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 just saved my life. Thank you!

Thanks for the explanation. Very easy to understand.

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.

@BensChemVideos

3 ай бұрын

You're very welcome! Best of luck to you 😀

thanks bro. You are the man.

very effective teacher. thanks

great video

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

@BensChemVideos

3 жыл бұрын

You're very welcome! Thanks for watching! 😀

Awesome video!

thank you very much

I can't thank you enough

@BensChemVideos

3 жыл бұрын

You're very welcome! Thanks for watching 😀

thanx alot :)

Thank you, the concept was very well understood after watching this!

@BensChemVideos

3 жыл бұрын

You're very welcome! Thank you for your feedback, and for watching so many of my videos! 😃

@dhivyamaalya

3 жыл бұрын

@@BensChemVideos I'm glad you uploaded them :)

very good video!

@BensChemVideos

3 жыл бұрын

Glad you liked it. Thanks for watching 🙂

Doing this lab tomorrow in Gen Chem 1, thank you sir!

when finding mass of solution, you didn't account for 2.4 g zn?

@seungjunbaek1494

3 жыл бұрын

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.

@thiphucphan9201

Жыл бұрын

Thank you so much for asking the question and also answering the question mate!

can u explain how u got 3.01 KJ?????

@Mayowa1994

8 жыл бұрын

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.