text:
How to choose a proper chromatography method to separate substrates?
I'm a freshman and I recently learned how to do TLC and column chromatography. I also know a little about other ways to separate things by chromatography, such as affinity chromatography, gel chromatography, ion exchange chromatography, and gas chromatography, but without many details.
I am facing problems in solving a question requiring us to judge how to separate these combinations of substances by chromatography (if there was no such confinement, this will be much easier):

N-phenylethanamide, N-phenylpropionamide
methoxymethane, ethoxyethane
Aβ36，Aβ42 (Aβ is a kind of peptide which is related to Alzheimer's disease, while the number, 36 and 42, refer to the amount of amino acids polymerized in the peptide)

Here are the specific issues I've met：

I've no idea how to deal with the first combination since their difference seems to not be significant, no matter in polarity, acidity (basicity), or molecular size. Is there any other chromatographic way, or did I get something wrong?
I think I can deal with the second combination by gas chromatography, since they are both volatile. But I'm not sure if it is a good idea. Is there a better way or not?
I've found some description for the separation of different Aβ, suggesting that the following ways are usually used: use antibodies (e.g. ELISA) or electrophoretic separation (e.g. PAGE). But seems that neither ways are related to chromatography. It is also said that reverse phase liquid chromatography after CNBr cleavage works, but this way is complex and time consuming. I'm wondering (a) if I can use antibodies or something else to deal with it by affinity chromatogrphy and (b) how CNBr works? (Now I know CNBr is used to cut proteins/peptides at where methionines are, but still not knowing how reverse phase liquid chromatography works.) Also I want to ask if there is a better way as well.

As you already mentioned, there are numerous types of chromatography, and they vary in several different ways. I can't tell you what would be best ways to separate each of these pair of substances, but I can give you some insights. Using chromatography always requires optimization of experimental procedures, and the first step is usually searching on papers and other references for information on how the specific separation is usually done.
N-phenylethanamide/N-phenylpropionamide and methoxymethane/ethoxyethane
Despite being very similar, the presence of one or two extra $\ce{CH2}$ group in each of the second molecules is enough to give the molecules different properties, specially considering they aren't big. For simple detection purposes, I'm sure it would be possible to find the appropriate solvent mixture to separate them using TLC or column chromatography.
A much more advanced technique, gas chromatrography is very powerful and could also be used to separate them, as long as the appropriate setup was used (injection and oven temperatures w or w/o gradients, column type and length, etc).
Aβ36/Aβ42
These would probably decompose in elevated temperatures, so gas chromatography is not a good idea. Peptides are, nowadays, mostly separated using liquid chromatography, usually in HPLC or UPLC systems using reversed phase columns. This type of chromatography will separate the peptides mostly by their polarity and size, and once again, reference search and experimental trials are essential to optimize conditions. HPLC and UPLC are used to separate mixtures of thousands of peptides (with varying efficiency, of course), but that should be a sign that two wouldn't be a problem, even if very similar.
Depending on the objective of the separation (identification and/or recovery), using CNBr or any other digestion step could be useful or not, but I don't think this would be the case for only two peptides.
PAGE and the variations of electrophoresis were for a long time the goto separation method for peptides and proteins, but recent developments in chromatography (and mass spectrometry) have made them fall off preference.
I'm not familiar with affinity chromatography (or the whole antibody thing, to be honest), so I can't say much about that, but, what I do know is that you won't always find or be able to manufacture the chromatographic column with the desired specificity.


text:
Tandem mass spectrometry: a1 and immonium ions
In tandem mass spectrometry of proteins one will usually end up with a mix of $b$ and $y$ fragments but also $a$ fragments and immonium fragments. My problem is that wouldn't an $a_1$ ion be the exact same as an immonium ion of the first amino acid in the peptide?

Some additional sites that have good info on $a$, $b$, $c$, $x$, $y$, and $z$ ions: http://www.matrixscience.com/help/fragmentation_help.html and http://www.sepscience.com/Information/Archive/MS-Solutions/244-/MS-Solutions-9-Peptide-Sequencing-with-Electrospray-LCMS-Part-1-Ion-Types-and-Nomenclature, if you haven't already seen them.

I don't know if you are still interested in this question. But yes, the a1 ion would have exact the same m/z as the immonium ion of all amino acids of the same type in the pepitde. Peptides typically fragment in low energy fragmentation methods such as CID by charg-directed mechanisms, which mainly produces y and b series ions. These ions can fragment further to lower b ions, a ions, or internal fragments and immonium ions. y and b ions are much easier to understand and we therefore mostly use these ions for interpretation of mass spectra.

source: Stack Exchange

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

sort: votes


text:
Autodock Vina flexible metals and water molecules
I have some problems with water and metals in binding site when using Autodock Vina.
Is it possible to make these flexible? They unfortunately remain where they are what leads to wrong interactions.
My other idea would be to delete it before docking, reinsert them and minimize everything.
Does anyone have a solution for metalloenzymes? (e.g. PDB:3NJY)

For the water molecules, you can try hydrated ligand docking in which they are added in the ligand file.
The documentation here: http://autodock.scripps.edu/resources/autodock-hydrated-docking.
I am not sure if the same approach can be used for the metal atom.

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

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


text:
Is there any acid or base which has a pH value under 0 and over 14?
what is the range of strong acid and base in pH scale.I have much confusion.

Please be more specific and do some research next time.

The known pH range goes from -28 to +22.

The solution with lowest pH value (to the left of pH scale) will be the strongest acid while the solution with highest pH value (to the right of pH scale) will be the strongest base.
In technical words, the acid which dissociates completely into H+ ions in water will have the lowest pH value, while the base dissociates completely into OH- ions in water will have the highest pH value.
EDIT : Strongest acid has pH of 1 while strongest base has pH of 14.
(0 - 6.99 are acids. 7 is neutral. 7.1 - 14 are bases.)
I hope that helps.

It can actually go between -3 and 16. But for most acids and bases you deal with, they will be in the range of 1-14. There are different variables you can adjust, ie temperature, etc, but it's no longer a useful measure.

@Jun-GooKwak They can go beyond the limits of pH scale?

@hardik See http://pubs.acs.org/doi/pdf/10.1021/ed083p1465

@Jun-GooKwak Wow, thanks for adding that to my knowledge. :)

Pure sulfuric acid has pH -12.

Usually the range of a $pH$ scale is $0-14$. A higher number on the $pH$ scale means that the substance is Basic, a lower number means it is Acidic.
Another thought to keep in mind is that there can be a $pH$ less than $0$ and greater than $14$. But let's not get into that right now. The pH scale was made for our convenience and cannot be a totally accurate representation of the real world.
Have you done some research on this? Be more specific in what you do not understand.
Possible duplicates:

What is the difference between the titration of a strong acid with a strong base and that of the titration of a weak acid with a strong base?
Why do strong acids have weak conjugate bases?
What is a strong base?

Check this out for more information.


text:
Merge rigid parts and flexible results after docking (Vina)
If I use flexible residues in the binding pocket, they are cut from the file.
The protein is broken and the amino acids are unbound in the room.
How can I get my results and the protein together again?

A Useful Script for AutoDock4 and Vina That Merges Flexible and Rigid PDBQT Output
« Prosciens.
http://prosciens.com/prosciens/a-useful-script-for-autodock4-and-vina-that-
merges-flexible-and-rigid-pdbqt-output/
Link to the script: http://prosciens.com/prosciens/oldproscienssarl/files/flexrigidpdbqt2pdb_template.sh


text:
Pyrolysis of canola oil
This is a topic that I cannot find much information about. I would like to know what compounds are in the products formed from pyrolysis of canola oil. I found out by accident that when trying to ignite some used canola oil in an aluminum can without a cover, the oil had become a mix of black solids and black liquid. I have not tried heating the oil in a container that has no exposure to oxygen. I could probably get better results by limiting all exposure to oxygen. I recently tried this experiment again and the oil is slowly turning black and looking like crude oil. I did find one article on this subject that says there are alkenes in the oil. Does anyone have more information on this?

Basically canola oil is obtained from a cultivated form of Rapeseed. Canola oil is just what rapeseed oil is called in North America.  From a marketing perspective it was thought that having the term "rape" in the product name would not help sales.
If you look for articles on rapeseed oil pyrolysis you'll find a lot of information since rapeseed (canola) oil is used in biodiesel production.
This article states that the major products produced from pyrolysis in a reactor include

gases: $\ce{CO2, CO, CH4}$ and $\ce{H2S}$
oils: oleic acid, 1H-indole, 2,3,5-trimethoxy toluene, toluene, (Z)-9-octadecanamide, psoralene, phenol and phenol derivatives

I didn't know what "psoralene" was, so here is a picture and a link


text:
Derivation spectrum
My book doesn't seem to properly explain the importance of derivating the Gaus spectral line, neither the meaning of it, so my questions are :
1.What do the spectral lines of 1st, 2nd, 3rd and 4th derivates of the basic Gaus specter tell us exactly ?
2.Why are the second and the fourth derivated spectral lines more used in practice ?
3.How does the amplitude of those spectral lines indicate in our results ?
Sorry for the lack of image.

Derivative spectroscopy you refer to is a senior technique deployed often in UV-Vis spectroscopy, yet there are examples in the field of infra-red and RAMAN spectroscopy and cyclic dichorism.

The theme in common is that at the resolution of the typical
experiment, while changing the wavelength of observation the
measured signal alters gradually and continuously and you observe
for example absorption bands -- and not, like in atom absorption
spectroscopy, for example, discrete absorption lines.  The following example sticks to UV-Vis absorption spectroscopy.
Often, absorption bands observed consist of several contributions, i.e. there are more than one individual absorptions differing slightly only in terms of their corresponding energy of absorption -- and instead of a pure Gaussian shape of your absorption band, the enveloping absorption band observed features shoulders, for example.
To determine the derivative of the spectrum recorded helps to determine the centres of the individual peaks, to ease the discrimination of the individual components of one absorption band.  This may lead to spectral deconvolution. A pictural introduction including examples and limits of application is provided here.


text:
Compare the ionization energies of C and Cl?
I saw a test question today that basically boiled down to comparing the ionization energies of $\ce{C}$ and $\ce{Cl}$. I know that in the periodic table, ionization energy generally increases as we move right across a row (because nuclear charge increases), and decreases as we move down a column (because atomic radius increases, so valence electrons feel less attraction to the nucleus). Note that $\ce{Cl}$ is in the row below $\ce{C}$, but is also to the right of $\ce{C}$, so it is not obvious how to compare the two. So, without looking up the ionization energies and using only the periodic table, how can I answer this question?

It's the other way around,

moving left to right across a row increases the IP due to increasing nuclear charge, and
moving down a column decreases the IP, due to increasing shielding of the nuclear charge by core electrons.

Chlorine is only one row below carbon, but it is three columns to the right.  As the following chart illustrates, moving to the right by one position generally has a similar or larger effect than moving down one position.

(image source)
Therefore, in this case moving further to the right by 3 positions (and increasing the IP) strongly outweighs moving down by only one position (and decreasing the IP).  Hence, in this case the IP of chlorine would be predicted to be greater than the IP of carbon.  Indeed, the IP of chlorine is reported to be 13.0 eV, while that of carbon is only 11.3 eV.

Sorry I knew it was the other way around I made a mistake when typing. But I don't think you have answered my question because how do you know that moving further to the right 3 times wins out over moving down one time. What if moving to the right affects the ionization energy less than moving down? This is not solid reasoning.

There are no what if's, this is how it is. The reasoning is provided in the chart, it evidently shows the reasoning behind this concept. ron clearly explains this.

@JoshuaBenabou  I've edited my answer in an attempt to address your comment.

@JoshuaBenabou I think you or your lecturer would have needed to analyse this graph in some detail, prior to the test, to be able to answer this question. I don't see how a qualitative argument from principal quantum number, effective nuclear charge or another periodic trend would provide a clear answer in this case, as the effects are opposite. This is the sort of test question I do not appreciate; when multiple qualitative arguments produce opposing conclusions, quantifying which one "wins out" often quickly becomes untenable in Chemistry, and it becomes a toss-up whether you were right.

I agree I don't like this kind of question because deciding qualitatively which factor wins out always comes down to sloppy reasoning. Thanks ron, I understand now.


text:
Mass Spectrum and Molecular Ion peaks
If a molecule contains four chlorine atoms, how many molecular ion peaks will this compound show in its mass spectrum?

The major isotopes of chlorine are 35 and 37.  If 4 chlorines are present in a molecule, then the following isotopic combinations of chlorine are possible

35, 35, 35, 35
35, 35, 35, 37
35, 35, 37, 37
35, 37, 37, 37
37, 37, 37, 37

So there will be 5 molecular ion peaks.  You could calculate their relative heights by factoring in the relative isotopic proportions, along with the number of permutations of each of the above combinations.

If a molecule contains four chlorine atoms, how many molecular ion peaks will this compound show in its mass spectrum?

Limitations of existing answers
Ron's answer is good for molecules that consist solely of chlorine, i.e., a hypothetical $\ce{Cl4}$ molecule.  But if there are other atoms [of elements that have more than one stable isotope] as well, then there will be more peaks.
An example using difluorotetrachlorobenzene
Suppose that the molecule is an asymmetric difluorotetrachlorobenzene, $\ce{C6Cl4F2}$.  Then all of the fluorine atoms will be $\ce{^{19}F}$, but every one of the six carbon atoms could be either $\ce{^{13}C}$ or $\ce{^{12}C}$, and every one of the four chlorine atoms could be $\ce{^{35}Cl}$ or $\ce{^{37}Cl}$.
A complete list of all isotopologues
Since there are six carbon atoms, there are 7 possible values for the total mass of carbon in each molecule (0 $\ce{^{13}C}$ atoms, 1 $\ce{^{13}C}$ atom, ... up to 6 $\ce{^{13}C}$ atoms).  Similarly there are 5 possible values for chlorine, so there are $7\times 5=~~$ 35 different molecular ions!  Here is a complete list:
       mass log10(abund) 12C 13C 35Cl 37Cl 19F
1  249.8722   -0.5102721   6   0    4    0   2
2  251.8693   -0.4031194   6   0    3    1   2
3  253.8663   -0.7219355   6   0    2    2   2
4  255.8634   -1.3929341   6   0    1    3   2
5  257.8604   -2.4899015   6   0    0    4   2
6  250.8756   -1.6980651   5   1    4    0   2
7  252.8726   -1.5909124   5   1    3    1   2
8  254.8697   -1.9097285   5   1    2    2   2
9  256.8667   -2.5807271   5   1    1    3   2
10 258.8638   -3.6776945   5   1    0    4   2
11 251.8789   -3.2660693   4   2    4    0   2
12 253.8760   -3.1589166   4   2    3    1   2
13 255.8730   -3.4777327   4   2    2    2   2
14 257.8701   -4.1487313   4   2    1    3   2
15 259.8671   -5.2456987   4   2    0    4   2
16 252.8823   -5.1070748   3   3    4    0   2
17 254.8793   -4.9999221   3   3    3    1   2
18 256.8764   -5.3187382   3   3    2    2   2
19 258.8734   -5.9897368   3   3    1    3   2
20 260.8705   -7.0867042   3   3    0    4   2
21 253.8856   -7.1979577   2   4    4    0   2
22 255.8827   -7.0908051   2   4    3    1   2
23 257.8797   -7.4096212   2   4    2    2   2
24 259.8768   -8.0806198   2   4    1    3   2
25 261.8738   -9.1775871   2   4    0    4   2
26 254.8890   -9.5618420   1   5    4    0   2
27 256.8860   -9.4546893   1   5    3    1   2
28 258.8831   -9.7735054   1   5    2    2   2
29 260.8801  -10.4445040   1   5    1    3   2
30 262.8772  -11.5414714   1   5    0    4   2
31 255.8923  -12.3059456   0   6    4    0   2
32 257.8894  -12.1987929   0   6    3    1   2
33 259.8864  -12.5176090   0   6    2    2   2
34 261.8835  -13.1886076   0   6    1    3   2
35 263.8805  -14.2855750   0   6    0    4   2

Why the complete list isn't always observed

Resolution limitations.  Many types of mass spectrometers won't be able to resolve the very small mass differences between some of these ions.  For example, $m/z=251.8693$ for $\ce{^{12}C6~^{37}Cl1~^{35}Cl3~F2}$ is only 0.0096 Daltons away in mass from $m/z=251.8789$ for $\ce{^{12}C4~^{13}C2~^{35}Cl4~F2}$.  In many instruments such peaks would not be well-resolved.
Sensitivity limitations.  Additionally, many of the 35 distinct isotopologues are very rare in abundance, well below practical limits of detection.  The abundance of $\ce{^13C6~^37Cl4~F2}$ is about fourteen orders of magnitude lower than the most abundant mass isotopologue, $\ce{^{12}C6~^{35}Cl3~^{37}Cl1~F2}$.

A theoretical spectrum at unit-mass resolution
The end result is that in most mass spectrometers, more than 5 but many fewer than 35 peaks will be observed for our example molecule of difluorotetrachlorobenzene.
Here are approximate results for an instrument with ~0.5 Da mass resolution, such as most quadrupole detectors.

How I have been calculating these results
I am a big fan of an R package called ecipex.  With that package these calculations are easy.  Here is my code to generate the list above.
require(ecipex)
df <- ecipex('C6Cl4F2', id=T, limit=1e-16)[[1]]
dfordered <- df[order(df[,'12C', decreasing=True]), ]
dfordered$abundance <- log10(dfordered$abundance)
names(dfordered)[2] <- 'log10(abund)'
dfordered

And to generate the graph:
require(ggplot2)
ggplot(df, aes(x=mass, weights=abundance)) +
  geom_bar(stat='bin', binwidth=0.5, color='black', fill='gray') +
  theme_bw() +
  xlab('m/z, Da') +
  ylab('abundance') +
  scale_x_continuous(breaks = 248:258, limits = c(248,258))

Both to shed complementary light on the lesser obvious limitation that comes along with first answer, as well as adding how to compute the expected pattern are appreciated.

This answer simply aims to add to the one by Curt F. that there is software available to do these calculations, too.  Beside the ones with the mass spectrometers, and assuming Ali Naqvi participates in initial lab classes, the Chemical Calulator of the Gnome Chemistry utilities may be worth a trial to get familiar with typical isotope patterns of organic molecules.  As a training, I found this beneficial to recognize them in a routine analysis with ease.
To return to the initial question, a $\ce{Cl4}$ would appear as this -- note there are five peaks.

On contrast, Ron's example of C6Cl4F2 -- nine peaks -- were this one:

Its always nice to see two independent means of calculating something give the same results.  Thanks for your answer Buttonwood.

$\ce{C6Cl4F2}$ was Curt F's example, not mine.


text:
Ink / formulation chemistry: What lab requirements
Little open information can be found on the requirements of the manufacture, development and testing of gel and printing inks.
It seems that most ink manufacturers have their own specific lab configuration.
Would you know of a recource, or could you list me a possible lab setup for such an endeveavor? While I have found books on the topic, few of them are of very practical nature.
Of specific interest to me are

analytic methods
synthetic apparatus
testing procedures and protocols

Thank you very much!


text:
Plotting ΔG vs Temp for Electrochemical Reaction
I’m doing an electrochemistry lab that requires me to construct various galvanic cells in an attempt to solve for $\Delta H$, $\Delta G$, and $\Delta S$ values of an unknown. During one part of the experiment, I had to heat a cell to a temperature of 70 °C and record its voltage in 15 degree increments until the temperature reached 10 °C.
The cell I am using to perform this step on is represented by the equation:
$\ce{Zn + Sn^2+ -> Zn^2+ + Sn}$
After performing the procedure, I’m getting concerned that my data may be inaccurate. Theoretically, I was using 0.1 M concentrations of each solution in the cells. According to the Nernst equation, it seems that the voltage or potential of the cell should not be affected by a change in temperature if the concentrations are equal because $\ln(0.1/0.1) = 0$.
However, according to my professor, I should be seeing a linear relationship when I plot $\Delta G$, calculated from the voltage measurements I took, vs the temperature in K.
My question is, theoretically, what should happen to the voltage as I heat the cell, assuming that the concentrations of the solutions were off? I’ve calculated the $\Delta S$ value based on the standard values listed for these entities and I’m getting a value of −85.18 J/(K mol).
Calculation: $\Delta S = (-112.1 + 51.55) - (41.63 + -17.0) = -85.18\ \mathrm{J/(mol\ K)}$
Based off of this result, it would appear that the slope of this plot must come out to be a positive value because $m = - \Delta S$. Does this mean that I should be seeing an increase in voltage as temperature decreases? Regarding the $\Delta H$ of this reaction, which is represented by the Y-intercept (b) of the equation of the best fit line for my data, it appears I must also have a negative value of −145090 (in units of J) based upon the calculation I did for this, also using standard values.
Calculation: $\Delta H = (-153.89 + 0) - (0 + -8.8) = -145.09\ \mathrm{kJ/mol}$
I’ve tried to be as detailed as possible in describing by question, and hopefully it makes sense. I’m just getting confused on what the data should look like. Unfortunately, its not possible for me to redo the experiment. Please note that this is for my own understanding, I will not submit the assignment with fixed data. I appreciate any assistance that anyone can offer me.

My question is, theoretically, what should happen to the voltage as I heat the cell, assuming that the concentrations of the solutions were off?

This question is tough to answer because I don't know what you mean by "off".  So let me ask (and answer) a few related questions that I think will help you.
What should happen to voltage as a function of temperature in an electrochemical cell
The electrical potential of electrochemical cells varies with temperature, even when the cell is at standard conditions.  A 1985 NIST document explains the situation well.  Only the standard hydrogen electrode has a standard potential that does not vary with temperature.  (The SHE potential is defined as 0 volts at every temperature.)  For other cells, usually the assumption of a linear variation with temperature is usually very good.  That is,
$${\cal E}_T^\circ = {\cal E}_{298}^\circ  + (T-298.15)\cdot \frac{d{\cal E}_{298}^\circ}{dT}$$
Note that this variation is completely unrelated to the Nernst equation.  The same reference gives values for $\frac{d{\cal E}_{298}^\circ}{dT}$ for zinc(II)/zinc metal of 0.119 milliVolts per Kelvin and for tin(II)/tin metal -0.32 milliVolts per Kelvin.
Don't be confused by the $T$ in the Nernst equation
The Nernst equation is
$$ {\cal E}= {\cal E}^\circ + \frac{RT}{nF}\ln Q$$
It does contain a $T$ in there, but don't be mislead into thinking that the $T$ you see is the sole means by which a cell potential ${\cal E}$ can vary with temperature.  ${\cal E}^\circ$ also varies with temperature (see above).

According to the Nernst equation, it seems that the voltage or potential of the cell should not be affected by a change in temperature if the concentrations are equal because ln(0.1/0.1)=0.

Thus, hopefully my answer makes clear that this statement is wrong, because ${\cal E}^\circ$ varies with temperature.


text:
pH of a weak base calculation
Calculate the $\ce{pH}$ of a $100$ $\ce{mL}$ solution containing $0$$.0375$ moles of the weak base, Sodium Benzoate, $\ce{C6H5COONa}$.

$$\ce{C6H5COO + H2O -> C6H6COO- + OH-}$$

$$\frac{0.0375 ~\mathrm{mol}}{.100~ \mathrm{L}} = 0.375~M$$
$$\ce{pOH} = 0.43$$
$$\ce{pH} = 13.57$$
Is this right? Thank you.

Benzoate is a weak base, so you can't assume it reacts completely to form $\ce{OH^{-}}$. What's the data you need in this case? Also, your reaction is written a bit incorrectly.

Would you need Kb?

Welcome to Chemistry SE!
To do this specific problem you need to use the $K_{b}$:
$$\ce{\frac{x^{2}}{.375-x}}=K_{b}$$
Finding the $K_{b}$ and solving for x will give you the concentration of $\ce{OH-}$. Remember to take the $-log$ of that and solve for the $pH$.
Also next time please show some work and take your time to format these questions so people can try to reply back to you.


text:
pH of dilute HCl solution
What is the pH of a solution of $3.4 \times 10^{-10}$ moles of $\ce{HCl}$ in a volume of $\pu{150.7L}$?
This is dilute so I expect pH to be high.
$$\frac{3.4 \times 10^{-10}\ \mathrm{mol}}{150.7\ \mathrm L} = 2.26 \times 10^{-12} \frac{\mathrm{mol}}{\mathrm L}$$
Then pH is 11.65. Is this correct?

What's the pH of pure water? If you add an amount (however small) of strong acid to pure water, what do you expect the new pH will be?

It should be ~7

Like you guessed, the pH should be around 7, since the concentration of HCl is very low. Obviously, it can't be 11.65, since it's still an acid. The reason for this discrepancy is that, for low concentrations, you must consider the self-ionization of water in order to get the correct concentration of $\ce{H3O+}$ ions.
Consider the equilibrium:
$$\ce{2H2O <=> H3O+ + OH-}$$
$$K_w = \ce{[H3O+][OH^{-}]} = 1 \times 10^{-14} \tag 1$$
and also the dissociation of the acid:
$$\ce{H2O + HCl -> H3O+ + Cl-}$$
Then
$$c_0 = \ce{[Cl^{-}]} = 2.26 \times 10^{-12} \tag 2 $$
and
$$\ce{[H3O+] = [Cl^{-}] + [OH^{-}]} \Rightarrow \ce{[OH^{-}] = [H3O+] - [Cl^{-}]} \tag 3$$
If we replace $(2)$ in $(3)$ and that in $(1)$, we end up with
$$ K_w = \ce{[H3O+]} (\ce{[H3O+]} -c_0) $$
$$ \ce{[H3O+]} (\ce{[H3O+]} -c_0) - K_w = 0$$
using $ x = \ce{[H3O+]}$
$$x^2 - c_0x - Kw = 0$$
This can be easily solved. Taking only the positive value of $x$:
$$ x = \frac{c_0 + \sqrt{c_o^2 + 4K_w}}{2}$$
Inserting the values of $x$ and $K_w$, the result is:
$$ x = 1.000023 \times 10^{-07} = \ce{[H3O+]}$$
So
$$pH = -\log\ce{[H3O+]} = -\log{1.000011 \times 10^{-07}} = 6.999995 $$
Which is not surprising at all, but is the calculated answer.
For a less diluted solution of $\ce{HCl}$ with concentration of $1 \times 10^{-8} \frac{mol}{L}$, which still can't be calulated directly, the result is $pH = 6.98$.

Thanks but I don't get how you came up with Equation 2 relating [H3O+] to [OH-] and [Cl-].

I think you meant equation 3. That's what we call charge balance. The only species in our system are $\ce{H2O}$, which we may consider with constant concentration because of how high it is, and the mentioned ions. The total system charge must be zero, so the concentration of cations has to equal the concentration of anions. Another way to look at it is noting that you have two sources of $\ce{H3O+}$, the water self-ionization and the acid's dissociation, and each one of these processes will lead to corresponding concentrations of their counter ions.

As you may have noticed, your answer of 11.65 is counter-intuitive as it is higher than 7, meaning that dilute hydrochloric acid is somehow more BASIC than water.
The key here is accounting for the $1.0 \times 10^{-7} M$ H+ ions, or $1.507 \times 10^{-5}$ moles of H+ ions for 150.7L, already present in the water due to self-ionization. An addition of $3.4 \times 10^{-10}$ more moles of H+ ions would be practically insignificant (5 orders of magnitude difference). Hence, the pH would be near 7.

Just for curiosity, an analytical solution can be found for such a dilute acid, but the hydrogen ion concentration is the same as in pure water to at least six significant digits (one part in a million), so for all practical purposes the resulting pH is identical to pure water.


text:
How to determine the acidity or basicity of dissolved salts?

One kg of each substance below is dissolved in a bathtub full of water. Will it be acidic, basic, or neutral?
  $\ce{KNO2}$
  $\ce{HF}$
  $\ce{(CH3)3NHBr}$
  $\ce{HO-CH2-CH2-O-CH2-Cl}$

$\ce{KNO2}$, $\ce{K+}$ $\ce{NO2-}$ $\ce{K+}$ is inert and $\ce{NO2-}$ is the base of strong acid, so it should be basic overall.
$\ce{HF}$, $\ce{H+}$ $\ce{F-}$ $\ce{H+}$ is strong acid and $\ce{F-}$ is strong base so it should be neutral overall.
$\ce{(CH3)3NHBr}$, Molecule acts like a $\ce{H3NHBr}$ or $\ce{NH4+}$ $\ce{Br-}$ so $\ce{NH4+}$ is acidic $\ce{Br-}$ is neutral, so overall solution is acidic.
$\ce{HO-CH2-CH2-O-CH2-Cl}$, No idea.
Can someone please explain them, if they're wrong? Thank you.

$\ce{NO2^{-}}$ is the conjugate base of a weak acid, nitrous acid ($\ce{HNO2}$). Also, $\ce{F^{-}}$ is a weak base in water.

So KNO2 is still basic overall. HF is basic because of F-. I'm lost on the last two.

I'm assuming the last one is a weak acid.

What is the last molecule comparable to? HOCl?

The $\ce{KNO2}$ is basic, the HF is acid, the $\ce{(CH3)3NHBr}$ being an ammonium salt is acid too, the last one is an alcohol, I would consider it neutral for your purpose. But you are missing one step to get your answer quite right.
Last part, in the pH world, everything turn around two ions H+ (or H3O+) and OH- and the equilibrium between those two. A substance is considered acid when it either give a H+ as disolving or takes an OH-. On the other hand a substance is considered basic when it give an OH- or takes a H+. Like HF is an acid (H+, F-) and NaOH is basic (Na+, OH-). For KNO2, it can take an H+ to go back to HNO2, so it is basic. For organic compounds, alcool groups are usually considered inerte as they don't ionate in normal conditions.
Now, maybe I'm wrong but your question sound like if you put all those in a bathtub what'S the final result? Is it right or not?

I don't get what I'm supposed to do?

(CH3)3NHBr isn't ammonium salt, but bromoamine and and alcohol hydrolyzes to HCl @Asker123

So @mithoron could you tell me the difference between those two molecules BrNH2 vs NH4Br?

https://en.wikipedia.org/wiki/Chloramine, These two last are much harder than first ones. Major difference is covalent vs ionic.

Oops, I thought it's tert-buthyl there, but it's trimethyl ammonium... But still this organic molecule is chloroether, which hydrolyses and therefore solution will be acidic.

Assuming your question is how to determine the acidity or basicity of salts.

Acidic Salt: A salt formed between a strong acid and a weak base is an acid salt.

$$\ce{HCl + NH4OH -> NH4Cl + HOH}$$
  The $\ce{NH4Cl}$ is an acidic salt and if it dissociates it will be acidic.

Basic Salt: A basic salt is formed between a weak acid and a strong base.

$$\ce{NaOH + HAc -> NaAc + HOH}$$
  The Sodium Acetate the basic salt that dissociated into acetate ions and takes an H+ in water.

Applying this logic to your question gives us: $\ce{KNO2}$ is a basic salt, the $\ce{HF}$ is an acid, the $\ce{(CH3)3NHBr}$ being an ammonium salt is acidic too, the last one is an alcohol which has no effect.
Thanks.

OK. That makes sense. But can you also figure out whether each anion/cation is a product of a strong/weak acid/base and determine the acidity/basicity of the salt using that? Ex: KNO2, The NO2- comes from HNO2, a weak acid, so it is basic overall (and K+ is an inert ion)?

@user15831 That is exactly what you are supposed to do. But I doubt that will help with hydrocarbons.


text:
Why does a weak acid and a salt containing its conjugate base form an acidic buffer?
I read this today:

If you mix equal amounts of $1 M$ $\ce{H_3PO_4}$ and $1 M$ $\ce{NaH_2PO_4}$ you get a buffer at $pH<6$.

The reason given is that a weak acid plus a salt containing the conjugate base forms a acidic buffer.

If you mix equal amounts of $1 M$ $\ce{NH_3}$ and $1 M$ $\ce{NH_4Cl}$, you get a buffer at $pH>8$.

The reason given is that a weak base plus a salt containing the conjugate acid forms a basic buffer.
I don't understand why this is true. Let's take the first statement for example. If you have equal amounts of a weak acid and its conjugate base, then by definition you have a buffer. Whether the buffer is acidic or basic shouldn't depend on whether the salt contains the conjugate acid or the conjugate base, right?

Try to find pH by yourself :) The first acidity to $\ce{H_3PO_4}$ is a strong acidity $\ce{pKa_1}=2,15$ and $\ce{pKa_2}=7,20$ and the third $\ce{pKa_3}=12,42$

Do you know the Henderson-Hasselbalch equation? The answer comes from a straightforward analysis.

Yes I know the Henderson-Hasselbach equation, it gives that the pH equals the pKa which for an acid is low, but this is just a mathematical trick and I want a conceptual explanation.

The statement "a weak acid plus a salt containing the conjugate base forms an acidic buffer." is not correct. More properly, a mixture of equal amounts of a weak acid (but not too weak: with a pKa < 7) with its conjugate base (salt) does indeed give an acidic buffer. To get a basic buffer you would use equal amounts of an even weaker acid (with a pKa > 7) and its conjugate base (salt). Similar considerations apply to formation of buffers from a weak base plus a conjugate acid salt.

This answer is correct. Not sure why you got a downvote.

Your suspicion is correct: the buffer pH depends on the $\rm{pK_a/pK_b}$ of the acid/base in question, not whether the buffer salt contains a weak conjugate acid or conjugate base. We can test this out via the application of the Henderson-Hasselbalch equation.
Consider acetic acid ($\rm{pK_a = 4.76}$) mixed with the same concentration of sodium acetate. Via the Henderson-Hasselbalch equation, $\rm{pH = pK_a + log \frac{[Ac^{-}]}{[HAc]}}=pK_a + log\ 1 = pK_a = 4.76$. Thus, the acid/conjugate base buffer is acidic.
Now take hydrocyanic acid ($\rm{pK_a = 9.21}$) mixed with the same concentration of sodium cyanide. Again, via the Henderson-Hasselbalch equation, $\rm{pH = pK_a + log \frac{[CN^{-}]}{[HCN]}}=pK_a + log\ 1 = pK_a = 9.21$. The acid/conjugate base buffer is now basic. A similar analysis can be made for bases mixed with their conjugate acid salts.
So why does this happen? It's quite simple. If an acid has a $\rm{pK_a < 7}$, then it is more acidic than its conjugate base is basic ($\rm{pK_b > 7}$), so when mixing an equal amount of the acid and its conjugate base, the acidic character wins and the buffer is acidic. If the acid has a $\rm{pK_a > 7}$, then it is less acidic than the conjugate base is basic ($\rm{pK_b < 7}$). Mix an equal amount of both, and as you might expect, the basic character wins, and the buffer ends up alkaline. Again, a similar analysis can be made for buffers made with bases and their conjugate acid salts.

I am a pretty elementary level student myself, so please excuse and ignore if you find me saying something wrong.
By definition a buffer solution resists the sudden change in $\mathrm{pH}$ on addition of a small amount of strong acid or strong base or due to dilution.
One possible way for this to occur (in case of acid buffer) is for the strong acid introduced into the system (buffer) to be converted to a weaker acid (that is why we require both a weak acid and its salt), same for a strong base.
Now, consider ethanoic acid and sodium ethanoate:
$$\ce{H+ + CH3COONa -> CH3COOH}$$
where $\ce{H+}$ – proton from strong acid; $\ce{CH3COOH}$ – a weaker acid than a one introduced, hence no considerable change in $\mathrm{pH}$.
Also,
$$\ce{CH3COOH + OH- -> CH3COO- + H2O}$$
where $\ce{OH-}$ is from introduced base (mildly reversible reaction).
The acetate here lacks a hydroxyl group as opposed to the species you introduced.
For a basic buffer
$$\ce{NH4OH + H+ -> NH4+ + H2O}$$
where $\ce{H+}$ is from acid; $\ce{NH4+}$ is weakly acidic; and
$$\ce{NH4Cl + OH- -> NH4OH + Cl-}$$
where $\ce{OH-}$ is from base; $\ce{NH4OH}$ – weak base.
However, both basic and acidic buffers serve the same purpose and in nearly the same way.. they just have different compositions and consequently different $\mathrm{pH}$ and $\mathrm{pOH}$.


text:
If the temperature of a solution decreases, what is the sign of the enthalpy change?
Here is what seems to be a straightforward question: you dissolve some ammonium chloride in water at $25\ \mathrm{^\circ C}$, lowering the temperature of the solution. Is the change in enthalpy for the dissolving process positive or negative?
My reasoning: if the temperature of the solution decreases, it means heat was released into the surroundings, so the enthalpy change is negative.
Apparently the answer is the enthalpy change is positive. What am I doing wrong? Whatever it is, it must be stupid, but please forgive me for its late here.

Identify the system and the surrounding here; I think your confusion stems from that.

When I was doing enthalpy, this was one of the tricky concepts that I still sometimes get wrong. Here is how you are supposed to remember it.

Identify the System

The system is where the reaction happens. You cannot physically monitor the change in heat here. The change you always measure happens in the surroundings.

Identify the Surroundings

This is the surroundings where usually the change is expressed as.

Now that we got the basics cleared away, let us look at your problem.

My reasoning: if the temperature of the solution decreases, it means
  heat was released into the surroundings, so the enthalpy change is
  negative.

This is what I usually think of, but now you must understand that as I've said before the change always happens in the surroundings and is associated with the surroundings. So the temperature of the solution decreases, which means that the surroundings temperature decreased. The solution is part of the surroundings. This means that the temperature of the system increases. Thus making the $\Delta H$ positive and the reaction endothermic.

Good Question, always remember that you cannot, in most situations, physically measure the change in the system.
Thanks.

Can you be more clear about what EXACTLY is the system and what exactly is the surroundings. Also please explain why the thermometer is measuring the surroundings and not the system. Then I will accept your answer.

There are certain cases where the boundary between the system and the surroundings is kind of blurred. System would be the aqueous ammonium chloride; surroundings would be the solution/water solvent. Energy from the surroundings (water) is being added to the system to break the ionic bonds.

It might help to think about it with an example. $\ce{NH_4Cl}$ is the 'active' ingredient in cold packs (ice packs). The system in this case, consists of the individual ions that separate when placed in solution ($\ce{NH_4^{+}}$ and $\ce{Cl^{-}}$). The solution makes up the surroundings - since the temperature of the solution decreases, the surroundings lose heat, and the system gains heat. Enthalpy is defined by a loss or gain of the system's heat. Since the system gains heat, the enthalpic value is positive (endothermic).
In general, for dissolving salts in solution, this is the case.


text:
Why isn't dry CaCl₂ used to dry HCl gas instead of concentrated sulfuric acid?
In our chemistry books, it is said that concentrated sulphuric acid is used to dry $\ce{HCl}$ gas.
But, concentrated sulphuric acid is a strong oxidising agent so, why isn't $\ce{HCl}$ gas oxidised to $\ce{H2O}$ and chlorine gas?
Why dry $\ce{CaCl2}$ (Calcium Chloride) isn't used as drying agent?
[Note: $\ce{CaCl2}$ have no reaction with $\ce{HCl}$ as they both have chloride ion in common]

cacl2 is a solid (amorphous mostly) and it is hard to handle with amorphous powdery solid, because they get mixed with the solution to be dried

Does the book mention more details about the process? Specifically, is it talking about drying the $\ce{HCl}$ gas, or producing it from $\ce{HCl}$ solution? For the $\ce{CaCl2}$ part, I'd start with the physical state: bubbling a gas through a liquid seems a lot more efficient than through a solid.

Sulphuric acid isn't oxidizing enough to do this.

Why does concentrated $\ce{H_2SO_4}$ oxidise? Due to the nascent oxygen it gives:
$$\ce{H_2SO_4}\rightleftharpoons\ce{SO_2 +H_2O +[O]}$$
Now, if you want $\ce{HCl}$ to be oxidised, the following reaction must exist:
$$\ce{HCl} +\ce{[O]}\rightarrow\ce{HClO}$$
But, you can observe that the reaction$$\ce{HClO}\rightarrow\ce{HCl} +\ce{[O]}$$
can't be backward under normal conditions (you probably have read this reaction in the bleaching action of bleaching powder). So, fortunately, you can safely use $\ce{H_2SO_4}$ in this purpose...and take its advantage to $\ce{CaCl_2}$, as agha rehan abbas and Molx mentioned.

Oxygen is more electronegative than chlorine so H2O is more stable compound than HCl .Hence,it is very possible to form H2O by oxidising HCl

There are several reasons:

Efficiency
When a stream of wet hydrogen chloride is run through a tube filled with coarse calcium chloride "rocks", interaction with the drying agent takes place at at its surface.
When the same wet $\ce{HCl}$ ist passed through sulfuric acid, the contact area is given by the surface area of the bubbles. Moreover, bubbles mean mixing. The trapped water is distributed over the drying agent.
Safety
Wet calcium chloride will clog, which might result in a pressure buildup in the drying tube.

This is because calcium is higher than hydrogen in the activity series of metals  due to which, the objective of the reaction will be destroyed. This is why concentrated sulphuric acid is used.

Can you elaborate your bit more?

Calcium metal is higher in the activity series; calcium chloride and hydrogen chloride do not react at all, as mentioned at the end of the question.


text:
Why would hexacyanidocobaltate(II) be uncommon while pentacyanidoaquacobaltate(II) is common?
In Huheey's Inorganic Chemistry it is written that hexacyanidocobaltate(II) is uncommon but pentacyanidoaquacobaltate(II) is common. Something is mentioned about Jahn-Teller distortion, but I don't understand how it applies here.

We use it in the lab all the time, I don't think it's uncommon.

that is artificial. i think huheey said not common in nature

In Inorganic Syntheses (1946), a synthesis of $\ce{K3[Co(CN)6]}$ is described, which boils down to the following three equations:[1]

$$\begin{align}\ce{CoCl2 (aq) + 2 KCN (aq) &-> Co(CN)2 v + 2 KCl (aq)}\tag{1}\\
\ce{Co(CN)2 + 4 KCN (aq) &-> K4[Co(CN)6] (aq)}\tag{2}\\
\ce{2 K4[Co(CN)6] (aq) + H2O &-> 2 K3[Co(CN)6] (aq) + 2 KOH (aq) + H2}\tag{3}\end{align}$$

The procedure requires precipitation of the cobalt(II) cyanide by carefully adding a $\ce{KCN}$ solution to a cobalt(II) solution. This precipitate is filtrated and washed, then redissolved in excess cyanide solution giving a deep-red solution. This in turn is heated to reflux for some 15 minutes, whereupon the colour changes to yellow indicating the formation of the hexacyanidocobaltate(III).[1]
I personally doubt equation $(3)$. Instead, I would assume that the oxidation of hexacyanidocobaltate(II) to the corresponding cobaltate(III) proceeds unter oxygen consumption, not hydrogen liberation. That would give equation $(\text{3'})$:
$$\ce{4 K4[Co(CN)6] + O2 + 2 H2O -> 4 K3[Co(CN)6] + 4 KOH}\tag{3'}$$
In any case, however, the presence of $\ce{[Co(CN)6]^4-}$ or hexacyanidocobaltate(II) as a stable intermediate in solution suggests to me that your premise is false and thus the book you cite from incorrect.

Reference:
J. H. Bigelow and J. C. Bailar (1946) Potassium Hexacyanocobaltate(III), in: Inorganic Syntheses, Volume 2 (ed W. C. Fernelius), John Wiley & Sons, Inc., Hoboken, NJ, USA. DOI: 10.1002/9780470132333.ch72.


text:
Is cyclopentadiene or cycloheptatriene more acidic?

Which of the following cyclic compounds are more acidic?

The most acidic proton is positioned on the carbon that is at the top of the above drawings (the methylene hydrogens) on each of the two species, as deprotonation allows resonance.
I am aware of Hückel's rule, which states that an aromatic species has $4n+2$ π-electrons. Only the five membered ring would fulfil this requirement. However, I can draw resonance structures where the seven membered conjugate base has the double bond at each position on the ring. I ask why is it not aromatic? Which conjugate base is more stable? Essentially it's a case of aromaticity vs number of resonance structures.

Essentially it's a case of aromaticity vs number of resonance
  structures.

Well put!
Aromaticity is a very strong driving force so aromaticity wins out; Huckel's rule is more important than the number of resonance structures.  Therefore cyclopentadiene is more acidic than cycloheptatriene.  Even though we can draw 7 resonance structures (one with a negative charge on each carbon atom) for the tropylium anion and only 5 for the cyclopentadienyl anion, Huckel's rule tells us that the cyclopentadiene anion ($4n+2$ π electrons, $n=1$) is aromatic while the tropylium anion with 8 π electrons (doesn't fit $4n+2$) is not.
Another example is cyclooctatetraene.

Like benzene, we could draw resonance structures by shifting the double bonds in this molecule too.  But this molecule is not aromatic, it does not fit the $4n+2$ rule.  In fact, Huckel says with 8 π electrons it is antiaromatic.  To avoid this destabilization cyclooctatetraene adopts a tub-like conformation.

(image source)
Again aromaticity trumps resonance structures.
Edit: Huckel's Rule: Aromaticity - Antiaromaticity
If we look at the energetic positioning of the molecular orbitals (MO's) in a cyclic, conjugated polyene, we can quickly understand the basis for Huckel's rule.
We can use Frost diagrams to determine the energetic positioning of these MO's.  Here is the diagram for benzene, we see that all of the pi electrons are in bonding MO's, the molecule is a stable aromatic that fits the $4n+2$ rule ($n=1$).

Here is the diagram for cyclooctatetraene, and we see that not all of the electrons are in bonding MO's, two π electrons are in non-bonding MO's.  They don't contribute to bonding or stabilization.

This is a $4n$ π electron compound ($n=2$) and accordingly would be classified as antiaromatic by Huckel's rule.  As mentioned above, the compound adopts a non-planar conformation to avoid this destabilization.

What are the origins of this anti aromaticity and why is it specifically when there are $4n\pi$ electrons?

I learned it as part of Huckel's rule: cyclic systems with 4n+2 pi electrons are stabilized (aromatic) while thiose with 4n pi electrons are destabilized (antiaromatic).  It comes from the molecular orbital diagrams.  See these earlier SE Chem questions, ref1, ref2.

Apparently $4n\pi$ rule does not work after n=2: the molecules try to escape anti-aromaticity and get into boat / twisted shapes.


text:
Disproportionation of plutonium species - Latimer's diagram
Why (according to provided answers) $\ce{Pu(IV)}$ disproportionates  into adjacent species whilst $\ce{Pu(V)}$ does not in acid aqueous solution.
$$\ce{\underset{(+6)}{Pu}O2^2+ ->[\pu{+1.02 V}] \underset{(+5)}{Pu}O2+ ->[\pu{+1.04 V}]\underset{(+4)}{Pu}^4+ ->[\pu{+1.01V}] \underset{(+3)}{Pu}^3+}$$
I thought it is the other way - the more positive potential of $\pu{+1.04V}$ on the right of $\ce{PuO2+}$ species will drive the disproportionation while in $\ce{Pu^4+}$ the higher value of $\pu{+1.04V}$ to the left renders the species by itself thermodynamically stable.

In acidic solutions ($\mathrm{pH} = 0$), the values of the relevant reduction potentials of $\ce{Pu^3+/Pu^4+}$ ($\pu{+1.01 V}$), $\ce{Pu^3+/PuO2+}$ ($\pu{+1.03 V}$), $\ce{Pu^3+/PuO2^2+}$ ($\pu{+1.02 V}$), $\ce{Pu^4+/PuO2+}$ ($\pu{+1.04 V}$), $\ce{Pu^4+/PuO2^2+}$ ($\pu{+1.03 V}$), and $\ce{PuO2+/PuO2^2+}$ ($\pu{+1.02 V}$) are all very similar.
Therefore, plutonium has the remarkable feature that all four important oxidation states can be present simultaneously in aqueous solutions.
In aqueous solution, $\ce{Pu^4+}$ disproportionates into a mixture of oxidation states:
$$\ce{2Pu^4+ + 2H2O <=> Pu^3+ + PuO2+ + 4H+}$$
$$\ce{Pu^4+ + PuO2+ <=> Pu^3+ + PuO2^2+}$$
Thus, the net reaction of the disproportionation of $\ce{Pu^4+}$ is approximately:
$$\ce{3Pu^4+ + 2H2O <=> 2Pu^3+ + PuO2^2+ + 4H+}$$
Nevertheless, $\ce{Pu^4+}$ may still be the predominant oxidation state in the mixture.
Besides, the equations imply that the disproportionation of $\ce{Pu^4+}$ dcreseases as the acidity of the solution further increases.
Concerning pentavalent plutonium ($\ce{PuO2+}$), your expectations are correct. In typical acidic solutions, $\ce{PuO2+}$ indeed quickly disproportionates into a mixture of oxidation states:
$$\ce{2PuO2+ + 4H+ <=> Pu^4+ + PuO2^2+ + 2H2O}$$
$$\ce{Pu^4+ + PuO2+ <=> Pu^3+ + PuO2^2+}$$
However, the given values for the reduction potentials apply to pH = 0, and the various reduction potentials of plutonium depend on pH. Furthermore, the stability of individual oxidation states can be strongly influenced by hydrolysis, or the formation of complexes or compounds with low solubility products. Therefore, the predominant oxidation state can be changed by means of multiple factors. Furthermore, the valence of plutonium solutions may change as a consequence of radiolysis.

Tetravalent plutonium as the ion $\ce{Pu^4+}$ does not disproportionate according to the equation:
$$\ce{3Pu^4+ + 2H2O -> 2Pu^3+ + PuO2^2+ + 4H+}\label{a}\tag{1}$$
The products on the right-hand side of Eq. $\ref{a}$ inevitably react according to Eq. $\ref{b}$.
$$\ce{Pu^3+ + PuO2^2+ -> Pu^4+ + PuO2^+}\label b\tag2$$
The true disproportionation equation contains all the disproportionation products in their stoichiometric proportions. Those products are $\ce{Pu^3+}$, $\ce{PuO2^+}$ and $\ce{PuO2^2+}$, and $\ce{H+}$.
Equation (1) is the limiting form of the disproportionation equation as the acid concentration increases. The limiting form of the disproportionation equation as the acidity decreases is Eq. $\ref{c}$.
$$\ce{2Pu^4+ + 2H2O -> Pu^3+ + PuO2+ + 4H+}\label{c}\tag3$$
The disproportionation equation at intermediate values between the limiting cases of "high" and "low" acidities is more complicated. It contains $\ce{Pu^4+}$ on the left-hand side and the three species $\ce{Pu^3+}$, $\ce{PuO2^+}$, and $\ce{PuO2^2+}$ on the right-hand side. The coefficients in the disproportionation equation are integers followed by decimal fractions.


text:
The hydrogen ion in acidified potassium manganate (VII)
Recently, I am playing with redox reaction and I come over a compound acidified potassium manganate (VII) ($\ce{KMnO4/H^+}$). I get confused that why they put $\ce{H^+}$ beside $\ce{KMnO4}$. Doesn't the hydrogen ions flow freely in the solution or something else.
And so, when we are writing a chemical reaction with occur $\ce{KMnO4}$, we sometimes exclude the $\ce{H^+}$. Why that so?

Permanganic acid ($\ce{HMnO4}$) is a strong acid, so it's not easy to form via protonation of  $\ce{MnO4^{-}}$. When adding permanganate ions to very very acidic mixtures, for example pure liquid $\ce{H2SO4}$, you can consider that the reaction $\ce{MnO4^{-} + H^+ <=> HMnO4}$ happens to an appreciable extent, producing permanganic acid. In the specific case I mentioned, however, $\ce{HMnO4}$ is a short lived entity, as it is quickly dehydrated by the sulfuric acid to form the anhydride $\ce{Mn2O7}$, a very reactive species.

I think you have known that the permanganate(VII) ion is a strong oxidizing agent because manganese is in the highest +7 oxidation state.
But in different pH conditions, $\ce{MnO_4^-}$ Will be reduced to different product:

In a acidic solution, it is reduced to the colourless manganese(II) ion, $\ce{Mn^{2+}}$.
$\ce{8H^+ + MnO_4^{−} + 5e^{−} → Mn^{2+} + 4H_2O}$
In a strongly basic solution, it is reduced to the green manganate ion, $\ce{MnO_4^{2−}}$.
$\ce{MnO_4^{−} + e^{−} → MnO_4^{2−}}$
In a neutral solution, it is reduced to the brown manganese dioxide $\ce{MnO_2}$.
$\ce{2H_2O + MnO_4^{−} + 3e^{−} → MnO_2 + 4OH^{−}}$

But the more understandable way to say, I think, is that $\ce{MnO_4^-}$ can be reduced to different oxidation states; however, it is an idea about chemical balance. If there is $\ce{H^+}$, the first reaction dominates (since we put in more reactant). While if there is $\ce{OH^-}$, neither reaction 1 nor 3 happens (1: lack of reactant; 3: too many product), so the third dominates.
Hope this helps.
Addition: in those different reactions, there are also different tendency to take place and make $\ce{MnO_4^-}$ reduced. In fact, the first reaction (happening with acidic solution) have a much larger tendency to take place. Thus, if $\ce{H^+}$ presents, $\ce{MnO_4^-}$ will have a larger tendency to oxidate other compounds. Thus some reactions require $\ce{MnO_4^-}$ with $\ce{H^+}$ instead of only $\ce{MnO_4^-}$.


text:
Ksp calculation of Cr(OH)2
At $25~\mathrm{^\circ C}$, $10.24~\mathrm{mg}$ of $\ce{Cr(OH)2}$ are dissolved in enough water to make $125~\mathrm{mL}$ of solution. When equilibrium is established, the solution has a $\mathrm{pH}$ of $8.49$. Estimate $K_\text{sp}$ for $\ce{Cr(OH)2}$ (Ans : $1.47\times 10^{-17}$)
I have calculated it but not have the same answer as this.
I tried it this way:

$\mathrm{pH}$ is $8.49$, then $\mathrm{pOH}$ is $5.51$. I use this formula: $\mathrm{pOH} = -\log[\ce{OH}]$
The $[\ce{OH}]$ is $10^{-5.51}$
Find molarity $\ce{Cr}$: $(10.24\cdot(1/1000)\cdot(1/86))/(0.125)=9.525\times 10^{-4}$
Substitute into $K_\text{sp} = [\ce{Cr}][\ce{OH}]^2$

But after I substitute value of $[\ce{Cr}] = 9.525\times 10^{-4}$ and $[\ce{OH}]$ of $(2\times 10^{-5.51})^2$, I get answer of $3.6\times 10^{-14}$.
Where have I gone wrong? Any Suggestions?

Your calculation of $[\ce{OH-}] = 10^{-5.51}\ \mathrm{mol\ l^{-1}}$ is correct.
The pH is well above 7. Therefore, the present $[\ce{OH-}]$ is mainly a result of the dissociation of $\ce{Cr(OH)2}$:
$$\ce{Cr(OH)2 <=> Cr^2+ + 2 OH- }$$
Thus, you know
$[\ce{Cr^2+}] \approx \tfrac{1}{2} [\ce{OH-}] = \tfrac{1}{2}\cdot10^{-5.51}\ \mathrm{mol\ l^{-1}} $.
$K_\text{sp} = [\ce{Cr^2+}][\ce{OH-}]^2 = \tfrac{1}{2}\cdot10^{-5.51} \cdot \left(10^{-5.51}\right)^2 = 1.48\cdot 10^{-17}$

Why question give me how many gram of Cr(OH)2 have. Why does it not relevance to this question? @Loong

@ChampMuangwong As long as the solubility product is exceeded, the total mass of $\ce{Cr(OH)2}$ is not relevant. However, you may use the total mass of $\ce{Cr(OH)2}$ to check whether the solubility product really is exceeded.

how to check that solubility product is exceeded?

2OH when calculate isn't it a 2*10^-5.51 then power of 2? @Loong

@ChampMuangwong You may calculate $[\ce{Cr^2+}]$ and $[\ce{OH-}]$ assuming that all $[\ce{Cr(OH)2}]$ dissolves. If this $[\ce{Cr^2+}] \cdot [\ce{OH-}]^2 > K_\text{sp}$, then the solubility product is exceeded.

2OH when calculate isn't it a 2*10^-5.51 then power of 2? @Loong

@ChampMuangwong The expression for the solubility product is $K_\text{sp} = [\ce{Cr^2+}] \cdot [\ce{OH-}]^2$. Thus you really use the value of $[\ce{OH-}]^2$ and not $(2\cdot[\ce{OH-}])^2$.


text:
Reference for UV/Vis Spectroscopic Data request
I am looking for a complete reference of UV/Vis spectroscopic data for organic materials ( mostly interested in λmax ) .
I have the CRC handbook but it only has data about some organic solvents in water and very limited data on dyes.
Can anyone suggest a comprehensive reference that includes the λmax of a wide variety of molecules in different solvents ?

To know the chemical name, molecular structure, or best: the unique registry number provided by the American Chemical Society (CAS-number) is beneficial in your search as trade names may vary depending on the supplier of the material you are interested in.  The index number of the Colour Index (like C.I. 73000 for indigo) frequently used in the business of dyes potentially is of lesser use here.  In decrease of data coverage:
There are commercial sources like Reaxys (by Elsevier) allowing to search for absorption data.  As this (in part) emerged from a much senior database about chemical reactions, Beilstein (back to 1771) this is a typical database used in Departments of Chemistry at a University or Industry. SciFinder, based on the data indexed by the American Chemical Society equally allows you to find such data -- for both it might be helpful to request assistance / initial training to fully benefit from these sources.  Note: there is so much published data, none of both huge data bases may index / point back to all of the original journal articles, patents, conference proceedings, etc. pp.  Depending on your search strategy, typically you find the data you look for in both data bases--at different places.  Sometimes "The Two" are complementary to each other, sometimes an other (third) search engine finds the data on the net that neither ACS nor Reaxys indexed (yet).
Personally, because of the design "how the numerical data are managed" I would suggest Beilstein as the better choice here: if you find your compound, the data base will list in a table wavelength and molecular absorption coefficient altogther with the bibliographic reference -- often including the solvent of the analysis.  And the thought occurs to me, it is easier to query with Boolean logic, like "find a compound of molecular mass in the range of xyz to x'y'z with an $\epsilon$ greater than 20000", for example.
Less in coverage -- yet worth to point to are the compilations Sadtlers Handbook of ultraviolet spectra (today part in Bio-Rad's "KnowItAll" database) and the handbook of stains, dyes, and indicators by Sigma-Aldrich, too mention some of the printed references usually found in the section of spectroscopy/chemistry in university libraries.  Here you find not only one $\epsilon$ at $\lambda_{\mbox{max}}$, but a complete print of the spectrum.  (Be aware, the elder the sources sometimes use wavenumbers or kilo-Kaysers instead of nanometers to reference the wavelength.)
Taking into account your user profile related to quantum chemistry, perhaps you want to train your quantum chemical model with known experimental data, to evolve from a descriptive to a predictive -- and small molecules may serve you well, too.  If so, take a look at NIST's webbook based on these references.

Last but not least: Take these data with a grain of salt, often there is some limitation: the same compound / material often exhibits a slightly different UV-Vis absorption spectrum if analyzed once in solvent A, and subsequently in solvent B.  Bathochromic / hypsochromic shifts (solvatochromism) may still occur.  Solvent polarity is among the  parameters in this play.

Thank you for such a great answer. I am in University of Colorado and here we have not Beilstein or Reaxys in our library database ( that's such a shame ! ).  We have access to SciFinder but I think it has not consistent data on UV/Vis. For example 75-07-0 ( acetaldehyde ) in "substance identifier " part gives spectra of everything but UV/Vis ! ( Or maybe I am missing something in my search ?)

@Aug 1) The old Beilstein database (about compounds considered as organic) and the old Gmelin database (for what was considered organometallic) were blend into Reaxys with addition of some bio-related fiels. 2) SciFinder should be the same in its results regardless of the location of access -- searching "acetaldehyde" and then clicking on "spectra" offered at least two references to UV-Vis spectra.

@Aug Perhaps I mix University of Colorado in Boulder with the University in Denver (sorry), yet the currently seen list under http://ucblibraries.colorado.edu/research/subjectguides/chemistry.htm (still?) mentions both Scifinder/ACS side-by-side with Beilstein.  Albeit the later offers numerous more entries regarding UV-Vis of acetaldehyde, I equally see it being listed under "Beilstein Crossfire", a name used till 2009 (before the merge of Beilstein and Gmelin database into Reaxys).


text:
Why does sunlight cause colors to fade?
If you leave something outside, its colors seem to inevitably fade or bleach due to exposure. Is this due to UV absorption? What sort of mechanism causes this - is it that man-made dyes deform on a molecular level? Are there notable materials that are exceptions?
I have a full undergraduate background in physics but not so much training in chemistry. A technical explanation would be cool, though.
To turn this string of questions into a focused query, let me refer back to the title:
Why does sunlight cause colors to fade?

Here's an interesting observation I hope an answer touches on: red paint is noticeably the fastest colour to degrade, while blue is the slowest. Multicoloured posters exposed to enough sunlight eventually become completely blue; even the mixed colours become a faded blue, as if the "RG part" of the RGB colour space were almost entirely removed.

Dyes and pigments work by absorbing certain wavelengths of light and reflecting or transmitting the rest. When a dye molecule absorbs a photon, an electron is excited to a higher energy state. Most of the time (neglecting fluorescence), the molecule de-excites by giving off heat and returns to the ground state intact. However, because the excited state is a high energy state, it has the potential to undergo a chemical reaction, breaking a covalent bond or otherwise irreversibly reacting with another molecule. This changes the electronic structure of the molecule which changes its absorption properties: e.g. many dyes that absorb visible light have large systems of conjugated double bonds and if these are broken, the absorbance can shift to much shorter wavelengths. How likely this kind of destructive chemistry is depends on the nature of the dye. Organic dyes tend to be more susceptible to photobleaching than things like quantum dots and inorganic pigments.
Touching on Nicolau's comment, red paint can often degrade faster because it absorbs higher energy (shorter wavelength) light. The more energy that is available, the more likely it is that reactions will occur. This is more pronounced as we enter UV wavelengths, which is why UV-B can cause thymine dimers in DNA, which can lead to skin cancer. For outdoor signs and such, UV-blocking coatings are often applied to extend the life of the dyes.
It's also not necessary for the light to interact directly with the dye molecules to cause bleaching: depending on what other molecules are present with the dye (solvents, etc), these other molecules may produce reactive species that then react with the dye.


text:
How to synthesize 6-methylhept-1-ene

What methods are there to produce this compound? What are its properties as a component of automobile petroleum? How does it impact on detonation resistance of automobile petroleum?

I think that the title compound might be produced from catalytic cracking of petroleum. But I don't have a clue of what methods are there to produce exactly this specific compound. I also think that it helps to increase the detonation resistance of automobile petroleum but I can't explain why. So I'm seeking for the explanation. Surprisingly, there's nothing on the Internet, so this site seems to be the only place where I can possibly find some information.

There are a few widespread misconceptions about this topic. The original question, edits, and comments give the impression that such misconceptions may also contribue to the confusion in this case. Therefore, we should like to clear up some of the misconceptions first.
High-octane fuel does not contain a high concentration of octane.
The composition of gasoline varies widely; anyway, typical gasoline contains less than 1 % octane.
The octane number of gasoline does not describe the composition of gasoline. It refers to the measurement of the knock resistance of gasoline, which is usually expressed by their research octane number (RON) or motor octane number (MON). Both parameters are determined in single-cylinder four-stroke test engines with an adjustable compression ratio according to standardized test procedures, e.g. EN 25164 or ASTM D 2699 for RON, and EN 25163 or ASTM D 2700 for MON. During the measurement, the compression ratio is raised until a certain standardized level of knocking is reached. The corresponding octane number is determined by finding a reference mixture of heptane and “octane” that gives the same knock level. The octane number is the fraction of “octane” (in %) in the “octane”-heptane mixture. For example, gasoline with an octane number of 95 has the same knocking performance as a mixture of 95 % “octane” and 5 % heptane; however, this does not mean that the gasoline contains 95 % “octane”.
Nevertheless, the wrong idea that gasoline would contain a high concentration of octane is widespread. For example, in the first relevant Google hit, Ritter, S. Chem. Eng. News 2005, 83 (8), 37:

Octanes are an important component of gasoline because they help provide smoother combustion in the car’s cylinders and prevent knocking.

Also remarkable is the fact that many textbooks and lectures select the combustion reaction of octane of all the possible combustion reactions of gasoline to show the principle of combustion:

$$\ce{2C8H18 + 25 O2 -> 16 CO2 + 18 H2O}$$

The reference substance for the octane number is not octane.
The octane number refers to a mixture of heptane and “octane”. However, the “octane” used for this mixture is not the unbranched alkane octane. It is actually a branched isomer of octane with the traditional name “isooctane”. Thus, gasoline with an octane number of 95 actually has the same knocking performance as a mixture of 95 % “isooctane” and 5 % heptane; however, this also does not mean that the gasoline contains 95 % “isooctane”.
The octane number of pure octane is actually very low. Its knock resistance is worse than the performance of pure heptane, which has a octane number of zero by definition.
The “isooctane” that is used as a reference substance for the octane number is not isooctane.
The name isooctane was in accordance with older nomenclature rules. The corresponding systematic name according to current IUPAC recommendations is 2-methylheptane. However, the “isooctane” that is used for the determination of the octane number is not isooctane, i.e. it is not 2-methylheptane. This “isooctane” actually is 2,2,4-trimethylpentane, which is another isomer of octane. Thus, gasoline with an octane number of 95 actually has the same knocking performance as a mixture of 95 % 2,2,4-trimethylpentane and 5 % heptane; and once again, this does not mean that the gasoline contains 95 % 2,2,4-trimethylpentane.
By way of comparison, the octane number of pure isooctane (2-methylheptane) is only about 23.

Irrespective of which compound is really meant in the original question (octane, 2-methylheptane, or 2,2,4-trimethylpentane; the question and comments also mention the similar alkenes 6-methylhept-1-ene and 2,4,4-trimethylpent-1-ene, which might be a mistake), such components of gasoline are usually not selectively synthesized in a pure form.
Gasoline consists of hundreds of different compounds, and the exact composition can vary widely. It is produced by blending of different product streams of oil refining. Today, only a small part (so-called straight-run gasoline) is directly obtained by distillation of crude oil. Other important product streams are thermally cracked gasoline, catalytically cracked gasoline, catalytic reformate (so-called platformate), isomerate, alkylate, polymer gasoline, and oxygenates.


text:
How to make hydrogen peroxide into gel
How to make hydrogen peroxide into a gel or just simply really thick and it has to be able to be in your mouth?
The only thing i have came up with is (Gelatins).
What should i use?

There are a number of vegetable gums, such as carageenan, pectin and gum arabic, and man-made methyl cellulose which might be compatible with low concentrations of $\ce{H2O2}$, which you can determine easily. Any impurities, particularly transition metals, may make $\ce{H2O2}$ decompose, so it might be best to keep the gelling agent and peroxide separate until just before use.
If for oral use, this site suggests 0.75% to 1.5% solutions, one quarter and one half the strength of the common antiseptic N.B. $\ce{H2O2}$ bleach is much stronger and is not for oral use. Note that there are dangers with use of peroxide as a mouthwash.

gels are of different types and shelf lives. If you are thinking of a true gel (water gel) then natural gums like guar, carrageenan, pectin, gum Arabic, etc are not useful. The gel will break down within no time. There are no material that can give hydrogen peroxide a good viscosity with shelf life more than 7 days.

So that sounds reasonable then, to mix the gel at the time of use.  So making a thicker gel with pectin or xantham gum and applying it should be fine.


text:
How do auxochrome groups work?
Functional groups like $\ce{OH}$, $\ce{NH2}$, attached to unsaturated compounds (chromophores) are called auxochromes. These groups tend to shift the absorption wavelength to the infrared region, which is called a bathochromic shift.

Why do these groups shift the light absorption wavelength?
Why exactly to the infrared region; that is, why is the absorption red-shifted or lowered in energy?

Generally speaking, the typical auxochromic groups possess (at least) one pair of non-bonded $n$-electrons and -- if taken alone -- do not absorb in the UV.  If attached, for example to a benzene ring, their pair(s) of $n$-electron interact(s) with the $\pi$-electrons, and the delocalization of the $\pi$-system is extended.
Now the catch:

Bathchromic auxochromic groups (like the mentioned $\ce{-OH}$ and $\ce{-NH2}$) increase the electron density in the $\pi$-system. This is typically caused by a strong $+M$-effect. As they lower the excited $\pi^*$-level, a longer wavelength is needed for the excitation along $\pi{} \rightarrow {} \pi^*$, hence the redshift.
In contrast to the former, substitution equally may alter the $n \rightarrow{} \pi^*$ transition, inducing the opposite effect, moving the absorption towards higher absorption (blue) and a hypsochromic shift.  One example is the to compare the absorption of indigo yielding "jeans blue" with 6,6'-dibromoindigo, main constitutent of tyrian purple. A second, that heteroatoms next to a $\ce{C=O}$ carbonyl group induce a hypsochromic shift like in the case of acetaldehyde example here in comparison to acetic acid here

Note the use of the terms of "bathochromic shift" and "hypsochromic shift" along solvatochromism.

I wasn't aware that electron density played a role.  I thought the basic premise was that any extension of the conjugation path leads to a smaller HOMO-LUMO separation and consequently a bathochromic shift (particle in a box).  Looking at the Woodward-Fieser rules even bromine and chlorine lead to a bathochromic shift since they extend the conjugation path.

@ron I altered the second part of my answer and provided some examples to support this section.  Provided unsaturated compounds often provide both $\pi \rightarrow{} \pi^$ and $n \rightarrow{} \pi^$ transitions, I now speculate the Woodward-Fieser rules empirically base on the former ($\pi \rightarrow{} \pi^*$).

Thanks for the clarification!  I think you're right about the W-F rules.


text:
Single displacement reaction
Will sodium metal, if mixed well with iron oxide, displace the iron in a chemical reaction? I know thermite involves the mixture of aluminium powder and iron oxide, and the single displacement reaction results in molten iron and aluminium trichloride. The other day I melted some sodium metal to the point of melting but reacting with the air. I wonder if either mashing up sodium with iron oxide and then igniting, or adding melted sodium to iron oxide will act like a thermite reaction.

Sodium metal will react with iron (III) oxide to produce sodium oxide and iron.
$\ce{6Na +Fe2O3 -> 3Na2O + 2Fe}$
This is also like a thermite reaction, so try to use the soam\e method as the reaction.


text:
Mercury into mercury oxide
How can I turn mercury metal into mercury oxide? I know that many metals will react with oxygen if enough heat is supplied in the presence of oxygen. If I heated up mercury enough and also supplied oxygen from my oxygen tank, would that result in mercury oxide? Also how about lead? Thank you

Which lead oxide do you want to make ?

If you give enough oxygen to let it react to mercury, it still won't work just like copper and sliver. Because they are unreactive metals.
However, it will form mercury oxide when mercury and oxygen react at 350°C as the chemical equation below.
$\ce{2Hg + O_2 -> 2HgO}$
P.S: The mercury oxide will decompose back to mercury and oxygen at 400°C.
Lead is a rather more reactive subatances. So, try burning it. If it don't work, means you need a higher temperature.

Thank you very much! So how much oxygen would I need? Could I just use a bernzomatic oxygen tank and blow oxygen onto the mercury while heating the mercury in a crucible? And would lead work the same way?

Remember that one mole of oxygen gas react with two mole mercury atoms. If you are thinking about how much to use, try to learn a bit about stoichiometry.

For an advise, use a boiling tube instead of a crucible. The mercury oxide might escape when you use a crucible.

I would strongly advise against this mercury(II) oxide synthesis. Metallic mercury is highly toxic, and its low vapour pressure at room temperature is already an exposure hazard. Heating it to 350°C means heating to its boiling point, making it a much larger containment problem. A better idea seems to be reacting the mercury with excess concentrated nitric acid, creating $\ce{Hg(NO3)2}$, which can decompose thermally to $\ce{HgO}$. This is also hazardous, however, as $\ce{Hg(NO3)2}$ is highly soluble in water. Why do you even want to perform this synthesis?


text:
Chemical formula and heat
In my Physical Science class, we are learning about chemical formulas and how to balance them. I balanced this equation:
$$
\ce{Fe2O3 + Al -> Fe + Al2O3}
$$
One of the other students asked me if this chemical formula releases or absorbs heat.
Could anyone help me give a great explanation for this question?

https://en.wikipedia.org/wiki/Aluminothermic_reaction

Check $\Delta H$ tables and calculate if the reaction is endothermic or exothermic.

@Sydney U:  Please note currently (22:10 UTC time), your reaction equation is not yet balanced correctly.

Look up thermite for more information on this displacement reaction. My students enjoyed seeing it demonstrated.

The best way to solve this problem is to look at the $\Delta H$ charts found online. Using the simple equation below will give you the answer.

$$\sum \Delta H_{\text{Products}} - \sum \Delta H_\text{Reactants} = \Delta H_{\text{Reaction}}$$
  If the $\Delta H$ is positive in sign then the reaction is endothermic and if it is negative in sign then it is exothermic.


text:
How is drift due to static electricity eliminated from an analytical balance?
I've recently encountered a problem when using an analytical (0.1 mg accuracy) scale to determine the mass of a sample of ferrocene powder. The reading in the balance seems to drift either upwards or downwards when I drop some of the powder in an erlenmeyer flask. The reading stabilizes when I add water to form a solution.
I've heard that static electricity might cause this kinds of problems when measuring powders so I suspect this is the case here, because the problem only appears when I add the ferrocene powder. I've try reaserching some ways to get rid of the problem, a static gun seems to work in some cases (no idea if it would work with a powder though) and I've heard that air ionisers also work, but don't have access to either of these.
Although this is the first time I've encountered this situation, I've heard its pretty common, so what do people usually do to get rid of the problem in a cheap manner? Is there an easy solution?
I'm using gloves to operate the scale, which I've been told to do in order to keep the flasks clean of oil from my fingers. Can the gloves be worsening the problem? What can I do about that?

Static charge can affect the measurements made by an analytical balance, as can magnetic materials because the mechanism in modern analytical balances involves a feedback loop between the load that is measured and an electromagnet. Insulating weighing vessels (plastic or glass) tend to accumulate static charge, especially when handled with another insulator (gloves) by the triboelectric effect.
You should first make sure that the body of the balance is grounded properly according to the manual for your model.
As a short-term solution to static charged glass, you might try arranging the sample so that it is father away from the balance. One way might be to place the sample on top of an empty sample bottle on the balance. As you mentioned, a more robust solution for discharging static build-up is to have a commercially available air ionizer nearby. The device will either produce an equal amount of positive and negative ions in the air at high voltage AC or DC electrodes, or with an ionizing radioactive material such as is found in anti-static brushes used for cleaning vinyl audio records.
If you are frequently measuring magnetic materials, many analytical balances can be configured in a bottom-loading arrangement. In this setup a platform is hooked beneath the balance, and the sample is supported sufficiently far away from the force-restoring motor so that it does not influence the measurement.

Wearing gloves may contribute to the static charges you observe, especially if you have to work with gloves not only to protect yourself, but per se you can't do otherwise (glove box) and a special, palm size ionizer is indeed a remedy.

While I do not pretend "it works every time", these are my suggestions, that sometimes work alone, sometimes in combination:

Electric discharge of yourself.  Sometimes over the day I become "charged", too, and a simple tip of my hand to the (air dry) water pipeline or heating pipe / radiator removes some of the charges that can't reach earth while walking in firm (but nonconducting) lab shoes over polymer-coated surfaces.
Electric Discharge the clean spatula.  Prior to transferring material to the balance, tip the metallic spatula against something that is both conductive to the earth and clean.  Clean and air dry water pipeline, heating pipe (vide supra) are handy.
Attempt the mass transfer with a spatula made of a different material rather than metal.  Beside glass, different plastics may be represent a surprising  alternative.  Krischenbaum et al. reported recently how to deploy drinking straws as disposable spatulas in Journal of Chemical Education.

Do you have any specific suggestions for weighing inside a glovebox? It has frustrated me plenty in the past, but I didn't have access to anything fancy like a static gun or ionizer. Can analytical mass measurements still be performed in this situation?

@Nicolau Saker Neto: No doubt about your observation.  The normal gloves + the ones to protect them against the chemicals while weighting-in milligram-quantities of catalyst.  It is possible thanks to the existence of such devices, I'll edit accordingly.

Picture reference: http://www.gloveboxes.com/access/static-eliminator.php

Would it help to use a different vessel to put the powder on?  Like a metallized weigh boat or a piece of aluminum foil instead of plastic or paper?   I'm not an expert in this area but I feel like I've seen similar products advertised as helping in cases like this.

@CurtF.  Changing the material the weighting boat is made may influence the situation when the material intentionally leaves the spatula (and there are boats made of glass, too).  I recall an experience to transfer a sub-mg quantity of Pd catalyst in the glove box, where once the ionizer was switched on (then visible plasma near its tips) the transfer on paper was eased, yet till cumbersome and these little pans behaved much better.

I have been using gloveboxes for several years in different labs across the world and I find that Ar gloveboxes completely eliminate the problem of static electricity commonly experienced with N2 gloveboxes. I have no explanation for this explanation but I am pretty sure N2 is the culprit because we once switched an N2 box to Ar and the problem disappeared. Before that we spent our time fiddling around the balance with static guns, air ionisers and such, without much success (the only exception being the Po ionisers which are pretty efficient). On the other hand, when purging an Ar box with N2, static electricity will usually appear for a couple of days. In my opinion the price difference between N2 and Ar is simply not worth the trouble.

The largest part of the answer does not really handle the question.


text:
How many hydrogen atoms are in a sample of NH3?
How many hydrogen atoms are contained in 8.66 moles of $\ce{NH3}$?
The answer is given after the log has been taken, so the final answer is 25.19.
What I have been doing is multiplying $8.66\text{ by }6.02\times10^{23}=5.21332\times10^{24}$ and when I take the log of that I get 24.71. I'm not sure what to do to get the correct answer.

Before you use log you have to multiply the answer of $8.66~\mathrm{mol}$ by $6.02 \cdot 10^{23}$ by 3 because the answer of $8.66~\mathrm{mol}$ by $6.02 \cdot 10^{23}$ gives you the number of ammonia molecules and in each molecules there are 3 hydrogen atoms present.
So it is:
$$8.66 \cdot 6.02 \cdot 10^{23} = 5.21332 \cdot 10^{24}$$
$$5.21332 \cdot 10^{24} \cdot 3 = 1.563996 \cdot 10^{25}$$
$$\log(1.563996 \cdot 10^{25}) = 25.19$$


text:
UV/Vis spectroscopy transition reference
There are mainly fundamental transitions happening during excitation and UV/Vis spectroscopy (as n- > σ* ,π->π* ,n- > π ,..).  I am looking for a good reference that shows these transitions for molecules (for example shows n- > σ* for acetone , π- > π* for benzene,... ).
Actually I am talking about tables. I can do guesswork for most of the molecules based on their structures but definitely I need references for publication.
Update: Mostly interested in these molecules: Acetone, Acetaldehyde , Benzene , Naphthalene, Anthracene , Methylene Blue, Phenanthrene , 1-Butanol

Here are a number of resources that may be of use for your research.
The UCLA website presents tables and data in their webpage Theory of Ultraviolet-Visible (UV-Vis) Spectroscopy, including reference tables of common functional groups, aromatics and compounds often used in the laboratory.  (Many compounds you listed are in the tables).
A reference that may be of use is Spectroscopy of Organic Compounds (Kumar, 2006) which include several tables that include information of the compounds you list.
The book Spectroscopy of Organic Compounds (Kalsi, 2007), from page 12 provide several tables of information related to your listed organic compounds, including details of the electronic transitions. The chapter "Ultraviolet (UV) and Visible Spectroscopy" (starting page 9) has a subsection "The Designation of Various Transitions" which details the electronic transitions using examples in context (from page 11, summarised in a diagram on page 12).  Examples of solvents are discussed from page 23 onwards.  (All available online).
Transitions are explored in a bit of detail in chapter 4.3 of the UC Davis ChemWiki The Dynamic Chemistry E-textbook, Section 4.3: Ultraviolet and visible spectroscopy


text:
What an aqueous solution really contains?
I want to know what an aqueous solution is. For example, let's consider a $1~\mathrm{M}$ aqueous solution of $\ce{MgCl2}$. Does it also contain $\ce{MgOH2}$? Though it is a simple question, I feel it is important to understand something as vital to chemistry as this.

It's alright, everyone has trouble with this at first. :) Aqueous means that the specific substance has been dissolved in $\ce{H2O}$.

Then aqueous solution of MgCl2 means MgCl2 is dissolved in water

Not really a simple question at all, but a topic that can cause difficulty.  Good question!

water as solvent --> aqueous solution

If you are going to answer, please make sure that your answer adds something to discussion and does not rehash what others have already written.

Uh... I think the insight you need is about chemical equilibrium.
An aqueous solution is exactly the mixture formed when something dissolved in water. This is easy to understand, but here is a more difficult question: what's the component of this mixure?
To answer this is not easy.
In a glass of pure water, there are still a little $\ce {H+}$ (to say presicely, $\ce {H3O+}$) and $\ce {OH-}$. This is because sometimes proton ($\ce {H+}$) will "jump" from one $\ce {H2O}$ molecular to another one.
$\ce {2H2O <=> H3O+ + OH-}$.
(Since we usually do not use the expression $\ce {H3O+}$, this process can be rewrited as $\ce {H2O <=> H+ + OH-}$ to avoid disruption.)
This phenomenon (called the autoionization of the water) is very weak. The concentrations of $\ce{H+}$ and $\ce{OH-}$ in "pure water" is $\ce{[H+]=[OH^{-}]=}1.0 \times 10^{-7} M$ when the temperature is 25°C (298K).
Then you put something in this glass of water. As other people mentioned, most of covalent compounds, such as glucose, will not break apart in water since the intermolecular force (covalent bond) do not break. Water moleculars may surround those covalent compounds like a cage to make them dissolved. So there won't be other things.
But sometimes it breaks: when acid compounds are dissolved into water, like $\ce {HCl}$ and $\ce {H2SO4}$, they will give their hydrogen atoms as proton to water:
$\ce {HCl + H2O -> H3O+ + Cl-}$
$\ce {H2SO4 + H2O -> H3O+ + HSO4-}$
This is because they are more acidic than $\ce {H3O+}$, here "acid" means "who gives proton(s)", while "base" means "who accepts proton(s)" (This kind of definion is called as Brønsted-Lowry theory, and as you see, in autoionization of water, one of the $\ce {H2O}$ is acid and another is base).
Noticed that acid like $\ce {HCl}$ and $\ce {H2SO4}$ will nearly definitely give their $\ce {H+}$ to water compounds (they are too “willing” to do it), we call those acid as strong acid. But some acid are not: protons sometimes jump back.
For example, $\ce{HSO4-}$ will meet an equilibrium:
$\ce{HSO4- + H2O <=> H3O+ + SO4^2-}$
This kind of acid will be called as weak acid. In an aqueous solution of a weak acid, most of the weak acid compounds keep unchanged.
Yet we haven't mentioned what you concerned: how about aqueous solution of ionic compounds? They usually break their intermolecular forces (ionic bond) via hydration, i.e. combination with water compound. Water compound have some charges on them, so they will attract ions and break them apart. Especially, we will see bases we are familiar with -- those who will ionized as cations and $\ce {OH-}$ -- first.
In fact, the structure of hydrated cation will be complex ("$\ce {H3O+}$" is also not the actual situation), so we usually just write them as unhydrated form, such as $\ce {Fe^2+}$ or $\ce {Na+}$. As you know, there are weak bases and strong bases.
So bases like $\ce {NaOH}$ are quite "willing" to accept protons:
$\ce {NaOH -> Na+ + OH-}$
But $\ce {Fe(OH)3}$ may not, and it also meet an equilibrium:
$\ce {Fe(OH)3 <=> Fe^3+ + 3OH-}$
Again, in an aqueous solution of a weak acid, most of the weak base sturcture keep unchanged. And for this reason, weak bases are usually insoluable in water, since water "have no good idea" to deal with the whole relatively large ion sturcture. But in fact, the word "insoluable" means that there is another equilibrium like this:
$\ce {Fe(OH)3_{(aq)} <=> Fe(OH)3_{(s)}}$
and the solid form dominates.
So here we meet the hardest part: what about adding some salts into water, like $\ce {MgCl2}$ you mentioned?
So far you have learned about ionization of the salt, and you know most of salt will completely ionized. But how about those ions?
A balance like $\ce {Mg(OH)2 <=> Mg^2+ + 2OH-}$ shows that $\ce {Mg(OH)2}$ is a  weak base. But if we see it as $\ce {Mg^2+ + 2OH- <=> Mg(OH)2}$ it means $\ce {Mg^2+}$ is a weak acid. And if you remember I have said in autoionization of water, the water compound can be both base and acid (we say this kind of substances are amphoteric). Then we can know that there will be a new equilibrium like this:
$\ce {Mg^2+ + 2H2O <=> Mg(OH)2 + 2H+}$
(We usually call this as hydrolyzation of salt.)
And because $\ce {Mg^2+}$ is a weak base, you may expect that there will be not a lot $\ce {Mg(OH)2}$ and it's still soluable. Also there will be not a lot $\ce {H+}$ so the solution is just a bit more acidic then water.

Addition: You can also think this comes from these two balances:

$\ce{H2O <=> H+ + OH-}$ (autoionization).
$\ce{Mg(OH)2 <=> Mg^2+ + 2OH-}$. in further calculation you may meet in future,  this kind of idea can make you avoid some mistakes.

So you may expect in a $1M$ aqueous solution of $\ce {MgCl2}$, there will be a lot of $\ce {Cl-}$ ($2M$) and $\ce {Mg^2+}$ (a bit less than $2M$), a little $\ce {H+}$ and a little $\ce {Mg(OH)2}$, and there will be $\ce {OH-}$  which is negligible in the most case. (Of course there is very much water!)
While you can see in the case of $\ce {NaCl}$ there won't be $\ce {NaOH}$ and $\ce {HCl}$ compounds in the solution: because $\ce {NaOH}$ and $\ce {HCl}$ are "strong", there won't be a hydrolyzation equilibrium. But yes, you can say there are $\ce {H+}$ and $\ce {OH-}$, as you have known, there are $1.0 \times 10^{-7} M$ of them. However, we usually not concern about this. And as you see, all those $\ce {H+}$ and $\ce {OH-}$ in an aqueous solution, in effect, come from water compounds. (This is why there will be "strong" acid and base: because they are all giving/receiving protons from water, even if they actually have different ability, compared to water, they are behaving undistinguishably. This is called as leveling effect of solvent.)
There are a lot of things I haven't mentioned here, because I think you are not able to calculate something about equilibrium, and maybe you are not familiar with thermodynamic idea yet. But here I just hope what I said is interesting enough for you.
Well, a simple conclusion:

To find out what will be there in an aqueous solution, you may consider:

If the solute ionize: Is it a covalent compound? Is it acid?
If yes, whether the ions are weak acid or base (in Brønsted-Lowry theory)? Remember, a cation which can be formed from a weak base will be weak acid; while an anion which can be formed from a weak acid will be weak base. (Or you may use an awful word: they are conjugate acid base pair!)

I think these two step are enough for you for the time being.

Ahaha! I think you really like explaining things! Keep up the good work! (But including a conclusion is very recommended with long answers)

Good job on taking the time to properly format your answer with LaTeX. Just a tip: you can write equilibrium arrows using \ce{<=>} instead of   \ce{<->}, so you get the correct $\ce{<=>}$ instead of $\ce{<->}$.

@Asydot>" ionic compounds? They usually break their** intermolecular** forces" or  Intramolecular?

Aqueous means that the specific compounds has dissolved in $\ce{H2O}$ or in other words, it dissociated into it's ions. When it is dissolved:

Ionic Compounds: $\ce{NaCl}$ for example dissociates in water to $\ce{Na+}$ and $\ce{Cl-}$ ions. Intramolecular breaking.
Covalent Compounds: Covalent compounds do not break apart but they do break their intermolecular force.

And one more thing, SOLIDS, LIQUIDS, GASSES aren't usually aqueous. I'm saying usually because there is always some exception to this and someone will point that out.
Hopefully this helps, comment for more information.

Then an aqueous solution of solution of NaCl is a mixture of Na+, cl-,OH-and H+.

Then do  Na+, cl-,OH-and H+ mixed and form NaOH and H2O?

Compare energy released in dissociation of NaCl and energy required to form NaOH/HCl . You will have your answer.

Sorry for this, but I don't understand the solid, liquid, gas thing ... If you are talking about the notations s, l, g and aq, then there shouldn't be exception. We use s, l, g for roughly description of the state of substances, but aq is accurate; it is, I think, refer to a phase. If you see aq and l in the same chamical equation, then they must be in different phases.

Exactly, that's what I meant.

Aqueous means your solvent for the solution is water. It's a term to differentiate from other solution like alcoholic or etheric, etc.
On the sol chem part of your question, a molecule can take many different form in solution and their amount can usually be calculated with a constant and equilibrium equation, but for $\ce{MgCl2}$, it is considered that most of the salt that dissolve turn into $\ce{Mg^{2+}}$ and $\ce{2 Cl-}$.

An aqueous solution is a solution in which the solvent is water. It is usually shown in chemical equations by appending (aq) to the relevant chemical formula. For example, a solution of table salt, or sodium chloride (NaCl), in water would be represented as NaCl(aq).

Source: wikipedia

So does it means NaOH is reacted with water and formed NaOH?

Not on this case, because $\ce{NaOH}$ is a very strong base and won't be formed in solution. But if you dissolve, for example, $\ce{NH3}$ in water (so, an aqueous solution of ammonia), you'll have the equilibrium $\ce{NH3 + H2O <=> NH4OH}$.

You can't directly copy paste from the wikipedia. If you do then you need to mention it as a source and add directly pasted stuff in blockquote. Please consider the editing.