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What is the difference between an angular and linear skeletal formula? |
When drawing a skeletal formula, what is the difference between an angular version and a linear version?
I was asked to draw the Z isomer of Resveratrol:
![(E)-Resveratrol][1]
For which I drew:
![My (Z)-Resveratrol][2]
However the markscheme states that:
> skeletal structure must be correct **and angular not linear**
I haven't come across the difference between the two before and can't find anything on google to suggest one. Their drawing of the correct answer is equivalent to mine, however I am concerned that I may have drawn the linear version as my benzene rings are in a line and theirs are not. Can anyone put my mind at ease?
Their Version:
![Their (Z)-Resveratrol][3]
[1]: https://i.stack.imgur.com/sKlxu.png
[2]: https://i.stack.imgur.com/5gRaA.jpg
[3]: https://i.stack.imgur.com/2qYhs.jpg |
Can derive the quantity of a substance from a [GC/MS](http://en.wikipedia.org/wiki/Gas_chromatography%E2%80%93mass_spectrometry) report if I know:
* the ratio to another substance in the data,
* the quantity of that second substance, and
* the molecular mass of both substances.
The site was very dense and vague regarding how GC/MS test work, but fairly upfront regarding how they derive their ratios.
Here are the numbers (slightly tweaked, but more or less in proportion):
Substance A:
* Quantity: $300 \:\mathrm{mg}$ (Know already, not part of GC/MS report)
* Mol. Mass: $500 \:\mathrm{g/mol}$ (found in external source)
* Peak Proportion: 3
Substance X:
* Quantity: Unknown
* Mol. Mass: $425 \:\mathrm{g/mol}$ (again, external source)
* Peak Proportion: 1
To be clear, the ratio of 1:3 means that substance X has a horizontal peak $\frac13$ of substance A. I'm not clear on what the x-axis is actually a measurement of, which is part of why I'm at a loss. The site does make it very explicit that the ratios are not directly proportional to mass ratio (so if it found a ratio of 3:1 of glucose to arsenic, this doesn't mean the substance is 75% glucose and 25% arsenic, only that the glucose "peaks" 3 times higher, which I've taken to mean "3 times the oomph", but that may be incorrect, as well).
So, if I know that the molecular mass($M$) of substance X is 0.85 of substance Y, can I derive the actual mass of substance X using the formula:
\begin{equation}\mathrm{qty}_X=\frac{\mathrm{qty}_A\times(M_X/M_A)}{\mathrm{peak}_A/\mathrm{peak}_X}\end{equation}
with data being:
\begin{equation}qty_X=\frac{300\:\mathrm{mg}\times(425/500)}{3/1}\end{equation}
Which simplifies to:
\begin{equation}qty_X=(300\text{ mg}\times 0.85)/3\end{equation}
and finally the result of $85 \:\mathrm{mg}$.
So I guess in the end there are 3 questions:
1. Is this even how GC/MS results work?
2. If so, is my assumption to derive the mystery quantity using the molecular mass correct?
3. Is the math itself in order? (I'm specifically worried that I should invert either the mass ratio or the peak ratio or both).
Of course, if the answer to the first questions is no, then my true question is: can I derive the quantity of substance X with the given data, and if so, what would be the right approach?
If anyone is curious for some context, I need to know the actual quantity of substance X as I know that, by mass, it has a threshold between harmless and toxic, so just knowing substance X is $\frac13$ "peak" of substance A doesn't let me know if I should let my dog/child/self ingest it. |
The ideal gas equation (daresay "law") is a fascinating combination of the work of dozens of scientists over a long period of time.
I encountered Van der Waal's interpretation for non-ideal gases early on, and it was always somewhat in a "closed-form"
$${\bigl(p + \frac{n^2a}{V^2}\bigr)(V - nb) = nRT}$$
with $a$ being a measure of the charge interactions between the particles and $b$ being a measure of the volume interactions.
Understandably, this equation is only still around for historical purposes, as it is largely inaccurate.
Fast-forwarding to the 1990s, [Wikipedia](http://en.wikipedia.org/wiki/Equation_of_state#Elliott.2C_Suresh.2C_Donohue_equation_of_state) has a listing of one of the more current manifestations (of Elliott, Suresh, and Donohue):
$$\frac{p V_m}{RT}=Z=1 + Z^{\rm{rep}} + Z^{\rm{att}}$$
where the repulsive and attractive forces between the molecules are proportional to a shape number ($c = 1$ for spherical molecules, a quadratic for others) and reduced number density, which is a function of Boltzmann's constant, etc (point being, a lot of "fudge factors" and approximations are getting thrown into the mix).
Rather than seeking an explanation of all of this, I am wondering whether a more "closed form" solution lies at the end of the tunnel, or whether the approximations brought forth in the more modern models will have to suffice?
|
When testing for pH for common household cleaning products, such as dishwasher fluid, I noticed that majority of them tend to be basic? Why is that? Do acid make good cleaning solutions as well? |
Why are the majority of cleaning solutions basic? |
I'll add to Aesin’s answer that in this case, the burden of proof rests on the side of an analytical (or closed-form) equation of state. Statistical mechanics explicitly guarantees that there is a relationship between $p$, $V$ and $T$, i.e. that they are not independent state variables. However, no further generic statement can be made about it, and only by making appropriate approximations can one actually write an actual equation of state (EOS). One such EOS is the ideal gas law, others are as you have cited, but if one were to prove the existence of a closed-form generic EOS for fluids, that would be a major upheaval of the current understanding of thermodynamics of fluids.
For an overview of the thermodynamics of fluids, you can refer to your favorite statistical mechanics or statistic thermodynamics textbook (my own preference goes to [McQuarrie](http://www.amazon.com/Statistical-Mechanics-Donald-Allan-McQuarrie/dp/1891389157)). For a introduction to issues specific to the liquid state (but also valid in general for fluids far from the ideal gas behaviour), I would recommend [Theory of Simple Liquids](http://books.google.fr/books/about/Theory_of_Simple_Liquids.html?id=Uhm87WZBnxEC&redir_esc=y), by Hansen and McDonald. |
Why are there no edge dislocations in an FCC lattice? |
Dishwasing agent Are mainly used to remove oil and protein based deposits from plates, glasses and other utensils
Dishwashing agents are usually anionic
Anionic detergents are alkylbenzenesulfonates.
They consist of two parts
1. Sulfonate part which is hydrophilic(water loving i.e. dissolves in water)
2. Alkylbenzene part which is lipophilic(fat loving i.e dissolves in fats and oils)
Thus the sulphonate bonds to water while alkylbenzene bonds to oil or protein or fat . Thus fats,oils,proteins & grease dissolve in water
Mechnism of a dishwashing agent$ ~~~~~~~~~~~$![enter image description here][1]
Ingredients as on a dishwashing liquid bottle(the one in my house)
![enter image description here][2]
As you can see **sodium dodecylbenzenesulphonate**(an alkylbenzenesulfonate) is one of the main ingredients.
----------
`Do acid make good cleaning solutions as well?`
Yes they can be, but not as a dishwashing agent.
According to [Wikipedia][3]
>Acidic washing agents are mainly used for removal of inorganic deposits like scaling. The active ingredients are normally strong mineral acids and chelants. Often, there are added surfactants and corrosion inhibitors.
[1]: https://i.stack.imgur.com/j30zD.jpg
[2]: https://i.stack.imgur.com/V3MWQ.png
[3]: http://en.wikipedia.org/wiki/Cleaning_agent#Acidic |
The [Hammett plot][1] is commonly invoked in organic chemistry to reason about the plausibility (or implausibility) of various reaction mechanisms. The vertical axis is essentially the logarithm of an equilibrium constant (or rate constant) measured relative to a hydrogen functional group, which I understand. However, I am at a complete loss for understanding what is being plotted on the horizontal axis, which conventionally are denoted $\sigma$ and $\sigma^\pm$. What are these parameters and how are their values determined for various functional groups? In practice it seems that people look them up in tables, but where do the values in these tables come from?
[1]: http://en.wikipedia.org/wiki/Hammett_equation |
How are $\sigma$ and $\sigma^\pm$ determined in Hammett plots? |
Dishwashing agents are usually anionic detergents ie they are alkylbenzenesulfonates.
They consist of two parts
1. Sulfonate part which is hydrophilic(water loving i.e. dissolves in water)
2. Alkylbenzene part which is lipophilic(fat loving i.e dissolves in fats and oils)
Thus when in water they dissociate to give $\ce{RSO3-}$ and $\ce{Na+}$ . As $\ce{Na+}$ is from a strong base($\ce{NaOH}$) while $\ce{RSO3-}$ is a weak acid thus the resulting solution is **basic**.
The sulphonate bonds to water while alkylbenzene bonds to oil or protein or fat . Thus fats,oils,proteins & grease dissolve in water
Mechnism of a dishwashing agent$ ~~~~~~~~~~~$![enter image description here][1]
Ingredients as on a dishwashing liquid bottle(the one in my house)
![enter image description here][2]
As you can see **sodium dodecylbenzenesulphonate**(an alkylbenzenesulfonate) is one of the main ingredients.
----------
`Do acid make good cleaning solutions as well?`
Yes they can be, but not as a dishwashing agent.
According to [Wikipedia][3]
>Acidic washing agents are mainly used for removal of inorganic deposits like scaling. The active ingredients are normally strong mineral acids and chelants. Often, there are added surfactants and corrosion inhibitors.
[1]: https://i.stack.imgur.com/j30zD.jpg
[2]: https://i.stack.imgur.com/V3MWQ.png
[3]: http://en.wikipedia.org/wiki/Cleaning_agent#Acidic |
When drawing a skeletal formula, what is the difference between an angular version and a linear version?
I was asked to draw the Z isomer of Resveratrol:
![(E)-Resveratrol][1]
For which I drew:
![My (Z)-Resveratrol][2]
However the markscheme states that:
> skeletal structure must be correct **and angular not linear**
I haven't come across the difference between the two before and can't find anything on google to suggest one. Their drawing of the correct answer is equivalent to mine, however I am concerned that I may have drawn the linear version as my benzene rings are in a line and theirs are not. Can anyone put my mind at ease?
Their Version:
![Their (Z)-Resveratrol][3]
[1]: https://i.stack.imgur.com/sKlxu.png
[2]: https://i.stack.imgur.com/5gRaA.jpg
[3]: https://i.stack.imgur.com/2qYhs.jpg |
What is the difference between melting and dissolving? I am looking some general features.
The answer should be adaptable to the melting/dissolving of ice cube (water) in a class of pure alcohol (ethanol) just below(or at the) melting point of ice, or similar phenomena.
I am now assuming that the ice is dissolving and melting at the same time.
**Edit:**
In other words which reaction energy is higher in the following reactions:
$H_2O(s) \to H_2O(l)$
$H_2O(s) + n EtOH(l) \to H_2O \centerdot EtOH_n$
Or, are there substances that release more energy in dissolvation than consume in melting?
**Edit 2**
To summarize the differences and similarities, here is a list (I will update the list, if new comments arrive):
**differences**
- melting is (almost) always exothermic, dissolving can be endo- or exothermic
- melting is within one substance, dissolvation is between solute and solvent that are different substances
**similarities**
- In both cases, bonds are broken by crossing an energy barrier
- phase of matter changes
- You can create a situation where it is difficult to say whether the substance is melting or dissolving.
|
What is the difference between melting and dissolving? I am looking some general features.
The answer should be adaptable to the melting/dissolving of ice cube (water) in a class of pure alcohol (ethanol) just below(or at the) melting point of ice, or similar phenomena.
I am now assuming that the ice is dissolving and melting at the same time.
**Edit:**
In other words which reaction energy is higher in the following reactions:
$H_2O(s) \to H_2O(l)$
$H_2O(s) + n EtOH(l) \to H_2O \centerdot EtOH_n$
Or, are there substances that release more energy in dissolvation than consume in melting?
|
QTPIE is a [fluctuating charge model](http://www.slideshare.net/acidflask/constructing-a-rigorous-fluctuatingcharge-model-for-molecular-mechanics) that substantially improves modeling of polarization and charge transfer. I was curious whether it has ever been applied to 1D metals such as polyacetylene to provide new perspectives on topics such as [Peierls' condensation and phase solitons](http://prb.aps.org/abstract/PRB/v29/i4/p2109_1)? |
Has QTPIE ever been applied to 1D metals such as polyacetylene? |
<h2> Basic dishwashing agents? </h2>
- Some dishwashing agents contain substantial amounts of NaOH, see below.
- Sometimes NaClO solutions are used as disinfectant for dishwashing.
<br/>$\require{mhchem}$
As $\ce{Cl- + ClO- + 2 H+ <=>> Cl2 ^ + H2O}$, they are kept basic (and $\ce{Cl-}$free, but $\ce{3 ClO- -> 2 Cl- + ClO3-}$ and $\ce{2 ClO^- -> 2 Cl^- + O2 ^}$) for stabilization.
- Soaps (chemical meaning) are salts of fatty acids. Na- and K-soaps are good for cleaning and as salts of weak acids and strong bases they are basic.
<h2> Dishwashing liquid chemistry? </h2>
If we approximate the food rests you want to clean away during the dishwashing as mixture of carbohydrate, proteins and lipids, then your dishwashing liquid needs that take care of those 3 substance classes in water.
- carbohydrates are hydrophilic - no problem here.
- amphiphils are needed to clean lipophilic substances like lipids, oils, greases as [Ashu explained already](http://chemistry.stackexchange.com/a/358/160). They can take care of quite a bit of protein as well.
- basic solutions usually cause a faster hydrolysis of the amide bonds in proteins than acids. Usually quite a bit faster: NaOH-solutions immediately give a "soapy" feeling while e.g. HCl doesn't.
- As far as I know here (Germany), basic agents for *household* dishwashing machines are uncommon/seldom/not used at all. They are less corrosive (for both the machine and your fingers if you more or less accidentally decide to use it for dishwashing by hand).
- For dishwashing liquids where you are thought/supposed to put your hands in, the pH is usually neutral to slightly acidic, so you need to scrub a bit more, but don't dissolve your skin.
<h2> Acids for cleaning? </h2>
Acids (usually acetic acid or citric acid, maybe HCl) are used in "household chemistry" to clean limescale (CaCO$_3$). But that usually isn't the problem of your dirty dishes.
Dishwashing machines use ion exchangers to soften the water (or phosphate in the dishwashing agent), so you don't get limescale or lime soaps inside the machine.
|
What kind of bonds silica bead forms with polystyrene? Silica is now as ~100 $\mu$m spherical particles in bulk polystyrene that is made by melting.
How can I bond the silica particles
1. more tightly
2. more loosely
to the polystyrene?
Can this be generalized to any nano/micro particle and plastics? I am thinking now metal oxides and other thermoplastics. |
How is silica bead bonded into polystyrene? |
I have to stretch my memory to remember how this goes (and read a bit from [here][1]).
First, how do you define burgers vector? You take the end point of the dislocation and make a circle around it as it were a perfect lattice, the extra (or missing step) from the full loop is the burgers vector.
Now, is the burgers vector also a lattice vector? If yes, then it is (edge)dislocation. If no, then it is only a **partial** (edge)dislocation (partial dislocations have their own name).
Now, how does this work on a fcc-lattice case. On quick read for link I pasted, it seems that fcc-lattice cannot have edge dislocations, because the burgers vector is not a lattice vector.
You may want to go this through by yourself and check that I didn't make any mistakes.
[1]: http://www.tf.uni-kiel.de/matwis/amat/def_en/kap_5/backbone/r5_4_1.html |
X-ray photoelectron spectroscopy ([XPS][1]) is a particularly useful technique for examining the surfaces of all sorts of materials. Organic monolayers are particularly difficult to characterize completely via this technique due to the fact that many carbon peaks do not differ much by chemical shift. Traditionally quantitative information is obtained via a lot of guess-work and forced peak fitting. It would be better predict peaks shapes and peak positions beforehand to aid in the interpretation and prevent a terribly mis-guided interpretation. How well can theory help the non-specialist with this? I wouldn't have the time to develop a program to do this either. Preferably one could point me to the right direction in terms of a research paper/software.
[1]: http://en.wikipedia.org/wiki/X-ray_photoelectron_spectroscopy |
X-ray photoelectron spectroscopy ([XPS][1]) is a particularly useful technique for examining the surfaces of all sorts of materials. Organic monolayers are particularly difficult to characterize completely via this technique due to the fact that many carbon peaks do not differ much by chemical shift. Traditionally quantitative information is obtained via a lot of guess-work and forced peak fitting. It would be better to predict peaks shapes and peak positions beforehand to aid in the interpretation and prevent a terribly mis-guided interpretation. How well can theory help the non-specialist with this? I wouldn't have the time to develop a program to do this either. Preferably one could point me to the right direction in terms of a research paper/software.
[1]: http://en.wikipedia.org/wiki/X-ray_photoelectron_spectroscopy |
What is the difference between melting and dissolving? I am looking some general features.
The answer should be adaptable to the melting/dissolving of ice cube (water) in a class of pure alcohol (ethanol) just below (or at the) melting point of ice, or similar phenomena.
I am now assuming that the ice is dissolving and melting at the same time.
In other words which reaction energy is higher in the following reactions:
$$\begin{align}
\ce{H_2O(s) &\to H_2O(l)\\
H_2O(s) + n EtOH(l) &\to H_2O \centerdot (EtOH)_{n}}
\end{align}$$
Or, are there substances that release more energy in dissolvation than consume in melting?
|
I have the following question to solve:
Tungsten (W) and chlorine (Cl) form a series of compounds with the following compositions:
Mass % W Mass % Cl
72.17 27.83
56.45 43.55
50.91 49.09
46.36 53.64
If a molecule of each compound contains only one tungsten atom, what are the formulas for the four compounds?
My answer is as follows:
For one gram of tungsten, chlorine has mass<br />
$\frac{27.83}{72.17}=0.3302$ g<br />
$\frac{43.55}{56.45}=0.7715$ g<br />
$\frac{49.09}{50.91}=0.9643$ g<br />
$\frac{53.64}{46.36}=1.157$ g.
Since $\frac{0.7715}{0.3302}\approx\frac{7}{3}$, $\frac{0.9643}{0.3302}\approx3$, and $\frac{1.157}{0.3302}\approx\frac{7}{2}$, the number of atoms of chlorine for a given mass of tungsten are respectively in the ratio $6:14:18:21$.
So if a molecule of each compound contains only one tungsten atom, the formulas are WCl$_{6}$, WCl$_{14}$, WCl$_{18}$, and WCl$_{21}$.
Is this correct? |
How to compute molecular formula? |
In the standard brown ring test for the nitrate ion, the brown ring is $$\ce{[Fe(H2O)5 (NO)]^{2+}}$$
In this compound, the nitrosyl ligand is positively charged, and iron is in a $+1$ oxidation state.
Now, Iron has stable oxidation states $+2,+3$. Nitrosyl, as a ligand, comes in many flavours, of which a negatively charged nitrosyl is one.
I see no reason why the iron doesn't spontaneously oxidise to +3 and reduce the NO to -1 to gain stability. But I don't know how to analyse this situation anyway. I think that there may be some nifty backbonding increasing the stability, but I'm not sure.
So, why is iron in +1 here when we can have a seemingly stable situation with iron in +3? |
Why is iron in the brown ring compound in a +1 oxidation state? |
In “periodic table”, the adjective is related to the noun *period*, and comes from Ancient Greek *περίοδος* through French *périodique*. In “periodic acid”, it is formed from the prefix *per-* and *iodic* (like *peroxide* and *permanganate*).
[Wiktionary][1] lists their respective UK pronunciations of as `/pɪə(ɹ).iˈɒdɪk/` and `/ˌpɜːraɪˈɒdɪk/`, markedly different: `pɪə` (as in *piece*) vs. `pɜː` (as in *perfume*); then `i` (as in *it*) vs. `aɪ` (as in *eye*).
However, is that distinction really made in practice? Would a native US/UK/Aussie speaker make the difference when talking in the lab?
[1]: http://en.wiktionary.org/wiki/periodic#Etymology_1 |
<h2> Basic dishwashing agents? </h2>
- Some dishwashing agents contain substantial amounts of NaOH, see below.
- Sometimes NaClO solutions are used as disinfectant for dishwashing.
<br/>$\require{mhchem}$
As $\ce{Cl- + ClO- + 2 H+ <=>> Cl2 ^ + H2O}$, they are kept basic (and $\ce{Cl-}$free, but $\ce{3 ClO- -> 2 Cl- + ClO3-}$ and $\ce{2 ClO^- -> 2 Cl^- + O2 ^}$) for stabilization.
- Soaps (chemical meaning) are salts of fatty acids. Na- and K-soaps are good for cleaning and as salts of weak acids and strong bases they are basic.
<h2> Dishwashing liquid chemistry? </h2>
If we approximate the food rests you want to clean away during the dishwashing as mixture of carbohydrate, proteins and lipids, then your dishwashing liquid needs that take care of those 3 substance classes in water.
- carbohydrates are hydrophilic - no problem here.
- amphiphils are needed to clean lipophilic substances like lipids, oils, greases as [Ashu explained already](http://chemistry.stackexchange.com/a/358/160). They can take care of quite a bit of protein as well.
- basic solutions usually cause a faster hydrolysis of the amide bonds in proteins than acids. Usually quite a bit faster: NaOH-solutions immediately give a "soapy" feeling while e.g. HCl doesn't.
- As far as I know here (Germany), basic agents for *household* dishwashing machines are uncommon/seldom/not used at all. The actually used solutions are less corrosive (for both the machine and your fingers if you more or less accidentally decide to use it for dishwashing by hand).
- For dishwashing liquids where you are thought/supposed to put your hands in, the pH is usually neutral to slightly acidic, so you need to scrub a bit more, but don't dissolve your skin.
<h2> Acids for cleaning? </h2>
Acids (usually acetic acid or citric acid, maybe HCl) are used in "household chemistry" to clean limescale (CaCO$_3$). But that usually isn't the problem of your dirty dishes.
Dishwashing machines use ion exchangers to soften the water (or phosphate in the dishwashing agent), so you don't get limescale or lime soaps inside the machine.
|
I think your question implicates another question (which is also mentioned in some comments here), namely: Why are all energy eigenvalues of states with a different **angular momentum quantum number** $\ell$ but with the same **principal quantum number** $n$ (e.g. $3s$, $3p$, $3d$) degenerate in the hydrogen atom but non-degenerate in multi-electron atoms?
Although `AcidFlask` already gave a good answer (mostly on the non-degeneracy part) I will try to eleborate on it from my point of view and give some additional information.
I will split my answer in three parts: The first will address the $\ell$-degeneracy in the hydrogen atom, in the second I will try to explain why this degeneracy is lifted, and in the third I will try to reason why $3s$ states are lower in energy than $3p$ states (which are in turn lower in energy than $3d$ states).
## $\ell$-degeneracy of the hydrogen atoms energy eigenvalues ##
The non-relativistic electron in a hydrogen atom experiences a potential that is analogous to the [Kepler problem][1] known from classical mechanics.
This potential (aka Kepler potential) has the form $\frac{\kappa}{r}$, where $r$ is the distance between the nucleus and the electron, and $\kappa$ is a proportionality constant.
Now, it is known from physics that symmetries of a system lead to conserved quantities ([Noether Theorem][2]).
For example from the roational symmetry of the Kepler potential follows the conservation of the angular momentum, which is characterized by $\ell$. But while the length of the angular momentum vector is fixed by $\ell$ there are still different possibilities for the orientation of its $z$-component, characterized by the **magnetic quantum number $m$**, which are all energetically equivalent as long as the system maintains its rotational symmetry.
So, the rotational symmetry leads to the $m$-degeneracy of the energy eigenvalues for the hydrogen atom.
Analogously, the $\ell$-degeneracy of the hydrogen atoms energy eigenvalues can also be traced back to a symmetry, the $SO(4)$ symmetry.
The system's $SO(4)$ symmetry is not a geometric symmetry like the one explored before but a so called dynamical symmetry which follows from the form of the Schroedinger equation for the Kepler potential.
(It corresponds to rotations in a four-dimensional cartesian space. Note that these rotations do not operate in some physical space.)
This dynamical symmetry conserves the [Laplace-Runge-Lenz vector][3] $\hat{\vec{M}}$ and it can be shown that this conserved quantity leads to the $\ell$-independent energy spectrum with $E \propto \frac{1}{n^2}$. (A detailed derivation, though in German, can be found [here][4].)
## Why is the $\ell$-degeneracy of the energy eigenvalues lifted in multi-electron atoms? ##
As the $m$-degeneracy of the hydrogen atom's energy eigenvalues can be broken by destroying the system's spherical symmetry, e.g. by applying a magnetic field, the $\ell$ degeneracy is lifted as soon the potential appearing in the Hamilton operator deviates from the pure $\frac{\kappa}{r}$ form.
This is certainly the case for multielectron atoms since the outer electrons are screened from the nuclear Coulomb attraction by the inner electrons and the strength of the screening depends on their distance from the nucleus.
(Other factors, like spin and relativistic effects, also lead to a lifting of the $\ell$-degeneracy even in the hydrogen atom.)
## Why do states with the same $n$ but lower $\ell$ values have lower energy eigenvalues?##
Two effects are important here:
- The **centrifugal force**${}^{1}$ puts an "energy penalty" onto states with higher angular momentum. So, a higher $\ell$ value implies a stronger centrifugal force, that pushes electrons away from the nucleus.
1. The concept of centrifugal force can be seen in the radial Schroedinger equation for the radial part $R(r)$ of the wave function $\Psi(r, \theta, \varphi) = R(r) Y_{\ell,m} (\theta, \varphi )$
\begin{equation}
\bigg( \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \mathrm{d}^{2} }{ \mathrm{d} r^{2} } + \underbrace{ \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \ell (\ell + 1) }{ r^{2} } } - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } - E \bigg) r R(r) = 0
\end{equation}
\begin{equation}
{}^{= ~ V^{\ell}_{\mathrm{cf}} (r)} \qquad \qquad
\end{equation}
The radial part experiences an additional $\ell$-dependent potential $V^{\ell}_{\mathrm{cf}} (r)$ that pushes the electrons away from the nucleus.
- **Core repulsion** (Pauli repulsion), on the other hand, puts an "energy penalty" on states with a lower angular momentum. That is because the core repulsion acts only between electrons with the same angular momentum${}^{1}$. So it acts stronger on the low-angular momentum states since there are more core shells with lower angular momentum.
1. Core repulsion is due to the condition that the wave functions must be orthogonal which in turn is a consequence of the Pauli principle. Because states with different $\ell$ values are already orthogonal by their angular motion, there is no Pauli repulsion between those states. However, states with the same $\ell$ value feel an additional effect from core orthogonalization.
The "accidental" $\ell$-degeneracy of the hydrogen atom can be described as a balance
between centrifugal force and core repulsion, that both act against the nuclear Coulomb
attraction.
In the real atom the balance between centrifugal force and core repulsion is broken,
The core electrons are contracted compared to the outer electrons because there are less inner electron-shells screening the nuclear attraction from the core shells than from the valence electrons.
Since the inner electron shells are more contracted than the outer ones, the core repulsion is weakened whereas the effects due to the centrifugal force remain unchanged. The reduced core repulsion in turn stabilizes the states with lower angular momenta, i.e. lower $\ell$ values. So, $3s$ states are lower in energy than $3p$ states which are in turn lower in energy than $3d$ states.
Of course, one has to be careful when using results of the hydrogen atom to describe effects in multi-electron atoms as `AcidFlask` mentioned. But since only a qualitative description is needed this might be justifiable.
I hope this somewhat lengthy answer is helpful. If something is wrong with my arguments I'm happy to discuss those points.
[1]: http://en.wikipedia.org/wiki/Kepler_problem
[2]: http://en.wikipedia.org/wiki/Noether%27s_theorem
[3]: http://en.wikipedia.org/wiki/Runge-Lenz
[4]: http://www.desy.de/~jlouis/Vorlesungen/QMII06/vortrag_09.pdf
|
In [a series of papers in the early 1980s](http://www.springerlink.com/content/g83871qw83n26152/abstract/), Michael Springborg explored an interpretation of the [Wigner phase space function](http://en.wikipedia.org/wiki/Wigner_quasi-probability_distribution) as an electron density in a six-dimensional $(q,p)$ phase space. He applied it with some success to several simple compounds.
Is the Springborg 6D phase model model used in modern molecular orbital modeling? If not, are there specific weaknesses that make it unsuitable for current computational approaches? |
Is the Springborg 6D phase space model used in modern molecular orbital modeling? |
I'm reading a book on carbon nanotubes (P.J.F. Harris, Carbon Nanotube Science, to be specific) at the moment and puzzling about Stone-Wales or 5775 defects. This is where two adjacent carbons undergo a 90-degree rotation about their mutual centroid and generate a defect consisting of two five membered rings and two seven membered rings, as I've illustrated below (This is only supposed to be qualitative, and is on a finite PAH fragment rather than a nanotube, but you should get the idea):
![Stone-Wales defect on PAH][1]
[1]: https://i.stack.imgur.com/cLFKf.png
I note that tropylium ($\ce{[C7H7]^+}$) is aromatic as a cation, and the ever famous cyclopentadienide ($\ce{[C5H5]^-}$) is aromatic as an anion, so what I'm wondering is: do the 5 and 7-membered rings in a Stone-Wales defect pick up either formal or partial negative and positive charges to maintain aromaticity in a fashion consistent with the rest of the structure? |
What does the charge distribution around a Stone-Wales defect look like? |
I have the following question to solve:
Tungsten (W) and chlorine (Cl) form a series of compounds with the following compositions:
Mass % W Mass % Cl
72.17 27.83
56.45 43.55
50.91 49.09
46.36 53.64
If a molecule of each compound contains only one tungsten atom, what are the formulas for the four compounds?
My answer is as follows:
For one gram of tungsten, chlorine has mass<br />
$\frac{27.83}{72.17}=0.3302$ g<br />
$\frac{43.55}{56.45}=0.7715$ g<br />
$\frac{49.09}{50.91}=0.9643$ g<br />
$\frac{53.64}{46.36}=1.157$ g.
Since $\frac{0.7715}{0.3302}\approx\frac{7}{3}$, $\frac{0.9643}{0.3302}\approx3$, and $\frac{1.157}{0.3302}\approx\frac{7}{2}$, the number of atoms of chlorine for a given mass of tungsten are respectively in the ratio $6:14:18:21$.
So if a molecule of each compound contains only one tungsten atom, the formulas are $\ce{WCl_{6}}$, $\ce{WCl_{14}}$, $\ce{WCl_{18}}$, and $\ce{WCl_{21}}$.
Is this correct? |
My Sister asked me this:
> "Why is it that when we chew gum, it is soft and mushy to begin with , but slowly gets firmer and firmer like after 20 mins of chewing?"
I think it is because when we initially chew gum it has plenty of sugar causing rapid salivation and hence mushiness and the rubber polymer is very flexible to begin with and slowly looses its elasticity hence becomes firmer.
Is this the reason?
What is the chemistry involved ? |
I'm reading a book on carbon nanotubes (P.J.F. Harris, Carbon Nanotube Science, to be specific) at the moment and puzzling about Stone-Wales or 5775 defects. This is where two adjacent carbons undergo a 90-degree rotation about their mutual centroid and generate a defect consisting of two five membered rings and two seven membered rings, as I've illustrated below (This is only supposed to be qualitative, and is on a finite PAH fragment rather than a nanotube, but you should get the idea):
![Stone-Wales defect on PAH][1]
[1]: https://i.stack.imgur.com/cLFKf.png
I note that tropylium ($\ce{[C7H7]^+}$) is aromatic as a cation, and the ever famous cyclopentadienide ($\ce{[C5H5]^-}$) is aromatic as an anion, so what I'm wondering is: do the 5 and 7-membered rings in a Stone-Wales defect pick up either formal or partial negative and positive charges to maintain aromaticity in a fashion consistent with the rest of the structure?
P.S.
----------
Big props to F'x, who attempted electronic structure calculations on the PAHs depicted, regrettably without SCF convergence. |
I think your question implicates another question (which is also mentioned in some comments here), namely: Why are all energy eigenvalues of states with a different **angular momentum quantum number** $\ell$ but with the same **principal quantum number** $n$ (e.g. $3s$, $3p$, $3d$) degenerate in the hydrogen atom but non-degenerate in multi-electron atoms?
Although `AcidFlask` already gave a good answer (mostly on the non-degeneracy part) I will try to eleborate on it from my point of view and give some additional information.
I will split my answer in three parts: The first will address the $\ell$-degeneracy in the hydrogen atom, in the second I will try to explain why this degeneracy is lifted, and in the third I will try to reason why $3s$ states are lower in energy than $3p$ states (which are in turn lower in energy than $3d$ states).
## $\ell$-degeneracy of the hydrogen atoms energy eigenvalues ##
The non-relativistic electron in a hydrogen atom experiences a potential that is analogous to the [Kepler problem][1] known from classical mechanics.
This potential (aka Kepler potential) has the form $\frac{\kappa}{r}$, where $r$ is the distance between the nucleus and the electron, and $\kappa$ is a proportionality constant.
Now, it is known from physics that symmetries of a system lead to conserved quantities ([Noether Theorem][2]).
For example from the roational symmetry of the Kepler potential follows the conservation of the angular momentum, which is characterized by $\ell$. But while the length of the angular momentum vector is fixed by $\ell$ there are still different possibilities for the orientation of its $z$-component, characterized by the **magnetic quantum number $m$**, which are all energetically equivalent as long as the system maintains its rotational symmetry.
So, the rotational symmetry leads to the $m$-degeneracy of the energy eigenvalues for the hydrogen atom.
Analogously, the $\ell$-degeneracy of the hydrogen atoms energy eigenvalues can also be traced back to a symmetry, the $SO(4)$ symmetry.
The system's $SO(4)$ symmetry is not a geometric symmetry like the one explored before but a so called dynamical symmetry which follows from the form of the Schroedinger equation for the Kepler potential.
(It corresponds to rotations in a four-dimensional cartesian space. Note that these rotations do not operate in some physical space.)
This dynamical symmetry conserves the [Laplace-Runge-Lenz vector][3] $\hat{\vec{M}}$ and it can be shown that this conserved quantity leads to the $\ell$-independent energy spectrum with $E \propto \frac{1}{n^2}$. (A detailed derivation, though in German, can be found [here][4].)
## Why is the $\ell$-degeneracy of the energy eigenvalues lifted in multi-electron atoms? ##
As the $m$-degeneracy of the hydrogen atom's energy eigenvalues can be broken by destroying the system's spherical symmetry, e.g. by applying a magnetic field, the $\ell$ degeneracy is lifted as soon as the potential appearing in the Hamilton operator deviates from the pure $\frac{\kappa}{r}$ form.
This is certainly the case for multielectron atoms since the outer electrons are screened from the nuclear Coulomb attraction by the inner electrons and the strength of the screening depends on their distance from the nucleus.
(Other factors, like spin and relativistic effects, also lead to a lifting of the $\ell$-degeneracy even in the hydrogen atom.)
## Why do states with the same $n$ but lower $\ell$ values have lower energy eigenvalues?##
Two effects are important here:
- The **centrifugal force**${}^{1}$ puts an "energy penalty" onto states with higher angular momentum. So, a higher $\ell$ value implies a stronger centrifugal force, that pushes electrons away from the nucleus.
1. The concept of centrifugal force can be seen in the radial Schroedinger equation for the radial part $R(r)$ of the wave function $\Psi(r, \theta, \varphi) = R(r) Y_{\ell,m} (\theta, \varphi )$
\begin{equation}
\bigg( \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \mathrm{d}^{2} }{ \mathrm{d} r^{2} } + \underbrace{ \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \ell (\ell + 1) }{ r^{2} } } - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } - E \bigg) r R(r) = 0
\end{equation}
\begin{equation}
{}^{= ~ V^{\ell}_{\mathrm{cf}} (r)} \qquad \qquad
\end{equation}
The radial part experiences an additional $\ell$-dependent potential $V^{\ell}_{\mathrm{cf}} (r)$ that pushes the electrons away from the nucleus.
- **Core repulsion** (Pauli repulsion), on the other hand, puts an "energy penalty" on states with a lower angular momentum. That is because the core repulsion acts only between electrons with the same angular momentum${}^{1}$. So it acts stronger on the low-angular momentum states since there are more core shells with lower angular momentum.
1. Core repulsion is due to the condition that the wave functions must be orthogonal which in turn is a consequence of the Pauli principle. Because states with different $\ell$ values are already orthogonal by their angular motion, there is no Pauli repulsion between those states. However, states with the same $\ell$ value feel an additional effect from core orthogonalization.
The "accidental" $\ell$-degeneracy of the hydrogen atom can be described as a balance
between centrifugal force and core repulsion, that both act against the nuclear Coulomb
attraction.
In the real atom the balance between centrifugal force and core repulsion is broken,
The core electrons are contracted compared to the outer electrons because there are less inner electron-shells screening the nuclear attraction from the core shells than from the valence electrons.
Since the inner electron shells are more contracted than the outer ones, the core repulsion is weakened whereas the effects due to the centrifugal force remain unchanged. The reduced core repulsion in turn stabilizes the states with lower angular momenta, i.e. lower $\ell$ values. So, $3s$ states are lower in energy than $3p$ states which are in turn lower in energy than $3d$ states.
Of course, one has to be careful when using results of the hydrogen atom to describe effects in multi-electron atoms as `AcidFlask` mentioned. But since only a qualitative description is needed this might be justifiable.
I hope this somewhat lengthy answer is helpful. If something is wrong with my arguments I'm happy to discuss those points.
[1]: http://en.wikipedia.org/wiki/Kepler_problem
[2]: http://en.wikipedia.org/wiki/Noether%27s_theorem
[3]: http://en.wikipedia.org/wiki/Runge-Lenz
[4]: http://www.desy.de/~jlouis/Vorlesungen/QMII06/vortrag_09.pdf
|
I googled a bit about what $\ce{AgNO3 + AlCl3}$ will become and found out the following:
$\ce{3AgNO3 + AlCl3 -> Al(NO3)3 + 3AgCl}$
But why? I know that $\ce{Ag}$ is higher up in the Reactivity serie than $\ce{Al}$ but that do not make sense to me in this problem... Is it because $\ce{Cl}$ is just one when $\ce{NO3}$ is multiple? |
For example — "high purity chloride salt of Zinc (ZnCl2.2H2O)" or "various concentrations of FeCl2.4H2O"
What does the number before H2O mean?
|
How to understand this form of writing the solution — ( Some salt • n H2O)? |
Le Châtelier's principle only states that a system previously at equilibrium will want to _stay_ at equilibrium - that is, if we perturb it, it will try to go back to equilibrium. This system is _initially not at equilibrium_, and therefore we don't need his principle.
I think the reaction proceeds because the formation of AgCl<sub>(s)</sub> is enthalpically favored. Molecular cohesion (enthalpy) is greater in AgCl than in AgNO<sub>3</sub>, thus promoting its formation. How do I know that? Well, it seems AgCl is less ionic than AgNO<sub>3</sub> since it is insoluble in water.
This reaction is a type of methathesis reaction. It can be called _double displacement reaction_ or _salt methathesis reaction_, depending on who you're talking with. |
For example — "high purity chloride salt of Zinc (ZnCl₂ . 2 H₂O)" or "various concentrations of FeCl₂ . 4 H₂O". What does the number before H₂O mean?
|
How to understand this form of writing the solution: (some salt • n H₂O)? |
Le Châtelier's principle only states that a system previously at equilibrium will want to _stay_ at equilibrium - that is, if we perturb it, it will try to go back to equilibrium. This system is _initially not at equilibrium_, and therefore we don't need his principle.
I think the reaction proceeds because the formation of AgCl<sub>(s)</sub> is enthalpically favored. Molecular cohesion (enthalpy) is greater in AgCl than in AgNO<sub>3</sub>, thus promoting its formation. How do I know that? Well, it seems AgCl is less ionic than AgNO<sub>3</sub> since it is insoluble in water.
This reaction is a type of metathesis reaction. It can be called _double displacement reaction_ or _salt metathesis reaction_, depending on who you're talking with. |
In this molecule, which is 3,5-dimethyl-4-carboxypyrazolate (unless I got it wrong):
![enter image description here][1]
I do not think there is free rotation around the bond indicated in red, because one can write a resonnance structure with significant reasonable weight where a double bond would be located between the two groups.
However, how can I go further than that and qualitatively describe the energy barrier for the rotation. I have found \[1,2] that in the case of benzoates, the barrier is in the range 3–6 kcal/mol. Should I expect it to be higher or lower in the above molecule?
---
\[1] [10.1021/jp0529467][2]<br>
\[2] [10.1021/jo991146n][3]
[1]: https://i.stack.imgur.com/CFC2U.png
[2]: http://pubs.acs.org/doi/abs/10.1021/jp0529467
[3]: http://pubs.acs.org/doi/abs/10.1021/jo991146n |
<h2> Basic dishwashing agents? </h2>
- Some dishwashing agents contain substantial amounts of NaOH, see below.
- Sometimes NaClO solutions are used as disinfectant for dishwashing.
<br/>$\require{mhchem}$
As $\ce{Cl- + ClO- + 2 H+ <=>> Cl2 ^ + H2O}$, they are kept basic (and $\ce{Cl-}$free, but $\ce{3 ClO- -> 2 Cl- + ClO3-}$ and $\ce{2 ClO^- -> 2 Cl^- + O2 ^}$) for stabilization.
<br/>
*@Ashu asked whether I'm really sure, so here's an example: with 15 - 30 % KOH + 15 - 30 % NaOCl, pH 14 "Liquid intensive cleaner for dishwashing machines" ([technical data](http://www.pack2go.de/media/uploads/G%20300%20Geschirrsp%C3%BClmittel%20fl%C3%BCssig%20Desinfektion.pdf), [saftety sheet](http://www.pack2go.de/media/uploads/G%20300%20Sicherheitsdatenblatt.pdf) both German, though)*
- Soaps (chemical meaning) are salts of fatty acids. Na- and K-soaps are good for cleaning and as salts of weak acids and strong bases they are basic.
<h2> Dishwashing liquid chemistry? </h2>
If we approximate the food rests you want to clean away during the dishwashing as mixture of carbohydrate, proteins and lipids, then your dishwashing liquid needs that take care of those 3 substance classes in water.
- carbohydrates are hydrophilic - no problem here.
- amphiphils are needed to clean lipophilic substances like lipids, oils, greases as [Ashu explained already](http://chemistry.stackexchange.com/a/358/160). They can take care of quite a bit of protein as well.
- basic solutions usually cause a faster hydrolysis of the amide bonds in proteins than acids. Usually quite a bit faster: NaOH-solutions immediately give a "soapy" feeling while e.g. HCl doesn't.
- As far as I know here (Germany), basic liquids for *household* dishwashing machines are uncommon/seldom/not used at all (I'm not so sure about the tablets). The actually used solutions are less corrosive (for both the machine and your fingers if you more or less accidentally decide to use it for dishwashing by hand).
- [This page from a New Zealand poison centre](http://www.poisons.co.nz/fact.php?f=26) talks about the *"dishwasher [...] powder or tablets [...] are often highly alkaline"*
- For dishwashing liquids where you are thought/supposed to put your hands in, the pH is usually neutral to slightly acidic, so you need to scrub a bit more, but don't dissolve your skin.
<h2> Acids for cleaning? </h2>
Acids (usually acetic acid or citric acid, maybe HCl) are used in "household chemistry" to clean limescale (CaCO$_3$). But that usually isn't the problem of your dirty dishes.
Dishwashing machines use ion exchangers to soften the water (or phosphate in the dishwashing agent), so you don't get limescale or lime soaps inside the machine.
|
In [a series of papers in the early 1980s](http://www.springerlink.com/content/g83871qw83n26152/abstract/), Michael Springborg explored an interpretation of the [Wigner phase space function](http://en.wikipedia.org/wiki/Wigner_quasi-probability_distribution) as an electron density in a six-dimensional $(q,p)$ phase space. He applied it with some success to several simple compounds.
Is the Springborg 6D phase model model used in modern molecular orbital modeling? If not, are there specific weaknesses that make it unsuitable for current computational approaches?
-----
ADDENDUM:
In a world dominated by Hilbert spaces, the orbitals of chemistry are interesting because they are one of the few places in physics where a 3D spatial image of the underlying quantum situation is consistently retained. (My recollection is that Schrodinger liked such low-dimensionality approaches, since they emphasized waves that could be visualized.)
Alas, the problem of course is that you need both position $q$ and momentum $p$ to get an accurate picture of the overall electron state. Using a space-only 3D $q$ representation forces the analysis of molecules to flip back and forth at rather arbitrary thresholds between localized (position space $q$) and delocalized (momentum space $p$) views of the electrons in the molecules or metal crystals in question.
That bothers me, both from a physics perspective and from a computational integrity perspective.
The problem is that there exists a very deep and profound physics symmetry between $q$ and $p$ in terms of how effects such as Pauli exclusion applies to fermions. I find that simple symmetry nothing short of amazing, because to me there's just nothing intuitive about the idea that half-unit spin would result in such extraordinarily similar behaviors in these quite different spaces. At some level it seems to be one of those cases of "that's just the way it works." So, it seems pretty reasonable that accurate modeling of orbitals of many diverse sizes likely requires this symmetry to be captured accurately.
If that's true, this older work by Springborg (he still heads a research group BTW) strikes me as a possible opportunity to for taming some models, making them smoother and more sane over a much broader range of molecular sizes. That's because Springborg's 6D electron density functions -- if they work OK in broader contexts -- might help eliminate any need for abrupt switches between "mostly position" or $q$ 3D views (classic "electron clouds," whatever those really mean) and "mostly momentum" or $p$ 3D views.
A good place to test whether 6D $(q,p)$ electron distributions might provide higher computational stability would be in modeling a range of lengths for a long-chain carbon polymer such as polyacetylene. (I always pick polyacetylene for its simplicity, but there are other good choices.) Springborg appears not to have explored that domain, since as best I can tell his papers addressed mostly simpler bonds. (I think; I don't have them.) The [instabilities of approximate density functional theories](http://chemistry.stackexchange.com/a/365/83) _may_ (or may not) be related to $(q,p)$ transition instabilities. That's not a question I can or should try to address, as there are folks out there who know that domain orders of magnitude better than I ever will. |
Le Châtelier's principle only states that a system previously at equilibrium will want to _stay_ at equilibrium - that is, if we perturb it, it will try to go back to equilibrium. This system is _initially not at equilibrium_, and therefore we don't need his principle.
I think the reaction proceeds because the formation of AgCl<sub>(s)</sub> is enthalpically favored. Molecular cohesion (enthalpy) is greater in AgCl than in AgNO<sub>3</sub>, thus promoting its formation. How do I know that? Well, it seems AgCl is less ionic than AgNO<sub>3</sub> since it is insoluble in water.
This reaction is a type of metathesis reaction. It can be called _double displacement reaction_ or _salt metathesis reaction_, depending on who you're talking with.
----------
**EDIT**
The other answers make an error, I think. Both suggest that the _solubility product_ of AgCl is of importance in determining _why_ the reaction takes place. My opinion is that the argument, **as presented**, is circular, in the following manner:
_The solubility product of AgCl in water is very low, therefore we observe formation of a precipitate._
vs
_AgCl is insoluble in water, therefore its solubility product is low._
It's easy to see that the fact AgCl is insoluble in water implies that it has a low K<sub>sp</sub> in water. If someone can _predict and quantify_ the solubility product of AgCl in water _from first principles_, then that's another story, but the explanations offered only rely on observation (experiment) to quantify the K<sub>sp</sub>.
My argument goes down the ladder. If you can answer why AgCl is insoluble in water, then you can definitely predict that Ag<sup>+</sup> and Cl<sup>-</sup> ions _from different sources_ (i.e. in different solutions) put into contact will spontaneously form a precipitate.
Now you maybe thinking this argument is also circular, but it's not.
We have access to experiment, which clearly informs us that AgCl is _insoluble_ in water. We can confidently assume (but we might be wrong!) that the AgCl bond is _less_ ionic than the AgNO<sub>3</sub> bond, since it _practically does not dissociate/has a low K<sub>sp</sub> in water_. A bond with a greater covalent character is more stable, since the electrons are _shared_ and not _transfered_.
Perhaps the crystal structure of AgCl makes it especially stable to solubilization from H<sub>2</sub>O. I just think it's obvious that the enthalpy of formation of AgCl is favorable, since it has to counter an unfavorable entropy loss in the formation of an ordered crystal lattice.
----------
As you can see, I cannot provide you with a _thorough_ answer, it would be extremely time consuming (for me) and I'd have to conduct many many calculations. I really hope someone can clarify the _why_. Maybe you can do your own research from the answers provided here and come back with a more "elaborate" answer. |
Oxygen is not a fuel by itself, it's only the oxidant. Combustion involves a fuel (acetylene, for example) and an oxidizer (here, O2). So if you have _no fuel_, but _all oxidizer_, then what's left to burn? |
Pure oxygen does not burn. Oxygen is a _supporter_ of combustion, not a combustible material(fuel) itself.
Combustion reactions are as follows:
$$\text{combustible stuff} \ce{+ O2 -> } \text{various oxides}$$
Note that nitrous oxide ($\ce{N2O}$) is also a supporter of combustion, which means that you can use it to carry out a combustion reaction without oxygen--but this is only because $\ce{N2O}$ generates oxygen itself.
Since combustion is a reaction with oxygen, what's left for oxygen to oxidise? Itself? Not likely. You need a fuel for a combustion reaction.
|
Oxygen is not a fuel by itself, it's only the oxidant. Combustion involves a fuel (acetylene, for example) and an oxidizer (here, O<sub>2</sub>). So if you have _no fuel_, but _all oxidizer_, then what's left to burn?
Combustion is just a name we give to certain reactions. The point to get here is that the burning and its associated temperature is dependent on the reactants' chemical nature and stoichiometry. When a flame has a higher temperature than another, it means that both reactions give products that have different enthalpies of formation.
The greater the difference between the enthalpy of the products and reactants, the greater the flame temperature. |
This is one of the questions with which I have puzzled over, and can arrive at no definite conclusion. Why are strong acids or bases, such as H2SO4, HNO3, HCl, and NaOH not suitable primary standards?
How will their higher or lower pH affect the accuracy of the results of a titration? |
Why are strong acids and bases not suitable as primary standards? |
This is one of the questions with which I have puzzled over, and can arrive at no definite conclusion. Why are strong acids or bases, such as $\ce{H2SO4}$,$\ce{ HNO3}$, $\ce{HCl}$, and $\ce{NaOH}$ not suitable primary standards?
How will their higher or lower pH affect the accuracy of the results of a titration? |
What's the difference between precipitate and turbidity? |
[Slush hydrogen](http://en.wikipedia.org/wiki/Slush_hydrogen) is a mixture of liquid and solid hydrogen **at the triple point** considered as a possible vehicle fuel. What is the need of having it at the triple point? Couldn't any other set of thermodynamic conditions consistent with liquid–solid equilibrium work as well? |
Why is slush hydrogen at the triple point? |
I am not planning to do these types of experiments but I would like to know the order of magnitude of the time it takes to run an LC-MS/MS experiment. Let's assume that I have the sample prep (protein) finished and I just need to load the sample into the LC. How long would I need to wait before I would get MS/MS data?
I know that for an typical HPLC experiment, the sample takes around 20 minutes to go through the column. |
> Ag is higher up in the Reactivity serie than Al but that do not make sense to me in this problem
- indeed this is **no** redox problem, oxidation states do not change.
- instead the equilibrium is <br/>
$\ce{Ag+ + Cl- <=> AgCl v}$
As you marked this question as homework:
- Edit the question and put the oxidation numbers to confirm that it is not a redox reaction.
- Look up the solubility constant of AgCl. Together with the total initial concentration of $\ce{Ag+}$ and $\ce{Cl-}$, you can then calculate the equilibrium concentrations.
- You'll see that the precipitation moves the reaction towards equilibrium.
----------
@CHMs comment:
> Of course, I can already predict that the solubility product will be low, precisely because I see a precipitate.
Sure. I've seen it as well. However, from the original question I'm not so sure that the OP did actually see it. I may be wrong, though.
And, by the way, I could predict the precipitate from free standard enthalpy as well as from solubility constant.
> Why is the precipitate favored? Because once it forms, it can't go back to solution - but then why can't it go back into solution?
But this is **plainly wrong**: it *can go back into solution* and this is not only a possibility, but actually happens pretty fast. There really is an dynamic equilibrium
$$\ce{Ag+_(aq) + Cl(aq)- <=>> AgCl_(s)}$$
Here are two experiments you can do to prove it:
1. If you disturb the equilibrium by removing either $\ce{Ag+}$ or $\ce{Cl-}$, the AgCl will dissolve again (now Le-Châtelier-Braun is coming in ;-) ).
<br/>
Here are some ideas:
- Complex (mask) the $\ce{Ag+}$, e.g. by $\ce{NH3}$ or $\ce{CN-}$<br/>have you ever done this? If not, you should really try. It is really fast.
- reduce the $\ce{Ag+}$
- oxidize the Cl-, so that $\ce{Cl2 ^}$ is removed out of the system.
<br/>
(you could also remove the AgCl, but that isn't practical, as you'd never be sure whether you observe formation of more precipitate or whether you didn't remove all of the AgCl)
2. Crystal growth in saturated solutions. You may have to wait quite a while for this one, though...
|
> Ag is higher up in the Reactivity serie than Al but that do not make sense to me in this problem
- indeed this is **no** redox problem, oxidation states do not change.
- instead the equilibrium is <br/>
$\ce{Ag+ + Cl- <=> AgCl v}$
As you marked this question as homework:
- Edit the question and put the oxidation numbers to confirm that it is not a redox reaction.
- Look up the solubility constant of AgCl. Together with the total initial concentration of $\ce{Ag+}$ and $\ce{Cl-}$, you can then calculate the equilibrium concentrations.
- You'll see that the precipitation moves the reaction towards equilibrium (unless the concentrations in the aqeous solution are not extremely low).
Here are two experiments that you can do to check that it is really a **dynamic** equilibrium (though much on the side of the precipitate)
$$\ce{Ag+_(aq) + Cl(aq)- <=>> AgCl_(s)}$$
1. If you disturb the equilibrium by removing either $\ce{Ag+}$ or $\ce{Cl-}$, the AgCl will dissolve again.
<br/>
Here are some ideas:
- Complex (mask) the $\ce{Ag+}$, e.g. by $\ce{NH3}$ or $\ce{CN-}$<br/>have you ever done this? If not, you should really try. It is really fast.
- reduce the $\ce{Ag+}$
- oxidize the Cl-, so that $\ce{Cl2 ^}$ is removed out of the system.
<br/>
(you could also remove the AgCl, but that isn't practical, as you'd never be sure whether you observe formation of more precipitate or whether you didn't remove all of the AgCl)
2. Crystal growth in saturated solutions. You may have to wait quite a while for this one, though...
|
In “periodic table”, the adjective is related to the noun *period*, and [comes from](http://www.etymonline.com/index.php?term=periodic&allowed_in_frame=0) Ancient Greek *περίοδος* through French *périodique*. In “periodic acid”, it is formed from the prefix *per-* and *iodic* (like *peroxide* and *permanganate*).
[Wiktionary][1] lists their respective UK pronunciations of as `/pɪə(ɹ).iˈɒdɪk/` and `/ˌpɜːraɪˈɒdɪk/`, markedly different: `pɪə` (as in *piece*) vs. `pɜː` (as in *perfume*); then `i` (as in *it*) vs. `aɪ` (as in *eye*).
However, is that distinction really made in practice? Would a native US/UK/Aussie speaker make the difference when talking in the lab?
[1]: http://en.wiktionary.org/wiki/periodic#Etymology_1 |
> Ag is higher up in the Reactivity serie than Al but that do not make sense to me in this problem
- indeed this is **no** redox problem, oxidation states do not change.
- instead it is a phase transition equilibrium
$$\ce{Ag+ + Cl- <=> AgCl v}$$
As you marked this question as homework:
- Edit the question and put the oxidation numbers to confirm that it is not a redox reaction.
- Look up the solubility constant of AgCl. Together with the total initial concentration of $\ce{Ag+}$ and $\ce{Cl-}$, you can then calculate the equilibrium concentrations.
- You'll see that the precipitation moves the reaction towards equilibrium (unless the concentrations in the aqeous solution are not extremely low).
Here are two experiments that you can do to check that it is really a **dynamic** equilibrium (though much on the side of the precipitate)
$$\ce{Ag+_(aq) + Cl(aq)- <=>> AgCl_(s)}$$
1. If you disturb the equilibrium by removing either $\ce{Ag+}$ or $\ce{Cl-}$, the AgCl will dissolve again.
<br/>
Here are some ideas:
- Complex (mask) the $\ce{Ag+}$, e.g. by $\ce{NH3}$ or $\ce{CN-}$<br/>have you ever done this? If not, you should really try. It is really fast.
- reduce the $\ce{Ag+}$
- oxidize the Cl-, so that $\ce{Cl2 ^}$ is removed out of the system.
<br/>
(you could also remove the AgCl, but that isn't practical, as you'd never be sure whether you observe formation of more precipitate or whether you didn't remove all of the AgCl)
2. Crystal growth in saturated solutions. You may have to wait quite a while for this one, though...
|
> Ag is higher up in the Reactivity serie than Al but that do not make sense to me in this problem
- indeed this is **no** redox problem, oxidation states do not change.
- instead it is a phase transition equilibrium
$$\ce{Ag+ + Cl- <=> AgCl v}$$
As you marked this question as homework:
- Edit the question and put the oxidation numbers to confirm that it is not a redox reaction.
- Look up the solubility constant of AgCl. Together with the total initial concentration of $\ce{Ag+}$ and $\ce{Cl-}$, you can then calculate the equilibrium concentrations.
- You'll see that the precipitation moves the reaction towards equilibrium (unless the concentrations in the aqeous solution are not extremely low).
Here are two experiments that you can do to check that it is really a **dynamic** equilibrium (though much on the side of the precipitate)
$$\ce{Ag+_(aq) + Cl(aq)- <=>> AgCl_(s)}$$
1. If you disturb the equilibrium by removing either $\ce{Ag+}$ or $\ce{Cl-}$, the AgCl will dissolve again.
<br/>
Here are some ideas:
- Complex (mask) the $\ce{Ag+}$, e.g. by $\ce{NH3}$ or $\ce{CN-}$<br/>have you ever done this? If not, you should really try. It is really fast.
- reduce the $\ce{Ag+}$
- oxidize the Cl-, so that $\ce{Cl2 ^}$ is removed out of the system.
<br/>
(you could also remove the AgCl, but that isn't practical, as you'd never be sure whether you observe formation of more precipitate or whether you didn't remove all of the AgCl)
2. Crystal growth in saturated solutions. You may have to wait quite a while for this one, though...
----------
F'x pointed us to a paper [Paradigms and Paradoxes: The Solubility of AgCl in Water](http://www.springerlink.com/content/g455t63482504217/). As it is behind a paywall, I'll try to get a few points out of it.
First of all, it's actually kind of an overdue homework question:
> It is well established in the pedagogical literature that AgCl is insoluble in water while NaCl and
KCl are soluble [...] What is usually left unsaid, however, is why AgCl is so much less soluble than NaCl or KCl.
Here are some interesting points for the question:
- Liebman concludes from thermodynamic data (and contrary to HSAB expectations) that $\ce{Ag+}$ is *better* solvatized than $\ce{Na+}$ or $\ce{K+}$
- Lattice energy for AgCl ($\ce{Ag_{(g)}^+ + Cl_{(g)}^- -> AgCl_{(s)}}$) is more negative than for NaCl and KCl.
- The paper estimates that the enthalpy of $\ce{AgCl_{(g)} -> AgCl_{(s)}}$ is comparable to that for the alkali chlorides.
- Liebman concludes from that, that the AgCl crystal is nothing special but is rather similar to NaCl and KCl
- So the point that is different is that the formation of diatomic AgCl is much more favorable than formation of diatomic NaCl or KCl.
*My personal conclusions:* It's a really though question. I'll stick to the experiment with measurable properies such as equilibrium constants. ;-)
|
> Ag is higher up in the Reactivity serie than Al but that do not make sense to me in this problem
- indeed this is **no** redox problem, oxidation states do not change.
- instead it is a phase transition equilibrium
$$\ce{Ag+ + Cl- <=> AgCl v}$$
As you marked this question as homework:
- Edit the question and put the oxidation numbers to confirm that it is not a redox reaction.
- Look up the solubility constant of AgCl. Together with the total initial concentration of $\ce{Ag+}$ and $\ce{Cl-}$, you can then calculate the equilibrium concentrations.
- You'll see that the precipitation moves the reaction towards equilibrium (unless the concentrations in the aqeous solution are not extremely low).
Here are two experiments that you can do to check that it is really a **dynamic** equilibrium (though much on the side of the precipitate)
$$\ce{Ag+_(aq) + Cl(aq)- <=>> AgCl_(s)}$$
1. If you disturb the equilibrium by removing either $\ce{Ag+}$ or $\ce{Cl-}$, the AgCl will dissolve again.
<br/>
Here are some ideas:
- Complex (mask) the $\ce{Ag+}$, e.g. by $\ce{NH3}$ or $\ce{CN-}$<br/>have you ever done this? If not, you should really try. It is really fast.
- reduce the $\ce{Ag+}$
- oxidize the Cl-, so that $\ce{Cl2 ^}$ is removed out of the system.
<br/>
(you could also remove the AgCl, but that isn't practical, as you'd never be sure whether you observe formation of more precipitate or whether you didn't remove all of the AgCl)
2. Crystal growth in saturated solutions. You may have to wait quite a while for this one, though...
----------
F'x pointed us to a paper [Paradigms and Paradoxes: The Solubility of AgCl in Water](http://www.springerlink.com/content/g455t63482504217/). As it is behind a paywall, I'll try to summarize a few points out of it.
First of all, it's actually kind of an overdue homework question:
> It is well established in the pedagogical literature that AgCl is insoluble in water while NaCl and
KCl are soluble [...] What is usually left unsaid, however, is why AgCl is so much less soluble than NaCl or KCl.
Here are some interesting points for the question:
- Liebman concludes from thermodynamic data (and contrary to HSAB expectations) that $\ce{Ag+}$ is *better* solvatized than $\ce{Na+}$ or $\ce{K+}$
- Lattice energy for AgCl ($\ce{Ag_{(g)}^+ + Cl_{(g)}^- -> AgCl_{(s)}}$) is more negative than for NaCl and KCl.
- The paper estimates that the enthalpy of $\ce{AgCl_{(g)} -> AgCl_{(s)}}$ is comparable to that for the alkali chlorides.
- Liebman concludes from that, that the AgCl crystal is nothing special but is rather similar to NaCl and KCl
- So the point that is different is that the formation of diatomic AgCl is much more favorable than formation of diatomic NaCl or KCl.
*My personal conclusions:* It's a really though question. I'll stick to the experiment with measurable properies such as equilibrium constants. ;-)
That is, insolubility of AgCl is no surprise to me: after all, I'm used to the *fact* for years. And given the insolubility of AgBr and AgI, I'd guesstimate unsoluble even though AgF is soluble (after all $\ce{F⁻}$ is usually different from the other 3). But I find the particular situation far away from any textbook kind of reasoning that should introduce general chemical concepts to students. Unless the message is supposed to be that reality is complicated... (which OTOH allows for the beautiy of reality and makes it interesting). |
> Ag is higher up in the Reactivity serie than Al but that do not make sense to me in this problem
- indeed this is **no** redox problem, oxidation states do not change.
- instead it is a phase transition equilibrium
$$\ce{Ag+ + Cl- <=> AgCl v}$$
As you marked this question as homework:
- Edit the question and put the oxidation numbers to confirm that it is not a redox reaction.
- Look up the solubility constant of AgCl. Together with the total initial concentration of $\ce{Ag+}$ and $\ce{Cl-}$, you can then calculate the equilibrium concentrations.
- You'll see that the precipitation moves the reaction towards equilibrium (unless the concentrations in the aqeous solution are not extremely low).
Here are two experiments that you can do to check that it is really a **dynamic** equilibrium (though much on the side of the precipitate)
$$\ce{Ag+_(aq) + Cl(aq)- <=>> AgCl_(s)}$$
1. If you disturb the equilibrium by removing either $\ce{Ag+}$ or $\ce{Cl-}$, the AgCl will dissolve again.
<br/>
Here are some ideas:
- Complex (mask) the $\ce{Ag+}$, e.g. by $\ce{NH3}$ or $\ce{CN-}$<br/>have you ever done this? If not, you should really try. It is really fast.
- reduce the $\ce{Ag+}$
- oxidize the Cl-, so that $\ce{Cl2 ^}$ is removed out of the system.
<br/>
(you could also remove the AgCl, but that isn't practical, as you'd never be sure whether you observe formation of more precipitate or whether you didn't remove all of the AgCl)
2. Crystal growth in saturated solutions. You may have to wait quite a while for this one, though...
----------
F'x pointed us to a paper [Paradigms and Paradoxes: The Solubility of AgCl in Water](http://www.springerlink.com/content/g455t63482504217/). As it is behind a paywall, I'll try to summarize a few points out of it.
First of all, it's actually kind of an overdue homework question:
> It is well established in the pedagogical literature that AgCl is insoluble in water while NaCl and
KCl are soluble [...] What is usually left unsaid, however, is why AgCl is so much less soluble than NaCl or KCl.
Here are some interesting points for the question:
- Liebman concludes from thermodynamic data (and contrary to HSAB expectations) that $\ce{Ag+}$ is *better* solvated than $\ce{Na+}$ or $\ce{K+}$
- Lattice energy for AgCl ($\ce{Ag_{(g)}^+ + Cl_{(g)}^- -> AgCl_{(s)}}$) is more negative than for NaCl and KCl.
- The paper estimates that the enthalpy of $\ce{AgCl_{(g)} -> AgCl_{(s)}}$ is comparable to that for the alkali chlorides.
- Liebman concludes from that, that the AgCl crystal is nothing special but is rather similar to NaCl and KCl
- So the point that is different is that the formation of diatomic AgCl is much more favorable than formation of diatomic NaCl or KCl.
*My personal conclusions:* It's a really though question. I'll stick to the experiment with measurable properies such as equilibrium constants. ;-)
That is, insolubility of AgCl is no surprise to me: after all, I'm used to the *fact* for years. And given the insolubility of AgBr and AgI, I'd guesstimate unsoluble even though AgF is soluble (after all $\ce{F⁻}$ is usually different from the other 3). But I find the particular situation far away from any textbook kind of reasoning that should introduce general chemical concepts to students. Unless the message is supposed to be that reality is complicated... (which OTOH allows for the beautiy of reality and makes it interesting). |
Please take a look at the following video:
http://www.youtube.com/watch?feature=player_embedded&v=f95bNFVZkh0
I am working on a new project, and i need to find whats the best liquid to hold ferrofluid inside the glass (or maybe even plastic) container, so that ferrofluid (that easily stains everything) does not stain or stick to glass?
also, is there any special preparation for the glass needed?
If you had experience with this please advise.
|
Please take a look at the following [video][1]:
[![Ferrofluid][2]](http://www.youtube.com/watch?feature=player_embedded&v=f95bNFVZkh0)
I am working on a new project, and i need to find whats the best liquid to hold ferrofluid inside the glass (or maybe even plastic) container, so that ferrofluid (that easily stains everything) does not stain or stick to glass?
Also, is there any special preparation for the glass needed?
[1]: http://www.youtube.com/watch?feature=player_embedded&v=f95bNFVZkh0
[2]: https://i.stack.imgur.com/kV1dL.png |
Please take a look at the following [video][1]:
[![Ferrofluid][2]](http://www.youtube.com/watch?feature=player_embedded&v=f95bNFVZkh0)
I am working on a new project, and I need to find whats the best liquid to hold ferrofluid inside the glass (or maybe even plastic) container, so that ferrofluid (that easily stains everything) does not stain or stick to glass?
Also, is there any special preparation for the glass needed?
[1]: http://www.youtube.com/watch?feature=player_embedded&v=f95bNFVZkh0
[2]: https://i.stack.imgur.com/kV1dL.png |
> Related: http://chemistry.stackexchange.com/q/383/22
Experimentally, $\ce{AgCl}$ is insoluble in water, but $\ce{AgNO3}$ is soluble. They're pretty common in a lab (well, $\ce{AgCl}$ is a common precipitate)--so I think most of us know this.
By [Fajan's rules][1], on the other hand, larger anion $\implies$ more polarization/covalent character $\implies$ less solubility. But, $\ce{NO3-}$ is the larger anion, yet $\ce{AgNO3}$ is more soluble.
Is there any theoretical reason for this?
[1]: http://en.wikipedia.org/wiki/Fajans%27_rules |
Why is $\ce{AgCl}$ less soluble than $\ce{AgNO3}$? |
[Hematite](http://en.wikipedia.org/wiki/Hematite) is composed of $\ce{Fe2O3}$, and is paramagnetic, whereas [magnetite](http://en.wikipedia.org/wiki/Magnetite) is $\ce{Fe3O4}$ and is diamagnetic. Magnetite's nature is due to the presence of both $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$ (Wikipedia even goes so far as to call it $\ce{FeO}\cdot \ce{Fe2O3}$, but explains that this is not a "solid solution").
I can surmise that in the crystalline solid, there are divalent and trivalent cations dispersed within. I can also assume that there might be some sharing of electrons between the oxygens and each of these cations. One would assume this also happens within the hematite crystal, so what is it about "tossing" the divalent cation into the mix of the crystalline structure that makes the magnetite diamagnetic?
Does the charge differential between the two cations cause a permanent dipole, if so, why don't all of the small dipoles simply cancel each other out? |
What is it about the relationship between the $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$ in magnetite that makes it diamagnetic? |
I was told that "Sodium Tallowate, sodium cocoate and sodium palm kernelate are the 'natural' forms of [SLS][1]". Is this true? I thought "natural" SLS would be SLS derived from a natural product such as coconut oil.
[1]: http://en.wikipedia.org/wiki/Sodium_dodecyl_sulfate |
Is soap the natural form of Sodium Lauryl Sulfate? |
Dishwashing agents are usually anionic detergents—i.e. they are alkylbenzenesulfonates.
They consist of two parts:
1. Sulfonate part which is hydrophilic(water loving i.e. dissolves in water)
2. Alkylbenzene part which is lipophilic(fat loving i.e dissolves in fats and oils)
Thus, when in water, they dissociate to give $\ce{RSO3-}$ and $\ce{Na+}$ . $\ce{Na+}$ is from a strong base (making it a weak conjugate acid), while $\ce{RSO3-}$ is from a weak acid (making it a weak conjugate base), thus the resulting solution is **basic**.
The sulphonate ion bonds to water, while alkylbenzene bonds to oil, protein, or fat . Thus fats, oils, proteins, and grease dissolve in water
The following is the mechnism of a dishwashing agent:
![enter image description here][1]
These are the ingreadients mentioned on a dishwashing bottle in my house:
[![enter image description here][2]][3]
As you can see, sodium dodecylbenzenesulphonate (an alkylbenzenesulfonate) is one of the main ingredients.
----------
>Do acid make good cleaning solutions as well?
Yes they can make a good cleaning solution, but not as a dishwashing agent.
According to [Wikipedia][4] ,
>Acidic washing agents are mainly used for removal of inorganic deposits like scaling. The active ingredients are normally strong mineral acids and chelants. Often, there are added surfactants and corrosion inhibitors.
[1]: https://i.stack.imgur.com/j30zD.jpg
[2]: https://i.stack.imgur.com/V3MWQm.png
[3]: https://i.stack.imgur.com/V3MWQm.png
[4]: http://en.wikipedia.org/wiki/Cleaning_agent#Acidic |
(-)-sparteine, in my experience, decomposes relatively quickly in ambient conditions. What is the reaction taking place to cause it to decompose? |
In [a series of papers in the early 1980s](http://www.springerlink.com/content/g83871qw83n26152/abstract/), Michael Springborg explored an interpretation of the [Wigner phase space function](http://en.wikipedia.org/wiki/Wigner_quasi-probability_distribution) as an electron density in a six-dimensional $(q,p)$ phase space. He applied it with some success to several simple compounds.
Is the Springborg 6D phase model model used in modern molecular orbital modeling? If not, are there specific weaknesses that make it unsuitable for current computational approaches?
-----
ADDENDUM 2012-05-14:
In a world dominated by Hilbert spaces, the orbitals of chemistry are interesting because they are one of the few places in physics where a 3D spatial image of the underlying quantum situation is consistently retained. (My recollection is that Schrodinger liked such low-dimensionality approaches, since they emphasized waves that could be visualized.)
Alas, the problem of course is that you need both position $q$ and momentum $p$ to get an accurate picture of the overall electron state. Using a space-only 3D $q$ representation forces the analysis of molecules to flip back and forth at rather arbitrary thresholds between localized (position space $q$) and delocalized (momentum space $p$) views of the electrons in the molecules or metal crystals in question.
That bothers me, both from a physics perspective and from a computational integrity perspective.
The problem is that there exists a very deep and profound physics symmetry between $q$ and $p$ in terms of how effects such as Pauli exclusion applies to fermions. I find that simple symmetry nothing short of amazing, because to me there's just nothing intuitive about the idea that half-unit spin would result in such extraordinarily similar behaviors in these quite different spaces. At some level it seems to be one of those cases of "that's just the way it works." So, it seems pretty reasonable that accurate modeling of orbitals of many diverse sizes likely requires this symmetry to be captured accurately.
If that's true, this older work by Springborg (he still heads a research group BTW) strikes me as a possible opportunity to for taming some models, making them smoother and more sane over a much broader range of molecular sizes. That's because Springborg's 6D electron density functions -- if they work OK in broader contexts -- might help eliminate any need for abrupt switches between "mostly position" or $q$ 3D views (classic "electron clouds," whatever those really mean) and "mostly momentum" or $p$ 3D views.
A good place to test whether 6D $(q,p)$ electron distributions might provide higher computational stability would be in modeling a range of lengths for a long-chain carbon polymer such as polyacetylene. (I always pick polyacetylene for its simplicity, but there are other good choices.) Springborg appears not to have explored that domain, since as best I can tell his papers addressed mostly simpler bonds. (I think; I don't have them.) The [instabilities of approximate density functional theories](http://chemistry.stackexchange.com/a/365/83) _may_ (or may not) be related to $(q,p)$ transition instabilities. That's not a question I can or should try to address, as there are folks out there who know that domain orders of magnitude better than I ever will.
-----
ADDENDUM 2012-05-16
I mentioned "algorithmic opportunity" and wondered what I meant myself by that, since as @AcidFlask reminds the $q$ and $p$ basis sets are fully interchangeable in terms of representing any stationary wavefunction.
The issue is (again, I think) one of accumulating uncertainty. Picture doing some kind of calculation in the $q$ representation. For a highly localized electron, you should get a pretty solid answer, but for something like a delocalized electron in a conductive polymer, a higher level of error would be likely (not a necessity).
Next, Fourier transform the result into $p$, and do the next iteration of whatever it is you are trying to calculate. There the nature of the errors should change, but in the case of a delocalized electron, it's likely to work out a bit more accurately.
You reverse transform that and repeat, with the objective of converging to as solution that looks pretty stable in both spaces. That, I would argue, is the "algorithmic opportunity" part of the 6D argument: that is, that by incorporating both views, you can come to an _overall_ solution that diverges less from reality when applied to a broad range of molecular orbital sizes.
Are there problems with that analysis? Oh my yes. You'd have to be nuts to do complete Fourier transforms twice per full iteration of your convergence, and even then, the devil will be in the details about whether it even _works_.
But the operative word was "opportunity," that is, such an analysis hints that there _may_ be away to use the dual perspective to create more stable approximation methods. To really work the algorithm would necessarily "live" a lot closer to a continuous 6D representation with constraints than a per-iteration transform back and forth. Can that be done? I don't know; as I said, folks like AcidFlux know this area orders of magnitude better than I ever will.
And finally, for an non-mathematical readers (AcidFlux, shoo, go away now!) who may have wandered in: Imagine a very flexible membrane with many embedded holes (but the holes don't stretch). Label one side of the membrane "ordinary space" (abbreviated $q$, don't ask why), and the other side "momentum space" (labeled $p$, and I told you not to ask). Next, put water balloons with fixed quantities of water in them through each hole, glued to the holes.
The $q$ side of each balloon now represents the "real space" representation of the water balloon. If you squeeze that part very tightly, it will shrink down almost to a point, so it looks more like a particle than a balloon. You think you have "captured" that balloon n a small space, but beware! On the other side, the $p$ side of the membrane, the balloon has gotten comparatively huge! So your point-like "capture" of the balloon on the $q$ side was more illusion than reality, since in the broader view that includes both sides, all yo really did is shift a lot of the balloon from the $q$ world into the hidden $p$ side.
Note also that you don't really _need_ to see both sides at once, because each side of the balloon fully determines the size of the other side. From the amount of water on one side you can _always_ calculate exactly how much water is in the balloon on the other, hidden side. And the opposite is also true! That is, if you know how big each balloon is on the $p$ side, you also automatically know how much of the balloon is on the $q$ side.
That kind of equivalence between two views that may appear at first to be quite different is an example a "basis set." A basis set has enough "pieces" in its tool kit to allow you to represent something _completely_, in the sense that you have all of the information you will ever need to know about the system.
Alas, complete is not always the same as convenient. For example, if you try to stack together the balloons as they appear on the $q$ side, you will quickly find that the hidden $p$ side of the balloons can make it very difficult. The balloons just won't pack as easily together as their forms on the $q$ side seem to suggest they would. The problem, of course, is that you have those huge $p$ sides of the balloons interacting in a very different way over in the $p$ or "momentum space" side of the membrane.
It's not that any information is missing! After all, since the $q$ side provides a complete basis set for describing the balloons, you _know_ that when a balloon is small there that it must necessarily be large in $p$. The trick, then, is make sure that you don't have to recalculate the size on the balloons on the other side every single time you try to do something with them on the $q$ side. That's inefficient at best, and could easily turn into a significant source of errors.
So, it's a bit simpler and certainly a lot less calculation if you preserve a simultaneous image of _both_ sides of the membrane, and change them only when you have to, e.g. after squeezing some of the balloons on either (or both) the $q$ and $p$ sides. Doing that is inherently redundant, sure, since you only _need_ to have one side to recreate the other.
However, if you've ever done some serious algorithm design, you know that "caching" mostly persistent results can be a great way to reduce calculation times, sometimes hugely. More subtly, it can also help stabilize the problem if it turns out that you have complementary error modes across your different "views" (basis sets) of the final result.
So, put that all together, and that's what I'm trying to suggest _may_ also be true for the physics of electrons, which similarly like to slosh around (as in Planck's Theory of Slosh," and yes _of course_ I just now made that up!) and spend some of their time point-like in one space (called $q$ or position space; ordinary, real space), or alternatively in another more abstract space (called $p$ or momentum space). Interestingly, just as with the balloons, they bump into and push each each other in almost exactly the same way in _both_ of these two spaces. Particles that do that are called fermions.
So why are fermions so turf-protective in _both_ of these quite different spaces? I really don't think anyone really knows, at least not with the kind of depth that is as satisfying as when chemistry started coming together and making deep sense. It's just one of those little mysteries of how the universe works, one (there are several actually) that's still sitting there thumbing its nose at us and saying "ha ha, gotcha!"
And that's cool! What's a universe without a few mysteries still hanging around, after all? |
By what reaction does (–)-sparteine decompose in ambient conditions? |
Lithium and magnesium are Group 1 and Group 2 elements respectively. Elements of these groups are highly ionic, and I've never heard of them forming significantly covalent _inorganic_ compounds.
Yet these elements form a variety of organometallic compounds ($\ce{PhLi}$, the whole family of Grignard reagents, etc). Organometallic compounds have significant covalent character(i.e., the bond can be called covalent) in the carbon-metal bond.
What's so special about carbon that makes these elements form covalent bonds?
|
Why do Magnesium and Lithium form *covalent* organometallic compounds? |
I've been asked this question a few times, and while I _think_ I know the answer, I'd like to know more.
Graphite, as we know, is a sheet polymer. Since polymers are bound to be finite by physical considerations, graphite must have "edges". My question is, what happens at these edges?
I can cook up these possibilities:
- A graphene sheet "folds back" on itself, thus forming a tube and taking care of _most_ of the edge carbons. But this would make graphite less slippery, so I doubt this is the case.
- Random elements/compounds from the environment at time of synthesis latch on to the edges, taking care of the valency of Carbon
- The edge carbons form double/triple bonds amongst themselves
I'd like to know more about this. I'm also curious what happens to the resonance of graphene at the edges. |
Oxygen is a rather boring element. It has only two allotropes, Oxygen and Ozone. Oxygen has a double bond, and Ozone has a delocalised cloud, giving rise to two "1.5 bonds".
On the other hand, Sulphur has many stable allotropes, and a bunch of unstable ones as well. The variety of allotropes, is mainly due to the ability of Sulphur to catenate.
But, Sulphur does not have a stable diatomic allotrope at room temperature. I, personally would expect diatomic sulphur to be _more_ stable than diatomic Oxygen, due to the possibility of $p\pi -d\pi$ back-bonding.
So, why do Sulphur and Oxygen have such opposite properties with respect to their ability to catenate?
|
Why does Sulphur, but not Oxygen, catenate? |
In [a series of papers in the early 1980s](http://www.springerlink.com/content/g83871qw83n26152/abstract/), Michael Springborg explored an interpretation of the [Wigner phase space function](http://en.wikipedia.org/wiki/Wigner_quasi-probability_distribution) as an electron density in a six-dimensional $(q,p)$ phase space. He applied it with some success to several simple compounds.
Is the Springborg 6D phase model model used in modern molecular orbital modeling? If not, are there specific weaknesses that make it unsuitable for current computational approaches?
-----
ADDENDUM 2012-05-14:
In a world dominated by Hilbert spaces, the orbitals of chemistry are interesting because they are one of the few places in physics where a 3D spatial image of the underlying quantum situation is consistently retained. (My recollection is that Schrodinger liked such low-dimensionality approaches, since they emphasized waves that could be visualized.)
Alas, the problem of course is that you need both position $q$ and momentum $p$ to get an accurate picture of the overall electron state. Using a space-only 3D $q$ representation forces the analysis of molecules to flip back and forth at rather arbitrary thresholds between localized (position space $q$) and delocalized (momentum space $p$) views of the electrons in the molecules or metal crystals in question.
That bothers me, both from a physics perspective and from a computational integrity perspective.
The problem is that there exists a very deep and profound physics symmetry between $q$ and $p$ in terms of how effects such as Pauli exclusion applies to fermions. I find that simple symmetry nothing short of amazing, because to me there's just nothing intuitive about the idea that half-unit spin would result in such extraordinarily similar behaviors in these quite different spaces. At some level it seems to be one of those cases of "that's just the way it works." So, it seems pretty reasonable that accurate modeling of orbitals of many diverse sizes likely requires this symmetry to be captured accurately.
If that's true, this older work by Springborg (he still heads a research group BTW) strikes me as a possible opportunity to for taming some models, making them smoother and more sane over a much broader range of molecular sizes. That's because Springborg's 6D electron density functions -- if they work OK in broader contexts -- might help eliminate any need for abrupt switches between "mostly position" or $q$ 3D views (classic "electron clouds," whatever those really mean) and "mostly momentum" or $p$ 3D views.
A good place to test whether 6D $(q,p)$ electron distributions might provide higher computational stability would be in modeling a range of lengths for a long-chain carbon polymer such as polyacetylene. (I always pick polyacetylene for its simplicity, but there are other good choices.) Springborg appears not to have explored that domain, since as best I can tell his papers addressed mostly simpler bonds. (I think; I don't have them.) The [instabilities of approximate density functional theories](http://chemistry.stackexchange.com/a/365/83) _may_ (or may not) be related to $(q,p)$ transition instabilities. That's not a question I can or should try to address, as there are folks out there who know that domain orders of magnitude better than I ever will.
-----
ADDENDUM 2012-05-16
I mentioned "algorithmic opportunity" and wondered what I meant myself by that, since as @AcidFlask reminds the $q$ and $p$ basis sets are fully interchangeable in terms of representing any stationary wavefunction.
The issue is (again, I think) one of accumulating uncertainty. Picture doing some kind of calculation in the $q$ representation. For a highly localized electron, you should get a pretty solid answer, but for something like a delocalized electron in a conductive polymer, a higher level of error would be likely (not a necessity).
Next, Fourier transform the result into $p$, and do the next iteration of whatever it is you are trying to calculate. There the nature of the errors should change, but in the case of a delocalized electron, it's likely to work out a bit more accurately.
You reverse transform that and repeat, with the objective of converging to a solution that looks pretty stable in both spaces. That, I would argue, is the "algorithmic opportunity" part of the 6D argument: that is, that by incorporating both views, you can come to an _overall_ solution that diverges less from reality when applied to a broad range of molecular orbital sizes.
Are there problems with that analysis? Oh my yes. You'd have to be nuts to do complete Fourier transforms twice per full iteration of your convergence, and even then, the devil will be in the details about whether it even _works_.
But the operative word was "opportunity," that is, such an analysis hints that there _may_ be away to use the dual perspective to create more stable approximation methods. To really work the algorithm would necessarily "live" a lot closer to a continuous 6D representation with constraints than a per-iteration transform back and forth. Can that be done? I don't know; as I said, folks like AcidFlux know this area orders of magnitude better than I ever will.
And finally, for any non-mathematical readers (AcidFlux, shoo, go away now!) who may have wandered in, the section below provides a simple, easy-to-imagine visual analogy to explain why location and momentum both play a major role at the level of molecules.
Imagine a very flexible membrane with many embedded holes (but the holes don't stretch). Label one side of the membrane "ordinary space" (abbreviated $q$, don't ask why), and the other side "momentum space" (labeled $p$, and I told you not to ask). Next, put water balloons with fixed quantities of water in them through each hole, glued to the holes.
The $q$ side of each balloon now represents the "real space" representation of the water balloon. If you squeeze that part very tightly, it will shrink down almost to a point, so it looks more like a particle than a balloon. You think you have "captured" that balloon n a small space, but beware! On the other side, the $p$ side of the membrane, the balloon has gotten comparatively huge! So your point-like "capture" of the balloon on the $q$ side was more illusion than reality, since in the broader view that includes both sides, all you really did is shift a lot of the balloon from the $q$ world into the hidden $p$ side.
Note also that you don't really _need_ to see both sides at once, because each side of the balloon fully determines the size of the other side. From the amount of water on one side you can _always_ calculate exactly how much water is in the balloon on the other, hidden side. And the opposite is also true! That is, if you know how big each balloon is on the $p$ side, you also automatically know how much of the balloon is on the $q$ side.
That kind of equivalence between two views that may appear at first to be quite different is an example a "basis set." A basis set has enough "pieces" in its tool kit to allow you to represent something _completely_, in the sense that you have all of the information you will ever need to know about the system.
Alas, complete is not always the same as convenient. For example, if you try to stack together the balloons as they appear on the $q$ side, you will quickly find that the hidden $p$ side of the balloons can make it very difficult. The balloons just won't pack as easily together as their forms on the $q$ side seem to suggest they would. The problem, of course, is that you have those huge $p$ sides of the balloons interacting in a very different way over in the $p$ or "momentum space" side of the membrane.
It's not that any information is missing! After all, since the $q$ side provides a complete basis set for describing the balloons, you _know_ that when a balloon is small there that it must necessarily be large in $p$. The trick, then, is make sure that you don't have to recalculate the size on the balloons on the other side every single time you try to do something with them on the $q$ side. That's inefficient at best, and could easily turn into a significant source of errors.
So, it's a bit simpler and certainly a lot less calculation if you preserve a simultaneous image of _both_ sides of the membrane, and change them only when you have to, e.g. after squeezing some of the balloons on either (or both) the $q$ and $p$ sides. Doing that is inherently redundant, sure, since you only _need_ to have one side to recreate the other.
However, if you've ever done some serious algorithm design, you know that "caching" mostly persistent results can be a great way to reduce calculation times, sometimes hugely. More subtly, it can also help stabilize the problem if it turns out that you have complementary error modes across your different "views" (basis sets) of the final result.
So, put that all together, and that's what I'm trying to suggest _may_ also be true for the physics of electrons, which similarly like to slosh around (as in Planck's Theory of Slosh," and yes _of course_ I just now made that up!) and spend some of their time point-like in one space (called $q$ or position space; ordinary, real space), or alternatively in another more abstract space (called $p$ or momentum space). Interestingly, just as with the balloons, they bump into and push each each other in almost exactly the same way in _both_ of these two spaces. Particles that do that are called fermions.
So why are fermions so turf-protective in _both_ of these quite different spaces? I really don't think anyone really knows, at least not with the kind of depth that is as satisfying as when chemistry started coming together and making deep sense. It's just one of those little mysteries of how the universe works, one (there are several actually) that's still sitting there thumbing its nose at us and saying "ha ha, gotcha!"
And that's cool! What's a universe without a few mysteries still hanging around, after all? |
Lithium and magnesium are Group 1 and Group 2 elements respectively. Elements of these groups are highly ionic, and I've never heard of them forming significantly covalent _inorganic_ compounds.
Yet these elements form a variety of organometallic compounds ($\ce{PhLi}$, the whole family of Grignard reagents, etc). Organometallic compounds have significant covalent character (i.e., the bond can be called covalent) in the carbon–metal bond.
What's so special about carbon that makes these elements form covalent bonds?
|
First, a note: while oxygen has fewer allotropes than sulfur, it sure has [more than two][1]! These include O, O<sub>2</sub>, <sup>1</sup>O<sub>2</sub>, O<sub>3</sub>, O<sub>4</sub>, O<sub>8</sub>, metallic O and four other solid phases. Many of these actually have a corresponding sulfur variant. However, you are right in a sense that sulfur has more tendency to catenate… let's try to see why!
Here are the values of the single and double bond enthalpies ([source](http://faculty.ncc.edu/LinkClick.aspx?fileticket=L2sV5SDzIhE%3D&tabid=1871)):
O—O 142 kJ/mol
S–S 268 kJ/mol
O=O 499 kJ/mol
S=S 352 kJ/mol
This means that O=O is stronger than S=S, while O–O is weaker than S–S. So, in sulfur, single bonds are favoured and catenation is easier than in oxygen compounds.
It seems that the reason for the weaker S=S double bonds has its roots in the size of the atom: it's harder for the two atoms to come at a small enough distance, so that the $p$ orbitals overlap is small and the $\pi$ bond is weak. This is attested by looking down the periodic table: Se=Se has an even weaker bond enthalpy of 272 kJ/mol ([source](http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html)).
[1]: http://en.wikipedia.org/wiki/Allotropes_of_oxygen |
To analyse such situations, you first must realize that nearly everything is present as ions. For example, if you took the reaction $\ce{NaCl +KBr->NaBr +KCl}$, the "direction" it goes in has no real meaning since $\ce{Na+}$, $\ce{Cl-}$, $\ce{K+}$, $\ce{Br-}$ are present as ions. There is no actual _reaction_ going on here, just a mixing of ions.
In this case, we have something slightly different. $\ce{AgCl}$ precipitates here:
$$\ce{3AgNO3_{(aq)} + AlCl3_{(aq)} -> Al(NO3)3_{(aq)} + 3AgCl v}$$
Similar to the $\ce{NaCl}$ case above, $\ce{NO3-}$ and $\ce{Al^{3+}}$ are present as ions on both sides--so nothing to worry about there. We can safely remove them from the equation. This reduces the reaction to:
$$\ce{3Ag+ + 3Cl- -> 3AgCl v}$$
Now, we have reduced the question to "why does $\ce{AgCl}$ precipitate?". The easiest way to decide is by looking at thermodynamic stability--solid silver chloride is more stable. We can experimentally prove this, or use standard enthalpies.
Reactions proceed in the direction of more stability, so the above reaction goes forward.
---------
Of course, I sort of "assumed" that $\ce{AgCl}$ precipitates in the beginning--this was so that I wouldn't have to complicate stuff. Nevertheless, we can, in a similar way, experimentally prove that the others are found in ionized form and generally do not precipitate. There are some nice arguments why $\ce{AgCl}$ is less soluble than $\ce{AgNO3}$ [in this post][1].
[1]: http://chemistry.stackexchange.com/questions/410/why-is-ceagcl-less-soluble-than-ceagno3 |
> Is dissolving always exothermic?
No, it is not always exothermic. For example, dissolving $\ce{NaOH}$ is an exothermic reaction, while dissolving $\ce{NH4NO3}$ is an endothermic reaction (This is from my personal experience)
> Is melting always endothermic?
Yes, it always is (except for the exception F'x gave). This is because during melting the body to be melted takes in "latent heat of fusion" from the surroundings, thus bringing down the temperature. The latent heat of fusion is required to break the intermolecular attraction forces that define a solid. Breaking such bonds requires energy, hence the reaction is endothermic.
>Can alcohol dissolve ice?
Lower alcohols like methanol and ethanol may be able to dissolve ice. They can effectively form hydrogen-bonds with water, but higher alcohols may not be able to do that. |
Nitrile gloves are made of [**nitrile rubber**][1], or **poly(butadiene/acrylonitrile)**. This polymer is [highly soluble](http://www.polysciences.com/Core/Display.aspx?pageId=98&categoryId=114&productId=454) in chloroform, with some papers I found indicating that one can dissolve up to 18% in mass of nitrile butadiene rubber in chloroform. Moreover, it [permeates easily](http://onlinelibrary.wiley.com/store/10.1002/ceat.200900268/asset/97_ftp.pdf;jsessionid=3685250CB0326943BAD0364311FC7D5D.d04t03?v=1&t=h2bwx35v&s=7690887daf7950a04640bf1898f100a03673ee4b) through NBR, meaning we can expect the dissolution to be fast in addition to thermodynamically
favourable.
Finally, I would not expect the mechanism here to be any different from that of any polymer dissolution by a good solvent. The solvent will permeate through the polymer, intercalate between polymer chains, and solvate them (inducing swelling). Once solvated, the network of polymer chains looses its mechanical properties and they can fully separate. *(I wish I could find a good existing illustration for that part, but I can't right now… If anyone can, feel free to edit!)*
[1]: http://en.wikipedia.org/wiki/Nitrile_rubber |
[Ortho- and parahydrogen][1] are two forms of the $\ce{H2}$ molecule that are distinguished by their pairing or antipairing of nuclear spins, giving rise to metastable singlet (ortho-) and triplet (para-) states. This constitutes an example of nuclear spin isomerism.
Whilst the thermal properties of these two species are different, my understanding is that their chemical properties are essentially identical. Are there any examples of this kind of isomerism having nontrivial implications for chemical reactivity?
[1]: http://en.wikipedia.org/wiki/Spin_isomers_of_hydrogen |
Are there any examples of nuclear spin isomers having consequences for chemical reactivity? |
The qualifier *'free'* seems to be ubiquitously attached to discussion of radicals as highly reactive species with unpaired spins. What, precisely (or imprecisely, as the case may be) does *'free'* really mean?
*<sub>Free as in free speech? Free beer? Freedom of movement? Kinetic accessibility?</sub>* |
What makes a radical 'free'? |
I'm attempting to perform atomistic optimal structure alignment on an inorganic species. Can anyone point me in the right direction to software or a technique for performing this?
There exists a wide range of structural alignment software, but this really appears to be geared towards aligning protein backbones, whereas I'm interested in aligning unlabelled Cartesian coordinates of small molecules. |
How can I perform structure alignment on an inorganic species? |
The [R/S naming system](http://en.wikipedia.org/wiki/Chirality_(chemistry)#Naming_conventions) for stereocenters relies on the [Cahn–Ingold–Prelog priority rules][1] to rank the four substituents of a stereocenter; the R or S name is then attributed depending on the spatial orientation of the ranked substituents. While this is mostly used for asymmetric carbon atoms, it holds true of other stereocenters with four substituents.
However, nitrogen and phosphorus atoms can also act as stereocenters if the inversion of their configuration is blocked (e.g. because they're in a cyclic molecule) or slowed enough (e.g. because of bulky substituents). Apparently, these stereocenters are also labelled in R/S fashion. Examples I found include [Tröger's base][2]:
![Tröger's base][3]
(above is the (5S,11S)-enantiomer, below is the (5R,11R)-enantiomer),
and [DIPAMP][4]:
![DIPAMP][5]
---
Having said that, I can't find any reference as to how the R/S naming scheme is applied to these tetrahedral stereocenters which don't have four substituents, but a lone electron pair and three substituents instead. I assume that the same scheme can be followed if one considers the lone pair as the lowest-ranked substituent, but I can't find any canonical reference as to that.
So, how are such stereocenters named? What authoritative reference can I find on this topic?
[1]: http://en.wikipedia.org/wiki/Cahn%E2%80%93Ingold%E2%80%93Prelog_priority_rules
[2]: http://en.wikipedia.org/wiki/Tr%C3%B6ger%27s_base
[3]: https://i.stack.imgur.com/KQNxz.png
[4]: http://en.wikipedia.org/wiki/DIPAMP
[5]: https://i.stack.imgur.com/WaclT.png |
How are non-carbon stereogenic centers named (S/R)? |
I'm attempting to perform atomistic [optimal structure alignment][1] on an inorganic species. Can anyone point me in the right direction to software or a technique for performing this?
There exists a wide range of structural alignment software, but this really appears to be geared towards aligning protein backbones, whereas I'm interested in aligning unlabelled Cartesian coordinates of small molecules.
[1]: http://en.wikipedia.org/wiki/Structural_alignment |
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