My goat frfr, if Krug had a million fans, I would be one of them. If krug has a thousand fans, I would be one of them. If krug had a single fan, I would be that fan. If krug had no fans, I wouldn't exist. Jeremy krug forever my chemistry goat
In the last question where you got 120 i got a 116.67 from my calculator, does that still count as correct as long as i show my steps and calculations ( revised them like 10 times 😭 ).
For the rate law experiment at the start, what is the third column, the initial rate of what, the cio2 or the oh or the rate of appearance of one of the profucts
Excellent question! We usually calculate the initial rate in terms of whatever substance has a coefficient of 1. This isn't always clear in most textbooks, but in this case it would be the rate of appearance of ClO3-, the rate of appearance of ClO2-, etc, or the rate of disappearance of hydroxide divided by 2, etc. This open-source textbook demonstrates this better than most that I've seen: chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/12%3A_Kinetics/12.04%3A_Rate_Laws
the math is a little tricky in kinetics... ln, log, e. if only I was good at math, I get the concept but the math here is super tricky. do future unit's get harder with the math?
Since the time in this problem is measured in seconds, we need to express our rate in 'change in concentration per second' and our rate constant k in s-1 so the units cancel out.
Personally I'd call it a poorly-written question. However, the question can still be solved, as long as one of the concentrations stays constant for at least one of the reactants at one point. For example, if there are two reactants, both are doubled, and the rate quadruples, this could mean two possible things: (1) one reactant is 2nd order while the other is 0th order, or (2) both reactants are 1st order. We'd need a little more information to pin down which is which. I hope this helps!
On the right side of the equation we have (0.020 M) squared times (0.030 M) to the first power. M squared times M gets us M cubed. Thanks for watching!
helped me more in 11 mins more than my chem teacher ever did in a month🙏🙏
Thanks for watching! Sometimes it helps to see a quick summary of something to put it all in perspective.
My goat frfr, if Krug had a million fans, I would be one of them. If krug has a thousand fans, I would be one of them. If krug had a single fan, I would be that fan. If krug had no fans, I wouldn't exist. Jeremy krug forever my chemistry goat
Thanks for the kind words, and for being a fan! Enjoy your break!
Actual lifesaver over here
Glad I'm able to help!
Thank you so much for your lifesaving videos!!
You're very welcome. And thank you for watching!
Thank you for explaining this! we took notes on this in class but this explained it better and now I understand
I'm glad my video helped you understand this concept. Thanks for watching!
god bless these videos they will save my AP chem grade 🙏
I'm so glad my videos are helpful to you. And thanks so much for watching and commenting!
In the last question where you got 120 i got a 116.67 from my calculator, does that still count as correct as long as i show my steps and calculations ( revised them like 10 times 😭 ).
Yes, my answer is 120 because the answer in this case should have only 2 significant figures.
omg I forgot to consider sig figs... I thought I forgot how to do basic algebra LOL
2:38 how did the rate inc by 9??
bcs 0.248/0.00276= approx 9
For the rate law experiment at the start, what is the third column, the initial rate of what, the cio2 or the oh or the rate of appearance of one of the profucts
Excellent question! We usually calculate the initial rate in terms of whatever substance has a coefficient of 1. This isn't always clear in most textbooks, but in this case it would be the rate of appearance of ClO3-, the rate of appearance of ClO2-, etc, or the rate of disappearance of hydroxide divided by 2, etc.
This open-source textbook demonstrates this better than most that I've seen: chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/12%3A_Kinetics/12.04%3A_Rate_Laws
Very very helpful video but how did you get 120, i've tried so many times but i keep getting like 116.66
Thanks! And excellent question. I rounded off to 2 sig figs
@@JeremyKrug ah i see, thank you
same question, thanks for askin
what would you do if in all 3 experiments, the initial molarity of an element is never the same
You divide rate laws
mr krug u need to make a patreon ill donate my life savings to you
Thanks for the kind words! A Patreon might be in the works in the upcoming couple of months.
I love you Mr Krug
Thanks!
the math is a little tricky in kinetics... ln, log, e.
if only I was good at math, I get the concept but the math here is super tricky.
do future unit's get harder with the math?
The kinetics unit is about as ‘mathematical’ as it gets. Thanks for watching, and best wishes in your chemistry course!
for the last question how did we get s-1?
Since the time in this problem is measured in seconds, we need to express our rate in 'change in concentration per second' and our rate constant k in s-1 so the units cancel out.
Mr Krug what would you do if they didn't keep one of the concentrations constant and changed both instead
Personally I'd call it a poorly-written question. However, the question can still be solved, as long as one of the concentrations stays constant for at least one of the reactants at one point. For example, if there are two reactants, both are doubled, and the rate quadruples, this could mean two possible things: (1) one reactant is 2nd order while the other is 0th order, or (2) both reactants are 1st order. We'd need a little more information to pin down which is which. I hope this helps!
why is m cubed?
On the right side of the equation we have (0.020 M) squared times (0.030 M) to the first power. M squared times M gets us M cubed. Thanks for watching!
my goattttttt
Thank you!
thx i got 95 in chem
Congratulations! You've got this!
If I defeat you in battle would I gain your knowledge of chemistry and pass the ap test
lol I don't think so, but I'm always glad to share my knowledge and help you pass either way!
if it wasnt for you ima fail ts
I know you're going to keep working and keep improving this year. Don't give up!
lock in gang