text:
Preparation of ferric hydroxide sol
I have read that before the preparation of ferric hydroxide sol, the beaker must be cleaned by the steaming out process, while it is not necessary for starch sol. Why is that?

It might be because starch sol is quite stable to impurities, whereas ferric hydroxide sol isn't. I am quite sure that is also mentioned in your book.


text:
Why do asymmetric stretches occur at higher wavenumbers than symmetric stretches?
I learned that the $\ce{N-H}$ bonds in a primary amine lead to two signals the infrared spectra at $\sim \pu{3400 cm^-1}$, one for symmetrical stretching and another for asymmetrical stretching. For example, here is the IR spectrum of butylamine, taken from Pavia & Lampman's Introduction to Spectroscopy (5th ed.):

The same is seen in, for example, carbon dioxide, where the asymmetric stretch ($\Sigma_\mathrm u^+$) occurs at $\pu{2348 cm-1}$ and the symmetric stretch ($\Sigma_\mathrm g^+$) at $\pu{1373 cm-1}$ (from ChemTube 3D; the NIST Webbook gives similar values).
Why is this so?

source: Stack Exchange

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

sort: votes


text:
How does common salt speed up the setting of the plaster of Paris?
We use plaster of Paris (POP) to cast and lift foot prints/tyre prints from the crime scene for evidence purpose. While preparing cast, we follow standardized protocol and add a pinch of common salt to the POP solution. It catalyzes thickening of the POP. I am keen to know chemistry behind it. How does salt react with POP?

Interesting question. Dissolving POP in salt water will make POP set faster too, so it isn't solid NaCl that increases the set rate. An interesting paper is found here (I have no idea if it the current state of the art...) https://ia801704.us.archive.org/17/items/someaspectsofset755clif/someaspectsofset755clif.pdf

As a side comment I'll point out that exactly how crystals form, and how they grow into different forms is still an active area of research. Many minerals have multiple forms influenced by temperature, pressure concentrations, flow rates and so on. So as you peel the onion on this question it quickly gets very complex.

I tried using brine instead of just a pinch of salt and the gypsum plaster didn't set at all. Not sure why just an observation. I guess too much salt could cause it to set too soon whilst still particulate and so the particles can't set together

My guess would be that it absorbs the water in the plaster of paris. Plaster of paris normally has what's called 'water of crystallization' in it. However, it sets too fast with just this, so extra water is added. The salt absorbs that extra water, thereby making it set faster.
Here's a website with details about plaster of paris chemically: http://collections.infocollections.org/ukedu/en/d/Jg16foe/7.4.html
and a book called 'Science in the Real World: A simplified story of how technology using chemistry and physics is used in the real world of industry' has good information on it as well.

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


text:
Is this asking for the equivalence point?
Is this question asking for the equivalence point of the titration curve?

At what point does the concentration of acetic acid equal the concentration of the acetate ion? What is the pH of the equimolar, ideal buffer solution at this point.

At first, I thought this question was asking about the point right before the titration curve shoots up. However, I don't recall ever learning about such a point.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. This appears to be a homework question, please share your thoughts and attempts towards the solution - although I am not sure if it is just a very basic question. Have a look at the wikipedia page of buffer solutions, I am certain it will address this issue.

The equivalence point is would not be what you are looking for. You are trying to find the half-equivalence point, where instead of all the acid being neutralized and converted to acetate, only half is. If you have a titration curve, this occurs at half the equivalence volume. Based on MaxW's formula below, what would the pH be when the concentrations of acetic acid and acetate are equal?

If you were titrating acetic acid with a base, then the equivalence point would be where as much base has been added (i.e. moles of base vs moles acid) to the solution as there was acetic acid. This is essentially the point where all the acetic acid has been neutralized.
In general the "pKa" is the term used to define the point at which the protonated and unprotonated forms are equal. The pKa is from the negative log of the acid dissociation constant as given by the reaction:
$$\ce{HA <=> H+ + A-}$$
and the equation: $$ K_\mathrm{a} = \dfrac{\ce{[H+] [A- ]}}{\ce{[HA]}}$$
where $\ce{[H+] = [A- ]}$ then:
$$K_\mathrm{a} = \dfrac{\ce{[H+]^2}}{\ce{[HA]}}$$
and $$ \ce{pKa} = -\log \ce{Ka} = -\log \dfrac{\ce{[H+]^2}}{\ce{[HA]}}$$


text:
Why can't antiaromatic compounds just escape planarity and become non-aromatic to avoid destabilization?
Why can't just all anti-aromatic systems lose planarity (and hence conjugation) and be non-aromatic like cyclooctatetraene ?
Would they not be more stable that way? Because anti-aromatic systems tend to be highly destabilized.
What can be a good working rule to predict whether an anti-aromatic system will remain anti-aromatic or whether it will somehow manage to lose planarity and become non-aromatic?

All the anti-aromatic systems I can think of undergo distortions to avoid it. Can you provide some examples of the systems you are thinking of?

@bon so you are saying that practically anti-aromatic systems do not exist.

I think that there is exactly one antiaromatic system (and its derivatives) known to mankind and that is cyclobutadiene — which is too strained to deplanarise in any way so it has to be planar. All other systems distort themselves as far as I know to avoid antiaromaticity — or in some cases such as the cyclopentadienyl cation, don’t even allow formation.

@Jan Apparently some cyclobutadiene derivatives are non-planar. http://onlinelibrary.wiley.com/doi/10.1002/anie.198002111/pdf

@Jan Doesn't cyclobutadiene avoid being antiaromatic by distorting from a square structure to a rectangular structure with alternating single and double bonds?

@ron Well yes of course, the structure contains localised bonds. But correct me if I’m wrong: even with differences in bond length, it still has a single pi system and the only reason why it is not a biradical is the lowering of symmetry from $D_\mathrm{4h}$ to $D_\mathrm{2h}$. That still makes it formally antiaromatic imho.

https://chemistry.stackexchange.com/questions/57678/is-compound-aromatic-if-it-has-also-anti-aromatic-ring

https://chemistry.stackexchange.com/questions/44743/aromaticity-of-1-1-bicycloprop-2-en-1-ylidene-and-1-4-dioxine

What about going a bit further, why would an anti-aromatic systems that can distort form in high enough concentrations to be observable to begin with?  I assume that if the energies of two molecules are similar and one is anti-aromatic there will be some Boltzmann distribution of states?

They do try to escape planarity if possible. As a simple example, consider cyclooctatetraene. It has 8 electrons, which is a non-Huckel number, considering it's a multiple of four, and would correspond to an antiaromatic system, if the ring was planar, and allowed delocalization. However, cyclooctatetraene isn't planar. It's rather "tub-shaped", as seen in this structure here:

Click the image for interactive 3-D model
The electrons aren't delocalised either. Cyclooctatetraene is known to add bromine, and generally behave like a cyclic polyalkene in many aspects.
The only cases where molecules don't escape planarity is when they cannot. This happens in the case of cyclobutadiene. Such a molecule is so small, it cannot be distorted to such extents. As a result, it remains planar, antiaromatic, and immensely unstable:

Click the image for interactive 3-D model


text:
Negative numbers for ionic radii?
I was looking for a good table of ionic radii for an experiment on the van't-hoff factor when I stumbled upon R. D. Shannon, Acta Cryst. 1976, A32, 751.
In table 1, it lists 1 of the ions of carbon with as so:

C +4 1S 2 III .06 -.08

The bold numbers are of concern here. The first number, 0.06 Å is what the paper refers to as the crystal radius, while the second number, -0.08 Å, is referred to as the effective ionic radius. I was also unable to find any good distinction between the two in the paper, except that the radius of $\ce{O^2+}$ for the crystal radius is 1.40 Å, while the effective ionic radius for $\ce{O^2+}$ is 1.26 Å.
I also doubt that this is a typo since it agrees with this statement made later in the paper:

crystal radii differ from traditional radii only by a constant factor of 0.14 [ Å].

Interestingly $\ce{H+}$ is listed as having both a negative crystal radius as well as a negative effective ionic radius. Only light, highly positively charged ions appear to give negative values ($\ce{H+}$, $\ce{D+}$, $\ce{C^{4+}}$, $\ce{N^{5+}}$). It is likely that these values are calculated in a way which cannot handle the extremely high demand for at least some covalency from these strongly polarising cations, so the model breaks down.

Though  I'm not 100% sure about this, the paper seemed to not be a model, but instead was directly measured. I couldn't find a real model anywhere so I can only assume that it must have been directly measured. These values were also not discussed anywhere in the paper as far as I can tell.

Also, that about negatively charged ions: why are they not affected?

Is it possible that the radii were measured in referential unit like amu?

I'm going to take a much closer look at the paper and then come back to this question.

The articl;e with the mathematical model seems to be: R. D. Shannon and C. T. Prewitt "Effective ionic radii in oxides and fluorides" Acta Cryst. (1969). B25, 925-946 doi:10.1107/S0567740869003220 // Seems a single best fit radius was calculated

Ionic radii are not measured on individual ions by some sort of mini-micrometer, but bond distances can be measured and then radii are assigned to the individual atoms/ions. The fuzziness of the assignments can be seen by the variation in ionic radii as the coordination number changes: if you give the ion more room (greater coordination number), it will fill the space - if you assume that the ligands have the same size as usual (whatever that is).
(There are materials (ferroelectrics) in which the coordinated ion does not completely fill the space allotted. In $\ce{BaTiO3}$, the $\ce{Ti^{+4}}$ ion is so small that it can shift its position in the octahedral cage of oxygen anions in response to an external electric field.)
The data in the Shannon paper are extensive, certainly a credit to the Central Research Department of DuPont in 1976, but the trends reported seem to be much more valuable than any specific datum.
Nicolau mentioned covalency as one factor in determining radii. Purepani brought up negative ions not having an effect of negative radii. Assigning a radius to a proton other than zero suggests that you are allowing it to have some electron density surrounding it, able to repel an anionic ligand. Electrostatically, a positive ion will attract some electron body (anion or atom) and come closer; depending on your initial assignment of radii, you can adjust to keep your idea of proton radius constant, or adjust, even into the negative region (because we assume you have a good imagination). Negative ions have a larger size than protons, and are less hard, so they can dent a little. One reason to use negative radii for the cations is that the anion may be dented only at one site, but has a "normal" radius everywhere else around it.
My approach to handling a negative radius is to conclude that the proton (or other mini- cation) is so electrostatically demanding that it can begin to bury itself in the outer shell of the anionic ligand, if the coordination number allows such a deviation in distance to be measured. And since we don't want to redefine the anionic radius, a negative radius is assigned to the proton, and the mathematics works out OK.
I guess you could say atoms and ions are somewhat mushy, and mathematics can deal with it.

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


text:
Zero dipole moment of p-dinitrobenzene
Why is $\mu=0$ for p-dinitrobenzene ? How can the dipole moments of the two Nitro groups exactly cancel each other ? Shouldn't it have a non-zero, albeit feeble, dipole moment ?

They exactly cancel each other because they are exactly similar.

Why would it be non-zero? The molecule is planar and the nitro groups are symmetric.

@bon So both the N-O bonds in a nitro group are of exactly equal length.

@Abhirikshma Yes that is correct.

The nitro groups most probably undergo free rotation and so on average are symmetrically disposed about the molecule, so no dipole. If there was some restricted rotational motion, due to say interaction with adjacent H atoms then the dipole could be very small but still approximate to zero.

@porphyrin The two nitro groups will statistically rotate at a different frequency. The higher rotational states have different bond angles. Plus the still and rotating nitro group have different interaction with the aromatic ring. Both clearly lead to a small, nonvanishing dipole moment.

@Karl, I am sure that  both nitro's will have the same average behaviour  over time, this time being no more than a few nanosec, and so they will be equivalent (or interchangeable) in any normal dipole measurement. However, my point was that if there is a potential partly restricting rotation,  then it might be that there is a preference for nitro's to be at some angle with the ring and this may generate a small dipole. I say this only as it is reported below that there is a small dipole and so seek an explanation. I do not find this explanation very convincing, however.

@porphyrin You mean that the rotational states of the two nitros vary so quickly that it can have no influence? In the liquid, possibly, but think about the gas phase. Also: A dipole that fluctuates very quickly is still not a vdW interaction, which are very high frequency. I doubt you can neglect it.

@Karl, By 'quickly' I mean that both nitros exhibit average values of their motion  because of the relatively long time it takes to make a measurement i.e. >> nanoseconds.  Naturally it is assumed that the vapour or solution is dilute enough so that each p-dinitrobenzene is isolated from others and that any  buffer gas or solvent is inert. I don't see what vdW has to do with it.

We can measure bulk dielectric properties in the gigahertz range, there's no need to do that in nanoseconds. And put dinitrobenzene in an inert solvent and tell me that it has totally the same induced-dipole-only interaction with it than the solvent molecules with themselves.

This paper (J. Sep. Sci. 2011, 34, 1489–1502) lists 1,4-Dinitrobenzene as having a very small dipole moment (0.01 Debye). No error is given on the dipole moments, but they do list compounds with exactly zero dipole moment (naphthalene for example).

Let the two nitro groups have a different rotational excitation against the aryl core, and i'm sure you can construct a small dipole moment. Firstly, higher rotational states have different bond angles.
Or easier to imagine: Let one stand in plane with the aryl ring, and one perpendicular or rotating. Different interaction with the aromatic ring, clearly leads to slightly different electronic structure, ergo: a dipole moment.
The dipole moment should vanish at low temperatures, but i guess the stuff crystallises before.


text:
ChemDraw: Automatic Caption with Number for Molecules
When working on some reaction schemes in ChemDraw, I sometimes forget to add text/captions with numbering and align it with my molecules before placing all the molecules in a reaction scheme and aligning that.
Now that I already have a scheme, adding the numbers to the molecules will either mean I have to place the numbers by hand (which would mean they are not exactly centered) or aligning the captions to the molecule. The latter option would work, but would also mean I have to redo a lot of the work aligning my scheme.
The question is then: Is there a way to automatically place a number under a molecule, so I do not have to worry about realigning everything?

There is no fully automatic method for creating labels. However, you can manually generate property labels, which are automatically placed and centred without disturbing any existing alignment, as follows.
Select the entire molecule, right-click the selected structure, then click Analysis, and then click one of the available options (Name, Chemical Formula, Exact Mass, Molecular Weight, m/z, or Elemental Analysis):

Use the Text tool (on the tool bar) to modify the text of the label (after your edit, you will receive the warning “This change to the text has caused auto-updating to be disabled.”):

You can use the Text menu, the Style toolbar, or shortcut keys (e.g. CTRL+B) to change the text format (structure labels are usually written in bold). Repeat this procedure for all relevant structures:

Oooooh, I never knew that, great feature!

Usually (personally) I manually create the textboxes and then use Object > Align... to get the desired alignment. Of course this works too.

@orthocresol OK, however Object -> Align also moves the molecule, and OP wants to avoid that.

Oh, yeah, oops, didn't see that.

Just obey these steps from the current manual (according ChemDraw V19.1 documentation):

Welcome to Chemistry! A screenshot or picture of an exercise is not searchable. Please consider rewriting it, so that it can be of help for future visitors.


text:
Is Cold water heavier than hot water?
I have come across this Quora Answer where it was written that cold water is heavier than hot water.
Are there any proofs which support it?

You understand what mass and density are? Colder objects usually have higher density.

@Mithoron Yeah, I understand that. However, I have seen a counter argument against that which confused me.

So about mass see http://chemistry.stackexchange.com/a/20103/9961 Yes higher energy means higher mass, but this effect is usually negligible.

It's confusing because you are confusing mass and density.  One molecule of hot water is slightly heavier than one molecule of cold water because of the extra energy.  However, one liter of cold water is heavier than one liter of hot water because many more cold molecules will fit in the liter.  Therefore if you take a liter of cold water and heat it the mass will increase, but so will the volume, for a net decrease in density.

@James: Excuse me for basically saying the same thing as your comment, in more words.

@DevSolar: No problem at all.  I didn't have time to create a good answer.  Hopefully giving a short answer in a comment as I did is not against etiquette here.

I'm voting to close this question as off-topic because it is about physics, not chemistry.

Consider migrating this question onwards to physics. It is not within what is usually defined as ‘chemistry’.

The "counter argument" you link to cites E = mc^2 and the higher energy content of hot water.
First off, note the "c^2" in that formula. That's the speed of light, squared. It is a huge number, meaning that you need huge differences in energy to get tiny changes of mass.
The other way around, you only need little mass to generate lots of energy, something nuclear physicists have been exploiting for quite some time. ;-)
So, while that effect is "true" to some extend, it is tiny. But yes, one molecule of H2O has more mass when hot than the same molecule when cold.
But when we talk about "weight", we usually mean "density", i.e. weight per volume. And here, a different, much more pronounced effect comes into play: Most substances expand when heated, reducing their density. Each molecule might be a little bit heavier, but the molecules are much farther apart, more than offsetting the E = mc^2 effect.
Cooling down increases the density -- the same amount of matter takes up less volume. An effect that home preservers have exploited for generations when canning things: You put hot produce into a container, and close it. The contents cool down and contract, producing underpressure in the container, sealing it. That's the "pop" you hear when opening a glass of marmelade the first time.
So, per molecule hot H2O is (a bit) heavier, but per volume hot H2O is (much) less dense.
At least down to about 4 degrees celsius, where a peculiar quirk to the H2O molecule makes it become less dense again with further cooling. That's why a pond starts freezing at the surface, not the bottom, where the 4 degree water is up to the point where the whole pond is frozen solid. An effect that fish have been exploiting for quite some time now. ;-)

This was the complex answer I was hoping for when Googling a similar question, but doesn't quite answer the question I had.  I supposed I should ask it as a separate question:  Where is the coldest water in a glass of ice water?  Is cooling fairly uniform?

@NDGeek: The coldest water in a glass of ice water would be the ice in the ice cube, in solid state. ;-) But you're right, that's better asked as a separate question (as you are actually asking about the coldest liquid water, I presume).

Relativistic effects (which is what that answer talks about) are tiny compared to chemical effects (that's why it took us so long to understand them). Hot water has a lower density than cold water like most other substances.
Note that there is also an anomaly concerning cold water and ice that causes ice to float, but for the most part of the temperature range, water behaves like any other substance that expands with heat.

Water below 4 degrees is lighter than water at 4 degrees. So that means within that range that cold water is lighter than hotter. However when you compare water above 4 degrees, it gets less dense as it gets hotter.

A proof will be the fact that water is most dense at $4^o$C and therefore sinks to the bottom and supports aquatic life even when a water body is frozen.
State of a matter or the state of aggregation as the name suggests gives us that state in which matter wants to collect or aggregate. As the temperature increases the atoms gain energy and start to move around and occupy more space as compared to when their temperature was lower hence their density decrease and hence becomes lighter.
This is a true but not completely since this is would be true if water was just what we imagine it to be. Which it's not. Due to presence of hydrogen bonding in water we get to see many amazing phenomenons.
While $\ce{H2S}$ is a gas $\ce{H2O}$ is liquid all because of H-Bonding.
Water is most dense at $4^o$C. This contradicts my original explanation but is supported by the fact that the explanation is very broad and not completely correct. The anomaly can be explained by presence of H-Bonding.


text:
In PCl2F3, why are two Cl atoms not on axial position?
According to these sites:

Chemistry Help

Molbase

in $\ce{PCl2F3}$, both $\ce{Cl}$ and one $\ce{F}$ atom are on equatorial position, and other 2 $\ce{F}$ atoms are on axial position as follows:

The first site mentions that to minimise repulsions of $\ce{F - F}$ bonds. But this structure is in fact increasing the repulsions, and decreasing stability, since the bond angle of $\ce{F - P - F}$ bond will be about $\mathrm{90^\circ}$
If instead all the $\ce{F}$ atoms are on equatorial position, and the 2 $\ce{Cl}$ on axial positions, then the repulsion may be further minimized.
So according to me, the structure should be:

Question: Why is my structure wrong? Please explain. Thank you!

See, the repulsion is not only between fluorines.

@IvanNeretin But $\rm F$ on equitorial position is a menace for all. It will repel the two $\rm F$ on axial very strongly, and also repel the chlorine.

True, but chlorine would repel chlorine even stronger.

@IvanNeretin Thats why i placed the $\rm Cl$ atoms on axial position. This also gives symmetry to molecule. The site I linked said that Fluorine's bond $e^-$ repel more strongly due to its greater electronegativity

Molecules do not care about symmetry. As for the repulsion, well, looks like the site was wrong on this one.

Strongly related: http://chemistry.stackexchange.com/questions/18427/why-does-f-replace-the-axial-bond-in-pcl5/18544#18544

I am VTCing as a duplicate. The sole difference is the number of chlorines.


text:
Do SN2 reactions always invert absolute configuration?
I know that an $\mathrm{S_N2}$ reaction proceeds through complete inversion of configuration, but is it necessary that the absolute configuration (R/S) of the chiral centre change after an $\mathrm{S_N2}$ attack ?

No, it is not.Why do you think it should?

Short answer: No

For the ease of thinking about this, let's imagine $\mathrm{S_N2}$ attack of $\ce{CN-}$ on bromochlorofluoroiodomethane, where iodide is being displaced.

Absolute stereochemistry is retained in this case because the highest priority substituent is becoming the lowest priority substituent. If you were to find the absolute stereochemistry of $\ce{2 ->3->4}$ in (R)-bromochlorofluoroiodomethane, which becomes $\ce{1->2->3}$ after attack, you would see that it is S and after attack becomes R. You can see that here:

The assigned configuration is based on the substituents at the stereocenter.  If what is entering has a different prority than what is leaving relative to the other groups, there may be no change in R/S.
For example:
The SN2 reaction of cyanide with MeCClHCH(SEt)2; in this case S alkyl chloride-->S nitrile (or R-->R) because the -CH(SEt)2 substituent is higher priority than -CN. The -Cl was 1st priority in the starting material.

To determine whether the $S_N2$ substitution inverts $R/S$ configuration, just do some simple arithmetic.
Let $Nu$ be the nucleophile, $L$ be the leaving group, and $R_1,R_2, R_3$ be the remaining groups.  $L$ will be higher in priority than 0, 1, 2, or 3 of the groups, and similarly for $Nu$.  You work out the difference between the two correct numbers for your specific reaction.  If it's even, you invert absolute configuration, if it's odd, you don't.  For example, in @Ringo's example above the leaving group, iodide, is higher than 3 of the retained groups but cyanide is higher than 0.  These numbers have an odd difference --> absolute configuration is retained.


text:
Is there any type of liquid, other than mercury, that PTFE teflon is known to "float" to the surface in?
The specific gravity of PTFE teflon is roughly 2.3. I have a 3/8" diameter sphere made up of entirely PTFE teflon, and I would like for it to float to the surface when dropped into a liquid.
Does anyone know of a liquid, other than mercury, that will allow this feat to be achievable?
(I prefer that the liquid be both non-toxic and inexpensive)

Gallium as well, presumably.

https://en.wikipedia.org/wiki/Clerici_solution - unfortunately toxic and expensive. That aside, I know that caesium chloride is used for centrifugation purposes although I'm not sure what's the highest density it can reach (lazy to calculate it right now).

Tetrabromomethane - Density: 3.42 g/cm³

$\ce{CHBr3}$ or $\ce{CH2I2}$ would do as well.

I don't think cesium chloride is quite dense enough (in aqeeous solution).  Cesium bromide might do the trick.

Various liquids are denser than $2.2\rm~\frac{g}{cm^3}$. $\ce{CBr2F2}$ is one of them with fairly low toxicity. Check out halons.


text:
Does mercury (I) chloride react with HCl?
If solid $\ce{Hg2Cl2}$ is placed in concentrated $\ce{HCl}$, will it react?
If so are the products $\ce{Hg2^2+}$, $\ce{H2}$, and $\ce{Cl2}$ ?
$$\ce{Hg2Cl2(s) + 2HCl(aq) -> Hg2^2+ + 2Cl2(g) + H2(g)}$$
I am teaching predicting products on a Chemistry I level, and I am not familiar with the subtleties of mercury chemistry. The short answer is "no reaction" because our "rules" state that a double displacement must occur with both reactants in aqueous solution. But I am seeking the reality behind the theory.
The entropy change would favor the reaction since it yields 4 mole of gasses from 1 mole of solid and 2 mole of solution.

I think you didn't thought-out well this reaction...

Your equation is not charge-balanced. Please edit your post to indicate what reaction you're interested in.

I think the reaction would be: $$\ce{Hg2Cl2(s) + 2HCl(aq) -> HgCl4^{2-} + 2H^+ + Hg(s)}$$ Heating would definitely give this reaction.


text:
Is peroxoborate and perborate the same thing?
Recently I came across the following question:-

Preparation of Sodium Peroxoborate- Equations

The term peroxoborate came from the fact that it has the peroxo bond $\ce{-O-O}-$. So, it should be strictly called peroxoborate. But some books says  peroxoborate while other books say perborate. Is it true?
I googled sodium perborate.  Wikipedia recognizes the compound as sodium perborate. Then, I googled sodium peroxoborate, the following message appear:-

So, is peroxoborate and perborate the same thing? If so, why some book says  peroxoborate while othe books perborate? It should be strictly called peroxoborate since it contains the peroxo bond $\ce{-O-O}-$.

Update
This is what written in Weller, M.; Overton, T.; Rourke, J.; Armstrong, F. Inorganic Chemistry, 6th ed:-

Sodium perborate is used as a bleach  in laundry powder, automatic dishwasher powders, and whitening toothpastes. Although the formula is given as $\ce{NaBO3.H2O}$ or $\ce{NaBO3.4H2O}$, the compound contains th peroxide anion, $\ce{O2^{2-}}$ and is more accurately given by $\ce{Na2[B2(O2)2(OH)4].6H2O}$.

But this is what written in Lee, J.D. Inorganic chemistry

$$\ce{2NaBO2 + 2H2O2 + 6H2O -> Na2[B2(O2)2(OH)4].6H2O }$$
  Sodium peroxoborate is used as a brightener in washing powder. It is compatible with enzymes which are added to some biological powders.

So, what is the correct name:- sodium perborate or sodium peroxoborate?

unfortunately, there is no consistency between individual cases in the matter. Sometimes  "per- " prefix denotes peroxo compound (like per-sulfate, per-borate); and sometimes it doesn't (per-chlorate, per-bromate).

Yes, this is the question that came in my mind. But I modified the question. Why there are certain flaws in the suffixes or prefixes or  any other rules in the naming of compounds though it is accepted universally?

TL; DR: Yes, perborate and peroxoborate are both the same thing. Perborate is just more popular, but peroxoborate is a more literate name for the $\ce{[B2(O2)2(OH)4]^2-}$ ion.

A perborate is a name attributed historically in XIX century to the compounds containing $\ce{BO3-}$ anion (see e.g. an overview [1]). In 1900s it quickly gained a foothold in the market and remained there since, as sodium perborate quickly became a popular additive for cleaning products.
For instance, the name of the popular laudry detergent Persil, which Henkel introduced in 1907, is a syllabic abbreviation formed combining German words "Perborat" (sodium perborate, bleaching agent) and "Silikat" (sodium silicate, dirt remover).
If we take another look at Google Ngram Viewer listing exclusively for peroxoborate, we'll notice that this term emerged only after 1961:

The reason is that in 1961 Hansson et al. [2] published a refined crystal structure of a controversial sodium perborate which was previously denoted with either $\ce{NaBO2 · H2O2 · 3 H2O}$ or $\ce{NaBO3 · 4 H2O}$ formula, and it was demonstrated that the correct formula unit is $\ce{Na2B2(O2)2(OH)4 · 6 H2O}$ and the name should indeed be sodium peroxoborate due to the presence of "the double peroxo bridge connecting two boron atoms" (I suggest having a look at this short communication from 1961: it's a great example of a concise and literal data presentation with effective illustrations).
A comment left in section "Per-acids of non-metals" in Wells' Structural Inorganic Chemistry [3, p. 422] is also quite peculiar (reference number updated; emphasis mine):

The demonstration that the sodium peroxoborate, originally written
  $\ce{NaBO3 · 4 H2O}$ or $\ce{NaBO2 · H2O2 · 3 H2O}$, is in fact $\ce{Na2[B2(O2)2(OH)4] · 6 H2O}$ [2] containing the cyclic ion of Fig. 11.2(a), with $\ce{O-O},$ 1.47 Å as in $\ce{H2O2},$ suggests that speculation about the structures of these compounds is less profitable than X-ray crystallographic examination.

More recent textbooks also often make a note that it's more correctly to use peroxoborate, yet perborate is a more popular name due to above-mentioned historical background of borates in cleaning industry.
For example:

Büchel's Industrial Inorganic Chemistry [4, p. 24]:

Sodium perborate (more correctly sodium peroxoborate) …

Structure and Bonding, vol. 105 [5, p. 35]:

Peroxoborate salts, commonly referred to as perborates, …

References

Tanatar, S. Perborate Und Ihre Konstitution. Zeitschrift für Physikalische Chemie 1898, 26U (1). https://doi.org/10/ggbdvm. (in German)
Arne Hansson; Fredga, A.; Sjöberg, B.; Dam, H.; Sjöberg, B.; Toft, J. On the Crystal Structure of Hydrated Sodium Peroxoborate. Acta Chem. Scand. 1961, 15, 934–935. https://doi.org/10/bhnnth. (Free Access: PDF)
Wells, A. F. Structural Inorganic Chemistry, 4th ed.; Clarendon Press: Oxford, 1975. ISBN 978-0-19-855354-0.
Büchel, K. H. Industrial Inorganic Chemistry, 2nd, completed rev. ed ed.; Wiley-VCH: Weinheim ; New York, 2000. ISBN 978-3-527-29849-5.
Group 13 Chemistry III; Roesky, H. W., Atwood, D. A., Eds.; Structure and Bonding; Springer: Berlin, Heidelberg, 2006; Vol. 105. https://doi.org/10.1007/3-540-46110-8.

This is necessarily  not  a chemistry answer but is a continuation of aventurin's answer.

Google ngram viewer is an online search engine h  that charts frequencies of any word or short sentence found in the sources printed between 1800 and 2012 in American English, British English etc.

So, here is the Google ngram for peroxoborate and perborate.

So, you can see that perborate is more often used than peroxoborate. Though, the correct name (IUPAC name) is tetrahydroxidodi-µ-peroxido-diborate, people often mention the compound sodium perborate. Sodium peroxoborate seems to be obsolete.

No, the names refer to different compounds. Attached below are their Lewis representations.

I should add that these structures were arrived at using Chemdraw's name to structure conversion. And it appears that this representation of perborate violates the octet rule. So this may be incorrect.

Yes it's very much incorrect. Read https://en.wikipedia.org/wiki/Sodium_perborate

Yes. I've seen that. But take a look at this compound. There seems to be conflicting nomenclature in the literature. http://www.chemicalbook.com/ChemicalProductProperty_EN_CB7214578.htm

"simplistic NaBO3·nH2O-type formulas are just a convenient way to express the average chemical composition" Seems  site you're citing didn't care

Here's another example illustrating the seemingly conflicting nomenclature associated with the compound. http://www.orgsyn.org/demo.aspx?prep=CV9P0522

The drawn lewis representation of 'perborate' is grossly incorrect.

The correct IUPAC name is disodium tetrahydroxidodi-µ-peroxido-diborate.
This name describes the structure with two bridging peroxido groups.

However, perborate or peroxoborate are well known and more often used.

Yes, but which is more often used: perborate or peroxoborate?


text:
When are the prefixes bis, tris etc. used in nomenclature of complexs?
Are the prefixes bis-, tris- etc. used only for ligands that are bidentate in complex nomenclature?

For naming coordination compounds, the multiplicative prefixes ‘di’, ‘tri’, ‘tetra’, ‘penta’, etc., are used in the case of simple ligands, for example:
$\ce{[PtCl4]^2-}$  tetrachloridoplatinate(2−)
The alternative multiplicative prefixes ‘bis()’, ‘tris()’, ‘tetrakis()’, ‘pentakis()’, etc. are not only used for bidentate ligands. They are generally used with composite ligand names or in order to avoid ambiguity, for example:
$\ce{[Pt(PPh3)4]}$  tetrakis(triphenylphosphane)platinum(0)
The corresponding section in the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) reads as follows:

IR-9.2.2.2 Number of ligands in a coordination entity
Two kinds of multiplicative prefix are available for indicating the number of each type of ligand within the name of the coordination entity (…).
(i) Prefixes di, tri, etc. are generally used with the names of simple ligands. Enclosing marks are not required.
(ii) Prefixes bis, tris, tetrakis, etc. are used with complex ligand names and in order to avoid ambiguity. Enclosing marks (…) must be placed around the multiplicand.

The use of multiplicative prefixes is also explained in the IUPAC Technical Report Brief guide to the nomenclature of inorganic chemistry. Pure Appl. Chem. 2015, 87(9–10), 1039–1049 as well as in the corresponding four-sided lift-out document, which is available as supplementary material.
Composite ligand names include ligand names which are substituted or which themselves contain multiplicative prefixes or locants.
For example, the normal prefix is used in ‘diammine’ $(\ce{(NH3)2})$, but the alternative multiplicative prefix is used in ‘bis(methylamine)’ $(\ce{(CH3NH2)2})$ to make a distinction from ‘dimethylamine’ $(\ce{(CH3)2NH})$.
Note that the use of the alternative multiplicative prefixes is not limited to coordination compounds. They are generally used for compositional names in inorganic chemistry when a simple multiplicative prefix may be misinterpreted. For example,
‘tris(iodide)’ is used for $\ce{(I- )3}$ (rather than ‘triiodide’, which is used for $\ce{I3-}$);
‘bis(phosphate)’ is used for $\ce{(PO4^3- )2}$ (rather than ‘diphosphate’, which is used for $\ce{P2O7^4-}$.

+1 for the link to the new technical report and the Red Book.

Are bis tris and tetrakis considered in alphabetical order?

@Yashwini No, ligand names are taken in alphabetic order. E.g. $\ce{[CrCl2(H2O)4]+}$ is named tetraaquadichlorochromium(III). Notice, aqua comes before chloro. The prefix doesn't matter.

Both the prefixes bis-, tris-, tetrakis- etc. and the prefixes di-, tri-, tetra- etc. are used to describe how many times a certain single ligand attaches to a central metal — and this usage is no different here than in typical molecule nomenclature. The ‘standard’ form one would use is di-, tri-, tetra- etc.
However, sometimes this can lead to ambiguity or confusion. Consider for example the triphosphate ion $\ce{P3O10^5-}$. Its structure is described well as $\ce{(O3P-O-PO2-O-PO3)^5-}$ — it is a single molecule consisting of three phosphorus atoms with oxygens in between and each phosphorus having a total of four oxygens surrounding it. Triphosphate can form stable chelate complexes with calcium ions and is therefore often added to washing powder. Let us assume a complex of calcium and triphosphate described by the formula $\ce{[Ca(P3O10)]^3-}$. (I don’t know if this is an actually existing one, but it can serve well as an example.) If we had to name it, we would arrive at triphosphato-calciate.
Now, of course, simple phosphate ions can also coordinate metals. It is totally conceivable that three phosphate ions coordinate a single calcium ion to give a complex represented by the following formula $\ce{[Ca(PO4)3]^7-}$. (Actually, this in itself is rather unlikely due to the high negative charge — using hydrogen phosphate would be a much more likely example but wouldn’t serve well enough to highlight the problem. Please note again that I don’t claim any of these complexes to actually exist, I am merely using them as convenient examples.) We now have three phosphato-ligands so naively we could name the complex triphosphato-calciate again.
This is where the ambiguity-prevention squad makes their glorious entry. To make it clear that we are dealing with three distinct phosphato ligands, we can switch to the prefix tris- which is reserved for multiple distinct occurrances of a single species. So a better name for this second complex $\ce{[Ca(PO4)3]^7-}$ would be tris-phosphato-calciate.

After ambiguity, let’s turn to confusion. Triphenylphosphane $\ce{(C6H5)3P}$, often abbreviated $\ce{PPh3}$, is a common ligand for catalysts used in organic chemistry. It consists of three phenyl rings connected to a phosphorus atom; the donated lone pair is phosphorus’. Consider Wilkinson’s catalyst $\ce{[RhCl(PPh3)3]}$. Naming this one would result in chloridotri(triphenylphosphano)rhodium(I) as an initial guess. Note the double-tri in the middle causing possible confusion. Instead, the prefix is changed to tris- which again means three occurrances of the same distinct species — in this case, triphenylphosphane. So the commonly used name for Wilkinson’s catalyst is chlorido-tris(triphenylphosphano)-rhodium(I).

As you can see in the final example, triphenylphosphane is a monodentate ligand not able to chelate. Yet tris- (or in other compounds: bis-, tetrakis- etc.) is used. So it is wrong to think of that set of names only for polydentate ligands. In fact, trisphosphato explicitly refers to three single phosphate ligands rather than a much more polydentate triphosphate moiety.


text:
Are there any gases more dense than liquids?
Inspired by Is there any type of liquid, other than mercury, that PTFE teflon is known to "float" to the surface in?, are there gases that are more dense than liquids? (So bubbles would sink down instead of bubbling up?)

The critical point of Xenon is at a temperature of 16,6 °C and a pressure of 5,84 MPa (~ 58 bar) with a density of 1.1 $\mathrm{g, cm^{-3}}$. Water is liquid under this conditions.
I guess that water drops would float up in gaseous Xenon slightly below the critical point ($p < p_{crit}$, $T < T_{crit}$).

@aventurin thanks, this gives rise to another question, but that would be physics: would liquid droplets form and raise, or gas bubbles form and sink?

What about sulphur hexafluoride? It can allow a solid boat to float on it, but I'm not sure about how a liquid would work with it. Also, Just shake up some oil and water to see how the two separate. My guess would be that the gas separating from the water would do the same.

@NA A boat can float on SF6 because the boat shape holds air. A solid block of the same material would sink. At standard temperature and pressure, there is no known liquid less dense than SF6 or even WF6.

@aventurin Amazingly, YouTube user Cody Don Reeder (of Cody's Lab fame) managed to implement this after some trial and error. Watch this video to see liquids floating in gaseous xenon at room temperature. Mind blown!

Very nice indeed!

You kinda interrupted me on related question you deleted, I was going to say if you changed it to "what polymers are soluble in supercritical CO2" than I think it would be OK.

It depends on the conditions. Let's start decomposing your question in two related questions:
Denser gasses at SATP?
There are indeed some gasses that are quite dense. Sulfur hexafluoride has a density of 6.17 g/L  while tungsten hexafluoride of 12.4 g/L. But usually, they are not so dense compared to the density of liquids.
Lightest liquid at SATP?
The density of liquids largely depends on the atomic mass of the compounds. Hydrogen that is the compound with the lowest atomic mass has a density of 70.85 g/L, which is probably the lowest density you can find.
So at room temperature is not possible but if you increase the pressure the density of the gasses will increase while the density of the liquid won't increase much because liquids are not appreciably compressible.  It is hence theoretically possible to achieve a gas with a greater density compared to that one of a liquid, and also the coexistence of the two phases should be possible in certain conditions but the pressure required won't allow you to observe it.
Bubbles sinking down?
That won't happen also if there was a denser gas compared to a liquid. Bubble formation is quite a different process in this case inverting the gravity would be better! ;-)


text:
Does adding mass to a liquid (not water) increase the volume
The Question sounds stupid but i have a
legit problem
I mixed 1000 mL coconut water and 225 g of agar-agar to make a coconut water [CW] agar. It gave me more than 1000 ml. the medium went beyond the 1000 ml mark. the CW agar was also too soft to use
then i tried mixing 775 mL coconut water and 225 g agar-agar it gave me little below the 1000 mL mark. the CW agar was just right not too soft and not to hard :|
(I did heat after mixing during those two trials but the volume remained the same after cooling to room temp.)
Seems to break rules taught to us.  i though when you add mass to a liquid it won't increase the volume but the density will increase or does this only apply to water?

You "understanding" of "the rule" is wrong. In general if 100 ml of a solution is needed then the solute should be dissolved in some water and then diluted to 100 ml final volume. Even mixing 50 ml of ethanol with 50 ml of water doesn't yield exactly 100 ml of solution.

so any principles explaing why this happen? i searched everywhere in the net and all of it are about water and some other chemicals mix with it. Even my teachers can't give answers

https://en.wikipedia.org/wiki/Partial_molar_property
Read this.

I tested using water and agar -agar the volume didn't change at all or there was no noticeable change.  i  did it at a small scale though. i added 25 g of agar- agar to 75 ml water but the volume remained the same while doing it to coconut water the volume increase. The Link kinda help but still trying to get a grasp on it might read more about it thanks for the reply.

"The rule" is that an ideal mixture will have the volume of the sum of its components and the density will be the molar averaged density of the components. Real mixtures deviate from this rule to varying degrees.
The name of the property you're looking for that describes this deviation is "excess volume". Excess volume can be either positive (1L of A and 1L of B creates >2L of AB mixture) or negative (1L of C and 1L of D creates <2L of CD mixture). The deviation from the rule depends mostly on how the molecules of the components interact with themselves compared to with the each other. Without knowing much about this specific case, I would guess that the big agar polymer molecules interrupt the local organization of the water molecules. This causes the density to decrease.
For many fluids, such as ethanol and water, the densities have been well studied and you can look up tables of excess volume. On a quick google search, it doesn't appear that anyone has ever done this for coconut water and agar. Unfortunately, that means more work for you. Fortunately, it means you get to break new scientific ground!

thank you! that cleared up things i can finally rest in peace XD


text:
Are there phosphorescent/fluorescent materials that absorb near infra-red waves and emit higher wavelengths?
This SO discussion here lists some chemicals (and interestingly chlorophyll) that absorb visible radiation and emit infra red. I was wondering if there were chemicals that absorbed near infra-red and emitted much larger wavelengths (near microwave).

It does work that way. Imagine that you put 10 energy units into creating an excited state. When that excited state decays you can't get more than 10 energy units back in the emitted photon, and you generally get less.

@MaxW do you know chemicals that absorb near infra-red and emit microwaves?

Microwaves would be too low in energy for most (all?) electron transitions. There might be some molecule that would do that but it would be really, really weird. There are molecules (chemicals, minerals) that emit in the infrared. Absorbing in high energy infrared, and emitting a low energy infrared photon would be possible. (Don't know one off the top of my head.)

What do you mean by "higher" wavelengths? Do you mean longer wavelengths (with lower energy) or did you mean higher frequencies (shorter wavelengths, higher energy).

All molecules emit in the microwave region. They all have rotational energy levels and they absorb and emit all the time as energy flows between these levels.  But this not what you are asking I suspect. If you blast thin Gallium Sulphide wafers with a femtosecond laser then it is possible to generate terahertz radiation.


text:
What did I do wrong with my lotion bar that it was grainy at the bottom of the mold?
I want to create a with cocoa butter bar and coconut oil, that emulsifies quickly when rubbed in your hands. I made a tiny batch with a 6:1 ratio (6 tbsp CB and 1tbsp CO) and they have held together just fine in a variety of temperatures. (The only thing I did different with these besides the smaller ratio of coconut oil is melt both oils together instead of separately).
I went ahead and made a larger batch and tried a tad bit more coconut oil 11(oz):2.5(oz) and I got bars that were totally grainy (I assume it was the coconut oil?). Interesting, the last 5 bars of the batch were fine when I popped them out of the mold. I am assuming the coconut oil is lighter and rose to the top and was poured out first, leaving very little for the last few molds. I"m having difficulty attaching a photo to this message but I can try again.
I heated separately both the cocoa butter and coconut oil in glass measuring cups and hot water on the stove. max temp 170F for 20ish minutes ( is that too hot perhaps?).
Then I mixed the two oils together and mixed them with a blender stick for 40 seconds. Scooped bubbles off that rose to the top.
As I was making sample sized bars, I poured 5/8 oz of mixture into individual silicone molds. But outside in 32F for 15-20 minutes. Then room temp 65-69F for 24 hours.
Ingredients used:
Cocoa Butter - Unrefined, Organic
Coconut Oil - Virgin, Cold pressed, Organic
Questions:
Should I use less coconut oil? I am looking for a bar that emulsifies quickly in your hand that is not too sticky.
If not coconut oil, might another natural product help soften the cocoa butter to that end?
Am I heating the oils too high or too long?
Is there a way to cure the bars successfully?Make sure there are no tiny bubbles in the mold?
Thank you so much for your time and expertise.

This question might be better suited for the cooking stack exchange


text:
Kinetic isotope experiment used to distinguish radical and concerted mechanisms for the Diels Alder reaction
In the Pericyclic Reactions Oxford Primer it is written that:

the deuterium
  substitution of the hydrogen atoms in the diene and dienophile that
  participate in a possibly Diels-Alder reaction, gives rise to inverse
  secondary isotope effect (as the carbon atoms change from trigonal to
  tetrahedral geometry):

If both bonds are forming at the same time, the isotope effect when both ends are deuterated is geometrically related to the isotope
  effect at each end.
If the bonds are being formed one at a time, the isotope effects are arithmetically related.

I don't understand the principle behind this. Why does the isotope effect occur and how does it tell us whether the mechanism is radical or pericyclic? I need an answer much more concise than what is presented in the book.

What Fleming is likely referring to is work by Edward Thornton (J. Am. Chem. Soc. 1972, 94, 1168) in which he attempted to make use of secondary kinetic isotope effects (KIE) to prove whether the Diels Alder reactions being studied were indeed pericyclic, or occurred via a radical pathway.*
The phrasing in your post (which I assume is from the book itself) isn't particularly helpful. By 'geometric', Thornton was actually talking about the geometric mean and the arithmetic mean as ways of considering the isotope effects.
The reaction they actually studied was the retro Diels-Alder of 9,10-dihydro-9,10-ethanoanthrace, with varying levels of deuteration (none, di, tetra) as shown in the figure below.

In order to discuss whats going on, we need to define some terms:

In the reaction, two bonds are being broken, lets call those bonds A and B. Breaking bond A has a KIE of fA and breaking bond B has a KIE of fB
Each molecule (no deuteration, di-, tetra-) has an associated rate constant, lets call these k0, k2, k4 (with the subscript indicating how many deuteriums we have).

We can now consider the calculation that Thornton went through to determine if the reaction is indeed a concerted one:#

In a completely concerted reaction, intuitively fA = fB, since the TS is symmetrical. If the reaction is non-concerted (aka stepwise), fA cannot equal fB unless there is symmetry to the TS.
Considering the di-deuterated molecule, two pathways are possible (breaking A or B first) to give a non-symmetrical intermediate. Thornton defines rate equations for these two as fAk0/2 and fBk0/2. We can do this since k0 is the same no matter which bond is broken in the undeuterated starting material (with the f modifying it to take into account the KIE).
k2 (the rate constant for the retro Diels-Alder of the di-deuterated starting material) can therefore be thought of as the sum of two pathways (break A then B or B then A). Mathematically: k2 = k0(fA + fB)/2. Crucially, this relationship is valid independent of whether the reaction is concerted or stepwise.
k4 (the rate constant for the retro Diels-Alder of the tetra-deuterated starting material) is able to be obtained similarly. Mathematically: k4 = fA fBk0.
At this point, Thornton provides the result of solving the two previous equations simultaneously, which provides a relationship as below:

$$f_\mathrm{A},f_\mathrm{B} = \frac{k_2}{k_0} \pm \left[\left(\frac{k_2}{k_0}\right)^2 - \frac{k_4}{k_0}\right]^{1/2}$$

It is from this relationship that we are able to consider whether the reactions are concerted or stepwise:

if concerted, fA = fB = k2/k0, which means that (k2/k0)2 == k4/k0. This now allows us to measure the rate constants, plug in the numbers, and determine if the reaction is concerted.

*: There are many examples of Diels Alder reactions that are stepwise, via radical mechanisms. The KIE's measured are necessarily small, meaning its often difficult to know if what you're looking at is actually correct. One thing that does hold up generally is that the reactions aren't ionic, due to the lack of any real solvent dependance (if the reactants are soluble in the solvent and you heat it hot enough, they generally proceed at the same rate, suggesting no stabilisation of 'intermediates' in polar solvents).
#: I'm paraphrasing the original paper, which will hopefully name it more digestible, but I still encourage you to read the actual thing, despite it being fairly dense

Sorry for the lack of MathJax, it takes me too long to look up the various things I need to make the right equations, and it slows Safari down. Feel free to edit!

By the way, were you intending to have one bullet point at the end on how to tell if it's stepwise?

I was going with 'if it doesn't conform to this, then it ain't concerted'. I'm less convinced by their argument for the stepwise case (as it could take many different forms).


text:
How can I search molecules by constituent atoms
Given a set of atoms, I would like to find (some / the most common) molecules that include only (some/all) atoms of the set. Is there any online search engine that can do this?

E.g.
  Atoms={C,H}
  -->
  Molecules={CH$_4$, C$_2$H$_6$, ...}

Note:

I am looking for known molecules.
It is OK if I can only find chemical formulae, so without all possible configurations.

Personally, I doubt so because there are literally millions, if not billions, of compounds that contain C and H (or any combination of atoms that you specify, unless it's something esoteric like Zr, N, Se).

Is there some database where you could enter a formula like $C_aH_bO_c$ and get molecules ( compounds) that fit (where a,b and c are given integers)? Certainly by the time that you get to 100 carbons there are more unknown possible compounds than known ones.

Pft, you can even check Wikipedia for compounds containing specific elements.

You can search ChemSpider using a formula for one... http://www.chemspider.com/Search.aspx?rid=40b3a5ef-3142-406c-bbb0-c05e69d10a6e

I'm pretty certain the question here is about searching known molecules, not theoretical ones. There are millions of connected chemical-like graphs with formula N3O4P5 yet I doubt very much that many of them correspond to stable chemical structures!

@gilleain - Yes, I am looking for known molecules.

You can use ChemSpider's full search:
http://www.chemspider.com/FullSearch.aspx
and click on the 'Elements' section. Alternatively, there is a direct link here:
http://www.chemspider.com/ElementsSearch.aspx
This gives you two periodic tables that you can select elements from for your search. I think the upper one is 'allowed' and the lower one is 'forbidden', but I haven't tested this.
Of course, as pointed out in the comments, the set {C, H} will give you most of the database! However, for rarer elements this can be a better filter.


text:
Choosing a vacuum pump for fractional distillation
I have a fractional distillation setup. I would like to purify a small  mixture ($\pu{\approx 100-150 mL}$) of high boiling point compounds (between $200$ and $\pu{300^\circ C}$). The one that interests me has a b.p. of $80\rm~^\circ C$ at $\pu{13 mmHg}$. I don't know what this mixture is composed of.
Now the tricky part and the one that give me the most headaches is the vacuum pump. I need one that has enough vacuum power to be able to distill at $\pu{13 mmHg}$ and one that is not expensive too.
I also need a system to be able to monitor and control the vacuum pressure too, so the vacuum pressure remains constant.
I figured that a diaphragm pump is not suitable since the ones that have that kind of vacuum power are way more expensive than rotatory vane pumps. Can anyone help?

Could you correct your orthography? Also I suggest not to thank in questions, but in comments.

Have you tried if a water aspirator works? It can go as low as 10mmHg (I think) and should be OK for the scale you want

Thank K_P i have thought about a water aspirator but the water temperature would need to be a very low to reach 13 mmHg, i have thought about using a water pump connected to a water aspirator, the water pump would use a precooled water in a container and then put it back there, but i assume i need i strong water pump but how much strength do i need ?

If you are keeping your hands on you can't beat a water aspirator for cost, simplicity, and chemical compatibility.
Here is an aspirator assembly with a gauge and valves I put together for about $50.  Connecting it to a standard hose bib it can pull up to 27mmHg.  The water valve controls the vacuum rate, and if you have a closed system once you reach the desired vacuum you can shut one of the other valves to hold it.

Note: This was inspired by Lysander's answer to my similar question last year, which has additional useful details.

I dont know what materials you'll be working with, but Ace usually has a pretty cheap 1.5 cfm two phase pump. They work alright. It pulls too fast in my opinion and not a very deep vacuum. It still hits -29.5 inHg (at the altitude im at), but you can regulate the pressure easily with a second cheap valve already placed on the pump itself. You can also add a inlet and exhaust trap for pretty cheap.
If your pretty good with tools, spend an extra 200USD and buy a deep vacuum off Ebay. Something that just requires an added belt drive motor and you can build yourself a pretty diesel, multi thousand dollar rotary vane pump for about $200 dollars.


text:
How to calculate galvanic cell spontaneity?
I was wondering how to calculate cell spontaneity for a galvanic cell. Do you just look up the enthalpy values of electrode reactions? Does the electrolyte solution you choose have an effect on cell spontaneity?

To determine spontaneity of a reaction, one looks at the sign of $\Delta G$ for the process. And you have the following relation between the change in standard Gibbs free energy and the (standard)EMF of the cell:
$ \Delta G^0 = -n\,F\,E^0$
where, $n:=$ no. of electrons exchanged
$F:=$ Faraday constant (Charge on 1 mole of electrons. About $96,500\,\mathrm{C}$)
So, it is sufficient to compute the $E$ and look at it's sign. A positive sign for the EMF would ensure a negative value for $\Delta G^0$ and thus, you can conclude that your cell reaction is spontaneous.
The EMF of the cell is given as follows:
$$E^0_{\mathrm{cell}} = E^0_{\mathrm{cathode}} - E^0_{\mathrm{anode}} $$
Where, $E^0_{\mathrm{cathode}}$ and $E^0_{\mathrm{anode}}$ are the standard reduction potentials for your cathode and anode half reactions, and you can easily find tabulated values for most metals, ions.
An illustrative example:
Consider the following cell reaction:
$$\ce{Zn_{(s)} + Cu^2+_{(aq)} -> Zn^2+_{(aq)} + Cu_{(s)}}$$
Your cathode half reaction is:
$$\ce{Cu^2+_{(aq)} + 2e- -> Cu_{(s)}}$$
and your anode half reactions is:
$$\ce{Zn_{(s)} -> Zn^2+_{(aq)} + 2e-}$$
the tabulated reduction potential values for the two:
$$E^0_{\mathrm{anode}}= -0.763\,\mathrm{V} \qquad E^0_{\mathrm{cathode}}= +0.342\,\mathrm{V}$$
Plugging this into
$$E^0_{\mathrm{cell}} = E^0_{\mathrm{cathode}}-E^0_{\mathrm{anode}}$$
we obtain,
$$E^0_{\mathrm{cell}} = +1.103\,\mathrm{V}$$
Thus, this cell reaction is spontaneous.


text:
How do I get the molarity of solution 1 given volume 1, molarity 2, and volume 2?
For example:

What is the molarity of a $\ce{NaOH}$ solution if $25\rm~mL$ of the solution is exactly neutralized by $44\rm~mL$ of $0.32\rm~M~HCl$?

I know that the abstract function is
$$\text{Given Volume of Known Molarity} \times \text{Molarity Factor} \times \text{Mole Ratio} \times \text{Per Volume Factor} \times \text{Conversion Factor Per Liter}$$

@ A.J what are your workings out so far? The policy on this site is to not straight out answer homework questions without going through the OP's workings out first

The reaction that occurs: H+ + OH- --> H2O
When you multiply the volume of HCl by the molarity of HCl you get the total amount of moles of H+ that reacted (mol/L x L = mol).
This means that the same amount of OH- was needed for the reaction (1:1).
Volume of NaOH is known and thus yields the answer.

This doesn't answer the question or really make sense.

@Joel I disagree. It does in a way answer the question and it makes perfect sense. However, it would have been better suited as a comment since it is neither complete nor does the question itself comply with our homework policy.

@Martin , then why disagree? It's not the same amount , it's the same molar ratio. From there you can calculate the amount.


text:
Why are there 2 equivalents of Ca in an example stated in Wikipedia?
Applying what's been said in the Wikipedia page for equivalents, if $1\ \mathrm{mol}$ of $\ce{NaCl}$ and $1\ \mathrm{mol}$ of $\ce{CaCl2}$ are dissolved in a solution, there is $\rm1~eq~Na,~2~eq~Ca$, and $\rm3~eq~Cl$ in that solution.
Is this a misstatement in Wikipedia?

The Wikipedia article uses a funny defintion of equivalent that I have not heard yet even though I use the term extremely frequently in the lab. But they define it well so here goes:
One equivalent of ions is defined as $1~\mathrm{mol}$ of charge. So if we dissolve $1~\mathrm{mol}$ of $\ce{Na+}$, we have exactly $1~\mathrm{mol}$ of positive charges in solution, so $1~\mathrm{mol} \cdot 1~e = 1~\mathrm{eq}$ ($e$ being the elemental charge). The $1~\mathrm{mol}$ of $\ce{Cl-}$ contributes $1~\mathrm{eq}$ of negative charges, balancing the entire thing.
Calcium, however, has a $\mathrm{+II}$ oxidation state. So $1~\mathrm{mol}$ of $\ce{Ca^2+}$ gives us $1~\mathrm{mol} \cdot 2~e = 2~\mathrm{eq}$ of positive charge.

Sir I meant this only in my answer. Actually it's not funny but we see equivalent as the number of eq. electrons bringing out the reactions. I did not write this because it might have confused the asker.

You see we have 2e = Weight of metal/ its equivalent weight( from electro-chemistry).

@AkshayPratapSingh There is no need to comment on an unrelated answer to answer to a comment I gave under your answer. Instead, comment in the same spot and ping me with @ followed by the username.

The link in the original post may refer to the equivalent of the number or ratio of electrons shared by each of the specific ions, according to their chemical formula, in the solution, rather than the ratio of specific ions that are present in the solution.

No the statement is absolutely correct.
We can write number of equivalents = Weight of the metal/ its equivalent weight. Since the equivalent weight of calcium in CaCl2 is 20 gm hence we get the number of equivalents of Ca as 2.

Reread the Wikipedia article, please.


text:
What is the chemistry behind black colour of platinum black?
Platinum is a silvery white metal and is highly unreactive then why it's powdered form "platinum black" is platinum metal but is black in colour. why would a metal such as platinum become black as a powder?

Might be related to light scattering from small particles of platinum (platinum black is a fine powder) vs. bulk, but not sure. Good question!

@ToddMinehardt is right, it's not about the chemistry, it's about the physics of how light interacts with it.  As good as a question as it is, I'm flagging it as off-topic for Chem.SE, as it more properly belongs on Physics.SE.

@ToddMinehardt For reference, this answer on Physics.SE has a short paragraph about a related modification of $\ce{Pt}$ that also leads to a black color.

I wrote a relevant answer in a past question.

While the answer might be that's it's a physical effect rather than a chemical one, the question is still about a particular compound and the user wouldn't have known that it was a physical effect or else they wouldn't have asked here, so I'm glad the hold was removed.

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


text:
I am confused about the stability of carbanion III and IV in the image

I am confused about the stability of carbanion III and IV in the image, I think III is more unstable because of electron donating effect (+R) of  $\ce{-OCH3}$ group but my book says III is more stable than IV. In compound III, $\ce{-OCH3}$ would donate its lone pair and would make the carbanion unstable but no such effect happens in compound IV. $\ce{-OCH3}$ in compound III would also have -I effect to stabilize the carbanion but I have read that for $\ce{-OR}$ group, +R effect is more than its -I effect i.e., its electron donating nature is more than its electron withdrawing nature through inductive effect, +R means electron donating by resonance (In $\ce{-OR}$, $\ce{R}$ is any alkyl group)

To answer this question, I have been trying to find the pKa values of the four parent acids, but I have only found acetone. It is the only compound of the four listed in the well known Bordwell and Evans tables. Ethyl acetate (not methyl) is listed, so I supposed that could be used, but I cannot find an authoritative source for acetaldehyde and methyl vinyl ketone. I am frustrated by my inability to easily search for papers that report pKa values.

The picture clearly says that it is abput resonance stability. So drawing all possible resonance structures for every substance should help probably.

I think in 3, carbanion is resonance stabilised. While in 4, C=C-C=O are in conjugation, thus will not stabilise carbanion.

Why does conjugation prevent stabilisation? I see no reason for this.

I think in both (III) and (IV), Cross Conjugation is present which generally prevents resonance from taking place.

In (III), Both CH3O and CH2minus can give electrons to the C in middle.
In (IV), After resonance, CHminus(on left) and CH2minus(on right), can give electrons to the C in middle.

Supposedly, in (III), O will not prefer to give it's LP electrons because of high Electronegativity.
In (IV), CHminus(on left) can comparatively give electrons more easily than O.
Also +M effect of CH2minus is greater than O(with LP).
Therefore, CROSS CONJUGATION EXTENT in (III) is less than in (IV).
Hence, (III) is more stable than (IV).


text:
Rutherford gold foil experiment
This question may sound silly.
What actually I understood from this experiment is that the positively charged nucleus which is in the center is responsible for the bending and reflection of the alpha rays.Here is my question:Why cant there be any small holes in the gold foil which might have caused this kind of behavior?

Why would they cause deflection of the $\alpha$ particles? Bear in mind that they did not appreciate that $\alpha$ particles might diffract when this experiment was done.

Does that mean alpha particles cannot be diffracted by surfaces and can pierce through all kinds of surfaces the same way..

I have never actually thought about this before. I would be interested to see if $\alpha$ particle diffraction is at all significant in this situation. I might look into the experiment a bit more later and do some calculations. $\alpha$ particles may still be diffracted by appropriate sized apertures if they have the right energy. It's a question of whether it's relevant here. Obviously in a historical context they didn't know about diffraction so they came to the only (correct) conclusion.

If you calculate the wavelength of an MeV alpha particle, you will find that diffraction is not in the cards at all. For neutron diffraction, at 1/4 the mass, you are down in the thermal regime (<<MeV) for diffraction. And, if surfaces were a real problem, then composition analysis by Rutherford Backscattering Spectrometry (RBS) would be significantly harder than it actually is. So, by simple physics you need to "hit" some entity in the solid that is heavy to be able to have elastic backscattering. Voids will not do at all - it has to be a nucleus.

"hit some"  ...If the alpha particles are travelling at high speeds even hitting on the like charged particles(electrons) would reflect....Am i not right??

No the experiment showed that the alpha particles interacted with a very small highly positively charged nucleus.

An alpha particle can certainly scatter off an electron, but it cannot backscatter - simple kinematics shows that. The deflection of the electrons leads to so-called electronic energy loss, distinct from the nuclear energy loss term in ion-solid interactions.

@JonCuster You should write an answer based on this.

Reflect means to bounce back, as in a ball bouncing off a wall. The average density of an atom is very low, so the observed reflection was startling... like throwing many billiard balls at a mass of fluffy cotton  candy and every now and then a ball comes bouncing back!
The explanation offered was that there must be some "hard", dense object inside that fluff... the nucleus. Electrons are effectively spread out in their orbitals, and tend to act individually (bound by electromagnetic forces). An alpha particle, massing $\pu{4 \times 10^{9} eV}$, could not be reflected by an electron massing ~$\pu{5 \times 10^{5} eV}$ -- that would be like bouncing a baseball off a ping-pong ball.
Since both alpha particle ($\ce{He}$ nucleus) and gold nucleus are positively charged, and the gold nucleus is held together by powerful nuclear forces (so all its particles' masses are effectively combined to ~$\pu{2 \times 10^{11} eV}$), it explained the reflection. Since reflection occurred only rarely, it implied that the nucleus was a small, dense target.

This doesn't actually answer the question, which is about whether $\alpha$ particles will diffract in this situation.

bouncing a baseball off a ping-pong ball.....why cant a fast moving baseball  change its direction when hit by a ping-pong ball???

The baseball can change direction. However, for it to bounce backwards in the center-of-mass frame from an elastic collision requires that the mass of what it is bouncing off of has to be larger. Now, conversion from center-of-mass to the lab frame throws a bit of uncertainty in to the process, but not much. Consider this - for an electron to displace an atom in a crystal lattice (i.e. transfer ~5-10eV to the atom) requires a near 1MeV incident electron (which is relativistic).

Consider that to get diffraction effects you generally need the wavelength of the diffracting particle to be related to a length scale of importance in your sample. For the moment, take it to be something on the order of an inter-atomic spacing, roughly an Angstrom ($\pu{\mathring A}$).
Taking an alpha particle at $\pu{3MeV}$ (the ones originally used were more on the order of $\pu{5MeV}$, depending on the source used). The de Broglie wavelength is straightforward to calculate, and turns out to be $\pu{8 \times 10^{-15} m}$, or $\pu{8 \times 10^{-5} \mathring A}$, 4+ orders of magnitude smaller than an inter-atomic spacing. So, one can rule out diffraction as a cause of energetic alpha particle scattering.
Now, plenty of folks do use neutron scattering to measure structure in materials. The trick is that they have to be very slow neutrons, or 'thermal' neutrons. Going through the de Broglie equation for a $\pu{25 meV}$ neutron (millielectron Volt, or basically room temperature) you get about $\pu{1.8 \mathring A}$ - which is right there with inter-atomic spacing. Now you have diffraction from solids!
So, energetic alpha particles are not diffracting off of anything related to crystal structure. They are interacting with a strong $1/r^{2}$ central force to give the observed differential cross section for scattering. And that force is centered on something heavy so as to result in backscattering. This shows the existence of the nucleus. (Another effect can be real interactions between nuclear energy levels and the alpha particle, but that is getting in to nuclear physics, a topic for another day.)
As a follow up to some of the comments - alpha particles do indeed interact with electrons in the sample, but their mass is so small that standard kinematics shows that an alpha will not backscatter off of an electron (unless it is highly relativistic - an alpha has a rest mass of about $\pu{3.6 GeV}$, an electron has a rest mass of about $\pu{511 keV}$).


text:
Can you make a magnetic fluid that's drinkable
I need to make a ferrofluid that is drinkable so I obviously can't use the traditional method of using iron as it is harmful. Are there any chemicals that are not harmful that can take on magnetic properties or and magnetic metals that are not harmful to consume?

‘as [iron] is harmful’ — a simplification at best. We would die without iron.

Read Nails for Breakfast? and see if breakfast cereal manufacturers agree with your assumption "I obviously can't use the traditional method of using iron as it is harmful"

Most people know that breakfast cereals are fortified with iron and other essential minerals and vitamins. What they don’t
  know is that in some cases the iron is actually added in the form of metallic, elemental iron—nails for breakfast!

Like permeakra commented, only a small amount iron, whether metallic or as a compound, is safe to consume.  See iron poisoning for details.

FeCl3 LD50 = 450 mg/kg . Iron IS harmful in large quantities.

Toxic effects begin to occur at doses above 10–20 mg/kg of elemental iron https://en.wikipedia.org/wiki/Iron_poisoning

@permeakra everything is harmful in large quantities.


text:
Before the CIP system was created, how did chemists ensure that the D/L system matched the R/S assignments?
The ᴅ/ʟ system I have referred to is the one chemists used before the Cahn-Ingold-Prelog (R/S) system where they arbitrarily assigned the (+) and (-) isomers of glyceraldehyde,  'ᴅ' and 'ʟ' respectively and related other carbohydrates via reactions with retention of configuration. They were eventually found to match the (R) and (S) assignments (for the lowest asymmetric carbon in Fischer projections) of those molecules today (particularly biomolecules), and that is where my question arises.
Even though the molecules were related to glyceraldehyde through reactions that proceeded with retention of configuration, it is also known that the resulting molecules may not always retain the absolute configuration of the original molecules (which depends on priorities of the new groups). And since it was before CIP's (R) and (S) system came into place, there was no way to know if the molecules had changed absolute configuration during the reactions. So how did the chemists manage to retain the configuration of all those molecules with respect to the glyceraldehyde isomers at all?

Are you only concerned with carbohydrates? Other compound classes can't necessarily be related to glyceraldehyde, unless you are using a convention that I'm not familiar with.

Yes, I am talking about carbohydrates or any other molecule that could be successfully related to the Glyceraldehyde enantiomers, for that matter...

I'm a little confused by the question as the d,l assignment is a physical property and there is an implication (through the mention of "retention of configuration") that an assignment convention is also being used (like the Cahn-Ingold-Prelog system). These assignment systems do not align.

Chemists actually still use D- and L- prefixes today, especially for carbohydrates where using it saves a lot of characters: Compare D-ribose to (2R,3R,4R)-2,3,4,5-tetrahydroxypentanal. The nomenclature is also in common use for amino acids, since all proteinogenic amino acids (except for glycine) in animals are L-configured (Bacteria, smart as they are, have many exceptions), while cysteine has the opposite absolute configuration to all other amino acids due to sulfur’s high priority. But your question revolves around history.

When Emil Fischer first devised the D- and L- nomenclature in 1891, not much was known about stereochemistry at all. It was known that carbon took four bonds and what the elemental composition of monosaccharides was. And methods were known to find out what a compound was, but much less sophisticated than nowadays. You would essentially derive it back to known products via known reactions and measure melting point (solids) or refractive index (liquids) for additional confirmation. Optical activity had been known since 1850, so that was probably performed in Fischer’s lab, too. At the same time, Pasteur performed his experiments, separating a racemic mixture of tartrate crystals into the enantiomers.
The choice Fischer made, when he assigned the D-configuration to (+)-glycerinaldehyde (the plus sign meaning that linear polarised light is rotated clockwise), coming from Latin dexter meaning right, was arbitrary. Not even essentially arbitrary, it was completely arbitrary. He probably knew he had a $50~\%$ chance of getting it wrong but in no way was he going to get proven wrong anytime soon. He decided it fitted nicely: the glycerinaldehyde that polarised to the right (clockwise) shall be the one with the asymmetric hydroxy group pointing to the right. He used that as a basis for all further sugars, knowing that all reactions he performed would take place on the aldehyde and leave the hydroxy group unchanged.
There was a fair bit of understanding how reactions worked even then. The first step to most sugar syntheses starting from glycerinaldehyde is the addition of a cyano group to give a cyanhydrine: Attack to the aldehyde, forming a new carbon-carbon bond and nothing else.
$$\ce{HOH2C-CHOH-CHO + N#C- + H+ -> HOH2C-CHOH-CHOH-C#N}$$
In any case, once a reaction like that was performed, derivatisation back to known compounds was done, and they would have realised if something had gone wrong with the hydroxy group. So we can be pretty sure they knew what wouldn’t influence it.
It wasn’t until half a century later and the X-ray crystal structure of glycerinaldehyde that one actually found out what the configuration of its carbon atom was. When it was proven that he had backed the lucky horse and the hydroxy group of D-glycerinaldehyde indeed pointed rightwards according to his convention, Fischer probably smiled a big smile from up in heaven. But if he had been wrong, he would have shrugged and thought ‘better luck next time.’

Now how does all this connect to the absolute configuration? Not at all! The absolute configuration is assigned on a per asymmetric atom basis. You need to know what the atom’s surroundings are like and can then apply a descriptor directly to it. Most notably, it was not the molecule as a whole that served as the reference but only the neighbouring groups and their priorities (read: atomic charge values). Any similarities to other stereodescriptor schemes, in use or not, was entirely accidental and unintended.
You have to remember that the retention of configuration in atomic terms does not automatically mean that the absolute configuration is retained. Take for example the Mosher acid as depicted below (image was taken from Wikipedia, where a full list of authors is available):

(R) and (S) Mosher acid.
Assigning the (R) configuration to the structure on the left is trivial. Assume you were to turn the (R) Mosher acid into the corresponding acid chloride. Note that although the asymmetric carbon was not touched the acid chloride turns out to be (S)-configured! So absolute configuration is something far more ‘wobbly’ than stereocentres are; make a remote change and influence an absolute configuration although that part of the molecule wasn’t touched.

Incidentally, that is the beauty of the Fischer nomenclature, especially when you consider the age in which it was invented. You don’t need to know which oxygen is pointing where you don’t need to figure out a long chain of priorities. You know you started at D-glycerinaldehyde so your sugar is now D-ribose. And you can take a pen and draw the lowest hydroxy group pointing to the right.
Do note make the mistake and try to correlate Fischer nomenclature with the sign of optical activity or with absolute configuration; all attempts will fail. Each has their use, their place to shine and none treats all similar molecules in the same way. Use wisely.


text:
Does internal potential energy affects temperature or only kinetic energy?
If I take an extreme case, where a body has only an internal potential energy with zero internal kinetic energy, does this body have a temperature? Another question related to it: if two objects A and B having different temperature: A: having only internal potential energy and B having only internal kinetic energy, can heat flow from A to B ?(temperature of A > B)

You know these cases aren't really possible?

yes I know. but I want to see if potential energy affects temperature and if potential energy can cause transfer of heat

Well, I guess no in both cases.

I believe you know the simple answer to both of your questions to be no. Though you should realize that heat tends to flow from the object with more kinetic energy to that of lower kinetic energy. So why would you think an object with no kinetic energy would transfer heat? It has none to transfer.

Ok but in general, internal potential energy does not affect temperature?


text:
What does the subscript n on methylene groups in a cyclic alkene mean?
I've recently come across an unfamiliar notation on a cyclic alkane. One carbon had parentheses around it with a subscript $n$. Does this refer to rings of variable sizes (i.e.  one or more $\ce{CH2}$ groups are located in this position)? If not, what does this refer to?

Yes, n denotes the number of $\ce{CH2}$ groups between the parentheses.
Example:

n = 1, 2 or 4 means cyclopentane, cyclohexane, or cyclooctane.

So, just to clarify, this means that the ring (drawn as cyclohexane) could actually represent a cyclopentane or cycloheptane, etc. as indicated by the subscript? Therefore, this would not represent a group attached to the ring that is not, itself, part of the cyclic structure?

The parenthesized group with the subscript is repeated n times. So if cyclohexane is drawn, the repetition of a $\ce{CH2}$ migth result in cyclohexane (n = 1) or bigger rings. I've added an example to illustrate this.


text:
Explaining the Relative Energies of Various Vibrational Modes
Specifically, I'm wondering why it is that the asymmetric stretch in most (although there are exceptions to this I believe) molecules is higher in energy than the symmetric stretch.
Similarly, I would like a nice, physical (ideally quantitative/mathematical) explanation of why bending modes are usually lower in energy than stretching modes.
I can rationalize this second question on the basis of electrostatic forces. That is, any distortion from the minimum of the potential energy well will either cause more coulombic-type attractive forces or more coulombic-type repulsive forces. These forces should not be as large for a lateral displacement (bending) than for a stretch of some kind. This makes sense because the change in radius will not be as large when simply moving to the side from an equilibrium position. Of course, the change in radius is significant because electrostatic potential energy goes as $\frac{1}{r}$.
Anyways, that's a pretty classical answer, so I wonder how both of those questions can be explained more quantitatively as well as quantum mechanically.

I don't know how to explain the bending mode, but there is a nice mathematical explanation for the difference in stretching modes.
Take $\ce{CO2}$, carbon dioxide, as an example. I'll use B3LYP/6-31G(d,p) values. $\nu_{1}$, the symmetric stretch, is 1372 cm$^{-1}$, and $\nu_{3}$, the asymmetric stretch, is at 2436 cm$^{-1}$. Now consider $\ce{CO}$, a single carbonyl group (the molecule won't work for this example). Looking at an IR table, the average stretch will be about 1750 cm$^{-1}$. $\ce{CO2}$ looks a lot like two $\ce{CO}$ bonds that are linear and share a carbon. But, clearly $\ce{CO2}$ doesn't have a doubly-degenerate stretch mode of any kind at that frequency, so something is causing the local modes to not be identical to the normal modes. Because both local modes have the same symmetry, they are allowed to mix with some coupling between them, just like molecular orbitals. A local mode is a vibration that is completely uncoupled from all other modes; its motion does not influence the motion of other atoms and vice versa.
Set up a Hamiltonian for a degenerate two-level system with some coupling between them. Keeping with a Huckel-like picture, I'll use $\alpha$ for the energy level of each degenerate local mode and $\beta$ for the coupling between them.
$$
\hat{H} = \begin{pmatrix}
\alpha & \beta \\
\beta & \alpha
\end{pmatrix}
$$
Obtaining the normal modes and their energies is equivalent to diagonalizing this Hamiltonian, where the eigenvalues are the normal mode energies and the eigenvectors are the normal modes themselves as linear combinations of the local modes. Doing so gives the normal mode energies:
$$
\alpha - \beta, \, \alpha + \beta
$$
and the normal modes themselves:
$$
\begin{pmatrix}
-1 \\
1
\end{pmatrix},\,
\begin{pmatrix}
1 \\
1
\end{pmatrix}
$$
By convention in Huckel theory, $\beta$ is usually negative, so the 2nd eigenvalue is the one lower in energy.
If you take the dot product between each normal mode eigenvector and the vector of local modes:
$$
\begin{pmatrix}
\phi_{1} \\
\phi_{2}
\end{pmatrix}
$$
out falls the symmetric and antisymmetric combinations leading to the symmetric and asymmetric stretches. Leaving out the normalization factor of $1/\sqrt{2}$:
$$
(-\phi_{1} + \phi_{2}), \, (\phi_{1} + \phi_{2})
$$
Setting $\alpha = 1750$ cm$^{-1}$ and $\beta = -500$ cm$^{-1}$ gives a symmetric stretching frequency of 1250 cm$^{-1}$ and an asymmetric stretching frequency of 2250 cm$^{-1}$. Not bad for a model with a single parameter.
This is how the energy levels (eigenvalues) split with increasing the coupling between the initially degenerate local modes into distinct normal modes:

Here is a link to the Jupyter Notebook containing the math I did. This is a simplified (math-wise) version of what we did in a paper, since the conceptual point is identical.
This sort of description is how any two-level time-independent system works. See how the two spin energy levels of a free electron change in energy when an external magnetic field is applied (Zeeman effect):

@pentavalentcarbon Comparing bending and stretching frequencies.. A stretching mode means that you will be significantly changing the orbital coupling of a bond (i.e., $\beta$ in Hückel theory). This clearly will change the energy of the system significantly. While overlaps change slightly during bending motions, the effect is much less - the distortion is not along a bond axis.

@GeoffHutchison We showed in the paper that charge transfer into the sigma bonding orbital of carbon dioxide causes a geometric distortion, which then causes frequency shifts. So, for a qualitative understanding of vibrations, the simple geometric picture is still effective, and any non-linear behavior in the energy shifts requires a better description of orbital behavior than $\beta$.

Ahh... This is more what I expected the math to look like. I've even seen this mathematical process before so I should've known better. Am I correct to guess the local mode description would work better for a stretch with a light atom. Like a C-H stretch?

Its already a while ago but still: What else than the force constants (energies) effects on $\beta$? Is there an "overlapp"-like term? What happens when two CO ligands are attached to a big molecule (complex)?


text:
How does hot air remove guests from cyclodextrin?
Cyclodextrin, as I know, has a hydrophobic cavity. Therefore, it would absorb hydrophobic organic compounds. But the organic compound can also be removed with hot air since the interaction between the organic compound and the cavity is an induced dipole attraction. How much heat would be enough to break the attraction and how would it break the attraction?


text:
Effects of a non-inert electrode in a half cell reaction involving hydrogen gas as the electrolyte ion
If there was a galvanic cell that had a copper cathode electrode and with a sulfuric acid electrolyte, will the copper ions be reduced and react with the electrolyte or will the H+ ions be reduced?
Normally, with a gaseous electrolyte, an inert anode material is used. I was just curious as to what happens when copper is used as supposed to an inert electrode.

Wait, where did those copper ions come from?

so there is a copper electrode submerged in a sulfuric acid electrolyte. That was my question: would copper just act like an inert electrode like platinum and transfer the electrons, or would the copper electrode be reduced and react with sulfuric acid?

Electrode can never be reduced. It consists of metallic copper; it is the reduced form already, you can't reduce it any further.

so the hydrogen in sulfuric acid would be reduced to H+ ions?

If you started with a full cell with two inert electrodes immersed in sulfuric acid and applied a large enough voltage between them, a current would flow. At one electrode, the easiest to oxidize species would be oxidized (in this case, that would probably be 2 H2O -> O2 + 4 H+ + 4e-). At the same time, the easiest to reduce species would be reduced at the other electrode (probably 2H+ + 2e- -> H2). If you switched the cathode electrode from some inert metal to copper, the easiest reduction would still be 2H+ + 2e- -> H2 (Remember, there is no Cu2+ in the system to reduce). If, on the other hand, you changed the anode from inert metal to copper, the easiest oxidation would probably be Cu -> Cu2+ + 2e-, so the net reaction would change.

what if the sulfuric acid electrolyte is part of a galvanic cell. What would happen then?


text:
Hydrogenation of cycloalkanes
Why can we not hydrogenate or brominate etc. on cycloalkanes to convert into linear molecule? And are there any exceptions to this?

Well, the $\ce{C-C}$ bond in cycloalkanes is the same as $\ce{C-C}$ bond in alkanes - that is, quite solid. Except when it is strained like in cyclopropane, which you indeed can brominate, hydrogenate and do whatever else.


text:
Is nucleus at the centre of the atom? If so why?
In postulates of Bohr's model of the atom, it is said that the nucleus is at the centre of the atom. In Rutherford’s α scattering experiment he said that out of 20,000 alpha particles only 1 particle is deviated by $180^\circ$. If the particles in the atom are moving continuously (like an electron), why is that nucleus at the centre of the atom? If that is so, then when we focus all the α particles at the nucleus (the point where an α particle got deviated by $180^\circ$) all should get deviated by $180^\circ$ right?
What I mean is that why is the nucleus at the centre of the atom?

Forget the Bohr's model. The particles in the atom are not really moving; also, they are not really particles.

I am a 12th class student. Can you please explain?

I believe this has been discussed a dozen times before. Wait a bit, surely someone is going to post a link to some earlier explanation.

Read this, for example: http://chemistry.stackexchange.com/questions/36987/does-the-shared-electron-in-covalent-bonds-revolve-around-nucleus

@IvanNeretin I have a feeling you read too much quantum stuff ;) At least from OP's point of view. From very basic pov it's similar situation as with Solar System. Nucleus is highly charged and massive analogously to Sun.

First, Rutherford did not show that the nucleus was at the center of an atom, only that the majority of the mass of an atom was concentrated into a small volume.  Now, given a small nucleus with all the positive charge, the electrons will, to first order, groups around the nucleus, not sit off to one side.  For that last bit - how, exactly, do you plan to focus all (or any!) alpha particles so that they will all hit one nucleus? For historical accuracy, the original experiment used alpha particles from natural decay processes, which were collimated with apertures - no focusing.

Why is the nucleus the centre of the atom?

Definition. Whatever the electron does, chemists define the position of the nucleus to be the centre of the atom meaning that electrons can be polarised away.
Quantum mechanics. When studying an isolated atom, the only thing that influences electron behaviour is the nucleus. You can calculate the ‘path’ of an atom‡ relative to the position of the nucleus with quantum mechanics and your result will be that quite a few electrons will occupy a perfect sphere whose centre is the nucleus.

The second point may require elaboration. But it really is quite simple. You have a very lightweight, very fast particle with a charge of $-1$ (your electron) and you have a very heavy, very slow particle with a charge of $+1$ (your proton). By the time the proton has moved an inch, the electron has moved a mile so we can approximate that only the electron really moves. (This has the fancy name Born-Oppenheimer approximation.) Now remember that both are attracted to each other. The proton doesn’t move, but the electron feels the ‘desire’ to ‘move towards’ the proton. (Inverted commas because this is a macroscopic view that only gets us so far when dealing with quantum effects.) If you average this over time, you will find that the electron will have spent the same amount of time on one side of the nucleus as on the other and the mean electron position will be in the nucleus. Spherically outwards from the proton, the chance of locating the atom decreases monotonously with distance; the angle does not matter. We have a perfect sphere with the proton being the centre.

There are so many different approaches to this question, I am sure. But, if you look at it strictly from a very simplistic perspective, the nucleus has a net positive charge and the electron cloud has a net negative charge. The electron cloud gets as close as it can to the nucleus, but not too close due to inter-electron repulsion. That makes the most probable distribution of the electron cloud to be a uniform distribution with the nucleus at the center. Because the position of the nucleus is only a probability, not all focused alpha particles would back-scatter by 180 degrees experimentally.


text:
How to check percentage yields in a given reaction?
For a general reaction , if it is asked to predict and not given by theoretically, how would we determine the percentage yield of the products based on their mechanism , stability , path taken for the reaction etc. ?
For example , where my main problem lies is  :
According to Solomon and Frhyle it is said that chlorine is highly unselective. But in a problem I was asked percentage yields of the two products:

So how do I go about determining the percentage yields in this reaction ?

They are determined experimentally. There isn't any possible way you can determine percentage yield by looking at a reaction (unless there is only 1 possible product).
What the text is trying to explain to you is that when comparing analogous reactions of $\ce{Cl2}$ and $\ce{Br2}$, there will be higher percentage yields of the more highly substituted (and stable) product in the $\ce{Br2}$ reactions. For example, in the reaction between propane and $\ce{Cl2}$, the yields of 1-chloropropane and 2-chloropropane are 45% and 55% respectively. In a reaction between propane and $\ce{Br2}$, the yields of 1-bromopropane and 2-bromopropane are 3% and 97% respectively.
The exact reason for these selectivities has to do with the differences in energies in the transition states for the formation of the free radicals involved in the mechanism for these reactions. If you're interested in learning more you can see for yourself here.

Yeah the transition state energies helped quite a lot!