text:
Methods to prepare lactic acid from acetaldehyde
I got one method [here](using acetaldehyde and carbon monoxide)1.
But is there any alternative method (using specific catalyst)?

What is the downvote for?

I'm not the downvoter or the upvoter, but it would be really nice if you add a summary of what's written there here. I could assume the downvote is because someone wasn't inclined to click on an external link to find out what your question really is.

A fairly simple two step synthesis involves nucleophilic addition of $\ce{HCN}$ to the carbonyl group and then aqueous hydrolysis of the nitrile to yield the carboxylic acid product.

Acetylene + $\ce{HCN-Ba(CN)2}$ → vinyl cyanide
Vinyl cyanide - hydrolysis → acetaldehyde cyanohydrin
Acetaldehyde cyanohydrin - dil HCl → lactic acid


text:
In Water Solution will there be an interaction between Ozone Gas and Carbonic Acid
Original Question Posted 8/31/15
We are trying to predict what will happen when Carbon Dioxide gas is introduced into a contact chamber that serves as an Ozone Contact Chamber.  These Contact chambers presently oxidize compounds found in Municipal Pool Water along with Chlorine at below 5ppm. there are 3 contact chambers (one for each pool) and the Temperature range is from 104 to 80 F. We are switching from Hydrochloric acid to Carbon Dioxide to keep the Total Alkalinity from dropping when the acid is introduced.
"Predict what will happen" is exceptionally broad. If you can narrow your question to specifics, that will help a great deal. For instance, we don't know what the compounds are in the municipal pool water. See this link for a guide to asking questions here on chemistry.SE. –  Todd Minehardt yesterday
Ozone and carbon dioxide will be pretty much indifferent to each other. Indeed, ozone is known for its ability to oxidize almost everything, but CO  2   is already a final product of oxidation. "What is dead may never die", they say. Likewise, CO  2   may never be oxidized. –  Ivan Neretin yesterday

Thank you for taking the time to respond Todd and Ivan. The other compounds in the municipal pool water would include body oils, salt from perspiration and whatever else a particular bather might introduce. We are using Calcium Hypochlorite as a chlorine source and, depending on the PH may also leach Calcium from the Plaster of Pool Walls. The calcium harness varies in the Three separate bodies of water from 400 to 1200 and we endeavor to keep the PH between 7.2 and 7.6.  The Total Alkalinity is around 20 in the 2 bodies of water in which the PH is controlled by adding Hydrochloric acid. We have started using Carbon Dioxide in our smallest body of water and are introducing it in the water returning to the pool perhaps 150 ft away the Total Alkalinity in that pool has risen to 120 ppm which helps to maintain a more stable PH.  Because the piping to the other 2 pools is not as long we are thinking about introducing the Carbon Dioxide into the line the ozone is injected into before the contact chambers. Ivan stated that the Carbon Dioxide would not interact with the ozone.  My research implies that both Carbonic Acid (the objective of injecting Carbon Dioxide) and Ozone are very unstable, and Carbonic Acid has a very short life.  Is it likely that the Carbonic Acid would interact with the Ozone before the Carbonic Acid can effect a PH change (in effect neutralizing both objectives)?        Thank you  for any feedback you may provide.

source: Stack Exchange

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

sort: votes


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Tollen's Test and Werner's Theory
The production of silver ions in Tollen's test violates Werner's theory of
non - ionizable secondary valency since we are using the diamminesilver(I) complex as the reagent. How does one reason this??

i.e the silver ion inside the complex should not be precipitable or ionisable according to Werner's Theory.

Ionizable and non-ionizable in this context refers to the observation that the counterions of a complex ionize in solution and can thus be precipitated and the ligands of the complex cannot and are thus non-ionizable.
There are two things at work here:

All complexes exist in an equilibrium with the free ligands. In the case of diamminesilver(I), $$\ce{Ag+ + 2NH3 <=> [Ag(NH3)2]^+}$$ the equilibrium is overwhelmingly toward the complex and no AgCl will be precipitated if chloride is added. However, if we add iodide instead, we will actually observe an AgI precipitate because AgI is far less soluble than AgCl and there is sufficient $\ce{Ag+}$ that we see some precipitate. Therefore, there is actually always a little bit of $\ce{Ag+}$ free in solution.
The Tollen's test is a redox reaction. Unlike the case above where we're precipitating an ionic compound and thus need to form an ionic bond, electron transfer to the silver is quite easy and once reduced, it is unfavourable for the silver to remain in solution. Even more extensive ligand coverage in something like $\ce{[Fe(CN)6]^{4-}}$ doesn't prevent redox reactions from occurring. Even some metalloporphyrins undergo redox reactions at the metal centre.

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What causes the "white water" smell?
I can detect a distinct smell when I am near water going over a weir. I could tell that I was near "white water" if I was blindfold. Can anyone tell me what causes this odour?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center. || Um, I've never heard of this terminology. What's "white water"?

@inɒzɘmɒЯ.A.M It means rough water that is turned white by the foam - like the type you might get in a fast flowing, rocky river.

As in 'white water rafting' - ah, I got it now - interesting question!

Perhaps you could describe the smell in a bit more detail.

related http://chemistry.stackexchange.com/questions/19495/does-water-have-a-smell, http://chemistry.stackexchange.com/questions/25076/what-is-the-smell-of-a-glass-of-water-in-contact-with-fresh-air

Can you clarify the limit of this sense?  E.g., does it only apply to freshwater in a natural river?  Or can you also tell with pool water, or in a lab with distilled water?  (Because river water is chemically vastly more complex than just "water"!)

See:
What makes rain smell so good

I don't have any journal article citations to back me up, but I've always understood that this distinct smell is from water aerosol.  You can detect the same smell during a fine misting rain as well as near water sprinklers (if the sprinklers produce a very fine mist of water).  But you don't get this smell from water vapor (steam) or bulk water, so there is clearly a specific range of water droplet size that triggers the olfactory response.
I've personally always wanted to know what the specific size range of these aerosol droplets needs to be to trigger the olfactory response.  I'm confident someone has studied this, but I don't know which journals to search to find the answer.

Possibly moss and foliage, along with damp dirt. Moist banks shaded by medium trees next to the white water areas could be a great growth site, and in my experience, this is true. If you are referring to white water on blank rocks, with no sign of organic matter, feel free to correct me.

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Lab safety - Long vs. Short Sleeves on Lab Coats
I have three lab coats, two with long sleeves, and one with short sleeves.
I saw that the concepts about lab safety vary from country to country; in some places it looks like the short-sleeve coat aren't allowed or they tell you to avoid it.
I'm a little confused about it: When should I use long-sleeves and when should  I use the short-sleeve coat?
Are the short-sleeves really wrong in a chem lab?

The short sleeve has no place in a chemistry lab.
Considering the coats purpose as personal protective equipment (PPE) is to reduce or eliminate the potential or risk of exposure to materials then the long sleeve coat would be the correct PPE to wear in a chemistry laboratory along with extra PPE which should have been assessed prior to performing the procedure.
Even in a microbiology or molecular biology lab, where a lot of the solutions might not seem harmful, the cellular material in suspension is classed as a bio hazard, which the long sleeve will provide better protection.
Hazardous chemicals are still commonly used in these labs, too, and will most likely be present whether you using them or not. It's not just for chemical exposure either also for general protection.

To add to @Joel's excellent answer, one of the reasons that short-sleeved lab coats are even used at all, is according to a HealthCare Business and Technology article Short-sleeved lab coats just as germy as long-sleeved, due to

Some hospitals (and individual health pros) have switched to short-sleeved lab coats under the belief that the reduction in the amount of fabric would also lead to a drop in the amount of bacteria picked up during the day — and potentially passed on to patients and co-workers.

However, based on the Journal of Hospital Medicine paper Newly cleaned physician uniforms and infrequently washed white coats have similar rates of bacterial contamination after an 8-hour workday: A randomized controlled trial, has found that there is

no statistically significant difference between the amounts of the bacteria on the long- or short-sleeve coats.

So, asides from the protection to your arms from chemical or biological splashes, glass splinter etc, there is no additional protection from the spread or pickup of bacteria with short sleeve lab coats.

At my university we are all required to wear long sleeve coats in chemistry labs. It simply protects more of your body compared to a short sleeve coat.
As an example, in one of my labs a guy had his sleeves rolled up and then he decided to rest his forearm on the fumehood while he was working with sulfuric acid. It turned out he spilled some acid earlier so he ended up with a burn all along his arm. This could also happen if pouring chemicals - if you spill some, your arm won't be covered.
Long sleeves are also important when handling liquid nitrogen with cryogenic gloves. You need to have your sleeves going over the ends of the gloves, to prevent any liquid nitrogen falling into the gloves where it can't evaporate easily off the skin and could cause burns.
I suppose one hazard with long sleeves is that spilling a chemical on your hands could cause it to roll down your sleeves and increase the contact time of the chemical with your skin as you then need to take the time to remove your lab coat before you wash your arm, but ideally your coat will be tight at the cuffs to minimise this risk.

authors: Илья Маковоз


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Does Photoelectric effect occur in a micro-oven while heating food?
Does the photoelectric effect occur while heating food on a metal plate in a microwave oven? If not, then why doesn't it occur?

I'm voting to close this question as off-topic because it is purely about physics and so would be more appropriate on physics.SE.

But Photoelectric effect is written in my chemistry book  . Even most of the  books .

No, the photoelectric effect does not occur in a microwave oven.
The workfunction for a metal is the minimum photon energy required to eject an electron from the metal via the photoelectric effect.  Workfunctions are typically in the range of several electron volts (eV).  Potassium, for example, has a workfunction around 2 eV, see here for a list..
The energy in a microwave photon can range from 0.00001 to 0.001 eV; much to low to eject an electron from a metal.
For more background on the photoelectric effect see this earlier SE Chem Q&A


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What are the products of the Grignard reaction with methyl-magnesium iodide?
I want to react my ketone with methylmagnesium iodide (the Grignard reagent). I then need to use a dilute acid solution to protonate the $\ce{MgI-O-R'}$ product to produce my alcohol: $\ce{H-O-R'}$
Now I was trying to figure out what acid to use. The first thing that came to mind was sulfuric acid. What would the products be?
In general the reaction should look like: $$\ce{R'=O + XMgI + H-A -> X-R'-OH + MgI+ + A-}$$
Now if I use sulfuric acid the conjugate base reacted with the Grignard products is something like $\ce{MgIHSO4}$. Is this correct?
This molecule doesn't look very stable so my thinking is that the $\ce{Mg}$ donates an electron to $\ce{I}$ in $\ce{MgI+}$ to form elemental iodine and $\ce{Mg^2+}$, then this reacts with $\ce{HSO4}$ to produce $\ce{Mg(HSO4)2}$ and $\ce{MgSO4}$.

Your product is that X-R'-OH. What's the point in worrying about byproducts?
Also, that Mg-something is an ionic salt and hence is not made of molecules. Also, you are not going to crystallize it anyway, so there is not even a definitive formula. It's just such-and-such ions in the solution.

I would like to recycle as much of my byproduct as possible since the iodine is expensive. I agree about the Mg salt solution... Although I am not sure about the elemental iodine presence, I guess I'll just see when I do the experiment.

Your reaction scheme doesn't make much sense. $\ce {R'=O}$ is not a ketone. Is $\ce X$ supposed to be $\ce Me$? This is confusing since $\ce X$ is most commonly used as the symbol of a halogen.
I would rewrite as:
$$\ce {R'RC=O + MeMgI +H_2O-> R'RC(Me)OH + Mg(OH)I}$$
$\ce {R'(R)C=O}$ is implied.
Make sure you use very dilute aqueous sulphuric acid since your product is a tertiary alcohol. I would rather use saturated aq. $\ce {NH_4Cl}$. Even water will do but I suspect that the work-up might not be as smooth (a less water soluble magnesium salt will be formed). There will be no iodine formed and in any case I don't find the idea of recycling iodine cost effective. It will cost you more to isolate iodine that can be pure enough to work in the notoriously capricious Grignard reagent formation. In fact I would suggest you to buy a commercially available solution of $\ce {MeMgX \ (X=Cl, Br, I)}$ since preparing a Grignard can be very challenging and you will also have to handle toxic and volatile $\ce {MeI}$, have access to dry ether/THF etc. I assume you follow a detailed experimental procedure from the literature.

I like the idea of using a weak acid so I don't need to worry about how dilute my acid is. Doesn't the Mg(OH)I ionize in the solution? And in order to form Mg(OH)I, doesn't the NH4+ -> H+  + NH3 where then the NH3 +H2O -> OH- +NH4? But then that process would self eliminate?


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Determining reaction formula from optical absorption graphs

The rate of a reaction between A + B was followed spectroscopically at three different wavelengths, and the percent absorption at each wavelength was plotted as a function of time.  Which of the following mechanisms is consistent with the spectral data?

(a) $\ce{A + B -> C}$
(b) $\ce{A + B -> D, D -> E}$
(c) $\ce{A + B -> F, A + B -> G}$
(d) $\ce{A + B -> H, H + A -> I}$
(e) none of the above

The correct answer is (b), which I can believe, but not sure why.
So, it can't be (a), because in $\lambda_2$, the curve went up then down, indicating the presence of an intermediate. However, I'm not sure why this can't be (d) or (c) as well. As long as F, G, and H are active at $\lambda_2$, then wouldn't all of the reactions show a similar curve?
I'm not sure what to do with $\lambda_1$ and $\lambda_3$, as I think they just tell me that A + B is consumed, and that the product and/or intermediates aren't active at either $\lambda_1$ or $\lambda_3$.

Let's see. You are right in that it can't be (a). Now consider (c): A and B are consumed, while F and G accumulate. Where is that intermediate that goes up, then down?
As for (d), I believe there is no way we can tell that from (b) based on the picture alone, without looking at the numbers. See, suppose $\lambda_1$ actually measures B, and A is in huge excess; what would be the qualitative difference between the two? I can't see any.


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What happens when acetone reacts with hydrogen cyanide?

Which of the following is the major organic product when acetone is allowed to react with hydrogen cyanide in the presence of a catalytic amount of sodium cyanide?

The correct answer is (B).
However, I got this answer without using the catalytic sodium cyanide.  In my proposed mechanism, the dissociated proton from $\ce{HCN}$ activates the carbonyl, which facilitates the nucleophilic attack of $\ce{CN-}$.
What role does the $\ce{NaCN}$ play?

The acetone carbonyl is attacked by the CN$^-$ that comes from the NaCN.  This tetrahedral intermediate has a oxygen anion, which then grabs a proton from the HCN, regenerating the CN$^-$.

The product that is formed is cyanohydrin.

Cyanohydrins can be formed by the cyanohydrin reaction, which involves
  treating a ketone or an aldehyde with hydrogen cyanide ($\ce{HCN}$) in the
  presence of excess amounts of sodium cyanide ($\ce{NaCN}$) as a catalyst:
$$\ce{RR’C=O + HCN → RR’C(OH)CN}$$

Since, the reactant is acetone both R and R' are $\ce{CH3 -}$ group.

In this reaction, the nucleophilic $\ce{CN^−}$ ion attacks
  the electrophilic carbonyl carbon in the ketone, followed by
  protonation by $\ce{HCN}$, thereby regenerating the cyanide anion.

Now, what is the function of catalyst sodium cyanide?

For successful cyanohydrin formation it is important to have free
  cyanide ions available to react with the ketone or aldehyde. This can
  be achieved by using a salt (e.g. $\ce{KCN}$ or $\ce{NaCN}$) or a silylated
  (e.g. $\ce{Me3SiCN}$) form of cyanide under acidic conditions or by using $\ce{HCN}$
  with some base added to produce the needed $\ce{CN^−}$ nucleophile.(Here)


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What is the difference between convection and conduction?
I was having a conversation with my dad the other day and he described something as happening due to convection, and it dawned on me that I have no real clue what the difference between and convection and conduction is.
I started to do some reading on this and haven't really found any good clarification.
Here is what I do understand:
Convection:

Involves transfer of heat within a single substance by mass flow
Does not take place in solids due to a solid's inability to behave like a fluid
The heat flow is governed by diffusion of higher energy molecules and thus a distribution of that energy evenly, or by advection which basically seems to be bulk diffusion to me

Conduction:

Transfer of heat within a body by diffusion
This diffusion differs from convection because convection supposedly happens through a material carrier
Can take place in solids

Confusion:
No doubt, some of the points I just put up there are wrong or else I basically see no point in making a distinction between convection and conduction when the processes can be described more easily by diffusion which redistributes energy by principle of maximizing entropy.
So why define these two different terms when they are essentially the same process perhaps just happening in different mediums?
Contradictions:
I was looking at the Wikipedia pages, which I usually find to be reliable, for convection and conduction and was only confused more.
For instance, from the conduction page:

"In gases and liquids, conduction is due to the collisions and diffusion of molecules during their random motion."

Now, from the convection page:

"Convective heat and mass transfer take place both by diffusion – the random Brownian motion of individual particles in the fluid – and by advection, in which matter or heat is transported by the larger-scale motion of currents in the fluid."

Those two statements seem to be basically the same. The mechanism for the heat transfer is the diffusion of randomly moving moving particles.
On the other hand, the conduction page explains that conduction takes place in solids through,

"In solids, conduction is mediated by the combination of vibrations and collisions of molecules, of propagation and collisions of phonons, and of diffusion and collisions of free electrons."

That seems like a perfectly good explanation of what takes place when heat is being distributed towards thermal equilibrium in a solid.
Questions:
What about the above statement that conduction in solids is mediated through collisions of electrons and vibrating atoms makes that process conduction rather than convection?
What of the above points are incorrect?
On a molecular level, (i.e. using typical chemistry ideas) what makes a distinction between convection and conduction?
Finally, what is meant by the statement that convection happens through a material carrier?
Does that imply that conduction does not happen through a material carrier? If so, how?

I hope you can see my confusion between these two terms. They seem very convoluted to me.
I would point out that good and correct answer need not address each of my sub-questions specifically because a good answer may clear up my confusion and hence eliminate one of those sub-questions.

Can I have an explanation from whoever downvoted this? The criteria for a downvote is, "This question does not show any research effort; it is unclear or unuseful." I did my best to present this question as clearly as possible and I believe it's useful for anyone interested in  thermodynamic phenomena.

I am not the downvoter, but I think one reason could be that this question is only very vaguely related to chemistry. It might serve a better purpose on [physics.se].

@Martin-マーチン It's more chemical engineering than chemistry -- perhaps it should be on Engineering.SE?

Convection is a mechanism that enhances conductive heat transport, rather than being a separate physical phenomenon.  It operates by bringing hotter parcels of fluid and colder parcels of fluid into closer physical proximity with one another so that conduction can take place more easily and rapidly.  One way this is accomplished is by shearing the fluid near a heated wall in such a way that the colder faster-moving layers further from the wall flow over the hotter slower-moving layers closer to the wall to physically cause the temperature gradient normal to the wall to increase.  This enhances the rate of conductive heat transfer.  Another situation is where the fluid is being mixed by a stirrer to create thin striations of hotter fluid and colder fluid throughout the fluid.  Conduction between these hotter and colder striations increases, again because of the much larger temperature gradients that have been created.  This is typical of how convection works.
What I'm saying is that the deformations experienced by the fluid while flowing are such that they cause the temperature gradients in the fluid to get higher, and this increases the rate of heat transfer.
Imagine a deck of cards split in half, with all the red cards (hot) in one half and all the black cards (cold) in the other half. If you bring these two slabs of hot and cold material together and wait, heat transfer will occur very slowly between the hot and cold regions. However, if you shuffle the deck so that there is a hot (red) card sandwiched in between every pair of cold (black) cards, and a cold (black) card sandwiched in between every pair of hot (red) cards, the heat transfer between the hot and cold layers will be much more rapid. This shuffling effect is what convection does.

Haha you'll notice I edited the question so as to not be presumptuous about engineers. Wasn't trying to be offensive. Frankly I'm just jealous I'll never make as much money as any of you guys :) So, you're saying that convection is essentially the mass transport (where the mass being transported has a temperature differential with the rest of the system) and conduction is the actual heat transfer taking place?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. While it may seem strange at first we do not use salutations in our Questions and Answers. Hence I removed it.

this has alluded me as well, in high school physics books you typically have heat transfer divided into 3 categories. convection surely seems little bit out of frame for me. as you said it's not the samke kind of "thing" as radiation and conductivity.


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If you can smell something does that mean that this thing loses molecules?
When you can smell something there must be some molecules in your nose which signalize the smell in your nose. Does that mean, that e.g. a cheese is losing molecules which fly into your nose?
And does that effect the weight or amount significantly?

Yes, the thing you smell surely does lose some molecules, and consequently some weight. It can be said that it slowly evaporates or sublimates. But the strength of smell varies greatly in different compounds, as does the rate of evaporation (sublimation). Some compounds may be gone completely in a few minutes, and some would stay for years, or even decades.


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How to prepare glacial acetic acid from sodium acetate?
What would be the most practical process to prepare glacial acetic acid from sodium acetate?
I'm thinking to react the sodium acetate with an acid to convert the solution to acetic acid and sodium salt but I'm not sure of the most practical way to proceed from here without distillation (which seems impractical).
I think solvent extraction (using a protonated solvent such as $\ce{CH2Cl2}$) then removal of the solvent would be a way but I'm not sure.
I would like to hear from someone who's done it before.

I do not think there is an insoluble sodium salt and distillation of glacial acetic acid just sound incredibly dangerous and horrifying.

I don't see anything wrong with distillation of acetic acid per se, so it's a way to go as long as it happens under fume hood. Distillation of bromine or nitric acid is way more dangerous and still is used.

It forms the azeotrope with water which is difficult to remove.

@Joel Same with nitric acid. Still, it is prepared by distillation from reaction mixture of sulfuric acid and potassium nitrate. Same is probably applicable to acetic acid. Or, in case you are not brave enough, the traditional way is a way to go (freezing the azeotrope)

I'll take that on board thanks , I'd like to find a way to do bit by solvent extraction.  I'd rather distill a suitable solvent than the actual acid it's much quicker. It's not a matter of being brave enough I understand the variables its about practicality.

Why would you want to prepare acetic acid from sodium acetate anyway? It’s cheap as sand, isn’t it?

No not really ! I have a lot of sodium acetate and instead of throwing it out I wanted to see if I could to something with it...

@Joel even so, the ingredients and equipment you need to covert sodium acetate back to glacial acetic acid will probably cost you more than just buying some glacial acetic acid. Don't waste time, effort and money on futile stuff unless it is fun (which this won't be).

@matt_black. I know what your saying but it's not helping . I don't want to buy it that's the whole point I don't really need it.

I would oxidized ethanol if I really needed to synthesize it for something.

There is a rumor around my lab that if you freeze vinegar at a very specific temperature, than get rid of the water ice, than thaw it, you could in theory extract Glacial Acetic Acid, but again, its just a rumor, and I haven't tested it.

You would only have to get the acetic acid down to below 15c and above 10c to prevent any water from freezing. Even though the melting temp for water is 0c, it is always safer because of the azeotrope to just keep it above 10c.

Dichloromethane is not a "protonated" solvent.

Matt_black lol think I meant polar


text:
Raise the pH of a solution and prevent build up of carbonates
I need to raise the pH of a solution so the easiest approach is to add alkali hydroxides (NaOH, KOH, etc). But as most of you know, the CO$\ce{_2}$ in the air can react with these hydroxides to form carbonates which usually  precipitate out the other metal ion dissolved in the original solution. So what else are my options?

How about $\ce{NH4OH}$? This does not for carbonates as readily, but it does form complexes with many metals. It depends on your intended use.
It might be easier to keep things away from the air:

Use a closed container.
Float an inert liquid (e.g. paraffin oil) atop your solution, or just a layer of pure water, which might take a while to mix.
Use an inert gas over the container to displace air.


text:
Titration of H3PO4 and H2SO4 with methyl orange and phenolphtalein as indicators
There are some problems I don't understand how they are solved (I have the answer). Since I can't find anything related to these on the internet, I'm asking about the problem and the answer here.
Problem:

For the titration of $\mathrm{25~mL}$ solution which has $\ce{H3PO4}$ and $\ce{H2SO4}$ in it, there are spent $\mathrm{37.2~mL~0.1008~M~NaOH}$ in the presence of methyl orange as an indicator. The other $\mathrm{25~mL}$ of the solution spent $\mathrm{49.6~mL~NaOH}$ in the presence of phenolphthalein as an indicator. Find the concentration of both acids.

Answer:
In the presence of methyl orange, will happen these reactions:

$\ce{H2SO4 + 2 NaOH -> Na2SO4 + 2H2O}$
$\ce{H3PO4 + NaOH -> NaH2PO4 + H2O}$

And there are spent $\mathrm{37.2~mL~NaOH}$.
In the presence of phenolphthalein, the following reactions will happen:

$\ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O}$
$\ce{H3PO4 + 2NaOH ->  Na2HPO4 + 2H2O}$

And there are spent $\mathrm{49.6~mL~NaOH}$.
Question:
The procedure to find the concentration of the acids is obviously easy, but what I don't understand is the volumes of $\ce{NaOH}$ we should take to make the calculations.
The volume of $\ce{NaOH}$ needed for $\ce{H3PO4}$ is $\mathrm{49.6-37.2=12.4~mL}$.
The volume of $\ce{NaOH}$ in presence of methyl orange, needed for $\ce{H2SO4}$ is $\mathrm{37.2-12.4=24.8~mL}$, and in the presence of phenolphthalein is $\mathrm{49.6-24.8=24.8~mL}$.
There are also more similar problems with different compounds, but I totally don't understand the logic how these volumes are calculated! How do you know which one to deduct?
If you also know any website with these problems please send me the link, thanks.

Subtracting volumes, while 100% legitimate, is a peculiar and unobvious thing to do. Personally, I'd rather go with molar quantities. Let's say your 25 mL sample contains a moles of $\mathrm{H_3PO_4}$ and b moles of $\mathrm{H_2SO_4}$. How many moles of NaOH you would then need for the first titration? What about the second?

You say that it is obviously easy and that you have the answer. But you don't show the obviously easy way and you don't tell about the answer. Could you please add those things?

The formula for the calculation is C (H2SO4) x V (H2SO4) = C (NaOH) x V (NaOH)

For all its worth, here are the answers : C(H2SO4)= 0.05 M and C(H3PO4)=0.05 M

I for myself like to watch titration curves, because they often give hints about where to look. So the figures below show you the theoretical titration curves of your both acids$^{\ast 1,2}$ each and also titrated together with an aqueous solution of NaOH$^{\ast 3}$.

But ... how exactly can this be helpful?

You can see that at the "first" inflection point, which methyl orange would detect, both acids lost one to two protons. So how many equivalents of $\ce{NaoH}$ was needed?
At the "second" inflection point, which is detected by phenolphthalein, only phosphoric acid has another inflection point. And how many equivalents of $\ce{NaoH}$ was needed now?

With those thoughts in mind, the question can also be answered without my fancy titration curves.

 You need to solve this equation system to get both concentrations: \begin{align}3 c_B V_B &= 2 c_{S1} V_i+c_{S2} V_i\\4 c_B V_B &= 2 c_{S1} V_i+2 c_{S2} V_i\end{align} where $V_i$ is the initial volume of both acids ($\mathrm{25~mL}$).

$^{\ast 1}$ $\mathrm{pK_a}$ of $\ce{H2SO4}$: $-3,~1.9$ (according to the german wikipedia)
$^{\ast 2}$ $\mathrm{pK_a}$ of $\ce{H3PO4}$: $2.16,~7.21,~12.32$ (according to the german wikipedia)
$^{\ast 3}$ the increasing volume is neglected and both acids are assumend to have the same concentration

Thanks, but i won't have time for curves in the exam !

As I said, you don't need the curves.

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text:
Mechanism for oxidative cleavage of tertiary alcohols by NaOCl
I came across this today

An ester $\ce{C4H8O2}$ on reaction with $\ce{CH3MgCl} $ and hydrolysis gives an alcohol. The alcohol on reaction with $\ce{NaOCl}$ and $\ce{H3O+}$ gives acetic acid. Identify the structure of the ester.

But what I'm facing a problem with is the role of $\ce{NaOCl}$. My teacher said it can cleave and oxidize a tertiary alcohol, say tert-butyl alcohol, to acetic acid.
But I couldn't understand the mechanism. Any help will be appreciated.

There are not so many possibilities for your ester that yield an alcohol which could be oxidized to acetic acid. Which esters could you imagine from $\ce{C4H8O2}$?

My problem is not with the ester...but how will NaOCl oxidize the the alcohol obtained...the mechanism..

I don't see any evidence in SciFinder that this is an actual transformation. Can you confirm with your teacher that a simple tertiary alcohol can be oxidized to an acid?

I actually think you are misinterpreting the problem, but I'll wait for confirmation about the tertiary alcohol oxidation.

@jerepierre Maybe i misinterpreted what my teacher said.But it will take atleast a week before i can confirm with him.But what why did you say i'm "misinterpreting" the problem?What do you think will be the solution?What is the particular ester C4H8O2 according to you?What will be NaOCl mechanism according to you?Maybe i'm having a conceptual problem...but its always good to discuss :-)

Reaction of an ester with a Grignard reagent gives two products: (1) a tertiary alcohol coming from the carbonyl carbon and two equivalents of the Grignard reagent and (2) an alcohol liberated from the ester. I believe it's this liberated alcohol that is reacting with bleach, since a tertiary alcohol is typically resistant to oxidation. However, the liberated alcohol might be primary, secondary, or tertiary depending on the identity of R2.

The information given says that after treatment with bleach, acetic acid is formed. This tells us two things: (1) that liberated alcohol is a primary alcohol, since the oxidation goes all the way to an acid and (2) that liberated alcohol has two carbons, since acetic acid has two carbons. That implies that the liberated alcohol is ethanol. The starting ester has four carbons total, which leaves two carbons for the carboxy fragment. The original ester was ethyl acetate.


text:
Mechanism of reaction between substituted aniline and potassium thiocyanate in presence of bromine and glacial acetic acid
Im currently doing this project where in we are synthesizing 2-amino-6-substituted benzothiazole by using p-substituted aniline with potassium thiocyanate in presence of bromine and glacial acetic acid.
My question is: how does the cyclization take place?
I know that the amino group of aniline is ortho and para directing so the phenyl thiocyanate is formed as the intermediate but then this intermediate reacts with bromine in presence of acetic acid to give 2-amino-6-substituted benzothiazole ... I do not get that step.
Please provide me with the mechanism as to how the cyclization takes place with those two reactants.

I think the mechanism is similar to the ones described in this paper (chapter 4.7.1): http://shodhganga.inflibnet.ac.in/bitstream/10603/9231/9/09_chapter%204.pdf. With the difference that your intermediate is brominated with $\ce{Br2/H+}$ instead of iodinated.


text:
Why does soap reduce surface tension of water
One explanation I read is that soap's long non-polar end will seek the air at the interface between water and air, and that the ionic end will form ionic-dipole bonds with the water. In the process of moving to the surface, the soap molecules force apart the water molecules, and hence the water molecules no longer have hydrogen bonds with each other, and the surface tension is weakened. However, wouldn't the ionic heads of the soap molecules just form stronger ionic-dipole bonds with the water, increasing inter molecular forces, and therefore increasing surface tension? What is the true explanation for why soap reduces surface tension of water?

Apoorv did not give the right explanation, because this would imply, that surface tension goes especially down, when micelles are formed, but the opposite is the case. Compare e.g.: first chapter of The Colloidal Domain by D.Fennell Evans and Håkan Wennerström.
First thing to note is that pure water has a higher surface tension than hydrocarbon chains which is due to the strong dipole dipole interaction.
If there is a very dilute solution of surfactant in water no micelles are formed but you have monomers "floating around". In contrast to ions in water they are not homogenously distributed, but they concentrate especially on the surface where their polar head is directed to the water and their hydrophobic chain is directed to the air. In the end you substituted water molecules with hydrocarbon chains which leads to a lower surface tension.
If you increase the concentration of surfactant, they start to form micelles in the water (and not on the surface). So after reaching this critical micelle concentration (called CMC) the surface tension stays quite constant. This constant behaviour is used experimentally to estimate the CMC.
See the figure below Equation 11:
http://infohost.nmt.edu/~jaltig/SurfaceTension.pdf
Please note that Apoorv described some kind of reversed micelle. The normal way is that they stick their polar head to the water and have their hydrocarbon chain inside the micelle.
In the picture the white group is the polar head:
https://en.wikipedia.org/wiki/Micelle#/media/File:Phospholipids_aqueous_solution_structures.svg

The polar end of the surfactant is attracted to the polar water molecule. When enough surfactant molecules attach to a water molecule, it gets covered in surfactant and forms a unit. These units have a weaker force of attraction because only the non polar head of the surfactant is exposed thereby lowering the surface tension.


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Calculating concentration of dissolved ions
A sample of $\mathrm{2.00~g}$ of iron(III) sulphate is dissolved in water to give $\mathrm{100~cm^3}$ of aqueous solution. What is the concentration of $\ce{SO4^2-}$ ions in $\mathrm{mol~dm^{-3}}$?
My attempt:
I found the number of moles which is $\mathrm{0.005~mol}$.
Then, 1 mol of iron(III) sulphate has 3 mol of $\ce{SO4^2-}$
So the concentration is $\mathrm{\frac{0.005}{0.1} \cdot 3 = 1.5 \cdot 10^{-1} ~mol~dm^{-3}}$
But the given answer is $\mathrm{1.5 \cdot 10^{-2}~mol~dm^{-3}}$. Why?

I have improved the formatting of your post using $\LaTeX$. For more information on how to do this yourself please see here and here. Please avoid using Latex in titles due to searching issues

Now it looks to me that you are right, and the given answer is wrong.

You're right and the given answer is wrong. Here's the math:
$$\left(1\;\mathrm{mol}\;\ce{Fe2(SO4)3}\over {399.88\;\mathrm{g}\;\ce{Fe2(SO4)3}}\right)\left({2.0\;\mathrm{g}\;\ce{Fe2(SO4)3}\over 100\;\mathrm{cm}^{3}}\right)\left({3\;\mathrm{mol}\;\ce{SO4^2-}\over 1\;\mathrm{mol}\;\ce{Fe2(SO4)3}}\right)\left({10\;\mathrm{cm}\over 1\;\mathrm{dm}}\right)^{3} =\ce{[SO4^2- ]} $$
$$\ce{[SO4^2- ]}= 1.5\cdot10^{-1}\;\mathrm{mol}\;\mathrm{dm}^{-3}$$


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Is water stream deflecting in an electric field really based on water's polarity?
A common highschool chemistry example for the polarity of water is the experiment where stream of water is deflected under the influence of a charge (electric field).
Now I have come across some condradicting claims, namely this Youtube video from Veritasium (from around 1:50) and more importantly a publication by Ziaei-Moayyed et. al..[1]
These sources suggest that this phenomenon is not based on the polarity of water but charged impurity ions in the water stream. They also claim that the electric field could also only re-orient the molecules, but not deflect them as the electric field also has an opposite effect on the other side of the dipole. This has been bothering me and it is hard to find concluding evidence either way. What are your thoughts?
References:

Ziaei-Moayyed, M.; Goodman, E.; Williams, P. Electrical Deflection of Polar Liquid Streams: A Misunderstood Demonstration. J. Chem. Educ. 2000, 77 (11), 1520. DOI: 10.1021/ed077p1520.

At first I thought there was no way this could be hard to answer, but it seems that there is actually debate about this phenomenon to an extent. I would lean towards agreeing with that journal article because I don't think it would be published without being reasonably sound. I also just found another paper which reports distortion of the surface of a nonpolar liquid using a Van de Graaf generator, which seems quite surprising. Read about that here.

Very interesting. Seems to be caused by induced polarization of the nonpolar molecules. Vertical offsetting requires stronger attractive force to overcome gravity, cohesion and rejecting coulomb forces (?), thus the Van de Graaf generator was needed. So is the distance between dipole sides great enough to cause noticeable acceleration due to coulomb's law?

Since impurities are charge neutral in sum (so much negative charge in ions as positive), how would electric field deflect them ?

If electric field cannot deflect electric dipoles, magnetic field cannot deflect magnetic dipoles (same force laws, only letter is changed) - which is off course not true..

An electrostatic field will attract both conducting and nonconducting materials even though those materials have no initial charge through electrostatic induction. That is why you can pick up lint or aluminum foil with a charged comb or balloon. See this demo with a neutral aluminum can.

Good point. The video was about inducing charge to a conductive material (aluminum) but polarization works for nonconducting materials as well. Why then the hexane stream is not deflected in  this video? Is it because the attractive force is simply too weak to notice?

Exactly! Water and alcohols are polar and are, but given sufficient voltage, even non-polar hexane would exhibit some attraction due to electrostatic induction. The degree of polarity corresponds roughly to the dielectric constant: hexane=~2; isopropanol=~20; water=~80 (http://macro.lsu.edu/howto/solvents/Dielectric%20Constant%20.htm). BTW, thanks for listing that video!

The polarity of water?
Rory Geoghegan
The deflection of a stream of water by an electric charge is often cited as a demonstration of the polarity of water molecules. The idea has even found its way into science text books.
What’s wrong with the explanation?
Water molecules are polar; this is true. However it is also true that they electrically neutral and that they are very small. If a negatively charged rod is held 10 cm away from a water molecule then the attractive force on the hydrogen side and the repulsive force on the oxygen side are, for all practical purposes, equal and opposite.
Easily disproved
Here are some simple demonstrations that clearly show that the ‘polarity argument’ is unsatisfactory, the first being the most obvious.

No deflection occurs if the water becomes a stream of separate drops, unless the charged rod is sufficiently close to the point where the stream becomes discontinuous.
(This refutes the polarity argument.)

Get a 500 ml plastic drinks bottle. Using an awl or nail make a hole (about 1 mm in diameter) in the side of the bottle a few centimetres from the bottom. Place the bottle on a large insulated block (e.g. styrofoam container) near a sink. Fill the bottle with water and arrange it so that a stream of water flows into the sink. If a charged rod is brought near the stream it will not show a continuous deflection. (It may show a slight deflection initially but it will not last.)

Clamp a salt cellar on a retort stand so that the salt can flow out. Bring a charged rod near the salt stream. It does not deflect. (The charge separation in NaCl is greater than that of water; so are the charges. NaCl is about five times more polar than water but it is not deflected.)

(Don’t try this one.) A continuous stream of mercury would be deflected by a charged rod even though mercury atoms are not polar.

A stream of iron filings is not deflected by a charged rod.
These results cannot be explained by the polarity argument.

The right answer
The right answer is perfectly simple. Continuous conductors may be deflected by an electric charge. Water is a conductor and so if, for example, a negatively charged rod is brought near an unbroken stream of water coming from a tap then a positive charge can be induced on the part of the stream near the charged rod. The force of attraction between the two will deflect the stream. This cannot happen if the stream is broken into separate drops.
Rory Geoghegan

Ordinary water isn't that strong a conductor. So I doubt that this explanation (roughly the equivalent of the magnetohydrodynamic effect of a moving conductor in a magnetic field) works. I suspect that the effect is the same as would happen if non-conducting polystyrene beads were dropped through an electric field. We know they deflect but polarity has nothing to do with it. A key part, BTW, is that the field is not uniform.

G. W. Francis "Electrostatic Experiments" 1844 showed that charge separation in water can be created by electrostatic induction (experiment 153).
William Thomson (Lord Kelvin) later demonstrated that charge separation in water occurs even in the usual electric fields at the earth's surface and developed a continuous electric field measuring device using that knowledge.  He then developed the water-dropper electrostatic high-voltage generator using the same concept.  Others have since shown experimentally that the amount of charge separation in water is proportional to the electric field strength.
I believe that a stream of water is deflected because the electric field induces macroscopic charge separation in the water and not because water has polarisable molecules that have reoriented.


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Accurate textbook about synthesis of (organophosphorus) chemical warfare agents
Lately I've been interested in the chemistry of nerve agents due to reading Patrick's Introduction to Medicinal Chemistry in which they were only briefly discussed. I bought the Handbook of Toxicology of Chemical Warfare Agents by Ramesh C. Gupta which appeared to me as a comprehensive work discussing the physical chemistry, biological action, history and legislation of a whole range of chemical agents (in particularly organophosphorus agents).
However it lacks experimental information (i.e. synthesis). I found a book that detailed (to the somewhat questionable extreme) on this information, having the rather obscure title The Preparatory Manual of Chemical Warfare Agents (it's a Google preview link, quite handy to see if a book is worth the money). Now since I'm not an expert in organophosphorus synthesis (else I wouldn't be interested in it, right?) I cannot fully appreciate the level of accurateness. And while you may argue that, since it's only of theoretical use to me, the level of accurateness is unimportant, I wouldn't really feel satisfied learning about something from a textbook that finally turns out to be based on completely wrong information. Because then what knowledge have I gained?
Now for the question: the book raised my suspicion because the author seems not to be related to any serious research unit or university for that matter. The poor lay-out, language, and somewhat too practical advice given here and there, screams Uncle Fester type bull shit book. So, is this book based on legitimate research? And if not, where can I read about (organophosphorus) chemical warfare agents synthesis from a reliable (preferentially written by well known and respected chemists) source, that goes into enough detail on a chemical level (and does not necessarily contain Uncle Fester type practical information in which I'm obviously not very interested)?

The Ledgard book looks like a practical manual for terrorists. I browsed through some of the procedures and they look solid, they are taken from published patents. Of course some of the little details that separate the experienced chemist from the dead chemist are left to be discovered. Personally I would enjoy reading this book, it's 10 times more scary than any Stephen King novel.


text:
Does alcohol hydrogen in the enol tautomer of a β-dicarbonyl have sp² character?
My reasoning for the somewhat implausible "$\mathrm{sp^2}$" character of the hydrogen between these two oxygens on the enol structure of the right comes from recognizing that if it were, then we would have 6 $\mathrm{sp^2}$ atoms with 6 π-electrons, so maybe that extra bit of instability from having a higher orbital on hydrogen could be sort of taken care of by being stable like an aromatic compound and forming a hydrogen bond.
So how could I go about figuring out if this is true, either with MO theory or something else?

(source: ucdavis.edu)

So are you asking whether the hydrogen $2p$ orbitals are involved in bonding?

Yes, I think so, or if they are occupied ever, for instance would I expect this compound to experience a chemical shift when put into a magnetic field similar to an aromatic compound.

Not a chance. 2p are just too far away. Look at the energy levels diagram.

As @IvanNeretin points out, the hydrogen is s-hybridized like normal.  This $\ce{O..H..O}$ structure is just another example of a 3-center-2-electron bond involving hydrogen.  Diborane ($\ce{B2H6}$) is another example.  See here, and links therein, for more information on such bonds involving hydrogen.

In reference to the possibility of downfield NMR shifts, the C-13 and H-1 spectra from SDBS show no signs of aromaticity.

Wait a minute. Diborane is electron-deficient, but this thing is not. See, we had a normal 2-electron O-H bond, then another O went in with its lone pair (2 more electrons). If anything, this should be a 3-center-4-electron construction.

@IvanNeretin Yes, I agree, I should have written "3-center-4-electron" in my earlier comment.  Thanks for catching that.

Yes, it has 3c-4e character, but it is limited, as here's asymmetric h-bond - one of contributors much more dominating


text:
carbon electrodes have potential difference even with no electrical input?
What is happening in this photo? I set up a simple electrolytic cell today of copper (II) sulfate solution with carbon electrodes. It was connected to a powerpack (set at ~8.0V), and a variable resistor with the resistance cranked up to its maximum as in the image below. You can also see the ammeter and the voltmeter in the photo.
When the powerpack was switched off, the current fell immediately to zero, as expected, but the voltmeter still read a non-zero number (initially 0.80V, then later seemed to decrease to 0.18 V). I turned the electrodes upside-down so that the clean ends were in solution (rather than the end covered in copper), but reading still read the same. Removed the electrodes to a beaker of water (but not deionised water), and it still read 0.18V or close to.
Tried the exact same thing with pure copper electrodes, but it read 0.00V immediately upon turning off the power supply.
Tried completely taking the leads out of the power supply (which was off anyway), and the carbon rods still read a non-zero value even with the circuit obviously incomplete.
What's happening here? Copper electrodes did not show the same effect. Is there something about carbon electrodes in particular that can make them act as a capacitor? The current stayed at 0.00A the whole time. This effect was NOT seen when the electrodes were lifted out of the solution/water.

Edit, to respond to the answer below:

This would be the most interesting answer.
I did consider initially this, so first I tried up-ending the
electrodes so the clean ends were in solution, but the reading
remained the same. Second, it showed no such behaviour with the pure
copper electrodes.
I think it's unlikely to be a case of the voltmeter left floating.
Surely if that had been it, regardless of which electrodes (copper or
carbon) were used, random static charges would've been present? Plus,
the phenomenon was reproduced when I swapped out different electrodes
and then swapped back in the funny-acting ones, so the behaviour
wasn't exactly random.

However, I did forget to mention, that I also tested this with a
second clean set of carbon electrodes and got 0.00V! Support for 2nd
explanation (but if so, how does that work exactly?). I still would've
liked it to have been the 1st explanation.
I also note that I left the apparatus overnight, and by next morning
the reading was 0.02V. I've unfortunately have had to clear the
materials away now, but if I get a moment I may try replicating the
result.

Some possibilities:

Porous carbon electrodes do act as a capacitor, the basis for supercapacitors (aka ultracapacitors).
You are electrolytically depositing copper on the negative electrode when you run current through the solution, even though it might not be enough to see. This creates an electrochemical cell.
Simply disconnecting the power supply leads will leave a high-impedance voltmeter floating, so it might be detecting random static charges. You might need to bypass the voltmeter with a high resistance (10 or 20 megohms) to reduce static sensitivity.

@Colm Prunty, though as you say answer 3. is not likely because of the control with copper electrodes, I'm inclined to think that impurities in the carbon electrodes, whether from initial fabrication or from electrolytically deposited copper inside a porous electrode, might be responsible for the EMF.  What is the purity of the electrodes? You might experiment with high-purity graphite from http://www.baycarbon.com/products.htm, http://www.coidan.com/electronics.htm or other source (but I don't know the cost, which might be prohibitive). If you do, handle them with fresh polythene gloves.

Actually I meant point 2 as the answer. Number 3 is not likely not only because of the control with copper electrodes, but from my edit, the result was not replicated with fresh carbon electrodes. It may be you are right about the impurities such as the electrolytically deposited copper. Don't know the purity of graphite, but it probably isn't great.


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How to name tricyclic hydrocarbons?
Recently, I read about the nomenclature of bicyclic compounds, by mentioning the number of carbons between two bridgeheads. But, then I came across the structure of a weird type of compound, whose IUPAC name is tricyclo-[2.2.1.02,6]-heptane. As the name suggests it must be a polycyclic compound, having three alicycles. 2.2.1 suggests that there are 2, 2 and 1 carbons between the bridgeheads. But what about the next part? What is the actual rule for naming tricyclo compounds?

The main ring of a polycyclic hydrocarbon ring system is selected so as to include as many skeletal atoms of the structure as possible. The main bridge is the bridge that includes as many of the atoms as possible that are not included in the main ring. Bridges other than the main bridge are called secondary bridges. Thus, tricyclic hydrocarbons consist of the main ring, the main bridge, and one secondary bridge.
The nomenclature of tricyclic hydrocarbons is based on the nomenclature of bicyclic systems. The additional steps for the naming of tricyclic hydrocarbons are described in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) as follows.

P-23.2.5 Naming and numbering tricyclic alicyclic hydrocarbons
P-23.2.5.1 Tricyclic hydrocarbons having an independent secondary bridge are named on the basis of a bicyclic system (…). Rings not described by the bicyclic system are defined by citing the number of atoms in the independent secondary bridge as an arabic number. The locants of the two attachment points of the independent secondary bridge to the main ring are cited as a pair of superscript arabic numbers (lower number is cited first) separated by a comma.
The name of the tricyclic system is then constructed by citing:
(a) the prefix ‘tricyclo’, in place of ‘bicyclo’, indicating the presence of three rings in the polyalicyclic system;
(b) numbers indicating the bridge lengths, starting with the two branches of the main ring (…), followed by the main bridge (…), and the secondary bridge (with superscript locants separated by commas indicating its points of attachment to the main ring), all separated by full stops and placed in brackets,
(c) the name of the acyclic hydrocarbon having the same total number of skeletal atoms.

For example tricyclo[2.2.1.02,6]heptane

(The main ring is shown in red, the main bridge is shown in black, and the secondary bridge is shown in blue.)

But then, if the numbering of the secondary bridge is done on the basis of its attachment to the main ring, why isn't it 1,4?

@Aneek As you can see from the structure diagram, the secondary bridge (length: 0; shown in blue) is attached to the atoms 2 and 6. This is not to be confused with the main bridge (length: 1; shown in black), which is attached to the atoms 1 and 4.


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Experimental Validation of Schrodinger's Electron Cloud theory
I am currently doing a report for school on the electron cloud atomic structure theory. One of the major points on my report is the experiments performed by the scientists (Schrodinger and Heisenberg) to validate their hypothesis.
I cannot find any evidence of any experiments performed by them to validate their claim. Did they perform any experiments or was their theory purely based on mathematics? (There seems to be a lot of that surrounding this topic)
Thanks in advance, Sam.

Schrodinger & Heisenberg were theoreticians, they didn't do experiments, but a lot of experimental data were obtained already by experimentalists that required new theory. There are myriads sources on that matter. And of course, myriads experiments were done after the invention of quantum mechanics as well to test it validity.

Sorry to have to ask but could you link some of the sources please? Thanks


text:
Can I use a Liebig condenser for condensing sulfuric acid?
I have a Liebig condenser but I read that the air cooled condenser is preferred for high temperature boiling points. My question is whether I can still use my Liebig with minimal risk... Should I just run it without coolant or even with warm water? There shouldn't be pressure build up if I just run it dry. If I used warm water though, I still have a huge temperature differential between the jacket and sulfuric acid vapour, not so much with just running it without coolant since there will be a gradient between the air temp from inner tube to outer tube.


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Iron(II) oxide on stainless steel?
I noticed after some use that a stainless steel spoon that I have started forming $\ce{FeO}$. Stainless steel is not supposed to rust like this. I don't know of any black iron hydroxides, chromium oxides, or chromium hydroxides so that lead me to deduce that it is Iron(II) Oxide.
But why would stainless steel oxidize? I mean it does not make any sense since I was told that stainless steel does not oxidize.
And how can I get rid of the oxide? Putting the spoon in phosphoric acid perhaps since phosphoric acid is a really good rust remover?

But the spoon is only weakly magnetic and it has this black oxide. It even says "Stainless steel".  Water must have done this because it wasn't there before and now it is and it never happened from exposure to the air. And if the chromium oxidized I would see a green oxide on my spoon instead of black. I know it can't be from carbon because it would form carbon dioxide. I don't know of any black nickel oxides and nickel isn't likely to oxidize in the first place. So it has to be the iron that oxidized.

I'd really like to see a photo before drawing any conclusions.

Iron(ii) oxide is unstable below about 560 or 570°C, disproportionating to form the mixed oxide Fe3O4, and thus is not expected to form under ambient cobditions.  If your black rust is an iron oxide at all, a big if, then I would bet on Fe3O4.

It was probably a poor grade of stainless steel. The chromium oxidizes before the iron therefore you obtain a thin layer of protection. In poor grades of stainless steel, I have seen rust form over a period of time or during exposure to non ideal conditions such as high humidity. This is probably because the manufacturer added the minimum chromium required to meet standards, and so you get a smaller ensemble of chromium oxide and more iron oxide.
You can treat the metal with phosphoric acid to remove the rust. You can also treat the stainless steel with concentrated sulfuric acid to produce a phosphate layer for additional protection with minimal loss of material (they do this in industry)


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Does H2O2 break down into H2O + O or H2O + O2?
In hydroponics, Hydrogen Peroxide is used to sanitise nutrient solution from pathogens. Allegedly it does not take sides, oxidising aerobic as well as anaerobic, including damaging the roots, so I was told. I believe this argument, since O as a free radical is highly reactive. I would appreciate your input on this. However, this isn't the main question.
The main question is that manufactures also claim that it will provide DO, dissolved oxygen, to the solution. I find this hard to believe. Reason being, DO refers to dissolved $\ce{O2}$ (Dioxide) not Monoxide. When $\ce{H2O2}$ breaks down it breaks down to $\ce{H2O + O}$ (free radical), right?
Can you please put some clarity into this?

Whenever the free radicals have nothing to react with, they (being highly reactive) quickly recombine with each other to produce $\mathrm O_2$. So both claims are true.
And by "quickly", I mean really quickly.

Also, it is extremely likely that your nutrient solution contains metal ions that will readily act as catalysts to decompose the peroxide you put in the solution.  Even a small quantity of iron ions will do the job, for instance.  It's likely that by the time you use the solution, the peroxide has already largely decomposed.

$\ce{H2O2}$ -> $\ce{H2O +1/2O2}$
This might make it easier to understand.
$\ce{H2O2}$ will homolytically cleave for form two $\ce{.OH}$ radicals. Radicals are very reactive and will start a chain reaction, but ultimately you will end up with water and oxygen products from $\ce{H2O2}$.
(Why does it form two hydroxy radicals? because the O-O single bond is weak and unstable - see MO theory)
How do we know this? Look at a bottle of $\ce{H2O2}$ that has been left out of the fridge for too long. There will be a slight pressure build up of $\ce{O2}$ gas and the % concentration $\ce{H2O2}$ will be down due to formation of $\ce{H2O}$and loss of $\ce{H2O2}$.

For sake of completion,
You can alternatively balance the equation thus:
$\ce{2H2O2 -> 2H2O + O2}$

Hydrogen peroxide is unstable under standard conditions and breaks down into $\ce{H2O + O2}$. Although you are not left with a free radical, diatomic oxygen is still an extremely powerful oxidizing agent, which is the same reason rust forms on metals. So even though it is not a free radical, it still has the ability to oxidize your organic compounds.

If it would be that simple, hydrogen peroxide would have no advantage over $\mathrm O_2$, which we already have in the air; then why would anyone use peroxide at all?
Also, $\mathrm O_2$ is a radical, though that is hardly relevant for the present discussion.

But where did you find the O2? I see 2 atoms Hydrogen and 2 atoms of Oxygen. In your example there would need to be 3 atoms of Oxygen. Do you understand my confusion? Thanks


text:
How to inhibit Mn2+ autocatalysis in the oxalate-permanganate reaction?
It is well known that permanganate in acidic solution can oxidize oxalate ions (or oxalic acid). The reaction produces $\ce{Mn^2+}$, which catalyzes this very reaction.
$$\ce{2MnO4^- + 16H^+ + 5C2O4^2- -> 2Mn^2+ + 8H2O + 10CO2}$$
As a part of a demonstration on chemical kinetics, I want to inhibit this autocatalysis somehow. (I intend to perform three reactions: one with normal autocatalysis, one with $\ce{Mn^2+}$ added, which will run faster, and one preventing the $\ce{Mn^2+}$ from catalyzing the reaction.)
At first I thought I might use EDTA to complex the manganese ions. Alas, it didn't help. Maybe the environment is too acidic. (The reaction was actually faster with EDTA, I don't know why.)
Perhaps I should try a different chelating agent? Perhaps I could precipitate the $\ce{Mn^2+}$ ions? What do you suggest?

At first the reaction is slow. But when we heat up, the reaction gets catalyzed due to formation of $\ce{Mn^2+}$ ions. It happens due to the fact that $\ce{Mn^2+}$ gets oxidised to $\ce{Mn^3+}$ ions. (Source).
$$\ce{4Mn^2+ + MnO4- + 8H+ -> 5Mn^3+ + 4H2O}$$
So, we have to inhibit the formation of $\ce{Mn^3+}$ ions so as to inhibit the autocatalysis reaction. For this,  we have to use  phosphonate and carboxylate based chelating agent that solubilize (hydr)oxide-bound $\ce{Mn(III)}$ . The following is some relevant information from the site:

phosphonate-based chelators that solubilize (hydr)oxide-bound
  Mn(III) via ligand-promoted and/or reductive dissolution at
  circum-neutral pHs.
pyrophosphoric acid (PP), methylenediphosphonic acid (MDP), and
  phosphonoacetic acid (PAA), ligand-promoted dissolution is
  predominant: from pH 6−8, initial dissolution rates and plateau
  concentrations for dissolved Mn(III) decrease in the order PP > MDP >
  PAA, and at pH 5, MDP reacts equally well (with birnessite) or more
  efficiently (with manganite) than PP, and PAA remains the least
  reactive chelator. For manganite reacting with an excess concentration
  of aminophosphonate/carboxylate-based chelators, the
  aminophosphonate-containing iminodimethylenephosphonic acid and
  glyphosate yield appreciable amounts of dissolved Mn(III), but the
  aminocarboxylate-based methyliminodiacetic acid yields solely
  dissolved Mn(II) via Mn(III) reduction.


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Why and when are bonds broken
I am new to chemistry (I have only done a GCSE in it) so forgive me if I am missing something obvious. But I am trying to very roughly simulate some chemical reactions. My background is in Artificial Life and Adaptive Systems rather than computational chemistry or chemistry in general. My stumbling block is that I am unsure exactly why bonds break and when they break.
Here are the things I am confused about:
Is the breaking of bonds by using energy a prerequisite for new bonds to be made with the involved atoms? If so, is this because the more energy there is, the more the atoms will move/vibrate, eventually pulling the bond apart? Otherwise, can bonds be broken by a new bond sort of pulling it apart in some way?
Is the strength/length of the bond what determines the amount of energy needed to break it? I'm also unsure of what the strength or length of a bond is based on. Am I correct in thinking that bond order is about the number of shared electrons in a covalent bond? But what about ionic bonds? Is it the amount of electrons "taken" by the more electronegative atom?
I know that there are multiple questions here. But they all answer the bigger question. Why and when do bonds break exactly? I have had a hard time trying to figure this out by searching on the internet alone. Often I just get pages explaining the energy requirements of a reaction. Thanks in advance, any help would be appreciated.

Is the breaking of bonds by using energy a prerequisite for new bonds to be made with the involved atoms?
A: No, it is possible for a bond to form between two species without first breaking a bond. But it is more common for reactions to occur with a tradeoff between breaking old bonds and forming new ones.
If so, is this because the more energy there is, the more the atoms will move/vibrate, eventually pulling the bond apart?
A: Adding energy in the form of heat can lead to breaking of bonds, which in turn can lead to formation of new bonds.
Otherwise, can bonds be broken by a new bond sort of pulling it apart in some way?
A: You can have a "displacement" reaction where one atom replaces another in a structure: the bond to the new atom forms at the same time as the other bond is broken. That process has a lower energy barrier between reactants and products, relative to breaking an existing bond first and then forming a new bond later.
Is the strength/length of the bond what determines the amount of energy needed to break it? I'm also unsure of what the strength or length of a bond is based on.
A: The strength of a bond is the same as the amount of energy needed to break it (in the absence of other reactants). Bond length tends to correlate inversely with bond strength, as for example in going from a single to a double to a triple bond between carbon atoms. Bond strength depends on a number of factors--bond strength tends to be greater between two atoms of about the same size. Bonds between two electronegative atoms like oxygen or chlorine are weaker than a bond between two carbon atoms.
Am I correct in thinking that bond order is about the number of shared electrons in a covalent bond? But what about ionic bonds? Is it the amount of electrons "taken" by the more electronegative atom?
A: You are correct here in your understanding of bond order. But the term "bond order" does not apply to so-called ionic bonds -- for example in sodium chloride -- where an electron is completely donated from sodium to chlorine and does not participate in keeping the atoms together. Rather, it is the electrostatic attraction between the sodium and chloride ions that keep them together.

Actually, I find it somewhat hard to come up with an example where a bond would form without first breaking a bond. See, here is your newly formed molecule, now let's point at any atom in it and ask: where did this atom come from? It must have originated from another molecule, or maybe bulk metal, or ionic crystal - in short, something held together by bonds, which had to be broken. Atoms do not float around all by themselves. Well, noble gases do, and some of them also form compounds ($\ce{XeF_2}$, etc.) But then there was that bond in $\ce{F_2}$, which we broke to make this...

So let's ask the question, what two chemicals, each of which can be stored in a bottle in pure form, will combine together via a process of only bond formation and not breaking. Like boron trifluoride and trimethylamine.

Boron trifluoride has bond order 4/3, for each fluorine is linked to boron with 1 regular sigma bond and 1/3 dative bond. In the resulting complex, the B-F bond order is 1, as illustrated by pyramidal geometry and increased bond length. Hence that 1/3 is broken. Is we disregard bond orders altogether, then the examples are plenty. Any Diels-Alder addition would do. Also, trimerization of acetylene to benzene, why not.

So, Ivan, what you are holding out for is a reaction between two atomic species that results in bond formation. For a reaction between two molecules, the formation of a new bond will cause a change in the existing bonds, no matter how slight.

Exactly, and that's the problem with molecules. As for atomic species, there are not too much of them around. Well, we might imagine how a lone hydrogen atom, lost in the immense vastness of outer space, meets another such atom. Not that it does not happen in nature, but the example looks kind of far-fetched.


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O-N-O bond angles in oxides of nitrogen

If we look at the bond angles of O-N-O in figure (c), (D) and (F), there not too much change in bond angle. Even though they are notable.
So my question is,
Why do we observe very small change in angles?

This seems to be two entirely separate questions so you should ask them as so.

A simplistic explanation would be to say, that in all relevant oxides nitrogen exists in $sp^2$ hybridization state. The reason for it is that it have to form n+1 bonds with n atoms, forcing it to form a double bond, that requires it to adopt said hybridization. || 2. When one needs to quickly estimate geometry of a molecule, quantum chemistry calculation may help. For precise experimental measurements diffraction methods are suited, say X-ray diffraction for solid state and gas electron and gas neutron diffraction for gas phase.


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How to predict the effusion/diffusion for a mixture of gases?
Suppose a a container contains a gaseous mixture of methane and hydrogen. The mixture effuses out through a hole in the container. Now hydrogen is lighter gas so it must effuse faster and all of hydrogen should come out of the flask earlier than methane (before all methane effuses out) and a stage should come when there is only methane left in the container.
Is this how it happens?

Sort of, but not quite.
If our hole is large (say, as if pierced by a toothpick), then the gases will go out in pretty much the same composition that is found inside, so their ratio in the mixture will stay nearly the same till the very end.
If the hole is small, about the size of free path of a molecule in a gas, then it's a different story. BTW, these conditions are called Knudsen flow. Methane weighs 16, and helium weighs 4 (I know you've said "hydrogen", but let's assume it was helium, just to make the calculations easier). So helium particles are 4 times lighter than those of methane, hence they fly 2 times faster. So there would be two times as many of them in the outgoing stream - that is, if the mixture inside was 1:1. See, it's not like all helium flows outside, while methane molecules stand politely in the line, saying "After you".
As the story progresses (which is not going to be fast with just one tiny hole), the mixture inside gets enriched with methane. So does the outgoing flow, but still it remains richer in helium relative to the mixture inside, and hence continues to shift its composition in the same direction. Yet still, pure methane is a limit which is never actually attained.
That being said, pressing gas mixtures through porous membranes is a valid way to separate the components by molecular mass regardless of their chemical properties, and as such, was used for isotope enrichment.

@ShashankKumar There wouldn't be a situation with pure methane as concentration of lighter gas would decrease limiting the outflow


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Iron (II) dioxygen reaction mechanism?
Ok so I'm confused about the mechanism for this process:

Now the idea here is that iron first becomes oxidised to Fe(III) by placing an electron in the antibonding orbital of dioxygen. This is followed by the addition of 2 electrons from an external source. This causes the splitting of the O-O bond resulting in the formation of a hydroxide ion. In addition, iron loses another electron to oxygen forming the Iron(IV) oxide species shown.
Now what I do Not understand is the overall charges on the species at the end. I feel as if the original O-Fe(III) bond was considered purely ionic hence the lack of a formal charge on oxygen, yet at the end it seems as if the O=Fe(IV) bond is covalent, otherwise there would be a negative 2 charge on oxygen!
Is there a switch between ionic and covalent formalisms in the middle of this process?
If anyone could help explain the charges with perhaps a curly arrow mechanism to indicate the numbers of electrons on each atom then that would be most appreciated!

Quick look into available articles made me think that the first stage is one-electron transfer forming superoxide anion. Since superoxide is extremely reactive, the further steps depend highly on environment, as it would react with first available candidate.


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How are partial molar Gibbs excess functions correctly defined?
I think I've found a mistake in my thermodynamics textbook, Chemical Thermodynamics for Process Simulation, but as thermodynamics is hard and I'm a relative novice, I wanted to check here.
The textbook, in talking about partial molar Gibbs excess functions of species in solutions, writes that we can express the partial molar Gibbs free energy of a species $\bar{g_i}$ in solution, as the sum of the partial molar Gibbs free energy in an ideal solution, $\bar{g_i}^\mathrm{id}$, and an excess function, $\bar{g_{i}}^\mathrm{ex}$:
$$\bar{g}_i = \bar{g}_i^\mathrm{id}+\bar{g}_i^\mathrm{ex}.$$
It doesn't specify whether we are thinking of evaluating $\bar{g}_i^\mathrm{id}$ at the system pressure or at a reference pressure, but immediately after it 'expands' the above expression into the form:
$$\bar{g}_i(T,P) = \Big(\bar{g}_i^\mathrm{pure}(T,P_0) + \mathcal{R}T\ln x_i\Big) + \left(\mathcal{R}T\ln \frac{f_i}{x_if_i^0}\right)$$
Clearly we're comparing $\bar{g}_i$ to the ideal value, $\bar{g}_i^\mathrm{id}$, at standard pressure $P_0$. However, they then go on to define the activity $a_i=f_i/f_i^0$ and the activity coefficient $\gamma_i = a_i/x_i$ and substitute these in to get
$$\bar{g}_i(T,P) = \bar{g}_i^\mathrm{pure}(T,P_0) + \mathcal{R}T\ln x_i  +  \mathcal{R}T\ln \gamma_i.$$
Now, I'm all fine up to here. The first part, $\bar{g}_i^\mathrm{pure}(T,P_0) + \mathcal{R}T\ln x_i $, is the ideal solution Gibbs free energy at standard pressure, and the final term, $\mathcal{R}T\ln \gamma_i,$ is a correction in two ways: It corrects for the non-ideality of the solution, and it corrects for the fact that we may be at a different pressure $P$.
But the book then says:

For an ideal mixture the excess part is equal to zero, since the activity coefficient is $\gamma_i=1$.

I think this is ridiculous: we can only make this conclusion if we have an ideal solution at $P=P_0$.
Is this an error in the textbook? Also, I can see two possible routes to an error: Either $\bar{g}_i^\mathrm{id}$ is actually to be evaluated at system pressure, in which case their last statement is correct but the formula they give is wrong, or else their definition of excess functions is the standard one (actual value minus ideal value at ref. pressure) but they made a mistake in concluding $\gamma_i=1$ for all ideal solutions, independent of the pressure.

Could you please provide the complete reference to the textbook you are using, i.e. authors and chapter/pages.

Hi Martin, it's Chemical Thermodynamics for Process Simulation by Jurgen Gmehling et al, p. 161-162.

The confusion is perhaps not uncommon. However, whichever book it is has jumped from discussing the free energy of pure materials, and their variation with various things (T, P, V, ...), to discussing mixtures.  In this case, all variation of the 'pure' substances is present in the ideal terms - that is because those variation exist without regard for any mixing terms.  What you are focused on here are the impacts of making a solution (or binary alloy, ...).
So, to make the mixture, you take a certain amount of substance A, and a certain amount of substance B, and mix them together. Before mixing, the total free energy of the system is $x_A G_{A} + x_B G_{B}$ where $G_{A,B}$ are the Gibbs free energies of those substances at whatever the conditions you specify before mixing. So, what does mixing do? Well, at the very least you get the increase in entropy, $RT(x_{A}\ln{x_{A}} +x_{B}\ln{x_{B}}$. If there is any interaction between A and B, you now have non-ideal terms which can be represented in any number of different ways. However, the point is that they are "all the rest that happens" during the mixing, and those things may be functions of concentration, temperature, pressure, ...
But, to come back to your question - no, this is not an error in the textbook (whichever it is). But, the did not make it clear that they were leaving behind the explicit dependencies of free energy on the various parameters - they figured you had that part down.
I would also note that, when one calculates a phase diagram (ultimately one hopes the book might get to that, and miscibility gaps and whatnot), you are free to choose the reference points or phases of A and B yourself - that choice makes no difference in the phase boundaries and tie lines. And, this comes back to just what $G_{A,B}$ means.

"But, the did not make it clear that they were leaving behind the explicit dependencies of free energy on the various parameters - they figured you had that part down." I'm a little confused by your meaning here. And to clarify my question, It seems clear to me that if you define $\gamma_i$ as a correction factor to correct  for both non-idealities and non-standard pressures, you can't claim $\gamma_i=1$ whenever non-idealities are present: you also need to be at the standard pressure the activity coefficient is correcting for :)


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Molecular Orbital Sequence
Two highly respected books: Atkins, Chemical Principles 5th ed. and Clayden Organic Chemistry 2nd ed. don't agree on the orbital arrangement of the $2p$ orbitals of $\ce{N2}$. Clayden on page 94 states that it is $\sigma_{(2p)}, \pi_{(2p)}, \pi^*_{(2p)}, \sigma^*_{(2p)}$ and Atkins on page 118 states that it is $\pi_{(2p)}, \sigma_{(2p)}, \pi^*_{(2p)}, \sigma^*_{(2p)}$.
I am a retired engineer with a consuming interest in how the brain works and am trying to give myself a good grounding in organic chemistry. This may be a trivial issue, but to a novice it hardly seems so. This kind of division within the chemistry community is very concerning.
Any ideas as to what's going on here?

Atkins is right. Clayden probably didn't want to go to deep into the matter. Refer to this question: http://chemistry.stackexchange.com/questions/14417/what-is-the-origin-of-the-differences-between-the-mo-schemes-of-o%E2%82%82-and-n%E2%82%82

Thanks for the clarification, Marco. I think you're being a little too generous towards Clayden though. If you're lucky enough to own both books and are studying by yourself such inconsistencies are massively distracting.


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Detecting fake gold
I am thinking of buying gold. If necessary, one gram at a time.
Is there a surefire way of checking if the alleged gold is real and of the stated caratage without a lab?

Buy from a trustworthy vendor.

I can't believe no one mentioned Archimedes.

There are lots of test for detection of real gold.

Visual inspection: Inspect the piece for official markings. And look for noticeable discoloration. If the gold seems to be wearing off and showing a different metal beneath it, you probably have a piece that is only gold plated.
Bite test: Bite down on your gold with moderate pressure. Examine your gold for any markings. In theory, real gold will show indents from your teeth; deeper markings indicate purer gold. This is actually not a recommended test, as you can damage your teeth.
Magnet test: Hold a magnet up to the item. Gold is not a magnetic metal, so if it pulls towards, or sticks to the magnet, it’s fake.
Density test: Weigh your piece of gold. Fill a vial with water. It’s helpful if the vial has millimeter markings on the side, since that will make it easier for you to read the measurements for this test. Place your gold in the vial. Use the following formula to calculate density: Density = mass/volume displacement. A result close to 19 g/ml indicates either real gold, or a material with a density similar to gold.
Ceramic test: Find an unglazed ceramic plate to use. Drag your item across the surface. A black streak means your gold is not real, whereas a gold streak indicates your item is genuine.
Nitric acid test: Place your piece of gold in a small stainless steel container. Put a drop of nitric acid on your gold and watch for any resulting reaction to the acid. A green reaction indicates your item is either a base metal or gold plated. A gold reaction indicates your item is gold-plated brass. A milk-colored reaction would indicate gold-plated sterling silver. If there is no reaction, you mostly likely are dealing with real gold.

Click here for detailed steps.
Try to perform all or at least 4 steps to confirm (with ceramic test compulsory).

This is an excellent answer but a Google search away. Can a fake alloy fool some or all of these tests?

@JJ_Jason density test, ceramic test and nitric acid test cannot be fooled by a fake alloy, according to me.

Density is fooled with tungsten

@Mithoron But tungsten is much more harder than gold.

Color would be bigger problem, but it can be simply put inside bars.


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Good software to perform simulations on polymer (absorption spectra)?
Sorry guys,I'm totally new in this field,for personal interest I'd like to start some calculations on conducting polymers ,could someone suggest a good software to perform simulations on polymer (absorption spectra)?

I echo @Martin-マーチン. While there are many groups that do similar work, my group certainly runs these types of calculations - but your question  is currently too broad to answer well.

https://en.wikipedia.org/wiki/List_of_quantum_chemistry_and_solid-state_physics_software || For solid state check programs with PW-style basis sets. Beware, that calculating adsorption specra may be much harder than for common gaussian/gamess.

It depends very significantly on what you want to accomplish.
I'm not going to suggest software, since there are many programs that can (and are) used to predict optical absorption spectra of conjugated polymers.
First, you can either approximate the properties of the polymer by running several calculations on oligomers (e.g., dimer, trimer, tetramer, etc.) and extrapolating the polymer band gap. Because conjugated polymers are similar to the particle-in-a-box model, you can roughly plot the HOMO-LUMO gap or excitation energy vs. 1/(# of repeat units) and get an approximately linear fit.
If you want more information, you could instead run a calculation on a polymer in a periodic unit cell, which will give you the electronic structure of the infinite system (but no information on disorder).
As suggested in the comment above, you can use multiple programs and different computational approaches depending on:

What kind of polymer
What kind of information you want
What computational resources you have
How accurately you want the information

The two most common methods for predicting optical absorption energies are the ZINDO/S semiempirical method, and TDDFT. The latter is more accurate particularly for predicting absorption intensity from the oscillator strength of the excitation energies.

Yeah,thank you!Geoff. I'm not totally new..well I've worked already in research several years,I've used always CASSCF,CASPT2,TDDFT,and interface for using Molecular Dynamics. I've always used Gaussian,Molcas,Turbomole always with University Licence. Never work with polymers before...and now I was just wondering if someone can suggest some free software suitable for polymers. Any suggestion is welcome! even a website to download some .xyz geometries, because as fr protein there are PDB files, I was wondering if there is some similar database for polymers.Thank you!

I'm biased, since I'm the creator, by I suggest http://avogadro.cc/

For example, if you want to set up a unit cell for a polythiophene. But you can also build a monomer and copy/paste to grow the chain.


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Why is Ionization Energy definition restricted to gaseous atoms?
Here's the definition of ionization energy that I am learning at this level:

energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of uni positively charged gaseous ions.

I would like to know, why did they specify that the removal of electrons have to be at gaseous state and the ions that form are gaseous as well? Does it have to do with the amount of energy supplied that causes the elements to vapourize?

In fact, it is not restricted at all. You may compare the energies needed to remove one electron out of a solid; these are also meaningful, albeit in a different way, and known for a wide range of substances. But when you are talking about atoms, you want to measure atoms, and the only way to have an undisturbed lone atom is to put it in a gaseous state. And of course, the resulting ion should be in that state, too; after all, you can't make a solid out of positive ions alone, and if you want to add negative ions to counterbalance them, an ambiguity arises as to which ions should you use.

The reason why ionisation energy is measured as the energy required to remove 1 mole of electrons from 1 mole of substance is that it allows the ionisation energy of different substances to be compared. This is because the ionisation energy of a substance is going to be different in different states.Therefore you can't compare the ionisation energy of a substance in a solid state to the ionisation energy of another substance in the liquid state. For this reason, the ionisation energy of all substances are measured in the gaseous state. It is like the standard solubility of a substance is taken at 25 degrees. It doesn't make sense to compare of the solubility of one substance at 25 degrees and the solubility of another substance at 100 degrees.
Now you might ask why did they choose to measure the ionisation energy of substances in the gaseous phase, why not the solid or liquid phase? The reason why ionisation energy is measure in the gaseous phase is because in the gaseous phase, there is very little attraction between other particles which effects the ionisation energy. Therefore, if the ionisation energy was measured for a substance in the solid, since the particles are close together, the intermolecular forces will distort the value of the ionisation energy.


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How is Negative Temperature Hotter than Infinite Temperature?
Recently I have read articles stating that negative Kelvin has been achieved. I was a bit speculative at first as I though how can you get cold than 0 Kelvin, but after doing some research it makes sense on how this is possible.
However the one thing that I still not sure is that it stated that the temperature scale can't be thought as being linear but rather as a loop which goes in the following order, "+0 K, … , +300 K, … , +∞ K, −∞ K, … , −300 K, … , −0 K".
How can negative Kelvin be hotter than infinite temperature?

Related: Negative Kelvin Temperature

I'd rather not think of negative absolute temperature as real; true, in some sense it has been achieved, but this sense is very different from what we commonly use. Think of it as a metaphor of sorts.

The other answer given is correct, but for completeness, I will give the thermodynamic, rather than statistical mechanics, explanation.
Thermodynamics gives us that $$\frac{\partial E}{\partial S}=T$$
This means that  when entropy--as a function of energy--increases, the temperature must also increase.
On a macroscopic level, this makes perfect sense because you're never going to think about a system reaching a maximum temperature because energy and entropy are always going to increase together.
Of course, it must be remembered that this definition of temperature was being given long before people were thinking about quantum mechanics. Even more, they were only concerned with temperatures because they realized that they could use a temperature difference to get work out of a system. A good example of this is the Carnot heat engine. The reason I reference the Carnot heat engine is that it demonstrates that when temperature was being defined mathematically, which was around this time, they were concerned with heat flow. We know that heat flows from a hot body to a cold body, so this means, almost exclusively, that heat flows from high energy to low energy.
For instance, in a system with only two energy states containing $N$ particles, the state of maximum entropy is when $\frac N2$ particles are in both states 1 and 2. The state of maximum energy, however, is when all $N$ particles are in energy level 2.
That means, if we push beyond the state of maximum entropy, energy will increase as entropy decreases, hence, $\frac{\partial E}{\partial S} < 0$.
So, negative kelvin can be "hotter" than infinite temperature because our definition of temperature has nothing to do with "hotness" as we feel on the macroscopic level, it is merely an exchange between energy and entropy.
Furthermore, negative kelvin is hotter than infinite temperature, not because it would feel hotter, but because heat flows from the negative temperature object to the positive temperature object because we can demonstrate mathematically that the negative temperature object is in a higher energy state. It makes sense to think about it that way because 90% of chemistry is just looking for the low-energy state.
In regards to the issue of our temperature scale going in a loop, there is a solution to this using the thermodynamic beta, $$\beta=\frac{1}{kT}$$Consider this graph which shows that temperature as a function of energy is asymptotic and thus discontinuous, while $\beta$ is continuous as a function of energy.

Also worth pointing out is that $\beta$ has units of energy, and is thus more physically relevant than temperature because energy is fundamental to all matter, while temperature is basically nothing more than a mathematical construct to describe "hot" and "cold".

It is all about the definition of temperature itself.
All atoms of any compound with a temperature T have different energies associated with them, therefore temperature does not make sense on molecular levels. So when you look at the energy distribution of all atoms that form a compound, you will find its temperature on the spot of the energy E axis, that has the highest probability associated with it.
The probability for a specific energy value (or state) can be calculated with:
$$P \propto e^{\frac{-\Delta E}{KT}}$$
$K$ being the Boltzmann constant.
In positive temperatures, you will always find more atoms in a lower energy state. So your probability P for finding atoms in higher energy states will decrease because $\Delta E$ is becoming bigger and bigger as you go up. But when you can get some atoms to stay in a higher energy state, which they would usualy not populate, then your probability has increased for that higher energy state. And the only way to fit this into the equation is to make T negative.
When you now bring a second compound B of any positive temperature to the original compound A, which has negative temperature, then energy will always flow from compound A to compound B. Which means A is hoter, regardless of the temperature of B. This is because this unusual energy state is very unstable and can only be maintained by an "atom trap".
Reference: http://m.youtube.com/watch?v=yTeBUpR17Rw

It may be worth mentioning that the whole population inversion thing refers to some specific energy levels; it isn't (and can't be) applicable for all of them.
Also, a body with this kind of "negative temperature" (like laser medium) does not have to be hot in everyday sense, i.e. hot to the touch.

In the simplest terms,
At negative temperatures the further addition of heat (or energy) decreases the entropy instead of increasing it.
Hence heat flow occurs from lower (or rather more negative temperature) to higher (or less negative) temperature.
So we say that bodies at negative temperature are hotter than those at positive temperatures.


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Synthesis using Cationic Hetero-Cope Rearrangement
By searching for the reference of the synthesis below I arrived at an example of Cationic Hetero-Cope Rearrangement, but I can't access the paper. How does it work here, and what happens in the first step of the synthesis (the addition of iBuOH and TsOH at 80 degrees)?

I assume the "numbers" correspond to the order of the reaction steps.  The first step forms the beta-keto enol ether.  That and a look at the abstract  might make the rest of the sequence understandable. I don't see any Cope reaction involved in the formation of this product, just a Diels-Alder.

What is reduced by DIBAL-H?

After 1,4 addition of the vinyl grignard, the DIBAL-H reduces the remaining ketonic carbonyl.

Strange selectivity. Anyway, when does on OH group eliminate? When BuLi is added (one eq eliminates one OH group, the other deprotonates the second)?

I'm just guessing that it's when the molecule is heated in the last step.

Seems more reasonable.

I see you're a first year undergrad.  Are they covering stuff like this and trans-cyclooctene chirality in a class, or are you studying these things on your own?

I am studying on my own. I find the university pace not so demanding. By the way, this problem is from an organic chemistry competition (to put the numbers that correspond to the order of the reaction steps).

Impressive!  You ask good questions too.  If you stay with it you may one day find that you have a real gift for chemistry.  Good luck in your studies!

BTW Marko don't forget that self-answers are highly encouraged!

This would be the reaction sequence:


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Why the "Diverse Ion Effect" increases Ksp?
I've read in chemwiki about the diverse ion effect. Assuming I have an aqueous solution of $\ce{NaCl}$ and an associated $K_\mathrm{sp}$. If I add $\ce{KI}$ there is an associated $K_\mathrm{sp}'$ for $\ce{NaCl}$ where $K_\mathrm{sp}' > K_\mathrm{sp}$. This actually means that I can dissolve more $\ce{NaCl}$ than I could before. Why does this happen? Shoudln't the new solution actually be closer to saturation and $K_\mathrm{sp}' < K_\mathrm{sp}$?

To illustrate the "diverse ion effect", we will study how the solubility of $\ce{AgCl}$ changes in absence and presence of $\ce{NaNO3}$, given that the molar solubility product of $\ce{AgCl}$ is $1.76 \times 10^{-10}$.
The exact definition of the solubility constant is: $$K_{sp}= a(\ce{Ag+}).a(\ce{Cl-})=[\ce{Ag+}].\gamma(\ce{Ag+}).[\ce{Cl-}].\gamma(\ce{Cl-})$$
Where $a(\ce{Ag+})$ and  $a(\ce{Cl-})$ are the activities of these ions, and $\gamma(\ce{Ag+})$ and $\gamma(\ce{Cl-})$

In dilute solutions, i.e. absence of $\ce{NaNO3}$:  $$\gamma(\ce{Ag+})=\gamma(\ce{Cl-})=1$$
So, the solubility of $\ce{AgCl}$ is equal to $$[\ce{Ag+}]= \sqrt{K_{sp}}= 1.33 \times 10^{-5}$$
In presence of $\ce{NaNO3}$ of a concentration of $0.1 \mathrm{mol/kg}$, then
$$[\ce{Ag+}]= \sqrt{K_{sp}}/ \sqrt{(\gamma(\ce{Na+}).\gamma(\ce{NO3-})}
= 1.33 \times 10^{-5}/ (0.769)= 1.73 \times 10^{-5} $$
i.e. in the presence of a soluble salt, the molar solubility of a slightly soluble salt increases.