Katie Cheng It's so intuitive yet effective Some sandwiches are just Bologna(O) and bread(H), others are double or triple decker and have a different medium like... peanut butter (Ammonia), or have different things that act as bologna but aren't or dishes that don't need it (a turkey sandwich or literally just table salt). It's brilliant!
The best teacher i never tell my friends to be the excell in class. I already speedrun math to linear algebra with professor dave videos and now i came to learn chemistry
I actually didn't understand this video, so I went to the khan academy video after reading your comment.. to see if it would be any different! Strangely enough, I understood their explanation! KA video is more detailed than this one is, I don't understand why you guys didn't like it! In any case ,I'm greatful to both these educators !
@@adarsh65kumar Perhaps learning the same thing from multiple teachers helps better, just a hypothesis though, it happens to me too. If I learn something from one video and don't get it clearly enough, I watch a different one on the same topic and it gives me a different perspectives on the same topic, making it a bit easier to understand.
+abdirahman abdilahi because more people watching this channel, which happens in my experience to be the best of its kind on youtube (yes, more appealing and discrete and comprehensive than Khan Academy), means more scientifically literate people, means better world 🙏
Limiting Reagents and Percent Yield are one of the most confusing topics for me in Chemistry. Thanks to you, I understand it better. This is really helpful for students like me. Thank you very much!
literally speaking, i just stumbled across this channel looking for revisions before mah test but i was bowled over by the content and clarity of the video. u literally have earned a new subscriber... :)
Thanks to this video, I am seeing some light. The graphics add a transitioning touch to the relationships of the numbers we just can't quite see with erase markers. You are making a great difference.
رقم افوغادرو 6.022×10²³ رقم ٦ هذا المقصود به كاربون . الكتلة المولية : العنصر × كتلته . C⁶ × 12 . عندما يحصل تفاعل كيميائي العنصر يتغير من جهة اليسار إلى اليمين اي النواتج ، لذلك يجب حساب مول جديد وهذه هي الطريقة : - قبل التفاعل الكيميائي البروبان × عدد المول / الكتلة المولية = المول . - بعد التفاعل الكيميائي المول × عدد المول الجديد / عدد المول القديم المول المحصول × عدد البروبان / عدد المول .
New subscriber here. Big thank you proffesor Dave for sharing these. It is indeed a big help to us students who are having hard time in understanding the world of chemistry. Peace Be with you.
Can I just say, you`re extremely helpful!! Thank you so much!! Keep up the great work, and know that you`re helping so many people, like myself. G-d bless you.
الكاشف : هو الذي ينفد اولًا بالتفاعل . الفائض : هو الذي تبقى منه كميات زائدة بعد انتهاء الكاشف . لكي تستخرج المول بدون تعب : لكل عنصر او مركب معادلته ، خذ القانون : غرام العنصر • واحد مول / المستخرج من غرام العنصر . الكتلة المولية قبل التفاعل • واحد مول / العدد الذي بجانب الصيغة . بعد هذه التفاعلات الاقل هو الكاشف والأكثر هو الفائض . بعيدًا عن العدد المضروبة به الصيغة . لا تستخدم الكاشف للمعادلة التالية ان شاء الله لأنه سيظهر الناتج خطأ . الظاهر الاخير • العدد بجانب الصيغة الظاهرة بعد التفاعل / العدد بجانب الصيغة قبل التفاعل • المستخرج غرام / العدد بجانب الصيغة بعد التفاعل . هذه هي المعادلة المذكورة 2:35 هنا في هذه الدقيقة والثانية .
رقم افوغادرو 6.022×10²³ رقم ٦ هذا المقصود به كاربون . الكتلة المولية : العنصر × كتلته . C⁶ × 12 . عندما يحصل تفاعل كيميائي العنصر يتغير من جهة اليسار إلى اليمين اي النواتج ، لذلك يجب حساب مول جديد وهذه هي الطريقة : - قبل التفاعل الكيميائي البروبان × عدد المول / الكتلة المولية = المول . - بعد التفاعل الكيميائي المول × عدد المول الجديد / عدد المول القديم او للاختصار : كتلة العنصر • مول ( العنصر ) / الكتلة المستخرجة • العدد الذي بجانب الصيغة مول ( مع العنصر ) / مول الصيغة ( العنصر الثاني ) • غرام الماء / واحد مول ماء = الناتج 20.0 g • 1 mol \ 44.0 g • 4 mol \ 1 mol • 18 g \ 1 mol = 10 g \ 11 g • 4 \ 1 • 18\1 = 40 \ 11 g • 18 \ 1 = رقم / 11 ازلنا جميع g ومول تقريبًا . الي انعقدت عليك هي الغرام للأول.
The final percent yield 89.3% is a little off because of a rounding error: 19.1 ÷ 21.42857 ≈ 89.1333% ≈ 89.2%. Don't round to sig figs too early in the process and you should be fine 😊
quick question, when we getting the theoretical yield for CH3COOH, why we taking 21.4g as our final answer, not 21.42g. I went back and rewatch the sig figs chapter, but it still couldn't help me with the result. Thank you in advance for helping !
Sir if u have time please try to make a video on equivalent concept in chemistry really hard to get feel and intiution of that topic I would be very very thank ful to you ( from India Telangana)
This is great work, but unfortunatley it would be nice if the problem showed how the understanding of dimensional analysis played a role. But other than that it helped me get to an answer!
Great video but I wish the calculations were explained more in-depth; you just say, "so we will now calculate how much product we will get," then just throw up a formula without explanation. I would have appreciated it if you explained why there was 60.1g of CH4N2O in the calculation, so I , for example, wouldn't have had to guess why. I realize now that is the molar mass of CH4N2O, but it would have been nice for that to be pointed out. 😫
when converting to moles in the ex. in 1:45, why can't I convert between the substances after I have just the first substance in moles? and also why do I get different result while doing it in comparison to just calculating it separately? 10g NH3* 1 mole NH3/ 17g NH3= 0.588 moles NH3 0.588 moles NH3* 1 mole CO2/ 2 moles NH3= 0.294 moles CO2 --------while calculating separately will give me 0.227: 10g CO2* 1 mole CO2/ 44g CO2= 0.227 moles CO2
ok so the answer is because if I do that, I will get the amount of the other substance in moles that will theoretically react flawlessly with the first substance and not the actual amount of the substance. I don't know if anybody needs it but there you have it
if during the experiment, you only included 97% of grams of 1 of the reactants, would you still get the same substance you’re expecting? only that, you wont get the theoretical yield amount in grams? OR the whole experiment would be wrong because the expected substance wont happen?
oh wait!!! i think, we can still get the same substance for as long as we have the reactants necessary. the change in grams in the reactants does not affect the expected substance, only the amount in weight afterwards. please tell me if im right or wrong, Prof Dave. :) if im right or wrong, why or why not
Hi, professor Dave. May I ask you why the answer of the question is 89.3, not 90? I thought sig fig would be 1 since the one with the fewest sig fig is 100 (1 sig fig). Sorry if the question is too silly :( and I very apprieciate your well-made videos.
professor dave whippin out the bologna sandwich analogy makes him my favorite person ever
Fr
He’s a living legend.
Katie Cheng
It's so intuitive yet effective
Some sandwiches are just Bologna(O) and bread(H), others are double or triple decker and have a different medium like... peanut butter (Ammonia), or have different things that act as bologna but aren't or dishes that don't need it (a turkey sandwich or literally just table salt).
It's brilliant!
The best teacher i never tell my friends to be the excell in class. I already speedrun math to linear algebra with professor dave videos and now i came to learn chemistry
I always knew Jesus saves, but I didn't think he'd be saving my Chemistry grade
👀💪💪💪✊✊✊🦁🦁🔥🔥🦁🦁🙏🙏♥️❤️❤️🔥🔥🔥🔥💯💯💯💯💯
Well,he kind of looks like Jesus😂
@@denicagovender5029that was the point of the joke 🤦♂️
.. can we not pls😅
Came here from Khan Academy. Much clearer presentation than the current Khan Academy video on limiting reagents. Thank you so much!
I actually didn't understand this video, so I went to the khan academy video after reading your comment.. to see if it would be any different!
Strangely enough, I understood their explanation! KA video is more detailed than this one is, I don't understand why you guys didn't like it!
In any case ,I'm greatful to both these educators !
i like khan academy more.. they explain it better (my opinion)
@@adarsh65kumar Perhaps learning the same thing from multiple teachers helps better, just a hypothesis though, it happens to me too. If I learn something from one video and don't get it clearly enough, I watch a different one on the same topic and it gives me a different perspectives on the same topic, making it a bit easier to understand.
I love you professor Dave, you know all about the science stuff and you made everything as easy as a piece of bologna sandwich ☺
yeah yeah!!!!
That's why "He knows lot about science stuffs Professor Dave explains" 🎶🎵
Bhai is bande ki shakal ranveer kapoor se milti h😂😂😂🤣🤣
Bas bal jangli wale h isko to animal movie me bhejo shikar karega soor ka sala ranveer kapoor ka bhai lagta h ye angrez😂😂😂
personally you deserve more subscribers and people need to see this.
i tell myself that every day! pretty please tell your friends :)
I agree!! Don`t worry, 1 quarter is worth more the 10 pennies. The people that subscribe seem to be very fond of you. Thanks a bunch.
+Professor Dave Explains i don't understand
+abdirahman abdilahi because more people watching this channel, which happens in my experience to be the best of its kind on youtube (yes, more appealing and discrete and comprehensive than Khan Academy), means more scientifically literate people, means better world 🙏
Limiting Reagents and Percent Yield are one of the most confusing topics for me in Chemistry. Thanks to you, I understand it better. This is really helpful for students like me. Thank you very much!
1 week of not understanding class into 4 minute video. thank you chemistry jesus
THIS IS AN AWESOME CHANNEL!! Please for the love of god don't stop making these videos! I need this so I don't fail chemistry!!!
It's been four years. Did you pass?
@@samisiddiqi5411 Yes, Yes I did. XD
@@rafyraffee yayyyyyyyyy
Thank you so much! I wasted so much time on Kahn Academy confused as all hell, and you summed it up perfectly in less than 5 minutes!
exactly!!!!!!!
Thank you Professor Dave for explaining this so well, I've got a final in a few days and need someone to actually teach me well. YOU ARE THAT MAN
This is perfect for studying for my grade 11 chemistry exam. Thank you Professor Dave!
How is college 😂😂
@@mgrn7106 no cap you be doin this on the daily in intro chem classes
I was here for my retake for a chemistry test, and I fell in love with these videos! Thank you professor Dave
literally speaking, i just stumbled across this channel looking for revisions before mah test but i was bowled over by the content and clarity of the video. u literally have earned a new subscriber... :)
you are saving my life gr11 Chem exam tomorrow and 0 idea what I'm doing
howd it go?
i need answers how did it go?
Thanks to this video, I am seeing some light. The graphics add a transitioning touch to the relationships of the numbers we just can't quite see with erase markers. You are making a great difference.
Hello Sir I am From Pakistan .Your videos are always helpful for me .Biology and chem lectures are admirable .Keep it up 😊
Indeed your way of teaching is extremely easy to understand you actually deserve 10m or even more than that
Professor Dave has been my favorite so far, am enjoying everything.
Thank you for teaching me in 4 minutes what my teacher could not teach for the past few months
I have an A in Chem thanks to you!! my prof honestly gives me no help
رقم افوغادرو 6.022×10²³
رقم ٦ هذا المقصود به كاربون .
الكتلة المولية : العنصر × كتلته .
C⁶ × 12 .
عندما يحصل تفاعل كيميائي العنصر يتغير من جهة اليسار إلى اليمين اي النواتج ، لذلك يجب حساب مول جديد وهذه هي الطريقة :
- قبل التفاعل الكيميائي
البروبان × عدد المول / الكتلة المولية = المول .
- بعد التفاعل الكيميائي
المول × عدد المول الجديد / عدد المول القديم
المول المحصول × عدد البروبان / عدد المول .
thank u i love u i was getting rly stressed out about my chem exam but this video SAVED ME all the other vids i watch sucked
Khan academy took 20 minutes to explain what you did in 4 and yours is still better 💀
New subscriber here. Big thank you proffesor Dave for sharing these. It is indeed a big help to us students who are having hard time in understanding the world of chemistry. Peace Be with you.
This video is life saving, thanks tons for the explicit explanation.
I'm a freshman in college and your videos are helping me so much!! :D thank you for these awesome videos!!!
Can I just say, you`re extremely helpful!! Thank you so much!! Keep up the great work, and know that you`re helping so many people, like myself. G-d bless you.
For the Question at 4:00 . Can you please quickly tell me how many moles of CH3OH will react in this rxn?
limiting reagent=number of moles/stoichometric coefficient💗💗
This is enormously helpful!
Probably the best thing I have ever subscribed to.. thank you.
sir can you please a separate video for difficult questions of limiting reagents and tricks to solve them.
prof dave literally saves my life
Thank you for this. I now fully understand how to compute for the theoretical yield.
haha great! I use a similar example when explaining the limiting reactant to my students; sandwiches make for great examples!
Bruh how did I understand this in 5 minutes but not in a 2 hour class.😩❤️
I am come from Bangladesh. Really your lecture helped me
Thank you professor dave. Because of your teaching skills my exam was easy
thanks professor dave!
If I had $1 per everytime I called someone a stupid face I'd be rich
@@ChemistryTeacher-qz6vpI think your the stupid face here 🤣🤓
Thanks for the brief refresher. I love you!
الكاشف : هو الذي ينفد اولًا بالتفاعل .
الفائض : هو الذي تبقى منه كميات زائدة بعد انتهاء الكاشف .
لكي تستخرج المول بدون تعب :
لكل عنصر او مركب معادلته ، خذ القانون :
غرام العنصر • واحد مول / المستخرج من غرام العنصر .
الكتلة المولية قبل التفاعل • واحد مول / العدد الذي بجانب الصيغة .
بعد هذه التفاعلات الاقل هو الكاشف والأكثر هو الفائض . بعيدًا عن العدد المضروبة به الصيغة .
لا تستخدم الكاشف للمعادلة التالية ان شاء الله لأنه سيظهر الناتج خطأ .
الظاهر الاخير • العدد بجانب الصيغة الظاهرة بعد التفاعل / العدد بجانب الصيغة قبل التفاعل • المستخرج غرام / العدد بجانب الصيغة بعد التفاعل .
هذه هي المعادلة المذكورة 2:35 هنا في هذه الدقيقة والثانية .
رقم افوغادرو 6.022×10²³
رقم ٦ هذا المقصود به كاربون .
الكتلة المولية : العنصر × كتلته .
C⁶ × 12 .
عندما يحصل تفاعل كيميائي العنصر يتغير من جهة اليسار إلى اليمين اي النواتج ، لذلك يجب حساب مول جديد وهذه هي الطريقة :
- قبل التفاعل الكيميائي
البروبان × عدد المول / الكتلة المولية = المول .
- بعد التفاعل الكيميائي
المول × عدد المول الجديد / عدد المول القديم
او للاختصار :
كتلة العنصر • مول ( العنصر ) / الكتلة المستخرجة • العدد الذي بجانب الصيغة مول ( مع العنصر ) / مول الصيغة ( العنصر الثاني ) • غرام الماء / واحد مول ماء = الناتج
20.0 g • 1 mol \ 44.0 g • 4 mol \ 1 mol • 18 g \ 1 mol =
10 g \ 11 g • 4 \ 1 • 18\1 =
40 \ 11 g • 18 \ 1 = رقم / 11
ازلنا جميع g ومول تقريبًا .
الي انعقدت عليك هي الغرام للأول.
Great! with you Chemistry is so easy! thank you for your tutorials!
this is such a great channel, thank you prof. dave!!!
woooooo hooooo I said my friends about your channel and they are just loving it!!
fun mix of puzzles and math! this is fun! im getting the answers right as well!! woo!!! thanks dave!!!
Awesome analogy...
love ur expalnation so much..u help me a lottt, tq prof
Thanks you helped me a lot !
thank you 🌸
Thanks a lot for your explanations.
Great teacher
thank you, professor you gave us, a clear explanation
This is ez stupid face try takin an art degree then tell me what's hard
you are a very good chemistry teacher
Thank you, chemistry jesus
Shut up stupid face
The final percent yield 89.3% is a little off because of a rounding error: 19.1 ÷ 21.42857 ≈ 89.1333% ≈ 89.2%. Don't round to sig figs too early in the process and you should be fine 😊
Wow❤u teach well...I now understand chemistry ⚗️
quick question, when we getting the theoretical yield for CH3COOH, why we taking 21.4g as our final answer, not 21.42g. I went back and rewatch the sig figs chapter, but it still couldn't help me with the result. Thank you in advance for helping !
its 3 sig figs so you wouldn't add the 2
Thanks for your explanation :)
Your explains is more than great Mr Dave 🖒🖒🖒🖒
I still don’t know how did he get the stuff that appeared in 1:39
Update: I got it
Thanks Prof. Dave..
thank you 🌷
"if you don't understand stoichiometry clap your hands"👏👏
👏👏
Thank you very much professor dave 💓
Guys pls someone explain how to find the actual yeild if not given
thank you so much.....
You are amazing!!!
Thank you very much Sir.
sophmore taking ap chem with no chemistry background. you made this much easier to understand, thank you!
Am I missing something or is the first conversion factor at 2:37 totally not equal to one?
Where were you the whole year good sir...
How to get the 0.294 mol CO2
0.588 mol NH3• 1mol CO2/ 2mol NH3
fantastic...How can I thank you .professor Dave
Thank you for your vids prof
Do you HAVE to find out how much of both reactants can react with each other to subtract and find the excess.
Well like the amount of excess
Dammit Dave, that’s a glorious flowing mane you have
Me singing along to the Intro song and then feeling stupid...😂😂😂🤣😭😭
Sir if u have time please try to make a video on equivalent concept in chemistry really hard to get feel and intiution of that topic I would be very very thank ful to you ( from India Telangana)
This is great work, but unfortunatley it would be nice if the problem showed how the understanding of dimensional analysis played a role. But other than that it helped me get to an answer!
i have a separate clip on dimensional analysis, take a look!
عاد يا بروفيسر ديف انت الزيك منو 😍..تتعلى ما تتدلى
Thank you sooo much.😊☺️
thank you ssso mmuch
thank u chemistry jesus
Great video but I wish the calculations were explained more in-depth; you just say, "so we will now calculate how much product we will get," then just throw up a formula without explanation. I would have appreciated it if you explained why there was 60.1g of CH4N2O in the calculation, so I , for example, wouldn't have had to guess why. I realize now that is the molar mass of CH4N2O, but it would have been nice for that to be pointed out. 😫
How did you get the mole in the problem???
when converting to moles in the ex. in 1:45, why can't I convert between the substances after I have just the first substance in moles? and also why do I get different result while doing it in comparison to just calculating it separately?
10g NH3* 1 mole NH3/ 17g NH3= 0.588 moles NH3
0.588 moles NH3* 1 mole CO2/ 2 moles NH3= 0.294 moles CO2
--------while calculating separately will give me 0.227:
10g CO2* 1 mole CO2/ 44g CO2= 0.227 moles CO2
ok so the answer is because if I do that, I will get the amount of the other substance in moles that will theoretically react flawlessly with the first substance and not the actual amount of the substance.
I don't know if anybody needs it but there you have it
Thanks!
Very nice explanation......in a small amounta time.... :)
Very good!
Molecular mass of ch4n2o is 16+24+16= 54 g/ molecular but how did 60.1g is multiplied with .227 moles ?? Please explain how did 60.1 comes
it's the molecular mass, you miscalculated it.
Legend...
if during the experiment, you only included 97% of grams of 1 of the reactants, would you still get the same substance you’re expecting? only that, you wont get the theoretical yield amount in grams?
OR the whole experiment would be wrong because the expected substance wont happen?
oh wait!!! i think, we can still get the same substance for as long as we have the reactants necessary. the change in grams in the reactants does not affect the expected substance, only the amount in weight afterwards. please tell me if im right or wrong, Prof Dave. :) if im right or wrong, why or why not
I am confused how you got 60.1 grams of CH4N2O. I got 61.04 grams of CH4N2O by calculating the molar masses of NH3 and CO2.
60.1 is the molar mass, man!
CH4N2O is 60.056 grams rounded to 60.1 grams of molar mass.
Cool I was doing the wrong calculation.
Thanks, man!
Hows the limiting CH3OH not CO?
Idk
holy shit bro ty
Hi, professor Dave. May I ask you why the answer of the question is 89.3, not 90? I thought sig fig would be 1 since the one with the fewest sig fig is 100 (1 sig fig). Sorry if the question is too silly :( and I very apprieciate your well-made videos.
if anyone else could reply to me, I would also very apprieciate.
How do you do the last question professor dave?
thanks
Nicely done, you did not show which was the limiting reactant but it's CO
i put a box around it!
but both ch3oh & co equate to 0.357 mol. im confused why you chose co as the LR. need clarification on this one prof. dave.
thanks in adv.
He looks like Jesus himself if Jesus knew Chemistry.