text:
The Lewis structure of HNO3
So I was checking the structure of nitric acid in Wikipedia, however I couldn't quite fathom why it looked like that because it seemed to contradict the following statement:

A Lewis structure with small or no formal charges is preferred over a Lewis structure with large formal charges.

If we draw it like the one on the right, we get rid of formal charges and the structure is said to be more "stable". Why does this concept not work in this case?

The left one satisfies 8-electron rule (something like that...)

Yeah but let's take a look at another example: H3PO4
Although the octet formation is not fulfilled, the structure with no formal charges is preferred. Here is the comparison:
http://i.imgur.com/XLpgIjn.png

How do you know which one is prefered?

I'm using Wikipedia.
http://en.wikipedia.org/wiki/Phosphoric_acid

Well, if some reference does not provide a justification for their claim, there is no serious reason to trust it :). Everyone can copy/paste here and there

It is quite strict that the number of electrons around N should not exceed 8.  Having lower than 8 electron is not ideal but still OK.  Nitrogen has only 4 valance orbital, enough to hold 8 electrons.   It is not a matter of preference-- the second structure is simply impossible. It is just like you cannot draw a double bond to a hydrogen atom.

It is very important to know that Lewis octet rule is one thing and the Lewis Formula another. That does not mean that they may not be combined. The main flaw of Lewis formula is that it can only describe covalent bonds with two electrons each. Hence a Lewis structure is never a complete representation of the bonding picture.

Nitrogen is in the second row with no $d$ orbital in the valence shell.  It obeys the octet rule and cannot have more than 8 electron.
There are exceptions to the octet rule.  Having less than 8 electron is less preferable but still possible, and is commonly seen in free radicals and cations. On the other hand, having more than 8 electron is extremely unfavorable for second period atoms. Such electronic structures can be found in extremely unstable species or excited states, such as the CH5 radical.
As a comparison, first shell atoms obey the duet rule while atoms in the 3rd shell and beyond may obey 18-electron rule, 12-electron rule or 8-electron rule.   However, 18-electron rule and 12-electron rules are much less strict than 8-electron rule and violations are common place.
According to valence bond theory, the electronic structure of a molecule is a combination of all possible resonance structure that you can write down, including structures with all possible formal charges and strange electron counts.  However, their contributions are not even.  some of them are more favorable than others.
For HNO3, in order to satisfy the octet rule, the nitrogen atom would form 1 double bond and 2 single bonds.  Based on octet rule alone, there are 3 possible resonance structures that are favorable.

However, the first two resonance structures are significantly more favorable than the third, because they have smaller amount of formal charges.  As a result, we usually only write the two dominant structures. The bond between OH and N is close to a normal single bond. The other two N-O bonds have bond order near 1.5.
It is also common to write a mixture of resonance structures as the hybrid form

Note that this representation is not a single Lewis structure, but a convenient way to represent many resonance structures in the same figure. It gives no information about the exact bond order or formal charges on individual atoms.  The dotted lines indicates that in some structures where that bond is a single bond and in others a double bond, and the order is somewhere between 1 and 2.

Well! here is a short quick reply..!!
I was pretty sure that the lewis stucture of $\ce{HNO3}$ would be the one with 0 formal charge and than I found this. Though this link only shows how to calculate formal charge on $\ce{HNO3}$ but it gives a little hint.
Secondly a thought came to my mind that if you will focus on Resonance, there will be double bond character on both bonds between N & O.

Historically, there was nothing wrong with structures like the one you drew on the right. Iwan Ostromisslensky had no problems drawing 4-chloro-1,2-dinitrobenzene with a pentavalent nitrogen in 1908.[1] (Yes, a nitro group is not nitrate or nitric acid, but after some time of searching I just took what I found to prove my point.)

Staudinger and Meyer drew dimethylanilinoxide similarly in 1919:[2]

I wasn’t able to find the point in time when preferences changed so I gave up. It must have been sometime around when the quantum chemistry of orbitals was becoming better and better understood, and it was realised that there are only four orbitals nitrogen has access to for bonding (2s and three 2p). Henceforth, people described structures with nitrogen in the centre with four bonds only, better reflecting reality.
For elements of higher periods, most notably sulfur and phosphorus, many people still choose to draw too many bonds rather than write charge-separated structures. This is often explained with ‘d-orbital participation’ — but from a practical point of view, the 3d orbital has a very similar energy to 4s yet nobody suggests 4s participation. All ‘extended octet’ structures can be drawn in a way that conforms to the octet rule, so maybe it is only a matter of time until $\ce{P=O}$ bonds in phosphate disappear.

As for rules to determine the likelihood of Lewis structures, this is a better set:

Lewis structures in which all atoms have an octet (doublet for hydrogen) are preferred.
If 1. cannot be fulfilled, Lewis structures which have the least number of elements with sub-octet structures are preferred.
If 1. or 2. generate a set of possible structures choose one that has a minimal number of formal charges.
If 3. leaves a set of possible structures, choose one where the formal charges are distributed according to electronegativity (electronegative elements having negative formal charges).
If 4. leaves a set of possible structures, choose one in which the formal charges are closer together.
If your final structure has an expanded octet on a main group element, start again at 1.

References:
[1]: I. Ostromisslensky, J. Prakt. Chem. 1908, 78, 263. DOI: 10.1002/prac.19080780121.
[2]: H. Staudinger, J. Meyer, Helv. Chim. Acta 1919, 2, 608. DOI: 10.1002/hlca.19190020161.

I have even seen trifluoroamine oxide drawn with a double bond from nitrogen to oxygen making 10 electrons around nitrogen.  The nitrogen-oxygen bond actually does have double-bond character, but this comes from ionic contributing structures of the form $\ce{(O=NF_2^+)F^-}$, not anything with ten valence electrons on nitrogen.

Even though you may have a filled octet, if the central atom has a positive formal charge, we generally form double bonds until the formal charge is reduced to 0 as close as possible as that will be the more stable configuration. However, there are some exceptions.
Please take a look at the chlorate ion. For chlorate, you would expect to leave four single bonded oxygens to the chlorine, however, we are left with a formal charge of +3 on the chlorine and -1 on each of the oxygens. Therefore, we form double bonds until the formal charge is removed and are left with only a formal charge of -1 on the single bonded oxygen.
edit: looking at your question the structure on the left is preferred even though the one on the right has a lower formal charge.
Perhaps Wikipedia has the wrong structure for the Nitric acid page? It looks like all the oxygen atoms are single bonded in the figure to the top right.
https://en.wikipedia.org/wiki/Nitric_acid
Looking for images, Wikipedia does have the right resonance structures with the double bonded oxygen, however, there is only one double bond contrary to what we expect which is two double bonds.
https://commons.wikimedia.org/wiki/File:Nitric-acid-resonance-A.png

Yes, indeed. Doesn't this principle contradict with the structure of HNO3 though? The formal charge of the center atom (nitrogen) is said to be +1.

@ZaferCesur Chlorine and nitrogen are fundamentally different.  Nitrogen atom has no d orbital because it is the in second period.  It has only s and p and the maximum number of electron it can hold is 8.  Chlorine is in 3rd period.  It can hold up to 18 electrons.

Okay, I got it now! You should maybe post an answer so that I can upvote or something.

@XiaoleiZhu Yep, in this case it is not possible to form two double bonds as extended octets are formed only by atoms with vacant d-orbitals in the valence shell (p-elements
from the third or later periods). Thanks for the clarification.

The Wiki picture is a graph that is just a bond and angle mapping. Even in chlorine (esp., and lower elements) d-orbitals have no role in bonding (@XiaoleiZhu). The concept of hypervalency (see also goldbook) is still under criticism and one usually refers to four-electron-three-centre bonds, rather than including d-orbitals in main group elements.

Three oxygens are bounded to one nitrogen atom.Oxygen  being more electronegative attracts the pair of electron more strogly.One of the oxygen binds with an atom of hydrogen and share an electron with nitrogen to complete its octet and the remaining oxygen atom share an other electron with nitrogen atom. In this way electron pairs are partially donated and coordinate covalent or dative bond is formed


text:
Sources for raw 1H-NMR data?
Where can one find examples of raw data for an 1H-NMR spectrum?
This may sound as an odd request, so I'll give some context.
I have found lots of images of spectra such as this:

where you can see an image of the peaks. What I am interested is a spreadsheet
of the $(x,y)$-values used to draw the spectre. The data is needed for a student mathematics/chemistry project (high school). The mathematics section will be on calculating the area under a curve using numerical methods. To get a connection to the chemistry part, I'd like the student to use realistic data.
(I am aware that it is the ratio of the areas of the peaks that are interesting, so the actual scaling of the y-axis is unimportant.)

Any links or references are welcome - I was not able to find anything via Google.

Sadly, this is not an easy problem to solve.  Do you have access to Excel or Mathematica?

I have access to both.

I might suggest that you write a short program to simulate the spectra.  The problem with using real data is the amount of points (upwards of 32k) that make publishing CSV files of raw spectra unreasonable.  Below is a quick attempt to give you what you are looking for.  I apologize for those who do not use Mathematica; however I tried an Excel version and it became unwieldy very quickly.
I have three functions:
Clear[feature];
feature[shift_, n_, coupling_, intensity_] := Module[{},
  Plus @@
   Table[Binomial[n, k]*intensity/(n + 1)*
     Cos[(x)*(shift + coupling (k - n/2))], {k, 0, n}]]
Clear[fid]
makefid[feature_, broadening_] := Module[{},
  Exp[-broadening x] Plus @@ {feature}]
Clear[transform]
transform[fid_] := Module[{resolution = 4096, $y, $x},
  $y = Re[Fourier[Table[fid, {x, 0, resolution, 0.1}]]];
  $y = Take[$y, Round[Length[$y]/2]];
  $x = Range[Length[$y]]/(resolution/(2 Pi));
  Transpose@{$x, $y}
  ]

The function feature creates a peak with a chemical shift as the first argument, the splitting as the second, the coupling constant (use 0.1 for starters) and the intensity (its integral).  makefid creates a simulated FID based on the feature that it is passed and transform makes the spectrum.  It can be used like this:
a = feature[1.2, 2, 0.1, 3];
b = feature[2, 0, 0.1, 3];
c = feature[4.1, 3, 0.1, 2];
fid = makefid[Plus @@ {a, b, c}, 0.01];
ListLinePlot[transform[fid], PlotRange -> {{0, 5}, All}]

The features {a,b,c} are simulated ethyl acetate yielding the following spectrum:

The data you want is in transform[fid] which can be exported as a CSV file:
Export["myfile.csv",transform[fid]]

Which will make it accessible to other software packages and you can perform your math lesson plan.  To make sure this is working properly in Mathematica, we can make a function out of our data and numerically integrate the three peaks:
data = Interpolation[transform[fid]];
NIntegrate[data[x], {x, 0.9, 1.5}]
NIntegrate[data[x], {x, 1.5, 2.2}]
NIntegrate[data[x], {x, 3.7, 4.5}]

This produces areas of 0.32, 0.25, and 0.33, respectively, which is what we expected.

Thank you for your Mathematica code. It was very instructive to see how you generate the peaks from the features.

Raw NMR data is usually stored in proprietary format, which is vendor specific, and even though most NMR data processing software can handle multiple formats, it is not a straight forward process to get data into an x-y format for use with other software (Excel, Matlab, Origin etc).
However, to answer your question:
There are a few sites where you can access raw 1H NMR data.
Possibly the most useful is the Spectral Database for Instructors, which has a small database of 1H, 13C and 2D spectra available for download. The data are in Bruker format (fid, acqus for 1D and ser acqus, acqu2s for 2D). This site has some useful information about how to access and process this data, for that is another question altogether!
You can trawl your way through Chemspider; you may get lucky with an odd spectrum or two, but I don't think this is worth the effort
There are a few old sources of NMR data in jcamp format that still exist. Jcamp was/is the universal data format proposed for many data sources (widely used for IR, UV for example). All major NMR software has the potential to read and write in jcamp format, although you can run in to problems using different jcamp version. Examples of sites with jcamp data include: AcornNMR, Polymer Science Learning Center, JCamp test data at the University of West Indies (with links to what jcamp is all about) are just some. You can probably google nmr jcamp to find more. I can't vouch that these data are in the public domain, and I have omitted a couple of databases that I know are not.
If you want real examples of real molecules run by real researchers (and not just for teaching purposes), you probably can't go past Open Science researchers like the Open Source Malaria project. You can follow the links to the compounds and save the raw NMR data.
Of course, a lot will depend on the types of molecules you would like to get data for, and how many different structures you need. It may be easier to approach a friendly NMR Facility Manager at your nearby College or University and ask very nicely for some data for some simple molecules. I'd be happy to provide further information about that if you need.
Once you have your data, you will need to be able to convert it into a format that you can make sense of. Again the best approach here will depend on how many data sets you need and how long you are likely to use the data for. You can download demo versions of NMR processing software like Mestrenova or Spinworks which should allow to read in raw data formats and export in a more desirable (ASCII X,y) format.

Thank you for your thoughtful answer. I found some real world data in the Spectral Database for Instructors. If I could I would have accepted both answers.

Sadly it seems that "Spectral Database for Instructors" has closed?
The advice that Mestrenova can save the data as xy-coordinates are spot on. To find data files one can google for, say, "aspirin fid file".

This is an old thread, but The Open Spectral Database osdb.info is starting to have some nice spectra available for download in several formats.

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text:
How do I find molar mass using gas laws?

A gaseous sample of a compound has a density of $\pu{0.977 g/L}$ at $\pu{710.0 torr}$ and $\pu{100.00 ^\circ C}$. What is the molar mass of this compound?

When I worked the problem I got $\pu{29.9 g/mol}$ is that right?

A gaseous sample of a compound has a density of $\pu{0.977 g/L}$ at $\pu{710.0 torr}$ and $\pu{100.00 ^\circ C}$. What is the molar mass of this compound?

The ideal gas law states $$pV = nRT\tag{1}\label{id-gas}$$ where $p$ is defined as the pressure, any units, $V$ is the volume in liters, $n$ is the amount of substance of gas, $R$ is the universal gas constant, and $T$ is temperature in kelvin.
The information given to us is the density of $\pu{0.977 g/L}$, a pressure of $\pu{710.0 torr}$, and temperature of $\pu{100 ^\circ C}$.
Knowing that amount of substance, $$n = \frac{M}{m},$$ where $M$ is molecular mass in $\pu{g/mol}$, and $m$ is the mass of gas in gram, we can see the unit gram cancels out and we are left with the unit mole.
However, we are given a density, $$d = \frac{m}{V} = \pu{0.977 g/L}.$$
Plugging into $\eqref{id-gas}$ the molecular mass and mass for $n$ we get:
$$ pV = \frac{M}{m}RT .$$
Isolating $m$ in density, $m = d\cdot V$, and isolating $m$ in $n = m/M\Leftrightarrow m = n\cdot M$, we can combine those two equations for $m$:
$$n\cdot M = d\cdot V$$
Isolate $n$, and plug back into our ideal gas equation $\eqref{id-gas}$:
\begin{align}
 n &= \frac{dV}{M}&
 \implies&&
 pV &= \frac{dV}{M}RT
\end{align}
Volume cancels out, and we isolate for $M$:
$$ M = d \cdot \frac{RT}{p}. $$
Plug in our known values, knowing that $\pu{1 torr} = \pu{1 mm Hg}$ and $\pu{760 mm Hg} = \pu{1 atm}$:
$$ M = \pu{0.977 g//L} \cdot \frac{
  \pu{0.082057 L * atm // mol * K}\cdot\pu{373.15K}}{
  \pu{710 torr}\cdot \pu{mmHg//torr}\cdot\pu{{1atm}//760mmHg}}
     = \pu{32.0 g//mol}$$
And our answer has three significant figures.

No! The answer $\pu{29.9 gm/mol}$ is wrong.
The best way to solve these type of problems is Ideal Gas Equation. You need to do a little modification as provided in answer by Rauru Ferro, i.e.
$$P V = n R T$$
Since $ n = m / M $, we can also write above as
$$P V = \left( \frac{m}{M} \right) R T.$$
Since density $d = m/V$,
$$P = \left( \frac{d}{M} \right) R T.$$
Now you can easily apply this modified equation in your problem.
Before applying this, all you need to do is to convert the pressure from torr to atm:
$$\pu{1 atm} = \pu{760 torr}.$$
And also you will need to see the units of gas constant and other quantities as these type of problems seems much easier, but not-focusing-on-units can lead you to the wrong answer.
The answer for your problem will be $\pu{31.9 g/mol}$ approximately.

From gas equation:
$$ P V = n  R  T$$
As $n = m / M$
and  density $d = m / V$,
$$P = d  R  T / M,$$
so
$$M = 0.977 \cdot 0.082 \cdot 373 / (710/760) = \pu{32.1 g/mol}.$$

Density is $\pu{0.977 g/l}$ at specified conditions.
What happens to the density if you increase the pressure from $710$ to $760$?  This tells you to multiply by $\frac{760}{710}$.
What happens to the density if you cool it from $373$ to $273$?  This tells you to multiply by $\frac{373}{273}$.
If $\pu{1 l}$  has this mass, what is the mass of $\pu{22.414 l}$?

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text:
Gas Law problems with stoichiometry
Consider the following reaction : $\ce{H2 + Cl2 -> 2HCl}$
A) If you have 15.00 liters of hydrogen reacting with 10.00 liters of chlorine at STP, how many liters of HCl would you produce?
B) What would be the limiting reagent?
C) How many moles of excess reagent would remain?
Please explain.

Molarity = moles/Liters
to check for $\ce{H2}$: M = 2/15.00 = .13334
to check for $\ce{Cl2}$: M = (35.45 x 2)/10.00 = 7.09
But I do not think that is correct. can somebody show me the correct way to do it.

Please see my comment on your other question.

Avogadro's law says that $V/n = \text{constant}$ at constant temperature and pressure.   That means your volumes can be used "as if" they were moles in the chemical reaction stoichiometry.   So, if the mole ratio is 1:1 for H$_2$ and Cl$_2$ reacting, then the limiting reagent is obviously the smaller volume of 10.0 L for Cl$_2$.
For (b) you can easily calculate the volume of HCl since 1 mole of Cl$_2$ produces 2 moles of HCl.  So, you get 20.0 L of HCl.
For (c) the excess reagent is the H$_2$.  Since it reacts with 10.0 L of Cl$_2$ in a 1:1 mole ratio, then you have 15.0L - 10.0 L = 5.0 L of H$_2$ left over.   Since this all ends up at STP, you can use the ideal gas law to calculate the moles of H$_2$ left over:
$n_{H_2} = \displaystyle \frac{p V}{R T} = \frac{(1 \text{atm})(5.0 \text{L})}{R \times 273.15 \text{K}} = 0.2230749373505515$ mol
Don't forget sig. figs., which in this case is determined by the 2 sig. figs. in the volume of 5.0 L, giving you
$n_{H_2} = 0.22$ mol
For Review check out:
gas laws: http://www.grandinetti.org/Teaching/Chem122/Lectures/Gases
If you have a mac, check out PhySyCalc. It will automatically do all the unit conversions in your calculations.
https://itunes.apple.com/us/app/physycalc/id644105995

You do not need the molar masses to do this question. The ideal gas law includes a relationship between volume $V$ and moles $n$ assuming pressure and temperature are constant (they are at STP).
$$PV=nRT$$
$$n=\dfrac{PV}{RT}$$

At STP, 1 mol of any gas occupies 22.4L. Using this knowledge, you can then convert each volume of gas to moles and then determine the limiting reactant using the 2:1 molar ratio of product to reactant.
As for part a, the amount of HCl produced would be dependent on the limiting reactant.
Part c could then be answered by subtracting the number of moles of limiting reactant from the excess reactant.

You only need to know that at STP $V_m=24.4\:\mathrm{L\:mol^{-1}}$, and that $n$ (number of moles of of a compound or molecule) is equal to $\frac{V_{gas}}{V_m}$.
Find limiting and excess reagents using stoichiometric ratios (in this case it is 1:1), therefore $\ce{H2}$ is limiting the limiting reagent.
Use stoichiometry to find amount of $\ce{Cl2}$ reacted, and then subtract the initial number of moles from the final number of moles.


text:
Gas law and moles

A sample of solid magnesium reacts with excess $\ce{HCl}$ solution to form hydrogen gas. The sample of gas measures $1.628\ \mathrm{L}$ at $22.0\ \mathrm{^\circ C}$ and $0.930\ \mathrm{atm}$ of pressure.
  a) how many moles of gas were collected?
  b) how many grams of magnesium did you start out with?

First I stated by writing the chemical equation
$\ce{Mg + HCl -> H2 + MgCl}$
then convert the temperature to kelvin and pressure to atm and volume to L
$V = 1.628\ \mathrm{L}$
$T = 295.15\ \mathrm{K}$
$P = 0.930\ \mathrm{atm}$
$PV = nRT$
$0.930\ \mathrm{atm} \cdot 1.628\ \mathrm{l} = n \cdot 0.08206\ \mathrm{l\ atm\ K^{-1}\ mol^{-1}} \cdot 295.15\ \mathrm{K}$
which will give that $n = 16.0\ \mathrm{mol}$. Did I do this correctly?

Although it doesn’t affect the result, your equation is incorrect. Magnesium tends to form $\ce{Mg^2+}$ ions and chlorine tends to form $\ce{Cl-}$ ions. That’s why magnesium chloride compounds are found in 1:2 proportions. After balancing the coefficients your equation should look like this:
$\ce{Mg + 2HCl -> MgCl2 + H2}$
Then using the formula $pV = nRT$ we get $n\approx0.063\ \mathrm{mol}$
Since $n(\ce{Mg})=n(\ce{H2})$ we have $0.063\ \mathrm{mol}$ of magnesium, which is approximately $1.531\ \mathrm{g}$.

For (a)
$$n_{\ce{H2}} = \frac{p V}{R T} = \frac{(0.930\ \text{atm}) (1.628\ \text{L})}{R \times (22.0\ \text{K} + 273.15\ \text{K})} = \frac{(0.930\ \text{atm}) (1.628\ \text{L})}{R \times 293.2\ \text{K}} = 0.06251389252288155\ \text{mol}$$
With the correct number of significant figures
$n_{\ce{H2}} = 0.06251\ \text{mol}$.
For (b)
the balanced reaction tells you that 1 mole of $\ce{Mg}$ makes 1 mole of $\ce{H2}$, so $n_{\ce{Mg}} = 0.06251\ \text{mol}$.  To get the mass of $\ce{Mg}$ use the atomic weight
$m_{\ce{Mg}} = n_{\ce{Mg}} \times M_{\ce{Mg}} = 0.06251\ \text{mol} \times M_{\ce{Mg}} = 1.519400157768636\ \text{g}$.
Again, don’t forget we only have four significant figures
$m_{\ce{Mg}} = 1.519$ g


text:
Gas law with molecules at the same time
How many grams of oxygen are present at 25.0 C in a 5.00L tank of oxygen enriched air? The total pressure is 30.0 atm and the only other gas is nitrogen with a pressure of 17.0 atm.
After converting the temperature to Kalvin, I do not really know what to do.
How can I find the grams of oxygen with another molecule in the container?

What do you know about the pressure of oxygen?

The pressure for oxygen shoul be 13.0 atm. Now, apply equation gas.

We have 2 gases in the container, oxygen ($O_2$) and nitrogen ($N_2$). Now we are given that $N_2$ has a pressure of $17$ atm out of $30$ atm, the total volume of the container is $5.00$ L and that the temperature is $25^o$C or $298$ K. So somehow, we need to find out the volume that oxygen occupies and from there we will be able to find out the grams of oxygen present in the container.
First let's find out the mole fraction $N_2$.
You can find mole fraction by the following formula,
 $$Mole\, Fraction = \frac{Partial\,Pressure\,of\,Gas}{Total\,Pressure}$$
So,
Mole fraction of $N_2$= $\frac{17}{30}$=$0.6$
Therefore mole fraction of $O_2$=$1-0.6$=$0.4$
Now we need to find out the total no. of moles by the formula,
$$PV={nRT}$$
Putting the values given for the whole system,
$$30\times5=n\times0.0821\times298$$
$$\Rightarrow\,n=6.1$$
So moles of oxygen in container = $0.4\times6.1$ = $2.5$
Consequently, grams of $O_2$ present in container = $2.5\times32$ = $78.4$
So that is how you get the grams of oxygen. Hope you understood everything. If not, feel free to ask by commenting. Good Luck!


text:
Memorize Transition Metal Electronic Ground State
In the course of my undergraduate studies it has become clearer and clearer that it is quite helpful to know the ground electronic state of transition metals by heart, if only to speed up the process of analysing a reaction or to evaluate a compound in respect to its number of valence electrons.
Edit: Note that when I say ground electronic state, I don't mean the state in vacuum, but when the metal is bonding to ligands. As a consequence all the electrons are found in the $d$ orbitals.
My Method:
I have now put the $d$-Block of the periodic table as a wallpaper on my screen background so I get constantly exposed to the arrangement of the elements.
What is your way to memorize what metals have which $d^x$ ground state?

Out of curiosity, is any example that the difference between Fe 3d$^6$4s$^2$ and Ru 4d$^7$5s$^1$ produces significant consequence for analysing reaction?

@user26143 I don't know if there's a direct chemical consequence from the differences in atomic ground states, but there certainly are consequences from the reason the ground states are different in the first place - the anomalous compactness of 3d orbitals relative to 4d ones.

Question edited for clarification in this regard.

I don't know if it is worthwhile for perfectly memorizing electronic configuration, since there are many subtleties in transition metals. More important, the electronic configuration of metal in compound will depend on ligand, like the high-low spin cases (which is not covered in periodic table wallpaper?), http://en.wikipedia.org/wiki/Crystal_field_theory, and http://www.academia.edu/847281/Icon_of_chemistry_the_periodic_system_of_chemical_elements_in_the_new_century "4.1. Free and Bound Atoms"

I think OP merely wants to know an efficient way to associate each transition element to their group? i.e. knowing by heart that ruthenium is from group 8, alongside iron and osmium.

I see, that would be the position of element in periodic table.... perhaps after staring at the wallpaper of periodic table for a couple of days, it enters the memory...

We use a periodic table exactly for that purpose (ie to NOT memorize things just for the sake of memorizing things).

Wikipedia has a comprehensive list on chemistry mnemonics, with which I'm now memorizing the position of the transition metals.
For example, for the first two periods of the elements (excluding neon):

H He Li Be B C N O F
Ha. Healthy Little Beggar Boys Catching Newts Or Fish.

Or my favourite for the first period of transition metals:

Sc Ti V Cr Mn Fe Co Ni Cu Zn
Saint Thomas Very Carefully Made Five Completely New Copies of Zarathustha.

Marked as accepted for the time being, if someone comes along and has a good option that works for them I'd be very glad to hear about it!

The asker could well create their own mnemonic more suited to their objectives by reciting the transition elements group by group instead of period by period (i.e. Sc → Y → La/Lu → Ac/Lr , Ti → Zr → Hf → Rf , V → Nb ... , with period 7 transition metals optional). Chain them all together for extra points! The very act of challenging yourself to create your own mnemonics will help you learn, so it's doubly effective.


text:
Why do amorphous substances have different values of physical properties in different directions?
Today my teacher taught about amorphous and crystalline substances. Crystalline substances are said to have different values for physical properties (like speed of light, electrical resistance, etc) in different directions. Crystalline substances are said to have orderly arranged constituents, however this orderly arrangement is not said to be same in different directions. Thus, they are said to show different values for physical properties in different directions.
  Amorphous are said to be isotropic in nature, i.e they have same value for all the physical properties in different directions. I was stuck here, amorphous substances have disordered arrangement of constituents. So, they should strictly have different values in different directions. Is it we speak about average values of physical properties in case of amorphous substances? or is it that, they strictly have same values for the physical properties in all directions?

Yes, you’re exactly right that we are talking about an average. If we could somehow look at a small enough portion of an amorphous solid, we would see differences between the properties in different directions. Indeed, a small enough portion of, say, amorphous ice is a single water molecule, which is certainly not isotropic: we can talk about the direction perpendicular to the molecular plane or whatever.
But when we talk about real, physical materials, as usual in chemistry, we are talking about Avogadro’s number (plus or minus a few orders of magnitude) of molecules. So taking the spatial average of an amorphous structure makes perfect sense, and the law of large numbers means that the tiny differences between properties in different directions will be utterly undetectable.
(As an aside, it’s worth noting that the perfect crystalline structure we imagine is also an average over every unit cell in the material – it’s just that because each unit cell is oriented in more or less the same way, different directions are not “mixed” as part of this average, so crystals can have anisotropic properties.)

There is probably some misunderstanding here... As an macroscopic example, take a book. And then split it in half. There is only one way you can do it with small effort. That's anisotropy. Then run the book through paper shredder and you obtain isotropic material (http://www.compatibletoner.org/wp-content/uploads/2009/09/shredded-papers.jpg). You can split it easily in half in all directions and also other properties are independent of the direction (compared to the book).


text:
A full verion of the periodic table (For computer game)
I am looking for a full set of data on every element on the periodic table. (and if possible those only speculated about)
When I say full I mean everything. Electron configurations, Valence electrons, melting point, Atomic mass, Isotopes and what they decay into and all the rest.
Is there such a database anywhere?

My nephew knew the first 150 or so first-series Pokemon, and all about them, by the time he was 7. I often thought that if there was a game with characters named after the chemical elements, he'd be streets ahead by the time he went to senior school. I hope your game will be as exciting (and profitable) as Pokemon!

It is answered here.

You can find all these information on Wolfram Alpha.  You can use the website directly to obtain these information, and you can call their API to directly fetch the data online from your own program.  Of course, you can also use Mathematica to write your program and query the database with WolframAlpha function.

Learned the api in about an hour and now I can query their database from c#. This is just amazing. Thank you.


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What are ketoximes?
I am reading a book which is saying about the Syn-Anti nomenclature of oximes. It is showing this molecule to be Syn-methyl-ethyl ketoxime.

I can't understand what is a ketoxime and how is there nomenclature done. Can somebody explain please ?

An oxime is a chemical compound belonging to the imines, with the general formula $\ce{(R^1)(R^2)C=N-O H}$,

How is the nomenclature of methyl-ethyl ketoxime done?
The answer is :
Look at your photo.
1. On the left side you can see "$\ce{CH3-CH2 -}$" (ethyl).
2. On the right side you can see "$\ce{CH3 -}$" (methyl).
3. On the middle you can see "$\ce{C=N-O-H}$" (ketoxime).
Now assemble (1 & 2 & 3 ) ..then you will have the name "methyl-ethyl ketoxime".
For more you can  see  : https://en.wikipedia.org/wiki/Oxime

According to the old IUPAC recommendations (1979 as well as 1993), compounds having the general structure $\ce{R-CH=N-OH}$ or $\ce{RR'C=N-OH}$ are called generically ‘oximes’. They are further classified as ‘aldoximes’ and ‘ketoximes’, respectively. They are named by placing the functional class name ‘oxime’ as a separate word after the name of the corresponding aldehyde $\ce{R-CH=O}$ or ketone $\ce{RR'C=O}$, respectively.
Therefore, the systematic name of your example according to the principles of functional class nomenclature is butan-2-one oxime.
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), functional modifiers are still acceptable for general nomenclature purposes, but the preferred IUPAC names (PIN) are substitutive names for oximes. They are named substitutively as ‘ylidene’ derivatives of hydroxylamine and not by functional class nomenclature as in previous recommendations.
Therefore, the preferred IUPAC name of your example is (butan-2-ylidene)hydroxylamine.


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Chemical shift value and proton NMR spectroscopy
I am a bit confused about what chemical shift value signifies in proton NMR spectroscopy. If the proton of the standard substance TMS absorbs X Joules of energy to go into resonance, than a substance with a chemical shift value of 2.5ppm means it absorbs:
energy absorbed= X- (2.5X/10^6)?
Could someone please tell me if this is what chemical shift value is mathematically. In my book, it says the substance absorbs "2.5 millionth less amount of energy than TMS to go into resonance"... but I don't really understand this.

https://en.wikipedia.org/wiki/Chemical_shift
has a pretty good explanation (see the Chemical Shift Referencing section).

The chemical shift is a measure of how much energy it takes to flip a nuclear spin relative to a standard.  For 13C and 1H that standard is tetramethylsilane(TMS).  long's answer below talks about Hz, and you can calculate the energy differences using Planck's constant.  That being said, this topic is almost never used chemically beyond a basic level understanding.  All that matters is that different chemical shifts correlate to different types of chemical environments.  Unless you are really grooving on the physics don't dwell on it.

Firstly, the ppm scale is a dimensionless scale that describes the energy level of a signal relative to a reference (usually TMS). Significantly, it is magnetic field independent, and so allows comparison of signals recorded on different systems (not just a 300MHz vs a 600MHz spectrometer, but between all different 600MHz spectrometers, because no two will have been set up at identical Bo fields).
The chemical shift scale is defined as:
$$\delta =(v-v_r )*10^6/v_r$$ , where $v_r$ is the spectrometer reference frequency (of TMS, say)
Now, for a 600MHz NMR spectrum, each ppm represents 600Hz. So, if the resonance frequency of a signal at 0ppm is 600,000,000Hz, the signal at 1ppm represents an energy of 600,000,600Hz. At 2.5ppm, the frequency is 600,001,500Hz.
The chemical shift scale is often confusing in its labelling, but the thing to remember is that as you go to higher chemical shift values, you go to higher frequencies.
Higher ppm == downfield == Deshielded == higher frequency
Lower ppm == upfield == Shielded == lower frequency

downfield and upfield are outdated and should no longer be used, since NMR no longer scan the spectrum. Other then that, excellent answer.

Upfield and downfield are fine, NMR always scans the spectrum, it just does so by Fourier transform these days.

Yes, upfield and downfield are accepted terminology, and this is unlikely to ever change. A metal hydride at -20ppm is upfield of an aromatic proton at 8.3ppm. A methyl amine at 3.5ppm resonates downfield from the TMS signal at 0ppm


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Formation of coordination complexes?
I'm having trouble understanding how exactly coordination complexes are formed.
Compounds such as $\ce{[Cu(H2O)6]++}$ are the ligands filled into the empty 3 of 4p orbitals+ 1 of 5s orbital + 2 of 4d orbitals or how?

If you want to analyse a compound, it is best to start with the $d$-electron count of the metal. For $\ce{Cu^{2+}}$ we get $d^9$, since $\ce{Cu^0}$ would be $d^{11}$.
Now, consider the following (simplified) molecular orbital diagram for a $d^1$ compound:

As you can clearly see, the ligand electron pairs mix with some of the $d$, the $s$ and $p$ orbitals of the metal.
Now, instead of $d^1$ we have $d^9$, so we just start filling in 8 more electrons into the MOs. This gives us a full $t_\text{2g}$ MO and 3 electrons in the $e_\text{g}$ MO. ($t_\text{2g}$ and $e_\text{g}$ are ways of describing the symmetry of the orbitals. This comes from applied group theory.)
So you see, the $5s$ orbital never comes into play here, only the $3d$, $4s$ and $4p$ orbitals of the metal take part in bonding.
You might want to look at Ligand Field Theory and Molecular Orbital Theory to further your knowledge in inorganic chemistry. Not everything is as simple as the "fill in the boxes" method.


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ions proportion
I have a precipitation reaction of this form:
$\ce{Ca^{2+} + CO_{3}^{2-} → CaCO3}$
How many mole of $\ce{Ca^{2+}}$ ion is contained in a 200ml solution?
I'd like to know how to calculate this, thanks for the help!

Please see the link below in the close reason for information on asking your homework questions on the site.

Carbonates are usually insoluble in water except for when combined with group 1 elements and and ammonium ions, however, as you may already know, almost everything is slightly soluble in water.
Generally in order to see a precipitate form, a solution of sodium carbonate and aqueous calcium chloride are mixed. The calcium ions and carbonate ions are then in contact and can form the calcium carbonate precipitate. The other product is sodium chloride which is known to be soluble in water.
Please keep in mind that $ M = \frac{moles \; solute}{liter \; solution}$
Which is expressed as mol per liter.
The solubility of calcium carbonate is very dependent upon the pH of its environment. Assuming that this reaction occurs in water, the pH will be very close to 7. Tap water generally contains many ions, chloride for instance that will interfere with our calculation in the moles of calcium ions that we have.
First, we need the solubility constant product denoted as $K_{sp}$.
http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Solutions/Solubilty/Solubility_Product_Constant,_Ksp
Taken from the UC Davis ChemWiki "the Solubility Product Constant, Ksp is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution. A more a substance dissolves, the higher the Ksp value it has."
The expression for the solubility product of calcium carbonate is $$[Ca^{2+}][CO_3^{2-}] = 8.7 * 10^{-9} $$
The concentration of each ion is 1:1 as shown in our equation, so there is no need to square the concentrations.
If S mol of $CaCO_3$ dissolved in 1L, the equilibrium concentration of $Ca^{2+}$ will be $[Ca^{2+}] = S$. The concentration of $CO_3^{2-}$ ions will be $[CO_3^{2-}] = S$ again, because each mol of $CaCO_3$ each produces one mol of $Ca^{2+}$ ions and one mol of $CO_3^{2-}$ ions. Therefore,
$$[Ca^{2+}][CO_3^{2-}] = S * S = S^2 = K_sp = 8.7 * 10^{-9}$$
Solving for $S^2$ gives
$$ S^2 = 8.7 * 10^{-9}$$
Taking the square root on each side of this equation gives
$$ \sqrt{S^2} = \sqrt{8.7 * 10^{-9}}$$
$$ S = 9.327*10^{-5}$$
The equilibrium concentrations are therefore
$$ [Ca^{2+}] = S = 9.327*10^{-5} M $$
$$ [CO_3^{2-}] = S = 9.327*10^{-5} M $$
Since we have our concentration of $Ca^{2+}$ ions, we can then calculate the number of moles of ions that we have.
$$ M = \frac{moles \; solute}{liter \; solution}$$
$$9.327*10^{-5} M = \frac{moles \; solute}{1 L}$$
$$ mol = 9.327*10^{-5} M * L^{-1} * \frac{1L}{1000mL} * 200 = 1.8654*10^{-5} \; mol \; of \; Ca^{2+} ions $$

For completeness sake, lets take a look of the number of moles in grams.
Because the molar mass of $CaCO_3$ is $100.0869 g/mol$, using our our definition of molarity from above, we can calculate the gram solubility which is
$$ gram \; solubility = (9.327*10^{-5} \; mol \; L^{-1})(100.0869 \; g/mol) = 0.009335 \; g/L$$
Converting this to our 200mL solution,
$$ 0.009335 \frac{g}{L} * \frac{1L}{1000mL} * 200mL = 0.001867 g$$

Last note:
When $$CaCO_3(s) \rightarrow Ca(aq) + CO_3(aq)$$
some of the carbonate hydrolyses to produce HCO3.
$$CO_3(aq) + H_2O(l) \rightarrow HCO_3(l) + OH(aq)$$
In this instance, the hydrolysis of carbonate decreases the concentration of carbonate, which pulls the solubility equilibrium to the right making calcium carbonate more soluble.
http://www.colby.edu/chemistry/CH142/CH142B/SolubilityCalciumCarbonate.pdf
Please take a look at the link provided, as the actual number of moles will vary due to the pH of the environment. They nicely derive the formula at eq. 11 which you can then find the number of moles of $Ca^{2+}$ ions.

Thanks so much for your extensive answers!  However, it is best to prompt the user asking the question with hints before offering a full answer.  If you see questions of this type (just a homework question with no effort towards a solution), please flag them as "off-topic".

@jonsca My apologies. I am a little new here. :)


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Hydrogen ozonide existence
Does the hydrogen ozonide exist? I mean if this compound could be found in Nature
$$HO_3$$
I am interested in the theoretical or phenomenological of such a chemical compound...

Yes.  It exists in the atmosphere, is proposed to be intermediate in various atmospheric reactions and has been experimentally detected.
See
Science Vol*285* No.5424(1999): p81
Note:  Full text is available after free registration.

Could you provide me some alternative paper (free online available) to study the subject? Chemistry is cool! And ozonides are pretty interesting and generally not very well known compounds by the general public (like xenon chemistry or non-stoichometric compounds).

http://www.chem.ucla.edu/~ana/pdf/3.pdf


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Molarity exercises
Can anyone direct me to a good book with lots of molarity exercises, i really want to practice this and would like to find as many types of exercises involving molarity. I of course did google this and there are some, but usually they are three or four exercises per resource.  Thanks in advance.

This type of question might not cut it on Chemistry.SE because it is primarily opinion based, and as you mentioned can be answered by a google search.

It's also essentially a service recommendation. However, I have a better Google search for you. The first site have 16 problems, the second 4, the third has 25, the four 24, etc.

U can try rc mukherjee,an indian author for preliminary level.and schaum publishers for higher level.


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What is hydrogen-rich water?
My question is: Is hydrogen-rich water true? I am aware of the alkaline water. I have read several articles in the Internet regarding this kind of water but I cannot seem to be convinced. Wikipedia isn't even talking about it.
I am puzzled with how can a water be enriched with hydrogen. It's definitely $\ce{H2O}$ and everybody knows it. It's not water anymore if it's not $\ce{H2O}$.
This link is one of the examples of websites that I read about hydrogen-rich water.

Questions about any of these "ionated" waters are okay. A link to the claim about hydrogen rich water would improve this question.

IMPORTANT: Read http://www.chem1.com/CQ/ionbunk.html and the linked http://www.chem1.com/CQ/wonkywater.html#HYD. Note that the published 'studies' are almost invariably by authors from the countries most afflicted by related scams.

Hydrogen water is a solution of molecular hydrogen $\ce{H2}$ in water, which shouldn't be confused with acidic or basic solutions.  The first scientific reference to hydrogen water appears to be in the journal, Nature Medicine which can be accessed here if you have a personal or institutional subscription to the journal. The claim is that molecular hydrogen can serve as a mild reducing agent and therefore has antioxidant properties; namely, it can reduce hydroxyl radicals:
$$\ce{H2 +2 OH\cdot -> 2H2O}$$
The article has generated a considerable amount of research, with well over 300 journal articles citing this work originally published in 2007.  There seems to be some merit to the claim although in my brief scan of the abstracts, there does not appear to be a large body of evidence suggesting that commercial products are beneficial.  One such article, which is publicly accessible is here.  As mentioned in another answer, the commercial product approach to generating hydrogen water appears to be the oxidation of magnesium in water, which generates $\ce{Mg^2+}$ ions.  As with any other nutritional supplement, I would be very wary of entering this type of treatment without consulting a professional.  Magnesium is an important nutrient, however at high doses it does show signs of toxicity.

Hydrogen in water is also not that pleasant thing, as seen on Apollo missions: http://books.google.de/books?id=ExkCU3lnBhIC&lpg=PA166&ots=ogiwrFFiD9&dq=apollo%20hydrogen%20%20water%20flatulence&pg=PA166#v=onepage&q=apollo%20hydrogen%20%20water%20flatulence&f=false
I came across the signs of heart problems due to hydrogen as well on Apollo missions, but cannot find an article now.

@bobthechemist I am actually drinking one right now. Therefore, according to what you've said, this hydrogen water is not a recommended supplement for us. If this considerable amount of research is not supported by many, then there might be some issues regarding that research. Why would someone keep a secret for good health, if it actually benefits us?

@LesterNubla I am neither recommending nor not recommending its use; I am recommending caution and that you should consult a health professional.  I do not believe that someone is keeping a secret health benefit from the general public, but this science has only been studied for a few years, and that is what makes me cautious.

I was completely unaware of the concept of hydrogen-rich water, and although there is a lot of misinformation and quackery around it, a quick search on Google afforded an article (whose quality I cannot vouch for). It would seem that hydrogen-rich water is nothing but regular water with dissolved hydrogen gas. You could simply pump $\ce{H_{2(g)}}$ into some water, but there is a smarter way to do it, by producing the hydrogen in situ via the addition of reactive metals such as magnesium, according to the following reaction:
$$\ce{Mg_{(s)} + 2H2O_{(l)} → Mg(OH)2_{(aq)} + H2_{(g)}}$$
Producing the hydrogen directly inside the water also seems to temporarily supersaturate the solution and increase the amount of hydrogen dissolved, because the equilibrium solubility of hydrogen gas in water is tiny, hydrogen gas is only soluble at around 1.6 mg per kg of water at 20°C. The exact mechanism by which hydrogen-rich water has medicinal properties I do not know, and that would probably be best answered on Biology.SE if you're curious. It is conceivable that the dissolved hydrogen gas could act as an antioxidant, though relying on the distribution of a gas across the body seems a bit strange (there are gaseous substances known to be important for our metabolism, such as $\ce{NO}$). I wouldn't rule out the possibility of it being nonsense before reading into it a lot more, though.

Great info above, I'm glad you took time to look into H2! The initial ? here was basically the difference between Alkaline water & H2 Water.
Alkalinity can be easily manipulated by the addition of alkaline minerals although this does not increase the molecular Hydrogen content. Much research on Hydrogen gas as a powerful antioxidant and its endogenous antioxidant enzyme enhancing mechanism (see chart #4 in the 2nd study posted below) exists today. Hydrogen is unique in that it has the rare ability to make it past the BBB / Blood Brain Barrier making especially useful for Neuro-degenerative & neurodevelopmental conditions.
Magnesium sticks can create H2 water in small amounts. Most common today is the process of ionization or Electrolysis more commonly recognized in scientific terms as ERW / Electrolyzed-Reduced Water made by quality ionizers. Some call it Kangen Water, a trade name given to the ionized water by one particular Japanese manufacturer whose machines are recognized as home use medical devices.  Here are just 2 of such studies of which the 2nd study continues on to explain the potential of molecular H2 in various prominent disease conditions.
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3285010/ and http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3660246/ and because of the great importance this ERW has had in my child's life as it greatly reduced her seizure activity, Autism & related issues which is why I study it.


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Electrolysis Anode equation
In the electrolysis of $\ce{Na2SO4}$, is it valid to write the anode half-reaction (oxidation) as
$$
\ce{OH-} \rightarrow \ce{H+} + \frac{1}{2}\ce{O2} + 2\ce{e-}
$$?
My book says it's
$$
\ce{H2O} \rightarrow 2\ce{H+} + \frac{1}{2}\ce{O2} + 2\ce{e-}.
$$
Other than the additional $\ce{H+}$, there's no difference between the two.

This depends on the medium in which you do the electrolysis. In the case of a basic medium your equation would be more correct, but in the case of an acidic medium your answer would be wrong, since the concentration of $\ce{OH-}$ ions would be far too small.
Generally: In acidic medium you balance your redox equations using $\ce{H+}$ and $\ce{H2O}$, in basic medium you balance with $\ce{OH-}$ and $\ce{H2O}$.
Note: Since you generate $\ce{H+}$ ions it is possible to argue that starting with a solution at $\text{pH}=7$ gives you an acidic medium, resulting in the equation in the book.

The book does not define the medium. It only asks for the half reaction(s).

You generate a base from an acid; while this is not necessarily forbidden, there's a certain ugliness to it in this case, since we're talking about water and proton exchange reactions are very fast. Maybe this lead the authors of the book to that equation.

"[...]Base from an acid": Are you referring to OH- -> H+? Would you not mean it the other way around? Ok, so you're saying the ugliness is to write it as if H+ is generated from OH-, right?

Of course, the other way around :) And yes, that's exactly what I meant with that comment.


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Best consumption-safe liquid for caffeine solubility?
The question is simple:  What human-consumption safe liquid has the best caffeine solubility?
It's 200mg/10mL of water.  That's a lot of water for one dose!
I found some ramblings on a fourm stating it's "low" solubility in propylene glycol and glycerine (glycol) but no sources or exact numbers, so that doesn't mean much.
In boiling water, it seems much higher, 66g/100mL, which is more of the range I am hoping.  So what happens if I boil a liter of water, dump 100g into it, and then let it cool?
I am hoping for something in the range of 200mg in 1mL-3mL, but as low as possible, and relatively easy to acquire.
All help very much appreciated.

Why do you think 200mg/10mL water is a lot of water for one dose?  Also what do you mean by consumption-safe? Water is the only liquid that is safe to take for a large quantity.  Glycol is moderately toxic.   The safest liquid besides water is likely alcohol, but you do not want to take it with caffeine.

It probably goes without saying, but please be cautious when ingesting anything that you've created in a lab.

Thanks for the tip with glycol.  I was just hoping to get it down to 1-3mL so it could fit in a smaller dropper bottle.  It's easier to carry 10mL around than 30mL.

I hope you are also aware that caffeine itself is moderately toxic. It's not usually a problem in caffeinated beverages because it is present in rather small amounts, but it's all too easy to ingest a couple of grams of the pure substance. Your body also may develop some resistance to the effects of caffeine after some time, and increasing the dose to compensate may lead you into unsafe consumption territory.

For reference sake the LD50 of caffeine is estimated at ~150 mg/kg. Since there's a slow process of removing caffeine from your system, this can also be exceeded by consuming caffeine through the day. Reference

I tried this a years or so ago and found that the caffeine went into solution at 66g/100mL quite readily in boiling water but when it cooled to room temperature it came out of solution and solidified. I wish I had known at the time that a little acidic acid would have made a difference in maintaining a saturated solution. I may have to get out the old test tubes and give it another go. It would be interesting to know what the threshold of a saturated solution is using a little acid as part of the solvent.

Could always go with decaffeinated coffee as the solvent. ;-)

Safety Warning
Caffeine is a toxic substance. When following the advice below, always keep in mind the factors that will help you kill yourself if you are not careful! For more details you can start out on Wikipedia.

Supersaturation
Your idea of preparing a supersaturated solution is not that bad, but it might be that you have some nucleation sites in the bottle where you want to store it. Furthermore, it is probably not that easy to transfer the supersaturated solution from the container where you prepared it to the vial.
Acidic Medium
I've come across this PDF that mentions something interesting:

Solubility in water is increased by adding dilute acid (e.g. HCl or citric acid).

It would be very easy to prepare an acidic solution (I'd aim at a pH of about 3 or so, in the general vicinity of energy drinks and Coca Cola, beverages known to contain large amounts of caffeine) from some citric acid¹ and dissolve caffeine in there. Maybe you could even conduct a study about the solubility of caffeine at different pH values!
Water/Ethanol Mixture
There is one dissertation (PDF) that touches on the solubility of caffeine in a mixture of water and ethanol. I wouldn't trust it too much, since it was written using MS Word and contains several formatting errors (it says $K_{ow}$ for example, instead of $K_\text{ow}$). The graphic containing the data can be found on page 52, but for your convenience it is reproduced below:

(Seriously, the figure caption is useless, and the plot is so bad I'm tempted to redo it just for this answer.)
Never mind the linear relationship that is indicated by the line, the diamonds are the important part here. As you can see, there seems to be a solubility maximum at around $x=0.5$, where $x$ is the volume fraction of ethanol in the water-ethanol mixture. So that's a whopping 50 vol.-% of ethanol, about equivalent to a cask-strength whisky!
However, it is probably not a good idea to use this knowledge to its fullest, because that will be quite a lot of ethanol, and as already mentioned in the comments it is probably not very wise to combine the consumption of caffeine with ethanol.
Does it have to be dissolved?
There are also caffeine tablets and pills that you can take. They come with the additional guarantee that some FDA-like agency has looked into the production of these products, so you can assume they are safe to ingest. Also, they come with instructions and more than enough warning labels so you can educate yourself further.
Conclusion
If you really want a caffeine solution, it is probably your safest and best bet to create an acidic solution where you can dissolve above-normal (i.e. pH=7) amounts of caffeine.

¹ Though please do not ask here how to calculate the amount you need, that would be considered a homework question and get deleted pretty quickly...


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Why are carbocations with sp3 hybridisation highly unstable?
Why is it that sp3 hybridized carbonium ions do not exist? Or why are they very difficult to form?
For instance, nucleophilic substitution in bicyclo compounds at the bridge head carbon is not very easy. I don't understand the explanation given in the book. Thanks for any help.

When you hybridize towards $sp^3$, you use one $p$ orbital more compared to the $sp^2$ hybridization.
The additional $p$ orbital raises the energy of the 4 $sp^3$ orbitals higher with respect to the energy level of $sp^2$-hybridized orbitals ($p$ orbitals are higher in energy than $s$ orbitals), which in the energy-minimizing world of chemistry is not favoured by the system.
The energy is minimized by hybridizing in a $sp^2$ fashion (in which all the $\sigma$-bonding electrons go) and adopting a planar geometry.


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How do I remove free biotin after biotin-protein conjugation?
I want to quantify a biotin-protein conjugate after chemical biotinylation. After biotinylation, the solution contains both biotinylated protein and free biotin.
I intend to use a HABA-avidin biotin quantitation kit like this one from Pierce. Since the biotinylated protein and the free biotin will both displace HABA and decrease the absorbance, I need to remove the free biotin so that I only quantify the biotinylated protein.
How do I separate the free biotin from the solution?

I did some bioconjugation with NHS and maleimide biotins to proteins, peptides, and oligosaccharides. I removed the free biotin using spin filters or dialysis. Just be sure to use a MWCO well below the mass of your protein. I even used prep scale HPLC to remove biotin from the oligosaccharide. The HABA kit works well, I think I used that same one to quantify peptide - avidin linking. But try to keep your biotin protected from light and oxygen, it oxidizes fairly easily, I was getting +16 and +32 masses on LC-MS and couldn't explain it until I found some 50 year old paper on biotin sulfoxide. The oxidized biotin doesn't bind avidin nearly as well as normal biotin.

You could precipitate the protein with alcohol (or acetone) (of course it will probably denature the protein), centrifuge the mixture and redissolve the protein to determine the amount of biotin bound to the protein.


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Non-stoichometric compounds
I am looking for references (review articles, reports) about non-stoichometric compounds. Any idea of where should I look?

http://en.wikipedia.org/wiki/Non-stoichiometric_compound

Wikipedia articles are "light", I am searching for something more "hard" and professional.

The Wikipedia article at least links to some book references. Maybe that's useful...

There are several resources available for this, for example, searching in Google Scholar yields over 37,000 articles (this is without any filters for age or specifics).
Organisations such as the American Chemical Society have a searchable publications link; related, CAS has a searchable reference list and ChemWorx which is a way to manage your research.  A similar set up is available with the Royal Society of Chemistry.
Individual universities also have searches/tutorials - for example, The University of California, Davis have a 'ChemWiki'.  Additionally, university libraries also are often accessible to the public and have search functions for journal articles, books.  Some universities, such as the University of Connecticut have Research Database Locator.
As mentioned in the comments, Wikipedia, particularly if a page is written well, have several links to journal articles and books etc.


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What's up with SiO₂ being tetrahedral?
If silicon dioxide tends to form a crystal lattice with four $\ce{O}$'s around a central $\ce{Si}$, why isn't the molecular formula $\ce{SiO4}$ then? I'm confused why it's unique in that its molecular formula doesn't match up with its geometry.

The problem you are having with the formula is based on a misunderstanding about how far you can take molecular and structural formulae.
For compounds that form distinct molecules it is often worth writing the molecular formula in a way that helps you understand the structure of the molecule. But this is a convenience not a generalisation that can apply to all possible compounds.
It doesn't apply to $\ce{SiO2}$ because there is no such thing as an silica molecule. Silica, like many other minerals, is a 3D network of bonds with no discrete molecular components. In silica each Silicon is bonded to 4 Oxygens, but each oxygen is shared with two silicons. This gives the $\ce{SiO2}$ formula. This also explains why mineralogy is harder than chemistry.

Every $\boxed{^{14}\textrm{Si}}$ atom is surrounded by 4 $\boxed{^8\textrm{O}}$ atoms.   On the other hand, every $\boxed{^8\textrm{O}}$ atom is surrounded by 2 $\boxed{^{14}\textrm{Si}}$ atoms.
Suppose that the number of $\textrm{Si-O}$ bonds is $n$, then the number of $\boxed{^{14}\textrm{Si}}$ atoms is $n/4$, while the number of $\boxed{^8\textrm{O}}$ atoms is $n/2$.  Therefore you arrive at the formula $\textrm{Si}\textrm{O}_2$.

As a short answer, chemical formulae are only really meant to show the proportions between different atoms in a compound, and in general they do not carry any information regarding the way the atoms are bonded. In quartz, there are simply two oxygen atoms for each silicon atom, hence $\ce{SiO_2}$.

Your answer appears to describe the empirical formula, the simplest whole number ratio of atoms. But not the molecular formula, which confusingly for silica is also $\ce{SiO2}$ according to PubChem. My question is getting at why silica does not follow convention with having a $\ce{SiO4}$ molecular formula since that would be a better description of its tetrahedral structure.
Take butane for example. There are 2 carbons for every 5 hydrogens so its empirical formula is $\ce{C2H5}$. But its molecular formula is $\ce{C4H10}$ since that describes its actual structure.

Structure of one molecule that is. Is there not 4 oxygen atoms in one molecule of silica? If not, then why is it tetrahedral? I was taught that the molecular geometry is derived from the molecular formula, at least that's what we use all the time in class. Are you saying molecular formula can't be used across the board like this? Are there are other exceptions?

@reflexiv First of all, notice that if you accept $\ce{SiO2}$ is the empirical formula, then the "molecular formula" (a misnomer since quartz does not form molecules; it is an network solid) cannot be $\ce{SiO4}$, as the latter is not equal to the former times an integer. The following sentence from the Wikipedia article for quartz may also help: "[Quartz] is made up of a continuous framework of $\ce{SiO4}$ silicon–oxygen tetrahedra, with each oxygen being shared between two tetrahedra, giving an overall formula $\ce{SiO2}$."

Also, I reiterate that molecular formulae do not in general carry structural information. You may know the structure of simple molecules from their formula, such as for $\ce{NH3}$, $\ce{H2SO4}$ or $\ce{IF5}$, but you may be hard-pressed to figure out that by $\ce{C_{14}H_{12}O2}$ I am referring to biphenyl-4-ylacetic acid. Sometimes we write out formulae in such a way that helps us figure out the structure. Phosphorous pentoxide is often written as $\ce{P4O_{10}}$ because the latter contains a hint to the structure of each molecule.

Similarly, though ethanol and dimethylether are funcional isomers and therefore have the same molecular formulae ($\ce{C2H6O}$), we often write $\ce{C2H5OH}$ for ethanol and $\ce{H3COCH3}$ or $\ce{O(CH3)_2}$ for dimethylether. However, this is not the general case.

Because the oxygens are shared with more than one silicon,so in large crystal structure the ratio of silicon to oxygens will be 1:2 not 1:4

As one oxygen is not wholely solely with one silicon atom but it is bonded with two silicon atoms thus contribution of one oxygen is  1/2 towards one SiO2 molecule.. As one silcon is bonded to 4 oxygen atoms so 4(1/2) = 2 thus contribution of oxygen towards silcondioxide is 2 and we have one silicon so formula becomes SiO2.


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Why does the excitation and emission spectrum of a fluorescent molecule have overlap?
Question is rather self-explanatory. Came up during a lecture without a concrete answer.
I understand that the differences in emission wavelengths is due to relaxation to the lowest energy level of S1, but why do fluorescent molecules necessarily overlap in their excitation and emission spectra?

If you are asking why there is no overlapping of lines in emission or excitation spectra, the answer to this question also explains why Rutherford model of atom was rejected. If lines were overlapping, it mean that atoms emitted energy (in the form of radiation) in continuous manner, if this was the case, electrons would collapse into the nuclei giving Thomson model of atom. But this never happens. So, atoms can emit energy in discrete manner, which was explained by Bohr. Thus, there will be no overlap of lines in emission or excitation spectra.

Energy-Level Diagram
There is overlap because between the excitation and the emission spectra because there exists a Range of orbitals at different energies. You'll always notice that the energy for excitation is always higher, or the lambda value is always lower (these say the same thing since E=planck's constant * frequency).

I've attached a HE-Ne laser schematic from Verdeyen's Laser Electronics book. For a better visual. What you need to know is let's say your gain medium, in this case He-Ne was excited at it's lowest wavelength (we'll call it Lambda1) then it's emission wavelength may be only 2nm higher (or red shifted), Lambda2.

However, as you'll notice in this schematic there are many wavelengths at which this gain medium (He-Ne) can emit light. These other possible excitation orbitals can have lower energy levels that correspond to higher wavelengths and thus overlap on the spectrum.

This is most easily seen in solution. The overlap in molecules occurs only when the electronically excited states of a molecule have rather similar geometries and so transitions between vibrational levels in the two states occur and the same wavelength.  After excitation the excited state rapidly looses energy until only the v=0 level is populated. This takes about $10^{-13}$ sec, far faster than fluorescence.
If the ground state and excited state have very different geometries then there is almost no overlap between absorption and emission. The picture tries to explain the mirror image overlap and features ( the picture is not to scale unfortunately as the length of the red and blue dotted lines should be the same for a couple of transitions)

 In an isolated molecule in the gas phase when only one vibrational level is excited then emission comes only from this level to varoius levels in the ground state; only 1 emission transition overlaps with the  absorption line.


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What can cause artesian well water to smell?
Recently, our water from the artesian well started to smell (again). It only smells bad for only few seconds than the smell disappear.  We have sulfur-iron removal system, water softener, UV lamp. Only the water from the system I listed above smell, because we also have a Reverse Osmosis installed in the kitchen and the water filtered from it does NOT smell.
We did a hydrocarbon test that was negative. We also did a normal "drinkable water" analysis and it passed with nothing ususual. My question is what are the possibilities of that smell?
PS. I can confirm the smell is not sulfur (different from rotten egg)
See this post for our water results before installing the system if it may be useful

I suspect the UV lamp is creating small amounts of ozone. It is an unstable, highly reactive strong oxidant, which is used in low concentrations for some water purification procedures as it readily destroys organic matter (such as bacteria). The strong-smelling gas can readily be produced from oxygen by turning on a high intensity UV lamp in air. You should always let rest for a couple of minutes after being collected from an ozonizer, to let the residual ozone escape. Once the smell disappears, it should be okay to drink.

As mentioned in the comments, water is often treated by ozonation in order to sterilize it and ensure it can be drank with no ill effects. Ozone is a very reactive gaseous allotrope of oxygen, produced in situ by exposure of atmospheric oxygen to UV-B light (in fact this happens naturally in the stratosphere, generating a thin ozone blanket covering the world). Ozone can be produced artificially by using a mercury lamp, which emits large amounts of UV light. The water is then brought into contact with the ozone in order to be purified.
When water is collected after being ozonated, typically a small amount of ozone is still dissolved in the water. If the water is allowed to rest for a few minutes after being collected, the residual dissolved ozone diffuses out. Ozone has a very strong, characteristically sharp odour, which can be detected even at very low concentrations. This is likely why the asker's water had an odd smell. Ozone is toxic, though, so exposure should be avoided.

In my opinion, ozone is highly reactive in water forming radicals (like hydroperoxyl which is a precursor to hydrogen peroxide, and other reactive oxygen species, that should reduce odors). Here is a mechanism per a source:

The first reaction that takes place is accelerated ozone decomposition by a type of initiator. This can be an OH-molecule, see reaction 1:
1: $\ce{O3 + OH- -> •O2- + HO2• }$
This radical has an acid/ base equilibrium of pKa = 4,8. Above this value, this radical no longer splits, because it forms a superoxide radical, see reaction 2:
2: $\ce{HO2• -> •O2- + H+ }$ (pKa = 4,8)
Radical chain-reaction
Now, a radical chain-reaction takes place, during which OH-radicals are formed. The reaction mechanism is as follows:
3: $\ce{O3 + •O2- -> •O3- + O2}$
4: $\ce{•O3- + H+ -> HO3• }$ (PH < ≈ 8)
The OH-radicals that have formed react with ozone according to the following reaction mechanism:
5: $\ce{OH• + O3 -> HO4• }$
6: $\ce{HO4• -> O2 + HO2• }$
During the last reaction, HO2• radicals are formed, which can start the reaction all over again (see reaction 2). As a result, a chain-reaction develops, which is maintained by so-called promotors.

My guess on what is actually occurring is much more subtle. You now have less of something (as a consequence of your water processing) that is instrumental in the transient formation of an H2S smell. The explanation is perhaps reduced nitrate concentration.
Per this source, 'Controlled dosing of nitrate for prevention of H2S in a sewer network and the effects on the subsequent treatment processes', you may now have had a reduced nitrate concentration due to current water processing. The consequence is sporadic regeneration of the sewer smell.
A partial quote from the cited source:

A process for the elimination of septicity based on controlled addition of nitrate to sewer networks is developed...The results showed that dosing of nitrate was very effective for suppression of hydrogen sulphide in a rising main. During nitrate dosing the average level of hydrogen sulphide at the works inlet was 0.24 mg/l, with a daily variation within the range of 0–0.4 mg/l. Without nitrate addition the average background level of hydrogen sulphide was 4.2 mg/l, with a daily variation of 1–10 mg/l. The dosing was accomplished without significant breakthrough of nitrate to the treatment works. Dosing of nitrate resulted in an increased removal of soluble BOD across the rising main, but no significant change in total BOD or COD was registered at the treatment plant during the nitrate dosing. The nitrification process in the biological filters at the treatment plant was improved during the dosing period.


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Difference between reversible and irreversible thermodynamic process
I have been reading up (on this website) about the difference between reversible and irreversible thermodynamic process and it defines them saying that reversible processes are ideal processes while irreversible processes are the natural processes. Are these definitions correct?
If not how are they defined?

The simplest answer in thermodynamic terms is "a reversible process is one in which the entropy generation term is zero."

Yes, I understood that but can you explain what difference is there in the change in enthalpy, internal energy, heat, work, etc. if the process is reversible and irreversible.

For any process, all of these would be the same whether it was reversible or not. The only one that would differ would be change in entropy.

Sorry to dig up something this old but this is wrong. The heat transferred and work done will typically be different in a reversible and an irreversible process. U, H, S are state functions, so as long as the initial and final states are kept constant, $\Delta U, \Delta H$ and $\Delta S$ are independent of the reversibility of the process.

@orthocresol - you are right in that heat and work will probably be different for an irreversible process compared to a reversible one. You are also right that the $\Delta$'s of the state functions will be the same if the starting and ending states are the same. My point was that the only difference we know for sure is that the entropy generation term will be zero - that's different from $\Delta S$. I was assuming that op is talking about terms in a thermodynamic equation, and you wouldn't change anything other than $dS_{gen}$ if the process is reversible.

See my recent article on this very subject at the following link:
https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/
The conclusion is that, mechanically at least, the difference between reversible and irreversible processes is the result of viscous dissipation of mechanical energy in the irreversible case.
Chet

These are not definitions of reversible and irreversible processes, but it is a description of one of the differences between the processes.
In real world situations, however, for some processes, the system undergoing the changes responds much faster than the applied change, the deviation from reversibility may be negligible. In a reversible cycle, the system and its surroundings will be exactly the same after each cycle. (see here)

A (...) definition of a reversible process is a process that, after it has taken place, can be reversed and causes no change in either the system or its surroundings. In thermodynamic terms, a process "taking place" would refer to its transition from its initial state to its final state. again, wikipedia

Yes, I understood that but can you explain what difference is there in the change in enthalpy, internal energy, heat, work, etc. if the process is reversible and irreversible.


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Mass spectrometry: trouble with acetic acid fragmentation
I'm trying to do compound optimization for acetic acid on a EI-MS.
Later on it would be also run via HPLC to the MS.
But I do only get fragments of 53/54 and 32.
Also I need a high concentration to be able to do a compound optimization for the 32 fragment (...around 1mM solution).
I looked up other mass spectra of acetic acid online.... and they got nice fragments around 43/45 and 29
...what am I doing wrong?
Edit: for clarification of the question

We work with ESI, also tried it with APCI
Acetic acid as a liquid, with mobile phase of water and formic acid
yes, it is acetic acid
in the meantime we did a blank (thanks for the suggestion!) and we also got the 32 fragment... but non at 60/61 (for positiv mode)
so it should actually work, if we first make Q1 for 61 and then look for fragments
we re-run a fresh dilution with UV-water ... and got some nicer fragments... but unstable turbo spray signal.

Are you using a solvent? If so, what is it?

I'd expect fragments with m=60 (M), 45 (M-15) and 43 (M-17) - just like you ;) For m=54, i can imagine C4H6 from the retro-Diels-Alder reaction of cyclohexene.

I ran this question past our mass spec. guy, and he offers the following suggestions:

Have you run a blank to ensure you are not just measuring some other random compound; Or contaminants from the instrument; Or gas system.
Be aware when measuring ions less than 45m/z that you are not just measuring ions from air leaking in. For example: 18/19 is H2O; 28 is N2; 32 is O2; 40 is Ar; 44 is CO2. There are probably other ions to watch as well.
Is the ionization energy set to 70eV and it is measured in positive ion mode as that is what most online libraries use
Is too much sample being injected in the instrument as you do measure different fragment patterns when a lot of sample is injected as the molecular ion may be a charge bound cluster (similar to what is seen in ESI).
Is the acetic acid reacting with anything in the sample introduction path which is giving another compound? If it is a water injection into a GCMS, not only are you destroying your column but the solvent will expand in the injection port and goes back into your gas system as water has a very high expansion when going from liquid to gas.
Are you sure, really, really sure you are using acetic acid and not something else?


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Critical separation at which molecule breaks
I have the following question.

The potential energy of two atoms, a distance $r$ apart, is: $$U = -Ar^{-2} + Br^{-10}$$ Given that the atoms form a stable molecule at a separation of $\pu{0.3 nm}$ with an energy of $\pu{-4 eV}$, calculate $A$ and $B$. Also find the force required to break the molecule, and the critical separation at which the molecule breaks.

Molecule is stable at the distance when the first derivative is zero. But how to find the critical separation at which the molecule breaks?

Please don't deface the question, it could be useful to someone in the future

Have you found the solution to your own question? If so, we'd appreciate if you could write up an answer.

... As far as I know, in the absence of an external influence, the molecule never actually breaks. I mean, most people wouldn't call two atoms a metre apart a molecule as such, but there's no negative curvature stationary point in a potential like this at which you could say "if $r > x$, this molecule is broken and will not reform".

The potential energy curve is rather unusual though, so maybe it was created to "fit the problem" rather than to fit reality.

@NicolauSakerNeto: Well... It's kind of unusual, but it's a lot like a Lennard-Jones potential in effect.

I thought that maybe changing the powers would have somehow changed the result even though I couldn't see it, but after graphing it in Wolfram Alpha, evidently not.

But there is always an external influence. Thermal energy from the environment matters as other molecules will randomly buffet the two atoms in question meaning that, in reality, the atoms will separate.

@matt_black: While this is true, in the absence of stated external conditions the statistics of that process are undefined.

I start from the expression $$\ce{U = - Ar^{-2} + Br^{-10}} ............. (1)$$ If this expression ($1$) is derived with respect to r, it gives ($2$) $$\ce{0 = -A(-2)·r^{-3} + B(-10)·r^{-11}} ... ........(2) $$ And this can be rearranged to give ($3$) :  $$\ce{A = 5B·r^{-8}} ... ........(3) $$ If A from ($3$) is introduced into ($1$), the following expression is obtained :
$$\ce{U = - \frac{5 B r^{-8}}{r^{2}} + B r^{-10} = - 4B·(r)^{-10}} ............ (4)$$
Now, introducing U = $4$ eV, and r = $0.3$ nm, the following expression ($5$) is obtained : $$\ce{4 eV = - \frac{5 B (0.3 nm)^{-8}}{0.3 (nm)^{-2}} + B (0.3nm)^{-10} = - 4B·(0.3 nm)^{-10}} ............ (5)$$ This can be simplified and yield the final value of B in eV ($6$) :  $$\ce{B = (0.3 nm)^{10} ...}..........(6) $$ Then from ($3$) and ($6$) the final A value in eV is ($7$) : $$\ce{A = 5 B·r^{-8} = 5 (0.3 nm)^{10}·(0.3 nm)^{-8} = 5(0.3 nm)^{2}}.........(7)$$ I have the strange impression that this solution is too simple.....

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


text:
Mole fraction of solute of a dilute solution
According to our book, when the solution is dilute, the mole fraction of solute can be given as:
X2 = molality (m) x mass of solvent (M1) / 1000
Actually it is written as:
X2 = m/(1000/M1)
How is this true?
The only reason i can think of is that as the solution is dilute, X2 should almost be equal to zero. As the solution is dilute so, molality is very low and it is also divided by 1000 which will give us a value near to zero (for X2). But i know its a poor explanation. So, how does this equation work?

The formula is correct, being M1 the molar mass of the solvent. I'll call it below $ M_{solvent} $.
Starting from the definition of molar fraction:
$ X = \frac{mol_{solute}}{mol_{solvent}+mol_{solute}} $
For diluted solutions, we can simply write:
$ X = \frac{mol_{solute}}{mol_{solvent}} $
To introduce molality into the defintion of molar fraction, one has to consider the following:
$ mol_{solvent} = \frac{g_{solvent}}{g_{solvent}/mol_{solvent}}=\frac{g_{solvent}}{M_{solvent}} = \frac{kg_{solvent}}{M_{solvent}}1000$
Replacing the latter into the definition of molar fraction, you get:
$ X = \frac{mol_{solute}M_{solvent}}{kg_{solvent}1000} $
Considering that $mol_{solute}/kg_{solvent}$ is the molality $m$, you get your formula:
$X= \frac{mM_{solvent}}{1000} $

to convert g(solvent) into KG(solvent), you have multiplied g(solvent) by 1000. shouldn't it be divided?

I haven't done any conversion: in the definition of mole, mass must be expressed in grams. The use of kg is further imposed by the definition of molality. So, if you have $mol_{water}$=1 that means that you have 18 g of water or 0.018 kg * 1000 g/kg (notice that this last product still gives you grams).


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Melting points and boiling points of primary alcohols do not follow the same trend
If one considers boiling points (in °C) of primary alcohols, one finds the following:

methanol: 65
ethanol: 79
1-propanol: 97
1-butanol: 117
1-pentanol: 138

This trend is due to Van der Waals forces increasing with molecular weight.
Now if one focuses on melting points, I would expect the same trend, but these are the experimental values (in °C):

methanol: −98
ethanol: −117
1-propanol: −127
1-butanol: −90
1-pentanol: −79

the lowest value being the melting point of 1-propanol.
The general idea stating that with all else being equal, more massive molecules have higher boiling and melting points, does not fully apply for this set of alcohols.
So my question: why melting and boiling points of primary alcohols do not follow the same trend?

Melting points are a bit trickier to compare than boiling points, especially when you're looking at the lightest examples of a group of molecules. This graph for the unbranched alkanes exemplifies nicely how the trends differ in smoothness.
As you increase the weight of molecules, the observed trend depends on how you increase the weight. For example, if you start with methanol and increase the weight by adding more $\ce{-HCOH -}$ units (methanol, ethylene glycol, glycerol, and so on) you would find a very clear increase in both melting and boiling points. In some sense this is a more legitimate trend to analyse, because the relative amounts of different types of intermolecular interactions stays approximately constant (in particular, all molecules have one hydroxyl per carbon atom, and hence they all have about the same amount of hydrogen bonding per atom in the molecule). In your sequence of alcohols, the strong hydrogen bonding allowed by the hydroxyl group becomes "diluted" as the molecule gains an ever larger alkyl chain, which can only sustain much weaker van der Waals interactions. In the limit of alcohols with very large alkyl chains, the alcohols' melting and boiling points do not differ tremendously from their parent saturated hydrocarbons. Compare the melting and boiling points of 1-hexadecanol (49°C and 344°C) and hexadecane (18°C and 287°C).
Also, the temperature at which a substance freezes is dependent not only on the strength of intermolecular interactions, but how the solid packs. Even substances with strong interactions may only freeze at very low temperatures if they pack very poorly when solidified. Extreme examples to which this effect contributes are ionic liquids. They tend to have a very low vapour pressure at reasonable temperatures (as you would expect of many salts), but some melt even below 0°C. The freezing of such bulky entities would entail lots of ordering and therefore a very large decrease in entropy, so it is disfavoured (or equivalently, their melting is highly entropically favoured, so the free energy of the liquid phase becomes lower than the free energy of the solid phase even at relatively low temperatures). The exact geometry of the molecules in the solid is also important, as some favourable interactions may be suppressed in the solid state due to steric hindrance which may be less present in the more freely-moving molecules of a liquid. Furthermore, molecules with a high amount of symmetry tend to have high melting points, because freezing is entropically favoured due to molecules more readily fitting in the solid lattice.
Yet another effect to consider is how large a surface area each molecule has available for intermolecular interactions. This is often brought up to explain differences between melting and boiling points of branched and unbranched organic compounds; when comparing structural isomers, branched compounds have a lower available area for interaction than unbranched compounds, so the latter tend to have higher melting and boiling points (branching may sometimes create a combination of opposing packing and surface area effects, though).
So why does your sequence of alcohols behave the way it does? The fact that the trend in boiling point is monotonic while the melting point is not suggests that even though there is a relative shift in importance of types of intermolecular interaction as the alkyl chain increase (from hydrogen bonding to van der Waals interactions), it is probably not the source of the melting point drop. Hence it likely has to do with geometric factors in the solid. The alcohol's alkyl chain probably disrupts the hydrogen bonding network of the solid as it grows from methyl to propyl in a very severe fashion, by making the solid pack less well or by partially hindering the number or strength of hydrogen bonds in the geometry of the solid. Perhaps looking at the crystalline structures of the solid alcohols may provide further insight.
Edit: I wrote a wrong answer initially. I've updated it, but now I realize I'm still not quite sure it's correct. I originally meant to write just a comment, and as I kept writing more I forgot that my very first thought was actually incorrect, so the whole text wasn't built as well as it should be. Anyone is free to pick apart or repurpose what I wrote!
Edit 2: User Uncle Al posted an interesting list of compounds and their melting/boiling points, showing the importance of molecular symmetry and solid packing.


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Determine the position of an equilibrium process

For the exothermic reaction of nitrogen gas with iodine solid to produce nitrogen triodide gas. Write the balanced equation including heat and select the overall direction of shift in the reaction.

I have balanced the equation: $\ce{N2 (g) + 3 I2 (g) <=> 2 NI3}$
Is this equation correct?
I am confused on how to find the overall direction of shift. Should I use the Le Chatelier's principle? If so, then I think the overall shift is a "shift to the left"/reverse reaction. Am I correct?

You are right, you can use Le Chatelier's principle to predict the shift of the equilibrium when temperature or pressure is in- or decreased. It is stated that the reaction is exothermic, and this tells you on which side of the reaction equation the heat goes. And as both products and educts include gases, it would also be easy to predict how the equilibrium shifts when pressure is varied. I presume that the question, in its current form, is asking for the shifts that happen in all possible combinations of in- or decreased pressure or temperature.


text:
Does minimizing ANY type of energy ALWAYS predict a state of equilibrium?
My intuition says no, but I am having trouble coming up with concrete examples.
I know that minimizing Gibbs free energy predicts a state of equilibrium, while minimizing kinetic or internal energy does not correspond to equilibrium. However, I am not so sure about potential energy--- when PE is at a minimum, is the system also at equilibrium?

If you write a wavefunction of H$_2^+$ molecule concentrating at nuclei, e.g. $\delta(r-R)$, the potential energy is minimized. By the uncertainty principle,  the kinetic energy is quite large. This wavefunction neither corresponds to the ground state of H$_2^+$ nor the ground state/equilibrium of an ensemble H$_2^+$ molecules.

If you minimize Gibbs free energy, than you have equilibrium at constant pressure and temperature, etc... so the type of energy you choose corresponds to given ensemble. Roughly speaking.

No. None of the thermodynamic potentials is at its minimum at the equilibrium point of ANY system. This includes G.

When you find a minimum in the thermodynamic potential energy of a system, you will have found an equilibrium point.
The key is the thermodynamic potential part. Depending on the constraints of the system (i.e. which thermodynamic state variables are held constant) the thermodynamic potential energy function will be different.
For example, in a system of a fixed number of particles at constant temperature and pressure (we would call it NPT), the Gibbs free energy is the thermodynamic potential, and it will be minimized at equilibrium.
For NVT (constant volume and temperature) the Helmholtz free energy is the thermodynamic potential, and that is what will be minimized at equilibrium.
In a system consisting of a rock dropped from a height above the ground, gravitational potential energy is the thermodynamic potential (I am abusing the language a little bit here) and it will be minimized at equilibrium (when the rock hits the ground).


text:
What is pre-concentration (context: Breath analysis)?
What is the meaning of pre-concentration both in general and specific to this context of
a) "In addition to consistent sampling protocols,.. pre-concentration, and analysis of breath samples require standard methodology, calibration standards..."
b) "Offline breath analysis involves some form of pre-concentration of analytes followed by a separation step using high-resolution GC-MS based detection"

Pre-concentration just refers to the process of concentrating a sample before analysis, so that trace components won't be overlooked. If you're looking for something that's only present at a few parts-per-million, you could easily miss it if you used the raw sample, so ideally you want a way to separate off either large, common sources of noise/error, or to just extract the exact components you're looking for.
A quick look suggests, for example, that a key first component to remove in breath analysis is moisture, as some volatile components will dissolve readily in the water and make it harder to detect them.
Anyway, this paper on a few pre-concentration and detection techniques might be informative: Mochalski P, Wzorek B, Sliwka I, Amann A; Chromatogr B Analyt Technol Biomed Life Sci. 2009 Jul 1;877(20-21):1856-66.

Preconcentration is the process by which the concentration of an intended species is increased to gain sensitivity for the subsequent determination.


text:
What's the mechanism for this isotopic substitution reaction?
I'm curious to know what's the mechanism of:
$$\ce{HF + D^+ \rightarrow DF + H^+}$$

Maybe you should include your own thoughts on this?

@tschoppi. I have an idea (probably stupid) but I would prefer not to contaminate the discussion with it.

Well, you got my answer. I'd be curious to hear about your approach.

The use of a double arrow is appropriate in this case.

To answer this question you should understand the kinetic isotope effect.
Since you deliver no real data about reaction rate constants I assume that the dissociation of $\ce{HF}$ is faster (has a larger $k_\text{dis}$) than the dissociation of $\ce{DF}$. I postulate the following mechanism:
$\ce{HF + D+ <=>[k_\text{dis}^\ce{HF}] H+ + F- + D+ <=>[][k_{\text{dis}}^\ce{DF}] DF + H+}$
Essentially it is a pretty simple reaction mechanism, but it tends to forming $\ce{DF}$ over $\ce{HF}$ because the deuterium dissociates more slowly.

I was thinking about an intermediate (HFD)+ based on some quantum mechanical calculations I performed.

Hm, given the high strength of the HF bond that is also entirely possible. Depending on the medium, of course.

@ClaudeLeibovici That is probably a good description for a gas phase reaction. However, when inspecting the reaction in a condensed phase, calculations must be done much more carefully as bare hydrogen nuclei are extremely polarizing and will drag along one or more solvating entities with them.

Actually, $\ce{F-}$ isn't that stable, and one would expect a concerted mechanism.

You know this happens in water all the time...what do you mean by mechanism? this is just proton jumping or hopping. The kinetics are are a side issue.


text:
Chemical bond-covalent-valence electrons OH-
I would like to ask how many valence electrons took part in creating chemical bond in $\ce{OH-}$? Is the minus related to $\ce{O}$ or to $\ce{H}$? Are there any spare valence electrons which didnt take part in creating chemical bond in $\ce{OH-}$?

In the photo the orange electrons are non bond electrons.
And the black electrons are bond electrons.
There are 2 electrons took part in creating chemical bond in $\ce{OH}$.
The first one is from oxygen.
The second is from hydrogen.
This kind of bond is called "Covalent bond".
For more about Covalent bond :
https://en.wikipedia.org/wiki/Covalent_bond
The minus is  related to  the black and white electrons that are free (in Oxygen).
The white electron is come from other element.


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Is the proper formula for phosphorus oxide(V) P4O10 or P2O5?
I would like to ask about formula of phosphorus oxide(V). Should it be $\ce{P4O10}$ or $\ce{P2O5}$? Which is right when I have to write the molecular formula?

Which do you think is correct and why?  Please add more details as to what you have attempted in doing the problem already.

I have no idea. I tried to check it somewhere online and I found that P2O5 is empirical and more suitable for everything is P4O10 but still not sure enough.

Note that empirical formula and molecular formula are different.  Ambiguous use of "formula", which can mean two different things, causes your confusion.  P2O5 is the empirical formula and P4O10 is the molecular formula.

The simplest types of chemical formulas are called empirical formulas, which use only letters and numbers indicating atomic proportional ratios (the numerical proportions of atoms of one type to those of other types). Molecular formulas indicate the simple numbers of each type of atom in a molecule of a molecular substance, and are thus sometimes the same as empirical formulas (for molecules that only have one atom of a particular type), and at other times require larger numbers than do empirical formulas. (Wikipedia)

For phosphorus pentoxide the molecule seems to be an adamantane-like compound with 4 phosphorus atoms.
So the correct molecular formula of the compound would be $\ce{P4O10}$.