Watching this during my chemistry unit test 🙏

@lolae.4562

2 жыл бұрын

same haha

@aniketdesale494

Жыл бұрын

Savage

@Joseph-hf9mi

Жыл бұрын

Exam*

@rod2662

Жыл бұрын

@Joseph they called it a unit test I wrote unit test

@heelylife

Жыл бұрын

How did you do

short, straight to the point video. helped me out within 5 mins

Wow. English is not even my language, and yet you're explaining way better than some french people💯thanks and good job👌🏽👍🏽

Dude! You are the freaking MAN! Thank you so much, helps like you wouldn't believe!

This was perfect. Understood it immediately

Love u sir o finally understood excess reagent

Thank you good sir, my chemistry teacher has been KZread and you are now one of many KZreadr teachers

a bit late but he got 1.49 by multiplying 3.5 g of O2 by 1 mol of O2 then dividing it by 32 g of O2 then repeat (remember to cancel out everything except the top right which is the 17.031 g of NH3 for you to come up with grams of NH3 in your final answer)

Can you pls explain gram to gram conversion properly

thank you for refreshing my memory!

Thankyou so much, you saved me 1 day before my test

Thank you very much, really appreciate it

How did you get 1.49 ?????I'm so lost there

@cakemonster3983

3 жыл бұрын

I was confused too but I found out that you should follow the formula. You multiply and divide the numbers regardless if it is grams or mol

Easy and helpful, thanks

How'd you get 1.49 g

omg bless this video is everything

Very helpful, thank you

How did u get 1.49 ?

thank youuuuu for this simple explanation ,

love you sir🥺 tomorrow will be my chemistry final test and you help me a lot!

@SI.COYG6

Жыл бұрын

how was it

THANK YOU SOOO MUCH!!!!

Watching this while doing my assignment due in the next hour. Tnx

amazing job i got the same chemical reaction 😂❤️❤️

How did you get the 1.49g, Please I dont know how?

thankyou sir you made it easy

ALL THE OTHER VIDEOS DO SOME WIERD SHIT AND DONT EXPLAIN< BUT YOU, YOUR A LEGENDDDD

How did you get the 1.49g

Explanation is just a quick but amazingly did it well and more informative than I expected!!

thanks this was easy to follow!

What if it’s in moles and not grams?

Thank youuuu

God bless you sir

Oh how did you get the 17.031?

From India🇮🇳

life saver

Is it possible to calculate for the excess amount of excess reagent if the problem only include the mass of only one reactant?

@jaysondebrah8359

2 жыл бұрын

Find the mass of the other one

U saved meeeeeeeee

1:02

Ty

I finawy figuwed out how to do dis

can I directly cut from their mols instead? I mean by the mol of excess minus the mol of limiting, to find the excess mass...?

@senthilpuliadi6599

2 жыл бұрын

No u can't

how do you calculate on the calculator to get 1.49 im lost

@carelessbear800

11 ай бұрын

multiply straight across and divide by the bottom 3.5g O2 / 32g O2 = 0.107 (0.107 x 4 )/ 5 = 0.0856 0.0856 x 12.031 = 1.49g NH3 how i remeber its like a zig zag. diviide given mass by its molar mass. Then multiple and divide by the mole ratio (multiple by numerator first then divide by denominator) and finally multiple by the grams. (dont really need to divde by 1. its a given what ur anwser would be) hope this helps anyone

30 minutes till my chemistry midterm

Thank you so much 🤍🌷

i wonder what youre up to these days? engineer? MD?

Everything in blue here is unnecessary. The only thing it can be used for is determining WHICH is the excess/limiting reactant. It gives no data of any real value later on. You can also determine WHICH is the excess reactant by just starting with everything that he wrote in purple. Compare the amount of NH3 "Consumed" when reacting with 3.5g O2 (1.49g NH3) to the amount of NH3 given in the experiment (3.25g NH3). The amount consumed is less than the amount given in the experiment, thus some of the NH3 would be left over. Thus it would be the excess reactant. This saves time because it identifies which reactant is in excess, as well as giving you real numbers that can be used to determine how much in excess the reactant actually is. Edit: I do realize that for someone being introduced to the topic, your way is probably better for them to learn. On a test however, especially a test on stoichiometry, this would be burning time that you really don't have.

watching 5 hrs before Chem exam fml

3.5 x 1/32 x 4 /5 x 12.031 / 1 = 1.053g NH3 am I wrong here?

@bendyaerial5940

4 жыл бұрын

sorry, 17.031 actually looks like 12.031

@meddlemedley740

4 жыл бұрын

@Bendyaerial he meant 17.031 but prob wrote the number like 2 on accident

@Manuxe

2 жыл бұрын

@@meddlemedley740 17.031 is a molar mass of what?

Goat

Great video. I was lost at the point where you had the 1.49g of NH3. That calculation is supposed to give 1.053g of NH3

@friendlytutor8891

2 жыл бұрын

I think you mistook 17.031 as 12.031. he was using 17.031 because that is that molar mass of NH3 :)

Very poor explanation

How did you get the 1.49g, Please I dont know how?