hypervalent_iodine

! Moderator Note Over 9000, if you can't post in this thread without resulting to insults then you will find that you will not be able to post at all. Furthermore, resorting to calling someone asking you to back up your claims stupid is rather disingenuous of you, and a particularly poor way to hold discussion. Such actions are also prohibited on this forum, so I would ask you to refrain from them in the future. Do not reply to this note within the thread. And just to be clear: you will be suspended if your attitude doesn't make a rapid change.

! Moderator Note There have been a number of posts subsequent to Phi's warning that have proffered direct medical advice to the OP. I have removed these (as well as a reply or two to said posts). We are not in the business of diagnosing people over the internet, and we expect our members to respect this and refrain from such posts. Since it seems that the framing of this thread is too difficult to navigate without going against this directive, I am closing it.

You've quite possibly made caclium or magnesium phosphate, or calcium sulphate (most likely) by mixing those three salts together. None are very soluble.

! Moderator Note As amusing as this is, I think we're done here.

Doesn't the question tell you the Ka of acetic acid is 1.8 x 10-5 ? As for telling you to use a specific method, I wouldn't worry about it. The question is clearly primed for use in the HH equation. Anything else (particularly if you've been taught HH) is overcomplicating it and wasting your time. Since this is through Wiley Plus (which I assume is being used as a supplementary form of assessment with questions not written by your course providers), I wouldn't stress about the way they tell you to approach it and instead focus on methods you've learned yourself in class. That being said, some notes on the method you originally took and that they suggest. You were actually quite close, but I think you were killed by rounding errors resulting from not including the + and - x terms. The equation does get a bit tricky to solve, but it's best to keep them in these cases, unless you're using the HH equation as you probably should be.

You probably don't want to do LC or GC with metal complexes. ESI might help. I would certainly try that if you can. Could you give more information on what you are doing? It might help us identify potential products.

I think the question is worded a bit funny. I think the fact that you don't get the anhydride is taken as a given. The question is why you get that regioselectivity. The reason is simply that the anhydride that would form if the carboxylate reacts with the acetic anhydride is less stable than the ester that forms when the phenoxide reacts. Under basic conditions the phenoxide is also going to be a stronger base and more likely to react with the anhydride.

I honestly don't care if you stay or go. That's up to you. I'm merely suggesting to you that your reasons behind leaving don't make any sense.

Right, so when you substitute SFN with something else equally as unhelpful, what then? Are you just going to keep asking to be banned from places? I can't see how that helps.

We can ban you, but it honestly doesn't sound like your problem is your ability to log on to SFN. IMO, it wouldn't help your situation one bit.

I'm not a math expert, but of course it's possible. The likelihood would (I imagine) be very small, but it's still within the realm of possibility.

I suck at rote learning as well. Much like you, this section of general chemistry didn't fully click for me until I had something tangible and visual to relate all the information too. I'm glad it helped.

The mechanism would be similar to that of aromatic sulfonation, yes.

The l number describes the shape of the orbital, so the orbital it refers to can be figured based on how many orientations the l number can have, which is just simply how many ml numbers are possible. So: For l = 0 there is only one possible ml number (0). There is only one type of s orbital, therefor l = 0 corresponds to the s orbitals. For l = 1 you can have 3 ml numbers (-1, 0, +1). These correspond to the three p orbitals - px, py, and pz. For l = 2 you have 5 ml numbers, which correspond to the 5 d orbitals.

This is starting to get off topic, but I do feel like you're shifting the goal posts a bit here. You're initial claim was that most chemistry doesn't deal with anything beyond basic numeracy. I don't believe that this is a fair assumption. The math involved in quantum modelling, spectroscopy, docking, kinetics and other areas of physical chemistry, some types of analytical chemistry, etc., might not be what you would consider high brow, but surely it's beyond basic numeracy. OP, I'm sure you can find space to get an MSc of PhD in math and chemistry, but you will need to have some understanding of the chemistry and physics involved. You will also probably need to look abroad for an appropriate program and supervisor. I would look into the areas suggested and pick a few things you think you'd be interested in, perhaps look into some recent papers in the area, and then look to universities and institutions with groups that meet your needs.

Ajb, with respect, you can't both claim that chemistry is outside your area of interest and then make assumptions about what a typical chemist does or does not know.

Sensei, my main point was that I wasn't sure if what you suggested would be of any interest to current research. It seems a bit simple to me, though it touches on more interesting kinetic modelling.

Just to perhaps clarify some of Sensei's post, he is including the D block elements in his group numbering. Some places will teach that there are 8 groups, not 18. In that case, the group number corresponds to the number of valence electrons in the neutral species (e.g. Fluorine is group 7 and has 7 valence electrons). As for your second question, you can't determine the identity of the atom based on the quantum numbers of one of its electrons. That makes far too many assumptions and flies in the face of pretty basic principles. You firstly have to assume that the electron it is describing is the last one to be placed into an orbital, meaning it occupies the highest energy position. You can determine that this electron is in a 3d orbital, since n = 3 (the energy level) and l = 2 (this describes the type of orbital; l = 0 for s, 1 for p and 2 for d). The ml numbers tells you the orientation along the Cartesian axes, but since this is an atom, the d orbitals are degenerate and you have no way of knowing which of those 5 is filled in what order. That means your electron could have been 1st d electron, or it could be the last. I'm not sure how you're expected to solve that question. Are you given more information?

There is an entire discipline called mathematical chemistry. I think it's mostly to do with topology and graph theory. Perhaps look into that? Sensei also touches on the first area that I thought of, though his explanation / example is a bit oversimplified. Computational chemistry - which deals with a lot of modelling of various systems - requires a great deal of math and programming skills, which you might find to be in your area of interest. It encompasses a pretty wide range of sub areas, so you may wish to look up the discipline yourself. Kinetics and other aspects of physical chemistry are good also, but these areas have huge wet lab components and a solid understanding of practical techniques. In either case, you will also need some understanding of chemistry. Not having this would put you at a disadvantage, since you wouldn't fully understand the systems you're working with. I do not believe that mathematical chemistry would rely on such a contingent.

In that case, the problem may be that you're trying to import a drug listed as perscription only. From what I'm reading, in Europe it's only used in special cases, meaning that it might be tricky for your common person to buy it over the Internet.

! Moderator Note kingJack, Please do not use the website to garner funding for your project. This is a discussion forum.

I once had a package leave Chicago, arrive in Australia (where I live), then after 2 weeks pack up and make its way to London, back to Chicago and finally 7 weeks later, to my doorstep. I've had a few items never make it past customs. One time I paid $70 to have the item treated, and another time where that wasn't an option I left it to be destroyed. Australia has very strict biosecurity laws, so it's not surprising, it just makes buying certain things difficult. Perhaps a blessing in disguise for you, anyway. Chinese herbal powders and the like are notorious for not being what they should be, never mind the fact that many of their purported medical benefits are bogus.

! Moderator Note You should know by now that reopening closed threads is against the forum rules. Start abiding by them, or we will be forced to take further action.

! Moderator Note Mike, you are once again talking nonsense and refusing to listen and take on any of the advice given to you in a meaningful way. This seriously has to stop. Thread closed.

! Moderator Note Your thread attempts to refute something accepted by mainstream science. That it is your opinion is literally in the title. It belongs in Speculations. If you have an issue with this, please voice it via PM or the report feature. I will be removing any more off topic replies. Edit: others have mentioned the issues with your so-called support. Please address those claims.