text:
Why would 1kW power supplies fail around Nitric acid (HNO3) gas?
I experienced a situation where some electrical power supplies (rated over 1kW) were exposed to nitric acid ($\mathrm{HNO_3}$) fumes at roughly 9ppm for half a day; these power supplies intermittently failed (perhaps 2 to 3 times per hour) in the presence of the fumes.  Replacing the power supplies restored proper functionality to the equipment they were powering.
At first, I thought that perhaps there was arcing inside the power supply; however, nitric acid doesn't seem to have a flash point.  I'm trying to come up with a reasonable explanation for why these particular components would fail.  Maybe the fumes increased atmospheric conductivity, which shorted across components?
If there isn't a compelling reason from a chemistry perspective, that's also a fair response.

Looking at the title, I can't help but think the more reasonable question would be "Why would they not?"

You're missing the reality that other equipment next to it did not fail.  The difference was actually air flow rates through the equipment in question.  The exact reason for failure is important to nail down, so we can prioritize what things to replace first

Nitric acid corrodes copper. Drop a penny into some nitric acid (under a hood!) and you'll see that your power supplies really don't stand much of a chance in that environment.
This happens with dilute nitric acid:
$$\ce{3 Cu + 8 HNO3 -> 3 Cu^{2+} + 2 NO + 4 H2O + 6 NO3^{−}}$$
When you open up one of the failed power supplies, do you see bluish crusty white stuff anywhere? That's nitric acid corrosion.
Addendum: Here are some quantitative corrosion rates:
Karl Hauffe and Roman Bender: Copper. In Corrosion Handbook. Wiley-VCH, 2008.
Addendum 2: Even though the concentration of nitric acid in the air is low, your power supplies are sucking a lot of air through them in 12 hours.
Let's suppose you have a power supply that dissipates 300 W of heat and you can tolerate a 10 degree Celsius temperature rise. Then according to this reference you need an airflow rate of about 53 cubic feet per minute.
That's 1.5 cubic meters of air per minute.
The density of air in a warmish (300 K) room is about 1.177 kilograms per cubic meter. So you're pulling
$$\frac{1.5~\mathrm{m^3}\ \text{air}}{1~\mathrm{min}} \frac{1.177~\mathrm{kg}\ \text{air}}{1~\mathrm{m^3}\ \text{air}} \frac{9~\mathrm{mg}\ \ce{HNO3}}{1~\mathrm{kg}\ \text{air}} = 15.9~\frac{\mathrm{mg}\ \ce{HNO3}}{1~\mathrm{min}}$$
which over 12 hours is
$$\frac{15.9~\mathrm{mg}\ \ce{HNO3}}{1~\mathrm{min}} \frac{60~\mathrm{min}}{1~\mathrm{hour}} (12~\mathrm{hours}) = 11.4~\mathrm{g}\ \ce{HNO3}$$
If all of that reacted with copper in the power supply, this is how much copper would be converted into corrosion:
$$11.4~\mathrm{g}\ \ce{HNO3}\frac{1~\mathrm{mol}\ \ce{HNO3}}{63.01~\mathrm{g}\ \ce{HNO3}}\frac{3~\mathrm{mol}\ \ce{Cu}}{8~\mathrm{mol}\ \ce{HNO3}}\frac{63.546~\mathrm{g}\ \ce{Cu}}{1~\mathrm{mol}\ \ce{Cu}} = \fbox{4 g Cu }$$
...that's a lot of copper to lose (a penny is 2.5 g!). Even if only 10% of all the $\ce{HNO3}$ that's being sucked through the power supply reacts, it's still 0.4 grams of copper, certainly enough to significantly corrode electrical contact surfaces.

There is a nontrivial difference between 9ppm and the concentrated solution / penny in your answer. I'm not suggesting that this is wrong, but is 9ppm enough to reasonably explain failure within an hour of first exposure?

All it has to do is spoil a contact surface somewhere. And it'll work on other metals, too.

I did a back-of-the-envelope calculation of the amount of nitric acid your power supply fans might pull over 12 hours and the amount of copper that would be corroded and added it to my answer.

Wow, now you're going to make me drop half my chemistry.se rep on a bonus... Good work, thank you :-)

This is a very amazingly thorough answer! I have updated your math formulas, as I noticed you are still using the deprecated \rm (see meta for more details). We also use mhchem for chemistry markup, which is easy to use. Another helpful tutorial can be found  here.

Indeed a remarkable answer!+1


text:
Why does water boil vegetables, but oil does not?
When we add vegetables to hot water (boiling), it boils the vegetables, but when we add vegetables to hot oil (boiling) it fries the vegetables. Why?

Related: Why does hot oil fry things, but hot water does not?

The maximum temperature for water boiling is around 100 Celsius and there is a great deal of water present.  Oil can reach greater temperature and the absence of water allows other reactions, such as the Maillard reaction, to occur.  That's why you never see browning in boiled food.

Brinnb is correct, but the other big difference between boiling and frying is that vegetables already have a lot of water in them. Boiling in water doesn't really change that because everything is already surrounded in water, but frying in oil is hot enough to boil the water inside the food, so in addition to the Maillard reaction, deep frying tends to dehydrate food. This is one of the reasons the outsides of french fries get crispy, and taken to the extreme, this effect is used to dehydrate ramen noodles in the production of instant noodles.

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How to calculate the heat capacity of a calorimeter?
I need to find the heat lost of an unknown metal dropped into a calorimeter with $70~\mathrm{g}$ $\ce{H2O}$.
The initial temperature and final temperature of the $70~\mathrm{g}$ $\ce{H2O}$ and the calorimeter are $21~^\circ\mathrm{C}$ and $34~^\circ\mathrm{C}$. I already know that the heat gained by the water is $3807.44~\mathrm{J}$.
The metal's starting temperature and mass are $100~^\circ\mathrm{C}$ and $180.45~\mathrm{g}$, but that didn't help me much as I don't know the $C_{sp}$ of the unknown metal.
What I don't know is how to find the capacity of the calorimeter, any thoughts?
I already tried a number of solutions that didn't work:

$C_{sp} \times \ce{H2O} = q(\mathrm{calorimeter})$
$q(\ce{H2O})/m(\ce{H2O}) \times \Delta T = q(\mathrm{calorimeter})$
$q(\ce{H2O}) = q(\mathrm{calorimeter})$
I knew the unknown metal was one of three metals (lead, aluminum, copper),     so I tried finding $q(\mathrm{metal})$ using the 3 heat specific heat capacities, but it didn't work out since the data provided were only approximations.

Since all you were given is the temperature and mass of the metal, the temperatures and mass of the water, I would say that you ignore the calorimeter and use the figures to calculate the heat capacity of the metal (or whatever they are asking for).

If the question isn't specifically about measuring anything about calorimeter, I suggest you ignore it. But if by any means the examiner (or the teacher) may insist on the calorimeter having any special effects on the outcome then the question gets kinda unanswerable.

This is impossible to answer. Usually you have to assume that when no calorimeter heat capacity is given, then it negligible (i.e. you only use the heat capacity of the 70g $\ce{H_2O}$). You know the temperature drop of the metal and the energy increase of the water, combine both to obtain the heat capacity of the metal.

You know the heat capacity of the water given by:
$$C=4.148\ \mathrm{\frac{J}{g\ K}}\ \times 70\ \mathrm g = 290.36\ \mathrm{J/K}$$ Since the water gained $3807.44\ \mathrm J$, you know that the metal loss $3807.44\ \mathrm J$. Using $Q = mc\Delta T$ you get: $$c = \frac{Q}{m\Delta T} = \frac{{-3807.44\ \mathrm J}}{180.45\ \mathrm g\times -66\ \mathrm K} = 0.320\ \mathrm{\frac{J}{g~K}}$$
Using the Dulong–Petit law, which states that the molar heat capacity of a metal is approximately $3R$, we get:
$$\frac{3R}{c} = \frac{3 \times 8.314\ \mathrm{\frac{J}{mol~K}}}{0.320\ \mathrm{\frac{J}{g~K}}} = 78.01\ \mathrm{g/mol}$$
 $M=78.01\ \mathrm{g/mol}$ is closest to copper so I would guess that is the answer.

q=m X c X (T2-T1)
set heat lost by metal equal to heat gained by water.  any difference is due to the calorimeter absorbing heat.  hopefully you have values for the specific heats (c) of your various metalS??  if not, hopefully other answers get you where you need to be!

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How to calculate the molecular weight for a volatile substance introduced into a Dumas bulb?

In a Dumas bulb a volatile substance is introduced. After a few minutes when the liquid has evaporated, the bulb is sealed. It is known that the initial weight of the bulb with air is $12.0468~\mathrm{g}$, $12.4528~\mathrm{g}$ with the volatile substance and $350.6264~\mathrm{g}$ with water. Calculate the molar mass of the substance if all these measurements were done at $25~^\circ\mathrm{C}$ and $1~\mathrm{atm}$ of pressure.

The correct answer is $29~\mathrm{g/mol}$.
What I've done is the following.
I've constructed a system of two equations to know the mass of the bulb and the volume it can contain.
$m_b+\rho_\mathrm{air}V = 12.0468~\mathrm{g}$
$m_b+\rho_\mathrm{water}V = 350.6264~\mathrm{g}$
The solutions are $m_b=11.6144~\mathrm{g}$ and $V=0.339012~\mathrm{m^3}$
With this I can find the molar mass of the volatile substance by knowing that its true mass was $m=0.8384~\mathrm{g}$ and the volume it occupied was $0.339~\mathrm{m^3}$ and using the ideal gas equation.
The ideal gas law, can be used to say
$M=\frac{m}{pV}RT=0.060~\mathrm{g/mol}$
Which is wrong. I've done all sorts of other things and can't get $29~\mathrm{g/mol}$. Maybe this is the craziest thing I've done until now.

There are a couple of things wrong.
First, the mass of the volatile gas is not $0.8384\ \mathrm g$. It's the difference between $12.4528\ \mathrm g$ and $12.0468\ \mathrm g$.
Second, the volume is incorrect. $11.6144\ \mathrm g + 999.97\ \mathrm{kg/m^3} \times 0.339012\ \mathrm{m^3} \ne 350.6264\ \mathrm g$

Could you show me how to solve a problem like this? Thanks.

The difference between 12.4528 g and 12.0468 g is 0.4060 g. The 2 equations are $m_b+1275.4g/m^3 \times V=12.0468g$ and $m_b+999970g/m^3 \times V=350.6264g$. That should get you to the right answer.


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Gibbs Free Energy : What is it trying to say actually?
"Gibbs Energy is the useful work that can be extracted from the heat of a reaction or a thermodynamic process." I understand how it predicts the feasibility of a chemical reaction , considering the Entropy statement of the Second Law. But how does it relate with the useful work that can be extracted? (This definition is implicitly used in my textbook in deriving from Nernst Equation, the useful work that can be extracted from an electrochemical cell and I am missing the point why this actually works.) Is there an intuitive explanation? Also if this much of energy is "free", what happens to the energy that is "not free"? Why can't we extract the whole energy that is released?

Also I am interested to know if there is any direct relation between Gibbs Function and Kelvin-Planck Statement, which apparently expresses the inherent inability to convert the whole heat to useful work, or am I mixing up two very different aspects?
Thanks.

You can prove that the Gibbs free energy is equal to the maximum amount of work that a certain reaction (or any physical process in general) can perform, i.e. $\Delta G = w_{max}$. Such a proof can be found in any decent Physical Chemistry textbook (such as Atkins'). The remaining energy must be "lost" as heat to basically account for the observation that is essentially stated with the Clausius inequality $\Delta S \geq 0$ (i.e. the entropy of the universe is always increasing) . Of course the definition of entropy $\Delta S = \frac{q_{rev}}{T}$ and the Clausius inequality that is derived from it (and hence the Gibbs free energy) are inherently connected to Kelvins (and Clausius') statement of the second law. You can see from the definition (don't forget the change of entropy of the surroundings!) that (since $\Delta U = q + w$) the entropy is decreasing when heat is completely converted to work.
Ask yourself the following: if heat could be completely converted to work, then the collision of an inelastic ball with the earth (or any other object) would be reversible, have you ever seen a tennis ball suddenly jumping back from the ground?

If the remaining energy is lost as heat, then knowledge of free energy could somehow determine the efficiency of the process, e.g, if a chemical reaction is the source of energy in an internal combustion engine, we can directly determine the engine efficiency. Is it true? @Jori

@Sagy you can determine the maximum theoretical efficiency - the actual efficiency is given by $\frac{w_{out}}{E_{in}}$. Note that this only applies to processes occurring at constant temperature and pressure - so an internal combustion engine wouldn't fit.

What if we assume that temperaure and pressure varies in infinitesemal steps and hence integrate the Gibbs Function along any thermodynamic process ? Then, arent the constant pressure/temperature restriction relaxed ? @thomij

No, that won't work - you need to use the appropriate thermodynamic potential for each step. For an internal combustion engine, the cycle you want is the Otto cycle:

What are you meaning when you say "appropriate thermodynamic potential" ? Please elaborate kindly, and also, consider the problem that we have to consider the maximum efficiency from a combustion gasoline engine. The efficiency for the Otto cycle is fixed obviously, but, by free energy energy considerations, is there an upper bound for this efficiency?

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What causes the really bad smell of rotten potatoes?
Rotten potatoes have a distinctive unpleasant odour, and I can't find any definitive answer as to what the chemical involved is. I was thinking it might be Butyric acid, from the fermentation of the starch, but according to Wikipedia:

Butyric acid is present in, and is the main distinctive smell of, human vomit.

And those puppies don't smell like vomit. They're more like rotting flesh – the reason I'm asking is that I've spent half the day looking for a dead animal that the cat brought in or something, only to discover that the horrible smell was coming from the potato box.

Just as a side note, you can delay the putrification of your potatoes by putting them in the refrigerator crisper drawer. They will literally last at least 10 times as long in the refrigerator. And trust me I have tested this Theory multiple scientific ways & the refrigerator does the job..! Please do not get the mean idea of messing with your friends by putting a potato under the driver seat or another enclosed area where they might like to hang out in. You will not be able to hold in your Chuckles and you will get busted by your friends.!

My first guess was that it was putrescine and other polyamines---but the smell of putrefying potatoes actually comes from methyl mercaptan,  dimethyl sulfide, and dimethyl trisulfide. Check it:

Source: A. Kamiya, Y. Ose, Study on offensive odor (Report IV): A Consideration of Putrefaction and Offensive Odor of Solid Waste, Journal of Japan Society of Air Pollution, 18(5), 1983, pp  453-463.
Notice that the headspace composition changes quite a bit with time. Most of the headspace chromatography studies I found dealt with early detection of disease organisms in infected potatoes, rather than potatoes in full putrefaction mode. You'll also find many references to solanine poisoning from potatoes; solanine is a toxic glycoalkaloid, is nonvolatile, and has nothing at all to do with the foul smell and toxic gas produced by putrid potatoes.
Methyl mercaptan ($\rm CH_3SH$) has an odor described as "rotting cabbage" by ATDSR; it's one of the major contributors to the smell of farts (oh, sorry, "flatus" if we're being polite). It has an odor detection threshold as low as 1 ppb.
Dimethyl sulfide ($\rm (CH_3)_2S$) is responsible for the smell of the sea (in low concentrations); it too has a cabbagy smell.
Dimethyl trisulfide ($\rm CH_3SSSCH_3$) is present in relatively smaller amounts but it has an even stronger odor. The detection threshold is around 1 part per trillion, and it is apparently a strong insect attractant.
The gases produced by rotting potatoes are quite toxic (see Rotting potatoes in basement kill four members of Russian family).
(To keep the notes below in context, in my original answer I said that a plot point in "The Walking Dead" was that zombies couldn't smell delicious humans if they were wearing coats smeared with rotting flesh. I suggested that when the zombie apocalypse arrives, packing your pockets with putrid potatoes might work, too. Now I'm not so sure. Can zombies distinguish between polyamines and sulfur compounds? Perhaps they're stench connoisseurs.)

There is a fact which is important when describing/discussing strong odors of sulfurous compounds: the odor can vary very much by concentration and with time spent in an atmosphere with such odor. It can switch from bad to pleasant at some concentration, or the odor vanishes completely. A common example is the extreme pungend stink of craft cellulose plants, which vanishes at the second or third breath takin. Otherwise such plants were not possible even in rural areas.

@Fred Hi, can you please share the links to those paid research papers that you were talking about, which are actually working on the early stages of potato rot. I am studying these papers for acedamic purposes.
Also, I couldn't comment on the above thread as I am new here.

It is a combination of potato and a fungus or bacterium that causes a potato to rot and produce the offensive odor.  A major contributor to the odor is dimethyl disulfide which has been identified as a key component of the emitted volatiles (see here)

Although the above reference doesn't mention it (at least not in the abstract) dimethyl disulfide is often accompanied by hydrogen sulfide, dimethyl sulfide and dimethyl trisulfide, the trisulfide being particularly odiferous.
Not only is the odor of rotting potatoes extremely unpleasant, but it has been reported to be lethal.  Certainly eating rotten potatoes can be lethal (solanine poisoning).

Thank you for this! I wonder, though, if this study wasn't geared towards early detection of disease rather than advanced decomposition? They seemed to be focusing on headspace gases that were produced directly by the organisms. I couldn't read the paper from where I am right now because it's behind a paywall.

@FredSenese  Yes, these authors have published several papers on early potato rot detection methodologies.  My assumption was that rot is a relatively simple process involving a simple pathogen and a potato, and composition of the volatile mixture won't show much variation with time, but I don't "know" that.  I was also wondering about your answer. Everything living contains putrescine, more in animals, less in plants; but not everything dead smells like putrescine.  Do you have any data or reference that supports your answer of putrescine vapor in the air around rotting potatoes?

True. And alas, I don't. I haven't done a thorough search. My answer was just a hypothesis, not a definitive answer.   Maybe I'll pop a rotten spud into our headspace GC sometime.  All of the papers I've seen in a quick go with Google Scholar focus on early detection, rather than advanced decomposition. For example, this one http://onlinelibrary.wiley.com/doi/10.1046/j.1365-3059.2001.00594.x/full makes no mention of sulfur compounds at all over innoculated samples.

@FredSenese  wrote, "makes no mention of sulfur compounds at all over innoculated samples"??  Benzothiazole was the largest signal seen in the headspace of the tubers inoculated with pathogens, it was not present in the sterile water inoculated tuber headspace.

Oh, oops! I meant it makes no mention of your sulfur compound.

@FredSenese  My reference was published in 2007, yours in 2001; your authors probably didn't even know or think to look for it.

Never mind, ron, I found a reference that actually deals with putrefying potato (rather than just newly infected potato) and you're closer to being right than I was. I'll edit my answer accordingly.


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Methanol skin exposure serious or not?
I have just minutes ago been splashed with industrial grade methanol. It seemed like a small amount (1-3mL) on my skin. I already washed the affected area thoroughly with soap and water. Online I've reviewed SDS's for methanol and they all seem to say that immediate medical attention is needed, and that the person exposed can be asymptomatic for a while. Does a small splash like that warrant medical evaluation?

Washing your skin was the best you could do. You will be fine!

@everyone Thanks everyone for adding a little humor to my panic attack :) I know that such small volumes of toxic chemicals can often be taken care of by your body, but 1) SDS's are horrible at describing the amount of exposure (in human-intuitive units) that cause those scary effects they outline, and 2) I work with stuff that's 1 x 10^6 more dangerous than methanol (say like methyl vinyl ketone), where one drop on your skin can hospitalize you. I guess I wasn't sure where on the spectrum this one landed. Thanks though!

http://chemistry.stackexchange.com/questions/21915/why-is-methanol-toxic?rq=1 - relevant info

"[A] factory worker who spilled a gallon of methanol down his trouser leg, was dizzy on the following day, took a short nap, and woke up total blind" Biochemical aspects of Methanol Poisoning, Biochemical Pharmacology, vol. 11, page 405.
So spilling methanol on your skin is serious, but the amount in your case is relatively small.
The article also explains that ingestion of as little as "a teaspoon" (5 mL) has caused permanent blindness.
In any case, the internet is no place to get medical advice.  If you are concerned you should call your local poison control center.

But I agree with you in the part "if you're concerned you should call your local poison control center."

I get it now. You wanted to emphasize not to take these things a joke next time. Point taken and I do agree with you.

+1 for an in-depth answer after my initial comment. Concerning permanent blindness, this case reported in the British Journal of Ophthalmology might be interesting too.

For any poisonous material we have special terms defined. In your case biochemists defined PEL (permissible exposure limit).
PEL is mainly important when it comes to deciding about for example "will the mineral water of that source be applicable for everyday-use of humans?" Many things affect the PEL for humans; such as age (kid, adult, etc.), type of exposure (ingestion, dermal exposure, breathing etc.), the medical situation of the exposed (other ailments or symptoms that can aggravate the harmful effect of a special kind of chemical) and so forth.
As your profile says, you're in your thirties, and thus are considered an adult. According to here an ingestion of up to 500 milligrams in an adults diet is no problem, among other trivia. And according to here, (in which you're considered to be supposedly exposed short-termly) if you didn't have headaches there couldn't be any problem.
Addendum :
You were exposed to methanol in the dermal way, not ingestion. As a result, as the second reference mentions, the liver will be able to turn methanol into formaldehyde and then formate. Formate is toxic. As some of this methanol will evaporate and the rest will eventually be absorbed by the skin you might not be in serious trouble. However, a specialist's advice is the best in this case. Human biology isn't really something anyone will be able to give statements about without empirical observation.

What do you mean by "A single diet soda exposes you more than this"? The reference in your answer says a diet soda can produce 20 milligrams.

Aspartame in that specific diet soda was reported to have metabolised into 20 mg of methanol, and the guy had a 3 mg dermal exposure to it. I don't know Dave, but in your reference, the doctor "has proven some cases to have become blind in an ingestion of teaspoonful". Maybe those cases were special. Still, we have very little info about the guy's medical situation so both of our answers can be incorrect, or either can be correct!

3mL is 3000mg, not 3mg.

There is serious difference between ingestion and short, dermal contact

Oops, my bad @DavePhD. Actually it's a funny mistake!

@Mithoron Yes, for dermal contact amount that enters the body would depend upon surface area and duration of contact.

@MARamezani it wouldn't be funny if you were a doctor or nurse or pharmacist.  Somebody was off by a factor of 1000 at Christus Spohn Hospital and: http://www.cnn.com/2008/HEALTH/07/09/heparin.babies/

It might be worth mentioning that ethanol is an antidote for methanol poisoning.

Might be, but kinda gets "old". If anything bad was to happen to OP, it would've happened by now. But, you have a good point in there.

I'm still around... for now :)


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Electrode for methanol
Could you advice an electrode for electrophoretic deposition in methanol which won't spoil methanol and also wont be spoiled by methanol. I googled a bit and found that
1 copper, platinum and gold are catalysts of methanol oxidation
2 aluminum reacts with methanol
3 silicon has poor conductivity...
What can you advice?

Glassy carbon may be a good choice. Looking at this, electrooxidation of methanol doesn't seem to be possible at reasonable potentials on glassy carbon, and it's a common working electrode material, relatively inexpensive, durable, and available in many shapes and sizes. Graphite could also work, though it may need to be replaced periodically as graphite electrodes tend to be less robust.


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UV curable adhesive
I'm looking for a UV curable adhesive that can be removed without peeling or burning.  Ideal removal would be some kind of solvent over a relatively short period of time. I also need it to have a viscosity of 50,000 - 70,000 cP.
Right now I'm using this but it's not ideal.  Someone has to be there to remove it.  I'm doing this with a robot and robot's aren't very good at removing it.
Anyone out there have any ideas?
Edit
I've been looking at UV curable investment casting wax.  I think that this could be melted off.  The only problem with what I have found is the viscosity.  They are made for 3D printing and have a relatively low viscosity.

Dentists use UV cured adhesives all the time. Try looking up bite block and what it's made of.


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Elimination in α,β-dibromohydrocinnamic acid

The above is a picture of α,β-dibromohydrocinnamic acid (2,3-dibromo-3-phenylpropanoic acid). In the lab today I carried out an elimination reaction on this molecule using potassium carbonate giving the carbonate anion as a base. I thought about the mechanism and I initially thought that the base would attack the hydrogen on the leftmost carbon atom (out of the two with the bromine atom on it) as I thought that this would be the most acidic proton of those two carbon atoms because of resonance available with the aromatic ring. However, the proton on the carboxylate group is surely the most acidic. How does this effect things in terms of the mechanism and the product?

Question: was the product a carboxylic acid or bromostyrene?

I have literally no idea. I am sorry.

Just theoretically though, if you were to try and predict what would happen, what would you say?

google://dibromocinnamic+acid+elimination gives links for decarboxilating elimination, but the procedure I found uses $\ce{AgNO3}$ in addition to carbonate. The product is bromostyrene. Given that I would say, that carboxylic group deprotonates first, and then both $\ce{Br^-}$ and $\ce{CO2}$ leaves simultaneously.

Dibromohydrocinnamic acid can eliminate in a number of different ways as shown in the following diagram. I'm not sure what conditions lead to cinnamic acid, but I know it can happen.  You have recently been asking questions about E1 and E2 eliminations, so my guess is that you are running a well-known lab experiment that generates cis- and trans-bromostyrene and allows the student to study the effect of reaction conditions on the mechanism of an elimination reaction.

Usually, the experiment starts with the bromination of trans-cinnamic acid.  The bromination is run under conditions that lead to trans-addition of bromine so that only the erythro (see here for diagrams and mechanism) or (S,R) [and its mirror image (R,S) enantiomer] dibromocinnamic acid diastereomer is formed.  The other dibromohydrocinnamic acid diastereomer, the threo isomer [(R,R) and its (S,S) enantiomer], which can only be formed by cis-addition of bromine is not produced.
If you build a model or draw a Newman projection it will make it easier to understand what follows.
Using the erythro [(S,R) and (R,S)] diastereomer obtained from the above bromination, running the elimination reaction in a solvent like acetone (a solvent that won't stabilize a carbonium ion as much as protic solvents like water or methanol) you drive the mechanism towards the E2 elimination side.  With your model or drawing, put the carboxylate group (the base has removed the acidic carboxylic acid proton to produce the corresponding carboxylate anion) and the bromine on the next carbon in an anti-periplanar arrangement (you've discussed anti-periplanar in your earlier questions on the E2 mechanism) and see how in a one-step, concerted elimination of both carbon dioxide and bromide you wind up with the cis-bromostyrene isomer.
If you ran the reaction in a protic solvent (like water or methanol) that stabilizes carbonium ions, you push the reaction towards the E1 mechanism involving loss of the bromine adjacent to the benzene ring to produce a stabilized benzylic carbonium ion.  The stable carbocation ion has an appreciable lifetime and will rotate so 1) the soon-to-be-departing carboxylate group is lined up with the carbocation p-orbital on the adjacent carbon and 2) that the benzene ring and remaining (bulky) bromine are as far apart as possible (the lowest energy conformation).  Elimination of carbon dioxide from this conformation produces trans-bromostyrene.
Typically, the student is given several test tubes and the reaction is run under a variety of conditions where solvent and reagents are changed and then the amounts of cis- and trans-bromostyrene are measured by nmr.  From this information the student can see how solvent and reagent can shift the mechanism.  See this link for a bit more detail on the actual experiment along with some good Newman projections.
BTW, the lab should smell nice since the trans-bromostyrene produces a hyacinth-like odor.

Two question Ron :-) --> 1) Why do protic solvents stabilize carbocations better than aprotic solvents? Both are capable of donating a lone pair. 2) Is the bromine more bulky than the carboxylate group?

@Jori 1) protic solvents, particularly water and methanol, have high dielectric constants.  High dielectric solvents lower the energy (stabilize) charge better than low dielectric solvents (see here).  Also protic solvents can hydrogen bond with the leaving group and stabilize it, making carbocation formation a lower energy pathway. 2) The first atom in the carboxylate is a (relatively) small carbon atom, bromine is larger than carboxylate.

It seems that also electron-donating substituents on the phenyl ring can interfere with the mechanism, driving it more towards an E1-type elimination. At least that’s what happened to somebody who worked on my project before I did and who had a phenol derivative.


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Why are triamino compounds so rare?
I was doing a reaxys substructure search for the compound shown on the left below, with the 3 amines on 1 carbon. However, the closest I found were all trinitro compounds like on the right. I would have thought the nitro compounds would be less stable and likely to explode. But given the lack of triamine compounds, they must be hard to make. All I can figure is that when all three amines are protonated you end up packing a lot of postive charge in a small place.

Now that I look at that nitro structure I'm pretty sure I got the charges wrong. Just ignore that.

You may be familiar with the equilibrium that exists between a gem-diol and the corresponding carbonyl compound, as shown in the figure below.  The carbonyl double bond is very strong, so in most cases the equilibrium lies far to the carbonyl side.

The same type of equilibrium exists in the case of bis- and tris-amino compounds where all of the amino groups are attached to the same carbon atom.  In the case of such a tris-amino compound (see figure) the imino analogue of a carbonyl is generated, but since a third amino group is present, another equilibrium between the imino compound and a nitrile also exists.  Just like in the carbonyl case, the nitrile triple bond is very strong.  So when a tris-amino compound is generated an equilibrium with a very stable nitrile will be set up and the equilibrium will (usually) lie far to the nitrile side.  Most tris-amino compounds prefer to exist as nitriles.  If the amino group is fully alkylated (all of the amino hydrogens are replaced by alkyl groups), then this equilibrium cannot occur and such alkylated tris-amino compounds should be isolable.

And the trinitro compounds are relatively stable because they can't convert the imino form, even if they might explode. What if we replaced the primary amines with tertiary or quaternary amines? (Assuming that the steric hinderences could be overcome)

That's correct.  The nitro analogues can't tautomerize, so while they might be unstable (and explode) they won't rearrange.  Nice idea, if you removed all of the amino hydrogens and made tertiary or quaternary amines you would again block the tautomerization and should be able to isolate those compounds.


text:
Dissociation of different compounds?
What would be the dissociation of Hydrosulfuric acid in the presence of water?
I was thinking that Hydrosulfuric acid or $\ce{H2S}_{(aq)}$ would form $\ce{2 H+}$ and Sulfur ions.
With this logic, I keep getting the wrong answer :(
The right answer of this question is it would not dissociate at all.
Shall appreciate some help on explaining this. Thanks!

$\ce{H2S}$ has a mostly covalent character. You can decide for it by the crude way textbooks give: Calculate the difference in electronegativity to decide whether it's a non-polar covalent compound, a polar one or has ionic characters.
For the dissociation part (which is usually endothermic overall) it'll need more "ionic character". And it does dissociate in water, resulting in a weak acid named  sulfhydric acid. (The spelling varies with citable sources) Your textbook likes to think about it as not dissociable, only because it's a weak acid.
Addendum:
The "ionic character" plays its role of a rule a good one as explaining acid 'strength'. However, it acts almost flawlessly in the same period, but has exceptions when it gets to groups. As Dave mentions, $\ce{HF}$ is weaker than for instance, $\ce{HCl}$ and that's because the atomic radius changes more dramatically than electronegativity difference. (See here ) But, in the $\ce{H2S}$ molecule, this exception hasn't occurred. Dave pointed out a good example of when the rule isn't working that well, but as I said, we can ignore the other factors in this very case.

so all weak acids do not dissociate?

No, but "weak acids", as the name fully suggests, dissociate partially. At times, lesser than 1% of the acid dissociates in water, so it's better to think they don't dissociate at all. It's like "Don't bother me with details".

verry helpful mannnn

@MARamezani I don't think what your are saying about ionic character and strong versus weak acids is accurate.  The series HF, HCl, HBr and HI has respective ionic charaters of 45%, 17%, 12% and 5%, and HF is a weak acid while the others are strong.

@DavePhD I'll edit my answer to cover your part.


text:
Are all strong acids the same strength?
Since the six strong acids dissociate completely into their ions, and the reaction goes to completion, does this mean that acids such as $\ce{HI}$ and $\ce{HCl}$ have the same acidic strength?
I know that 0.10 M $\ce{HI_{(aq)}}$ and 0.10 M $\ce{HCl_{(aq)}}$ both contain 0.10 M $\ce{H3O+_{(aq)}}$, which leads me to believe they are equally strong acids. Is this correct?

There's something called the "solvent leveling effect". In short, there's a lowest pKa in a particular solvent, based on the basicity of the conjugate base.
In water, you're limited by $\ce{OH-}$. So given the same concentration of $\ce{HI}$ and $\ce{HCl}$ in water, they will indeed have the same pH.
In other solvents, you may have a difference between acids, depending on the pKa.

I don't think there is a lowest pKa in a particular solvent.  This study http://dx.doi.org/10.1063/1.1739994 quantifies the pKas of nitric and perchloric acid in water by NMR.  There is also the method of measuring HCl vapor pressure http://pubs.rsc.org/en/content/articlepdf/1936/tf/tf9363200743 . Maybe there is a lowest pH or at least a lowest p[H+] just due to the fact that there are only so many H3O+ ions that can fit in a given volume.  No way p[H3O+] can be less than -log 55.

No, not all strong acids are the same strength
There are more than just 6 strong acid.
The acid dissociation constant ($K_a$) quantifies the strength of an acid.  pK is negative log base 10 of $K_a$.
Trifluoromethanesulfonic acid pk <-13
Perchloric acid pK = <-9
HI pK = -9.3
HBr pK = -9
HCl pK = -6
sulfuric acid pK = -3
chloric acid pK = -1
nitric acid pK = -1

This estimation for triflic acid isn't correct http://www.nrcresearchpress.com/doi/pdf/10.1139/v78-385


text:
Could I get a hint on the structure of this molecule based on HNMR spectra?
Chemical formula is $\ce{C7H6O3}$. The IR spectrum has a peak at $\pu{3238 cm-1}$, which I believe could be due to an alcohol, phenol, or carboxylic acid. There's an additional peak at  $\pu{3015 cm-1}$, which I believe is due to either an alkene or aromatic group. The presence of aromatic and phenol peaks makes me think there's a phenol group in this compound.

What I'm stuck on are the four protons in the 7-8 ppm range. Especially the one split into a quartet. If I'm right about the IR, I just can't think of a way to get one proton with three neighbors to split with in a phenol.

That is not a quartet, but probably two overlapping signals. You always have to pay attention to the signal ratios and not just the number of split signals. The ratio here is 2:2:2:1, that is not a quartet ratio. A quartet would be 1:3:3:1.
There is a certain distortion in strongly coupled systems, the "roof effect" you can also see in this spectrum. This distorts the ratios a bit from the ideal ones, but not so much as to explain the ratio in the pseudo-quartet here.

Surprisingly, the three left peaks of the quartet end at the exact same position. What signal would it be then? I think the person who scaled that spectrum cut off the quartet so that the roof effect is still identifiable as well as the very broad signal at 11.5 ppm. If the quartet would be fully shown, the peak at 11.5 ppm would not be visible. Additionally, the doublet and and triplet build the left part of the roof and the quartet (or what's left of it) builds the right part of the roof.

@Matthew You're right, it looks a bit cropped. I still don't  think it is a real quartet, even with the roof effect it doesn't look right. The second peak from the left also looks like it has a small shoulder, I would guess this  is a doublet and a triplet overlapping. Individual integrals for each signal would be nice, but we don't have that here.

Does the IR have any stretches near ~1700? This would be a great way to rule in (or out) a carbonyl. It looks like there is one from the above graph, and you have a broad peak ~ 11 ppm, which should stand out as a very specific functional group.
Mad Scientist is correct, that is not a quartet at 7 PPM, but is a doublet and a triplet partially overlapped.
Here's your hint: you have TWO oxygen-containing functional groups and an aromatic system (It looks like you already knew that from your sketches in the photo). If the aromatic system contains 4H, and is there are two functional groups, it is likely a di-substituted benzene ring. Which of the three possible types (ortho, meta, para) could give rise to two doublets and two triplets in the splitting?

P.S. I believe the spectrum is correctly referenced. The tiny peak near 7.26 ppm is likely chloroform from the solvent - it is much smaller than the other peaks and chloroform is the solvent of choice for most organic NMR.

Looking at your picture you also seem to have annotated an IR peak at around $\mathrm{1700~cm^{-1}}$ which is indicative of the carbonyl group. Together with the IR peak at $\mathrm{3238~cm^{-1}}$ and the very broad NMR peak far downfield this supports your supposition that there is a carboxylic acid present.
Looking at the molecular formula you now only have one $\ce{O}$ and one $\ce{H}$ unaccounted for so it must be a phenol group. As mentioned by Mad Scientist the pseudo-quartet on the NMR spectrum is in fact an overlapping doublet and triplet. You have two doublets and two triplets in total so you should think about which of ortho, meta and para substitution would give you this splitting pattern.

I once had a similar example in which benzene was used as solvent but the spectrum was not referenced to its peak at 7.16 ppm. If the quartet is really a quartet you might think about the meaning of that singlet at 7.26 ppm.


text:
What does it mean that an electron pair "spends more time" nearer an atom?
In a book I am reading, it is said

In the $\ce{HCl}$ molecule, the shared pair of electrons spend more time nearer the chlorine atom. In the $\ce{HF}$ molecule, the shared electrons spend more time nearer the fluorine atom.

What does "spend more time" mean here? Has this been experimentally validated? If yes, how?

When you come to the quantum level (looking at one electron) things don't behave anymore like we normally think. To make it anyway understandable those expressions were invented such as "spending more time". Electrons are less described as a dot but as a smearing (electron density).
Chlorine and fluorine have higher electronegativities than hydrogen. The electron density therefore is not equally distributed between the two bonding partners but drawn towards fluorine/chlorine. To illustrate this, you can say "the electron spends more time at F/Cl".
This thinking was established as molecular orbital theory (s orbitals, p orbitals, ...). There have been other models before, e.g. you might know the shell model (k shell, l shell, ...). For experimental evidence you might want to take at look at this -- but most of all the citations therein: http://en.wikipedia.org/wiki/User:Chem507f10grp4/sandbox#Experimental_Evidence_Supporting_Molecular_Orbital_Theory
Basically, the shell model ran into a series of properties it did not calculate correctly or precisely. Molecular orbital theory was found to predict many properties and came to well calculated values.

There is now a technique that can visualize chemical bonds vis a vis their electron density and how that electron density interacts with a carbon monoxide molecule on the tip of an atomic force microscope. http://www.rsc.org/chemistryworld/2013/09/first-pictures-hydrogen-bonds-unveiled-afm

It would also be notable to say that what the textbook said was to clarify the meaning of electronegativity, along with MO basics.

This is terrible terminology. Especially in MO theory, the approximations are based on the time-independent Schrödinger equation. It very much is expressions like these, which produce inaccuracies and misunderstandings. And to be fair here to the students as well, when they learn about MO theory, they probably already know something about probability; at which point it becomes just stupid to dumb it down again.

On the question: 'What does "spend more time" mean here?', from a theoretical vantage, why explore approximations that are based on the time-independent Schrödinger equation?
Perhaps instead, this could be viewed as an application of the nonrelativistic version of the time-dependent Schrödinger equation for the wave function in position space of a single particle (a shared electron) subject to a potential (from an electric field from the respective cases of the chlorine and fluorine atoms).
Conceivably (for those more of an expert in this field), the product could be a theoretically based graphic depiction of the expected position over time. Visually, does the shared pair of electrons spend more time nearer the chlorine atom with HCl? In the case of the HF molecule, does the shared electrons spend more time nearer the fluorine atom?
As an example of one of the more readable discussions on the time-dependent Schrödinger equation in a position basis, complete with graphs, please see, for example,this source. Here is a good educational video describing the probabilistic nature of the quantum realm at here and also this. A related video is found here.

This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review

To spend more time in a nonrelativistic sense means literally in a probabilistic answer, one is more likely to be in one place (near an atom, for example) than away. This what precisely what the time-dependent Schrödinger equation for the wave function in position space addresses for a single particle (like an electron). Please review all my sources, and explain to me why this is not correct. It is actually, in my opinion, the most appropriate, albeit advanced and erudite, explanation presented. I also have an advanced degree in Statistics and Applied Mathematics (ask advanced questions).


text:
How paper-tag survive in the dry-cleaning process?
This prevent the clothes from being ruined by ink on the paper tag.
To put the garments attached with this paper through the whole process of dry-cleaning,
I have the questions listed below

paper type?
ink type?
Is paper need to be coated?
if need >> what is this coating technique called?
printing technique (refer to http://en.wikipedia.org/wiki/Printing)

Note that, the dry-cleaning solvent is not just water.

Does that sound scientific?
I don't know the more appropriate place to ask this kind of question. If there is, just let me know.
Thanks,
pakorn

By the way, welcome to chemistry.SE! You have taken the tour, so I assume you know what this is about.

Thanks MARamezani. I think, this question is some kind of technology/chemistry-related. I was looking for the tech SE and you can guess what I found. no tech!
SE does it the narrower way to categorize the sites. Narrow enough to make me not to know where I am supposed to post this. Or I should tell SE in the area51.

This Q is enough chemistry for a good answerer to answer it right here. No need to panic! :)


text:
How does HF dissolve glass?
By what mechanism does HF proceed in dissolving glass? Why is it the only acid that has this capability? Is it because of the small size and high electronegativity of fluorine?

But what exactly do you mean by "only this capability"?

@Matthew, it is  why does only it have this capability or why is only HF capable of doing so

$\ce{HF}$ reacts with glass ($\ce{SiO2}$).[1]
$$\ce{SiO2 + 4HF -> SiF4 + 2H2O}$$
$\ce{SiF4}$ is not a solid that consists of vertex-connected tetrahedra like $\ce{SiO2}$ but is a gas at room temperature. Technically, $\ce{HF}$ is not a solvent since in this case it reacts with the glass vessel.
According to Spierings:[2]

$\ce{HF2-}$ ions are adsorbed on surface silanol groups, the $\ce{HF}$ molecules on vicinal silanol groups and $\ce{H+}$ on surface bridging oxygens in siloxane units. [...] These are transformed into surface groups such as $\ce{\bond{#}Si-F}$ and $\ce{\bond{#}Si-O-SiF3}$. The adsorption of $\ce{HF}$ and $\ce{HF2-}$ increases the electronic density on the bridging oxygen in the siloxane unit. This in turn makes these oxygens more basic, so more $\ce{H+}$ ions are adsorbed, which leads to more siloxane bonds being broken per time unit, i.e. a kind of catalytic effect. [...] The catalytic action of $\ce{H+}$ ions on breaking siloxane bonds also occurs in the dissolution of glasses in acidic and weakly alkaline solutions.

Hollemann, A. F.; Wiberg, E. Hollemann–Wiberg: Lehrbuch der Anorganischen
Chemie, 34th ed.; de Gruyter: Berlin, 2007; Vol. 102.
Spierings, G. A. C. M. Wet chemical etching of silicate glasses in hydrofluoric acid based solutions. J. Mater. Sci. 1993, 28 (23), 6261–6273. DOI: 10.1007/BF01352182.

I know the reaction, but my question is about mechanism how does this happen, does fluorine go interstities of the glass or h+ or the kinetic energy is responsible for it. Technically, HF doesn't dissolve glass since a solvent does not react with your compounds.; thats why I said does it disintegrate glass then because of what: Energy or size of ions or both

What do you mean by the kinetic energy of the fluoride ion?

@Dissenter, I am reffering to maxwell velocity distribution in liquids, which is related to collision theory of reactions

I've no idea what "a solvent does not react with your compounds" means especially in this context. Hydrofluoric acid clearly does react with glass. And that is what matters since most other things do not.

Oh boy... thank you for pointing that out. I mixed things up. Now it should be correct.


text:
How is "F2/sig(F2)" the data quality in SXRD?
When I recieve a dataset from single crystal X-ray diffraction, I can determine the quality from a graph in which F2/sig(F2) is displayed in dependence of the frame.
But I forgot the basics... I have found that the $R_{int}$ value is calculated as such:
$$R_{int}=\frac{\Sigma|F^2-F_{mean}^2|}{\Sigma|F^2|}$$
Reference: http://www.canadiancrystallography.ca/cccw/files/CIF-file%20and%20validation.pdf
But what exactly is F2/sig(F2) now?

Would "sig" be short for sigma $\sigma$ (i.e. the standard deviation).

That could be as well as the sum $\Sigma$. Maybe F2/sigma(F2) is part of the formula by which $R_{int}$ is defined. But would that make sense?

$R_{int}$ considers in this summation symmetry relevant reflections, e.g. in your equation the term $\Sigma|F^2-F_{mean}^2|$ takes an intensity of reflection and subtracts from this number the mean intensity of all its symmetry equivalents. If redundancy of your data collection is high (the case for all area detectors), then with this number you can tell how close in intensity are the equivalents, and thus how good is your crystal.
Another measure of quality is $R_{\sigma}$
$$R_{\sigma}=\frac{\Sigma[{\sigma}(F^2)]}{\Sigma|F^2|}$$
It tells you how close are the estimated standard deviation of equivalents (e.g.: were all equivalents collected with the same precision?)

$F^2/\sigma(F^2)$ is basically the signal-to-noice ratio, a simple measure of how well your crystal diffracts. It is hardly related to $R_{int}$ at all. Indeed, to calculate $R_{int}$ you need to have the whole dataset, otherwise it just would not make sense. As for $F^2/\sigma(F^2)$, you have it as soon as you've collected a little bit of data.


text:
Should the liquid be included in Kc expression?
Should the concentration of liquid water be included in the $K_\mathrm{c}$ expression for the reaction
$$\ce{2 NO2_{(g)} + 7 H2_{(g)} -> 2 NH3_{(g)} + 4 H2O_{(l)}}$$
if water is liquid and all the other compounds are gasses?

I attempt to improve formatting. Yours was a bit "wierd".

In the equation that you stated initially, water would be omitted from the equilibrium constant expression.
Your second question (about esterification that you posted in a comment to one of the answers) is extremely interesting.
I have given it some thought: in the standard case of esterification water should be included  because it is responsible for causing the reverse process ( hydrolysis of ester formed).
But in the case of zeolites, which by virtue of their structure, probably act as shape selective catalysts..ester hydrolysis by the water would be affected. (in some cases, I have read that selectivity towards ester formation is nearly 100%) In a case such as this, one can safely remove water from the equilibrium constant expression.
(I am not be a 100% correct, please excuse me, since I am just a student myself. Perhaps, someone else can provide a better/clearer explanation)

No. If the water condenses it is effectively removed from the reaction equilibrium. Just as with solids that precipitate.

What about the water produced during an esterification process? With zeolites it doesn't participate in Kc, but without it does, right?


text:
reference states for activities
I wanted to ask clarifications about a passage in Atkin's physical chemistry book, chapter 9, in the paragraph 9.2 description of equilibrium.
For studying equilibrium of a reaction, where $\nu_i$ are the stechiometric coefficients it imposes $\sum_i \nu_i \mu_i=0$ with $\mu$ the chemical coefficients. At this point it writes the chemical coefficients as a function of the activities $\mu_i=\mu_i^\circ+RT\ln(a_i)$ and identifies the term $\sum_i \nu_i \mu_i^\circ$ with the standard reaction Gibbs energy. I do not understand this identification. I thought that the standard state for the definition of activities (even dependent on wether the substance is a solute or a solvent) was not the same reference state for computing the reaction Gibbs energy (and does not depend on the role played by the substance in the reaction). I'm surely getting something wrong...

$\mu_i=\mu_i^o+RT ln(a_i)$ is the definition of activity ($a_i$).
Therefore, the standard state for standard chemical potential ($\mu_i^o$) is necessarily part of the definition of activity.
For a pure liquid or solid, the standard state is the pure substance at 1 bar.
For a liquid solvent, the standard state is also the pure liquid at 1 bar.
For a gas, the standard state is the fictitious state of an ideal gas at 1 bar.
For solutes, there are multiple conventions, for example the fictitious state of the solute having a molality of 1 mole/kilogram and exhibiting the behavior it does at infinite dilution.
$\Delta G^o = \sum_i \nu_i \mu_i^o$ by definition.
So there is no one standard state for $\Delta G^o$, because the various substances may have different standard states.

Actually Atkins says that for a solute the $\mu_i^0$ is your equation is defined as $\mu_i^0=\mu_i^+RT ln(K_B/p^_B)$, where $\mu_i^$ is the chemical potential of the pure solute at the same temperature, $p^_B$ the vapour pressure of the pure solute ad that pressure, and $K_B$ the constant given bu Henry's rule. I think this is the chemical potential of the pure substance when the vapour pressure is $K_B$ (not sure) but anyway I do not see here a link with the standard conditions (1 bar) that are used to define the standard Gibs free energy of reaction...

I wanted to say that $p_B$ is the vapour pressure at that temperature (I do noto think the external pressure plays a role).. I also used i and B ad the same symbol I'm sorry bit I'm noto able to edit the message...

@tirrel The way you and/or Atkins have $\mu_i^o =$ (function of temperature) doesn't make sense. Should $\mu_i^o$ just be $\mu_i$? The closest equation that I know of to what you have written is $\mu_i = \mu_i^* +RTln(x_i)$ for an ideal solution and $P_i = K_ix_i$ for Henry's law and $P_i = x_iP_i^*$ for Raoult's law.

DavePhD: this is what I understand from Atkins, which is a well respected book, so maybe I do not understand it. The section is 7.2, if you want to have a look.
By the way in your equation $μ_i=μ^∗_i+RTln(x_i)$, which also appears in Atkin's book how do you interpret the symbols? I though that $μ^∗_i$ was the chemical potential of the pure substance at the temperature given (so it is temperature dependent). And by the way if $x_i=1$ we get $μ_i=μ^∗_i$, which suggests to me that the reference state must indeed be temperature dependent.

@tirrel I agree Atkins is a great book, but I don't have it at home, only Levine.  The superscript in $\mu_i^o$ means "standard state" and the superscript in $\mu_i^$ means "pure state".  The equation $\mu_i=\mu_i^+RTln(x_i)$ doesn't say anything about the standard state, and isn't generally true, it is an ideal-law type of equation.  Make sure you check if the equations are true generally, or just in some ideal condition, and whether Atkins says $\mu_i^o$ or $\mu_i$

As far as I understand, Atkins says that: $μ_i =μ^∗_i +RTln(p_i/p^_i)$ is the equation valid for every real solution where $p_i$ is the partial pressure at a certain composition and temperature $T$ and $p^_i$ is the vapor pressure of the pure substance. For solvents activity is defined as $p_i/p^*_i$ and therefore we get $μ_i =μ^∗_i +RTln(a_i)$. This equation is generally true and $a_i$ tends to $x_i$ for $x_i$ that tends to one (Raoult's law). This for solvents. For solutes we have Henry's low and I already wrote the equations (of general validity) that I've understood...

Maybe we can just concentrate on solvents for the time being and then we pass to solutes, which seem to be more complicated... Do you agree with the definition I posted or do you have an other one?

For solvent, the standard state is the pure solvent at 1bar (according to IUPAC), and $\mu=\mu^*+RTln(a)$, where activity ($a$) could be written as $\gamma x$, $\gamma$ being the activity coefficient and $x$ being mole fraction (according to Levine).

According to your definition DavePHD $\mu^*$ depends on temperature. The only condition is to be at $1 bar$, which actually I think that at normal conditions is not very important, since the vapour pressure of a pure substance should depend very little on external pressure

Anyway, what about solutes, which are more complicated? And then, if the standard conditions are different in the two cases, as I'm getting to believe (what does Levine says?), how do I know from a table that a certain substance is a solvent or a solute (according to the experimental conditions, it can be the former or the latter...)?

Now I'm looking at parts of Atkin too through google books.  For solvent, Atkins equation 5E.1 is exactly the same as what I posted from Levine.  The equation you posted in your very first comment with $K_B/p^*$ is equation 5E.7, but Atkins makes clear that it is for ideally dilute solution.  Then Atkins introduces real solutes and $\mu_i=\mu_i^0+RT ln(a_i)$ is equation 5E.9, and explains that Henry's Law behavior is only approached in the limit of infinite dilution and "activity" captures the deviation from ideal behavior. (continued below)

(continued from above) Then Atkins introduces the mole fraction convention and says the standard state would be that same as that discuss in an earlier. Then Atkins discusses the molality convention and says the standard state would be different from the previous one.  Levine discuss these two convention, plus two other convention (concentration scale and treating the solute exactly the same as solvent).

Thanks DavePHD! Just a last comment: the same substance therefore can have multiple standard Gibbs chemical potentials reference states, depending if it is acting as a solvent or a solute. Right?

Yes, as Atkins puts it, "The selection of a standard state is entirely arbitrary, so we are free to choose one that best suits our purpose..." Section 5.11(c)

So is the point of contention here "how does the standard state for the activities relate to the standard state for the Gibbs free energy change"?
To see how choice of standard state for an activity relates to the standard state chosen for the chemical potential, write the equation as
$$\mu_i = \mu_i^\circ + R T \ln a_i = \mu_i^\circ + R T \ln \frac{\gamma_i c_i}{c_i^\circ}$$
where the standard states are consistent.The standard state is being defined so that
when $\gamma_i c_i = c_i^\circ$, $\mu_i = \mu_i^\circ$. If we're using molarities, we can choose $c_i^\circ = 1\ M$.
Now suppose you want to define a standard state that's different from 1 in the units you're using for concentration; e. g, suppose we have a reaction that includes $\rm H^+$ and we want to define the biological standard state ($^\oplus$) as $c_{\rm H^+}^\oplus = 10^{-7}\ M = 10^{-7} c_{\rm H^+}^\circ$, while using molarities as our concentration unit.
We can compute a new chemical potential:
$$\begin{array}{rcl}\mu_{\rm H^+} &=& \mu_{\rm H^+}^\circ + R T \ln \dfrac{\gamma_i c_{\rm H^+}}{c_{\rm H^+}^\oplus}\\
&=& \mu_{\rm H^+}^\circ + R T \ln \dfrac{\gamma_i c_{\rm H^+}}{10^{-7} c_{\rm H^+}^\circ}\\
&=& (\mu_{\rm H^+}^\circ + 7 R T\ln 10) + R T \ln \dfrac{\gamma_i c_{\rm H^+}}{c_{\rm H^+}^\circ}\\
&=& \mu_{\rm H^+}^\oplus + R T \ln \dfrac{\gamma_i c_{\rm H^+}}{c_{\rm H^+}^\circ}\\
&=& \mu_{\rm H^+}^\oplus + R T \ln a_{\rm H^+}
\end{array}$$
Now at unit activity $\gamma_i c_i = c_i^\circ$ and $\mu_i = \mu_i^\oplus$. Your chemical potential (and your Gibbs free energy changes) are now defined in terms of this new standard state. And note that the chemical potential at this new standard state ($\mu_{\rm H^+}^\oplus$) is indeed temperature dependent. To convert between the two standard state Gibbs free energies in this particular case,
$$\Delta G^\oplus = \Delta G^\circ + 7 \nu R T \ln 10$$
where $\nu$ is the stoichiometric coefficient of $\rm H^+$ in the reaction.
TL;DR: The standard state you choose for the concentration at unit activity becomes the standard state for the standard state Gibbs free energy change.


text:
Totally confused about the location of "s,p,d,f" Orbitals inside the atom
Here is the other crazy question... I know that electrons revolve around nucleus in their orbits.But my question is "how the orbitals are located in the orbits? Do they really posses Spherical, dumb bell, double dumb bell shapes inside the orbits?? Totally confused....

I know that electrons revolve around nucleus in their orbits.

You should "unknow" that, because that is the failed Bohr Model of the atom.
The are no fixed orbits for electrons in atoms.
The Bohr Model has been superceded by the Schrodinger model.
The atom is described in terms of probabilty density: for each point in space the probability that an electron will be at that point (when the atom is in a given state such as 1s, 2s, 2p, etc).

my question is "how the orbitals are located in the orbits?

There are no orbits

Do they really posses Spherical, dumb bell, double dumb bell shapes inside the orbits??

The shapes you seen in drawings that are spherical, dumb bell, etc. are meant to show regions of higher electron probabilty density.  For example, someone might draw a sphere on the basis that there is a 90% probablity of the electron in that orbital being within the sphere at a given moment.  However, in reality, for example in the 1s state, the electron could be anywhere in space. Other states, such as 2s, have nodes, with the electrons possibly being anywhere except at the nodes.
So think of the shapes as showing where the electrons in that orbital are often, but remember that they can really be anywhere excpet at the nodes.

Thanks for the answer... But I have one more doubt.. What about degeneracy of an orbital I mean every orbital is having a fixed amount of energy and the electrons present in those orbitals will possess the same energy..If there are no orbitals (fixed path for the electrons to revolve) Then how can you account on the constant energy of an orbital???

In the Schrodinger theory, the Schrodinger equation is taken as a postulate, and only discrete values of energy are found to be solutions. Total energy is kinetic energy plus potential energy.  Various distances from the nucleus have different potential energies, but kinetic energy must be considered also.


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What is an example of a very low temperature endothermic reaction?
Shakashiri describes the reaction of solid Ammonium Thiocyanate and solid Barium Hydroxide Octahydrate resulting in an endothermic reaction leading to subzero temperatures. I've performed this experiment and measured temperature as low as about $-10\ ^\circ\text{C}$.
Are there known chemical reactions that can lead to even lower temperatures?

I attempt to change the title so this question won't seem "primarily opinion-based", hopefully. As for the reaction, I can suggest the formation of ATP form ADP. But since you didn't mention the $\Delta H$ for that reaction, it's really hard to say that to what degree should your reaction be endothermic.

it is claimed that temperatures up to -90 may be achieved with precooled ice and magnesium chloride hexahydrate.

@M.A.R i think op meant low temprature spontaneous endothermic reactions.

Yes there are. In Vogels famous textbook on practical organic chemistry there is a section about the cooling effect of certain salt (mixture) solutions in water.

If ice is temporarily not available, advantage may be taken of the
  cooling effect attending the solution of certain salts or salt
  mixtures in water. Thus a mixture produced by dissolving 1 part of
  NH4C1 and 1 part of NaNO3 in 1-2 parts of water causes a reduction in
  temperature from 10 to — 15 °C to — 20 °C; 3 parts of NH4C1 in 10
  parts of water from 13 to -15°C; 11 parts of Na2S2O3.5H2O in 10 parts
  of water from 11 to - 8 °C; and 3 parts of NH4NO3 in 5 parts of water
  from 13 to — 13°C

(Taken from Vogels 5th edition, p. 70)


text:
Do chemicals undergo granular convection?
Do chemicals of an average weight, such as Polyethylene Glycols, undergo granular convection (the Brazil nut effect)?

Good question; is ultracentrifugation, http://en.wikipedia.org/wiki/Ultracentrifuge, affected by size of molecules, as well as mass?

@DrMoishePippik is it?


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Polyprotic acids and bases
I just have a question on how you handle polyprotic bases.
Question:

Show how oxalate ion can be a polyprotic base

My answer:

My question is whether we stop doing the reaction until the oxalate ion has no charge or do we keep going on as below:

Another question I have is ;why hypochlorite ion would be a monoprotic base, rather than a polyprotic base?

If you have a molecule or ion with multiple sites that can be protonated (as you do here), and the sites can interact with each other (as they can here), the affinity for protons drops rapidly as more sites are protonated.
The proton affinity of the ion in your second reaction is a couple of orders of magnitude less than in your first reaction because of this effect.
Piling a third proton on there would be far less favorable than that---oxalic acid is a weak acid, and you're forcing it to act as a base by making it accept a proton1. You should stop with the two reactions you have.
It's really the same thing with hypochlorite ion; putting one proton on is fine (you get HClO, which is a weak acid). Putting a second proton on is tough, because you'd be making an acid act as a base.

1. This kind of thing does happen, though such protonated acids will quickly lose their extra proton to the surrounding water molecules. If you're using glacial acetic acid as a solvent, your acid ion is $\rm CH_3COOH_2^+$ instead of $\rm H_3O^+$,
and $\rm CH_3COO^-$ instead of $\rm OH^-$.

oh ok, so for any polyprotic acid, you should stop at two hydrogen right?

You should stop when you've added enough protons to its ions get a neutral molecule. For example, with $\rm PO_4^{-3}$ you'd add 3 protons one at a time to get neutral $\rm H_3PO_4$.

okay that makes sense. Thank you so much Fred for clearing my doubts


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The Verb for Polymorphism
How can you elegantly describe that a polymorphic substance changes its structure?

It polymorphs
It morphs
It undergoes polymorphic transition
It experiences a morphic change

Or what would be even better terms?

Chemical ELL.SE! Please propose! :D

It "transitions from one polymorph to another"
It "transitions to another polymorph"
It "transition to polymorph beta"
It "transforms to another polymorph"
It "transforms to polymorph beta"
I like "transitions" better than "transforms", but I've seen both

I see "transition" as the more common form

Does "polymorph" have a different definition in chemistry than colloquial English?  Because here you're using polymorph as a form/shape rather than as an object that changes shape.  A thing doesn't change to a polymorph, a polymorph changes shape to a thing.  http://en.wikipedia.org/wiki/Polymorphism_(materials_science) doesn't seem to disagree with me, though my knowledge of chemistry is minimal so I may be dead wrong here.

This is what puzzled me in the beginning and is the reason why I ask. Originally, a polymorph is a substance which can have different shapes at the same conditions. But apperently each form of this polymorph is called "polymorph". Ugh...

In chemistry, the different polymorphs have different arrangements of atoms or molecules, but there is no macroscopic change in the shape of an object.

@DavePhD I would like to object that there were never a macroscopic change of the crystal.  If atoms and molecules rearrange to yield a new polymorph with a unit cell of different volume, the density of the sample will change.  If these changes exceed the Young moduli, the crystal may become brittle and shatter into pieces.


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What determines humidity limit / dew point of the air? - Why can air only hold a certain amount of water?
From Wikipedia:

The dew point is the temperature at which the water vapor in a sample of air at constant barometric pressure condenses into liquid water at the same rate at which it evaporates. At temperatures below the dew point, water will leave the air.

I am wondering what is is about the air which determines how much $\ce{H2O}$ vapour can be stored in a set volume of air at a set temperature without the water condensing back to liquid water.
Also, why is it that at higher temperatures the air can hold a greater amount of water vapour?

One way to understand this is if you think of air as bunch of atoms and molecules moving around in random fashion. The speed at which they move is measured by the temperature of air. (This is true even if we are not talking about air, but we don't need to bother about that here.) So when we say higher temperature we mean that the speed at which these particles move is higher. Furthermore the faster they move the less likely they are to stick together.
Now water molecules like to stick to each other and when too many of them are together they "condense". But if you were to increase the temperature of the air (and hence the water molecules), they end up moving around faster and less likely to stick together. This way you can add more water molecules for the same amount of air.
The other thing to think about is pressure. In some sense it describes how crowded the place is. The higher the pressure, the more crowded it is. So when you increase the pressure, you increase the chances that the water molecules will stick together and hence "condense".

Another way to think about this is through Le Çhatelier's principle, which states that when a system in equilibrium (especially chemical equilibrium) is disturbed it tries to adjust in a way which minimizes the disturbance.
So now think of a box with some water and air. If you leave it long enough the system will reach an equilibrium, wherein no more water will evaporate or condense. But say now you increase the temperature of the air, the system will try to readjust so as to reduce the change in temperature. To do this it will cause the water to evaporate (as evaporation requires/consumes heat) and thereby reducing the change in temperature.
Similarly when you increase the pressure which is similar to decreasing the volume of the box, the air tries to account for this by condensing out some of the water (since liquid water occupies less volume than gaseous water), thereby reducing the effective decrease in volume seen by the air.

You may wonder why other particles in air do not "condense." They do, but only at much lower temperatures and/or higher pressures.


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Slater determinant as an unperturbed atomic wave function
I've deduced following postulates from studying my chem books.

1) Slater determinants are eigenfunctions of an unperturbed atomic Hamiltonian, which contains kinetic and central potential energy parts of each electrons only, since spin orbitals constituting the determinants are originated from one-electron Hamiltonian eigenfunctions.

Here, atomic Hamiltonian = (kinetic part of each electrons) + (central potentials between atomic nucleus and electrons) + (interelectronic potentials)
Spin orbit interaction and the other effects are neglected.

2) Slater determinants or their linear combinations are eigenfunctions of total spin angular momentum and total orbital angular momentum operator($S^2$ and $L^2$) simultaneously.

In addition to them, I would refer to information from Quantum Chemistry, 6th Ed., written by I. N. Levine.
(p. 312) Since $S^2$ and $L^2$ commute with the atomic Hamiltonian and with the exchange operator, the zeroth order functions should be eigenfunctions of $S^2$ and $L^2$.
(The zeroth order functions Levine mentioned above is indicating the single or linear combinations of Slater determinants in a same configuration)
Well, I can accept the fact that the atomic Hamiltonian, $S^2$ and $L^2$ commute.
However if the single or linear combinations of Slater determinants in a same configuration are the zeroth order functions (eigenfunctions of the unperturbed atomic Hamiltonian), how the fact that $S^2$ and $L^2$ commute with the atomic Hamiltonian makes the Slater determinants be the eigenfunctions of the atomic Hamiltonian??
For example, a Slater determinant corresponding to the one of the Helium-first excited states, $|1s\alpha~2s\beta|$, is an eigenfunction of the unperturbed atomic Hamiltonian(this determinant is a zeroth order function), $S^2$ and $L^2$, but not of the atomic Hamiltonian.

However if the single or linear combinations of Slater determinants in
  a same configuration are the zeroth order functions (eigenfunctions of
  the unperturbed atomic Hamiltonian), how the fact that S2 and L2
  commute with the atomic Hamiltonian makes the Slater determinants be
  the eigenfunctions of the atomic Hamiltonian??

Actually, you go it right, but let us spell this out step-by-step. First, forget about angular momentum operators, since you do not need them to understand that a single Slater determinant is indeed not an eigenfunction of the atomic Hamiltonian (as you called it). That is it, you are right. A Slater determinant is an eigenfunction of the unperturbed Hamiltonian which describes a system of independent electrons, but not of the exact one.
Nevertheless, a Slater determinant can be used as a trial wave function in a variational procedure. That is the whole point of the Hartree-Fock method: a Slater determinant $\Phi$ is not an eigenfunction of the atomic Hamiltonian, but we could evaluate it energy $\left\langle \Phi \mid H \mid \Phi \right\rangle$ using Slater rules, and consequnetly, we could minimize it to find an upper bound to the ground state energy. The resulting Slater determinant obtained by minimizing the energy is indeed only an approximation to the ground state wave function.
Note also that the exact wave function (which is an eigenfunction of atomic Hamiltonian) can be expressed as a linear combination of Slater determinants for the various possible electronic configurations, not just a single Slater determinant. And this is the theoretical basis for configuration interaction (CI) method.

Now back to all this business with angular momentum operators. A single Slater determinant is not necessarily an eigenfunction of neither $\hat{L}^2$ nor $\hat{S}^2$. However, as you mentioned, the Hamiltonian indeed commutes with all these operators (in non-relativistic approximation), and thus, the exact wave function is also an eigenfunction of $\hat{L}^2$ and $\hat{S}^2$ (as well as $\hat{L}_z$ and $\hat{S}_z$). To be more precise, the exact wave function can be chosen to be a simultaneous eigenfunction of all these commuting operators $\hat{H}, \hat{L}^2, \hat{S}^2, \hat{L}_z, \hat{S}_z$. And since it is really desirable, we require the same from out trial wave function: we would like it to be an eigenfunction of these angular momentum operators.
So we want to construct our trial wave function being an eigenfunction of angular momentum operators $\hat{L}^2, \hat{S}^2, \hat{L}_z, \hat{S}_z$, but a single Slater determinant would not necessarily do it, so we form what is called a spin-adapted configuration state function (CSF) - a linear combination of Slater determinants which is an eigenfunction of angular momentum operators.
Note, however, that this is not always possible.
For instance, for the case of unrestricted determinants using a linear combination would not help in making the trial wave function an eigenfunction of $\hat{S}^{2}$.


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Is a triplet state or singlet state more stable?
Is a triplet state or singlet state more stable? How can an atom have a long lifetime in a triplet state even though two electrons have the same spin?

Whether or not the triplet state is the ground state depends on the molecule. There is no general answer along the lines of singlet is always preferred. Could you narrow your question a little, asking for more specific examples?

You question is correct. May be you should ask it like that How can an atom stay long time in an excited triplet state though the electrons have same spin??

The key issue here that we are comparing excited states. So let's begin by defining Singlet and Triplet Excited States.
Singlet and Triplet Excited States:  A singlet or a triplet can form when one electron is excited to a higher energy level.  In an excited singlet state, the electron is promoted in the same spin orientation as it was in the ground state (paired).  In a triplet excited stated, the electron that is promoted has the same spin orientation (parallel) to the other unpaired electron. The difference between the spins of ground singlet, excited singlet, and excited triplet is shown in the illustrative Figure below. Singlet, doublet and triplet is derived using the equation for multiplicity, $2S+1$, where $S$ is the total spin angular momentum (sum of all the electron spins). Individual spins are denoted as spin up ($s = +1/2$) or spin down ($s = -1/2$).  If we were to calculated the S for the excited singlet state, the equation would be $2[(+1/2) + (-1/2)]+1 = 2(0)+1 = 1$, therefore making the center orbital in the figure a singlet state.  If the spin multiplicity for the excited triplet state was calculated, we obtain $2[(+1/2) + (+1/2)]+1 = 2(1)+1 =3$, which gives a triplet state as expected.

Illustrative Figure: Spin in the ground and excited states The difference between a molecule in the ground and excited state is that the electrons is diamagnetic in the ground state and paramagnetic in the triplet state.This difference in spin state makes the transition from singlet to triplet (or triplet to singlet) more improbable than the singlet-to-singlet transitions. This singlet to triplet (or reverse) transition involves a change in electronic state. ***For this reason, the lifetime of the triplet state is longer than the singlet state by approximately 104 seconds fold difference.***The radiation that induced the transition from ground to excited triplet state has a low probability of occurring, thus their absorption bands are less intense than singlet-singlet state absorption. The excited triplet state can be populated from the excited singlet state of certain molecules which results in phosphorescence. These spin multiplicities in ground and excited states can be used to explain transition in photoluminescence molecules by the Jablonski diagram.

What difference in the spin state makes the transition from singlet to triplet rather than singlet to singlet??

It's purely a probabilistic phenomenon. But the probability of the transition singlet to singlet is bigger than the transition singlet to triplet.

If you consider stability simply in terms of energy and not lifetime, then a triplet state is definitely more stable than a singlet state, i.e., a triplet state is of lower energy than a singlet excited state.
This can be understood when we look at the energy contributions in a singlet and a triplet. Consider a simple case with two electrons labeled as $1$ and $2$ and two spin orbitals $a$ and $b$. In the singlet state electron $1$ occupies orbital $a$ with $1/2$ spin and electron $2$ occupies orbital $b$ with $-1/2$ spin. In the triplet state, $1$ occupies $a$ and $2$ occupies $b$, both with the same spin. The energy of the singlet state is given as:
$E_S = h_{11}+h_{22}+J_{12}$
And the energy of the triplet state is given as:
$E_T = h_{11}+h_{22}+J_{12}-K_{12}$
Here $h_{nn}$ is a contribution to energy irrespective of the spin of the electron. $J_{nm}$ is the coulomb term which exists between every unique pair of electrons. $K_{nm}$ is the exchange term, which only exists for an electron pair of the same spin. The exchange term has no easy classical interpretation. But importantly, it is a positive number, which implies that $-K_{nm}$ always decreases the energy.
Therefore, $E_S$ is greater than $E_T$, meaning a triplet state is more stable compared to a singlet state.

hmm... but is there any reason to say that exchange contribution be positive? For the coulomb contribution, it must be positive since it contains the square of wavefunction term only. But it's not for the case of exchange $K_{12}$.

Is triplet state stable or singlet state ??

It depends upon the spin system (atom, molecule, etc.).
Diamagnetic compounds have singlet ground states, so for many compounds the ground states are singlets.
Examples of triplet ground states include dioxygen.  Singlet dioxygen is a excited state.
For dihydrogen, the singlet state is the ground state, but at room temperature, there is an equilibrium concentration of 75% triplet state and 25% ground state. Near zero kelvins, the singlet state dominates, but the energy difference between the states is so small that, in accordance with Boltzmann statistics, at room temperature the states are populated in proportion to the number of microstates, the triplet having 3 and the singlet having 1. But in this spin system, the spins are proton spins not electron spins.
Methylene also has a triplet ground electronic state and singlet excited state.

How can an atom stay long time in a triplet state though the electrons have same spin??

Firstly, electronic "triplet state" doesn't exactly mean the electrons have the same spin.  The three possibilities referred to by the word "triplet" are 1. both spins are up 2. both spins are down and 3.  $1/\sqrt2$(up,down + down,up)
In any case, the lifetime of any excited state is not fixed even for a particular spin system.  Mechanisms for the system to transition to the ground state need to be considered.  In the gas phase the excited state may be long lived but it could be many orders of magnitude less in a liquid solvent.
Just to give an example, for atomic helium in the triplet 1s2s state, the lifetime to decay by radiation to the 1s1s singlet state is 7859 seconds.

Stability is about the energy level which is a thermodynamic factor while lifetime is a kinetic factor. That is why they are not necessarily agree with each other.
Triplet live longer simply because it is kinetically harder for the excited electron to jump back to the original orbital and at the same time flip its spinning state. It is not really connected to the stability of it. Stability is more complicated depending many other aspects.


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Iodine ion vs caesium ion size
An iodine ion is the larger of the two, but why if they have the same amount of electrons? Also, $\ce{Li}$ is larger than $\ce{Li+}$, don't more electrons mean larger radius?

Iodine is having 53 electrons and cesium is having 55 electrons in general. Iodine ion can be made by adding or by removing electrons from it. Here you added two electrons to iodine and made it an iodine ion in which number of electrons of iodine ion became equal to the number of electrons of cesium but the composition of nucleus(number of protons) is different in both iodine ion and cesium ion. Obviously cesium is having more number of protons than iodine, so cesium nucleus can hold it's electrons more closer in cesium than iodine nucleus in an iodine ion which results in larger radius of iodine ion than cesium..
Same number of electrons doesn't mean that their radii are same. Same logic is applicable for lithium ion and lithium.. Lithium ion is having less electrons and more protons compared to lithium.. So radius of lithium is larger than radius of lithium ion


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Analytical Chemistry predicting the order of elution of compounds in a reverse phase HPLC

Predict the order of elution for a mixture containing caffeine and benzoic acid, using a reverse phase HPLC. Explain your answer.

My try:
Since benzoic acid is more acidic that caffeine, which means it has a lower pH, it is hence more polar. Since for HPLC, the mobile phase is the polar phase, elution of polar compounds will be favoured hence benzoic acid will be eluted first. Is my answer and explanation correct?

The logic is wrong. Lower $\mathrm{pH}$ (actually you mean $\mathrm{p} K_\mathrm{a}$ here) does not mean more polar.
It is more complicated to decide the polarity then you think. For reverse phase HPLC, the more polar and more soluble compound come out first. For reverse phase HPLC, it is common to have acid (TFA for example) in the eluent. It this case, caffeine can be protonated and become much more polar because it is ionic now. Solubility is another factor to consider. So if you want to predict the order, you have to provide more information.


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Is tetraammine dichloro cobalt (III) an optically active compound?

Is
$$ \text{Co}[{(\text{N}{\text{H}}_{3})}_{4}{\text{Cl}}_{2}]$$
optically active? If so, write its stereo-isomers.

I know that for a compound to be optically active, it should rotate the plane of Plane Polarized Light, but is there a way to do this without experiment?
Thanks!

Please do not use formatting in the titles; because of url slug issue. (I know the Q seems a bit scary now, but it's better than the version with all those "mark-ups" in the face. :D)

Co(II) or Co(III)? Edited title is now in disagreement with formula in question!

$\ce{Co[(NH3)_4Cl2]}$ can exist as cis and trans isomers as shown in the following figure.

image source
If a molecule contains a plane of symmetry it cannot be optically active or chiral.
In the cis isomer there is a plane of symmetry containing the $\ce{Cl-Co-Cl}$ atoms, therefore this isomer cannot be optically active.  With the trans isomer there are several planes of symmetry (one contains the $\ce{Co}$ atom and the four $\ce{NH3}$ groups, and there are four more planes of symmetry containing the $\ce{Cl-Co-Cl}$ atoms and either containing or bisecting the pairs of amino groups), therefore the trans isomer is also optically inactive.

The 6 ligands are at the 6 vertices of an octahedron.
There will be cis and trans isomers, but neither isomer is optically active.
For the trans isomer, there will be a central axis containing Cl-Co-Cl, and the 4 N atom lie in a plane of symmetry, bisecting the Cl-Co-Co line segment.
For the cis isomer, there is a 90 degree Cl-M-Cl angle, with the line segment Cl-Cl being an edge of the octahedron.  The plane perpendicularly bisecting the Cl-Cl segment is a plane of symmetry. The plane passes through the Co and 2 of the N atoms as well.


text:
Ion-selective electrodes
Some ion-selective electrodes can be expensive (starting from $500). Is there any relatively inexpensive do-it-yourself alternatives that would still give very accurate results? How does one make an ion-selective electrode?

Ion selective electrodes are a group of electrodes that are very different from each other.
The common bits are, a selective membrane, a "sensing" electrode and a reference electrode to measure against.
The details of the membrane and sensing electrode depend highly on what selective electrode you are trying to build.
Here is an old patent on a Cl selective electrode for reference:
http://www.google.com/patents/US3740326
Regarding getting accurate results with a DIY: this really depends on which electrode you are trying to build. In the Cl example, I don't see any reason why a DIY would perform worse.


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Is there a name for the intersection of the graphs of first and second order reactions?
Is there a name for the point where a first order rate of reaction meets a second order one.

This point highlighted above.

Welcome to chemistry.se! I don't really understand your question. The kinetics of a given reaction is fixed. You can not have a first order and a second order kinetics  for the same reaction. Any way, at the point of intersection, the rates of reactions are equal.

As far as I know there isn't a specific term for that point.
If you have a one-step first order reaction
$$\rm A \xrightarrow{k_1} B + C, \quad \text{rate} = k_1[A]$$
that can run backwards as a second order reaction
$$\rm B + C \xrightarrow{k_{-1}} A, \quad \text{rate} = k_{-1}[B][C]$$
and the rates of those reactions are equal, you can call that point "equilibrium".
You can't really plot the rates separately the way you have unless you have a common concentration, e. g. plot a change in concentration $x$ where $\rm [A] = [A]_0 - x$, $\rm [B] = [B]_0 + x$, and $\rm [C] = [C]_0 + x$.


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What is actually a redox reaction?
I need to know what a redox reaction is. How could a redox reaction possibly be identified? Also, I have a confusion whether,
$$\ce{H -> H+  +  e-}$$
$$\ce{e- + Cl -> Cl-}$$
$$\ce{H + Cl -> HCl}$$
is a redox reaction.
Furthermore, could anyone please tell me if ionization and reduction and gaining electron and oxidation are referring to the same phenomena?

A redox reaction is any reaction in which oxidation and reduction are occurring. The word redox is a portmanteau of REDuction-OXidation. So, if we can correctly identify reduction and oxidation, we can identify redox reactants. In practice, this can be a little challenging, so I will focus first on the second part of your questions (definitions) and then go back to the reaction.

Ionization is any process that produces ions.

The conversion of a hydrogen atom to a hydrogen ion by loss of an electron is an ionization.
$$\ce{H -> H+ + e-}$$
So to is the reaction of hydrochloric acid and water, which produces hydronium ions and chloride ions.
$$\ce{HCl + H2O -> Cl- + H3O+}$$

Oxidation is a loss of electrons.

The ionization of hydrogen atoms to hydrogen ions is an oxidation.
$$\ce{H -> H+ + e-}$$
So to is the conversion of carbon monoxide to carbon dioxide:
$$\ce{2CO + O2 -> 2CO2}$$
Note that this type of reaction is what gave us the word oxidation - compounds are reacting with oxygen and atoms end up with more oxygen atoms bound to them.

Reduction is a gain of electrons.

The ionization of chlorine atoms to chloride ions is a reduction.
$$\ce{Cl + e- -> Cl-}$$
So too is the conversions of acetylene to ethane:
$$\ce{C2H2 + 2H2 -> C2H6}$$
Not all oxidations and reductions are ionizations. There is a simple mnemonic in English for remembering oxidation and reduction: OIL RIG (Oxidation Is Loss Reduction Is Gain).

Oxidation and reduction happen together.

In principle we can write an equation that shows only oxidation:
$$\ce{H -> H+ + e-}$$
but oxidation cannot happen on its own. That electron that was lost needs to be picked up by something else. This reaction that shows only oxidation is called a "half-reaction" and needs to be coupled with a reduction half-reaction to show a valid chemical reaction:
$$\ce{H -> H+ + e-}$$
$$\ce{Cl + e- -> Cl-}$$
$$\ce{H + Cl ->HCl}$$
In the other examples I gave, both processes are occurring. In the oxidation of carbon monoxide, the oxygen molecules are being reduced.
The surest way to identify oxidation and reduction (and thus redox) is to compare the oxidation states of the reactants and the products. Determining oxidation states/numbers is more complex that I want to put here, so start with the Wikipedia article.
For the carbon monoxide example:
                  reactants    products
C                 +2           +4
O (bonded to C)   -2           -2
O2                0            -2

The carbon atoms are being oxidized (oxidation number increased) and the oxygen atoms in $\ce{O2}$ are being reduced (oxidation number decreases). The oxygen atom already attached to carbon did not change its oxidation number.

A redox reaction involve transfer of electrons between species and the term redox comes indeed from the combination of terms "oxidation" and "reduction".
The species which is oxidized loses electrons whereas the species which is reduced (with higher oxidation potential) gains electrons. An oxidant (oxidating agent) is a species that can oxidize other species, whereas reducing agents can be oxidized, thus transferring electrons to other species (subsequently reduced). While being reduced, the oxidation number of a species decreases and vice versa, oxidation implies an increase in the oxidation number.
One example of everyday life redox reaction is rusting, which consists in oxidation of iron by oxygen thus forming $\ce{Fe(III)}$:
$$\ce{4Fe + 3O2 -> 2Fe2O3}$$
Another example is the cellular respiration, a complex pathway of oxidative degradation of sugars producing energy (stored as chemical energy in the acid anhydride bonds of ATP). Although the process is quite complex involving several enzymes and many steps, the overall reaction is quite simple:
$$\ce{C6H12O6 + 6 O2 -> 6 CO2 + 6 H2O}$$


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Making three membered ring using different reactants

I am struggling with the above question. It will be the conjugate base of trichloromethane that is involved. The same is true of the carboxylate. However, I have no idea about the reaction mechanisms. Please help.

You are correct that the first step with both reagents involves the formation of the trichloromethyl anion ($\ce{CCl3^{-}}$).  This anion can eject $\ce{Cl^{-}}$  and irreversibly form dichlorocarbene ($\ce{:CCl2}$).  Carbenes are very reactive intermediates, reactive enough to add to a double bond and form a strained 3-membered ring (a cyclopropane).
The reaction you've shown is said to be stereospecific.  That is, you started with cis-2-butene and wound up with a cis-dimethyl-dichlorocyclopropane; the cis olefin geometry is preserved in your cyclopropane product.  This indicates that the reaction involved singlet dichlorocarbene.  Carbenes can exist as singlets or triplets.  As the following figure indicates, singlet carbenes add stereospecifically (your case), triplets do not.  The "slow spin flip" step in the triplet carbene reaction is slow enough that rotation can occur about what was the olefinic bond and this rotation gives rise to cis and trans cyclopropane isomers.

I think that both reactions involve carbene as common intermediate which then is the species involved in the subsequent cyclopropanation reaction.
These are the two equations which lead to the formation of the carbene:
$$\ce{KOH + CHCl3 ->  :CCl2 + KCl + H2O}$$
$$\ce{Cl3CCO2Na ->  :CCl2 + NaCl + CO2}$$
Subsequently, the alkene undergoes cyclopropanation by dichlorocarbene.
In the second case, heat is required in order to promote decarboxylation, which is entropically favored.

It will be the conjugate base of trichloromethane that is involved.

Only as the initial intermediate! $¸\ce{CHCl3 + OH- ->Cl3C^- + H2O}$
The carbanion will lose chloride and yield dichlorocarbene $\ce{Cl3C- -> Cl- + Cl2C:}$
The latter then adds to the $\ce{C=C}$ double bond.

The same is true of the carboxylate.

No! In this case, decarboxylation (loss of $\ce{CO2}$) is the first step. The fate of the carbanion is however as described above.

Why does decarboxylation occur? Aren't carboxylate ions stable? Is that why heat is required?

@RobChem You're right. Usually, they are stable. Decarboxylation may occur when a carboxylate is oxidized to an acetoxy radical, which is not the case here, or when when a resulting carbanion is stabilized by electron-withdrawing substituents. And yes, in your case, heat is necessary for the decarboxylation of the trichloroacetate.