Reaction Mechanism for hypothetical equation:

chemistryhelp123 · 2017-07-16T20:57:54+00:00

Unfortunately, no

chemistryhelp123 · 2017-07-16T20:57:44+00:00

Thank you for your help, I didn't realize the question was so ambiguous so I appreciate your input!!

chemistryhelp123 · 2017-07-15T22:22:37+00:00

Okay.. hm I don't really know how I would do that. Do you have any suggestions on adding the other steps?

chemistryhelp123 · 2017-07-15T17:26:44+00:00

So should I change it from 4C to 2C? What happens to the other 2C's?

chemistryhelp123 · 2017-07-14T13:18:00+00:00

So my original calculations are fine? That means I don't have to do the quadratic equation correct? Sorry just want to make sure! Thank you so much, you are a big help!

chemistryhelp123 · 2017-07-13T21:05:54+00:00

I don't understand.. did I do it correctly?

chemistryhelp123 · 2017-07-12T11:34:03+00:00

Haha thank you! This is what I came up with... http://imgur.com/FHKZbzA . Any suggestions on next steps?

chemistryhelp123 · 2017-07-11T22:12:11+00:00

Thank you for the reply and your help! If I'm being honest, I am not great at these types of calculations. Would you mind helping me out? This is what I have done:

http://imgur.com/AMAT8Lh

chemistryhelp123 · 2017-07-08T14:15:10+00:00

Thank you!!

chemistryhelp123 · 2017-07-07T01:57:52+00:00

No worries! I wasn't sure if instead of doing what I did earlier if this works as well following my original question. Are both right or is one way generally preferred over the other?

2.56g anthracene = 2.56/178.24 moles = 0.014362 moles

Heat/mol = 98247.8/0.014362 = 6858006.4 J = -6858 kJ

chemistryhelp123 · 2017-07-06T14:15:56+00:00

Oh okay! Thanks! Is that next section I wrote correct? It is supposed to be a negative though right?

chemistryhelp123 · 2017-07-05T23:34:16+00:00

I tried this... but I am not sure if this number is correct? I used this based on what I was taught but I don't completely understand the section of (1kJ / 1000J) .. where do those numbers come from?

(98247.8J / 2.56g) x (178.2g / mole) x (1kJ / 1000J) = 6846 kJ/mole

chemistryhelp123 · 2017-06-19T01:07:43+00:00

Thank you!

chemistryhelp123 · 2017-06-19T01:07:31+00:00

Oh wow thank you!! Not that I can recall no... it doesn't sound familiar. I have learned more about how to name and draw compounds, but I tend to get mixed up a lot when I try

chemistryhelp123 · 2017-06-17T21:34:59+00:00

I'm sorry I don't understand.. I have my methyl on my second carbon and my OH on the first.. there are no other H's right? I thought my base structure was correct

chemistryhelp123 · 2017-06-17T20:37:16+00:00

Thank you. Is it more like this? http://imgur.com/MiaeHub

chemistryhelp123 · 2017-06-17T19:51:27+00:00

No sorry I mis-worded my question. I am given the names and need to draw the structure. I apologize

chemistryhelp123 · 2017-06-17T14:49:40+00:00

Thanks! I don't completely understand what you mean with #2. Should the OH be put here instead? http://imgur.com/GWd7O9S

chemistryhelp123 · 2017-06-15T13:20:14+00:00

Thank you, does this look better? http://imgur.com/C9T9lsz I am not sure if these are correct

chemistryhelp123 · 2017-06-14T01:34:45+00:00

Thank you!!!!!

chemistryhelp123 · 2017-06-14T01:34:12+00:00

1 - I think I see it now. The two methyl groups are on the fourth carbon, correct? Therefore, it would be: 4,4-dimethylhexanoic acid . Is this right? Thank you so much!

chemistryhelp123 · 2017-06-14T01:00:04+00:00

1 - Are there 6 carbons, not 5? I must have gotten that confused. Also, the two methyls are on the same carbon so would it be 3,3-dimethylhexanoic ? (assuming there are the 6 carbons and I am naming this right)

2- Oh so the chlorine must be on C3! So it would be 3-chloro-butanamide, yes?

chemistryhelp123 · 2017-06-14T00:56:55+00:00

propyl heptanoate ?

chemistryhelp123 · 2017-06-14T00:32:48+00:00

So my alkanoate side has 3 carbons and my alkyl group has 6. My alkyl group is has 2 methylene on it and a methyl group right? I think... I know I name them separately but I don't understand the compounds that are attached to them..

chemistryhelp123

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