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How would mixing sea salt and honey affect the water activity of honey?
Honey has around 20% water content which is one mechanism of it's antibacterial action. If I mixed equal parts of honey and salt how would this affect the water content/activity of honey? Would it lower the content of water and thus provide a stronger antibacterial effect?
Also how would mixing salt with honey affect its other properties?
Thank you.

When you mix the salt with the honey, does the salt dissolve?  If so, then the water activity of the honey is going down.  If not, then it may not be.

If it consists of 20% water I am assuming the salt would dissolve, no?

The solubility of salt in honey is probably much different than salt in water because of a large number of other solutes such as sugars, etc.  It's very difficult to predict what would happen so in this case I think we need some experimental data.

Thank you for your quick responses. For the sake of the argument let's assume the salt would dissolve in honey. How much salt would I need to add to the honey to reduce the water content to 0%?

Reducing water activity is not the same as reducing water content.  2. It would not be possible to reduce water activity to solely by adding salt.  3. You could reduce water "content" to 0 by using an infinite amount of salt.

Is there a way to reduce the water activity of honey?

Why, you may just evaporate water away (say, by vacuum drying). The problem is that honey would crystallize, and thus cease to be honey. But it still would taste better than salted honey, that's for sure.

Is there any other way that I could reduce bacteria through osmosis similar to how a salt solution does this?

Any antibacterial effect of honey would need to be determined without salt, first, and then compared with the salt added.
It would be hard to state that there is any antibacterial effect without evidence. It would be interesting to see the results of a careful controlled experiment, with varying concentrations of honey on different bacteria. For example, simply covering the growth medium with honey might just prevent growth of aerobic bacteria, as it excludes oxygen, so should not be considered an "antibacterial effect". Comparing the honey with an equally-concentrated sugar solution would help determine if osmosis (or sugar concentration) is the cause of antibacterial actoin.

The salt will fall out because the 20% water is not enough to dissolve it. Plus many fine salts have anticaking agents such as TriCalcium Phosphate, which are not soluble in water.


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Programs for chemical structure visualisation and drawing 2d and 3d
I want to create list of simple and useful software to draw molecules 2d and 3d with small description. Not necessary only freeware software. Also web-based services are interesting.
Your help in improving this list is greatly appreciated. I will continue this list and add more information with some pictures.
If you know some usefull and easy software, please, write about it!

Web-based freeware site chemicalize.org
Free. Cross-platform.
It does not require installation of java or flash, as output/input can be drawn by hand, SMILES srting, IUPAC name, some molecular geometry file formats.

Web based freeware site chemspider.com
Free. Cross-platform.
It searches by IUPAC name, SMILES, can be drawn by hand, but there is HUGE limitation - it searches only existing molecules, but shows some useful information about compound.

http://molview.org/ web-based
VMD - freeware open-source program from University of Illinois
Free. Cross-platform.
This program mostly for visualization, but has some building features. Can be run under any OS. Opens a lot of different formats. GUI is not very friendly for the beginners. Used mostly for visualization large molecules (proteins, membranes), molecular dynamics simulations (especially from NAMD).

Mercury - Crystal Structure Visualisation, from The Cambridge Crystallographic Data Centre (CCDC)
Free. Cross-platform.
New freeware version has some limitations - not possible to calculate distance to the plane (now it is feature for paid version). It can open/save a lot of chemical structure format files, all OS are supported.
Chemcraft
Free trial. +Academic discount. Windows/Linux (works fine under wine).
Basically is non-free version (there is quite long trial period and there is old free versions through internet without any restrictions). Is one of the simplest and easiest 3d drawing programs. There are versions for Windows/Linux and under Linux/Mac OS it is possible to run under wine.
BIOVIA Draw (old name Accelrys Draw - successor of ISIS draw)
Free  for Academic and Non-commercial Use. Windows.
Only windows (can't say yet about working under wine). Mostly 2d drawing software, can wok with a lot of different formats, also supports 3d visualization.
Avogadro
Free. Cross-platform.
Molecular builder/editor. Drawing is not the best, but has built-in optimisation tool. Opens/saves a lot of formats, useful for preparing inputs for quantum-chemical calculations. Fast and simple.
Openbabel
Free. Cross-platform.
Read, write and convert over 110 chemical file formats. Free cross-platform format converter.
Marvin Suitefrom ChemAxon
Free. Cross-platform.
Very good 2d/3d drawing software supports a lot of formats. Easy and useful.
ISIS/Draw
Free. Windows.
Very old drawing software. Support most common formats. Incompatible with some new drawing formats. Possible to download in the internet.
ChemDoodle
Non-free. Cross-platform.
Has a lot of features bult-in. 2/3d drawing. Has applets for phones.

I am asking about software you are using - I haven't tried all possible software.

http://chemistry.stackexchange.com/questions/28997/software-for-viewing-a-structure-and-altering-bond-angles

I think this is much too broad a question. Not only are there dozens of such programs, but there are multiple websites with such listings, e.g. here and Wikipedia, among others.

@Mithoron Yes, Avogadro also was on my list.

@GeoffHutchison Yes, but what about our edition of such list?

You could post this in last part of this list: http://chemistry.stackexchange.com/questions/37303/resources-for-learning-chemistry

My understanding is that such shopping questions are beyond the intended scope of this forum.

Indeed, with rare exceptions, such a question is beyond the scope. Asking about "I need a 2D drawing program for Linux" is fine, but this is a Q&A forum. As I mentioned, there are multiple lists, including Wikipedia - better suited for such things.

I've been involved in the development of
Speciolab - a drawing app for 2d chemical structures. It is tailored for students and researchers needing a simple tool to draw structures or reactions - and export to a report or paper. Available for Mac and Windows. It is currently in beta. Try it for free at https://specio.com

source: Stack Exchange

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

sort: votes


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What concentration (M) of citric acid should I use in a titration with 0.02M LiOH?
I am using an automatic titrator that records the amount of acid/base added and the pH of the solution in the form of a graph. I am trying to titrate citric acid to capture all 3 pKas in this graph. I will be titrating 40mL of a ???M citric acid solution made up in 0.01M LiNO3 as a background electrolyte. The software's titration schedule begins by adding increments of 0.1M HNO3 until the pH reaches 2.5, then it adds increments of 0.02M LiOH until the pH reaches 10. What concentration (M) of citric acid should I use in this titration in order to capture its pKas?

We need to know how big the increments of are to answer this question.  Are they 1 mL?  0.01 mL?

The largest increment is 0.1mL. However, the titration software decreases the mL of base added as the titration reaches equivalence point. So at the beginning of the titration, the increments are 0.1mL LiOH added, but as the pH begins to increase rapidly with each addition, the increments decrease non-continuously (down to a minimum increment of 0.01mL), then after the equivalence point has passed the increments increase again. Hope this helps.

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


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Why is silver nitrate used for cyanide titration?
Or more to the point, since silver isn't exactly cheap: Why can't cuprous nitrate ($\ce{CuNO3}$) substitute for the pricier ($\ce{AgNO3}$) in cyanide titration?
I assume that free cyanides bind eagerly with all Group 9-11 metals, since they're used industrially to leach the heavier ones.
I'm guessing that there's something relatively quirky about the Period 4 metals Co, Ni, and Cu that prevents their use as a titrant of aqueous cyanide: E.g., it appears that nickel nitrates are very hygroscopic.

Out of curiosity (it kind of affects my answer), does this work with $\ce{HCN}$? Or just the $\ce{CN^-}$ ion?

@DGS - AFAIK cyan salts (which I would assumed to include HCN) are always titrated in aqueous solution.  However the pH of the solution is not specified, which leaves me wondering whether it is assumed to be neutral or whether that is irrelevant to the titration....

@DGS - After further thought: The titration is for "free cyanide."  My guess is that CN ions that have bound to H in solution are not "free" and won't bind to other metals.  (Therefore, if someone is titrating for cyanide reagent quality they probably want to do it in a very alkaline solution?)  In fact, hydrolysis of cyan salts is notorious for producing HCN (en route to the usual irreversible nitrile hydrolysis).

In that case, I'm sticking with what I've already said below. And yes, they would have to adjust the pH to significantlt higher than 9.2 (pKa of HCN in water). Cool question.

Due to the hardness of copper ions relative to silver ions, copper ions will bind hydroxide more readily in the presence of cyanide than silver ions do. Once ligated by $\ce{OH^-}$ ions, $\ce{Cu^{1+}}$ will oxidize to $\ce{Cu}^{2+}.$
The softer acid, $\ce{Ag^{1+}}$, doesn't do this nearly as quickly, especially not in the presence of a softer ligand like cyanide.


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Hazards of Keeping Freshwater Fish in a Container that was Used for Adipic Acid
I have an Intermediate Bulk Container (IBC) which is a large 3' x 3'x 3' plastic cube. The label on the container described it as storing Adipic Acid.
Is it safe to use this container to grow plant life and harvest fish life for aquaponics? Or is there a possibility that this acid has leached into the plastic?
My quick research has shown that adipic acid is a mild skin irritant.

If you clean it out well with soap and water, the fact that it was used to store adipic acid shouldn't matter. Not knowing the grade of the reagent (i.e. the impurities, may be some less water-soluble ones) or even what plastic the container is made of might be a bigger risk here.

Would it help if I provided more information that was on the label? Are there perhaps some simple home tests that I could do to measure the pH of the water? Perhaps if I did a pH test on water I filled the container with, and then tested the pH of the water again 1, 7, 21, and 35 days later?

You must monitor and adjust the pH of fish tanks.

Adipic acid occurs naturally in beets and sugarcane, common foods, so it is not likely to be very toxic to fish if sufficiently diluted by washing. As @DGS and @ghost state, normal cleaning of the container followed by adjusting the pH for the fish should be OK. If you need to use an electric in-tank heater, though, check that it is safe to use in a plastic container.


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Calculation of an equilibrium constant when data for reactants and products is given
Lets say that I want to calculate the equilibrium constant $K$ for a chemical reaction where $\ce{A -> B}$
$$K = [\ce{A}]/[\ce{B}]$$
where $[\ce{A}]$ is the concentration of $\ce{A}$ and $[\ce{B}]$ is the concentration of $\ce{B}$ at equilibrium
I have measured the quantities $[\ce{A}]$ and $[\ce{B}]$ and thus have two sets of data.
To calculate $K$ should I:

Take the ratio of $[\ce{A}]$ and $[\ce{B}]$ at each data point and then calculate a mean average of all of the $K$ values?
OR Plot a graph of $[\ce{A}]$ against $[\ce{B}]$ and use the gradient of the best fit line as the value of $K$?

My instinct tells me it is the former as the straight line plot on excel has a non-zero intercept (which is not reasonable). If it is the first option, can you explain to me why it is so?

How do you know you're at equilibrium?  We need a better description of your experiment to answer your question.  Why are there different concentrations of [A] and [B]?  What did you vary to get these different concentrations?


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Nitrogen oxides produced in diazotization
My A level exam board describes an experiment to produce methyl 2-hydroxybenzoate (methyl anthranilate) from methyl 2-aminobenzoate (methyl salicylate).
I did the experiment as described (including keeping the temperature below 5 degrees), but when I added the sodium nitrite solution, a significant amount of brown gas was produced, which I identified as nitrogen dioxide. My teacher said this shouldn't happen, and wasn't sure why it was happening. What is wrong with the plan?
Edit: I think too much sodium nitrite is used; calculation suggests about 3g of it should be added, so I think a decimal point is missing. Am I right in thinking this?
Here is the plan:

Requirements:
  2-aminomethylbenzoate 5 cm3 (6.5 g),
  concentrated sulfuric acid 5.6 cm3 (10 g),
  sodium nitrate(III) solution 60 g in 100 cm3 water,
  urea (solid).
  Measure out 45 cm3 of distilled water into a 250 cm3 conical flask. Slowly and carefully add 5.6 cm3 (10 g) of concentrated sulfuric acid to the water. Add to this warm solution 5 cm3 (6.5 g) of 2aminomethylbenzoate. (You may have to warm this first as its melting point is 24 ºC). Place the flask in an ice bath and add about 45 g of ice to the solution in the flask. Slowly add 20 cm3 of sodium nitrate(III) solution to the flask making sure that the temperature does not rise above 10 ºC. Now warm the solution slowly to 25 ºC and hold at this temperature for 5 minutes. Add about 4 g of solid urea to remove the excess sodium nitrate(III). Warm the solution to 80 ºC and then boil it for no more than 5 minutes. Pour the mixture into a 250 cm3 beaker that contains about 50 g of ice cubes. Separate the oily lower layer and dry using anhydrous sodium sulfate. Re-distill the oil. The boiling point of pure 2-hydroxymethylbenzoate is 223 ºC. Think about how you can check the identity of your product.

Possibly. Also, you have to keep the reaction mixture cool (below about 5°C) otherwise the nitrous acid decomposes, producing nitrogen dioxide.

Also, perhaps you could copy-paste the relevant experimental protocol into your question rather than posting a long, ugly link.


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Poly(3-hexyltiophene-2,5-diyl) (P3HT) reactions with inorganic salts in aqueous solution
I would need for my research at least as a starting point to know what kind of reactions occur at the interface of P3HT and an electrolyte. The electrolyte is simple $\ce{KCl}$ solution, maybe with $\ce{HCl}$ or $\ce{KOH}$ added. The reactions of polytiophene with the same type of electrolyte will do, if there is no information on P3HT. The two molecules don't seem to be that much different to me from the point of view of electrolytic reactions.
What I really need is information about reaction kinetics, at least in the form of some reaction constants, but simple stochiometric equations are still better than nothing at all, so I'm interested in those, too. I have been searching for some time now, but I'm not a chemist, and I don't really know where to look to find something like this.

Assuming the electrolyte is inert (e.g., won't oxidize or reduce the P3HT), then there are no likely reactions. I would look up electrochemical potentials of KCl and P3HT, but I don't expect much would happen.

@GeoffHutchison If you measure the characteristics of a transistor made as such that you put two electrodes under a P3HT thin film, drop a drop of electrolyte on the top, and insert an electrode into the electrolyte, the characteristics of the transistor drastically change. These transistors are sensitive to both changing the pH or salt concentration. The sensitivity comes from capacitive moulation, but also there is a continuous drift in the characteristics, that might be due to surface reactions. My first impression also was that it doesn't react, but seemingly experiment shows otherwise.

I'll point to a very recent paper on oligothiophenes for solution-gated FET sensing that suggests there are no chemical reactions. (This article popped up in my feed - there are many others on solution-gating organic electronics.)
$\alpha,\omega$-dihexyl-sexithiophene thin films for solution-gated organic field-effect transistors. Appl. Phys. Lett. 108, 073301 (2016)
The authors look at many factors influencing the device performance:

Finally, excellent transistor stability is confirmed by continuously operating the device over a period of several days, which is a consequence of the low threshold voltage of DH6T-based SGOFETs. Altogether, our results demonstrate the feasibility of high performance and highly stable organic semiconductor devices for chemical or biochemical applications.

The devices in this article use KCl and phosphate buffer. They vary the pH and ionic strength.
I haven't seen anyone suggest chemical reactions, and I'm not sure how it would occur. Certainly the oligo- or polythiophene could degrade over time, but most articles argue these are stable for long-term use.
The authors do note a small drift on the time scale of days:

both changes may possibly be attributed to the passivation of trap states at
  the semiconductor-electrolyte interface by water molecules penetrating into the thin film

I can't speak to the drift you see, but I'd make sure that:

You aren't seeing evaporation of solvent over time (e.g., from device heating) that's changing your electrolyte concentration.
You use a proper reference electrode
You try removing water via vacuum and repeat the experiment (i.e., you drive off any water penetrating the semiconductor film)

Certainly organic semiconductors are porous, and while mostly hydrophobic, it's not unreasonable to expect that after days of use, some water may migrate into the film.

@GeoffHutchinson Thank you very much. To be honest, I'm not doing the experiments, but the simulations, and my experimentalist conterpart told me this one, and that it would be nice to have a model of the drift. I'll definitely talk with her, and she promised to show me how she fabricates and measures these devices. However, she is no amateur, and I don't suppose she would talk about a drift that comes from experimental error. What might happen however, is that the P3HT thin film itself degrades when operating the device (I read it in a Master's thesis though, so it's not 100% sure).

@GeoffHutchinson I didn't specify it in the question, that is true, but there is a fundamental difference between the transistor that they have made and ours: ours is solution processed, thus the P3HT film is disordered, while in the article they have used molecular beam deposition which results in single crystals. I could also imagine that the disordered P3HT film is less stable, than a single crystal (although right now I don't know why). Maybe not because of surface reactions then, but some internal change. Can this be possible?

@user3237992 - I'd be happy to consult more, but find my e-mail. Definitely if you're doing simulations, see how they're making the devices. I would not be surprised solution-deposited P3HT has differences - it's likely much more porous than the crystalline films. But I doubt it's from a chemical reaction - more likely some sort of water interpolation or rearrangement of the film. As I said, please send me an e-mail.


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nomenclature of citric acid

I named Citric acid : "3-hydroxypentan-1,3,5-trioic acid"
but when I searched , I found that the right name is "3-carboxy-3-hydroxypentane-1,5-dioic acid"
Can anyone explain to me why my nomenclature is wrong?

Note that, although the name “3-carboxy-3-hydroxypentane-1,5-dioic acid” is unambiguous and describes the correct structure, it is not in accordance with current IUPAC recommendations. The preferred IUPAC name is “2-hydroxypropane-1,2,3-tricarboxylic acid”. Furthermore, the name “citric acid” is also retained for use in general nomenclature, but no substitution is recommended; however, the formation of salts and esters is allowed.

First you must find the longest chain, which incidentally does not contain one of the carboxylic acid groups. The longest chain is 5 carbons long (hence pentane), there are 2 carboxylic acid groups on this chain, one on each end (hence pentane-1,5-dioc acid), there is a hydroxy group on the third carbon in the chain (hence 3-hydroxypentane-1,5-dioc acid), and finally there is also a carboxylic acid group on the third carbon (hence the full systematic IUPAC name: 3-carboxy-3-hydroxypentane-1,5-dioic acid).

The "-oic" ending means to convert the carbon into a COOH group, not that there's an additional COOH group added to that carbon.
Look at your main chain. You have five carbons, of which carbons 1 and 5 are fully oxidized, hence "pentane-1,5-dioic acid". If the "COOH" group was counted as an independent entity adding an additional carbon, then it would be a three carbon chain, to which was added two COOH groups.
That's why it can't be a 3-oic acid - that would imply that the third carbon of main chain has been oxidized to a COOH group (which is preposterous), not that there's a COOH group that's been added. Instead, the COOH group (carboxy) has been appended to the three position, and the molecule is named accordingly.

If you would name it as an 1,3,5-trioic acid then the carbon atom of the 3-carboxy group had to be part of the of carbon skeleton from which the name of the acid derives, namely 3-methylpentane, not pentane.
So the name is 3-carboxy-3-hydroxypentane-1,5-dioic acid, or, named as carboxylic acid, 2-hydroxy-1,2,3-propanetricarboxylic acid.

2-hydroxypropane-1,2,3-tricarboxylic acid is the correct name. For tricarboxylic acids we are finding longest carbon chain with excluding carboxyl groups. Then we are naming it with all fuctional groups but without carboxyl groups. At the end position carboxyl groups and say tricarboxylic acids.


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Nontraditional career/grad school path
I'm a math major (got my BA in May 2014) currently working in finance. I really don't feel any motivation for my current job except that it pays the bills. I would like to know if anyone has made a jump from a non-chem background into something related to chemistry, be it a lab position, grad school etc.
My rationale for choosing chemistry is that I've enjoyed working in a lab for two summers and that physical sciences are very interesting for me.
Right now, my only option seems to be UCLA extension evening/weekend classes, maybe getting a part time lab gig at a local CC (or UCLA if I am very very lucky) and see if that will get me a job or a grad school.
Here are my questions:

How would I transition from having no chemistry background?
Do I have good reasons to switch to chemistry? What else should I ask myself before I make this change?

Thank you.
Best regards,
squelchy451

It's rather for Academia SE...

Concur -- these sorts of career track questions are off-topic for this particular Stack Exchange site, per the site definitions.  Great question for Academia.SE, though.

I'm voting to close this question as off-topic because it is more appropriate on academia.SE.


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Carbonic Acid Titration with Sodium Hydroxide
For my school lab, I'm determining the solubility of carbon dioxide at different temperatures. I will determine the amount of dissolved carbon dioxide by titrating carbonated water with sodium hydroxide. I'm confused as to how I can determine the pH at equivalence point (for indicator)? I don't know the concentration of the carbonated water so how would I go about determining the appropriate indicator?

Why would the concentration affect the indicator that you use? You know you need to pick an indicator that covers the end point? The pH of the end point isn't affected by the concentration.

How do I determine the endpoint then?

The two pKa values for carbonic acid are given as 6.37 and 10.25, so you may want to pick the midpoint between these (pH 8.31) as your equivalence point. At that pH, a tiny amount of base will change the pH a lot, so the amount of base needed should not depend strongly on the exact value you select for the equivalence point. But the pH range is rather narrow, so I would recommend that you use a pH meter instead of a visual indicator for the titration.

The pH would be basic at the equivalence point. This makes the standard phenolphthalein an even more ideal choice. It goes from clear to purple at around a pH of 8.


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How is dimerization done?
Nitrogen dioxide is dimerized to form dinitrogen tetroxide according to the equation
$\ce{2NO2 -> N2O4}$. How do two molecules of one compound fuse to form another?

I think you can answer your question easily. Draw the Lewis structures for NO2 and you'll see how the dimer is obvious solution to some problem.

The same way as any other reaction takes place?

The electronic structure of NO2 leaves an unpaired electron on the nitrogen. This is one of the reasons it is quite reactive. But it is also the reason why it can easily dimerise to N2O4 as the two unpaired electrons in two molecules of the gas can pair up creating a weak bond. The dimer is favoured at low temperatures when the thermal energy isn't large enough to shake the dimer apart (I'm simplifying a bit). This is why the colour of the gas changes depending on temperature as the dimer is colourless but the NO2 is brown.

This is very correct. However, when reading the questions it appears to me the person asking it (on a rather basic level) did not understand what the dimer would look like. There seems to be a misconception about the binding process, given the phrasing of "fusing". The crucial bit here is to understand that the atoms of the 2 nitrogen dioxide molecules are not broken apart and refused to form something new, but that the molecules stay intact and bind together. See also my answer.

Maybe the confusion arises because you're unclear about what a dimer is. The molecules don't "fuse". They form a bond.
It's better to think of the dimer as $(NO_2)_2$ rather than $N_2O_4$.

I think it would be even clearer if you were to include the Lewis structure.


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Number of hydrogen bonds of organic compounds?
How to count the organic molecules ' hydrogen bonds in average?
What are the numbers of hydrogen bonds formed among alcohols ,alkanoic acids ,aldehyde , ketones , amine ,amide , and ether when they are dissolved in water?

Well, that's simple: (number of hydrogens linked to highly electronegative atoms like N or O) + (number of lone pairs).

Hint you could write down general formula for eg for alcohols it $C_nH_{2n}O$ and also  due to covalent bonds the hydrocarbons arent dissolved in water.

but like water ,when i count the H bonds in average , it should be two ,but if i add up the number of H linked to O and its lone pair ,it becomes four ,why does this occur ??


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Why is tin(IV) iodide bright orange while elemental iodine is violet?
Does it have anything to with iodines polarizability? Is there a way to explain this using sigma bonding or an MO diagram?

Well, they're completely different compounds. Is there any reason you'd expect them to not have different colours?

Why is $\ce{KI}$ white? As @NicolauSakerNeto said there is no reason why they should have the same colour, simply because they contain a common element.

I understand that. I was hoping someone could explain in terms of MO theory or bonding symmetry. For example, is the I-I bond much lower energy than the Sn-I bond so it would absorb more lower frequencies of light?


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E1 and E2 elimination reactions of isotopically labeled bromocyclohexanes

Question
  a) What would the product of the reaction be for an E2 mechanism
b) What would the products be for an E1 mechanism

The Br coming out of the page is the 79 isotope while the Br going into the page is the 81 isotope.

There is no mention of reaction conditions.
I am not totally sure what the effect of the isotope will have on bromine's ability to leave. I am guessing that the bromine-81 will leave since it is bigger and can stabilise the negative charge better.
I am also not sure how the products for the E1 and E2 mechanism will be different. Won't they be the same?

By any reasonable chemical definition of "size" the isotopes should be the same, so that can't be it. The first thing to think of when looking at isotopes is that the heavier isotope makes slightly stronger covalent bonds due to reduced zero-point energy of bond vibrations. For such high atomic masses, though, the difference must be extremely slight, so I don't know if that's the right path to go down.

@NicolauSakerNeto Thanks for that. I wasn't aware of the fact that heavier isotopes made stronger covalent bonds. This might be the way to go because I can't think of anything else that will effect its leaving ability.

This is commonly called a "kinetic isotope effect".  The wikipedia article is decent..

I have a feeling that the isotopes are only there to distinguish between the two bromine atoms. Although there is an isotope effect, with the high mass bromine the reactivity difference will be negligible. Think about the difference in mechanism between E1 and E2. I would also guess that the stereochemistry is given to you, and that will play a critical role in an E2 reaction. If you provide the full structure (and reaction conditions, if possible) I can give a less speculative answer.

By way of comparison, the kinetic isotope effect for the E2 elimination of $\ce{HCl}$ from (1-chloropropan-2-yl)benzene is only $k(\ce{^35Cl}):k(\ce{^37Cl}) = 1.0059$. The relative mass change for $\ce{^79Br}/\ce{^81Br}$ is even smaller.

Under E1 conditions (typically hot, polar protic solvent), one of the bromines will leave as bromide to give a carbocation. Ignoring the isotope effect, there's about an equal propensity for either of the isotopic bromines to leave, giving about a 50/50 ratio of the two carbocations. Deprotonation at the beta-position gives an alkene. I've only shown the most substituted possible alkene since that will be the most stable and major product.
(Note, the carbocation could rearrange, but I'm ignoring that for this level of analysis.)

Under more strongly basic conditions an E2 reaction is more likely. There is a strong preference for an anti-periplanar arrangement of the leaving group, carbons where the alkene will form, and the deprotonated hydrogen. This stereochemical requirement is more easily visualized in the chair forms, which are not equivalent. The structure on the left is more stable since the methyl group is placed equatorially. In this conformation, there are two hydrogens that are anti-periplanar with the bromine-81 atom (green). A bulky base would react with the least hindered hydrogen (purple) and eliminate the bromine-81 to give the disubstituted alkene shown. A smaller base would be more likely to produce the most stable product, which results from deprotonating the hydrogen in blue.
Just for completeness I've shown the elimination product of the less stable chair conformation. In this case, there is only one anti-periplanar arrangement. If the orange hydrogen is deprotonated, then the bromine-79 atom can be eliminated to give the cyclohexene shown.

Thanks  for the great answer. So basically,  for elimination reactions in general, if the atom is heavy, different isotopes won't effect the end product. The end product will just depend on its stereochemistry, right ? While for smaller molecules the lighter isotope is more likely to leave. Am I right?

@Nanoputian The kinetic isotope effect is related to the ratio in the masses of the isotopes. The only atom that I would worry about difference in product distribution would be hydrogen. If isotopes are specified for heavy atoms, they are more useful as tags to determine mechanism.


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pka from molecular dynamics
How can one calculate pKa values from first principle calculations, static or dynamics?
I have encountered some works using CPMD where people try to find pKa values from -(i) constrained Molecular Dynamics (MD) and (ii) metadynamics. These are somewhat more intricate ways, I believe. Is there any other simpler ways (theoretical/computational)? Can not we do it from simple MD?
Please suggest some references as well, if possible.
Thanks in advance.

Initially I tried to use the basic formula for pKa, given in wikipedia..                      $$pK_a=-\ln \left[ \frac{[A^-][H_3O^+]}{[HA][H_2O]}\right].$$For my case, I used this formula as follows, $$pK_a=-\ln \left[ \frac{69\times 2}{31\times 58}\right]=1.12$$ against experimental values of about 2.0. Here the numbers that I used represent not the molar concentration, but rather the proper number of molecule/ion. How well do you think the result is reliable?

The calculation you suggest would be valid if the given concentrations were at the system's equilibrium. If you don't intend on directly simulating a grand-canonical ensemble, feel free to accept the (only) answer below.

Have you considered a free-energy calculation?
That is, calculate the free-energy of the dissociated species and of the neutral one, and calculate the pKa via the formula for activity.
There was this GROMACS tutorial, but it seems the link is broken.

That's what I'm trying to do now, as you have suggested! I would not be able to use gromacs as my calculations are based on AIMD simulations. For free energy calculations, as statistical physics says, the formula is $F=-kT \ln Z$, where Z be the partition function. How can I calculate Z from MD trajectory?

If you can't be bothered to look up a 'smart' way, you can do this directly by binning position and momentum space and build a histogram from real-space trajectories (gotten from a MD simulation, of course).


text:
What are the chemical formulae for galvanic corrosion of iron (rusting)?
The conditions for corrosion, I am searching equations are as follows:

The other metal is copper
The temperature is $25~\mathrm{^\circ C}$
The electrolyte is $\ce{NaCl}$
pH 7
A limited amount of oxygen

Which hydroxides and oxides are formed when iron undergoes galvanic corrosion? Which ones are dominant?

You might want to provide information about the conditions - what is the other metal, what is the medium, what is the temperature, and so forth - to narrow the scope. As it stands, there are a lot of 'it depends' sorts of answers that may or may not apply to your specific situation.

Details added to the question.

@Chemobot 1) use \mathrm for units 2) no need to put every number in math mode - it only makes the page load more slowly and does not help much with presentation


text:
Is it practical to extract zymase from yeast for ethanol production?
Suppose I wanted to make some ethanol fuel. Normally, I would get yeast, mix with water, add sugar and wait. However, there is an enzyme (zymase) in yeast that does this. Looking at the Wikipedia article for zymase, someone managed to extract it using only quartz and diatomaceous earth.
The advantages of this is that it could potentially produce large amounts of concentrated ethanol, whereas yeast could only produce less than 20% due to it dying. So, do people use such a method, and if not, why not?

There is a huge industry converting grains into ethanol for fuel. 20% efficiency would be very unproductive. So there is some better process. (Evidently with removal of tetraethyl lead, to reduce lead exposure,  ethanol is a effective and cheap antiknocking compound.)

@MaxW I think not that only 20% of sugar is converted, it only produces 20% ethanol solution. At least, according to my understanding.

So, do people use such a method, and if not, why not?

No, people do not use such a method.  There are several reasons:

In fact, conversion of glucose to ethanol requires many enzymes, not just one.  The crude extract originally known as zymase was not characterized until later to consist of so many different molecules.  The molecules that comprise zymase include all the enzymes of glycolysis (e.g. glucose-6-phosphate isomerase, triose phosphate isomerase, phosphofructokinase, triose phosphate isomerase, etc.)
The other component of "zymase" as it was understood historically included dialyzable components.  Later these were identified as coenzymes, and are now understood to consist of molecules such as NAD+, ATP, and coenzyme A.
Live yeasts can synthesize all the enzymes and co-enzymes they need to convert glucose to ethanol.  But yeast extracts such as zymase can't. As they convert sugar to ethanol, eventually cofactors are used up, lost to processes such as dilution, oxidative damage, or enzyme promiscuity.  When a zymase extract loses its activity because of these reasons, it can never get them back.  But live yeasts are always synthesizing more coenzymes (and enzymes) in a programmed response to any sensed shortage.
Even zymase would not go to 100% ethanol concentrations.  Enzymes denature and lose their effectiveness at high alcohol concentrations.  The concentration where this happens is probably higher than the concentration that is lethal for yeast, but I don't think it is close to 100%.


text:
If for reversible process, ΔS= 0 for universe (system+surroundings), why would there be non-zero change in gibbs energy?
As Second Law says, the entropy must increase for universe
$$\Delta S_\textrm{universe}\ge 0$$
Now, we know, $$\Delta G= -T\Delta S_\textrm{universe}\le 0^\dagger$$
For reversible process, $$\Delta S_\textrm{reversible, universe} = 0$$
That would mean $$\Delta G_\textrm{reversible}= 0$$
But does it happen so?
Peter Atkins, in his book writes:

At constant temperature and pressure, for a reversible process: $\mathrm dG = \mathrm dw'_\rm{rev}\;.$

Now, isn't it contradicting that $\mathrm dG\ne 0$ for reversible process? If $\Delta G= -T\Delta S_\textrm{universe}$, then wouldn't $\mathrm dG= 0$ for $\mathrm dS_\textrm{universe}= 0$ during reversible process?

\begin{align}^\dagger \Delta S_\textrm{universe}&=\Delta S_\textrm{system}+ \Delta S_\textrm{surroundings} \\ &= \Delta S_\text{system}+ \frac{-\Delta H_\text{system}}{T} \\ \implies   -T\Delta S_\text{universe} &= \Delta H_\text{system} - T\Delta S_\text{system} \\&=\Delta G  \end{align}

Forgot about enthalpy H?

@Mithoron: Huh?

G=H-TS so your equations look wrong...

@Mithoron: Edited.

Think more about it and you'll see why you're wrong.

@Mithoron: Periodic Table?

The proper equation is
$$\mathrm dG=\mathrm dH-\mathrm d(TS)$$
Since $H=U+pV$, we rewrite the above equation as
$$\mathrm dG=\mathrm dU+(p\mathrm dV+V\mathrm dp)-T\mathrm dS-S\mathrm dT$$
Furthermore, we know from the first law that
$$\mathrm dU=\delta Q + \delta W$$
and
$$\Delta S=\int{\frac{\delta Q_\text{rev}}{T}}$$
which is equivalent to writing
$$T\mathrm dS=\delta Q_\text{rev}$$
so, putting it all together and remembering that temperature is constant, we have
$$\mathrm dG=T\mathrm dS+\delta W_\text{rev}-T\mathrm dS$$
Then,
$$\mathrm dG=\delta W_\text{rev}$$
where the pressure-volume work went to zero because, as you point out in one of your comments, all of this is only valid for zero expansion work.
I hope this kind of answers your question, but you seem to have a pretty good grasp on this already.

Also, it probably shouldn't be written that
$$\mathrm dG=\mathrm dW_\text{rev}$$
because this seems to imply that integrating this expression will have you take an integral with respect to work, but this cannot be done because work is not a state function. You would figure this out, however, when trying to figure out what values of work to use as the boundaries in that integral.

$\delta Q\ne 0,,\textrm{but},,=  T\mathrm dS$

For a reversible process the heat transfer does equal zero.

No, no; this does happen only for adiabatic process.

Ya you're right about that. I think there shouldn't be heat transfer in this case because I believe this is how Atkins gets to the result you're talking about.

@jheindel It is not correct to say that "for a reversible process, there must be no heat loss from the system."  What about an isothermal reversible process?

@jheindel Also, your expressions for dG are incorrect.  dG=dH-d(TS), which leads to dG=VdP-SdT

Oh I see where it's going wrong. I'll edit it later. And ya you're right. At the time I couldn't figured out what was thing wrong so I reasoned very poorly.

The equation in Atkins' book refers to the system only, and not to the universe, and, moreover, in this equation, w' specifically refers to non-PV work, like electrical.

G at constant temperature and pressure doesn't talk about the entropy of the universe?

I thought dG would be zero as $\mathrm dG= -T\mathrm dS_\rm{universe}$ and $\mathrm dS_\textrm{universe}= 0$ for reversible process?

Doesn't Atkins' book specify these constraints when it talks about the change in G for a system experiencing a process at constant temperature and pressure?  What exactly does the book say leading up to your highlighted statement?  How exactly does it define w' (with the prime)?  You seem to have omitted important details in your question.

He actually doesn't elaborate anything; only he does mention the term maximum non-expansion work at const. temp. and pressure

I guess he should have been more precise.  I don't have a copy of Atkins, but I can tell you the Denbigh (Principles of Chemical Equilibrium) has a development that spells this out with more precision.

https://books.google.co.in/books?id=BV6cAQAAQBAJ&pg=PA117&lpg=PA117&dq=non+expansion+work+gibbs+free+energy+peter+atkins+physical&source=bl&ots=N4rC6dkgPO&sig=kUlgR5ig24YI3D8V4-dbYtqBfd8&hl=en&sa=X&ved=0ahUKEwjs5_GA047LAhUQC44KHQfVBuUQ6AEIHDAA#v=onepage&q&f=false

Two Thermodynamics books I highly recommend are Fundamentals of Engineering Thermodynamics by Moran et al, and Introduction to Chemical Engineering Thermodynamics by Smith and Van Ness.

Thanks for the recommendation; but I can't immediately buy either of the two now ;/

Don't you think Gibbs energy is related to non-expansion work?

In the online version you provided, the Justification 3.3 is similar to the development in Denbigh, although Denbigh is more precise. I do think that, for a reversible change at constant temperature and pressure, the change in the Gibbs energy is equal to the non-expansion work.

Oh! But reversible process means the change in entropy of the universe is 0, isn't it?

So?  What is your point?

As $\Delta S_\rm{universe}= 0$, the equation (valid at const. temp and pressure) $\Delta G= -T\Delta S_\textrm{universe}$ yields 0  but also the derivation that Atkins shows gives $\Delta G= w'$ anon-zero quantity. So, isn't it contradictory?

It's not contradictory.  The equation we have been discussing applies to a specific closed system, and not necessarily to the universe.  If you want to apply the equation to the universe, then, for the universe, w' = 0, because the universe is not doing any reversible non-PV work on anything.  The reversible non-PV work that a closed system does on the surroundings is exactly equal in magnitude and opposite in sign to the reversible non-PV work that the surroundings does on the system.  So, if you include both system and surroundings, the reversible non-PV work is zero.

Are you saying $\Delta G= -T\Delta S_\rm{universe}$ (at const. temp. and pressure) is wrong?

I'm saying that, for a reversible process at constant temperature and pressure, the non-PV work done by the universe is zero, and $\Delta G$ of the universe is zero.  But please understand that, the real applicability of the equation in Atkins is for a closed system (of lesser extent than the entire universe) experiencing a reversible change at constant temperature and pressure.  I've run out of ideas on how to explain this more clearly.  Maybe someone else can help out.  I need to go to bed now.

Good night ;) So, at reversible process, the non-expansion work of the universe is zero?

I think it is a very bad idea to think of this in terms of the universe.  The focus should be on a finite closed system.  If you include both the system and surroundings, then this combination can't do work on anything (either reversible or irreversible).  The work the system does on the surroundings is just the negative of the work that the surroundings does on the system, so the sum of the two (the net work) is zero.  Try to picture in your mind the universe doing work on anything, given that work only occurs at the boundary of a system.

Let us continue this discussion in chat.

I don't think so.  I'm pretty much out of ideas on how to explain this.

@user36790 In the future, it's better to formulate some of these into new questions rather than having a back and forth in chat between just two people.

@Chester Miller : Have you visited the chat-room & read my comment?

At constant temperature and volume, \begin{align}W_T & \le (U_1- U_2)- T(S_1-S_2)\ \implies W_T & \le (F_1 -F_2) \end{align} Now, if reversibility exists then $$W_T= F_1- F_2$$ though $dS_\textrm{universe} = 0$

Generally it is written that $$\mathrm dS_\textrm{surr}= \frac{-\mathrm dH_\textrm{sys}}{T}$$ but it is valid only when the non-expansion work is zero. Generally $$\mathrm dH = \underbrace{w'}_{\textrm{non-expansion work}} + \delta q$$

So, when one uses $\mathrm dH $ to denote heat transferred (or absorbed) in the surroundings, then he is definitely taking $w' = 0,,$ isn't it?


text:
Is a galvanic cell only called that if copper and zinc are used?
I know other metals can be used to get the same effect but I'm interested in the name. Is it only called a galvanic cell if copper and zinc are used? Or is the name "voltaic cell" used only when copper and zinc are used? I'm confused because I've heard them being used interchangeably.

Maybe you are thinking of a Daniell cell, which consists of a copper electrode and a zinc electrode.

Voltaic cell and galvanic cell are both synonyms for the same thing: An electrochemical cell, which produces an external electric current. So the two terms can be used interchangeably.
Check out also a site from Georgia state University, where it says:

Electrochemical cells which generate an electric current are called voltaic cells or galvanic cells, and common batteries consist of one or more such cells.

Addendum: Nevertheless there might be some galvanic cells, like the Daniell Cell, which have a defined electrode material and defined electrolyte concentrations to supply a specific voltage.
For the Daniell cell these would be:
$$\begin{array}{lc}
\text{Electrode material:} & \ce{Zn|Zn^{2+}||Cu^{2+}|Cu}\\
\text{Electrolyte concentrations:} & c(\ce{Zn^{2+}}) = c(\ce{Cu^{2+}}) = 1\ \mathrm{mol\ l^{-1}}\\
\text{Voltage:} & U(25\ \mathrm{°C}) = 1.10\ \mathrm{V}
\end{array}$$


text:
What does a molecules color have to do with its bond/orbital energies?
For example, elemental iodine is deep violet. Its sigma bond or perhaps the lone pairs are capable of absorbing all visible light frequencies except violet which is why we see it as that color.
Lithium, on the other hand, is red which means its valence electrons must be reflecting red frequencies.
So, does that mean the valence/bonding electrons of $\ce{I2}$ are higher energy than the lithium valence electrons?
In other words, is there a relationship between an orbitals energy and the type of light it can reflect?
You could perhaps compare $\ce{I2, Sn^{II}I2}$, and $\ce{Sn^{IV}I4}$, and which are violet, yellow, orange (in decreasing order of visible light energy).
Perhaps it has to do with symmetry? $\ce{I2}$ is $D_{\infty\mathrm{h}}$, stannous iodide is $C_{2\mathrm{v}}$, and stannic iodide is $T_\mathrm{d}$.

Bonds do not absorb light, neither do lone pairs. It is a transition between certain orbitals that actually works.

You could have edited earlier question http://chemistry.stackexchange.com/questions/46785/why-is-tiniv-iodide-bright-orange-while-elemental-iodine-is-violet

related http://chemistry.stackexchange.com/questions/16633/why-is-gold-golden

@IvanNeretin: It is even worse, the absorption of light corresponds transition between states.

Would you leave out iodine, there could be some trends observed. Lot of information can be gained from  https://en.wikipedia.org/wiki/Ligand_field_theory and subsequently https://en.wikipedia.org/wiki/Tanabe%E2%80%93Sugano_diagram

Probably interesting too, although more organic: http://chemistry.stackexchange.com/questions/9493/what-is-the-origin-of-the-colour-of-azo-dyes/9495

I understand there is absorption going on here, but WHICH electrons (as in WHICH orbitals) are the ones doing the absorbing of red, orange, yellow, and green to give Iodine its purple color. Is there any way to know this information?

Rather than give answers to your specific question I try to explain what happens in general as there is some confusion in your question. You should really consult a book on spectroscopy, of which there are many, for a thorough explanation.
In a molecule the arrangement of electron in orbitals is called a state, hence ground state, first exited state etc. When all the electrons a molecule has fill up all the orbitals with no gaps left, this is the ground state. They will fill up half of all the orbitals, two to an orbital.  When a photon is absorbed and it has sufficient energy (usually in the visible or UV part of the spectrum) an electron is promoted from one of the bonding orbital to a anti-bonding orbital which has higher energy. This arrangement of electrons is called an excited state. This transition between states is what gives molecules and atoms their colour.
In iodine the difference between the lowest excited state and the ground state occurs in the visible part of the spectrum. (for simplicity, I'm ignoring any vibrational transitions accompanying the electronic excitation). Benzene has a larger energy gap between ground and excited state than I$_2$ and so absorbs only in the UV and looks transparent to us.
The same basic idea applies to all atoms and molecules. Where the molecules or atoms absorb is a function of the exact electronic energy difference between ground state and excited state of the particular species. An understanding of the nature of the bonding and so energetics allows us to predict trends in absorption spectra, but exact values can only be obtained from quantum calculations.


text:
Why do reactions of sodium halides with sulfuric acid produce sodium hydrogen sulfate instead of sodium sulfate?
In the reaction of sodium halides with sulfuric acid, sodium hydrogen sulfate is produced \eqref{NaHSO4}. Why would this be produced instead of sodium sulfate \eqref{Na2SO4}?
\begin{align}
\ce{NaX + H2SO4 &-> NaHSO4 + HX}\tag1\label{NaHSO4}\\
\ce{2 NaX + H2SO4 &-> Na2SO4 + 2 HX}\tag2\label{Na2SO4}
\end{align}
Is this because of lower energy requirement for this reaction compared to forming sodium sulfate?

Related: Why do sodium halides react so differently with sulfuric acid?

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text:
Does dissolving a solute lowers the activation energy of a reaction of that solute?
I'm formulating a reason why the use of solvents in some reactants are preferred rather then mixing just the pure components together. My gut feeling tells me that it lowers the activation energy of the reaction but, I can't find explicite answers to confirm my guess. Can someone help me with this :)?

Dissolution may or may not affect the activation energy, but there is a factor more important than that. See, you can't really mix two solid compounds on molecular level, even if you grind them together very thoroughly.


text:
Is hydroquinone polar
Which of the following molecules is/are polar

$\ce{B3N3H6}$
Hydroquinone
Benzene
$\ce{IF7}$

Molecules (4),(3) are non-polar. Molecule (4) has pentagonal bipyramidal structure and the dipole moments of \ce{I - F}$ Bonds cancel out.
I thought the answer was (1) (inorganic benzene). But the answer given is (2). How is hydroquinone or para-hydroxyphenol polar?

I find the term "polar" somewhat confusing. They all are polar, except benzene, but some have zero dipole moment.


text:
What is the patina formed on copper pennies?
I'm currently taking high school chemistry and I don't quite understand my results from my experiment.
I had two pennies where both were put into a mixture of salt and vinegar. One of the pennies was dried and cleaned and the penny was like brand new. While the other penny Is left it in the mixture for 2 and 3 days with water.

Once I took it out, the penny had patina  forming. I'm not quite sure what type of reactions are occurring here and what elements/compounds are coming into play.

Do you know the years of the pennies?

Not really but why is that of any importance?

Pre-1982 pennies are 95% copper. Post-1982 pennies are 97.5% zinc. Zinc oxidizes more readily than copper, but with the 95% copper penny, that may not be an option, leading to copper "rust," i.e. oxidation.

The opposite may be true, though. If the zinc has no exposure to the oxidant inside the penny (new pennies are a copper shell around a zinc core), then the copper might have to oxidize, where the 95/5 Cu/Zn pennies may have been an alloy, allowing the zinc access to the oxidant.

Oh wow seems like you have profound knowledge on this. Let's assume the penny is very recent for this problem.

...and in 1943 pennies were a zinc coated steel. Those actually do rust.

Maybe rust was a bad way of describing the surface of the penny. It looks like this: http://cdn2.buggyandbuddy.com/wp-content/uploads/2013/06/Pice.jpg

Patina - that would be the right word.


text:
Why do some gases have lower value of Z for a particular pressure?

In the above graph,the minima of the curve for methane is more than that of nitrogen. Also, for a given value of pressure, the value of $Z$ for methane is less than that of nitrogen. They seem to meet at high pressure. Why do some gases have lower value of Z?
The same thing is observed if you compare methane and carbon dioxide. But they don't seem to meet at high pressure. This is because the $Z$ for methane increases at a faster rate. Why does Z increase faster for some gases?
The real gas equation is $$\left(P+\frac{a}{V^2}\right)\left(V-b\right)=RT$$
I tried expressing $Z$ in terms of $P$ and $T$ by eliminating $V$.
$$V=\frac{ZRT}{P}$$
$$\left(P+\frac{aP^2}{Z^2R^2T^2}\right)\left(\frac{ZRT}{P}-b\right)=RT$$
$$ZRT-Pb+\frac{aP}{ZRT}-\frac{aP^2b}{Z^2R^2T^2}=RT$$
Multiplying by $Z^2$ on both sides,
$$Z^3RT-Z^2(Pb-RT)+\frac{ZaP}{RT}-\frac{aP^2b}{R^2T^2}=0$$
$$f(Z,P)=0$$
(I am assuming constant temperature)
Should I find $\frac{\partial f(Z,P)}{\partial P}$ and make it equal to zero to get information about the minima?

The trend you've noticed here is a very interesting way of studying the interactions of a gas with itself.
You are certainly right about how to find the minimum of these curves, but the important thing here is what that minimum actually means.
$$Z=\frac{PV}{RT}$$By analyzing this, we expect that $Z$ will tell us something useful about the way the volume of a gas changes as pressure changes (holding $T$ constant just gives better information on a graph). Specifically, we see that if pressure increases but volume decreases more, we will have a number less than 1 and if volume decreases less than pressure increases, we will have a number greater than 1. So what does that number tell us?
Well, if pressure increases and the volume gets even smaller than we would expect for a real gas, these gases must be attracted to each other. Then, we can interpret the minimum of this graph as being the pressure (for a given temperature) at which a gas molecule is attracted to itself most strongly. Beyond this point, repulsive interactions become more and more important. Notice that there is also no maximum. Or, there is no point where attractive interactions dominate repulsive interactions again which is what we would expect.
Now, to explain why we see the behaviors we do from certain molecules, we have to imagine what is causing there to be attractive and repulsive interactions at all. This is due to polarizability and the amount of van der waal's forces.
Indeed, if we look up value of the parameter $a$ for the molecules present (I have them in a textbook) we find:$$a_{\ce{H2}}=.24646$$$$a_{N_2}=1.3661$$$$a_{\ce{CH4}}=2.3026$$$$a_{\ce{CO2}}=3.6551$$where $a$ is a measure of how attracted the molecules are to themselves (in the horrible unit of $\frac{dm^6*bar}{mol^2}$).
These values do indeed follow the trend we see in the graph above. That is, larger value for $a$ has a larger attractive deviation initially. $\ce{H2}$ is something of a special case because it is effectivly unpolarizable, so we see a linear increase in the compressibility factor with pressure. This basically means that hydrogen will deviate from the ideal gas precisely the same at all pressures.
My best guess at why the slopes then increase at different rates has to do with the size of the actual molecule. Again, looking up the values for $b$ which represents the excluded volume of a molecule, we have:$$b_{\ce{H2}}=.026665$$$$b_{N_2}=.038577$$$$b_{\ce{CH4}}=.043067$$$$b_{\ce{CO2}}=.042816$$with $b$ in units of $\frac{dm^3}{mol}$.
So, this predicts methane will have a slope increasing more quickly than both nitrogen and carbon dioxide which is what we see and that hydrogen will have the shallowest slope which is again what we see.
Hope that helps.

So the equation $f(Z,P)=0$ will not work properly because it will give a Maxima and a minima.

I assume there's one compound which reaches the lowest value of Z at a given temperature. I wonder what it is, and what makes it so. Wikipedia has a list which suggests iodobenzene is up there, which I didn't quite expect.

@NicolauSakerNeto cubic functions can also have a Maxima. But Z vs P curves don't have a Maxima. Is there a way to get the curve equation using my approach?

Does $f(Z,P)=0$ even make sense? It was derived from the real gas equation.

Can you comment on the role of a temperature in this graph? If the temperature is changed, keeping P constant, what changes will be observed?
And will the line for dihydrogen remain straight?

Also, tallying your data from this page, I think there is a typo in the unit of van der waal constant $a$ and $b$ that you have stated. The magnitude of the values matches, but the unit doesn't.

@Guarang regarding the units, a $1/ dm^3=1/ L$.

Considering Z for the van der Waals equation (upper part of the image), deriving this equation with respect to Pressure(P), the slope at P = 0 (initial slope of the curve) can be negative or positive. If "b" is greater than the term "a / RT", the effect of molecular size governs the behavior of the system, the slope is positive and Z>1. If b is less than a / RT, the attractions prove that the gas has a volume less than that calculated through the ideal gas equation, the slope is negative and Z<1.

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text:
Why are Erlenmeyer flasks shaped the way they are?
The title says it all. We could conceive a good number of other possibilities. Why did Erlenmeyer arrive at this design?

I'm sure it was by experimenting with various shapes.

Imagine a time when magnetic stirrers were unknown. You, or your lab assistant, would mix solutions in suitable glassware, shake manually, put it aside and then observe a change.
An Erlenmeyer flask is much more suitable than a simple beaker

the flat bottom gives it a stable stand on the bench
the narrrow neck allows to hold and shake a large flask with one hand
the conical shape prevents spills when the flask is shaken circularly

This is a great answer for provenance.  However, you also get some reflux action when boiling solvents which tends to minimize crystal cakeup along the sides of the vessel, and as such tends to make resulting product somewhat more pure.


text:
Balancing molecular and net ionic equations for reactions
Source:'Chemistry: The Central Science 11th edition' by Brown. Ch.4, 4.53:
Question:

Write balanced molecular and net ionic equations for the reactions of:

chromium with hydrobromic acid;
tin with hydrochloric acid.

My attempts:

$\ce{ Cr(s) + 2HBr(aq) →CrBr2(aq)~ or~ CrBr3(aq)? + H2(g)}$
I got the left part of the reaction correct, but I can't figure out what charge chromium should have because it can have either $\ce{+2}$ charge or $\ce{+3}$ charge when it ionizes.  The answer says that it should be $\ce{CrBr2}$, but why can't it be $\ce{CrBr3}$?

same thing for this option. Why can't $\ce{SnCl4}$ can be an answer?


text:
What is the maximum homogeneous nucleation rate of supercooled liquid quartz?
I read in College Physics, Volume 1 (by John Oren Reed, Karl Eugen Guther) that fused quartz is just supercooled liquid quartz. Is it really supercooled liquid quartz?
As liquid quartz is cooled below its melting point, does it eventually get to a temperature where achieves such a high homogeneous nucleation rate that it can't be supercooled any further.
Is it the case that the homogeneous nucleation rate only goes up to a rate that's still low enough to avoid nucleation at the macroscopic level with slow cooling and then starts going down again as a result of bonds not getting broken and reformed as often due to the lower temperature?


text:
How Fittig reaction gives good yield if phenyl cations are unstable?
I am confused,after reading from NCERT text that phenyl cations are unstable due several reasons like the carbon in phenyl group is sp2 hybridized,resonance structures do not stabilizes the phenyl cation.But then,in Fittig reaction we do have phenyl cations as intermediate.For example in a reaction between two molecules of chloro benzene and sodium metal we get a diphenyl molecule.Here we have two phenyl cation generated (if i am right).Then what factor stabilizes the phenyl cation?If there is any error in the theoretical aspect of the question please help me correct it as i am not quite sure about the mechanism.
I would be very much glad to accept an answer with some thermodynamical data like the bond enthalpies of C-Cl bond and newly formed product's bond enthalpy that is C-C bond enthalpy(in the case of diphenyl)and the total realease in energy.

Then, probably, we don't have phenyl cations as intermediates?

@IvanNeretin  so what you are saying is that phenyl cations are not being generated?right? –

The Sandmeyer reaction involves the formation of aryl cations. Although they are very high energy they are formed in this reaction because nitrogen (the leaving group) is also formed in the process. Since nitrogen is an extremely stable molecule it drives the formation of the phenyl cation..

@ron yeah that would be correct.But in wurtz reaction what you are saying would be that the formation of NaCl releases lot of energy and the C-Cl sp2 hybridised bond will broken as the energy released  by formation of NaCl is greater than the bond enthalpy of C-Cl bond.correct?Would you provide me with the bond enthalpy of C-Cl bond in chloro benzene as i didn't get any result from internet.

I don't know if the Wurtz reaction involves cations, but the reaction involves solid sodium and complex things can take place on a metal surface.

@ron in some internet  sites i found that a phenyl anion being generated and in some it says it proceeds with free radical mechanism.But i feel that if in any case the C-Cl bond is broken a phenyl cation would be generated as chlorine is more electronegative.

The Wikipedia page for the Wurtz-Fittig reaction suggests the alkyl halide is reduced twice to the organosodium species, which undergoes electrophilic attack on the aryl halide.

The most probable mechanism for Wurtz reaction or Fittig reaction involves formation of organic free radicals and not cations. Like in Wurtz reaction, phenyl free radicals are formed in Fittig too.

I checked out the site you mentioned.But it doesn't have any data regarding    the mechanism for halobenzene.Rather it has only data regarding aliphatic halides (that is wurtz reaction and not fittig reaction),which probably won't be the answer as both involves completely different type of molecules.One is aromatic and other is aliphatic and C-X bond these two have different bond enthalpy(if i am right).

Expecting different mechanisms for different molecules is good. However, what is important is in both the cases, Na is used. It is a good reducing agent and can provide electron equally well to aliphatic and aromatic compounds. One more aspect is the C-X bond in aromatic compounds is stronger also implies, X cannot leave as X-


text:
Reactivities: interhalogen compounds versus halogens
My textbook says that interhalogen compounds are more reactive than halogens, because the $\ce{X-X}$ bond in halogens (for example, dichlorine or $\ce{Cl2}$) is stronger than the $\ce{X-X'}$ bond in interhalogens (for example, chlorine monofluoride or $\ce{ClF}$).
(In the symbols that I have used, $\ce{X}$ and $\ce{X'}$ have been used to denote dissimilar halogen atoms.)
So my question is: why is the bond between dissimilar halogens weaker than the bond between atoms of the same halogen?
I am aware that the bond between two identical atoms will be a completely covalent bond due to no difference in the electronegativities of the participant atoms. So, the $\ce{X-X}$, bond as in $\ce{Cl2}$ will have no ionic character.
In case of an interhalogen compound, however, the difference in the elecronegativities of the (dissimilar) participant atoms will cause the bond to become polar and thereby acquire ionic character.
So is it the ionic character of the interhalogen compounds that causes them to be more reactive than halogens? But then, this means that ionic bonds are weaker than covalent bonds.
But, the answers here as well as here say that an ionic bond is stronger.
But this link says the very opposite. Okay, the people writing the answers there have written it to suit an eight grader's understanding, but even so, they have stated that covalent bonds are stronger than ionic bonds.
So which alternative am I to believe? I know that most bonds are neither purely ionic not purely covalent, but intermediate between these two extreme characters, but this only serves to increase my bewilderment.
Can somebody please resolve this, or tell me where I have gone wrong in my reasoning as regards the reactivities of the interhalogen compounds in comparison with halogens?
And finally, please tell me the reason why interhalogens are more reactive. Thanks!

See this:https://in.answers.yahoo.com/question/index?qid=20090728202251AA0E7oV

This question makes less than little sense. Some covalent bonds are stronger than some ionic bonds. And some ionic bonds are stronger than some covalent bonds.

Here is the answer for which bond (ionic or covalent) is stronger.
There is no direct answer to this question as it depends on atoms that take part in bond formation.For example you might be knowing that a Fluorine- Fluorine bond is weaker than a chlorine-chlorine bond due to high inter electronic repulsion.In another example iodine-iodine bond is weaker than chlorine-chlorine bond.Similarly there is certainly this sort of variation in ionic bond as well.
So if you compare the weakest ionic bond and strongest covalent bond then covalent bond would be stronger.And if you compare stongest ionic bond and weakest covalent bond, then of course ionic bond would be stronger.Also in many substances its not only the bond strength that matters but also the crystal structure.
An example where covalent bond is stronger:Carbon-Carbon bond is stronger than NaCl bond.
Whereas NaCl ionic bond is more stronger than Br-Br covalent bond.
So it depends on the atoms that participate in bond formation.
visit https://labs.chem.ucsb.edu/zakarian/armen/11---bonddissociationenergy.pdf for bond enthalpy.


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How is it possible that meta xylene exists?
Methyl group is ortho para directing so it should direct the
 second methyl group to ortho or para position giving rise to ortho or para xylene but how can meta xylene be formed?

Related: http://chemistry.stackexchange.com/questions/5660/how-does-toluene-react-at-higher-temperatures-and-why. As @Klaus mentioned, there are many other routes to m-xylene that do not involve Friedel-Crafts reactions.

Friedel Crafts alkylation of toluene is far less regioselective than you might think.
The meta-substituted products are thermodynamically favoured. At higher temperatures, the alkylation yields more 1,3-dialkyl substituted benzenes.
On a larger scale, Dimethylbenzenes are typically not synthesized by alkylation of toluene, but stem from catalytic reforming processes using alkanes as feedstock.

I don't know about the practical issues, but theoretically benzoic acid can undergo Friedel-Crafts alkylation to form 3-methylbenzoic acid which can be reduced to m-cresol and further to m-xylene


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Is there an easy way to test for the amount of iron (Fe) in water?
I have a rusty metal rainwater tank, and I'd like to find the amount of iron in the water. (The water I've collected is perfectly clear.)
How would chemists in the 1800's have gone about this? What chemicals/apparatus would I need to purchase?
So far, I've soaked a tissue in the water and am letting it air dry to see if any rust discoloration becomes apparent.
The following info from the WHO might be helpful:

Iron (as Fe2+) concentrations of 40 µg/litre can be detected by taste
  in distilled water. In a mineralized spring water with a total
  dissolved solids content of 500 mg/litre, the taste threshold value
  was 0.12 mg/litre. In well-water, iron concentrations below 0.3
  mg/litre were characterized as unnoticeable, whereas levels of 0.3–3
  mg/litre were found acceptable (E. Dahi, personal communication,
  1991).
In drinking-water supplies, iron(II) salts are unstable and are
  precipitated as insoluble iron(III) hydroxide, which settles out as a
  rust-coloured silt. Anaerobic groundwaters may contain iron(II) at
  concentrations of up to several milligrams per litre without
  discoloration or turbidity in the water when directly pumped from a
  well, although turbidity and colour may develop in piped systems at
  iron levels above 0.05–0.1 mg/litre. Staining of laundry and plumbing
  may occur at concentrations above 0.3 mg/litre (4).
Iron also promotes undesirable bacterial growth ("iron bacteria")
  within a waterworks and distribution system, resulting in the
  deposition of a slimy coating on the piping (4).
  http://www.who.int/water_sanitation_health/dwq/chemicals/iron.pdf

Iron particles can precipitated by fine filter paper and visually checked

In a rainwater tank in contact with air, dissolved iron is probably oxidized to iron(III) ($\ce{Fe^3+}$).
A simple and sensitive test for $\ce{Fe^3+}$ in water uses thiocyanate ions ($\ce{SCN-}$, also known as rhodanide), which form the blood-red coloured complexes $\ce{[Fe(SCN)(H2O)5]^2+}$, $\ce{[Fe(SCN)2(H2O)4]+}$, and $\ce{[Fe(SCN)3(H2O)3]}$.
$$\ce{[Fe(H2O)6]^3+ + SCN- <=> [Fe(SCN)(H2O)5]^2+ + H2O}$$
Take a precise volume of the water (e.g. $10\ \mathrm{ml}$), acidify with dilute sulfuric acid (e.g. $2\ \mathrm{ml}$), add a thiocyanate solution (e.g. $5\ \mathrm{ml}$ $\ce{KSCN}$ or $\ce{NH4SCN}$, $c =0.5{–}2\ \mathrm{mol/l}$), and fill up with distilled water to a precise volume (e.g. $25\ \mathrm{ml}$).
An orange to red colour should appear.
The colour of the solution can be compared to that of standard solutions that contain known concentrations of $\ce{Fe^3+}$, e.g. prepared from iron(III) nitrate ($\ce{Fe(NO3)3.9H2O}$).
In a laboratory, the intensity of the red colour could be measured using a photometer.
Similar simple test kits are commercially available, e.g. for aquariums, garden ponds, or other water or soil samples.

I agree with colorimetry being the method of choice. It cannot be very precise by definition, but you just cannot make anything wrong. It is very robust and does not need precise scale. The standards can be prepared by further dilution in known ratios. You could also use your camera as a very imprecise photometer.

My stock method is to use 2,2'-bipy and iron to make the red complex for measurement, my method would require you to add a reducing agent such as hydroxylamine to get all the dissolved iron into the form of iron(II). But the thiocyanate method would work,

I would disagree with Ssavec, when used with care colorimetry can be used to get very precise results

If you want to test for the presence of iron(II) in solution, you could add $\ce{CO3^{2-}}$ ions to form a green precipitate.  This could be extracted and measured, although it could be a very small amount unless you concentrated the sample water by evaporating it down.  The main tool would be an accurate balance, but also some liquid measuring equipment.
An alternative method that would be more analytically accurate would be a redox titration using something like potassium permanganate or dichromate.  This would require an accurately known concentration of the reagent, and some more specific equipment like a burette and a stand, and a volumetric flask for precise liquid measurements.

Very small amounts of $\ce{Fe^{2+}}$ ions are quite improbable in water, because they are quickly oxidized by atmospheric oxygen, and transformed into $\ce{Fe^{3+}}$ ions. And this $\ce{Fe^{3+}}$ ion does not make a green precipitate with $\ce{CO3^{2-}}$ ions. It makes a brown precipitate. The same critics could be made for the second method, using potassium permanganate, which only reacts with $\ce{Fe^{2+}}$ ion and not with $\ce{Fe^{3+}}$ ions

There is a problem which I think that the other people might not be aware of. It is a problem which is well known in nuclear chemistry. If you take a sample of lake water from some random lake such as a Swedish lake then if you measure the plutonium content of the water by digesting the sample and then filtering then you get a higher value for fallout plutonium than if you filter it first and then digest.
While iron is very different to plutonium, I think that the same problem could exist. The amount of iron which is present as a fine (colidal solid) might be important. I would suggest the following.
Take a large sample of water from the tank, now filter part of this through the finest filter you can find. The other part you should combine with a known volume of hydrochloric acid and boil it up.
After cooling the boiled acidic water should be filtered. I would take both filtrates and then measure the iron by means of colourmetric tests.
My normal test for iron uses 2,2'-bipy to form a complex with the iron(II), this test can be used to determine what fraction of the iron is present as iron(II).
I would make up four volumetric flasks with the filtered iron solutions, I would put in each flask ammonium aceate (to buffer the mixture), 2,2'-bipy solution and the iron sample. I have found that this method works down to ppm levels of iron. To two of the flasks I would add hydroxylamine solution to reduce all the iron to iron(II). I would also make up a control (reagent blank) flask and a set of standards in the range 0 to 10 ppm of iron. I would then use a UV / vis machine to measure the red complex [Fe(bipy)3]2+
I think that a late victorian chemist would be able to make 2,2'-bipy from pyridine, but if you want a method which an early victorian could have used then use potassium thiocyanate to form the red complex with the iron(III).
For that method you would need to put in the flasks potassium thiocyanate, you might want to read about the Fricke dosimeter which is used in nuclear chemistry to measure very high radiation doses. It is based on the conversion of iron(II) in acidic media which is oxygen saturated. The conversion to form iron(III) is under some conditions proportional to the radiation dose. Some people use thiocyanate in the iron(III) determination to make it more senstive.
I suggest you read
The International Journal of Applied Radiation and Isotopes
Volume 29, Issue 3, March 1978, Pages 151-157
https://doi.org/10.1016/0020-708X(78)90137-0
If you want to dig deeply into the problem of how to measure iron(III) with thiocyanate


text:
Does glycerin promote bacterial growth the same as water?
A sodium chloride + water solution draws the water out from bacteria through the process of osmosis.
Would a sodium chloride + glycerin solution have the same antibacterial effect?
If I understand correctly if the concentration of water is greater on the outside of a bacterial cell the water will go inside the bacteria allowing it to multiply or grow faster. Does glycerin have the same effect?

If I understand correctly if the concentration of water is greater on the outside of a bacterial cell the water will go inside the bacteria allowing it to multiply or grow faster. Does glycerin have the same effect?

Usually the point of adding salt to water is to lower its activity, in order to make the water concentration outside the cell lower than on the inside. This leads to water wanting to escape the cell, causing osmotic pressure to swell the cell.  With strong enough salt, the osmotic pressure causes water to flow out of the cell, so much that the cells dry up and aren't able to function.

Water is miscible with glycerol, which makes glycerol an osmolyte.  That means that glycerol would have the same effect as salt.  Water mixed with glycerol is lower in concentration than pure water.  Putting a happy aqueous bacterium into 100% glycerol will cause immense osmotic shock, probably resulting in death for most bacteria, due to too much of their intracellular water trying to escape to the extracellular, water-free (but glycerol rich) environment.

The downside of using glycerol is that at low concentrations (say 10 g/L or even 100 g/L) it is a great food for many types of microorganisms.  Sure at >1000 g/L it is a deadly osmolyte, but at low concentrations, its a yummy food.  Salt would merely lose its effectiveness as it gets diluted, but glycerol on dilution becomes a potent agent that would enhance microbial growth.  For this reasons, and others (costs, viscosity) it is not often used (outside of laboratory settings)

Thank you, very insightful. That would make homemade soap an antiseptic due to its high glycerine content (in commercial soaps, manufacturers remove glycerine from their faux "soap" to sell separately at a higher margin to the cosmetics industry). Make your own soap, people.

I know this is an old answer, but you've got it backwards. The water coming out of the cell does not cause it to burst. It flows out easily and the cell shrinks to the point that it's too dried out to function. To burst a cell, you need water to flow in, so you need water that is not salty. If there is no cell wall or only a weak cell wall, the cells may burst. Most bacteria are hardy enough to maintain their structure, but treatment with compounds like penicillin can make some of them susceptible.

Thanks for the correction!  Edited.  Let me know if the current version seems OK to you.


text:
What is the exact meaning of ionic and covalent character?
In my textbook, it is written that 'Even in case of covalent bond between two Hydrogen atoms there is some ionic character.'
Didn't get that.

Sounds like nonsense to me. I mean, they may have had in mind things like polarizability, induced dipoles... but still, this just sounds ugly. If H-H is not a 100% covalent bond, then what is?

What the book means is that the canonical forms $\ce{H-H <-> H^{-} ... H^{+} <-> H^{+} ... H^{-}}$ all contribute to the true picture of bonding in $\ce{H2}$, even if we often imagine the bond as purely covalent. This is true. It is possible to prove that calculations which ignore the ionic resonance structures result in a worse approximation of experimental energy levels. One of the computational/theoretical chemistry users around here can probably give a great answer regarding this.

Resonance like thing in H2 molecule?

Hmm I think you are right. So is this London Forces?


text:
Compound with two functional group
Compound X has 3 carbon atoms in one molecule. It contains a primary alcohol and a carboxylic acid functional group.
Draw the displayed formula of Compound X.
Help! Condensed formula is fine for me!

Welcome to Chemistry.SE!  This appears to be a homework question, please share your thoughts and attempts towards the solution.

Draw a chain of three carbon atoms.
Make one of them $\ce{COOH}$. You have two carbon atoms to choose from.
Attach $\ce{OH}$ to one of the remaining carbon atoms to make it a primary alcohol. You have one carbon atom to choose from.

So I have OH on one end and COOH on the other end, am I correct?
OH-CH2CH2-COOH

@Benedict Based on your description, I came to the same conclusion :) We learn: If in doubt, draw structures! If you found my hints helpful, please mark my answer as accepted. Btw, welcome to Chemistry.SE!

Hi, I would like to ask one more thing. Is it possible for Compound X to go through self-polymerization? A condensation polymerization?

@Benedict Yes :) You're right, the condensation will form chains with a repeating ester motive.


text:
Deviation of Ytterbium from general trend
What is the reason for deviation of ytterbium from the general trend of lanthanides in terms of hardness, melting point, density etc.?

You can find many 'deviations' in the periodic table - nobody said it would be a set of monotonic series, so don't be surprised when things are not simple. One thing to note for the three properties you listed is that Yb is a fcc crystal at room temperature, while the immediate neighbors (Gd to Hf) are all hcp (although many, including Yb, go bcc at higher temperatures). Why fcc? Well, the electronic structure happens to be happier in that crystal structure based on a number of competing factors.

To compliment Jon Custer's comment, we can likely trace this deviation to the electron configuration of ytterbium. Note that ytterbium is not the only deviation among the lanthanides - Europium deviate as well:

Why is that? Note the electron configurations of europium and ytterbium:

Europium - $\ce{[Xe]} 4f^7 6s^2$
Ytterbium - $\ce{[Xe]} 4f^{14} 6s^2$

In europium, the $4f$ subshell is half-filled, an arrangement that is particularly desirable from a quantum mechanics standpoint. In ytterbium the $4f$ subshell is completely filled, another 'happy' arrangement. These configurations lead to changes in bonding behavior, for example noting that most lanthanides have the hexagonal close pack or the double hexagonal close pack crystal structures. The only ones that are different are samarium, which is rhombohedral, europium which is body-centered cubic, and ytterbium with is face-centered cubic.
The link between electron configuration and physical properties exists elsewhere in the periodic table, but the trends are not as clear. However, transition elements with filled or half-filled $d$ subshells are likely to have lower melting points than their neighbors.


text:
Comparing pKa of Alpha and Beta Amino Acids

The $\mathrm{p}K_{\mathrm{a}}$ of the functional groups in L-glutamic acid (shown below on the right) from left to right are $4.1$, $9.5$, and $2.1$. Predict the $\mathrm{p}K_{\mathrm{a}}$ values of B-glutamic acid (below on the left).

I am not sure on the answer and I am just guessing here.
The $\mathrm{p}K_{\mathrm{a}}$ value for the $\ce{COOH}$ group on the right will increase at it won't be stabilised by the amino group any more. The $\mathrm{p}K_{\mathrm{a}}$ of the $\ce{COOH}$ on the left won't really change.I am guessing that the $\mathrm{p}K_{\mathrm{a}}$ of the amino group will decrease. That is a complete guess.
Can anyone please explain how exactly the amino group effects the stability of the $\ce{COO-}$ group and also how the $\ce{COOH}$ group effects the stability of the $\ce{NH3+}$ group.


text:
What is a prosthetic group of protein? What are prosthetic groups?
I have found this term while I studying the topic of ETS in respiration in plants.

A prosthetic group is a smaller organic molecule strongly bound to a protein. Typical examples are

haemoglobin (bound to haeme)
biotin (bound to carboxylases)
flavines (bound to flavoproteines)


text:
Is butanoic acid more soluble in water than ethyl acetate?
I was asked which of the two following isomeric compounds, a carboxylic acid and an ester, was more soluble in water.

I believe both molecules should be somewhat soluble in water as they have polar C=O and C–O bonds, which should allow them to form dipole–dipole interactions with water. However, I'm not sure how the difference in the structures would affect the solubility.

Try to solve the question by analogy. Imagine that you perform a couple of experiments on different glasses of water:

Add vinegar to one glass with water. Does it mix? Yes, it does!
Add olive oil to another glass with water? Does it mix? No, it does not.

Vinegar is a carboxylic acid, whereas olive oil is made of esters. This should suggest to you that carboxylic acids in general are more soluble than esters.
Carboxylic acids, $\ce{R-COOH}$, are acids, that partly dissociate in water
$$\ce{R-COOH + H2O -> H3O+ + R-COO-}$$
to form ions. Supposing that the alkyl chain isn't extremely long, even the undissociated part of the acid added is solvated in water, a polar protic solvent, by hydrogen bonding interactions. In other words, the carboxylic acid is highly soluble in water.
The solubility of ethyl acetate in water is 8.3 g/100 ml (Wikipedia), whereas butanoic acid is completely miscible in water. That is to say, butanoic acid completely dissolves in water, regardless of the proportion of acid to water.


text:
Are there any compound lists searchable by elemental content?
As I say in the question, I am looking for lists of compounds that I can search based on elemental content. Essentially, I want to find candidate compounds for boron doping liquid scintillator, we have a few likely candidates already but we want to see the available chemical space and see if there are better/cheaper candidates. I want to avoid high electron densities, so essentially I want to find compounds that only have boron, carbon, and hydrogen. I will filter later for hazard/reactivity/flammability.
Does such a list / service exist? If not, is there a general list of compounds with their formulas that I could then parse with a python script or something?

Check out Matweb. It's free and might be of use to you.

A bit of searching shows little of use there sadly, mostly they list things like boron nitride, boron carbide, and alloys with boron in them which would not dissolve well at all. Thanks though, and it could be quite useful for other purposes later.

After some more searching, followed by asking reddit, I got a few good suggestions.
ChemSpider's advanced search method allows you to, among other things, search for elements containing specific elements and lacking other elements.
Also, for large lists of compounds I can use python tools to extract the tables from the CRC Handbook pdfs and then, search that data.


text:
Why do atoms bond? Why are unfilled valence shells energetically unstable?
I teach school Physics. And am trying to get a reasonable explanation of the root cause of the energetic instability of unfilled outermost electron shells that leads to atoms bonding. For most of what I teach, I am able to tie things back to the four fundamental interactions/forces. But unfilled valence/outer shells being the reason why atoms bond beats me.  I get Pauli exclusion. I think I get Schrodinger's equation. Yet, to me these don't provide a fundamental explanation of why a shell 'filled' by bonding is more energetically favorable than neutral atom with an unfilled outermost shell. Do I just accept this as fundamental property of our universe?

At the level of fundamental interactions there are no shells. Dozens of approximations has to be introduced starting from the fundamentals on the way to orbitals and shells. And understanding these approximations is essential to understand the stability of species with filled valence shells and instability of species with unfilled valence shells.

I'm not sure this question 'why' can actually be answered.  It just 'is'.  Wildcat's comment does not help (while true) and Spontification's answer speaks to myriad exceptions.

@Lighthart I object that my comment does not help. Once a person understand the orbital approximation it is rather trivial to understand why it is energetically more favourable to fill a particular shell before starting to fill the next one.

The comment it fails to actually address 'why', but rather discusses what is needed to answer the original question.  This is why I find the comment unhelpful.

@Wildcat I get that it is energetically more favorable to fill a particular shell before starting to fill the next one.  My question is "in terms of chemical bonding, why is a filled outermost shell more energetically favorable than an unfilled shell?" Or simply put, why do atoms bond? Lighthart are you saying I just tell my students "It is"?  In which case, from the perspective of grade-school science I should teach them that to understand the physical sciences they need to understand there are four fundamental interactions, the atomic/standard model, and the "unhappiness of atoms."

@SantoshZachariah, until now I didn't thought your question is about bonding... Well, anyway. The simplest explanation of a chemical bonding is that, loosely speaking, if occupied orbitals of few atoms are collectively higher in energy than occupied orbitals of a molecule, then its energetically favorable to form this molecule. Fundamentally, one might say that bonding is driven by the principle of minimum potential energy.

Unfilled shells aren't "filled by bonding". When a bond is formed, atomic orbitals are fully reorganized into molecular orbitals. This a much more complicated process than a simple filling of unfilled shells. And besides, all this is just a model of reality. In reality, all you have is electron density being redistributed in energetically more favorable way.

But using the orbital model, you can explain why atoms with unfilled outer orbitals usually bond while those with filled do not as follows. For a couple of atoms with unfilled outer shells, you will populate only the bonding MO which is energetically favorable. But when you have two atoms with fully filled shells, not only bonding MO, but antibonding as well will be populated, which is energetically unfavorable. See for details this question.

The idea of a "full" outer shell is a bit of a misnomer, for instance 3rd row has 3d orbitals but doesn't bother to occupy them until 4th row. Does Argon have a full outer shell, even though it has a $3d^0$ configuration?
Generally there's a large energy gap between the p-orbitals in one row and the s-orbitals in the next row, because of the increase in shielding and such factors, which acts as a kind of natural barrier, or a step, to adding any more electrons than what appears to be a 'full' valance shell.
But within any row, the amount of shielding and associated factors (penetration, nuclear charge, etc) remain largely similar, so if the valence shell isn't 'full' there is still sufficient attraction between the electron and a neutral atom that a negative ion may be stable.  Adding more than this, the energy gap is too large to overcome under usual circumstances so it stops at a 'full' shell.
There are always exceptions.  Noble gases lower down can form compounds even though they are "stable" "full" outer shells, and quirks like some transition metals having $s^1d^{10}$ configuration go to show that some of the ideas about electron filling and stable shells are rules of thumb, each individual case really needs to have its energy levels calculated to explain them.

Alright but why is it energetically favorable for a halogen to gain an electron. Your explanation doesn’t work because it isn’t going down a principal quantum number by gaining an electron.


text:
Gamma Spectroscopy: 511 keV peak?
Given a source with no gamma rays at 511 keV, what are two reasons for a peak at this location in a gamma spectrum (on HPGe detector)?
I know the annihilation peak from pair production appears at 511 keV and if a gamma ray had an energy of 1533 keV then the double escape peak (1533- 1022 = 511) would also appear at this energy.
Is there another reason that a peak could appear at this location, or are these the only reasons?

(3) sum peak from two 505.5 keV gamma rays.

A true annihilation peak at $E = 511\ \mathrm{keV}$ is caused by the detection of individual annihilation photons. The source of such annihilation photons can be:
1) Pair production in the surrounding material caused by high-energy gamma radiation
If the gamma energy is large enough to make pair production in the surrounding material (typically in the high-$Z$ shielding material surrounding the detector) relevant, the gamma quantum (a high-energy photon) may disappear and be replaced by an electron and a positron. The electron and positron may travel a few millimetres before losing their kinetic energy to the absorbing medium. When its kinetic energy becomes low, the positron may combine with an electron in the absorbing medium. Then both disappear and are replaced by two annihilation photons of $E = 511\ \mathrm{keV}$ each.
2) Positron emission of the source
If the source consists of a radionuclide that decays by positron emission (β+ decay), and the positron is stopped and undergoes annihilation before it reaches the detector (i.e. the positron is stopped within the source volume or in the material that is encapsulating the source), this region acts as a source of annihilation photons of $E = 511\ \mathrm{keV}$ each.
If one annihilation photon reaches the detector and is absorbed within the detector, an annihilation peak appears in the spectrum at an energy of $E = 511\ \mathrm{keV}$.
Furthermore, a peak at $E = 511\ \mathrm{keV}$ could be caused by mere coincidence due to
1) Coincidence summing
If two gamma rays are detected approximately at the same time, the spectrum will show only one count at the sum of their gamma energies. If the sum amounts to approximately $E = 511\ \mathrm{keV}$, the resulting peak would look like an annihilation peak.
2) Double escape
If the gamma energy is large enough to make pair production in the detector material relevant, the photon may disappear and be replaced by an electron and a positron. The electron and positron may travel a few millimetres within the detector before losing their kinetic energy and finally combining with an electron. Then both disappear and are replaced by two annihilation photons. If one annihilation photon escapes from the detector without interaction within the detector, a single escape peak appears in the spectrum at an energy of $511\ \mathrm{keV}$ below the full-energy peak. If both annihilation photons escape, a double escape peak appears at an energy of $1.02\ \mathrm{MeV}$ below the full-energy peak.
If the remaining energy is approximately $E = 511\ \mathrm{keV}$, the resulting escape peak would look like an annihilation peak. This effect is not possible for single escape peaks since their energies are always well above $E = 511\ \mathrm{keV}$, but it is theoretically possible for double escape peaks.

Wonderfully comprehensive answer!  Thank you so much!


text:
Oxidizers Cabinet
I am currently reorganizing the chemical inventory in our lab and grouped all the oxidizers together. Would it be safe to put the following chemicals together in the fume hood cabinet? I have checked the SDS for any incompatibilities/storage and everything seems to be in the clear.
Oxidizers Cabinet
Sodium chlorate
Potassium dichromate
Sodium dichromate
Potassium permanganate
Potassium persulfate
Aluminum nitrate
Ferric nitrate
Magnesium nitrate
Nickel nitrate
Sodium nitrate
Silver nitrate
Strontium nitrate
Potassium nitrite
Sodium nitrite
Sodium cobalt nitrite
Sodium peroxide
Thanks in advance,
C

Sodium chlorate tics me a bit. It is highly hygroscopic, but if dry it may produce nasty mixtures with most organics of any sort and many nonorganics. Persulfate has low active oxygen content and probably may be moved to a less secure location.

As with your previous question on azides, the same regulation on storage classes (LGK), namely TRGS 510, applies.
You neither seem to have ammonium nitrate (LGK 5.1C), nor organic peroxides (LGK 5.2).
If your oxidants belong to the storage classes 5.1A and 5.1B, you may store them together.


text:
What are reducing sugars?
I found that from many definitions reducing sugars are those which contain free aldehyde or ketone group which can reduce reagent like Tollen's Reagent.But I have also read that ketones get oxidized only with very strong oxidizing agents like alkaline KMnO4 or acidified potassium dichromate under vigorous condition.So according to what i read,then i could say that fructose is a non reducing sugar as it won't be able to reduce weak oxidizing agents like Tollen's Reagent.
But its also said in my textbook that all monosaccharides are reducing sugars.But fructose is a monosaccharide.Then how can it be a reducing sugar.Please explain me if there is any misconceptions from my side.

Not a duplicate per se, but the answer is there nonethelss

I din't see that question when i typed it SE should work upon this.

It's more tricky example, but if Community marked it then who knows... Still, searching for dupes may be AI complete ;)

@Mithoron  i agree with that.Somebody must develop an efficient AI :-p.

http://meta.chemistry.stackexchange.com/questions/3071/searching-guidelines-for-newbie