could you check the next page where I replied to the post from yesterday pertaining to nylon calculations.
Thanks
C
I partially answered the question.
Oh..I'm sorry that your brain is cooked but I greatly appreciate you helping me out.
Thanks Dr.Bob =)
Good grief...I actually did the problem correctly HOWEVER it doesn't figure in with what I found as the theoretical yield =(
Problem (copied)
I added in the lab:
10ml 5% solution of Hexamethylenediamine +
10 drops NaOH (20% solution) +
10ml 5% aq solution of adipoyl chloride (slowly pouring to form a top layer over the hexamethylenediamine and NaOH solution)
_______________________
what I did...based on what you said...even though it ended up the same as what I did on my own.
0.5g hex (1mol/116.21g)= 0.004302mol hex
0.5g adi (1mol/183.03g)= 0.002731mol adi
Limiting: 0.002731mol adi(1mol nylon/1mol adi)(226.319g/1mol nylon6.6)= 0.61807 g Nylon
Supposed to have theoretically: 0.61807 g Nylon
What I had when I weighed it:
0.818g Nylon
I think it was a little wet on one spot about 1/2 a inch but I don't think it was enough to account for the 0.2g difference.
Thus I am throughly confused and have a headache right now..
(I spent the whole day looking up what polymer would make the best pipe and how to make it and I really didn't get anywhere)
Thanks Dr.Bob
above
Remember it doesn't take very much water to weigh 200 mg; i.e., only about 0.2 mL (about four (4) big drops) and your product could easily absorb that much water if it wasn't completely dry. I didn't understand exactly how you converted the mols of nylon to grams; i.e., isn't the molar mass of nylon in the big numbers because it has polymerized.
As an AI text-based model, I don't have the capability to check specific web pages or posts. However, I can provide you with general instructions on how to navigate to the next page and find your reply in most common scenarios. Please follow these steps:
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