For problem 1 you can also do 1.107/13=8.2~8 which is our number of carbons 2. 107-13(8)=3 3+8=11 which is the number of Hydrogens so we Have C8H11. 4. From the nitrogen rule we know there has to be a nitrogen since we have an odd m/z 5. 1 CH2 = N..so then we have C8H11 - CH2 = C7H9N as our answer

in my class, we call the peak with a relative intensity of 100% the "base peak," which indicates the molecular weight of the most stable carbocation. the" m peak" is the peak furthest to the right, indicating the molecular weight of the parent compound

Thank you soo much señorita this was a great help 💐💐💐💐

it helps a lot, thank you for the detailed explanation

this was very helpful thank you🐱

thank you very much, that was very helpful.

Where 1.1% came from?

Thank alott ❤❤❤

good good good

Good explanation.

Very helpful, thanks

For problem 1 apply rule of 13 within 1 step u got ans😅

Thank you 😊

Thanks

super helpful, thank you!

How to see and calculate the relative abundance more than 100 scale.

Thanks 😊

Please where did you get the 1.1% from

nicee! thx

How old are you

dump lecture. dont see it ........waste of time.