This video was excellent. Provided in the context of mcat Q's and clear, procedural, and succinct. Deserves far more views.
@seanoconnor3112 жыл бұрын
This was fire. When's the merch dropping?
@e-27584 ай бұрын
Lifesaver. You are so concise and clear in the way you teach. Thank you!
@tristenmelvin18555 ай бұрын
This freaking video is awesome. Was so confused on this topic and came back to relearn it!! Gonna be returning to this playlist
@AlishaSlomers2 ай бұрын
such a great explanation! thank you
@getitdone001010 ай бұрын
Deserves far more views
@LindseyDeaver2 ай бұрын
Thank you!!
@dreamchaser51737 ай бұрын
This was excellent. Thank you so much!!
@MedSchoolCoachMCATPrep7 ай бұрын
Glad it was helpful!
@emilybernal47368 ай бұрын
thank you! this was incredible!
@MedSchoolCoachMCATPrep8 ай бұрын
Glad you liked it!
@ANNALERCH4 ай бұрын
Do you combine the rate laws of steps 1 and 2 to get an overall rate law for the entire reaction?
@dikigurung2222 Жыл бұрын
thank you for saving my life
@Obamnaz2 жыл бұрын
Can you do a video on the integrated rate laws please?
@dwl26018 ай бұрын
When determining rate law from experiment data, do we also only use gaseous and aqueous reactants in the rate law?
@denissemedina46197 ай бұрын
In that case it does not matter.
@carinaashcraft16272 жыл бұрын
Best explanation
@misaestudiando2 жыл бұрын
Do the trials you pick to determine rate always have to be next each other? For e.g. lets say Trial A has x2 change between trial 1 & 3, and Trial B & C remain unchanged. Because there was a "jump" would it still be okay to calculate reaction order with those values?
@MedSchoolCoachMCATPrep2 жыл бұрын
Nope, you can always pick any two trials to work with!
@rachel39321 Жыл бұрын
At the end you wrote/said that reaction order is the stoiciometric coefficients but that is not actually the case. even it happens to be the same number occasionally, they are not related.
@diannenewcomb9781 Жыл бұрын
Yes. The rate order usually isn't the coefficients. Rate order aka exponents need to be determined experimentally.