text:
How is dioxygen stabilised by coordination to haemoglobin?
I wonder how the binding of $\ce{O2}$ to a molecule of hemoglobin stabilizes it to the extent that I can carried around the body with the risk of it being involved in oxidation reactions minimized. The same applies to hemocyanin.
I am looking for an answer invoking orbital interactions, back-bonding etc.

Note how I formatted your molecular formula for oxygen using MathJax formatting. This is how we should format chemical equations on this site, and there is a good introduction here. It's really pretty simple to get the hang of it and it makes your posts look much more professional.

Are you  asking about the bonding of oxygen molecules to the Fe in heamoglobin (and to Cu in hemocyanin) and about the different structures of the protein in different states ? As to the latter see the chapter on this in Stryer 'Biochemistry' and in Kuriyan, Konforti & Wemmer , 'The molecules of Life'.

@porphyrin Thanks! I'm primarily looking at the bonding with the iron-containing haem. It seems like oxygen would be a superoxide, so I wonder if there are any orbital interactions involved that would stop it from reacting with molecules around it. If not, I wonder if the stability is to do with the whole protein (e.g. the oxygen-containing harm being buried deep within its core).

related http://chemistry.stackexchange.com/questions/33780/why-does-carbon-monoxide-have-a-greater-affinity-for-hemoglobin-than-oxygen?s=1|0.8430

Your question incorrectly assumes that oxygen molecule is unstable or reactive at room temperature. No orbitals, sorry

I think you have a misconception. The binding of oxygen to haem and its release are processes that are entirely equilibrium-controlled by the equilibrium shown in $(1)$.
$$\ce{haem + O2 <=> haem{-}O2}\tag{1}$$
Meaning that wherever high partial oxygen pressures are observed, oxygen preferentially binds to haem but likewise it is released where low partial oxygen pressures are observed. Thereby, Le Chatelier’s principle nicely controls where oxygen is released from the red blood cells’ haem groups — it will always be in those areas where oxygen is required. Haem itself has evolutionally acquired the role of binding oxygen rather well but only to make sure that it actually reaches the distant parts of the body.
Oxygen in its $\ce{O2}$ modification is also not as bad as your initial post seems to assume. Yes it can react with quite a few substances and many of its reactions have a low activation barrier — but the concentrations in which it would be released into the blood stream ‘undesiredly’ are concentrations the body has learnt to cope with over millenia of evolution. So while minor damage may occur, most damage is quickly repaired by dedicated enzymes. Most importantly, any liberated oxygen will always be $\ce{O2}$ and never superoxide or peroxide.
Only while being bound to haem do oxygen’s (and iron’s) electronic properties change. On a microscopic level, the oxygen molecule causes a one-electron oxidation of iron converting itself to superoxide. This also causes a spin-flip in the iron centre which goes from high to low spin. The combined $\ce{\overset{\mathrm{+III}}{Fe}-O2}$ system can finally undergo antiferromagnetic coupling to give an overall observed diamagnetic ground state. Details can be found in this answer. However, while it is superoxide the $\ce{O2}$ moiety will never leave the iron centre — or, more precisely described, as soon as $\ce{O2}$ leaves, the entire process is reversed again. Thus, even though the bound state includes a superoxide ion no superoxide can ever be directly released into the surrounding blood stream.
$$\ce{[\overset{\mathrm{+II}}{Fe}(por)] + ^3O2 <=> [\overset{\mathrm{+III}}{Fe}(O2^{.-})(por)]}\tag{2}$$

Thank you! So while bound to Fe, oxygen (superoxide) isn't in contact with the surrounding solution at all, even with water molecules diffusing to the binding cleft that it can oxidize or surrounding amino acid side chains?

@substitutedbutadiene Well, it is somewhat close enough to the surrounding solution to be liberated as neutral dioxygen but it does not show the reactivity of free superoxide.


text:
How does biodiesel have so much energy density?
I find it fascinating that something like cooking oil can be turned into fuel. I know it can't have that energy density off the bat. So what do they do to make fuel for cars? Do they have to add other ingredients, and if so, why?

Could maybe get some better answers at the  chemistry SE site?

It's just a bunch of carbon-carbon and carbon-hydrogen bonds. That's what you have in gasoline... Cooking oil is just not as volatile so is less efficient as an engine fuel.

You cross-posted this and after migration there's an exact duplicate

Where you go wrong is in thinking that cooking oil has less energy density than biodiesel. The Wikipedia page for biofuel energy content list a number of different types of fuel. Along with a long list of other fuels, it lists the following:
Fuel Type   Specific energy (MJ/kg) Energy Density (MJ/L)
Methanol    19.9 – 22.7             15.9
Fat         37.656                  31.68
Sunflwr oil 39.49                   33.18
Olive oil   39.25 - 39.82           33 - 33.48
Biodiesel   37.8                    33.3 – 35.7
Diesel      48.1                    40.3

As you can see, both the specific energy (MJ/kg) and energy density (MJ/l) are about the same as biodiesel and close to those of ordinary diesel fuel. To convert the oil into biodiesel, some methanol is usually added, but that does not increase the energy density.

The OP is correct in the statement that cooking oil does not quite have the energy density of regular diesel fuel. On an energy per volume basis, vegetable oils have on the order of 3/4 of the energy available for combustion as does regular diesel fuel. But that's OK, particularly if it were not going to be reused or recycled in some other way.
There one problem with the physical characteristics of vegetable oils, namely that they are too viscous to use unmodified in most stock diesel engines.  That leaves two possibilities: 1) modify the fuel or 2) modify the engine.
1) When the oil is modified so as to be used in a standard diesel engine it is called biodiesel fuel.  According to  this Wikipedia article:

Biodiesel refers to a vegetable oil - or animal fat-based diesel fuel
  consisting of long-chain alkyl (methyl, ethyl, or propyl) esters.
  Biodiesel is typically made by chemically reacting lipids (e.g.,
  vegetable oil, soybean oil, animal fat (tallow)) with an
  alcohol producing fatty acid esters.
Biodiesel is meant to be used in standard diesel engines and is thus
  distinct from the vegetable and waste oils used to fuel converted
  diesel engines. Biodiesel can be used alone, or blended with
  petrodiesel in any proportions. Biodiesel blends can also be used
  as heating oil.
The National Biodiesel Board (USA) also has a technical definition of
  "biodiesel" as a mono-alkyl ester.

2) If the vegetable oil is not converted to biodiesel fuel, then it will be too viscous to form a fine enough mist in the engine to ignite.  The simple solution here relies on the fact that viscosity is a function of temperature. So, the mechanical solution is to simply add a pre-heater to the fuel line to heat the unmodified oil such that it's viscosity will be low enough to form a fine enough mist to ignite, and no further modification is necessary.

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sort: votes


text:
NaOH CO2 trap troubleshooting
I was wondering if anyone could help troubleshoot an experiment I'm working on.
I'm running an experiment to measure the microbial respiration of compost samples using a $\ce{NaOH}$ trap.  Each sample is in a jar with 20 ml of 1M $\ce{NaOH}$ and is placed in an incubator for 24 hrs.  After 24 hrs, 20ml of .5N BaCl2 and an indicator are added to the $\ce{NaOH}$ and it is then titrated with $\ce{HCl}$.  I run a few blanks each time, with an $\ce{NaOH}$ trap in a jar without a compost sample. This is repeated over 4 days In theory, I should need less $\ce{HCl}$ to neutralize the $\ce{NaOH}$ which has been sitting with the compost samples then the $\ce{NaOH}$ just by itself.  Sadly, after hours and hours of work in the lab, this has not been the case!
I have 45 different samples of different types of compost. Each time I run the test, about a third of the samples require more $\ce{HCl}$ to neutralize than the blank.
The blanks are all consistent in the amount of $\ce{HCl}$ needed, I ran three last night and one the night before.  The test also seems to be picking up samples which are respiring heavily- when there are moldy samples, they require only a small amount of $\ce{HCl}$.  I can't really decipher a trend with the ones that require more $\ce{HCl}$ to titrate than the blank other than that it's somewhat consistent. 41% of the samples which needed more $\ce{HCl}$ than the blank on day 2 also needed more on day 1, and of the day 2 "error samples", 88% were from a treatment which had an "error sample" on day 1.
I thought perhaps this had to do with ammonia volatilizing from some of the samples, but this doesn't necessarily make sense with the samples which are taking more $\ce{HCl}$ than the blank. I also noticed that on the samples which require more $\ce{HCl}$ to titrate than the blank, there seems to be more precipitate (barium carbonate?) that settles out after the titration.
Do you have any idea what could be happening or what I am doing incorrectly?  It's definitely a long process to do this measurement for 48 replicates over 4 days - I'm hoping the data isn't totally worthless.
Thanks
Edit: Pictures

On the left are samples which required less HCl to neutralize than the blank.  (I forgot to add an extra 20ml of BaCl2 to the second from left, which I had added an additional 10 ml NaOH to, so that one is an error).  In the middle is the blank and a sample which required the same amount of HCl to neutralize as the blank.  Then on the right are samples which required more HCl than the blank to neutralize.

re: "Each sample is in a jar" I assume that the jar is sealed? If so the oxygen in the jar will be depleted quickly. Basically you need oxygen to make the CO2.

Is the jar sealed? Have you tried stirring during the incubation? Have you computed the theoretical change associated with your experiment?

Hmmmm interesting. The method says to "close the lid tightly" of the incubation vessel.  Otherwise I believe the trap would pick up CO2 being respired by all the other samples.  I'm expecting about 300 mg CO2 respired on average per day? Doing some rough back of the envelope calculations there would be about .016 moles O2 in the jar and with 300 mg CO2, that would be  .006 moles used.  If the CO2 respiration was a lot higher, that could definitely deplete all the O2.   Even if all O2 was depleted in a sample, shouldn't that NaOH trap need less HCl to titrate than the blank?

Jar is sealed.  I would worry about them not being air tight but my blanks all require the same HCl to titrate.  I haven't stirred during the incubation.  As for the theoretical change I believe it would go like this: CO2 + 2NaOH = Na2CO3 + H2O, then Na2CO3 + BaCl2= BaCO3 + 2NaCl, then the HCl + NaOH = NaCl + H20

Filtering BaCO3 out?

The methods instruct me to go straight to titration with HCl after adding the BaCl2. I think the BaCO3 is supposed to be precipitated and so it shouldn't react with the HCl?  Perhaps what's going on here is the solution is not being adequately mixed and the Na2CO3 isn't fully precipitating out as BaCO3??  Thanks for the help!

The BaCO3 must be insoluble enough to not have much carbonate in solution. With HCl you're just titrating to solution is barely acid, not like conc. HCl.

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text:
Can bromine water be used to compare the reactivity of liquid 1-heptene and liquid heptane?

Describe how you would carry out a lab test to compare the reactivity of liquid 1-heptene and liquid heptane. State clearly what the possible results of the test are and what they would indicate.

Would you carry out the lab test using bromine?
Add about 1 mL (about twenty drops) of 1-heptene to one
container and about 1 mL of heptane to the other container.
Add an equal volume of bromine water to each of the liquids then
shake each container to the same extent.
Leave the containers to settle, then observe.

Would this process work for heptane and 1-heptene? Would the colour of the solution change?

According to this Wikipedia article:

"In electrophilic halogenation the addition of elemental bromine or
  chlorine to alkenes yields vicinal dibromo- and dichloroalkanes
  (1,2-dihalides or ethylene dihalides), respectively. The decoloration
  of a solution of bromine in water is an analytical test for the
  presence of alkenes."

Alkanes tend to be much less reactive toward bromine, so yes, your bromine water test should allow you to distinguish between heptane and 1-heptene.

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text:
Conditions for free radical bromination using NBS and peroxides?
In the free radical bromination of 1-methylcyclohexene using NBS and peroxides a very important consideration to get the appropriate product is:
A. The wavelength of light used for initiation
B. The polarity of the solvent
C. Temperature
D. None of the above
You would need a high enough temperature / low enough wavelength to form the radicals but he said nothing about polarity of the solvent, and there can only be one answer.

From practical experience, I can tell you that the answer is "All of the above".
First, your peroxide initiator needs some energy to undergo homolytic cleavage. This can either be achieved through UV photolysis (A) or heating (B).
$$\ce{RO-OR <=>[h\nu][\mathrm{or\ }\Delta] RO. + .OR}$$
The amount of energy needed will correlate to the strength of the O-O bond. The energy available from UV irradiation is dependent on wavelength because $E=hc/ \lambda$. The amount of energy available from heat is dependent on temperature.
I've initiated NBS reactions both ways, but I prefer heat since I can more easily heat a larger vessel.

What about solvent polarity?

Solvent is the tricky point, and depending where you are in your studies, you may not have covered the other things that NBS can do. Under the right circumstances, NBS is also an electrophilic source of bromine, especially in polar protic solvents.
Here are two examples from my personal experience using the same substrate with different conditions:
Using methanol as a solvent without a radical initiator, I got electrophilic bromination of the arene.

Using 1,2-dichloroethane as a solvent with heat and a radical initiator, I got radical bromination of the benzyl position and electrophilic bromination of the arene.

Using 1,2-dichloroethane as a solvent with UV irradiation and a radical initiator, I got radical bromination of the benzyl position.

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text:
If fluorine and cerium ion have more oxidising power than KMnO4, why is KMnO4 more popularly used as an oxidising agent?
It is quite clear from the electrode potential values that $\ce{F2}$ and $\ce{Ce^{4+}}$ are better oxidizing agents that $\ce{KMnO4}$.

Have you considered price, safety, toxicity, stability to store..? Also, oxidizing agents are not interchangeable in many reactions - electrode potential is not the only important parameter when you want to prepare specific products.

Using $\ce{F2}$  instead of $\ce{KMnO4}$ is much like using a hand grenade to kill a mosquito.

KMnO4 you find in glass bottles in most chem labs. For a strong oxidising agent it's relatively non-toxic and won't inflame and destroy a good number of common materials too quickly, such as glass, metals and human beings. Fluorine will. Fluorine is nasty, very nasty, and no sane person uses it unless necessary. Ce4+ I don't know, but you do come across in labs at the undergrad level once in a while, and once even at school. My guess is that it is much more expensive then KMnO4.

Lanthanides are pretty expensive. There's a reason why there popularly called as ''rare earth metals".


text:
How to calculate the hybridization of ethane with H=½[V+M-C+A]?
I wanted to calculate the hybridization of this molecule $\ce{CH3-CH3}$. I noticed a shortcut formula given in my class notes for calculating the hybridization. It was
$$H=\frac{1}{2}×[V+M-C+A].$$
With

$H$: Hybridization
$V$: Number of valence electrons of central atom.
$M$: Number of monovalent atoms (i.e hydrogen or halogen) around central atom.
$C$: Cationic Charge.
$A$: Anionic Charge.

When I tried to calculate the hybridisation of $\ce{CH3-CH3}$ by using above formula, where for any one of the carbon atom $V=4$, $M=3$, $C=0$, $A=0$ then $H=\frac{7}{2}$. But as we all know hybridization of every carbon in ethane is $\ce{sp^3}$, i.e. $H$ should be $4$. I know other methods of calculating $H$ like $H= \text{no. of sigma bonds} + \text{no.of lone pairs}$, which gives me the right answer, but why does the above formula give something different? And also for any molecule if using that formula gives fractional answer then what does that mean?

Are you sure you don't mean formal charge?

But what difference formal charge will create as it is also zero.

I have not come across this formula before and I cannot make any sense of it. Please be aware that hybridisation itself is just a model, it is just a visualisation and interpretation technique. Simple approaches to that will only give you a very approximate answer to what would mathematically be correct. In ethane the $\ce{C-C}$ bond would have more p-character than the $\ce{C-H}$ bonds.

Consider CH3 as one of the monovalent group for first C to calculate the value of H.

The shortcut formula cannot be used
to predict hybridized state of an atom in compounds having
cyclic rings or catenation. Reference: https://www.worldwidejournals.com/indian-journal-of-applied-research-(IJAR)/recent_issues_pdf/2014/March/March_2014_1492755966__17.pdf

The mistake that you have made is incorrect understanding of M. From what I remember, M was number of single bonds, which would be only with a monovalent atom, in the context you learnt tbe formula. This was taught in chemical bonding where only atoms are bonded to central atom.
Ex- PCl5, PCl3
You should modify this definition to get the answer  here. Any monovalent group (-CH3) should be there in M.


text:
What is the name of C(OOOH)₄?
I've been practicing organic nomenclature and I was wondering if anyone could show me the proper way to approach naming this molecule. My guess would be methane-quadra-dioxyanol.

I would call it boomane, since if you attempted to synthesize it something would probably go boom.

This molecule doesn't have any carbon to hydrogen bonds, so it likely wouldn't even be properly named using organic nomenclature. Tetrachloromethane, for example, is correctly named carbon tetrachloride. If you wanted to throw technicalities to the wind, however, along with the fact that this compound would likely be so incredibly unstable that no one would even bother attempting to synthesize and isolate it, it would probably be named tetratrioxidanylmethane.
You might notice the name bears some similarities to tetrachloromethane. That's because it's four of the same group, the trioxidanyl group (from trioxidane), bonded to a single carbon.

Under this entry, PubChem hosts even a $\ce{CH4O22}$ with an surprisingly systematic name (after all) of "bis(hydroperoxyperoxyperoxyperoxy)-bis(trioxidanyl)methane"

So far, however, experimentally determined properties are still to add to this entry created on February 23rd, 2016.  And the absence of a CAS registry number suggests (maybe for the better) that this maybe high-energy material was not not yet isolated.

I think, it’s surprisingly unsystematic. I would have expected a more systematic name labelling the longer oxygen chain as a chain similar to the all-carbon compound. But $+1$ for the finding!

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)), the preselected name for the compound $\ce{HO-O-OH}$ is trioxidane. According to rule P-68.4.1.1, such compounds with three or more contiguous identical chalcogen atoms are treated as parent structures. For example, the preferred IUPAC name of the compound $\ce{CH3OOOCH3}$ is dimethyltrioxidane (note that the locants of 1,3-dimethyltrioxidane are omitted in the preferred IUPAC name).
Therefore, the parent structure for the preferred IUPAC name of the compound that is given in the question should be trioxidane and not methane. However, this compound contains four individual trioxidane structures. In order to name such assemblies of two or more identical parent structures that are linked by a bi- or multivalent substituent group, multiplicative nomenclature can be used (e.g. 1,1'-peroxydibenzene). In this case, four trioxidane parent structures (“tetrakis(trioxidane)”) are linked by a tetravalent methanetetrayl substituent, which leads to the preferred IUPAC name 1,1',1'',1'''-methanetetrayltetrakis(trioxidane).


text:
Measuring or calculating the pH of 4M NaOH? I prepared the solution in Milli Q water
I prepared $4\ \mathrm{M}$ solution of $\ce{NaOH}$, and the glass electrode I use cannot measure it the $\mathrm{pH}$ correctly. What are the other options do I have to exactly know the solution's $\mathrm{pH}$?

This link explains the problem. http://www.all-about-ph.com/acid-and-alkaline.html The following link is for a high pH electrode but I couldn't find any specifications for it. :-(  http://www.awedirect.co.uk/ph/ph-electrode/high-alkaline-ph-electrode-9037-10b.html

Do you really need to know pH or just that solution is 4.00 molar NaOH? Can you assume that the solution is "pure" NaOH? No NaCl, no KOH. I'm wondering about a sodium electrode, or conductivity. Carbonate from the air will be a problem.

Did you use tap water or distilled water?

thanks ringo, I had calculated the same way, and I had made it in Miili q water, pH 5.7, and using AR grade NaOH.

For $\mathrm{pH}>12$ and $\ce{Li+}$ or $\ce{Na+}$ concentrations greater than $0.1\ \mathrm{M}$, glass electrodes experience alkali error. Because the concentration of $\ce{H3O+}$ ions in solution is so low, interference from alkali metal ions (like $\ce{Na+}$) becomes noticeable and causes the electrode to read lower than the true $\mathrm{pH}$ of the solution. $\ce{K+}$ typically cause less error than $\ce{Li+}$ or $\ce{Na+}$ due do its larger size.
Since sodium hydroxide is a strong base and dissociates completely into $\ce{Na+}$ and $\ce{OH-}$ ions, though, it's actually quite simple to just calculate the $\mathrm{pH}$.
$$K_\mathrm{w}=\ce{[H3O+][OH-]} \Rightarrow$$
$$ 1 \times 10^{-14} \ \mathrm{M^2}=\ce{[H3O+] \cdot 4\ \mathrm{M}} \Rightarrow $$
$$\ce{[H3O+]}=2.5 \times 10^{-15}\ \mathrm{M}$$
$$\mathrm{pH}=–\log(2.5 \times 10^{-15}\ \mathrm{M})=14.60$$
If you are worried about whether or not you truly have a $4\ \mathrm{M}$ solution of $\ce{NaOH}$, you should try titrating it with a known concentration of acid (in the past I always used potassium hydrogen phthalate). This process is called standardization in acid-base parlance.

I am pretty sure that is not 'exact'. If the electrode cannot measure it, there is a decent chance, that the final solution does not behave like an ideal solution.

@Martin-マーチン Addressed this


text:
Is the total entropy change of all isothermal processes 0?
We define  total entropy as: $$\Delta S_\mathrm{total}=S_\mathrm{system}+S_\mathrm{surroundings}$$
I know that for an isothermal AND reversible expansion: $S_\mathrm{total}= 0$
But what about an isothermal expansion at a constant pressure? Does $S_\mathrm{total}$ also equal 0? If not, then would the equation look something like this:
$$\Delta S_\mathrm{total}=\frac{q_\mathrm{rev}}{T}+\frac{q_\mathrm{surroundings}}{T}$$
where $q_\mathrm{surroundings}$ is the heat lost or gained by the surroundings by a process and $q_\mathrm{rev}$ is the heat lost or gained by the system IF that process was to become reversible.

Are we talking about reversible processes only? Entropy is conserved in all reversible processes.

@Greg Please read my question

Please describe in more detail what you are referring to when you say isothermal expansion at constant pressure.  Are you saying that you add heat to the system while you hold the pressure constant at the original value, or are you saying that you hold the system in contact with a constant temperature bath while, at time zero, you drop the external pressure to a lower value and then hold the external pressure at that constant value until the system re-equilibrates.

Is the total entropy change of all isothermal processes 0?

No. Only for reversible processes.
However your equation is correct.
For an irreversible process in which an ideal gas expands isothermally against a constant external pressure
$$\Delta S_\mathrm{total}=\frac{q_\mathrm{rev}}{T}+\frac{q_\mathrm{surroundings}}{T}$$
becomes
$$\Delta S_\mathrm{total}=\frac{q_\mathrm{rev}}{T}+\frac{w}{T}$$
where $w=-p \Delta V$, since $q_\mathrm{surroundings}=-q_\mathrm{system}$ and for the isothermal process the first law stipulates that $w=-q_\mathrm{system}$.
For a free expansion $p_\textrm{ext}=0$ and no work is done. No heat is exchanged and the entropy of the surroundings does not change. The change in entropy of the system then drives the process.

Isothermal expansion at constant pressure is NOT a reversible process and hence ∆S(universe) is always greater than zero.
I.e. $$\Delta S_\mathrm{univ}=\frac{Q_\mathrm{actual}}{T}-\frac{\int PdV}{T}$$
$$i.e. ~ \Delta S_\mathrm{univ} >= 0 $$
Equality holds only for reversible processes.
\n
Only reversible processes are iso-entropic.

$\Delta S_\mathrm{universe}$ is not always positive.

You might want to take a look at this discussion of how to properly format mathematical and chemical equations. It's really not hard to learn and it will make your answers much more readable.


text:
Does iron in water become visible when being heated?
I'm using ground water to provide my daily needs: Laundry, bath, toilet, ... I don't use it as drinking water, but I do cook with it.
What I noticed is that when using the water cold it appears as water from a bottle, but when I start to heat the water it turn orange as if the iron becomes more visible? It also is visible from water coming out of the boiler.
Does iron in water become visible when the water is being heated?
Another fun fact, just for information, when it rains a lot the orange color is more dense.
Greetings.

Hello and welcome to Chemistry.SE. If you have any questions about the site I would start by taking the short [tour]. Regarding your question, do you know anything else about the water, like it's pH or if it's pretty hard water, etc. You may be on the right track as orange/redish colored water can be an indicator of iron. Anyway, good luck with solving your mystery!

Iron becomes visible as it gets oxidized to $\ce{Fe^3+}$, hydrolyzes all the way to hydroxide, and precipitates from the solution. All this is somewhat facilitated by heating.

Question already asked here : http://outdoors.stackexchange.com/questions/5894/can-groundwater-still-be-consumed-if-after-boiling-the-water-turns-brown-with-an

In pine forests, freshly-dug soil (podzol) is often a pale blue-green from reduced iron, but after an hour or two exposed to air, it turns reddish brown as the iron oxidizes.  This is the same effect you've noticed.


text:
Entropy- Judge my knowledge
Entropy? — So basically I’ve been trying to understand what it is and what I’ve come to realize is. Entropy basically is energy dissipated. What I mean is this — so we know what enthalpy is — it measures the useful energy in the inicial state and in the final state. Heat — measures the energy being transfered during the process, as such, it’s the total energy. Entropy — consists of the amount of energy that was dispersed. As such we can say that a spontaneous reactions occurs, when the entalpy variations, [useful energy] is higher than the energy dissipation [entropy] can we not?

Welcome to Chemistry StackExchange. It will help to get a clear answer if you are more careful in formatting your question and be specific in what you want an answer to, rather than asking if your opinion is correct.

Please never use acute accents as a substitute for apostrophes, it looks absolutely horrible. Please also note that full stops and commas are followed by a space and that capital letters are reserved for the first word in the sentence, the pronoun I proper nouns and a few terms derived from proper nouns. Most importantly, energy and entropy are not capitalised but I is always.

@Jan, indeed, this should be in EnglishSE

Your idea of energy dispersal is heading in the right direction
Entropy is a thermodyanmic potential describing the distribution of energy states available to the system.
Assuming that there is a finite number of degenerate (equivalent energy) states $\Omega$, then the absolute entropy $S$ of a system is:
$$S= k_\mathrm{B} \ln \Omega$$
where $k_\mathrm{B}$ is the Boltzmann constant.
Let's consider a very simple system of some particles in a fixed volume. If there are four spots that particles can occupy, we can construct states and count them.
In the case where there are no particles, there is only one state:
$$\begin{array}{|c|c|}\hline
\ &\  \\ \hline
\ &\  \\ \hline
\end{array}\\
S=k_\mathrm{B} \ln 1 = 0$$
When there is one particle $x$ in the system, there are four states:
$$\begin{array}{cccc}
\begin{array}{|c|c|}\hline
x &\  \\ \hline
\ &\  \\ \hline
\end{array} & \begin{array}{|c|c|}\hline
\ &x  \\ \hline
\ &\  \\ \hline
\end{array} & \begin{array}{|c|c|}\hline
\ &\  \\ \hline
x &\  \\ \hline
\end{array} &
\begin{array}{|c|c|}\hline
\ &\  \\ \hline
\ &x  \\ \hline
\end{array}
\end{array}\\
S=k_\mathrm{B} \ln 4$$
When there are two particles there are twelve states (we have to pretend the two particles are different to get there: $x_1$ and $x_2$)
$$\begin{array}{cccc}
\begin{array}{|c|c|}\hline
x_1 & x_2  \\ \hline
\ &\  \\ \hline
\end{array} & \begin{array}{|c|c|}\hline
x_2 & x_1   \\ \hline
\ &\  \\ \hline
\end{array} & \begin{array}{|c|c|}\hline
x_2 &\  \\ \hline
x_1 &\  \\ \hline
\end{array} &
\begin{array}{|c|c|}\hline
x_2 &\  \\ \hline
\ &x_1  \\ \hline
\end{array} \\
\begin{array}{|c|c|}\hline
x_1 &\  \\ \hline
x_2 &\  \\ \hline
\end{array} & \begin{array}{|c|c|}\hline
\ &x_1  \\ \hline
x_2 &\  \\ \hline
\end{array} & \begin{array}{|c|c|}\hline
\ &x_2  \\ \hline
x_1 &\  \\ \hline
\end{array} &
\begin{array}{|c|c|}\hline
\ &x_2  \\ \hline
\ &x_1  \\ \hline
\end{array} \\
\begin{array}{|c|c|}\hline
x_1 &\  \\ \hline
\ &x_2  \\ \hline
\end{array} & \begin{array}{|c|c|}\hline
\ &x_1 \\ \hline
\ &x_2  \\ \hline
\end{array} & \begin{array}{|c|c|}\hline
\ &\  \\ \hline
x_1 &x_2  \\ \hline
\end{array} &
\begin{array}{|c|c|}\hline
\ &\  \\ \hline
x_2 &x_1  \\ \hline
\end{array} \\
\end{array}\\
S=k_\mathrm{B} \ln 12$$
For similar reasons, three particles would generate 24 states, and so would four particles. When we get to bigger systems (especially macro systems), include different types of particles and interactions between particles, the situation gets more complex, but there are statistical methods that can handle it.


text:
Direct conversion of amides to esters
Rarely does the literature describe the synthesis of esters from amides. I understand that amides are a relatively unreactive functional group but they can easily be hydrolysed to give carboxylic acids under basic or acid conditions. Why can't they be converted to methyl esters under similar conditions using methanol in place of water?
Surely the mechanism would be identical to hydrolysis? Additionally, the pKa of water is very similar to methanol so I cannot justify it on the basis of nucleophilicity/basicity. The only thing I can think of is steric hindrance but methanol isn't that bulky compared to water.

A google search gave the following reference doi:10.1038/nature14615 which should give you a clearer understanding.

Thanks but I've read that paper already and it doesn't answer my question. It just makes the usual statement that amides are unreactive due to resonance stabilisation. If they are so stable why do carboxylic acids readily form but not esters?

You are right about the rarity of amide to ester conversion. Amide to ester conversion has remained a challenging and underdeveloped synthetic transformation. It is often due to the fact that amides are sufficiently stable because of its resonance-stabilized $\ce{C–N}$ group that imparts a planar geometry to amides and as such, esterification is difficult and requires the use of harsh acidic or basic conditions, while employing a large excess of nucleophile (e.g. using alcohol nucleophile as solvent). In contrast, esters are generally reactive under a variety of mild condition and therefore the opposite conversion of ester to amides is quite normal.
The most promised method of achieving amide to ester conversions is Keck’s methylation/hydrolysis although this method is limited to the synthesis of methyl esters and thus researchers has developed some method to expand this conversion's feasibility and efficiency.

Hie et. al. has achieved amide to ester synthesis using nickel complex catalyst. They studied the conversion of various aryl and heteraryl amides. One studied reaction is conversion of benzamide to methyl benzoate using $\pu{10 mol{\%}}$ $\ce{Ni(cod)2}$, $\pu{10 mol{\%}}$ SIPr, 2 equivalents of methanol, and toluene as solvent at $\pu{110 °C}$ for 12 hours. (Ref. 1)

An efficient two step transformation of 1° and 2° amides to methyl and ethyl esters has been developed using trimethyl- and triethyloxonium tetrafluoroborates, and dilute acid. This method was applied to 1-benzyl-4-phenylamino-4-piperidinecarboxamide, a precursor in the synthesis of carfentanil, to produce the methyl ester in $\pu{60{\%}}$ yield and the ethyl ester in $\pu{80{\%}}$ yield. (Ref. 2 and 3)

Conversion of amides to ester has been achieved using $\ce{CeO2}$ catalyst and HY zeolites. A variety of metal oxides as catalyst and ester yields has been discussed. (Ref. 4)

References

Hie, L., Fine Nathel, N., Shah, T. et al. Conversion of amides to
esters by the nickel-catalysed activation of amide C–N bonds. Nature
524, 79–83 (2015). DOI: 10.1038/nature14615 (PDF link)
Amide to ester conversion: a practical route to the carfentanil class of analgetics
Douglass F. Taber and Mohammad Rahimizadeh, The Journal of Organic Chemistry, 1992, 57 (14), 4037-4038, DOI: 10.1021/jo00040a061
Anthony J. Kiessling & Cynthia K. McClure (1997) The Conversion of Amides to Esters with Meerwein'S Reagent. Application to the Synthesis of a Carfentanil Precursor., Synthetic Communications, 27:5, 923-937, DOI: 10.1080/00397919708004212
Conversion of Amides to Esters via Selective Cleavage of Amide C–N Bonds over a CeO2 Catalyst by Shimizu et. al. (PDF)


text:
What is the role of size in formation of complex compounds?
In my textbook, these reasons are given as of why transition metals form a large number of complex compounds:-

Smaller size of metal ions
High ionic charges
Availability of d orbitals for bond formation

What is the use of the first reason, i.e small size? Shouldn't it be opposite? In interhalogen compounds too, in XYn, X should be of greater size than Y. So why opposite in case of complex compounds?

The 3d electrons are attracted strongly to the nucleus (positive charge) and this overalls 'pulls' the electron cloud closer to nucleus making the atomic radii much smaller, which is why there is a general trend of decreasing atomic radii across the period.
Since in transition elements series the 4s orbitals are filled first before 3d orbit, this is turn makes the transition element very reactive and readily form compound as the 4s electrons are released easily, (in addition 3d orbitals provide a screening effect) and excited to a higher energy level during typical reaction to form  complex salts.


text:
What caused this reaction between chlorine and brake fluid?
I recently saw a video on Facebook of a couple of people pouring brake fluid into a bottle containing pool chemicals (chlorine pellets, I believe). Now this was a rather strong exothermic reaction, resulting in a fire. At one point during the fire, fire appeared to be vapors burning under pressure (like a propane tank fire). I have included a link here. Unfortunately, it is not the original video that I watched.
My question here is what in the brake fluid is reacting so strongly with the chlorine? And what are the chemical products of this reaction (what's in the smoke)?

Pool Chlorine is calcium hypochlorite and brake fluid is polyethylene glycol. Here is a proposed mechanism for the reaction.

Thank you for your response! If no one else responds, I will likely accept this answer. :)

This reference looks like a good find. Writing out at least the key parts of the reaction would greatly increase the quality of your answer, and probably earn you more up-votes ;)


text:
Does mercury chloride (ii) evaporate at room temperature?
if you leave mercury chloride (ii) exposed to air in a room does it evaporate to any significant level which can cause toxicity?

According to this 1911 Journal of the American Chemical Society article (Johnson, F. M. G, JACS 1911, 33, 777), the vapor pressure of $\ce{HgCl2}$ is 3 mmHg at 152 degrees Celsius. Since the article is behind a paywall, I will reproduce the data below:
$$\begin{array}{|c|c|}
\hline
T/ \mathrm{^\circ C} & P/ \mathrm{mmHg} \\ \hline
152 & 3\\
195 & 20 \\
231 & 82 \\
256 & 198\\
262 & 237 \\
275 & 325\\
278 & 421 \\
283 & 481\\
302 & 754\\
\hline \end{array}$$
The authors did not report values for lower temperatures since their method focused on setting a pressure in the system and then measuring the temperature of sublimation/boiling point. However, using the data and some help from the Clausius-Clapeyron relationship, an estimate for the room temperature vapor pressure is $2.4\times 10^{-4}\ \mathrm{mmHg}$, which assuming you are at 1 atm, is 0.32 parts per million when the atmosphere is saturated. Consult a safety data sheet or other source for toxicity considerations at that exposure level.


text:
Spontaneity of Reactions- Gibbs Energy or Entropy change
I just wanted to check if I am thinking right:
The reason why we consider the Gibbs energy to be the measurer for spontaneity of a reaction is because it does not depend whether the reaction was exo- or endothermic.
Is it arising because:
 if the change in entropy were less than zero, the Gibbs change in energy is less than zero? So if that were true, then entropy may not be a measurer for spontaneity, correct?

The reason why we consider the Gibbs energy to be the indicator for spontaneity of reactions is because it is such an indicator. Let's distinguish facts from interpretations. Entropy is an indicator too, but in different conditions.

The only determining factor for spontaneity of a process is the total change in entropy of the universe. This must be non-negative.
For a general process within a system, we can decompose this into:
$$\Delta S_{\mathrm{univ}} = \Delta S_{\mathrm{sys}}+\Delta S_{\mathrm{surr}} \geq 0$$
The computation of the system's change in entropy $\Delta S_{\mathrm{sys}}$ is not trivial, but doable. On the other hand, there is no practical way to directly measure or compute the corresponding change for the surroundings. We generally take shortcuts by measuring something else and deriving this value.
For example, for a isothermal, isobaric process, we assume that
$$\Delta S_{\mathrm{surr}} = -\frac{\Delta H_{\mathrm{sys}}}{T}$$
Substitution into the first equation gives:
$$\Delta S_{\mathrm{univ}} = \Delta S_{\mathrm{sys}} -\frac{\Delta H_{\mathrm{sys}}}{T} \geq 0$$
$$-T\Delta S_{\mathrm{sys}} + \Delta H_{\mathrm{sys}} \leq 0$$
This value is $\Delta G$.
For a constant volume process, you use the Helmholz free energy $\Delta F$ instead and just use internal energy change $\Delta U$ instead of $\Delta H$.
From this perspective, free energy is a convenient way to capture the total entropy change of the universe while only considering your system, but not surprisingly, there are assumptions you make in order for the magic to happen.

Neither enthalpy nor entropy alone can decide the spontaneity of a process.
For example, evaporation of water from oceans is an endothermic process $(\Delta H \gt 0)$, which clearly violates the fact that a system aims for minimum energy, yet this process is spontaneous (or naturally occurring).
To determine the spontaneity of a process the Gibbs-Hemholtz equation was introduced. This combines the effects of both enthalpy and entropy for determining the spontaneity of a process.
$\Delta G=\Delta H-T\Delta S$
A system generally aims for two things; minimum energy and maximum probability to be found in the universe (or more entropy). Therefore, for a spontaneous process, $\Delta H \lt 0$ and $\Delta S \gt 0$, Which gives an overall negative change is Gibbs energy.
For a spontaneous process, $\Delta G$ is negative.

There is more to it. $\Delta G$ determines the spontaneity of processes at constant $T$ and $P$. Otherwise we have to use something else.


text:
Why is thallium(III) oxide stable while thallium(III) chloride unstable
Why is thallium(III) oxide stable while thallium(III) chloride unstable
Both the compounds have +3 oxidation state of thallium and the electronegativity difference between chlorine and oxygen is not very large. Then why is there such a large difference between the stability of the 2 compounds?

The difference between standing 1 inch from a cliff and 1 inch over the cliff is not very large either, yet it may mean a great deal in the end.

@IvanNeretin so is this difference basically due to electronegativity? How can we predict the stability of such species in the absence of experimental data?

This difference is due to oxygen and chlorine being different elements. They differ in size, in weight, in all aspects of chemistry and in pretty much everything else. It's not like we can take an element and tune its electronegativity to a different value, while leaving all other properties intact.

As for predicting stability, I'd take a look at the oxidation potentials ($\ce{Tl^3+/Tl+, Cl2/Cl-,\text{ and }O2/OH-}$). True, they are measured in water solutions, and hence not applicable directly to solid compounds, but still they may give us a vague idea of what to expect.

@IvanNeretin basically I came across this question which asked me to compare the stability of these 2 and a few other compounds but I didn't know what basis to use to do so in the absence of any data given to me. The oxidation states were same. So how can I compare their stabilities using periodic trends and no other data?

The stability of thallium(III) oxide and thallium(III) chloride can be compared by observing two properties:

Size of corresponding ions

Size of $\ce{Tl^3+}$ is 0.95 Å, size of $\ce{O^2-}$ is 1.44  Å and that of $\ce{Cl-}$ is 1.8  Å. So, it pretty seems that difference in size between thallium and oxygen atoms is less than that of difference in size between thallium and chlorine atoms. So, thallium(III) oxide is stable than that of thallium(III) chloride. (data)

Difference in electronegativity

Electronegativity of thallium is 2.04, oxygen is 3.44 and that of chlorine is 3.16. So, difference in electronegativity of thallium-oxygen is more than that of thallium-chlorine. So, thallium(III) oxide is stable than that of thallium(III) chloride. (data)

Number 2 is too small a difference to be convincing.  Might we consider lattice energy effects?


text:
Are resonating structures real or not?
We say that hybrid structure is real. It has distinct properties as compared to its Resonating Structure. Is it because that all the experiments we carry out are on a sample, lots of molecules are present. Are they hybrid or are they lots of Resonating Structures in an equilibrium state that they collectively show properties of the hybrid. Also, some structures are said to contribute more to a hybrid structure. What does that mean?
Also, in charged molecules like carbonate ion, the charge is fractional. Won't that mean that electrons get "divided"? I know that partial charges can arise in covalent bonds due to a difference in electronegativity, but this seems different from that.

They are hybrids, individual resonance structures do not exist. Search for resonance on this site there are lots of writeups.

All the experimental proofs in our books are on a large sample. Can't all the structures contribute to effects that are similar/ expected from hybrid?

Look at it this way. A horse is the average of the resonance forms Pegasus and Unicorns.

Electron-density cloud diagrams are your friends when it comes to resonance

Resonance is a concept that had to be invented by man because the previous concepts to describe reality were insufficient. Resonance is sometimes wrongly taught as swinging back and forth between the different resonance structures, but that is an idea that people thought up and that has little to do with the actual reality.
It is very hard if not impossible for humans to actually understand the reality of chemistry at a molecular or atomic level. At these scales, quantum mechanical descriptions must be considered and any observed object can be both a wave and a particle — I think, at one point interference patterns could be observed with Buckyballs, confirming the wave-particle dualism up to $\ce{C60}$; just to give you an idea. Considering this, it is very understandable that concepts are hard to grasp and frequently misinterpreted, even by nobel laureates.
The key to correctly understanding resonance is that there are not many resonance structures that co-exist, but that overlaying all the resonance structures together into one gives us a better clue of the chemical reality at atomic scales. Taking the benzene molecule, we do not have double bonds on one side half the time or for half the molecules we’re observing. We have six identical bonds with a bond order of $1.5$. Localised double bonds, if present at any time, should react with halogens in an electrophilic addition reaction, yet that never happens with benzene. (Other experimental proof can also be used.) It is the fact that writing this $D_\mathrm{6h}$ symmetric structure on paper with our Lewis model formalisms is impossible that inspired chemists to think of resonance structures to better explain what is going on.
You also mention fractional charges and the possibility that ‘electrons get “divided”.’ Rest assured that fractional charges are possible even without electrons getting divided. A computationally well-studied system is the allyl anion[1]. Two canonic resonance structures as shown below are typically drawn.
$$\ce{H2C=CH-\overset{-}{C}H2 <-> H2\overset{-}{C}-CH=CH2}\tag{1}$$
Each of the terminal carbon atoms has an effective charge of $-\frac12$. How is this possible, you may ask? Molecular orbital theory does the best job of explaining it. The key orbital is the π HOMO which looks as shown in the image.

Notice that it is a wave function; it has four lobes, two centred on each of the terminal carbon atoms. If you try to understand this MO with conventional mechanics, you may be inclined to propose some kind of tunneling mechanism for the electrons to get from one lobe to the other. That would be wrong. They are not in one lobe and then go somewhere else, the two electrons are in all four lobes simultaneously. So we can have fractional charges (happens if you add up the electrons on each carbon; proof left to the readed) even without having to split electrons simply because of the wave-like nature of electrons.
Many fascinating observations await you as you enter quantum chemistry and

[attempt to] understand whatever
  Binds the world’s innermost core together[2]

Be prepared to forget what you think you may know if you decide to dive into this world.

Note and Reference:
[1]: Taken from Professor H. Zipse’s webpage (LMU Munich). For some reason, the allyl anion system shows N/A. Instead, I linked to the allyl cation which is identical except for one orbital without electrons.
[2]: J. W. von Goethe: Faust. Der Tragödie erster Teil. 1808.


text:
Do polar compounds travel further or less far than unpolar ones in hexane/ethyl acetate TLCs?
Would ethyl acetate/hexane be considered nonpolar or polar overall? I know that the ethyl acetate is polar and the hexane is nonpolar. I am trying to understand what if polar or nonpolar analgesics would travel further in TLC. I know that if  ethyl acetate/hexane is polar than a more polar one will travel further, and if it is nonpolar then nonpolar will travel further.
I was looking here and I see that caffeine is the most polar (out of aspirin and acetaminophen) and traveled the least distance (lowest $R_\mathrm{f}$) in my experiment. I thought that ethyl acetate/hexane was polar so I would have thought caffeine would be the least polar since it traveled the shortest.
So am I wrong that  ethyl acetate/hexane is polar or are they wrong that caffeine is the most polar?

For normal phase TLC the more polar compound always travels slower.

The link seems to have disappeared. Ethyl acetate is polar. Hexane is non-polar. So by using a mixture of the two you can get solutions of varying polarity. Adjusting polarity that way is easier than trying to find 10 different solvents of varying polarity.

I used a silica gel for the plate... Does that make it normal phase TLC?

Basically using a polar stationary phase (which a silica plate is) makes it normal phase TLC.

Never have I seen a normal phase (i.e. silica gel or alox) TLC where a more polar compounds travels further than a less polar compound — and I have used some extremely (considering normal phase TLC) polar eluents in the past.
The top-running compounds are always the least polar and the slowest are always the most polar in normal phase TLC.

How about using aniline vs glacial acetic acid to develop amines versus carboxylic acids? // Grand dogmatic statements in chemistry always bother me. It seems that there is always some weird exception.  // I do get that you're talking about "normal expectations"...

@MaxW Haha~, but then we have the anion in one case and the neutral molecule in another. I consider those different species for the sake of the argument.


text:
Side chain chlorination of substituted toluene
I know that we can convert toluene to benzaldehyde using side chain chlorination in presence of light, followed by hydrolysis.
Is it possible to use the same method to convert a substituted toluene?
An example would be to convert p-fluorotoluene to p-fluorobenzaldehyde.

Hello and welcome to Chemistry.SE! If you want to learn more about how the site works, this short [tour] is a good place to start.

Check the literature. The issue is that radicals are electron deficient, so this is probably not going through a very stable intermediate...

In general, this is the basis of organic chemistry’s notion of functional groups: a reaction will occur if a given functional group (here: benzylic hydrogen that can be turned into a benzyl radical easily) is present regardless of any other groups in the molecule. Thus, in theory, the reaction should occur just as well.
In practice, functional groups are never fully innocent. They can either disturb the entire reaction by interfering with the mechanistic pathway (e.g. you cannot perform Paterson aldol reactions in the presence of alcohols) or they can subtly affect the electronic and steric parameters of a molecule to ease or inhibit a reaction. For this reason, journal articles that establish novel transformations often ‘synthesise a table’ — i.e. they take multiple related compounds and perform the reaction for each to show whether it works with a given functional group at all and how the yield is affected.
In this case, the benzylic radical intermediate is an electron-deficient species. Fluorine is an electronegative substituent that destabilises electron-deficient species. However, fluorine is unlikely to interfere with the generation of radicals or react with them. So without having consulted the literature I would predict the reaction to work but require longer reaction times, higher temperatures or give lower yields.

Yes, I also know about Glorius’ robustness screening, but after having talked to a guy in method development I no longer consider it any more interesting than the typical synthetic tables.


text:
Penta-valent ion in water?
I would like to have a solution with a pentavalent positive ion. I need something that, preferably in water:  $\ce{XY -> X^5+ + Y^5-}$. What can XY be?

I don't think this is too broad, considering that as far as I know, no such compound exists.

If you consider a small polymer chain with eg 5 amine groups it can exist, but 5+ is a very high charge, so you need some really strong stabilization mechanism if you want to concentrate it to a single group or small molecule.

You won't get a +5 charge on a single atom (or even a small molecule) in water. It will be heavily coordinated by water/hydroxo/oxo ligands.

Just to back this up there aren't a huge amount of single atom tetravalent simply hydrated ions  that come to mind, let alone pentavalent, if it is single atoms you are looking for. For the cation Cerium, Thorium and Uranium are possibilities that come to mind, but without looking them up I can't be sure.

Even the tetravalent cations (most of them) are heavily hydrolyzed and barely exist in water solutions. Pentavalent cation is even more of a challenge. Protactinium might be a possibility, but you can't have it anyway.
Speaking of anions, there is not even a trivalent single-atom anion that would not hydrolyze instantly and completely upon touching water.
On the other hand, polyatomic anions (and cations, for that matter) can have pretty much any charge you want.

One five-charge anion you might see is tri(poly)phosphate, as in the sodium salt $\ce{Na_5P_3O_10}$.  See https://en.wikipedia.org/wiki/Sodium_triphosphate.


text:
After a unitary transformation, is Koopmans' theorem still valid?
Koopmans' theorem says that the energy of the HOMO of the Hartree-Fock orbitals is equal to the first ionization energy of whatever system is being studied. This is only approximate because it assumes no changes in the orbitals in the ionic state, no relativistic effects, and no electron correlation. Ignoring those approximations, there is another feature of this which I don't understand.
The wavefunction as determined using HF can be unitarily transformed so that the total energy and wavefunction are preserved, but the "orbitals" themselves are not. Under many of these transformations, I assume that the orbital energies change. Is this true? If it is true, I assume Koopmans' theorem is still valid, but is now more complicated so that the ionization energy is some combination of orbital energies. Can anyone expand on this and let me know if my thinking is correct?
Also, what is it that is so special about the Hartree-Fock orbitals that Koopmans' Theorem has such a simple interpretation, while, if what I say above is true, it seems to be quite convoluted for other sets of orbitals?

As a note, I asked a question which more or less contained this question, but was a bit broader. The question I'm asking right now is the question I am really more interested in having answered.
You can find that other question here: Observability of Orbitals and Orbital Energies

HOMO has a well defined energy only when it corresponds to an eigenvalue from HF. After such a unitary transformation the orbitals have no well defined energies any more

I don't think that the orbital energies ever correspond to the eigenvalues of the Fock operator. Rather the eigenvalues are the sum of these orbital energies, but I'm pretty sure these orbital energies are not unique because the orbitals themselves are not unique. The second half of that sentence depends on the answer to this question which I'm not completely sure of.

They do, the Fock operator is an effective one electron operator and this is how you find the orbitals and the correspond energies.

Ah sorry. I was confusing the Fock matrix and the Fock operator. The eigenvalues of the Fock matrix are the energy of the whole system which corresponds to the sum of eigenvalues of the Fock operator which are indeed the orbital energies. Sorry to suggest otherwise. What I'm pointing out in the question is that the set of orbitals and their eigenvalues given by the Fock operator is not the only solution we could come up with. So, I'm basically asking what is so special about the HF orbitals that Koopmans' theorem should be so simple when that doesn't seem to be the case for other orbital sets.

No, again, the eigenvalues of the Fock matrix (which is the Fock operator in some basis) are the canonical orbital energies. Moreover, the sum of orbital energies is NOT the total system energy. Check out this paper for more info: http://pubs.acs.org/doi/abs/10.1021/ed200673w

@levineds Thanks for the reference. I see what you're saying and what they're saying. I should have been more precise when saying total energy. I have also since cleared up my confusion about the Fock operator and Fock matrix. Things were pretty jumbled but I'm glad to straighten them out!

Disclaimer: I like Tyberius' answer, but I'd like to go a little further.
$%
\newcommand{\ll}{\left\langle}\newcommand{\rr}{\right\rangle}
\newcommand{\lb}{\left|}\newcommand{\rb}{\right|}
\newcommand{\op}[1]{\mathbf{#1}}$

The beauty of Koopmanns' theorem is its simplicity. Quoting from the gold book:

Koopmans' theorem
Directly relates experimental ionization potentials with energy levels of molecular orbitals. The theorem states that the ionization potential required to remove an electron from the orbital $\Psi_i$ is given by the negative value of the energy of the orbital, $−\varepsilon_i$, as calculated within the Hartree–Fock approximation. The theorem is not applied to localized molecular orbitals, which are not eigenfunctions of the effective hamiltonian.

While Koopmans' text (in German: Physica 1934, 1, 104-113.) is a bit hard to comprehend, due to the ancient language I never learned, it basically boils down to two important facts.

There is a unitary transformation of the Lagrange multipliers which diagonalises them. These form a characteristic set of elementary wave functions, matrix, which in turn have a characteristic eigenvalue each. If all eigenvalues are different, then all elementary wave functions are specified through the Hamiltonian. (If there are equal eigenvalues, a freely chosen unitary transformation can be performed.)
In other words: There is one set of canonical orbitals that diagonalises the Fock matrix; each canonical orbital has an eigenvalue. For non-degenerate systems those orbitals are specified through the Hamiltonian.
This choice of functions has physical meaning, as the eigenvalue (disregarding a small correction) can be equated to the ionisation energies of the corresponding electron.

The second statement implies that there is no relaxation of the orbitals when removing an electron. This is oftentimes referred to as "frozen MO" approximation. Koopmanns himself appreciates that there should be a contraction of orbitals when removing an electron, however, he doesn't go into detail.
Nowadays we explain Koopmans' theorem in some simpler terms, where we remove the $k$th electron:
\begin{align}
&& E_N &= \sum_{i=1}^N H_{ii}
        + \frac12\sum_{i=1}^N\sum_{j=1}^N (J_{ij} - K_{ij})
        + V_{\mathrm{nuc}} \tag1\\
&& E_{N-1}^k &=\sum_{i=1}^{N-1} H_{ii}
        + \frac12\sum_{i=1}^{N-1}\sum_{j=1}^{N-1} (J_{ij} - K_{ij})
        + V_{\mathrm{nuc}} \tag2\\\hline
&& E_N - E_{N-1}^k &=  H_{kk}
        + \frac12\sum_{i=1}^N (J_{ik}-K_{ik})
        + \frac12\sum_{i=1}^N (J_{kj}-K_{kj}) \tag{$1-2$}\\
\therefore&&
E_N - E_{N-1}^k &=  H_{kk} + \sum_{i=1}^N (J_{ki}-K_{ki})\\
\therefore&&
E_N - E_{N-1}^k &= \varepsilon_k\\[2ex]
\text{with}&&
H_{ii} &= \langle \phi_i(\mathbf{x}_1)|
          \mathbf{H}^\mathrm{c} | \phi_i(\mathbf{x}_1)\rangle\\
&&
J_{ij} &= \langle \phi_i(\mathbf{x}_1) \phi_j(\mathbf{x}_2) | r_{12}^{-1}
          | \phi_i(\mathbf{x}_1) \phi_j(\mathbf{x}_2) \rangle\\
&&
K_{ij} &= \langle \phi_i(\mathbf{x}_1) \phi_j(\mathbf{x}_2) | r_{12}^{-1}
          | \phi_j(\mathbf{x}_1) \phi_i(\mathbf{x}_2) \rangle\\
\end{align}
As you can see, Koopmans' theorem is not limited to the HOMO, but can be used for any occupied MO.
There is a very important point to be always be kept in mind when working with any Hartree-Fock based approaches:
Even though the equation
$$\op{F}_i\phi_i = \varepsilon_i\phi_i
     \tag{3}\label{fock-pseudo}$$
suggests an eigenvalue problem, it is not. Remember the definition of the Fock operator and the operators contained
\begin{align}
  && \op{F}_i &= \op{H}^\mathrm{c} + \sum_j (\op{J}_j - \op{K}_j),\\
\text{with}&&
     \op{J}_j\lb \phi_i\rr &=
       \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
       \lb \phi_j(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr,\\
\text{and}&&
     \op{K}_j\lb \phi_i\rr &=
       \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
       \lb \phi_i(\op{x}_1) \rr \lb \phi_j(\op{x}_2) \rr.
\end{align}
As you can see, the "one-electron" Fock operator depends on the solution of all "one-electron" Fock operators (ref. Szabó-Ostlund p. 115). The Hamilton operator is not the sum of all Fock operators, and the total HF energy is not the sum of all orbital energies. As a result of that, the canonical orbitals are actually unique solutions.
The Fock operator is associated with the whole wave function, and while unitary transformations will keep the wave function and its energy equivalent, everything else will fall apart.
Another important consideration is the fact that the Fock operator is only well defined for occupied MO. You sometimes find the statement
$$E_{N+1}^{l>N} - E_N = \varepsilon_l,$$
which can not be applied in the same way; it is very basis set dependent.
From all that above it is obvious, that Koopmans' theorem only works for systems where a single determinant approximation is reasonable. It also explains, why it only works for HF; although there are generalisations for DFT.

Is Koopmans' theorem still valid after a unitary transformation?
No. As stated above, the Fock operator is associated with the $N$-electron wave function. After a unitary transformation and then removal of an electron, the energy of the $N-1$-electron wave function is not conserved.
As an illustrative example:
\begin{align}
&& \sum_{i=1}^{N} H_{ii} &= \sum_{i=1}^{N} H_{ii}'\\[2ex]
\text{with}&&
H'_{ii} &= \langle \phi'_i(\mathbf{x}_1)|
          \mathbf{H}^\mathrm{c} | \phi'_i(\mathbf{x}_1)\rangle\\
\text{and}&&
|\phi_i\rangle &\color{red}{\neq} |\phi'_i\rangle\\[2ex]
&& \sum_{i=1,i\neq k}^{N-1} H_{ii} + H_{kk}&=
   \sum_{i=1,i\neq k}^{N-1} H_{ii}' + H_{kk}'\\
\text{for } H_{kk} = H_{kk}':
&& \sum_{i=1,i\neq k}^{N-1} H_{ii} &=
   \sum_{i=1,i\neq k}^{N-1} H_{ii}'&&
\implies|\phi_i\rangle \color{red}{=} |\phi'_i\rangle\\
\end{align}
I have indicated the contradiction in red. Therefore it follows:
$$|\phi_i\rangle \neq |\phi'_i\rangle
  \implies H_{kk} \neq H_{kk}'
  \implies
  \sum_{i=1,i\neq k}^{N-1} H_{ii} \neq \sum_{i=1,i\neq k}^{N-1} H_{ii}'
$$
You can follow that trough for the other terms and will see that the expectation value of energy of the $N-1$-electron wave function needs to be different from the expectation value of energy of the $N-1$-electron wave function after unitary transformation.

I'm basing my attempt at an answer off of Modern Quantum Chemistry by Szabo and Ostlund, p.119-122.
The general development of Hartree-Fock is done via functional variation of the ground state energy $E_0=\left<\psi_0\right|\!\hat{H}\!\left|\psi_0\right>$. At the end of this, we obtain the matrix form of a differential equation
$$
f\left|\chi_a\right> = \sum_{b\,=\,1}^N \epsilon_{ba}\left|\chi_b\right>
$$
where $f$ is the Fock operator and the sum is over all $N$ occupied spin orbitals. By appropriate unitary transformation (that which diagonalizes the matrix $\epsilon$), we obtain the equation in its canonical form
$$
f\left|\chi_a'\right> = \epsilon_{a}\left|\chi_a'\right>
$$
where the $\chi_a'$ are the canonical Hartree-Fock orbitals and the elements along the diagonal of $\epsilon$ are interpreted as molecular orbital energies.
My understanding of why it would be difficult to use Koopmans' theorem with other orbitals is that the canonical Hartree-Fock orbitals are unique in putting the above matrix equation in diagonal form. In the noncanonical forms, there isn't a clear value of the orbital energy for a particular $\left|\chi_a''\right>$ (i.e., they don't return an eigenvalue when acted on by the Fock operator).

Correct, I would expect the "energies" to be linear combinations of the canonical Fock matrix eigenvalues, with the linear coefficients determined by the linear combination of eigenvectors from the canonical basis that form each new unitary-transformed (?) eigenvector. Sorry that's worded so awkwardly...

@pentavalentcarbon I'm not sure if that's actually valid. It apparently works for occupancies, as in, e.g., the fractional occupations that result in the construction of natural orbitals. But if it were valid for energies, I would think calculating the transformed orbital energies this way would be used all over the place, and AFAIK it's not.

@pentavalentcarbon Or, well, maybe I should take that back. I'm reading Weinhold's NBO book, and apparently GENNBO does attribute energies to the individual NBOs. So, apparently there is some method for rationally assigning such energies -- and the method you describe seems sensible.

@hBy2Py I put energies in quotes because it isn't an energy. Not all eigenvalues are energies. Transformed orbitals may look more interesting physically but are more ill-defined, even more so for their eigenvalues. NBO eigenvalues may be interesting for comparison within a set but there is probably no quantitative meaning behind them at all.

While we are busy arguing about the validity of eigenvalues from unitary transformations, I wonder what the connection is to actual natural orbitals, that is, the eigenvectors that diagonalize the canonical density matrix (rather than the canonical Fock operator). Though, one could use a DM from a correlated method but let's not go there.

@pentavalentcarbon Aren't the the eigenvectors that diagonalize the density matrix the same as those that diagonalize the Fock matrix for HF? That is to say, aren't the canonical HF orbitals also the natural HF orbitals?

I don't think so (would love some confirmation of this). For the Fock matrix we have $C^{\dagger}FC=E_{\text{canonical}}$ and for the density matrix one would have $U^{\dagger}DU=E_{\text{natural}}$ (just using $U$ to differentiate, I don't know what the convention is); I don't think $C$ and $U$ are necessarily the same.

@pentavalentcarbon I might be misunderstanding this page and something I was told by a professor, because I had thought that diagonalizing the Fock matrix had to diagonalize the density matrix because the Fock Matrix is a function of the density matrix.

@Tyberius To my best understanding, the "density matrix" used in the course of a HF calculation, as referred to by your link, is very different from the first-order reduced density operator, $\Gamma(\mathbf r_1)\equiv\int{\Psi\Psi^*,d\mathbf r_2d\mathbf r_3\ldots d\mathbf r_n}$, whose matrix representation is diagonalized to generate natural orbitals. Personally, I would be inclined to call the former "density matrix" an "overlap matrix," instead, as it's what is used to probe the orthonormality of the canonical MOs.

@hBy2Py I believe you are correct. I'm still somewhat in the early stages of learning quantum chemistry and I had only seen the one instance of the density matrix  in doing HF calculations.

I'm planning on writing up a question to ask about this difference, and post it here and on the ORCA forum. Intuitively, it seems like these have to be different, but I'm not at all sure.

@hBy2Py I know you said you would make a question on this but: 1) yes, the orbitals that diagonalize the Fock matrix also diagonalize the density matrix (in HF), 2) as such the natural orbitals are the canonical orbitals for HF, 3) the 1RDM is different from the n-particle density matrix. The 1RDM is the density matrix integrated over all deg. of freedom except 1 electron. HF is the approximation that all higher density matrices are direct products of the 1RDM and so the HF density specifies the full n-particle DM. 4) the density matrix is NOT any kind of overlap matrix

@levineds If you would be interested, I could post that question soon. I think people on the site (and myself personally), would benefit from seeing a fleshed out version of your answer.


text:
Storing oxygen generated via water electrolysis
I want to create a somewhat pure oxygen gas from electrolyzing water and using sodium bicarbonate as an electrolyte.
How can I harvest the oxygen once it is released from the water, and store it in a container without adding oxygen to the existing air in the container? Meaning that I want pure oxygen in the container and not normal air.

Hello and welcome to Chemistry.SE! I suggest taking the short [tour] if you have any questions about how this site works. Regarding your question, the simplest approach to this may be to displace the water from a full, inverted container like a test tube. If you could rig some kind of inverted funnel over the anode (in the water) with a hole at the top for a short straw that goes into the inverted container, that might do what you need. This is really a mini-engineering question as you've already done the chemistry in creating the oxygen, but maybe someone will have a good idea. Good luck!

I think the simplest approach is to fill a test-tube with your electrolyte solution, seal the open end with your latex-glove covered thumb, then immerse the sealed end in your cell. If you can tilt your anode several degrees from the perpendicular, you should get a stream of oxygen bubbles that you can then catch with the test tube. You certainly won't get them all but it shouldn't be a problem to get 3/4 of them anyway. If you have trouble inverting the test tube without having a little air left, just slide some flexible tubing to the top and suction out the air. Take care to watch the level of your electrolyte as the solution that is displaced from your test tube will of course be filling your cell.
After you have collected all of the oxygen you need, seal the open end of the test tube with a piece cut from a latex glove and a rubber band. You will then have a test tube filled with oxygen gas and any residual electrolyte solution.


text:
Is recycling introducing toxins into products that come in contact with food?
Paper products like napkins and paper plates made from recycled material might have unsafe levels of toxins if the original recycled materials contain toxic inks, etc. How big of a health concern is this?
EDIT: Not sure how this is too broad -- my goal is to start a discussion about a problem that I have not heard mentioned before.

Surprised but such products do exist. http://letsgogreen.biz/pages/paper.html

All toilet paper is made at least partially from recycled material.

None, if you can trust your producer, and the safety agency controlling him.
The problem is more the large water and bleach consumption of the recycling factory, that makes sure your new recycled toilet paper is safe, and the environmentally safe waste disposal of said factory.

I am not convinced that if toxins start out in the original paper that it can be effectively removed at least in an economical way. So part of the question is, how toxic are, for example, the inks and dyes in a colorful laundry detergent box? At one time, I am pretty sure such a box would have had a lot of lead; at one time the labels on Wonder Bread were printed with lead ink.

@Jeff  Such inks have long been forbidden in the more civilised parts of the world.

I hope so. Are you familiar with the Wonder Bread incident? A label printed inside-out so bread was in direct contact with ink caused acute lead poisoning in kids. BTW perhaps recycled paper may not come from "civilized" parts of the world.

@Jeff Well, that's true. Plastic toys with toxic paint etc. regularly make it into our newspapers. Horray for globalised production without retraceble quality control.


text:
How to write mass balance equation for the scenario when molecules are getting generated inside an sphere?
I am working on the physical understanding of diffusion through a nano machine(a shell with hollow interior in which molecules are produced) and have taken few assumptions:

There is rate $r_T$ with which molecules are produced inside this machine, and  we are not concerned with the production process.
The machine is spherical shell and emitting molecules through pores on its body.
We are diffusing in an infinite radius sphere where the center of the nano machine coincides with the center of sphere and is stationary.
Let the radius of the spherical be $r+\delta r$ where r is the radius of inner side region where mass is getting produced with rate $r_T(t)$ and $\delta r$ the width of shell. $j_1$ is flux per unit area.
Assuming that there is a porous outer through which molecules are diffusing with Partition coefficient 'H'.

Now I am trying to write mass balance equation for it so as to get diffusion flux in terms of concentration.For this I apply this formula:
input +generation-output-consumption=accumulation
Taking its rate/differential form, I know that rate of accumulation in $\delta r$ will be zero.
so, by somewhat same approach as when we have a spherical solute dissolving sparingly in water I got:
$$r_T(t)+(4\pi r^2j_1)_r - (4\pi r^2 j_1)_{r+\delta r}=0$$ and when I try to solve it taking $\delta r$ to be very small and dividing both sides by the volume of the shell, I got this equation:
$$\frac {r_t(t)}{4\pi r^2 \delta r}=\frac{1}{r^2} \frac{d (r^2 j_1)}{dr}$$
which seems to be unbalanced, further applying it to Fick's first law doesn't make any sense. How should I proceed in my case? Is my mass balance equation wrong?
$$$$

You're pretty close.  Is $r_t(t)$ supposed to be the total rate of generation from r = 0 to r = R, or is it supposed to to be rate of generation per unit volume?

Thanks for your interest, $r_T(t)$ is the total rate of generation from r = 0 to r = R.

Then the left hand side of the equation should read:  $\frac{3r_t(t)}{4\pi R^3}$

Incidentally, is this problem time-dependent, or are you assuming quasi steady state?

$r_T(t)$ is the rate of production of particles with respect to time. I think quasi steady state is an easier thing to do. You are asking that because in latter case we have equilibrium established?Then yes  quasi steady state.

I was just wondering, for clarification, whether you were trying to treat this as a true transient problem, or were just assuming that you had allowed enough time to pass for quasi steady state to be established.

" had allowed enough time to pass for quasi steady state to be established" that should be fine for me, as this will simplify the problem a little bit( I feel so) , and once done I can move for transient part too.


text:
How can you calculate a new Keq when a reaction is reversed?

Consider the following equilibrium:
  \begin{align}
  \ce{H2 (g) + I2 (g) &<--> 2HI (g)}& K_\mathrm{eq} &= 50.0
\end{align}
  What is the value $K_\mathrm{eq}$ for the reaction rewritten as:
  \begin{align}
  \ce{2HI (g) &<--> H2 (g) + I2 (g)}& K_\mathrm{eq} &= \mathbf{?}
\end{align}
  Choices
  \begin{align}
  \mathrm{A}&:-50.0 &
  \mathrm{B}&: 0.0200 &
  \mathrm{C}&: 25.0 &
  \mathrm{D}&: 50.0
\end{align}

The answer is $0.0200$, but I'm confused as to why? How can you figure it out?

Do you know how to write the equation for the equilibrium constant in terms of the concentrations of products and reactants? Doing that might shed some light on what's going here. Hint: notice that 0.02 = 1/50

The first Keq equation would be [HI]^2/[H2][I2]= 50. The reaction written backwards is the reciprocal, so would Keq just equal the reciprocal of 50?

Yep, you got it ;)

For homework like questions, please use the tag homework-and-exercises

First let's use a conceptual approach, as this multiple choice problem is answerable in that way.
$K_\mathrm{eq}$ cannot be negative.  Therefore $\mathrm{A}$ is incorrect.
$K_{\mathrm{eq}}$ is greater than one when the $\ce{left -> right}$ reaction goes farther; it is less than one when the reverse reaction goes farther (by going farther I mean end in greater concentration at equilibrium).
The
$$\ce{H2 + I2 <=> 2HI}$$
reaction has $K_{\mathrm{eq}}$ over $1$.  Therefore the
$$\ce{H2 + I2 -> 2HI}$$
happens faster than the
\begin{align}
&&\ce{H2 + I2 &<- 2HI}\\
\text{or}&&
\ce{2HI &-> H2 + I2}.
\end{align}
So in
$$\ce{2HI <=> H2 + I2}$$
or the reverse, the $\ce{left->right}$ reaction occurs slower; the $K_{\mathrm{eq}}$ must be less than $1$.  $\mathrm{B}$ is the only answer.
Another way to approach the problem is using algebraic reasoning.
Initially,
$$K_{eq} = \frac{\ce{[HI]^2}}{\ce{[H2][I2]}} = 50.$$
Then, when you reverse the reaction,
$$K_{eq} = \frac{\ce{[H2][I2]}}{\ce{[HI]^2}}
\frac{1}{\left(\frac{\ce{[HI]^2}}{\ce{[H2][I2]}}\right)}= \frac{1}{50.0} = 0.0200.$$
In general when you reverse the equation
$$K_{\text{eq (new)}} = \frac{\text{products (new)}}{\text{reactants (new)}} = \frac{\text{reactants (old)}}{\text{products (old)}} = \frac{1}{K_{\text{eq (old)}}}.$$

It would be easier, more conceptual, I feel to do in terms of gibbs free energy. The log term is exactly negative of initial, indication new K is inverse of original.


text:
Question on organic chemistry where we are required to find enol content

Which of the following compounds is expected to  have maximum enol content in ethanol?
(A) $\ce{CH3COCH2CO2CH3}$
(B) $\ce{CH3COCH2CO2C2H5}$ ✔(I think this is the correct answer)
(C) $\ce{CH3COCH2COCH3}$
(D) $\ce{CH3COCH(CH3)COCH3}$

I think the ketoesters will have greater enol content than the diketones because they have stronger $-I$ effect resulting in greater positive charge on the hydrogen atoms of the $\ce{CH2}$ group in between the ester and ketone group. This leads me to believe the hydrogen will be lost easily. Now between option one and two I think option (A) makes no sense because it will be converted to option (B) via trans esterification, so I have marked option (B). Please let me know where I am going wrong and the correct concept involved if what I think is not right.

Thermodynamically, the greater enol content can be associated with greater acidity. Thus, the question can be rephrased as ‘which is the strongest acid of these four?’ We can ignore the interconversion of the methyl ester A into the ethyl ester B — it will happen because we are in ethanolic solution, but extremely slowly as we do not have any strong bases present. (And even if we had strong bases, these would deprotonate A more further reducing the susceptibility of nucleophilic attack on the carboxy group.)
For reference and posteriority, these are the compounds we are talking about:

methyl acetoacetate (A)
ethyl acetoacetate (B)
acetoacetone (C)
3-methylacetoacetone (D)

Simply by looking at the names we can see how highly related they are and how much subtle differences must matter. We can predict all four of them to be rather acidic since if the inter-carbonyl hydrogen is released and an enolate is formed, this enolate will always be in contact with an electron-withdrawing π system on the right-hand side: either another keto-group or an ester group.
That immediately gives us our first hint to compare the four: an ester group is less electron-withdrawing than a ketone, so we predict A and B to be less strong acids than C and D. This is convenient as it means we do not have to concern ourselves with the acidity difference of a methyl versus ethyl ester.
The difference in acidity between C and D is also subtle but more obvious. In both cases, the corresponding enolate anion would be planar. However, the methyl group is much larger than a hydrogen atom meaning that the enolate of D is subject to 1,3-allylic strain — which cannot be lessened without destroying planarity and thereby reducing conjugation. Thus, the enolate of C is more stable than the enolate of D.
This is backed by acidity data from the Bordwell tables — see table.
$$\textbf{Table 1: }\text{acidity data of the four compounds}\\
\begin{array}{lcccc}\hline
\text{solvent} & \mathrm{p}K_\mathrm{a}(\textbf{A}) & \mathrm{p}K_\mathrm{a}(\textbf{B}) & \mathrm{p}K_\mathrm{a}(\textbf{C}) & \mathrm{p}K_\mathrm{a}(\textbf{D}) \\ \hline
\ce{H2O} & -/- & 10.7 & \phantom{0}9.0 & -/-\\
\text{DMSO} & -/- & 14.2 & 13.3 & 15.1\\ \hline\end{array}$$

Thanks a lot . That was really helpful. I would like to ask you ,can't we explain on basis of inductive effect of methyl group ,the second part as to why c is chosen over d

@Ambarish Not really. Remember that the anion — $\ce{Me-C(-O^-)=CMe-C(=O)-Me}$ is oxygen centred, thus pretty far away. In fact, I was initially inclined to assume a greater acidity for D since it leads to the higher-substituted double bond.

I believe that C is the correct answer. If you draw out the enol structure from one of the ketones, you can see that the C-C double bonds is conjugated with the neighboring ketone which provides a stabilizing effect. Furthermore, either ketone can form the enol form, so there is double the chance for the enol to form.
You would not see the transesterification reaction happen here because it is thermodynamically unfavorable for ethanol to displace the methyl group without the presence of a catalyst. You are not wrong about the inductive effect of the ester, but since the solution we are in is ethanol losing the hydrogen would be unfavorable. In water, the ester might win out, as the enol could freely exchange its hydrogen.

If you draw out the enol of A , B and D, you should notice that they, too are conjugated to a pi system. Also, if you say C is strongest you should also consider why not D which is identical to C save an additional methyl group.

I think Jan , the methyl group by its +I effect will decrease the positive charge density on the hydrogen atom ,thus making it difficult for Ethanol (already a weak base) to abstract it . Also you can think like this ,that the carbanion obtained after abstraction of hydrogen will be destabilized by the +I of methyl group .

And regarding ester showing resonance ,I am still doubtful with the answer given . That is primarily the reason I thought the answer is b ,because resonance is present in the esters also , on addition the ester via its -I effect ,I think , creates a stronger active methylene group ,from which the hydrogen can be lost more easily ...... So that doubt still remains


text:
What are potential sources of low recovery in this acid-base extraction experiment?
I did an acid-base extraction experiment. First extracted a solid mixture containing benzoic acid, p-nitroaniline, and naphthalene into three separate solutions:

NaOH extract
HCl extract
3.organic with the neutral compound.

Then the HCl and NaOH extracts were neutralized and underwent vacuum filtration to isolate crude p-nitroaniline and crude benzoic acid, while the organic solution underwent rotary evaporation to get crude naphthalene.
Crude benzoic acid was purified by recrystallization using hot filtration, and then benzoic acid crystals were isolated by suction filtration. The melting range of the benzoic was relatively pure.
How can I improve this experiment? My recovery of benzoic acid was less than 40%
And as for sources of error, extraction processes always lose some compound during each step. What are other probable sources of error?

Do you know the recoveries of the other 2 compounds?

What was % recovery for crude p-nitroaniline and crude benzoic acid?  What was your final % recovery for p-nitroaniline and naphthalene? // Zone refining

How many NaOH extracts did you use? How many back extractions did you do? What was the back extraction solvent? Did you use a single component solvent for recrystallization? Please provide more experimental details.

Jason, what organic solvent did you use to extract the aqueous solution. Have you tried using different organic solvents?

I can only assume that you we following some recipe for the purification of the benzoic acid. Frankly if I were doing this I would have done it differently.
Wikipedia shows that the solubility of benzoic acid is 1.7 g/L (0 °C) and 56.31 g/L (100 °C) in water. So I'd added benzoic acid to a test tube and put in a boiling water bath. I'd added just enough water to get the benzoic acid to dissolve and then let the liquid cool in air. When down to room temp, I'd put it into an ice bath. After 10 minutes in ice bath I'd pour off water. I'd wash with a few drops of ice cold water. I'd test for purity and if not good enough I'd repeat.
0.34 g of crude benzoic acid should dissolve in about 6 ml of boiling water.
Each water cycle should have an efficiency of close to (56.31 - 1.7)/56.31 = 97%

Jason I don't understand your comment. If purity wasn't an issue why were you recrystallizing the benzoic acid? // (1) Did you "invent" hot filtration purification method? How much water did you use? // If you dissolve 0.34 g in 5 ml of boiling water then a small mass won't dissolve. But the benzoic acid that doesn't dissolve doesn't really get purified much.

@Jason - How much water did you add?

@Jason -  Did you "invent" the hot filtration purification method or were you following the experimental procedure given?


text:
In which solvent is AgBr least soluble?

The solute $\ce{AgBr}$ is least soluble in

A $\ce{H2O}$
B $1.0~\mathrm{M}\ \ce{FeBr3}$
C $1.0~\mathrm{M}\ \ce{CaBr2}$
D $1.0~\mathrm{M}\ \ce{AgNO3}$

I'm confused as to why the answer is B. Is it because $\ce{Ag}$ has low solubility with $\ce{Br}$, and the higher the concentration of $\ce{Br-}$ ions the less $\ce{Ag}$ will dissolve?

While solubilities are typically quoted in $\mathrm{g/l}$ or related units, the actual physicochemical idea behind it is an equilibrium constant known as the solubility product. In general, dissolution and precipitation can be considered two directions of the following equilibrium:
$$\ce{m A^n+ (aq) + n B^m- (aq) <<=> A_mB_n (aq) <=>> A_mB_n (s) v}\tag{1}$$
The outer sides are both preferred with respect to the central one; thus, the central compound is usually ignored, giving the following equilibrium constant:
$$K_\mathrm{sp}^{-1} = \frac{a(\ce{A_mB_n})}{a(\ce{A^n+})^m a(\ce{B^m-})^n}\tag{2}$$
Notes concerning equation $(2)$:

technically, this would be a precipitation equation; the solubility product is derived from the opposite equation hence the superscript $-1$.
activities should be used here, and not concentrations. This makes a difference as it allows for an activity of a solid (which has no concentration in solution — being solid outside of the solution).

Going the dissolution pathway, the activities of $\ce{A^n+}$ and $\ce{B^m-}$ are in the numerator and the solid’s activity is in the denominator. Thankfully, the activity of any solid compound is always $1$ per definition. That gives us $(2')$:
$$K_\mathrm{sp} = a(\ce{A^n+})^ma(\ce{B^m-})^n\tag{$2'$}$$
Which is the way the solubility product is typically represented. Now since this is an equilibrium process, Le Chatelier’s principle influcences it. Most importantly, you can add any one reactant to generate more of the product and vice-versa. If you take a look back at $(1)$, you’ll notice that the dissoluted ions enter the equation separately. Thus, we can use higher concentrations of one ion to reduce the overall solubility since then less of the other ion can remain dissolved.
This already tells us that all of options B through D result in a lower solubility since bromide ions (or silver ions) are dissolved. But how do we compare the three? The key here is that a $\pu{1M}$ solution of a salt does not necessarily mean that all its ions have a concentration of $\pu{1M}$. Instead, for $\ce{CaBr2}$ two moles of bromide are in solution for each mole of salt so if $c(\ce{CaBr2}) = \pu{1M}$, the concentration of bromide will be $c(\ce{Br-}) = \pu{2M}$.
Therefore, the solubility of $\ce{AgBr}$ will be lower in $\ce{CaBr2}$ (twice the concentration of bromide) and lower still in $\ce{FeBr3}$ (thrice the concentration of bromide). Hence B is your answer.

I would claim the opposite, I think. Solubility is fundamentally how much substance $\ce{A}$ dissolves in $\ce{B}$ for specified conditions, and one way (not unique) of expressing analytical composition has dimensions $\mathsf{M\ L^{-3}}$. The solubility product is a method of estimating solubilities, but a number of criteria must be satisfied in order for $K_\mathrm{sp}$ to be useful. // 10.1021/ed018p126; solubility in Gold Book

AgBr would not dissolve anymore if the solubity product is greater than Ksp.In option B, that thing is equivalent to saying 3M Br-. Let AgBr dissolve by x moles/litre.

$x(x+3)=Ksp$

In option D , you have 1M Ag+, so if y moles/lit dissolve then

$y(1+y)=Ksp$

Clearly x is less than y .so answer is B


text:
Gas laws, finding pressure
What is the resultant pressure if 0.6 mol of
ideal gas at 273 K and 1.18 atm in a closed
container of constant volume is heated to
563 K?
Answer in units of atm

We have PV=nRT
(where symbols have there usual meaning.)
You see V is constant as the container doesn't change or deform ( as you mentioned)
so thinking that way PV/nRT=constant
as moles are constant and not leaking
R is universal gas constant .
so ;
P/T is constant
then $P_1/ P_2 = T_1/T_2$

What about the moles though?

they would just be constant provided there's no leakage

Oh I see. So would you just plug it in to be: 1.18/273 = P2/563

So would it be 2.43?

hmm; just math , just calculation :)

Hello and welcome to Chemistry.SE! At this point I suggest you do a couple things. First, take the short [tour] to familiarize yourself with how this site works. Secondly, this is not a "homework" site. You will need to show some work and effort to start off with before others will put their efforts into helping you. How far can you get? What specifically don't you understand? This post give some good guidance for asking homework questions. Best of luck with it!

yes I knew that should be a comment but didn't had any option ; sorry for that

@ airhuff I have edited the answer accordingly


text:
Why don't all electrodes produce the same results in the electrolysis of potassium iodide?
While in lab, I accidentally used two paper clips while trying to electrolyze potassium iodide and all I saw was was phenolphthalein turning pink at the cathode and no brown at the anode.
I asked my instructor to allow me to repeat the experiment with pencil lead, which produced brown and pink, aluminum foil, which only produced pink, silicon which produced brown and pink, and copper, which produced brown and pink.
Then I used one paperclip at the cathode and a piece of pencil lead at the anode and I found that the anode turned pink and the cathode turned brown. Then I flipped the pencil lead and paper clip (anode to cathode and vice versa), and only the cathode (pencil lead) turned pink and there was no brown at the anode (paper clip).
My original hypothesis was that the increased resistance in the silicon and graphite lowered the current since the battery provided a constant $\pu{9V}$, but being able to get brown and pink with copper invalidated that hypothesis.
This has been bugging me all day, and my instructor says she does not know why this is happening.

Obviously, more active metallic anodes oxidize themselves instead of oxidizing iodide, hence no brown.

To distill down your observations: you always saw pink at the cathode, but only some materials produced a brown color when used as the anode. This means that the anodic materials that didn't produce brown didn't oxidize the iodide to iodine.
So, the question is what was happening at the anode when the iodide was not being oxidized. The answer is simply that the anode material itself was easier to oxidize than the iodide in those cases, so the anode itself was oxidized and the iodide was not, so there was no brown color produced.


text:
Various meanings of Ka, Kb etc
A little while back, I introduced the feature to mhchem to properly format $K$ constant variables, like $\ce{Ka}$ rendering as $K_\mathrm{a}$, $\ce{pKb}$ rendering as $\mathrm{p}K_\mathrm{b}$, etc. (besides $\ce{K_a}$ and $\ce{pK_b}$).
I was inspired by seeing quite some usage of Ka, pKb inside \ce, here at Chemisty.SE. I also checked if there were other meanings of these terms and found none.
But now, I am getting doubts how future-proof this is. What if the next new chemical element's name will have the 'Ka' abbreviation? What if some field will invent the 'Ka' 'nickname'?
Do you know any other meaning of Ka, Kb, Kw, Kf, Kc, Kp, Ka1, Ka2, Ka3, Kb1, Kb2, Kb3, Keq, Ksp other than $K$ constants?
How likely is it that alternative meanings for these will emerge?
Or, to phrase it differently: How likely is it (or will it be) that somebody typing \ce{Ka}, \ce{Keq}, ... does intend other output than $K_\mathrm{a}$, $K_\mathrm{eq}$, ...?

Ha? These are just arbitrary symbols of mathematical, chemical constants, and have been used long before you were even born.  Except $pK_a$ and $pK_b$, which are clearly defined symbols.

Ooops, some grammatical mixup in the last sentence. I rephrased it (and added one more). Does it make sense now?

Much better. I don't understand how the system works, though. Is the list you give above complete?

There are a number of two-letter symbols in use. Deborah number $De$, Raleigh number $Ra$, Weissenberg number $We$, and quite a few more in other fields. https://de.wikipedia.org/wiki/Dimensionslose_Kennzahl The Knudsen number $Kn$ might collide with your system, perhaps.

I can't imagine an element name have 3 letters (Ksp Keq), having a number would be really really weird. Ka seems the only one in some danger. Hopefully the powers to be would realize the problem and choose something else.

Do you have any statistics about who uses your "abbreviations"? It seems a bit strange to add the \ce tag to a simple mathematical constant.

@MaxW All the temporary element names (systematic element names) were/are/will be three letters. Unu, Unb, Unt, Unq, Unp, Unh, ...

@Karl. It is a very, very low percentage. You can easily scan the list yourself. Therefore, it is not a big deal if I would remove support again. But also consider all the posts that were entered as Ka and got fixed by others because it did not render properly.

I would rather not have $\ce{Ka}$ render $K_\mathrm{a}$. \ce{} is meant to typeset chemical expressions such as compounds or reactions. $K_\mathrm{a}$ is not a chemical expression, it is a physical/mathematical one. It makes practically zero sense to include these in the same command. Remember that a core principle of $\mathrm{\LaTeX}$ is to use semantic markup — having a chemical markup for a physical constant violates that rule.

I think that most of the uses of $\ce{Ka}$ on site were people thinking ‘Ah, when my last post using CO2 was corrected, they wrote $\ce{CO2}$. So now that I have Ka it should probably be $\ce{Ka}$. And in the next line $\ce{m=5g}$.’ Do you really want to account for every semantic misuse?

But since these two do not answer the question, they shall remain comments.

@Jan I understand your "not every misuse" argument. But could you explain why you think the acid dissociation constant is not a chemical expression?

@mhchem Simply because it is used in physical (or physicochemical) equations. The same is true for pressure, partial pressure, mole fraction, amount of substance, volume and any other physical constant/variable that can be attributed to a single compound.

I endorse what Jan says.

Let's move over to Meta: About the scope of mhchem

@Jan The acid dissociation constant is a mathematical object used in the course of describing a chemical phenomenon. It is not itself a chemical expression.

@hBy2Py My point exactly, except I used mathematical expression rather than object.

That's nitpicking. $K_\mathrm{a}$ is a symbol for a ... "thing" ... mostly used in chemistry. Wikipedia puts $K_\mathrm{a}$ in the chemistry category. Following your arguments, one could could state that the + in a chemical equation is not chemical, but a mathematical character describing a chemical phenomenon. Or one could say that oxidation states should not be part of \ce because they are a "dimensionless number of chemistry" just as $K_\mathrm{a}$ is. I am very open for arguments to remove $K_\mathrm{a}$ support from \ce, but please don't tell me $K_\mathrm{a}$ is not chemical.