Also I might be wrong but when you did -log[.0725] in the last problem, you should have gotten 1.1397 instead of .1397 which would change the final answer.
This helped me sooo much THANKS CAT!! love how its simple clear and to the point thankyou!!! "new sub":) pls keep making videos and you are wayy to underrated
So for example lets say the molarity of Ca(OH)2 is .01M. We would first do -log [.02] because there are 2 OH ions, We get around 1.70. Then we would just do 14-1.70 resulting in an answer of 12.30. Sig figs might be off
I’m watching this video because I don’t fully understand the basics. So I really wish you would have written the “base 10” model 🥴 …. Because that’s where you lost me 😔
oh that just means "multiplied by ten to the power of -POH" as in [OH-] = 10 ^-POH. I hope that helped, I know I'm a year late to your comment though haha
You really explained this basic concepts very well.i would rate you 100%.
If I had you as my teacher, I wouldn’t miss one day and get 100 on every exam.
real
we can only dream
Also I might be wrong but when you did -log[.0725] in the last problem, you should have gotten 1.1397 instead of .1397 which would change the final answer.
Agreed
She speaks so calm
Suddenly i realized i love chemistry
This helped me sooo much THANKS CAT!!
love how its simple clear and to the point thankyou!!!
"new sub":)
pls keep making videos and you are wayy to underrated
if she was my high school teacher my attendance is 100%
Shouldn't there be two minuses in SO4? 1:37
Yes
this was so incredibly well explained! Thanks! :)
Hello
The last question:
c=n/v
c= 0,029 mol / 0,4 L
c= 0,0725 mol/L
pOH = -log (0,0725 M)
pOH= 1,139
14 - 1,139
pH = 12,86
THX YOU
YOU ARE MARVELOUS and so nice how you explain in order, step by step...
Thank you so much for your video !
im a grduate in chemistry and i donno why am i watching this but that feels good... and btw great explaination
Thank you , that was a very useful explanation that saves me from reading 20 pages :)
Very great explanation. My concepts are clear. thank you madam. Big Fan.
You are the best and thank you very much
In last question
POH=1.139
PH=12.86
True
what if we need the pH of Ca(OH)2?
So for example lets say the molarity of Ca(OH)2 is .01M. We would first do -log [.02] because there are 2 OH ions, We get around 1.70. Then we would just do 14-1.70 resulting in an answer of 12.30. Sig figs might be off
nice thats helps a lot
Informative
i love this
Thanks
I’m watching this video because I don’t fully understand the basics. So I really wish you would have written the “base 10” model 🥴 …. Because that’s where you lost me 😔
oh that just means "multiplied by ten to the power of -POH" as in [OH-] = 10 ^-POH. I hope that helped, I know I'm a year late to your comment though haha
Wish u were my chem teacher
Tq🤗
Well explain mam
Thanks :)
is she recording with a landline phone lmao
Bro😂😂😂
Thanks from Pakistan
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You should show without using a calculator.
🙏
❤👍
Do you also teach in onlyfans?
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hello mam