hypervalent_iodine

You won't find a crystal structure for it. I already gave you a list. Please read back on the thread.

ben, you already have a thread in which I have answered these questions. No, you will not find a crystal structure for either. As with fibre glass, there are many polymers they use for memory foam. I gave you a list in your original thread if you'd bothered to read it.

Since this is homework, you'll have to show us what working you've done and specific points where you need help or simple clarification. We aren't here to do your work for you; that part is entirely up to you.

From the Oregon Board of Psychologist Examiners: There are exceptions to this, but what they boil down to is, 'you cannot call yourself a psychologist without a PhD'. A PhD you admit not to having. So this? Is now very nearly an appeal to authority disguised by a blatant misrepresentation of who you actually are - a bricklayer who pretends he's a psychologist.

Could I get some clarification as to what you're calling dry? To me, it means you are after anhydrous Cu(OH)2, which I was under the impression was not impossible (I may be mistaken in that). Removing water from a compound in the manner I think you are suggesting doesn't fully work on account of there being water in the atmosphere. Usually, if you were completely drying something out it would imply the use of vacuum and / or elevated temperatures. Definitely not. If you leave it open to atmosphere, then yes, it will decompose to give this. The question now is do you understand why leaving it open to atmosphere will do this? It's all well and good telling you that you're correct, but it amounts to nothing if you were simply throwing answers out with no real logic behind why you chose them. Hardly. There's nothing wrong with asking questions if you don't understand something.

Topic split from here. So far as I can tell, the big bang and the age of the universe has nothing to do with swansont's credentials or how and when he chooses to share them.

Well, it does, it's just a weird thing to ask you. And yes, I was hinting towards what you've managed to work out.

Each point is a carbon with a certain number of hydrogens attached to them (each carbon will have four bonds, so you should be able to work out how many hydrogens are attached to each carbon). The hexagons with the double bonds are called phenyl rings and yes, the O is for oxygen. The OH group is called a hydroxyl group (a hydrogen bound to an oxygen). How to read the above structure properly is a lot for me to write up. Lucky for us, other people have already taken the liberty to writing it. Have a read of this website. I'm sorry, I don't really know. As I said, I'm an organic chemist and this is all very much outside of my area. I suppose a prudent question would be, compared to what?

This is where it starts to get to the limit of my knowledge. I don't know of any advantages, per say. They are chiral, which means that their mirror images are non-superimposable onto on another the same way that your hands are (i.e. your hands are mirror images of each other and if you try to place one hand on top of the other, it won't be superimposable). This means that they can rotate the plane of polarised light to some degree and are in that sense, optically active. That's my best guess, though it doesn't answer your question very well. Nope. Fibre glass consists polymer reinforced glass fibres; usually they use co-polymers like the epoxy one I linked you.

If you can tell us what answers you're working our them, it would help us see if you're doing anything wrong. The best hint I can give you is to make sure you pay attention to which orbitals have the highest energy when you are removing them to generate your ions. This should be easy for the bromonium ion (which is a weird example to have given you, by the way; are you sure that one is written down correctly?), but the cobalt and zinc being transition metals are slightly different, although the logic is the same; electrons are removed from the highest energy orbitals first.

Yes, but be aware that the structure actually looks like this: Image drawn myself. Drawing conventions in organic chemistry are designed so we don't have to spend our lives writing out the letters C and H over and over again (also to reduce clutter). You'll often see images that are literally just a bunch of zig-zag lines. They might not look like much, but believe it or not, these do actually hold quite a lot of chemical information in them. In you above image, the ends of the lines coming out of the parentheses represent CH3 groups.

The atomic structure would not change, it would still be SiO4, but the way they arrange them selves (i.e. the crystal structure) might. It can change it quite a lot. The best example of this is carbon. In one arrangement you get graphite, which consists of layers of carbon sheets and which is flakey and easy to break. In another arrangement, you get diamond. As for the quartz, I don't really know. I would say not very much. The only difference between the two is that one part of the crystal structure is twisted; they way they are linked is exactly the same. In diamond vs. graphite, they way they are linked/bonded together is very different, giving rise to the huge differences in their physical properties.

It tells you in the question how many copper atoms are present, so all you have to do is solve for x the same way I did in my previous post. The net charge of the molecule is 0, so again we have: 0 = 3x -3 (3x because there are three copper atoms in the formula) 3 = 3x Which gives us x = 1, i.e. Cu+. So we have 3 coppper atoms, all with a +1 oxidation state, and a central nitrogen with a -3 charge. As much as I hate to promote this sort of learning, there are some things that you just have to remember. Unless in their elemental form, hydrogen exists in a +1 state, oxygen is -2 (unless it's in a peroxide, in which case it is -1) - there are a few. Take a look at this wiki page. It should help. Again, you would treat this question algebraically: 0 = x + 2(-2) 0 = x - 4 Giving x = 4, which makes our lead Pb (IV), or Pb4+.

Have a look at the crystal structure. It just occurred to me that you might not have realised this, but the crystal structures show ball-and-stick models of your compound. Each ball represents an atom and each stick is a bond. They are usually colour coded; carbon is usually grey, oxygen is red, nitrogen blue, etc. In the quartz crystal structures shown on the wiki page, the Si atoms are represented in grey and the oxygen in red.

Is this a homework question?

Geez, off-topic much? Wolfram to the rescue! The square root of 17 to 26 decimal places: 4.12310562561766054982140986

Oh, feel free to keep asking more. I'm bored as anything stuck here in bed.

What you are referring to is called the octet rule. And yes, that is exactly what it means. If you look at the crystal structure you had of it, you'll see all of the SiC molecules are all joined together; each C is bound to 4 Si atoms and each Si atom is bound to 4 C atoms. Wikipedia has some good images. I think I linked it a few posts back, but if not, you can find them here. Yeah, most likely. As I said on the last page somewhere, you can make fibreglass from a number of different polymers - either a thermoplastic or a thermosetting plastic (eg. epoxys, polyesters or polyvinyl esters). It is usually a combination of a couple of different polymers, so my recommendation would be to find an example of just one polymer and draw that. That's alright, I enjoy helping where I can. I'm sorry I haven't been more attentive, I've been quite sick the past four or so days and haven't been very active.

The I and II describe the oxidation states of the copper. Copper (II) is Cu2+ and copper (I) is Cu+. You can tell the oxidation states by treating the whole thing as an algebraic expression. In the case you've listed, the net charge of the complex is 0, which means that the sum of the charges of the constituents must sum to 0. We can assume that the sulphur has an oxidation state of 2-, which makes the equation 0 = x - 2 (where x is the oxidation state of the copper) Rearranging that gives x = 2, so the copper in CuS must have a +2 oxidation state. For Cu2S, we have: 0 = 2x -2 2 = 2x Which then gives us x = 1, so in this case our copper must have an oxidation state of +1. Ligands such as water, ammonium, cyano groups, etc., all have fairly well-defined defined charges. Water is neutral, ammonium has a +1 charge, cyanide has a -1 charge, etc., etc. Sulphur tends to be a bit tricky in cases like this, as it can exist in quite a number of oxidation states due to its ability to form hypervalent bonds; in this case, however, the copper monosulphides will have S2-. Also, as an aside, I'm not a gentleman

I didn't read what he had in brackets. Yes, you're right. Banks, when you write posts you'll notice that the bar that gives you options to bold, underline, etc., also gives you the function to sub and super script. They are the X2 and X2 buttons, respectively.

You would usually write it with the charge in brackets, though it is sometimes not necessary. Your example makes it clear that the oxidation state is +2, so you wouldn't have to write (II), but it is just as correct if you do.

Alpha is trigonal, beta is hexagonal. I don't really know, sorry. In truth, I'm an organic chemist, so this is a little out of my depth. If you want them the same as in your last post, then yes, but it would be difficult and convoluted. Alpha is correct. The beta one is cubic; more specifically, it has a crystal crystal structure similar to diamond, called zincblende. You could draw them like that, but I would only do it if you absolutely had to.

As the others have said, you might like to look up basic naming conventions. The big hint as to how you were meant to know there was a double bond was the 'en' (for alkENe) as opposed to 'an' (for alkANe) as Fuzzwood pointed out. Similarly, if it were a triple bond, you would see 'yn' for alkYNe. IUPAC conventions can get confusing with large, non-linear molecules, but basic ones such asnyoursnarenfairly easy to figure out once you get a feel for the prefixes, etc. And of course the best way to get a feel for it is to practice. Start small and work your way up and you'll soon start to find it comes quite naturally to you.

You're really going to claim that the pictures you've posted here are of you? Unless your name is actually Izabella Carr or Andrea Nocturna, I can say that they most certainly are not. On the contrary. By copy pasting something that is not yours and not acknowledging the actual author of your post, you are taking the credit. This is blatant plagiarism, which you should already know.

Damn, Phi for all. What have I told you about harnessing the new members? Only if I'm invited.