text:
Why do the protons in pentacarbonylmethylmanganese show up at –0.21 ppm?
The $\ce{^1H}$ NMR of $\ce{Mn(CO)5CH3}$ shows a resonance at –0.21 ppm, which has to be due to the methyl group. Why are the protons so shielded?
I would have thought that all the pi backdonation from the Mn to the carbonyl would reduce the electron density on the $\ce{CH3}$, hence increasing the chemical shift.

Agostic interactions can also shift H resonances upfield in OM compounds with alpha protons

http://www.ilpi.com/organomet/agostic.html

I thought of an agostic interaction, too. However, to me, Mn(CO)5Me doesn't seem to be a prime candidate for an agostic interaction, being coordinatively and electronically saturated. That said, I wouldn't know.

Unfortunately, you don't give an indication of what you expect the chemical shift for the -Me to be, other than higher than is observed. However, you are correct in your assumption; the π back donation from the Mn centre to CO ligands does decrease the shielding, and therefore does increase the chemical shift, but this is a good case of how chemical shifts are all relative.
The dominating factor in shielding in d-orbital transition metal complexes comes from the paramagnetic ring currents within the incomplete valence d-shell. The circulating electron in the d-orbital produces a much greater angular momentum than for a p-orbital complexes (typically organic molecules, say). The influence of the d-orbital ring current induces a very large diatropic shift for ligands close to the metal centre. These can by very large for hydrides; up to -50ppm in some cases, and still significant for methyl ligands.
So, your starting point for the chemical shift of the -Me ligand is something that has significant shielding, and would be largely negative. What you would expect, based on your argument, and indeed what is observed by experiment, is that as you increase the number of π accepting ligands around the metal centre, you get an increase in the chemical shift of your -Me ligand. For example, as you go from M(PR3)5(CO)1 to M(PR3)1(CO)5, there would be a general trend to deshielding and an increase in the observed chemical shift.
Relative changes in chemical shift in transition metal/ligand chemistry are a great tool for determining stereochemistry and relative positioning of other ligands; a trans substitution will have a greater influence on the shielding than cis, for instance, however the overall dominating factor in the local shielding of your observed -Me still remains the diatropic shift from the metal centre.

Going to assume the solvent is CDCl3.
It is true that pi Back-donation withdraws electron density from the Metal Center to deshield it, but also note that carbonyl groups are still electron donating and the Methyl group can be treated as negatively charged (additional shielding).


text:
What is the purpose of Gaussian, itself?
I have been doing some research into computational chemistry, having never looked into the area before, and have read about a popular, paid-for software "Gaussian", and free alternatives such as "GAMESS". From what I understand they are both capable of employing methodologies such as PM3, but I am struggling to understand the point of the software itself; how is a Gaussian PM3 calculation any different to a GAMESS calculation? There must be a reason one would want to pay for it.

related http://chemistry.stackexchange.com/questions/27599/gamess-and-gaussian-what-is-the-maximum-efficient-number-of-supported-nodes

One difference (that has little to do with your query but is relevant) is that software developed by a commercial, for-profit operation in some cases will be better maintained and more portable than similar software forged in the open-source community: you're not paying for functionality as much as you're paying for support, documentation, development by professional software engineers, and other tangibles that reduce your own costs in terms of time and money.

GAMESS is not open source, it is free for academic use. Beyond that, there are a lot of implementation that Gaussian has that GAMESS does not, and vice versa. Researchers in the computational chemistry branch often utilise many different programs. A lot of them are not free. Gaussian also is often a good choice for beginners, since it comes with a more comprehensible user interface and documentation. Also the user community is much larger. And try to search for problems with GAMESS online - it's a nightmare because of actual games.

Apart from the above, I hardly see any way of objectively answering this question. It will most likely be based on personal preferences and/or opinions. And then we can throw a couple of more software packages into the mix and it gets even more complicated.

@Martin-マーチン, thanks for your comments; I've edited the question to reflect the "free" nature of the software as opposed to the open-source one cited earlier.

@Martin-マーチン Gamess is open-source (i.e. you can read the code), but it is not free.

Gaussian includes a lot more methods within one nicely done package and, frankly speaking, is faster. Paid developers do better job and have paid time to implement more advanced techniques in the product. However, since gaussian is proprietary, it is not always beneficial to employ it, and for many  simpler tasks investment into hardware is more reasonable. When I checked last time, gaussian allowed calculation of analitical frequencies for dft methods, while gamess didn't. It meant, that for any large molecule gamess couldn't produce IR spectra in reasonable time using DFT  approach.

@permeakra if you pay enough you can read the source code of Gaussian, too. Both programs are not open source software, because that would mean that you can modify and redistribute the code to anyone. Both are proprietary software, they come with serious limitations in licencing.

@permeakra GAMESS allows analytic DFT frequencies with some reasonable restrictions. While it is true that DFT in GAMESS may be slow, that isn't where it excels, and Gaussian tends to be among the slowest for common tasks.

There are several interrelated reasons (disclaimer: I am a GAMESS developer).
Gaussian has a lot of features and interoperability.  To take your PM3 example. Yes, the results of a PM3 calculation are identical in GAMESS and Gaussian. But until recently you couldn't do a PM3 calculation in solution using the PCM method in GAMESS.  Also, what if you want to do a PM6 calculation instead. In GAMESS this is currently only possible only for elements up to F.
Because of this Gaussian has been the program of choice for many people, which also means they have trained a lot of people to use Gaussian.  Learning to use a new (and perhaps free) program well can be very time consuming, so it might be more cost effective to pay for a program you know how to use well.  And chances are you probably have to learn how to use several new programs to get all the functionality of Gaussian.

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text:
Do you include the "10 H2O" in the formula for borax?
When you calculate the mass of borax needed to make up a standard solution, do you include $\ce{\cdot 10 H2O}$ in the formula for borax? Why or why not?

In general, you include water when it is there, and you don't when it is not, much like you would do with anything else in a chemical formula. That dot ($\cdot$) in the formula does not (I repeat, does not) make everything after it "insignificant" or "kinda not real".

You are talking about the waters of hydration in crystal structure $\ce{Na2B4O7 \cdot 10H2O}$, and the answer is invariably—yes.
These water molecules account for 47.24% of the crystals weight! If you wanted to make $1\: \mathrm{L}$ of $1\: \mathrm{M}$ of $\ce{Na2B4O7}$, and you did not account for these waters of hydration, you would end up making a $0.4818\: \mathrm{M}$ solution. Less than half the desired concentration!
This is the result of a double effect. Not only are you not adding as much salt as you thought you were, but because the salt contains water it is diluting the solution even further. In the case above $95.05\: \mathrm{mL}$ of water are actually added to the solution just in the salt!

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text:
Suggestions for solvents capable of efficiently performing alpha-bromination and nucleophilic substitutions
We performed the one-pot preparation of bupropion in class where the bromination of the carbon adjacent to the carbonyl group was performed using DCM as solvent, and upon completion the solvent was switched to n-methyl-2-pyrrolidone and was then reacted with an amine. I know that the nucleophilic substitution won't occur in DCM, the reason for which was also explained here on this site, so what alternate solvent could be used that allow for both bromination and substitution to occur without a solvent switch, AND still allow this procedure to still be conducted within a single class session? I asked the professor, but he said that he would have to think about it and get back to me... yeah, I have a feeling I am being blown off.
I believe that water would work for the bromination as long as the mixture is kept acidic, but then that complicates things with the amine as it will need to be in its basic form, unless that doesn't matter because by the time you add the amine, the bromination has already taken place and you would just neutralize first. However, aren't these reactions very slow in aqueous conditions? I know that water lowers nucleophilicity, so would that explain the hindered reaction times?
What about acetonitrile? I know this can be used in bromination, as I have read procedures that claim increased selectivity for monobromination.
In the event that you are not familiar with the preparation I am referring to, I am providing a link to the ACS paper that we used as the basis of the experiment.
http://pubs.acs.org/doi/abs/10.1021/ed077p1479
EDIT: There is also a follow up question regarding the usage of more than an equivalent of the amine during the substitution step as an acid scavenger. What is up with that? Why isn't the acid being formed in the substitution step being neutralized with an appropriate base that won't hydrolyze the bromoketone first? This just seem rudimentary to me, but I lack the encompassing understanding at this time. Is there a procedure I could reference as an example of what I just proposed?

As a general comment published procedures have been optimised. I don't have access to the paper you link so I don't know if the procedure is from process chemistry (where you expect a high degree of optimisation) or from med chem (where the priority is to get the compound out fast but there is still some optimisation).
But your questions are still valid. So I agree that doing the bromination - amination in one pot using the same solvent is a reasonable experiment to try and actually acetonitrile is a good choice. Why do you think acetonitrile is nucleophilic? it is not (unless you deprotonate it with a strong base). So it is very possible that the first researcher tried your approach initially and possibly found that the bromination was not very selective or the amination step was too slow, or low yielding and switched to the current procedure.
I also agree that you can use a base instead of excess of amine, for example an amine base like triethylamine or inorganic bases like $\ce{K2CO3}$ are standard in this case and this is what they would have used if they had to do the reaction with a different amine (as the nucleophile) of higher cost.

You are absolutely right about acetonitrile. I had been cramming my head with strong base deprotonations that night and acetonitrile had been one of the substrates in the literature I had been reading. It somehow transferred to this question! Thank you for your response. I will take a look at this since my exam is over and post a follow up response if I have one after thinking about it.

Acetonitrile is likely not used for student procedures as this since it is more expensive than DCM. That is my initial response, but since I was looking to increase my knowledge I will look into this from the perspective of acetonitrile.


text:
Why do liquids freeze into a polycrystal solid when they freeze fast?
I don't see any possible explanation. If there's a crystallite at one orientation, why would a crystallite at another orientation start growing instead of more molecules just building onto the surface that's already there in the same orientation? Does the size of the crystallites vary as the reciprocal of the speed of movement of the boundary between the solid state and the liquid state?

Another crystallite would start growing for the very same reasons that the first one. It is not like the first crystal to form would immediately send a telepathic signal throughout the whole volume of liquid to the effect "it's OK, I'm here, stop forming new crystals, wait for me to grow and reach you".

Embryo crystals spontaneously and independently form in a supercooled liquid. A crystal successfully growing in one location doesn't prevent others from growing elsewhere. As the temperature decreases, the rate of embryo formation and chance of successful growth increase, so at a low temperature you're likely to get multiple successful crystals.

Actually, usually, only a very small part of a liquid gets supercooled before it begins to freeze if the heat is sucked out of it quickly and then after that, you get a migrating boundary between the liquid and the solid moving as a result of continuous suction of heat and the solid still forms in polycrystal form. I want to know why that happens.

My Masters dissertation is on the control of supercooling by proteins. Are you talking about water or liquids in general? Water has unique properties.


text:
Why even the most volatile solvents leave a residue after evaporating?
After a solvent evaporates off of a surface there's always some residue left in the form of a strain.
Why it is the case even for highly volatile solvents such as acetone, toluene etc?

I don't experience this with dichloromethane. Ether can leave a little water, maybe, but I haven't noticed that, either. Maybe it's the quality of chemical you're using? Of course, wash acetone is often lower quality, and I definitely wouldn't call toluene highly volatile. It probably has some petroleum contamination most of the time, too.

The most common reason why evaporating solvents leave a residue is because they are good solvents and some of the surface or contaminants on the surface are dissolved in the solvent. This is very common for most non-inert surfaces with solvents like acetone.
Another explanation when the surface is very unreactive (glass for example) is that the evaporation of a volatile solvent cools the surface and causes other things (like water vapour) to condense onto the surface. This is certainly true for dichloromethane and ether which are so volatile that evaporating a large amount can freeze nearby water.

Just 1 nitpick to what I think is otherwise a good answer: your first sentence says "...some of the surface or contaminants on the surface are dissolved in the solvent.". This certainly explains the appearance of what looks like a residue, but is another phenomenon entirely, and is not actually a process of leaving a residue from the solvent, as stated earlier in the sentence, "...evaporating solvents leave a residue...". While that is a bit of a nitpick, I think that it kind of confuses the issue of the question, whether the solvent is leaving a residue. Still, +1 for the answer as a whole.

If the surface material is clean and neither soluble nor reactant in any way with the solvent, then it is probably a solvent purity issue. If you have a good micro-balance, you could pour a known volume of the solvent into a pre-weighed light-weight, i.e. aluminum foil, container, dry it, and re-weigh. From that you can calculate the non-volatile solids content of your solvent and compare it to the specifications.

In my experience most highly volatile compounds like acetone & pure ethanol do not leave any residue at all unless they have already dissolved something. If you are using them for cleaning purposes, the residue that you are seeing Left Behind is not from the acetone, but rather from the dissolved substances. Several successive wipes using a clean acetone solution each time will ultimately yield a perfectly clean surface with zero residue of any kind.

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text:
Wavy baseline in carbon NMR
I realize anything NMR-related is essentially physics, but I figured this was a better place to ask this. I have this issue fairly regularly (with one instrument in particular) where my baseline ends up looking like the one in the image below. The (inverted) sine wave-shape is what I'm referring to, if it's not obvious.

Is there something specific that causes this baseline shape?  Is there a way to fix it? Preferably I'd want to use something that fixed the signal/input, but if I have to resort to a processing method, I'd settle for that.

This baseline roll seen in this spectrum is an example of acoustic ringing, and is quite normal for many low frequency NMR experiments that are recorded over wide sweep widths. It occurs as a consequence of slight breakthrough of the excitation pulse, as it rings down to zero. Ideally, you would like to start collecting signal at your receiver immediately after your excitation pulse, however it does take a small amount of time (microseconds) for the pulse to decay to zero. If your pre-scan delay (typically configured for your spectrometer during the configuration process) is too short, you can get this ringing artifact. It is more pronounced for weaker samples, but usually nothing to worry about.
To correct this as an experimental acquisition parameter, you would make your pre-scan delay longer; for Topspin, this is the parameter DE in your acquisition parameter list. It is normally defined during spectrometer configuration, and needs to be as short as practically possible to maximise your signal detection - typically less than 10 microseconds. For some nuclei, much of the signal has already decayed during this pre-scan period, and require some backward linear prediction to help. I'd speak to your local friendly NMR spectroscopist before mucking about with changing parameters like DE.
It is perfectly normal, and you should be able to get rid of it by applying a simple baseline correction algorithm - for Topspin you should use the commands abs or absd. This should work for your spectrum. As a word of caution when applying baseline corrections to your 1H spectrum, this will also apply an automatic integration for you, overwriting any integration you may have done previously. You can avoid this by using the optional n switch (abs n). For severe cases of baseline roll, apply a partial basline correction using absf between the limits of absf1 and absf2. Your Topspin manual can provide a fuller description, including other baseline correction modules.
In recent versions of TopSpin (4 onwards) there is also the apbk command, which uses a new algorithm to perform simultaneous phase and baseline correction. This can yield better results than the traditional apk + abs combination, where phase and baseline correction is carried out separately.

Awesome, this is really helpful. I'll try the baseline correction and see how that goes. If I have to do anything beyond that, I'll ask one of our NMR guys to help me out before moving on to other measures.

This seems to happen alot when you try to auto phase spectra with poor signal to noise.  Longer acquisition times more concentrated samples and more delay time to relax between pulses will quiet down the baseline and allow for better phasing.

I tried phasing this one myself. Any time it flattened out, I had signals at ~90°. This wasn't finished, but when I let these things all night even, I come back to less noise but the same kind of shape. Also, that thing was pretty concentrated. RG was like 25 for the proton.

Is this in vnmrj? When was the last time the carbon channel on the instrument was tuned?

This is Bruker Topspin. I hadn't tuned it at this point. I did check it, though. It looked pretty good. I might have gotten it slightly better but not much. This NMR isn't as good for X nuclei as it is for H, it's a BBI. I can always just turn around and use one of the BBO's, but that fixes the results and not the problem.

Did you filter the solution after you made up the nmr sample?  Particulates in the sample will also cause this,  as will any biphasic or immiscible liquids.  This used to happen to me all the time with air sensitive organometallics that would decompose and precipitate out during longer acquisitions.

I did the filter it, but it had just come off a column. Could still be an issue, these compounds seem to have solubility issues. Great in DCM, terrible in everything else, including chloroform. I could have tried DMSO, but then I couldn't recover it very easily.

Small amounts of paramagnetic material can also really mess up the baseline and even shift peak resonance frequencies. Not likely in typical organics but a concern if there may be metals in there. The problem is they are usually NMR silent (apart from messing things up). How about chlorobenzene-d5?

Haven't tried it, but with a benzyl ester and a tosyl group, it would probably be soluble. It is pretty soluble in toluene. I'll look around for some. Might mess with my aromatic region, though.

What about degassing solvent? I know NOE stuff is affected by normal levels of oxygen. Could that be an issue here?


text:
Why are there different values for enthalpy of combustion, depending on the calculation method?
Why are there different values for enthalpy of combustion, depending on the calculation method?
Take for example the combustion of ethanol:
$$\ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O (l)}$$
If I calculate this using enthalpy of formation:
\begin{array}{c|c}
\text{Substance} & \Delta H^\circ /\pu{(kJ/mol)}  \\
\hline
\ce{H2O(l)}    & -285.8 \\
\ce{CO2(g)}    & -393.5 \\
\ce{C2H5OH(l)} & -277.0
\end{array}
\begin{align}
  \Delta H &=
    \left[2\cdot(\pu{-393.5 kJ/mol}) + 3\cdot(\pu{-285.8 kJ/mol})\right]
    -\left[\pu{-277.0 kJ/mol} +  3\cdot(\pu{0 kJ/mol})\right]\\
  &= \pu{-1367.4 kJ/mol}
\end{align}
If I calculate this using bond energy:
\begin{array}{c|c}
\text{Bonds} & \Delta H^\circ /\pu{(kJ/mol)}  \\
\hline
\ce{C-C}      &  347 \\
\ce{C-H}      &  414 \\
\ce{O-H}      &  464 \\
\ce{C-O}      &  351 \\
\ce{C=O(CO2)} &  803 \\
\ce{O=O}      &  498
\end{array}
\begin{align}
  \Delta H &=
    \pu{347 kJ/mol} + \pu{351kJ/mol} + \pu{464 kJ/mol}
    + 5\cdot(\pu{414 kJ/mol}) \\
  &\quad
    + 3\cdot(\pu{498 kJ/mol})
    + 4\cdot(\pu{-803 kJ/mol}) + 6\cdot(\pu{-464 kJ/mol})\\
  &= \pu{-1270 kJ/mol}
\end{align}
Why is there such a big difference between them?

Average bond enthalpies (note the word "average") are calculated in a different way from formation enthalpies. Formation enthalpies are well-defined and precise, meaning that two different people could measure them the same way and get the same answer. Average bond enthalpies are averages over many different types of bonds.
For example, there is only one way to make $\ce{CH4}$, but there are many different molecules with $\ce{C-H}$ bonds. The enthalpy of dissociation of a $\ce{C-H}$ bond in $\ce{CH4}$ is different from that in $\ce{C2H6}$, and in $\ce{C6H6}$, and really in any other molecule you can think of. To find average bond enthalpies, a selection of a large number of those bonds in many different molecules is taken, and the enthalpy of dissociation for each is measured. Then, the average (mean) is recorded. If you look at some different charts of bond enthalpies, you will notice two things: First, there are some numbers marked "exact" - $\ce{H-H}$, $\ce{Cl-Cl}$, $\ce{F-F}$, for example. This is because there is only one molecule with each of those types of bonds. The other thing you will notice is that depending on which chart you are using, many of the numbers will be slightly different. This is because there is no universal, unchanging standard describing which molecules are used to determine each bond - it depends upon what the people making the chart decided to use.
Because of this difference, when making predictions, average bond enthalpies are less accurate than formation enthalpies. In practice, they are used for quick estimations, or as a last resort when formation enthalpies are not available.
When I taught chemistry, I would have my students measure reaction enthalpies in lab, then calculate them using formation enthalpy tables and average bond dissociation enthalpy tables. The goal was to show them that average bond enthalpies can be reasonably close for rough estimates, but that formation enthalpies are much better in practice. It also showed them that accurately measuring reaction enthalpy in the lab is difficult!

Because calculation with bond energies are notoriously imprecise.
Those bond energies that you use do not reflect the exact C-C bond energy in ethanol, for example. It is merely an "average" C-C bond energy taken from a wide variety of organic compounds. Obviously if you use it to calculate the enthalpy change of a reaction specifically involving ethanol, your answer is not going to be correct.
Using bond energies is often only good enough for a ballpark estimate. You should never place much trust in those numbers - certainly not as much as you seem to have in your question.

It's because bond enthalpy calculations apply to molecules in their gas phase!
That is, you would have to account for enthalpy change of vaporization of ethanol.
Enthalpy change of formation:
$$\ce{C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O (l)}$$
Gas phase reactio for bond energy calculations:
$$\ce{C2H5OH(g) + 3O2(g) -> 2CO2(g) + 3H2O (g)}$$
If we take the enthalpy change of vaporization of water and ethanol as $41 \ \mathrm{kJ\ mol^{-1}}$ and $42 \ \mathrm{kJ\ mol{-1}}$, we get $\Delta H_f \approx -1270 - 41 - 42 \approx -1350 \ \mathrm{kJ\ mol^{-1}}$
Close enough!
I made a similar mistake here:
Discrepancy between calculated and experimental enthalpy change of formation? Hydrogen Iodide

Note that the bond enthalpy change is the major but not the only part of the reaction enthalpy, what can be easily seen in reaction enthalpy temperature dependence.

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text:
Physicist notation for spatial orbital
In quantum chemistry, the two-electron integrals often denoted as physicist's and chemist's notations.
For spin orbital, the physicist's and chemist's notations are
$$ \langle pq | rs \rangle = \int \int p(1)^* q(2)^* \frac{1}{r_{12}}  r(1) s(2) d^4 x_1 d^4 x_2 $$
$$ [ pq | rs ] = \int \int p(1)^* q(1) \frac{1}{r_{12}}  r(2)^* s(2) d^4 x_1 d^4 x_2 $$
respectively.
For spatial orbital, I have only found the chemist notation
$$ ( pq | rs ) = \int \int p(1)^* q(1) \frac{1}{r_{12}}  r(2)^* s(2) d^3 r_1 d^3 r_2 $$
Is there any physicist's notation for spatial orbital?

As far as Szabo/Ostlund is concerned, there's no other notation (Table 2.2, p 68).

yep, that's the reason for asking this question..

I guessed that that might be the case :) but that's all the info I have anyway, thought it was worth a mention anyway since you didn't mention the book.

Yes, there exist "physicist's notation" for integrals over spatial orbitals and symbolically it is looks the same as "physicist's notation" for integrals over spin orbital: the good-ol' Dirac bra-kets is what is usually called "physicist's notation".


text:
Anodized Aluminum...danger?
I recently got an iMac, but have been feeling anxious about using it as  the keyboard is made of aluminum as is also the actual computer.
the keyboard and computer are made from Anodized aluminum, while the
base is made from aluminum and plastic,
How long does the anodization last? Like how long does it take for erosion to take place? I read somewhere salts and I think acids could erode it, but I have to check about that. Should I be concerned of amount of aluminum that could be reabsorbed into my skin over several years typing on the keyboard? Granted I could get another keyboard, though I really like the keyboard it came with. In addition, its difficult psychologically  because I guess I worry about just touching the on button to put the computer on..I could also use something else to touch it...but yeah thats just too much trouble over a computer.
The fears stem from a possible but apparently not proven link of alzheimers to increased aluminum in the body.
Considering this is not the same as dietary intake, I would hope that it should not be an issue, and also because its anodized aluminum, nevertheless I am seeking looking for assurance as the internet can be quite brutal and adamant about the position it takes. You know there are these other "health" sites supporting claims that it could cause Alzheimers.
I literally when to look for publications concerning aluminum alzheimers link. But it was not helpful..some of what I saw just perhaps scared me. Granted I did not literally read the whole article mainly just the Abstract. Nor did I commit my self to a period of time to look over the matter.
Anyway, yeah I have to make a decision soon to keep or return the iMac like less than two weeks perhaps; the anxiety is painful. If anyone could give me insight
concerning aluminum and anodized aluminum, if they can be transferred dermally and if so, how much is absorbed and if is significant. I read here that just a small amount is transferred dermally.
http://www.atsdr.cdc.gov/phs/phs.asp?id=1076&tid=34
In addition now to think of it the aluminum could be transferred to the mouth if you were to eat with it without washing hands.yeah not the kind of way I'd want to live worrying about this.
I think soda cans have aluminum s in the soda, but a computer is something I may use nearly everyday, for perhaps hours so..

It isn't a problem. You're in more danger driving back to the store than you are from the keyboard.

You might wish to try at [skeptics.se].

I'm voting to close this question as off-topic because it just is.

ok thanks orthrocresol..and thanks MaxW for responding...perhaps people who are in the chemist field may know something about anodization..and how long it may take to wear off..nevertheless it does stray off into other areas and the question I guess doesn't totally  fit in this category .so if you want the question closed then

You are aware that a lot of cookware is made from aluminium or anodised aluminium? So your food will often be in contact with it and in a harsh environment. Also the link between aluminium and alzheimer's  is now thought to be spurious: an artefact of the analytical technique used and not a real link to the disease.

[…] How long does the anodization last? Like how long does it take for erosion to take place? […]

This depends on the thickness of the oxide layer, the cleanliness of the aluminium surface before anodization, the curvature (deposition might be different at edges), the mechanical and the chemical stress. In summary, the question might be, at least, very difficult to answer.
But since you're more worrying about possible health issues rather than about the quality or look of the computer, you might take the following into consideration:

A lot of antacids still do contain large amounts of aluminium salts, such as magnesium aluminium silicates. These product are used orally! To my knowledge, higher frequencies for Alzheimer disease among patients that frequently use these antacids has never been observed.
The German Alzheimer Gesellschaft does not see a correlation between aluminium and Alzheimer disease, see this note in German.
In a more recent report, the Bundesinstitut für Risikobewertung (BFR),  a German federal institution for risk assessment has addressed the question in a FAQ (in english). Let me cite a paragraph:

Is there a connection between aluminium intake and Alzheimer’s disease?Various studies have tried to demonstrate a connection between aluminium intake and Alzheimer’s disease. However, due to the patchy data situation, they do not provide irrefutable scientific evidence for such a connection.

In summary, I agree with the comment given by MaxW. Don't worry about about intoxication from touching the keyboard, but pay attention when you cross the road.


text:
Model of a particle in a linear box
I'm studying the model of a particle in a linear box when we have a chain of unsaturated carbons. I have a chain with 3 bonds and 4 atoms of carbon.
I'm applying the expression
$\lambda = \frac{8cmL^2}{ n^2 h}$
(instead of n I'm calculating the variation of n) where $c$ and $m$ are the light speed and the mass of the electron, respectively and $L$ is the length of the box ($=(3+1) \times d_\ce{CC}$, where $d_\ce{CC}=\pu{1.41E-10 m}~$ is the carbon-carbon distance), and I assume the transition with less energy is $3^2-2^2$.
I calculated 210 nm and the correct answer should be 207 nm. Am I taking an incorrect approach?

Ok, first of all you need to be way clearer in what you are saying. It took me a while to figure out what you were even trying to calculate, because you didn't say what $\lambda$ represented. For a while I thought you were calculating the de Broglie wavelength of the electron in the orbital. It's easy to use maths to write "the variation in $n$". All you need to do is:
$$\lambda = \frac{8cmL^2}{(n_2^2-n_1^2)h}$$
where $n_2$ is the upper level and $n_1$ is the lower level. Then say $n_1 = 2$, $n_2 = 3$. Anyway, my answer is 209.76 nm.

What is a linear box? How carbons come into picture? Where does the 207nm as an answer come from?

210 instead of 207? I'd say that's an excellent agreement for such a crude approximation.


text:
Raman Spectrum of carbon chains: How to identify the vibrational modes
So I am using NWChem to calculate the raman spectra of C2H2 and C4H4. For example for the first I obtained the following frequencies:
#  freq [ 1/cm]  S [Ang**4/amu]
------------------------------------------------------------
  1   644.63        0.0087
  2   679.18        0.0076
  3   792.83        0.0001
  4  2146.53        8.0917
  5  3506.05        0.0007
  6  3645.57       37.8189

and for example, when comparing with the expected results from http://www.chem.purdue.edu/gchelp/vibs/c2h2.html , I could say that my mode number four corresponds to C-C stretching.
So now, when I perform the same calculation for C4H4, I get the following:
#  freq [ 1/cm]  S [Ang**4/amu]
------------------------------------------------------------
  1   216.77        2.3873
  2   305.29        0.7775
  3   548.05        0.8095
  4   643.94        1.7042
  5   677.47        0.5218
  6   700.29        6.8906
  7   893.88        0.1890
  8   928.98        0.5379
  9  1033.68        0.1554
  10  1134.09        1.5802
 11  1357.84       16.9774
 12  1504.93       31.2415
 13  1715.33       17.1094
 14  2269.08       38.4659
 15  3293.55       19.5975
 16  3313.05       78.1132
 17  3449.94       44.9757
 18  3578.50       24.6781

My question is this, how could I identify the vibrational mode corresponding to C-C stretching now?
And if I performed the same calculation for C6H6, how could I identify the C-C vibrational mode?
Is there a way of identifying what each mode corresponds to? not only in calculation but also in experiments?
thank you

When you perform a IR or Raman calculation, the primary step is the diagonalization of the mass-weighted Hessian:
$$
\begin{align}
H_{ij,ab}^{\mathrm{mw}} &= \frac{1}{\sqrt{m_{i}m_{j}}} \frac{\partial^{2} E}{\partial a_{i} \partial b_{j}} \\
\mathbf{Hq} &= \lambda \mathbf{q}
\end{align}
$$
where $H_{ij,ab}^{\mathrm{mw}}$ is a $3N\times3N$ Hermitian matrix, $i,j$ are indices running over the atoms, $m_{i}$ is the mass of atom $i$, and $a,b \in x,y,z$ (the $3N$ Cartesian molecular coordinates). The eigenvalue $\lambda_{i}$, multiplied by some conversion factors, is a frequency in the harmonic oscillator approximation, and the corresponding eigenvector $\vec{q}_{i}$ is the normal mode, the collective atomic displacements, for that frequency.
(A calculation of Raman intensities involves extra work, namely the calculation of $\frac{\partial^{3} E}{\partial \varepsilon_{\alpha} \partial \varepsilon_{\beta} \partial a_{i}}$, but the normal modes are identical.)
Aside from performing literature searches for normal mode frequency assignments on related systems, a good alternative is the visualization of these normal modes. Many outputs from frequency calculations can be opened in Avogadro:

Clicking on a frequency in the right-hand side pane (only IR intensities are displayed) can display both the displacement vectors and an animation of the displacement (see the lower right corner box). Avogadro is able to read this portion of the NWChem output file:

          -------------------------------------------------
          NORMAL MODE EIGENVECTORS IN CARTESIAN COORDINATES
          -------------------------------------------------
             (Projected Frequencies expressed in cm-1)
...
                    7           8           9          10          11          12

 P.Frequency      252.66      366.45      606.76      773.89      790.45      808.06

           1     0.06344     0.00000    -0.02127     0.00000     0.00000     0.00096
           2    -0.10753     0.00000    -0.09540     0.00000     0.00000    -0.00247
           3     0.00000    -0.03639     0.00000    -0.01614     0.01014     0.00000
           4    -0.08820     0.00000    -0.09326     0.00000     0.00000     0.00127
           5     0.02291     0.00000    -0.01064     0.00000     0.00000    -0.00238
           6     0.00000     0.00615     0.00000     0.14073    -0.07465     0.00000
           7    -0.13091     0.00000     0.17533     0.00000     0.00000     0.05512
           8     0.01443     0.00000     0.09744     0.00000     0.00000     0.01777
           9     0.00000     0.19419     0.00000    -0.08894     0.11256     0.00000
          10     0.11294     0.00000    -0.04324     0.00000     0.00000    -0.12834
          11     0.07728     0.00000     0.04956     0.00000     0.00000    -0.02839
          12     0.00000    -0.10132     0.00000    -0.04711    -0.12489     0.00000
          13     0.29675     0.00000    -0.23722     0.00000     0.00000     0.83429
          14     0.12532     0.00000     0.00136     0.00000     0.00000     0.22209
          15     0.00000    -0.33078     0.00000     0.53191     0.69392     0.00000
          16     0.24449     0.00000     0.20104     0.00000     0.00000     0.01418
          17    -0.04794     0.00000    -0.02443     0.00000     0.00000     0.00175
          18     0.00000    -0.37449     0.00000    -0.49495     0.26769     0.00000
          19     0.03319     0.00000    -0.05962     0.00000     0.00000    -0.00066
          20    -0.29762     0.00000    -0.33571     0.00000     0.00000    -0.01410
          21     0.00000     0.25757     0.00000     0.27323    -0.15332     0.00000
          22    -0.06568     0.00000    -0.11340     0.00000     0.00000    -0.00250
          23     0.13575     0.00000    -0.12891     0.00000     0.00000    -0.02550
          24     0.00000    -0.29795     0.00000    -0.17367     0.10673     0.00000
...

So, to find the C-C stretching mode (or any vibration of interest), clicking through the list in Avogadro is more convenient than eyeballing the displacements in the output file. With the advent of accurate computational methods, the assignment of peaks in experimental spectra is usually aided by calculations identical to these. It's even possible to plot the computed spectra with some artificial broadening:

I am looking at http://avogadro.cc/wiki/Spectra , along with some other pages related to avogadro, and there seems to be no way of reading nwchem's output from a raman task. How did you do it?

I used a NWChem IR job, because I couldn't get a Raman job to run for some reason.
On the page you've linked, when it says what software is supported, only the latest development versions support Raman intensities, and only for GAMESS. The point I was trying to make is that the NWChem frequencies and normal modes will still be read, which is what you want. If you're interested in Raman intensities, you'll have to match them in the output file directly.

What o you mean by match them to the output files? How exactly?

If you're interested in Raman spectra, find the frequency/mode you want to identify in Avogadro, then look at the NWChem output file, find that frequency, and read the Raman intensity.

The problem I am having is with getting the information from nwchem to avogadro. Could you provide a bit more detail on how to get the IR frequencies from nwchem and visualize them on avogadro, please?

Open the output file in Avogadro using the Open dialog box under the File menu. Please come to our chat room so we don't pollute this comment thread: http://chat.stackexchange.com/rooms/3229/the-periodic-table

Which one of the output files?

Let us continue this discussion in chat.

The solution to the problem of Avogadro not reading the output file is that it ended in *.txt, when Avogadro requires quantum chemical output files to end in *.out.


text:
Is the standard enthalpy of formation always non-positive?
If the standard enthalpy of formation is defined as

the energy change when 1 mole of a substance is formed from its elements in their standard states

and forming chemical bonds is an exothermic process, then does this mean all standard enthalpies of formation are non-positive?
Why do some compounds such as $\ce{NO}$, $\ce{NO2}$ have positive standard enthalpies of formation? Is this because to form $\ce{NO}$, we must first break the bonds in $\ce{N2}$ and $\ce{O2}$ (endothermic) to form the bond in $\ce{NO}$ (exothermic, smaller energy produced than the energy required to break the bonds of original standard state elements)?

Well, you've pretty much answered your own questions.

Yes i agree right answer is no

All combustion reaction are???????

Another good example is ozone (O3), whose $\Delta H^{\circ}_f = 142.7 ,\mathrm{ kJ/mol}$.  It's also interesting to consider monatomic N(g), which has $\Delta H^{\circ}_f = 473 ,\mathrm{ kJ/mol}$.  That's because, for every mole of N(g) you form, you have to break a half mole of N≡N triple bonds, which have a bond dissociation energy of 2 x 473 = 946 kJ/mol, and you don't form any bonds to compensate.

@Fatima Please note your comments did not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review

You'v answered the question yourself.
In general, any enthalpy of reaction is the measure of the energy remainder after breaking the bonds (positive) and what energy we got back from the formation of the bonds (negative).
The net sum can be positive or negative, and it depends on what the reactants and products were. As you correctly said yourself, $\ce{NO}$ happens to have a positive standard enthalpy of formation because, as experimentally determined, more energy went into breaking the bonds of reactants than what we got back from forming the products.
So,

Is the standard enthalpy of formation always non-positive?

The answer is no.


text:
Does benzene structure stand for a single resonance form or the whole molecule?
It is known that benzene has two main resonance forms

Often we still draw benzene molecule as one of them

My question is, actually when we draw the lower figure, do we refer to the whole benzene molecule or one of the resonance structure or depend on the context?

Since neither of the two resonance structures truly exists as shown, the bottom is taken to represent the resonance hybrid of the top unless suggested otherwise (I'm not sure I've ever seen it suggested otherwise, but it could happen?).

https://chemistry.stackexchange.com/questions/50005/does-benzene-structure-stand-for-a-single-resonance-form-or-the-whole-molecule

I approach this question from the opposite direction.
Benzene is commonly drawn as a ‘cyclohexatriene’ corresponding to the Kekulé structure, i.e. with three single bonds and three double bonds, despite the fact that the six bonds of benzene are actually indistinguishable from each other.

This graphical representation of benzene is in accordance with Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) as well as Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008).
Therefore, when you see such a structure diagram in the literature, it usually represents a real benzene ring and not a particular resonance form.
However, if a structure diagram is explicitly meant to show the individual resonance forms, they are commonly drawn with a two-sided arrow.

me thinks you over sold the use of the double bond forms. See page 104 of the pdf file (page 380 of printed copy) http://lcso.epfl.ch/files/content/sites/lcso/files/load/IUPAC%20drawing%20rules.pdf

@MaxW Certainly, the shorthand form shown on page 380 which indicates all resonance forms of benzene by a circle is acceptable, especially to show the equivalence of the six bonds. However, the preferred style is the ‘cyclohexatriene’ structure diagram (see almost all other structure diagrams of benzene rings in that document, and “Conventionally, benzene is commonly drawn with three single bonds and three double bonds, (…)” on page 379). Anyway, the question is not about how to draw benzene; it is about how to interpret a given ‘cyclohexatriene’ structure diagram.

"Anyway, the question is not about how to draw benzene; it is about how to interpret a given ‘cyclohexatriene’ structure diagram" +1 That's exactly what I mean.

I'm shocked, but section GR-6.5 on page 105 of pdf file goes into more detail and reinforces the idea that the double bond form is preferred unless resonance is specifically being discussed.  Weird. I really like the circle form better.

The answer is you are referring to neither of them. That is because resonance structures don't actually exist in reality. We only use them to give us a rough idea what the actually molecule and bonds look like. A common way to explain resonance structures is this:
An explorer from a far distant land travels to a new continent and sees a strange animal that he has never seen before, a rhinoceros! When he comes back home, not really sure how to describe this strange animal, he tells everyone that it looked like an unicorn and a dragon mixed together.

So what the explorer did was he tried to explain some weird, not easily imagined object (a rhinoceros) by describing it as a mixture of two mythical, non-existent objects, but which are easily imagined by people.
That is exactly what the resonance structures of benzene (also for any other molecule) are. They don't actually exist, but rather benzene exists as a weird mixture of these resonance structures. The reason why we use resonance structures is that they give us an idea how to imagine what benzene looks like.

What a great analogy! I've never seen this before, but I'll be sure to use this to explain resonance in the future!

The analogy seems to originate from "Resonance in Organic Chemistry" by George Willard Wheland - 1955, page 4. In an earlier version (The Theory of Resonance and Its Applications to Organic Molecules, 1944) of the monograph he used a mule as an intermediate between a horse and a donkey which are both real creatures.

@MaxW my organic lecturer explained tautomers using the above analogy. Where you explain a mule as a mixture of two real animals, a horse and a donkey (the individuals tautomers in this case).

This is veering off-topic but I strongly disagree with the use of this analogy for tautomerism. Tautomerism is not the same as resonance. Tautomers independently exist and are interconverted via a rapid process. The relative abundance of each tautomer can be directly measured, e.g. via NMR spectroscopy. Resonance forms do not exist and are not interconverted.

I've seen the following drawing used instead of the three double bonds. The circle signifies that the ring is aromatic. It's an alternative that is used to show that all bonds are the same.

This. It reinforces the notion of the pi electrons all being in a weird circular flux of no fixed position.

The way I would put it is that we are compensating for the inability of Lewis structures to satisfyingly depict electron delocalization, so instead we pick whichever "resonance" structure is most appropriate, with the understanding that these are only poor approximations of the real situation. (FWIW, I don't quite like the "circle" notation, as this is not very useful for mechanistic "electron pushing" analyses. How many electrons are there in a circle, anyway?)

Electron pushing is where the evil starts; the circle has 6 electrons.

If electron pushing is evil, my soul is lost for all eternity and destined to burn in the deepest and darkest six-electron circle of hell … @Mart


text:
Mechanism of Diels-Alder reaction
I have a question about the mechanism of the Diels-Alder reaction. Is this a two-electron or one-electron process as described by Fig (a) or (b) respectively or does not matter? How do we know that?

A Diels-Alder reaction is a [4+2] cycloaddition, which is formally shown as a two electron process. I suspish that both ways of drawing the mechanism might be equivalent thouh. I was recently having a confusion about MS fragmentation products. Generally, I draw McLafferty rearrangements with  three 2 electron arrows just like a Diels-Alder, but my book uses six 1 electron arrows. I don't think it really makes much of a difference. For Diels-Alder, though, I've never seen the six 1 electron arrow mechanism in a book, but that doesn't really mean it's wrong.

I'm don't think the mechanism on the right would necessarily require adherence to the Woodward-Hoffman rules.

It's a little like with your question with benzene - both are only lame ways to try to represent what's really happening.

First of all, full arrow represents a movement of 2 electrons. Half arrows represent movement of 1 electron.
Diels-Alder is a 6π electron pericyclic reaction. The diene provides 4π electrons and the dienophile 2π electrons to the formation of the bonds. The Diels-Alder reaction belongs to [4+2] cycloadditions according to the Woodward-Hoffmann nomenclature.
A purely formal way of determining the participating electrons consists of simply counting the electron flow arrows that are used to illustrate the mechanism and multiplying by two. Each arrow moves two electrons.
You have 3 arrows representing a reaction, so 3*2=6 electrons
The number of electrons participating in the cycloaddition determines wether the reaction follows either a supra-supra (s+s) or a supra-antara (s+a) mechanism.
This is a nice source for Diels-Alder mechanisms

This doesn't answer the question. Beyond that, it doesn't even point out anything useful. It mentions a lot of things that are related but definitely off-topic. The structure on the left uses three full arrows to represent six electrons. The structure on the right uses six half arrows to represent six electrons. Maybe the phrasing in the original isn't perfect, but look at the images and you clearly see that the question was asking if each electron acts alone vs each electron being part of a pair that moves togethers.


text:
Relationship between Gibbs free energy, enthalpy, entropy and rate of reactions
I'm having a little trouble constructing the bigger picture of chemical reactions.
From the collision theory we know that molecules react if: 1) they collide 2) they collide bearing a sufficient amount of energy to surmount the activation energy and collide at a specific geometric orientation.
From the relationship between Gibbs free energy, enthalpy, entropy and temperature we know: $\Delta G = \Delta H - T\,\Delta S$,
and that a reaction is only thermodynamically favored when $\Delta G < 0$. And even when $\Delta G < 0$ the reaction may still not occur at a measurable rate because it has too high an activation energy.
From the Arrhenius equation we know: $k = A ~\mathrm e^{-E_\mathrm a / (RT)}$
Linking all these together:
1. Reactions only occur when particles can collide with each other.
2. Colliding particles will only react if they bear sufficient energy (can be increased through temperature and/or adding other external energy sources) and have correct orientation (chances can't be increased?).
3. Only particles in thermodynamically favored reactions have enough energy to overcome $E_\mathrm a$ and so will spontaneously occur without external energy applied. But they may occur extremely slowly if they have a very high $E_\mathrm a$ (according to the Arrhenius equation).
Is this reasoning correct?

You're forgetting about equilibrium ΔG=-RTlnK

I'm having a little trouble constructing the bigger picture of chemical reactions.

Let's go step by step.

From the collision theory we know that molecules react if: 1) they collide 2) they collide bearing a sufficient amount of energy to surmount the activation energy and collide at a specific geometric orientation.

Yes, in order to react, molecules taking part in a (bimolecular) reaction need to:

be in contact;
be correctly aligned to each other (a random alignment may not react);
have enough energy to overcome the activation barrier ($\Delta G^\neq$).

From the relationship between Gibbs free energy, enthalpy, entropy and temperature we know: $\Delta G = \Delta H - T\,\Delta S$,
  and that a reaction is only thermodynamically favored when $\Delta G < 0$. And even when $\Delta G < 0$ the reaction may still not occur at a measurable rate because it has too high an activation energy.

What you're describing is the Gibbs reaction energy: the Gibbs energy between products and reactants.
This indeed tells us how thermodynamically favourably the reaction is ($\Delta G < 0$).
But the phrase in bold above is wrong.
The reason is that the Gibbs activation energy ($\Delta G^\neq$) is the Gibbs energy difference between transition state and reactants.
As Mithoron mentioned, Gibbs reaction energy would give you the equilibrium constant for the reaction:
$$K = e^{-\tfrac{\Delta G}{R T}}$$

From the Arrhenius equation we know: $k = A ~\mathrm e^{-E_\mathrm a / (RT)}$

Generally, from Eyring theory,
$$k = \frac{k_B T}{h c_0} e^{-\tfrac{\Delta G^\neq}{R T}}$$
Comparing to Arrhenius equation, see:
$$k = \frac{k_B T}{h c_0} e^{-\tfrac{\Delta H^\neq - T \Delta S^\neq}{R T}}
= \frac{k_B T}{h c_0} e^{\tfrac{\Delta S^\neq}{R}} e^{-\tfrac{\Delta H^\neq}{R T}}$$
Thus,
$$A = \frac{k_B T}{h c_0} e^{\tfrac{\Delta S^\neq}{R}} \\
E_a = \Delta H^\neq$$
That's what you would find from an Arrhenius's plot.

Linking all these together:
  1. Reactions only occur when particles can collide with each other.
  2. Colliding particles will only react if they bear sufficient energy (can be increased through temperature and/or adding other external energy sources) and have correct orientation (chances can't be increased?).
3. Only particles in thermodynamically favored reactions have enough energy to overcome $E_\mathrm a$ and so will spontaneously occur without external energy applied. But they may occur extremely slowly if they have a very high $E_\mathrm a$ (according to the Arrhenius equation).
Is this reasoning correct?

The reasoning is correct, except for point three above (in bold), for which the same error is found as before: reactions may have a big $K$ (small $\Delta G$) but small $k$ (big $\Delta G^\neq$) or vice-versa.

In the second point, that happens because in Arrhenius equation, $e^{-\tfrac{E_a}{RT}}$  actually corresponds to the fraction of molecules that have an energy which more than $E_a$. (Part of Arrhenius' theory which explains his equation). So as $T$ increases, the negative exponent of $e$ decreases and hence the value of the number increases. And no, the chances can not be increased. All you can do is use a positive catalyst to increase the amount of colliding molecules and hence increase the rate of reaction.
All others things are correct and so is the reasoning.


text:
Increasing the length of a copper wire via electrolysis
I'm currently experimenting with copper coating using electrolysis. I'm using copper wire from my home, like the one you use for grounding. I want to electrolysis with longer copper wire (say 5 times the height of the container) to find out if the process will be shortened. My question here is about how I should arrange the wires in the container and if one arrangement has different effect than the other.
Here is what I have in mind:

all horizontal wires connecting to one vertical wire(this will be 5 separate pieces same length connected to the vertical wire)
bent like snake up and down (in one piece)

Do you think these two arrangements would work the same or differently?

The two arrangements would produce identical results, because the copper wire conducts electricity so much better than the electrolyte solution.
However, the second arrangement is more practical because no joins are needed so it would be more reliable.
Generally, the surface area of the copper anode should be similar to the cathode which is to be coated to avoid concentrating the current unevenly, so increasing the anode area would probably help shorten the coating time somewhat, as it would avoid oxygen being evolved from the anode which can happen if the anode area is too small.

Thanks for your comment. You just raised out my other question. Is avoiding oxygen to evolve a bad thing? Please brief.

You should always try to avoid electrolyte breakdown in electroplating. It can have many bad effects. For copper electroplating, the voltage needed is very small, around 0.5 volts, which is much less than the breakdown voltage of water. If the voltage is higher than 1.2V then you will get oxygen from the positive (source of copper) electrode and hydrogen from the negative (the item being plated with the copper). Excess oxygen can cause black copper oxide sludge which can cause plating to be rough and not bright. Excess hydrogen at the negative causes hydrogen embrittlement, lowering ductility.

I am aware of the volt but dont know that it has to be around 0.5 volts. Coming back I do remember see one or two bubble on the plate. I dont know about black copper oxide sludge but I do get very little amount of blue type sludge sometime. Do you happen to know what that is?

The blue sludge is probably a hydrated copper hydroxide or hydroxysulphate. If you only see a small number of bubbles, then the voltage is probably correct. I hope you get good results. I enjoy copper plating - it is a most rewarding experience with many possibilities for experimentation.

I actual like the blue sludge and have collected about 1cm thick radius of 3cm. I think I would like to make a bottle out of it but cant product it every time. Is there a way to produce it consistently?

Sure. It's a bit sensitive to the electrolyte concentration, but if you search for synthesising copper hydroxide by electrolysis there are lots of how to guides and YouTube vids on doing it.


text:
Enzyme Specificity vs Selectivity
Background:
I can see a difference in stereoselective and stereospecific reactions in organic chemistry. However, both terms are used interchangeably as synonyms in a biochemistry class.
Question:
What differs selectivity from specificity in enzymatic reactions. What is meant when we talk about enzyme specificity and enzyme selectivity?

Specificity is the term used to define the selectivity of enzymes
for their substrates. The selective qualities of an enzyme are collectively recognized as its specificity.
Other texts have synonymised enzyme selectivity with substrate specificity:

The non-covalent forces through which substrates and other molecules bind to enzymes are similar in character to the forces
that dictate the conformations of the proteins themselves : Both involve van der Waals, electrostatic, hydrogen
bonding, and hydrophobic interactions.
Molecules that differ in shape or functional group distribution
from the substrate cannot productively bind to the
enzyme; that is, they cannot form enzyme–substrate complexes
that lead to the formation of products.

The extraordinary ability of an enzyme to catalyse only one particular reaction is a quality known as specificity.
Specificity means an enzyme acts only on a specific substance,
its substrate, invariably transforming it into a specific product. That is, an enzyme
binds only certain compounds, and then, only a specific reaction ensues.
Some enzymes show absolute specificity, catalyzing the transformation of only one
specific substrate to yield a unique product. Other enzymes carry out a particular
reaction but act on a class of compounds. For example, hexokinase (ATP-hexose-6-
phosphotransferase) will carry out the ATP-dependent phosphorylation of a number
of hexoses at the 6-position, including glucose.

Pioneering enzyme specificity studies at the turn of the 20th century
by the great organic chemist Emil Fischer led to the notion of an
enzyme resembling a “lock” and its particular substrate the “key.”
The specificity of an enzyme (the lock) for its substrate (the key) arises from their geometrically complementary shapes. This was later
modified to a more accurate induced fit hypothesis given the high
flexibility, conformationally dynamic properties of enzyme molecules.

References

Voet and Voet Biochemistry 4th ed: Introduction to Enzymes
Biochemistry Grisham & Garret How Can Enzymes Be So Specific?

Selectivity is used when talking about the substrate, for example an enzyme can be more selective for D- sugars compared to L-sugars.
Specificity is used when talking about the product. For example an enzyme that only accepts a single compound could specifically hydroxylate the 2 position, even though there are be more positions that could be hydroxylated.


text:
How much steric hindrance does a phenyl group offer?
I've come across a question in my revision in which I am asked to compare the steric hindrance provided by a number of substituents. Those relating to phenyl are ranked relatively low, despite it being what I thought was a rather large substituent. Is this due to the planarity of the ring, or is it just much smaller than I originally thought; does a tertiary alkyl substituent provide more steric hindrance because it is larger, or because it is spread in all directions?

despite being what I thought was a rather large substituent

I agree, generally speaking the phenyl group is considered a rather large substituent.
One metric commonly used to compare "steric size" is the cyclohexane A-Value.  In the cyclohexane ring system, bulky substituents prefer the equatorial position rather than the axial position. This is due to steric (this can also be described in electronic terms since it is due to the repulsion between interpenetrating electron clouds) destabilization. For example, in methylcyclohexane, the methyl group can exist in an axial or equatorial position. In the axial position, it is destabilized by the repulsive steric (electronic) interactions it has with the axial hydrogens on carbons 3 and 5.

The larger the A-Value, the larger the substituent and the greater its preference for the equatorial location ($\mathrm{\Delta G = -RTlnK_{eq}}$ where $\mathrm{K_{eq}}$ represents the the axial/equatorial ratio at equilibrium).
Here is a link to a Wikipedia article on A-Values.  Look at the table of A-Values in this article.  A methyl group has an A-Value of 1.7 kcal/mol. a phenyl group has an A-value of 3 kcal/mol.  The only substituent in the table with an A-Value larger than phenyl is the t-butyl substituent with an A-Value > 4 kcal/mol. The phenyl group is a relatively large substituent.


text:
Change in gas phase reaction equilibrium when not all product and reactant are increased
$$\ce{2NOCl <=> 2NO + Cl2}$$
What will be the effect on equilibrium concentration of $\ce{NOCl}$ when equal moles of $\ce{NOCl}$ and $\ce{NO}$ are introduced in the mixture at constant temperature?
Now this is what I think
Increasing the moles will result in increased concentration of both $\ce{NOCl}$ and $\ce{NO}$. But there also lies a second product $\ce{Cl2}$. Since there's an increase in $\ce{NO}$ and $\ce{NOCl}$, there should be an equal increase in the concentration of $\ce{Cl2}$ to keep the equilibrium-constant constant. But since there's no addition of $\ce{Cl2}$ molecules there should be a decrease in $\ce{NOCl}$ concentration which will decompose further to increase the molar ratio of $\ce{Cl2}$ and hence achieve equilibrium. Please enlighten me where my logic is flawed.

From comments (edit by BuckThorn): The reported answer is that the position of equilibrium remains unchanged and hence the concentration of $\ce{NOCl}$ remains constant.

Given the reaction $$\ce{2NOCl <=> 2NO + Cl2}$$  we can formulate the equilibrium as $$\mathrm{K} = \dfrac{\ce{[NO]^2[Cl2]}}{\ce{[NOCl]^2}}$$
The problem then asks "What will be the effect on equilibrium concentration of NOCl when equal moles of NOCl and NO are introduced in the mixture at constant temperature?"
Now if we add equal amounts of $\ce{NOCl}$ and $\ce{NO}$ we don't have to do it at the same time to end up with the final equilibrium.
So let's add the $\ce{NOCl}$ first. We now have two possibilities (assuming that it is very unlikely that K=1 exactly).

If K>1 then more $\ce{NO}$ and $\ce{Cl2}$ form.
If K<1 then less $\ce{NO}$ and $\ce{Cl2}$ form.

Now let's add the $\ce{NO}$. This has to push the reaction to the left.
So the net result is it depends. You have too have enough data to plug into the equilibrium expression and other equations to work the net result out.

Your logic is true.
When you increase the concentration of NOCl and NO,  Chlorine become limiting compound. So it will affect the equilibrium.
Increase in amount of NOCl,  will leads to more decomposition but it will also affected by concentration of NO in product. So in decomposition process, NO is controlling compound.
And new equilibrium will achieved with more concentration of NOCl and NO and slightly more of Chlorine compare to initial concentration .


text:
Can I get some pointers on what to do in calculating the hardness of water?
Information provided: $\pu{1.64 g}$ of $\ce{Ca(NO3)2}$ and $\pu{0.94 g}$ of $\ce{MgCl2}$ are dissolved in $\pu{10 L}$ of water, calculate the hardness of water.
I have used these two equations: $m = n \cdot M$ and $c = \frac nV$.
What I have done is simply insert the molar mass of both $\ce{MgCl2}$ and $\ce{Ca(NO3)2}$ in the first equation, isolated $n$, then used that answer to isolate $c$ in the next equation. But I need to find the mass for $\ce{Ca^2+}$ and $\ce{Mg^2+}$ only, how do I do that? Have I found the concentration of the wrong things?
Please explain in simple language. Thanks!

The hardness of water is defined as the sum of the dissolved $\ce{Ca^2+}$ and $\ce{Mg^2+}$ species and can be reported in different ways (moles/L, g/L, and so on). The one I have seen most commonly is mg/L, so that is the calculation presented below.
You need to determine how many milligrams of $\ce{Ca^2+}$ and $\ce{Mg^2+}$ are present and then divide that number by the solution volume (here, 10 L).
For $\ce{Ca^2+}$:
$$1.64\;\mathrm{g}\;\ce{Ca(NO3)2}\cdot{1\;\mathrm{mol}\;\ce{Ca^2+}\over 164.09\;\mathrm{g}}\cdot{40078\;\mathrm{mg}\over 1\;\mathrm{mol}\;\ce{Ca^2+}}
 = 400\;\mathrm{mg}$$
For $\ce{Mg^2+}$:
$$0.94\;\mathrm{g}\;\ce{Mg(Cl)2}\cdot{1\;\mathrm{mol}\;\ce{Mg^2+}\over 95.2\;\mathrm{g}}\cdot{24305\;\mathrm{mg}\over 1\;\mathrm{mol}\;\ce{Mg^2+}}
= 240\;\mathrm{mg}$$
Thus, you have 640 mg $\ce{Ca^2+}$ and $\ce{Mg^2+}$ in 10 L of solution so the hardness is:
$${640\;\mathrm{mg}\over 10\;\mathrm{L}} = 64\;\mathrm{mg}\cdot\mathrm{L}^{-1}$$


text:
How to compare the Heat of Combustion qualitatively?
I have been doing practice problems in General Organic Chemistry. I got some questions that asked me to compare the Heat of Combustion of some molecules. I couldn't get any of those questions and the logic behind their answers. I tried googling but it didn't help. The search results were not direct. So, I'd like to know the factors on which the heat of hydrogenation depends. I am not looking for an answer that describes it quantitatively but rather qualitatively.
Here are some example questions:

Which one of the following has the smallest heat of combustion?

The answer is (c). But why is that?

Rank the following substances in order of decreasing heat of combustion (maximum $\rightarrow$ minimum)

The answer is 2>4>1>3. But again, why is that?
I knew that the heat of combustion depends on the number of carbon atom. But in these two  examples the number of carbon atoms are same. Is there anything I need to watch out for?

I guess you're asking about effect of delocalisation on the heat of hydrogenation?

@Mithoron Not just that. But some other common dependence of heat of hydrogenation on other things too.

That may be too broad, it would help if you'd rather specified this.

@Mithoron I added some examples. I hope it's alright now. :)...And oh..I also changed the heat of hydrogenation to heat of combustion.

Och no now it looks like homework :D Rollback, quickly!  ;)

related http://chemistry.stackexchange.com/questions/35684/reactivity-of-propellanes http://chemistry.stackexchange.com/questions/32450/comparing-heat-of-hydrogenation-per-mole-of-butenes-and-1-3-butadiene

@Mithoron Ok ok...I get it. :p

For isomers differences in the heat of combustion are due to differences in the internal energy of the educts. If intermolecular interaction (lattice energy etc.) can be neglected it reflects differences in the internal energy of the molecules.
In your first example C has the smallest heat of combustion since it has the lowest internal Van der Waals strain.
In the second example the order is given by the ring strain, and the Van der Waals strain between the two methyl groups at the cyclobutane ring.


text:
Do all covalent nitrates on heating decompose to give nitrogen dioxide?
Being covalent we can conclude that they will on heating decompose completely, but will they give only $\ce{NO2}$ and not any other oxide is the question.
It is observed in $\ce{Pb(NO3)2, AgNO3,}$ and also $\ce{LiNO3}$ (the anomalous behaviour of $\ce{Li}$) that they decompose on heating to give $\ce{NO2}$ and the above mentioned nitrates possess good amount of covalent character. So as a rule can I remember that all covalent nitrates decompose to give $\ce{NO2}$ or there are any exceptions to it?

I would not classify those as 'covalent' nitrates.

Xenon nitrate has unusual mechanism of decomposition.

What criteria do you use to state that the lead and silver nitrates possess good amount of covalent character ?


text:
Enthalpy change vs. change in internal energy in a system
I've been trying to get my head around the differences between enthalpy change and internal energy change for a system. Here's my understanding of it so far. I'd appreciate any feedback on my understanding so far on the concepts - especially if there are misconceptions (and there probably are)

Internal energy is the 'intrinsic' energy of some system, can be measured with reference to something within the system, e.g. chemical bonds, vibrational energy states, etc.
Enthalpy change is the heat (i.e. energy transferred due to temperature difference) change at constant pressure due to some chemical process.

$\Delta H = \Delta U + P \Delta V$
$\Delta H$ is characteristic of a given chemical process - i.e. this value is the same regardless of what conditions it is run under (e.g. varying pressures, volumes, starting temperatures). As long as it's the same reaction, $\Delta H$ will be the same. It's the overall energy input required to make a process work, and is split between influencing $U$ and performing $PV$ work.
$\Delta U$ is a function of the specific conditions (i.e. volume, pressure) in which a specific instance of a chemical process is run.
So it's theoretically possible to set up a chemical process (say vaporisation) to run in such a way so that $\Delta U$ is zero, and all the enthalpy change is accounted for in the $P\Delta V$ term?
And what about cases where P is not constant? What happens then? Can we calculate the value of $\Delta H$? And does $\Delta H$ even have a meaning in this case?

Both $U$ and $H$ are not directly related to any particular process.  They are physical properties of the material being processed, and are unique functions of the thermodynamic state of the material, and not the process.  They are directly proportional to the amount of material, and its thermodynamic state is characterized by its temperature and pressure (and composition for a multicomponent system).  The internal energy is a fundamental property, but the enthalpy is a defined property given by $H = U + PV$, and is just a convenient parameter to work with is solving many kinds of thermodynamics problems.
For a process carried out at constant pressure in a closed system, the amount of heat $Q$ added to the system turns out to be equal to the change in enthalpy between the initial state of the material and the final state of the material.  For a chemical reaction carried out at constant pressure, the amount of heat $Q$ that must be added to the system in order for the final temperature of the products to equal the initial temperature of the reactants turns out to be equal to the enthalpy of the pure products minus the enthalpy of the pure reactants at that temperature.  Both these results follow from the first law of thermdynamics.
For a chemical reaction, generally both the internal energy and the enthalpy change between the initial and final equilibrium states.

And what about cases where $P$ is not constant? What happens then? Can we calculate the value of $ΔH$? And does $ΔH$ even have a meaning in this case?

Since $H$ is just defined as $U + PV$, we can always calculate the change in enthalpy between two thermodynamic equilibrium states of a system.  $\Delta H$ certainly has a value in this case, even if it is not equal to the amount of heat added for a process between the two states.
The critical thing to remember is that $U$ and $H$ are physical properties of the materials comprising a system, and are independent of any specific process.


text:
Hydrolysis of Sucrose over Sn1Ac or Sn2Ac
In "March's Advanced Organic Chemistry" I have read about the general mechanism of the acetal hydrolysis. It names the acid-catalyzed SN1 or SN2 as a possible mechanism for a Acetal-hydrolysis.
Now I am interested to know if SN1 or SN2 is mainly going on at the Hydrolysis of Sucrose to Glucose and Fructose. So I compared the different parameters which influence SN1/SN2.

Polarity of the solvent: Water -> really polar -> SN1 preferred.
Stabilization of a possible Carbokation-Intermediat: Primary (Anomeric C!), so not really stabilized by Hyperconjunction -> SN2 preferred. (Is there Hyperconjunction also in the ether-O?)
Mesomeric stabilization There is also a mesomeric stabilization between the resulting carbocation and a Oxonium-Ion in the SN1-Mechanism.
Steric Hindrance: In my opinion, through the planar region around the anomeric C (planar ether), there is no steric hindrance -> SN2 possible.
Temperature: high temperature -> SN1 preferred.

So to conclude: Many different aspects. Because of the main points "steric hindrance" and the destabilized primary carbocation intermediate I would conclude that the hydrolysis of Sucrose goes mainly (maybe a little bit over SN1) over SN2.
Have I forgotten something important? Are my assumptions "No steric hindrance" and primary carbocation-intermediate correct?
I have attached my possible SN1-mechanism.

Unstabilized primary cations never undergo $S_N1$ reactions so yes you are correct.

@bon So the mesomeric stabilization isn't enough stabilization?

I would not call the intermediate a primary carbocation. Yes the primary carbocation is one of the resonance forms but it is an extremely minor one. The oxonium ion resonance form is a much larger contributor. In any case, both pathways are possible. The SN1 mechanism you have drawn is perfectly correct. In addition to the factors you considered, the SN1 pathway is particularly favoured for the β-anomer because of the anomeric effect (n -> σ* donation from ring oxygen lone pair into C-O σ*).

@orthocresol Tanks for your answer! So the oxonium-resonance-structure stabilize the whole SN1 intermediat and so we can assume that also the SN1-mechanism happens also at a "big" amount?

@orthocresol To the anomeric effekt: Sucrose has the Alpha-D-Glucose-monosaccharid, so the the anomeric OH-group makes the whole intermediat not more stable. But if the Glucose-rest is the leaving group, and Fructose works as the intermediat (is that the case?), there your spoken Anomeric effect works?

Yes the SN1 pathway does occur significantly in carbohydrate reactions. Also, oops... sorry. The anomeric effect works for the anomer which has the leaving group placed in an axial position, which in this case is the α-anomer. My bad.

@user2365846 Primary cations that are otherwise unstabilised do not undergo SN1. However, this one is highly stabilised by the neighbouring oxygen so it will. Sorry I wrote my previous comment without properly looking at the structure.

The mechanism supplied in the question statement is wrong according to this paper, where the evidence favored Carbon-Oxygen bond cleavage on the fructosyl-side instead of the glucosyl-side of the bridge Oxygen. As that reference states, "the initial products of the hydrolysis are a-D-glucose and a fructosyl carboxonium ion"

But don't feel bad, many textbooks have shown the same mechanism (detailed in the Discussion section of the paper). It's possible that the reaction mechanism changes with conditions, but I know of no evidence for that.
However, the same paper discusses that the mechanism was thought to depend on the type of sugar undergoing bond cleavage:

Results from a computational study show that SN1 mechanisms via a carbocation on fructose and glucose are both feasible, with the one on fructose slightly more likely.
Some other points:
1) The question statement refers to "primary" carbocations twice, and so do three of the comments. But in this reaction the 2 alternatives are a secondary or tertiary carboxoniun ion (for this terminology see Olah (1998) Onium Ions or ref 11 of the above paper).
2) It makes sense that an SN1-type mechanism would take place on the fructosyl-side of the bridge Oxygen since the resulting carboxoniun is tertiary rather than secondary.
Now, the real question is: How would you experimentally determine the mechanism? That is where the science comes in. The fact that so many textbook authors have presented this classic reaction with different mechanisms should be a big lesson to all students.  That expert authors would go to the extent of drawing detailed reaction mechanisms in a textbook that are not backed up by experimental reality, shows how mistrustful we should be of any non-experiment-based opinions. And this is for a classic, textbook example.  How much more unreliable "expert opinions" are for much more complicated things!


text:
Should carbon tetrachloride really be considered an organic compound?
Wikipedia says:

Carbon tetrachloride […]  is an organic compound with the chemical formula  $\ce{CCl4}$.

From an organic point of view, carbon tetrachloride can be called the tetrachloro derivative of methane and is described in many organic textbook along with $\ce{CH3Cl}$, $\ce{CH2Cl2}$ and $\ce{CHCl3}$.
From an inorganic point of view, carbon tetrachloride can be considered chloride of a group 14 element (along with $\ce{SiCl4}$, $\ce{GeCl4}$).
Shouldn’t carbon tetrachloride be considered an inorganic compound? It neither contains any $\ce{C-H}$ bond nor is it found in any living organism.
Tetrachloromethane is its organic name. But it is generally not called by this name. Carbon tetrachloride is its better name and inorganic compounds are named like that.

The distinction organic/inorganic is (often) not a very useful one, except to put the word on books, lectures and institute names. ;) All (in)organic molecular chemistry people speak basically the same language. Chemistry of inorganic solids and ionic melts is something much different, for example.

As Karl said, does it really matter?

Related: Is carbon dioxide organic or inorganic?

There are a number of questions here on Stack Exchange discussing the distinction between organic and inorganic compounds, such as:

What is the definition of organic compounds?
What does 'organic/non-organic molecule' means exactly? (Check out ron’s answer there for different contemporary definitions of organic compounds.)
What is the first organic compound to be discovered ? Urea or alloxan? (Check out my answer there for one of the oldest definitions.)

The distinction between organic and inorganic chemistry is one that is upheld mainly because of traditional reasons. It is also upheld because some more specialised fields are often well included into one of the two traditional definitions: solid state chemistry is usually inorganic while natural product synthesis is almost entirely organic chemistry by all applicable definitions (save the very old historical ones).
But remember that there are a number of fields that straddle the boundary. Coordination chemistry often involves organic and inorganic ligands around an inorganic metal centre. It depends on the research being performed whether the group wishes to be considered organic or inorganic. See for example the groups of Professor Klüfers and Professor Knochel at the LMU Munich: The former classifies himself as an inorganic chemist and investigates carbohydrate-metal complexes among other things; the latter classifies himself as an organic chemist and investigates organometallic reagents and reactions.
So in a way, the organic/inorganic distinction compares well to the border between Europe and Asia. Everyone will agree that France is in Europe and China is in Asia but what about Georgia and Turkey?

Note that your dualism is short-handed. $\ce{SiHCl3}$, trichlorosilane or silicochloroform is produced in an industrial scale and trichlorogermane $\ce{GeHCl3}$ is also known. On the silicon side of things, apart from tri- and tetrachlorosilane the di- and monochlorosilanes $\ce{SiH2Cl2}$ and $\ce{SiH3Cl}$ are known just like their carbon counterparts. And of course, silane $\ce{SiH4}$ is known, too. (I am not sure exactly how much is known about the corresponding germanium compounds. The English Wikipedia didn’t have entries for them but didn’t contain trichlorogermane either, which was only present in German and Russian.)
So beware when drawing analogies. They could go into a different direction that you may think.

Speaking of surprising facts: Did you know that chloromethane is the most-produced secondary metabolite by mass? So let’s add chlorosilane into the organic realm by analogy!

Finally, as an aside note: I actually use tetrachloromethane much more than I would carbon tetrachloride; that goes for both languages in which I do chemistry.

Its been a while that you answered some question here....

@NilayGhosh That’s true. I needed to take a break to concentrate on practical work ;)

Simple carbon compounds such as the tetrachloride blur the boundary between organic and inorganic. You should know that presence in natural systems is not an indicator of a compound being organic; many organic compounds are completely artificial, though they may be derived from natural feedstocks.
There is no strict criteria for organicity (is there a better word?). Consider perfluorinated carboxylic acids, they have no C-H bonds but are considered organic.

Or HCN, definitely inorganic, right? But what about cyanogen NC-CN? Ethine is organic, but acetlyides?

Simple carbon ions are considered inorganic.

I would assert HCN is definitely an organic compound.

@Karl I agree with Lighthart's comment because HCN is found in fruits like apples,  cherries, almonds etc. and also in mammals. See wikipedia page of HCN(occurence).

@Lighthart Yes, HCN  is an organic compound.

That's debatable: HCN is also a pseudohalogenic acid, cyanogen the corresponding pseudohalogen.


text:
Calculating temperature after adding heat to ice

If $53.2~\mathrm{kJ}$ of heat are added to $15.5~\mathrm{g}$ of ice at $-5^\circ\mathrm{C}$, what will be the resultant state of matter in which water is present and also calculate its final temperature.

I have done this:

Step 1: Energy required to make ice at 0 Celsius
Step 2:Energy required to make  water at 0 Celsius
Step 3: Energy required to make steam at 0 Celsius
Step 4: Energy required to make steam at 100 Celsius
Step 5:Energy required to make steam at 'x-100' Celsius

Please leave a comment stating if any errors are present. The answer I got is x = 808 Celsius. The correct answer is 303 Celsius.

You don't make steam at 0°C. You heat water to 100°C and only then evaporate it.

So i have to make water from 0 Celsius to 100 Celsius to 100 Celsius steam and then continue step 5?

Yeah, like that.

It is an enthalpy balance that requires knowledge of the enthalpy change at each step: (The steps you have listed are not correct.)

Solid heat capacity of ice of the form e.g. $C_p^\mathrm{solid}(T)$. This allows you to calculate the amount of heat (or energy) required to warm the sub-cooled ice from $-5~^{\circ}\mathrm{C}$ to $0~^{\circ}\mathrm{C}$.
(Latent) heat of fusion, $\Delta H_\mathrm{fus}$ then must be added to transform the solid ice to liquid water at $0~^{\circ}\mathrm{C}$.
Next we must warm the liquid water from $0~^{\circ}\mathrm{C}$ to its boiling point at $100~^{\circ}\mathrm{C}$, which requires knowledge of the liquid heat capacity of water e.g. $C_p^\mathrm{liquid}(T)$.
Now at the boiling point, we must account for the required amount of energy to transform the liquid water into its vapor state at $100~^{\circ}\mathrm{C}$ using the (latent) heat of vaporization of water $\Delta H_\mathrm{vap}$.
The final step then involves super-heating the vapor from $100~^{\circ}\text{C}$ to $T_\mathrm{final} = ?~^{\circ} \mathrm{C}$, which requires knowledge of the vapor heat capacity of steam e.g. $C_p^\mathrm{vapor}(T)$.

You know the overall enthalpy change ($53.2~\mathrm{ kJ}$) and you know the starting temperature ($-5~^{\circ}\mathrm{C}$), so all you have to do is add the results from each of the steps and solve for $T_\mathrm{final}$ e.g. $303~^{\circ}\mathrm{C}$.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help].


text:
Partial charges in a covalent bond
My professor stated the following statement:

In a covalent bond, there is a presence of a partial charge in the atoms that combine to form a compound. For example in $\ce{H2O}$, since $\ce{O}$ is more electronegative, therefore there is a presence of a partial negative charge around $\ce{O}$. Similarly, as $\ce{H}$ is less electronegative, it has a presence of a partial positive charge around it. These partial charges are very feeble.

I found this statement to be little incomplete. So, I decided to find the reason of this partial charge. One of the video that talked about partial charges stated this:

Since $\ce{O}$ is more electronegative than $\ce{H}$, therefore the shared electrons tend to spend more time around $\ce{O}$ than $\ce{H}$ and this is the reason of the partial charges.

Now, my questions are:

 If the shared electrons spend more time around $\ce{O}$ then it means that for some time $\ce{O}$ will be negatively charged and for some time $\ce{H}$. If this holds true, then why do we say that there is presence of  partial charges  around the atoms instead of saying that there is a presence of complete charge around the atoms but the charges keep on changing?
 What does my professor mean by 'feeble' charges? Is the charge reduced or something else?

If the question is unclear, ask me to edit it in comments.

"For some time" should not be taken literally. It is just another awkward attempt to wrap our head around the quantum paradigm. In fact, the charges do not keep on changing.

@Ivan Neretin but a covalent bond experiences sharing. Doesn't it mean that electrons need to change the atom with which they are? And wont it affect charges?

Well, yes, some change does occur when the bond is formed, but not after that.

@Ivan Neretin Why is there no change later on if they  are literally 'sharing'? Also, can you please write an answer to the question please? My previous question also had no answer and only comments. Thanks

It seems you (quite wrong) assumed that an electron actually spends some time at some region of space. It is wrong. An electron can be detected at some point of space, yes, but in reality it doesn't work this way. It is much better to imagine an electron as a cloud with density following some weird and contr-intuitinve rules. Though it is also incorrect on fundamental level, electron gas approach does give a good basic idea. After all, DFT-LDA methods is surprisingly  adequate for such a simple idea.

We describe an object, which does not exist, but is very real, called and 'orbital'.  It is a region of space where an electron is 'likely' to be.  It is the result of charge field interactions between positively charged nuclei.  Normally we only discuss orbitals in the context of  1-2 Angstroms distance between atomic nuclei.  Some orbitals have a character which we call 'bonding'.  A bonding orbital, occupied by one or two electrons might be thought of as a 'bond'.  However, bonds also do not exist, even if they are very real.
It is somewhat nonsensical to describe where in an orbital the electron is.  If you must, the electron may be described as having a location of the entire orbital simultaneously, with vary weights to all of those locations, which are averaged.  This needs to be thought of in the concept of 'wavefunction', which is a mathematical function describing the distributions of these weights.  In the case of a polar covalent bond, the wavefunction more heavily favors locations near the more electron negative nucleus, and so the orbital 'concentrates' more of its charge near one end of the 'bond' and results in what we call a partial charge.
The reason we get squeamish about describing 'when' and 'where' is that the Heisenberg uncertainty principle renders the uncertainties energy and momentum of the system to be absurdly large when dealing with systems that are roughly the size of a chemical 'bond'.
At this point, nothing should be cleared up with regard to the original question.
The reality of the situation is that using when and where in this context is very likely to lead to predictions that will lead one astray when predicting physical phenomena.  As bizarre as quantum mechanics is, it has been shown to be highly predictive of the bizarre world of atoms and molecules, and so you have to try to figure out how to not ask questions from a classical  physics framework.

What do we actually mean by describing the charge as partial or feeble? Does it mean that the charge is reduced?

A partial charge means that the physical interactions are described as if portions of the molecule bear less than one elementary charge unit.  The actual magnitude is somewhat difficult to measure (and frequently dynamically changing due to molecular motion), but their existence is used to describe non-covalent bonding interactions between molecules such as London forces and Hydrogen Bonding.

The study of chemistry requires one to build up every more complicated models to explain the behavior of the atoms. Dividing bonding into "pure" ionic bonds and "pure" covalent bonds is one such model that is wrong. No bonds are purely ionic or purely covalent. The bonds are always something in between.
The problem with the bonds in $\ce{H2O}$ being purely covalent is that water has a dipole moment. The way to explain the dipole is to note that the bonds are not purely covalent but have some ionic character too. Thus the oxygen has a partial negative charge but not a full -1 charge. The two hydrogen atoms must then each have a small positive charge. Now with partial charges the water molecule can have a dipole moment.
So the gist is that ionic bonds and covalent bonds are labels that we put on molecules to predict behavior. Such models are useful in limited circumstances. But if we try to overextend the model it leads to wrongful conclusions.


text:
can metamers contain N as differing functional group?
Are $\ce{Ph - NH - CH3}$ and $\ce{Ph - CH2 - NH3}$ metamers?

This term is so old, I'd never even heard of it, had to look it up. Hooray for wikipedia
https://en.wikipedia.org/wiki/Metamerism
. As i understand it, this applies to the two compounds you give.
The proper modern term is "isomer", a substance having the same molecular formula as a second one, but in different structural arrangement. (And yes, the two are isomers, although you would only rarely come in a situation where you want to discuss them as such, because they are chemically very different.)


text:
Calculate the flow rate in the steady state of a dynamic equilibrium
I have the following situation:
I am interested of the reactions rate of an enzymatic reaction, depending of the substrate concentration. The thing is, that I don't add the substrate directly to the enzymes; the substrate has first to pass (I think diffuse) through a membrane.
I assume an infinite amount of substrate (outside of the membrane), so it won't run out. Further I assume that the product of the enzymatic reaction is eliminated immediately.
$Substrate_{outside} \leftrightarrow Substrate_{inside} \rightarrow Product $
The first reaction ($\leftrightarrow$) is k1 and the second ($\rightarrow$) is k2 (I didn't figure out how to name the arrows, sorry)
k1 is the diffusion trough the membrane which I think should be proportional (fick's law) to the difference of $Substrate_{outside} - Substrate_{inside}$, so it is:
$k1 = D * Substrate_{outside} - Substrate_{inside}$, where $D$ is some constant.
k2 is Michaelis-Menten:
$k2 = \frac{V_{max} * Substrate_{inside}}{K_m +Substrate_{inside}}$
when I solve eq. 1 to $Substrate_{inside}$ and insert it to eq. 2, I got :
$k2 = \frac{V_{max} * (D *Substrate_{outside} -- k1)}{K_m + D *Substrate_{outside} -- k1}$
in the steady state is k1 = k2 so:
$k = \frac{V_{max} * (D *Substrate_{outside} -- k)}{K_m + D *Substrate_{outside} -- k}$
but if I solve it, I get some strange function, which does not looks like I think it should look.
In short words: I am looking for $k(Substrate_{outside},D,K_m,V_{max})$, whereby k is the speed of the reaction.
I am not really into this topic, so sorry if I was unclear or not 100 % correct. Thanks for your help.


text:
1,2-dibromopropane reaction with OH and NaNH2
We have the reaction of the alkyl halide $\ce{CH3-CHBr-CH2Br}$ with 1) $\ce{OH-}$ 2) $\ce{NaNH2}$ The answer is propyne.
I am unable to understand why we have performed an elimination reaction here. Why can't it be a substitution reaction instead? Also, how do we decide which two carbon atoms would lose their hydrogen atoms? (as in elimination, there are two choices for losing the $\ce{H}$ atom - from left and right carbon respectively)

Related http://chemistry.stackexchange.com/q/41888/16683

You may get some substitution, but the dominant products are likely those formed from elimination reactions.

Substitution, meaning displacement reactions are Sn1 or Sn2. These reactions are dependent on the leaving group, nucleophilicity vs base of attacking group, and solvent. Currently you have two good leaving groups (Br) and strong bases. Strong bases (have localized electron density, no resonance etc.) favor proton attack due to partial + charge on H, no spatial constraints. So the protons are attacked preferentially. It would be much easier sterically and kinetically compared to either Sn2 (recall l, these also requires specific orientation of attack) or Sn1. These reactions will occur but at much much small amounts, such that on time of reaction they are minimal.
The reason for the -yne product is because there are two adjacent leaving groups. Once the first comes off (which is aided by partial resonance by the other Br), the electron densities around the double bond, the large radii of Br, along with the positive charge character on Br due to resonance, it easier to remove the next Br, but this requires a strong base. The amine is definitely strong enough, but so is the OH ( less so due to higher electronegativity of Oxygen which doesn't want to share its electrons so to speak )
Because you end up with a -yne it doesn't matter which Br is removed first, regarding the final regiostereochemistry.