Constant Pressure Calorimetry
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- เผยแพร่เมื่อ 12 ม.ค. 2025
- This video is a basic discussion on how constant pressure calorimetry works, and how it is used to calculate the change in enthalpy associated with a chemical reaction.
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I know this video is officially old, but due to the fact that you single-handedly enabled me to solve an enthalpy of reaction performance assessment problem for my college Chemistry class, I owe you a huge thank you.
You have explained this perfectly, I do love chemistry a lot and solving these problems is not a big problem
@@erastuskihali4298 thanks bud!
Thank you Ben!!!! you saved my life!!!! I was trying to understand and process all of the new information online :( I really need to go back to school cuz learning by myself is being very challenging
You're very welcome! Thanks for watching! 😀
Thanks for the explanation. Very easy to understand.
Thank you, the concept was very well understood after watching this!
You're very welcome! Thank you for your feedback, and for watching so many of my videos! 😃
@@BensChemVideos I'm glad you uploaded them :)
great video
Awesome video!
very good video!
Glad you liked it. Thanks for watching 🙂
thanks bro. You are the man.
thank u my final is in three days and i got this one wrong during the year. Makes sense right now, I pray it will on the test.
You're very welcome! Best of luck to you 😀
very effective teacher. thanks
I can't thank you enough
You're very welcome! Thanks for watching 😀
when finding mass of solution, you didn't account for 2.4 g zn?
bruh read the question carefully plz. It literally says that it makes 50.0mL of the solution meaning it takes into account of the mass of zn, therefore meaning that you don't need to add the mass of zinc.
Thank you so much for asking the question and also answering the question mate!
thank you very much
Doing this lab tomorrow in Gen Chem 1, thank you sir!
thanx alot :)
Is deltaH the same as deltaE?
No
can u explain how u got 3.01 KJ?????
he multiplied the mass of the soln, the specific heat of the soln, and the change in temp and got 3,009 J. He then rounded up to 3 sig figs = 3.01 X 10^3 J
I was just wondering why we don't take work into account in constant pressure calorimetry.