text:
What was the lithium concentration in 1940's 7-Up?
I just came across this article in the New York Times, where lithium (or more correctly, lithium salts), a known anti-depressant had been added to beverages in the 1940's. They also mention that a study found negative correlation between the suicide rate of a county and the amount of lithium in its tap water, despite the fact that the concentration is relatively low ($\pu{<0.170 mg L^-1}$), especially compared to the doses administered in therapy.
Now I was wondering what the lithium content in those beverages was (obviously, they are not sold anymore), but couldn't find anything?
The article mentions that you can still buy lithia water ($\pu{0.50 mg L^-1}$), but I'm more interested in the 1940's 7-Up mentioned, as those probably had different food standards at the time they were made.

Related: Do higher rates of lithium in groundwater lead to a decreased suicide rate?

Have you considered contacting 7-up (owned by Dr Pepper in the US and Pepsi everywhere else) and asking them?  There's a chance they might give some sort of answer if you say it's for a school project or something like that. As far as I have seen, the general internet only seems to know that there was lithium citrate in the drink rather than how much was in it.

You did not account for serving sizes. You gave an estimate for 12 oz. A typical serving of Coke in the 40s was 6 oz, and one would think 7up was similar. Serving sizes have increased dramatically since the Super Size era started.

Given that the formula was a trade secret and that lithium was removed from the product ~70 years ago, the likelihood of this question being answered by providing a detailed formula seems remote. Note in addition that the formula might have evolved to reduce cost or improve market reception. You might therefore want to settle for an estimate of the upper limit on the lithium citrate concentration, which is relatively easy to provide.
A basic estimate can be made based on (1) the historical limits of sodium concentration in sodas; (2) the current composition of 7-up; and (3) the relative saltiness (taste) of lithium citrate versus sodium.
First one can attempt to determine what range of sodium concentration makes for a flavorful soda. Here is a quote:

The amount of sodium in soda varies depending on the type. A well-known brand of cola has 46 milligrams of sodium in a 12-ounce can. The same sized can of root beer has 71 milligrams; a can of lemon-lime soda has 64 milligrams; a 12-ounce can of all-natural soda contains 36 milligrams. These are just examples, and the amounts in particular brands of soda may differ. It's best to check the label of your favorite soda to make sure.

This leads to a range of ~150-200 mg/L sodium concentration.
Next, consider the current formula for 7-up (in the USA):

in 12 fl oz (355 mL):
    40-50 mg  sodium,
    60 mg potassium

The current formula contains ~110-140 mg/L sodium and ~170 mg/L potassium.
Finally, consider the saltiness of lithium citrate versus sodium. Lithium citrate was the salt form of lithium in the 7-up recipe  according to a number of sources including the Wikipedia:

The soft drink 7Up was originally named "Bib-Label Lithiated Lemon-Lime Soda" when it was formulated in 1929 because it contained lithium citrate. The beverage was a patent medicine marketed as a cure for hangover. Lithium citrate was removed from 7Up in 1948.[Ref 4]

and  this one (which cites Ref 1):

As to why “7Up,” C.L. Grigg never explained how he came up with the cryptic name. Several theories exist about its origin:
     "7Up was the product of seven ingredients. (Which, in a way, was at least true with regard to the classes of ingredients in that original formulation: sugar, carbonated water, essences of lemon and lime oils, citric acid, sodium citrate, and lithium citrate.)"

Lithium and potassium are among few cations with a salty flavor similar to that of sodium, albeit weaker. Citrate reduces the saltiness further. According to Ref 2, the saltiness of lithium and potassium can be estimated to be ~0.4 and 0.6 that of sodium, repectively, while Ref 3 explains that citrate has an inhibitory effect.
Consider now the potassium concentration (~170 mg/L) in the current formula of 7-up. Potassium is added as the citrate salt, and could be replaced with the lithium salt with little change in taste perception, according to what is reported on the relative saltiness of the cations and citrate salts. This suggests that if the taste of the beverage has remained reasonably constant since inception, it is not unreasonable to estimate an upper bound of ~200 mg/L in the concentration of lithium citrate (replacing the potassium salt) in the orginal recipe.
Note this is based entirely on the taste perception and not on any medical implications of consuming lithium at such concentrations. Lithia water (a brand of spring water sold in the USA) contains only ~0.5 mg/L of lithium. For reference, an online medical advice site suggests that a dose for treatment of depression is ~60 mg of lithium daily. So a dose of ~60 mg lithium citrate per serving certainly seems high relative to lithium-rich spring water but is just about a typical clinical dose.
References

Rodengen, Jeffrey.   The Legend of Dr. Pepper/7Up.   Fort Lauderdale: Write Stuff Syndicate, 1995.   ISBN 0-945903-49-9.
McLaughlin, S., Margolskee, R.F.. The Sense of Taste. American Scientist, 82, 538-545 (1994).
C.M.D. Man. Technological functions of salt in food products. in Reducing Salt in Foods, Woodhead Publishing, 2007. ISBN 978-1-84569-018-2.
Gielen, Marcel; Edward R. T. Tiekink (2005). Metallotherapeutic drugs and metal-based diagnostic agents: The use of metals in medicine. John Wiley and Sons. p. 3. ISBN 0-470-86403-6.

Thanks for the answer, that is a very interesting take on how to estimate the lithium content. Since you would likely drink more than just one serving per day the lithium content appears quite high, but googling a bit it also appears that the doses in therapy seem to be a bit higher (300-900 mg lithium carbonate). Anyway, it seems that 7Up contained non-negligible amounts of lithium.

@snurden Yes, if the concentration in 7-up really was that high you wouldn't want to dring too much :-)  The Li concentration would be ~1/5 of total carbonate concentration, so ~60-180 mg/L. Anyway, on a related topic I was just reading that Mt Fuji water contains vanadium: https://www.ncbi.nlm.nih.gov/pubmed/15663202


text:
What is the major product in the reaction of 2,3-dichlorobutane with sodium amide in liquid ammonia?
I am not able to decide which among 1,3-butadiene and 2-butyne is the major product of this reaction. Will there be even a major product out of the two or will there be a mixture?
According to me, the product formed from the 1st mechanism should be the major product as it is more stable (two double bonds in a conjugate alkene are more stable than a triple bond). But I also think that product formed from mechanism II can be formed faster as the hydrogen being picked up in the first step of mechanism II is more acidic (due to the electron withdrawing effect of electronegative chlorine atom) as compared to the hydrogen being picked up in mechanism I.
Another factor, which seems to favor the formation of product by mechanism I, is the picking of hydrogen atom in the 2nd step which is easier from an sp3 carbon than from an sp2 carbon (in mechanism II).

These are both reasonable mechanisms, and the question outlines well the factors favoring each. In favor of mechanism I:

Low temperature suggests kinetic deprotonation
Statistically more terminal hydrogens than internal hydrogens

In favor of mechanism II:

Small base suggests thermodynamic deprotonation

In these cases where there are conflicting factors, I agree with @Lighthart that the best answer is to actually look at the experimental results. That will give an indication of how to weigh the various factors.
In this case, I do not find an exact hit in SciFinder, but the reaction of  2,3-dibromobutane does come up in two older reviews by Bergstrom and Franke. I do not have access to Franke's review, although SciFinder has a note that "Bromine represents any halogen."
The SciFinder result for the Bergstrom review is deceptive. There is a comment "generalized reaction, halogen and R-groups can vary," and I don't find the exact result (or reasonable analogs) in the review.
More broadly, there are seven results for an internal, vicinal (1,2) dibromide reacting with sodium amide to give an alkyne. One reliable example is the synthesis of stearolic acid from Organic Syntheses.
Similar SciFinder searches for the 1,3-diene product come up empty.
Based on this research, I believe the alkyne would be the predominant product.

A further conclusion is that I wouldn't try to make either of the possible compounds by this route.

I agree. These kinds of questions are interesting from a teaching standpoint to compare various factors that influence the outcome of a reaction that can occur by competing pathways. The experimental version would be an interesting lab exercise (except for the liquid ammonia) to demonstrate the principal. If I needed a conjugated diene or an internal alkyne in high yield, I would something else.

It would be a good experiment to run because teaching how to handle liquid ammonia is a worthwhile exercise in itself

Liquid ammonia is cold, this means you are going to form the 'kinetic' product.
Kinetic products are the ones that are most likely to occur via molecule collision, generally from the least sterically hindered deprotonations.  There is some hinderance for forming the first intermediate on the way to product II.
I would expect product I to form.

Consider however, that in very strong base, product II can isomerize into an acidic alkyne: $$\ce{CH3 -C#C -CH3 <=>[\ce{NH2-}]CH3CH2 -C#C -H} \ \ce{CH3CH2 -C#C -H + NH2- -> CH3CH2 -C#C- + NH3}$$

True, under thermodynamic conditions.(read: not liquid ammonia temperatures).  Less sure about this under kinetic conditions.

On the other hand, NaNH2 is an extremely small base so the steric effect will be minimized.

Agree on this point except that the likely molecular collisions are more likely to happen on the terminal hydrogens, of which there are three times as many.  My answer is a weighted guess, I would defer to an actual experiment to resolve this.

@Lighthart - isn't the product II the kinetic product as it is easy to pick more acidic hydrogen? I say this because I think the activation energy for picking up more acidic hydrogen in the first step of mechanism II will be less than that for mechanism I. Moreover isn't the more stable product known as the thermodynamic product? which makes the product I as the thermodynamic product and product II as the kinetic product?

More acidic hydrogen is a thermodynamic consideration.  Your point about energy of activation is a good one, but with 3 times as many hydrogens in a vastly less sterically crowded environment, the preexponential may win this argument and favor product I. Also, I get the acetylene as more stable owing to 6 sp3 CH bonds versus 6 sp2 CH  bonds outweighing any loss in conjguation or pi-bond energy.

If you look at the heat of hydrogenation of both compounds then you will see that 1,3-butadiene is more stable than 2-butyne.  http://www.chem.ucla.edu/harding/IGOC/E/enthalpy_of_hydrogenation.html

Then I shall defer to the experiment.

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text:
Which is more electronegative: hydrogen or astatine?
Which is more electronegative - hydrogen or astatine? I'm a bit confused because my teacher taught me that for an atom to be electronegative it has to be on the top of the periodic table and/or on the right. But hydrogen is on the top-left and astatine is on the bottom-right of the periodic table.
How does one determine which is more electronegative?

Check out an electronegativity chart. A few of them say that Astatine is more electronegative.

@PrittBalagopal I agree with you that looking up a chart is the best option, but I believe the OP is looking for a theoretical approach if they were asked such a question in exam (where such EN charts are generally not allowed).

Astatine does not quite exist, so why bother. As for hydrogen, it is not on the top-left. It is on the top-everywhere.

As alluded to by @IvanNeretin, the placement of hydrogen in the periodic table is a bit of a special and controversial case. There is indeed a decent argument to be made that it should really be a wide bar extending all the way across the top row, from above lithium to above fluorine.

An interesting way to determine which of $\ce{A}$ or $\ce{B}$ is more electronegative is to take their binary compound $\ce{A_xB_y}$ and observe its dissociation. Note that it can be $$\ce{A_xB_y -> xA^y+ + yB^x-} \tag{1}\label{a}$$ or $$\ce{A_xB_y -> yB^x+ + xA^y-} \tag{2}\label{b}$$
Reaction $\ref{a}$ indicates that $\ce{A}$ is less electronegative than $\ce{B}$, because in the dissociation of $\ce{A_xB_y}$, $\ce{A}$ carries the positive charge, while $\ce{B}$ carries the negative charge. Reaction $\ref{b}$ instead indicates that, for similar reasons, $\ce{A}$ is more electronegative than $\ce{B}$ 1

Going by the same logic, we may tend to observe a hydrohalo acid in this case, specifically, the hydroastatic acid/hydrogen astatide. Quoting from its Wikipedia page,

This chemical compound can dissolve in water to form hydroastatic acid, which exhibits properties very similar to the other four binary acids, and is in fact the strongest among them. However, it is limited in use due to its ready decomposition into elemental hydrogen and astatine (my emphasis) Because the atoms have a nearly equal electronegativity, and as the $\ce{At+}$ ion has been observed, dissociation could easily result in the hydrogen carrying the negative charge. (my emphasis) Thus, a hydrogen astatide sample can undergo the following reaction:
  $$\ce{2HAt -> H+ + At- + H- + At+ -> H2 + At2}$$

which also explains the spontaneous decomposition into elemental hydrogen gas and the astatine precipitate.
If you also have a look at their electronegativities, both have a Pauling electronegativity $2.2$, further complicating the comparison.
Therefore, based on these evidences, I can say, without the least hesitation, that theoretical comparison of the electronegativities of hydrogen and astatine is not possible.

[1]: a shrewd observer would notice that the apt way to write the second reaction's reactant would be $\ce{B_yA_x}$ instead, since $\ce{A}$ is more electronegative

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text:
How to transfer precise volumes of viscous substances from a container to a beaker?
Pipetting doesn't work with viscous materials because the suction/pressure is not strong enough and because relatively high volumes of the liquid remain at the external walls of the pipette tip when withdrawing from the liquid.
So how can I transfer precise volumes (hundreds of microliters +/- 10%) of viscous substances?

Working in mass and not volume is usually more accurate with a good balance. Is it possible to cool/freeze the liquid, break it up and mass out what you need? This worked with some of the oils I used as a grad student,  but might not with biologics where the temp change might damage them.

Alternatively maybe heat the liquid if viscosity falls with temperature increase

Good comments from Stevie D, as a follow on, warm it up and measure by mass. If you can only do a large quantity,  measure into a volumetric flask and dilute with a solvent to give a specific concentration. Then pipette this.


text:
Synthesis of indigo from anthranilic acid - reaction not taking place
I was trying to synthesise indigo via the Heumann indigo synthesis (II.), which involves heating antrhanilic acid and chloroacetic acid, both dissolved in isopropanol. After heating up, a hydroxide or another strong base is added to form indigo
I have slightly altered the synthesis process, added trichloroacetic acid instead of chloroacetic, and also added more reactants to tilt the equilibrium towards the product.
The problem is, that any signs of reaction have taken place after more than 3 hours
I have discussed the issue with my chemistry teacher, and he said that the trichloroacetic acid had been of poor quality (made 1990, majority of the original crystals dissolved, turned to brownish liquid). I also think could be simply stopped by the forming of salt (sodium trichloroacetate) and that being unable to interact with anthranilic acid.
I'm very desperate about this. Do you have any thoughts on how to get that reaction going?

http://www.chemexplore.net/indigo.htm

If you do not want to do the Heumann synthesis for some reason or you are having trouble, the page you link gives several friendlier alternatives. The von Baeyer synthesis from 2-nitrobenzaldehyde and acetone is easy (but the yields are low).

Chemistry is not about playing with chemicals in the lab. Read carefully the link that you provided and it will be clear why you cannot replace chloroacetic with trichloroacetic acid. I'm surprised that your instructor thought this was a good idea.

Trichloroacetic acid was definitely the wrong way to go.
The reason is the chloroacetic acid is alkylating your anthranilic acid, adding the missing carbons Ben Norris comments on.  Ttrichloroacetic acid is not a great alyklating agent, owing to steric bulk.
If you are dead set against working it up and separating your anthranilic acid, you might try just adding the chloroacetic acid.
Otherwise clean it up and start over.

I concur. Anthranilic acid has 7 carbon atoms. Indigo has 16, which is more than can be accounted for by two equivalents of anthranilic acid. The extra carbon atoms come from chloroacetic acid (but cannot come from trichloroacetic acid).

Doesn't the indigo precursor need a reaction temperature in the region of 300 degrees C to convert to indigo?


text:
Can you synthesize phenol from salicyladehyde?
Can you synthesize phenol from salicyladehyde? Well for starters I know that salicyladehyde can be obtained from phenol through Reimer–Tiemann reaction but I was wondering if the reverse was possible.

Oxidation and decarboxylation.  But this is an expensive way to make phenol.

An even more expensive way: stoichiometric rhodium decarbonylation. Apparently, though, that is now possible catalytically.

It is possible to convert salicylaldehyde into phenol in a two-step process. Firstly, salicylaldehyde is oxidized to salicylic acid with a standard reagent like $\ce{KMnO4}$ (see the answers to this question).
In the second step, salicylic acid is thermally decarboxylated to phenol in a simple distillation apparatus. Solid salicylic acid is strongly heated in a round-bottom flask equipped with a short path distillation head with thermometer. Heating is maintained to keep a reasonable distillation rate of the desired product (phenol passes over at a temperature range from 145 to 165 °C). Most by-products of the reaction, like phenyl salicylate, have higher boiling points and remain in the residue. The crude product can finally be purified by vacuum distillation, for example, at a pressure of 30 Torr and a temperature range of 85 - 90 °C (reference).


text:
Finding value of equilibrium constant
I tried to find the value of equilibrium constant several times.
I failed..
the value of equilibrium constant at 500K
has enthalpy 250kJ/mol and entropy 48J/mol*K
the answer is $ 2.45*10^{-24} $
which formula do I have to use?

What formulas have you tried?

@Lighthart I have used $lnK=-delta/RT + deltaS/R$

$\Delta G=\Delta H - T \Delta S = 250-48 \times 10^{-3} \times 500 = 226 KJ/mol$
$K=exp(-\frac{\Delta G}{RT})=exp(-\frac{226 \times 10^3}{8.314 \times 500})=2.45 \times 10^{-24}$

Oh... my problem was that I did not multiply 10^-3 to entropy. Thank you.


text:
What happens to the shielding of the proton when an alcohol is oxidized?
The full question:

For H-NMR, the magnetic field experienced by a hydrogen atom can be
  externally applied or internally induced or some combination of the
  two. When the applied and induced fields act in concert, a proton in
  these fields is said to be deshielded. Which of the following
  processes will result in the greater deshielding of the proton? Please
  choose from one of the following options:

Removing nearby carbonyl groups
Adding electropositive atoms like silicon
Oxidation of primary alcohols with $\ce{KMnO4}$
Reducing terminal alkyne groups

This is what the question is asking me, but I'm confused because if you oxidize a primary alcohol, this gives an aldehyde or a carboxylic acid (depending on reaction conditions), and the proton in question goes into forming water. So if an H disassociates, does this equate to a total deshielding, or how is this explained with respect to proton NMR?
I realize that water does show up on NMR spectra, but I'm unclear what I am being asked. It also obviously depends if this is primary, secondary, or tertiary alcohol.
The answer given is (3) Oxidation of primary alcohols with $\ce{KMnO4}$. I don't understand why.

Is the question really asking about the shielding of a H attached to the carbon that is alpha to the alcohol?  Then you'd be comparing that H to an aldehydic H.

thanks, I think this is the problem , that i dont think its asking about that specific hydrogen. I think it makes sense now. (C.A.'s and ald have alot of deshielding)

Can you give some background on the question? It seems like a follow-up question, a question about a particular proton on a particular molecule, or at least a question with some information leading up to this question. Maybe supply a screenshot or photo of the entire question?

The alcohol proton becomes an acid proton. Potassium permanganate oxidizes primary alcohols to carboxylic acids. In the process, the hydroxyl proton goes from 2-5 ppm to >10 ppm (generally). Therefore, since it is shifted to a higher ppm, it is being more deshielded.


text:
Ammonium Hydroxide Name
Why is an aqueous solution of $\ce{NH3}$ (ammonia) often referred to as $\ce{NH4OH}$?
I know that dissolved ammonia gas reacts with water in the following equilibrium:
$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
However, $K_\mathrm b = 1.8\times10^{-5}$. Since only a small amount of $\ce{NH3}$ dissociates, why is the whole solution called "ammonium hydroxide"?
Is ammonium hydroxide soluble, or is it an actual aqueous species?

It's not always called ammonium hydroxide. I call it "ammonia" or "aqueous ammonia." It's just a preference thing. It's an easy way to convey "aqueous ammonia."

The fictitious substance “ammonium hydroxide” $(\ce{NH4OH})$ was suggested in accordance with the Arrhenius theory. In 1882–1883, Svante Arrhenius had researched into the conductivity of electrolytes. His theory of electrolytic dissociation was discussed in his dissertation in 1884 and published in refined form in 1887. In 1903, the Nobel Prize in Chemistry was awarded to Svante Arrhenius “in recognition of the extraordinary services he has rendered to the advancement of chemistry by his electrolytic theory of dissociation”.
According to the Arrhenius definition, acids are compounds that dissociate and release hydrogen ions $(\ce{H+})$ into the solution:
$$\ce{HCl -> H+ + Cl-}$$
Bases are defined as compounds that dissociate and release hydroxide ions $(\ce{OH-})$ into the solution:
$$\ce{NaOH -> Na+ + OH-}$$
The need for a hydroxide group in bases according to the Arrhenius definition led to the proposal of $\ce{NH4OH}$ (i.e. “ammonium hydroxide”) as the formula for hydrated ammonia in water. (Note that this problem does not exist in the definition of bases according to the Brønsted–Lowry theory, which was proposed in 1923.)
However, ammonium hydroxide cannot be isolated; the fictitious solid compound does not exist. Nevertheless, the misleading traditional name “ammonium hydroxide” is still widely used for solutions of ammonia in water.
Actually, a solid compound (melting point: −79 °C) with the correct atomic composition (molecular formula) can be crystallized from aqueous solutions of ammonia. But even this compound is not ammonium hydroxide $(\ce{NH4OH})$. It actually is a real hydrate $(\ce{NH3.H2O})$ in which the molecules are linked by hydrogen bonding.

A review (https://chemistry.stackexchange.com/review/close/116067, about https://chemistry.stackexchange.com/questions/131455/are-these-two-reactions-the-same) brought me here.  With the school closed, I found (at least) this OA entry in the COD: http://www.crystallography.net/cod/2310926.html which is, given the time of publication (10.1107/S0365110X59002419, http://scripts.iucr.org/cgi-bin/citedin?a02661, OA(?)), lacking the hydrogens, but -- contrary to the more recent doi.org/10.1021/ja9052569 --, a model at ambient pressure.

Since only a small amount of ammonia dissociates, why is the whole solution called "ammonium hydroxide"?

Well the operative word is dissociates.  Just because it dissociates doesn't mean ammonium hydroxide isn't in solution.  There are many salts that dissolve in water without a great deal of dissociation.  Further ammonia is incredibly hydrophillic and a more appropriate equilibrium of the solution is represented as:
$$\ce{NH4OH_{(aq)} <=> NH4+_{(aq)} + OH-_{(aq)}}$$

That's a good point. There is definitely going to (essentially) always be a stabilizing hydrogen bond, i.e. it will exist short of as a complex between the two.

It definitely isn't more appropriate.

So is the $\ce{NH_4OH}$ an actual molecule?


text:
why is yield of a reaction negative?
Suggest an explanation and method of improvement when a reaction has negative yield i.e. the mass of container with product is less than the mass of empty container. This happened to me twice in an orgo lab and I have no idea whats going on.

There is some error in your measurement because this isn't possible.

Backfilled an empty flask with hydrogen?

Error in calculation as this is not possible. Negative yield would be like saying you went out and destroyed someone else's product

How large was the difference as compared to the weight of the container and/or the expected weight of the product? ? Did you use the same container? Back in the days, round-bottom flasks sometimes had little glass hooks attached to the neck (for springs). Did one of the hooks break off in between two measurements? Did you use a cap/lid on the empty container?

In all seriousness, I've had to deal with negative yields before because I was weighing glassware off the shelf before concentrating down a very small amount of product on high-vac. I still had product by NMR, but it was very dilute despite being all I had. What I do now when I have very, very small amounts of product is pump on my flask before getting a tare weight to remove any moisture on the inside of the flask. That small amount of moisture evaporating was enough to throw it off, apparently, because it doesn't happen anymore. This is not likely to be what happened to you, though.

If the mass of the container with product is less than without then there are some possible mistakes:

you wrote down/used the wrong mass/container
you weighted the empty container with plug/screw cap and afterwards without
you changed the plug/screw cap
you had the container filled with a gas with higher density than air (common with argon), which can be a difference of some hundred mg depending on the size.
the temperature wasn't the same, don't weigh hot/warm flasks

From my experience as TA in synthetic labs: it's most likely the second one.

From my experience during my PhD student time, it’s likely the last.

At the risk of stating the obvious, one possible explanation is the release of a volatile product. Thus the use of thionyl chloride to produce an acyl chloride releases Sulphur dioxide which release takes mass with it.

That is not a negative yield... that's called "I don't know how to calculate a theoretical yield".


text:
Litmus paper - turning red, blue and even bleached
I have blue litmus paper, and if I put it in an acid it turns red. I also have red litmus paper, and if I put it in a base, it turns blue.
I know the question about litmus was asked and answered and I like the answer:
How does the litmus pH indicator work?
However, damp litmus paper also becomes bleached in presence of chlorine gas. I believe it has nothing to do with the mechanism described in aforementioned question. I wonder what the mechanism is. Wikipedia actually suggests:

For instance, chlorine gas turns blue litmus paper white – the litmus dye is bleached, because of presence of hypochlorite ions. This reaction is irreversible, so the litmus is not acting as an indicator in this situation.

I also found an equation which describes formation of hypochloric acid (source):
$$\ce{Cl2 (g) + H2O (l) -> HClO (aq) + HCl (aq)}$$
The problem that I have now is as follows: We know that litmus paper gets discolored because of contact with hypochloric acid. But what is actually the chemical reaction for that?

I'm not sure about this (plus it doesn't give sufficient details) so I am not posting this as an answer: Litmus is a mixture of dyes extracted from lichens, and most of these dyes/indicators have colour due to a extended pi conjugated system. The acid-base environment alters the pi conjugated system and causes a shift in absorbance, causing colour change. The HOCl is the bleach in the reaction, and it probably functions by destroying the extended pi conjugated system.

Azolitmin is known to be one of the major components of litmus: azolitmin I have no access to SciFinder at the moment, but I do recall hydantoin imides to be exquisitely sensitive to oxidizing agents, and I don't see azolitmin being an exception.

@tovok Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. I have converted your answer into a comment for now. With a bit more rep, you will be able to post comments on any question/answer.

I just came across a mention of the easiness with which the double bond is disrupted by halogens, in the 10th grade textbook by Rudzitis (Russian).  Nice.

I can supply some details now, and hopefully this ought to qualify as an answer.
As I mentioned earlier litmus is a mixture of 10-12 dyes (CAS number: 1393-92-6).
The acid-base indicator properties of litmus are primarily due to 7-hydroxyphenoxazone chromophore (pictured below)

The answer you linked to discusses the acid-base indication mechanism in some detail, so I shall skip over that.
Anyway, what this serves to establishing that it is indeed a extended $\pi$-conjugated system that we are dealing with in the chromophore.
Now, $\ce{HOCl}$ would bring about halohydrination (basically an electrophilic addition) across the $\pi$ bonds, thus disrupting the conjugated system.
Halohydrins are compounds that contain an $\ce{-OH}$ and $\ce{-X}$ groups on adjacent carbons.
This image describes the general mechanistic scheme in a simpler case:

Anyway, after said addition reaction has taken place in the chromophore, the $\pi$-conjugated system is disrupted, and thus, we see an absence of colour.

Thank you, getafix! Did I understand you correct, that there is no reaction with HOCl, but instead there is a reaction with Cl2, followed by reaction with H2O?

Well, the chlorine+water is furnished by the HOCl (so to speak). The reaction of chlorine you put down with water in your question is reversible.

@SleepyHollow - the reaction is probably with the hypochloride ion (ClO minus), since HClO dissociates in water. Here's a nice description of electrophilic addition, showing how the breakdown of the pi-bond occurs and what follows that.


text:
Is LiOH a weaker base than NaOH?
Is $\ce{LiOH}$ a weaker base than $\ce{NaOH}$?
Note: I'm not interested in "why," but rather what the "real" $\mathrm{p}K_
\mathrm{b}$ values are.
$$
\begin{array}{lcc}
\text{Data for Alkali Metal Hydroxides} \\
\hline
\text{Cation} & \mathrm{p}K_\mathrm{b} & \text{Solubility} & \text{Solubility} \\
\text{} & \text{} & \pu{g/100 ml} \text{ at } \pu{20 ^\circ C} & \pu{moles/L} \text{ at } \pu{20 ^\circ C} \\
\hline
\ce{Li} & -0.36 & 12.8 & 5.34\\
\ce{Na} & 0.2 & 111 & 27.75\\
\ce{K} & 0.5 & 112 & 20.0  \\
\ce{Rb} & ? & 180 & 17.6\\
\ce{Cs} & -1.76 & 200? & 13.3?\\
\hline
\end{array}
$$
The $\mathrm{p}K_\mathrm{b}$ data is from Wikipedia ($\ce{Li}$, $\ce{Na}$, $\ce{K}$, and $\ce{Cs}$). Wikipedia doesn't have a value for $\ce{Rb}$ and I couldn't find one.
The solubility data is from the Wikipedia Solubility Table. Wikipedia doesn't have a solubility value for $\ce{CsOH}$ at $\pu{20 ^\circ C}$  and I couldn't find one, About $\pu{200 g}$ would be a reasonable guess, which would be $\pu{13.3M}$.
One point is that the Wikipedia article for lithium hydroxide says "It ... is the weakest base among the alkali metal hydroxides."
Another point is an earlier question "Why is $\ce{LiOH}$ a weaker base than $\ce{NaOH}$". User Lighthart references a paper "Ions in water: The microscopic structure of concentrated hydroxide solutions" by Imberti et al. but the abstract at least doesn't seem to address the strength of the alkali hydroxides directly (I don't have access to the paper itself since it is behind a paywall).
Is this one of the Bertrand Chord paradox problems where it depends on how we define "stronger"?

With a $\mathrm{p}K_\mathrm{b}$ of -0.36, $\ce{LiOH}$ would seem to be a stronger base than either $\ce{NaOH}$ or $\ce{KOH}$.
Certainly more moles of $\ce{NaOH}$ dissolve in a liter of solution at $\pu{20 ^\circ C}$ than any other alkali hydroxide.

EDIT 3/22/2016 11:31 pm
Found a different table of $\mathrm{pK}_\mathrm{b}$ data for the alkali hydroxides: Source-2
EDIT 3/23/2016 12:12 am
Source-3 CRC Handbook of Chemistry and Physics, 92nd edition, "Dissociation Constants of Inorganic Acids and Bases. Only $\ce{Li}$ $(\mathrm{p}K_\mathrm{a}= 13.8)$ and $\ce{Na}$ $(\mathrm{p}K_\mathrm{a}=14.8)$ are listed as $\mathrm{p}K_\mathrm{a}$ of the cations. It is the same data as in this pdf file which is from another edition.
$$
\begin{array}{lcc}
\hline
\text{Cation} & \text{Wikipedia} & \text{Source-2} & \text{CRC}\\
\hline
\ce{Li} & -0.36 & 0.18 & 0.2 \\
\ce{Na} & 0.2 & -0.56 & -0.8 \\
\ce{K} & 0.5 & -1.1 & \\
\ce{Rb} & ? & -1.4 & \\
\ce{Cs} & -1.76 & -1.76 & \\
\hline
\end{array}
$$

"Hence, amongst the alkali metal ions, lithium salts are the most hydrolyzable salts and the base strength increases from LiOH (pKb = -0.7) to NaOH to KOH (-1.7) G.S. Manku "Theoretical Principles of Inorganic Chemistry" - Page 260 // https://books.google.com/books?id=be0T64ZWp9EC&pg=PA260&dq=LiOH+pKb&hl=en&sa=X&ved=0ahUKEwiVhZibgtbLAhUMRyYKHXt_DBgQ6AEIHDAA#v=onepage&q=LiOH%20pKb&f=false

Just a guess- Lithium cation is smaller than the sodium cation, so the size of LiOH must be smaller than NaOH. Hence, the electrons will be pulled strongly, and it will be harder for them to leave. Hence, it would be a weaker base. Also, Lithium compounds are largely covalent, which could again be a possible reason.

What.... is going on here?

@Brian - I'm trying to figure if pKb of LiOH is really less than NaOH or not. Not "why" just "if". There seem to be conflicting values. I don't know what the "real" values are.

@MaxW So it's a research project, work in progress?  "Research scratchpad" is not one of the typical use-cases for Chem.SE that I've seen thus far.... More I think about it, the more comfortable I'm getting with it, but it's...different.  I would say: once you're satisfied, you should probably edit the question down, remove all but one answer, and edit that answer into a short review article.  Or, perhaps convert the whole thing to community wiki?

@Brian - I'm long retired from chemistry. I don't have access to journals. I was just trying to document what I could. I was hoping that someone who had access to journals could supply some reference to actual research papers and settle the question.

@MaxW Ahhh, gotcha. Might want to edit that into the question, that you're seeking literature-search assistance from those with better library access.

In Proton Affinities of the Alkali Hydroxides J. Am. Chem. Soc., 1969, 91, pp 2810–2811, it is shown that gas phase affinities in $\pu{kcal/mol}$ are:

$\ce{LiOH}$: 240.7
$\ce{NaOH}$: 247.6
$\ce{KOH}$: 262.6
$\ce{CsOH}$: 269.2

So, yes, $\ce{LiOH}$ is weaker, at least in the gas phase.
In aqueous solution, ACIDITY FUNCTIONS FOR STRONGLY BASIC SOLUTIONS Chemical Reviews volume 66, pages 119-131:

the order of basicity for the same molarity of aqueous solution, $\ce{LiOH < NaOH < KOH}$.

For example, for a $\pu{1M}$ solution, the acidity function (H) is:

$\ce{LiOH}$: 13.48
$\ce{NaOH}$: 14.01, 14.16 (two different measurement techniques)
$\ce{KOH}$: 14.17

So in both the gas and aqueous phases, $\ce{LiOH}$ is a weaker base than $\ce{NaOH}$.

Can there be something that is stronger/weak acid/base as a gas, than as a liquid? Hmm..

The reference
  Acids and Bases (Essential Chemistry)
  Kristi Lew
  Publisher: Chelsea House Publications; Library Binding edition (December 1, 2008)
  Language: English
  ISBN-10: 0791097838
  ISBN-13: 978-0791097830

from page 42
I clipped the image from Goggle books. So far as I know the pKb values for LiOH, NaOH and KOH do not have a primary source listed.

Wikipedia references
LiOH, pKb -0.36
The reference listed is Kristi Lew, Acids and Bases (Essential Chemistry)
Note: I changed Wikipedia to value of -0.36 from what had been -0.63 which seems to be a transpositional error.
NaOH, pKb 0.2
Reference is website Polyprotic Acids & Bases. UC Davis ChemWiki
Website does not have source of values listed. Of the alkali metal hydroxides only NaOH and KOH are listed. NaOH and KOH values agree with Kristi Lew, Acids and Bases (Essential Chemistry)
KOH
Wikipedia lists a pKa value of 13.5 for which pKb = 14.0 - pKa = 0.5
There is no source listed.
RbOH
No pKb value given.
CsOH, pKb -1.76
Reference given is http://www.periodensystem-online.de/index.php?sel=wert&prop=pKb-Werte&show=list&el=4&id=acid which is Source 2.
No idea where website got its data.

  Quantitative Chemical Analysis, Eighth Edition Edition
  Daniel C. Harris
  Publisher: W. H. Freeman; Eighth Edition edition (April 30, 2010)
  Language: English
  ISBN-10: 1429218150
  ISBN-13: 978-1429218153

To my knowledge no reference is given for the complex constants. I clipped the problem from Google books.
Given the complex constants cited, for 1.00 molar solutions:
       [M+] = [OH-]   [MOH]
   Li      0.65        0.35
   Na      0.85        0.14

So it would seem that indeed that NaOH is a stronger base than LiOH based on this data.

"The Aqueous Chemistry of the Elements" by George K. Schweitzer, Lester L. Pesterfield, Oxford University Press; 1 edition (January 14, 2010), ISBN-13: 978-0195393354 // page 100 "LiOH*H2O and LiOH are very soluble, completely dissociated in HOH, and act as a strong base."  // page 104 "NaOH is very soluble, completely dissociated in HOH, and acts as a strong base."

Why did you post three answers?

I'm curious as to what the "real" value is since various sources have different answers. I was trying to document some of the various sources.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Stoichiometry and iodometric titrations

$1.25~\mathrm g$ of a sample of bleaching powder is dissolved in $100\ \mathrm{mL}$ of water. $25\ \mathrm{mL}$ of it are treated with $\ce{KI}$ solution. The iodine so liberated required $12.5~ \mathrm{mL}$ of $\frac{\mathrm{M}}{25}$ hypo solution ($\ce{Na2S2O3}$) for complete titration. Find percentage of available chlorine from the sample of bleaching powder.

My attempt:
In $25~\mathrm{mL}$ sample:
mEq of $\ce{Cl2}$ in bleaching powder $=$ mEq of $\ce{I2}$ liberated $=$ mEq of hypo solution
$\mathrm{mEq}$ of Hypo $= NV = \left(1\times\frac{1}{25}\right)\times12.5 = 0.5$
This is where I have a problem, if we are to go by the solution provided in my book, then the n-factor for the hypo solution is 1, whereas, when I write the balanced redox reaction, that is,
$\ce{2S2O3^2- -> S4O6^2- +2e^-}$
You can see that n-factor should be $2$.

No. The $n-$factor is still $1$ because one mole of $\ce{S2O3^2-}$ furnishes 1 mole of electrons.
In your balanced reaction, 2 moles of $\ce{S2O3^2-}$ gives 2 moles of electrons. To find n-factor, you need to calculate the amount of electrones given by one mole of $\ce{S2O3^2-}$.


text:
Theoretical Value of Activation Energy of Potassium Iodate and Sodium Bisulfite?
I investigated the Arrhenius equation using the reaction between potassium iodate (containing the IO3- iodate ion) and sodium bisulfite (containing the HSO3- hydrogen sulfite ion. The ionic equations for the reaction’s two stages are as follows:
IO3- (aq) + 3HSO3- (aq) -----> I- (aq) + 3SO42- (aq) + 3H+ (aq)
5 I- (aq) + 6H+ (aq) + IO3- (aq) -----> 3I2 (aq) + 3H2O (l)
I rearranged the Arrhenius equation to give:
ln(k)=(-Ea)/R)(1/T) + ln(A)
I changed the temperature of the reaction and timed the speed of the reaction, which I used to calculate the values of k. I then graphed the relation between ln(k) and 1/T , with the slope giving –Ea/R.

My graph ended up having a slope of -0.0003, so I obtained an activation energy of 0.00024942 J/mol.
I was wondering if anyone had a theoretical value of the activation energy of this reaction I can compare my experimental value to?

You need to plot the other way round - lnK on y axis...

Thank you, I got the much more reasonable 31200 J/mol this time.


text:
Volume change inside a balloon upon decreasing the the outer pressure

A balloon is filled with hydrogen at room temperature. It will burst
  if pressure exceeds 0.2 bar. If at 1 bar pressure the gas occupies
  2.27 L volume, upto what volume can the balloon be expanded?

Now i can easily figure out that the pressure is 0.2bar at a volume of 11.35L. However, when i check the answer in my book, it says that the volume of the balloon should be less than 11.35L.
However, if you decrease the volume from 11.35L, wouldn't the pressure become greater than 0.2bar, hence bursting the balloon?

Is this the REAL problem - exactly as stated? The problem doesn't make sense to me.

@MaxW I copy pasted the problem from a pdf of the excercise book (Although the book itself doesn't make sense. Before teaching any chemistry it dives into Quantum Mechanics (The physics companion is still on units/measurements and classical mechanics..) and starts randomly pointing out facts from QM without deriving or explaining them. That's India for you)

The relationship between pressure and volume is given by Boyle's Law.
$$\mathrm{P}_1\mathrm{V}_1 =  \mathrm{P}_2\mathrm{V}_2$$
      P       V
     0.199    11.41
     0.20     11.35
     0.201    11.29

     1.0      2.27

So yes, you're right. Using Boyle's Law, the volume should be more than 11.35L so the pressure stays below 0.20 bar.

However this doesn't make real world sense. When you blow up a balloon, as the pressure increases the volume increases.
Edit - Thanks @Mithoron here is a version of the problem that works...

A balloon is filled with hydrogen at room temperature (25 C) and pressure (1.00 bar) to a volume of 2.27 L in a chamber. The pressure in the chamber is then reduced isothermally. The balloon will burst when the pressure in the chamber is reduced to 0.200 bar. What will the volume of the balloon be just as it explodes?

It does make sense: if you gradually lower pressure around ballon it will burst.

@Mithoron - Ok, let's assume that the balloon starts at 1 bar. If it pops at 0.2 bar then you're already 0.8 bar over bursting strength. // I tried to think of relative pressure vs absolute pressure, but that just doesn't seem to work either.

No, no, it's the upper bound! The lower the pressure, the bigger balloon. When it's too big, it breaks.

@Mithoron - ah... So the pressure in the room is 1 bar when the balloon has a volume 2.27 L, then you evacuate the air pressure in the room. That does work better. Not sure how you work the 0.2 bar pressure into that. It seems the statement should be "if the pressure is lower than 0.2 bar."

And that's where authors made mistake...

When you will decrease the volume, surely the pressure will increase. But this pressure will not act on the balloon's body. The picture shows two types of pressure, one is inside the balloon and another is outside the balloon. Now imagine, if the volume increases, the size will also increase and as the size increases the area will also increase. We also know that pressure is directly proportional to force. Therefore the increase in pressure will result the balloon to explode. I hope this will help  .

If you consider the pressure as force per unit area, the force will also decrease when increasing the volume..

Sorry for the mistake in my last answer. Now, I want to say that this anomalous answer is coming because the  Boyle's law is only applicable on ideal gases. But in reality gases don't show ideal gas behavior and thus this this unexpected thing occurs.


text:
Is it possible to have an airborn fluorescent particle that reacts to lightning?
I want to measure the expansion of the atmosphere when it is heated and impacted by lightning, to have an idea of the propagation of sound waves made by lightning.
To do this, the ideal would be to have a box of static air which changes color or causes a durable fluorescence indicating the reaction of the electricity with the air and showing the patterns of expansion of air that has been heated by electricity.
Please can y ou suggest some chemicals which can be used to make the effects of lightning on air visible?

Air itself emits photons after it has been excited by lightning, e.g. $\ce{N2^{$*$} -> N2 + $h\nu$}$. However, the life times of the excited states are probably too short for your purpose.

Sound and air expansion are two different things

I think you mean thunder but I could be wrong.

Maybe something that fluoresces in the presence of ozone? A quick search gave this article. Lightning is known to convert oxygen into ozone, so I think this should work or at least give a good starting point for future development.


text:
Why do steel and iron not passivate the way aluminum does?
When aluminum is exposed to oxygen a micro layer or film of oxide appears on the surface and oxygen cannot penetrate any further.
With iron or steel, however, oxygen continues to reach into the metal, rusting it deeper over time. When iron is heated the oxide ("scale") forms much faster and the oxygen can deeply penetrate into the metal.
What is the reason that aluminum and iron are different in this way?

Are you talking about air, oxygen, or air/oxygen in presence of water?

It is all about conditions. Iron passivates e.g. in the concentrated sulphuric acid.

Related: Pilling-Bedworth Ratio

The main reason is nature of the oxide layer formed on the surface. Iron oxides are not remotely like aluminium oxide. They tend to be flaky and porous meaning that they offer no protection to the surface when formed. And iron oxides tend to be lower density and higher volume than the iron they are made from. Aluminium oxide is not at all porous and is very strong.
Bulk aluminium oxide is corundum, which is one of the strongest minerals (it is just about the hardest other than diamond). Corundum occurs as the gemstones sapphire and ruby when it has the right impurities and they are known for their hardness as well as their beauty. It is also unusually dense for a mineral made up of relatively light elements. So even a thin layer provides strong protection.
It is also worth noting that the aluminium oxide layer can be disrupted (e.g. by gallium or mercury), in which case aluminium does the equivalent of "rust", in some cases quite spectacularly.
It is also worth noting that some steels can be made to have a passivation layer which is why the are "stainless". This is achieved by alloying with metals such as chromium that facilitate the creation of a stronger oxide layer on the surface.

Aluminium oxide does not dissolve in gallium or mercury. Metallic aluminium does.

@Karl True. But if the film has any flaws at all, the ingress of gallium or mercury will prevent it reforming as it normally does in air. and that is enough to destroy its usefulness.


text:
Molar concentration of water and pKa value
By defintion, $\mathrm pK_\mathrm a$ value is defined as
$$\mathrm pK_\mathrm a = -\lg ([\ce{H2O}] K_\mathrm{eq})$$
Consider the reaction:
$$\ce{HCl + H2O <=> H3O+ + Cl-}$$
The Eq constant of $\ce{HCl}$ is defined as
$$K_\mathrm{eq} = \frac{[\ce{H3O+}][\ce{Cl-}]}{[\ce{H2O}][\ce{HCl}]}$$
Why is molar concentration of water fixed at 55.5? Wouldn't it depend on the extent of ionization?

What does HCl have to do with the pKa of water?

@jerepierre edited to be more specific.

For most practical purposes, we assume that the aqueous solution is dilute and that the concentration of water is constant, that is, $[\ce{H2O}] \approx 55~\mathrm{M}$. This is the standard that is used to define the ionization constant of water, $K_\mathrm{w}$.
If the equation $1\times 10^{-14}~\mathrm{M^2} = [\ce{H3O+}][\ce{OH-}]$ didn't hold (at $25~\mathrm{^\circ C}$) then we would have to come up with other equations for acid-base equilibrium. Even for strong acid/base dissociations we approximate the equilibrium equation with full dissociation (i.e. one forward arrow).
I recommend reading pp 12-15 of the following if you want a more theoretical treatment of strong acid/base dissociation where activities are used: http://www.chem1.com/acad/pdf/c1xacid2.pdf

Please use \ce{} for chemical formulae. (See edit)

The truth is convoluted. In reality the equilibrium "constant" isn't really a constant. The value is really a function that depends on temperature, pressure, and ionic strength of the solution. The gist is that in concentrated solutions, where the concentration of "free" water molecules isn't 55 molar, the equilibrium constant needs such a significant correction due to ionic strength that making the correction to water alone doesn't improve the accuracy of the formula.


text:
Calculating molar mass from gas data

34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure. What is the molar mass of phosphorus?

I tried to solve the above question and got 1249.8g. The book's answer sheet lists the answer to be 1247.7 g.
I don't understand how the heck that happened? I'm pretty sure a mole of phosphorous doesn't weigh more than 1.2 kg.

https://en.wikipedia.org/wiki/Garbage_in,_garbage_out - that's how. (I mean, there is nothing terribly wrong with your calculations.)

Well done recognizing that the answer you found was unreasonable!

To start your book didn't use significant figures properly. There is no way that the weight can have 3 significant figures in the mass and then conjure 6 significant figures in the molecular weight. To make matters worse the pressure is only given to 1 significant figure. YUCK!
Use PV = nRT to determine the moles of gas present.
$\mathrm{n =}\dfrac{\mathrm{PV}}{\mathrm{RT}} = \dfrac{0.1 \text{ bar} \times 0.03405 \text{ L}}{\text{0.08314 L bar K}^{−1} \text{mol}^{−1} \times 819\text{ K}} = 5.000 \times 10^{-5}\text{ mol}$
$\text{MW} = \dfrac{\text{m}}{\text{mol}} = \dfrac{ 0.0625\text{ g}}{5.000 \times 10^{-5}\text{ mol}} = 1,249.8\text{ g/mol}$
This is of course wrong. It would seem that the pressure should have been 1.00 bar. That would have given an answer of 125 g/mol. Using 31.0 g/mol as the atomic weight of phosphorous would reveal that the phosphorous vapor was $\ce{P_4}$.


text:
Canonical structures of (1E)-prop-1-en-1-yl acetate
I have to find all of the possible canonical structures of this molecule

And here's what I did

Are they correct and if yes, did I miss any possibility?

These are correct, and I see one more fully octet structure that is closely related to a non-octet structure.

Would it be this one that I've added on the top row?

I added it in the question.

No, in that drawing oxygen has 10 electrons. I would only draw two, to be honest. You're already delocalising a lone-pair on oxygen in one direction. You could also delocalise it in the other direction.

That structure isn't possible... Sort of on the right track though. Think more about what you have on the bottom line and apply it to the top line.

what about now?

Is it correct like this?

If oxygen is neutral in the last structure you drew, it would still have 10 electrons. We only observe elements capable of an extended octet in the 3rd period. However, if it has 8 electrons, its charge would be 2+. Oxygen is very electronegative, so that doesn't sound so likely, right?
I will post a more detailed answer, or at least resources, a bit later. Meanwhile, please edit your post, so that it doesn't look confusing to someone new coming in. As you did in your original post :-)!

It is correct, the oxygen in the last structure has 8 electrons and one positive charge on it. No idea how 3 bonds and 1 lone pair will add to 10 in other comments.

No I took it away, there was an oxygen making four bonds and having one lone pair.

If my oxygen makes 4 bonds and has one lone pair, is it neutral?

I just did it again and I changed the results in the question so please take a look and tell me if something is wrong.

May it look like this? In those two possibles ways?

Your top row is not possible. Look at the last three rows: there are octet violations

I don't see how.

And I have done the same things in the bottom row

Oh, you are correct! You have an octet rule violation in the bottom row also.

But how? Is it because of the triple bond of the oxygen?

On your image, you have two structures with oxygen atoms that have one double bond, one single bond and two lone pairs = 10 electrons. You also have two structures with oxygen atoms that have two double bonds and one lone pair = 10 electrons.

So the oxygen with one double bond and a single bond should have one lone pair and the oxygen with two double bonds should have none?

And the formal charges work. For the oxygen with one double bond, one single bond and two pairs it really have a charge of -1 since 6-3-4=-1

You have also missed one along the allyl ether moiety


text:
derivation of enthalpy internal energy relationship
A question asks us to derive a relationship between the enthalpy change and the change in internal energy for a reaction.
Now, I know the relationship is:
$\Delta H = \Delta U + V \Delta p + p \Delta V$
But i thought enthalpy was conveniently defined this way ( not derived) so that enthalpy change equals heat change at constant pressure. So which

First a brief note on the mathematics of Legendre Transforms
Consider a function in two independent variables: $ f(x,y)$
We right the differential of this function as follows:
$df(x,y) = \Big( \frac{\partial f}{\partial x}\Big)_ydx + \Big( \frac{\partial f}{\partial y}\Big)_xdy$
Now, I shall define $u := \Big( \frac{\partial f}{\partial x}\Big) $ and $w := \Big( \frac{\partial f}{\partial y}\Big)$
so, $df = udx + wdy$ [equation 1]
Here, u and x (and similarly w and y) are called conjugate variables. (i.e $ux$ and $wy$ will have the same units as $f$)
[Think of it this way: In thermodynamics, PV, TS have the same units as H, U,G,A]
Now, our goal is transform our original function $f(x,y)$ to a function $g(x,w)$. We wish to replace y, with it's conjugate variable and the function $f $ and $g$ will have the same units.
so, we take the differential of the product $wy$ : $d(wy) = ydw + wdy$
We subtract this from [equation 1] and get $d(f-wy) = udx - ydw $
and can deduce that our Legendre Transformed function $ g(x,w) := f-wy$
Now, we can easily apply this method to Internal Energy (U) to obtain Enthalpy (H)
We have $U(S,V)$ and $dU(S,V) = \Big( \frac{\partial U}{\partial S}\Big)_VdS + \Big( \frac{\partial U}{\partial V}\Big)_SdV $
Now, here we seek $H(S,P) \equiv g$
a)  $f \equiv U$
b) $S \equiv x$ is the variable we hold fixed
c) $V \equiv y$ is the variable we switch
d)  $\Big( \frac{\partial U}{\partial S}\Big)_V := T \equiv u $(the conjugate of the variable we are holding in place
e) $\Big( \frac{\partial U}{\partial V}\Big)_S := -P \equiv w$ (the conjugate of the variable we change
$ g =f-wy $
so, $H(S,P) = U-(V)(-P)$
Or, $H(S,P) = U + PV$
From here, you can proceed to write down: $dH = dU + PdV + Vdp$
and for macroscopic changes: $\Delta H = \Delta U + P\Delta V + V\Delta P $

You can define enthalpy through its difference, so in this sense you are right. But maybe the question that have been asked to you refers to an specific kind of system or specifics restrictions. I am just guessing.
For example, if it is about reaction between ideal gases at constant pressure and temperature, you'll have the general expression
$$ \Delta H = \Delta U + \Delta (pV)= \Delta U + p \Delta V $$
which for this particular case, as $pV= nRT$, so $\Delta(pV)= \Delta(nRT)$, and under this ligatures $p \Delta V = RT \Delta n$, so
$$ \Delta H = \Delta U + RT \Delta n$$
Other way the question makes not sense.


text:
Formulas of synthetic polymers
I have a question regarding synthetics. $\ce{(C8H8)}_n$ is the formula for polystyrene. What exactly does the $n$ stand for?
Thanks

n is the number of molecular units in a chain, also called the degree of polymerization. This varies based on the molecular weight of the polymer. The molecular weight is the degree of polymerization times the molecular weight of a monomer.


text:
Is carbonyl-enol tautomerization only intermolecular or can it be intramolecular?
During workup of a conjugate addition reaction, it is likely that protonation occurs at the oxygen of the enolate formed. See discussion here: Do enolates get protonated at the carbon or oxygen atom? In that discussion, it is essentially said "The oxygen is probably protonated, but it really doesn't matter." However, I think this could be important in select instances, specifically in the protonation of enolates with diastereotopic faces (which isn't all that rare). But this mostly depends on if tautomerization can occur intramolecularly.

An example of what I'm getting at:
This is an example I just made up. This reaction likely would not actually happen like this. In actually, I could see some γ-deprotonation happening.
In this reaction, one chiral center is formed, and an enolate is formed, and the most likely atom to be protonated upon addition of an acid is the oxygen:

That enol could go on to form one of two products. The intramolecular reaction, I think, would likely give much more of the thermodynamic product (shown below) than an intermolecular reaction, though I'm not certain on that. The reactions here are simplified; I realize both would most likely make some mixture of the two products:

My thoughts are that the four-membered intermediate required for the intramolecular reaction will just be too strained for this to occur. I realize that in basic or acidic solution, catalysis will occur, as the Wikipedia page mentions. I guess the best way to phrase my question is this: In the hypothetical situation that a single ester enol molecule (like that shown as the product of O-protonation above) is dissolved in pure hexane, would that molecule be stuck in in the enol form or would it be able to rearrange? I realize this is more feasible with an dienol, but I'm specifically trying to figure out if the enol can do this.
For the sake of this question, ignore the basicity of the nitrogen. Maybe I should have used the diphenylmethane anion or something. Doesn't matter, it wouldn't reach the proton, either.

I guess it is also possible that intermolecular would lead to more thermo product because a smaller acid could fit in on the less accessible side.

Instead of hexane, I think gas phase is more what you're suggesting. And if it's gas phase, then someone can probably calculate it...

@jerepierre That would be fine. I considered that, but I'm an organic chemist. Haha, I don't do anything in the gas phase. Essentially, I was thinking "in a vacuum" but gas phase is to chemists what a vacuum is to physicists, right?

I'm inclined to agree with you about ring-strain.  However, we are back to the 'do the experiment' approach at this point, I can't really give an answer

@Lighthart I can't really think of how to come up with an experiment for this one. Or is that what you meant?

I mean synthesize compound and see what you get.

@Lighthart I think I'd have to adjust the setup a bit. Plus, I'm not sure that there would be a significant difference in product distribution unless I used a particularly hindered acid.

Can you specify what you mean by intramolecular reaction? Do you refer to a dimerisation kind of mechanism? How do you O-protonate without the lactone getting cleaved? The re-site of the alpha carbon is pretty much shielded with the bulky NPh2 group. I would suspect that any reaction happening there is virtually impossible. That would leave the si-site for an external attack. The intramolecular reaction is probably impossible, too. At least I cannot imagine any four-membered transition state for that where the proton has access to the pi-bond.

@Martin-マーチン  If you look at the top image, the middle image is the lithium enolate of a lactone. There is no doubt in my mind that this would be protonated in the presence of some acid. The lactone wouldn't be cleaved by forming the enol form of the ester unless you think the it is likely to form a ketene and an alcohol. Also, as far as protonation of the  re vs si sides, si should, I believe, be the side we are viewing, so I would expect that to be the harder side to protonate, despite its protonation leading to the kinetic product.

Also, I specifically mention several times and show that the reaction that I am talking about being inter- or intermolecular here is the tautomerization, and I show and mention right in the title: carbonyl-enol tautomerization. I don't think it is unreasonable to think that, if the enolate is quenched with something like aqueous ammonium chloride, that the small $\ce{NH4^+}$ could lead to some of the cis-bicyclic isomer being formed, especially because the side opposite the one we are seeing is somewhat concave.

So I guess, to specify, the intramolecular reaction I'm referring to is just that - an intramolecular migration of the enol proton to the alpha carbon. There are instances of these things being hypothesized, though I'm not sure what has come of those hypotheses (see this discussion).

@DGS I'm sorry, the first question should have been: Can you specify what you mean by inter-molecular reaction? I admit I failed to see the cleaved products, so thanks for clearing that up. Also I messed up the faces. I think protonation on the face we are looking at, si, is quite a challenge, probably due to the bulkyness of the phenyl rings. However, if you add NH4Cl to that mix, you can't ignore the amine. I don't know how the last link relates to this question though.

@Martin-マーチン Oh, okay, that makes sense. Yeah, a dimerization could occur in the way it does with a carboxylic acid -- this would be hard with a bulky enolate, but makes sense to me with a more accessible enolate. Using ammonium chloride was just one example (how I usually protonate my enolates), but yes, I agree that the amine would have to be considered in that case. The link provided deals with gamma-interaction between a proton and a negatively-charged atom, which I feel would be similar to the proton on an enolate being attracted to the alpha-carbon of the enolate.

@DGS I don't want to extend that into a discussion in the comments sections, so if you would like to continue to talk about it, I invite you to come to [chat]. Just very briefly now, how I think the case is here very different. You have to access the pi-bond which is perpendicular to the OCC-plane. In HONO you have an in-plane lone pair which could be accessed by the proton, hence giving the four membered transition state. If you consider the amine, then a nice six membered transition state could lead to the thermo product.

@Martin-マーチン Good point on the HONO thing stereoelectronics. That makes perfect sense. I think this molecule is too rigid for the hydroxyl proton to reach the amine, but it certainly would in a less rigid molecule.

Nice thought experiment but I might suggest using something like trimethylsilyl chloride. That won't hope off the oxygen, and Id bet you can induce stereo-selectivity that way.

I understood that letting the tautomerisastion step run for a longer time would give more of the thermodynamically controlled product. What would one do if s/he wants to get the kinetic product? Preferrably without chiral catalysts?

@EashaanGodbole, in the appropriate circumstances, you could have some kind of trapping agent present that will react more quickly than interconversion.

Trialkyl boranes tend to not add to aldehydes due to the strain of the 4-membered TS, though allyl boranes do, due to the presence of a 6-membered TS, in this case, the intramolecular addition is no different, so I would expect it not to occur.

tl;dr: The intramolecular variant is an impossible rearrangement. Only intermolecular proton transfer can be observed.

This is actually a very nice example to discuss sigmatropic rearrangements, because the desired [1,3]-rearrangement would be one. In general, these can proceed in four possible ways, depending on the configuration of the migrating group and whether it remains on the same side of the π system or not. We would have to draw orbital schemes to determine whether or not the orbital of the migrating group is able to interact both with the point of origin and the point of arrival favourably.
With regards to the configuration, the two possibilities are:

migration under retention
migration under inversion

A migration under retention is always possible because the corresponding orbital’s lobe is already pointing towards the π system it is migrating along. A migration under inversion is only possible if the (atomic) orbital whose bond is broken and later reformed has lobes on two different sides; typically for migrating atoms, this means a $\mathrm p$ orbital.
In the case of a proton shift, we must consider a hydrogen atom. Its only occupied orbital through which it bonds to the carbon chain is $\mathrm{1s}$. This orbital does not have two lobes, therefore a rearrangement mechanism under inversion is impossible; we need only consider the retention mechanisms.
With respect to the π system, the two possibilities are:

suprafacial (remaining on the same side)
antarafacial (switching sides)

Again, a suprafacial migration is practically always possible. For an antarafacial migration to be considered, the chain must be flexible enough so that both sides can interact with the migrating group; ideally at the same time.
Since the thought experiments are rather hard to perform, here are some depictions to make it easier.

Figure 1: Suprafacial [1,5] (left) and antarafacial [1,7] (right) hydrogen migrations. The depicted π orbital corresponds to the HOMO of a full π system including the $\ce{C-H}$ bond.
Generally, the full π system and the neighbouring $\ce{C-H}$ bond that is to be broken must be considered as I have done in figure 1. This π system can be drawn up with the standard rules for linear π systems. In the [1,5]-rearrangement, the system is penta-2,4-dien-1-yl while in the [1,7] case it is hepta-2,4,6-trien-1-yl. The $\ce{C-H}$ bond is considered part of the π system with the proton being the migrating group. Therefore, the HOMO (the orbital of importance) will be the one in question. (If the $\ce{C-H}$ bond is considered part of hydrogen, we must consider the π system’s LUMO, meaning that the orbital will remain the same. We must always have an occupied and an unoccupied orbital for consideration.)
The image shows us that the [1,5]-migration — over two double bonds — can only occur in a suprafacial manner with hydrogen. If we had a group capable of inversion, an antarafacial manner would be possible given the correct geometry. In the [1,7] case — over three double bonds —, hydrogen can only migrate in an antarafacial manner under retention; the only possible suprafacial manner would be under inversion which hydrogen cannot do.
This has been turned into a general rule of thumb: If your migration is [1,$4n+1$], it may proceed suprafacial under retention but if it is [1,$4n+3$] it must proceed in an antarafacial manner. In the [1,3]-case, only antarafacial migration would be possible, but there is no way that the π system can twist so it would allow for antarafacial migration. The π orbital in question is shown in figure 2 below.

Figure 2: The π orbital relevant for a [1,3]-migration.
Therefore, an intramolecular proton transfer is impossible and an external source would have to protonate.

One thing to note: the molecule itself is able to deprotonate/reprotonate another molecule of its type intermolecularly. So the compound would not be stuck in the enol form but a second molecule would provide the much desired proton to allow the keto form to be regenerated, even in pure hexane (if it is even soluble in that).

This is a very hypothetical situation, but let's say that you obtain that enol, and you dissolve it in an aprotic solvent, and that the enol is higher energy than the ketone, I would think a second molecule would act as the base and send your product to the lower energy state. There is no need for this to be intra-molecular, as the Pka of one enol is exactly the same as the Pka of the other enol, they will freely exchange protons up to the point where you will get protonation at the carbon.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Is there a way to force C-protonation of an enolate?
In another question, I mentioned that O-protonation of an enolate followed by tautomerization may lead to increased amounts of the thermodynamic product over the kinetic product (when those are different species).
The idea that the oxygen is likely protonated first has been discussed previously.
The way, I think, to get the most kinetic product is to use a bulky acid that would do C-protonation rather than O-protonation (assuming sterics explain the difference in kinetic/thermo product). The proton would have to be a softer electrophile, i.e. less positively charged than a proton from a stronger acid.
To continue with the example I gave in the previous question, I think it's likely that C-protonation would look like this, where the major product would be the kinetic product:

So my question is: are there acids that are known to specifically protonate the carbon of an enolate rather than the oxygen? I was thinking maybe BHT would work. Or maybe the thiol analogue of BHT - yes, it would be more acidic than BHT, but it would also be a softer nucleofuge than the BHT anion. This reasoning works with alkyl halides, i.e. softer leaving group = more C-attack, but I'm not sure if the same goes here.

I guess $\sigma ^{} _{\ce{S-H}}$ would be lower in energy than $\sigma ^{} _{\ce{O-H}}$, so even though BHT would a weaker acid than the equivalent thiol, the thiol would still have better HOMO-LUMO overlap with carbon and lower $\delta +$ on the hydrogen.

I think it's reasonable to extend the logic of O- vs. C-alkylation to protonation. I don't see any good way of proving the results of the experiment though.

@jerepierre What do you mean you don't see any good way of proving the results? You mean that even if the results came out as I expected, it wouldn't be definitive that it happened the way I described it?

what technique would you use to prove that the oxygen or carbon is protonated first and the observed product is not the result of equilibration?

@jerepierre Probably nothing would prove it, but an increase in kinetic product would at least lend evidence to it. How about... Get it good and cold, then shoot in the deuterated you choose, wait a small period of time, then shoot in the non-deuterated acid and if you get purely deuterated material back, it means you did the deuteration without tautomerization (so C-deuteration), but if you get a mix, you probably still had some enol tautomer in there. This, of course, assumes that protonation/deureration is faster than tautomerization which may be naive of me.

Trap it as the silyl ketene acetal and then protonate that, eg. https://doi.org/10.1016/S0040-4039(00)84541-4

You might use tert-butyl bromide as your "acid" source via an E2 elimination.


text:
Is there an equilibruim in an unsaturated calcium hydroxide solution?
Imagine an unsaturated(No Ca(OH)2 (s) just Ca2+ (aq) and OH- (aq))  aqueous solution of Calcium Hydroxide.
Is the following equilibrium present in that solution and does the transparency of an unsaturated Calcium Hydroxide vanish after a while?
$$\ce{ Ca^2+_{(aq)} + 2OH- _{(aq)} <=>  Ca(OH)2_{(s)}}$$

No and no (well, in real world the solution might get cloudy, but that's for entirely unrelated reasons).

Equilibria in solutions are given as solubility products, $K_{\text{sp}}$. Your reaction is:
$$\ce{Ca(OH)_2 <=> Ca^2+ + 2OH-}$$
for this, $K_{\text{sp}}=5.5\times 10^{-6}$ (from Wikipedia). What this means, is that in a saturated solution, at equilibrium, you will have:
$$K_{\text{sp}}=[\ce{Ca^2+}]  [\ce{OH-}]^2=5.5\times10^{-6}$$
However, for an unsaturated solution, you will have:
$$[\ce{Ca^2+}]  [\ce{OH-}]^2 <K_{\text{sp}}$$
As mentioned the solution does get cloudy for different reasons, namely the formation of $\ce{CaCO3}$ by reaction with $\ce{CO2}$.


text:
Dependency of Standard Gibbs free energy (using electrode potential) on stoichiometric moles of electron exchanged
I have a question regarding the definition of the number of moles $n$ when using the equation-
$$\Delta G=-nF\Delta E$$
I understand that $n$ is the number of moles transferred in the overall redox reaction, however, it seems to depend on how you write the equation.
If I write an equation as-
$$\ce{Ag+ + e- -> Ag}$$ $$\ce{\frac{1}{2} Cl2 + e- -> Cl-}$$
Then, we have $n$=1.
But if I wanted one mole of chlorine and I write:
$$\ce{Cl2 + 2e- -> 2Cl-}$$
then I get $n$=2.
Which one is preferred and why?

It does not depend on how you write the reaction. You were right in saying that, $n$ is the mole number transferred in the reaction. So, it is not the number of electrons you write arbitrarily in the equation that represents the reaction, it is rather proportional to the number of other substances, which, even being all arbitrary always satisfy that proportional relation.
Think of $n$ as the factor needed to relate one intensive property with one extensive property.

As @user1420303 said, $\Delta G$ is an extensive quantity, and $E$ is an intensive quantity.
First you should write $\Delta G$ as $\Delta G_m$ which means that $\Delta G$ relating to $\pu{1 mol}$ extent of reaction.
$$\ce{Cl2 + 2e- <=> 2Cl-}$$ $\Delta G_{m1} = -n_1FE_1, n_1=2$
$$\ce{2Cl2 + 4e- <=> 4Cl-}$$ $\Delta G_{m2}= 2 \Delta G_{m1}= -n_2FE_2$,
where $n_2=4=2n_1$.
Hence, $E_1=E_2$


text:
Internal energy calculation

The temperature of 1 mole of a liquid is raised by heating it with 750 joules of energy. It expands and does 200 joules of work, calculate the change in internal energy of the liquid.

I want to use the expression:
$$\Delta U = \Delta Q + \Delta W$$
so that:
$$\Delta U = 750\,\mathrm J- 200\,\mathrm J = 550\,\mathrm J$$
but it strikes me that it can't be that simple (first year college exam paper). Also, of what significance is the '1 mole' of liquid?

You proposed the right solution. Nothing to do with the amount of matter nor agregation state.

Yes. Couldn't leave it at that though, a comment has to be longer than three characters. "1 mole of liquid" is of no significance.

It is $Q$ and $W$ not $\Delta Q$ or $\Delta W$

Your calculation is correct. The standardized definition of the change in internal energy $U$ for a closed thermodynamic system is
$$\Delta U=Q+W$$
where $Q$ is amount of heat transferred to the system and $W$ is work done on the system (provided that no chemical reactions occur). Therefore, heat transferred to the system is assigned a positive sign in the equation
$$Q=750\ \mathrm J$$
whereas work done by the system on the surroundings during the expansion of the liquid is assigned a negative sign
$$W=-200\ \mathrm J$$
Thus, the change in internal energy is
$$\begin{align}
\Delta U&=Q+W\\
&=750\ \mathrm J-200\ \mathrm J\\
&=550\ \mathrm J\\
\end{align}$$
However, the question is a bit flawed since the given values are not typical for a liquid. By way of comparison, realistic values for water are shown in the following table.
$$\textbf{Water (liquid)}\\
\begin{array}{lllll}
\hline
\text{Quantity} & \text{Symbol} & \text{Initial value (0)} & \text{Final value (1)} & \text{Change}\ (\Delta) \\
\hline
\text{Amount of substance} & n & 1.00000\ \mathrm{mol} & 1.00000\ \mathrm{mol} & 0\\
\text{Volume} & V & 18.0476\ \mathrm{ml} & 18.0938\ \mathrm{ml} & 0.0462\ \mathrm{ml} \\
 & & 1.80476\times10^{-5}\ \mathrm{m^3} & 1.80938\times10^{-5}\ \mathrm{m^3} & 4.62\times10^{-8}\ \mathrm{m^3} \\
\text{Pressure} & p & 1.00000\ \mathrm{bar} & 1.00000\ \mathrm{bar} & 0 \\
 & & 100\,000\ \mathrm{Pa} & 100\,000\ \mathrm{Pa} & 0 \\
\text{Temperature} & T & 20.0000\ \mathrm{^\circ C} & 29.9560\ \mathrm{^\circ C} & 9.9560\ \mathrm{^\circ C} \\
 & & 293.1500\ \mathrm{K} & 303.1060\ \mathrm{K} & 9.9560\ \mathrm{K} \\
\text{Internal energy} & U & 1\,511.59\ \mathrm{J} & 2\,261.58\ \mathrm{J} & 749.99\ \mathrm{J} \\
\text{Enthalpy} & H & 1\,513.39\ \mathrm{J} & 2\,263.39\ \mathrm{J} & 750.00\ \mathrm{J} \\
\hline
\end{array}$$
When $1\ \mathrm{mol}$ of water with an initial temperature of $T_0=20\ \mathrm{^\circ C}$ is heated with $\Delta H=Q=750\ \mathrm J$ at a constant pressure of $p=1\ \mathrm{bar}$, the resulting expansion is actually only
$$\begin{align}
\Delta V&=V_1-V_0\\
&=18.0938\ \mathrm{ml}-18.0476\ \mathrm{ml}\\
&=0.0462\ \mathrm{ml}\\
&=4.62\times10^{-8}\ \mathrm{m^3}
\end{align}$$
The corresponding pressure-volume work is
$$\begin{align}
W&=p\Delta V\\
&=100\,000\ \mathrm{Pa}\times4.62\times10^{-8}\ \mathrm{m^3}\\
&=0.00462\ \mathrm J
\end{align}$$
which is clearly below the value given in the question $(W=200\ \mathrm J)$.
The values given in the question are appropriate for a gas. For example, realistic values for nitrogen are shown in the following table.
$$\textbf{Nitrogen (gas)}\\
\begin{array}{lllll}
\hline
\text{Quantity} & \text{Symbol} & \text{Initial value (0)} & \text{Final value (1)} & \text{Change}\ (\Delta) \\
\hline
\text{Amount of substance} & n & 1.00000\ \mathrm{mol} & 1.00000\ \mathrm{mol} & 0\\
\text{Volume} & V & 24.3681\ \mathrm{l} & 26.5104\ \mathrm{l} & 2.1423\ \mathrm{l} \\
 & & 0.0243681\ \mathrm{m^3} & 0.0265104\ \mathrm{m^3} & 0.0021423\ \mathrm{m^3} \\
\text{Pressure} & p & 1.00000\ \mathrm{bar} & 1.00000\ \mathrm{bar} & 0 \\
 & & 100\,000\ \mathrm{Pa} & 100\,000\ \mathrm{Pa} & 0 \\
\text{Temperature} & T & 20.0000\ \mathrm{^\circ C} & 45.7088\ \mathrm{^\circ C} & 25.7088\ \mathrm{^\circ C} \\
 & & 293.1500\ \mathrm{K} & 318.8588\ \mathrm{K} & 25.7088\ \mathrm{K} \\
\text{Internal energy} & U & 6\,081.06\ \mathrm{J} & 6\,616.83\ \mathrm{J} & 535.77\ \mathrm{J} \\
\text{Enthalpy} & H & 8\,517.87\ \mathrm{J} & 9\,267.87\ \mathrm{J} & 750.00\ \mathrm{J} \\
\hline
\end{array}$$
When $1\ \mathrm{mol}$ of nitrogen with an initial temperature of $T_0=20\ \mathrm{^\circ C}$ is heated with $\Delta H=Q=750\ \mathrm J$ at a constant pressure of $p=1\ \mathrm{bar}$, the resulting pressure-volume work is
$$\begin{align}
W&=p\Delta V\\
&=100\,000\ \mathrm{Pa}\times0.0021423\ \mathrm{m^3}\\
&=214.23\ \mathrm{J}
\end{align}$$
The corresponding enthalpy balance
$$\begin{align}
\Delta H&=\Delta U+W\\
750.00\ \mathrm{J}&=535.77\ \mathrm{J}+214.23\ \mathrm{J}
\end{align}$$
is quite similar to the values of the question $(\Delta H=Q=750\ \mathrm J,$ $\Delta U=550\ \mathrm J,$ and $W=200\ \mathrm{J}).$


text:
What are dipole moments in a molecule supposed to act upon?
In a $\ce{CO_2}$ molecule, a total of four electron pairs are shared between the carbon and oxygen atoms, such that 2 pairs are shared between the carbon atom and each oxygen atom. Oxygen has a greater electronegativity hence the atoms of oxygen will spend greater time with shared pairs and are partially negative. Since both oxygen atoms will exhibit this same behavior, the carbon atom will become partially positive.
I was reading about the dipole moment of $\ce{CO2}$ (e.g., as here), in which it said that the net polarity of the $\ce{CO_2}$ molecule is zero because if we add the individual dipole moment vectors directed from the positive pole (carbon) to the negative poles (the oxygen atoms), the resultant will be zero which implies the polarity of the molecule is zero.
My problem is this:
If these individual dipole moments (vectors) are showing the opposite and equal pull of shared pairs of electrons toward more electronegative atoms (oxygen atoms), then the carbon atom should stay partially positive with the oxygens remaining partially negative.
Given this, though, it seems that the molecule should remain polar on whole.  This is because these dipole moment vectors are acting on different shared pairs: the first vector acts on the two shared electron pairs between the carbon atom and one of the oxygen atoms, and the other acts on the two shared electron pairs between the carbon atom and the other oxygen atom. According to the following law, then, these vectors cannot cancel each other and thus $\vec u \neq\vec 0$:

Two equal and opposite vectors do not cancel each other out when acting on different bodies.

My best attempt to resolve this contradiction:
Since in reality it is the case that the net polarity of $\ce{CO_2}$ is null, I think that both vectors (viz., the 'pull' from both oxygen atoms) must be acting individually on all electrons of carbon atom and therefore they cancel each other out resulting in molecule to be non-polar on whole. This explanation isn't all that satisfying to me, however.

Also, see here.

Your confusion appears to arise from trying to consider dipole vectors as representing forces, which is incorrect.
Dipole vectors don't "act on" anything; they describe the displacement of the "center-of-charge"‡ of a given region of space (including both the electron cloud and the nuclei) from a particular reference point. In particular, I believe you are confusing dipole moment vectors with electric field vectors:

The electric field vector at a point describes the force experienced there by a point charge
The dipole moment vector at a point describes the displacement of the "center of charge" of the system from that point

Thus, whereas the electric field contributes to the forces present in a system, the dipole is simply a post facto descriptor of a system.

Further, in this case the law you quoted is not applicable:

Two equal and opposite vectors do not cancel each other out when acting on different bodies.

It only makes sense to calculate an overall dipole moment if you use the same reference point for all of the 'subsystem' dipole vectors. Thus, in this example of $\ce{CO2}$, the two $\ce{C-O}$ 'subsystem' vectors point in exact opposite directions from the same reference point, since $\ce{CO2}$ is linear.  Thus, since the outside atoms are both oxygens, the magnitudes of both 'subsystem' vectors are equal and their sum is the zero vector.

‡
For a non-electroneutral region of space, by center of charge I mean the charge-weighted centroid of the system: (1) Choose a reference point $\vec r_0$. (2) At each point $\vec r$ in the region of interest, multiply the charge at that point, $\rho\!\left(\vec r\right)$, by the distance from the reference point, $\vec r - \vec r_0$. (3) Integrate the resulting product, $\rho\!\left(\vec r\right)\cdot\left(\vec r - \vec r_0\right)$, over all space. (4) Divide by the net charge of the region of interest. Dipole moment calculations are more complicated for electroneutral systems, but the principle is similar.

In that case, carbon atom should become more positive in $\rm{CO_2}$ as compared to the carbon atom in $\rm{CO}$ molecule. And I think the dipole moments are forces is not all wrong because they can be thought as the forces exerted by somewhat negative pole(s) to shift the electron density towards themselves. This is the idea I got from Khan academy and one more lecture on youtube.

@SufyanNaeem Dipole moments are not forces. Period. They reflect the outcome of forces, yes, but they are not themselves forces. They are effect, not cause.

Well I think I am closer but please tell me more on what is displacement of center-of-charge or suggest me a book that could tell me about that. I thin, I have tou understand this thing only.

@SufyanNaeem Please see my further edits

More than once I have complained about how ill-defined the concept of polarity is. One of the most common ways to attest whether a molecule is polar or not is to judge it form its dipole moment. When you look at the equilibrium geometry of carbon dioxide, which has $D_\mathrm{\infty h}$ symmetry, you will see, that it is impossible to have an overall dipole moment due to the inversion centre, i.e. the carbon centre.
Due to the simple picture it paints, we often decompose the total dipole moment into bond dipole moments, because this gives us a handle on how polarised these bonds are. Often enough we can estimate an overall dipole for a molecule from bond dipole moments of similar molecules. However, the overall dipole moment is just one way to express how the electron density and the field of the nuclei differ from each other, and the decomposition into different vectors is arbitrary, a mathematical model if you wish.
You can look at the dipole moment $\mathbf{q}$ as the expectation value of the dipole operator $\mathbf{P}$ from the total wave function $\Phi_0$.
$$\langle\mathbf{q}\rangle = \langle\Phi_o|\mathbf{P}|\Phi_0\rangle$$
But let's have a closer look at carbon dioxide and what you have written.

In a $\ce{CO2}$ molecule, a total of four electron pairs are shared between the carbon and oxygen atoms, such that 2 pairs are shared between the carbon atom and each oxygen atom.

This is not true. This is the misconception you have when looking at a Lewis type bonding picture. If you employ the molecular orbital theory, you can see that there are actually twelve electrons in bonding orbitals, and four electrons in anti-bonding orbitals. Not counting core electrons. All of those electrons are delocalised over the whole molecule.

(Taken from an earlier post of mine. There is an error, MO 8 should be MO 7. BP86/cc-pVTZ level of theory. Click images to enlarge.)

If these individual dipole moments (vectors) are showing the opposite and equal pull of shared pairs of electrons toward more electronegative atoms (oxygen atoms), then the carbon atom should stay partially positive with the oxygens remaining partially negative.

This is true. This charge separation is also the reason for the quadrupole moment. Taken from BP86/cc-pVTZ:
 Traceless Quadrupole moment (field-independent basis, Debye-Ang):
   XX=              1.3594   YY=              1.3594   ZZ=             -2.7187
   XY=              0.0000   XZ=              0.0000   YZ=              0.0000

Probably a more known approach is to look at the partial charges on the same level of theory using different analysis tools.
      Mulliken     Natural        Hirshfeld     Atoms in
                   Population                   Molecules
1  C    0.336381     0.91266        0.295332      2.111095
2  O   -0.168190    -0.45633       -0.147682     -1.055548
3  O   -0.168190    -0.45633       -0.147682     -1.055547

You see, you are quite right. There is charge separation. Which mainly stems from the lone pair electrons at the oxygen atoms, i.e MO 6 and 7 (8 in the picture).
This is why I am very hesitant of calling a molecule like carbon dioxide unpolar.

Given this, though, it seems that the molecule should remain polar on whole. This is because these dipole moment vectors are acting on different shared pairs: the first vector acts on the two shared electron pairs between the carbon atom and one of the oxygen atoms, and the other acts on the two shared electron pairs between the carbon atom and the other oxygen atom. According to the following law, then, these vectors cannot cancel each other and thus $\vec u \neq\vec 0$:

Two equal and opposite vectors do not cancel each other out when acting on different bodies.

Electrons are indistinguishable, so are electron pairs. You cannot treat the electrons of a molecule as different bodies. You can only treat the molecule as a whole. The positive charged nuclei act on all electrons at the same time, all electrons are correlated. In the Born-Oppenheimer approximation you have a cloud of electrons action upon a fixed field of attractors (nuclei).

The dipole moment results from the center of "gravity" of all negative charges and the center of gravity of all positive charges being spatially separated. Its actually the product of the positive times the negative net charges times the distance vector between their centres.
Thus a dipole moment of zero actually means that the centres of gravity coincide (unless there are no charges at all), and it does not mean that there are no net (partial) charges around. So in deed in CO$_2$ the center of gravity of all positive charges is at the location of Carbon nucleus, but for reasons of symmetry so is the center of gravity of all the negative charges (electrons).
In contrast to a zero dipole moment CO$_2$ has a substantial electric quadrupole moment, this is maybe what your intuition suggests.


text:
Reactions of ketones and aldehydes

Our topic was about ketones and aldehydes (carbonyl group) including hemiketals and hemiacetals and he asked us to do a homework. I cannot find an answer to this one.Please explain me the reaction and the product formed.
PS: The figure at bottom is my best effort to do this question.

Have you tried to write a mechanism for this reaction?

How did you end up at the structure on the bottom?

Would it help to know that $\ce{R'-C(OR)3}$ is an orthoester?

There are 5 oxygens at the bottom and three at the top.  The only way you get more oxygens in these types of reactions is if you lose a double bond.  Therefore, something is amiss at the bottom.

But.....6 member lactones are super stable so I wouldn't be surprised if only 1 ring opened/ some recyclized

I'm not sure I would say "super stable" necessarily. Delta-valerolactone always (for some aggravating reason) comes as liquid lactone over polymerized material. See impurities listed here.

Let me rephrase since 'super stable' is not a technical term (though probably assessable and appropriate for those with beginning organic chemistry homework problems). Molecules like this with 2 nucleophilic atoms which if they attacked the carbonyl  would make a 6 membered ring would likely react in acidic media to some extent. Please post the solvent, acid used, concentration, and temperature of your homework such that we can guess at the product distribution.

Also of note; 4 membered transition states of proton transfer are unlikely in reality, but often used in these sorts of mechanisms. The proton is likely transferred first to a water molecule, a solvent molecule, or some other base in solution before it ends up on the indicated position. Chemists do this (and take it with a grain of NaCl) since it doesn't overly complicate the real mechanism in question, and could be any or all bases present.

You mentioned ketals, acetals, hemiketals and hemiacetals.
Let's have a look at some structures (1 to 5).

What do they have in common?

 Apparently, 1 to 5 each have one carbon atom with two single bonds to oxygen atoms. In 1, this central carbon atom bears a hydrogen atom (omitted in the drawing). Do you recognize that 1 is a "masked" aldehyde? 2 and 3 are apparently different. The central carbon atom, the one with the two oxygen atoms, bears single bonds to two other carbon atoms. 2 is made from cyclopentanone, 3 is made from cyclohexanone. When the protecting groups are split off in 1 to 3, they are gone and no longer a part of the molecule with the rebuild $\ce{C=O}$ group. But it mustn't always be like that. Take a closer look at 4 and 5 and figure out what will happen to them under acidic conditions. Will there be $\ce{C=O}$ and $\ce{OH}$-groups in the products?

Now have a look at the structures 6 and 7.

Are these different from 1-5?

 Apparently, they are! Each has one carbon atom with three single bonds to oxygen atoms! You will conclude that these are not the acetals that we worked on above. Under aqueous acidic conditions, the central carbon will not be converted to an aldehyde or ketone, but to a […]. On the other hand, 4 and 7 do have something in common when it comes to the question whether the "protecting groups will still be found in the product ;-)

actually what reaction is this?(i.e the one mentioned in the question).


text:
Is there any process to determine what elements a physical sample is made from?
I'm wondering if anyone knows of laboratory services that can analyze a sediment sample to tell what it's made of.
Background: our drinking water has small black flakes appearing, and our city water department cannot tell us what they are, and they say they have no equipment to test it. Their guess is either manganese, or sediment.
I'm not sure if there's a way to test an arbitrary sample to determine what it is. I've tried a few labs but none of them have the ability to do something like that.

You would need several techniques. An  atomic absorption spectrometer would be good for manganese, but for "sediment" (very broad), there is no telling. Chromatography of some sort + mass spec would be my first guess for general dirt, but maybe not. You could always do elemental analysis, but if it's not a pure sample, that won't help with IDing a specific compound, but elements would be easy enough. Honestly, without more than "sediment," you're going to have a hard time getting a concrete identification.

There are certainly techniques that can analyze one or a small number of particulates. A scanning electron microscopy with an x-ray detector (SEM/EDS) would be my first thought. X-ray diffraction (XRD) and perhaps x-ray fluorescence (XRF) are others. // I'm sure that after Flint everyone is nervous about their water. In general I'd be more worried about the stuff in the water that I couldn't see. Pb+2 would be invisible to the eye.

I could suggest checking with you neighbors to see if they have the same problem. Also a simple particulate filter in the kitchen would keep such particulates out of your food. // I'd wonder if you have an old iron supply line coming into your house that has corroded. The water flow through such a line is strong enough to break of chunks of the crud.

Hi all, thanks for the info, it's been a big help. It seems like it might be "scale" build up in our tank-less hot water system. Called the manufacturer as a guess, and they said it can happen when the water gets heated and reacts with whatever particular elements are in our water source. I wouldn't have guessed anything could solidify in this manner while passing through a tankless system, but that's what they said. Sorry to transform the chem board into a home improvement board.  Thanks for your help!

Well if the tankless hot-water is to blame that is easy to test. Filter some cold water through a coffee filter and filter some hot water through a coffee filter - may need to run for a while. I'd guess that the "shock" of going from no flow to full bore open would be more likely to dislodge crud. // PS - Did you know that you get crude forming a ppt in the bottom of a hot water tank? You're suppose to flush the ppt out once in a while.

Look for labs with ion chromatography (metals), electron capture chromatography (halogenated compounds), and gas chromatography/mass spectrometry (organic compounds). They should be common.


text:
Estimating for atomic number for an "unknown element"
A questions tells me there's this element with 3 isotopes (270.51 amu, 34.07% abundance; 271.23 amu, 55.12%; and 269.14 amu, 10.81%), and to solve for average atomic mass. I have done so and got 271 (270.758) for its atomic mass. Then it asks me to estimate its atomic number, and its location on the periodic table. How do I exactly "estimate"?
ps. I dont think this element actually exists...

Note that, in general chemistry, the atomic mass unit (amu), which referred to the atomic mass of oxygen, is obsolete since 1961.

You need at least four significant figures not 3 for the average atomic weight, but I'd give 5 in this case. // I know that there is some curve relating atomic number and atomic weight but I can't remember the name right now. // Peeking at the periodic table, Sg with atomic number 106 seems to fit.

See here: https://en.wikipedia.org/wiki/Semi-empirical_mass_formula#Examples_for_consequences_of_the_formula the curve is shown here: https://en.wikipedia.org/wiki/Nuclear_binding_energy#Nuclear_binding_energy_curve

@MaxW there is also a good atomic number vs. atomic weight answer here on chem.se: http://chemistry.stackexchange.com/questions/32158/is-there-a-function-to-approximate-atomic-mass-from-the-atomic-number/32159#32159

@ph13 - Thanks for the link! I upvoted Geoff's answer. I'm sure he wasn't the first to play with the equations, but he gave a really good answer. // It doesn't seem that the power law equation has a particular name which somewhat surprises me.

Well I found it out. It is seaborgium atomic number 106, mass number 271.
To be found in period B6, group 7.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. At the moment this reads more like a comment than an actual answer - could you elaborate a little more, for example how you found it, because that is the actual question. With a bit more rep, you will be able to post comments on any question/answer.