text:
Partial Atomic Charge in GAMESS
In GAMESS, a calculation of optimisation gives data about partial atomic charges (like Mulliken or Lowdin charges). Can you please let me know how to calculate the other types of partial atomic charges (like Hirshfeld, Maslen or Politzer charges)?
FYI: I have used B3LYP and MP2 for optimization.

Unfortunately GAMESS does not support any other than the named population analyses.
You will need to use a different program. For that matter I suggest MultiWFN, which comes with a variety of different population analyses and is still in active development. From the website:

Population analysis. Hirshfeld, VDD, Mulliken, Löwdin, Modified MPA (including three methods: SCPA, Stout & Politzer, Bickelhaupt), Becke, ADCH (Atomic dipole moment corrected Hirshfeld), CHELPG, Merz-Kollmann and AIM methods are supported.

The currently developed version MultiWFN 3.4 should be able to even read GAMESS output directly as stated in the update log:

GAMESS-US output file now can be used as input file (not comprehensively tested, currently only single point task at HF/DFT level is formally supported). The suffix of output file should be changed to .gms so that Multiwfn can properly recognize it

As a workaround for older versions, and a reliable way to import this data, you can export a .wfn file with GAMESS. I will give a brief summary for this, since I am still using MultiWFN 3.3.8. From the corresponding manual:

For example, the wavefunction represented by GTFs is enough for Hirshfeld population, so you can use .fch/.molden/.31~.40/.wfn/.wfx file as input, but .pdb, .xyz, .chg, .cub and .grd files do not carry any wavefunction information hence cannot be used.

As I have previously stated, .wfn files are technically outdated, especially when it comes to larger basis sets. However, for the time being, they are our only option to make it work.
I am using GAMESS version % DEC 2014 (R1) and the following input to demonstrate the procedure. I am using a Linux environment, so you might have to adjust this guide accordingly.

 $CONTRL SCFTYP=RHF            ! Restricted calculation.
         RUNTYP=OPTIMIZE       ! Geometry optimisation.
         COORD=ZMT             ! Z-matix specifies coordinates.
         AIMPAC=.TRUE.         ! Requests wfn file to be written.
         $END
 $BASIS  GBASIS=STO NGAUSS=3   ! STO-3G minimal basis
         $END
 $GUESS  GUESS=HUCKEL          ! Start with Hückel MO.
         $END
! Following group contains molecule specification.
! Blank lines are important.
 $DATA
Water
Cnv  2

O
H  1 rOH
H  1 rOH  2 aHOH

rOH=1.09
aHOH=110.0
 $END

Note that the indentations before the keywords are important. ! indicates a comment. Any additional output will be written to the punch file. This is dependent on how you set up GAMESS, but it normally is located in ~/scr/<rootfilename>.dat.
To extract the .wfn file you can use the following little script, or extract it manually from the .dat file.

#Script to extract wfn file from GAMESS dat (punch) file
#! /bin/bash

[ ! -z $1 ] && inputfile="$1" || exit 1
outputfile="${inputfile%.*}.wfn"

startpattern="----- TOP OF INPUT FILE FOR BADER'S AIMPAC PROGRAM -----"
endpattern="----- END OF INPUT FILE FOR BADER'S AIMPAC PROGRAM -----"

sed "/$startpattern/,/$endpattern/!d;//d" "$inputfile" > "$outputfile"

This should give you the following .wfn file for analysis:

Water
GAUSSIAN              5 MOL ORBITALS     21 PRIMITIVES        3 NUCLEI
  O    1    (CENTRE  1)   0.00000000  0.00000000 -0.14560943  CHARGE =  8.0
  H    2    (CENTRE  2)  -1.43257773  0.00000000  1.05592484  CHARGE =  1.0
  H    3    (CENTRE  3)   1.43257773  0.00000000  1.05592484  CHARGE =  1.0
CENTRE ASSIGNMENTS    1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  2  2  2  3  3
CENTRE ASSIGNMENTS    3
TYPE ASSIGNMENTS      1  1  1  1  1  1  2  2  2  3  3  3  4  4  4  1  1  1  1  1
TYPE ASSIGNMENTS      1
EXPONENTS  1.3070932E+02 2.3808866E+01 6.4436083E+00 5.0331513E+00 1.1695961E+00
EXPONENTS  3.8038896E-01 5.0331513E+00 1.1695961E+00 3.8038896E-01 5.0331513E+00
EXPONENTS  1.1695961E+00 3.8038896E-01 5.0331513E+00 1.1695961E+00 3.8038896E-01
EXPONENTS  3.4252509E+00 6.2391373E-01 1.6885540E-01 3.4252509E+00 6.2391373E-01
EXPONENTS  1.6885540E-01
MO  1                     OCC NO =   2.00000000 ORB. ENERGY = -20.25158070
  4.22735195E+00  4.08851077E+00  1.27421022E+00 -6.18804850E-03  8.27701479E-03
  6.24678653E-03  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00
  0.00000000E+00  0.00000000E+00  6.97684832E-03  4.38722945E-03  6.95010854E-04
 -1.54630881E-03 -1.49552256E-03 -4.66089059E-04 -1.54630881E-03 -1.49552256E-03
 -4.66089059E-04
MO  2                     OCC NO =   2.00000000 ORB. ENERGY =  -1.25754111
 -9.93961353E-01 -9.61316148E-01 -2.99600254E-01 -2.02173727E-01  2.70423693E-01
  2.04092795E-01  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00
  0.00000000E+00  0.00000000E+00  2.05800516E-01  1.29412887E-01  2.05011755E-02
  4.30889918E-02  4.16737970E-02  1.29879022E-02  4.30889918E-02  4.16737970E-02
  1.29879022E-02
MO  3                     OCC NO =   2.00000000 ORB. ENERGY =  -0.59384261
  0.00000000E+00  0.00000000E+00  0.00000000E+00 -0.00000000E+00  0.00000000E+00
  0.00000000E+00  1.02654522E+00  6.45519180E-01  1.02261083E-01  0.00000000E+00
  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00
 -1.24405583E-01 -1.20319664E-01 -3.74983838E-02  1.24405583E-01  1.20319664E-01
  3.74983838E-02
MO  4                     OCC NO =   2.00000000 ORB. ENERGY =  -0.45973131
 -4.42352599E-01 -4.27824176E-01 -1.33334109E-01 -1.28843153E-01  1.72338126E-01
  1.30066155E-01  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00
  0.00000000E+00  0.00000000E+00 -1.26642108E+00 -7.96359559E-01 -1.26156734E-01
 -8.17262213E-02 -7.90420432E-02 -2.46339524E-02 -8.17262213E-02 -7.90420432E-02
 -2.46339524E-02
MO  5                     OCC NO =   2.00000000 ORB. ENERGY =  -0.39261686
  0.00000000E+00  0.00000000E+00  0.00000000E+00 -0.00000000E+00  0.00000000E+00
  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00  1.67545020E+00
  1.05356804E+00  1.66902879E-01  0.00000000E+00  0.00000000E+00  0.00000000E+00
  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00
  0.00000000E+00
END DATA
RHF      ENERGY =      -74.9659012162   VIRIAL(-V/T)  =   2.00600297

You can now use this file and load it with MultiWFN. After the start-up process you should find the key values printed to the screen like this (I abridged it a little bit):

 Multiwfn -- A Multifunctional Wavefunction Analyzer (for Linux)
 Version 3.3.8, release date: 2015-Dec-1
[...]
 System energy:     -74.965901216200 Hartree,   Virial ratio:  2.00600297
 Total/Alpha/Beta electrons:     10.0000      5.0000      5.0000
 Net charge:     0.00000    Expected multiplicity:    1
 The number of orbitals:     5,  Atoms:      3,  GTFs:     21
 This is restricted close-shell single-determinant wavefunction
 Title line of this file: Water
 Formula: H2 O1
 Molecule weight:        18.01528

 Loaded water-sto3g.wfn successfully!

                    ------------ Main function menu ------------
[...]
 7 Population analysis
[...]

You may now choose 7 and enter the following menu:

                 ============== Population analysis ==============
 0 Return
 1 Hirshfeld population
 2 Voronoi deformation density (VDD) population
 10 Becke atomic charge with atomic dipole moment correction
 11 Atomic dipole corrected Hirshfeld population (ADCH)
 12 CHELPG ESP fitting charge
 13 Merz-Kollmann (MK) ESP fitting charge
 14 AIM charge

Follow the guidance of the manual and the on-screen instructions to produce the required charges. For comparison here are the computed Hirshfeld charges, written to water-sto3g.chg with the following command sequence 7; 1; 1; y.

  O     0.000000    0.000000   -0.077053   -0.296017
  H    -0.758088    0.000000    0.558771    0.147959
  H     0.758088    0.000000    0.558771    0.147959

The element identifier is followed by the Cartesian coordinates, which are followed by the charges.
In principle all other implemented charges are equally as easy to compute. Keep in mind, that basin based charges (e.g. AIM) need more computing power and depend heavily on molecule and basis set size.
Apart from this procedure, you can obtain natural charges with the NBO6 program, which conveniently interfaces with GAMESS. Unfortunately you have to buy it.

Partial charges are tricky because they are not observables, therefore it is difficult to define them rigorously, however in same cases it can be done.
I usually never use analysis of the population-based charges since they can be very arbitrary in assigning electrons to a given atom.
I know with Gaussian you can obtain Hirshfeld charges (therefore fitting the ESP) but honestly in GAMESS I do not know.
A more rigorous way of dividing the electron density is to use the Quantum Theory of Atoms in Molecules. I cannot go into details since it would take really a lot, however it is possible to obtain all kind of atomic properties partitioning the molecular electron density in basins of charge containing the nuclei, following rigorous topological arguments. From the atomic densities it is possible to obtain the expectation values of observables (atomic dipole/quadrupole moments, energy, magnetic properties etc...) and integrating the atomic density it is possible to know "how many" electrons are contained in the atomic basis, therefore the atomic partial charge.
The free software AIMALL can definitely help you.
Using GAMESS it is possible to generate wavefunction files (file.wfn) that contain the information needed to AIMALL to perform the decomposition of the molecular density. It is very easy and free, I strongly suggest you this approach.
In case you were dealing with planar molecules, there is another rigorous way to obtain information about the partial charge of atoms via the Atomic Polar Tensors, defined as
$$P_{\sigma I}=\frac{\partial \mu_{\sigma}}{\partial r_{I}}$$
where sigma can be the x,y,z molecular dipole component and I is one of the $3N$ cartesian coordinates of the atoms defining the molecule. This tensor is analytically calculated by almost every Quantum Chemistry software, GAMESS included.
I know it is difficult to be absolutely exhaustive within an answer, there are many more things and details to say, however I give you some useful link where you can pick up more theory if needed
AIMALL web page: http://aim.tkgristmill.com/
Paper of Dinur of the charge in planar molecules: http://pubs.acs.org/doi/abs/10.1021/j100169a030
I really hope it can be helpful!

As far as I know, the standard (free) mode of AIMALL supports only a few atoms and basis functions and is in my eyes utterly useless. (It's also such a slow user interface that I would not recommend buying a professional version.) This post reads more like a commercial than an actual answer, and I disagree that QTAIM is to complex to be explained here.

No one said it is too (not to) complex, I wrote "it would take really a lot". I would be pleased to see you explaining the atomic virial theorem and the formalism of bond/ring/cage critical points in one answer! I think I would learn a lot! Moreover I expressed my enthusiasm for AIMAll since I used it a couple of times and I think it is really a nice tool..of course if you work with many atoms it makes no sense to perform QCT. So I cannot understand what's wrong with my naswer..He ansked and I replied with a possible way to solve his problem.

Obviously the full derivation of QTAIM would be too (I'm very sorry for my typo) long for a post. However, the basis on what the electron density is analysed, divided and what criteria are used to obtain the partial charge are not. I can understand that you like AIMAll, however, it is proprietary software (quite expensive actually) and therefore not free. The standard edition has a limitation of 12 atoms, which is barely enough to cope with benzene. If you are well versed with QTAIM, maybe you can answer my question.


text:
Speed at which methanol dissolve in water?
Suppose that in a container of volume V we pour V1 of methanol and V2 of water (V2>V1), supposing that V2 is big enough that all the methanol will be dissolved, do we have to mix them mechanically or the diffusion will be enough if we want them to mix in seconds?

Methanol is infinitely soluble in water and vice versa. If you mean how long does it take for the solution to be relatively uniform, it will be more than several seconds. You should stir.

Is their a formula calculating the exact time and rate?

No. The way you pour it is going to matter. The size and shape of the container. You're basically modelling diffusion which can be done, but it's pretty dependent on initial conditions...

As already mentioned by @Zhe, the rate of the mixing of methanol in water (or the other way around) depends (among other parameters) on the rate of addition, the geometry of the vessel, if / how the solutions are mixed by a stirrer.  In addition, this process diffusion is dependent on temperature and gradients of concentration.  One approach to quantify diffusion is based on the differential Fick's laws by determination of the diffusion coefficient.
As difficult it may appear, the diffusion of methanol in water was determined experimentally, as here and here, for example.  If you do not have access to a university library, NIH / HHS made a more recent reference available here, and a more general value is given here

source: Stack Exchange

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

sort: votes


text:
Why is a cationic trialkylammonium functional group inductively withdrawing in nature?
The alkyl groups on the nitrogen ion should donate electrons (alkyl groups are electron donating in nature). So, irrespective of the nitrogen ion being a cation (which is an electrophile), the alkyl groups should compensate the electrophilicity of the nitrogen cation, and the ion as a whole should be electron donating in nature.
Then why is $\ce{R3N+}$ considered to be electron withdrawing?

The inductive effect is very weak; the inductive effect will reduce the charge on the nitrogen atom but it does not make it zero. Nitrogen is quite electronegative and it hates to be positively charged (even partially). It tries to pull electrons towards itself wherever possible.

I don't know the reason, but $\text{R}_3\text{N}^+$  is considered highly de-activating and meta-directing for electrophilic substitution on Benzene. I wanted to ask this question!

By far, the overall positive charge of the $\ce{[-NR3]+}$ group is its dominating characteristic when considering inductive behavior.  Bear in mind that the magnitude of the charge variations due to the inductive effect is much smaller than $q$, the electron charge. These variations are written as $\delta^+$ and $\delta^-$, after all, and it's implicit in the notation that $|\delta^\pm| \ll 1$.
So, no matter how electron-donating the alkyl chains might be relative to the hydrogens in a simple amine group, $\ce{[-NH3]+}$, the overall functional group remains strongly electron withdrawing.

It's worth pointing out here a common misconception about ammonium cations; quoting Fry$^\dagger$ (emphasis added):

[One must] appreciate that while the central nitrogen atom of the [tetra-n-butylammonium] cation is customarily written with a positive charge, quantum chemical computations at all levels of sophistication all agree that in tetraalkylammonium ions each of the hydrogen atoms actually carries a small positive charge and all of the carbons and the central nitrogen atom carry a negative charge.

Thus, even though we usually write tetraalkylammonium cations $\ce{(R4N+)}$ and trialkylammonium side groups $\ce{(-R3N+)}$ as though the positive charge is localized on the central nitrogen atom, the strongly-electronegative $\ce N$ will in fact always carry a $\delta^-$ charge.  It would thus be more accurate to write these species as $\ce{[NR4]+}$ and $\ce{[-NR3]+}$, respectively.

$^\dagger$ Fry, A.J. Electrochem Soc Interface, Summer 2016 issue, pp. 37-40. doi:10.1149/2.F04162if
       (free article through 9 Apr 2017!)

In the first part of the answer, you've written that alkyl groups will direct some electron density towards the nitrogen atom (although very feebly). If that's the case then why $NR_3+$ is even more electron-withdrawing than $NH_3+$? Shouldn't alkyl groups, with their superior +I ability than H atoms, decrease the -I effect in $NR_3+$?

@AumkaarPranav Very good question! I don't actually know the reason for that. I'd suggest asking it as a brand new question...it does seem to be a hole in the logic of my answer.

Nitrogen is a very electronegative element and the fact that nitrogen bears a positive charge in the ammonium ion makes it even more electrophilic(as you pointed out) . Alkyl groups are only weak electron donating groups and can't help much about nitrogen's positive charge.Hence the ammonium cation is inductively withdrawing in nature.

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text:
Pesticide Edolan U
I have recently become aware that old woolen blankets from the 1940:s to the 1970:s often contain the pesticide "eulan U 33" or "Edolan U" and wonder what is the best way to dispose of these. Ordinary burning is apparently not enough.

It can be found that Edolan-U (Eulan-U-33) is

CAS 69462-14-2
1-chloro-N-[4,5-dichloro-2-(2,4,5-trichlorophenoxy)phenyl]methanesulfonamide sodium salt, or
sodium 1,4',5'-trichloro-2'-(2,4,5-trichlorophenoxy)methanesulfonanilide
(PubChem doesn't know it yet, only its free sulfonanilide "acid", CID 13932360),
according to the PAN PesticideInfo.org database record (and combination with some other chem. DB searches), where you can find potentially useful regulatory info.

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


text:
Is there such a thing as an acid without a hydrogen?
I'm in AP Chemistry and we are learning about the Brønsted-Lowry model and my teacher mentioned that "for the most part" acids have hydrogen, could there be such a thing as an acid that has no hydrogen in it?

Well, one can buy gas bottles filled with anhydrous HF (nasty stuff, not to play with at home). Now, you can ask if that is an 'acid' in the bottle. Once exposed to air or water, it certainly is.

By definition, Bronsted-Lowry acids (proton donors) must have a proton, so they must have hydrogen. Lewis acids are a different matter altogether though, and chances are you'll come across it soon enough. PS @JonCuster question was changed, OP meant "without hydrogen" not "anhydrous".

@orthocresol - So it goes. Still an interesting question - is pure HF an acid or not without the presence of water.

@JonCuster HF is a stronger acid than F- is. Acidity is a continuous scale, not an either/or or yes/no binary category.

@JonCuster Would methane still be considered a flammable gas if it's stored in an anoxic environment?

@R.M. - certainly one could say that the reactions/equations that determine the hydrogen ion concentration are different between anhydrous HF and aqueous HF.

Not totally related, but most students don't realize that a hydrogen ion is simply a proton. You'll benefit from that knowledge in college level chem.

@gogators my teacher actually taught that to us at the beginning of the chapter, and it makes some of the actions of acid base reactions a lot more intuitive

OP, I appreciate that you called it a "model". In my opinion that is a much better way of describing it than "theory". Arrhenius, Bronsted-Lowry, Lewis... After all, they are only human conceptions to aid us in categorizing our observations of how different chemical species react. There was never some hypothesis that "acids release protons" that became a theory, rather, Bronsted and Lowry defined that "acids release protons", in their very useful but ultimately arbitrary categorical model.

$\ce{AlCl3}$ and $\ce{BF3}$ are acids without hydrogen. :)

It depends on which definition of acids and bases you are using.
According to the Arrhenius theory, acids are defined as a compound or element that releases hydrogen (H+) ions into the solution. Therefore, there are no Arrhenius acids without a hydrogen atom.
According to the Brønsted–Lowry acid–base theory, an acid is any substance that can donate a proton and a base as any substance that can accept a proton. Hence, there are no acids without a hydrogen atom according to this theory.
But according to the Lewis theory of acids and bases, an acid is any substance that can accept a pair of nonbonding electrons. In other words, a Lewis acid accepts a lone pair of electrons. According to this theory, acids without a hydrogen atom can exist. (A coordinate bond is formed between the Lewis acid and base. The compound formed by the Lewis acid and base is called a Lewis adduct)
A great example of this would be $\ce{BF3}$. It is neither an Arrhenius acid nor a Brønsted–Lowry acid but it is a Lewis acid. The boron atom accepts a pair of nonbonding electrons from another atom or ion to complete its octet. Here $\ce{BF3}$ is a Lewis acid as it accepts a pair of nonbonding electrons.

The fluorine ion here is a Lewis base as it donates a pair of electrons.
If you want to get more rigorous about the definition of Lewis acids: "a Lewis acid is a type of chemical that reacts with a Lewis base to form a Lewis adduct".
More about Lewis acids and bases here.

Just out of curiosity, would such thing form acid-like solution in high pressure liquid form of oxygen difluoride? (assuming it didn't have access to anything else to react with, I know I'm probably showing just how little I know by asking this question)

How could fluorine make a dative bond with the bromine atom when fluorine has a higher electronegativity than bromine?

There are quite a number of theories regarding acidity and basicity, but in this case, will explain the Lewis acid.
The Lewis Theory of acids and bases
This theory extends well beyond the things you normally think of as acids and bases.
The theory

An acid is an electron pair acceptor.
A base is an electron pair donor

The Lewis acid-base theory explains why $\ce{BF3}$ reacts with ammonia. $\ce{BF3}$ is a trigonal planar molecule because electrons can be found in only three places in the valence shell of the boron atom.
As a result, the boron atom is sp2 hybridized, which leaves an empty 2pz orbital on the boron atom. $\ce{BF3}$ can, therefore, act as an electron-pair acceptor or Lewis acid. It can use the empty 2pz orbital to pick up a pair of nonbonding electrons from a Lewis base to form a covalent bond.
$\ce{BF3}$, therefore, reacts with Lewis bases such as $\ce{NH3}$ to form acid-base complexes in which all of the atoms have a filled shell of valence electrons, as shown in the figure below:

Consider another scenario:

Water is an example of a lewis base. Carbocations are examples of lewis acids. When water reacts with a carbocation, as shown above, one of the electrons pairs from oxygen is used to form a new sigma bond to the central carbon in the carbocation.
References

The Lewis Theory of acids and bases
The Lewis Definitions of Acids and Bases
Lewis acids Chemistry

As noted by others, there is the concept of Lewis acids, and other answers to this question explain that concept well. To reiterate: A Lewis acid is a chemical that will accept electron pair donation from another chemical (which is called a Lewis base in this context). It is also true that Brønsted-Lowry defined acids as $\ce{H+}$-donors and bases as $\ce{H+}$-acceptors. As such, a BL acid without hydrogen is strictly not existent.
I note two important exceptions that sort of defy the strict definition. Boric acid $\ce{B(OH)3}$ does not detach one of its original protons but accepts an electron pair from $\ce{OH-}$. This raises the concentration of $\ce{H+}$ through protolysis of water. $\ce{FeCl3}$ reacts likewise. Of course, these reactions require water, thus, in the strict sense, a complex of water and the substances noted constitutes the BL acid, not the substances themselves.

Plenty of answers here focused on the definitions of Brønsted-Lowry and Lewis acids, which is great. But there is an important point I saw mentioned only in passing (in TAR86's answer), and I think it comes closer to answering your question that reciting the somewhat arbitrary acid-base definitions of our forefathers (useful as they are).
Could there be such a thing as a Brønsted-Lowry acid that has no hydrogen in it?
Yes! At least in a pragmatic sense. As elaborated here, highly-charged metal cations such as $\ce{Al^{3+}}$ and $\ce{Fe^{3+}}$ can liberate protons in aqueous media. When such cations are placed in water, water molecules  coordinate/associate with the metal ion:
$$\ce{Al^{3+} + 6H2O -> [Al(H2O)6]^{3+}}$$
Here is a structural cartoon of the hexavalent aluminum hydrated complex:

This complex easily releases three protons:
$$\ce{[Al(H2O)6]^{3+} -> Al(H2O)3(OH)3 + 3H+}$$
So, not only is $\ce{Al^{3+}}$ a Lewis acid, but practically speaking it is also a Brønsted-Lowry acid (and triprotic, nonetheless!) Such highly-charged metal cations are as close as you're going to come to a Brønsted-Lowry acid without hydrogen. While they are not acids in a literal (single-step) sense, if we are considering aqueous solutions, then functionally speaking they are indeed Brønsted-Lowry acids because they effect the release of protons and thus lower the solution's pH.
I'm not sure if your teacher had this in mind, or was just trying to cover their arsenic in case there were exceptions they had forgotten about, but there is your answer.

I just wish to clarify the other answers. The definition of a "Lewis Acid" is a chemical species which reacts with the electron lone pair of a "Lewis Base" (and a Lewis Base is a species which has a reactive lone pair of electrons). This is a generalization and abstraction of the more common Brønsted-Lowry theory.
Consider that Science is full of 'levels' of abstraction. We have positive integers (the natural numbers), but then we add zero, then negative integers, then rationals, finally we might think we're done when we get to real numbers, but then along come complex numbers, and quaternions, octonions, vectors, fields, and on and on.
While historically a fact, so you need to be familiar with the concept of Lewis Acid/Lewis Base/Lewis Adduct, you should also be aware that all chemistry has to do with the (partial or total) donation and accepting of electronic charge, so while lone pair chemistry is a large enough area to justify its own terminology, it's just the same old, same old "chemistry is about the electrostatic interactions of atoms" story.


text:
Is there an intramolecular hydrogen bond in guaiacol?
Is there a hydrogen bond between the $\ce{H}$ atom of $\ce{-OH}$ group and the $\ce{O}$ atom of the $\ce{-OCH3}$ group in ortho-methoxyphenol? I was told no. But I want to know why.

I recommend looking up the relevant bond lengths and seeing if the H and the O "match up" at the right distance to form a hydrogen bond, as dictated by the geometry of the molecule. H-bonds tend between 1.6 and 2 angstroms.
Also note that the most effective hydrogen bonds have all three atoms co-linear (X-H~X); the hydrogen bond becomes weaker as the angle deviates from 180°. You could find the theoretical angle your three H-bonding atoms will share (assuming the distance is reasonable for an H-bond to occur in the first place), and consider how much this angle deviates from 180°.
This brief article mentions models that include the H-bond energy's dependence on the H-bond angle, where the H-bond energy is proportional to some power of the cosine of the angle:
$$U \propto \cos^n(\theta)$$
By this model, when the angle becomes a right angle (a "side-on" H-bond), the potential energy rises to zero. An H-bond in this perpendicular orientation is no longer a stabilizing interaction. So you will want your angle to be between 180° ($\pi$) and 135° ($3\pi/4$) or so to have an interaction worth calling a hydrogen bond.

So why is there at least some intramolecular hydrogen bonding in ethylene glycol?  https://www.ncbi.nlm.nih.gov/pubmed/16316257

I have seen plenty of examples of molecules completing a six or five membered ring through H bond. How does it happen then?

According to a quick calculation on the DF-BP86/def2-SVP level of theory, there is an intramolecular hydrogen bond with a strength of about $\ce{-26.6 kJ/mol}$ according to a quantum theory of atoms in molecule (QTAIM) analysis.

For an explanation of the evaluation of the bond strength see: E. Espinosa, E. Molins, C. Lecomte. Chem. Phys. Lett. 1998, 285, 170-173.
Analysis done with MultiWFN 3.3.8. Calculations with Gaussian 09 D.01.

Ortho-methoxyphenol actually can form an intramolecular hydrogen bond, but this can be altered by protic hydrogen acceptor solvents such as DMSO.  See https://www.ncbi.nlm.nih.gov/pubmed/19473035 (abstract available in front of the paywall).


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Plasma cleaner Increasing pressure in vacuum
We have a plasma cleaner, in order for the plasma to be created there has to be around 200-300 mtorr.
As soon as we turn the plasma, the mtorr instantly goes up but the vacuum line is still active.
The question is: What is the plasma in the chamber doing to make the pressure increase?
If this is the wrong section of stacked exchange can someone redirect me to the right section?

Do you observe the same increase in pressure if a classical McLeod gauge is used? (Caution, some labs officially banned any mercury...) Agreed, 200 mtorr is not the typical working range for a Pirani, but what if the charges or/and frequency generated by the plasma cleaner interfere with the electronics of your pressure sensor?  The nature of the plasma (Argon, Nitrogen, Oxygen; different glow) needn't be the principal cause.

You can also tell if the militorr is increasing because the plasma turns from purple to white, also at around 500 mtorr the plasma cuts off.


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Dichromate-thio titration
Can dichromate in acidic media be titrated directly with buffered thiosulfate solution using $\ce{H2O2}$ as an external indicator instead of  using $\ce{KI}$ which is a standard procedure?
$\ce{H2O2}$ is used as an external indicator because dichromate in the presence of excess acid reacts with $\ce{H2O2}$ to yield a transient blue color due to formation of $\ce{CrO5}$.


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How does the steric effect affects the electronic density of the central atom?
For example, if you have Nitrogen attached to some methyl groups, and other attached to some ethyl groups. Why the electron releasing effect on the ethyl will be lower than for the methyl?

Both methyl and ethyl groups have +I effect and its obvious that the ethyl groups have more electrons than the methyl groups, so in this case the present of more electron density on nitrogen atom with ethyl groups lessen the -I effect of the nitrogen atom toward the ethyl groups compared to that on the methyl groups. I think this should be the reason.

No, no, no, no.

Oh! So what's reason?

Read title of question. Also you wrote that having stronger inductive effect gives weaker effect...  Clever making it c.w. though, but it won't help I'm afraid.

Okay. I'll pounder on it ;)

CW shouldn't be used like this. If it's your answer, then it's your answer - CW is for posts where you want the entire community to pitch in.

I'll take note of that next time.

Maybe the question wasn't so clear, but thank you for your response. I was trying to understand why larger alkyl groups will have a lower electron releasing effect, you know? Why the electronic density of Nitrogen will be more negative with the Methyl than with the Ethyl or Propyl...


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Supercritical fluids polarity
Water is a polar solvent, but in the supercritical fluid state, it is much less polar, I presume because of the cleavage of hydrogen bonds. On the other hand, carbon dioxide is non-polar, but as a supercritical fluid it is more polar. Why is this so?

Geez, again this mixing of net dipole and solvent polarity, it doesn't matter much that CO2 has no net dipole.

It has a quadrupole moment, I guess it is related to it's polarity.


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Why is carbon dioxide bent in its first excited state?
R. H. Crabtree writes in The Organometallic Chemistry of the Transition Metals (p 145, 6th ed.) that

$\ce{CO2}$ is linear in the free state but bent [...] in the first excited state...

Why is this so? (The context refers to an electronically excited state.)

If you think about how linear molecules would deform (vibrational modes) as they become excited you can think of the carbon atom and the oxygen atoms shifting up and down w.r.t each other making a bent shape (i.e. a "v" and inverted "v" with the tip of the "v" being the carbon atom). Unfortunately I can't quote you anything, I'm digging back to what I learned in University in one class.

This effect is the so-called Renner–Teller effect and is a consequence of a coupling between vibrational and electronic motion and thus a breakdown of the Born–Oppenheimer approximation. You might be familiar with the Jahn–Teller effect which is related to the Renner–Teller effect. In degenerate states of linear molecules with three or more atoms, the total energy of the molecule can be reduced by bending the molecule. Consider a three atomic linear molecule in a degenerate state (such as $\ce{CO2}$ in its excited $\Pi$ state). When the molecule bents, the degeneracy is lifted as it can bend in the $x$ or $y$ direction. Because of symmetry requirements, the potential has to be an even function of the bending coordinate. Ignoring the lifting of the degeneracy we may approximate the zero-order potential with a Taylor expansion as
$$
V^0=ar^2+br^4+\cdots,
$$
where $r$ is the bending coordinate and $a$ and $b$ are constants describing the potential ignoring the vibronic interaction. When we do consider the vibronic interaction, the degenerate potential splits in two, but the splitting must still be an even function of $r$.
$$
\Delta V=\alpha r^2 + \beta r^4 + \cdots
$$
If the interaction is so large that $a<\alpha/2$, the potential will not have its minimum at $r=0$, but will have two minima at $\pm r_e$ and the molecule has a lower energy when it is bent.
See also Herzberg Vol 3, p 26.

Okay, the J-T effect applies to non-linear molecules while the R-T effect applies to linear.

That is correct :)


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Why "initial" corrosion studies only, rather than longer time-spans?
Some papers say they study the identification and quantification
of species involved in initial atmospheric corrosion. Consider the case of this article in Corrosion - The Journal of Science and Engineering, for example, it provides an exceptionally good model, but I am interested in the copper corrosion process as a whole, not just the beginning. What exactly is understood as "initial corrosion" and what's the relevance of studying such a short time-span? That paper's tests reach no longer than 300 minutes, while other studies measure time in weeks, not minutes.

I don't think this can really be answered in a general way - it depends on the metal, the environment, how much liquid is around, how salty it is, etc.

Here's an attempt at an answer -
Initial corrosion may be taken to mean when the electrochemical reactions occurring are the main driving force for the material to corrode, meaning changing chemical and/or physical properties. As corrosion continues, physical factors (stresses, strains, grain defects, crystal defects, etc.) can contribute and dominate the reason why a material might fail. Studying the initial mechanism can lead to solutions that inhibit or prevent corrosion since it is currently not likely you can salvage a heavily corroded piece of metal and return it to its original service.
As for the duration variance of the studies, it could be that mechanisms change with time or that some systems are slow to corrode but still need characterization to allow for the material in question to be more accurately used. This, however, is more my opinion than actual experience.


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Why does crystalline solids have well defined 3d geometry
Crystalline solids have well defined geometry because the intermolecular forces are uniform throughout the solid unlike amorphous solid where it varies from place to place.
But still the question hold why does some solids have this property like why $\ce{NaCl}$ forms a crystalline solid but not glass or something else.
I thought it is because crystalline solids are ionic but diamond is an counterexample among many.
Question :
Why does some solids have crystalline structure while others don't ?

Keep two concepts separate: the solid crystalline state is not bound to occur on ionic compounds only.
In a simple view, crystalline describes a regular arrangement of matter, with predictable order both in short, as well in long range.  For the crystalline state, it is unimportant if the constituents of the crystal are atoms (like metal atoms), charged particles (like your example of NaCl), or neutral molecules (like benzoic acid) or even a mixture of neutral and charged molecules (many protein crystals contain water).
Some materials are easier packed in such a regular pattern, than others.  In the instance of metals, their (on first approximation) sphere like atomic geometry and similar radius eases such a settlement.  Inorganic salts tend to be a bit more complicated, as often smaller ions are localized in gaps built by larger ions (octahedral / tetrahedral gaps).
Organic molecules may equally build crystals, yet their shape and conformational flexibility around single bonds (for example in alkyl chains) and multiple intermolecular interactions may keep them over a long time in a glassy state (like supercooled benzoic acid, or some brands of honey).  More than for the aforementioned examples: formation of a nice crystal may be thermodynamically favourable, but it is not enough "just to cool down" below the melting temperature to reach this energetic minimum.  There must be seeds of crystallization, and typical conditions of their formation (temperature, concentration, etc.) are not necessarily the same where nicely shaped crystals growth further.
It is not surprising crystals attract attention, for example in material science and crystallography, and that there are journals with focus on the relationship between molecular properties, and macroscopic / crystallographic ones; like CrystEngComm or Crystal Growth & Design.

"some brands of honey", is honey a solid ?

@A---B While "honey" is a mixture of several compounds, it is save to assume it largely consists of sugars, with fructose and glucose as the most important contribution among them.  These, if pure, are solids.  The ease honey crystallizes out correlates with the ratio of fructose : glucose, because yes, if not solid, honey is a supercooled liquid (cf. https://en.wikipedia.org/wiki/Honey#Phase_transitions, for example).  It is one of the examples, outside the chem lab, that illustrate how thermodynamics is about energies, and kinetics about the rates reactions and phase transitions occur.


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Stereochemistry of Gabriel Synthesis
I read that the primary amine product of Gabriel Synthesis is a racemic mixture. However, when analyzing the reaction the stereochemical step(s) are all $S_N2$. Thus, I figured if you are careful you can track the stereochemistry of the (or both) $S_N2$ reaction(s). Can anyone clarify this for me?

As far as I remember, the Gabriel reaction transforms  primary haloalkanes into primary amines.  How do you attribute absolute configuration on a primary haloalkane of type $\ce{R-CH2-Br}$ to the Carbon atom just adjacent to the one of Bromine?

@Buttonwood It works for secondary haloalkanes as well. For example, people use it to make amino acids which are secondary amines (with the exception of glycine).

@Cocmos So far, my association with the synthesis of amino acids was the Strecker synthesis.  Revisiting the Gabriel synthesis again, I found indeed an entry relevant to your welcomed comment, used in an answer below.  The preparation of glycine with phtalimide, as outline in the Polish wikipedia about the same topic, indeed differs in its procedure.

The lower part of the Dutch wikipedia about the Gabrielsynthese indeed displays an example where this reaction is carried out to yield a secondary amine:

The original publication (DOI: 10.1021/jm9804376, J. Med. Chem., 1999, 42, 593-600) equally includes similar conversions with racemic secondary substrates like (R/S)-($\pm$)-1-methylpentyl bromide to yield (R/S)-($\pm$)-N-(1-methylpentyl)phthalimide which in turn was then converted into (R/S)-($\pm$)-(1-methylpentyl)amine hydrochloride.


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What is the temperature of evaporating water (not boiling)?
The vapor pressure of water at $\pu{70 ^\circ C}$ is $\pu{233.7torr}$.
If I were to put a temperature probe directly above the liquid water, what would it read?
In other words, does a water molecule need to have $\frac32kT$ (where $T=\pu{373 K}=\pu{100 ^\circ C}$) to escape the liquid?

While it's a liquid water can escape what contains it due to the difference of chemical potential and it will escape until the thermodynamic equilibrum has not been reached. This is due to phenomenas of diffusion. You'll find some explaination there

No, the average thermal energy doesn't need to be the one corresponding to 100C.
Solo water molecules can exist in vacuum or in a gas phase with whatever kinetic energy. Liquid water condensation starts when the concentration of the water molecules is enough high (there is a tipping point), i.e. when the partial pressure of water reaches the vapor pressure for the given temperature.
From the classical thermodynamic point of view (similar argument to Hexacoordinate-C's comment), there is a chemical potential of water in the gas phase which is dependent on the partial pressure of water, and there is a chemical potential of condensed water for a given temperature. When the partial pressure is low, the gas phase chemical potential is lower than it would in the condensed phase -> it is stable in gas phase. When the partial pressure reaches the vapor pressure, the chemical potential is equal to the gas phase and condensed phase.

I'm sorry, this answer doesn't make much sense to me. Is there a resource that discusses the relationship between chemical potential and partial pressure that you recommend?

Any thermodynamics textbook which discusses chemical topics (concentration, pressure etc dependence of chemical potential, generally around concepts like fugacity)


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Entropy change of an ideal gas undergoing adiabatic expansion into vacuum
For the adiabatic expansion of an ideal gas into an empty chamber, I am told that the entropy change of the system is not zero, and that the entropy change of the surroundings is zero.
But if the expansion is adiabatic, then it should be isentropic, so should not the entropy change of the system be zero? And how can you infer anything about the change in the surroundings?

What makes you think that the entropy change of the system should be zero?  Do the surroundings know that anything even happened inside the container?


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Mechanism for the synthesis of 2-pyridone from pyridine N-oxide
I'm specifically interested in the pathway from pyridine N-oxide, where there is some kind of reaction and rearrangement with acetic anhydride.
I'm assuming there is first a substitution (by the negative oxygen) at the carbonyl. The -OAc group can then attack at the alpha-position on the ring.
We could continue and say that the loss of H and the -OAc group on nitrogen happens next to restore aromaticity, but then we're stuck with that OAc group instead of the carbonyl.
From what I could find, it doesn't seem that there's a radical mechanism here, but if there is some intramolecular rearrangement, I cannot seem to find it.

Ester cleavage would give the hydroxypyridine which would tautomerise to the pyridone. Something like this: https://i.stack.imgur.com/1ncpx.jpg If there's a funky rearrangement, I can't see it - suspect it's more of a question of what nucleophile is cleaving the ester. [Edit: nitrogen in the pyridone is missing a hydrogen. Classic error...]

@orthocresol This makes a lot of sense; I was drawing parallels here to what happens in a similar reaction with an alkyl group in the alpha-position, where you do get a sigmatropic rearrangement, but it did make me blind to the last step...

Pyridine N-oxide itself is generally fairly resistant to nucleophilic attack as the ring is already 'electron rich' and doesn't want to accept additional electron pairs.
When treated with acetic anhydride, however, the carbons at the 2 position become electrophilic (the acetate is electron withdrawing making the C=N bond more like an iminium), hence it is possible to have acetate attack, with subsequent elimination to re-aromatise the ring, leaving us with a 2-substituted pyridine.
If the acetate is hydrolysed (water, or mild aqueous base), the 2-hydroxy pyridine is formed, which may tautomerise to the more stable 2-pyridnone.

It's worth noting that this mechanism differs significantly from the Boekelheide reaction (the initial pyridine has an additional methyl group at the 2-position), whereby the acetate bound to nitrogen is migrated by means of a [3,3] sigmatropic rearrangement. This rearrangement isn't possible/favourable in this case, as the [3,3] requires a 6 membered transition state, whereas without the methyl group only a 5 membered TS exists.


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Is it true that heavy water is not blue?
I believe I saw this claim somewhere on the internet a long time ago. Specifically, it was claimed that the difference could be observed by filling one long, straight tube with light water and one with heavy water, and looking through both tubes lengthwise (so that light has to travel through the tubes' lengths before reaching the eye), whereupon the light water would appear blue as it does in the oceans, and the heavy water would not. The explanation given was that heavy water has a different vibrational spectrum because of the greater mass of the $^2$H atom, which seemed perfectly plausible.
However, I am no longer able to find a source for this claim, which is strange because if it were true, surely it would not be so difficult to find a source?

I didn't even know that normal water is blue. But it seems to be so.

@Mindwin Very interesting article.  To me, though, this begs the question "How many meters of a given substance must light penetrate through before we can determine its 'true colour'?"

@Steve-O - the absorption length of water depends a lot on wavelength. At the minimum (418 nm) it takes 227 m to reduce the intensity of the light to 1/e. See this answer on the physics site. Funny that questions about the light absorption of water appear almost simultaneously on both sites. Coincidence?

A five-hundred-gallon polyethylene tank of pure [RO] water is a distinct strikingly blue color under room lighting

Based on your description, I may have found the article you originally saw, or at least one very similar.
Researchers from Dartmouth College published a paper$\mathrm{^1}$ in which they report, among other things, the results of viewing sunlit white paper through two 3 meter lengths of plexiglass; one filled with $\ce{H2O}$ and one with $\ce{D2O}$. Sure enough, because of the lower frequency of the maximum absorption of $\ce{D2O}$ in the red to near IR wavelengths, the blue color that is characteristic of $\ce{H2O}$ is far less pronounced in $\ce{D2O}$. This website is based on the published paper and additionally shows a photograph of the blue colored $\ce{H2O}$ on the left with the far less colored $\ce{D2O}$ on the right:

"Why is Water Blue", Charles L. Braun and Sergei N. Smirnov, J. Chem. Edu., 1993, 70(8), 612

Comments are not for extended discussion; this conversation has been moved to chat. In any case, the comments have been rendered obsolete by David Richerby's edit (thanks, David). I simply left them in a chat room in case anybody still wanted to debate it.

Just a note: this website (the one linked in Tyberius' answer) contains a version of the same picture shown here, but the description there suggests that the right tube is actually empty.

@GOTO0 , that is definitely the same picture, thought the fact no source is given is disturbing. The site I linked to is a reproduction of their article published in the Journal of Chemical Education. They fully cite the article, though it only links to an abstract and I don't have free access to the whole thing. The bottom line though is that assuming the Dartmouth College group is not flat out plagiarizing something, their article and descriptions of their work (including the photographed water set up) must be the correct one. They probably wouldn't be happy about the use in the other link.

@GOTO0, I have notified the original authors of the link you cited. I will post any response they have regarding the photograph. Thanks again for bringing this to my attention.

@airhuff I completely overlooked that the image I would be looking for was in my link. However the description in both the reproduction and the original paper(I have access through school) suggests that the right tube is empty. "The large tube volume and a limited budget precluded checking to see if light transmitted through a D2O filled tube was indeed white, as expected.". Unless the image was obtained after the original article (there are no pictures in the original article) it seems as though it is prohibitively expensive to attempt.

Another quote "D2O is colorless, at least for path lengths on the order of meters, because a minimum of 6 stretching quanta would be required for any transition in the visible region.". So a lot of D2O would be required to actually get an image like the tube on the right. The reproduction you posted seems to be word for word the original article, plus the few pictures thrown in.

@Tyberius , thx so much for your efforts looking into this. I did think that picture looked pretty low quality for a publication. A good high res picture would have made all this pretty obvious. The text sure seems clear about it though. I do hope to hear back from the primary author from Dartmouth, but I'm not holding my breath.

Your "This website" link is now dead.

I'm a little surprised at the claims that the experiment is prohibitively expensive to perform properly. Nearly pure D2O is used routinely as a solvent for NMR experiments, so it is obtainable is reasonable quantities, and since this experiment does not require dissolving anything in the D2O or otherwise "using it up", it could be recovered and repurposed as a solvent.

This does seem to be the case. I don't have images of the different types of water, but I did find this overlaid IR-visible spectrum of water and heavy water:

As you stated, the presence of deuterium shifts the absorbance spectrum of heavy water further into the IR region, rendering it colorless.
The website I found this on (http://www.webexhibits.org/causesofcolor/5B.html) goes into great detail on why water, and various other compounds, exhibit particular colors.

The spectra show very clearly why $\ce{D2O}$ is preferred over $\ce{H2O}$ in certain NIR experiments, it is simply more transparent in the respective spectral region, or in other words, has a different color.

If you saw it during the 1990s online era, it might well have been reference links on my old website, from this article on physics misconceptions in grade school.
I'd been doing some textbook-consulting back then, and this page was my effort to stir up trouble/perform viral education/convert innocent children into critical thinkers.  I was camping out on appropriate Newsgroups as well as many 1990s physics forums, and encountering controversy from those who refused to accept that water could have a blue color (since after all, no grade-school science text ever mentioned this.)  And yes, the non-blue of D2O was a continuing topic of the time.
Besides JCE journal and the Dartmouth paper, another reference from the article is Water Absorption Spectrum, from LSB Univ., which takes an in-depth look at H2O vibrational absorption, and specifically points out the flat spectrum of D2O, and its coloration entirely due to Rayleigh scattering.

Unforunately nobody maintains heavy water in kiloliter, white-interior, open-top tanks.  :)  If you're going to set one up yourself, I recommend a white-painted stepped-depth setup, to display the color effect as with this example with H2O, or another.

Aren't the walls of the pool blue in your "example with H2O", invalidating the proof?

@Ruslan in what way?  If the water was removed, would the top of the lower step still be blue?  Let me find white marble or white tile swimming pools https://i.pinimg.com/564x/f3/00/cc/f300ccbf3870a4c07833693fb95cfe75.jpg or https://media.architecturaldigest.com/photos/5cdf2d7f711f00f6203fc2f1/master/w_1600,c_limit/Claesson-Koivisto-Rune-Pool.jpg

Yeah, these are much more convincing examples. I suppose you may be right about the top step, but still it's better to clear any possible doubts and use something without blue background (as you do now).

The problem with the answer given in both of the other two responses (so far) is that they claim that water has the unusual property of having transmitted red light filtered/absorbed so that the transmitted light is blue. I have zero problem with that. The authors of the J.Chem Ed. piece go on to talk about the looking at reflected light from a Colorado lake or the Caribbean. There, you have to be careful, reflected light is clearly not the same as transmitted light and the inclusion of those scenes is a distraction. The two tubes are clearly indicated to be left = water, right = air (due to the cost of D2O being above their budget). I also have zero problem with the assertion that the visible absorption spectra of heavy water is, essentially, flat (although I'd prefer a higher resolution spectra as well as a baseline which is level). (This statement ignores non-linear optical effects, which is quite reasonable.) However there's a clear difference between saying there's little absorbance and that the heavy water is colorless. Scattering will occur. Even in ultra-pure D2O, scattering will occur. see Rayleigh Scattering. It is inherent in the electrical nature of matter. So, blue wavelengths will be more scattered than the reds, regardless of other specific interactions. Heavy Water is colorless and the sky is not blue. If you're ok with that statement, then ok, we're done but perhaps you should take a look outside on a clear day, if you are lucky enough to live somewhere where we haven't fouled the air with pollution. Now, it takes miles of sky, I'm not sure how much liquid water it would take to have a noticeable scattering effect. Anyway, a reductive approach to the question of color should, sooner or later synthesize all of the various causes of color back into the (gesalt) whole. If X is a function of A,B,& C, then showing that one material's X is dominated by A while another material's X isn't doesn't say squat about what X is for the second material. You need to include all contributions. For what it's worth, being completely colorless would be a balancing act amongst all the factors (A, B, C, ..) and would be quite unusual. We usually just consider the color to be relevant to some expected optical path length, so color is typically context dependent. Reflected, transmitted, short path length, long, white light, colored light, etc., etc.

agreed the "color of water" is just like the "color of the sky" mainly due to Rayleigh Scattering, no reason to think that D2O would have markedly different properties on the scattering.

I actually agree that this is a good answer, though I think you could format it better and try to limit your answer to one cohesive point to make it better.

"'color of water' is just like the 'color of the sky' mainly due to Rayleigh Scattering" @DanS not true.  However, that was the conventional explanation from pre- ~1930s, never mind that the deep underwater environment isn't colored sunset-red!  The situation is non-ideal, since thick water is a blue-colored filter and often a white sand bottom is present.  Also we usually don't encounter utterly silt-free water. E.g. famous blue Crater Lake contains large-particle "glacier milk," and would appear milky white if not for the water's inherent blue color.

@alphonse  Separate the examples? Caribbean beach, white sand & 0-10M depths: we see brilliant turquoise/blues against the submerged white background. Rayleigh colors are overwhelmed by water's blue-filter effect. Next, deep ocean, which appears black if no scattering. But Rayleigh scattering returns light from +10M depths,  which light is deep blue from water's absorption spectrum.  Next, deep D2O oceans appear sky-blue from Rayleigh scattering, and divers encounter a sunset-red-lit environment.  D2O Caribbean beach scenes might appear aerogel-gray: shallow D2O over a white sand reflector.

Pure water in large quantities is a definite striking blue color. The first time I saw a large PE tank of reverse osmosis water I was startled by the blue color. It is blue because of the second overtone of the O-H stretch at about 3400 wavenumbers absorbs near 700nm removing some red light. The O-D stretch is about 2400 wave numbers so its overtones will still be in the near infrared and does not affect the visible spectrum.


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Cyanamide mesomeric contributors
Which form of the cyanamide ion is more stable?
As per the resonance rules, I find both of them equally stable as number of bonds is the same, more electronegative atom has negative charge and so on. Is my conclusion right?

Can you rotate the picture anticlockwise?

The original picture is oriented properly. It got twisted when I uploaded it here, and I've got no idea of how to make it straight


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What is the significance of Sphingolipids in human?
I want to know the significance of sphingolipids in human. I have learnt that sphingomyelin is the most significant type of sphingolipid in human. Also that the sphingomyelin serves as a structural constituent of myelin sheath of nerve fibres.
Is that all? I need to clarify this. Please help me. Thank you

cross-posted http://biology.stackexchange.com/questions/57945/what-is-the-significance-of-sphingolipids-in-human

Don't cross-post between the sites, not getting answer quickly is not justification.

There are multiple known functions of sphingolipids and their metabolites in humans, although the roll played in the structure of myelin might be the most important and well understood.
However, according to this Wikipedia article, sphingolipids also play a roll in cell recognition and signalling, among other specialized functions. Additionally, some of it's metabolites have been shown to perform numerous functions. From the above article:

Recently, simple sphingolipid metabolites, such as ceramide and sphingosine-1-phosphate, have been shown to be important mediators in the signaling cascades involved in apoptosis, proliferation, stress responses, necrosis, inflammation, autophagy, senescence, and differentiation.

The bottom line answer to your question is that there are many other functions of sphingolipids in humans, particularly if you include it's metabolites.  The article cited above is well referenced and should serve as a good starting point to investigate (or just enumerate) the other functions of sphingolipids in humans.


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Why does aluminum foil "electrocute" dental fillings?
This is probably one of the weirdest questions on here. When I was a kid, I put aluminum foil in my mouth (maybe it was still attached to some food or something, I don't remember.)
Anyway, it made my mouth feel like I'd licked a 9-volt battery. Being the natural born scientist, I repeated the procedure with similar results.
Why does this happen? I really think it has something to do with my dental fillings since the sensation occurred near them.

It's basically setting up a galvanic cell that involves a nerve root in your tooth. The term for it is actually "galvanic shock"!

Ouch...  actually a known phenomenon.  Between the foil of aluminum/Al, and the metal filling of amalgam in your tooth, you created a little electrochemical cell.  Your saliva served as an electrolyte and your nerves around sensibly detected the current. This already "works" with little splinters of Al foil wrapping chocolate (obviously, it was not a small Hershey bar), yet is less powerful than the said 9 V block.
(one reference)

Slightly off topic, but possibly based on similar electrochemical interactions in the mouth, some people have even reportedly picked up radio signals with their teeth & dental fillings! http://skeptics.stackexchange.com/questions/3638/is-it-possible-to-pick-up-radio-signals-from-dental-fillings

@theo I'm unaware if the gate-shaped metal detectors at the airport may be attuned to catch passengers with metal fillings in their teeth, yet this comment prompts me to think if other objects then the metallic filling of a tooth may be susceptible to electrochemical interaction. What about braces and dental bridges?  What about voluntary piercing (across tongue, lip, cheek, etc), equally made of metal? Simply I do not remember if people invested in the later remove this decoration / attire, potentially acting as an electrode, prior to eating, drinking, kissing because of electrochemistry.


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Gas spectra in the 400 nm - 1000 nm range
Disclaimer: My background is computer science.
Is it possible to detect any gas using a visible light and near infrared (400 nm - 1000 nm) hyper spectral imaging camera? This can be related to gas leaks, pollution, gasses released in fires... From what I found when googling, this is not really possible, but I though to check here to be sure.

There are gases that absorb in the visible region. Some are nasty and not things you'd run into every day like bromine. The only thing that comes to mind that could be rigged up with a camera like that would be an NO2 detector. NO2 is red to brown colored and is major industrial pollutant. Accurately quantifying NO2 with a single modded camera doesn't seem likely or useful though.

The answer is that you can detect all types of gases if you have the correct light source and use Raman spectroscopy. Raman is more adaptable than absorption or emission spectroscopy as it is a scattering process and the laser does not have to operate at the wavelengths that the molecules absorb, thus one laser can be used for all samples. A suitable source is going to have to be a narrow band laser (although when Raman discovered this effect he used sunlight). You will also need notch -filters to remove the laser wavelength and prevent it from saturating your detector. Not cheap.

This is a broad question and the answer depends on the type of gases and the type of spectroscopic technique.

Absorption :The spectral region described by you refers to the visible region and some part of the infrared region. Hence, all colored gases can be observed. Some gases which absorb light in the lower infrared region 700 nm to 1000 nm region can also be detected.
The principle of absorption is completely different in different regions of the electromagnetic spectrum (400-700 nm - electronic absorption, 700-1000 nm = vibrational absorption) and so the intensities will also be different. In this case  a broad band light source is used as a source of incident light and the absence of certain wavelengths due to absorption is signal of some gases.  (Colorless  gases will not be detected in the visible region, though possible in the infrared region).
Emission : Other spectroscopic techniques might be possible involving some excitation (using light in far UV - UV region), and then emission in the 400-1000 nm region and this emission is detected by the detector. Colorless gases can be easily detected by emission approach but requires expensive (and dangerous) UV light source(s).

Specifically which gases is being detected (by any approach) needs careful investigation by a chemist and probable comparison with already existing literature on absorption and emission of possible gases.


text:
How to synthesize 1,3-pentadiene from butanal?
How do you synthesize 1,3-pentadiene from butanal?

I have the (possible) last steps of bromination of 1-pentene at the allylic position and then an E1 reaction to give the final product. I don't understand how to get from butanal to 1-pentene though. Also if you know another way to do the synthesis then feel free to share.

I believe one way would be to start with butanal participating in a Wittig reaction to form 1-pentene.
$\ce{CH3CH2CH2CHO + H2C=PPh3 -> CH3CH2CH2CH=CH2}$
Then allylic bromination, as you said.
$\ce{CH3CH2CH2CH=CH2 ->[{NBS}][{hv}] CH3CH2CHBrCH=CH2}$
Dehydrohalogenation with a hindered base like t-butoxide (to favor the elimination reaction) should give the diene. The only thing I'm not sure about is whether it would be E1 or E2. It isn't very hindered, but it is a secondary halide. With those factors, plus the alkoxide, I think it will be an E2 mechanism.
Let me know if I made a mistake somewhere.

How about :
Step 1:
Add $CH_3MgX$ to butanal to get pentan-2-ol.
Step 2:
Dehydrate the alcohol to pent-2-ene
Step 3:
Add $Br_2$ to pent-2-ene to get 2,3-dibromopentane
Do a double elimination to get 1,3-pentadiene


text:
Why is it wrong to calculate the amount of oxygen directly from the products of a combustion reaction?
I've been stumped by a question in my textbook concerning combustion analysis.

A $\pu{0.5438g}$ sample of a liquid consisting of only $\ce{C}$, $\ce{H}$, and $\ce{O}$ was burned in pure oxygen, and $\pu{1.039g}$ of $\ce{CO2}$ and $\ce{0.6369g}$ of $\ce{H2O}$ were obtained. What is the empirical formula of the compound?

Firstly, I calculated the amount of substance of carbon and hydrogen by dividing the masses of each product by their respective molar masses.
\begin{align}
(1.0390/44.01) = n(\ce{C}) &= 0.02360\\
2\cdot(0.6369/18.02) = n(\ce{H}) &= 0.0706\\
\end{align}
My textbook then states that I must find the sum of the masses of carbon and hydrogen, and then deduct this sum from the mass of the liquid compound to get the mass of oxygen:
\begin{align}
0.02360 \cdot 12.01 &= 0.283535\\
0.0706 \cdot 1.01 &= 0.28637\\
0.5438 - (0.283535 + 0.28637) &= m(\ce{O}) \\
m(\ce{O}) &= \pu{0.1888g}\\
\end{align}
I have no qualms with this approach, but why can't I find the amount of oxygen directly from the products?
\begin{align}
\text{I.e.}&& 2\cdot(1.0390/44.01) &= n(\ce{O}) &\text{in }\ce{CO2}\\
\text{and} && (0.6369/18.02) &= n(\ce{O}) &\text{in }\ce{H_2O}.\\
\end{align}
$$\text{Amount of oxygen in liquid compound} = n(\ce{O})\text{ in }\ce{CO2} + n(\ce{O})\text{ in }\ce{H2O}.$$
I know this doesn't work, but my question is why? It seems perfectly intuitive to me, and yet it doesn't work.

Let's first write the chemical equation for the reaction (I will leave it unbalanced because the stoichiometry is irrelevant):
$$\ce{C_xH_yO_z + O2 -> CO2 + H2O}$$
To do this question, we need to find the mole ratio of carbon, hydrogen and oxygen in $\ce{C_xH_yO_z}$, then we can find its empirical formula.
From the above equation, it is clear that all the carbon in $\ce{CO2}$ originated from the compound and that all the hydrogen $\ce{H2O}$ originated from the compound. However, all the oxygen in $\ce{CO2}$ and $\ce{H2O}$ didn't just come from the compound, it came from the compound and gaseous oxygen. Therefore, from your method, you are just finding the total amount of oxygen in the reactants, not the number of moles of oxygen in the compound.

While it is certainly unnecessary for the argument, I believe a balanced reaction equation would make it a lot stronger.

Both (at the time of writing) other answers are absolutely correct, but I'd like to approach it from a slightly different angle. I think it is one of the most important traits of a chemist to analyse the initial situation correctly. Read the question very carefully again. Below (in the spoiler) I highlighted the part which is an important information that you might have overlooked.

 A $\pu{0.5438g}$ sample of a liquid consisting of only $\ce{C}$, $\ce{H}$, and $\ce{O}$ was burned in pure oxygen, and $\pu{1.039g}$ of $\ce{CO2}$ and $\ce{0.6369g}$ of $\ce{H2O}$ were obtained. What is the empirical formula of the compound?
 In most of these texts you might alternatively read  [...] was burnt in excess oxygen, [...]

I will always suggest writing the balanced equation first, as usually the chemistry becomes more obvious converting a text into a mathematical construct:
$$\ce{C_xH_yO_z + 1/2(y/2 + 2$x$ - $z$) O2 -> x CO2 + y/2 H2O}$$
from there you can see that as long as $z \neq \frac{y}2 + 2x$ you will need more oxygen for a complete combustion. It will later also make it easier to find the actual empirical formula.
One final advice: Always write down the units when calculating values. It'll help you find mistakes easier. For example, it should have rather been:
\begin{align}
\frac{\pu{1.039g}}{\pu{44.01g/mol}} = n(\ce{C}) &= \pu{0.0236mol}\\
2\cdot\frac{\pu{0.6369g}}{\pu{18.02g/mol}} = n(\ce{H}) &= \pu{0.0706mol}\\
\end{align}

Oxygen is found in both reactants, and the amount of O2 used for combustion (pure oxygen) is unknown and usually just given as 'excess' in order to have a complete combustion, i.e. the only products are H2O and CO2.
You are interested only in the oxygen that was originally part of the molecule, but you cannot tell which oxygen atom in each molecule of CO2 or H2O comes from the molecule being tested, and which comes from the pure oxygen you used for combustion. Therefore, you have to calculate it indirectly.
(You COULD have determined the molecule's oxygen content directly if the oxygen atoms were marked in some way either in the molecule or the pure oxygen gas. As a theoretic example, you could use a certain isotope of oxygen which can be detected easily in some way.)
Reference:
General Chemistry: Principles and Modern Applications; Petrucci, Herring, Madura, Bissonnette; Pearson, 2011; ISBN13: 9780136121497 - pp. 82-83 (for owners of older editions, it should be in chapter 3, combustion analysis section)


text:
Calculating stoichiometry of a reaction between xenon and fluorine

Xenon and fluorine were mixed in a 1:1 ratio by volume and sealed in a vessel attached to a barometer. After some time, crystals were found to have formed on the walls of the vessel and the barometer showed that the pressure had dropped to 70% of the original pressure. Analysis of the gases remaining indicated 4 parts xenon to 3 parts fluorine. Construct a balanced equation for the reaction. Assume constant temperature throughout.

Equal volumes of gas at same temperature and pressure have the same number of molecules, thus I deduce that the mole ratio of the 2 gases initially is 1:1. Since $pV = nRT$ then the number of moles of the two gases combined would be 70% of the total number of moles of the 2 gases at the start of the experiment.
I am unsure of how to proceed from here.

Xenon and fluorine were mixed in (1) a 1:1 ratio by volume and sealed in a vessel attached to a barometer. After some time, crystals were found to have formed on the walls of the vessel and the barometer showed that (2) the pressure had dropped to 70% of the original pressure. Analysis of the gases remaining indicated (3) 4 parts xenon to 3 parts fluorine. Construct a balanced equation for the reaction. Assume constant temperature throughout.

You have properly analyzed (1) and (2). It remains to analyze (3). We started with ten parts of gas, 5 parts xenon and 5 parts fluorine. From (2) and (3), we see that 4 parts xenon and 3 parts fluorine are left in the gas phase. Hence 1 part xenon ($\ce {Xe}$) and 2 parts fluorine ($\ce {F2}$) must have reacted to form crystalline $\ce{XeF4}$.
The balanced reaction is left as a simple exercise.

The choice of starting with ten parts of gas (as opposed to fifty parts, or a hundred parts) is arbitrary. Since we're given only ratios, we can choose an arbitrary number of parts of gas and work with that throughout. Choosing ten makes the math simple.


text:
why oxidation states of suphur atoms of Disuphur-monoxide is +2 and 0?

I) According to my logic, sulfur-sulfur linkage (linkage between same atoms) result in no charge separation.So, the oxidation number of the sulfur atom linked to only sulfur should be 0(zero) and other sulfur being +4 as it is connected to 3(more electronegative) oxygen atoms.
Thus, the oxidation number of sulfur atoms should be +4,0.
II)However, Using X-ray absorption to probe sulfur oxidation states in complex molecules claims that oxidation states are +6 and -2.
III)Again,Determination of Changes in Sulfur Oxidation States
in Prostate Cancer Cells proved that that oxidation states are +5,-1.
These three sets of results left thoroughly confused.Which one of them is correct and why?
Following the same sequence of logic:
My counter-question for the claims (I) and (II) is
"Why the oxidation number of sulfur atoms of Disulfur monoxide is [ +2 and 0 ] instead of [+4 and -2] or [+3 and -1] ?"

Related: Oxidation states of the sulfur atoms in the thiosulfate ion

orthocresol .Sir, I went to the link you have sent me but it doesn't answer my question and   specifically my counter question . What is the absolute answer of the oxidation states of sulphur?

Is it +6 and -2 or +5 and -1?

There is no absolute answer. Oxidation states are human invention and not facts of nature.

Couldn't you just treat terminal S as non-innocent ligand and give us a break?

Ivan Neretin. Sir , I know chemistry is a subject based on more or less hypothetical things such as electrons which nobody has seen but has great deal to be talked about.      Can  you please help me out from the dilemma of oxidation states?

Milthoron. Sir can you please explain in detail.


text:
Why is ZnFe2O4 a normal spinel?
Considering that $\ce{Zn^2+}$ is $\mathrm{d}^{10}$, $\ce{Fe^3+}$ is $\mathrm{d^5}$, and $\ce{O^2-}$ is a weak field ligand, the CFSE values of both the normal and inverse spinel structures should be $0$.
Why is it then that the normal spinel, with $\ce{Zn^2+}$ in tetrahedral holes and $\ce{Fe^3+}$ in octahedral holes, is formed?

The reason stated in the literature is fairly straightforward. However, actually tracking down the original source is harder than it looks.
Going back to Phys. Rev. B22, 115 (1980) one finds in the opening paragraph the statement

Mixed Zn ferrites of the type Zn$_{x}$M$_{3-x}$O$_{4}$ where M
  is a magnetic ion have been the subject of numerous
  investigations in the past, especially with regard to
  their magnetic properties.$^{1}$ The metallic ions occupy
  the tetrahedral A or octahedral B sites in the spinel
  crystal structure. The Zn ions preferentially occupy
  the A sites because of their tendency to form covalent
  bonds involving sp' orbitals.

Reference number 1, on the first sentence, is a book [J. Smit and H.P.J. Wihn, Ferrites (Wiley, New York, 1959)]. The statement on $sp3$ bonding is not backed up by a reference. I can only assume that it was considered common knowledge in the spinel community in 1980.


text:
Does Ni(H2O)6 have Oh symmetry or Th?
I optimized a geometry of $\ce{Ni(H2O)6}$ using quantum chemistry and got a molecule which is symmetrical for the $\ce{Ni-O}$ distances as shown
here:

In the literature it says that this complex has an $\ce{O_h}$ point group, but this complex looks like it is $\ce{T_h}$. Do I need to revise my group theory or is what I am looking at actually a $\ce{T_h}$ point group?

Welcome to Chemistry StackExchange. It look like $T_h$  as drawn (ignoring bonds  to H atoms, and considering all atoms are fixed), but without any H it is $O_h$. Additionally the OH group is going to rotate more or less freely so its average will be the same as if H atoms are ignored.

In addition to the Hydrogen atoms swirling around, wasn't the tetrahedral symmetry $T_h$ a subgroup of the octahedral $O_h$, too?


text:
Why does magnesium prefer to insert into C–Br bonds over C–Cl bonds?
In the formation of a Grignard reagent from a compound possessing both bromine and chlorine, magnesium preferentially inserts into the C–Br bond over the C–Cl bond. For example, the formation of a Grignard reagent from 1-bromo-2-chloroethane proceeds with good selectivity:[1]

How can this be explained? Can it be rationalised using a reaction mechanism?

Simon, P. Y.-R.; Oreal, H.; Audran, G.; Schulz, M.; Joly, J.-P.; Siri, D.; Siri, A. Proteasome Inhibiting β-Lactam Prodrugs Useful for the Treatment of Cancer and Neurodegenerative Disorders. World Patent WO 2018/115497 A1, June 28, 2018.

Simply put the C-Br bond is weaker than C-Cl, so oxidative addition occurs much faster. Oxidative addition is treated in more depth in any textbook on organometallic chemistry (the usual context is transition metals but it applies equally well to Mg).

As noted earlier, the determining factor is the greater strength of the carbon-chlorine bond versus carbon-bromine, which leads to a higher activation energy for the magnesium to react with the former.
The strength of a carbon-halogen bond, and therefore the activation energy required for magnesium to form a Grignard reagent, is greatest with fluorine and decreases progressively with heavier halogens. The reactivity sequence between carbon-halogen bonds and magnesium may therefore be extended:
$\ce{C-F < C-Cl < C-Br < C-I}$
We might amend the first Inequality to read $<<$ rather than just $<$, as carbon-fluorine bonds are completely unreactive with normal procedures. Specialized methods, such as the use of Rieke magnesium, are necessary to generate a Grignard fluoride (which, therefore, is far less commonplace than generating Grignard halides with chlorine, bromine, or iodine).
The hydrocarbon group, too, can impact reactivity. Chlorine, bromine and iodine can all couple their nonbonding pairs with an aromatic ring, strengthening the bond of the halogen to the ring. Thus aryl halides react less rapidly than alkyl halides with a given halogen. Encyclopedia Britannica, combining the substrate and halogen effects reports:

Organohalogens vary greatly in their rates of reaction with magnesium. For example, alkyl iodides generally react very rapidly, whereas most aryl chlorides react very slowly, if at all.

This may be used to advantage, for instance if an aryl compound with both chlorine and bromine is exposed to magnesium, the chloroarylmagnesium bromide is formed with good selectivity.

There are a few reasons:

C-Br bond is weaker than and thus more easily cleaved than C-Cl bond.

Mg-Br is more stable than Mg-Cl. The electron cloud of Br is more polarisable than that of Cl, due to Br having not just a larger number of electrons but also larger atomic radius, over which its electrons are distributed. The Mg-Br covalent bond formed will thus have a greater partial ionic character than Mg-Cl bond, which confers additional stability by electrostatic attraction.

I'd stick with (1).

Unfortunately I can only upvote @OscarLanzi once

Mg-Br covalent bond? Highly suspicious

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


text:
Should I have a pair of shoes that I wear only in the lab?
I bought some very nice heavy leather boots when I was working in a solid state synthesis lab, mainly with selenium and sulfur compounds, a few years back. I probably did spill small quantities of some very basic/acidic solutions with some selenium/selenate content on them a few times.
I don't work in a synthetic lab anymore, but I wore them again this weekend. It feels strange though, perhaps just from ingrained lab safety habits, to touch my face after putting them on etc. My gut feeling is that a) rain has washed away anything which was on them b) most chemists I knew as a research student wore their normal day shoes in the lab, so it should be fine. But what are the standard safety rules on this? Is it generally okay to wear the same shoes in the lab and in daily life?

Frankly I'd bee more worried about biological contamination from handling my belt. You can't really wash your hands until after you pull up your pants...

Mostly it won't matter unless you are handling very nasty or smelly compounds (and are a bit messy in your lab technique).

Standard safety rules (from a pharma and semiconductor processing POV):
To follow industrial hygiene practice absolutely to the letter, you should have dedicated outer layer garments that remain in a quarantined area that would not pose a gross cross contamination risk to other things that cannot be controlled for exposure risk. This means that you should not have been able to take those boots home with you after you wore them in the lab even one time. I do not think any company enforces such a rule, in fact, it is the employer's responsibility to provide PPE to workers and the worker's responsibility to use and maintain PPE (if applicable.) If the risk of cross contamination was assessed and determined to be moderate or higher, you should have been provided shoe covers at your solid state systems lab or dedicated work shoes that had to be kept in a quarantined area.
At the lab bench level, it is my experience that people wear their daily shoes unless you work at a larger site that has a safety shoe/shoe covering policy.

Thank you for this excellent answer; this addresses my concerns. I think it's hard for me to put into perspective without this sort of info, as my lab had a typically relaxed "academic" attitude towards safety policies.

You're welcome! I should note that my statements about employer responsibilities is true in most cases in the US. I can't assume you're from the US so your organization may not have been negligent if they didn't follow this program I am describing; even in the US R&D facilities either have exemptions to some safety and environmental rules or are too small to be within the scope of industry guidelines. But I digress...

I think it's not a good idea to carry them home if you were handling dangerous chemicals, this could pose a great danger to you if you get to consume them. some chemicals like Pb undergo bio accumulation and can not only affect you but also the entire ecosystem in your vicinity. so companies should take care of their employees by providing safety clothing in their areas of work only


text:
What is the standard molar weight of Carbon, and why?
The mole is defined as the amount of a chemical substance which contains as many representative particles, e.g., atoms, molecules, ions, electrons, or photons, as there are atoms in 12 grams of carbon-12 (12C), the isotope of carbon with relative atomic mass 12 (from Wikipedia).
However, when I go to websites such as http://www.webqc.org/molecular-weight-of-C.html or http://www.convertunits.com/molarmass/Carbon and I ask what is the molar weight of Carbon, I get answers such as:

Molar mass of C is 12.01070 ± 0.00080 g/mol

When I asked on a forum why this isn't precisely 12, I was told that typically carbon is a mixture of isotopes which have relative atomic masses of 12, 13 and 14. With the ratio of each, the average value comes out to 12.0107.
But where does that number come from?  Is it by definition?  Is it by convention?  Is it a practical number?  Where do the error bars come from?  For instance, I would expect to see different isotope distributions at different geographic locations (or what about the moon, asteroids, Jupiter, etc.).   I tried googling for "standard carbon isotope ratio" but didn't find anything definitive.  Of course then I start to wonder how molar weights of other elements are defined, with varying isotopic combinations, and I have the same question.

It is a number determined experimentally, hence the error bars. As for the abundance of different isotopes, these were mixed pretty thoroughly during the formation of the Solar System, though some minor variations do exist.

As you stated the molar mass is a weighted average based on isotopic abundance, specifically abundance on Earth. We can obtain the molar mass of carbon by averaging the mass of $\ce{^{12}C}$ (about $98.89\%$ abundance) and $\ce{^{13}C}$ (about $1.11\%$ abundance). $$(\mathrm{mass\space of\space \ce{^{12}C}})*(.9889)+(\mathrm{mass\space of\space \ce{^{13}C}})*(.0111)=(12)*(.9898)+(13.00335)*(.0111)=12.0206\text{g}$$ This deviates from your value above because the values I used for the abundance aren't very precise.
In terms of where these abundances comes from and how the error bars are determined, the short answer is that its complicated. The long answers is here in a technical report by IUPAC, particularly sections 1.2-1.4.


text:
Do catalysts increase the rate of exothermic reaction more in a reversible reaction?
I'm just learning chemistry for school and I've learnt that adding a catalyst reduces the activation energy for the forward and backward reaction by equal amounts, therefore the rate of forward and backward reaction increase by the same amounts so equilibrium stays the same but is reached more quickly.
I'm having some trouble understanding this. I know that the activation energy of the exothermic reaction is always smaller than that of the endothermic reaction. So thinking about the Maxwell Boltzmann distribution, a change in $E_a$ near to $0$ will have a greater effect on the rate of exothermic reaction than the endothermic reaction. Therefore, a catalyst should cause equilibrium to shift to the exothermic side?

Look at the form of the Arrhenius rate equation (this is fine since they mostly all have this form):
$$k = Ae^{-\frac{E_{a}}{RT}}$$
If you make the substitution, $E_{a}->E_{a}+\Delta E$, then you have:
$$k = Ae^{-\frac{E_{a} + \Delta E}{RT}} = Ae^{-\frac{E_{a}}{RT}}e^{-\frac{\Delta E}{RT}}$$
This means that you're just scaling the rate constant $k$ by another constant. This scaling applies to both the forward and reverse rate constants. If the concentrations were such that they were in equilibrium before, they're still in equilibrium now.
You'd have to prove this in general for complex reactions with complicated rate laws, but for simple reactions, this should suffice to help you see that the equilibrium is unaffected.

So the reaction rates don't increase by the same amounts, the increase by the same proportion?

Because rate dependence on energy is exponential, it is by proportion. In the end, the laws of chemistry work pretty well, so you can try to rationalize backwards from rate equivalence at equilibrium.

Oh, I see what i'm missing. There have to be 2 distributions, one for the reactants and one for the products. This will allow for both areas to become the same

wait but then it should favour the side of the reaction with the greater concentration of reactants?

Again, chemistry works, so no.

ok i guess i can take that. You see I have a very mathematical approach in scientific problems. I attempt to be very picky about every detail and different ways of coming to the conclusion. It helps so much in understanding physics but not so much in chemistry

Notice that what you're missing from the Maxwell-Boltzmann distribution is how this relates directly to the activation energy? Is it just a function of $v$? If so, how does decreasing the barrier translate to changing entire equation? You'd have to perform a total substitution. After that you need to integrate from some cutoff $v=v'$ (how does this depend on catalyst?) to $v=\infty$.


text:
What are the effects of heating paraffin wax above it's melting point
I'm working with a paraffin embedding station that we recently learned was overshooting it's set point (60C) and heating paraffin to 100C. The problem has been corrected, but I'm curious what the mechanical/physical effects of paraffin being heated to almost twice the melting point. This is for a microtome setup (slicing at 5 microns) so micro features can and would be effected.

The mechanical effects on the paraffin? Or on the apparatus generally? Also, could you explain in more detail how the apparatus works? I've never worked with one; the phase "paraffin embedding station" gives me a hazy idea of what it does, but more detail would be greatly helpful.

The paraffin is inserted as paraffin chips (solid at room temperature)  heated to a set point, and melted via a heating element. The melted paraffin comes out of a nozzle. The cassettes, molds, and sample are then doused with the liquid paraffin, then cooled until they can be separated. (looks like this  http://drp8p5tqcb2p5.cloudfront.net/uploads/tx_templavoila/1029219956.jpg

the effects on the paraffin*

In short: A brief heating to 100 C (instead to 60 C) probably did not destroy your paraffin.
It is not too surprising the paraffin accidentally was warmed to 100 C, instead to 60 C. Especially in the solid state it does conduct heat not very well, and has a considerable heat capacity, too.  If it melts around 60 C,  heating to 100 C won't lead to vigorous boiling (like boiling water at this temperature).  In addition, I assume once you noticed the temperature higher than necessary / suitable for the embedding, you rapidly down-regulated the heater, too.
Maybe the manufacturer of the paraffin provided some properties of the batch delivered to you, beside melting point an indication of the decomposition temperature, too; which however should be much higher than 100 C (say 150...175 C).  I think it is unlikely that you observed fumes similar to the ones by an overheated frying pan, which were indicative for decomposition of the paraffin.
The reasoning: Many of the contemporary candles are equally made of paraffin, and when the children make new ones, chunks of old candles and new paraffin are equally thrown into the same melting pot, surrounded by (gently) boiling water, too. And while they do so for hours, I still have to see one new candle that became brown or black because the vessel was heated for so long time.


text:
Where is the more reactive double bond?
I was thinking that the addition reaction would occur more easily in the double bond (marked with an arrow) where the carbons are less substituted, because they are more unstable.

You have to specify exactly what addition reaction you are talking about, see my comment on the answer. Reactivity is not an absolute scale as it strongly depends on the exact conditions. Compare the reactivity of C=O and C=C with $\ce{NaBH4}$ and $\ce{H2, Pd/C}$ for example

I would like to agree with you somewhat.
The stability of substituted alkenes is due to two factors:

Stabilizing interaction between the C=C (pi bond) (hyperconjugation)
Adjacent C-H (sigma bonds) on substituents.

The more substituents that are present in the double bond, the more hyperconjugation and the stable the alkene.
Consider the heats of of hydrogenation of these alkenes:

Heat of hydrogenation of alkenes is a measure of the stability of carbon-carbon double bonds.
Under similar conditions, the smaller the numerical value of heat of hydrogenation of an alkene, the more stable the double bond therein
To sum up,it is much is easier to create a new bond where  the bond is weak/less stable,which is the reason why alkenes (esp polysub.) easily undergo addition reactions.
Please note that, alkenes behave as nucleophiles (Lewis bases) in polar reactions, donating a pair
of electrons from their electron-rich C=C bond to an electrophile (Lewis acid),So in this case they need stabilisation (tert. carbocation more stable)
Update
Electrophilic addition of alkenes
As previosly noted alkenes behave as Lewis basis in polar reactions.
One classic example is reaction with HBr

The reaction occurs in two steps, protonation and bromide addition,
and involves a carbocation intermediate.
A hydrogen atom on the electrophile HBr is attacked by  electrons from the nucleophilic double bond, forming a new C–H bond. This leaves the other carbon atom with a + charge and a vacant p orbital. Simultaneously, two electrons from the H–Br bond move onto bromine, giving bromide anion.
The carbocation intermediate that results is itself an electrophile, which can accept an electron pair
from nucleophilic Br- ion to form a C-Br bond and yield a neutral addition
product.

Markovnikov’s rule

In the addition of HX to an alkene, the H attaches to the carbon with
fewer alkyl substituents and the X attaches to the carbon with more
alkyl substituents.
In the addition of HX to an alkene, the more highly substituted
carbocation is formed as the intermediate rather than the less highly
substituted one.

References
Image and some content from McMurry Organic chemistry

You don't know what addition it is, if it is polar addition of HX it is going to react on the more substituted double bond (tertiary carbocation). If it is hydroboration it is going to be less substituted (sterics), etc.

Better to ask the asker for clarification in such cases.

Sorry about that. If it is a Eletrophilic addition, then it wouldn't on the one that i pointed, right?

I'm not downvoting "just because I want to". The answer is wrong because it overgeneralises alkene reactivity.

@orthocresol, Im sure you have to answer the rest, since you brought that up. But seriously lets be FAIR!

Downvotes can be retracted easily after the post has been edited. I still think it would be good if there was a clear statement that the relative reactivity depends on the circumstances.

@xavier_fakerat Thank you for clearing that out. I really feel that i did the mistake here on not being specific, and Xavier gave the answer from what i gave. I should've be more specific, so i'll keep my upvote...


text:
Is it true that when hydrogen reacts with oxygen to form water, that if first forms an hydroxyl radical?
When I took chemistry in high school, the inevitable question as to whether or not water has a chemical name came up. Our instructor said no, but if it did it ought to be called hydrogen hydroxide. His reasoning was that when hydrogen reacts with oxygen, it first form an hydroxyl radical that then reacts with another hydrogen molecule to finally form water; since the reaction that converts hydrogen and oxygen into water is so chaotic, that hydrogen molecule will almost certainly be a different molecule than the one that contributed the first hydrogen atom. Was our chemistry instructor correct? Is this actually how hydrogen and oxygen react to form water?

This wanders from one question to a second. Could you edit it to focus on one or the other?

$\ce{OH.}$ isn't called hydroxide, it's hydroxyl; and don't take that as me advocating for "hydrogen hydroxyl". See also: Does water have a chemical name? and Do chemists refer to water as “dihydrogen monoxide”?

@orthocresol, dang! I know that! Stupid mistake.

Edited the question to better reflect the question posed in the body.