform: Flex AI - Your personal assistant for all your inquiries!
Become a Member

Other questions asked by students