Thanks Zuffyy and I am having a wonderful day! I'd rather be working with students in person rather than via Zoom or by making videos in the confines of my home office, but it's nice to still be able to provide value in this time. Happy Studying!
You have been blessed with such a great gift! God bless, thank you for taking the time to create this content!!! You have to be one of the best teachers I have seen.
seriously such a great help, I finally understood titrations today from your video and I find the way you teach so clear and understandable, thank you for what you are doing.
Thank you so much for this amazing lesson! I first got confused by the values of SRP when calculating E of the cell, when given 2 SRP values to construct a galvanic cell, I can't tell what metal is being oxidized or reduced, but since you emphasized that in galvanic cells E > 0, I can write down the 2-half reactions as long as it satisfy that criteria. You made it clear as day and I can't thank you more for that.
You're very welcome Deya! Unfortunately I didn't return to ASU this Fall. I decided to focus on making more online resources available to everyone during the pandemic. We'll see what happens next year. Best on your finals!
I've been subscribed for a while but never really watched any videos. I watched one for the first time now and I have to say, I was killing my grades not using you as a resource, but it's a mistake I won't ever make again!
Love the hyperbole Samira! You're very welcome! And did you figure out the answer to whether or not a molecule with no chiral centers can have hydrogens that are diastereotopic? I saw the question come in but it disappeared on me.
@@ChadsPrep yes thanks a lot for the reply, I rewatched the video, sometimes we don't have chiral carbon at the beginning but once we substitute an element with hydrogen, the achiral carbon will have 4 different groups attached and appear as chiral carbon, and make the hydrogens diastriotopic. Thanks again, Sir.
Hi how do you exactly know from where the electronsmare coming? From the reductor aide or the oxidator side? And how do you know how mamy H+ and e- you put in?
High Prof. Chad! How did you know that @14:57 Pb was oxidation state +2 in both reactant and product sides? Because surfer can have many oxidation states. It’s because of the sulfate ion right? Could this possibly have been any other oxidation state? I guess my question came from trying to break down the individual oxidation state of each atom in PbSO4. Oxygen gets -2, but then which one gets assigned next and how? Thanks!
Sulfur can have oxidation state ranging from (-)2 to +6 in other compounds, but as this is lead sulfide, the sulfur anion has a typical oxidation state of 2(-)
hey chad good video but you have got the signs wrong in1.04 it should be + 1.08 V as cromium is being oxidised so you take the opposite value for your table.
Hey Chad! This was uploaded 1 month ago. Firstly, I love that the whole lesson is in one shot and also, you are talking to us, the viewers. Secondly, are you offering courses anywhere?
Glad you like them! I was a little worried they wouldn't be as engaging as with students present (and they probably aren't), but I feel they more than make up for that in the efficiency of the presentation (and there's no other options during a pandemic!). I have online courses at chadsprep.com that include the study guides as well as hundreds and quiz questions and practice exams. Happy Studying!
Which chemical solution clean black mercury II sulfide (metacaniber) pigment. Thin metacaniber layer coating on paper. How to clean that pigment on paper
Back with another question! @32:55 16 H+ + 16 OH = 16 H2O…why is this when the ratio of H to O is 2:1 and not 1:1. Don’t we have a limiting reagent situation with H?
Sulfur is a monatomic ion and is always -2 as a monatomic ion (it takes on other oxidation states when part of a polyatomic ion). We figure this out based on it having 6 valence electrons and needing to gain 2 to have a filled octet. And if sulfur is -2 then Pb must be +2. Hope this helps!
@@ChadsPrep Perfect Thanks so much for all your videos. Im a member of your site and will be using it when i get to O Chem. you make things ridiculously easy to understand!
Potassium is +1 as an ion. Nitrogen is more electronegative than carbon so we'll assign him first. His typical oxidation state is -3 (being 3 electrons short of a filled octet). This leaves carbon to balance out the overall charge to zero since KCN is a neutral compound. Carbon will therefore have to be +2. Hope this helps!😊
Hey Luna - so each oxygen has -2 oxidation state and there are 4 oxygens giving a total of -8. Since N2O4 is neutral, we have to balance out this -8 with +8 from the N2. So each of the 2 N atoms has +4 giving +8 total. Make sense?
Chad I hope you have a wonderful day knowing you are helping so many struggling students achieve their dreams!
Thanks Zuffyy and I am having a wonderful day! I'd rather be working with students in person rather than via Zoom or by making videos in the confines of my home office, but it's nice to still be able to provide value in this time. Happy Studying!
You have been blessed with such a great gift! God bless, thank you for taking the time to create this content!!! You have to be one of the best teachers I have seen.
Thanks so much for the lovely comment Yesenia - and glad you enjoyed the video :)
Really appreciate your channel, your teaching is really easy to follow.
Aweseome JD! Glad you find it helpful!
Currently trying to keep my A in General Chem 2 and your videos are making this possible! Your teaching style is so so good!
Glad to hear it, Sarah - work hard and earn that grade!
You're the best chem teacher on UA-cam
Thank you
THIS IS SO TIMELY!! Bless you Chad.
Awesome Gracee! Very glad to hear it!
seriously such a great help, I finally understood titrations today from your video and I find the way you teach so clear and understandable, thank you for what you are doing.
You're welcome - Glad to hear it!
Thank you so much for this amazing lesson! I first got confused by the values of SRP when calculating E of the cell, when given 2 SRP values to construct a galvanic cell, I can't tell what metal is being oxidized or reduced, but since you emphasized that in galvanic cells E > 0, I can write down the 2-half reactions as long as it satisfy that criteria. You made it clear as day and I can't thank you more for that.
You're welcome and Thank You!
Legendary overview, as always, Chad.
Thanks!
Chad THANK YOU! You deserve so much in life, bless you!
You're welcome and Thank You!
GOD BLESS YOU CHAD! This is great!!!
Thanks Iron Ed! Glad you found this helpful! Happy Thanksgiving and Happy Studying!
Chad if I ever see you at ASU its going to feel like Im meeting a celebrity. Thank you for the videos!
You're very welcome Deya! Unfortunately I didn't return to ASU this Fall. I decided to focus on making more online resources available to everyone during the pandemic. We'll see what happens next year. Best on your finals!
Thank you, Chad, so much for doing these!
You're very welcome Kayla! Glad you find them helpful!
I've been subscribed for a while but never really watched any videos. I watched one for the first time now and I have to say, I was killing my grades not using you as a resource, but it's a mistake I won't ever make again!
Glad you are subscribed - Happy Studying!
thanks from Italy, wish u all the best
You're welcome, Stefano - thank you!
This man is a genius
Genius? Maybe, maybe not. But doing this for 20 years has definitely helped!
my life is complete now. special thankssss
Love the hyperbole Samira! You're very welcome!
And did you figure out the answer to whether or not a molecule with no chiral centers can have hydrogens that are diastereotopic? I saw the question come in but it disappeared on me.
@@ChadsPrep yes thanks a lot for the reply, I rewatched the video, sometimes we don't have chiral carbon at the beginning but once we substitute an element with hydrogen, the achiral carbon will have 4 different groups attached and appear as chiral carbon, and make the hydrogens diastriotopic. Thanks again, Sir.
@@26d8 Exactly right! Awesome!
@@ChadsPrep Thank you so much, Sir, I posted a file with 3 spectra on your website, I really appreciate your help. :)
thank you , you changed my life::)
Wow and Happy Studying!
Hi how do you exactly know from where the electronsmare coming? From the reductor aide or the oxidator side? And how do you know how mamy H+ and e- you put in?
High Prof. Chad! How did you know that @14:57 Pb was oxidation state +2 in both reactant and product sides? Because surfer can have many oxidation states. It’s because of the sulfate ion right? Could this possibly have been any other oxidation state? I guess my question came from trying to break down the individual oxidation state of each atom in PbSO4. Oxygen gets -2, but then which one gets assigned next and how? Thanks!
Sulfur can have oxidation state ranging from (-)2 to +6 in other compounds, but as this is lead sulfide, the sulfur anion has a typical oxidation state of 2(-)
@@ChadsPrep oh yeahhh rule# 5, duh lol Thanks!!!
@@JossinJax 👍👍👍
At 34:53, to make it basic, can I just change the H+ to the H2O that's on the other side. and then switch the H2O that' on the other side to OH-?
hey chad good video but you have got the signs wrong in1.04 it should be + 1.08 V as cromium is being oxidised so you take the opposite value for your table.
Hey Chad! This was uploaded 1 month ago. Firstly, I love that the whole lesson is in one shot and also, you are talking to us, the viewers. Secondly, are you offering courses anywhere?
Glad you like them! I was a little worried they wouldn't be as engaging as with students present (and they probably aren't), but I feel they more than make up for that in the efficiency of the presentation (and there's no other options during a pandemic!). I have online courses at chadsprep.com that include the study guides as well as hundreds and quiz questions and practice exams. Happy Studying!
Which chemical solution clean black mercury II sulfide (metacaniber) pigment.
Thin metacaniber layer coating on paper.
How to clean that pigment on paper
Back with another question! @32:55 16 H+ + 16 OH = 16 H2O…why is this when the ratio of H to O is 2:1 and not 1:1. Don’t we have a limiting reagent situation with H?
Omg never mind haha, it’s 16 moles of OH not of O! Ugh, okay lol
Glad you're all sorted :)
1:01:16
Thank you so much
God bless you 🙏🏻
You are welcome, Rita - to you as well.
Hi Chad, @1:22:30 can you just combine the two half reactions and then just write the reaction quotient from that? Thanks!
Precisely what you should do!
How did you figure the oxidation state for PbS in the first balancing example?
Sulfur is a monatomic ion and is always -2 as a monatomic ion (it takes on other oxidation states when part of a polyatomic ion). We figure this out based on it having 6 valence electrons and needing to gain 2 to have a filled octet. And if sulfur is -2 then Pb must be +2.
Hope this helps!
@@ChadsPrep Perfect Thanks so much for all your videos. Im a member of your site and will be using it when i get to O Chem. you make things ridiculously easy to understand!
THANK YOU THANK YOU THANK YOU
You're welcome 3 times!
This is awesome 🙌
Thanks!
2hr vid. Lets gooo!
Longer than typical - settle in and Happy Studying!
You are the best Thank You!!!!!!!
You're welcome, alexa - Glad the vids help!
Hi - Please explain the oxidation state of KCN?
Potassium is +1 as an ion.
Nitrogen is more electronegative than carbon so we'll assign him first. His typical oxidation state is -3 (being 3 electrons short of a filled octet).
This leaves carbon to balance out the overall charge to zero since KCN is a neutral compound. Carbon will therefore have to be +2.
Hope this helps!😊
keep it up sir, good job
Thanks!
@@ChadsPrep you r welcome sir
@@Chemcrown 👍 👍 👍
I still don't understand why you put a +4 on N_2 in N_2O_4
Hey Luna - so each oxygen has -2 oxidation state and there are 4 oxygens giving a total of -8. Since N2O4 is neutral, we have to balance out this -8 with +8 from the N2. So each of the 2 N atoms has +4 giving +8 total. Make sense?
THANK YOU
You're welcome, Sam!
Nice sir
Thanks.
1:32:45
1:16
Yes!