text:
Difference between single point energy and dimerization energy
I am investigating potential energy surface characteristics of some electrostatic monomeric interactions e.g. $\ce{F-H}$...$\ce{F-H}$.
I'm using DFT and SCS-MP2 to perform the necessary calculations for single point energies. Is it possible to compare single point energy with the electrostatic dimerization energy of the monomers?
Concepts or references will be highly appreciated.

How is this question different from your previous one?

Extremely related: http://chemistry.stackexchange.com/a/38/14  Also: http://chemistry.stackexchange.com/a/24376/14

@Wildcat I actually saw people using dimerization energy to represent such interactions, so I was a bit curious on the role of 'orientation of atoms' in both form of energies.

The question also doesn't make sense, since OP doesn't seem to understand what a SP calculation is.

Definitions
Rather than use the phrase "dimerization energy", which is the energy difference specific to two chemically identical species coming together, such as $\ce{2 BH3 -> B2H6}$, it is better to use the phrase interaction energy. See here and here for a detailed distinction between interaction energies and binding energies. I will simplify a bit some of the text there and in this answer.
The single point energy (SP energy, SPE) is the total energy of a system at one and only one configuration of the nuclear positions. In the case of two monomers (sometimes called fragments), the interaction energy will be the energy difference between the complex (sometimes called the supersystem) and the sum of the individual fragments or monomers at the same nuclear positions as in the complex. This gives rise to the equation
$$
\Delta E_{\text{int}}(\text{AB}) = E(\text{AB}) - \left[ E(\text{A}) + E(\text{B}) \right]
$$
where there are two molecules, A and B, leading to three calculations, one for each of the monomers and one for the AB complex. In the dimerization of borane from above, or your specific example of the $\ce{HF}$ dimer, assuming that the internal coordinates (bond lengths, bond angles) of each monomer are identical, then the equation above can be simplified as A and B are identical:
$$
\Delta E_{\text{int}} = E_{\text{dimer}} - 2 E_{\text{monomer}}
$$
To take your specific example of the $\ce{HF}$ dimer, you mention the role of the "orientation of atoms" in both the single point and dimerization energies. For a single monomer, assuming again that internal coordinates are not changing, the relative positions of the nuclei are not changing, and the total energy is invariant to translations and rotations. However, in the dimer or complex geometry, the relative orientation of one monomer to the other will change the total energy of the dimer, leading to a change in the interaction energy. For example, in addition to the linear form $\ce{F-H}\cdots\ce{F-H}$, you might be interested in the "flipped" versions $\ce{F-H}\cdots\ce{H-F}$ or $\ce{H-F}\cdots\ce{F-H}$, or forming a T-shape, etc. The interaction energy of each will tell you about the relative stability of that nuclear configuration relative to the other configurations.
The most important variables then in an interaction energy calculation are the relative orientation of the interaction species and the distance between each of the fragments. How do you decide this? Usually some intermolecular coordinate is scanned, such as the distance between closest atoms on the monomers, still while leaving the monomer internal coordinates fixed to give an unrelaxed potential energy surface scan.
Calculation Examples
For example, here is the interaction energy for the $\ce{F-H}\cdots\ce{F-H}$ configuration with a variety of methods using the cc-pVTZ basis set and an intraatomic distance of 0.927 angstroms, scanning the $\ce{H}\cdots\ce{F}$ distance:

From a plot like this, we see that MP2 overestimates the interaction and CCSD underestimates it compared to CCSD(T); the shape of the B97-D potential is probably incorrect. HF is missing some repulsive interaction at long-range that the other methods include. It isn't dispersion, as the total dispersion interaction is always attractive and the minimum of the potential is qualitatively correct. You could certainly use any density functional or other method of your choice here (SCS-MP2, SOS-MP2, CISD, ...).
You could similar plots for other configurations, or choose a single method and place multiple configurations on the same plot (here, CCSD(T)/cc-pVTZ):

From a plot like this, it's pretty clear that only the $\ce{F-H}\cdots\ce{F-H}$ configuration is ever bound; it would be interesting to compare the T-shaped configurations to the linear ones.
Interaction Energy Decomposition
You mention "electrostatic monomeric interactions", and we also see some interesting long-range behavior in plot 1. One of the most powerful techniques for understanding intermolecular interactions is symmetry-adapted perturbation theory (SAPT), which gives a decomposition of interaction energies in terms of electrostatics, Pauli exchange, polarization (called induction), dispersion, and charge transfer (though the definition of the latter is often ambiguous).
Here are each of the interaction energy components (SAPT0/jun-cc-pVTZ, monomer-centered basis):

Combining all the electrostatic (1), exchange (1), induction(3), and dispersion (2) terms gives

This makes it pretty clear that all points, $\ce{F-H}\cdots\ce{F-H}$ is bound mostly by electrostatic attraction, and dispersion is negligible. Maybe you knew that, but I didn't. My guess is that the presence of dynamic correlation in all the wavefunction methods damps the long-range electrostatics compared to Hartree-Fock.
References
Interaction energies

J. Řezáč and P. Hobza, Chem. Rev., 2016, 116 (9), pp 5038–5071

SAPT

B. Jeziorski, R. Moszynski, and K. Szalewicz, Chem. Rev. 94, 1887 (1994)
E. G. Hohenstein and C. D. Sherrill, J. Chem. Phys. 132, 184111 (2010)
E. G. Hohenstein, R. M. Parrish, C. D. Sherrill, J. M. Turney, and H. F. Schaefer, J. Chem. Phys. 135, 174107 (2011)
T. M. Parker, L. A. Burns, R. M. Parrish, A. G. Ryno, and C. D. Sherrill, J. Chem. Phys. 140, 094106 (2014)

Technical notes
All of these calculations were run with Psi4. None of the interaction or SAPT energies are CP-corrected.


text:
Can the structure of an azide ion contain a coordinate bond?
Can the structure of an azide ion ($\ce{N3-}$) be $\ce{N#N\bond{->}N^-}$?
I know the actual structure but was asking whether the single bond can be shifted to a N to make it a triple bond and a coordinate bond with other Nitrogen.
This is not a duplicate of Hybridization in azide ion? because I am asking whether it can have a coordinate bond or not while the other question asks what the hybridization of centre nitrogen in $\ce{N3-}$ is.

I have improved the formatting of your question using $\LaTeX$. For more information on how to do this yourself please see here and here.

possible duplicate of Hybridization in azide ion?

Jan is right - it's usually written as single bond with formal charges, but you can write it as an arrow - it's the same.

related http://chemistry.stackexchange.com/questions/32528/help-with-lewis-structure-ch2n2

Of course it can be a duplicate, as ron answered your question - wrote all mesomeric structures for azide, only used straight line and formal charges instead of arrow.

@satyajeet Do you understand what mesomeric structures are? You wrote one of structures of the ion needed to describe it - it's not other molecule!

No.
The issue with using your proposal is very simple: $\ce{N_2}$ sucks at being reactive. It doesn't want to give up or share its electrons with any other atom. In fact, the azide ion is usually explosive because it keeps trying to form nitrogen gas. $\ce{N_2}$ has no real polarity and no particular motive to coordinate to anything, a single N included.

Sorry. you're wrong - structures with additional formal charges are more important in azide than those without them.

Formal charge of the azide has nothing to do with the stability of gaseous nitrogen. Azide is not formed directly from diatomic nitrogen and azide ions are terribly explosive. Sodium azide is used in airbags because it explodes so well, mercury azides detonate if you do nothing with them, etc.  Nitrogen gas is extremely stable and will not try to coordinate to anything. Azide can adopt formal charges, yes, but that's distinct from a nitrogen coordinating to a nitride.

Thanks that was pretty much good answer but is ther any other way to think about its formation rather than just thinking it will be unstable.

see annswers to questions I linked, also one pair of N2 electrons can certainly be donated.

Can be? Sure, technically, I suppose an electron pair can usually be coordinated. But it's energetically unfavorable to the point of ridiculousness. The second question you linked has no answer, and I'm already very aware of the resonance in an azide (in which the most stable contributor is N=N=N).

This is the second important contributor and it is there in answer to hybridizatio.. - http://chemistry.stackexchange.com/a/19911/9961

The resonance structure is distinct from the coordination shown here. That resonance structure is a direct covalent bond. This structure proposes that nitrogen gas will directly coordinate with $\ce{N^-}$.

yes , yes sorry thank you Breaking Bioinformatics now I deeply realized that how can $\ce{N2}$ being thermodynamically extremely stable be co-ordinating with N , it will be a freaking dangerously unstable molecule then and will not actually represent the $\ce{N3-}$ molecule which is lot different from the one I mentioned above.

As @Jan points out in comment it's the same Satyajeet simply wrote another mesomeric structure and probably thought it's another compound

That definitive no is just very wrong. If one accepts donor-acceptor-bonds to be part of the Lewis formalism,  then the indicated structure is indeed a resonance contributor. There is no such thing as a most stable resonance structure, as they are only present when coexistent. It is possible to have a highest contributor, but it is not the same thing. I am sorry, but since this is wrong (and accepted) I had to down vote it.

You can’t really tell a dative bond from a ‘normal’ covalent bond until you break the bond. A dative bond will then dissociate heterolyticly while a covalent bond will dissociate homolyticly.
Just using that simplified way of putting it, you could say that there are many reactions of azides wherein $\ce{N2}$ is liberated leaving a nitrene or related structure; just one of many examples would be the Curtius rearrangement. However, that is not what is meant in the definition above; the definition talks about supplying the bond dissociation energy.
The BDE is not typically supplied in chemical reactions except in coordination compound ligand exchange and radical generation. If you were to apply it to $\ce{N3-}$, I predict homolytic dissociation. Thus, the bond between the nitrogens is more of a typical single bond.
Of course, aside from the chemical implications of dative versus covalent bond you may just have decided to use an arrow notation rather than formal charges. Technically nobody can stop you but still you shouldn’t call it a true dative bond.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
How to distribute the Boltzmann energy to internal and external motion?
If we have an isolated pentane molecule at room temperature (RT) the Boltzmann energy $(E=k_\mathrm{B}T)$ is approxately $0.59\ \mathrm{kcal/mol} \overset{\wedge}{=} 207\ \mathrm{cm^{−1}}$. There is not enough energy for an electronic or vibrational excitation, so the molecule is in its electronic and vibratory ground state. However usually 10–20 rotational states are populated at RT. Can we attribute mathematically (to a first approximation) this $207\ \mathrm{cm^{−1}}$ amount of energy to $x\:\%$ translational energy ($E_\mathrm{kin} = \frac{m v^2}{2}$) and to $(100-x)\:\%$ rotational energy?
This question is related to the question about the internal degrees of freedom for a molecule: Rotational degrees of freedom (3N-5 and 3N-6)

The translational energy of a mole of molecules is $\tfrac{3}{2}RT$ which corresponds to an average kinetic energy of $\tfrac{1}{2}M\langle v^2\rangle$.  The average reflects a Boltzmann distribution of speeds
$$f(v)=4\pi\left(\frac{M}{2\pi RT} \right)^{3/2}v^2\operatorname{e}^{-\tfrac{1}{2}Mv^2/RT} $$
i.e. some molecules have very low velocity and some have very high velocities.
Kinetic energy is constantly being transferred, so if you could monitor its velocity over time its average kinetic energy would be $\tfrac{1}{2}M\langle v^2\rangle=\tfrac{3}{2}RT$
In addition one mole of gas has $\tfrac{3}{2}RT$ of rotational energy ($RT$ for linear molecules), which also reflects an average rotational kinetic energy.  For individual molecules at any given time, the amount of translational and rotational energy is independent.  For example, a molecule can move very slowly while rotating very rapidly – until it collides with another molecule and things change.  However, on average there is an equal amount of translational and rotational energy (for non-linear molecules).

Is $f(v)$ equal to $\langle v^2\rangle$? I'd like to obtain numbers for pentane for the rotational energy and for the translational energy. Can we do this with formulas or do we really have to measure this values in experiments?

$f(v)$ is a function, $\langle v^2 \rangle$ is a number, so no. The rotational and translational energies for pentane and any other non-linear molecule are $\tfrac{3}{2}RT$

If there is only a single molecule (very very low gas pressure), which cannot collide, will its translational energy change during the time coordinate? If there is no collision I expect no Brown'sche molecular motion, so it will fly along a one-dimensional coordinate (if we neglect grativational forces) with the same energy as it rotates. I guess the molecule tumbles along the different rotational axis a,b,c. Or will the molecule rotate only along one axis?

The usual approximation is that translational, rotational, and vibrational motion is uncoupled. So if the molecule doesn't collide with anything then the translational kinetic energy (and the velocity) will be constant.  However, for an individual molecule that never collides the translational and rotational energy will not be equal.  That's only true on average and after many, many collisions.  Most likely, the rotational motion will reflect rotation along 3 distinct axes.

What does $f(v)$ tell me?

$f(v)$ tell you the probability of a molecule having velocity $v$.  Also $\langle v^2 \rangle=\int_0^\infty v^2 f(v) dv$

As you said the translational and rotational energies are different if there is no collision. Is there any formulae which relates the two energy values. I mean (with no collision), if I indice a dipole transition with light (and there is not enough energy for vibration or electronic transition) how many % will roughly go into rotational motion and how many % into translational motion in the case of isolated pentane?

The absorption of light depends on the wavelength. Sub-IR radiation typically means microwave, which excites rotational modes.  None of this internal energy can be transferred to translational motion.  The photon also has momentum ($p=h/c$), which can be transferred to the molecule. Everything we've talked about previously assumes thermal energy transfer and does not apply to light-induced excitations.

Ok, that's a very important point for me what you mentioned (that there is no internal to external energy transfer)! I guess (for our case) only the total energy and the total angular momentum is conserved. If we heat an isolated molecule it's thermal energy = kinetic translational energy will increase. Is it possible for the molecule to lower its translational energy by going to some higher rotational levels (= increasing the internal energy)?

In the case of a supersonic expansion the expanded gas is cooled down and gets to the ground state of rotational and vibrational levels. I'm thinking if the translational motion is also decreased (not only the internal degrees of freedom) because the internal degrees of freedom might not have enough energy for this adiabtic expansion.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Why starch (amylose) and cotton (cellulose) are so different?
Both amylose and cellulose have the same "monomer" structure (glucose), so what makes them look/form so differently?

Your structures don't clearly show the different configurations between α- and β- linked glycopyranosides. Cellulose is a non-branching (poly) β-glycopyranoside. Amylose (a component of starch) is a non-branching (poly) α-glycopyranoside. The enzymes in your mouth and digestive system are good at breaking apart the α-glycopyranoside linkage, but not the β-glycopyranoside (it doesn't fit into the enzyme).
Amylopectin (the other component of starch) is a branched (poly) α-glycopyranoside. It branches with a 1-6 linkage. Glycogen is another branched (poly) α-glycopyranoside.
Cellulose is not the only (poly) β-glycopyranoside; there is also β-glucans.

You can also add glyocogen to that list, a glycan found in us The difference is that cellulose has a linear structure, allowing parallel chains to lie in close proximity and for hydrogen bonding to predominate, strengthening the structure.. Starch, on the other hand, has side branches preventing as many hydrogen bonds from forming.
Note also that the site above shows very similar structures for starch and cellulose. The illustrations you've shown are using different methods to display the structure, emphasizing the cyclic nature in the first image, and the bond angles in the second. It would be interesting to see these molecules in atomic force microscopy, rather than in our conventional diagrams.
BTW, consider the different structure of carbon in diamond and graphite, where there is little bonding between layers of graphite, but diamond has three-dimensional strength.

Amylose is not really branched very much, although the other main component of starch, amylopectin, is.  The difference in amylose and cellulose is the geometry of the 1,4 glycosidic linkage and the effect of on the resulting crystal structure of the two polymers.  Inter-chain H bonding is greatly facilitated by the cellulose crystal structure, making it exceptionally resistant to hydrolysis, relatively insoluble, and a tougher material than amylose, all because it's 1,4 linkage is beta instead of alpha.

@CurtF., could you please explain (or attach a link explaining) the difference between alpha linkage and beta linkage?

There is some info here: http://www.chem.ucla.edu/harding/ec_tutorials/tutorial43.pdf


text:
pH effect in the hydrophobic interactions between two polypeptide chains
The rise in pH can change the protein conformation by changing the ionic interactions and hydrogen bonds between two polypeptide chains. This is a fact that I understand very well because the change in amino-acid ionization with pH. I would like to know if by changing the pH value the hydrophobic interactions between two chains can also be changed. If yes, what is the mechanism?
Article: a pH Induced increase in hydrophobicity...

Here is one possible mechanism:
In tightly folded, compact, "globular" protein structures, the most hydrophobic bits of the protein tend to be in the core of the structure, away from the surface.
If the pH changes, then by mechanisms you mention in your question, this structure can be disrupted.  If it is side chain interactions among surface-oriented, polar residues that "hold together" the globular structure, then a pH change can disrupt those interactions, allowing the previously isolated hydrophobic "core" to become  exposed.  This core can now bind more surfactant molecules, aggregate with itself, or do any number of other "hydrophobic" things.


text:
How to determine the worst resonance structures out of a given set?
So I was "happily" doing organic chemistry homework when I came across this question:

For 1-methoxy-1,3-butadiene, which of the following resonating structure is the least stable?
$$\begin{align}(\mathbf{a})\ &\ \ce{H2\overset{+}{C}-\overset{+}{C}H-CH=CH-O-CH3} & (\mathbf{b})\ &\ \ce{H2\overset{+}{C}-CH=CH-CH=\overset{+}{O}-CH3} \\
(\mathbf{c})\ &\ \ce{H2\overset{+}{C}-\overset{-}{C}H-\overset{+}{C}H-\overset{+}{C}H-O-CH3} & (\mathbf{d})\ &\ \ce{H2C=CH-\overset{+}{C}H-CH=\overset{+}{O}-CH3}\end{align}$$

I decided to take a look through my textbook to see what I was missing. The rules to compare resonance structure energies (and therefore determine relative contributions) were in my textbook.  It said:

The following structures are considered to contribute most:

Structures having filled octet a for second row elements (C, N, O, F) are stable.
Structures having minimum number of formal charges and maximum number of bonds.
Structure in which negative charge appears on the most electronegative atom (C < N < O).
Structure in which there is minimal charge separation while keeping the formal charges closer together.

Instead of answering my initial question, two more questions came to mind. They are (in order):

Why do the octets of only the second row elements have to be filled in for greater relative contribution?
I suppose it has something to do with the greater electronegativity values of the elements, but would still like some clarification.
From option (a) and (c) in my question, how did the two positive charges get next to each other?

Resonance structures, as stated in my textbook, can be represented by two or more Lewis structures that differ only in the positions of the electrons.
The options clearly do not just redistribute the electrons. Or at least I don't think so.
So if I had to guess, the answer would be (b) because the terminal carbocation is worse than an internal carbocation.
But it turns out, the correct answer is (c). And I think that it shouldn't even be considered a resonating structure.

Possibly there's an error in the question.  None of the 4 structures represent 1-methoxy-1,3-butadiene; all of the structures represent the dication of 1-methoxy-1,3-butadiene.  I suspect someone accidentally put a (+) sign instead of a (-) sign in each structure.  Had they been drawn correctly (with an equal number of positive and negative charges), then (c) would be the least stable since it has 4 separated charges, all of the others only have 2 (rules 1 and 2 in your list).

I'm afraid your textbook sucks ;)

Wow! I spent at least an hour on this question. I guess I should be more careful with my choice of text next time. @ron I can't believe I didn't see the mistake. Thanks.

Resonance structures are not real, they are only a description. Therefore there cannot be a most stable resonance structure. The premise of the question is wrong, regardless of the error in the formulae. See also here.

Can inductive effect due to nearby atoms not be considered as a point for stability?

(C) is technically a valid resonance structure. You can derive it from (B) by moving the electron pair from the double bond between the carbons 2 and 3 (numbering from the left: this may not be correct from a nomenclature standpoint, but I'm using it for this answer) onto carbon 2 and moving the electron pair from the double bond between the oxygen and the carbon onto the oxygen.
(C) is least stable because it is, simply put, absolutely disgusting. Charge separation (carbons two and three) is gross, you have two cations in close proximity (carbons three and four), a terminal carbocation (carbon 1), and three atoms with empty valence shells (carbons 1, 3, and 4). So while the resonance structure is valid enough for (C) to be considered a resonance structure, that resonance structure happens to be the spawn of the devil.
That said, the compound in the answers isn't the same as the compound named in the question, like ron said.
As to the first part of your question, the filled valence shells are more strictly necessary for the second row elements because they don't have d orbitals to confound the picture.

I find the second paragraph not only offensive, but also wrong. There are no such things as most/least stable resonance structures; the whole premise of the question is void. While charge separated structures in organic molecules are often minor contributors, that changes quickly, if you go to other elements than carbon and co. They are very often essential contributors, which explain reactivity towards certain mechanisms. It may look odd to you, but it most certainly is not the spawn of the devil. And your last sentence hints at octet expansion, which has been disproved multiple times.

As ron stated in the question’s comments, none of the resonance structures display 1-methoxybuta-1,3-diene. In fact, they show an extremely unstable (if generateable) dication. Choose a random $+$, replace it with a $-$ (don’t choose the one on the oxygen) to get resonances for the structure the question asks for. That should also semi-answer your question about the two positive charges.
As for the question on octets: It is the more traditional viewpoint that elements of the third and higher periods such as sulphur could expand their octets (e.g. in $\ce{SO2}$). This was generally explained with contribution by d-orbitals (and Pauling’s dislike for charge separation).
Nowadays, however, these structures are more generally considered as abiding by the octet rule invoking charge separation where necessary and multicentred bonds where applicable. The reason for this is that d-orbitals are too far removed energetically from s and p of the same shells to actively take part in bonding.
Finally, there are sub-octet structures to consider (heptets in radicals or sextets in carbocations, carbenes and the like). These exist and are undebated. This violation of the octet rule is ‘less bad’ (by a mile) than a super-octet. Why? Well, you’re not filling electrons into high-energy contributed orbitals but rather removing them from stable ones.
That said, when considering the first of these octet paragraphs, it is a lot harder for me to think of compounds that have sub-octets for elements of higher periods — most examples I know are carbon or nitrogen centred.

On the topic of the ugliness of (c), Breaking Bioinformatics has answered that nicely. But you could have arrived at the same conclusion by checking your book’s rules:

Does not apply here. No resonance has an octet on every carbon.
(c) has four formal charges, all the others have two. Stop here, mark (c) and grab a piece of cake.

Following rules are 100%-accurate and can be used to determine stability:

more the no. of π-bonds more stable the contributing structure is.
if the no. of π-bonds are equal then … the molecule with no charge or with no charge separation is more stable.
if the molecules are not containing any charges i.e neutral then … the molecule in which all atoms have a complete octet is more stable.
if the atoms in a molecule have an incomplete octet or have charge separation then … the negative charge is more stable on a more electronegative atom (like N,O) and positive charge on a less electronegative atom.
if above the condition is satisfied, too, in a molecule then … if in a molecule unlike charges are closer then it is more stable than the molecule which has unlike charges far away.
And if like charges are closer to each other then it is more unstable than in which like charges are far away.

According to inductive effect OCH3 is electron withdrawing group and positive charge next to this carbon make it relatively least unstable .So c is correct answer.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Do crystals have stronger intermolecular forces than amorphous solids?
Do crystals have stronger intermolecular forces than amorphous solids.
Are the properties of amorphous solids vs. crystals explained by their structures?

I didn't edit, but note, it's rather intermolecular forces than bonds.

The structure, of course, does lead to many of the properties.  Now, for an example to ponder: amorphous silicon is a four-fold coordinated, tetrahedrally-bonded semiconductor, just like crystal silicon.  Well, except that it has no long range order. It is also a separate thermodynamic phase exhibiting a first order phase transition to the solid. The band gap of the amorphous phase is similar to that of the crystal. The average bond length is similar. The phonon energies are similar.  The average atomic forces are slightly in the crystal's favor, but it is the lowest free energy allotrope.

Diamond crystal has much stronger bonds than amorphous red phosphorus. Amorphous sodium/calcium silicate, known as glass, has much stronger bonds than crystalline metallic sodium. (Also, in all four cases the word intermolecular doesn't really apply, because these compounds are not made of molecules.)
You see that the answer depends primarily on the particular compound, rather than on its being crystalline or amorphous. In cases when something can be either crystalline or amorphous with the same chemical composition, the bonds in both are pretty much the same, as explained in the comment by Jon Custer.
Oh, and yes, chemical and physical properties of pretty much anything are explained by its structure.

For some materials that can exist in the amorphous or crystaline state like quartz or corundum, the crystalline state is harder, that is, requires a harder substance to even begin to scratch it. For other other materials like diamond, the amorphous state is harder. However, for quartz and corundum, the amorphous state is less breakable than the crystalline state once they have already been scratched. That's because the cracks in the crystalline form introduced by scratching are less likely to loop around into a chip to when the crystal undergoes a collision with a hard surface, the tension is greatly magnified at the tip of the crack, which causes it to propagate and if the collision is fast enough, the crack will propagate through leading to fracture.


text:
Why does phenol not undergo Friedel-Crafts alkylation and acylation reactions, whereas anisole does?
Friedel–Crafts reaction of phenol
The answer to the question at the above link states that : "Like aniline, phenol too reacts to a very less extent during Friedal-Crafts reaction.
The reason being that the oxygen atom of phenol has lone pair of electrons which coordinate with Lewis acid."
But isn't the case with anisole, the same? However, my textbook states that anisole does undergo Friedel-Crafts acylation and alkylation.

You and the textbook are both correct. Anisole does undergo FC reactions but the O can coordinate (more weakly than in the case of phenol),  leading to demethylation.  It all depends on your reaction conditions and Lewis acid.


text:
Nomenclature of a cyclohexyl ether and locant position enumeration

As far as I know, 'ethoxy' should be given the locant '1' as it is a functional group. The compound should thus be named 1-ethoxy-2,2-dimethylcyclohexane.
However, I was told that the correct name is 2-ethoxy-1,1-dimethylcyclohexane. Can anyone please explain why this is so? My teacher explained this to me in terms of the lowest locant sum rule, which I just discovered doesn't even exist.

The smallest sum of locants rule does not exist in the IUPAC recommendations. The application of this ‘rule’ can lead to wrong results in many instances.
The corresponding section in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) actually reads as follows:

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.
(…)

With regard to numbering of locants, simple prefixes (simple substituent groups consisting of just one part that describes an atom, or group of atoms as a unit, for example methyl and ethoxy) are considered together with equal seniority:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(…)

Therefore, the example is named as 2-ethoxy-1,1-dimethylcyclohexane rather than 1-ethoxy-2,2-dimethylcyclohexane since the locant set ‘1,1,2’ is lower than ‘1,2,2’.

"...simple substituent groups consisting of just one part that describes an atom, or group of atoms as a unit, for example methyl and ethoxy". Could you please give more such examples. Also, "... low locants are assigned to them in the following decreasing order of seniority: ...(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;" Could you mention the order as well?

@ag_dhruv In this context, ‘simple substituent groups’ are substituents that are cited as prefixes before the name of the parent structure and do not modify the structure  (e.g. ‘bromo-’, ‘chloro-’, ‘methyl-’, ‘ethyl-’, and ‘nitro-’).

@ag_dhruv The ‘series of increasing numerical order’ mentioned in P-14.4 (f) corresponds to the order that is applied in P-14.3.5 above; e.g. ‘5-bromo-8-hydroxy-4-methylazulene-2-carboxylic acid’ has the locant set ‘2,4,5,8’ in a series of increasing numerical order.

Slightly irritating technicality, but according to P-14.4(f), the locant set under consideration is that of the "detachable alphabetized prefixes", hence in the example '5-bromo-8-hydroxy-4-methylazulene-2-carboxylic acid' it is '4,5,8' and not '2,4,5,8' (this is corrected in the errata).

@orthocresol .... Is 4-Ethoxy-1, 1-dimethylxyclohexane correct or 1-Ethoxy-4, 4- dimethylcyclohexane?

@Vishalprabhulawande Neither is correct, because chemical names should not be capitalised at the beginning (unless they are at the start of a sentence). Once you fix that, though, (4,1,1) is correct.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Electrochromic Materials, electrode setup
I've read a lot about electrochromism. This article on electrochromic materials has also been additionally helpful.
I was wondering, for these electrochromic materials, can you simply attach a small electrode to either side of the film like in setup 2, then apply a small voltage? Will the electrochromism only occur at the edges in this case?
Does anyone know how material dependent this effect is? I was hoping to use a polymer film in particular.

Trivia: a slice of potato with some sodium or potassium iodide solution sprinkled on it makes a primitive electrochromic "display". Apply a few volts to wires immersed in the electrolyte; iodine released at the positive electrode turns the potato purple or black. If the reacti0n hasn't gone to far, it's reversible by changing polarity. [Sorry, it's not really an answer to your question, but perhaps someone will make a starch display.]

I will answer your last question first, this is very material dependent. Not all materials will be electrochromic. It is a very specific effect where the material has different oxidation states that have different absorbance profiles(colors if you will). The most common ones are conducting polymers (https://en.wikipedia.org/wiki/Conductive_polymer) since these materials have different oxidation states that have vastly different electronic structure.
Two electrode electrochemistry can get pretty hard to understand. Two compare your setups 1 and 2, the main differences are the electrode area and the electrode distance. Other than that, the two are the same. So, yes, electrochromism will happen in setup2 as well. However, it is hard to predict exactly where it will happen and to what extent. It all depends on the material you select, it's resistance and doping characteristics. In either setup, it will start at the electrode and work it's way inward, obviously, setup 1 is more advantageous with the larger electrode area and shorter inter-electrode distance.


text:
Why are amino acids named amino acids?
When learning about amino acids, the primary things that were pointed out were that they had an amino group and a carboxyl group, and side chain that varies from one to the other.
The amino group of one can attach to the carboxyl group of another, making a peptide bond and water.
Both groups are clearly important, but what is it about the amino group that that's what goes in the name "amino acid"? Is the amino group in some way more important? Is it just something about the way/order in which they were discovered/identified that the name just came about?
Why amino acid? Not carboxyl acid, or some mix of the two, or something else?

I think you are forgetting that a carboxyl group is acidic.

When learning about amino acids, the primary things that were pointed out were that they had an amino group and a carboxyl group...
Why amino acid? Not carboxyl acid, or some mix of the two, or something else?

You stated why it is called an amino acid, the name comes from the amine (amino) and the carboxylic acid (carboxyl group).
Amines are weak bases, so the acid portion of the name must come from the carboxylic acid.

Ohhhhhhhh... yea. I just literally did the whole facepalm thing over the fact that I didn't....... wow. Thanks guys! XD

@Caesium-133 Don't worry! I HAD to read on to learn why as well. Facepalm here too!!!

You might think that the term "amino acid" comes from the more systematic names of these molecules, where the amine is a substituent.
Common name        systematic name
glycine            2-aminoethanoic acid
alanine            2-aminopropanoic acid
valine             2-amino-3-methylbutanoic acid
cysteine           2-amino-3-mercaptopropanoic acid
serine             2-amino-3-hydroxypropanoic acid

In all of these names "amino" precedes "acid".

You would be wrong!

"Amino acid" apparently entered the English lexicon in 1898, long before IUPAC reared its systematic nomenclature head. It is likely then, though I was not there, that "amino acid" came about from two considerations:

We were already probably using the prefix "amino-" to denote molecules with nitrogen groups in them.
"Amino acid" was probably easier to pronounce than other possible combinations.

By the way, interestingly, most of the amino acids in the normal conditions are not acids nor amines, but rather “inner salts”, i.e. dipolar ions, i.e. ‘zwitterions’, $\ce{H3N+-CH(R)-COO-}$, i.e. “inner ammonium carboxylate salts”, rather than aminocarboxylic acids.


text:
The difference between the different types of rhodamine
What is the difference between different type of rhodamine, for example rhodamine red and rhodamine green?
are these two type of rhodamine (red & green) have a good Raman signal?
We would like examples of Raman signals of these two types of rhodamine.

Welcome to chemistry.se! I think your first question can be answered by wikipedia. It would be helpful if you could provide the structures of the compounds in your post. Please also have a look at the homework policy which applies to your question.

Rhodamine green
CAS: 189200-71-3

Rhodamine Red
CAS: --


text:
Which one is the best notation for sodium acetate
I have seen two different notations for sodium acetate. The first one is:
$$\text{NaCH}_3\text{COO}$$
The second one is:
$$\text{CH}_3\text{COONa}$$
Now I'm confused, which one is the best to use?

I like the second one better because it implies the correct location of the ionic interaction (between sodium and the carboxylate). I cannot readily point to something that makes the first one "wrong".

First one is "inorganic"; in organic chemistry the second is usually used

I think NaOAc is the best, but that's because I'm lazy.

There is nothing wrong with either formula. And you can use even more:

$\ce{NaC2H3O2}$
$\ce{C2H3NaO2}$

It really depends on which point you want to bring across.
$$\ce{NaCH3COO}$$
This formula, being analogous to formulae like $\ce{NaCl}$ stresses the inorganic salt view more. It shows that there is a cation ($\ce{Na+}$) and an anion ($\ce{CH3COO-}$) which form a salt crystal together much like the anion $\ce{Cl-}$ would do with the same cation. Inorganic nomenclature prefers cations to be written first.
$$\ce{NaC2H3O2}$$
Is basically the same except for saying ‘I don’t care how those seven atoms combine to form the anion, all I care for is what’s in there.’
$$\ce{CH3COONa}$$
This one stresses a more organic-chemical point of view where it’s relevant where the cation is actually bound to within the molecule. Oftentimes, organic chemists would even write structural formulae with a bond between oxygen and sodium as if it were hydrogen. The reasoning behind this is ‘I don’t care if the structure ends up being a salt, all I need to know is that the sodium somehow connects to the carboxyl group.’
Note that you can combine the first’s and the third’s idea to give $\ce{NaOOCCH3}$ which works, but is a lot less used than either of them.
$$\ce{C2H3NaO2}$$
This is the Beilstein-type lookup formula. You just know the elemental composition of your compound and now want to look it up — you can’t know whether it’s sodium acetate or sodium hydroxyethanalate (if that name is even correct). Or maybe you just don’t care. The reasoning behind this is $\ce{C}$ first, $\ce{H}$ second, the remaining elements following in alphabetic order.

Finally, there are also shortened formulae. The most common (the one I actually use most) would be $\ce{NaOAc}$. Here, $\ce{Ac}$ is an abbreviation for $\ce{H3C-C=O}$ where whatever follows or precedes is bound to the carbonyl carbon. Note that this formula is not considered standard, and would need to be included in a list of abbreviations or defined otherwise if you decide to use it.

I'm glad you included NaOAc, that is also my preference

Ok , both of the formulas are absolutely correct.It depends on what you want to signify , basically if you want to look from the point of view of organic chemistry then you take the second formula. And If you any to look from just the point of view of just ions then you can render the positive ion($\ce{Na+} $) to shifted first and then the negative ion($\ce{CH3COO-}$)
.I think the second formula looks much more better because it is the usual notation for sodium acetate and my vote goes to the second.But if you want to show people something different you can obviously use the the first one.

Both are correct, but the second one is more commonly used. Actually, the way of writing a molecular formula signifies the structure of the molecule.
Let me explain: There are compounds that have the same molecular formula but have different structures. Such molecules are called structural isomers. An example for a carbon compound:

As you can see, even if both have the same formula they have different structures (they are structural isomers to say it precisely). Generally formulae of many compounds or molecules are written in a way so that they can signify the structure or arrangement of atoms in molecules.
Now look at that image again. If I want to write chemical formulae of both compounds that can signify their structures, I would write:
$\ce{CH3CH2CH2Cl}$ for the first one, and the second one is $\ce{CH3CHClCH3}$.
I guess now you can understand that the way of writing a formula defines the structure of the compound, so always write the formula in a way that can define or give some hint about the structure.

I disagree with this answer because it implies that only one of the two structures ($\ce{CH3COONa}$) be correct — or if not the only correct one, at least being one strongly favoured. I must say that that is definitely not the case. -1.

This has nothing to do with isomerism.-1


text:
Cleaning silver - which method is better?
What is the advantage of this approach over polishing the silver with an abrasive?
"Silver tarnish is an outer coating of silver sulfide, $\ce{Ag2S}$, formed when silver reacts with trace quantities of air-borne hydrogen sulfide, a smelly gas produced naturally by the digestion of food in living animals.  Silver atoms in silver sulfide have lost electrons to sulfur atoms in creating ionic bonds.  You can remove this tarnish with home electrochemistry.  Convert silver atoms back to their elemental state by restoring their electrons.  Sulfur atoms won't relinquish electrons to silver, but with the proper connection, aluminum atoms will.  Add about a liter of water with several heaping tablespoons of baking soda to an aluminum pan that has been scoured clean.  If you don't have an aluminum pan, use several large strips of aluminum foil at the bottom of another pan.  Bring the water to a boil and remove it from the heat source.  Slowly immerse a tarnished piece of silver; you'll see an immediate effect as the silver and aluminum make contact.  (Add more baking soda if you don't see this.)  Also, as silver is brought back to its elemental state, hydrogen sulfide is released back into the air.  You'll smell it!  The baking soda serves to remove a thin and transparent coating of aluminum oxide, which normally coats aluminum and thus prevent direct contact.  The baking soda also serves as a conductive ionic solution to permit the passage of electrons from the aluminum to the silver."

https://www.youtube.com/watch?v=vGCx9HZwYBo is a neat video on the topic.

Abrading the silver sulfide removes it (and some silver underneath the tarnish). Electrolytic cleaning actually restores the silver previously part of the tarnish. In addition, it cleans crevices inaccessible to manual polishing.
There are disadvantages, though:

If the article has niello inlay, it could be damaged.
If other materials, such as parchment or glue, are part of the article, they would be damaged.
If the article is a valuable antique, cleaning may reduce its value.


text:
Pauling's equation on the relationship between Electronegativity and Percentage Ionic Character
In Paulings eqn:
$$ \%~\ce{ionic~character} = \left[ 1−e^{−(Δχ/2)^2}\right] \times 100$$
What is $e$?
I for the love of me can't find it anywhere

Related to the answer: http://chemistry.stackexchange.com/questions/27535/find-percentage-ionic-character-if-electronegativity-is-given/27571#27571

'e' is a natural exponential function, so not an actual measurement.
The equation is correctly written as:
% Ionic character $= (1 − e^{−(\frac{Δχ}{2})^2})×100$
where the $−(\frac{Δχ}{2})^2$ term is raised as a power of the natural exponent 'e'.
Source: The University of the West Indies page Lecture 3. Polarizability, you'll need to scroll down the page a bit to the section labelled 'Bond character based on electronegativity differences'.

Thank you, that is what I thought. just needed some confirmation. That lecture is exactly what I used


text:
What is "Steel Reinforced Epoxy"
No, it's not epoxy for reinforcing steel. It's "steel reinforced epoxy."
http://www.jbweld.com/products/j-b-weld-twin-tube
How in the world does steel reinforce epoxy?
I can understand how carbon nanotubes dispersed in epoxy can help strength it, quote from paper below:

The mechanical behavior of multiwalled carbon nanotube/epoxy
  composites was studied in both tension and compression. It was found
  that the compression modulus is higher than the tensile modulus,
  indicating that load transfer to the nanotubes in the composite is
  much higher in compression. In addition, it was found that the Raman
  peak position, indicating the strain in the carbon bonds under
  loading, shifts significantly under compression but not in tension. It
  is proposed that during load transfer to multiwalled nanotubes, only
  the outer layers are stressed in tension whereas all the layers
  respond in compression.

But back to steel - as far as I know, steel is a crystalline material and it doesn't possess any property akin to CNTs. Can someone explain how this particular epoxy works?

I have no idea why you have problem here - you don't understand what reinforcing means or sth? This link leads to normal glue...

The link leads to a product that claims to have steel in it's composition as a structural element.

Hmm, it's strange but only in the case of glue.

I suspect the main thing reinforcing this product is jbweld's marketing department.

There's no "steel" in that stuff. It is simply filled with inorganic particles, probably (partly?) iron dust. I'm using jbweld too. Quick test with a bar magnet showed the tube with the "steel" component as well as the cured epoxy are indeed slightly ferromagnetic.
Most importantly, the filler reduces macroscopic shrinking during curing. That's a general principle: The density of the filler particles stays constant, while that if the glue component goes down, remarkably. The filler material type is not really important, only its surface must be compatible with the epoxy.
Plus the filler particles impede the propagation of cracks, and reduce the permeability to water, oxygen, etc. Again, that's a general property of fillers, not specific to this one.
It's hard to tell what the specific advantage of using iron/steel powder is, instead of some inorganic oxide (vulgo "sand"). A metal might catch environmental oxygen that diffuses in, increasing the temperature stability of the epoxy. If that's the case I don't know, but this stuff is remarkably temperature-resistant.


text:
Is there a known chemical that makes it possible to control the opacity of a large surface?
I have this idea of creating a large domed ceiling/roof for a building that can be made clear to see the actual sky, or made fully opaque/dark grey/black to project an artificial sky onto.
I know that things similar to this effect have been chemically achieved such as UV-activated sun glasses and remote controlled window tint, but I haven't been able to find much information that explains how this kind of thing is done chemically.
What options, chemically speaking, exist for making a large glass surface transform between being clear and opaque?
I'm looking for any known chemical that can do this in the form of a tint applied to a glass surface, or a liquid filling a hollow glass surface, or even a gas filling a hollow glass surface.
To be absolutely specific, I'm describing a chemical that can be either modified or added/removed to/from/within a glass surface to systematically control nearly full opacity or nearly full clarity of a layer applied to or contained within the surface of a curved glass panel.
Note: The opacity transition should be efficiently repeatable, meaning, the effect shouldn't deteriorate significantly with repetition.

There are

thermochromic materials that change color in response to heat
photochromic materials that change color in response to light
and electrochromic materials than change color in response to an applied electric field.

All of these materials are sub-categories of Smart Glass. I suspect that electric field sensitive Smart Glass materials are what you are looking for.  There are several different types of such Smart Glass materials.

Those with suspended particles that are normally randomly aligned, but when an electric field is applied they align an alter the light transmission properties of the material.  These materials have already found commercial application (airplane) and are moving down to lower cost applications (autos)
Electrochromic devices where a molecule undergoes a chemical change between colored and non-colored states when a field is applied.
Liquid crystal materials dissolved in a polymer matrix are normally randomly aligned, but, when a field is applied they align thus altering the light transmission properties of the material.

There are also some non-electrochromic smart glass technologies that are also discussed in the Smart Glass Wikipedia article linked above.  These are generally lower tech and therefore lower cost.  They work based on

Having several aligned windows, but with the alignments in different directions and then manually moving the windows into place to adjust the opacity.
Having two reflective windows with a space in between.  When the space is filled with a material having the same refractive index as the reflective windows, the assembly becomes transparent.
Using multiple polarized panels and moving them into or out of place to adjust opacity.

Which method you choose will depend upon how quick you need the materials to change color, the level of manual intervention you're willing to tolerate and how much your willing to spend.  If you google "Smart Glass" you'll find that this field is already well advanced.


text:
What is that process called in which we add water molecules to an ion to form a complex?
An example would be $\ce{[Fe(H2O)6]^2+}$ in which it has 6 water molecules associated with it. What is the process of adding these water molecules called?

What makes you think there exist some specific term for this particular kind of complexation?

Are you referring to "hydration"?

@ron thank you but i am confused is this called hydration ?Then what is  that called $\ce{FeSO4.5 H2O}$? The water of crystallization is also said to be a result of hydration reaction.Please clarify!!

$\ce{[Fe(H2O)6]^{2+}}$ and $\ce{[FeSO4(H2O)5]}$ are both examples of hydrated iron.  In the latter compound, one of the water ligands has been replaced by a sulphate ligand.

@ron what i am thinking is the first ion actually has coordinated $\ce{H2O}$  bonded to $\ce{Fe^2+}$ and in the second case the water of crystallisation is surrounding the compound FeSO4 by electrostatic attractions.

@Satyajeet Well, no; both are essentially the same thing only that one occurs in solution, the other has happened in crystallised form. If you're interested in the difference between hydration in crystals and hydration in solution, you should reword your question to reflect that.

I believe it is normally just called "addition of water".
There is also the term "hydration" but this can also refer to the addition of water to alkenes or alkynes which forms alcohols (in the case of alkene; I am not sure about alkynes).
For example, $$\ce{CH2=CH2 \text{(ethene/alkene)} + H2O -> CH3CH2OH \text{(ethanol/alcohol)}}$$
I personally think "addition of water" is just fine to use and people will get what you are saying.

I would definitely use 'hydration'.


text:
Dissociation reaction notation
This equation has been described as an equilibrium in an article, how would you explain it to someone with not a lot of understanding of it.
The equation is:
$\ce{CH3COOH <-> CH3COO- + H+}$
Although the description is of an equilibrium, it can be clearly seen that it is a resonance arrow. So how would you describe it?

No, no it's not a resonance arrow my friend, it has to be an arrow for equilibrium because it is a reaction equation of $\ce{CH3COOH}$ in its aqueous solution. It will dissociate feebly into its constituent ions because its a really weak acid. Thus establishing equilibrium quite a lot first.
Moreover in a resonance arrow we describe the same compound with different structures (which are all correct) and that is why we establish a resonance hybrid to denote all such possible resonating diagrams in a single diagram fulfilling the properties of the compound.
Nevertheless a resonance arrow can't anyway turn into a dissociation reaction arrow.


text:
Alcohol in the bloodstream modeled by differential equations
Does anyone know any good references which use partial differential equations to model the alcohol concentration in the bloodstream? I'm a mathematician, so I'm searching for a rigorous mathematical model.
If you have any other suggestions of models in chemistry that uses partial differential equations, feel free to post them here.

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

"If you have any other suggestions of models in chemistry that uses partial differential equations, feel free to post them here." that is pretty broad... It covers about all physical chemistry with thermodynamics and quantum chemistry

One such article that I found was published by the University of South Florida. The abstract for which reads:

Given a set of differential equations describing blood alcohol content as a function time, we integrated the
  equations to obtain a general solution. The general solution equation depends on three free parameters: the initial
  concentration of alcohol in the stomach after ingestion, the rate of alcohol absorption into the blood stream and the rate
  at which the alcohol is metabolized by the liver. We fitted our solution to experimental data to determine the unknown
  parameters for a particular subject.

If this sounds useful to you, it can be found here.


text:
R32 Space Group (Crystallography)
Why R32 Space Group is so named? I mean, if R stands for Rhombohedral and 3 for Threefold Axis, what 2 stands for?
I don't see any Twofold Axis in the cell:
http://img.chem.ucl.ac.uk/sgp/large/155az1.htm
EDIT:
What does it means 1/2, 2/3 and 1/6 on the arrows?
R32 have two cells.
This one: http://img.chem.ucl.ac.uk/sgp/large/155bz1.htm
And this other: http://img.chem.ucl.ac.uk/sgp/large/155az1.htm
Are they equivalent?
EDIT:
I have traced some lines to mark the screw-axes and the twofold-axis, but i can't see any point that follow the simmetry operator and the element of simmetry. Maybe i don't understand how it works.
Which is the height of the green line? And, which are the points of the screw-axis of the yellow line?

The 2-fold axes are parallel to the "paper plane ", i.e. perpendicular to the three-fold. They are depicted as arrows, the half-arrows indicate screw-axes. The numbers next to it show the "height" of the axis, i.e. where the plane parallel to the paper-plane cuts the unit cell: 1/2 would be in the middle, 1/3 would be a third from the bottom/top, etc. (obviously one only gives relative numbers here, as the absolute unit cell length varies depending on the real-life crystal you are looking at).
Also have a look here for the symbols: http://img.chem.ucl.ac.uk/sgp/misc/guide.htm
Otherwise, the two links you provided are equivalent, as you can see from the pictogram in the lower lefthand corner - the first link is in the "tri-/hexagonal", the second link in the "rhombohedral" setting. The symmetry for both is obviously the same (otherwise they would not be the same space group), it is just a different way of depicting the spacegroup by changing the point of origin.
See here how they relate to each other: https://www.uwgb.edu/dutchs/symmetry/rhombo-lattice.htm
This one should also be quite enlightening - but I am not quite sure whether it might be behind a paywall: http://it.iucr.org/Ab/ch7o1v0001/sgtable7o1o155/
(Alternatively, have a look at the hardcopy of the International Tables of Crystallography, Volume A and look up spacegroup 155 - any decent crystallography lab or library should have it).


text:
What is the difference between an amorphous solid and glass?
They are both formed when a liquid is supercooled rapidly, no free energy, and they both have irregular structures. What defines a glass other than how it is amorphous, transparent, and has a glass transition state?

While a glass is generally considered to be a supercooled, configurationally frozen liquid, not all amorphous solids are glasses. For example, amorphous silicon is a four-fold coordinated semiconducting solid, much like crystal silicon.  Liquid silicon is 8-12 fold coordinated with metallic bonding. Amorphous silicon has been shown to display a first-order phase transition to both the crystal and the liquid.
So, all glasses are amorphous solids, but not all amorphous solids are glasses.

So how can you tell the difference between glass and a supercooled liquid or any other amorphous solid?

The supercooled liquid will have an x-ray diffraction pattern similar to the liquid.  The amorphous solid may be different, and will be different if it is not a glass. Note above the difference in coordination number between metallic silicon and semiconducting crystal or amorphous. The first order phase transition from amorphous to crystal solids is shown in calorimetry.

While I am going to give this an upvote, I do object to "glass is generally considered to be a supercooled, configurationally frozen liquid" although I think it is technically (statistically) true, if a million people say a foolish thing it is still a foolish thing.

@A.K. - fair enough, and that could also be sub field specific. The sub field I was in some time ago (dealing with amorphous silicon amongst other materials) agreed on that distinction.

@A.K. when objecting to something it makes sense to explain in detail why one is objecting and the counter-reasoning if possible ;)

@Thomas The counterexample would be crosslinked polystyrene at RT. A glass, virtually undistinguishable from ordinary glassy PS, but cannot melt. Or any other rubbery substance below $T_g$.


text:
Advantages of thioesters over esters in fatty acid catabolism
In several metabolic pathways, Nature (live systems) chooses to make esterification using thioesters (CoA) and not esters. I would like to know what is the main aim of the Nature with this choice: faster reactions or more extensive reactions?

what is the main aim of the Nature with this choice: faster reactions
  or more extensive reactions?

Nature is frugal, its main aim is to spend as little energy as possible in a chemical reaction.  Nature will always evolve so as to use the lowest energy pathway.
If we compare the resonance structures for an ester and a thioester we see that we can draw the same three resonance structures to describe each structure.

However, the resonance structure on the right, the one with the double bond between the carbonyl carbon and the ester oxygen or the thioester sulfur, is less important in the sulfur case.  Why?  Because a double bond between second and third row atoms is not nearly as strong as a double bond between two second row elements (much better electronic overlap results between atoms of similar size).
As a result the thioester carbonyl is not as stabilized by resonance as the ester carbonyl.  Said differently, the thioester carbonyl is more reactive (lower activation energy to react) than the ester carbonyl.  The thioester motif provides nature with a more reactive carbonyl. This way nature doesn't have to spend as much energy to bring about reaction at this more reactive carbonyl.

Also worth mentioning that hydrogens that are beta to thioester carbonyls are more acidic than beta hydrogens in oxo-esters.  This fact is very important for fatty acid biosynthesis and polyketide biosynthesis.

The reason for that is the electron contribution from the alkoxy group which is bigger than that of the thiolate, due to different sizes of the atoms, right?

authors: Αντώνιος Κελεσίδης


text:
Is there a food-safe hydrophobic liquid with a substanially higher density than water?
Requirements
In short, I am looking for a fluid that meets these criteria:

Hydrophobic
Denser than water
Low viscosity
Food safe

Background
We are working on adding a pool and designing a backyard recreational area.  One design element that we would like to include in multiple places is waterfalls of varying sizes and styles; I love flowing water.  As an extension of that I started wondering if it would be possible to create an underwater waterfall.
Initially I was thinking this could be accomplished by creating a regular waterfall in the pool with normal water, which could then be capped with a clear glass or acrylic watertight cover.  This certainly works as an underwater waterfall, but the cover will be visible, and will likely cloud over time, not to mention the condensation issues.
So I'm wandering now if some sort of fluid exists that could actually flow underwater.  If it's dense enough it will flow down a rockface even under the water, and if hydrophobic will not mix with the surrounding water.  As it will be within a pool that humans will be in for extended periods it must be externally safe, and I'd like something food-safe as kids swallow pool water and it's possible this stuff might get kicked up.  And to add to that tall order, the wrong viscosity will ruin the visual effect.
Does such a fluid exist in a form that a normal person can purchase for a reasonable cost?  I live in the US (Pennsylvania).  Another question, Is there a readily available liquid that's more dense than water and insoluble?, seemed to be looking for something similar but the answer doesn't meet my safety requirements.

The only thing that comes to mind is brominated vegetable oil, but over the last few years this too has come under scrutiny for possible toxicity.  https://en.wikipedia.org/wiki/Brominated_vegetable_oil

Sorry to put a damper on your idea, but even though the fluid might be immiscible with water, natural saponins from plants, detergent residue in the pool (0r even on skin) and even lecithins from insect carcasses or other sources would eventually create an emulsion with some of the fluid and water. How about a gallium fountain (Ga is mildly toxic, though)? It would need to be kept above 30 C, though, which might be OK for a swimming pool.

What visual effect did you have in mind? Even if there exists a liquid that meets all your criteria, I can't imagine it'll make an interesting waterfall. At best, it might slowly flow down the side of the pool like molasses (EDIT: Not quite as slow as molasses, but I don't know if it would have any striking visual effect)

@DrMoishePippik That's a good point; I hadn't considered the effect of contaminants.  I wonder if that could be circumvented by replacing the fluid often enough.  A gallium fountain is an awesome idea, but is unfortunately economically impractical.  :(

@chipbuster I do realize the waterfall will be slow.  My 'low viscosity' requirement was to try and address this issue.  I imagine molasses would make a poor waterfall because of its goopy nature, but a more free-flowing liquid could result in what appears to be a waterfall in slow motion; a very striking effect.  Part of the appeal of this idea, of course, is that it's unusual, and will take some creativity and personal effort to bring to life.  It will help to personalize the space for me and provide an interesting conversation piece.

@CurtF. That's an interesting suggestion.  Wikipedia provides some information about toxicity of ingestion, but not much regarding topical contact.  I'll need to do more research on this.  Thank you.

The gallium fountain was meant in jest, though it could be created: 100 grams is currently ~US $60 at Amazon. An indium-gallium eutectic, m.p.  16 C, might be better... but it's ~US $1,600/kg at http://galliumsource.com/ (or through Amazon, again). Perhaps avery small fountain of molten metal?

@DrMoishePippik Possible, yes.  But even a relatively small fountain, accounting for the piping/tubing and pump volume, will probably require in the realm of a kilogram of gallium.  At $60/gram you'd need to trade in a couple of cars.

You could always go in the reverse direction, creating an underwater fountain using a liquid that is less dense that water.  This would significantly broaden your choices.  Note, however, that an underwater waterfall or fountain will attract the attention of children, who will play with it and try to disrupt it, thus scattering your hydrophobic liquid across the pool.

Hello I know I went late to the party, but I have a solution to this. You all already have the right idea, but you need to put your ideas together.
Double layer your pool
Whether it's gallium, water, oil, or some other liquid it will all work and be safe under glass.
Their are hydrophobic resins that you can treat the glass with, and if you double paned the area that you're waterfall would be under, it would not cause a health problem because there would be no contact.
For example if you had made window panes in your pool (let's just say in a t-shaped stretching halfway under your pool), and you coated the window panes with a hydrophobic resin, and you kept it under let's just say a temperature controlled environment, that would one solve your pool problem of being boring and two, keep your waterfall from fogging up.
So in short
1 Removable hydrophobic panels for cleaning
2 Heated/ chilled pool or temperature controlled environment
3 Heated / cooling pump for waterfall
4 Well designed lighting ( you'd be surprised what a little bit of light in the right spot can do)
I do not know if it's available to the public yet but there is a coolant used for computers that is one consumable and two low maintenance called Coriant. It has a lighter density than water, if you're  panes are hydrophobate this would not present a problem.
Colored mineral oil or antifreeze would also work here.
If you use antifreeze and we're creative with its use, it may assist in some other mechanical uses as well as being a display piece.
Please note that any hydrophobic material or substance is going to be toxic to breathe or have extended contact with.
This should not be a problem if it is under a air / water tight encasing.
I apologize that I'm so late in responding to this but I hope someone reads this and finds it useful thank you for your time.


text:
Activity of different positions in phenol with respect to electrophilic aromatic substitution
Which of the statements below is correct?

a) The ortho and para positions are more active than the meta position.

or

b) The ortho and para positions are activated while the meta position is deactivated.

Neither of the statements is incorrect. Both statements are stating that the electrophilic substitution occurs at the ortho and para positions. The first statement just tells you where the reaction will occur, but does not provide any reason. The second provides a reason, but doesn't provide a cause.

As the figure below illustrates, substituents on aromatic rings that are electron donating like the methyl and t-butyl groups (and also the hydroxy group [e.g. phenol]) increase the reaction rate at all positions (ortho, meta and para) relative to benzene.
On the other hand, electron withdrawing substituents (the chlorine and ester carbonyl examples) decrease the reaction rate at all positions relative to benzene.
So for phenol, or any other electron donating substituent, while the meta position does react more slowly than the ortho and para positions, it still  reacts faster than benzene.  Therefore, it would be incorrect to say that the meta position is deactivated. Only when the rate at a given position is less than the rate for a single position in benzene could one say that the position is deactivated.
Statement "a" in your question is correct.

(image source)
For future reference a note on terminology. The above figure tells us that nitration of toluene proceeds 24.5 times faster than the nitration of benzene; that is to say that the relative rate for nitration of toluene is 24.5 times greater than benzene.  The meta position in toluene reacts 3 times faster than a single position in benzene, this is referred to as the partial rate factor.

While the hydroxy group increases the electron density via mesomeric effects, wouldn't it decrease the electron density at meta position via inductive effects?

Yes, but 1) resonance effects usually outweigh inductive effects in aromatic systems and 2) inductive effects drop off rapidly with distance.

On drawing the resonance structures, isn't it only the ortho and para positions that get higher electron density? How does the meta position receive any resonance effect?

That is a very good question, and also a very different question from your original question; you might want to ask it as a separate question.  Basically, look at your resonance structures. Although the substituent does not interact with the meta position by resonance, it nonetheless stabilizes the overall meta intermediate because a substituent on a double bond stabilizes a double bond.  Therefore, the meta intermediate is stabilized compared to the benzene intermediate and will react slightly faster than the benzene case.


text:
Why do we say "approximation" in the dipole approximation in spectroscopy?
In the dipole approximation, the following relation holds:
$$
\hat{V} = -\hat{\mu} \cdot \vec{E}.
$$
When we say "approximation", I guess we want to point out that we are dealing with linear spectroscopy, where we have the polarizability of the molecule proportional to the electric susceptibility of a molecule:
$$
\vec{p}(\omega) = k\chi \vec{E}.
$$
If the wavelength of the light is not larger than the atom size, the dipole approximation is no longer a good approximation. This is probably also true for large molecules. I'm not sure if this has to do with large molecules no longer having symmetry, so transitions might be allowed which otherwise should be forbidden.
Is this correct?

As per your linked reference, it's an approximation because only the first term in the expansion (here, unity) is considered.
The issue with transitions that are allowed or forbidden arises as a result of using such an approximation: "Forbidden" transitions are not strictly forbidden, it's just that they are weak and, in the linear approximation of the transition dipole operator, are usually not observed.

Well, my question is more, if there is a relation between the "dipole approximation" formula on the top and the "linear spectroscopy" formula? (meaning: Are we dealing with non-linear spectroscopy when we don't make the dipole approximation?)

The term you have - interaction of a dipole in an electric field - is related to what is called "linear spectroscopy" in that it adequately describes the interaction of the system with a single source of the external electric field: That is, the allowed (and observed) transitions are those that arise from one field and not (as in the non-linear case) from more than one.

So for non-linear spectroscopy the dipole approximation $\hat{V}=-\hat{\mu}\overrightarrow{E}$ is no longer adequate?

I think it's adequate - but I believe if you want a model that describes the system being studied at a higher degree of accuracy, it would be wise to include higher-order terms when fitting data or modeling a system computationally.

If it's adequate: linear SpY and dipole approximation have not the same meaning/origin.

I'm afraid that's the extent of my knowledge - perhaps others who know more about the topic can contribute!

Yes, this would be interesting. However, many thanks at this point for your help. I like to communicate, this helps me to see how other scientists think about a certain question.


text:
What is a kg.cal per gram, as a heat of combustion?
I see heats of combustion given in kg [dot] calories per gram molecular weight (Handbook of Chemistry and Physics, 1969-1970). I don't know how that makes sense, as I don't know what a kg cal might be.  I would have expected units of energy per mass (calories per gram) or energy per mole.  What is a kg cal, and what is wrong in my thinking?

"gram" in the title of this thread should be "gram molecular weight".  I previously misread the units in the table in the Handbook.

$\mathrm{kg~cal}$ probably represents the kilogram calorie (especially given that it is an old book). Nowadays it is usually just known as the Calorie, usually abbreviated $\mathrm{kcal}$. It is roughly the amount of energy required to raise the temperature of $1~\mathrm{kg}$ of water by $1~\mathrm{^\circ C}$. Since this is temperature and pressure dependent, there are slight variations in the actual definition depending on the context in which it is used.

Okay, thanks.  Then a kg cal is the same as a "food calorie".  Back in the day, we just called a calorie a calorie and a kilocalorie a kilocalorie.  BTW I find that paraffin wax has about the same energy content as food fat.

I sincerely doubt that kilogram-calorie is "nowadays known as the Calorie"... the two don't have the same dimensions. Perhaps you meant to say that the kilogram-calorie-per-gram is now known as the Calorie?

The kilogram calorie is a unit of energy, not energy times mass, despite the name. It is named 'kilogram' to distinguish it from the 'normal' calorie which is the energy required to heat $1~\mathrm{g}$ of water by $1~^\circ C$. However, the kilogram calorie ($\mathrm{kcal}$) has become the 'normal' calorie due to its extensive usage in food labelling and is now colloquially referred to just as the calorie.

Presumably the "kilogram" in the name "kilogram calorie" refers to heating a kilogram of water, rather than heating a gram of water.

@JohnSurname That is correct.


text:
Can someone explain the equation for entropy?
Here's the equation:
$\Delta S = \frac{\Delta q}{T}$
The confusion I have with this equation lies in the variable for temperature, $T$.  Why is the temperature constant in this equation? Why can't the equation include $\Delta T$ instead of just $T$ and, surely, if $\Delta q$ is negative or positive the temperature must decrease or increase, respectively.

The temperature is not constant in this equation. This equation is just a simplification for cases when the temperature change is negligible and you don't need to take care of it. All the other cases, you should do the integration.

So, say the temperature change is not negligible, could one replace T with ΔT, or does a significant temperature change involve a much more complex equation?

The equation for $\Delta S$ will be very different depending on the conditions; however, all are derived from the infinitesimal relation in Shadock's answer, which happens to be the definition of entropy itself. Most commonly if temperature changes but pressure is kept constant one would use $\mathrm{d}q = C_p,\mathrm{d}T$ and integrate over the temperature range to get $\Delta S = C_p\ln\frac{T_2}{T_1}$ assuming $C_p$ is constant over the temperature range.

I would like to add one comment that by the Second Law of Thermodynamics, the infinitesimal equation only holds true if the heat transfer is reversible: $\mathrm{d}S = \frac{ \mathrm{d}q_{\mathrm{rev}}}{T}$. However, since $S$ is a state function, even if the entire process is not reversible, this has no effect on the value of $\Delta S$ since $\Delta S$ only depends on the initial and final states and not the process by which the system goes from the initial to the final states.

@kmcmillan You keep T, and you integrate the expression with this T changing along the process.

Your relation is not correct, so you must not understand something. If I were you I will asked the question in Physics SE but nevermind.
The good relation is : $$\mathrm{d}S=\frac{\delta Q}{T}$$
I let Richard Feynman explain it to you here (when you click on the link wait few seconds)
I hope it will help you, have fun, live long and do chemistry !

$dS=\frac{\delta q}{T}$ is just the infinitesimal version of $\Delta S =\frac{\Delta q}{T}$ and still implies that $T$ is constant or minimally changing in comparison to $q$. The OP is interested in knowing how this equation would be different if $T$ varied significantly (or perhaps that $q=Q(T)$.

And there is a big difference ...

Agreed, but it does not answer the OP's question (especially if the OP is only at the general chemistry level). It would also be helpful to summarize the content of your link to combat link rot.

I am french and don't speak english as well as i understand it so if you are you will do better than me :)

@BenNorris No, it doesnt imply that T is constant. To obtain the whole change of S, you have to integrate the infinitesimal expression, where T can vary at every point.

@BenNorris is right. I'm no where near this kind of stuff yet. I guess I just needed to see if this kind of question was beyond me or not (I kind of figured it was). Thus far, I've taken calculus I and general chemistry. Perhaps I should have mentioned that in the first place.


text:
Is there a type of detergent/surfactant that evaporates at STP?
Are there any type of detergent/surfactant chemicals that would be good for removing dirt and grease from fabric that also would evaporate from the clothes within, say, 24 hours leaving no residue?
Googling "volatile surfactant" turned up fluorocarbons and something called Surfynol 61, which is really 3,5-Dimethylhex-1-in-3-ol as possible candidates. I don't know if these would be safe for laundry.

Instead of looking for some magic chemical, why don't you tell us what you're trying to clean, and what has stained it.

Volatile surfactants
There are many molecules that are both surfactants (in water) and volatile, but of the ones I can think of, none can be used safely in home laundry applications.

Hydrocarbon derivatives.  The examples you found by googling are in this class.  3,5-dimethyl-1-hexyn-3-ol is the principal component of Surfynol 61 (probably named Surfynol 61 because it is a surface-active (surf) alkyne alcohol (ynol) with six carbons (6) and the alkyne moiety at carbon one (1).  Other examples would be medium-chain fatty alcohols like hexanol, octanol, cyclohexanol, etc.  These molecules reduce surface tension of water and have OK-ish cleaning power (not as good as larger but non-volatile alternatives like SDS), but their volatility is also their downfall: their use in home environments creates a strong fire and explosion hazard when used in automatic heated dryers.
Fluorocarbon derivatives.  I found the same article you probably did about volatile fluorocarbon surfactants.  The most powerful at reducing the surface tension of water seems to be nonafluoro-tert-butyl alcohol, which reduces water surface tension by 78%!  However this experiment required saturating the water with the vapor of this compound.  And 1 gram costs >$80.  And toxicity is a concern, despite this compound apparently being researched as a component of or precursor to artificial blood mixtures in the 1970s.
Other halocarbon derivatives.  These include solvents like perchloroethylene, chlorobutanol, etc.  These are all highly toxic, and with some of them, there is a flammability/explosion risk as with hydrocarbon derivatives.

Non-surfactant use
You might consider just using organic solvents that are not (very) surface active as a way to degrease fabric.  However, these may leach dyes out of the fabric or otherwise damage them.  Isopropyl alcohol (IPA) is accessible to home users, and is relatively safe.  Be sure to do the soaking in a well-ventilated area, and do not use heated dryers to remove the IPA!  Instead, rinse out the IPA thoroughly with water.  Depending on local regulations, you will have to dispose of all of the dirty IPA and IPA-contaminated rinsing water as solvent waste.
The best option
Pay to have your clothes dry cleaned!

Really, you want to rinse your surfactant out of your clothes. There's not much point in making an emulsion of your grease, then leaving it there till the surfactant evaporates. You want to emulsify the grease to render it water-dispersable, then disperse it in water, then remove the water as a liquid, then let the remaining water evaporate. To clean anything the surfactant has to leave in the aqueous/liquid phase of the cleaning solution. The use of a volatile surfactant would be to suspend an oily substance in water, and then leave it behind when the water/surfactant evaporate.
If you just want to wash out an oily substance without water, then as in dry cleaning, you just need a solvent, not a surfactant (and still want to remove it in the liquid phase). So if you want to clean a greasy stain, add the solvent, then blot it off. Acetone (or the previously mentioned isopropyl alcohol) would probably work well, with the stuff you don't blot off easily evaporating (or coming out in water).

May the force of additives be of aid to you!
Beforehand, choose an easier type of detergent:

"Surface-acting polyglucosides [which are] easily broken down by microbes, leaving no traces in the environment", is one (Quellen, 213).

What is slow to biodegrade may not be used as often-inform yourself, and go from there!
And please, be safe.
Source:
Why There's Antifreeze in Your Toothpaste: the Chemistry of Household Ingredients. P 213. Quellen, Simon. 2008.

Polysaccharides are not volatile.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. If you have a NEW question, please ask it by clicking the Ask Question button. If you have sufficient reputation, you may upvote the question. Alternatively, "star" it as a favourite and you will be notified of any new answers.


text:
Amending soil pH to change the color of hydrangea flowers
I would like to ensure that my hydrangeas are blue. I am told that adding ammonium sulfate or aluminium sulfate will cause the soil to be more acidic. I am also told that adding elemental sulfur will help. Will adding magnesium sulfate to the soil help?

Although it is true that soil $\mathrm{pH}$ plays a role in the color of hydrangeas, the presence of aluminium is actually what determines the flower color. Aluminium is a naturally occurring mineral that is found in most  soils, however, just because a mineral is in the soil doesn’t mean the plant is absorbing this mineral. This is where soil $\mathrm{pH}$ comes into play – plants are able to absorb aluminium when the soil $\mathrm{pH}$ is between 4.5 and 5.8 with the best range being between 5.0 and 5.5. When the $\mathrm{pH}$ moves to 6.0 and above then the plants will not be able to absorb the aluminium despite it being present in the soil.
Adding either sulfur or magnesium sulfate (epsom salt) to the soil will lower the $\mathrm{pH}$, but this will not help with the presence of aluminium in the soil, which is why many aluminium sulfate is so highly recommended.
This is not to say that magnesium sulfate would not be helpful in other ways, however, as magnesium is the central atom of pigment chlorophyll, which will help to improve the foliage quality of your hydrangea.


text:
Determination of carbon content of steel by liquid chemistry
It seems like the typical laboratory methods for determining the carbon content of steel are to cuppellate it in a stream of oxygen and then measure the volume of the evolved $\ce{CO}$ and $\ce{CO2}$, which involves many high-temperature complexities.
Why can't the carbon content of steel be measured by dissolving it in acid and reacting out the carbon at low temperature?

I am interested to know what happens the the carbon when steel when it is reacted with acid.  Would the reaction be different in nitric acid vs hydrochloric?

I don't know much about metallurgy. I think reaction of steel with acid leads to corrosion.  So does corrosion play any role here?

I think cuppellate is a foreign word meaning "burn" essentially.
The basic technique would be to have the iron sample in a small ceramic boat which is put into a hot stream of $\ce{O2}$ so the iron and carbon both oxidize. The iron oxide stays in the boat, and the $\ce{CO2}$ is sweep away and trapped/reacted somehow. We talking small samples 10 grams perhaps.
If you try to dissolve the steel in an acid solution then the $\ce{CO2}$ gas is formed in solution. It is really hard to get a small amount of $\ce{CO2}$ out of solution given that you are starting with a small amount to begin with. Also if you try to scale up such a reaction it gets to be a problem. A $2~\mathrm{cm^3}$ block of iron takes a long time to dissolve in acid.
So choosing the analytical method depends on a lot of factors. Time per sample, sensitivity, and overall cost being the main ones.

May I introduce you to the \ce{...} notation? Use that in MathJax to get upright chemical expressions. E.g. $\ce{C + O2 -> CO2}$ gives $\ce{C + O2 -> CO2}$. Learn more on meta


text:
Solvent effect on gold nanoparticle and silver nanoparticle in UV spectroscopy?
Can anyone explain to me that how solvent factor affect the wavelength and intensity of gold nanoparticle and silver nanoparticle in UV spectroscopy?

Attenuation of light related to the properties of material is governed by beer-lambert law Many molecules absorb ultraviolet or visible light. The absorbance of a solution increases as attenuation of the beam increases. Absorbance is directly proportional to the path length, b, and the concentration, c, of the absorbing species. Beer's Law states that
A = ebc, where e is a constant of proportionality, called the absorbtivity.
Different molecules absorb radiation of different wavelengths. An absorption spectrum will show a number of absorption bands corresponding to structural groups within the molecule. For example, the absorption that is observed in the UV region for the carbonyl group in acetone is of the same wavelength as the absorption from the carbonyl group in diethyl ketone.
here is beer law description:http://teaching.shu.ac.uk/hwb/chemistry/tutorials/molspec/beers1.htm

I believe this hardly answers the question. Lambert-Beer's law just states that transmittance is proportional to concentration and path length. It does in now way say anything about how a solvent will have an effect on that.

Mentioned in previous answer (beer law) explains differences in intensity because, apart from nanoparticles, each solvent absorbs differently. But in the most of the measurments of the nanoparticles suspended in solvent, spectrum originating from solvent itself is treated as baseline. Therefore final measured spectrum originates only from nanoparticles.
But medium in vincinity of the nanoparticle alters the spectrum of the nanoparticle itself.
UV-Vis absorption spectrum of the Au/Ag nanoparticles consists of more than absorption, because they manifest presence of LSPR (Localized Surface Plasmon Resonance).
First thing is interband absorption, in which nanoparticles absorbs light causing excitation of charges between bands in nanoparticle electronic band structure. Molecules and quantum dots (called artificial atoms because of their small size and similiar electronic structure) have discrete absorption spectrum and discrete energy levels. When we deal with larger structures e.g. nanoparticles, we must consider their band structure causing continous absorption spectrum). Nanoparticles are treated as 3D objects (which have parabolic energy-momentum electronic dispersion: Wikipedia - density of states). Density of states illustrates shape of absorption spectrum profile originating only from interband excitation and look like this: http://britneyspears.ac/physics/dos/images/Image441.gif.
Second thing is LSPR (https://en.wikipedia.org/wiki/Localized_surface_plasmon). It originates from oscillation of excited electrons in nanoparticles and it causes additional absorption and scattering events in nanoparticles solution. LSPR causes Lorentzian-type absorption profile with maximum peak placed by the wavelength corresponding to resonant oscillations of the electrons. This phenomenon occurs in the vincinity of surface of the nanoparticle and surrounding medium. Wavelength where LSPR peak is placed, depends strongly on the electrical permittivity of both the nanoparticle and the surrounding medium. So when You change between solvents, which have different permittivity, You can change placement of LSPR peak. That also counts for diffrent surface countings of the nanoparticles (because change of the permittivity surrounding metallic core), which also changes LSPR absorption.

There is one more additional effect originating from LSPR. When nanoparticles are well-suspended in solvent, they are "far" from each other and You cannot see effect from interaction between LSPR from two different nanoparticles. But, for example, in case you have nanoparticles with polar molecules attached the surface and have more polar solvent as medium, nanoparticles will tend to stick together and form agglomerates. Then they will be so close, that LSPR oscillations from each nanoparticle will interact wich each other - that will also change placement and intesity of LSPR peak.


text:
difference between absorption spectroscopy and extinction spectroscopy
can anyone tell me the difference between absorption spectroscopy and extinction spectroscopy in terms of experiment? and how to get extinction spectroscopy, how to get absorption spectroscopy? Thank you so much.

https://physicstoday.scitation.org/doi/abs/10.1063/PT.3.4547?journalCode=pto reports experimental distinctions between the two, and credits reference 14 with doing that work: Read https://iopscience.iop.org/article/10.1088/2053-1583/aab670 and https://aip.scitation.org/doi/10.1063/1.5047792 for experimental procedures. My guess is that you'd measure scattered light as well as transmitted. The scattered light would be the difference  - .

There should not be any difference in experiment, because the extinction is something you have to calculate by measuring the transmittance/absorbance of a given frequency.
The transmittance $T$ of a material is the ratio between the received intensity of a certain frequency to the transmitted intensity of the same frequency. $$T=I/I_0$$
The absorbance $A$ is defined as the decadic logarithm of the inverse transmittance. $$A=\log_{10}\left(I_0/I\right)=\log_{10}\left(1/T\right)$$
By using Beer-Lambert’s law $$I=I_0\operatorname{e}^{-\varepsilon~c~d}$$ one can calculate the extinction $\varepsilon$ from the measured transmittance/absorbance by $$\varepsilon=\frac{A}{c~d}$$ where $c$ is the concentration of the absorbing material and $d$ is the distance that the light has to travel in that given material.


text:
Equilibrium concentration, where does the 5.00 for iron thiocyanate complex come from?
Trial #1:
$$\ce{Fe^3+(aq) + SCN^- (aq)  ->   FeSCN^2+(aq)}$$
\begin{array}{|r|c|c|c|}\hline
\mathrm{Initial~}(M)&\phantom{-}6.00 &\phantom{-}10.00&\phantom{-}0.00\\\hline
\mathrm{Change}~(M) & –4.00 & –4.00 &+4.00  \\\hline
\mathrm{Equilibrium}~(M)&\phantom{-}2.00&\phantom{-}6.00&\phantom{-}4.00\\\hline
\end{array}
$$ K = 0.333$$
Consider the reaction represented by the equation:
$$\ce{Fe^3+(aq) + SCN- (aq) \longrightarrow  [Fe(SCN)]^2+(aq)}$$
Trial #2:
Initial: $10.0 ~M\ce{~Fe^3+(aq)}$ and $8.00~M\ce{ ~SCN- (aq)}$ (same temperature as Trial #1)
Equilibrium: $x~M\ce{~FeSCN^2+(aq)}$
I have tried putting $10.0$ and $8.00$ in an ice table, then when I have finished I start solving to get my equation as $$0=\frac{0.33x}{(10.0-x)(8.00-x)}$$ when I'm done I come out with $$0.33=\frac{x}{(x^2-18x+80)}$$ then I plug in for quadratic equation,
$$x=\frac{-6.94\pm\sqrt{6.94^2-4\times0.33\times26.4}}{2\times0.33}$$
my answers came out to be $-5$ and $-16.03$.  What am I doing wrong?

Welcome to Chemistry.SE. Take the [tour] to get familiar with this site. This appears to be a homework question, please share your thoughts and attempts towards the solution. It'll make us certain that we aren't doing your homework for you.

Apart from this, please consider copying the content relating to your question of the linked power point presentation. This site should be self contained, but links can change and then this question would not help anyone any more.

This is getting better - could you please format your post according to http://meta.chemistry.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here

Wow! That is a much improved edit - my retinas are saved!

Could you include your ICE table, I fail to see how you derived your formula.

Thank you LordStryker & Martin, I understand it now that LordStryker have helped me edit it, and thank you Martin for further explanation for this problem, so I use fractions instead of decimals, correct?

You are just making a small sign error:
\begin{align}
ax^2 +bx +c &= 0 & x_{(1;2)} &= \frac{\color{\red}{\pmb{-}}b \pm\sqrt{b^2 - 4ac}}{2a}
\end{align}
Taking $K=\frac13$:
\begin{align}
\frac13x^2 - \frac{21}{3}x +\frac{80}{3} &= 0 &
 x_{(1;2)} &= \frac{\color{\red}{\pmb{-}}(-\frac{21}{3})\pm\sqrt{\left(\frac{21}{3}\right)^2 - 4\cdot\frac13\frac{80}{3}}}{\frac{2}{3}}\\
&& x_{(1;2)} &= \frac{7\pm\sqrt{49-\frac{320}{9}}}{\frac{2}{3}}\\
&&           &= \left(7\pm\frac{11}{3}\right)\cdot\frac{3}{2}\\
&& x_{(1;2)} &= 16; 5
\end{align}
The first value obviously does not make any sense, hence $c(\ce{[Fe(SCN)]^2+})= 5~\mathrm{mol\, L^{-1}}$.


text:
What are the prerequisites in hydrogen bonding?
I know when a hydrogen bond is, but - can hydrogen bonds only occur between two molecules containing hydrogen? Or, only one of the molecules should contain hydrogen and the other one may not contain hydrogen, but should have a large electronegative difference between two of its sides?

Actually, theoretically speaking hydrogen bond is the attractive force between the hydrogen attached to an electronegative atom of one molecule and an electronegative atom of a different molecule. Usually the electronegative atom is oxygen, nitrogen, or fluorine, which has a partial negative charge. A hydrogen bond may occur between $\ce{H-O}$, $\ce{H-N}$, $\ce{H-F}$ (first two are more common). Apart from that, these hydrogen-bond attractions can occur between molecules (intermolecular) or within different parts of a single molecule (intramolecular)
See also the Wikipedia article about hydrogen bonding
A hydrogen atom attached to a relatively electronegative atom will play the role of the hydrogen bond donor An electronegative atom such as fluorine, oxygen, or nitrogen will be the hydrogen bond acceptor, irrespective of whether it is bonded to a hydrogen atom or not. An example of a hydrogen bond acceptor that does not have a hydrogen atom bonded to it is the oxygen atom in diethyl ether.
http://chemwiki.ucdavis.edu/@api/deki/files/4680/image115.png?revision=1

Hi Martin, as I said, I know th definition of a hydrogen bond.
My question is-  should hydrogen be present in both molecules participating in a hydrogen bond or not ? If not , will you please give me an example ? THanks a lot!

actually hydrogen bond is bond between one atom which must be hydrogen and other one can be oygen, nitrogen or fluorine it is not necessary to have hydrogen in both molecules involved in bonding one molecule must have hydogen( that one  is proton donor) here is example:https://www.google.co.in/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=8&ved=0CAcQjRw&url=http%3A%2F%2Fchemwiki.ucdavis.edu%2FPhysical_Chemistry%2FIntermolecular_Forces%2FHydrogen_Bonding&ei=9cCHVZDzEMeouQSb8LaICg&bvm=bv.96339352,d.c2E&psig=AFQjCNFisJ6bt_r2aigqvPOnP4LRP35CSQ&ust=1435046466119122