text:
Will fire corrode stainless steel?
I am considering building a stove out of stainless steel. I want to us stainless steel because the stove will have a double outer jacket to pass water through and heat it.
Will a wood fire sitting directly on stainless steel corrode it? Would a higher grade of stainless steel, say 316 instead of 304, prevent this?
Should the stainless steel be protected in some way, perhaps by a bed of fire bricks (at least) under the fire, to prolong its life? Or will stainless steel stand up to exposure to fire without corroding for a long time?

Maybe I should have posted this in Engineering?

Yes, fire does corrode steel over time, though it takes time. Somewhere around several mm per year of continuous exposure. Steel is not a fireproof material. If the steel is cooled, the process is slowed. There are special steels tuned for boiler use.

It its mm per year of continuous exposure, sound like I will be ok. This will only be used occasionally.

brief search revealed that if boiler tube is intended to boil, it should NOT be made of stainless steel. Special boiler steels should be used.

Interesting, thanks. In my case I'm only wanting to heat the water for a bath (outdoors), 50C would be my guess at the maximum temperature. I suppose next to the metal it may get hotter, then mix in. If it boils it gone wrong, as the bubbles tend to disrupt the convection flow that circulates the water.

A wood fire will not be hot enough to cause more than about 0.02" of metal loss in a year of continuous use. This is assuming about 1500 °F. It would make no difference if 410 (13 Cr - exhaust pipes) or 304 or 316 ss is used. If is has a complete water jacket, the life would be much longer.
Boilers are usually carbon steel; electric utility boilers have 1% or 2% Mo and maybe 1% to 2% Cr, not for oxidation resistance but for higher strength. With water in the tubes they are at a much lower temperature than the flame (roughly 700 °F to 1000 °F). However the wood will contain sulfur and other components that can increase oxidation rates.

Get a pad of steel wool.  This is stainless steel, because it's made for use in a sink, washing dishes.  Take a 9-volt battery and briefly touch both terminals to the steel wool.  It will catch on fire.  (Warning: don't try this at home, kids!  But it totally will.)  When the fire goes out, you'll be left with a pad of rusty steel wool.
Therefore, the answer is yes: fire can corrode stainless steel.
If your stainless steel is a big solid block rather than a super-high-surface-area pad of steel wool, the process will be a lot slower, but it will still happen over time.

Can you provide a source for the idea that most forms of steel wool are stainless steel?  I did a quick googling and most steel wool seems to be something called "carbon steel", not stainless steel.

@CurtF. the dishwashing sort might be but tends to be made of thicker "wire" (actually more like strips) than the sort you might find with the sandpaper (NB UK bias).  The latter doesn't burn with a battery, it gets red hot and melts, which can be use to ignite other things.

Non-sequitur. Steel wool has tiny filaments, and the metal is vaporizing because there's a large surface area through which to transfer heat into the steel wool, so it melts and vaporizes. This doesn't imply stainless steel corrodes, or is oxidizing - and in fact one of the reason stainless steel is used in engineering is, it is resistant to many corrosive industrial chemicals. The battery+steel wool just tells you that with enough energy per mass, steel will melt and vaporize - as will glass, gold, titanium, silicon, or just about anything else. This is not corrosion.

Here's a device that's been around for at least several decades, as far as I know:

(source: teploterm.ru)
I've been using something similar for around 15 years. The idea is you put firewood into the oven you can see at the bottom, and the chimney goes through a 90 liter water tank. The bottom is something like .5 - 1 cm thick cast iron, the rest is made from stainless steel sheet metal.
There are variants to the construction, but I think here are the two main points, coming back to your question:

This thing needs replacement around once a decade, being in use around 500-1000 hours a year.

If the stainless steel is thick, then rusting gradually becomes irrelevant. Does make the thing heavy, though.

Yes, there is corrosion, especially around the fiery hot parts.

Also, note that the plumbing within this thing is kind of special, as it needs to refill cold water while simultaneously mixing your hot and cold water. I remember that it was very damaging to run it empty.
And yeah, this is more engineery than chemicky...

Not easily from what I can find. But remember it is Stain-less not Stain-impossible.
Actually there are different type of stainless steel used for different stuff. Like for protect it from fire I feel A typical composition of 18% chromium and 10% nickel, commonly known as 18/10 stainless, is often used in cutlery and high-quality cookware will be good.
Source: Wikipedia
If you see here fire pits are also made up of stainless steel.

authors: Алёна Гутварева


text:
Did Berzelius unify the names of chemical elements using the Latin language?
I asked this question because the Chinese Wikipedia community is trying to decide whether it should use Latin names or IUPAC names for the first introductory sentence for its articles on chemical elements.
One of the central argument by those support Latin names is the statement that Swedish chemist Berzelius unified the names of chemical elements using the Latin language common to European countries.
They cite the source in Chinese language here:
http://www.cnki.com.cn/Article/CJFDTotal-ZHHJ198805015.htm
"Swedish chemist Berzelius first proposed to use Latin, a language common to European countries, to unify the naming of chemical elements, thereby changing the chaotic status of naming them." (My translation)
What I have found from the literature is that Berzelius's proposal helped unifying the symbols of chemical elements, but the names remain diverse in language sources (not all are Latin) and some remain disputed (not all unified). Thus, I gather that Berzelius's proposal helps, but it is far from the statement that he unified all the names of chemical elements using Latin. If so, it begs the question why IUPAC names for Gold, Silver, etc. are not Latin.
Thus, your help is needed to answer the question as stated in the title.

Are you saying the Latin name for gold should be 'aurum', and silver 'argentum' etc?

Yes. That is exactly what the Latin-note supporters believe in Chinese Wikipedia.

Somewhat related http://chemistry.stackexchange.com/questions/28387/why-do-some-chemical-symbols-contain-a-single-letter-while-others-contain-two-le/28496#28496

What I have found from the literature is that Berzelius's proposal helped unifying the symbols of chemical elements, but the names remain diverse in language sources (not all are Latin) and some remain disputed (not all unified). Thus, I gather that Berzelius's proposal helps, but it is far from the statement that he unified all the names of chemical elements using Latin.

This is essentially true. Berzelius was the first to suggest using symbols abbreviated from the names of the elements and this system, with some modifications, is still used today. Most of Berzelius' symbols were abbreviated from Latin names but a large number of the elements discovered since then have had symbols drawn from their common names and some of Berzelius' original symbols have been dropped in favour of more modern versions. Dalton had earlier used a system of drawn symbols to represent the elements but this was not widely taken up as the symbols were difficult to draw and memorise. (source)
Looking though this wikipedia article on the etymology of chemical elements, the most common language of origin appears to be Greek (argon, chlorine etc.), with Latin coming in second (mercury, rhenium etc.), including quite a large number of names which come originally from another language (often Greek) via Latin.
Additionally there are a large number of elements which are named after places (Swedish - ytterbium, yttrium, etc. ; English - americium, californium, etc. to name just a few) or people (mendelevium, einsteinium, fermium etc.) and have been given the Latinate endings -ium or occasionally -um (see this question for more discussion). Also note that according to IUPAC recommendations:

For linguistic consistency, the names of all new elements should end in “-ium”

The common, and IUPAC recognised, names of gold and gilver both have old Germanic origins and are recognised as official names because of their widespread usage. I have never seen the Latin names for gold, silver or lead (aurum, argentum and plumbum) used in a serious chemical context, although of course the elements' symbols (Au, Ag and Pb) are derived from them.
As such I would not recommend using the Latin names as the main recognized name in a Wikipedia page, although a side note about them might be justified.
EDIT: Thanks to @KlausWarzecha for pointing this out. There is one exception where the latin names (aurum, etc.) come into play: in anionic complexes, the latin version of the name is officially used. $\ce{[Cu(CN)4]^{3-}}$ is tetracyanocuprate(I), $\ce{[Ag(Cl)2]^-}$ is dichloroargentate(I), $\ce{[Au(Cl)4]^- }$ is tetrachloroaurate(III). However, this does not change the viewpoint that the Latin names should not be used when commonly discussing the element.

Thanks. It is very useful (though I cannot vote it up due to my newbieness here). Thank you for a very comprehensive and detailed answer.

There is one exception where the latin names (aurum, etc.) come into play: in anionic complexes, the latin name of the name is officially used. $\ce{[Cu(CN)4]^{3-}}$ is tetracyanocuprate(I), $\ce{[Ag(Cl)2]-}$ is dichloroargentate(I), $\ce{[Au(Cl)4]^{-}}$ is tetrachloroaurate(III).

@KlausWarzecha Good point, I forgot about this. Will edit it into the answer.

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text:
Heating of Ethers with HI and Cleavage of Ethers
Question:

Give the major product formed by the heating of the following ether with HI.
        CH3
        |
CH3-CH2-CH-CH2-O-CH2-CH3

My Attempt:
According to me, the product should be this due to delocalization of positive charge among the neighboring carbon atoms.

        CH3
        |
CH3-CH2-CH-CH2-I + CH3-CH2-OH

Answer:
I checked the answer in my book and it is completely different:

        CH3
        |
CH3-CH2-CH-CH2-OH + CH3-CH2-I

Now, I think they may have considered steric hindrance to be the deciding factor of site of attack of OH but will not delocalization be dominant over steric effect?
Also, what will be the effect of hyperconjugation on the reaction?

Your ether has $\ce{CH2}$ groups at both sides of the oxygen atom.
I would exclude a $S_N1$ reaction here! Instead, cleavage of the ether proceeds via $S_N2$. The oxonium ion resulting from addition of a proton is attacked by iodide at the least hindered carbon atom. Iodide is quite big ;-)

So that means that they indeed has taken steric effect to be the factor.

@SharadGautam Yes, absolutely. $S_N1$ reactions would give two primary cations with little stabilisation here.

How will Hyperconjugation affect the reaction?

According to me, The product should be this due to delocalisation of positive charge among the neighbouring Carbon atoms.

Even if you consider an $S_N1$ reaction pathway for this reaction, the positive charge could be more stable on the $\ce{-CH2CH3}$ branch as it has three hyperconjugation structures stabilizing the positive charge against one in the other branch. And gives the same products.
Moreover, as Klaus Warzecha said, steric hindrance comes into the play while considering $S_N2$ pathway which looks a more plausible pathway due to the relatively unstable primary carbocations that form in $S_N1$

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text:
Which allotrope of phosphorus is kinetically the most stable?
Which allotrope of phosphorus is kinetically the most stable?
I ruled out white and yellow as being quite reactive.
Red and black allotropes are both polymeric and comparitively less reactive.
Black looks the most stable one thermodynamically. But kinetically, I have no idea whether it should be red or black.
The answer provided to me is Red.
(Relevant chat discussion)

IIRC, Black is the thermodynamically most stable and Red is kinetically most stable. The reasoning, however, eludes me.

You can bookmark an entire conversation in chat, and link to it. I did it for you and inserted the link.

Thermodynamic stability is easy to define, but kinetic stability is not that easy to define. I would consider what it means to be kinetically stable.

I think the key to this question lies in the understanding of the difference between thermodynamic and kinetic stability. In this post, I find Thomij's answer the most rigorous and enlightening.
From this you should learn, that the thermodynamically most stable allotrope corresponds to the global minimum on the potential energy surface, while a kinetically stable allotrope, i.e. hindered through an activation barrier, corresponds to local minimum. From this it can easily be seen, that the thermodynamically most stable allotrope is also the kinetically most stable allotrope. This is due to the fact, that the barriers for interconversion of the different allotropes always have to be higher for the most stable one.
In the following two dimensional case you see that the thermodynamically most stable, i.e. the global minimum, is surrounded by two local minima. It must always be true, that $E_\mathrm{a}(1\to2)<E_\mathrm{a}(2\to1)$ if $1$ is a local and $2$ is a global minimum. The same applies to the other case. If this was not true, then $2$ cannot be the global minimum and hence not the thermodynamically most stable modification.

Having established that, it is therefore true, that black phosphorus is the thermodynamically and kinetically most stable form, see Hollemann-Wiberg, p. 680.
However, looking at a different reaction, the case might not be so simple any more. This is even true for different temperatures, as the following graphic will show (quoted from the linked book).

the thermodynamically most stable allotrope is also the kinetically most stable allotrope. Is this P specific? Since diamond/graphite doesn't show this.

@user223679 What do you mean with diamond/graphite doesn't show this?   This has to hold for any element/allotrope. If you are looking at different reactions and conditions, things might change, as kinetics is a function of those.

I am unable to conclude from the graphic that black is the answer. The paragraph comments on the thermodynamic stability while the flowchart seems to say that 550-620 degrees is required to convert red, which seems to say red is kinetically stable.

In response to your first comment, The classic textbook example is the conversion of diamond to graphite, which is thermodynamically favorable because the free energy of graphite is lower, but doesn't occur under ordinary conditions because the kinetics of the reaction (in the form of the immense activation energy required) are extremely unfavorable I meant this.

Of course red is kinetically stable, but kinetics is the only reason that it is stable, it is still unstable with respect to black and violet phosphor.

@user223679 Under normal conditions, diamond is kinetically stable and won't form graphite. It remains still less stable with respect to graphite.

In response to your 2nd comment, so red is kinetically the most stable?
it is still unstable with respect to black and violet phosphor. --> That seems to be unstability defined on the basis of thermodynamics and not kinetics.

3rd comment: Even in the case of P, under normal condition, wouldn't red be kinetically more stable, as your graphic's flowchart suggests that.

This discussion ended in chat. Look here for the bookmarked conversation.

@Martin-マーチン With regard to diamond, note that in the interconversion between graphite and diamond, assuming microscopic reversibility, the free energy barrier from graphite to diamond is higher than the reverse. Of course, graphite can under combustion much better, but again, it is not clear what "kinetic stability" even means...


text:
Standard Heats and Free Energies of Formation and Absolute Entropies of 2-Chloro-2-methylpropane
Does 2-Chloro-2-methylpropane have standard heats and free energies of formation and absolute entropies values in a liquid state?
http://www.wiredchemist.com/chemistry/data/thermodynamic-data
This website only provides the values in gaseous state and I can't seem to find the values for the liquid state anywhere.

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

Maybe you somewhere find a copy of CRC Handbook of Thermophysical and Thermochemical Data to have a look inside, if they have any values.

$\Delta H_\mathrm{f}^\circ = -211.3~\mathrm{kJmol^{-1}}$
Neither source provided entropy or free energy data though but this is a start.
http://courses.chem.indiana.edu/c360/documents/thermodynamicdata.pdf
http://webbook.nist.gov/cgi/cbook.cgi?ID=C507200&Units=SI&Mask=2#Thermo-Condensed

Couldn't you calculate those values using Hess's Law and the law related to free energy?


text:
Reaction of perchloric acid with cyclopropenol
Question:

Identify A and B.

My attempt
The first step will be the attack of $\ce{PhMgX}$ on carbonyl carbon, that is, 1,2 position followed by hydrolysis, yielding an alcohol.
But how does the reaction proceed with $\ce{HClO4}$?

you should read about cyclopropenyl cation in aromaticity section of advanced organic books.  HClO4 is a strong protonating agent and should protonate OH-group, that will leave. The remaining cation is known to be remarkably stable in organic chemistry and have stable salts.

@permeakra it is remarkably stable yes, but getting rid of the strained ring by oxidative cleavage (similar to hot $\ce{KMnO4}$) seems way more plausible. We shouldn't ignore the oxidative capabilities of $\ce{HClO4}$

The answer provided is wrong . Whenever cyclic conjugated ketones comes in question then the conjugation is the unsaturated part is the point of action. Which means double bond will be attacked.

Can you provide a reference?

The first step is indeed 1,2-addition of the Grignard reagent to the keto group, and product A is the alcohol:

Treatment of the alcohol with the strong acid $\ce{HClO4}$ cannot yield the usual dehydration product (alkene) in this case, as the corresponding alkene would have two double bonds in an already highly strained three-membered ring, which is highly unfavorable. Instead, after protonation of the $\ce{OH}$ group and subsequent loss of $\ce{H2O}$, the carbocation:

is obtained as the final product B (with $\ce{ClO4-}$ being the counteranion). The planar cyclopropenyl cation is a Hückel aromatic compound with $(4 \cdot 0+2) = 2$ $\pi$ electrons, and delocalization of the positive charge over the two phenyl substituents can provide further stabilization.

Carbocation as the final product, really interesting!

Yes this happens when the intermediate is highly stable with a stable ion like flouro borate, perchlorate etc. (Note that stable counter ion makes the ionic compound to be formed easily according to fajan's rules

@Archer I think this answer disregards $\ce{HClO4}$'s oxidative abilities. I think the stability gained by cleaving the alkene is far more than the resonance of the carbocation.

@napstablook $\ce{HClO4}$ is often used in dilute solution where it dissociates, and then the oxidation power of the perchlorate ion is kinetically squashed. Perchloric acid here is acting only as an acid.


text:
Identifying MgCl2 or CaCl2 in solution
I have an aqueous solution of either $\ce{MgCl2}$ or $\ce{CaCl2}$ in solution with $\ce{K2SO4}$. There is a white precipitate that forms on the bottom that does not dissolve when $\ce{HCl}$ is added (maybe it needs to be in excess?). Is there any way to find out if it's $\ce{Mg^2+}$ or $\ce{Ca^2+}$? I have already tried adding $\ce{NaOH}$, and the result was a not very cloudy, white precipitate, which is not really jelly like, but when compared to stock samples of $\ce{MgCl2}$ and $\ce{CaCl2}$ with $\ce{NaOH}$, they don't match. I tried a flame test but only could see the $\ce{K+}$ (violet flame).

First, it is no use of putting $\ce{NaOH}$ in the solution as both ion with form white percipitate and insoluble in excess of $\ce{NaOH}$.
Try aqueous $\ce{NH3}$. If the solution contains $\ce{Mg^2+}$ ion , it will white percipitate and insoluble excess of $\ce{NH3}$. However, $\ce{Ca^2+}$ ion will not form any percipitate when $\ce{NH3}$ it added.

$\ce{MgSO4}$, otherwise known as epsom salt, is soluble. $\ce{CaSO4}$ ($K_\mathrm{sp}=7.1 \times 10^{-5}$), is slightly soluble. It is extremely likely that unless there are some other unknown metal ions in this solution, the white precipitate you describe in the solution is $\ce{CaSO4}$.

Why not do the following, add an excess of sodium carbonate. Boil the mixture, cool and then centrifudge it down. Pipette off the supernatent. Next repeat the process several times to convert the sulfate into the carbonate.
Next wash the solid with water to remove the sodium carbonate. You now will have pure calcium carbonate or magnesium carbonate.
Next dry the solid on the filter in the following way, transfer into a preweighed crucible. Put it in an oven at 100 oC and dry to a constant mass.
Next when you have the mass of the metal carbonate, put it in a muffle furnace at 700 oC and ash it to a constant mass to get the mass of the oxide. You can then work out by gravimetric method if it was magnesium or calcium in it.
Another method would be to isolate the pure metal carbonate, add a weighed amount to a flask. Add some HCl to dissolve it and then adjust the pH with sodium hydroixde. Next titrate it with EDTA using Eriochrome Black T. This will allow you to determine the number of moles of metal in each gram of the dry carbonate solid.
Years ago an older chemist told me of how he used to measure zinc in mixtures of calcium, magnesium and zinc. He would do the titration with EDTA using Eriochrome Black T to get the total number of moles of Zn / Mg / Ca. He would then do the titration with KCN present to mask the zinc. By difference he would then be able to calculate the number of moles of zinc and the number of moles of calcium / magnesium.
You could always take the easy route and use AAS, ICPOES or even neutron activation, counting the calcium-45 as a method of determining the amount of calcium present in the carbonate solid.


text:
Capacitance and flammable liquid
I'm trying to attach a sensor to some tubes to measure the level of liquids in them.
see here
One tube contains toluene and the other contains Recosol R55 (shellite).
I'm thinking of using a capacitance level sensor like this or this.
But I am concerned about the safety of doing this. Could the capacitance sensor ignite these liquids? is this even possible?
Obviously I need to check with the suppliers of the device. However, most of datasheets for level sensors omit this information. Perhaps they are not game enough to make any claims about their sensor in this regard.
I don't really understand capacitance and whether it can ignite flammable liquids.

The capacitance sensor itself will certainly not present an ignition risk to your system. The sensor will be on the outside of the container, and it does not introduce any electrical potential that would cause a spark.
The way the device works is to measure the change in the sensor's capacitance depending on the dielectric effect of the liquid, which will increase as more liquid enters the electric field of the sensor (a capacitor).


text:
Which effect does adding hydrogen chloride have when determining the concentration of sodium hypochlorite in bleach via titration have?
Many of you might be familiar with this and I do understand this completely, however I do have a question.
If you are familiar with this, then this is the equation and this is the question for it.
$$\ce{ClO^{-} ~(aq) + 2I^{-} ~(aq) + H2O ~(l) -> Cl- ~(aq) + I2 ~(aq) + 2 {}^{-}OH ~ (aq)}$$
So for the equation above I understand everything except that in the titration process my instructions say to add $\ce{HCl}$ into the solution, I do not understand how adding $\ce{HCl}$ will have any impact.
I will provide an accurate reference later.

The reaction is in equilibrium and if you don't add any acid, then some of the $\ce {ClO-}$ does not react. Adding acid will change the equilibrium to the right so most of the $\ce {ClO-}$ reacts.


text:
Selecting between redox and neutralization reactions for a mixture of HCl, H2SO4 and NaOH
I know that you can have 2 acids reacting with 1 base. These 4 reactions are an example of this:
$\ce{3 HCl + H2SO4 + NaOH -> 3 H2O + SCl2 + O2 + NaCl}$
$\ce{6 HCl +  4 H2SO4 + 2 NaOH -> 8 H2O + 2 S2Cl2 + 5 O2 + 2 NaCl}$
$\ce{HCl + H2SO4 + 3 NaOH -> 3 H2O + Na2S + 2 O2 + NaCl}$
And
$\ce{HCl + H2SO4 + 3 NaOH -> 3 H2O + Na2SO4 + NaCl}$
I checked and all these equations are balanced. But will these reactions forming sulfur chlorides and ionic compounds with sodium and sulfur as well as $\ce{H2O}$ and for 3 of them, $\ce{O2}$, actually occur without any added heat or are they endothermic acid base reactions and thus require heat or will the stronger acid(in this case $\ce{HCl}$) be the one that reacts and the weaker acid (in this case $\ce{H2SO4}$) just make the pH of the solution lower?

Those first three equations might be stoichiometrically valid, but they look extremely weird from a chemical perspective, and I would very surprised if they happened. I suspect if you calculate the reaction free energy change using the free energy of formation for each compound, you will find that the last reaction is far more favourable than the rest.

As Klaus and I mentioned, the first reactions seem very strange from a chemical perspective, even if they are correctly balanced. This is where a chemist needs to exercise their intuition and experience.
Balancing reactions is nothing but solving a linear system of equations, so it is entirely possible to propose reactions which balance perfectly, but which are based on questionable chemistry. For example, in all but the last reaction, the sulphur atoms in $\ce{SO_4^2-}$ anions are being reduced to lower oxidation states. While this is possible in some conditions, it takes quite strong reducing agents, and none of the reagents shown are known to possess significant reducing character. Reactions which release gaseous oxygen are also not particularly common. Furthermore, two reactions propose the formation of sulphur chlorides, which would actually hydrolyse in contact with water.
Disregarding intuition, it should be possible to pick out which reaction is the most favourable by simply comparing the Gibbs free energy of reaction in each case; the most exergonic reaction, with the most negative value of $\boldsymbol{\Delta_\mathbf{r}G}$, should dominate (assuming no kinetic factors bar it from happening, which is unlikely in this situation). I did a very quick and dirty calculation, out of curiosity, by taking the standard Gibbs free energy of formation $\Delta_\mathrm{f}G^\circ$ for each compound and summing them together to obtain the standard Gibbs free energy of reaction:
$$\Delta_\mathrm{r}G^\circ = \sum\limits_\text{products}\Delta_\mathrm{f}G^\circ - \sum\limits_\text{reagents}\Delta_\text{f}G^\circ$$
This is only meant as a qualitative way to compare the reactions, as I am assuming standard state conditions for the reactions, which are not necessarily true. Furthermore, for a few species, I didn’t find the correct values in these two sources [1] [2]. $\ce{SCl2}$ and $\ce{S2Cl2}$ probably can’t be found tabulated for the aqueous phase because of their propensity to hydrolysis, so I used the free energies of formation for the gas and solid phases, respectively. I also didn’t find a source for aqueous $\ce{Na2SO4}$, so instead I used the data for the solid decahydrate $\ce{Na2SO4 . 10 H2O (s)}$. The results I got were:
$$ \small
\begin{array}{lcc}
\hline
\text{Reaction} & \Delta_\mathrm{r}G^\circ / \mathrm{kJ\ mol^{-1}}\\
\hline
\text{A} & +440  \\
\text{B} & +1940  \\
\text{C} & +590  \\
\text{D} &  \mathbf{-2610}  \\
\hline
\end{array}
$$
As you can see, even though the calculation was very rough, the huge disparity in free energy values suggests that the first three reactions are completely overwhelmed by the last one, so that of the four possibilities listed, only the last has any chance of occurring. Strictly speaking this doesn’t mean the last reaction happens, only that it’s much more favourable than the others presented; there could be an unlisted reaction which is even more favourable than all four mentioned in the question. However, empirical data confirms that the last reaction, a simple neutralisation reaction, is the major event when mixing $\ce{HCl}$, $\ce{H2SO4}$ and $\ce{NaOH}$ in practically any conditions.

Very good answer. This is a very good example of the value of teaching thermodynamics in chemistry.

So does that mean that these redox reactions are endothermic and thus require heat energy to be added in order for it to happen? Or would I need to use a different sulfur acid where the sulfur is either positively charged or negatively charged depending on whether the atoms it is bonded to other than hydrogen break off as cations or as anions(Which I think requires that it is a carbide since carbon doesn't mind if it is positive or negative when it is charged)?

If you want to make one of the first three reactions more likely, the problem is that you have to selectively favour them, while not accidentally favouring the last one too. Just increasing the temperature is a rather blunt tool, and could well make a favourable reaction even more favourable. One way to force the first three reactions is to consume a product which is only present there. For example, in this case, you could add a reagent to consume the oxygen gas. This doesn't necessarily mean one of the first three reactions would then take place; there are other reaction possibilities.

Fortunately, sulfurdichloride and disulfurdichloride (= sulfurmonochloride), which is the decomposition product of the former,
$$\ce{2SCl2 <=>> S2Cl2 + Cl2}$$
cannot be prepared that way.
I suggest to balance the neutralisation reaction without proposing redox reactions that lead to these compounds and I would stay with your fourth equation.


text:
How are RDM patterns for reactions in the KEGG database constructed?
With respect http://www.genome.jp/dbget-bin/www_bget?rp:RP00167
can anyone tell me how the RDM pattern were obtained??  Information what RDM is given here: http://www.genome.jp/kegg/reaction/
Note: This may be a silly question to all chemists, but not for me, i am a computer science student , and my research field in bioinformatics and so i have to study both biology and bit of chemistry, so please just dont close the question or be harsh on me, I have very little knowledge in chemistry and also googling doesn't help much. I asked this question in Biology forum, they closed the question and asked me to ask it here

Did you have a look at Computational Assignment of the EC Numbers for Genomic-Scale Analysis of Enzymatic Reactions by Minoru Kanehisa, the founder of KEGG (Kyoto Encyclopedia of Genes and Genomes)?

If you are anywhere near a college or university, you will probably be able to access the paper suggested by @KlausWarzecha. The Journal of the American Chemical Society is one of the most prominent chemistry journals in publication.

@jerepierre Definitely! In addition, the J. Chem. Inf. Model. paper suggested in my answer is available free of charge.

To my impression, there is no RDM value, but a RDM pattern!
This is about mapping atoms is a reactant and a product by assigning R (reaction center), D (difference atoms) and M (matched atoms) in both compounds. Pairs for R, D, and M are separated by colons. The pattern is thus given as
R(reactant)-R(product) : D(reactant)-D(product) : M(reactant)-M(product),
where a an asterisk marks a void centre.
In order to figure out how these patterns are retrieved, you definitely want to have a look at Modular Architecture of Metabolic Pathways Revealed by Conserved Sequences of Reactions, a freely available article by Minoru Kanehisa (KEGG founder), published in J. Chem. Inf. Model., 2013, 53,613-622 (DOI). Don't forget to read the Supporting Information, particularly the sections on KEGG atom types and  RDM chemical transformation notation.


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What is the meaning of scores when comparing reaction classes in the KEGG database
With respect to related R class in KEGG website:
http://www.genome.jp/tools-bin/relclass?RC00064  What is the biological or statistical relevance of the score? How is it computed? And does it signify the bonding strength of how probably it is for a pair of compounds to be an R pair?
Note: This may be a silly question to all chemists, but not for me, i am a computer science student , and my research field in bioinformatics and so i have to study both biology and bit of chemistry, so please just dont close the question or be harsh on me, I have very little knowledge in chemistry and also googling doesn't help much. I asked this question in Biology forum, they closed the question and asked me to ask it here

This question could also profit from more context and an actual title.

http://www.genome.jp/ >> KEGG >> KEGG Overview >> R number

And does [the score] signify the bonding strength of how probably it is for a pair of compounds to be an R pair?

No, it has nothing to do with that. It is not about how probably a particular reaction is.
When comparing RCs (reaction classes), the scores in the table of related reaction classes are just a measure on how similar these reactions are.
RC00064 apparently is an N-acylation or the formation of a peptide bond. RC00197 with a score of 1.0 is a N-formylation and considered very similar.
Change the cutoff to a lower value, such as 0.4, and compare the results at the end of the table with RC00064. You'll find dissimilar reaction classes like RC02518, the oxidation of a selenol to a selenic acid.


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Tap water leaving white residue after drying
There is no heating or cooling in this process.
We recently move to a new house and something strange is happening with the water in every tap of the house.
First thing I noticed is when you fill a bucket with the garden hose the bubbles in the water lingered around longer than they should have, ie. normal senario fill bucket with water and bubbles pop immediately, this senario the bubbles take a few good long seconds befor they pop.
Second thing I noticed was after waking away from the garden hose and returning after everything was dry there was white residue all over anything that the water had been on, so I wipe it down and did it again and same thing happened, so I went and tested all the taps in the house with the cold water and the same thing happened for all the taps, so all the taps are using the same water source.
I have heard of white residue being left behind after heating from a source like a kettle but never heard of white residue being left behind from a non heated source.
Is there any way I can find out what this is and should I be worried about this (more than I already am)?
Or alternatively does anyone here know what this is?

You either have hard water or a water softener system.
If you have hard water, then the residue is a combination of calcium / magnesium carbonate / chloride.
If you have a water softener system, then the residue is from the salt used to remove the hard ions.


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Why are the three most simple alcohols completely miscible?
I want to know why the three most simple alcohols (methanol, ethanol and propanol) are completely miscible in water?
I want to know what happens at molecular level.
I think hydrogen bonding plays an important role.

This happens because the intermolecular forces between different low molecular weight alcohols are very similar; as you say, they are dominated by hydrogen bonding. You can consider mixing on a molecular level to involve exchange of molecules from the two phases at their interface; if the energy changes involved are very small, mixing will occur as it is entropically favoured (mixed phase is more disordered) and the average thermal energy is far higher than the barrier to diffusion.

It would be great if you also add image explaining hydrogen bond. :-)

No-one really knows what the structures of hydrogen bonding clusters are in water, let alone water-alcohol mixtures. There's a good explanation of what hydrogen bonding is at http://chemistry.stackexchange.com/questions/5503/what-exactly-is-hydrogen-bonding-and-why-cant-it-exist-with-atoms-other-than-f.


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How common are automated pipetting systems
It's clear that automated pipetting is faster than doing it manually, but how common is it for a lab to have an automated pipetting system, and what usually are the circumstances that a lab have before they purchase such a system?

Are you asking about something different than what I think is an automatic pipette. Google seems to agree with me. The single channel versions of these devices run between 100 and 300 USD, which is hardly a big expenditure for most labs. We use the single channel devices in my general chemistry courses - every pair of students gets one. Even the multichannel pipettes are not that expensive.

@BenNorris http://en.wikipedia.org/wiki/Automated_pipetting_system

Ah! I have never seen such a thing, just the pipettes. I edited your question to put in a link to the device into it so that others like me who think that "auto pipetting" only refers to a hand-held device.

@Martin-マーチン: I see you welcome a lot of people here - that's nice :) Just so you know, the English expression is "consider accepting one", not "*consider to accept one" which sounds wrong to a native speaker. Hope you don't mind me giving you this feedback!

@psmears I don't mind at all, thank you for pointing that out. As a non native speaker I am always happy, when someone points out the mistakes I make. I have now edited my template, thanks again.

The circumstances in a lab that would make automated liquid handling appropriate would be a need to prepare large numbers of samples by a routine method, so that it becomes more cost and time effective to install and program a machine to do the job than to do it with handheld equipment.
This is clearly the case in high throughput drug discovery work and also in analytical labs that perform identical tests on large numbers of samples; it's less common outside these fields although high throughput techniques can also be used in the development and testing of formulations, for example for personal care products or paints and coatings.
Standard automated liquid handlers are designed to deal with volumes in the  microlitre to millilitre range and liquids that are not too volatile or viscous. Requirements outside this range, or extra constraints such as moisture or air sensitive reagents, will demand more complex and expensive equipment which may make automation less viable, although companies such as Zinsser and Chemspeed make devices for handling a wide range of liquid and solid substances.
To get a more specific answer than this you're probably going to have to narrow your question down to the particular industry and type of lab you're thinking of. If you're on LinkedIn you might get some useful responses from the Laboratory Robotics Interest Group discussion board.

Speed isn't everything. Reproducibility might be much more important!
Imagine that you have to transfer different volumes from five different stock solutions to a 1536 well plate and adjust the total volume by adding water or the common solvent of your choice. A programmable pipetting automat might be a blessing here.

Makes sense, how common is such a system? Does almost every lab have one? Or most labs? Or one third? I'm trying to get a rough estimate.

I agree that reproducibility is important, and I would think that of this reason they should be pretty common. My question is not only about why they're important or useful, but also how common they are in labs around the world.

@KristofferRyhl As usual, it depends. These machines, which are often used to transfer volumes in the sub-milliliter range, are definitely not standard equipment in the average organic synthesis lab. Typically, you'd rather find them with the microbiologists, or when high-throughput screening is the primary goal.

@KristofferRyhl - If you are worried about speed when filling well plates, you might be better off with a good multichannel pipette. The multi-pipette, while handheld, will probably be faster than a robot that can only fill one well at a time. The other benefit of a robot is that you can do something else while it is working.

I don't see how this answer addresses the question.

There is also something in between manual pipettes and a fully automated system. We use these guys for a lot of stuff. They can be programmed for a series of different volumes in one go for doing concentration series in well-plates and the like. They're still handheld, but much faster than normal pipettes.


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How does the temperature of water affect the reactivity of alkali metals acting on it?
There are a few experiments comparing the reactivity of alkali metals with cold water. Is there any purpose to use cold water? Does the temperature of water affect the reactivity of alkali metals?

Temperature affects all reactions.

How does it affect? What I'm thinking is, alkali metals should be more reactive with hot water because having higher energy. How do you think?

I suspect that they specify cold water so that the metals do not react so vigorously that they are unnecessarily dangerous to observe.  Yes, hot water should react faster than cold.

Increasing temperature does indeed increase the reactivity of alkali metals, or most things for that matter. The rate of reaction increases because the average kinetic energy of the molecules increases, and so more collisions can overcome the activation energy for the reaction to take place.
The reason for using cold water rather than hot water for the reactions of alkali metals is that alkali metals are already reactive enough at room temperature. Even fairly small lumps of potassium will burn if you put them in cold water so imagine what it would be like with hot water. This is not something you want to try without very careful precautions.

The rule of thumb is that each 10 Celsius rise in temperature doubles the reaction rate. So increasing the temperature certainly wouldn't slow the reaction. I doubt that the temperature rise would yield a very good correlation however to the overall reaction rate.
The metal is a solid so it is a different phase. The reaction is exothermic, so the temperature at the solution/metal interface is certainly higher than the temperature of the bulk solution. So the overall rate of reaction would be very complicated and depend on surface effects of the metal.


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ionising atom(s) with gamma rays
if a gamma ray hits an electron and transfers energy, does it hit that electron (ionising the atom), transfer all its energy and stop or does it pass through multiple electrons, transferring a portion of its energy each time?
i.e.) if a gamma ray has enough energy to ionise multiple atoms, will it, or will it just give all its energy to the one electron?
also if is has the energy to ionise multiple atoms, about how many electrons can it remove (for a range of ray energies)? (i know it is a bit general, but lets say for the first ionisation of potassium)
also, why is the dose equivalent so low? is it because it can only ionise 1 atom by +1, or because its ionizations (and hence energy) are dispersed throughout many atoms (making the meaning less energy transferred per ionisation)?
(i know this has a lot to do woth physics and so i posted it on the physics SE to, but i was wondering weather a fellow chemist could help me out, thanks)

I'm voting to close this question as off-topic because of http://physics.stackexchange.com/q/184826

Various mechanisms for interaction of gamma radiation with matter are possible. Depending on the energy and the composition of the absorbing material, the most important types for gamma radiation are photoelectric effect, Compton scattering, and pair production.
In absorption of gamma radiation of lower energy by the photoelectric effect, the photon is absorbed completely by the atom and an energetic electron is ejected. The energy of the emitted electron is the difference between the energy of the gamma-ray and the binding energy for that electron in the atom.
Gamma radiation of higher energy directly interacts with one electron by the Compton effect. Usually, this is the most probable interaction of gamma radiation. The incoming photon transfers a portion of its energy to the electron and is deflected through an angle with respects to its original direction. Since all angles of scattering are possible, the transferred energy varies from zero to a large fraction of the energy of the photon. The scattered photon may have sufficient energy to interact further by Compton effect or photoelectric effect.
If the energy of the gamma radiation is very large (i.e. it exceeds $1.02\ \mathrm{MeV}$, which is twice the rest-mass energy of an electron), the process of pair production is energetically possible. In the interaction, the photon disappears and is replaced by an electron and a positron. The process may be considered as the inverse of positron annihilation.
Your additional question about the equivalent dose is slightly misleading. The resulting equivalent dose from gamma radiation is not particularly low. By definition, the radiation weighting factor of gamma radiation is $w_\text{R}=1$. Thus, an absorbed dose of $D=1\ \mathrm{Gy}$ leads to an equivalent dose of $H=1\ \mathrm{Sv}$. However, you may ask why some other radiation types have a higher relative biological effectiveness.


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Purification of water with ferric chloride
Ferric chloride ($\ce{FeCl3}$) is a chemical used in water treatment. If ferric chloride is added to a water sample containing calcium bicarbonate, would the $\mathrm{pH}$ of the solution decrease or increase?

If I'd state that the pH drops, would you know why?
If you dissolve $\ce{FeCl3}$ in water, a hexaqua complex is formed. These complexes are known as cation acids:
$$\ce{[Fe(H2O)6]^{3+} + H2O -> [Fe(OH)(H2O)5]^{3+} + H3O+}$$

But where does the Cl3 went in the case?

@LoïcFaure-Lacroix There's no $\ce{Cl3}$ but three chloride ions $\ce{Cl-}$ acting as counter ions to the aquo complex. They just float around, happily solvated by water molecules.

authors: Генади Янков


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How photoluminescence (PL) spectroscopy works
Photoluminescence (PL) spectroscopy is used to find the energy levels in a compound but how does PL spectroscopy works? After exciting a material  with a particular wavelength whether PL machine looks for photons emitted with in a range of wavelenght?
The below image was taken from here.

To get this spectra the $\ce{CeO2}$ was excited at 345 nm and got peaks at 490 and 520 nm(approx) ie 2.53 and 2.38 eV respectively and its band gap is 3 eV. Whats happening here?

What is the question? The experimental procedure? The theory in general? The reason for the shift? The analysis of CeO2 in particular?

The explanation that follows is valid for compound dissolved in liquid solution where PL is understood either as fluorescence or phosphorescence.
PL is measured with a fluorimeter. The main optical components of this instrument are: a lamp that provides the light used to excite the sample, two monochromators (M), one for the excitation (Mex) and the other one for the emission (Mem) and a detector, usually a photomultiplier.
To measure PL the Mex is fixed at the excitation wavelength (in your case 345 nm) and the Mem does the scan, i.e. it let pass the luminescence (a narrow band) in a fixed wavelength range. This is then measured with the photomultiplier. At the end you obtain a spectrum, i.e PL vs wavelength.
From the spectrum that you show this range goes from 200 nm to 1000 nm. Usually the range starts after the excitation, this is to avoid to record the excitation itself that is not desired. Because in the spectrum that you show there is not the excitation band i suppose they use a filter (cut-off) to cut it out.

authors: Наталья Григорьева


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Solvent effects on acidity of ethanoic acid

Is ethanoic acid most acidic in DMSO, water or methanol?

I suspect that the conjugate base would be most stabilized by a polar protic solvent (meaning it's weakest in DMSO), however, water and methanol are both protic and polar. Which causes ethanoic acid to be most acidic lowest $\mathrm{p}K_\mathrm{a}$? This was asked in an exam question where the following $\mathrm{p}K_\mathrm{a}$ values were meant to be assigned to the acid in each of the solvents: $9.5$, $12.6$, $4.8$.

I agree with your reasoning regarding DMSO vs (water and methanol).  Now, between the two, which would be best able to stabilize free protons?  Alternatively, which is a stronger base, water or methanol, and thus which is more likely to form, H3O+ or CH3OH2+?

Is methanol more basic? inductive donation?

According to Evans' $\mathrm{pK_a}$ table, ethanoic acid has $\mathrm{pK_a}$ values of 4.76 and 12.3 in water and DMSO respectively. This answers the question of the ordering. As to why I might have something to say but I need to think about it for a bit and do some reading.

Ethanoic acid has $\mathrm{pK_a}$ values of 4.76, 9.63 and 12.3 in water, methanol and DMSO respectively (source here and here). Comparing this with the $\mathrm{pK_a}$ values given in the question we can see that ethanoic acid is most acidic in water and least acidic in DMSO. $$\ce{pK_{a}(H2O) < pK_{a}(CH3OH) < pK_{a}(DMSO)}$$
In general the acidity of a species in different solvents is affected by three factors:

The dielectric constant of the solvent. Solvents with larger dielectric constants favour charge separation because the force pulling the charges together as they are separated is lower. This is evidenced by the Born equation which estimates the Gibbs free energy of solvation of gaseous ions: $$\Delta G = -\frac{N_Az^2e^2}{8\pi\epsilon_0r_0}\Bigg(1-\frac{1}{\epsilon_r}\Bigg)$$
The strength of solvation of the conjugate base of the acid.
The strength of solvation of the conjugate acid of the solvent.

Dielectric constant: The dielectric constants of water, methanol and DMSO are 80.1, 32,7 and 46.7 respectively (source). This gives some support to the greater acidity in water but contradicts the observed acidity in methanol and DMSO. This suggests that dielectric constant may not be the major factor that determines the relatives acidities.
Anion solvation: As you suggested in the question, water and methanol are protic solvents but DMSO is aprotic and so the ethanoate ion will be more strongly solvated in water and methanol, shifting the acid-dissociation equilibrium in favour of dissociation. This is considered to be the main factor which explains the greater acidity of ethanoic acid in protic solvents as opposed to aprotic solvents. When comparing water and methanol the strength of anion solvation will be broadly similar because both form $\ce{O-H\bond{~}Anion}$ hydrogen bonds.
Cation solvation: The difference in the stability of the conjugate acid of the solvent may be the factor that causes the difference between water and methanol. The $\mathrm{pK_a}$ of $\ce{CH3OH2+}$ is approximately -2.5 (several sources, none giving exact answers) and the $\mathrm{pK_a}$ of $\ce{H3O+}$ is -1.7. Therefore protonated methanol is less stable than protonated water and so this disfavours dissociation of the acid in methanol. This page claims that this can be attributed to oxygen $p$-orbital donation into $\ce{C-H}~\sigma^*$ orbitals but I don't have enough knowledge of MO theory to verify this (maybe someone else can).
Overall the greater acidity of ethanoic acid in water than methanol seems to be down to the greater dielectric constant of water and the greater stability of the solvent cation formed.
This page says some good things about the subject (some of my ideas were developed from reading this).
Additional reading: This paper looks to have some detail on the anion solvation aspect but I can't access the full text so I'll leave it as a link.

Thermodynamically, solvation (including (de)protonation) relies on an energy difference: energy required to make a "hole" in the solvent for $\ce{CH_3COOH}$, as well as the energy required to make two holes - one for $\ce{CH_3COO^-}$ and one for $\ce{H^+}$ - compared to the energy released when stabilizing interactions form (dipole-dipole interactions, charge-dipole, cation-pi interactions, etc.) between solvent and solute molecules.
NOTE: When we say "holes" in the solvent, we literally mean the energy one would need to push aside the solvent molecules from each other to make a little hole big enough to put a single ethanoic acid molecule.

If the energy required to make a "hole" in the solvent is larger than the energy released by making all the stabilizing interactions possible, the molecule will not be effectively solvated (e.g. hexanes in water: hexane cannot form lasting stabilizing interactions with water.)
If the energy required to make a "hole" in the solvent is less than the energy released forming stabilizing interactions, the molecule will be solvated.(ethanoic acid in water)

It's about the ability of a solvent to stabilize the neutral and charged species.
Water: $\ce{CH_3COOH}$ and its acid and conjugate base are all polar molecules/ions. Water stabilizes $\ce{CH_3COOH}$ by forming excellent dipole dipole and dipole-charge interactions (and highly symmetrically because water is symmetric). The $\ce{H^+}$ from the acid forms $\ce{H_3O^+}$ in water (also highly symmetric) and is also stabilized by $\ce{H_2O}$, so often you read about "proton hopping" in acidic aqueous solution, which also happens in methanol, but to much less extent. You can also say water's lone pairs are larger and less hindered than methanol's, or you could say that the sigma-like MO for water overlaps better (and more symmetrically) with incoming molecules than methanol's. High symmetry allows much better orbital mixing (via more degenerate MOs), lowering the energy of water and hydronium significantly.
Methanol: Methanol is polar and has a $\mathrm{p}K_\mathrm{a}$ (so I've heard) around water's. Methanol $\ce{O}$ atom can accommodate a proton and when it does, it is stabilized partly by hyperconjugation (a type of orbital mixing; see below for visual) but is not stabilized to the extent water is in part because methanol is not as symmetric (neutral or protonated) as water is = less MO mixing. Yet, methanol forms good dipole-dipole and dipole-charge stabilizing interactions.
DMSO: DMSO is polar, but aprotic, suggesting it does not hold on to protons very well. It can form dipole-dipole interactions, but the sulfoxide $\ce{S=O}$ part likely doesn't overlap very well in a "sigma way" (head-to-head), but in a "pi-way" resulting in weaker overlap.
A note to illustrate hyperconjugation:
If protonated methanol is stabilized, it is likely stabilized intramolecularly  by hyperconjugation, which is electron overlap between the exposed (empty) $\mathrm{p}$-orbital on oxygen and the $\ce{C-H}$ sigma bond.

It will also be stabilized intermolecularly if surrounded by other polar electron-rich species (methanol, $\ce{CH_3COO^-}$ or $\ce{CH3COOH}$): These molecules will also temporarily donate electron density to the empty $\mathrm{p}$-orbital on the protonated methanol oxygen.
These effects combine to allow for methanol to harbour a proton, making it a little better base than one would think.

Can you address more of the issues that are asked about in the question?

Great job, that's perfect!

@user3158761 The oxygen p-orbital in $\ce{CH3OH2+}$ isn't empty though. It contains 2 electrons.

@bon p orb forms as a result of symmetry change about the O from bent to trigonal planar upon protonation. It contains some e- density, due to donation and mixing with other MOs, but is definitely relatively +charged because it represents part of the LUMO of $CH_3OH_2^+$, so is mostly "empty". I apologize for considering the p-orb apart from a MO. Thanks!


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Ketone to alkene regiochemistry
What are the reagents for the following reactions:

Typically, when thinking about ketone to alkene reactions, one thinks of the Wittig reaction. However, I don't see an appropriate form of that reaction here. This was posed as a question on my problem sheet and I suspect - after having given it some thought - that it's just a Hoffman/Saytzev question.
For reaction A, I think reaction with $\ce{MeMgBr}$ to make the tertiary alcohol. At this point, one has to think of a way to get the least thermodynamically stable product (the kinetic product). Perhaps sulfuric acid? It would protonate the hydroxyl so the transition state wouldn't have significant alkene character and the "base" is bulky. Correct?
For reaction B, I think it should be reacted with $\ce{NaBH4}$ to make the secondary alcohol then elimination can be performed using whatever. Maybe sulfuric acid again or just tosylate it then use $^t\ce{BuOK}$. Does this sound right?

Why can't you see a Wittig reaction working for reaction A?

How can it be done?

$\ce{Ph3P + MeI}$, then $\ce{NaH}$, then cyclohexanone. Unless you want to use $\ce{(Cp)2Ti=CH2}$ instead ;-)

B can be done either using the Shapiro reaction (tozylhydrazine, BuLi) or using the Bamford-Stevens rearrangement (tozylhydrazine,  somewhat less strong base eg. NaOMe). You can do Wittig for 1, or a Schrock carbene as Klaus suggests, or a Peterson olefination (TMS + base).

Part (a)
I agree with the previous comments that a Wittig reaction is the best way forwards especially in terms of regioselectivity of the double bond. PPh3, MeI and NaH would be an excellent combination. Perhaps draw the mechanism and you'll see how/why this work so well?
If you are interested, you could use (Cp)2Ti=CH2 also known as Tebbe's Reagent. Klaus Warzecha mentioned this already (above).
Part (b)
I like your idea of reducing, tosylating and using the non-nucleophilic base tBuOK


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Nomenclature of an organic compound
Which substituent should be placed first (in case of alphabetisation)?

isopropyl-
methyl-

I have come across two compounds: 1-methyl-3-methoxy-4-isopropylbenzene and 1-isopropyl-4-methylcyclohexane
So, is “iso” taken care of in case of alphabetisation? And, are there any special criteria for this operation?

The name “1-methyl-3-methoxy-4-isopropylbenzene” is not in accordance with IUPAC nomenclature. The systematic name is 1-isopropyl-2-methoxy-4-methylbenzene.

(Note that the prefix ‘isopropyl’ is retained for use in general nomenclature but the preferred IUPAC name is ‘propan-2-yl’.)
Simple prefixes (simple substituent groups consisting of just one part that describes an atom, or group of atoms as a unit, for example methyl, methoxy, and isopropyl) are arranged alphabetically disregarding any multiplicative prefixes. Any multiplicative prefixes are inserted later and do not alter the alphabetical order.
For example, ‘1,2-dibromo-’ is considered to begin with ‘b’; ‘isopropyl’ is considered to begin with ‘i’.
On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.5 ALPHANUMERICAL ORDER
Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
(…)
P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Therefore, the correct alphanumerical order in your first example corresponds to the name x-isopropyl-y-methoxy-z-methylbenzene.
In your second example, the correct alphanumerical order corresponds to the name x-isopropyl-y-methylcyclohexane.
Furthermore,

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule f takes precedence over Rule g.
Therefore, your first example is named as 1-isopropyl-2-methoxy-4-methylbenzene rather than 4-isopropyl-3-methoxy-1-methylbenzene since the locant set ‘1,2,4’ is lower than ‘1,3,4’.

Your second example is named as 1-isopropyl-4-methylcyclohexane rather than 4-isopropyl-1-methylcyclohexane since isopropyl is cited first as a prefix in the name.


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Theoretical versus empirical equations
I'm familiar with the fact that theoretical formulae are based on already known theory and are very frequently simplified (some equations don't include interactions between molecules for example) while the purpose of empirical equations (like polynomial equations) is just to follow experimental measurements even if it doesn't make any chemical or physical sense.
Which of these yield more accurate results (compared to reality)? What are the upsides and downsides of each group? Which are more universal? Which would require less compound-specific data (constants) to use it?
Edit: Removed van der Waals equation as an example of theoretical equations since, as pointed out by others, it's actually an empirical equation.

related question: http://chemistry.stackexchange.com/questions/342/why-are-equations-of-state-for-a-non-ideal-gas-so-elusive?rq=1

Note that the van der Waals equation requires two constants, in addition to the gas constant, that are unique to every gas and must be determined empirically. In general all theoretical models are attempts to generalize empirical models.

As pointed out already, the van der Waals (vdW) equation is also an empirical equation.  However, I think what you are saying is that the vdW equation is a physically motivated empirical equations where the empirical parameters in principle correspond to something "real" (e.g. molecular size and interaction strength) in contrast to a fit to an arbitrarily chosen fitting function such as a polynomial.
If you have lots of data points so that you can fit many parameters then a simple fit will be more accurate since it tends to have more fitting parameters than the physically motivated equation.  However, the simple fit will likely only be accurate for interpolation, and is still likely to be comparatively less accurate for extrapolation.

I would argue that e.g. a virial equation would be less physical, just because the vdW seems to (but in reality not) correspond something that is closer to a Chemist comprehension about small balls and big balls and their forces between them. From that point of view a viral equation is just a higher level of abstraction.

Yes, my point was that vdW equation comes from physical background even if I put it in the wrong category. Thank you for your answer.


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Thermodynamic vs Kinetic Sulphonation of Naphthalene

The left hand isomer is clearly the kinetically favoured product and the right hand isomer is the thermodynamically favoured product. However, I have no idea why. Can you explain?

In contrast to many other electrophilic aromatic substitution reactions, aromatic sulfonation is reversible, in other words it is an equilibrium. If you use a large excess of $\ce{SO3}$ you push the reaction to the sulfonation side; if you heat the sulfonated product in the absence of $\ce{SO3}$ it will de-sulfonate.  This makes sulfonation a nice method to 1) control where a second substituent will be introduced or 2) serve as a protecting group (block second substituent introduction) - and then when you are done, you simply de-sulfonate.
Now to your question
If we look at the intermediates leading to the 2 isomers we see that we can draw 2 resonance structures that preserve aromaticity in one ring for the transition state leading to naphthalene-1-sulfonic acid, but only one resonance structure where the other aromatic ring remains intact for the 2-isomer (draw the resonance structures to convince yourself).

(image source)
Therefore the transition state leading to the 1-isomer will be lower in energy and naphthalene-1-sulfonic acid will be formed fastest - it is the kinetically favored product.
However, naphthalene-2-sulfonic acid is a few kcal/mol more stable than its 1-isomer. This is due to an adverse 1,8 steric interaction present in the 1-isomer.

(image source)
Because the sulfonation reaction is reversible, if we run it long enough at high temperature (favoring reversibility), eventually we will wind up with the thermodynamically preferred (more stable) product, the 2-isomer.


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De-ionized water
Is de-ionized water more corrosive to metals than tap water? I have read that it "wants" the removed ions back and will aggressively take them from any available source.

Where did you read this claim?

santiago asks where I read this claim? Two places-Wikipedia-Purified water-reference 19 "Health risks from drinking DI water" Kozisek,F. (2005), and overclockers.com/pc-water-coolant-chemistry-parts 1 and 2.

The Wikipedia article merely states that sole consumption of purified water may deplete some electrolytes through normal excretion and lack of replacement,  not that it somehow "wants" the minerals.
As a matter of fact, pure, deionized water is much less corrosive to metals and is often used in closed systems to prevent corrosion because it helps prevent galvanic (electrolytic) corrosion.
Automotive coolant antifreeze contains specific corrosion inhibitors safe for aluminum and iron parts, but if water needs to be added, it should be distilled water, to prevent galvanic corrosion.
People living in northern climates often encounter corrosion of vehicles due to road salt, and learn to wash vehicles with tap water to remove the salt (high-purity or distilled water would be better, but not cost effective).

Deionized and distilled water are two very different things. Deionized is much less corrosive than distilled water, since deionized still contains many solute species, just not easy-to-ionize ones like sodium.

True... but for the purpose of removing road salt, either would work, but would be too expensive to be practical  (unless for washing a cincours d'elegance vehicle).


text:
Collisonally induced intersystem crossing
What makes a molecule a good initiator of collisionally induced ISC? $\ce{N2}$, $\ce{O2}$ and $\ce{CH4}$ have been mentioned as examples.
I imagine the most important factor is the ability to quench vibrational excitation so that the Franck-Condon factors for non-radiative decay are optimized.

There are two factors that I recall that can enhance intersystem crossing (ISC, the conversion of an excited singlet state to an excited triplet state, or vice-versa; $\ce{S_1 <=> T_1}$).  One is the presence of a "heavy atom" such as bromine or iodine in the colliding molecule (or the excited molecule itself).  The other is the presence of a paramagnetic species.
Both types of molecules enhance ISC by relaxing the quantum rules which forbid  singlet-triplet conversions. When a heavy atom or paramagnetic molecule collides with a molecule in its excited state it can facilitate the electron spin flip (e.g. converting paired electrons to unpaired electrons, or vice versa) through spin-orbit coupling.  It is this spin-orbit coupling that relaxes the rules forbidding direct conversion between $\ce{S1}$ and $\ce{T1}$ states.  In essence spin-orbit coupling leads to a mixing of the singlet and triplet wavefunctions, thereby allowing singlet-triplet conversion.


text:
Measuring Colloidal copper PPM & how to ignore the salt PPM in soluton
I made some colloidal copper with 9V battery. To make this water I had to add small amount of salt. However, the salt itself shows up in PPM count of my PPM reader. So how can I discount the salt PPM and come up with correct copper PPM in distilled water.

First method is to look for Faraday's law of electrolysis and put in the equation your data.
Second method is to add to TDS reading the initial reading of the salt solution before the electrolysis.

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center. || Could you be a little bit more elaborate? With a link to Faraday's law of electrolysis, explaining the method more thoroughly etc. Right now, this isn't unfortunately a very good-quality answer according to chem.SE's standards.


text:
Porous petri dish
I was wondering if there is a type of glass dish with small holes in it to use as a filter.  I've found a few but they are either fritted glass disks or a type of funnel.
I was looking for something similar to just a petri dish with no stems or inserts.

What scale of filtration are you talking about?  Do you want to filter tiny crystals of a precipitate from its parent solution, or is this for coarser purposes (like a tea  strainer, but glass and flat)?

I was looking for coarser purposes.  I have found some vented petri dishes but those seem to only be slits on the side.  I was looking for something more like a cocktail strainer where the openings are pin holes and covering the bottom.

Very large (i mean, ~ 0.5-1 l) filtering funnerls with glass filter do exist. Also, see this http://www.amazon.com/Kimax-Glass-Crucible-Kimflow-Fritted/dp/B004PLMF7G . There are some types of glassware, using fritted glass disk, but said glassware usually is used for some very specific task.

The closest I could think of are nozzle cover lids. Wouldn't such tool be ineffective at 'filter'-ing substances nonetheless?

what kind of nozzle cover lid?
Is it glass?

Most nozzle lids would be plastics or metal as glass would not be useful. However, I'm not sure whether a plastic lid could be transparent.

The glass part is more important, not the opacity.
I've seen borosilicate crucibles with the fritted glass insert but I was looking for something that I wouldn't have to change the strainer in the bottom.
Does something like that really not exist?


text:
Photoisomerization of Azobenzene
Rotation around the double bond of azobenzene is restricted because it would distort the P orbital overlap between the nitrogen atoms. However, in the $n \rightarrow \pi^*$ excited state ($S_1$), the rotation is no longer hindered and it can isomerize.
I don't understand why this is the case. You are taking an electron from the non-bonding orbital, and moving it into the pi-antibonding. However, the $\pi$-bonding is still present, so I see no reason why the rotation would be unrestrained. Is it something about the interaction between the electrons in the $\pi^*$ with the $\pi$ that weakens the double bond? And if so, how?

Azobenzenes are one of the most widely used moietes in photochemically  switchable systems. Nevertheless, the pathways for the E-Z isomerisation of azobenzene in the $S_1$ and the $S_2$ state have been in debate for decades and I'm not quite sure whether this is finally solved.

However, in the $n \rightarrow \pi^*$ excited state ($S_1$), the rotation is no longer hindered and it can isomerize.

Yes and no.
It is true that the $S_1$ state arises from a $n \rightarrow \pi^*$ transition in a weak, broad band with $\lambda_{\mathrm{max}} = 432\,\mathrm{nm}$ and $\epsilon \approx 400\,\mathrm{L\,mol^{-1}\,cm^{-1}}$ in hexane.
By the way, E-azobenzene exhibits another, much stronger absorption at $\lambda_{\mathrm{max}} = 319\,\mathrm{nm}$ with $\epsilon \approx 22800\,\mathrm{L\,mol^{-1}\,cm^{-1}}$ in the same solvent. This $\pi\rightarrow\pi^*$ transition to the $S_2$ state would indeed allow for an isomerisation through an one-bond-flip rotational mechanism.
(For the UV data, see 1 and 2).
But - you saw that coming after this lengthy introduction - the situation is much more complicated!
You have neglected the fact that E-Z isomerisations can proceed by other mechanisms than just the simple one-bond-flip that is usually postulated for alkenes!
Forgive me for not getting into all details on this, but for polyenes, to give an example, the so-called Bicycle Pedal (BP) and Hula twist (HT) pathways have been proposed and confirmed as space-conserving alternatives to the one-bon-flip mechanism. The latter can be visualized as "let's just flip the other side of the former double bond around like a rotor".
Back to the azobenzenes. Even from more recent works (2, 3, 4, 5), it seems that the E-Z isomerisation is still not fully explained, although a plain rotation around the former double bond can be excluded.
The two main fractions in the debate support mechanisms in which a contribution of inversion through a change of the $\ce{C-N-N}$ angle in the excited state is assumed. Whether this is the only effect and the reaction proceeds through a transition state with a linear $\ce{C-N-N}$ arrangement or whether an inversion-assisted torsional pathway provided the beter explanation is still unclear.
References
1 George Zimmerman, Lue-Yung Chow, and Un-Jin Paik, J. Am. Chem. Soc., 1958, 80, 3528-3531.
2  Alessandro Cembran, Fernando Bernardi, Marco Garavelli, Laura Gagliardi, and Giorgio Orlandi, J. Am. Chem. Soc., 2004, 126, 3234-3243.
3 Eric Wei-Guang Diau, J. Phys. Chem. A, 2004, 108, 950-956.
4 Eric M. M. Tan, Saeed Amirjalayer, Szymon Smolarek, Alexander Vdovin,    Francesco Zerbetto, and Wybren Jan Buma, Nature Communications, 2015, 6, Article number: 5860.
5 Junfeng Shao, Yibo Lei, Zhenyi Wen, Yusheng Dou, and Zhisong Wang,
J. Chem. Phys., 2008, 129, 164111

Aha! Thank you very much Klaus, this is very insightful. I will look into the papers and get further insight. I am curious to learn about the BP and HT pathways, and why they would occur in the S1 but not in the S0 state.

@Max BP and HT modes are usually discussed in the context of polyenes, think in double-bond homologues of  stilbenes. If you're interested in these, look for publications of Robert Liu, Jack Saltiel and Werner Fuß. As far as azobenzene is concerned, have a look at the references 2-4 and dig your way down from there. or, even, better, just use the azobenzene as building block in synthesis and don't care for the gory details ;-)

Thanks again for the advice! I'm looking into this for a molecular simulations class project, and the gory details are the most important part :)

Unless it is already among the references in the Nature article, http://dx.doi.org/10.1063/1.3000008 might be interesting too.

I would like to encourage you to give the citations in addition to the DOI, I had to click all the links, just to find out, that I already read two of the papers. I also think proper acknowledgement, at least a name, is a treat in sciences.

@Martin-マーチン No problem. Done.


text:
Change of electron spin in Carbon after Hybridization?
During bond formation in Carbon, one of the electrons from the 2s orbital jumps to one the empty p orbitals creating a hybrid orbital as seen in the figure below.

Now clearly one electron from the 2s orbital jumped to the 2pz orbital.
But what happened to the spin?
Before hybridization, one electron had an opposite spin, but after hybridization all electrons have the same spin.
Almost everywhere the same thing is shown.
Did the spin of the promoted electron get reversed?

That picture is a very crude oversimplification of many things. First of all, hybridisation is a function of geometry and will never happen in an atom by itself. The spin state is dependent on the excitation, it is possible, to excite from a triplet to a quintet state.

And that is why I want anyone who teaches chemistry to always emphasize that orbitals, hybridization, etc. are not real things and processes. They are just parts of modeling (and often oversimplified) description of the reality.

Short answer: No it didn’t. But hybridisation also never happened in the way it is drawn.
The typical hybridisation picture from textbooks that you have shown us, is not ‘the truth’, it’s not ‘an approximation of the truth’ and not even ‘somewhere near to the truth’. It is no more and no less than a set of drawings trying to use one concept to explain an experiment.
The experiment in this case is the structure of methane, which is found to be tetrahedral, while the concept is that of orbitals as standing waves and solutions of the Schrödinger equation. The latter says that in a carbon atom floating around in sheer nothingness, there should be five standing waves produced by the electrons, that were given the name 1s, 2s, 2px, 2py and 2pz. But this is an entirely hypothetical construct from our understanding — not many people have been to outer space and nowhere have people observed and experimented with single, unbound carbon atoms.
Carbon is only found in compounds (on Earth and in the near space, that is). But once it’s in a compound, you need to write down a whole new Schrödinger equation, that not only includes the carbon but also all the other atoms in the molecule. If you solve that equation, you will again get a set of standing waves, but waves, that are spread out across the entire molecule. You need to perfom quite a few mathematical tricks to reduce those waves back to the orbitals you calculated for a single carbon atom — but by doing so, you have lost the predictive power of where the electrons are to go. There is no longer a way to say if any electron is spin $\alpha$ or spin $\beta$.
For methane, calculations that there are in fact four bonding orbitals at two (!) different energy levels. If hybridisation occurred, one would expect only one energy level, because all carbon orbitals would have been made equal. The orbitals generally represent a carbon ground-state orbital added to the corresponding hydrogen orbitals of the same phase (carbon’s p-orbitals oriented so that the stab in between two hydrogens). The lower energy state of course corresponds to carbon’s 2s plus the four hydrogens. (See pictures in this German lecture script, part A, section 4.8)
Take a different molecule, e. g. formaldehyde. Drawing hybridisation schemes, you would expect the oxygen to be something like $\mathrm{sp}^2$ and have a ‘rabbit ear’-like pair of free electrons. But solving the Schrödinger equation does not give you any free electrons. Instead you find a multitude of orbitals, none of which exclusively surrounds the oxygen — excepting the one that represents oxygen’s 1s in simplified diagrams. Again, reducing this to any type of structure that would tell you anything about the ‘shape’ of the free oxygen electrons means that you would lose predictive power. (Luckily in that case, though, all electrons are spin-paired, so they will remain spin-paired.)

just a side note, Phil Skell studied the reactions of atomic carbon back in the 60's.  It reacts with ethylene to yield spiropentane.  The stereochemistry of the reactions with cis- and trans-butene is very interesting.

@ron Oh wow, didn’t know that. Got a doi/source for me?

Just google "p. skell, atomic carbon" and you'll find some.

authors: Масумех Никхах


text:
What is a non-classical carbocation?

What is a non-classical carbocation?
How is it different from a classical carbocation?

I am confused as I have come across this term many times on Chem.SE but there seems to be nothing for my level of understanding on the Internet!

Here is a picture of a "classical" carbocation, there is an electron deficient carbon bearing a positive charge.

There are many examples of "non-classical" carbocations, but the 2-norbornyl carbocation is among the best known.

Labeling experiments have shown that the positive charge resides on more than one carbon in the 2-norbornyl ion.  Early on, the data was explained by equilibrating classical ions, but soon another possibility emerged - one involving a single non-classical ion.
The problem comes down to: are the equilibrating classical ions ground state structures with the non-classical ion serving as the transition state, or is the non-classical ion the ground state? This debate went on for a very long period of time, but now most agree that the non-classical structure is the ground state in the 2-norbornyl system.  In fact, a recent, and difficult to obtain, crystal structure for the 2-norbornyl cation has been published proving that the ion exists with the non-classical geometry (thanks to Klaus for finding this reference, see his comment below).

A key difference between classical and non-classical structures is the bonding.  As illustrated above, a classical ion has a carbon with a sextet of electrons and 3 other bonds.  The non-classical ion, on the other hand, involves 3 carbons with 2 electrons spread over them.  This is called a 3-center 2-electron bond (hypercoordinate bonding) and is a clear marker for a non-classical ion.  Notice if you count all of the bonds to any of these 3 carbon atoms (solid and dashed lines) you count 5!  Sounds strange, but such "hypercoordinate bonding" is a permitted consequence of the 3-center 2-electron bond.

Here is a link to a simpler non-classical carbocation discussed here yesterday.

In addition to the link that you gave for the last image, the report in Science, 2013, 341, 62-64 might be worth to be linked too. The decade-long debate concerning the non-classical nature of the 2-norbornyl cation has been settled.


text:
Working with wave number equations, use trial-and-error?
I have a wave number question which I have partially solved:

What transition in $\ce{He+}$ ion shall have the same wavenumber as the first line in Balmer series of $\ce{H}$ atom?
  (a) $7\to5$ (b) $5\to3$ (c) $6\to4$ (d) $4\to2$

Here's how I solved it:
Note: symbol $Z$ is the atomic number
and $1/\lambda$ is wave number (I have written frequency by mistake)

At this point my teacher told me to continue using a trial and error method, is there any other way to solve this, perhaps using some other equation.

You have one equation with two unknown; there is no way to solve it without eliminating one of the unknowns (e.g. choosing a value of n1 or n2 and working through by trial and error) or needing more data on the transition.

What data about the transition would be sufficient to solve the equation?

I think 5/36 changed to 5/6 in your final steps. And you cancelled off $Z$ with $Z^2$. And the $Z$ is different in both cases. And in the final eqn, you probably forgot to mark them as n1 and n2. And you will have to visit these errors before going for the trial and error method. And thats it.

Yes, there is.

hydrogen: $$E=\frac{1}{hc}R\left(\frac{1}{(n_1)^2}-\frac{1}{(n_2)^2}\right)$$
helium ion: $$E=\frac{1}{hc}R(2)^2\left(\frac{1}{(x_1)^2}-\frac{1}{(x_2)^2}\right)$$
Since the energy is equal, we get: $$\frac{1}{(n_1)^2}-\frac{1}{(n_2)^2} = 4\left(\frac{1}{(x_1)^2}-\frac{1}{(x_2)^2}\right)$$
with simple rearragement: $$\frac{1}{(n_1)^2}-\frac{1}{(n_2)^2} = \frac{1}{\left(\frac{x_1}{2}\right)^2}-\frac{1}{\left(\frac{x_2}{2}\right)^2}$$

Here, one (obviously) possible answer of the equation is:
$$n_1 = \frac{x_1}{2}, n_2 = \frac{x_2}{2}$$

First balmer line is $n_1=3, n_2=2$. One possible answer is $x_1=6, x_2=4$
Ok, I know that with single linear equation and two variables, we can get multiple answers in the case of real numbers (actually, infinite). However, quantum number n can only be positive integer. I don't think there are other answers here.

[…] my teacher told me to continue using a trial and error method […]

But nobody told you to do it manually once you have understood how to adjust the Rydberg constant for $\ce{He+}$ and use the formula to get the energies for the given  transistions, in order to compare them with the $\ce{H_{\alpha}}$ line (656.3 nm):
import scipy.constants

transitions = ((7,5), (5,3), (6,4), (4,2))

def get_he_line_nm(n_upper, n_lower):
    '''calculate the energy for a line in a He+ spectrum and return
       it in nanometer'''

    energy = 4 * scipy.constants.Rydberg * (pow(n_lower, -2) - pow(n_upper, -2))

    return 1.0/(scipy.constants.nano * energy)
    # energy in nm

for upper, lower  in transitions:
    wavelength = get_he_line_nm(upper, lower)
    print('{0} -> {1} : {2:7.2f} nm'.format(upper, lower, wavelength))

will give the output

7 -> 5 : 1162.81 nm
5 -> 3 :  320.37 nm
6 -> 4 :  656.11 nm
4 -> 2 :  121.50 nm

This feels like much more effort than the pen and pencil approach

I agree :-) Once a student has understood the underlying concept and has actually done some calculations by hand to get a feeling for it, namely where to multiply and where to divide, one should encourage them to write a script to automate the task. That leaves more time to nag them with the next problem set :-D


text:
What is the exact meaning of GAMESS (US) SCF density convergence criteria?
In Gaussian the SCF convergence criteria is set to $10^{-N}$ by
SCF=(Conver=N)

Note that this criteria applies to the root mean square change in density matrix between two SCF cycles, while the maximum change convergence criteria is set to $10^{-N+2}$. For instance, for the default case of $N=8$ the output says

 Requested convergence on RMS density matrix=1.00D-08 within 128 cycles.
 Requested convergence on MAX density matrix=1.00D-06.
 Requested convergence on             energy=1.00D-06.

Now in GAMESS (US) the SCF convergence criteria is set to $10^{-N}$ by
$SCF CONV=1.0d-N $END

However, neither manual, nor the output do not explicitly specify does this criteria refers to the root mean square change in density matrix or to the maximum change in it? Judging from the default value (1.0d-05), it is the last one, the maximum change in density matrix, otherwise it would be too small for a reliable convergence. Thus, SCF=(Conver=N) in Gaussian corresponds to $SCF CONV=1.0d-(N+2) $END in GAMESS (US). But I'm not sure, so can anyone confirm that?

P.S. I tried to look at the source code, but it is an old FORTRAN mostly undocumented spaghetti mess with good-old six-lettered totally undescriptive variable names. :D

The difference is calculated in the subroutine DDIFF (scflib.src) and corresponds to the largest absolute change in the density matrix

Thanks! As expected, it is THE LARGEST ABSOLUTE CHANGE IN THE DENSITY MATRIX.


text:
Why are signals of carbon and hydrogen atoms (in NMR spectra) split by coupling to 19F?
Why signals of carbon and hydrogen atoms (in NMR spectra) are split by coupling to $\ce{^{19}F}$? Some spectra books only states that

$\ce{^{19}F}$ (natural abundance 100%) has a spin quantum number I of ½. The signals of carbon and H atoms up to a distance of about four bonds are split by coupling to $\ce{^{19}F}$.

but there is no explanation. Could someone please explain why $\ce{^{19}F}$ can couple with H or C and split their signals.

You always have a coupling if you have to NMR-active nuclei connected, there is nothing unusual about fluor here. But in the most common and simplest NMR experiments on small organic molecules you usually don't encounter such heteronuclear couplings.
The first important aspect is that many common nuclei in organic molecules are present mostly in an NMR-inactive isotope. Carbon and nitrogen for example are only present in the NMR-active isotopes 13C and 15N in tiny amounts. So you'll be able to see small satellite peaks sometimes, but no complete splitting. If only 1% of the nuclei is NMR-active, only 1% of the signal will be affected by the coupling.
The second aspect is that some experiments like 13C 1D experiments are typically recorded with decoupling. There is a coupling between 1H and 13C, but it is intentionally suppressed as it would complicate the spectra.

Thank you for the answer. It has solved my problem not only of splitting signals by 19F but also of some small peaks shouldering the main peaks in 1H-NMR. Now I know they're "satellite peaks". Excited to know that. Thank you so much.


text:
Carbon dating: What to compare the new value against?
Formula for calculating age by using carbon-14 (source: HowStuffWorks)
$$
t = \frac{\ln (N_\mathrm f/N_\mathrm o)}{-0.693}t_{1/2}
$$
The page doesn't explain what the $-0.693$ represents. What is $-0.693$?
Secondly, it says that $N_\mathrm o$ is the original level of $\ce{C-14}$ and $N_\mathrm f$ is the measured value. How do you get the original value? Is it the same for all living organisms?
Wikipedia says that the age of an object (e.g. a fossil) can be calculated by comparing the amount of $\ce{C-14}$ in the fossil and the amount of $\ce{C-14}$ in a sample of living tissue. The half-life of $\ce{C-14}$ ($\pu{5,730 years}$) can be used to find out how much time has passed since the death of the organism because you should be able to calculate how much time has passed by measuring how much $\ce{C-14}$ has decayed. But how can you calculate how much has decayed if you don't know how much was there in the first place?

Large elements of my answer is drawn from my own notes that I use for teaching General Chemistry II.

What is −0.693?
Radioactive decay processes follow first-order kinetics.  A first-order reaction is one where the rate depends only on the concentration of one of the reactants raised to the first power.  So consider the following reaction:
$$\mathrm{A} \longrightarrow \mathrm{products}$$
The rate can be expressed as the rate of change in the reactant concentration with a change in time such that
$$\mathrm{rate} = -\dfrac{\Delta[\mathrm{A}]}{\Delta t}$$
This simply means that [A] is being consumed (i.e. converted to products) as the reaction proceeds with time.  The corresponding rate law can be written as
$$\mathrm{rate} = k[\mathrm{A}]$$
where $k$ is a rate constant.  We can set these two expressions equal to each other to give
$$-\dfrac{\Delta[\mathrm{A}]}{\Delta t} = k[\mathrm{A}]$$
At this point we must apply a bit of calculus.  Let us write the previous equation in differential form to give
$$-\dfrac{d[\mathrm{A}]}{dt} = k[\mathrm{A}]~~~\mathrm{or}
~~~\dfrac{d[\mathrm{A}]}{[\mathrm{A}]} = -kdt$$
Integrate over $t=0$ to $t = t$ to give
$$\int_{[\mathrm{A}]_0}^{[\mathrm{A}]_t}\dfrac{d[\mathrm{A}]}{[\mathrm{A}]} = -k\int_0^t dt$$
which results in
$$\ln[\mathrm{A}]_t - \ln[\mathrm{A}]_0 = -kt$$
or
$$ \ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0}=-kt$$
where $[\mathrm{A}]_0$ is an initial concentration and $[\mathrm{A}]_t$ is the concentration at some time, $t$.
Now that we have our first-order integrated rate law, we can now discuss half-life.  Half-life (denoted as $t_{\frac{1}{2}}$) is simply the time required for some reactant concentration to drop to half its original value (where the original value is simply [A]$_0$.  So
$$t = t_{\frac{1}{2}}~~~\mathrm{when}~~~[\mathrm{A}]_t = \dfrac{1}{2}[\mathrm{A}]_0$$
Given this relationship, we can rearrange our first-order integrated rate law to solve for $t$ such that
$$\ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0} = -kt~~~\longrightarrow~~~
t = \dfrac{1}{k}\ln\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]_t} $$
Therefore, we can make some simple substitutions into our first-order integrated rate law at $t = t_{\frac{1}{2}}$ to give
$$t_{\frac{1}{2}} = \dfrac{1}{k}\ln\dfrac{[\mathrm{A}]_0}{\frac{1}{2}[\mathrm{A}]_0} $$
We can then simplify this equation such that
$$t_{\frac{1}{2}} = \dfrac{1}{k}\ln2\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]_0} ~~~\longrightarrow~~~
t_{\frac{1}{2}} = \dfrac{1}{k}\ln (2)
~~~\longrightarrow~~~
t_{\frac{1}{2}} = \dfrac{0.693}{k}$$
So to answer your question, the 0.693 value comes from the $\ln(2)$ term.  As ringo pointed out below (and I will include here to be thorough), you can now solve your first-order integrated rate law for $t$ and substitute in your half-life equation such that
$$\ln\dfrac{[\mathrm{A}]_t}{[\mathrm{A}]_0}=-kt
~~~\longrightarrow~~~
\dfrac{\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}}{-k}=t
~~~\longrightarrow~~~
\dfrac{\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}}
{
-\frac{0.693}{t_{\frac{1}{2}}}
}
=t
$$
or more cleanly...
$$t = \dfrac{\left(t_{\frac{1}{2}}\right)
\left(\ln\frac{[\mathrm{A}]_t}{[\mathrm{A}]_0}\right)}
{-0.693}
$$
How do you get $N_0$?
You could get $N_0$ by examining the activity of carbon-14 in a relatively recent object or plant.  Carbon-14 in a living plant is not expected to have decayed at all (since the half-life is over 5000 years!).  Therefore you could use its activity for $N_0$.  Of course this procedure relies on a few assumptions but carbon-14 dating is an approximation to begin with.

Very interesting about $\ce{^{14}C}$ dating is that there are calibration curves. Baffled me somewhat, but makes sense.

Carbon-14 dating works because in a living organism, carbon-14 is constantly decaying, but is also constantly being replenished by the organism, meaning its concentration in the living tissue essentially does not change. Though the concentration of the element carbon itself is different for all organisms, the percent abundance of carbon-14 in all living organisms will be $\approx \pu{0.00000000001\%}$ (one part per trillion). In a deceased organism, carbon-14 can no longer be replaced, so the supply of carbon-14 slowly diminishes over time ($t_{1/2}=\pu{5730 years}$). The concentration of carbon-14 in a deceased organism at any given time can therefore be used to find how long ago the organism stopped replenishing its supply of carbon-14.
The decay of $\ce{^14C}$ to $\ce{^14N}$ is given by the equation:
$$\ce{^14_6C -> ^14_7N + e-}$$
The equation you see is derived from the integrated rate law. This is a first order reaction with respect to both $\ce{^14C}$ and $\ce{^14N}$, meaning there is one of each atom in the rate determining step of the reaction. This means that the change in the concentration of each of these atoms can be used to determine the age of a sample which it decays can be modeled by the equation:
$$\ln[\ce{^14C}]=-kt+\ln[\ce{^14C}]_0$$
or as you have it in your question:
$$\ln[N_f]=-kt+\ln[N_0]$$
The rate constant for the equation, derived from the time taken for the concentration of the concentration to be half of the initial ($N_f=\frac{N_0}{2}$) is:
$$\ln[\frac{N_0}{2}]=-kt_{1/2}+\ln[N_0]$$
$$\ln[\frac{N_0}{2}]-\ln[N_0]=-kt_{1/2}$$
$$\ln[\frac{1}{2}]=-kt_{1/2}$$
$$\ln[2]=kt_{1/2}$$
$$\frac{\ln[2]}{t_{1/2}}=k$$
$$\ln(2)\approx 0.693$$
$$k=\frac{0.693}{t_{1/2}}$$
Solving the equation for any time $t$:
$$\ln[N_f]=-kt+\ln[N_0]$$
$$\ln[N_f]-\ln[N_0]=-kt$$
$$\ln[N_f]-\ln[N_0]=-kt$$
$$\ln[\frac{N_f}{N_0}]=-kt$$
$$\frac{\ln[\frac{N_f}{N_0}]}{-k}=t$$
Substitute in what we solved for:
$$\frac{\ln[\frac{N_f}{N_0}]}{-k}=t$$
$$t=\frac{\ln[\frac{N_f}{N_0}]}{-\frac{0.693}{t_{1/2}}}$$
$$t=\frac{\ln[\frac{N_f}{N_0}]}{-{0.693}}t_{1/2}$$

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Stereoselectivity solvolysis of cyclohexenyl halides
When a functionalized cyclohexenyl halide undergoes $\ce{S_{N}1}$, does a racemic product form (due to ring flipping) ?

The term ‘functionalized cyclohexenyl halide’ is ambiguous. Could you name one example?

The specific type of functionalization is not very important; I would be primarily interested to know the behaviour of the unfunctionalized molecule and secondarily how e.g. alkyl or hydroxy functionalization affects that.