text:
Why is mercury used in thermometers?
I know it expands when hot and grows compact when cool (that's about all I know), but I suppose many other substances do, too.
Also, and excuses for my ignorance, what the term used to describe the function of mercury in a thermometer? I mean, would you call it an "indicator" (insofar as it indicates the temperature), or is there some technical term?
Thanks!

In addition to the answer and the comment given, the use of mercury in thermometers had another advantage frequently used in the lab:
Mercury, as a metal, has a high electrical conductivity. Imagine two thin wires inside the capillary of your thermometer. One of them is fixed.
The other wire is can move along a threaded rod. Typically, this was realized via magnetical coupling to a freely rotating knob on top of the (hermetically sealed) thermometer. Upon heating, the mercury expands and finally reaches the upper wire previously adjusted to a particular working temperature. At this point, an electrical current may flow through the thermometer.

Source: http://upload.wikimedia.org/wikipedia/commons/4/48/Kontth.jpg
These contact thermometers, nowadays replaced by mercury-free electronic thermoswitches, once were the working horses in the lab when it came to controlled heating of oil baths or monitoring cooling circuits for uv lamps.

"In addition to the answer and the comment given..." -- There is no other answer or comment for me. Too bad, curious what this answer was made in addition to. Thanks!

Unfortunately, the comment by Michael D. M. Dryden was attached to an answer that was deleted by the owner. So the comment went with it. In the comment, it was pointed out that mercury can't be used for temperatures below -30 °C; kerosene and isoamyl acetate were mentioned as for low-temperature thermometers. Furthermore, it was mentioned that water would a possible but limited alternative for a narrow range (> 0 - < 100 °C.

Reasons:

It expands linearly (i.e a 1 kelvin temperature increase almost always produces the same change in volume over a reasonable range of temperatures). You might think that this is the norm - but not always. For example, water does not expand linearly. It has a maximum density at $4~\mathrm{^\circ C}$.
It can measure over a reasonable range of temperatures, from $-39~\mathrm{^\circ C}$ up to its boiling point.
It is quite dense in the first place. This is so that it can be used in configurations such as the Six's Thermometer — a type of thermometer that records the maximum and minimum temperatures of an area (I won't go into schematics now, although it is quite simple)


text:
Can you calculate the properties of a substance based solely on its atomic properties?
I'm trying to write some software that I can use to determine, roughly, what the physical properties of a pure substance are. I know I could just use a database of the known properties of each element, but that doesn't do much for me when I go to create molecules and combine substances. A database that lists how each element will react to another is likely impossible, so I got to thinking... there has to be a mathematical way of determining these properties!
For example, I understand that there are trends on the periodic table that show that the stronger the force between the molecules is, the higher the melting point will be. I also understand that we can calculate how strong that force is based on what we know about the atom. So why can't I find any information relating the two in some mathematical way? The same goes for color, hardness, malleability, etc. Do we really only know these values through testing? We know the values but we don't know why they are what they are?
I know this is deeply complex topic and may not even be computational possible. In those cases, I'm trying to do as much as I can to approximate.

Would you agree that the behavior of electrons and nuclei are atomic properties?  If so, with an appropriate operator, we can extract the properties you are talking about from the wavefunction.

While theoretically speaking we should expect 'yes' no general practical method is known. \

@permeakra, hmmm. The way I interpret this question, we should expect 'no' as the answer. The right way of determining properties of a molecule is to solve the Schrödinger equation for all the nuclei and the electrons. Strictly speaking, there is no even atoms in molecules in a sense which is given to them in this question.

@LordStryker, can you expand upon what the appropriate operator(s) may be? Also, can you link me to information regarding the wave function? I'm still pretty new to chem.

@Wildcat, how can I use Schrödinger's equation to determine molecule properties? I'm looking at the wiki page right now and... am quite lost. Pretty new to chem. Can you point me in a direction?

@Stradigos, sure, Google "quantum chemistry" and follow the white rabbit. :-D

@Stradigos One of the postulates of quantum mechanics is that all the information about chemical system can be represented (i.e. contained in) the wavefunction.  You may note that the Schrodinger equation uses the Hamiltonian operator to return the ENERGY of a system (energy being an example of a property that we want).  The Hamiltonian is an appropriate operator for the S.E. to obtain the eigenvalues  (i.e. energies) of the system (psi) we are representing.

@LordStryker How then do energies translate into physical properties? I read somewhere that the melting point of an element can be high because it takes more energy to break apart the bonds (especially covalent ones). Is that the connection I'm looking for or did I miss the point?

@Stradigos Energies can relate a lot of things.  For example, it gives insight into why water breaks NaCl bonds as opposed to the other way around (the energy of an OH covalent bond is larger than the energy of an NaCl ionic bond).  Energies tell us what molecules are likely to be observed experimentally.  They tell us the actual structure of atoms and molecules and their vibrational and rotational motions.  A lot of these pieces of information give us insight into physical properties like what you are inferring about.

For atomic and molecular-level properties (bond, angle, torsion energies, etc), quantum mechanics provides the equations for solving for these properties. However, there are a number of approximations that are employed, especially for larger, more complex molecules.
For bulk properties of a substance, thermodynamics and statistical mechanics provide equations for solving for a variety of material properties (critical properties, phase equilibria, various types of enthalpy, and so on). Read up on equations of state.
There is also the option of molecular simulation. In this case you are solving relatively simple, well-established equations (eg, Newton's equations of motion in molecular dynamics simulations) repeatedly in order to compute the (usually) equilibrium properties of a substance of interest.
In all cases, the accuracy varies, and validating results by comparing various methods is considered best practice. While there are "purists" out there who believe that all parameters in a model should be derived from first principles alone, in practice models are often fit to experimental data (or even data from quantum mechanical calculations) to improve accuracy.

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text:
Resonance contributors and IR stretching frequency
I am told that a carboxylic ester would have a higher carbonyl stretching wave number than a carboxylic amide due to there being more double bond character in the carbonyl group of the ester.
This makes sense prima facie; the resonance structures suggest that there is somewhat more double bond character in the carbonyl of the ester. And nitrogen and oxygen are nearly identical in mass.
Is this actually true and applicable?

And what's your question? Nitrogen is less electronegative than oxygen, so it gives its electrons in conjugation more easily, if that's your question.

Yes, it is true and applicable. In fact, the resonance structure you've drawn with the double bond between the carbonyl carbon and the amide nitrogen is so significant that one can actually observe restricted rotation about this bond on the nmr timescale.  That is to say, if a methyl group was attached to the amide nitrogen, you can see distinct resonances for the syn and anti isomers.  The rotational barrier about the C-N bond is around 20 kcal/mol, depending upon the substituents, etc.  In the case of the ester, for the reason you provided, the barrier to rotation about the carbonyl carbon - ester oxygen bond is much lower, typically in the range of 10-12 kcal/mol.
So in the case of the amide, the second resonance structure is more important than it is for the ester.  Consequently, the amide carbonyl bond is weaker than the carbonyl bond is for the ester.

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Ammonia to tri substituted nitrogen
I am told that an intramolecular proton abstraction can occur. For some reason this seems at least sterically unlikely. Is it possible sterically? Or is the proton abstraction best described as a equilibrium process between the nitrogen's hydrogens and water?

Until there's water present (I'm assuming that it's not the solvent), the proton can't come from water. The proposed intramolecular proton transfer would go through a 5-membered transition state, which is reasonable. If you don't like that mechanism, then two of the intermediates could transfer protons to each other. Depending on the concentration, one or the other or both could be occuring.
The exact nature of these proton transfers are difficult to prove and typically inconsequential, so you will often see "shortcut" mechanisms drawn which imply intramolecular proton transfer even when it is more unlikely than the one you show here.


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Predicting 260/280 of DNA or RNA from sequence?
I asked this in biology, but I think they don't want to deal with the chemistry behind UV-Vis.
My question has to do with the Poly A tails on mRNA. I've been doing in vitro transcription and tailing for a while now and have noticed that the untailed RNA has a 260/280 of about 2.0, like you'd expect, but successful tailing results in ratios from 2.15 to 2.45, with higher ratios correlating to better band shifts on agarose gels, implying a longer tail. I suspect the reason for this is because the adenine has a much higher 260/280 than the other bases, so if you add a lot of adenines to the 3' end of the RNA you add much more 260 absorbance than 280 absorbance.
After seeing this I tried to come up with a mathematical method to estimate tail length based on 260/280, since gels are annoying, while I can get the ratio directly from the nanodrop.
I started by using the 260/280 for each nucleotide, A: 4.50, G: 1.15, C: 1.51, U: 4.00, T: 1.47 source here: 1. I then took the sequence for the Luciferase gene I'm using for mRNA and counted the nucleotides, 449 A, 566 C, 470 G, 374 U. I tried to get the weighted average of the 260/280 for my RNA.
A: 449 x 4.50 = 2020.50
C: 566 x 1.51 =  854.60
G: 470 x 1.15 =  540.50
U: 374 x 4.00 = 1496.00
Sum:            4911.60

Nucleotides: 449 + 566 + 470 + 374 = 1859
Sum / Nucleotides : 4911.60 / 1859 = 2.64

2.64 is clearly not 2.0. Even accounting for the potential effect of pH and ionic strength on RNA 260/280 ( acidic solutions have lower ratios ) I can't explain why the calculated ratio is so high. Even if we assumed an RNA with 25% of each base, the weighted average is still (1.15 + 4.50 + 1.51 + 4.00)/4 = 2.79, even higher than for my real sequence.
Doing the same calculation for DNA, (1.15 + 4.50 + 1.51 + 1.47)/4 = 2.16.
So why don't my calculated ratios for RNA and DNA 260/280 match the "ideal" ratios of 2.0 and 1.8? It probably has something to do with base stacking interactions, but I don't enough to compensate for it.

You could try a different pH to get rid of secondary structure. It might change the absorbance spectra as well, though. Here is a paper where they measured spectra of DNA and its constituents: https://doi.org/10.1016/S0022-2836(61)80003-X
The abstract says the following (mμ is the same as nm):

The spectra of deoxyadenylic, deoxyguanylic, deoxycytidylic and thymidylic acids are given at pH 1, 2, 3, 5 and 7, from 220 to 310 mμ. The spectrum of thymus DNA denatured in 0·l N-acetic acid at pH3, at low concentration, is almost identical to the calculated spectrum of its constituent nucleotides.
The ratio of the absorbancies of DNA at 260 and 280 mμ reaches a constant value near pH 3, characteristic of the molar proportions of the bases. This appears to form the basis of a rapid method for estimating base composition.

The problem with doing this, especially with RNA, is that secondary structure can have a large impact on the absorption properties of the molecule. Interactions between bases will change the molar absorptivities in ways that may not be easily predictable. The "ideal" ratios are only guidelines and I wouldn't worry too much that a long string of a single base produces a deviation from them. If very accurate measurements are required, hydrolysis into nucleotides before measurement is usually done to avoid these interactions. For your specific case, it may be possible to construct a calibration curve of sorts to get an estimate if you can measure the tail length by some other means. So long as the tail is sort relative to the rest of the molecule, I would expect to see some predictable change in ratios.

A calibration curve would be a good way to do it, but precise measurement of PolyA tails is not easy. The best I've seen involve dense polyacrylamide gels and short RNAs, the gel is dense enough to differentiate single base differences. My RNA is too large, but I've thought about pairing the polyA with an Oligo(dT), then digesting the unpaired with a type of RNAse. However, that RNAse would also leave double stranded RNA intact, and mRNA has a significant amount of double strandedness due to its complex structures.

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What happens when 0.02 mole NaOH is added to a buffer solution?
I am asking this question for my son.
(This question is related but we still  have our confusions.)
We have one liter buffer solution of 4.74 pH which contains 0.1 mole $\ce{CH_3COOH}$ and
0.1 mole $\ce{CH_3COONa}$.
What will be the resulting pH when 0.02 mole $\ce{NaOH}$ is added to this solution? (Given $\ce{Ka} = 1.8 \times 10^{-5}$.)
As the definition of a buffer solution [A, B] goes, pH value should not change when small quantity of acid or base is added to it.

So, one way of answering the question  will be to simply comment that since
a small quantity of base has been added, the pH value will not change.

But, we find another solution (presented and accepted in some
moderately acceptable study-sheet) in the following line.
We will have a reaction like,
$\ce{CH_3COOH} + \ce{NaOH} = \ce{CH_3COONa} + \ce{H_2O}$
So, after the reaction, we have,
$\ce{CH_3COONa}$: $(0.1 + 0.02)\ \text{mole} = 0.12\ \text{mole}$
$\ce{CH_3COOH}$: $(0.1 - 0.02)\ \text{mole} = 0.08\ \text{mole}$
Now, $pH = pKa + \log{\text{[Salt]}\over\text{[Acid]}}$
Hence,
$pH = pKa + \log{{{n_\text{Salt}}\over V}
\over
{{n_\text{Acid}}\over V}}$
Which after plugging in the values gives,
$pH = 4.92$.

We are  confused about which of the above solutions is
correct.

pH does change while buffering

The second solution using the Henderson-Hasselbach equation is correct. While buffer solutions change pH very little near the pka of their constituent weak acid, it would not be correct to say that they do not change at all. Also, an addition of 20 mmol of NaOH is not insignificant to a 100 mM buffer—a 20% change in concentration. The Henderson-Hasselback equation in the form you've shown is an acceptable estimation of the buffer pH in a moderately concentrated buffer with an acid/base ratio close to 1. Deviations from it in that form can occur for very dilute or concentrated solutions or if it's well away from the pka.


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Calculating amount of ice required for heat loss
Question:

Given $400.0\ \mathrm{g}$ of hot tea at $80.0 °\mathrm{C}$, what mass of ice at $0 °\mathrm{C}$ must
  be added to obtain iced tea at $10.0 °\mathrm{C}$? The specific heat of the
  tea is $4.18\ \mathrm{J}/(\mathrm{g}\ \cdot°\mathrm{C})$ , and $\Delta H_{\mathrm{fusion}}$ for ice is $+6.01\ \mathrm{kJ\ mol}^{-1}$ .

$q$ is the amount of heat transferred. Since the solution is decreasing heat, then $q_{\mathrm{ice\ tea}}=-q_{\mathrm{ice}}$ We can then solve q using the information given.
$q_{\mathrm{ice}\,\mathrm{tea}}=\left(\dfrac{4.18\,\mathrm{J}}{\mathrm{g}\cdot ^{\circ}\mathrm{C}}\right)(400\,\mathrm{g})(-70\,^{\circ}\mathrm{C})=-1.17\times 10^5\,\mathrm{J}$
The next part cofuses me, this is what the solution manual does.
$q_{\mathrm{ice}}=1.17\times 10^{5}\,\mathrm{J}=(6.01\,\mathrm{kJ\ mol}^
{-1})\left(\dfrac{1000\,\mathrm{J}}{1\,\mathrm{kJ}}\right)\biggl(m_{\mathrm{ice}}\times\dfrac{1\,\mathrm{mol}\,\mathrm{H}_{2}\mathrm{O}}{18.02\,\mathrm{g}\,\mathrm{H}_{2}\mathrm{O}}\biggr)$
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+\left(\dfrac{4.18\,\mathrm{J}}{\mathrm{g}\ \cdot\ ^{\circ}\mathrm{C}}\right)(m_{\mathrm{ice}})(10.0\ ^{\circ}\mathrm{C})$
I'm not too sure where this equation is derived from or of its purpose. The top part is determining the the enthalpy of ice but why is it being added to the specific heat of ice?

HINT:
$\pu{4.18 J/g K}$ is the specific heat of water. Also question assumes that due to addition of tea leaves the specific heat doesn't change much and remains equal to that of water.
So if you mix ice and tea and want the final temperature at $\pu{10^\circ C}$ then what do you expect of ice. Shouldn't it must melt and reach to $\pu{10^\circ C}$ and that will require a further energy to raise it temperature from $\pu{0^\circ C}$ to $\pu{10^\circ C}$. Also the mass of ice will remain same even if it get converted to water


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Why are imines more susceptible to reduction than carbonyl compounds?
Why are imines more susceptible to reduction than carbonyl compounds? Imines react with $\ce{NaCNBH3}$ while the carbonyls don't.

Generally speaking, I think it is the reverse - carbonyls are reduced faster than imines with various sources of $\ce{H^-}$.  The reason usually cited (see pages 2-4 of this reference) for this behavior is that the imine double bond is less polarized than the carbonyl double bond.  In the figure below resonance structure II contributes less when X=N as compared to the case X=O, due to the decreased electronegativity of nitrogen compared to oxygen.

However, if the reaction is run under mildly acidic conditions, since the imine nitrogen is more basic than the carbonyl oxygen, more of the imine nitrogen will be protonated than the carbonyl oxygen.  This will make the protonated imine carbon much more electrophilic (positively polarized).  Consequently, under acidic conditions, an imine may be more reactive than a carbonyl to nucleophilic reduction (in addition to the above reference, also see this reference).

Oh, it's because imines are protonated in a bigger amount. Thank you.

You mention "acidic conditions".  But can there be hydride sources (that do not break down to H2) under acidic conditions?

@OscarLanzi Yes, take a look at the last reference (section 1.2.2.1)  in my answer.  They mention use of $\ce{NaBH3CN}$ and $\ce{NaBH4}$ \ $\ce{RCO2H}$ under mildly acidic conditions.

under acidic conditions, imines can be activated in a similar way to carbonyl compounds. Furthermore, since the imines nirogen is more basic than the carbonyl carbon, the equilibrium lies further towards species than it does i the corresponding carbonyl series. iminium ions are excellent electrophiles and reactive intermediates in synthesis.


text:
Do electrons only fill 'spin up' first? Or could it start filling 'down spins' first?
Due to Hund's rule, electrons start filling up the orbitals without pairing up. When this is happening, do the electrons all fill up the 'up' spin? Could they fill in the 'down' spin? Why do they prioritize the 'up' spin if they do?

No it does not matter whether you start with 'spin up' or 'spin down' electrons as long as you are consistent: If you start filling some orbitals with 'spin up' electrons you will have to keep doing this for all the orbitals you want to fill.
But: The convention is that you start with 'spin up' as this usually is taken to represent the $\frac{1}{2} \! \hbar$ eigenvalue of the $z$ component of the spin angular momentum operator and you should adapt to that in order to avoid confusion.

awesome. thank you! Is it then, fair to say that up/down of the first filling will be random?

@jkim Well, there are a few subtleties and I don't know how much you actually know about quantum mechanics. But if you only want to use the Aufbau principle and put some up/down arrows into boxes to describe the electron configuration of atoms and molecules then the choice between up and down should be random. But: The convention is that you start with 'spin up' as this usually is taken to represent the $+ \frac{1}{2} ! \hbar$ eigenvalue of the $z$ component of the spin angular momentum operator and you should adapt to that in order to avoid confusion.


text:
About possibly healthy salt substitutes
I noticed Heinz Ketchup has a salt substitute version that uses potassium chloride instead of the usual salt. If this is safe and tastes like salt could it be a better additive than sodium chloride? Since it has potassium wouldn't this be a benefit for Heart health or against stroke if not used too much?

The taste of potassium chloride is not close enough to sodium chloride so the salt substitutes usually contain a mix of potassium and sodium chloride.  This is healthier than pure sodium chloride.

The taste of $\ce{KCl}$ is a bitter and salty, but is not as salty as that of $\ce{NaCl}$, so the idea seems questionable.

Special Remarks on other Toxic Effects on Humans: Actute Potential
  Health Effects: Skin: May cause skin irritation Eye: Dust may cause
  eye irritation. Inhalation: Dust may cause respiratory tract
  irritation. Low hazard for usual industrial handling Ingestion: May
  affect behavior (coma, change in motor activity, listlessness,
  vertigo, mental confusion, paresthesias, general weakness, flaccid
  paralysis), metabolism, blood (change in clotting factor, electrolytic
  imbalance), cardiovascular (hypotension, circulatory disturbances,
  cardiac arrhythmias, heart block), and respiratory, gastrointestinal
  (irritation of GI tract, nausea, vomiting, diarrhea, abdominal
  discomfort, purging), and urinary(impairment of renal function)
  systems. Acute potassium intoxication by mouth is rare because large
  single doses usually induce vomiting, and because in the absence of
  pre-existing kidney damage potassium is rapidly excreted. Maximal
  nontoxic oral dose of KCl in man varies from 0.2g to 1 g of
  potassium/kg/day depending upon efficiency of individual excretory
  mechanism; lower doses sometimes cause impairment of renal function as
  shown by reduced inulin, and urea clearance. Chronic Potential Health
  Effects: May affect blood and cardiovascular system.

(c) $\ce{KCl}$ MSDS.
In general, If you are into this, I recommend to consult your doctor. $\ce{K,Na,Mg,Ca}$ ions are vital for cardiovascular and nervous system and affect kidneys. They are not highly toxic, and our body can deal with reasonable doses of each, but in case you have chronic condition, it may cause individual vulnerability.

These aren't really safety concerns at for normal uses in food. Many commercial products that are blends of NaCl and KCl are available, e.g. http://www.nutrasalt.com/Store/low-sodium-products/low-sodium-sea-salt.aspx . It is true that tons of potassium can cause problems (which is why it's used as part of the protocol for lethal injection), but it's not likely from ingestion. The MSDS for NaCl is similarly dramatic warning of causing abortions and has a comparable toxic dose level. KCl does, of course, taste terrible. To reduce sodium intake, I would think it better to just not use any salt.

@MichaelD.M.Dryden Let me put it this way: imbalanced KCl intake may cause extra stress onto cardiovascular system. While not lethal, it still may reduce expected lifespan a bit.


text:
Fuel Cell Chart - Anode, Cathode, Electrolyte
Can someone please check my table to make sure I have the correct information. I am especially unsure of the parts highlighted in yellow:

Have tried to search in google because all this is available on wikipedia.

I have, but I'm not sure if I have the correct information.

http://artsandsciences.virginia.edu/cchf/research/fuelcells.html


text:
Effect of water viscosity (increased due to NaCl concentration) on dissolution rate of CO2
We did an experiment involving different $\ce{NaCl}$ concentration(ranging, 1-4%) on 100 mL of water. Our aim of the experiment was to find the dissolution rate of $\ce{CO2}$in the brine solvent. We did some research and found that $\ce{NaCl}$ increases the viscosity of the solvent, which in-turn decreases the dissolution rate. We want to explain how the entire process works, in a microscopic level.
Answers with good source would be good.

By dissolving NaCl you increase the ionic strength of the solution, i.e. there is more electrostatic attraction between the layers of water increasing the viscosity. Carbon dioxide is non-polar, so by increasing the polarity of the solvent, the carbon dioxide tend to form less interactions with the solvent.


text:
Are phenylboranes susceptible to ipso or meta substitution?
Are phenylboranes susceptible to ipso or meta electrophilic aromatic substitution, i.e. what's the regioselectivity?

A SciFinder search turns up very little on this type of transformation.
Qui and coworkers run halogenations of phenyl boronate esters using N-halosuccinimide using gold(III) catalysis. The regioselectivity was essentially 1:1:1 for ortho:meta:para. When the reaction was carried out with other substituents as well, the other substituent's directing group preference dominated the regioselectivity.
Gold(III)-Catalyzed Halogenation of Aromatic Boronates with N-Halosuccinimides
There is one other reaction in SciFinder that is relevant, however I don't have access to the full text (and it's in Russian, so I wouldn't be able to read it anyway). The substrate is a benzene ring substituted with a borane cluster. The reported regioselectivity is 80:20 for meta:para.
Electrophilic substitution in the benzene ring of 9-phenyl-o- and -m-carborane. Zakharkin, L. I. et al Zhurnal Organicheskoi Khimii, 23(8), 1691-5; 1987.

I think that it's somewhat strange that this isn't examined more thoroughly. I am also unsure whether a boronate ester can be compared to a borane. The carborane looks more appropriate for consideration (excluding the steric effect contribution which doesn't allow ortho). The meta-direction would be in agreement with the resonance withdrawal. Seems like there is no evidence on ipso substitution.

I don't think there's been a real need to investigate that type of reaction. Boron relatively rarely appears in the target compound of a synthesis. Even then, it is probably introduced late in the synthesis.

One other note, SciFinder reports that phenyl borane is typically prepared through the organo mercury compound, something many people probably don't want to do. As a reactant, phenyl borane has been oxidized to the phenol, used in hydroboration, and to make boron complexes.

Practically speaking yes, but theoretically speaking I think that it would be interesting to examine this topic, as boranyl is the only group that gives electrons by inductive effect and withdraws them by conjugation.

9-(N-pyridine)-11-I-7,8-dicarba-nido-undecaborate are susceptible to ipso substitution and have substitution I-atom on Cl-atom. Read
Rudakov D. A., Potkin V. I.
Chlorination of 9-dimethylsulfonio-7,8-dicarba-nido-undecaborane and 9-(N-pyridine-)-11-iodo-7,8-dicarba-nido-undecaborane. pp. 69--72
Summary:
Chlorination of 9-dimethylsulfonio-7,8-dicarba-nido-undecaborane and 11-iodo-9-(N-pyridine)-7,8-dicarba-nido-unde-caborane with chlorine in dichloromethane produces 6,11-dichloro-9-dimethylsulfonio-7,8-dicarba-nido-undecaborane and 11-chloro-6-iodo-9-(N-pyridine)-7,8-dicarba-nido-undecaborane in 20% and 44% yields, respectively.

https://www.researchgate.net/publication/259363537_Chlorination_of_9-dimethilsulfonio-78-dicarba-nido-undecaborane_and_9-N-pyridine-11-iodine-78-dicarba-nido-undecaborane


text:
Can in any case the faster step of the reaction be rate determining?
I found a sentence in book which states the reverse step of this reaction ( I forgot what was it!)  has the faster step as rate determining .
Even Rate determining step-Wikipedia  states:

In chemical kinetics, the rate (or velocity) of a reaction mechanism with several steps is "OFTEN"  determined by the slowest step .

which I think  'OFTEN' means not 'everytime' so,
if faster step of the reaction is rate determining. Then why it is?

In the past I was also explicitly told that the slowest reaction is "often" the rate determining step, and like you I figured some day I would find a reaction where the fastest step determines the rate. Thinking over it a bit more now, however, I believe we may have misinterpreted what was meant. Rather than looking for an opposite situation, it is likely we were indirectly told that in some cases the rate-determining-step picture simply isn't applicable in the first place.
In general, reactions can't be modelled though such simple kinetic theory; in truth, when a bunch of reactants are brought together, all steps in every possible reaction route matter. It just happens that for several simple but relevant chemical systems there are few possible steps and few accessible reaction routes, and they have such wildly different rate constants that we can approximate by looking only at the slowest step of the fastest route. A more complex reaction with several steps, several side-reactions and similar rate constants will likely be poorly described by a RDS framework, so one can say that the reaction rate is not determined by the slowest step because the reaction rate won't be well determined by any single step.

You may also be interested in this past answer of mine.

Yes, that's the right interpretation.

I can provide you an example.
The oxidation of formate ion by peroxydisulfate in water solution:
$\ce{HCOO^{-} +  S_2O_8^{2-} -> CO_2 + 2SO_4^{2-} + H^+}$
has the following mechanism:

As you can see, the first step is the slowest, but by using the rate-determining step approximation you wouldn't arrive at the correct rate law which is: $r=k[\ce{HCOO^-}]^{1/2}[\ce{S_2O_8^{2-}}]$. The  first  reaction  is  very  slow,  so  most  of  peroxydisulfate  is consumed  in the third reaction. The correct rate law can be obtained by applying the steady-state approximation to the two radical species.
(This isn't the true mechanism but it is good for showing the point. A  more complex  mechanism  includes  the  formation  of  OH  radicals  and  several  chain termination reactions. That’s why the given rate law is valid only for a limited range of reactant concentrations.)

But how can peroxydisulphate ion be consumed in 3rd step because any co2- will be formed after first step only. I think!

But how can peroxydisulphate ion be consumed in 3rd step because any co2- will be formed after first step only. I think!

During the conversion of the very first peroxydisulfate anion, there wont be any CO2-, but steady state will be reached very fast afterwards.

This is not really a fair test, as reaction 1 is not needed  because $\ce{^.SO4^-}$ is produced is reaction 3 independently of reaction 1, i.e. it is probable that as the k quoted is made up of the rate constants one could put $k_1 = 0$ without it having much effect.


text:
How does lye appear in ashes?
One of the oldest recipes to create lye is this one:

Take well burnt ashes and put them in a barrel
Fill the rest of the barrel with water
Leave the suspension for a week or two
You can now try the lye concentration with egg - if 2/3 remain above the water, the concentration is sufficient for domestic uses
Separate the lye solution from ashes by carefully draining the water. You can also use filtration.

Now I know there are traces of various metals in us, as well in flowers and trees. But how can ashes accumulate such a big amount of sodium? And how is the sodium bound into it. How does it turn into hydroxide?

There is sodium and potassium in wood.  When the wood burns some of the sodium and potassium become bound to oxygen.  The oxides become hydroxides when exposed to water.

What are such amounts of sodium doing in wood? Are they in humans too? Does the potassium also end up in water as KOH?

Sodium chloride (common salt) is present a lot in most common organisms. Some Potassium ions for ionic balance.

Wood ashes are usually stated to be a source of potash, porassium oxide.  Rarely have I seen wood ashes stated as a source of lye.

@user2617804 And salt does burn into sodium oxide? Wouldn't than be more profitable to add salt into fire and then make lye of the remains?

I speculate that it would depend upon the heat of the fire.  A hot fire could drive off chloride and chloride would be replaced by oxygen as the fire cools.

Only a very small percentage of the original mass of wood remains after combustion.  The Combustion oxidizes the cellulose and other carbohydrates into CO2, CO, H2O, and other gases.
What's left over are the trace minerals, unburned creosote, tars, and other solids that couldn't be gasified.  Out of the first burn, much of the residue can be burned off with a higher temperature.
I have a large fireplace insert wood stove and I have burn through 10 cubic meters of wood to get less than 1 cubic foot of ash residue.  These are the minerals that include the sodium and potassium that cannot be vaporized.  Sodium and potassium are required by all life in trace amounts, including trees for the chemical processes involved in cells.


text:
Regioselectivity of Birch reduction (electron donating groups)
Why does Birch reduction on benzene derivatives with an electron donating group happen at ortho and meta positions? When the substitute is electron withdrawing, ipso and para reduction take place because the electron withdrawing group stabilizes the anion in these positions. What's the explanation for electron donating group directivity?

The explanation lies in the distribution of electron density in the intermediate radical anions that appear during the Birch reduction.
I use the following reactions as guiding examples:

Electron-withdrawing groups stabilize electron density at the ipso and para positions through conjugation and so the negative charge will mainly be found in these positions; subsequent protonation occurs para.

Electron-donating groups destabilize a negative charge (when compared to simple benzene) through conjugation and this effect would be strongest at the ipso and para positions.

Thus the negative charge will mainly be found at the ortho and meta position.

But why so? How does it destabilize ipso and para more than meta and ortho?

@Marko If you would have the same para/ipso pattern for a electron-donating group (EDG) as you have for an electron-withdrawing group (EWG) you would have more electron density next to the EDG or at the para position. But the EDG has a free electron pair, i.e. a high electron density, which in this case could be effectively engage in conjugation with the electron density at the ipso and para positions. And, loosly speaking, when two high electron densities interact they "repel" each other thus raising the energy.

@Marko With the ortho / meta pattern, the orbital interaction and overlap is different and the two high electron densities can better "go out of each others way". In the case at hand it is rather difficult to show this in terms of resonance structures. Orbital pictures would be more suitable.


text:
Removing HCl from water
I have water contaminated with dissolved chlorine:
$\ce{H2O + Cl2 <=>HCl + HOCl}$
During electrolysis, this is an issue. These are my products:
$\ce{NaCl + H2O ->T[electrolysis] NaOH + Cl2 + O2 + H2 }$
However, $\ce{NaOH}$ will be neutralised by $\ce{HCl}$:
$\ce{NaOH + HCl-> NaCl + H2O}$
I guess this can be considered hydrochloric acid. I want to remove the chlorine ions from water and prevent future dissolving.
Does increasing temperature help to remove the chlorine? Does boiling make the chlorine leave the mixture?
$\ce{H2O + HCl + HOCl ->T[???] H2O + Cl2(gas)}$
PS.: I understand that there are means of separating the electrolysis columns. But this question is really about removing $\ce{Cl2}$ or $\ce{HCl}$ from aqueous solution.

Can you let the mixture sit and ventilate for a while?  If the chlorine leaves as a gas eventually it will all be gone.  HCl should be less soluble in hot water so heating will drive it off faster.

I actually used to put an old PC ventilator on half of the glass, so air would forcefully strike the solution. But even after it smelled, as well as after boiling. Still thanks for info. I'll repeat the process.

1) Do NOT heat the solution, if you want chlorine out. Electrolysis of hot $\ce{NaCl}$ solution produces $\ce{NaClO3}$, then $\ce{NaClO4}$, not chlorine. You apparantly don't want this.
2) traditionally, in such setups some barrier between cathode and anode spaces is used. The easiest one would be unglazed ceramic pot (with no holes and no direct contact of anode and cathode solutions), though I'm not sure if it will work with chlorine ions. It is recommended to perform electrolisys of $\ce{MgSO4}$ solution first. In this case chlorine ions would migrate from cathode space. With barrier introduced you may expect chlorine to evolve from anode space and sodium hydroxide to accumulate in cathode space.
3) do NOT blow air atop sodium hydroxide solution. It reacts with $\ce{CO2}$ from air.

Thanks for warning me - apparently, I was totally unaware that I'm dealing with poison. Since the sodium chlorate decomposes at 300°C, it will be best to toss my current product.

@TomášZato I wouldn't call chlorates a potent poison. They are potentially dangerous in mixes with reducers, but not extremely toxic.


text:
Solution, hardener and metal required to cast key duplicates
I just watched this YouTube video for a DIY key duplication system and was trying to figure out what all the material components could be:

The "mold solution" (parts 1 and 2)
The "hardener"
The "low melting point ingots"

The mold gel has to be resilient enough to withstand molten metal (even if the melting point is low), but I can't figure out what materials would fit the bill. Any ideas?

depending on th alloy we are talking about temperatures 50(fifteen)-300 Celsium, so it may be something very common. My guess would be silicone and maybe polyurethane molds.

From the video, the material looks like some kind of silicone. There aren't many other materials that have that translucent elastomeric quality. If that's the case, the larger of the two mould solutions will be mostly some kind of silane precursor such as SiOH. The second part will contain the catalyst for the polymerization. Depending on the exact chemistry used it could be perhaps tin-based. My guess for the "hardener" is that it's simply more catalyst. Two part silicones with longer working lives are more common than very rapid cure ones so adding more/stronger catalyst as the third part may be just to make it cure quickly. Many silicones can withstand temperatures of 200+ °C.
As for the low melting point alloy, there are a number of things it could be. Given that the application is keymaking and people touch keys with their hands, I hope it's safe to assume that the alloy doesn't contain lead or cadmium. The video says its melting point is around 90 °C. The two closest common alloys without toxic metals are 58% Bi, 42% Sn @ 132 °C and 66.3% In and 33.7% Bi @ 72 °C, from the wikipedia page on fusible alloys, so it's likely some alloy with similar composition to one of these. Bismuth and tin are far cheaper than indium, so it wouldn't surprise me if the alloy was some mixture of those two elements.

I've seen dental alginate work as an excellent key mold before as well, and this material is readily available and cures very quickly. For the actual key, I would use a strong enough psi  2 part epoxy and reinforce it with strands of carbon fiber of just wire.


text:
2-methylpropene to tert-butyl alcohol Reaction Mechanism
Can someone explain what reaction is going on here?
I looked in my list of mechanisms learnt: I find nothing related!

If you've learned the E1 reaction, this is the reverse process, as detailed by Jannis below.

It is an acid-catalyzed hydration of the double bond. First, the double bond is protonated to yield the tert-butyl cation which is most stabilized by hyperconjugation (protonation at C2 would form a less stable primary cation). The cation is captured by nucleophilic attack of water, which after deprotonation yields tert-butanol and the regenerated proton.
$$\ce{(H3C)2C=CH2 + H3O+ <=> (H3C)3C+ + H2O} \tag 1$$
$$\ce{(H3C)3C+ + H2O <=> (H3C)3C-OH2^{+}} \tag 2$$
$$\ce{(H3C)3C-OH2^{+} + H2O ->~ (H3C)3C-OH + H3O+} \tag 3$$


text:
Why is carbon monoxide the product in reactions that produce high flame temperatures?
I was looking at substances that produce the highest flame temperatures upon combustion. I note that the equations for combustion are producing carbon monoxide rather than carbon dioxide.
$$\ce{(CN)2 + O2 -> 2CO + N2}$$
Why isn't carbon dioxide produced? Why doesn't the production of carbon dioxide give a hotter flame? It seems that a higher flame temperature would be generated by complete combustion. My thinking is that the heat of formation of carbon dioxide is around −400 kJ/mol while the heat of formation of carbon monoxide is around −100 kJ/mol so the additional heat released by producing carbon dioxide should account for carbon dioxide's greater heat capacity.

This is just a guess... But I wonder if this is due to entropic considerations. In the equation you have written, there are two molecules of gas getting converted to three molecules of gas. $\Delta S$ is positive, so it decreases $\Delta G$ as the reaction temperature increases. However, in the reaction $\ce{(CN)2 + 2 O2 = 2 CO2 + N2}$ you have an equal number of gas molecules, so the entropic component of $\Delta G$ will be small.
I'm not really sure why that would matter. At very high temps, the first reaction will presumably have a very large, negative free energy. Of course, the $\Delta G$ for the $\ce{CO2}$ reaction is most likely negative, so you might have to control the feed stoichiometry carefully to ensure that you are getting only the first reaction.
EDIT:
I got the $\Delta G$ values from the NIST-JANAF website, and it doesn't look like the free energy for these reactions is much different at 4000 K.
$\Delta_f G_{(CN)_2}^{0} = 136.036\ \mathrm{kJ\ mol^{-1}}$
$\Delta_f G_{CO}^{0} = -446.485\ \mathrm{kJ\ mol^{-1}}$
$\Delta_f G_{CO_2}^{0} = -393.183\ \mathrm{kJ\ mol^{-1}}$
$\Delta G_{rxn1}^{0} = -1029.005\ \mathrm{kJ\ mol^{-1}}$
$\Delta G_{rxn2}^{0} = -922.397\ \mathrm{kJ\ mol^{-1}}$
EDIT 2:
So, I just read one of the original papers on this type of flame, and it sounds like the thermal stability of the products is an issue as well. I take it that at these very high temperatures, some of the energy can be transferred to bond-breaking processes, which would lower the total flame temperature. $\ce{CO}$ and $\ce{N2}$ have very strong covalent bonds, so perhaps they do not readily dissociate at these temps.

mainly because in high flame areas, oxygen is used up by the fire more rapidly,thus there is less oxygen available,and thus CO carbon monoxide forms.


text:
What really happens atomically in an explosion?
Let's say a room is filled with butane, I then throw a cigarette into the room. What happens to the atoms/molecules of the butane when they are in contact with the heat from the cigarette?

The cigarette heats up a few oxygen and butane molecules, which causes them to combine, which releases heat, which heats up even more oxygen and butane molecules, etc, etc, boom.

It's obligatory to add that a room filled with butane wouldn't do much as there's no oxygen to react with.

What happens to the molecules is that they kill you.

The combustion of alkanes like butane is fearsomely complicated involving dozens of transient compounds and hundreds of different reaction. If you have a few spare hours there is a dissertation that presents a nice summary of the process here (this is a 1MB PDF).
A butane molecule is pretty stable and doesn't react with oxygen on contact so you need some way to get the reaction going. Typically the reaction is started by the generation of free radicals. These will react with stable molecules to split them up and generate more free radicals so in effect we have a chain reaction.
So the ignition process is basically the generation of free radicals, and that's what the lit cigarette does. The heat of the burning tobacco generates free radicals that then start the butane oxygen reaction.  You can also generate free radicals in lots of other ways e.g. using shock, from an electric spark, or by using a suitable catalyst such as the catalytic convertor on your car exhaust.

The butane and oxygen in the room are in a metastable state.  An example energy diagram of a metastable state is shown in the image below.  Before throwing in the cigarette, the butane and oxygen molecules are in the state labeled as 1.  This state is metastable because there is a lower energy state (which makes it statistically more probable) available if they were to combine to form $CO_2$ and $H_2O$.  This lower energy state is represented by the 3 in the image.
The problem is that there is an energy barrier which must be overcome for them to get to this more stable state.  This is where the cigarette comes in; it adds some thermal energy which pushes the system over the energy barrier (state 2 in the image) and allows it to fall into its more stable state.  In the process the excess energy gained from moving from state 1 to state 3 is given off as heat.  This excess heat causes more molecules in the area to combine, and you end up with a runaway reaction known as an explosion.

*Image taken from the Wikimedia Commons.

The butane and oxygen molecules' electrons gain enough kinetic energy to overcome the potential barrier that prevented them from disassociating from their respective molecules making the hydrogen and carbon atoms bond with oxygen – producing $\ce{CO2}$ and $\ce{H2O}$ and extra heat. Which would go to increase more electron's kinetic energy as well as increase the pressure of the air (since that what happens when gasses heat). This sudden increase in pressure is an explosion.


text:
Triple Bond Character in Acyl Chlorides
I am told that because of the poor overlap between a chlorine atom and a carbon atom and chlorine's relatively high electronegativity, there is a strong partial positive character on the central carbon in an acyl chloride molecule.
This makes sense. I am then told that because of this high partial positive character created through the inductive effect of chlorine and oxygen and the poor overlap between chlorine and oxygen, the oxygen's lone pair(s) actually participate in a sort of resonance with the central carbon, thereby giving the C=O bond in an acyl chloride partial triple bond character.
Is this true? Also what exactly is this called? Backbonding?

http://chemistrytextbookcrawl.blogspot.com/2012/12/infrared-spectra-of-acid-chlorides.html  The IR stretch is between 1775 and 1810 suggesting greater bond order.

The various resonance structures that one can draw for an acid chloride are shown below.

Structure IV, is the "triple bond" structure that you mentioned.  The acylium ion ($\ce{R-C#O^{+}}$) is very stable, so structure IV does contribute to the overall description of an acid chloride.  Note that the $\ce{C#O}$ bond in IV is not a real triple bond as the carbon remains $\ce{sp^2}$ hybridized, with the attendant 120 degree (non-linear) bond angles.  I've also drawn a fifth resonance structure involving hypeconjugation.  This is probably a reasonable structure as well since it places negative charge on oxygen and positive charge on the hydrogen.  It also helps explain why hydrogens on the carbon alpha to a carbonyl have increased acidity.
Usually the term "backbonding" is used when electron density is transferred from one atom to another through the use of antibonding orbitals.  See here for a better description.  In resonance structure IV, the lone pair from oxygen is being shared with an empty carbon $\ce{sp^2}$ orbital, the one that had been used to bond to the chlorine.  I've never heard that referred to as backbonding.

The ion is "stable" by itself or stable when solvated or stable in the gas phase?

Also @ron how can one calculate the percent resonance contribution for each resonance contributor?

@Dissenter The acylium ion is stable enough that it can be observed spectroscopically, both in solution and the gas phase, but it is still a reactive intermediate.  If you mix an acid chloride with $\ce{AlCl_3}$ in benzene, you can generate the acylium ion and it will react with the benzene via electrophilic aromatic substitution to produce a phenyl alkyl ketone.  There's no simple way to estimate the contributions of the various resonance structures, other than through the use of a computer, as far as I know.

Are there any programs I can use?

Martin or Philipp might know the answer to that.  I suspect that to get meaningful estimates you'd have to use something fairly advanced.

why does your triple bonded resonance structure have a ~120 degree bond angle?

Because nuclear positions can't change when we draw resonance structures and the carbonyl carbon was approximately sp2 hybridized.

So the bond angle should probably be between 120 and 180 degrees?

Probably closer to 120

@Dissenter To approximate the contributions of a resonance structure you would have to follow the valence bond theory approach. I am not aware of any free programs for this. Although this molecule is quite simple, it will still require a reasonably powerful computer and a day of calculating I guess.|| Another possibility to check if a resonance structure is well enough agreeing with the overall electron density, one can use the natural bond orbital theory approach. I am also not aware that there is free software for it around. Computing is much simpler though.


text:
How should I measure out a solution given in nanorods per ml to make a solution of a set number of mols?
I am trying to determine the number of nanorods that I would need to measure out to get a 1 mM solution.  For the gold nanorods that we ordered, it was mentioned that the single nanorods dimensions are 40 nm length by 10 nm in diameter.
The product's concentration is $6.3\times10^{11}$ nanorods per ml.
How would I measure out the solution to obtain a 1 mM gold nanorod solution?

It appears you would have 6.3E14 nanorods per liter, far less than needed to give a 1mM solution.  That is, unless they 1mM solution is referring to atoms of gold instead of nanorods.

In order to avoid any possible misunderstanding, we should first clarify that “nanorods” (and thus also “nanorods per millilitre”) is not the name of a unit. A nanorod is a very small (so-called nanoscale) object.
The number of nanorods in one millilitre of the given solution is $N_\text{nanorods}=6.3\times10^{11}$, i.e.
$$\frac{N_\text{nanorods}}{V_\text{solution}}=\frac{6.3\times10^{11}}{1\ \mathrm{ml}}=6.3\times10^{14}\ \mathrm{l^{-1}}$$
where $V_\text{solution}$ is the volume of the solution.
Since
$$c=\frac nV$$
and
$$N_{\mathrm A}=\frac Nn=6.022\,140\,76\times10^{23}\ \mathrm{mol^{-1}}$$
where $c$ is concentration, $n$ is amount of substance, and $N_{\mathrm A}$ is the Avogadro constant, the concentration of nanorods in the given solution is
$$\begin{align}c_\text{nanorods}&=\frac{n_\text{nanorods}}{V_\text{solution}}\\[6pt]&=\frac1{N_{\mathrm A}}\cdot\frac{N_\text{nanorods}}{V_\text{solution}}\\[6pt]&=\frac1{6.022\,140\,76\times10^{23}\ \mathrm{mol^{-1}}}\cdot6.3\times10^{14}\ \mathrm{l^{-1}}\\[6pt]&=1.0\times10^{-9}\ \mathrm{mol\ l^{-1}}\\[6pt]&=1.0\times10^{-6}\ \mathrm{mmol\ l^{-1}}\end{align}$$
The given nanorods are approximately cylinders with a length of $l=40\ \mathrm{nm}=4.0\times10^{-8}\ \mathrm m$ and a diameter of $d=10\ \mathrm{nm}=1.0\times10^{-8}\ \mathrm m$, i.e. a radius of $r=5\times10^{-9}\ \mathrm m$.
Therefore, the volume of one nanorod is approximately
$$\begin{align}V_\text{nanorod}&=\pi\cdot r^2\cdot l\\[6pt]&=\pi\times\left(5\times10^{-9}\ \mathrm m\right)^2\times4.0\times10^{-8}\ \mathrm m\\[6pt]&=3.14\times10^{-24}\ \mathrm{m^3}\end{align}$$
The density of gold is $\rho=19.32\ \mathrm{g\ cm^{-3}}=1.932\times10^4\ \mathrm{kg\ m^{-3}}$. Thus, the mass of gold in one nanorod is approximately
$$\begin{align}m_{\ce{Au}}&=\rho\cdot V_\text{nanorod}\\[6pt]&=1.932\times10^4\ \mathrm{kg\ m^{-3}}\times3.14\times10^{-24}\ \mathrm{m^3}\\[6pt]&=6.07\times10^{-20}\ \mathrm{kg}\end{align}$$
The molar mass of gold is $M_{\ce{Au}}=196.97\ \mathrm{g\ mol^{-1}}=0.196\,97\ \mathrm{kg\ mol^{-1}}$. Thus, the amount of gold in one nanorod is approximately
$$\begin{align}n_{\ce{Au}}&=\frac{m_{\ce{Au}}}{M_{\ce{Au}}}\\[6pt]&=\frac{6.07\times10^{-20}\ \mathrm{kg}}{0.196\,97\ \mathrm{kg\ mol^{-1}}}\\[6pt]&=3.08\times10^{-19}\ \mathrm{mol}\end{align}$$
Note that this amount is equivalent to only about $190\,000$ gold atoms since
$$\begin{align}N_{\ce{Au}}&=N_\mathrm A\cdot n_{\ce{Au}}\\[6pt]&=6.022\,140\,76\times10^{23}\ \mathrm{mol^{-1}}\times3.08\times10^{-19}\ \mathrm{mol}\\[6pt]&=1.9\times10^5\end{align}$$
Therefore, the concentration of gold in the given solution can be estimated as follows:
$$\begin{align}c_{\ce{Au}}&=n_{\ce{Au}}\cdot\frac{N_\text{nanorods}}{V_\text{solution}}\\[6pt]&=3.08\times10^{-19}\ \mathrm{mol}\times6.3\times10^{14}\ \mathrm{l^{-1}}\\[6pt]&=1.9\times10^{-4}\ \mathrm{mol\ l^{-1}}\\[6pt]&=0.19\ \mathrm{mmol\ l^{-1}}\end{align}$$
or
$$\begin{align}c_{\ce{Au}}&=N_{\ce{Au}}\cdot c_\text{nanorods}\\[6pt]&=1.9\times10^5\times1.0\times10^{-9}\ \mathrm{mol\ l^{-1}}\\[6pt]&=1.9\times10^{-4}\ \mathrm{mol\ l^{-1}}\\[6pt]&=0.19\ \mathrm{mmol\ l^{-1}}\end{align}$$


text:
How much energy does it cost to have electron configurations that are not in accordance with Hund's rules?
What is a ballpark figure for the difference in energy for an atom that follows Hund's rule vs one that has two electrons with opposite spins?  I'd be interested to know carbon and nitrogen.  Is there a bigger difference when there are two electrons with one spin and one electron with the opposite spin vs two electrons with opposite spins?

May I ask: What electron configurations were the ones you actually asked for?

Each energy level of a given electron configuration is described by a  Russell-Saunders term symbol, assuming LS coupling. The ground state term symbol is predicted by Hund's rules. Tables of atomic energy levels identified by their term symbols have been compiled by NIST. So, the only thing you have to do, is to compare the energies of the terms describing the electron configurations you are looking for with their respective ground states.
In the following pictures the notation $m_{S} = + \frac{1}{2} \overset{\scriptsize{\text{def}}}{=} \, \uparrow$ and $m_{S} = - \frac{1}{2} \overset{\scriptsize{\text{def}}}{=} \, \downarrow$ will be used for the spin quantum number $m_{S}$.
The data for carbon can be found here. The ground state for carbon is ${}^{3}\mathrm{P}_{0}$ - it is assigned the energy $0.00 \, \mathrm{cm}^{-1}$. From your question it's not 100 % clear what exact electron configuration you want. I assume it is either ${}^{3}\mathrm{P}_{2}$ or ${}^{1}\mathrm{S}_{0}$ therefore I will give the values for both of them (for comparison: the thermal energy at room temperature is about $200 \, \mathrm{cm}^{-1}$).

The data for nitrogen can be found here. The ground state for nitrogen is ${}^{4}\mathrm{S}_{3/2}$ - it is assigned the energy $0.00 \, \mathrm{cm}^{-1}$. From your question it's not 100 % clear what exact electron configuration you want. I assume it is either ${}^{2}\mathrm{D}_{5/2}$ or ${}^{2}\mathrm{P}_{3/2}$ therefore I will give the values for both of them.

How is a 3p0 not equivalent to 3p2? Is there a electron-spin coupling with the nucleus or something?  Just off the cuff I would think that these would be identical if considering only this representation of the electron configuration.

@LordStryker The energy difference is quite small compared to the other ones. Maybe it is a result of spin-orbit coupling.

@LordStryker The $^3P_0$ and $^3P_2$ differs in J, i.e. the mutual orientation of L and S. This energy difference is coming from the spin-orbit coupling (which is per definition proportional with $LS$). Since spin-orbit coupling is small for light elements that is why the energy gap is so small.

Question on carbon: First, in your diagrams, shouldn't singlet S zero be a singlet D two configuration? Second, in the link to your source for carbon transitions, why is there no transition for triplet S one in carbon - that is, for the valence carbon 2p subshell, an electron in ml=+1 and an electron in ml= -1? Thanks.


text:
Ionic Character
Can ionic character also be construed as a poor sharing of electrons rather than simply as the taking of electrons? For example, NaCl is ionic, and that's because the chlorine takes sodium's one valence electron. On the other hand, C-Br and C-I are sometimes referred to as ionic (at least informally).
Is it proper to refer to C-Br and C-I as ionic because of the poor overlap between carbon and the two respective halides rather than any electron-withdrawing effect per se; the EN difference between C and I is less than the EN difference in the "non-polar" C-H bond.

Is it proper to refer to C-Br and C-I as ionic because of the poor overlap between carbon and the two respective halides rather than any electron-withdrawing effect per se? No, they are certainly not ionic per se. Have a look here and here.

Good point, ionic things have a crystal structure.

Is it proper to refer to C-Br and C-I as ionic

I've never thought of those bonds as ionic.  For example, consider electrophilic aromatic substitution in bromobenzene. The sigma bond between the bromine and benzene carbon allows for inductive withdrawal of electron density from the benzene ring, hence bromobenzene typically reacts slower than benzene.  However, to explain the  ortho-para directing effect of the halogen you also need to invoke resonance interaction between the halogen and the benzene carbon p orbital. So there is both sigma and pi interaction between a halogen substituent and an aromatic carbon, that doesn't sound ionic to me.
BTW, even in the case of sodium chloride the electron is not completely transferred from sodium to chlorine.  Crude estimates using this methodology put it around 80%.

Perhaps polar covalent is a better term?

Yes, that's a good way to describe it.


text:
What is the formula for migration rate of DNA in agarose gel electrophoresis?
Let's consider a typical agarose gel electrophoresis experiment. An appropriate quantity (not so much that it is overloaded, but not so little that it cannot be visualized) of linear DNA pieces with identical sequence and chemical structure is loaded onto the gel, and constant voltage is applied for a given time. I suspect that there is a function:
$x = f(V, t, l, c, k)$
Where the variables are:

$x$, cm, distance from well
$V$, Volt, potential difference across the gel
$t$, minutes, time over which the voltage is applied
$l$, bp, length of DNA fragments
$c$, gr/100 ml, density of the gel given in grams of agarose per 100 ml of buffer
$k$, a constant which is empirically determined for a given running buffer and gel tank

Basically, I want a formula into which I can plug in the size of my DNA and a few other parameters, and figure out how long I have to run it to get decent separation (the output variable is x because I intend to plot $x$ over $t$ for a given planned electrophoresis).
Is such a formula possible? What is it?
I have previously worked with gel documentation software which fits linear, polynomial and exponential functions to the DNA (I couldn't figure out exactly what kind of fitting the software was doing). However, after training a best-fit model on my ladder (with ~12 bands) and applying it to my samples of known size, I found all of these fits to be very inaccurate (5-10% error). Therefore, I suspect the equation governing the migration rate of DNA is in fact not a simple exponential function (as is often taught in college) but more complicated. Unfortunately, I lack the background in physical chemistry to derive the equation myself.
If you have a good argument for why the relation should be a simple exponent, eg. $x = e ^ {\alpha \cdot l}$, that is also an acceptable answer.

I found a model based on the gamma distribution, which seems relevant. I'm not sure how to convert it to the formula I want, however.

I guess my answer is very old, but here it is for newcomers searching the internet for this formula:
$$D = a - b \log(m)$$
where

$D$: Distance traveled for the restricted fragment.
$a$ and $b$: Constants depend on the electrophoresis experiment.
$\log(m)$: The logarithm of the molecular weight ($m$)of the fragment.

Knowing $a,b$ and the distance traveled in your gel, you can easily determine the molecular weight of your fragment and its size by the following formula:
$$m = e^ \left(\frac{a-D}{b}\right)$$

This is a good approximation. The mobility of DNA in a gel is also temperature dependent, which is a parameter usually not well controlled and not constant across the gel. Therefor, a more complicated model would probably not be useful, and people use size standard on the same gel instead. For a reference, see e.g. https://www.sciencedirect.com/science/article/abs/pii/002196739180078U


text:
Photoelectrochemical (PEC) water splitting
I'm starting to prepare for my research project on photoelectrochemical (PEC) splitting of water into H and O using (F doped) titanium dioxide as the photoanode. I've been unsuccessfully searching in the literature for any particular reason(s) which could make PEC better than other methods of photocatalytic water splitting. What is, if anything, special about PEC water splitting compared to other approaches?

Splitting of water is an electrochemical reaction. Any photocatalytic process is by definition photoelectrochemical.  You need to clarify what you mean by the two different terms.


text:
Mixing of powders to make stock solutions
For a solution such as:
Luria-Bertani (LB) Broth:
10 g tryptone
5 g yeast extract
10 g of NaCl ($M = 58.44\ \mathrm{g/mol}$)
q.s. to 1 l, pH to 7.2, autoclave.
I would like to make a bulk amount of premixed powder, i.e. combine 1 kg tryptone, 500 g yeast extract and 1 kg NaCl in a big bottle, shake it and keep it on shelf. Whenever I need to make LB now I just weight 25 g of this premix and add to 1 l of water.
I’m concerned about a couple things thought. First how thoroughly it gets mixed, and second any settling of different chemicals over time. Either would cause many different batches to be messed up.
I’m wondering if anyone has experience with this and ways to avoid those two problems. I’m interested in doing this for other solutions as well. Do manufacturers who make these premixed powders re pulverize them or process them in a certain way to make it work?

Companies that sell such mixtures usually make them by dissolving everything into a solution, lyophilizing the solution, and then grinding the lyophilized residue.  This guarantees that the size of particulates in the mixture will not be related to the chemical identity of the particulates.
If you have to mix powders, I wouldn't recommend making such a large batch.  Why not make a smaller batch, enough for 10 - 20 bottles of LB, mix thoroughly by shaking (which will still not be as good as lyophilizing), and then immediately aliquot 25 g portions into a series of dry 1 L bottles?  Then you can store the dry bottles until you want to make some LB, and you don't have to worry about the powder settling etc. in each bottle once it has its own charge of powder.

Once the powders are mixed, it is unlikely that they will separate unless the container is subjected to vibrations.
As for mixing, companies that need to mix batches of material use a mixer to make their batches. The second photo in the article is similar to what I have seen.

I'm still hesitant, there's vibration everywhere, this mix will prob last 5 years and I don't have one of those awesome mixers. Especially with solutions where some components are very hydroscopic. I guess I'm stuck with aliquoting a single premix

It doesn't need to be so "awesome". Even a rolling bottle will eventually mix the material.

There is no guarantee that the equilibrium distribution of particles in the mixture is uniform.  Mixing powders is generally a bad idea.

@CurtF. Every tablet was a mixture of powders before it was compressed into a tablet. Are you saying that there is usually a large variation in the dose between tablets?

Yes, that is what I am saying.  It is actually a major problem in the drug industry and in the USA at least, the FDA is tightening quality control standards.  Pill to pill variation in dosage can be 10% for some products.


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Fluoride Color Reaction
I am trying to find visual ways to demonstrate water contaminants. For example, chlorine reacts to N,N-diethyl-p-phenyldiamine (DPD), forming a magenta color. The chemical doesn't need to be accurate, as this is more of pass-fail test of the water. What reacts with fluoride to provide similar results?

Not much reacts with fluoride.  Here is a fluorometric assay:  http://pubs.rsc.org/en/Content/ArticleLanding/2011/CC/c1cc10498e#!divAbstract Here is a review:  http://www.atsdr.cdc.gov/toxprofiles/tp11-c7.pdf

Fluoride has a high affinity for $Fe^{3+}$, and $Fe^{3+}$ forms deeply coloured complexes with phenols.  The decomposition of the latter by fluoride ion is the basis for colorimetric determination of fluoride.
Ligands used in this kind of assay include 5-phenylsalicylaldehyde and resacetophenone.  For occasional testing of well water I'd recommend any analytical lab.  I'd be very surprised if Merck didn't offer any fluoride test strips, at their price of course, although whether they go down to the 5 ppm range is a different question.  For those that really like to spend money, there's fluoride-sensitive electrodes.


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Determination of timescales using characterization methods
There are 4 different timescales in material characterization (as classified in the Atkins textbook):
a. Time during which a quantum of radiation or a particle can interact with a molecule.
b. Lifetime of excited state of a molecule.
c. Minimum lifetime species must have to be seen as a distinct species.
d. Total time of experiment in which the species is observed.
Now the question is how to determine each of these? Say I have a certain compound and want to know what is the time of interaction of a photon with the molecule (part a). Is there an experimental method or instrument to determine these timescales?

The absorption of a photon en route to an electronically excited Franck-Condon state typically typically proceeds in the femtosecond range.
The lifetime of an electronically excited singlet state mostly is in the range of several nanoseconds. In some cases, however, fluorescence lifetimes in the range of several tenth of nanoseconds have been observed.
In cases where the initial excited singlet state undergoes intersystem crossing to a triplet state, much longer phosphorescence lifetimes (seconds range) are observed.
In order to detect a transient, its lifetime must be longer than the laser pulse needed to generate it.
Once you've populated a state, you may either use

time-resolved absorption spectroscopy to record it's spectrum and the decay of the absorption at a particular wavelength over time or
use time-resolved emission spectroscopy to measure the decrease of fluorescence (or phosphorescence emission) intensity at a particular wavelength over time.

The time to interact with a photon is only a few femtoseconds (1 fs = $10^{-15}$ s) as this is the time for the excited state to be distinct from the ground state. Before that time the molecule is in a superposition of the two states with the photon.
Fluorescence is generally short lived as it is a spin allowed process, excited singlet state to ground singlet state. The fluorescence lifetimes range from a few picoseconds e.g. bilirubin, to several hundred nanoseconds, e.g. pyrene.
Intersystem crossing from a singlet state produced a triplet state and as the transition the the ground state is spin forbidden (triplet to singlet)then phosphorescence has a longer lifetime than fluorescence in the same molecule. Phosphorescence lifetimes vary from tens of nanoseconds (rare) to several seconds.
It is possible to measure decay times shorter than the light pulse (usually laser ) used to excite the molecules. This involves the mathematical process of convolution and is routinely performed in fluorescence decay measurements.
Measuring excited state decays by absorption is done by the pump-probe method (formerly called Flash-Photolysis). In this method a pump pulse excites the sample and a probe photon beam measures the change in absorption at a later time. The pump-probe process is repeated at different time delays until the time profile is mapped out.
Fluorescence decays are usually measured by time correlated single photon counting. In this method a short pulse excites the sample and the time delay after excitation at which a single photon is measured is recorded. This is repeated many thousand of times and a histogram of the recorded times represents the fluorescence decay time.


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pH Changes in primary cells

Primary Cells - pH changes during discharge

Alkaline Cell:
Manganese Dioxide Zinc Alkaline Cell
Anode (-)
$\ce {Zn(s) + 2OH^{−}(aq) → ZnO(s) + H2O(l) + 2e^{−}}$
Cathode (+)
$\ce {2MnO2(s) + H2O(l) + 2e^{−} → Mn2O3(s) + 2OH^{−}(aq)}$
Overall
$\ce {Zn(s) + 2MnO2(s) → ZnO(s) + Mn2O3(s)}$

$\ce {OH-}$ ions are consumed at the anode
$\ce {OH-}$ ions are formed at the cathode
$\ce {same[OH^{−}]}$
pH - no change

Is this correct? Do pH changes occur during the discharge process?

No, the reaction does not produce hydroxide ions. But the solution is already basic (probably between pH 12 and 14).

I fixed your edits - because I wrote it wrong at first... or have I corrected it incorrectly? Can you take a look again?

So was my answer.

so there is no change - but the solution remains basic?

Yes, the solution remains basic.


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Can the thermodynamic information of a compound be calculated or must it be found emperically?
Do attributes such as heat of formation have to be found through experimentation or can they be calculated based on the attributes of the bonds and the atoms involved?

They can be calculated with the help of computers. The area that is concerned with this stuff is Statistical Mechanics.

For gaseous compounds heat of formation can be calculated using quantum chemistry packages as a difference between energies, assigned by the package, though inaccuracies are possible. Entropy of formation is a bit trickier, it requires calculation of vibrational and rotational states, but still pretty straightforward. For condensed phase (liquids, solids) the situation is harder, as knowledge of structure of solid state must be known beforehand, or full exhaustive search of possible crystal structures must be performed.
In practice, however, both heat and entropy of formation are found experimentally.


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Why is pressure negative in force and work
The equation to measure work is $$w=F\cdot D$$
However, the equation for force is  $$F=(-P \cdot A)$$
Plugging this into the work equation we have:
$$w=-P\cdot A\cdot D$$
My question is, why is pressure negative in force?

In this context we are talking about the work done by a gas.
Rather than representing it as $w = -PAD$, think of the formula as representing an area (that the gas is pressing against) moving a distance $D$. Since this a change in volume, we can set $w = -P\Delta V$
If you are putting work into the system (say pressing a piston into a cylinder containing a gas at a given pressure), the volume is becoming smaller, so $\Delta V$ is negative.  The negative of the pressure times a negative $\Delta V$ is a positive value, which matches with the sign convention for putting work in.  Vice versa, an expansion of a compressed volume at a given pressure will push the piston back out of the cylinder and allow the gas to do work, so $w$ is negative which follows the sign convention for work output.

You have to represent the equation in the most physics sensible way. Work is a scalar and the dot product of force and displacement gives out a scalar. In the second line $F = -PA$ is not the correct representation. Actually, it is
$$\vec{F} = -P\vec{A}$$
See for a piston exerting pressure on the gas. The force vector acting on the gas is vertically downward but area vector is vertically upward and hence a minus implies that the two vectors are mutually opposite to one another.


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How to photolyze anything?
If you have read this question and you thought that what is this stupid question then sorry to ask this question to you. Hey reader if you want to increase one student's knowledge then please answer this. Hey reader please answer this in an easy way. I am only in 10th grade\class.

Very few people will like this question since it looks like you are asking us to do the work for you. If you have a specific question, or you can narrow you request to certain conditions, then you should have better responses.

Find a suitable solvent.
Record a UV-VIS spectrum to determine at which wavelength you need to irradiate.
Chose the right glassware. For $\lambda$ = 254 nm, quartz is mandatory. Irradiations at $\lambda$ = 300 nm and above may be performed in pyrex.
Chose the right light source. For $\lambda$ = 254 nm, use a low-pressure mercury vapour lamp. For irradiations at $\lambda$ around 300 or 360 nm, use low-pressure mercury vapour lamps with specific coatings. Medium- or high pressure lamps typically show much broader emission spectra. For experiments at $\lambda$ = 589 nm, use a sodium vapour lamp.


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Do dark lines on porcelain dishes indicate usage of aluminium for the color?
"I heard" that gray lines forming on porcelain indicate usage of aluminium in the coating. I would primarily like to verify this information.
Bowl #1 and bowl #2 have been used both for several months and were bought in different shops. The same observation goes for other dishes. Some don't feature lines, other dishes feature those after days already. The dishes who tend to display those patterns have been bought all at the same place (a large chain of stores specializing on furniture). The lines are most likely caused by cutlery as in cups the same lines are found forming ellipses at the bottom from the stirring. Again some cups don't feature those patterns others do.
Bowl #1 with no lines despite usage for a while:

Bowl #2 with lines:

Bowl #2 with lines - picture taken HDR mode:

Could it merely be due to differences in washing-up methods? In other words for one bowl a metal scourer was used to clean it as opposed to a washing-up cloth or brush?

If the two bowls look different although treated similar, the material must differ.
Your bowls are almost fully covered with a glaze - except for the ring on the bottom on which the ceramic was standing while fired in an oven.
Here, water can enter the ceramic core. If the ceramic material of one bowl is more porous, water can be taken up easier. Upon heating, as in a dishwasher, the water absorbed expands. The glaze can't withstand the tension build up and breaks.
This results in the cracks that you observe.

maybe I should have mentioned that somebody told me this aluminium-theory but I cannot track down a source for it

@Raffael I think it is a misunderstanding. There is aluminium in your dishware, but not in metallic form. It's just that kaolinite, the clay mineral from which porcelain is made, is a silicate with aluminium cations in the crystal lattice. (Aber jetzt: Fußball ;) )

that would mean that both bowls feature aluminium cations and hence the aluminium is not involved in forming the lines?

@Raffael Yes, exactly. Refining aluminium metal from alumosilicates is almost impossible.


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When is coal considered to be oxidized?
What amount of oxidization is required to refer to some coal as oxidized coal, and how long does it take for coal to become oxidized in temperatures of roughly 25-40°C?
I have found a research paper from 1940 that I don't have access to here. But other than that, I have had no luck. Does anyone have any ideas or references I can refer to.

It is a pretty specialized, industrial terminology I am not familiar with, so take this comment with caution, but after superficial reading of the article it seems: 1) not the carbon, but the other organic content of coal is oxidized and this oxidation is highly dependent both on the coal quality and external factors (temperature, moisture etc). 2)  there is no "oxidized coal", but more like a level of oxidation, without strict boundaries. The terminology mostly refer to the loss of heating value during storage.   ,


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How do we get rid of dangerous explosive acetone peroxide and nitroglycerine?
Imagine the police (or army or whoever's job is that) found out that some maniac placed a lot of primary explosives under a building. They have evacuated the building and deactivated the trigger.
What now? You have a barrel full of Acetone peroxide (terrorist's favourite explosive). You move it - it might blow.
Even if it doesn't - can you react acetone peroxide slowly, to avoid explosion?
What do factories do with this compound formed as a side waste?
Could you answer the same for nitroglycerine?

There are some good guidelines and recommendations here: http://www.bnl.gov/esh/cms/pdf/peroxides.pdf . There are ways to dilute such compounds in the laboratory before disposal, however I don't think this question is particularly answerable since what must be done in a given possible explosive situation depends on many factors that are not chemistry related. Consider scaling the question back by simply asking how to properly dispose or make safe the compound in question. It might help if you restrict it to one compound, since one solution might not apply to both.

I think it is not a chemistry question / mostly off-topic. The core part of the question  - how to move or discard some highly unstable / home made explosives - is offtopic. It makes only ontopic sense under lab conditions (how to avoid problems working with these materials, how to dispose them etc), which are rather different from the barel-under-the-bridge scenario both in scale and in constrains.  Also, acetone peroxide is not "terrorist's favorite explosive". It is idiots favorite explosive.

@Greg Where's written applied chemistry is off topic? Using science in practical (even if hypothetical) situations is also chemistry.

I don't know if it's practical to dissolve or otherwise stabilize acetone peroxide, but I could see absorbing nitroglycerine with sawdust or trapping it in gelatine, rendering it more stable, which is how dynamite is made. Certainly if it can be done safely, detonating in place is be the preferred solution for dealing with unstable explosives. It's done frequently with unexploded munitions if they're deemed too dangerous to move, e.g. WWII bomb in Munich. I also visited a potash mine once. Old dynamite tends to "sweat" nitroglycerine, but the explosives lockers are designed such that the contents can be detonated in place without causing damage, if this happens as it's safer than attempting to remove them.

In Dynamite, Kieselgur (diatomite) is typically used as an absorbent.

I think you are thinking of gelignite. But the principle is the same.

I can think of one way: Dilute it in ethanol (or other flammable solvent) and burn the mixture.
BTW, It would be very unlikely that either of these would be a generated in a side reaction.

Acetone peroxide is a by-product during ecstasy synthesis. But I guess that doesn't happen in factories. Does acetone peroxide dissolve in ethanol?

IDK, but acetone and ethanol are miscible so I would think that the peroxide is as well.

I'm definitely not going to test it any time soon and I wish for a complete and reliable answer.

It's a good thing that caution over rules recklessness.

TATP may be formed from acetone used as a solvent in the "classical" oxidation of isosafrole by aqueous performic acid, which is formed in situ from formic acid and hydrogen peroxide.

Add denatured alcohol or another soluble liquid. Even if its not completely dissolved the slurry becomes greatly insensitive. TATP gets its power when allowed to dry out; even the slightest moisture (water or solvent) will greatly hinder its power until allowed to dry out. Or wait 2-3 months for sublimation to occur. Nitroglycerin may be deflagrated in an open area.


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Why is a temporary name given to an element with an atomic number above 100?
All the elements with an atomic number more than 100 are given temporary names by IUPAC according to nomenclature rules. For example, element 101 was temporarily  named "Unnilunium" until they give it a permanent name (101 is now known as Mendelevium).
But why does the IUPAC give a temporary name to elements?

I doubt that IUPAC gave Mendelevium the name unniunium in 1955. In 1985, I know all 109 elements had assigned names and I thought it was weird that the new elements (after 109) were being given Latin numbers. I think I heard they wanted verification before assigning a name. It used to be that the discoverer of the element got to give the element its name.

https://chemistry.stackexchange.com/questions/16029/why-is-a-temporary-name-given-to-an-element-with-an-atomic-number-above-100/16051#16051

The systematic element names refer to a temporary name and symbol for hypothetical and newly synthesized elements. Because of naming controversies in the past, these names were introduced by the IUPAC to discuss the elements without ambiguity. After the discovery is confirmed, they receive a permanent name.
The IUPAC states in the red book this controversy as follows:

In the past, some elements were given two names because two groups claimed to have discovered them. To avoid such confusion it was decided in 1947 that after the existence of a new element had been proved beyond reasonable doubt, discoverers had the right to suggest a name to IUPAC, but that only the Commission on Nomenclature of Inorganic Chemistry (CNIC) could make a recommendation to the IUPAC Council to make the final decision.

As this text suggests, this is a highly political process, that can be dragged out quite easily. The systematic nomenclature and symbols for new elements are stated in IR-3.1.1 as follows:

Newly discovered elements may be referred to in the scientific literature but until they have received permanent names and symbols from IUPAC, temporary designators are required. Such elements may be referred to by their atomic numbers, as in ‘element 120’ for example, but IUPAC has approved a systematic nomenclature and series of three-letter symbols (see Table II).2 The name is derived directly from the atomic number of the element using the following numerical roots:
  0 = nil
  1 = un
  2 = bi
  3 = tri
  4 = quad
  5 = pent
  6 = hex
  7 = sept
  8 = oct
  9 = enn
  The roots are put together in the order of the digits which make up the atomic number and terminated by ‘ium’ to spell out the name. The final ‘n’ of ‘enn’ is elided when it occurs before ‘nil’, and the final ‘i’ of ‘bi’ and of ‘tri’ when it occurs before ‘ium’.
  The symbol for the element is composed of the initial letters of the numerical roots which make up the name.
Example:
  1. element 113 = ununtrium, symbol Uut

As for the transfermium elements, their names have been finalised only in 1994 (Pure Appl. Chem., 1994, Vol. 66, No. 12, pp. 2419-2421).
However, for Mendelevium the first publication already suggested the name and a symbol. It was in use before officially recommended by the IUPAC (A. Ghiorso, B. G. Harvey, G. R. Choppin, S. G. Thompson, and G. T. Seaborg, Phys. Rev. 1955, 98, 1518).

That what my teacher told today. IUPAC started keeping temporary name during cold war due to political reason as you have stated.

According to The Dissappearing Spoon, another issue is that once a name has been formally nominated to a particular numbered element, that name will never be recognized for any other.  If Mr. Smith discovered element 117 and named it Smithium, but someone else was deemed to have found it first, the name Smithium would be permanently deprecated.  Even if Mr. Smith later found element 118, the name Smithium would no longer be available for it.


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Is there any way to determine if the compound is oxide or peroxide or superoxide?
I know that in an oxide oxygen's oxidation number is $-2$, in a peroxide oxygen's oxidation number is $-1$ and in a superoxide oxygen's oxidation number is $-\frac12$. So, how will I be able to detect that any compound is oxide or peroxide or superoxide?
For example, to get the oxidation number of $\ce{Mn}$ in $\ce{KMnO4}$ how will I determine whether $\ce{KMnO4}$ is an oxide, superoxide or peroxide? Is there any way?

generally the most universal way to prove this is X-ray crystallography. It allows to obtain crystal structure of target compound directly. It has, however, an inherent flaw: it can't reliably detect position of hydrogen atoms, so in some cases some uncertainty may remain.
In case obtaining crystal is impossible, the second option is NMR, that allows to get various info about atom's neighborhood, but interpreting NMR data is really hard. In case of superoxide, EPR may prove direct evidence.
Generally there is no way to detect such things from brutto-formula. For example, compound $\ce{CH2O2}$ may be either dioxirane (methylene peroxide) or formic acid. In practice, however, peroxides are really unstable, so in most cases if compound is not overloaded with oxygen (like, $\ce{H2SO5}$) and is reasonably stable towards thermal decomposition, it is most likely oxide and not peroxide. Peroxides are usually formed either from another peroxide or electrochemically, or (in rare cases) via cold oxidation with molecular oxygen. And there are exactly three common superoxides.

If you want to distinguish oxide, superoxide and peroxide ions, generally you can use their X-ray structure well. Oxide is a single oxygen ion, superoxide and peroxide is $O-O$ with approximate bond distance of 1.33 and 1.49 A respectively. Superoxides also have an unpaired spin which can be often detected (even if it is coupling with some other spins).

If oxygen was with the metals of main group elements, the compound is peroxide, but if it was with nonmetals or other metals except for the metals of main group metals the compound is oxide. For example $\ce{KMnO4}$ is oxide it is composed of $\ce {K+}$ and $\ce{MnO4-}$ because there is manganese which is a metal that isnt among the main group elements group in the polyatomic ion. and the oxidation number of the $\ce{Mn}$ is $7^+$.


text:
Which is the bigger ion, F- or O-?
Well, according to the proton-electron ratio $\ce{O-}$ should be bigger than $\ce{F-}$
What about the charge/electron density in $\ce{F}$? Will it not affect the size of the atom of $\ce{F-}$?

I think $\ce{O-}$ is the bigger guy. Since $\ce{F-}$ have a complete octet and also have more protons than $\ce{O-}$.

@Siddharth Yadav, What about the charge/electron density in O?

definitely $\ce{O-}$
In $\ce{O-}$ 7 outer electrons are tied to nuclei with 6 positive charges (number 8 minus 2 electons of inner shell), while in $\ce{F-}$ 8 outer electrons are tied to nuclei with 7 charges. The difference, however, should be quite small.


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Book recommendations for protein structure refinement
I’m an undergraduate Chemistry student and I’m currently writing my dissertation about semi-empirical methods, particularly PM7, applied to protein structure refinement from X-ray structure. Could you recommend me any book/article about it?

I think this is clearly defined enough to get a reasonably-scoped answer.

I found an article here about PM6/7 which might be useful.


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Chemical structure of water-soluble polysaccharide
I don't know much about analytical chemistry, but I need to determine the chemical structure of a water soluble polysaccharide which is not pure and may have D-glucose, mannose and fructose. How do I actually know its chemical structure?
It's from a plant and I know it has fructan and a random polysaccharide which consists of glucose and mannose as the main part and it also has less than 20% impurities such as ash, fat, protein or other sugars. I need to know the arrangement between monosaccharides in the random polymer and the percentage of each polysaccharide.
I have read that it's amorph and linear but I'd rather to check if it is branched. The compound can be consumed with the testing.

Megazyme has several kits to analyze different polysaccharides. If you only need to do the analysis a few times, you can also download the manual for the kit, buy the necessary reagents locally and follow the procedure. http://www.megazyme.com/docs/default-source/artwork/product-list-for-web-may-2014.pdf?sfvrsn=2

Thank you but I want to know if I need to use NMR, IR, Mass spectrometry.... and how the procedure is.

Because you are dealing with a polymer, $^1H$ and $^{13}C$ NMR might be able to tell you the ratio of glucose, fructose and mannose (if the peaks separate), IR might allow you to determine the amount of branching (along with NMR), but MS won't be of much help. Depending on the impurities, they may need to be removed. Polysaccharides are slightly soluble in ethanol, so you can purify the polysaccharides by dissolving in a minimum amount of water, precipitating the proteins first with trichloroacetic acid, filter, then add 3 volumes of ethanol and filtering the precipitated polysaccharides.

Could you digest the polysaccharide into monosaccharides and run on HPLC and compare to known standards? Would get you the percent of each component, but not the sequence.

Depending on the impurities and the length of the sugars, LC-MS (Liquid Chromatography-Mass Spectrometry) would be an option. Comercial Mass Spectrometers can be coupled to a database to quickly find fractured mono sacharides. That way you know the composition. The order in which the polysacharides are made up is not as easy, but I can assure you it will be very chalenging using Spectroscopy to find the answer to that.


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Synthesis of 1-hexene from 1-octyne
This is my synthetic pathway. For one, is it plausible? For another, is it efficient?
1) Reduction of 1-octyne with hydrogen gas and Lindlar's catalyst to 1-octene.
2) Oxymercuration to install a hydroxyl group at the most substituted carbon. This creates 2-octanol.
3) Reaction with tosyl chloride and pyridine to create a sulfonate ester - a much better leaving group than an alcohol.
4) E2 elimination with sodium hydroxide and strong heating to create the internal, Zaitsev alkene.
5) Ozonolysis to create two aldehydes.
6) Reduction of longer aldehyde with hydrogen gas over transition metal to primary alcohol.
7) Dehydration of primary alcohol to alkene using a Lewis acid (aluminium trichloride in this case).

On the plus side, you've clearly thought about this problem and have worked out a detailed synthetic route.  I think most of the steps would work as desired.  The two steps that concern me the most are 4 and 7.  I suspect your step 4 will give a mixture of products.  Perhaps with further experimentation you could find reaction conditions that would give you the desired product without  other products from undesired side reaction (elimination in the other direction and perhaps some rearrangement products too).  I see your step 7 as the most problematical.  Aluminum chloride, heat and the alcohol will produce a carbocation and it will undergo significant rearrangement creating a variety of elimination and addition (polymer formation) products.  So while your reaction scheme is plausible, in my opinion it is not efficient.

Could I dehydrate the alcohol via heating and sulfuric acid?

That would still proceed through a (primary) carbocation intermediate and be subject to rearrangement and polymerization in addition to the desired elimination.  Perhaps dehydration under basic conditions might have a better chance of success.

You mean NaOH dehydration?

Any nucleophile that favors elimination over substitution.

I guess this is just a paper exercise, since trying to make 1-hexene from octyne means you forget efficiency. A very useful and interesting exercise though. I agree with Ron about potential issues with step 4. I think sulfuric acid is worth trying for step 7 though and if it fails you can try stronger dehydrating agents that are less/non acidic or an indirect method like your tosylate elimination or even selenium chemistry. For step 6 I would use the standard $\ce{NaBH4}$ or $\ce{LiAlH4}$ for reduction.
Alternatively, after step 1, you can apply your ozonolysis-aldehyde reduction - elimination cycle twice to get to the product. Generally you sacrifice 2 of your 8 carbons ie 25% "wasted" from your starting material.


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Activation Energy of a rotting apple
A slightly bruised apple will rot extensively in about 3.5 days at room temperature (22.0 degrees Celsius). If it is kept in the refrigerator at 0.5 degrees Celsius, the same extent of rotting takes about 12 days. What is the activation energy for the rotting reaction?

We do not do your work for you. Rephrase your question so we can guide you on obtaining the answer when you don't understand.

I have to say, among all the problems I've ever seen, I think this wins top place in the largest stretch of chemical kinetics theory...

Maybe so, but if such a question seems spurious, it might be quite nice to point out why it is so, and perhaps help the enquirer rephrase it to a question which lends itself to being answered by the theory.

Welcome to Chemistry Stack Exchange! Please add what you have attempted towards solving the problem into the body of your question.  Even a rough outline of what you know so far or an educated guess is fine.  For more information, see the site's homework policy for how to ask homework questions. Thanks!


text:
Is trans-2-butene IR active with regard to its C=C bond?
To be IR active, the vibration of a bond must result in a substantial change in dipole moment. Since trans-2-butene is symmetrical, will the C=C stretch show up on IR?
Something tells me no because if the C=C bond vibrates, there will be no net change in the bond dipole moment, unlike in cis-2-butene. Any polarization one way in trans-2-butene is cancelled out by a polarization in the opposite direction in trans-2-butene. Cis-2-butene, however, has a permanent dipole and any vibration only exacerbates this dipole.

What about the C-H bonds? They will show in the IR.

I was thinking specifically of the C=C bond.

How come something such as 2-butyne is IR inactive according to my book? Couldn't the methyls be moving in different directions here too?

Let us continue this discussion in chat.

Dissenter I think Martin has answered your first comment below.  If you would, add a separate question about what you and @LDC3 came up with in chat.  I'm going to leave the second comment for now.

@jonsca You can always click on the link since I didn't make it private: http://chat.stackexchange.com/rooms/16996/discussion-between-ldc3-and-dissenter

Not IR active, but Raman active. The measured vibrational frequency using Raman spectroscopy is $1682 \pu{cm^{-1}}$.

The $\ce{C=C}$ stretch will not show in IR. The frequency is $\nu_{\ce{C=C}}=1700~\mathrm{cm^{-1}}$ and has A gerade symmetry (DF-BP86/def2-SVP). There is no change in the dipole moment.

However, some bending frequencies involving the $\ce{C=C}$ double bond will show up. For example $\nu_{\ce{HC=CH}}=963~\mathrm{cm^{-1}}$ with a medium intensity.


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Index of Hydrogen Deficiency for para-Toluenesulfonylchloride
In para-Toluenesulfonylchloride, there are 5 pi bonds and 1 ring. This suggests a IHD of 6. However, the actual IHD is 4. How come?
Is it because the formula IHD = rings + pi bonds does not apply to compounds containing sulfur? If not, what formula do I use? I have this formula:
$IHD = \frac{2C + N - X - H + 2}{2}$
Using this formula yields a IHD of 4.

You have to start from the general formula for the degree of unsaturation
$$DU = 1 + \frac{1}{2} \sum n_i(v_i-2),$$
where $n_i$ is the number of atoms with a valency of $v_i$.
For para-toluenesulfunylchloride ($\ce{C7H7SO2Cl}$) you will therefore arrive at
\begin{align}DU &= 1 + \frac{1}{2}\Big[n_\ce{C}(4-2) + n_\ce{H}(1-2) + n_\ce{S}(6-2) + n_\ce{O}(2-2) + n_\ce{Cl}(1-2)\Big]\\
&=1+\frac12\Big[7(4-2) + 7(1-2) + 1(6-2) + 2(2-2) +1(1-2)\\
&=1 + \frac12\Big[14 -7 +4 +0 -1\Big]\\
&=6,
\end{align}
because sulfur uses all six valence electrons in this case.

Just to confirm though, you need to know the oxidation state of sulfur before using the equation.

@jerepierre Yes of course. I was just making the point that the OP used the wrong formula.

There are 5 pi bonds: three C-C in the ring and two S-O pi bonds. Add the ring, and the IHD is 6.
Because of it's ability to exist as higher oxidation states than the second row elements, I don't believe that there is any way to include sulfur in the IHD formula unambiguously. Consider the following two compounds, toluenesulfonyl chloride and an isomer.

The formulas are the same, but the IHD of the left structure is 6 but the right structure is 4.
I think the best way to handle third row (and beyond) is to treat the large atom as if it were it's second row analog (P->N, S->O, etc.), and use the result as the lower limit on the IHD.