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> but I can't find any canonical reference as to that. [Look harder][1] ;-) It seems to be that the lone pair has lowest priority. OK, I can't find any *authoritative* reference to that (I don't have access to the [original paper][2]), but I've got plenty of others: > In the Cahn-Ingold-Prelog priority system, things with lower atomic number have lower priority. Hydrogen has an atomic number of one, but a lone pair has an atomic number of zero. Therefore a lone pair is lowest possible priority attachment. ([source][3]--UCLA class notes) > Lone pairs have atomic number 0 ([source][4]--MIT OCW) > If a lone pair forms one of the groups, then it has the lowest priority. ([source][5]-- _Molecular Symmetry_ , David J. Willock) Unless it's a widespread hoax, I'd say we don't need to see the original paper. After all CIP is based on the "number of protons"--atomic number--with "mass number" being the tie-breaker. Lone pairs have 0 protons and nearly 0 relative mass, so it's natural to give them 0 priority. Now I wonder what happens to these rules if you have _two_ lone pairs, one of them [muonic][6] ;-) [1]: http://www.google.co.in/webhp?sourceid=chrome-instant&ie=UTF-8#hl=en&output=search&sclient=psy-ab&q=Cahn-Ingold-Prelog%20lone%20pair&oq=&aq=&aqi=&aql=&gs_l=&pbx=1&fp=c59d3fc9b9c4309b&bav=on.2,or.r_gc.r_pw.r_qf.,cf.osb&biw=1067&bih=548 [2]: http://onlinelibrary.wiley.com/doi/10.1002/anie.196603851/abstract [3]: http://www.chem.ucla.edu/harding/30A/30A_su09/stereo_30A.pdf [4]: http://ocw.mit.edu/courses/chemistry/5-43-advanced-organic-chemistry-spring-2007/lecture-notes/01_handout.pdf [5]: http://books.google.co.in/books?id=2Uj2Z1XfzFoC&lpg=PA303&dq=%22If%20a%20lone%20pair%20forms%20one%20of%20the%20groups,%20then%20it%20has%20the%20lowest%20priority.%22&pg=PA303#v=onepage&q=%22If%20a%20lone%20pair%20forms%20one%20of%20the%20groups,%20then%20it%20has%20the%20lowest%20priority.%22&f=false [6]: http://en.wikipedia.org/wiki/Exotic_atom#Muonic_atoms
> but I can't find any canonical reference as to that. [Look harder][1] ;-) It seems to be that the lone pair has lowest priority. <s>OK, I can't find any *authoritative* reference to that (I don't have access to the [original paper][2]), but I've got plenty of others:</s> Alright, IUPAC [says][3]: >Except for hydrogen, ligancy, if not already four, is made up to four by adding "phantom atoms" which have atomic number zero (0) and are thus always last in order of preference. >This has various uses but perhaps the most interesting is where nitrogen occurs in a rigid skeleton, as for example in a-isosparteine (14); here the phantom atom can be placed where the nitrogen lone pair of electrons is; then N-i appears as shown alongside the formula, and the chiraiity (R) is the consequence; the same applies to N-i6. Phantom atoms are similarly used when assigning chirality symbols to chiral sulfoxides (see example to Rule E—4.9). When ligancy is less than four, we obviously have some lone pairs hanging around (except in Hydrogen--which has been excluded, and some other cases like carbocations--which are planar anyway) So the lone pair becomes a "phantom atom" and has 0 priority. Now I wonder what happens to these rules if you have _two_ lone pairs, one of them [muonic][4] ;-) <sub>Check the [first revision of this post][5] for my original, non-authoritative references</sub> [1]: http://www.google.co.in/webhp?sourceid=chrome-instant&ie=UTF-8#hl=en&output=search&sclient=psy-ab&q=Cahn-Ingold-Prelog%20lone%20pair&oq=&aq=&aqi=&aql=&gs_l=&pbx=1&fp=c59d3fc9b9c4309b&bav=on.2,or.r_gc.r_pw.r_qf.,cf.osb&biw=1067&bih=548 [2]: http://onlinelibrary.wiley.com/doi/10.1002/anie.196603851/abstract [3]: http://www.iupac.org/publications/pac/1976/pdf/4501x0011.pdf [4]: http://en.wikipedia.org/wiki/Exotic_atom#Muonic_atoms [5]: http://chemistry.stackexchange.com/revisions/443/1
can an atom have more than 8 valence electrons? according I have reviewed in chemical texts, the max quantity is 8, but does not explained the why. then if is not possible, why an atom can not have more than 8 valence electrons? thanks in advance.
can an atom have more than 8 valence electrons? if not, why 8 is the limit?
Can an atom have more than 8 valence electrons? According to some chemistry textbooks, the maximum number is 8, but they do not explain why. If is not possible, why can't an atom have more than 8 valence electrons?
Can an atom have more than 8 valence electrons? If not, why 8 is the limit?
Can an atom have more than 8 valence electrons? According to some chemistry textbooks, the maximum number is 8, but they do not explain why. If it is not possible, why can't an atom have more than 8 valence electrons?
Chiral molecules tend to absorb one circular polarization of light more than the other. Is this ever used to isolate a particular enantiomer?
Is circularly polarized light ever used to isolate particular chiralities?
I'm familiar with the concept of a reaction coordinate from high school chemistry as some generic conformational parameter that all of the intermediate states of a chemical reaction lie upon. Recently, though, I've seen some presentations where chemists presented plots of the free energy with respect to two reaction coordinates, and my understanding is that the plot was of a measurement (I could be wrong, though). What exactly does the reaction coordinate mean? What does it mean to have more than one reaction coordinate for the same reaction? Is the free energy as a function of the reaction coordinate a measurable quantity?
What does "reaction coordinate" mean?
I liked this question because I had never thought much about it. However, it's not such a mistery because the answer is in the [Wikipedia “radical” page][1]: > Historically, the term radical was also used for **bound parts of the molecule**, especially when they remain unchanged in reactions. These are **now called functional groups**. For example, methyl alcohol was described as consisting of a methyl "radical" and a hydroxyl "radical". Neither are radicals in the modern chemical sense, as they are permanently bound to each other, and have no unpaired, reactive electrons. However, they can be observed as radicals in mass spectrometry when broken apart by irradiation with energetic electrons. So, the *free* in *free radical* was originally used to distinguish it from a radical bound to a carbon skeleton. --- Digging into historical chemistry texts reveals that “free radical” becomes really widely used around 1930: [![enter image description here][2]](http://books.google.com/ngrams/graph?content=free+radical&year_start=1800&year_end=2000&corpus=0&smoothing=3) However, some quotes for earlier works use it, though it does not appear so frequently that it may be considered a common expression… i.e. it is at that time simply the juxtaposition of the adjective *free* with *radical*, which is then synonymous with *group* (or today's noun *substituent*). For example: > “This hydrocarbon was formerly supposed to be the free radical of the methyl group, but no methyl compounds have in any way been derived from it, whilst by the action, of chlorine upon it, ethyl chloride is obtained as the product” (1870, *[Lessons in elementary chemistry][3]*) or > “Although, however, the free radical methylene is unknown, many of its compounds have been prepared, of which some, such as the haloid ethers, &c.” (1884, *A Treatise on Chemistry*, Henry Enfield Roscoe and Carl Schorlemmer) Though more quotes could be dug and provided, the ones I've read support this idea that free radical initially meant “the chemical group as a molecule by itself”. [1]: http://en.wikipedia.org/wiki/Radical_%28chemistry%29 [2]: https://i.stack.imgur.com/a8bMx.png [3]: http://books.google.com/books?id=8K8XAAAAIAAJ
While reading [this paper][1] (suggested by @jonsca [here][2]), I came across the following line: > Surface reconstruction is known to be extremely important for Si, the material of 20th century electronics. However, surface reconstruction has been much less studies for graphite. One of the reasons might be graphite's limited use in electronics because **fabrication of ordered graphite substrates is difficult.** (emphasis mine) I've always believed that graphite isn't much used in semiconductor tech since it isn't a metalloid and conducts in an entirely different manner, probably making doping not work. This reason is new to me, but quite interesting. I can sort of guess why "fabrication of ordered graphite substrates is difficult" --I'm not too sure though. At any rate, I'd like to know more about it. Exactly _why_ is such fabrication difficult? [1]: http://www.lehigh.edu/~svr204/newdata/mypapers/191-200.pdf [2]: http://chemistry.stackexchange.com/a/449/22
Why is fabrication of ordered graphite substrates difficult?
While reading [this paper][1] (suggested by @jonsca [here][2]), I came across the following line: > Surface reconstruction is known to be extremely important for Si, the material of 20th century electronics. However, surface reconstruction has been much less studies for graphite. One of the reasons might be graphite's limited use in electronics because **fabrication of ordered graphite substrates is difficult.** (emphasis mine) I've always believed that graphite isn't much used in semiconductor tech since it isn't a metalloid and conducts in an entirely different manner, probably making doping not work. This reason is new to me, but quite interesting. I can sort of guess why "fabrication of ordered graphite substrates is difficult" — I'm not too sure though. At any rate, I'd like to know more about it. Exactly _why_ is such fabrication difficult? [1]: http://www.lehigh.edu/~svr204/newdata/mypapers/191-200.pdf [2]: http://chemistry.stackexchange.com/a/449/22
This answer is intended to supplement [Manishearth's earlier answer](http://chemistry.stackexchange.com/a/445/83), rather than compete with it. My objective is to show how octet rules can be helpful even for molecules that contain more than the usual complement of eight electrons in their valence shell. I call it donation notation, and it dates back to my high school days when none of the chemistry of the texts in my small-town library bothered to explain how those oxygen bonds worked in anions such as carbonate, chlorate, sulfate, nitrate, and phosphate. The idea behind this notation is simple. You begin with the electron dot notation, then add arrows that show whether and how other atoms are "borrowing" each electron. A dot with an arrow means that that the electron "belongs" mainly to the atom at the base of the arrow, but is being used by another atom to help complete that atom's octet. A simple arrow without any dot indicates that the electron has effectively left the original atom. In that case the electron is no longer attached to the arrow at all, but is instead shown as an increase in the number of valence electrons in the atoms at the end of the arrow. Here are examples using table salt (ionic) and oxygen (covalent): ![salt and oxygen in donation notation][1] Notice that the ionic bond of $\ce{NaCl}$ shows up simply as an arrow, indication that it has "donated" its outermost electron and fallen back to its inner octet of electrons to satisfy its own completion priorities. (Such inner octets are never shown.) Covalent bonds happen when each atom contributes one electron to a bond. Donation notation shows both electrons, so doubly bonded oxygen winds up with four arrows between the atoms. Donation notation is not really needed for simple covalent bonds, however. It's intended more for showing how bonding works in anions. Two closely related examples are calcium sulfate ($\ce{CaSO4}$, better known as gypsum) and calcium sulfite ($\ce{CaSO3}$, a common food preservative): ![calcium sulfate and sulfite in donation notation][2] In these examples the calcium donates via a mostly ionic bond, so its contribution becomes a pair of arrows that donate two electrons to the core of the anion, completing the octet of the sulfur atom. The oxygen atoms then attach to the sulfur and "borrow" entire electrons pairs, without really contributing anything in return. This borrowing model is a major factor in why there can be more than one anion for elements such as sulfur (sulphates and sulfites) and nitrogen (nitrates and nitrites). Since the oxygen atoms are not needed for the central atom to establish a full octet, it is possible for some of the pairs in the central octet to remain unattached. This results in less oxidized anions such as sulfites and nitrites. Finally, a more ambiguous example is sulfur hexafluoride: ![enter image description here][3] The figure shows two options. Should $\ce{SF6}$ be modeled as if the sulfur is a metal that gives up all of its electrons to the hyper-aggressive fluorine atoms (option a), or as a case where the octet rule gives way to a weaker but still workable 12-electron rule (option b)? There is [some controversy even today](http://en.wikipedia.org/wiki/Hypervalent_molecule#History_and_controversy) about how such cases should be handled. The donation notation shows how an octet perspective can still be applied to such cases, though it is never a good idea to rely on first-order approximation models for such extreme cases. [1]: https://i.stack.imgur.com/tjb6K.gif [2]: https://i.stack.imgur.com/IIDSF.gif [3]: https://i.stack.imgur.com/aU9gk.gif
Just some while ago i was burning a box of cardboard in my backyard and I saw a very curious phenomenon . I saw that as the cardboard box started burning it started to curl. up after a while as i came back to check if the fire was out i saw that though the cardboard box wasn't a box any longer but individual sides/faces of the box had retained their ridges and texture though they had tuned greyish white and crumbly My question: * Why did the cardboard start to curl up as it began to burn * Why did it retain its texture if not shape and size?
Most molecules containing nitrogen atoms in trigonal pyramid configuration undergo a relatively fast process of [inversion](http://en.wikipedia.org/wiki/Nitrogen_inversion) at room temperature. On the other hand, the free energy barrier for [phosphines](http://en.wikipedia.org/wiki/Phosphine), [sulfoniums](http://en.wikipedia.org/wiki/Sulfonium) and [sulfoxides](http://en.wikipedia.org/wiki/Sulfoxide) are high enough that they are optically stable: the rate of racemization is slow at room temperature. I wonder what effect is responsible for this difference in behaviour. I expect that it has to do with the size of the central atom (N being one row higher than P and S), but is that all there is to it? And how does size impact the free energy barrier for the inversion: energetically or entropically (or both, of course)?
How come trigonal S and P compounds do not undergo inversion at room temperature?
Just some while ago I was burning a box of cardboard in my backyard and I saw a very curious phenomenon. I saw that as the cardboard box started burning it started to curl. Up after a while as I came back to check if the fire was out I saw that though the cardboard box wasn't a box any longer but individual sides/faces of the box had retained their ridges and texture though they had tuned greyish white and crumbly. My questions: * Why did the cardboard start to curl up as it began to burn? * Why did it retain its texture if not shape and size?
While reading [this paper][1] (suggested by @Janice [here][2]), I came across the following line: > Surface reconstruction is known to be extremely important for Si, the material of 20th century electronics. However, surface reconstruction has been much less studies for graphite. One of the reasons might be graphite's limited use in electronics because **fabrication of ordered graphite substrates is difficult.** (emphasis mine) I've always believed that graphite isn't much used in semiconductor tech since it isn't a metalloid and conducts in an entirely different manner, probably making doping not work. This reason is new to me, but quite interesting. I can sort of guess why "fabrication of ordered graphite substrates is difficult" — I'm not too sure though. At any rate, I'd like to know more about it. Exactly _why_ is such fabrication difficult? [1]: http://www.lehigh.edu/~svr204/newdata/mypapers/191-200.pdf [2]: http://chemistry.stackexchange.com/a/449/22
Given two crystal space groups, how can one determine if they are in a group-subgroup relationship? The specific case at hand is **P 21/m 2/m 2/a** (aka **P m m a**, [#51][1]) and **P m m 2** ([#25](http://img.chem.ucl.ac.uk/sgp/large/025az1.htm)), but knowing how to solve this question in a general manner will sure prove helpful in the future as well… [1]: http://img.chem.ucl.ac.uk/sgp/large/051az1.htm
Are these two space groups in a group–subgroup relationship?
According to some chemistry textbooks, the valence shell can only hold eight electrons (except for hydrogen). This is called the "octet rule", and is invoked often. But, many of the shells have a capacity of more than eight. (2,8,18,32.... are the capacities of successive shells). They are only restricted to eight when they are the _valence_ shell. Why is this so? Are there any examples where there are more than eight electrons in the valence shell?
Most molecules containing nitrogen atoms in trigonal pyramid configuration undergo a relatively fast process of [inversion](http://en.wikipedia.org/wiki/Nitrogen_inversion) at room temperature. On the other hand, the free energy barrier for [phosphines](http://en.wikipedia.org/wiki/Phosphine), [sulfoniums](http://en.wikipedia.org/wiki/Sulfonium) and [sulfoxides](http://en.wikipedia.org/wiki/Sulfoxide) are high enough that they are optically stable: the rate of racemization is slow at room temperature. I wonder what effect is responsible for this difference in behaviour (the larger energy barrier). I expect that it has to do with the size of the central atom (N being one row higher than P and S), but is that all there is to it? And how does size impact the free energy barrier for the inversion: energetically or entropically (or both, of course)?
Can an atom have more than 8 valence electrons? According to some chemistry textbooks, the maximum number is 8, but they do not explain why. If it is not possible, why can't an atom have more than 8 valence electrons?
I just bought a heat bag from store and it has special property. It is made up of transparent plastic bag and transparent unknown liquid inside when sold. There is a tiny circular metal piece in there. If you bend the metal piece, crystalization will occur and release quite a lot of heat. The crystalization will spread out and change all the liquid into crystals. When the hardened plastic bag cool down, you need to put it into boiling water and the crystals will dissolve and become liquid again. What is more interesting, I tried to heat it up but not letting all the crystal dissolve, take it out of the boiling water and assume it is ready for the next use. However, the half-dissolved liquid/crystal turns into all crystals again ( Without me bending the metal piece ). I am guessing the transformation was not complete and the process was reverted. Did Japanese invent this material?
What kind of material made up of such heat bag?
According to some chemistry textbooks, the maximum number of valence electrons for an atom is 8, but the reason for this is not explained. So, can an atom have more than 8 valence electrons? If this is not possible, why can't an atom have more than 8 valence electrons?
Can an atom have more than 8 valence electrons? If not, why is 8 the limit?
What material are heat bags made of?
I just bought a heat bag from store and it has special property. It is made up of transparent plastic bag and transparent unknown liquid inside when sold. There is a tiny circular metal piece in there. If you bend the metal piece, crystalization will occur and release quite a lot of heat. The crystalization will spread out and change all the liquid into crystals. When the hardened plastic bag cool down, you need to put it into boiling water and the crystals will dissolve and become liquid again. What is more interesting, I tried to heat it up but not letting all the crystal dissolve, take it out of the boiling water and assume it is ready for the next use. However, the half-dissolved liquid/crystal turns into all crystals again (without me bending the metal piece). I am guessing the transformation was not complete and the process was reverted. Did the Japanese invent this material?
I am currently searching for a portable dehumidifier but notice that most of them come with a indicator that actually using the Blue Silica Gel, which I believe that it may contain Cobalt Chloride. So, I would like to ask, <br> (1) if that is the case, would the **portable dehumidifier become a harmful product**?<br> (2) would a **Orange Silica Gel contain other harmful chemical** that is currently not found? Would like to hear expert advice from the community.
Blue Silica Gel more harmful than the orange ones?
I am currently searching for a portable dehumidifier but notice that most of them come with a indicator that actually using the Blue Silica Gel, which I believe that it may contain Cobalt Chloride. So, I would like to ask, <br> (1) if that is the case, would the **portable dehumidifier become a harmful product**?<br> (2) would a **Orange Silica Gel contain other harmful chemical** that is currently not found?
Magnetite, $\ce{Fe3O4}$ is able to capture $\ce{As(III)}$ and $\ce{As(V)}$ from drinking water through the following mechanisms. ![enter image description here][1] Bidentate-binuclear complex ![enter image description here][2] Monodentate-binuclear complex (both from [here](http://arsenic.tamu.edu/about/course/mod2/notes/pg7.htm)) I'm assuming these are standard reactions in inorganic chemistry, but are there known mechanisms (I don't have a text available)? (the first looks like an attack on the $\ce{As}$ by the $\ce{O}$, but the second seems to take place in two steps) Also, I've heard these referred to as "inner sphere" complexes. To what does that designation refer? [1]: https://i.stack.imgur.com/5HJGq.gif [2]: https://i.stack.imgur.com/PyORS.jpg
What are the mechanisms for capture of $\ce{As(III)}$ and $\ce{As(V)}$ by magnetite from water?
Magnetite, $\ce{Fe3O4}$ is able to capture $\ce{As(III)}$ and $\ce{As(V)}$ from drinking water through the following mechanisms. ![enter image description here][1] Bidentate-binuclear complex ![enter image description here][2] Monodentate-binuclear complex (both from [here](http://arsenic.tamu.edu/about/course/mod2/notes/pg7.htm)) I'm assuming these are standard reactions in inorganic chemistry, but are there known mechanisms (I don't have a text available)? The first looks like an attack on the $\ce{As}$ by the $\ce{O}$, but the second seems to take place in two steps. Also, I've heard these referred to as "inner sphere" complexes. To what does that designation refer? [1]: https://i.stack.imgur.com/5HJGq.gif [2]: https://i.stack.imgur.com/PyORS.jpg
Magnetite, $\ce{Fe3O4}$ is able to capture $\ce{As(III)}$ and $\ce{As(V)}$ from drinking water through the following mechanisms. ![enter image description here][1] Bidentate-binuclear complex ![enter image description here][2] Monodentate-binuclear complex (both from [here](http://arsenic.tamu.edu/about/course/mod2/notes/pg7.htm)) I'm assuming these are standard reactions in inorganic chemistry, but are there known mechanisms (I don't have a text available)? The first looks like an attack on the $\ce{As}$ by the $\ce{O}$, but the second seems to take place in two steps. [1]: https://i.stack.imgur.com/5HJGq.gif [2]: https://i.stack.imgur.com/PyORS.jpg
I've heard some people mention that this or that program isn't "up to date" with respect to the borders used in the Ramachandran plots to classify &phi;/&psi; as being in the most favorable/acceptable/disallowed regions. How *are* the accepted regions defined? What is the current accepted paper that covers this?
What are the border definitions in the Ramachandran plot?
In Cell Culture, DMSO is added to prevent the formation of ice crystals which may lyse the cells. Exactly how does DMSO act as a cyroprotectant?
How does DMSO serve as a cyroprotectant?
When mixing two fluids of different composition at constant pressure when no heat can escape, is the enthalpy additive? If not, how to compute the enthalpy of the mixture resulting from mixing two fluids with given compositin after equilibrium is reestablished?
Is enthalpy additive in mixing?
When mixing two fluids of different composition at constant pressure when no heat can escape, is the enthalpy additive? If not, how to compute the enthalpy of the mixture resulting from mixing two fluids with given composition after equilibrium is re-established?
As Ali said, unless the mixture is [ideal](http://en.wikipedia.org/wiki/Ideal_solution) the enthalpy of mixing (or excess enthalpy) is not zero. This is the basis for a whole field of thermodynamics (and in particular, statistical thermodynamics): **solution theories**. To quote only a few of the most famous ones: - VST (Vacancy Solution Theory) <sub>*[I don't have a good reference at hand right now, I'll try to come back and edit it in if noöne beats me to it]*</sub> - [NRTL][1] (Non Random Two Liquid) - SAFT (Statistical Associating Fluid Theory) and [its variants](http://www.sklogwiki.org/SklogWiki/index.php/SAFT_(statistical_associating_fluid_theory)): [original paper](http://pubs.acs.org/doi/abs/10.1021/ie00104a021), [a sort of review](http://pubs.acs.org/doi/abs/10.1021/ie0102201). In this literature, you'll frequently encounter the name of [John M. Prausnitz](http://www.cchem.berkeley.edu/jmpgrp/), whose book are considered classics (though I have not read them myself), including: - [Molecular Thermodynamics of Fluid-Phase Equilibria](http://books.google.com/books/about/Molecular_Thermodynamics_of_Fluid_Phase.html?id=VSwc1XUmYpcC) - [The Properties of Gases and Liquids](http://books.google.com/books/about/The_Properties_of_Gases_and_Liquids.html?id=9tGclC3ZRX0C) The above are for mixture of molecular fluids… polymer solutions are a bit of a special case of this, and specific models have been developed for them, the best-known being [Flory–Huggins][2]. --- Also, you did not specify what kind of *computation* of the mixing enthalpy you were considering, so I have to add that several molecular simulation techniques have been developped to addres this question. For a basic introduction on this topic, I would recommend Singh and Gubbins’ [review in *Molecular-Based Study of Fluids*](http://pubs.acs.org/doi/abs/10.1021/ba-1983-0204.ch004). [1]: http://en.wikipedia.org/wiki/Non-random_two-liquid_model [2]: http://en.wikipedia.org/wiki/Flory%E2%80%93Huggins_solution_theory
Portable dehumidifier require consumer to plug into a outlet to "re-generate" its crystal so that the indicator will change from pink color to its original blue color. However, I notice that there is usually a color-less gas (or smell - as it smell differently from the surrounding air). So, is the gas (or smell) from the portable dehumidifier (when plugged into a outlet for re-generation of its crystal) harmful?
Is the gas (or smell) that come out from a portable dehumidifier safe?
Can you explain what the crystal is? Based on the common colors you describe and the "Regeneration" I would think that the "crystal" you are describing is just a basic Desiccant (like a silica gel, its hygroscopic, it absorbs water). Most desiccants are pretty(never heard of any issues with common ones used in consumer goods) harmless. You really need to add more information to your description. What is the make (incase someone wants to look it up). The biggest danger I cant think of with a portable dehumidifier is bacterial/fungal growth. Both of these can give off a smell, it could be that when "regenerating the crystal" (which if it a desiccant like I think would mean heating it, is actually "cooking" the growth creating a smell. (But this is me trying to image what could potentially cause that smell, not what is happening)... Again, not enough information to reach any conclusion. (But I do think you are describing a desiccant).
A portable dehumidifier requires the consumer to plug it into an outlet to regenerate its crystal. When the crystal is thus "regenerated", the indicator will change from pink color to its original blue color. However, I noticed that there is usually a colorless gas (or smell - it smells differently from the surrounding air) accompanying this change. So, is the gas (or smell) from the portable dehumidifier (when plugged into a outlet for re-generation of its crystal) harmful?
In cell culture, dimethylsulfoxide (DMSO) is added to prevent the formation of ice crystals which may lyse the cells. Exactly how does dimethylsulfoxide act as a cryoprotectant?
How does dimethylsulfoxide serve as a cryoprotectant?
What software does the PDB used to generate these "cel-shaded" protein models with outlines around foreground atoms? ![Leptin](http://www.rcsb.org/pdb/education_discussion/molecule_of_the_month/images/149-Leptin_1ax8.jpg)
What software is used to generate the PDB molecule of the month images?
[Molprobity](http://kinemage.biochem.duke.edu) and some other protein structure validation tools report a *C&beta; deviation* statistic and offer plots for it (example below). Apparently if the C&beta; is greater than 0.25 &Aring;, some attention should be given to the residue. The statistic is given in units of angstroms, but I'm not sure what it's actually referring to. With the Rama plot, it's the peptide dihedral angles, but what is this? And why are the graphs 2-D despite the one dimension given? ![example C-beta graph][1] [1]: https://i.stack.imgur.com/cZBVI.png
What is Cβ (C-beta) deviation?
Can you explain what the crystal is? Based on the common colors you describe and the "Regeneration" I would think that the "crystal" you are describing is just a basic Desiccant (like a silica gel, it's hygroscopic, it absorbs water). Most desiccants are pretty harmless (never heard of any issues with common ones used in consumer goods). You really need to add more information to your description. What is the make (in case someone wants to look it up)? The biggest danger I can think of with a portable dehumidifier is bacterial/fungal growth. Both of these can give off a smell, it could be that when "regenerating the crystal" (which if it's a desiccant like I think would mean heating it, is actually "cooking" the growth creating a smell. (But this is me trying to image what could potentially cause that smell, not what is happening)... Again, not enough information to reach any conclusion. (But I do think you are describing a desiccant).
Is the smell coming out from a portable dehumidifier safe?
These illustrations are by [David Goodsell][1], and carry his famous look. Whilst I was not able to determine what specific software he uses, [this page][2] suggests that he probably creates his illustrations as a composite of renders and that the outlines and depth cueing are achieved by running a 'find edges' filter over the z-buffer of the scene, which can be output from many renderers. [VMD][3] is a fairly powerful program (that F'x has contributed patches to!) capable of [achieving this cel-shaded look][4] with minimal postprocessing by using a GLSL Fresnel shader. As F'x mentioned, QuteMol can do this style as well, however QuteMol has not been actively developed since about 2007 and is not terribly feature-rich, which is a shame as it's one of very few molecule viewers with competent ambient occlusion. How to do it with Blender and VMD --- In the first instance, I tried using the GLSL shader Goodsell style in VMD on the [PDB file 1RON][5], a neuropeptide. I was not satisfied with the results, but this would probably look decent on a protein SES as illustrated on the aforementioned VMD site. ![enter image description here][6] Note that I have used 2 VMD representations to separate out the style of the backbone and the sidechains, minus the hydrogens. I'm not going for a 1-to-1 clone of the referred style, just something thematically similar. I have shrunk this VMD output down because as F'x pointed out in chat, GLSL shaded spheres are not antialiased. This said, if you are very tricky, you might be able to force antialiasing using a shader method like [FXAA][7], which can be enabled in arbitrary programs in the very latest nVidia beta drivers (I know...). Alternatively, you can do what I did and manually oversample :D. Next step - Blender --- I output the VdW spheres as two separate VRML files, one for the backbone, one for the sidechains. These can be imported into [Blender][8]. I'm going to assume proficiency with Blender, however if you are not familiar, Blender is fairly easy to learn, especially versions 2.5 and later. ![enter image description here][9] I've combined the VdW sphere meshes of each component into a backbone and sidechain object, assigned materials for each (just changing the colours of the default diffuse shader) and have applied one level of Catmull-Clarke subdivision to the meshes, to make the spheres more spherical than the default. I've also removed the light source and turned on environmental lighting, which lights the scene isotropically and will generate some nice ambient occlusion while we're at it. I now render this at high resolution, obtaining a standard render and the Z-buffer for the scene. I've scaled them down here. **Note that if you want the spheres to be a flat colour, you should set the material to 'shadeless'**. I like the subtly shaded look better. ![enter image description here][10] Oh yeah, that looks nice. And here's the Z-buffer. Regrettably, the Z-buffer in blender has no antialiasing which is why I've rendered things huge and scaled down in Photoshop. ![enter image description here][11] I now run the Glowing Edges filter on the Z-buffer and threshold to remove faint spurious edges inside the spheres. ![enter image description here][12] Invert and level, set blendmode to multiply and you're done. ![enter image description here][13] Not perfect, but not bad. [1]: http://mgl.scripps.edu/people/goodsell/ [2]: http://www.bmc.med.utoronto.ca/molecule/ [3]: http://www.ks.uiuc.edu/Research/vmd/ [4]: http://www.ks.uiuc.edu/Research/vmd/minitutorials/glslgoodsell/ [5]: http://www.rcsb.org/pdb/explore/explore.do?structureId=1ron [6]: https://i.stack.imgur.com/xlAKk.png [7]: http://developer.download.nvidia.com/assets/gamedev/files/sdk/11/FXAA_WhitePaper.pdf [8]: http://www.blender.org [9]: https://i.stack.imgur.com/0N3Sy.png [10]: https://i.stack.imgur.com/50zHb.png [11]: https://i.stack.imgur.com/c5bTa.png [12]: https://i.stack.imgur.com/LoDSG.png [13]: https://i.stack.imgur.com/xUzcn.png
These illustrations are by [David Goodsell][1], and carry his famous look. Whilst I was not able to determine what specific software he uses, [this page][2] suggests that he probably creates his illustrations as a composite of renders and that the outlines and depth cueing are achieved by running a 'find edges' filter over the z-buffer of the scene, which can be output from many renderers. [VMD][3] is a fairly powerful program (that F'x has contributed patches to!) capable of [achieving this cel-shaded look][4] with minimal postprocessing by using a GLSL Fresnel shader. As F'x mentioned, QuteMol can do this style as well, however QuteMol has not been actively developed since about 2007 and is not terribly feature-rich, which is a shame as it's one of very few molecule viewers with competent ambient occlusion. How to do it with Blender and VMD --- In the first instance, I tried using the GLSL shader Goodsell style in VMD on the [PDB file 1RON][5], a neuropeptide. I was not satisfied with the results, but this would probably look decent on a protein SES as illustrated on the aforementioned VMD site. ![enter image description here][6] Note that I have used 2 VMD representations to separate out the style of the backbone and the sidechains, minus the hydrogens. I'm not going for a 1-to-1 clone of the referred style, just something thematically similar. I have shrunk this VMD output down because as F'x pointed out in chat, GLSL shaded spheres are not antialiased. This said, if you are very tricky, you might be able to force antialiasing using a shader method like [FXAA][7], which can be enabled in arbitrary programs in the very latest nVidia beta drivers (I know...). Alternatively, you can do what I did and manually oversample :D. Next step - Blender --- I output the VdW spheres as two separate VRML files, one for the backbone, one for the sidechains. These can be imported into [Blender][8]. I'm going to assume proficiency with Blender, however if you are not familiar, Blender is fairly easy to learn, especially versions 2.5 and later. ![enter image description here][9] I've combined the VdW sphere meshes of each component into a backbone and sidechain object, assigned materials for each (just changing the colours of the default diffuse shader) and have applied one level of Catmull-Clarke subdivision to the meshes, to make the spheres more spherical than the default. I've also removed the light source and turned on environmental lighting, which lights the scene isotropically and will generate some nice ambient occlusion while we're at it. I now render this at high resolution, obtaining a standard render and the Z-buffer for the scene. I've scaled them down here. **Note that if you want the spheres to be a flat colour, you should set the material to 'shadeless'**. I like the subtly shaded look better. ![enter image description here][10] Oh yeah, that looks nice. And here's the Z-buffer. Regrettably, the Z-buffer in blender has no antialiasing which is why I've rendered things huge and scaled down in Photoshop. ![enter image description here][11] I now run the Glowing Edges filter on the Z-buffer and threshold to remove faint spurious edges inside the spheres. ![enter image description here][12] Invert and level, set blendmode to multiply and you're done. ![enter image description here][13] Not perfect, but not bad. Bonus --- Here's the shadeless version, which is closer to Goodsell's ideal. ![enter image description here][14] [1]: http://mgl.scripps.edu/people/goodsell/ [2]: http://www.bmc.med.utoronto.ca/molecule/ [3]: http://www.ks.uiuc.edu/Research/vmd/ [4]: http://www.ks.uiuc.edu/Research/vmd/minitutorials/glslgoodsell/ [5]: http://www.rcsb.org/pdb/explore/explore.do?structureId=1ron [6]: https://i.stack.imgur.com/xlAKk.png [7]: http://developer.download.nvidia.com/assets/gamedev/files/sdk/11/FXAA_WhitePaper.pdf [8]: http://www.blender.org [9]: https://i.stack.imgur.com/0N3Sy.png [10]: https://i.stack.imgur.com/50zHb.png [11]: https://i.stack.imgur.com/c5bTa.png [12]: https://i.stack.imgur.com/LoDSG.png [13]: https://i.stack.imgur.com/xUzcn.png [14]: https://i.stack.imgur.com/xrTrc.png
These illustrations are by [David Goodsell][1], and carry his famous look. Whilst I was not able to determine what specific software he uses, [this page][2] suggests that he probably creates his illustrations as a composite of renders and that the outlines and depth cueing are achieved by running a 'find edges' filter over the z-buffer of the scene, which can be output from many renderers. [VMD][3] is a fairly powerful program (that F'x has contributed patches to!) capable of [achieving this cel-shaded look][4] with minimal postprocessing by using a GLSL Fresnel shader. As F'x mentioned, QuteMol can do this style as well, however QuteMol has not been actively developed since about 2007 and is not terribly feature-rich, which is a shame as it's one of very few molecule viewers with competent ambient occlusion. How to do it with VMD and Blender --- In the first instance, I tried using the GLSL shader Goodsell style in VMD on the [PDB file 1RON][5], a neuropeptide. I was not satisfied with the results, but this would probably look decent on a protein SES as illustrated on the aforementioned VMD site. ![enter image description here][6] Note that I have used 2 VMD representations to separate out the style of the backbone and the sidechains, minus the hydrogens. I'm not going for a 1-to-1 clone of the referred style, just something thematically similar. I have shrunk this VMD output down because as F'x pointed out in chat, GLSL shaded spheres are not antialiased. This said, if you are very tricky, you might be able to force antialiasing using a shader method like [FXAA][7], which can be enabled in arbitrary programs in the very latest nVidia beta drivers (I know...). Alternatively, you can do what I did and manually oversample :D. Next step - Blender --- I output the VdW spheres as two separate VRML files, one for the backbone, one for the sidechains. These can be imported into [Blender][8]. I'm going to assume proficiency with Blender, however if you are not familiar, Blender is fairly easy to learn, especially versions 2.5 and later. ![enter image description here][9] I've combined the VdW sphere meshes of each component into a backbone and sidechain object, assigned materials for each (just changing the colours of the default diffuse shader) and have applied one level of Catmull-Clarke subdivision to the meshes, to make the spheres more spherical than the default. I've also removed the light source and turned on environmental lighting, which lights the scene isotropically and will generate some nice ambient occlusion while we're at it. I now render this at high resolution, obtaining a standard render and the Z-buffer for the scene. I've scaled them down here. **Note that if you want the spheres to be a flat colour, you should set the material to 'shadeless'**. I like the subtly shaded look better. ![enter image description here][10] Oh yeah, that looks nice. And here's the Z-buffer. Regrettably, the Z-buffer in blender has no antialiasing which is why I've rendered things huge and scaled down in Photoshop. ![enter image description here][11] I now run the Glowing Edges filter on the Z-buffer and threshold to remove faint spurious edges inside the spheres. ![enter image description here][12] Invert and level, set blendmode to multiply and you're done. ![enter image description here][13] Not perfect, but not bad. Bonus --- Here's the shadeless version, which is closer to Goodsell's ideal. ![enter image description here][14] The Journey Continues! --- Poking around with the Toon shaders in Blender, I have found a super easy way to make Goodsell-esque figures! If you configure your material like so: ![enter image description here][15] You have a shader that will rapidly shade to black at angles far from normal with a lightsource. To make this work, you should disable environment lighting and position a bright point light behind the camera. You will note that I have adjusted the Toon size to 0.98, which will create a shadow just shy of perpendicular to the light source normal. You will also note that I have set the specularity to zero, which gives an approximately flat finish to the sphere. Render it up and you will get something like this: ![enter image description here][16] [1]: http://mgl.scripps.edu/people/goodsell/ [2]: http://www.bmc.med.utoronto.ca/molecule/ [3]: http://www.ks.uiuc.edu/Research/vmd/ [4]: http://www.ks.uiuc.edu/Research/vmd/minitutorials/glslgoodsell/ [5]: http://www.rcsb.org/pdb/explore/explore.do?structureId=1ron [6]: https://i.stack.imgur.com/xlAKk.png [7]: http://developer.download.nvidia.com/assets/gamedev/files/sdk/11/FXAA_WhitePaper.pdf [8]: http://www.blender.org [9]: https://i.stack.imgur.com/0N3Sy.png [10]: https://i.stack.imgur.com/50zHb.png [11]: https://i.stack.imgur.com/c5bTa.png [12]: https://i.stack.imgur.com/LoDSG.png [13]: https://i.stack.imgur.com/xUzcn.png [14]: https://i.stack.imgur.com/xrTrc.png [15]: https://i.stack.imgur.com/heCym.png [16]: https://i.stack.imgur.com/BmKnL.png
I know that $H^+$ is not possible in water and it is present as $H_3O^+$ . But later on i come to know that even $H_3O^+$ is not possible and that it is present as $H_9O_4^+$ Why does this happen? what give that compound so much stability that is not present in $H_3O^+$ or $H+$ ?
I know that H<sup>+</sup> is not possible in water and it is present as H<sub>3</sub>O<sup>+</sup>. But later on I come to know that even H<sub>3</sub>O<sup>+</sup> is not possible and that it is present as H<sub>9</sub>O<sub>4</sub><sup>+</sup>. Why does this happen? What give that compound so much stability that is not present in H<sub>3</sub>O<sup>+</sup> or H<sup>+</sup>?
I think that for this type of question a bit of context as to why you are asking the question, would be useful. If for instance you are in an Intro Chem class I would say the following: **All chemical systems tend towards their most stable state (in the absence of external energy being applied).** $H^+$ is just a single proton, a single positive charge without any negative charge to balance it out. Naked charges are very powerful forces, while it is possible for $H^+$ to exist in isolation (even to have a number of them together in a system), the presence of other matter which might be used to balance out the naked positive charge will quickly lead to a reaction. By forming $H_3O^+$ the single proton $H^+$ is able to spread its positive charge out over not just itself but the entirety of the bonded $H_2O$ molecule. (Note that because Oxygen is electronegative it can more easily stabilize that extra positive charge). Whats better than $H_3O^+$? Getting even more water molecules to share the burden of the extra positive charge and producing $H_9O_4^+$. Thus the short answer I would give to your question would be: There are two main factors: 1. Simply the distribution of that extra positive charge over a larger area/shared by more matter stabilizes the system. 2. The Oxygen in particular (being fairly electronegative) can "handle" the extra positive charge better than the lone proton ($H^+$) Hope that helps.
I think that for this type of question a bit of context as to why you are asking the question, would be useful. If for instance you are in an Intro Chem class I would say the following: **All chemical systems tend towards their most stable state (in the absence of external energy being applied).** $H^+$ is just a single proton, a single positive charge without any negative charge to balance it out. Naked charges are very powerful forces, while it is possible for $H^+$ to exist in isolation (even to have a number of them together in a system), the presence of other matter which might be used to balance out the naked positive charge will quickly lead to a reaction. By forming $H_3O^+$ the single proton $H^+$ is able to spread its positive charge out over not just itself but the entirety of the bonded $H_2O$ molecule. (Note that because Oxygen is electronegative it can more easily stabilize that extra positive charge). Whats better than $H_3O^+$? Getting even more water molecules to share the burden of the extra positive charge and producing $H_9O_4^+$. Thus the short answer I would give to your question would be: There are two main factors: 1. Simply the distribution of that extra positive charge over a larger area/shared by more matter stabilizes the system. 2. The Oxygen in particular (being fairly electronegative) can "handle" the extra positive charge better than the lone proton ($H^+$) Hope that helps. Because in my opinion, beyond this explanation the answer gets very complicated and as many have pointed out, broad.
What software does the PDB use to generate these "cel-shaded" protein models with outlines around foreground atoms? ![Leptin](http://www.rcsb.org/pdb/education_discussion/molecule_of_the_month/images/149-Leptin_1ax8.jpg)
I think that for this type of question a bit of context as to why you are asking the question, would be useful. If for instance you are in an Intro Chem class I would say the following: **All chemical systems tend towards their most stable state (in the absence of external energy being applied).** $H^+$ is just a single proton, a single positive charge without any negative charge to balance it out. Naked charges are very powerful forces, while it is possible for $H^+$ to exist in isolation (even to have a number of them together in a system), the presence of other matter which might be used to balance out the naked positive charge will quickly lead to a reaction. By forming $H_3O^+$ the single proton $H^+$ is able to spread its positive charge out over not just itself but the entirety of the bonded $H_2O$ molecule. (Note that because Oxygen can be seen in resonance to distribute the negative charge associated with its loan pairs it is particularly good at stabilizing the extra positive charge). Whats better than $H_3O^+$? Getting even more water molecules to share the burden of the extra positive charge and producing $H_9O_4^+$. Thus the short answer I would give to your question would be: There are two main factors: 1. Simply the distribution of that extra positive charge over a larger area/shared by more matter stabilizes the system. 2. The Oxygen in particular (being able to share the negative charge of its lone pair) can "handle" the extra positive charge better than the lone proton ($H^+$) Hope that helps. Because in my opinion, beyond this explanation the answer gets very complicated and as many have pointed out, broad.
[Molprobity](http://kinemage.biochem.duke.edu) and some other protein structure validation tools report a *C&beta; deviation* statistic and offer plots for it (example below). Apparently if the C&beta; is greater than 0.25 &Aring;, some attention should be given to the residue. The statistic is given in units of angstroms, but I'm not sure what it's actually referring to. With the Rama plot, it's the peptide dihedral angles, but what is this? And why are the graphs 2-D despite the one dimension given? [![example C-beta graph][2]][2] [2]: https://i.stack.imgur.com/hxzOd.png
Assume a cubic diamond crystal with hydrogen-terminated surface bonds. The terminating hydrogen atoms will form pairs that are geometrically close together in alternating diagonal pairs (think of the rectangles on corrugated walking steel). Next, picture a formaldehyde molecule being pushed oxygen-first into one of these terminating hydrogen pairs. If the oxygen atom can be made to react with the hydrogen pair, a water molecule would be expelled and the remaining $\ce{CH2}$ group would presumably bond to the two bonds of the diamond lattice. If you repeat this process over the entire cubic face, the result will be a new layer of diamond that is terminated with the same paired hydrogen atoms as the original layer. The entire layer-growth process thus in principle could be repeated indefinitely, resulting in a diamond grown at or near room temperature from an aqueous or other low-temperature solution. Question: Can anyone think of a fundamental reason why the above process, or some variant of it, could _not_ be used to grow diamond crystals at room temperature? (A "variant" might start with carbohydrate units larger than $\ce{OCH2}$, for example.) One or more geometrically sophisticated catalyst molecules would be required. Each catalyst molecule would snag, carry, and orient a formaldehyde molecules onto some part of the textured surfaces provided by the cubic diamond face, such as the face itself or along a growth edge. I did this a long time ago in high school, but my recollection is that the reaction should be mildly exothermic. But even if it is not, energy-source molecules (think of the role of ATP in living systems) could provide energy to enable an endothermic reaction. So... thoughts, anyone? Is there any deep reason why this approach to room-temperature or near-room-temperature diamond synthesis is clearly _not_ possible?
Is it possible that diamond can be grown from aqueous solutions?
Are there endothermic dissolution reactions?
Why is calcium fluoride insoluble? NaCl is soluble, but CaF2 is not. Why is this?
Why is calcium fluoride insoluble?
Why is calcium fluoride insoluble? $\ce{NaCl}$ is soluble, but $\ce{CaF2}$ is not. Why is this?
It's possible to select a perfluorinated hydrocarbon that is immiscible in another organic solvent, forming a distinct phase boundary. Such systems may become miscible at elevated temperature which is exploited in the design of some catalytic processes<sup>1</sup>. Why is it that many fluorous solvents are immiscible with organic solvents? Are their polarities sufficiently different or is there another reason? ----- **(1)** Housecroft, C.E.; Sharpe, A.G.; *Inorganic Chemistry 2e*; Pearson Prentice Hall; **2005**, pp. 798-799
Why do biphasic systems of fluorous and organic solvents form?
It seems from a brief search of the literature that 1-ethyl-3-methylimidazolium tetrafluoroborate (EMI-BF4) is a prototypical room temperature ionic liquid (RTIL) that has been studied extensively. However, numerous -- almost limitless -- combinations of cations and anions that give rise to ionic liquids are possible. Why does it seem that the imidazolium cations are so extensively studied? Is one possible reason the fact that $\text{EMI}^+$ is aromatic and thus affords the ionic liquid a high degree of stability? Similarly, why does it seem that $\text{BF}_4^-$ (along with, perhaps, $\text{PF}_6^-$) is **the** prototypical ionic liquid anion? I have seen studies of other anions (for example, $\text{CF}_3 \text{SO}_3^-$ and $(\text{CF}_3 \text{SO}_2)_2 \text{N}^-$), but $\text{BF}_4^-$ seems to be one of the most popular.
Why is 1-ethyl-3-methylimidazolium tetrafluoroborate (EMI-BF4) often considered a prototypical room temperature ionic liquid?
What is the specific heat capacity of *liquid* iron? The online handbook I've been using only lists specific heat for the solid and gaseous states. Is this a known property and I'm just looking in the wrong place, or is there a reason specific heat isn't listed anywhere for the liquid state? I only found one reference to its heat capacity by googling (611 J/kgK found [here][1]), but no sources are listed, so I'm skeptical. [1]: http://wiki.answers.com/Q/What_is_specific_heat_capacity_of_iron
Specific heat of liquid iron?
Does *liquid* iron have a constant specific heat capacity? Everywhere I've looked (save one sourceless online reference to Cp = 611 J/kgK [here][1]), I find no value listed for liquid iron. Is this because the heat capacity is temperature-dependent? Or is this related to a measurement problem? Or does an accepted constant value exist, but I'm just a poor researcher? [1]: http://wiki.answers.com/Q/What_is_specific_heat_capacity_of_iron
I am not a chemicist, I hope I will be enough specific. Suppose there are two chemical species A, B, with properties: - at temperature $t<T_r$, no reaction occurs between A, B (in any combination), - at $t\ge T_r$, A interacts with itself to create $A^2$, B reacts with itself to create $B^2$, and A, B are reacting to create $AB$, - $A^2, B^2, \text{and}\space AB$ are never reacting. In experiment, we first mix A and B in temperature $t<T_r$. Amounts of species mixed are $a$ for A, $b$ for B. Then, we add heat to obtain temperature $t\ge T_r$ and start the reaction. 1. What amounts of $A^2, B^2, \text{and}\space AB$ can be expected to be produced? 2. To obtain the amounts, should probability theory be used? E.g. amount of $AB$ equals to probability that species A, B will interact?
I am not a chemicist, I hope I will be enough specific. Suppose there are two chemical species A, B, with properties: - at temperature $t<T_r$, no reaction occurs between A, B (in any combination), - at $t\ge T_r$, A interacts with itself to create $A^2$, B reacts with itself to create $B^2$, and A, B are reacting to create $AB$, - $A^2, B^2, \text{and}\space AB$ are never reacting. In experiment, we first mix A and B in temperature $t<T_r$. Amounts of species mixed are $a$ for A, $b$ for B. Then, we add heat to obtain temperature $t\ge T_r$ and start the reaction. 1. What amounts of $A^2, B^2, \text{and}\space AB$ can be expected to be produced? 2. To obtain the amounts, should probability theory be used? E.g. amount of $AB$ equals to probability that species A, B will interact ("collide" or similar interpretation).
I am not a chemicist, I hope I will be enough specific. Suppose there are two chemical species A, B, with properties: - at temperature $t<T_r$, no reaction occurs between A, B (in any combination), - at $t\ge T_r$, A interacts with itself to create A<sub>2</sub>, B reacts with itself to create B<sub>2</sub>, and A, B are reacting to create AB, - A<sub>2</sub>, B<sub>2</sub> and AB are never reacting. In experiment, we first mix A and B in temperature $t<T_r$. Amounts of species mixed are $a$ for A, $b$ for B. Then, we add heat to obtain temperature $t\ge T_r$ and start the reaction. 1. What amounts of A<sub>2</sub>, B<sub>2</sub>, and AB can be expected to be produced? 2. To obtain the amounts, should probability theory be used? E.g. amount of AB equals to probability that species A, B will interact ("collide" or similar interpretation).
Does *liquid* iron have a constant specific heat capacity? Everywhere I've looked (save one sourceless online reference to $C_p$ = 611 J.kg<sup>–1</sup>.K<sup>–1</sup> [here][1]), I find no value listed for liquid iron. Is this because the heat capacity is temperature-dependent? Or is this related to a measurement problem? Or does an accepted constant value exist, but I'm just a poor researcher? [1]: http://wiki.answers.com/Q/What_is_specific_heat_capacity_of_iron
Is the specific heat of liquid iron constant?
Heat capacities are usually dependent on temperature except for ideal gases. You will generally find the values tabulated at specific temperature and pressure values. For instances for which one is interested in heat capacity as a function of temperature, this is usually approximated through a regression equation. The NIST JANAF Themochemical Tables use the Shomate equation with 5 coefficients, of the form: C<sub>p</sub>° = A + B t + C t<sup>2</sup> + D t<sup>3</sup> + E / t<sup>2</sup> where t is T (in K) / 1000, and A, B, C, D and E are regression coefficients. If one looks at the [Condensed phase thermochemistry data page for iron](http://webbook.nist.gov/cgi/cbook.cgi?ID=C7439896&Units=SI&Mask=2#Thermo-Condensed), one can find the relevant coefficients (A, B, C, D and E in the above equation) and thus calculate heat capacity of liquid iron in the 1809 - 3133 K range.
Okay, let's answer bit by bit: - Measurement of heat capacity of liquids: this isn't particularly hard to do, and in fact it's much easier to obtain $C_p$ experimentally than $C_V$… however, once you know one, the other can be determined from it and some other thermodynamic properties. Different types of calorimeters are used for liquids, but the best-known is probably the [Callender and Barnes apparatus](http://www.schoolphysics.co.uk/age16-19/Thermal%20physics/Heat%20energy/text/Specific_heat_capacity_measurement/index.html). - *“Does an accepted value exist?”* and *“Is the specific heat of liquid iron temperature-dependent?”* — Well, I found a good paper which includes measurements and comparison with previously published values: [*Thermophysical Properties of Liquid Iron*](http://www.springerlink.com/content/q636918771156554/), M. Beutl, G. Pottlacher and H. Jäger, *Intl. J. Thermophys.*, **1994**, 15, 1323.             ![Fig. 1][1] which confirms that considering that **heat capacity is roughly constant in the 1800–4000 K range**. The authors thus extract the constant pressure specific heat capacity as **825 J/kg/K**, apparently in reasonable agreement with previous work that they cite: ![enter image description here][2] [ ](http://insertingsomeverticalspacing.com) - *“Or am I a poor researcher?”* — Well, the above paper is the first result return by Google for a search for *"specific heat" and "liquid iron"*, so I'll let you be the judge of that ;-) [1]: https://i.stack.imgur.com/z5Glp.png [2]: https://i.stack.imgur.com/BAtoA.png
I am not a chemicist, I hope I will be enough specific. Suppose there are two chemical species A, B, with properties: - at temperature $t<T_r$, no reaction occurs between A, B (in any combination), - at $t\ge T_r$, A interacts with itself to create A<sub>2</sub>, B reacts with itself to create B<sub>2</sub>, and A, B are reacting to create AB, - A<sub>2</sub>, B<sub>2</sub> and AB are never reacting. In experiment, we first mix A and B in temperature $t<T_r$. Amounts of species mixed are $a$ for A, $b$ for B. Then, we add heat to obtain temperature $t\ge T_r$ and start the reaction. 1. What amounts of A<sub>2</sub>, B<sub>2</sub>, and AB can be expected to be produced? 2. To obtain the amounts, should probability theory be used? E.g. amount of AB equals to probability that species A, B will interact ("collide" or similar interpretation). **Update:** rates of the reactions are equal.
Heat capacities are usually dependent on temperature except for ideal gases. You will generally find the values tabulated at specific temperature and pressure values. For instances for which one is interested in heat capacity as a function of temperature, this is usually approximated through a regression equation. The NIST JANAF Themochemical Tables use the Shomate equation with 5 coefficients, of the form: C<sub>p</sub>° = A + B t + C t<sup>2</sup> + D t<sup>3</sup> + E / t<sup>2</sup> where t is T (in K) / 1000, and A, B, C, D and E are regression coefficients. If one looks at the [Condensed phase thermochemistry data page for iron](http://webbook.nist.gov/cgi/cbook.cgi?ID=C7439896&Units=SI&Mask=2#Thermo-Condensed), one can find the relevant coefficients (A, B, C, D and E in the above equation) and thus calculate heat capacity of liquid iron in the 1809 - 3133 K range. Since all coefficients except A are close to zero, for liquid iron heat capacity is nearly constant in this range.
This is a very good question. These compounds have been studied for long enough that most people working with them "know" the answers to these questions, but the answers are buried in physical chemistry articles from 20 or more years ago. The imidazolium tetrafluoroborate salts are popular RTILs because 1) they have the "right" properties, and 2) they are easy to make. Concerning these compounds having the "right" properties: Most RTILs require two characteristics in the their constituent ions to minimize the intermolecular order which would cause them to solidify. First, they need a cation-anion pair that cannot associate strongly, and second they need to have lower symmetry. Immidazolium tetrafluoroborate (BF<sub>4</sub><sup>-</sup>) salts have both of these. The aromaticity of imidazole is important. The positive charge of the imidazolium cation is distributed through the resonance of the aromatic pi system. The negative charge in BF<sub>4</sub><sup>-</sup> is distributed inductively over the four fluorine atoms. Additionally, imidazolium is planar and BF<sub>4</sub><sup>-</sup> is spherical. Both the diffuse charge distribution and poor shape match prevent these two ions from having a strong association. The two different alkyl groups (ethyl and methyl) on the imidazolium ring are enough to break the symmetry of the system and prevent more efficient packing, which would lead to the substance being solid. Concerning their ease of synthesis: The imidazolium salts are easy to prepare from imidazole and various alkyl halides, all commercially available and inexpensive. The predominance of the BF<sub>4</sub><sup>-</sup> anion is also a result of the synthesis. The synthesis of an imidazolium salt RTIL begins with imidazole and and an alkyl halide, say chloromethane. The product of this step is 1-methylimidazolium chloride. This salt is reacted with another alkyl halide, say chloroethane, in the presence of a base, say triethylamine. The products of the this step are 1-ethyl-3-methylimidazolium chloride and triethylammonium chloride. 1-ethyl-3-methylimidazolium chloride is a solid, because the chloride ion is small, with concentrated charge. The tetrafluoroborate salt is prepared by salt metathesis. 1-ethyl-3-methylimidazolium chloride is dissolved in water and an aqueous solution of silver tetrafluoroborate (AgBF<sub>4</sub>) is added. Silver chloride, AgCl, is virtually insoluble in water (_K_<sub>sp</sub> = 1.77 x 10<sup>-10</sup>) and precipitates. The solution is filtered, and the filtrate is concentrated to produce 1-ethyl-3-methylimidazolium tetrafluoroborate (EMIM BF<sub>4</sub><sup>-</sup>). AgBF<sub>4</sub> and AgPF<sub>6</sub> are commercially available, so you will see those ions more frequently. The other ions, while they have good properties for ionic liquids, are not usually available as silver salts, so those reagents have to be prepared. You are already doing three reactions to get to EMIM BF<sub>4</sub><sup>-</sup>. Who wants to do more synthesis just to make their solvents?
Fluorous solvents have the odd property of being both hydrophobic and lipophobic ([Ref][1]) and thus are not miscible with either aqueous or many organic solvents as you've noted. Fluorine, as the most electronegative element, does odd things to a molecule (and is often used in pharmaceutical compounds just because of some of these odd properties). In the perfluorinated hydrocarbons, the carbon-fluorine bonds are quite polar with the electron density higher toward the fluorine. The solvents themselves, however, are not polar due to free rotation about the C-C bonds, so they are not miscible with polar solvents and are hydrophobic. Because the F is so electronegative, the electrons are quite tightly held and the molecules have a low polarizability and unusually weak London dispersion forces, so they aren't miscible with non-polar solvents either. These weak London forces are illustrated by the fact that the perfluorinated hydrocarbons have nearly the same boiling point as their corresponding normal hydrocarbon, despite having more mass. ([Ref][2]) You mentioned heat as one way to render the organic and fluorous solvents miscible. Adding gaseous $\ce{CO2}$ to the biphasic system also makes the phases miscible in many cases due to the high solubility of $\ce{CO2}$ in both fluorous and organic solvents. ([Ref][3]) [This reference][4] suggest that shape may also be a consideration when considering flourous solvent's miscibility with alkanes, also through poor/weak interaction of London dispersion forces in the two liquids. [1]: http://en.wikipedia.org/wiki/Organofluorine_chemistry [2]: http://www2.chemistry.msu.edu/courses/cem958/FS04_SS05%5CStewartHart.pdf [3]: http://pubs.acs.org/doi/abs/10.1021/ie0308745 [4]: http://books.google.com/books?isbn=3527642137...
Fluorous solvents have the odd property of being both hydrophobic and lipophobic ([Ref][1]) and thus are not miscible with either aqueous or many organic solvents as you've noted. Fluorine, as the most electronegative element, does odd things to a molecule (and is often used in pharmaceutical compounds just because of some of these odd properties). In the perfluorinated hydrocarbons, the carbon-fluorine bonds are quite polar with the electron density higher toward the fluorine. The solvents themselves, however, are not polar due to free rotation about the C-C bonds, so they are not miscible with polar solvents and are hydrophobic. Because the F is so electronegative, the electrons are quite tightly held and the molecules have a low polarizability and unusually weak London dispersion forces, so they aren't miscible with non-polar solvents either. These weak London forces are illustrated by the fact that the perfluorinated hydrocarbons have nearly the same boiling point as their corresponding normal hydrocarbon, despite having more mass. ([Ref][2]) You mentioned heat as one way to render the organic and fluorous solvents miscible. Adding gaseous $\ce{CO2}$ to the biphasic system also makes the phases miscible in many cases due to the high solubility of $\ce{CO2}$ in both fluorous and organic solvents. ([Ref][3]) [This reference][4] suggest that shape may also be a consideration when considering flourous solvent's miscibility with alkanes, also through poor/weak interaction of London dispersion forces in the two liquids. [1]: http://en.wikipedia.org/wiki/Organofluorine_chemistry [2]: http://www2.chemistry.msu.edu/courses/cem958/FS04_SS05%5CStewartHart.pdf [3]: http://pubs.acs.org/doi/abs/10.1021/ie0308745 [4]: http://books.google.com/books?id=6MzGgfWZAIMC&pg=PA240&lpg=PA240&dq=solubility%20of%20organic%20solvents%20in%20fluorous%20solvents&source=bl&ots=0eS9j7ysiB&sig=XIpj4mJh2tawr9cVPdtvl9Rj3ZA&hl=en&sa=X&ei=zdC9T6jME6qsiAKM2_X1DQ&ved=0CEwQ6AEwBA#v=onepage&q=solubility%20of%20organic%20solvents%20in%20fluorous%20solvents&f=false
Can someone explain the nomenclature of 7-amino-4-methylcoumarin, pictured below ([from the Sigma-Aldrich website][1])? ![enter image description here][2] The alpha-carbons of the methyl- and amino-groups here are separated by a minimum of three C's, whereas the numbers 4 and 7 are separated by two digits. What am I missing? [1]: http://www.sigmaaldrich.com/catalog/product/sigma/a9891?lang=en&region=US [2]: https://i.stack.imgur.com/fCOKR.png [3]: http://www.acdlabs.com/iupac/nomenclature/
*(Assuming this is homework or self-education which should be treated homeworky).* > 1) why we are not adding the contribution of water to the H3O+ ions 1. Write down the dissociation constant equation with and without $\ce{H2O}$. What changes? 2. Can you find a formulation of the dissociation constant where it seems natural not to include $\ce{H2O}$? 3. Read up "activity". > 2) why we are assuming that the value 0.02x is dissociated from HF and not just x Write down & solve the equation with 0.02 x and with x only. What is the difference?
Ok, a little wierd one: In the title I am refering to silver spoons, which, when used to consume milk products (like yogurt), get a wierd taste. It could be described as umami/meaty, a little bit like onions. I came to the cunclusion that it must be the silver, because this taste is never observed when using steel spoons. My guess was: the silver (or maybe its oxide?) reacts with an ingredient of the milk product. My question is: Is this documented/known process/reaction or is this some kind of misperception on my part? If the former is true, what exactly is happening? (By)-products?
While trying to understand the solution of a problem given in my text book, I realized I'm having some difficulty with the solution. The problem is as folows: The ionization constant of HF is $3.2 \times 10^{-4}$. Calculate the degree of dissociation of HF in its 0.02 solution.Calculate the concentration of all species present $\ce{H3O+ , F+}$ and $\ce{HF}$ in the solution and its PF. In the solution of this problem, the equation is given as $$\ce{HF + H2O <-> H3O+ + F-}$$ The concentration of at the time of equilibrium are given as: $$\ce{HF} = 0.02 - 0.02x, \: \ce{H3O+} = 0.02x, \:\ce{F- }= 0.02x}$$ I have the following doubts: 1. Why we are not adding the contribution of water to the $\ce{H3O+}$ ions? 2. Why we are assuming that the value $0.02x$ is dissociated from $\ce{HF}$ and not just $x$.?
Turbostratic graphite is graphite in which there is quenched rotational disalignment between adjacent graphene sheets, i.e. one sheet is rotated with respect to its neighbor. I suppose this could be considered a crystallographic defect of sorts. How do these turbostratic layers form, and can they possess long range order, such as glide or helical symmetry (e.g. similar to a twisted nematic phase)?
While trying to understand the solution of a problem given in my text book, I realized I'm having some difficulty with the solution. The problem is as folows: The ionization constant of HF is $3.2 \times 10^{-4}$. Calculate the degree of dissociation of HF in its 0.02 solution. Calculate the concentration of all species present $\ce{H3O+ , F+}$ and $\ce{HF}$ in the solution and its PF. In the solution of this problem, the equation is given as $$\ce{HF + H2O <-> H3O+ + F-}$$ The concentration of at the time of equilibrium are given as: $$[\ce{HF}] = 0.02 - 0.02x, \: \ce{[H3O+]} = 0.02x, \:\ce{[F- ]}= 0.02x$$ I have the following doubts: 1. Why we are not adding the contribution of water to the $\ce{H3O+}$ ions? 2. Why we are assuming that the value $0.02x$ is dissociated from $\ce{HF}$ and not just $x$.?