text:
What made my solution cloudy?
I may know, but I would like an expert opinion and a possible solution?
I had an ethanol product from molasses, brown sugar and yeast which was clear. I then added (and feel all information is important):
1 vanilla bean
1 cinnamon stick
whole allspice berries
whole black peppercorns
1/2 star anise
1/8 tsp grated nutmeg
3, 2-inch pieces ginger
3-inch strips orange zest
I let that sit for 2 weeks and remained clear. I then poured that solution through a coffee filter into beaker and it remained clear.
Then I took tap water, boiled it and poured it into the bottle with the used spices and let that sit for a day. I then used that water to dilute the ethanol solution to a lower percentage. Again, running it through a coffee filter.
Final product, cloudy.
The only thing I can think of is my water. We get our water from a well so it could be high in mineral content.
Would water be the problem?
And solution to make it clear again?

You likely formed an emulsion. Many of the flavor compounds in the products you mentioned are negligibly soluble in water but soluble in ethanol. Once you add the ethanol extract to water the much of the ethanol from the extract will mix with into the water but the flavor compounds in the extract will not dissolve in the water, which leaves micro-sized spheres of flavor floating around the water.  these micro-sized spheres have a different refractive index of light and thus disperse the light making it cloudy.  The same thing will happen if you mix water and oil with soap.  If you look at Gatorade or PowerAde sport drinks all of them are cloudy because of the emulsion produced from micro-size pockets of flavor in the drink.

This is a well known phenomenon in certain alcoholic drinks called "louche" or the ouzo effect. It's often seen in the preparation of absinthe and ouzo.
It's caused by essential oils and other non-polar compounds from the botanicals (in your case the spices and orange zest - primarily the orange oils) which can be dissolved in the relatively non-polar high-ethanol solution, but are not soluble in a relatively polar high-water solution. When you add water to the alcohol these oils are no longer soluble, and they come out of solution in microdroplets, giving the solution a cloudy appearance.
To make the solution clear again, I see two options. The first is to add some sort of (edible) surfactant/detergent which will be able to promote the solubilization of the oils in water. There are a number of compounds that are used in food chemistry to do this, although few are easily accessible to home users. (Lecithin may be an option, although I don't know how well it would work in alcoholic drinks.)
Your other option is to simply increase the amount of alcohol in the solution until you reach a point where it's non-polar enough to dissolve the essential oils.

So I poured a couple ounces to a glass, then added some ethanol and sure enough. Cleared up!
SCIENCE!!


text:
Do surfactants exist in a gas phase?
I am wondering if there are surfactants that exist in a gas state that can be used to reduce the surface tension of a liquid.
I know its a broad question, so for example, are there any gases that could reduce the surface tension of water when it is exposed to it?

I am wondering if there are surfactants that exist in a gas state that can be used to reduce the surface tension of a liquid.

The most direct answer is "no".  Gases do not have surface tension and gas-phase molecules do not chemically affect the surface tension of liquids.

I know its a broad question, so for example, are there any gases that could reduce the surface tension of water when it is exposed to it?

Gases could, however, dissolve in the liquid in question.  Then, they aren't really gases, but solutes in the liquid.  And solutes can affect surface tension.  You can see this earlier question on volatile surfactants for some examples of water-soluble volatile surfactants for some examples.  A paper entitled Surfactant in the gaseous phase was brought up by MAR in the comments and is another good example.

What about this shiny thingy Curt? Does it suggest what I think it suggests?

Those are gases that dissolve in the liquid to behave as surfactants, as I refer to in the second part of my answer.  If my answer is confusing let me know how I can clean it up to reduce confusion.

Well, no, but all I could go for was reading the title of the journal, which wasn't that descriptive. :)

Thanks for the link, looks interesting although I it's $40 to buy the full article. It also mentions about the gas creating a think layer liquid at the surface, would would probably not be a good thing for my situation. I'm trying to come up with a novel way to reduce the surface tension of a top down resin 3D printer.

Surfactants are surface active agents.  So by definition, to work they need to create a thin-layer liquid at the surface.

Ah that helps, thanks. I was thinking about just using liquid surfactants but was worried they would dissolve into the resin and somehow change how the resin performs. I guess my question is, is it common to find surfactants that don't dissolve through the liquid, and are only located on the surfaces, even after being mixed?

Think about it like this: Does water and gasoline mix? - NO.  Now does water vapor and gasoline vapor mix? - Yes.  Gases do not need surfactants to mix as again there is no surface and thus no surface tension.

source: Stack Exchange

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

sort: votes


text:
Hydrolysis of 4-chloro-3,3-dimethylbut-1-ene forms which products much faster than primary alkyl halides?
Question

My Attempt
First a primary carbocation is formed and then methyl shift occurs to form a more stable secondary carbocation hence making (C) the major product, consequently I also arrived at answer (D) via hydride shift, and also option (A) would anyway be a possible product.
But the answer given is (B). Where am I going wrong?

Is the question wrong? Will (B) be the correct answer if it was asked much slower instead of faster?

SciFinder does not know any reactions starting from that reactant. So whatever the book says is based on theory, not experiment. Aka a bad question.

@Jan though i couldn't agree with you more, aren't all questions in competitive exams based on theory, is my explanation below legitimate?

No, not necessarily. In fact, good exam questions should be based on experimental evidence not on theoretical thoughts unless the idea is explicitly consider how it may happen. Your explanation below doesn’t seem bad but I am hesitant to vote on it.

Apparently, (B) is wrong for obvious reasons.

The product formed fastest will be kinetically determined. Shifting of bonds takes time so the fastest formed product will be the the one where minimum migrations occur hence (A).
As @orthocresol stated below, a stable compound similar to cyclopropylmethyl carbocation is forming here. "The double bond helps to kick out the chloride ion to give a stabilised carbocation, which makes it react much faster than usual". Hence product (A) will be favored thermodynamically too (refer).

I wouldn't say it's SN2 under those conditions. The double bond helps to kick out the chloride ion to give a stabilised carbocation, which makes it react much faster than usual. Go and read about the cyclopropylmethyl cation.

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


text:
What elastic polymeric material look like at molecular level?
I think there are few crosslinks in elastic material. So probably, the material would look like not so dense, maybe even transparent.
I received this question from my collegue and have difficulty in answering it.
I think this question is about know how the structure of an elastic polymer looks on the molecular level.
What does the elastic polymeric material look like at molecular level?

I don't quite understand the question. I could guess that you just want to know how the structure of an elastic polymer looks on the molecular level, but then I would say, that this question is too broad. Maybe you should consider adding more specific information to your post, otherwise I fear your bounty will turn into dust.

@Martin-マーチン I received the question from my collegue and I have had difficulty in answering it. So please add an answer about how the structure of an elastic polymer just looks on the molecular level. It may just be about that.

@Martin-マーチン I understand the question as interdisciplinary umbrella question perhaps most relevant to physical-chemistry. Generic answers outlining features such as repetitive units and common forces -- usual spatial measures such as the size of sub particles in nanoscale for typical particles ma be relevant answers. There must be some comparison of different polymers covered in some materials science book, researching.

Are you asking about the structure of elastomers?

Yes, I think. Elastomers.

Broadly, the elastic polymeric materials are structural with the mixture of networks of molecular chains and liquid. There may be a distinction between the terms "elastic polymeric materials" and "elastic polymers". Key features contain repetition of some parts so open to bioengineering such as bioprinting. Each specific elastic polymeric material differs at molecular level. For example some materials can have long repetitive carbon chains while some materials can have short repetitive carbon chains where each structure is attached together by some forces. There are different kinds of forces where their degree is relative to the molecule:
Brainstorming on the question "What look like?" What does it mean?

what is meant by "look like"?
intermolecular forces? Repulsive forces? Attraction forces?
EM fields between each charged particles? How are quantum phenomena in materials such as doping considered?
What does the electron current look like over the material?
What is the distribution of static charge?

Perhaps relevant

The book Edited by Xiang Yang Liu and Jing-Liang Li "Soft Fibrillar Materials Fabrication and Applications".

"As one of the most important classes of soft materials, supramolecular materials are of a mixture of networks of molecular
  chains/fibrils and a liquid. These self-assembled fibrous/ molecular
  architectures exhibit various functionalities, ie. superhydrophobicity
  or superior mechanical strength, etc. and consist of the controllable
  structures.
--
The book covers the most important soft functional materials, including small molecule physical gels, silkworm silk and spider silk
  fibers and functional fibers, with respect both to the fundamentals
  and to development and engineering methods. It provides the reader
  with the necessary knowledge on the chemical and physical formation
  mechanisms of these materials and demonstrates that one can rationally
  design and tune the fibrillar networks so that the resulting materials
  exhibit the desired functionalities.
This work is a must-have for all Materials Scientists, Polymer Chemists, Condensed Matter Physicists, and Biotechnologists working in
  this interdisciplinary field."

Structure = matrix/non-matrix + ... Energy of structures in EM fields and KE.

@hhh Is there any reason why this post should be community wiki?

The images below of vulcanized butyl rubber monomer and  chains should give some clarity.  Essentially elastic polymers (elastomers) consist of polymer chains that do not crystallize and are bound to each other by a cross-linking agent.
Butyl rubber [poly(isoprene); your typical elastic material] is a good material to look at for what makes a good elastomer.  As you can in Figure 1 see the isoprene monomer has a methyl pendant group on the molecule which makes the molecue asymetric and greatly inhibits crystallization of the polymer.  It also has two double bonds which you can see in Figure 2 creates molecules which form trapezoidal monomer patterns as opposed to the typical "zig-zag."  This special shape  and lack of crystallization is what allows the polymer to extend a greatly as it does and spring back to it origional shape.

Figure 1. Molecular Structure of Isoprene Monomer.

Figure 2. Molecular Structure of Natural Poly(isoprene).
Butyl rubber is often cross-linked with sulfur as shown in Figure 3.  Note the two distinct polymer strands and the sulfur atoms linking them.  This cross-linking disallows the chains from slipping by one another (like spagehtti strands being pulled with a fork) and reduces the ability of the material to permanently deform.

Figure 3. Molecular Structure of Vulcanized Butyl Rubber

How is the last image made? How do you know how to divide the structure that like? How do you know that the S-S-S is not part of either chain (green or blue)?

The S-S-S comes from mixing sulfur with the rubber and curing it in an oven at high temperature.  Sulfur is not a part of the origional chain, but rather bonds to the chains during curing.

How can you do this kind of coloring of polymers?

I took it from wikipedia.


text:
Difference between using Zinc Nitrate instead of Potassium Nitrate for Salt Bridge Solution?
I have a lab question which asks "Suppose that before you began this experiment you ran out of your salt bridge solution of Potassium Nitrate and the only substitute available was zinc nitrate, would this have an effect on the cell potentials you measured?"
Now I'm guessing the answer is yes, since Zinc will have a different reduction potential than Potassium, but I don't understand how to explain that in regards to the actual transfer of electrons and change in Voltage measurement.  Any ideas?
*Some background information, we used half cells of AgNO3, FeCl2, Zn(NO3)2, Ni(NO3)2, Cu(NO3)2, and measured the voltage of each combination of half-cells by connecting each to both positive and negative lead.
Thanks so much..


text:
What is the size of the ammonium ion?
I did an experiment in which I dissolved certain substances in water, measured the conductivity and compared that with the ion-radius. One of the things I dissolved was $\ce{NH4Cl}$. What should I use as the ion-radius  for $\ce{NH4}$, since it is not really one atom?

I finally found it :D It is 148 m^-12


text:
Interaction between energy and entropy in highly elastic polymers?
Consider the interaction of energy and entropy in the highly elastic materials of an ideal polymeric network.
Now Gibb's free energy cannot be used directly $\Delta H = \Delta G + T \Delta S$ where $\Delta S$ and $\Delta H=U+pV$ for the internal energy $U$ such that

$S(R) = -k_{B} \frac{ 3R^{2} }{ 2N l^{2} } + const$
$U_{eff}(R) = k_{B} T \frac{ 3R^{2} }{ 2 N l^{2} } + const $

of which I am unsure.
I think the Gibbs free energy may have some nonlinear behaviour with the material such as some correlations between different terms and more passive energy terms due to elesticity -- this may be described by things such as fugacity.
How can you describe the interaction of energy and entropy in highly elastic polymers?

If I remember correctly, the relationships you have presented are related to the configurational entropy of the polymer chains between cross links of the polymer network.  In these equations, R is the spatial distance between cross links, N is the number of chain segments between cross links, and l is the length of each chain segment.  The smaller the value of R, the greater the number of configurations that the chain can exhibit (i.e., the greater the entropy).  Thus, I believe that there should be a minus sign in your equation for S, because, as R increases, the fewer the number of configurations that the chain can exhibit, and thus the lower the entropy.  The parameter you call U is, I believe the Helmholtz free energy, and is a measure of the stored elastic energy of the polymer network.  As the R gets longer, the polymer chains have been stretched more, and more elastic energy is stored in the chains.  Conceptually, the polymer chains are like springs between the cross links.

Thank you for defining the terms +1. Do I understand correctly that fugacity/z-compressability factor only play role in applications with very high pressure profile or very high temperature? I moved my curiosity to new question in http://chemistry.stackexchange.com/questions/41977/when-fugacity-and-z-compressability-equations-instead-of-gibbs-free-energy-equat

Why are fugacity/z-compressability factors used as a substitute for Gibbs free energy criterion in certain applications?
High pressure profile or large heat in which polymers must be designed to work otherwise they can break. Applications contain polymer mud systems such as here related to petroleum industry. More in When fugacity and z-compressability equations instead of Gibbs free energy equations?.
Fugacity of a pure fluid

Fugacity criterion is often used as a substitute for the Gibbs free- energy criterion. The definition for fugacity comes from an analogue
with ideal gases that is derived for a closed system under isothermal
conditions. Eq. 6 for an isothermal process ($\mathrm{d}T = 0$) is

More in http://petrowiki.org/Equations_of_state.
Macromolecular features relevant to pressure

Fugacity: effective pressure which replaces the true mechanical pressure in accurate chemical equilibrium calculations

Compressability factor

The relationship between fugacity and compressibility is not as straightforward as you might think either. Starting with the expressions for the Gibbs free energy of the real and ideal gases, you can work out the following relationship:
$$\ln{\left(\frac{f}{P}\right)} = \int_0^P\!\frac{Z-1}{P}\,\mathrm{d}P$$
More in https://www.quora.com/Why-is-fugacity-or-z-always-less-than-1.


text:
Does CH3CHNH exist, and what is it called?
I am drawing different Lewis Structures of small compounds containing nitrogen, hydrogen, and carbon. One permutation is $\ce{CH_3CHNH}$, but I have never heard of this compound. Initially, and falsely, I thought this compound was acetonitrile. It isn't coming up in Wolfram Alpha, Pubchem, or Wikipedia.
Does this compound exist? Also, what would this compound be called regardless of its existence value?

https://en.wikipedia.org/wiki/Ethanimine

@Loong I found that searching 'CH3CHNH' did not get me anything, but Ethanimine does appear to be the right compound name for 'CH3CHNH'.

The -CH=NH functional group is called an imine (in this case a primary imine) your molecule would be called Ethan-1-imine

A.K. all your answers are one liners that, while a good answer somewhere like Yahoo answers, don't comply to our standards. Please refrain from posting such answers, as the goal here isn't quick answers, it's high-quality content.

The question was also about existance - this compound is very reactive.

I would like to see more information on the existence of the compound. I think @A.K. sufficiently answered the nomenclature part of the question.

It doesn't ocurr terrestrially as it is very reactive, but it can form in space.
Per Loong: https://en.wikipedia.org/wiki/Ethanimine

@Galen imines typically hydrolise very fast, unless stabilized by bulky substituents and $\pi$ conjugation. Typical example of stabilized imines are Schiff bases, oxyms, hydrazones, iminoesters and varios compounds with imine group included in aromatic system. like pyridine.


text:
Pentabromophenol Mass Spectrum
I am working on analyzing the mass spectrum of pentabromophenol but am struggling to determine the fragment at 60 m/z. It could possibly be 5 carbons however that is too much positive charge to exist at high abundance.
Below are the mass spec retrieved experimentally and from literature.
Literature Mass spec: NIST WebBook
Experimentally determined mass spectrum:

Can you give a zoom-in of the region around 55 - 65 m/z?

It seems to have the 1:1 $\ce{Br}$ ratio spaced by 1.
I think a reasonable assignment would be that it is $\ce{[BrCCOH]^2+}$
m = 120 and 122
m/z = 60 and 61

authors: علي ناجي طاهر


text:
Can there be gaps in autoignition temperatures?
I recently saw a request for solids that auto-ignite between 80C and 160C.  I can't find any, and none have yet been suggested.
Which leads me to the theoretical question: Is it possible for there to be gaps in auto-ignition temperature ranges that cannot be filled by any chemical?
More specifically: Supposed we're interested in auto-ignition of a chemical in a standard atmosphere (where the temperature is varied between storage and ignition).
First question: Is it true that any chemical that can auto-ignite in a given atmosphere at some temperature will not be shelf-stable when stored in that atmosphere at any temperature?  Or are there self-sustaining exothermic reactions that will not occur at all below a threshold (auto-ignition) temperature, and therefore would be indefinitely "shelf-stable"?
Second question: If shelf-stable auto-ignition is possible (or even typical), then are there any theoretical assertions that can be made about the possibility of auto-ignition temperature gaps?  E.g., if I know there is a chemical that auto-ignites at 20C, should it theoretically be possible to chemically dampen the reaction so as to raise the auto-ignition temperature arbitrarily?  Or is this just a "that's the way it is" property?  I.e., chemicals have the reactivity they have, and there's no reason to believe that you could pick an arbitrary temperature and find a compound that auto-ignites at that temperature?

+1. This is a GREAT question. If there's no answers I might put a bounty. I looked at your profile and saw you followed a proposal for a Materials Modeling Stack Exchange. If you don't mind another attempt at a Materials proposal, would you be so kind to commit to this?  https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=ZTUyZWY5OTQ1NDI4ZGVlMmU5MzM0ZThiZjNhOTRkYWFlMTA1NjM0OWEzOTZhYzlmMTBlNWQ4ZmIzYWE3MWFkN-4At28O9gekwyERVbYPhFIPwyHiajfB5BeMyrZKJaYV0 It started slow but lately it's been booming with 5 commitments/day and very soon it will reach 200.


text:
How to prove that chloroform is more acidic than fluoroform?

How to prove that chloroform is more acidic than fluoroform?

Using the concept that the inductive effect is dominant over the resonance (mesomeric effect) in the case of halide substituents: $\require{mediawiki-texvc}$

Since both have single hydrogen, let's compare stability of carbanions $\ce{CF3-}$ and $\ce{CCl3-}$.
Since fluorine shows more negative inductive effect than chlorine, $\ce{CF3-}$ should have been more stable than $\ce{CCl3-}$, so some greater cause must be acting here.
In $\ce{CCl3-}$ carbon has non bonded electrons in 2p orbital and chlorine has an empty 3d orbital, but since the energy difference is very high, $\ce{p\pi-d\pi}$ backbonding is barely possible. Worst case scenario, even if it's possible, then giving preference to backbonding will violate our initial rule (that inductive > mesomeric).
Bond lengths, $\ce{C-Cl}=1.75\AA, \ce{C-F}=1.35\AA$, but since chlorine is bigger than fluorine, they both will have approximately same electronic $\ce{p-\pi}$ repulsion (of lone pair of halogen and at anionic centre).

Which of my reasoning are wrong? How do I prove the fact that chloroform is more acidic than fluoroform?

Data: $\mathrm pK_\mathrm a (\ce{CHBr3})=13.7$ and $\mathrm pK_\mathrm a (\ce{CHCl3})=15.5$ and $\mathrm pK_\mathrm a (\ce{CHF3})=28$ (source - page 2)

My Views (I'm not entirely sure about these reasons)
(1) $\ce{CF3-}$ is highly unstable due to huge repulsions between the negative charge on carbon and the lone pair of fluorine.
(2)While in $\ce{CCl3-}$ chlorine has an empty d orbital thus the negative will be more dispersed into the this empty orbital  hence the repulsions will be quite less as compared to $\ce{CF3-}$.

I would $+1$ a self-answer, but I cannot $+1$ anything that assumes the usefulness of d-orbitals on chlorine.

I'm sorry @Jan i thought this question had went unnoticed, I'm leaving this question open again.

I don't think your reasoning is sound. You can promote electrons into orbitals but you cannot disperse charge into an orbital. However, you can disperse charge across the volume of the atom. Since chlorine has a larger atomic volume, you can say that the charge is dispersed across a larger volume of space in Cl.


text:
Finding the concentration of a solution
Q.) Determine the concerntration of a solution made by dissolving 10.0g of sodium chloride in 750.0ml of solution.
I did the following; 1mol of sodium chloride = 58.44

58.44*10 - 584.4g
Concentration = number of moles/volume
584.4/0.750 = 779.2 - and this is the wrong answer.

Any pointers would be appreciated.

Rather than doing operations on just the numbers, also pay attention to the units; carry them along in your calculations. Do you see any mistakes?

I should have divided - 10/58.44 and then divided that by .750 - thank you for your comment.

The molar mass $M$ of $\ce{NaCl}$ is $58.44\ \mathrm{\frac{g}{mol}}$ indeed:
$\displaystyle\text{molar mass}(\ce{NaCl}) = \frac{\text{mass}}{\text{amount of substance}} = 58.44\ \mathrm{\frac{g}{mol}}$
$\displaystyle\Rightarrow \text{amount of substance} = \frac{\text{mass}}{\text{molar mass}}=\frac{10.0\ \mathrm g}{58.44\ \mathrm{\frac{g}{mol}}} \approx 0.171\ \mathrm{mol}$
$\displaystyle\text{concentration} = \frac{\text{amount of substance}}{\text{volume}}=\frac{0.171\ \mathrm{mol}}{0.75\ \mathrm L}=0.228\ \mathrm{\frac{mol}{L}}$
Assumptions: Adding $\ce{NaCl}$ to these $750\ \mathrm{ml}$ does NOT change the volume noteworthy.


text:
Why are chemistry droppers bad at holding bromine?
Question in my chemistry test last week. The question was related to a previous question on bromine's density.

I imagine it has much more to do with bromine's vapor pressure (which is high). Pipettes have trouble with diethyl ether and other high vapor pressure liquids as well.

The diameter of the orifice of a "standard" dropper is "sized" to work well with water. But bromine has three factors that work against that large size. (1) Density of liquid (2) lower surface tension and (3) high vapor pressure.

Bromine seems not to wet glass which would be another reason that a dropper doesn't work well.  https://books.google.com/books?id=cm46AAAAMAAJ&pg=PA109&lpg=PA109&dq=bromine+%22wet%22+glass?&source=bl&ots=a7orFKoQ_q&sig=IEGZvJC4UqqJMmpBjbkheA9B-eM&hl=en&sa=X&ved=0ahUKEwjMg9TNpMjJAhVEdj4KHVQSBbYQ6AEIQDAH#v=onepage&q=bromine%20%22wet%22%20glass%3F&f=false

Since the standard dropper is designed to hold common liquid such as water or ethanol, the capillary size have some particular size. When it comes to bromine, the density is 3 times of water. In other words, the dropper need to hold a super heavy droplet, so you could expect that it won't be as efficient as usual. In addition, your previous question hinted on the density, so it is pretty obvious right?


text:
How to calculate the total concentration after mixing two solutions of differing dilutions?
If you mix $\pu{20 ml}$ of $\pu{3 M}$ sugar solution with $\pu{30 ml}$ of a $\pu{5 M}$ sugar solution, what solution do you end up with?
What I did,
\begin{align}
\text{volume} &= \frac{\text{amount of substance}}{\text{concentration}}\\
\dfrac{3}{0.02} &= 150\\
\dfrac{5}{0.03} &= 166.\overline{6}\\
150+166.\overline{6} &= 316.\overline{6}
\end{align}
This is the  wrong answer.
Any formulas which could help would be appreciated.

Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". (cc @safdar )

Please also note that descriptive terms or names of quantities shall not be arranged in the form of an equation; i.e. do not write "$\text{volume}=\frac{\text{amount of substance}}{\text{concentration}}$"; write $V=\frac nc$ instead.

3M and 5M mean 3 mol/L and 5 mol/L i.e. the concentration. 20ml and 30ml are the volume. You said you were calculating the volume but you actually divided the concentration by the volume which gives mol/L/L. And you don't need to calculate any volumes since they are given, and you can add them to get the final volume. They want you to use this to find the final concentration.

Ok, thank you, I had tried that. C=no of moles/volume. Following that I get (3/0.020)+(5/0.05) = 250. However the answer is supposedly 4.5M. I'm sorry to appear slow, but I haven't done this before!

I meant 4.2M instead of 4.5M as the answer given.

Ok, found an equation - (20/50) * 3 + (30/50)*5 = 4.2M

Remember 3M stands for 3 mol/L (the concentration), not 3 moles. So again you were actually dividing the concentration by the volume which gives moles/L/L. Your new equation is correct because e.g. for the first solution, you have diluted 20ml to 50ml, so the new concentration will be 2/5 of the original concentration. Alternatively you could calculate the number of moles of each, add them and then divide by the total concentration to gives the concentration in mol/L or M. I hope that helps you understand the formulas

Remember two solutions of different concentrations are mixed together, the total amount of substance of the solution is the sum of the amounts of the individual solutions. when you get the sum of the amounts, you add their volumes and use those two to determine the new concentration of the solution.
$$n_1=0.02\ \mathrm l\times3\ \mathrm{mol\ l^{-1}}=0.06\ \mathrm{mol}$$
$$n_2=0.03\ \mathrm l\times5\ \mathrm{mol\ l^{-1}}=0.15\ \mathrm{mol}$$
$$n_\text{total}=n_1+n_2=0.15\ \mathrm{mol}+0.06\ \mathrm{mol}=0.21\ \mathrm{mol}$$
New concentration:
$$c_\text{new}=\frac{n_\text{total}}{V_\text{total}}=\frac{0.21\ \mathrm{mol}}{0.05\ \mathrm l}= 4.2\ \mathrm{mol\ l^{-1}}$$

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


text:
Why are acids corrosive?
I'm in the second year of my biochem major, taking organic chemistry, and I feel like I've learned a lot about acids. The most popularly recognizable property of acids is their corrosive power. There are pop-culture depictions of acids everywhere, and all they seem to do is corrode stuff. Yet, with all I've learned about them, it's never been explained to me why they have this property, or how and when it works.

Not all acids are considered corrosive.  The strong acids which are corrosive are highly reactive so they react with most material they come into contact with and the by products start to dissolve in the acid corroding the material.

The reactivity of the anion towards the exposed material also plays a role in the acids potential to corrode.

Strong acids disassociate completely, weak acids do less so. When they do disassociate in water, they form $\ce{H3O+}$. This pulls the electrons from the thing it is reacting with. Lets take this to be Fe. ($\ce{H3O+}$ rips the electrons from the substance). So the reaction would go something like: $\ce{Fe + 2H3O+ -> Fe^2+ + 2H2O + H2}$. The $\ce{Fe^2+}$ formed is soluble in water. This also reacts with the other ion from the acid. This is what makes the acids corrosive.

So you're saying that acids have a tendency to take electrons from things, and things with fewer electrons tend to be more soluble? Like I know that metals seem to tend to do that, but is that sort of a general trend? Like maybe taking electrons like that tends to polarize things and the now-polar things tend to be more soluble?

Things don't neccassarily have to be soluble for the reaction. Even submersion in an acidic solution will work. I will talk mostly about H+ (as this is the only true acid according to bronsted-lowry theory (ignoring the proton acceptors as acids (lewis theory) as they function in the same way but a little less strongly)). This is a single proton. It is highly positive and this charge is what makes it so electronegative. Meaning that it has a very high tendancy to gain (simply rip off (force) from anything) electrons off of other things which makes them "vulnarable" for reaction with other ions


text:
What are inner core electrons and how do they influence chemical and physical properties?
There are six groups of p–block elements in the periodic
table numbering from 13 to 18. Boron, carbon, nitrogen,
oxygen, fluorine and helium head the groups. Their valence
shell electronic configuration is $\ce{ns^2 np^{1-6}}$(except for He).

The inner core of the electronic configuration may,
  however, differ. The difference in inner core of elements
  greatly influences their physical properties (such as atomic
  and ionic radii, ionisation enthalpy, etc.) as well as chemical
  properties

The above lines are from my school's textbook.
How does inner core electron create such differences and and what are inner core electrons?

Core electrons are electrons that are not valence electrons. I am sure your textbook has something about periodic trends going down a group. Go and read it.

Check out the differences between argon and krypton. Then extrapolate for other p-elements.

Inner core electrons do have a very strong affect on the chemistry of elements,for example inner d orbitals have a very poor shielding effect thus the outer electrons are bounded more strongly to the nucleus thus decreasing their reactivity and oxidation states.You will come across inert pair effect,stability of lower oxidation states down the group etc because of effect of inner core electrons.


text:
Equilibrium and Completeness of Reaction
In class we are currently learning about equilibrium and Le Chateliers principal. The principal makes sense but seems contrary to some other concepts. For example, using stoichiometry to calculate the mass of reactants for a certain amount of products. If we know the reaction is always going to equilibrium where the forward and reverse reactions are happening at the same rate, then how will the reaction ever be "complete" because according to the principal even if the reaction were to be complete and all the reactants were gone, the reverse reaction should be occurring. This is not the case in reactions like burning magnesium.
I'm guessing that not all reactions are reversible and that the principal is correct but in some cases it would take a long time to reach equilibrium.

Related answer that may be helpful: http://chemistry.stackexchange.com/questions/41597/equilibrium-constant-can-it-be-reached/41610#41610

The basic stoichiometry is done to understand the basic concepts of moles and ratios in reactions. Later on, as the reactions become more complex, alot of them don't go to completion and so for a lot of them, you need equilibrium constants. This being said, not all reactions end partway, some of them do go to completion. The most basic example is disassociation of salt ions in solutions. From basic stoichiometry, you learn that the ratio is one to one (like NaCl -> Na+ + Cl-). So Moles of NaCl added = the moles of Na and Cl ions in water. So basic stoichiometry is sometimes (not always) applicable but is essential for the learning process.


text:
Non-mathematical explanation for why the 3d orbital generally has higher energy than the 3p orbital
Why does the angular momentum of orbitals(l) affect the orbital's energy? Please provide a qualitative answer.


text:
Mechanism of reaction between triphenylmethanol and acetyl chloride
I am reading an old method for making triphenylchloromethane (Org. Synth. 1943, 23, 100):
$\hspace{30 mm}$
When I draw out the structures of the molecules, I see this:

But the $\mathrm{S_N1}$ mechanism doesn't make sense to me since $\ce{OH-}$ is a bad leaving group, and there is nothing to protonate it.

SN1 reactions are based on the stability of carbocation (R+), whereas SN2 reactions are based on the steric hinderance in R group.

I find it unlikely that the substitution proceeds via the hydroxyl group directly leaving. If there is any evidence that this reaction proceeds via direct $\mathrm{S_N1}$ of $\ce{OH-}$ by $\ce{Cl-}$, I would be very interested to see it. When an alcohol and an acyl chloride are mixed, I think it is much more probable that they will initially form the ester via nucleophilic acyl substitution:
$\hspace{27 mm}$

Presumably, this ester is rather unstable because of steric repulsions between the three phenyl groups and the ester group. So now, the $\ce{-OAc}$ group ($\ce{Ac} = \ce{COCH3}$) can leave as a relatively stable carboxylate ion to give the resonance-stabilised carbocation:
$\hspace{35 mm}$
The carbocation is then intercepted by chloride ions that were formed earlier to give the product you want. So, the second part of the reaction is a $\mathrm{S_N1}$ substitution of $\ce{OAc-}$ by $\ce{Cl-}$.
$\ce{OAc-}$ ($\mathrm{p}K_\mathrm{aH} = 4.76$) is a much better leaving group than $\ce{OH-}$ ($\mathrm{p}K_\mathrm{aH} = 15.7$).

There is another pathway possible, since $\ce{HCl}$ is formed from the initial reaction of the alcohol with the acyl chloride. If $\ce{HCl}$ is present in appreciable quantities, it can protonate the original alcohol and you could perhaps have a direct $\mathrm{S_N1}$ of $\ce{H2O}$ by $\ce{Cl-}$.
$\hspace{10 mm}$

It all came back to me!  Thank you very much for the detailed answer!

Yes, the hydroxide is a bad leaving group but look the structure if the hydroxide ion leave. It's a triphenyl cation! It's very stable carbocation (very high mesomeric effect). Then the reactive hydroxide react with acyl chloride and liberate Cl- that react with phenyl cation.


text:
Why does ozone reduce to oxygen?
The O's in ozone and in dioxygen all have oxidation states of 0. If this is the case, why is ozone considered to be reduced to form dioxygen (since there's no change in oxidation state)?
Would the half reaction for this reduction be $\ce{2 O3 -> 3 O2}$ ? Since there's no transfer of electrons, how would we combine this with an oxidation half reaction to cancel out the electrons transferred in the oxidation half?

Ozone is not considered to be reduced to form dioxygen like $\ce{2 O3 -> 3 O2}$. As you correctly pointed out, there's neither transfer of electrons, nor a change in oxidation state, so this can't be and isn't considered a redox reaction. On the other hand, there is a good deal of true redox reactions where ozone attacks something and leaves $\ce{O2}$ as a byproduct, but that's another story.


text:
Why is silver chromate red?
Copper(II) ions are blue and chromate(VI) ions are yellow therefore copper(II) chromate(VI) would turn out to be green because it's a mixture of blue and yellow.
In the same way, silver ions are colourless while chromate ions are yellow. So shouldn't silver chromate be yellow as well?

You can't just add colours like that. The ions interact with each other. Lead(II) and iodide ions are both colourless in solution but $\ce{PbI2}$ is yellow.

If so are there any ways to predict the colour of a compound?

http://chemistry.stackexchange.com/questions/2547/why-are-some-salt-solutions-coloured

You do not get green color by mixing a blue colored compound and a yellow colored compound. That's not how chemicals work.


text:
Acidity of o-halobenzoic acids

Rank the following compounds in decreasing order of acidity:

2-Fluorobenzoic acid
2-Chlorobenzoic acid
2-Bromobenzoic acid
2-Iodobenzoic acid

My approach
Since the ortho effect is proportional to the size of the substituent, it will be the greatest in the case of iodine and smallest in the case of fluorine. So, the molecule will be more non-planar in 2-iodobenzoic acid.
So resonance will be inhibited in all the above compounds. But what will happen to the acidity?

Acidity data:
The true order of acidity is $\ce{Br = I > Cl >> F}$. Relative to unsubstituted benzoic acid, all of the ortho-halogen substituted benzoic acid derivatives have a greater acidity, clearly having some stabilising effect on the resulting benzoate anion.

Data retrieved from Journal of Research of the National Bureau of Standards, 1958, 60, p569 - research paper 271. $\mathrm{p}K_\mathrm{a}$ values measured in water.

Rationalisation:

Polar effects All of the halogens exert a negative inductive (-I) effect, removing electron density from the ring. This makes the benzoic acid derivatives more acidic in all 4 cases since the resulting conjugate base is stabilised to a certain extent by the presence of the electron withdrawing halogen.
Steric effects As the halogen gets bigger, strain may force the carboxylate out of planarity with the aromatic ring. Whilst fluorine is roughly the same size as the hydrogen atom, bromine and iodine are both sterically more demanding.
Internal H-bonding This is suggested by another answer below, whilst this has been studied and shown to operate for protic groups such as alcohols and amines, in halogens it has little impact on stabilising the negative charge.

Overall, steric and polar effects operate in tandem to give an overall acidity for the benzoic acid. It's perhaps not intuitive which one will dominate (e.g. fluorine is far more electronegative, but also far smaller than iodine, so which is more acidic).
The table below, taken from the cited paper, attempts to separate out steric and polar effects through calculations on both the acid and its conjugate base. Sadly, they only look into $\ce{F}$ and $\ce{Cl}$ (no $\ce{Br}$ or $\ce{I}$), but it can be clearly seen that:

Hydrogen bonding is making no significant contribution to the acidity of the substituted benzoic acid
Polar and steric effects are both in operation, with sterics being far more important in the ground state (the carboxylic acid) and polar effects dominating for the conjugate base (the carboxylate).

Table copied from New. J. Chem., 2004, 28, p67

At least 3 (or 4) facts are required to be considered for the acidic strengths of halogen acids:

The resonance stabilization of carboxylate anion is poor in benzoic acid due to cross conjugation, wherein the pi system of phenyl ring also 'competes for' the same $\ce{C=O}$, where negative charge of one of the $\ce{O-}$ would have been shifted

Steric effects of any group, irrespective of its +M/-M effect, will allow $\ce{-COOH}$ to go perpendicular to the plane, destroying the cross conjugation; which give a more stable carboxylate, which means more acidic substituted $\ce{Ph-COOH}$
(This is what they say is, ortho effect)

Halogens have dominant negative inductive effect over their positive mesomeric effect.
H-bonding would have played some role in stabilizing carboxylate ion in halogen acid.

Could you please cite the source from where you got these images, or are these self-made? Thanks!

Gaurang, this is my work only

The intramolecular hydrogen bonding between the halogen substituent and the adjacent carboxyl group (as shown below) would decrease the acidity of the halobenzoic acids:
$\hspace{60 mm}$
Since the strength of the hydrogen bonding decreases in the order $\ce{F}>\ce{Cl}>\ce{Br}>\ce{I}$, we can arrive at the order of acidity $4 > 3 > 2 > 1$.


text:
Determining the order of basicity of acetate, phenoxide, hydroperoxide, and hydroxide ions

Arrange the following ions in order of basicity:
  (1) $\ce{CH3COO-}$ (acetate)
  (2) $\ce{PhO-}$ (phenoxide)
  (3) $\ce{HOO-}$ (hydroperoxide)
  (4) $\ce{HO-}$ (hydroxide)

The first thing I see is that all of them have the same donor atom, i.e. oxygen.
In both $\ce{CH3COO-}$ and $\ce{PhO-}$, the charge is stabilized by resonance. So, they are least basic, $\ce{PhO-}$ being the less basic among the two as you can see that in case of $\ce{PhO-}$, the number of resonance structures are more than $\ce{CH3COO-}$.
Next, comparing $\ce{HO-}$ and $\ce{HOO-}$ I find that the negative charge on $\ce{O-}$in $\ce{HOO-}$ maybe slightly more stable as the charge may be shared by the other oxygen, whereas $\ce{HO-}$ has no such option.
So I get the basicity order as 4 > 3 > 1 > 2. But the answer is 3 > 4 > 2 > 1. Where am I wrong?

You are partly wrong and partly right. The correct order is $4 > 3 > 2 > 1$ ($\mathrm pK_\mathrm a$ data taken from the Evans table):
$$\begin{array}{cc}
\hline
\text{Molecule} & \mathrm{p}K_\mathrm{a} \\
\hline
\ce{CH3COOH} & 4.76 \\
\ce{C6H5OH} & 9.95 \\
\ce{H2O2} & 11.6 \\
\ce{H2O} & 15.7 \\
\hline
\end{array}$$
Between acetate and phenoxide, phenoxide does have more (non-trivial) resonance forms, but all of them involve putting a negative charge on carbon which is hardly favourable. So, the stabilisation derived from resonance is not much at all.

On the other hand, in acetate, the other major resonance form has the negative charge on an electronegative oxygen. The moral of the story is that the actual number of resonance contributors is not the only deciding factor:

Between hydroperoxide and hydroxide, the $\mathrm{p}K_\mathrm{a}$ values show that you are right in saying that the second oxygen in the hydroperoxide ion stabilises the negative charge. This operates entirely via an inductive effect, though. There is no resonance involved, which partly explains why both ions are more basic than phenoxide and acetate.

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


text:
Short Cut Method to Calculating pH of Polyprotic Acid?
I am doing the following question:

Calculate the pH of a $4.00\ \mathrm{mol\ L^{-1}}$ solution of citric acid.
  $\mathrm{pK_{a1}} = 3.09~~~~~ \mathrm{pK_{a2}} = 4.75~~~~~ \mathrm{pK_{a3}} = 6.40$

Usually for polyprotic acids, we can assume that the second, third and other dissociations  of $\ce{H+}$ ions are negligible when compared to the first dissociation. However I believe that this is only valid when the pKa values are separated by around 4. So for citric acid, this would be an invalid assumption as the pKa values are pretty close to each other.
Now, I can find the pH of the solution by doing the long way by first finding the amount of $\ce{H+}$ ions formed by the first dissociation, then using that to find the amount of $\ce{H+}$ ions formed by the second dissociation and so on.
However the room that they have given for this question is pretty small and has only has enough room for a couple of lines of working out and I definitely wouldn't be able to fit all my working out in that box.
So am I missing some short cut method to finding the pH in cases like this, or is the above assumption that I have outline at the start still correct for citric acid?

Looks mighty like they want you to assume the said approximation, despite its being not that great for this particular case.

The assumption that the second (and higher) deprotonation steps are negligible for the calculation of the pH of a polyprotic acid is valid for citric acid.
You can compare this by calculating the exact solution and then stepwise neglect deprotonation steps.
(For the sake of simplicity I write $c(\ce{H+})=x$ and $k_{a,n}=k_n$.)
Exact: $$x=\frac{k_W}{x}+c_0(\ce{H3A})\frac{x^2 k_1+2x k_1 k_2+3 k_1 k_2 k_3}{x^3+x^2 k_1+x k_1 k_2+k_1 k_2 k_3}$$
Neglecting $k_3$: $$x=\frac{k_W}{x}+c_0(\ce{H3A})\frac{x k_1+2 k_1 k_2}{x^2+x k_1+k_1 k_2}$$
Neglecting $k_2$ and $k_3$: $$x=\frac{k_W}{x}+c_0(\ce{H3A})\frac{k_1}{x+k_1}$$
Solving those equations give the following pH values (with super unrealistic high precision):

Exact:               1.2469291050971560
w/o $k_3$:           1.2469291065371300
w/o $k_2$ and $k_3$: 1.2470654318802320

As a pH can only be measured precisely within something like $\pm \left(10^{-2}\ldots10^{-3}\right)~\mathrm{pH}$, the calculated pH can be rounded to be 1.247 in every singly case. No matter which deprotonation step was neglected.
So for your case it is negligible. For more general cases I have to look it up and will add it to my answer within the next days.

Thank you for the great answer! So is there any rule of thumb about knowing if you can assume the second dissociation is negligible? Before I thought the pKa values had to be different by 4 but that obviously doesn't work for this example.

As always it depends and there is no clear border ... the pictures will follow. But another point ... is this just a "normal" homework excercise from school or is it more "real" from the lab? 4 M is a very high concentration and unless this is not just a "normal" homework excercise - which I assumed - there is no simple concentration=activity assumption. Afaik, this can actually only be made if the concentration is below ~0.001 M.

Since the solution will be acid in water, the $\mathrm{pH}$ will be below all the $\mathrm{p}K_\mathrm{a}$ values.
Given that the $\mathrm{pH}$ is below all the $\mathrm{p}K_\mathrm{a}$ values, an upper limit is that the second $\mathrm{p}K_\mathrm{a}$ will contribute $10^{(3.09-4.75)} = 0.022$ protons for each proton that the first $\mathrm{p}K_\mathrm{a}$ contributes.
The effect of these 2% more protons on $\mathrm{pH}$ would be about $\log (1.022) = 0.0009$ $\mathrm{pH}$ units.
In real life, the problems calculating $\mathrm{pH}$ from the information given in the question, would be:

$\pu{4 M}$ citric acid means 768 grams of citric acid in $\pu{1 L}$ of solution.  Will that much really dissolve?

If that much did dissolve, and concentration of $\ce{H2O}$ is down to about $\pu{22 M}$ instead of the usual $\pu{55 M}$ , how do you account for that in the equilibrium calculations?

The 3 $\mathrm{p}K_\mathrm{a}$ values are calculated based upon extrapolating ionic strength to zero, what would these values be at $\pu{4 M}$ citric acid?

The $\mathrm{pH}$ of 50% citric acid is reported to be 0.6, and if you could make $\pu{4 M}$ citric acid, it would be still lower $\mathrm{pH}$.  It doesn't matter if you use 1, 2, or 3 $\mathrm{p}K_\mathrm{a}$ values, you wouldn't calculate the right value (to say 0.1 $\mathrm{pH}$ units) for a solution that concentrated.


text:
p-d sigma bond - Why not possible?
Why is a head on (sigma) bond between a p-orbital and a d-orbital not possible? Is it because extent of overlap becomes too little? Can you please illustrate?

I'm pretty sure your premise is wrong. I can't see why there wouldn't be tons of metallic compounds with p-d $\sigma$ bonds, such as $\ce{WCl6}$, which is undeniably a very covalent compound and whose only really available orbitals are the p orbitals in chlorine and d orbitals in tungsten.

Are you sure? Can you show me the overlapping diagram for such a compound?

Your premise is incorrect. $\unicode[Times]{x3c3}_\mathrm{p-d}$ are very much possible. Consider any hexahalidometal complex such as $\ce{[FeF6]^3-}$. Since fluoride is essentially unhybridised as iron would be, too, we can separate the atomic orbitals into irreducible representations of the $O_\mathrm{h}$ point group. We find that iron’s d-orbitals transform as $\mathrm{t_{2g} + e_g}$ and the 18 fluoride p-orbitals as $\mathrm{a_{1g} + e_g + 2t_{1u} + t_{1g} + t_{2g} + t_{2u}}$. If we separate out those 12 p-orbitals that are perpendicular to the iron-fluoride axis, the ones that remain are $\mathrm{a_{1g} + e_g + t_{1u}}$. You may immediately realise that $\mathrm{e_g}$ is an irreducible representation of both and thus that the fluorides’ p-orbitals can and will mix with the iron’s $\mathrm{d}_{x^2 - y^2}$ and $\mathrm{d}_{z^2}$.
Professor Klüfers included the calculation results of said ion in his internet scriptum of his coordination chemistry course. The ones of relevance are:

You may be slightly surprised at the shape since you cannot recognise the typical $\mathrm{d}_{z^2}$ shape. But as Brandhorst, Grunenberg and Tamm explain, these two orbital shapes are mathematically equivalent descriptions for the $\mathrm{e_g}$ orbitals.[1]
The molecular orbitals shown are both of antibonding type (formally $\mathrm{e_g^*}$). The corresponding bonding orbitals are much lower in energy and centred on the fluoride ligands. Both of the depicted ones are occupied only by a single electron in α spin.

References
[1] K. Brandhorst, J. Grunenberg, M. Tamm J. Chem. Educ. 2008, 85, 1692. DOI: 10.1021/ed085p1692.


text:
Nodal planes in a molecular orbital
How are nodes in a molecular orbital different different from those of atomic orbitals?
eg. For a Px-Px Pi bond, we would say that the nodal plane is the Y-Z plane (correct me). How about for a Pi* (antibonding) molecular orbital?
I think it should be similar to an atomic orbital, but please confirm.

The nodal planes in a $\pi^*$ orbital are analogous to the nodal planes in a $d_{xy}$ atomic orbital. Note the plural.

Why is that? Can you explain more?

They are not. A nodal plane is just a plane where the electrons’ wave function is zero. There is no distinction between ‘two types of zero’.

I see, so nodal planes are similarly defined as areas where probability of finding an electron is 0? That means nodal planes for all 3 cases - p-p atomic, p-p pi and p-p pi* will be same, right?


text:
Why are alkanes 'non-polar'
I am unsure with regards to the specific reason why alkanes are non-polar as I have heard different explanations. My teacher said that it is because, although there is a difference in electronegativity between carbon and hydrogen atoms, the carbon is shielded by the hydrogen atoms from reacting with the external environment and therefore, since only the hydrogen can react, it is non-polar (something along the lines of it is the difference between delta + and delta - ends that actually make a molecule polar). From searching on the internet, however, I have found the explanation that it is because the difference in electronegativity is so small (0.3) that it is non-polar, which makes sense, however it contradicts what my textbook states: That only bonds between elements with equal electronegativity are non-polar. Can someone clear this up for me? Thank you!

It is just a matter of semantics. The difference in electronegativity between C and H is just so small that, for all intents and purposes, the bond (and the molecule) is practically non-polar.

In addition, unsubstituted, symmetrical alkanes, e.g. methane, are even less polar (though there is still a small instantaneous dipole moment). See http://butane.chem.uiuc.edu/cyerkes/Chem104ACSpring2009/Lecture_Notes_104/Lect14c.html

so it does not have anything too do with the fact that carbon is 'on the inside' and therefore cant interact with the outside environment that it is not polar?

@frostedcake No, it doesn't. Also, there is quite a set of polar hydricarbons like azulene

This highlights a common theme throughout much of science. Very rarely are two things exactly equal. The question to ask is, "are they 'equal enough' that we can't really tell a difference between them?" In this case, the electronegativities of $\ce C$ and $\ce H$ are "equal enough" to make alkanes non-polar.

There are two reasons why molecules are polar or not: the symmetry of the molecule and the polarity of the bonds in it.
Bonds can be thought to have electric dipoles if the elements in the bonds are different. Bonds between more polar elements will have stronger dipoles. But, in a molecule with many bonds, you need to know the overall structure to know whether the molecule is polar or not. Imagine taking all the individual bond dipoles and adding them up (as vectors, in other words taking account of the direction they point as well as the size of the dipole).
So polar bonds are not enough to make a molecule polar. Carbon tetrachloride has 4 fairly polar bonds but is non-polar because the dipoles in each bond cancel each other out because of symmetry. Many simple hydrocarbons are non-polar for this reason: they have enough symmetry that any polarity in their C-H bonds cancel out. Many simple hydrocarbons are too symmetric to have any polarity, but some are not: isobutane, HC(CH3)3 is polar (dipole moment 0.13 D) because it isn't symmetric and the slight dipoles in its C-H bonds don't cancel out. But compare to chloromethane which is  asymmetric and has a very polar bond: it is fairly polar (dipole moment 1.9 D) as it is dominated by  just one polar C-Cl bond.
But even the non-symmetric hydrocarbons are not very polar compared to things like chlorinated hydrocarbons so are often grouped with completely non-polar compounds.

This is the correct answer, addressing both the (roughly) non-polarity of the bonds themselves and the non-polarity of the overall molecule.

It is because of the very small difference in electronegativity between carbon and hydrogen.This is the reason why alkanes are very less acidic,do not show electrophillic substitutions etc.

Strictly speaking, everything is polar unless it has a perfect spherically symmetric charge distribution, which in ordinary chemical settings means noble gas atoms. Otherwise, where there is not a dipole moment there can be higher order polarity and it can impact solubility and reactivity. The quadrupole in carbon dioxide is one such example.
However, in alkanes all the electrical polarities, not just dipoles, are so small that they have little effect on the properties of the material as a solvent or solute, so they act as a fully nonpolar material would.


text:
Bond order of HCO3- as compared to (CO3)2-
How can we say that the bond order of $\ce{HCO3-}$ is less than that of $\ce{CO3^2-}$? I see that the resonance in $\ce{HCO3-}$ is not as effective as in latter because in latter it is equi-energy resonating structures, and thus energy released in $\ce{HCO3-}$ would be less. But how does that relate to the bond order? The bond order of $\ce{CO3^2-}$ should be $\ce{4/3}$, but why can we say that $\ce{HCO3-}$ will be slightly less than $\ce{4/3}$?

Can we really?$%$

See previous post http://chemistry.stackexchange.com/questions/1302/bond-order-for-carbonate-ion-for-resonance

Yes but then can we talk about BO or Bond length of HCO3-?

You can view bicarbonate approximately as hydroxyl + carboxylate, with bond orders appropriate for them.


text:
Non-zero dipole moment of parahydroxy phenol
How can we explain the non-zero dipole moment of parahydroxy phenol? I know that it cannot be explained by inductive effect, and has something to do with resonance, but still how can we say it is non-zero? I mean, since it's symmetrical, even the resonance effect should cancel out, right?


text:
Explain the 3c 2e or 3c 4e bond in these compounds
Can someone please explain how these compounds are formed/how the 3c 2e comes into existence.

Polymeric structure of BeCl2 - 3c 4e
BeH2 - 3c 2e
CH5+

Please explain in terms of the hybridization of the central atom and how the electrons are shared etc. Thanks!
My approach -
I know that in B2H6 there is a banana bond (3c 2e), but in BH3 it makes more sense - both B go into sp3 hybridization, and one hybrid orbital of each is empty where some overlapping takes place due to the H bonded with the other B, thus causing the electron pair to jump from one B to another. Thus named banana bond. With CH5 I don't understand how C is able to hybridize in a way such that it can have a banana bond. I read that it is of the form CH3 and H2 and 3c 2e bonding in between them, but I don't understand how it takes place.
For BeCl2 there are 2 Be having 2 Cls each and also coordinate bond by other 2 chlorines. Is that right?
But for BeH2, how can there be a 3c 2e bond? I mean based on the same principle, there must be a bond between Be, H that can jump to the other Be with same H, but Be doesn't have enough electrons to do this. Thus my confusion.

This looks like a [tag:homework] question.

No, really, I am more interested in how a 3c 2e bond is formed than getting these answers. I want to know how the "banana bond" comes into existence in such compounds. It is the concept of 3c 2e that I'm interested in. Please edit the question if you want.

Basically any question with the wording your question has is considered homework; it needn't be literally one. Self-study questions, puzzles etc. also count as homework. Don't worry, they're not banned. But, we require a minimal effort. You can just tell us what you did that didn't get answers. We'll take it from there.

Ok I am editing my question now

related http://chemistry.stackexchange.com/questions/32319/hypervalency-in-elements-in-the-second-period http://chemistry.stackexchange.com/questions/4027/carbon-with-5-bonds http://chemistry.stackexchange.com/questions/22088/dimers-and-polymers-of-electron-deficient-compounds http://chemistry.stackexchange.com/questions/34752/beryllium-hydride-violates-the-octet-rule


text:
What is the major product formed on addition of Hydrogen Iodide to Pent-2-ene?

Orginal question:

(1) One of the carbocations formed (CC on the right) is only slightly more stable than the other (both are 2°).
(2) Iodine ion ($\ce{I-}$) is a slightly bigger ion and it faces more steric hinderance while attaching to one of the carbocation (CC on the right).
These two factors are contradictory hence how do i compare them and arrive at a major product?

I would say the reason that the iodine ends up in the 2-position is because at the 2 and 3 positions the charge of the carbo-cation is on a secondary position and has no thermodynamic driving force. The adjacent secondary hydrides may shift without much energy loss leaving the major product to be kinetically determined. Since there are two positions for an addition to the 2-position carbon and one for the 3-position carbon, your 2-position addition has a statistical advantage. Combine this with the fact that two adjacent ethyl groups would be more stericly hindered to attack than a methyl and propyl group and it makes reasonable sense that the addition occurs on the 2-position.

Thanks for the answer, i just have a little bit trouble understanding this part "Since there are two positions for an addition to the 2-position carbon and one for the 3-position carbon your 2-position addition has a statistical advantage".

This is because if the positive charge is on the 3-position, then hydride shift can come from either of the ajacent carbons (2)

Also your answer reminded me that Hyper-Conjugation (obviously >inductive effect) is making stability of 1st carbocation > 2nd carbocation. I would be pleased if you could mention that too in your answer.


text:
How to condense a melted wax in to small aggregate forms like stones of different sizes?
I am an Architect trying to do some experiments with wax. Normally when the wax is melted and poured on to a surface and it takes the shape of the surface with some strings formed . But instead I would like to condense the wax before it hits the surface so it becomes a small aggregate of varying sizes in the form of a stone . How can this be achieved ?
I want it to condense quickly before it hits the surface once its relieved from the melting point and gets in contact with the room temperature of 12-14 Degrees celcius.Should any solvent be added to the melted wax to achieve this state I wanted?
Your suggestions are most welcomed.

Pour it into the water.

large-scale version: https://en.wikipedia.org/wiki/Shot_tower

Part of your problem is that wax is an amorphous solid and a mixture of various compounds.  This causes it to have no specific temperature that can be defined as a melting point, but rather a range of temperatures where it transforms from a hard brittle solid to a soft plyable solid to a liquid.
The best I can recomend is that you first pour very hot wax into a much larger amount of ice cold water to form wax shot.  Then take the shot and a propane torch, torch the wax beads as they fall to partially melt the surface of the shot and that may get you your desired effect.


text:
What's the difference between "conversion" in condensation polymerization and in radical polymerization?
The conversion in condensation polymerization is defined as the fraction of functional groups reacted,
$$p=\frac{[M]_0-M}{[M]_0}$$
while in radical polymerization it's defined as the fraction of monomer that has reacted.
Why is it not the same thing?

Consider dimers. All the monomer has reacted, but the chain length is short.

It is the same thing as long as you have monomer with one double bond (functional group) in radical polymerization.

The reason they are defined differently is based on the differing mechanisms:
All of the monomer in step-growth (of which condensation is a subset) polymerization are reactive, and they are consumed very quickly, forming (mostly) dimers which then go on to react further.  At this point the average length of chains is 2-- not really a polymer, but also relatively monodisperse.  However this reaction is going to continue, and you'll get long chains after some time.  But the monomer itself is gone.
In addition polymerization (of which radical is a subset) there are very few reactive intermediates, and many monomers get added per chain. After 50% of the monomer is converted in addition polymerization, you have some really long chains, swimming around in the remaining monomer (half the original amount).  This contrasts the other situation.
Basically, you can't measure monomer concentration to reasonably monitor the progress of a step-growth polymerization, while you can do such for addition polymerization.  Conversely, you can generally monitor functional groups to monitor both kinds of polymerization.


text:
Why is benzoic acid a stronger acid than acetic acid?
Why is benzoic acid ($\mathrm{p}K_\mathrm{a} = 4.20$) a stronger acid than acetic acid ($\mathrm{p}K_\mathrm{a} = 4.76$), even though the conjugate base in case of benzoic acid is destabilized due to electron donation through resonance. Whereas there is nothing of that sort operating in acetic acid; only inductive and hyperconjugative electron donation.

The answer lies in the resonance energy of various resonance structures of both the carbanions (formed after donating H+).

related http://chemistry.stackexchange.com/questions/7309/resonance-of-benzoic-acid

Among various types of resonance effects equivalent Resonance makes the most stable molecules. Here the resonance with benzene ring causes the carboxylate ion to lose its precious equivalent Resonance and thus decrease its acidity.

@SujithSizon You're second comment is wrong and vide your first comment - we have oxoanions, not carbanions here.

@Mithoron hmm.. but isn't stability of oxanions proportional to carbanion stability here? Also isn't equivalent resonance in $\ce{COO-}$ the reason for the higher energy of separate resonance structures. Don't we use the same fact for proving that acidity: H-COOH > Ph-COOH?

@Mithoron the questions that you have cited are in no way same as what I had asked. Interestingly the two questions in the links even contradict each other.

Aromatic ring is inductive acceptor and mesomeric donor - together it makes benzoic acid stronger then  acetic which is only inductive donor. @SujithSizon I don't get your point

@Mithoron you cannot probably say that aromatic ring is inductive acceptor  straight away. What if we add a fluorine group to it ?

@Mithoron When we try to show that formic acid is more acidic than benzoic acid, all these points regarding Inductive and Mesomeric effect are immediately flawed. Sooner or later Abhirikshma will face the same situation, thats why i was trying to give him hint regarding resonance energies.

I found a solution on YouTube

https://youtu.be/wH3dVLxxoYs     (in HINDI)

It is all about the electron withdrawing/ donating nature of the acid's functional groups.  The more electron withdrawing the group, the stronger the acid will be.  The more electron donating the weaker the acid will be.  A phenyl ring is electron withdrawing, while a methyl group is electron donating.  I don't know why resonance would destabilize an organic molecule or ion.  if you are referring to the carboxylate donating electrons to the phenyl ring, I don't see how this could be possible and not violate the octet rule.

The phenyl ring is donating electron to the carboxylic acid group through resonance, that is what I am talking about.

The aromatic ring is so thermodynaically stable that the phenyl ring donating electrons to the carboxylate is not a major resonace structure. Though it is enough to dictate meta directed electrophilic aromatic substitution.

How is that the ortho effect in substituted benzoic acids increase the acidic strength ?

please elaborate.

When we add a bulky substituent at the ortho position it throws the carboxylic acid group out of plane and thus successfully inhibits the phenyl ring from donating electron through resonance, and increases the acidic strength.

@Abhirikshma http://chemistry.stackexchange.com/questions/7683/ortho-effect-in-substituted-aromatic-acids-and-bases Please refrain fro  prolonged discussions in comments. There's chat for that

@A.K.  I agree that because of the stability due to aromaticity, phenyl ring donating electrons to the carboxylate is not a major resonance structure and that electron withdrawing nature of phenyl group stabilizes the negative charge, but in that case how will we explain that formic acid(pKa=3.751) is more acidic than benzoic acid(pka=4.204). Before reading your comment, I believed that it was because of the destabilization caused by phenyl ring to caboxylate. But now how to explain it?

OP is on the point, to satisfy one condition(benzoic acid>acetic acid or formic acid>benzoic acid)  other will contradict instantly


text:
Stereoselectivity of dihydroxylation with osmium tetroxide

Dihydroxylation responds to steric effects and proceeds at less hindered positions. What will be the major product in the following reaction if the reagent used is $\ce{OsO4}$ followed by $\ce{NaHSO3}$?

Won't interactions between $\ce{- OH}$ bring $\ce{OsO4}$ on the same side as $\ce{- OH}$? Or will the $\ce{OH}$ sterically hinder the reagent from attacking the olefin from the same side?
Why I am I having this doubt is because I read that $\ce{m-CPBA}$ can be directed by hydroxy groups. For example, if the compound given below were to react with $\ce{m-CPBA}$, the product formed is:

The reason is because of interactions between $\ce{m-CPBA}$ and the substrate like this:

EDIT:

It's given in Classics in Stereoselective Synthesis by Erick M Carreira and Lisbet Kvaerno that $\ce{OsO4}$ with $\ce{NMO}$ does not respond to directing effects while $\ce{OsO4}$ with $\ce{TMEDA}$ responds to directing effects.

It's given in the same book that the reason why $\ce{OsO4}$ doesn't attack on the same side as $\ce{-OH}$ is because of electrostatic repulsions between oxygen atoms. These repulsions increase in E-alkenes. Fair enough. But won't the same reason apply for $\ce{m-CPBA}$? Why won't the oxygen atoms repel $\ce{m-CPBA}$ but only $\ce{OsO4}$?

The problem is OsO4 in NMO does not respond to directing effects. This is given in "Classics in stereoselctive synthesis" book by Wiley. No reason is given. Will there be a difference if NMO is replaced by sodium bisulphate?

Also, OsO4 in TMEDA responds to directing effects.

I don't know if this will help but in OsO4, oxygen atoms acts as oxide ligands and hence donates its electron pairs to the central atom (ie: Osmium) hence reducing electronegativity/ability to form H bonds (of oxygen atoms). Whereas in other cases the amino groups react with OsO4 and allow amine interactions...

I know that. What about NaHSO3? Does it respond to directing effects.

@AdityaDev NaHSO3 is just a normal reducing agent, infact it is reduced by adding two moles of H2O like http://i.stack.imgur.com/AtL7U.jpg.

See p 11 of this group meeting presentation: http://evans.rc.fas.harvard.edu/pdf/smnr_1993-1994_Dart_Mick.pdf There are different competing models with references to primary literature. It's a great question and I apologise that I don't have enough time right now to research and write the kind of answer it deserves. Hopefully someone else will do it.

In case of mCPBA a very favourable H-bond can be formed in the transition state/intermediate. The hydrogen in the acid is strongly polarized. I expect no good H-bond between OsO4 and the ROH alone.

Dihydroxylation responds to steric effects and proceeds at less hindered positions. What will be the major product in the following reaction if the reagent used is osmium tetroxide followed by sodium hydrogen sulfite?

The crux of the original question seems to be why standard Upjohn dihydroxylation conditions (cat. osmium tetroxide w/ NMO as the terminal oxidant) doesn't respond to hydrogen bonding based directing effects, whilst stoichiometric osmium tetroxide in combination with TMEDA does.
One of the earliest reports of diamine ligands on osmium (of which TMEDA is one) was by EJ Corey's group in 1996 (, who obtained a crystal structure of osmium tetroxide bound to, in this case, a chiral diamine in a bidentate fashion. (If you're familiar with the Sharpless Asymmetric Dihydroxylation you'll see why this was an important first step).

With the coordination of diamine ligands to the osmium tetroxide understood, several groups began to think about why this allowed for hydrogen bond based direction of the dihydroxylation. The most comprehensive reference I've found is a chapter in Organic Reactions by Donohoe et al. (DOI: 10.1002/0471264180.or076.01 sadly not widely available, but a .doc file missing the figures is available from the publishers website which may be somewhat helpful).
Donohoe's group (in collaboration with several others) also published a detailed study of the transition states of these directed reactions in 2002 (DOI: 10.1021/jo026161y hopefully more widely available).
The essential argument being made is that the TMEDA (or other chelating diamine) is acting as a Lewis-base, providing electron density to the osmium tetroxide, which in turn makes it more willing to hydrogen bond (more electron dense carbonyls are more willing to H-bond). This is perhaps a sensible conclusion given how electrophilic osmium tetroxide is, with 4 highly electronegative atoms all pulling for electron density.
In the paper Donohoe proposes two possible TS structures for the directing effect, with either the oxygen or the nitrogen being involved in coordination to the hydroxyl of the substrate. Through NMR and IR work it was eventually determined that TS-A was the true pathway in which the TMEDA is donating to the osmium tetroxide, with the carbonyl on osmium using its lone pair to coordinate to the hydroxyl of the substate.

To conclude. Standard Upjohn conditions give the product expected by a steric argument alone (opposite to the hydroxyl) whilst modified TMEDA conditions give the product formed through H-bond mediated direction.

But won't the same reason apply for m-CPBA

This part of the question should maybe be another question entirely, but I think its fair to say that osmium tetroxide bears zero resemblance to mCPBA, the electronics of a metal centre vs the electronics of a peracid are completely different and as such their relative ability to hydrogen bond and hence be directed will be significantly different.


text:
Reactions of alcohol with sulfuric acid
I knew two chemical reactions of alcohol with sulfuric acid

$\ce{CH3CH2OH + H2SO4 -> CH2CH2}$
Here product is having a double bond (ethene) and this reaction happens at 443 K temperature.
$\ce{CH3CH2OH + H2SO4 -> C2H5OC2H5}$
Here product is ether an happens at 413 K temperature.

But today I came across another reaction.

$\ce{C2H5OH + H2SO4 -> C2H5OSO2OH + H2O}$

Now my question is when will this third reaction happen? Any specific temperature?
The book from which I found this reaction doesn't describe anything. Just says: Alcohols react with inorganic acids (except halogen acids) to form inorganic esters.

Have you tried searching? https://en.wikipedia.org/wiki/Organosulfate

@Mithoron It doesn't say anything about this reaction.

You understand that these are equilibrium reactions, which happen concurrently and lots of things affect those equilibria?

@Mithoron You mean to say these are three different Major products at different temperatures?

Yes, but you can also influence concentrations directly on start and during reaction. Also pressure, additional compounds like solvent and other factor may be important.

The organic sulphate should be rather prone to solvolysis and loss of sulphate — at least colleagues of mine have gotten themselves quite a few grey hairs when trying to isolate a sulphate-containing secondary metabolite.

This article on Wikipedia says your compound, ethyl sulphate, can be synthesized, by adding sulfuric acid drop-wise to boiling ethanol and temperature should be kept below 413K which agrees with your data. I found also an article according to which authors conducted esterification at about 300K, but with starting presence of anhydrous sodium sulfate.
Also proportion of reagents should be right, if you add more than equimolar amount of acid, removal of leftover is problematic, as sulfuric acid has high boiling point and there's no suitable selective extractant.

See: "The Ethyl Sulfuric Acid Reaction", Evans, P. N. and Albertson, J. M., J. Am. Chem. Soc., 1917, 39, 456.

Actually this is the product produces when ethene reacts with sulphuric acid. You can say like that firstly ethyl alcohol coverts to ethene and futher it reacts with sulphuric acid to give $\ce{CH3-CH2-OSO3H}$. It futher reacts with water to give ethyl achohol again at $\pu{100 ^\circ C}$

You can add a mechanism to your answer: https://chemistry.stackexchange.com/questions/86760/what-would-be-the-correct-mechanism-for-hydration-of-ethene


text:
Predicting the products of reaction between ammonia and hydrogen peroxide
Like we do in organic chemistry is there any way to write the mechanism of the reaction between $\ce{NH_3}$ and $\ce{H_2O_2}$?
Can the products be predicted?

Majority of answer content came from this link and this link.
Strong solutions of $\ce{H2O2}$ with a few drops of $\ce{NH4OH}$ (aq. ammonia) is left for 24 hours without any nitrite formation occurring. But upon longer standing, even with a small amount of hydroxide then nitrite forms. Nitrite also forms when a dilute solution of $\ce{H2O2}$ is mixed with $\ce{NH4OH}$  and is evaporated over pure conc. $\ce{H2SO4}$ with a bell jar. $\ce{H2O2}$ forms (even in very dilute solutions) nitrite very rapidly, if the $\ce{H2O2}$ solution is mixed with a few drops of $\ce{NH4OH}$ and a little $\ce{NaOH}$  and this then boiled in a retort to a very small volume.
$\ce{2NO2- +  2H+(aq) <=> H2O + NO + NO2}$
$\ce{NO2- + NO3- + 2H+(aq) <=> H2O + NO2}$
Nitrogen dioxide is then able to attack ammonia.
$\ce{2NO2 + 2NO + 4NH3 -> 2NH4NO2 + 2H2O + 2N2~~~~~~~~~~~~~~~~~~~~(1)}$
In any case, solutions of ammonium nitrite decompose on heating to 60 to 70C.
Perhaps the very slow spontaneous reaction of aqueous ammonia with $\ce{H2O2}$ takes place the small equilibrium with amide anions, $\ce{NH2-}$.
$\ce{NH2- + H2O2 -> NH2- + 2OH-}$
Even in $\ce{NH4OH}$, there is a very slight equilibrium with amide ions.
$\ce{2NH3 <=> NH2- + NH4+}$
Amide rapidly and vigorously reacts with water, so the equilibrium of amide existence in water must be very low.
$\ce{NH2-+ H2O -> NH3 + OH-}$
In the base catalysed decomposition of hydrogen peroxide, the mechanisms is presumably:-
$\ce{H2O2 + OH- -> HOO- + H2O}$
$\ce{HOO- + H2O2 -> HOOOH + OH-}$
$\ce{HOOOH --> HOOO- + H+(aq)}$
$\ce{HOOO- -> OH- + O2}$
Also, ammonium nitrite decomposes into nitrogen and water.
$\ce{NH4NO2 → N2 + 2 H2O~~~~~~~~~~~~~~~~~~~~~~~~~~(2)}$
A combination of ammonia and hydrogen peroxide is often used in the process of bleaching hair. Hydrogen peroxide has the necessary properties to remove the color from hair - it oxidizes one of hair’s melanin pigments to a colorless substance. This results in the hair becoming lighter. However, it is not recommended by hair professionals (those who don't understand chemistry but want your cash), and can cause extensive damage to hair (even causing it to fall out).
Therefore, the overall reaction is:-
$\ce{2(NH3•H2O) + 3H2O2 ->[\Delta] N2 + 8H2O~~~~~~~~~~~~~~~~~~~~~~~(1) + (2)}$

As far as I can tell, the prior answer while posting several well known reactions, did not actually detail how nitrite is introduced.
To be honest, this is not an easy question to answer and as far as I know, even  the pundits have not agreed on the precise pathway. I suspect the actually kinetics starts with the action of light on an alkaline solution introducing the hydroxyl radical and possible solvated electrons:
OH-(aq) + hv --> •OH + e-(aq)  (see ‘Flash photolysis in the vacuum ultraviolet region of sulfate, carbonate, and hydroxyl ions in aqueous solutions’ by Elie Hayon, and John J. McGarvey, in J. Phys. Chem., 1967, 71 (5), pp 1472–1477, DOI: 10.1021/j100864a044 , link: https://pubs.acs.org/doi/abs/10.1021/j100864a044?journalCode=jpchax )
This could be followed by some radical reactions:
NH3 + •OH --> •NH2 + H2O  (See Eq (1) at "Kinetics and Mechanism of the Reaction of •NH2 with O2 in Aqueous Solutions", by B. Laszlo , Z. B. Alfassi , P. Neta , and R. E. Huie, J. Phys. Chem. A, 1998, 102 (44), pp 8498–8504, DOI: 10.1021/jp981529+, July 15, 1998, at https://ws680.nist.gov/publication/get_pdf.cfm?pub_id=831494 )
•NH2 + O2 --> •NH2O2  (Same source, Eq(2))
O2 + e-(aq) --> •O2- (Same source, Eq(3))
•NH2O2 + •O2- --> O2 + NH2O2- (Same source, Eq(7))
And also, some standard chemical reactions:
NH2O2- --> NO- + H2O (Same source, Eq(8))
NO- + H2O2 = NO2- + H2O  (Speculation, as NO- with O2 (not H2O2) forms peroxynitrite, but NO2- + H2O2 --> peroxynitrite, see http://www.jbc.org/content/289/51/35570.full.pdf, so NO- + 2 H2O2 = (NO2- +
H2O2) + H2O = aqueous peroxynitrite is a supportive argument for the speculated reaction as it suggests a consistent reaction system!)
NH4+ + NO2- = NH4NO2 .
The source of oxygen above follows from the instability of alkaline H2O2 to self-decompose through various pathways or from the presence of air in an open vessel. If dust (which can contain transition metals like iron,..) is allowed to enter the solution, a possible Fenton (or Fenton-like with Cu, Mn,..) reaction could proceed producing a source of hydroxyl radical (in place of the action of light) although this suggested path would be hindered by the addition of OH-, so not my preferred likely explanation.


text:
Racemic compound widely available
Can racemic compounds be found at the local drug store non-prescription? Where might one be found most readily/cheaply? Only one chiral center is needed. More may actually be unhelpful. The purpose here is to practice separating enantiomers.

I know that ibuprofen is sold as racemic mixture, it's a drug for fever treatment. It's very cheap in Brazil, one dollar each box (400mg per tablet).

I think ibuprofen is common and cheap everywhere, possibly with a few exceptions.


text:
Buffer Systems and Ion Reaction
If we have a buffer solution as follows where HA is a weak acid and $\ce{A-}$ is its conjugate base, $\ce{HA + H2O<=>H3O+ + A-}$, do we ever consider $\ce{A- + H2O<=>HA + OH-}$? This is because in all of the buffer system problems I've done we only use the first reaction to find pH and other values but, we never consider the reaction of the conjugate base with water to form hydroxide ions which should increase the pH. Why is this so?

If you are in an acid media don't exist a reason for consider the second equilibrium.

Yes the equilibrium certainly exists. In fact, there exists a third equilibrium, the self ionisation of water. Considering any of the two equilibria gives the concentrations of all the species involved, there is no need to write down the third equilibrium as the third equation will be a combination of the first two. The problem arises only when you invoke stoichiometry. For example, if $0.1\ \mathrm{mol}$ of the acid dissociated, it would certainly give $0.1\ \mathrm{mol}\ \ce{A-}$, but $0.1\ \mathrm{mol}$ would not be the amount of $[\ce{H3O+}]$ as it is part of another equilibrium. However, if we do make this approximation, ignoring the hydrolysis of $\ce{A-} $, we can justify it in some cases. Say if $K_\mathrm{a}$ is around $10^{-5}$ then the equilibrium constant for the hydrolysis would be $K_\mathrm{w}/K_\mathrm{b}$ which would be around $10^{-9}$. Thus it would not be a bad approximation to ignore the second equilibrium for finding the pH.


text:
What is the mole fraction of dihydrogen in a mixture, given the relative rates of effusion of its components?

A mixture of $\ce{H2}$ and $\ce{He}$ at $\pu{300 K}$ effuses from a small hole in a gas chamber. What is the mole fraction of $\ce{H_2}$ in the original mixture, if $3.0$ times as many $\ce{He}$ atoms as $\ce{H2}$  molecules escape from the hole in one minute?

The answer apparently relies on the fact that $\ce{H2}$ collides with the walls $\displaystyle\frac{1}{\sqrt2}$ as often as $\ce{He}$, and we therefore need $\sqrt{2}$ as many $\ce{He}$ atoms as $\ce{H2}$  molecules to yield an equal amount of substance escaping. Considering that the ratio is in fact 3, the final answer is then that the mole fraction of $\ce{H2}$ is
$$x(\ce{H2}) = \frac{1}{3\sqrt{2} + 1}.$$
Can anybody please explain how they got the $\displaystyle\frac{1}{\sqrt2}$ part?

If you are familiar with Graham's Law, you will know that the ratio of the rates of effusion of two gases is inversely proportional to the square root of the ratio of their molar masses.  That is:
$$\frac{R_1}{R_2}=\sqrt{\frac{M_2}{M_1}}$$
Here, $M_1 = M(\ce{H2}) = \pu{2 g//mol}$ and $M_2 = M(\ce{He}) = \pu{4 g//mol}$, so the lighter of the two, hydrogen, effuses at $\sqrt{2}$ the rate of helium, or conversely, the heavier, helium, effuses at $\frac{1}{\sqrt{2}}$ the rate of hydrogen.  That's where the fraction came from.
Since we know that helium is escaping three times faster than hydrogen, there must be three times more helium than would product an equal rate of flow for the gases.  An equal flow would occur when there is $\sqrt{2}$ times as much helium, and triple flow would occur with $3\sqrt{2}$ times as much helium.
The total amount of gas present is then:
$$\ce{H2} + \ce{He} = 1 + 3\sqrt{2}$$
and the mole fraction of hydrogen is:
$$x(\ce{H2}) = \frac{n(\ce{H2})}{\text{total amount of substance}}
= \frac{1}{3\sqrt{2} + 1}$$


text:
Is there an easy way to make PVC from PVA?
Say, I took Elmer's glue (or a similar Polyvinyl Acetate) and I inserted some Chlorine Tabs and warm it up while stirring.
Would the electronegative Chlorine turn some of it it into Polyvinyl Chloride (PVC)?
Would I get a Dioxin Compound too?
Would I get any other toxic Side-Product?

If it's a practical question, just don't do it. You won't get any PVC this way.

I want to tell you what the practical question is:
I mix together gelatin, casein, silicone and many others to try making bio-plastics from cheap home materials.
And some recipes also mention the use of PVA.
First of all: I do not want to accidently create toxic vapours, I was even afraid of mixing DMSO with EthylenAcetate.
Secondly: the PVA compound will always stay water solulable - could be nice to partly turn this compound into PVC.

Choring Tablets have positivly charge cholrine atoms you want negative, so chlorine tablets will not work.  And no you won't get a dioxin compound unless you burn it, then you may ge some.  If you want to make PVC from PVA you can use Hydrochloric acid but you may not get a good yield.

thanks for your answer!
I was unable to vote you up, somebody has voted you down and I have no clue why he did.

I really don't think that HCl + PVA will give much PVC. It is always going to be far easier to make PVC from the monomer.


text:
How to create different fluorescent colors using household items
This is a request:
I wanted to create fluorescent liquids that have different colors.
I'm going to dye my flowers in different fluorescent colors and I'd need a recipe for fluorescent colors that aren't too toxic.
Please bare in mind that the injection of the liquid can only happen by spraying or watering with the liquid.
The liquid needs to be able to handle being Waterdown by tap water.
I'd recommend using household items to make this more helpful to the average day user. Also any color will do but, the most wanted Colors are:

List item

Blue
Green
Red
Yellow

Welcome to chemistry.stackexchange!  This is an interesting question.  May I suggest that you edit your question to include a description of the research you have already done on the topic?  For example, what documents have you already read?  If you put that into the question, folks trying to answer it won't waste your time by directing you to reference materials that you have already seen.

http://chemistry.stackexchange.com/questions/41858/how-to-create-different-fluorescent-colors-using-household-items?s=1|0.5338

If you dismantle highlighter pens, you can find the felt material inside. This material contains fluorescent chemicals, such as pyranine. This felt like material is usually encased in a plastic film, remove the film and soak the felt in water.
If you use too much water then it may become too dilute and you may start to encounter an issue with the ink/water mixture sticking to the flower.

Are you using a black light to view the fluorescent colour?

it might be worth trying to cut the stems and keep the flowers in the ink containing water. Then as they absorb the water they will absorb the fluorescent ink as well and may take on some of the colour (and viewable under black light).

You might try some light-sticks for fluorescent chemical. Some of these chemicals are non-colored or pale  in visible light, but are brightly fluorescent under long-wave UV.
Note that though those I use are labeled "non flammable [sic] and non toxic [sic]... ingredients will not cause injury to the eye... do not drink or ingest," they do contain easily broken glass containers with a "luciferase". Use caution with any experiment.


text:
How to monitor (high) oxygen content in gas (air)?
How could I go about monitoring (semi-continuously, say, one measurement per 10 minutes) oxygen content in an oxygen-rich air mix (air with substantial amount of nitrogen removed with a Zeolite filter)?
There are questions (1, 2 ) about measuring oxygen content in air, using a candle and checking volume after burning. Them being cumbersome is less of a problem - worse that lighting a candle in 80+% oxygen atmosphere can a be quite... energetic reaction.
Are there any other reactions that can provide oxygen content measurements like that, preferably not involving open flame?
(Neither accuracy nor measurement time are very important, but the measurement should be easy to perform, and repeatable, no large amounts of expensive consumables. Something that could be automatized would be nice.)

Would not an oxygen sensor designed for such an application be the best bet? Googling is a bit tricky since many results point to oxygen sensors for automobiles, but there are multiple vendors for 0-100% oxygen meters.

http://www.systechillinois.com/en/paramagnetic-cells_54.html

I would say paramagnetic oxygen analysis or a typical four gas meter.