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is it 5.5 liters or atm??

Timestamp for Self: 14:40

slayyy

Thank you so much for breaking this down into easy steps! love the pancake analogy - super helpful!

Hey Doc, I can’t find the next lecture (35) for some reason, where can I see it? Your videos are genuinely amazing, never thought I’d actually get electron configuration this fast. Thanks and keep up the good work.

Uh oh…

Such a great teacher.

I am so happy that I found this video. This is very helpful for my exam prep. Thank you Dr. E 🙂

Well, okay. Still quite confused. I get everything that brings the electron configuration up to Radon. It is the last 6 electrons (really the last four) that confuse me as they do not seem to follow the literal aufbau principle. It makes sense that the 7s orbital fills first with a maximum of 2, so 7s^2. But the next and final four should all fill in the 5f orbital for 5f^4. Yet apparently the 6d orbital snags one of these despite that there is plenty room left in the 5f orbital to house it. So, instead of 7s^2, 5f^4, we have 7s^2, 5f^3, 6d^1. It gets even more confusing in that the next element--Neptunium--keeps the 1 electron in the 6d orbital while the 5f orbital picks up another one for a configuration of 7s^2, 5f^4, 6d^1. But then with the very next element--Plutonium--one more electron goes into 5f orbital but then also snags the one in the 6d orbital for a configuration of 7s^2, 5f^6. Then Americium picks up one more electron in 5f orbital for 7s^2, 5f^7 BUT then the next element--Curium--gets a single electron back into the 6d orbital again for a configuration of 7s^2, 5f^7, 6d^1. AAAAAAAARRRRRRRRRGGGGGHHHHHHH! How am I supposed to rationally or systematically KNOW this when doing the electron configurations???? I've been told to note the Hund rule but that doesn't seem to help--it's still not clear to me how I'm supposed to figure out that the 6d orbital snags just one electron preventing the 5f orbital to get it after latter's first three are there (and when it can hold up to 14). What am I supposed to know to fill this out correctly without just looking at my periodic table that gives me the configuration (as using the noble gas notation)??

First of all 5f3 is before 6d1 we know they have same energy but since 6d1 have a bigger radius we put it in front

I got this. I’m passing this test.

Thank you for this video, it's really helpful.

yeah im failing

the formula for Ecell = Ered - Eoxi

Thanks mam 🙏

Thanks mam

Thanks mam

This video reminder of my first lecture in lab, long time ago

Thanks for the video Cardi B❤

Thank you Dr. E

is it 5.5 liters or atm??

It helped me a lot. My concepts are now clear. Thank you very much for this!!!! God bless!!!

My favorite quote...,"what comes after pico? who cares. We'll probably be dead." Is this an acceptable test question answer?

Dr. E I love you so much, I don’t know how I would’ve passed Chem without you ❤❤❤

Thank you so much for your videos! I really appreciate how you break the concepts down for easy understanding.

you're the best teacher in the world.

Thank you so much mam

Love your video it's nice working out the problem with someone. Thank you!

thank you for this!! :)

Simply the least useful contrived exercise invented by teachers who never actually did research. Without a proper analysis of standard deviations and confidence intervals for sample sizes greater than 30, this whole is utter nonsense.. Every measure needs a calibration and understanding of accuracy and precision of each instrument. In short, nonsense built on stilts.

Best chemistry professor of all time!!! Thanks so much professor, you just make everything so easy to understand

In problem #4 the 3 subscript is missing from molecule B3(PO4)2 in the word problem/mw. Fortunately, however, I was able to determine the correct value without balancing the whole equation, simply by isolating mols of PO4 :)

wow. you deserve more views.

at 13:50, isn't initial [F-] = 0.45M? (the concentration of [NaF] that fully dissociates in the solution before adding KOH)

I feel enlightened

This professor makes me love chem!

Very nice video, thank you very much

YOU ARE AMAZING

I love this professor

Dr. E, how would I measure a 25 mL grad cylinder? Thanks

love your videos💕

epic

For Br, in literature says the oxidation numbers are 1- 3+ 5+ and 7+, how can I justify the 3+? Also for S the oxidation numbers are 2- 2+ 4+ and 6+, how can I justify the 2+?

A much easier way to look at this problem would be to conclude that after the first equivalence point there would be no more H3X left ,so only a third as much titrant would be required to reach the first E.P. The rest of the base is used to reach the 2nd and 3rd E.P. At these points you will only have H2X-,HX-. left .Another way to thing about this is 3moles of NaOH is needed to react with one mole of H3X.If you know the moles of NaOH used then one third as much H3X will be used.

#3 is also 0.0467 which is 3 sig fig

a) 234000 is what we have in our worksheet

33:39 :- You said we will round for -/+ if 4 was 6 so how about for x/% do we need to round or we will only take the sig fig and not change the decimal number?

Alkanes -1 , Alkenes -2 , Alkyne - 3. Cycloalkane - ring to rule them all