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Liquifying metal and toxic fume exposure
I just read a very entertaining Youtube comment thread on a video where TheBackyardScientist melts tin on the stove in his (enclosed) kitchen, the proceeds to cook a steak on it.
According to the mass of commentaries (and votes), what he did was idiotic due to the toxicity of the fumes released when melting the tin, and the fact that he was in an enclosed space without a respirator.
Reading about tin and organotin compounds a bit, nothing is mentioned about them being 'released' upon melting, but rather that they are a particular subset of compounds of tin, and some happen to be toxic.
Regarding 'metal fume fever' and its causes, it seems that it is a small subset of metals that when oxidized could cause nervous system damage, in which tin does not take part.
Aside from burning yourself horrendously, what dangers are associated with the melting of (we can assume nearly pure) tin, particularly in a closed space? Is what he did really that dangerous?

None of the organic compounds of tin would survive till the melting point of the metal. As for the rest, well, when compared to other heavy metals, tin is not all that dangerous.

The "pure" tin probably had more than a bit of lead. Such a stunt in a kitchen in which you're going to eat is just absolutely stupid.

Copper pots, kettles and food cans used to be TINNED because it was a relatively inert metal.  As a metal there is little risk, metal fume fever is usually transient but can be very bad from zinc, not sure of tin.  Tin dust would be the result of grinding or other ways of powdering it, seems not to be a significant issue here.  Fearmongering is not a word but it is a modern trolling sport.

You ask:

Aside from burning yourself horrendously, what dangers are associated with the melting of (we can assume nearly pure) tin, particularly in a closed space? Is what he did really dangerous?

According to the SDS for tin metal from Sigma-Aldrich, tin metal (at the purity level for the product mentioned here, which is <= 100% and may not be that of what was used in the video you mentioned) is not a hazardous substance or mixture.
There is some danger specifically relating to inhalation of tin dust or aerosols (see sections 6.1 and 7.1 in the linked document):

6.1 Personal precautions, protective equipment and emergency procedures
  Use personal protective equipment. Avoid dust formation. Avoid breathing vapours, mist or gas. Avoid breathing dust.
7.1 Precautions for safe handling
  Avoid formation of dust and aerosols. Further processing of solid materials may result in the formation of combustible dusts. The potential for combustible dust formation should be taken into consideration before additional processing occurs. Provide appropriate exhaust ventilation at places where dust is formed.

In section 8.2 (Exposure controls) it is noted that:

Respiratory protection
  Respiratory protection is not required. Where protection from nuisance levels of dusts are desired, use type N95 (US) or type P1 (EN 143) dust masks. Use respirators and components tested and approved under appropriate government standards such as NIOSH (US) or CEN (EU).

In section 11.1 (Information on toxicological effects, a subsection of Toxicological Information), information pertaining to toxicity can be found. Inhalation here will refer to tin dust or aerosols that can be inhaled:

Acute toxicity
LD50 Oral - Rat - male and female - > 2,000 mg/kg (OECD Test Guideline 423)
LC50 Inhalation - Rat - male and female - 4 h - > 4.75 mg/l (OECD Test Guideline 403)
Dermal: No data available

If the tin was/is at the near 100% purity level, the main risks associated with handling it are associated with the formation of dust (workplace inhalation limits are given in section 8). The HMIS (Hazardous Materials Identification System) ratings in section 16 are 1 for health hazard, 0 for flammability and physical hazard, and no value assigned for chronic health hazard.
Conclusion: Assuming the person in the video did not inhale or ingest tin dust (a proposition that largely becomes moot after it has melted, in any case), there appears to be little to no danger associated with the activity with respect to melting tin in a closed space (especially such a small amount of it).

:-( First I doubt that some knuckle head who wanted to try cooking with molten tin is going to spend hundreds of dollars for high purity tin. Second cooking with it is different than just melting it. Once you melt it is going to oxidize some and the oxide will form a skin on the surface. I'd be very worried that the oxide would concentrate the heavy metal impurities in the tin. So all in all cooking with molten tin is a stupid stunt that shouldn't be tried.

@Maxw tin is a bit more active than lead, and, more importantly, tin is widely used to cover tin cans. Thus, lead-free tin is not an issue. The stunt is no more stupid than consumption of $\ce{K4[Fe(CN)6]}$

I did XRF for years. I understand that "pure" tin used in such a stunt could be safe, but I doubt the ability of most to understand what "pure" tin is.  Very few people would have the ability to adequately test tin for impurities. Buying high quality tin (reagent grade) is very expensive. Cooking a steak with molten tin is just a stupid stunt that should not be repeated. Downplaying the risk is a disservice to the idiots who might be inclined to repeat the stunt.

I have re-tinned some old copper cookware; no fumes, but it is not easy to get a mirror surface. I believe all old copper cookware is tin coated (essentially pure tin).
"Tin" cans have less than a 0.001 in. coating of tin on the outer surface for corrosion protection; various organic coating are used on the interior. For salvage ,the tin is removed chemically.
Zinc is another story. It is impressive to see a "zinc boil" in brass overheated in a foundry ($\ce{Zn}$ vapor pressure > atmospheric).  I have had "zinc chills", you get over it in a day. But I do recommend avoiding $\ce{Zn}$ vapor (ie zinc oxide).


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Kinetic Isotopic Effect in EAS
It is observed that rate of reaction of two isotopologues of benzene, $\ce{C6D6}$ and $\ce{C6H6}$, is the same for electrophilic aromatic reactions, as the deprotonation is not the rate limiting step.  Instead the carbocation is usually limiting.
Do iodination and sulphonation somehow do show different rates with these isotopologues?  Is $\ce{C6H6}$ the faster of the two, due to stronger C-D bond (smaller size of D)? If this is correct what is the reason?
Does the protonation becomes the rate limiting step? if yes, then why in only with iodination and sulphonation?
Update : Found potential energy graphs for suphonation and nitration and it does seem like protonation either becomes the rate determining step or just smaller than carbocation hill, but why in sulphonation, the deprotonation curve was high is still a mystery.

The slow step is electrophilic attack. Deprotonation after electrophilic attack is likely not the slow stop in any of those cases.

-----Update----

In general, deprotonation of a $\ce{C-D}$ bond would require more energy than that for $\ce{C-H}$ bond because the zero point vibrational energy of the bond is lowered. By referring to our harmonic oscillator zero point energy equation,
$E_0 = \frac{1}{2}h \nu$
where frequency is,
$\nu = \frac{1}{2 \pi} \sqrt{\frac{k}{\mu}}$
and the reduced mass ($\mu$) being,
$\mu = \frac{m_1m_2}{m_1 + m_2}$
we see that as the mass for either atom increases, the zero point energy decreases, therefore, the bond is relatively more stable than its lighter isotopologue. However, swapping out isotopes in a molecule does not change the potential energy surface for the reaction [1]
1.) Sulphonation
Changes in the rate determining step may arise from the type of sulfuric acid used. Since fuming sulfuric acid has a higher concentration of $\ce{SO3}$ (the electrophile) then more dilute forms of $\ce{H2SO4}$, the rate limiting step is no longer forming the bond between benzene and $\ce{SO3}$, rather, it is re-forming the aromatic ring through deprotonation. Ergo, if you did a KIE study with different concentrations of sulfuric acid, you may see that only fuming $\ce{H2SO4}$ shows significant values of $\frac{K_\ce{H}}{K_\ce{D}}$.
2.) Iodination
This article [2] studied the iodination of benzene with the source of iodine being generated through the "oxidation of 0.5 g of $\ce{I2}$ in acetonitrile-tetraethylammonium perchlorate to 2.4 faradays/mol."

The article reported that the secondary kinetic isotope effect for $k_1$ and $k_{-1}$ is negligible, however, $\frac{K_\ce{H}}{K_\ce{D}} = 2.25$ for $k_2$.
Here is a thesis I found that discusses the iodination of aromatic rings [3]
The thesis, in short, suggests that whether the deprotonation is rate limiting has to do with the concentration of the reactants, along with the transition state of the molecule. The transition state is affected by the strength of the base, since a higher $\mathrm{p}K_\mathrm{b}$ may give a transition state where the hydrogen is partially bound to both the ring and the base, making this step in the reaction faster (therefore, not rate limiting). However, it did only investigate substituted aromatic rings, where substituents can play a significant role in the transition state, making the reaction with benzene a little less comparable.
3.) Nitration
This article reports no KIE observed in nitration of benzene [4].

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Volume of sucrose (solute) contributing to volume of solution? (solvent = water)
I'm trying to make stock solution of 3M sucrose in water.  In 25ml, this is 25.672g sucrose (FW 342.3).  However, that mass of sucrose, in the powder, occupies a pretty substantial volume, such that if I add 25ml of water to it, I end up with a dissolved solution ~42ml in volume (it is fully dissolved).  Puzzled by this, I've also prepared a second solution in which I am only adding water up to the 25ml line in my 50ml conical (so far only 9ml).  This is obviously struggling to dissolve as the saturation solubility for sucrose is like 5M or something like that.
So what's the deal?  Is my 40ml solution with the correct mass of solute and volume of solvent really 3M?  My head says yes but my gut is uneasy.

I've never made that particular 3.0 molar solution, but the correct method as you noted is to dissolve the 25.6 grams of sucrose in some water, and then dilute the whole solution to a final volume of 25.0 ml.

The English wikipedia page for sucrose notes a solubility of 2000 g/L at 25 degrees Celsius, corresponding to an approximately 5.8 M solution. 3 M shouldn't be a problem in that respect, but to make dissolving easier you could warm up your solution slightly if that is appropriate in your protocol.
Now, molar is a concentration equivalent to mole per liter and this volume is the total volume including solvent and the dissolved material. So 25 ml of a 3 M solution of anything, say x, should have a final volume of 25 ml, including the solvent and 75 millimole of x.
This means your second approach is correct (being 25.6725 g of sucrose, to which you add water until the total volume is 25 ml).

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How to calculate the volume of oxygen in a reaction between hydrogen peroxide and bleach using this apparatus
I know that the reaction between bleach and hydrogen peroxide produces oxygen but using this apparatus and information, I need to state how to calculate the volume of oxygen released:

At the start of the experiment, the measuring cylinder is completely filled with water. Hydrogen peroxide solution $(\pu{20 cm3})$ is added to the flask using the syringe. When the bubling stops, the volume of gas collected in the measuring cylinder is recorded.
How would the volume of oxygen given off be calculated?

From my initial thoughts, I thought that the volume of oxygen will simply be the volume of the gas collected, however, the answer is said to be the total volume of gas collected subtract $\pu{20 cm^3}$. I fail to understand this as I believe that the $\pu{20 cm^3}$ of the hydrogen peroxide reacts with the bleach completely and even if not, will still only produce oxygen and not any other gas, making the use of subtraction unnecessary and wrong for our answer.

This is how it works in my head:
If instead of peroxide in the syringe, you just had $\pu{20cm^{3}}$ of air, then injecting the air will cause a reading of $\pu{20cm^{3}}$ to happen.  The water gets pushed out of the way.
Injecting peroxide would do the same thing (approximately).  After the reaction, there is still saltwater and leftover reactants taking up space.  The oxygen produced will take up additional space (and push the water away) because it is a gas and it wants to expand.

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How does an air conditioner produce liquid
I always notice that when an air condition is working and cooling the room it, is heating outside the room and a Liquid is dropping from the other side of the air conditioner. how does this happen and is it drinkable?, what is the name, is it water(H2O)?

I'm voting to close this question as off-topic because (a) the answer is easily found using an internet search engine; and (b) I'm not comfortable advising anyone on the consumption of water condensate from an air conditioning unit.

Hmmm... I've very comfortable advising not to drink the condensate. The condensate is not at all like distilled water.  Quite possibly lead solder on the coils plus heaven knows what sort of biological gunk (bacteria, mold, algae...) has grown from the air pulled through the coils.

I'm voting to close this question as off-topic because it is not about chemistry.

Just because the condensate is contaminated post-condensation doesn't mean it is isn't being distilled.

At the risk of answering a question that may be better suited in Physics.SE, yes, it's water. It's condensation from the evaporation coil, which is where the refrigerant expands in order to cool the air drawn through the unit. That is how an AC unit also dehumidifies the air.
Chemically speaking, distilled water is drinkable, but it isn't great to drink. It can sap electrolytes from your body over time. From a unit like that, however, it may be contaminated with oils, rust, dirt, etc.


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Alternative to a proton
I know I might sound terribly stupid but the idea just hit me.
We define elements on the periodic table as "lumps" of matter with specific amount of specific particles (electrons neutrons and protons)
We use this table because every different "lump" or atom that has a specific amount of protons has a set of characteristics that is shared among all instances of atoms that have the same number of protons.
Now what if there is,or could be, another form of matter "lumps" that instead of protons have another type of particle (lets say deuterons),that could also give have a set of characteristics to each instance of those lumps with the same amount of "deuterons" , essentially creating another types of "elements"
Im not talking about antimatter here although that could be an answer and if someone has to point me into resources regarding if and what "anti-elements" there are out there I would be glad for that.
TL;DR : different type of elements ,using another particle instead of a proton

check out the exotic atoms: https://en.wikipedia.org/wiki/Exotic_atom  I think this is what you mean (using e.g muons to create atoms).

Your example with deuterons is unfortunate as they are composed of protons and neutrons and, as such, formally putting deuterons together would just reconstruct the periodic table we know. However, there are atoms with particles other than protons, neutrons and electrons. Two relevant terms to refer to such are exotic atoms and hypernuclei. One of their main properties is that they tend to be extremely unstable, so they have essentially no use in chemistry.
That said, there has been a relatively recent study on muonic helium, which is slightly different from a regular helium atom (two protons and two electrons) in that one of the electrons is replaced with an almost identical subatomic particle, the muon. It behaves like the electron in all aspects, except that it is about 200 times heavier, and decays within microseconds. During its brief existence, the muonic helium atom was observed to behave like a regular hydrogen atom. This is because muons and electrons are distinct subatomic particles and so Pauli's exclusion principle does not apply. The electronic configuration for muonic helium would be $\mathrm{(1s)^1}$ (like hydrogen), and its "muonic" configuration would also be $\mathrm{(1s)^1}$.
Furthermore, the much larger mass of the muon means it spends its time much closer to the nucleus than an electron of similar energy. The electron is so much further out, in fact, that it "sees" the nucleus and the muon as occupying almost the same position, and so the negative charge of the muon almost cancels out the positive charge of one of the protons (technically, the muon provides almost perfect electronic shielding to the valence electron). The lone electron then behaves almost as if it were attached to a single proton, i.e. a hydrogen atom.
So yes, exotic atoms can be qualitatively different from the elements we know, but their investigation and use is certainly limited by their instability.


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Low melting point of Manganeese
Melting point of manganese is lower than that of other 3d series except copper and zinc.
Melting point increases from scandium to chromium. The reason given in my book for this trend is the increase in number of unpaired d electrons which increases interatomic forces on account of covalent bonding.
But if thats the case, shouldnt manganeese have the highest melting point?

see this:-http://chemistry.stackexchange.com/questions/4766/melting-and-boiling-points-of-transition-elements

http://chemistry.stackexchange.com/questions/26534/why-is-enthalpy-of-atomization-of-mn-exceptionally-low


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Why does the activation enthalpy of ester hydrolysis increase with an increase in alcohol-chain length?
(Particularly concerning acid hydrolysis) Is there some sort of chemical reason as to why with increasing alcohol-chain length, there is also an increase in the minimum energy needed to cause the ester bond to break?
Does the molecule as a whole become more stable because of the large -R group of the alcohol section of the molecule, for example?
(I am thinking of esters like methyl and ethyl ethanoate).

You mean like CH3CO(OCH3) and C2H5CO(OCH3)?


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Why does the ink on my whiteboard move?
...or at least I thought so, when I saw that the ink that was on my whiteboard last night became really messed up the next day.
The image below shows my whiteboard before (on the left) and after (on the right). The time span was one night.

As you can see, the words became really squiggly, and they mostly travelled downwards.
When I woke up, a few hours before I noticed my whiteboard mess, I noticed lots of condensation water on my whiteboard due to my air-conditioner (my whiteboard is made of glass, if it matters).
My theory is that the water pushed the oil surfactant* of the marker ink mostly downwards. The oil surfactant in the ink coats the board to ensure easy erasing, so the oil should be the closest chemical to the board. Since oil and water are immiscible, the hydrophobic oil ended up being pushed away by the sliding condensed water (sliding due to gravity) and the ink travelled down.
However, my theory does not explain why the ink travelled upwards in some parts of the board. I have not shown it here due to sensitive information.

Is my theory lacking? What do you guys think about this phenomenon? And no, it wasn't a prank.
Can the ink travelling in directions other than downwards be explained?

*Regarding ingredients, all alcohol-based dry-erase marker inks contain 4
  basic ingredients:

Release agent- an oily (can be simply a mineral oil) surfactant or co-
  solvent which will coat the board to provide easy erase barrier (such that
  the colorant slides off of the board surface and does not penetrate or
  stain over time)

A nice thing to do , would be to watch it or record it overnight so you can actually see what is moving where and if you can find some pattern in those movements. If it doesn't happen when you're watching, then probably you should call Sam and Dean.

Draw a straight line, spray it with water and see what happens.

Start adding salt to your dry erase ink.

Is an air duct nearby?

This is a very unformed idea, so I'm hesitant to write it up as an answer (feel free to run with it if you like it, anyone).  What if the movement wasn't due to condensed water sliding around due to gravity, but instead variations in which parts of the board were slightly moistened varying with respect to time.  Downward motion would be favored (b/c of gravity) but you could still have patches of moisture develop above your writing that pulled the letters upward.  In any case, this is a really cool phenomenon, thanks for sharing and I hope you get your answer.

This is a wild guess, so I am open to criticism.
Brownian motion could explain this phenomena.

Water was on the board, so it is possible a film of water condensed on the board some time in the night.  This means the humidity was high while the board was still relatively cool. This film may have allowed the oil particles to suspend.
The ink bound tighter to itself than the non-porous glass board, and the ink itself may be buoyant in water.
The ink binding to itself explains why the letters stayed semi-intact instead of diffusing into a mess.
The water molecules are moving constantly which would allow the bigger particles, which in this case would be the letters, to do the random walk characteristic of Brownian motion.
(As a note the random walk characteristic of Brownian motion tends in the direction of gravity.)
The diffusion coefficient of the letters would be temperature dependent, so the temperature in the room may have briefly spiked just warm enough for the oil to diffuse some then it came back down so the ink did not run completely off the board.

Brownian motion does not apply to macroscopic objects of the size of the letters. Even a particle with a radius of 0.001 mm would move less than 1 mm under the above conditions. See A. Einstein, Ann. Phys. 322, Nr. 8, 1905, p. 559 (http://www.physik.uni-augsburg.de/annalen/history/einstein-papers/1905_17_549-560.pdf).

I remember Brownian drift entering into certain diffusion calculations, though the timescale was long.  You may be right though, it may be much easier to argue some form of covective drift as opposed to a Brownian drift.

i guess this movement is slightly affected by the motion of the air in your room.
or else it is possible that as long as your ink is wet it will slip down ,but as it is very slippery the gravity dosen't affect the ink that much and so the ink travels upward
.p.s. it's my wild guess


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Why doesn't H-O-O-O-O-O-O-O-O-H exist?
The question has its roots in why only carbon mainly shows catenation. I don't see why such a compound cannot be formed. Why don't we have a whole branch dealing with such compounds other than organic chemistry? If this compound can exist then such atoms will also be able to form a large number of compounds like the organic compounds.

Look at C-C and O-O bond strengths - and compare the O-O bond strength to that of O=O. Same goes for nitrogen (compare N-N single with N-N triple bond). Why can't you just have graphite-like layers of nitrogen?

Computationally it has been predicted that arbitrarily large $\ce{H2O_{x}}$ molecules are technically stable, though after more than a few they are presumably quite shortlived.

@Nicolau Saker Neto so you are saying that what sets carbon apart is its tetravalency and not catenation?

@NicolauSakerNeto Derek Lowe posted on peroxide peroxides. Seems the five-oxygen species is about as long a chain as you're likely to get, even if only fleetingly.

There are other elements which can catenate (boron and sulfur come to mind, iodine and mercury too in certain conditions), but carbon is pretty unique. I don't think any other element is expected to have a similar amount of covalent molecular chemistry

@NicolauSakerNeto If one is willing to allow chains composed of multiple types of atoms, one can do decently well with $\ce{Si}$ and $\ce{O}$ -- e.g., the siloxanes. (Though, even these generally use nicely stable organic carbon side/cap groups, as in polydimethylsiloxane.)

@Brian Ah yes, some element combinations can go quite far. Silicon and oxygen are definitely a good match with rich chemistry. Sulfur and nitrogen also catenate well. Then there's transition metal polyoxometalates. Surely many other examples. Here's a Wikipedia link which is quite relevant

A related, and surely under-upvoted, topic is here, dealing with the low strength of the peroxo bond. As well, this question addresses the thermochemistry of $\ce{H2O2}$ decomposition directly.

O is quite small. And it has lone pair and it has more tendency to go in -2 than in -1 state. It is like the second most electronegative element. Why?You know it better than me. Why do you think  such a compound would be stable? If it were we would already be studying about them. C on the other hand, has the best suitable properties for catenation.


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Fragmentations and rearrangements in n-(4-alkyl phenyl) alkanoic acids in mass spectrometry
I have this example exam question for which I don't have the answer. I have been working on it the whole morning but I can't figure it out. The question suggests three molecules, only one of which corresponds to a given fragmentation table and mass spectrum following EI-based MS.
The molecules are 4-(p-hexyl phenyl) butyric acid:

and two variants where the phenyl is shifted one to the left (pentylphenyl pentanoic acid) or one to the right (heptylphenyl propanoic acid).
The table and EI-MS spectrum are unfortunately not of ideal quality, but they are as follows:

The 177 and 188 were given to both be directly stemming from the molecular ion, and with one of them having an even and the other an odd mass, this suggests a simple cleavage and a cleavage following rearrangement.
With pentylphenyl pentanoic acid I got close, but the fragment following rearrangement is 192 Da instead of the 188 Da seen in the spectrum.

UPDATE - Let me add two other examples of what I have tried so far. Maybe there is something wrong with my approach or someone can see the obvious change that would get to an observed peak at 188 m/z from here. The fragmentation on the left (below), is where there actually are 188 and 60 fragments, but the 60 fragment carries the charge (this peak at 60 m/z can be seen in the spectrum). On the right, I tried to get to a similar fragmentation while starting by putting the charge on the phenyl.

On an unrelated note, please always upload images to the SE imgur using the create image button in the editor. (You can enter a link from the web there.) I know it is sometimes a pain, but it preserves the images in case their original source (here: a probably personal Dropbox, unlikely to withstand the test of time) gets deleted. We want questions and answers to be helpful to future visitors, that’s why =)

Presumably this is electron ionization (EI)-based mass spectrometry?  You may want to edit the question to add this detail.

The question doesn't specify it, but I suppose you are correct. I edited the question. May I ask how you recognised it, just for my learning process :)

In your update you nearly got to the point: the McLafferty rearrangement from the carboxylic acid can lead to two products, either the ionized acetic acid (ML product) or the ionized alkene (ML'). In this case, ionization energy of the enol tautomer of acetic acid is probably much larger than that of alkenyl substituted benzenes (for instance the closest I could find in structure is o-Allyl-Toluene for which the IE is 7.78 eV). Thus from application of the Stevenson-Audier rule, the charge should remain on the moiety with the lowest ionization energy, leading to the formation of the m/z 188 moiety (ionized p-pentyl-allylbenzene).
This hypothesis is further reinforced as it also explains the formation of m/z 117 from a oss of 71 ($C_5H_{11}^\bullet$) from fragmentation of the m/z 188 radical cation, through a pathway similar to that leading to m/z 177 from the molecular ion.
(This is likely oversimplified mechanism interpretation: I strongly suspect that some other rearrangements or cyclisation likely occur in the process. I did not run a complete database search for other EI spectra of similar molecules, but for instance 10-phenyldecanoic acid leads to only a very minor loss of 60. Thus I would suggest that the m/z 188 probably undergoes cyclization and / or hydrogen migration processes leading other products.)

I didn't think of inductive cleavage after the rearrangement (rather than the alpha cleavage I drew) but clearly, if the ionisation energy is lower it is the favoured option. All clear :)


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How to recycle 60 tons/hour of CaCO3?
I hope this is the right place to ask. I'm not a chemist so my question may seem obvious but it's not for me.
I work in the egg industry and we produce every hour 60 tons of dried egg shells as waste.
Those shells come in small grains (few millimeters) but can be further reduced to powder (like talco or marble powder).
This powder is $97~\%\ \ce{CaCO3}$ and $3~\%$ proteins.
What I'm interested in is an idea to recycle this product in a cost-effective way. At the moment a small part is used in cultivations but the rest is just stored in abandoned mines and I don't like the environmental impact that this solution has.

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. This question essentially boils down to how can we use calcium carbonate in my opinion, since the proteins are easily removed from ground egg shell.

@Jan no, it does not since the protein is a burden in most $\ce{CaCO3}$ applications

@permeakra Naively I’m going to suggest heating it to a few hundred degrees burning the proteins to smithereens and ending up with $\ce{CaO}$ … but I could be underthinking it.

@permeakra Thank you for the edit, I wasn't sure about the tags and also I've misspelled CaCO3..

@Jan Heating is a problem if you consider the cost of the energy required to heat all this powder.

@permeakra You are right about the protein, they are a burden for most applications. They are also the cause of the smell of this stuff and that alone limits the applications.

Out of the blue I can offer cement and lime production, but I'm unsure if organics is OK there. Both processes include heating to significant temperature, but in reducing atmosphere it won't help and I'm not aware of details. Both, however, are large-scale processes, 60 ton/hour may be too little for them. $\ce{CaCO3}$ is also used in metallurgy, again, unsure about organics and scale there.

To separate the proteins, why not NaOH to cleave all the amides and leave the salts in an aqueous phase, separable from the majority of the calcium carbonate.  Only if the proteins are an issue of course.

@Beerhunter thank you for the hint. The proteins are indeed a problem. Next week I'll meet with someone that have a better understand of the process involved than me and I'll ask him to test your idea.

Actually, one more idea is calcium carbide production. Technical grade calcium carbide is rather dirty and organics is not a problem here for sure.

Actually, finely ground calcium carbonate is one of the major fillers used in papers because they are cheaper than wood fiber.  Nowadays, paper contains about 10-20% CaCO3 as filler. And I don't think presence of protein will be a problem here.

60 tons on $\ce{CaCO3}$ per hour means half a million tons per year. That's quite some calcium carbonate.
Hens need $\ce{CaCO3}$ to produce eggs. So the most natural idea to recycle it is to produce feed lime.
Or consider producing agricultural lime. The annual consumption of agricultural lime is about 2.5 million tons in Germany, and about 25 million tons in Brazil. Your production is in this scale. The presence of 3% proteins is probably of little concern if not beneficial.

Feed lime - To avoid disease problems with feed it would seem that you'd have to lime the eggshells then convert back to calcium carbonate. That is just too complex when you could buy calcium carbonate to start with.

Agricultural lime - So much is used that the egg shells would be a very small source of calcium carbonate (assuming that the product is fine enough and dry enough to be mixed directly...). So the plant that makes agricultural lime would be near a calcium carbonate mine. So the problem would at least be the shipping costs between the two plant locations.

Thank you @aventurin. I accept your answer because in the end we are selling it 20% as feed lime and the rest of it as agricultural lime.


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What is the difference between carbon and charcoal filters?
I'm confused about the difference between carbon and charcoal filters.
Ultimately, charcoal contains carbon but I have come across an exam question that asks for their individual roles in a water filter.
I have done some research and found this page that states that 'activated carbon' and 'activated charcoal' can be used interchangeably.
Would any of you be so kind as to help me decipher the individual role of carbon filter and a charcoal filter in an ideal water filter?

Activated carbon is usually of higher quality than activated charcoal, but the words are often used interchangably.

First of all and I want to clarify something that it often misunderstood, which is that charcoal has carbon but it is NOT carbon.  Charcoal actually have a bit of residual Hydrogen and oxygen in it and is again NOT carbon.
Carbon is a pure element that occurs naturally in a variety of allotropes.  Charcoal is the charred product from pyrolyzing wood.  Carbon is found as graphite and diamond and has no intrinsic filtering properties.  Charcoal is a porous material that can burn or adsorb pollutants. Activated charcoal is charcoal which has been specially processed to have a very high surface area.  This surface is what pollutants adsorb (not a(b)sorb) to to be remove from air and water.  If you said activated charcoal or activated carbon anyone would know what you were talking about, but activated charcoal is more proper as the material is in fact charcoal, NOT carbon.

Carbon filters use activated carbon, which is a highly porous material with a large surface area. This makes it very effective at adsorbing impurities, such as chlorine, volatile organic compounds (VOCs), and heavy metals.
Charcoal filters use charcoal, which is a type of carbon that has been burned in the absence of oxygen. Charcoal is also porous, but it has a smaller surface area than activated carbon. This makes it less effective at adsorbing impurities.
Another difference between carbon and charcoal filters is that carbon filters are typically more expensive than charcoal filters. However, carbon filters are also more effective at removing impurities.

Property
Carbon filter
Charcoal filter

Material
Activated carbon
Charcoal

Surface area
High
Low

Effectiveness
More effective
Less effective

Cost
More expensive
Less expensive


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aluminum hydroxide as an electrolyte
A friend and I recently produced some aluminum hydroxide, $\ce{Al(OH)3}$ for a project, and we wondered how it functions as an electrolyte. We know it has a low conductivity and solubility, but we wanted to know for knowledge's sake. We put some aluminum hydroxide in a bowl and used a power supply running at 31.9 volts (The resistance was high enough for the current to be immeasurable). We also used aluminum electrodes (aluminum foil).
The electrodes were bubbling and a white precipitate was forming on the top, and it seems we have less hydroxide than we started with...
So what exactly did we do?
$%edit$

When you are applying 31.9V across I am assuming cm-ish distances, you are definitely causing a lot of electrochemistry to happen in the solution.
You are oxidizing one of your aluminum electrodes to $\ce {Al^3+}$, therefore precipitating $\ce{Al(OH)_3}$ and probably reducing water on the other electrode to evolve $\ce{H_2}$.Along with these, you are probably doing a lot of other reactions with any and every impurity around, but these two should dominate.
Measuring ionic conductivity properly requires doing an AC measurement since you cannot have a sustained DC current without doing electrochemistry of some sort. The electrochemistry on the electrode surface ends up the main factor influencing your measurement in any DC setup.


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Is sand a mixture or a compound?
I was given an assignment to find out if sand is a mixture or a compound I'm sure it can't be an element. Considering the fact that a mixture contains two or more substances that are not chemically combined and not proportionally fixed and a compound is contained substances that are chemical combined and are fixed proportionally. Sand contains a lot of divided small rocks and mineral particles.

Indeed you are correct. It is not an element. Now tell us about your thoughts on whether it is a mixture or a compound. Perhaps you could start by telling us your definition of mixtures and compounds and how this applies to sand.

Filter through this.

Sand is a mixture. Sand is classified as a heterogeneous mixture because it does not have the same properties, composition and appearance throughout the mixture. A homogeneous mixture has a uniform mix throughout.

I guess there are different kinds of sand.

This answer doesn't actually explain why sand is a mixture. It just states that it is.

Indeed there are different kinds of sand. White Sands National Monument has gypsum "sand" and the "black sand" on the Punaluu Black Sand Beach in  Hawaii is ground up volcanic rock.

The main component of sand is $\ce{SiO2}$, silicon dioxide. If you purified the sand, you could obtain a pure compound, but in general, natural deposits of sand have $\ce{SiO2}$ and a lot of other "stuff" (e.g. organic molecules, iron oxides, and even biological matter), which is called a mixture.


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How does salt concentration affect corrosion rate of iron?
I have conducted an experiment where I test the affect of $\ce{NaCl}$ concentration on corrosion rate of iron. I did this in an environment were all of the iron was submerged in water and the water was not aerated. The results I got are shown in the graph below.

Are there any research papers that could support or neglect my data? Can anyone explain this relationship? Is 1M just an outlier?

Run one more experiment, this time at 0.9 or maybe at 1.2. One data point can be a meaningful trend, or it can be an outlier, you can never tell. Well, if the experiment actually takes years to run, then things become tricky.

water was not aerated || in this case the corrosion is attributed solely to dissolved oxygen. Solubility of oxygen decreases with salt concentraion AFAIK, while chlorine ions promote corrosion of iron.

Are you completely sure that that point does not correspond to 0 mol/L? It would fit better there, if still a bit low. And it is an obvious point to include in the trials. If so, the interpretation would be that with a very small amount of salt the corrosion increases a lot, and with higher quantities, the corrosion still increases but not so fast.

100% sure that the point corresponds to 1 mol/L, as couple of trials were conducted

The involvement of water accounts for the fact that rusting occurs much more rapidly in moist conditions as compared to a dry environment such as a desert. Many other factors affect the rate of corrosion. For example the presence of salt greatly enhances the rusting of metals. This is due to the fact that the dissolved salt increases the conductivity of the aqueous solution formed at the surface of the metal and enhances the rate of electrochemical corrosion. This is one reason why iron or steel tend to corrode much more quickly when exposed to salt (such as that used to melt snow or ice on roads) or moist salty air near the ocean.
Water is the enabler of fast oxidation of iron so freshwater will also cause rust. However, salt water is a very good conductor (lots of dissociated ions) and so there are a number of electrolysis reactions that tremendously accelerate corrosion in salt water.
For example if you have iron in contact with salt water and also in contact with another metal such as aluminum (also in contact with the water) you effectively get a battery which drives very fast corrosion processes.
This effect can be reversed by using a metal (like zinc) which causes the current to be reversed and in effect the zinc corrodes rapidly, protecting the iron. This is the principle of a 'Cal-rod' which is used to slow the rusting of hot water tanks

Corrosion of iron requires oxygen which is a non-polar molecule.  If you add salt to water you increase the mixture's polarity making it more difficult for oxygen to dissolve in it.  This may contribute to the lack of corrosion in your sample.  I can tell you for certain though that this is NOT an outlier.  This observation is very common that corrosion is inhibited at higher salt concentrations.

The comments and other answers suggest that there is more than one variable affecting corrosion. Concentration of salt, aeration, buildup of corrosion product (stifling), oxygen solubility, stirring/liquid mobility...
A possible conclusion could be "Over the range from 0.2 to 0.8, the corrosion rate is affected proportionately by NaCl concentration..." and do another set of experiments over another range.
Note that the slope of corrosion rate does not intersect zero at zero concentration, indicating that the iron corrodes even without NaCl.


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How can I understand the Lewis acid-base interaction of iron(III) and water?
The interaction between $\ce{H+}$ and $\ce{NH3}$ as respectively a Lewis acid and a Lewis base is clear for me. $\ce{NH3}$ has a lone electron pair and $\ce{H+}$ has no electrons, or, saying politely, an incomplete duet, and can use nitrogen's lone pair to complete its duet, and a dative (coordinate) bond is formed in the process.
However when I think about $\ce{[Fe(H2O)6]^3+}$, I don't see where these six lone pairs of electrons (which are coming from water molecules) can go in the iron ion. The electron configuration of $\ce{Fe^3+}$ is: $\mathrm{1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^6\ 3d^5}$. It has five 3d electrons, and all (namely, 2) of its 4s electrons are gone. All five 3d electrons have the same spin, and they occupy five d orbitals. OK, I do remember that 3d orbitals are funny and in the transition metals 3d orbitals get divided into low and high energy level 3d orbitals (hence the color of transitional metal solutions).
But even with this, how do six water molecules position their lone pairs there? We have space for one pair of electrons (where 4s electron left) and then space for 5 more electrons, so in total, for 7 electrons. 6 molecules of water would give us at least six pairs of electrons, i.e. 12 electrons. So we are 5 places short to place all these electrons.
So the question is: how are they placed?

you're right that 3d can't be filled, now we need space for 12 electrons {all these are coordinate bond and each bond completely fills one orbital i.e. 2 electron donor} so, 4s complete, 4p complete and 4d's 2 orbitals are used hence the hybridisation sp3d2, if it was nh3 , it would force pairing {energy needed to pair after filling 3rd electron is less than energy gap created by strong field ligand between [dxy,dxz,dyz] and [dx2-y2,dz2] (pretty much everything stronger than H20 for 3d series and all ligands for 4d and 5d)} Fe would have 2 empty 3d ones, hence d2sp3 is hybridisation.

should i elaborate to an answer?

Thank you very much, ELiT! It makes sense for me, at least partially. Two questions: (1) where do I read about sp3d2 hybridization? do you know where I can find picture of such? (2) Do you know where I can find picture of Fe[H2O]6 +3  with this coordinate bonds graphically examplained?  Thank you again!

 here you go, posted an answer.

Coordination chemistry is that point in time where the simple main-group octet rule starts to loose its practicality. (In reality, not even main-group chemistry is as easy as it is often made in introductory courses, but the fact that d-orbitals of the same period are energetically too far removed to actively take part in bonding and that core orbitals are usually fully populated make things easier.)
The general principle is that electrons are represented by wavefunctions called orbitals that have an intrinsic energy value. These orbitals — being waves — can be mixed in an additive or subtractive way much like you can add or subtract sine waves to each other. When mixing orbitals, you always have one method which creates constructive intereference and one which creates destructive interference — the constructive interference will always have a lower energy than the original and the destructive one will always be higher. The caveat is, that the energy gain from constructive interference will always be less than the energy lost from destructive interference, so that mixing is only overall favourable if one of the resulting orbitals ends up empty (or filled by only one electron).
In the case of Lewis acids and bases, one side (the Lewis base) will always contain a pair of electrons in an orbital of relatively high energy (for filled orbitals) and the other side (the Lewis acid) will usually contain no electrons in an orbital of relatively low energy (for unfilled orbitals). These two can mix favourably: Lowering the populated ligand’s orbital and raising the unpopulated Lewis acid’s orbital creates an overall stabilisation of the system. Typically, for an octahedral complex like $\ce{[Fe(H2O)6]^3+}$, the overall picture will look something like this:

Figure 1: Orbital scheme of a basic octahedral complex. Image originally taken from Prof. Klüfers’ internet scriptum to his coordination chemistry course.
On the right-hand side in figure 1, you can see the six donating ligand orbitals that are each populated with an electron pair. The symbols underneath donate the orbitals’ symmetry point groups. On the left-hand side, you can see a central metal. The lowest-lying set of orbitals corresponds to 3d, the next to 4s and the final to 4p. You can see how mixing with these unoccupied orbitals gives the ligand orbitals an overall stabilisation. It is this that makes $\ce{[Fe(H2O)6]^3+}$ more stable than a ‘naked’ $\ce{Fe^3+}$ ion. The energy gained is largest for the octahedral geometry shown which is why exactly this arrangement is adopted.
There are higher-arching rules, too. For the broad area of metal-carbonyl complexes, one often finds arrangements that correspond to an electron count of 18 overall — and a stable number 18 is also true for many other complexes. 18, of course, corresponds to 10 d-electrons, 2 s-electrons and 6 p-electrons and thus a fully-populated ‘shell’. (Note that the d-electrons belong to the lower shell, but energetically they are all close together; much closer than the d-subshell of the same shell.) Therefore, $\ce{[Fe(CO)5]}$, pentacarbonyliron(0) is stable, iron having eight electrons and each carbonyl donating two more. Cobalt would need nine in addition to the nine it has in its ground state, so it assumes a dimeric structure of $\ce{[Co2(CO)8]}$. Nickel is satisfied with eight additional electrons, so $\ce{[Ni(CO)4]}$. However, you should never just look at the higher-arching rule and thereby decide whether a complex is more stable than another one or not.

So, effectively, the question is: How are electrons from the covalent bond between ligand and central metal atom or ion placed with regard to the electron configuration?
$\ce{Fe^{3+}}$ has the condensed electron configuration of [Ar]$3d^5$. When the ligands approach the iron ion from the X, Y, and Z, the ion forms six bonds. It would be incorrect to state that the remaining 3d electrons are involved in the bonding with the ligands. Instead, the iron ion uses 6 orbitals from the 4s, 4p, and 4d orbitals to accept the lone pairs of electrons from the ligands. However, before they are used, the orbitals are hybridized in order to create six orbitals of equal energy.
If you want to dig deeper into the subject, I highly recommend the lecture notes from this webpage here.

Thank you, Mattias! I read through ppt you recommended, it was very useful. However, there is something I am confused about. On page 7, bottom slide, it says that  Crystal Field Theory "Assumes ionic bonding between the metal and the ligand instead of covalent bonding". I started to address this topic from point of view of interaction between Lewis acid and Lewis base, which is, as far as I know, implemented via coordinate covalent bond. Point of view of ppt you recommended sort of removes Fe[H2O6]+3 from the area of discussion about interaction between Lewis acid and Lewis base.

It is curious that while Crystal Field Theory (CFT) describes the interactions between the metal ion and the ligand as electrostatic, the Ligand Field Theory (LFT) describes the bonding between the two parts as covalent (forming a dative bond). I have been researching a bit, and the Wikipedia article for [LFT] (https://en.wikipedia.org/wiki/Ligand_field_theory) states that LFT is used to describe octahedral complexes, such as the one in your example, while other complexes can be described with reference the the CFT.

I'm gonna give basic results of Crystal Field and related VBT, and I don't think the generalizations made on shape are good in that ppt like Ni+4 Co+3 etc..
Assumptions
1.Electrostatic interaction only between ligand's dipole or charge and positive charge of central metal ion
2.Ligands are point charges

(source: pgcc.edu)

Look up shape of an octahedral, x,y,z axes passing through vertices and central metal at its centre, ligands approach along this axis and increase the energy those orbitals which are along axes which are dx2-y2 and dz2 due to electronic repulsion

so, we have 3 "d" degenerate {equal energy} of one level of energy and two of a higher level, now FILL the orbitals {from start if you want to but just the d subshell is fine} by paulli,hund's rule . When there are 3 unpaired electron in lower energy level, its about to happen, if the energy gap between the levels {increases with strength of field of ligand, look up spectrochemical series for strength [chelating ligands have extra stability and need to be taken care of properly, their strength is more than specified in order as that order doesn't take ring formation entropy increase into consideration [if we made 2 ammonias into ethylene diammine, reaction's products are more so entropy increases and free energy becomes more negative]]} is more than energy needed to piss off hund (pair the electrons) then, pairing will start until the 3 orbitals have 6 electrons, now the next batch starts and start filling like you would normally do, when it's done, start to count orbitals left with no electrons with are lowest in energy and they (6 of them) hybridise to 6 equal energy orbital and as my comment said, each coordinate bond fills one orbital so, we needed 6 orbitals
things are a little complicated for tetrahedral and square planar with the energy gaps not really as simple as described above {to this day i can't convince myself of dz2 in square planar} so, just google it and learn it , and try to derive it by the same method as before
When we write d before, it means inner orbital d orbital is used ex- 3d for Fe+3
and when after one means 4d for Fe+3 i.e. the next shell's d

Always start with coordination number i.e. number of ligands
[Any one wants to make inner orbital complex as shorter and better bonds]

When it's
4 - Tetrahedral {sp3}[only if we don't get an empty d orbital left , even by hook or crook] or Square Planar {dsp2}
5(rare) -  Triagonal Bipyamidal {sp3d} and {dsp3}
6 - octahedral {sp3d2} or {d2sp3} with same logic, latter being preferred if possible
Other things include delocalized (resonance) coordinate bond and it will be specified on how many atoms electron cloud being donated is delocalized , colour [photon accepted and released's wavelength can be correlated withe energy gap between HOMO and LUMO and the electron gets excited to the next lowest energy orbital i.e. LUMO, and then dexcites , photon's energy =hv and can be used to measure crystal field splitting energy with is the energy gap we are discussing], magnetic moment = (n(n+2))^1/2 Bohr Magneton {it's a unit} where n is the number of unpaired electrons in the configuration etc, Also, that energy gap increases rapidly with oxidation number and size and charge density {size becomes smaller as we move down in d block due to lanthanoid/actinoid contraction} so, pretty much any ligand is strong for 4d and 5d metals and for 3d, anything after and including nh3 is strong and anything before and incuding H2O is weak except for Co+3, by strong and weak i mean gap is or is not big enough to cause pairing,
Extra :

If there is one unpaired electron and all ligands are strong, then transference will take place if inner orbital complex would be formed {meaning, the electron jumps to a higher level for the good of them team, and the emptied orbital can take part in bonding}

If there are 2 unpaired electrons and again, all ligands are strong, again they will be forcefully paired if dsp2 i.e. CN=4 is there. {meaning would be formed if they got paired}

Any metal with oxidation number > 4 , All ligands are strong for it like Ni+4.
and other factors like number of d electrons, proton number etc also influence it but basically, it increases with positive charge on central metal

from here ligand theory starts


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What are the advantages of eutectic solvents over aqueous solutions for electroplating?
I'm trying to electrically strip lead contamination from aluminum.
This answer suggests using a deep eutectic solvent (DES) as the conducting liquid for plating the lead onto a copper cathode.  But now that I'm thinking about it I can't see why.  Is a DES more conductive?  More inert with respect to (these) metals?
What advantage does a DES have over conductive aqueous solutions for electroplating?


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Solvation rates of crystalline and noncrystalline materials
Assuming equal surface area (and other variables), would a highly ordered crystal of some substance dissolve more slowly or quickly than (or at the same rate as) a disordered sample of the same substance?
I immediately thought it would be slower for a crystal to dissolve before thinking about why. Thermodynamically speaking, it should be entropically more favorable to break up the ordered crystal structure, but that doesn't determine kinetics. Kinetically, I think the energy required to break the intermolecular bonds in a crystal would cause the dissolution to be slower.


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Hiyama vs Sonogashira coupling
Why is Sonogashira's coupling more feasible than Hiyama's coupling when both could occur? I have seen many examples in syntheses, here is one (the penultimate reaction):

I'm not very familiar with the Hiyama coupling, but it seems to me that, to afford the Hiyama product, you would need $\ce{F^-}$ in there, or some base that would attack the silicon atom.

Meaning that the transmetalation is too slow otherwise. Thanks.

To be honest, I'm curious as to why they didn't do the Hiyama reaction in the first place to get the natural product. As I said, not real familiar. I'm sure there's a reason.

Probably because there is a bigger tendency of the reagent to undergo Sonogashira coulping, therefore the yield would be lower.

The Sonogashira is the standard reaction to try here first and it worked. Here is one page with alternatives and indeed there is a Hiyama example (but notice that Si and Br have been swapped i.e. they couple a alkynyl halide with a vinylsiloxane): http://www.organic-chemistry.org/synthesis/C1C/chains/enynes.shtm

Sonogashira with subsequent Hiyama coulping opens great synthetic possibilities.

So I will answer my question. In Organic Chemistry Portal it is written that Hiyama coulping requires an activating agent such as fluoride ion or a base because crucial for the success of the Hiyama Coupling is the polarization of the Si-C bond. Activation of the silane with base or fluoride ions (TASF, TBAF) leading to a pentavalent silicon compound is a first necessary step. The whole mechanism is given here:

On the other hand, Sonogashira coupling doesn't require any activation and it happens preferentially in the above mentioned synthesis.


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Hydrolysis of vinyl silanes
Alkynyl silanes can be easily hydrolyzed with TBAF like in the last step in the synthesis:

How feasible is the hydrolysis of vinyl silanes with TBAF? Does it actually happen or some other routine must be used?

According to This article, vinylsilanes are stable in TBAF solutions, allowing other silyl groups to be removed selectively. They can be activated in the presence of TBAF using palladium similarly to a halide, though.
This link shows C2-trialkylsilylalkenes being cleaved with potassium t-butoxide, 16-crown-8 and TBAF. I'll look more thoroughly when I get to a computer, though.
I also found this example, but it had no reference, so I'll try to find one:

How to remove the silyl group?

Oh, you wanted a method to do it, too? That I'm not sure about. Looking at references, it doesn't look simple. See edit made.

Interesting. Strong bases and mild acids can cleave them.

What search engine did you use? I couldn't find this.

I can't find it either, by using Google, I found the book.

Ah, ok. It's a good one.


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Is exposure to amyloid beta proteins dangerous?
Recently, I began working in a lab in which our main area of study is Aβ peptides. My concern is whether or not exposure to high levels of Aβ proteins is hazardous.
I suspect that the peptides cannot pass through one's dermis, but this is just speculation. Also, if one were to accidentally ingest Aβ, it would likely be broken down into its monomers by your digestive system before it could cause any harm.
I've tried to find resources covering my concerns but to no avail.

Do you speak german? I have a newspaper article that could lead you to some papers covering it. As far as I read it, it is still a question of research.

I am going to let someone with more expertise actually answer, but for anything you use in a lab that you don't prepare yourself, there's (by law, in the States at least) a MSDS available for it, and usually with a chemical hygiene plan that anyone can access.

No I don't speak German.

We synthesize these proteins in our lab. We work with the wild type protein and mutants of it. But I hadn't actually thought to consult an MSDS. Might be wise. Thanks.

This question is suited for [biology.se] as it does not discuss any aspect that particularly pertains to the chemical reactions underlying the process.

Amyloid proteins form insoluble aggregates and they cannot pass through the dermis. The aggregates can be (and frequently are, in the intracellular context) resistant to digestion but they would not be absorbed from the gut. Moreover, these peptides are not "toxic" per se. It is their property of forming aggregates and plaques that gives rise to the pathology observed in the associated neurodegenerative diseases. It is important to note that neural tissue does not regenerate like many other tissues in the body and that is one reason why neural tissue is vulnerable.
So unless you inject some of these proteins inside your cranium, it is unlikely that you will suffer from any problems beacause of Aβ exposure.


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Why do ethanethiol molecules have stronger London dispersion forces than those of ethanol?
I know that, molecules with higher molecular weight tend to form London forces. Is there any other reason which makes Ethanethiol to have strong London dispersion forces as compared to Ethanol?

Oooh, interesting question. Let me have a go at this and we'll see if there's someone else who can put it in better terms.
So, the way I tried to reason this out is to remind myself that every single bond, in each of these two molecules, is the same except the ones where C is bonded to O or S and the ones where H is bonded to O or S. We then pay attention to these bonds, particularly.
Now, at this point, we also have to remind ourselves that O is more electronegative than S and therefore, pulls electrons towards itself much more than S does and both of these atoms are a tad bit more electronegative than C or H. Therefore, the region of higher electron density in ethanol is O and the region of higher electron density in ethanethiol is S. We also have to remember that O has a more negative partial charge than S does. We also need to remind ourselves that S is larger than O which, by the way, is also a contributing factor as to why O is more electronegative than S.
I then uses these facts and try to make sure my reasoning is sound; If we look at the way london dispersion forces work, we realize that it is easier for electrons to fluctuate between molecules when they are within the bonds than when they are within the atoms constituting the molecules. In other words, when the electron density is concentrated in the bond, as is the case when there is 0 electronegativity difference, then london dispersion forces are stronger. (Electron fluctuation, by the way, is the basis for dispersion forces)
Using the reasoning above, it becomes natural to think that the dispersion forces in ethanthiol are stronger, since the electron density is more concentrated in the bonds than it is when you consider ethanol. Therefore, it is easier for the electrons to fluctuate between two ethanethiol molecules than between two ethanol molecules, which is why ethanethiol molecules have stronger dispersion forces between themselves as opposed to the ethanol molecules.
Note: However, ethanol molecules have a much more powerful intermolecular force available to them, which is hydrogen bonding. So, clearly, ethanol molecules have a higher boiling point or whatnot than ethanethiol. Just to make sure that this is clear enough.


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Migration of R group in Hofmann rearrangement
In the mechanism of Hofmann rearrangement, one step involves migration of $R-$ group from carbon to nitrogen as shown: This reaction proceeds even if the migratory aptitude of the $R$ group is poor (like in the case of ethanamide). What motivates the migration of $R$ group and makes the reaction proceed?

P.S.: I think the reason cannot be the negative charge on oxygen since it already delocalized by resonance, which shifts the electron cloud partially towards nitrogen, which is in turn partially shifted towards Bromine due to its -I effect. Here's my depiction:

Does this answer your question? Why does alkyl rearrangement happen in the Hofmann rearrangement?

I think it can be viewed (at least for "educative purpose" i.e. "lie for student") as intramolecular substitution - nitrogen electrophilic attack with CO... as leaving group.


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Qualititative inorganic analysis
Why is it that a specific order is followed while performing qualitative inorganic analysis? I mean, why are we told to first test for the anion in the given salt (and there is an order to be followed for which anion to test for first, which second and so on) and then for the cation?
Does it really matter which way we do it?
How does it affect the analysis if we don't follow the order of cations/anions?

It is just a matter of preference. If you wish to test for the cation first, you can do that and nothing is going to change. Generally the order is given to make the job easy for students, so they can remeber the tests (and also because the profs. know which salt is generally given and the given order is the order of probability. Like if there are 4 salts with sulfate ions, it is better to do the test for sulfates first).


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How to explain covalent bonds with Bohr's atom model?
I was reading about covalent bonding and got to know that in this type of bonding, the atoms of elements "share" electrons with each other to have an octet or duplet. But, I also have been taught that electrons keep on revolving around the nucleus of the atom, much like the planets around the sun. But, if this is true, how are the able to easily share electrons if these electrons keep on moving around their respective nuclei? I hope I have made myself clear. If not, then please tell in the comments.

electrons keep on revolving around the nucleus of the atom, much like the planets around the sun.

Electrons are quantum particles. Trajectories are not applicable for them. One have to use quantum wave equation to explain "movement" (note the quotes, you cant say anything about electron movement, you can only determine the probability of electron presence in some place) of electron. Planetary (Bohr) model of the atom can help to describe some effects for isolated atoms using Newtonian physics' equations with some amendments (Bohr quantum rules), but it does not describe bonding.
So for bonding you can use valence bond theory (which exactly the case with sharing electrons you described) or some more complex theories (molecular orbital (MO) theory, as widely used example) which consider quantum wave equations for describing electrons. But you surely can't mix in Bohr atom model in this case, because it is too simple to describe bonding.
Hope my explanation helps.


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Corey-Seebach reaction regioselectivity
Does Corey-Seebach reaction with conjugated enones undergo in a direct or conjugated fashion?

Can the two sulfur atoms soften the carbanion enough for it to undergo conjugated addition? If not, would addition of Cu(I) enhance the proportion of conjugated addition?

Are you talking about forming the dithiane? or the addition of a lithiated dithiane into a conjugated enone? (I'm guessing the former, but the Corey-Seebach reaction is usually described as the addition of the dithiane rather than its formation)

No, I've meant the latter. I bet on direct addition.

If I remember correctly, 2-phenyl-2-lithio-1,3-dithiane exclusively yields the thermodynamically controlled 1,4-addition product when reacted with cyclohexen-2-one in THF at room temperature. At -78 °C under otherwise similar conditions, the  the kinetic 1,2-addition product is predominantly (~ 2:1) obtained. Adding HMPA to the lithiated dithiane prior to adding the enone also favoured reaction in Michael fashion, even at low temperatures.

Okay, is't all about conditions. I could accept your comment as an answer if you provide a reference.


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Lucas test for alcohols
The Lucas test for alcohols is based on the way primary, secondary and tertiary alcohols react with Lucas reagent. If on adding Lucas reagent to an alcohol, the mixture immediately becomes turbid, then it is a tertiary alcohol.
I am aware that alcohols get converted to alkyl chlorides on reaction  with concentrated HCl, and that it the alkyl chloride that is insoluble and hence responsible for the turbidity.
But why do tertiary alcohols react the fastest with HCl?

Because the rate of the reaction is directly proportional to the stability of carbocation

The tertiary alcohol has more R groups compared to primary & secondary alcohols, there by removing the C-O Bond from a tertiary alcohol is more easier. (the tertiary carbocation is more stable)

In Lucas test, the alcohol acts as a polar protic solvent and the -OH group can get protonated to form water which is a good leaving group. So SN1 is the most likely mechanism and the order of reactivity is 3°>2°>1°.


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Wax layer destruction from seed coats
I'm working with some orchid species, and i encounter a specific issue. Some of those species deposits upon their seed coat a wax layer. This layer makes seeds behave in a very hydrophobic way. You cannot dispense them in a pour water.
That behavior has many impacts. One of the most significant is that it does not let water come in touch with the actual seed and then with embryo to make it sprout.
So i tried many ways to remove that layer but nothing actually seemed to work out.
I treated seeds with the following chemicals:
a) 1% NaOCl with some dish soap water solution for 10min. After throwing seeds into this solution you could see that they can be evenly dispensed in it and come in touch with the solution. But if you wash them up with pure water for example 3 times they are getting back their hydrophobic behavior.
b) 1% H2SO4 water solution for 10min. I thought that with this treatment waxes will be removed from the seeds but nothing like that happened. Maybe they had been better dispensed in the pure water comparing to the previous treatment but nothing significantly changed.
c) 93% ethanol solution for 1, 2, 3, ... 30min. results here were also very disappointed. While the seeds are immersed into ethanol, they were very well dispensed into the liquid without any hydrophobic behavior. After three washes with pure water, hydrophobic properties came back.
Treatment time seems not to play any significant role on the results.
My conclusion is that there is no destruction of the waxes with the above chemicals. The only thing that happened probably, is the creation of a Micelle around the seed, which make it for a while (the time that it was in the solutions with soap and ethanol) non hydrophobic.
Finally my question is, if there is any way to destruct chemically, to some grade the waxes. So even when you put the seeds in a pour water not be so much hydrophobic.
Thank you.

Solvents like hexane/pentane/heptane/toluene are excellent for getting rid of wax/grease in a chemical sense (like dissolves like), however I'd be cautious about using them with biological material like seed samples as theres the potential for the solvents to penetrate and kill the while thing.

Yes, i also thought of such organic solvents but i believe that they are gonna harm the living cells. Or maybe in a very small concentration do not harm them.

You might be better posting in Biology.SE, I'm sure there will be a common procedure for doing this that a plant biologist would know, rather than us chemists speculating about what chemicals would remove it without killing/dissolving the rest of the seed :)

Maybe you are right :). I will post it there too.

I'd wonder if you're going in the wrong direction. A drop of dish washer soap shouldn't cause any harm to the seeds and would likely be enough to disperse the seeds in the water.

Actually yes. Dish soap disperse very well seed in water but i don't want any molecules of soap later , to be in touch with seeds and that's why i wash them up with pure water. After the wash, they loose their  dispersion in water


text:
Why does mercury in mercuric acetate have a lone pair?
The configuration of mercury ends with $6s^{2}$. In mercuric acetate, mercury is in the +2 oxidation state, so does it still have a lone pair?

Of course it has lone pairs. 1s², 2s², …

TLDR:
Mercury in its (+2) salts does not have lone pairs readily accessible for covalent bonding. However, its 5d electrons do affect its properties,  resulting in a relatively high polarisability.
The long story:
Mercury is the last 5d element and has electron configuration $[\ce{Xe}] 4f^{14} 5d^{10} 6s^2$. The $6s$ electrons are the outermost electrons and readily available for covalent bonding. $4f$ electrons are virtually inert since they are very deep. $5d$ electrons belong to the previous shell, but are the outermost of them and once mercury looses its $4s$ electrons, $5d$ electrons become the outermost and define its properties. They are still too strongly bound to the nucleus to participate in covalent bonds under ordinary conditions ($\ce{HgF4}$ was only observed spectroscopically in inert gas matrix). , but are far enough to wiggle around and participate in non-covalent interactions, such as van-der-waals interactions. $6p$ orbitals are also available for covalent bonding, though extent of their use is often debatable and/or unobvious.

These are 5d electrons. They can participate in chemical bonds, e.g. in mercurinium cations.

You should elaborate.


text:
Dependence of carbocation stability with respect to the length of alkyl group

Which carbocation is more stable, the ethyl carbocation or n-propyl carbocation?
What is the effect of alkyl group length on the stability of a carbocation?

Your first question, which carbocation is more stable, ethyl or n-propyl, has already been answered (see here).

@ron actually i couldn't completely understand the answer (quite a high level for me).But what i found from your answer is ethyl cation is an exception right?.But i have a doubt,is it correct that only methyl group can show inductive effect (can -CH2- group show inductive effect?)?Also would you post a simple answer about the dependence on length of alkyl group.

Yes $\ce{-CH2 -}$ can show an inductive effect, all alkyl groups can.

@ron thanks.Would you please explain me the dependence of +i effect on length?Is there any dependency?Am I wrong about this concept?

There is likely a measurable difference between methyl and ethyl due to the hyperconjugation reasons mentioned in the answer below. Beyond that I would expect very small differences (for example between n-propyl and ethyl).

So what you are saying is that +I effect of pentyl group and hexyl group will be having less difference.right?

Also how will we explain the variation in pka of acetic acid and propanoic acid with respect to +i effect? pka value:http://crab.rutgers.edu/~alroche/Ch20.pdf

Carbocation stability is widely known to increase going from primary -> secondary -> tertiary.

In the absence of other factors (such as lone pair donation etc.), the major contribution to the greater stability of tertiary carbocations vs primary is due to hyperconjugation between the vacant p-orbital (carbocation) and the C-H sigma bonding MO's.

Based on hyper conjugation alone, you can predict that length of the alkyl chain will have no effect on the stability by itself, only the number of neighbouring C-H bonds with which hyper conjugation can take place.
In specific regard to ethyl vs n-propyl cation, neither are considered stable.

Then why is there a variation in the acidity of primary aliphatic acids say between hexanoic acid and butanoic acid ?though the ka value difference is small still what is the reason for the differencce?


text:
Why does benzaldehyde not respond to Fehling's test?
All aldehydes are supposed to respond to Fehling's test. So why doesn't benzaldehyde respond to it?
Does it have something to do with the fact that benzaldehyde is an aromatic compound? If so, then how?

Related: Why does formic acid give positive Fehling's test?

The rate-limiting step of the Fehling’s test reaction with aldehydes is the formation of the corresponding enolate:

The subsequent reaction of the enolate with copper(II) proceeds through a single electron transfer mechanism.
Aldehydes that lack alpha hydrogens, such as benzaldehyde or pivalaldehyde (2,2-dimethylpropanal) cannot form an enolate and thus do not give a positive Fehling’s test result under usual conditions.

From where did you find the mechanism of this reaction?

Could you kindly share the source/citation? Thank you!

In benzaldehyde, the carbonyl group is an electron withdrawing group so the carbonyl group pulls the electron from the electron-rich benzene ring. Due to this, the polarity of the $\ce{C-H}$ bond in the carbonyl group is reduced because the $\ce{C-H}$ bond now has a higher electron density. Thus, the $\ce{C-H}$ bond becomes stronger (the lesser the polarity of a bond, the stronger the bond ). So Fehling's solution (comparatively a weaker oxidizing agent than Tollen's reagent) can't oxidize benzaldehyde (an aromatic aldehyde). However, Fehling's solution can oxidize an aliphatic aldehyde.


text:
What is the total bond energy of CO2?
How can you calculate the total bond energy of CO2? I've searched for the bond energy of C=O and found it was 745, but sometimes 799. Which one is the right one to use? And do those numbers mean the bond energy of CO2?

there are many C=O bonds with different chemical environments, bearing slightly different energies. Usually the reported value is an average of many instances of that bond.

You're asking a pretty complex question. The short answer is that to calculate such an energy you have to utilize quantum mechanical calculations, or otherwise perform macroscopic experiments on bulk $\ce{CO2}$.
A quick ab-initio calculation of a O=C=O molecule with bond-length 1.164 Angstroems (quite close to experimental and theoretically deduced length) yields the following Hartree-Fock SCF energy (using the 6-31G* basis set):

-187.634176090515 Hartrees, which is
-492633.5668524824 kJ/mol

This is a huge negative energy, and rightly so -- it is how much energy is theoretically required to break the $\ce{CO2}$ molecule entirely into subatomic particles. In other words, it is the energy holding $\ce{CO2}$ together -- the energy that defines the molecule.
Better approximations can be made with higher levels of theory but don't deviate from these values by much.
The simplistic route is to add up collective bonds from a data table provided in your textbook, e.g. $2 \times$ C=O = total molecular bond energy. This is not the same thing as the total molecular energy though: the latter includes energetic contribution of the nuclear-electron interactions, not just atomic bonds.

I wouldn't worry too much about the slight differences in the bond enthalpy for the C$=$O. I will discuss why after the next paragraph.
The bond enthalpy is the amount of energy (in kJ) that is required to break 1 mol of that bond. Therefore, it is also a measurement of how much energy is released as 1 mol of that same bond is formed. The bond enthalpy is dependent upon the strength of the bond. The higher the strength, the larger the bond enthalpy.
The bond enthalpy is more of an average value than an accurate figure. For example, the energy required to break the first double bond between the central carbon atom and the oxygen is different from the energy required to break the second double bond. This is because the strength of the bond is dependent upon its environment. When you break off that first oxygen atom, it changes the environment.
So you can use any one of the those bond enthalpy values.
Source(s): Any good textbook in general Chemistry should discuss how enthalpy bonds are calculated. Personally, I used the Pearson Baccalaureate Textbook for IB Chemistry HL

Thanks! This helped. But what is the total bond energy of CO2?

In $\ce{CO2}$, the potential energy in the carbon-oxygen double bonds is equal to the number of such bonds in the molecule multiplied with the average bond enthalpy.
If you are in doubt, it is always useful to draw the Lewis Structure of the molecule to help define the bonds.

I know that at this point I'm basically necroposting, but none of these answers actually directly answered the question, which was disappointing since I came here looking for the answer to this same question, and yet after 7 years, nobody actually answered it.
Lets address the question actually being asked:
"What is the total bond energy of CO2" and lets notice that he's referring to chemical bonds, not nuclear/subatomic bond energies since he also said: " I've searched for the bond energy of C=O and found it was 745, but sometimes 799. Which one is the right one to use?"
Ok, so now we have a question that we can actually answer.
To start off, carbon forms 2 double bonds with oxygen atoms like this O=C=O which makes sense since carbon has 4 more valence electrons than a full shell and oxygen has 2 less than a full outer shell. I often like to informally write atoms with their valence electrons written like such: C+4 O-2 since it makes it easier to see where they are relative to a full shell.
Anyway, so to make any stable molecule with this notation, you just have to make it neutral, so this means 2 oxygens and one carbon: 2x(-2) + 1x(4) = 0.
Now, this molecule has 2 essentially identical double bonds O=C=O on either side of the carbon, and the energy of one of these bonds is what we are looking for, or more correctly, the average energy between both of the double bonds present. now technically speaking they have the same strength, but practically speaking, one of the double bonds will come off first when involved in a chemical reaction, meaning that the other double bond will experience a change in strength since the molecule as a whole is now significantly different. we would have to account for this, but this is usually what most online values will already be. generally they display the average value, that is as long as they say that collectively "C=O bonds in CO2 = X kJ/mol" with no further information on individual bonds.
another thing to mention is that C=O bonds in CO2 will have a different strength than C=O bonds in other molecules, and for the same reason as earlier, its because the molecules are significantly different.
and another thing we have to watch out for is whether the value we find online or elsewhere is for the molecule as a whole (both double bonds) or for only one of the double bonds, but usually that is easy to tell as it will either say "per mol of CO2" which is both, or "per mol of C=O" which is only one, and likely the averaged value.
Now, as I said at the start, I too am looking for the same answer, meaning that I don't have a definitive answer for you that I can whip out of my back pocket on demand. however, I can walk you through my logic on how to find a more accurate and satisfactory answer, which is precisely what I did.
When we search for the value online, we get a lot of the same 2 answers you got, one around ~750 and another about ~800. this difference is large enough to cause error in calculations, and so a more accurate answer is necessary. now if we look a little deeper we'll find that the sites offering the 750 value are primarily discussing carbonyl groups. such as: The Carbonyl Group - Chemistry Libre Texts
These are C=O groups within other molecules. and now if we refer to Wikipedia: Carbon-oxygen bond - Wikipedia (I know its not always perfectly reliable, but well known science data is usually accurately represented) we will find the following paragraph:

"Bond lengths of C=O bonds are around 123 pm in carbonyl compounds. The C=O bond length in carbon dioxide is 116 pm. The C=O bonds in acyl halides have partial triple bond character and are consequently very short: 117 pm."

Notice the bond lengths. they are 123pm for carbonyl compounds, 116pm for carbon dioxide, and 117pm for acyl halides. If you remember, the shorter the bond length, usually the significantly stronger the bond. CO2 actually has the shortest bond length here, even surpassing the acyl halides with their sort-of triple bond character, and triple bonds are some of the strongest in all of chemistry, such as N≡N (nitrogen gas) and C≡O (carbon monoxide). this means that of all the values for C=O bonds in various molecules, the one for CO2 should most likely have the highest bond strength of all of them. also, generally speaking, simpler binary compounds such as CO2 will have some of the strongest bonds, such as the C≡O from earlier (which was conveniently similar to CO2).
There is however, one last thing we should check, and that is whether the values from earlier refer to the individual C=O bonds in CO2 or the whole molecule of CO2. now, we already have a bit of a hint, since the carbonyl groups from earlier were said to have the 750 value, and carbonyl groups are individual C=O bonds, not CO2 molecules, this means that 750 must be the value for a single C=O carbonyl bond. we also know that the carbonyl bonds should be a little less strong than the CO2 bonds. now, if either the 750 or 800 value were for a whole CO2 molecule, then when you split that in half (~400) to look at only one of the C=O bonds, it would be less strong than the carbonyl bonds (~400 < 750-800) which we know isn't true. therefore, the 750-800 value is for only one individual C=O bond in CO2.
Time for conclusions:
We know that the 750-800 values are per bond, and we know that the 750 value belongs to the carbonyl groups, and we know that CO2 has a shorter bond length on average than the carbonyl groups, and therefore likely higher bond strength, meaning that the most likely answer to your question is that the 800 value belongs to CO2 and that the bond strength in kJ/mol of individual C=O bonds for CO2 is ~800 kJ/mol. that would be 1,600kJ/mol of CO2 molecules, that's 1.6MJ/mol of carbon dioxide, now that's a lot of energy!

CO2 is a too complex molecule to perform such a back of the envelope calculation. Unfortunately this post is very long, without saying much. I don't actually think it's answering the question. But then again the question is really hard to answer.

@Martin-マーチン thanks for the feedback, as I'm quite new here and would like to refine my ability to answer questions effectively. however I should point out that my post did not claim to definitively answer the question, it was only meant to point people in the right direction. I was also later able to find more definitive references which aligned very closely with my answer.


text:
Oxygen reduction - beginner level
I am embarrassed to ask but I don't get something really basic.
When molecular oxygen and hydrogen react to produce water, we say the oxygen has been reduced and the hydrogen has been oxidised. Reduction, I understand, is the gain of electrons.
So I look at the oxygen atoms prior to the reaction and it seems to me each has six of its own plus two more that are shared.
And I look at the oxygen in the water molecule and it has... six of its own and two that are shared.
So in what sense has it gained electrons?
Of course, before posting this question I have searched and read for several hours. All I can find is talk of half-reactions. The clue here seems to be that the oxygen is ionised to $\ce{O_2-}$ prior to reacting with $\ce{2H+}$. But this just confuses me. Maybe this intermediate step does involve reduced oxygen but the end point is still the same number of electrons as we started with, so if it does get reduced it must be oxidised again immediately afterwards, for no net effect.
Where am I going wrong?

You can look at the change in formal charges as the reaction takes place. In the diatomic species $\ce{H2}$ and $\ce{O2}$, the formal charge on oxygen and hydrogen is zero. In $\ce{H2O}$, the hydrogen atoms each have a formal charge of +1 while the oxygen has a formal charge of -2.

@ToddMinehardt Oxidation state not formal charge.

@Mithoron - Good catch, thanks! Oxidation state indeed.

This is a really good question which I've never given much thought about.
I'll try to explain the reason as much as I know it. There may be a better explanation.
The reaction can be represented as -
$$\ce{2H_2 + O_2 <=> 2H_2O}$$
Now in O2 you have a structure like
$$\ce{O=O}$$
And in H2O you have a structure like (ignoring the tetrahedral hybridisation and bent shape)
$$\ce{H-O-H}$$
We know that Oxygen is second most electronegative element after fluorine, which is because of it's small size and strong effective nuclear pull towards electrons.
Hydrogen on the other hand, is not as electronegative as Oxygen even after having a small size, because of very less effective nuclear pull.
When a bond between H and O is formed, O pulls the shared pair of electrons towards itself because it is more electronegative than H. This is the "gain" of electrons. The atoms do not actually take up electrons (when it is covalent bond) but pull it towards themselves, thus getting reduced. You will not find this "gain" or pull of electrons in the O-O bond because they have equal pulls, hence 0 oxidation state. But in water, Oxygen has +1 oxidation state.
Similarly, for hydrogen, it pushes the electrons towards Oxygen in water. Not exactly losing the electrons but density of electrons around Hydrogen atom decrease due to this push by hydrogen and pull by Oxygen. Thus it has "lost" it's electron, becoming positively charged which wasn't the case when it was in H2
Also, you'll notice that when more bonds are formed, there is more change in oxidation state of the atom.
Hope this helped.

"But in water, Oxygen has +1 oxidation state." What?? I assume you meant "Hydrogen has +1 oxidation state."?


text:
What is the IUPAC name of [Co(NH3)5H2O]Cl3?
I am not sure if it is Pentaammineaquo cobalt(III) chloride or Aquopentaammine cobalt (III) chloride.
Should the prefix be ignored while naming the ligands in alphabetical order? Besides how should the name be written with proper spacing?

Multiplicative prefixes are not considered in determining the alphabetical order of ligands.
According to the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book), the systematic names of coordination entities are generated by using the following principles:

IR-9.2.2.1 Sequences of ligands and central atoms within names
The following general rules are used when naming coordination compounds:
(i) ligand names are listed before the name(s) of the central atom(s),
(ii) no spaces are left between parts of the name that refer to the same coordination entity,
(iii) ligand names are listed in alphabetical order (multiplicative prefixes indicating the number of ligands are not considered in determining that order),
(…)

Therefore, the systematic name of the coordination entity in $\ce{[Co(NH3)5H2O]Cl3}$ is “pentaammineaquacobalt” since the ammine ligands are named before the aqua ligand according to alphabetical order. Note that, since 1970, the recommended name for water $(\ce{H2O})$ as neutral ligand in coordination complexes is “aqua” (formerly “aquo”).
The ionic composition of the complex salt $\ce{[Co(NH3)5H2O]Cl3}$ may be indicated in several ways:

IR-9.2.2.4 Charge numbers, oxidation numbers and ionic proportions
The following methods can be used to assist in describing the composition of a compound:
(i) The oxidation number of the central atom in a coordination entity may be indicated by a Roman numeral appended in parentheses to the central atom name (including the ending ‘ate’, if applicable), but only if the oxidation state can be defined without ambiguity. When necessary a negative sign is placed before the number. Arabic zero indicates the oxidation number zero.
(ii) Alternatively, the charge on a coordination entity may be indicated. The net charge is written in arabic numbers, with the number preceding the charge sign, and enclosed in parentheses. It follows the name of the central atom (including the ending ‘ate’, if applicable) without the intervention of a space.
(iii) The proportions of ionic entities in a coordination compound may be given by using
multiplicative prefixes. (See Section IR-5.4.2.1.)

Therefore, correct systematic names for the compound $\ce{[Co(NH3)5H2O]Cl3}$ are:

pentaammineaquacobalt(III) chloride
pentaammineaquacobalt(3+) chloride
pentaammineaquacobalt(III) trichloride
pentaammineaquacobalt(3+) trichloride

Note that multiplicative prefixes are usually not used in binary names if there is no ambiguity about the stoichiometry of the compound.


text:
Acceptable levels of HCl?
For a science/engineering project we are taking pool water (disinfected), filtering to 0.1 microns, and pumping to the kitchen.  Of course there is still 5 ppm of chlorine present so we want to neutralize the chlorine.  Asorbic acid in the form of orange juice neutralized the chlorine,  but I understand that it also creates some amount of HCl. This method is used to neutralize chlorine in water being introduced into ponds, rivers, etc. so that the chlorine does not kill the fish.
Could the asorbic acid actually create enough HCl to be dangerous,  how can I test this,  and what is acceptable?  I used a standard pool testing kit to test the chlorine.

pH of lemon juice and cola is around 2-3. As long as you don't go below this point, drinking should be fine.
It is different if you consider fish living in such water. pH from several ppm would not drop below 5, which is close to rainwater, but significantly lower than normal river or sea water (6-7 and 7-8 respectively). You should consult specialized literature in the case.

HCl is completely dissociated in aqueous solutions, so the protons and chloride ions are independent from each other.
As long as the acidity (pH value) of the water is acceptable you should be fine (note that sea water contains about 20000 ppm chloride).
If the fish is sensible to chloride you should also take into account that the water has a higher chloride concentration than the water originally used to fill the pool. So you might want to measure chlorine, too.

HCl is completely dissociated in aqueous solutions, so the protons and chloride ions are independent from each other. What exactly do you mean by this? The concentrations are obviously not independent because they must be equal.

No, they must not be the same as long as other ions are present which is certainly true for pool water. To determine if the acid is hazardous to fish it should be sufficient to measure the pH. I've added a remark about higher chloride concentration.