The Periodic Table

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user123733

20:01

@Jan what more should I do

Loong

The first step (mass of solution) is a good idea. However, you really should add the missing units. For the next step, you may want to find out what ppb means and include this information into the question.

M.A.R.

I edited your question. Now you must be able to understand what is false in your reasonning. Why don't you use the infomation about "parts per billion" ?????????????????????????????????????????????????????????? — Hexacoordinate-C 1 hour ago

@Hexa's case is an unfortunate case of questionmarkitism

It's a very dangerous ailment, and it's contagious.

Loong

@M.A.R. contained

M.A.R.

I'm not sure ??????????????

Achoo ?????????????

Loong

cleanse the chat room!

preferably with fire

M.A.R.

20:11

Loong to cure the disease, can I PLEASE go and flag so you edit that comment?

Loong

@M.A.R. hm, maybe you should refresh that page

M.A.R.

All hail Loong

Jan

@Loong live @Loong!

Loong

do you mean "mod abuse"?

Jan

Q: Urgent: Similarities in Vitamins

Focus on the structure, formula, bonding type and physical properties of water soluble vitamins (B series, and C) and fat-soluble vitamins (A, D, E, and K). Identify structural similarities found in the two categories.

URGENT!!!11!!11eleven

@Jan @M.A.R. Is exclamationmarkitis as contageous as questionmarkitism?

M.A.R.

20:30

@Jan They all use English alphabet letters, and that's unfair.

Why doesn't any vitamin use Persian letters?

@Jan Yes, and more lethal.

Jan

@M.A.R. Because 19th century (I think)

M.A.R.

!!flip/19th century

Chemobot

◟(`ﮧ´ ◟ )І6ʇɥ ɔǝuʇnɹʎ

hBy2Py

20:43

@ortho, I piled on:

A: What is the significance of the kink in the [A] vs T graph for a 2nd order reaction?

Summary As orthocresol correctly notes, the presence of this kink is simply a feature of the $x$-shifted hyperbola that is the solution to the second-order kinetics problem, and the presence/characteristics of the kink will vary depending on the particular parameters of the problem. However, th...

orthocresol

20:55

I saw that. :)

But am a bit busy right now. I just browse front page to see if there's anything that needs to be dealt with.

M.A.R.

@hBy2Py Show-off

Eat that upvote

hBy2Py

@M.A.R. .... isn't that what answering on SE is all about?

M.A.R.

@hBy2Py It's also about getting repz. Ask Jan.

Jan

@M.A.R. No, badges and hats!

And also for ban-Mjölniring ~~homework dumpers~~ people who post their homework assignments here.

DSVA

21:11

good evening everyone =)

frostedcake

21:45

hey guys, quick question, do carbonyl groups attract each other? I wrote down in my notes that this is a factor that increases boiling points in them, but I haven't been able to find it on the internet after searching

Jan

A neglegible amount of pi(e)-stacking, imho. And some dipole-dipole-interactions. But for all intents and purposes they don’t attract each other more or less than two other polar groups that cannot hydrogen bond.

frostedcake

so basically the same as alkanes?

"Both aldehydes and ketones are polar molecules because of the presence of the carbon-oxygen double bond. As well as the dispersion forces, there will also be attractions between the permanent dipoles on nearby molecules."

"That means that the boiling points will be higher than those of similarly sized hydrocarbons - which only have dispersion forces."

so there are van der Waals, permanent dipole to dipole attractions, (which give them a higher boiling point than alkanes), but no hydrogen bonding which means they have a lower boiling point than alcohols?

Jan

Yeah, the C=O bond is polar, so you will get dipole interactions, but they are not the strongest of dipole interactions. No hydrogen bonds between carbonyls.

frostedcake

22:02

ok cool thanks

Jan

Noprob =)

Dovah-king

22:29

Jan

Hello other person.

I presume you have a question, too? ;)

Dovah-king

I'm having a very hard time studying organic reactions

I'm an engineering physics student, so i feel a lot comfortable with math or physics. But organic reactions seems like just loads and loads of information that i cant connect. Any tips studying it?

Im sorry if this is a vague question

Jan

Try and find the most nucleophilic position and the most electrophilic position. Usually, they will react.

Or, look out for possible pericyclic reactions that don’t follow the nucleophile/electrophile scheme (e.g. Diels-Alder).

Dovah-king

I have one question

In E2 reactions, why does the two atoms that will "leave" the alkene must be in periplanar position?

I think this also applies for E1

Jan

anti-periplanar.

It’s the best orbital overlap you can get.

It does not apply for E1 which is stepwise (leaving group leaves first, proton follows later).

Dovah-king

22:38

Yes, anti periplanar, sorry.

Jan

Although it is true that the proton, before it leaves, will align with the empty cationic p orbital.

Dovah-king

Why is it the best?

Another thing: when a E1 reaction is happening, will it always be also a Sn1 reaction happening together?

Jan

@Dovah-king Typically, both will happen if there is both a nucleophile and a base (and many nucleophiles are weak bases, too).

For anti-periplanar, if you arrange the orbitals in that manner, the pi bond your forming ‘is basically already formed’, just needs to get shifted a little.

For syn-periplanar, this is not the case, you would have a triangular shape.

towc

hope y'all don't mind if I ask noobish questions that could easily be googled, but finding weird varied answers

Dovah-king

@Jan i dont get why "is basically formed"

Jan

22:44

Well, compare the pi-orbital on the right to the orbitals on the left. If you ignore the differently sized lobes and the skewing, all features present in one are in the other, too. Most importantly, they are already parallel. So you only need to change small things to get from one side to another.

@towc Nah, fine. Just don’t be sad if nobody answers.

towc

yh sure thanks

Jan

I can feel the sensation of an @orthobreath down the back of my neck … O__O''

orthocresol