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How much CO2 is created during water electrolysis with a graphite anode?
I'm experimenting with water electrolysis and wanted to use non-toxic materials, so I settled on pure graphite rods for both electrodes and Epsom salt (magnesium sulfate) as the electrolyte.  The system works reasonably well, but upon further reading I found that some of the $\ce {O2}$ is likely reacting with the graphite and being converted into $\ce {CO}$ and $\ce {CO2}$ at the anode.
How much $\ce {O2}$ am I probably losing to $\ce {CO}$ and $\ce {CO2}$ ?  Just a small percentage, or a majority of the potential $\ce {O2}$ production?  How can I tell?  I'm capturing the resulting mixed gas product in soap bubbles and igniting it but it seems to burn somewhat less vigorously than I was expecting, which makes me think there is rather more $\ce {H2}$ and less $\ce {O2}$ than optimal.

Just out of curiosity, I would love to see your experimental setup. Could you take a picture?


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What is the Chemistry behind "good" and "bad" cholesterol & their relation to Cis/Trans fatty acids?
I have searched Google for this but cannot seem to find an explanation. My question is as follows - what is the Chemistry/Biochemistry behind cis and trans fatty acids resulting in "Bad" LDL and "Good" HDL cholesterol? Is one synthesised favourably with cis/trans fatty acids as precursors?
I am under the impression that LDL results in reduced cell permeability (how?), but other than that I am struggling to find information.

Your question is better suited for biology site, and actualy almost answered there: http://biology.stackexchange.com/questions/5152/hdl-and-ldl-cholesterol

source: Stack Exchange

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How to produce fumes in chemistry lab?
In chemistry lab we perform experiments to find the unknown solution. I just want to produce some fumes in lab. We are only allowed to use few common dilute solution like $\ce{NH_4Cl}$, $\ce{NH_4OH},$ $\ce{HCl}$, and few more which are generally allowed to students.
Is it possible to form fumes in lab?

Why would you want to produce nasty smelling and potentially harmful gasses in the lab? This just sounds very insane to me.

@Martin frankly speaking, my practicals are over in 15-20 minutes but my lab teacher wants that we should be there for at least 45 minutes.But after this she will not force us!

Aqueous ammonia and hydrochloric acid are solutions of gases.
In a fume hood place two beakers with aliquots of these two solutions (one in each beaker, don't mix) side to side.
Did I mention the fume hood?
You'll see the formation of solid ammonium chloride from the reaction of the two gases.
$$\ce{HCl(g) + NH3(g) -> NH4Cl(s)}$$
Don't forget to clean up the mess!

I wanted to know with liquid solution, is it possible?

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Where does the energy in the exothermic reaction of CaCl2 with water come from?
Today I created $1~\mathrm{M}$ aqueous solution of $\ce{CaCl2}$ and realized that the water before the mixture had a temperature of 23.4 Celsius degrees. Then I added the $\ce{CaCl2}$ and the temperature started growing until it reached 27.5 Celsius degrees. So we have an exothermic reaction. What does that actually mean and how does it work in my example? By the book, an exothermic reaction occurs, when the energy stored in reactants is smaller than the energy stored in products.
Here we have (with my opinion):
$$\ce{CaCl2 + H2O <=> Ca^{2+} + 2Cl^{-} + H2O}$$
From where did that energy come from? Is it just released from the fission of $\ce{CaCl2}$ reactant or from the fusion of some molecules of $\ce{CaCl2}$ again in products? Also, is it possible to have in the products side $\ce{CaCl2}$ molecules or only ions of $\ce{Ca}$ and $\ce{Cl}$?

Actually, exothermic means that energy is released so that the initial energy is higher than the final one. That's say the contrary of your assumption.  This energy is contained in the atoms states themself and in the bondings. So when you break the bonds to form another compound, you change the energy state of your atoms, releasing energy if the final state is more stable (less energetic). Since in your reaction it's the case, the reaction happens then spontaneously.
Since you should keep in mind that nothing is lost, everything is transformed, this extra energy remaining is transformed in heat.

Actually, exothermic just means that the overall change in the enthalpy is negative. That usually means heat is transferred to the system. When you are talking about the spontaneous reaction, you are talking about an exergonic process, wich can still be endotherm. End exothermic reactions could be endergonic, but this is just a hypothetical case.


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Symmetry representations for acetonitrile (CH3CN)
Can anybody share the symmetry labels for acetonitrile ($\ce{CH3CN}$), please?
I mean to assign the symmetry labels to orbitals of acetonitrile

I am afraid I disagree with Geoff's assignment of acetonitrile to the point group $C_{\infty v}$ and "approximation" with $C_{2v}$, so I will make another attempt to answer this question. The way I have structured this answer, the resources I used, and the level of detail included follow a similar question on this site.
Molecule structure
First you need to know the geometrical structure of your compound. This can be found easily from Wikipedia. The image is reproduced here.

Assigning the point group
Then you need to find the point group of acetonitrile. This can be done by following flow charts, one of which is available here. While it may be a little difficult at the beginning, with some practise you can look at a molecule and determine its point group accurately, so it is worth working through the flow chart for a range of molecules until you are confident with this. There are other arrangements of this flow chart, but all flow charts should give you the same result, so use the one provided in your course if possible.
With the flow chart I have linked to, we go through these steps (if you have a flow chart organised differently, you will encounter the same steps, but likely in a slightly different order - check this is the case!):

molecule is not linear
molecule contains only one $C_3$ axis
molecule DOES contain a proper rotation axis which is $C_3$ (rotation over methyl group)
the highest rotation axis is $C_3$ but there are no perpendicular $C_2$ axes
the molecule does NOT contain a horizontal reflection plane ($\sigma_n$). This can sometimes be confusing -- in general, a horizontal plane is perpendicular to the highest rotation axis.
the molecules DOES contain 3 vertical reflection planes ($\sigma_v$).
therefore the point group is $C_{3v}$.

"Orbitals"
In your question, you said that you need to "to assign the symmetry labels to orbitals of acetonitrile". This is a little unclear, because you can mean that you want to assign symmetry labels to atomic orbitals of acetonitrile, or molecular orbitals of acetonitrile. I am going to assume that you need the molecular orbitals here. But how do we find the molecular orbitals?
Molecular orbitals are constructed by making combinations of atomic orbitals. Which means that, a logical next step is to consider all the valence atomic orbitals (VAOs) on all the atoms of acetonitrile.
Atomic orbitals and symmetry sets
To help us talk about VAOs, I have put numbers on all the atoms.

Carbon, atoms 1 and 2: $2s$, $2p_x$, $2p_y$, $2p_z$
Nitrogen, atom 3: $2s$, $2p_x$, $2p_y$, $2p_z$
Hydrogen, atoms 4, 5 and 6: $1s$
When we try to tackle these types of problems, it is common to construct symmetry sets of atomic orbitals that would transform in the same way under the operations of the point group of the molecule. You will also notice that you cannot use any of the operations of the group to transform certain atomic orbitals on the same atom into each other (cannot transform $2s$ to $2p$, cannot transform $2p_x$ to $2p_z$, etc), so these should be considered as separate sets. The $p_x$ and $p_y$ orbitals are mixed by the rotations and reflections of the group, so we should consider them as one symmetry set.
In acetonitrile, there is only one nitrogen, so we have to consider as separate sets:

$2s$
$2p_z$
$2p_x$, $2p_y$

Carbon 1 and carbon 2, though both carbons, cannot be transformed from one to another, and your chemical intuition should also tell you they are different carbons because they have different bonds attached to them, so they are also in separate sets of their own. The three hydrogen atoms, on the other hand, are related by rotation (a $C_{3v}$ rotation on the principal axis, so they can form one set of 3 hydrogen 1s orbitals.
Assigning symmetry labels to each set of valence atomic orbitals
The usual trick at this point is to simply read off the orbitals that have corresponding symmetry labels, e.g. the $p_x$ orbital would transform as $x$ on the character table. This definitely works for any atom that lies on the origin point of the point group, but it is not completely obvious where you would take the origin point for putting a (right handed) axis system on acetonitrile. The "middle" of this molecule would appear to be somewhere between the two carbons, and that may seem a little awkward.
Remember that in all cases, the choice of origin and axis system is actually arbitrary, however you should pick the most convenient one. Now, for acetonitrile you can pick any point on the principal rotation axis ($C_{3}$) and it would work, because that is an inherent feature of the $C_{3v}$ point group. If this is not obvious, I suggest looking at the answer I referenced previously, and also working through any other examples that you can find.
Luckily for us, $C_{3v}$ has quite a small character table (i.e. it does not have many symmetry elements), so we can work through it and check. The character table of the point group $C_{3v}$ is shown below and can also be found here. The character tables for many common point groups can be found here.

$s$ orbitals that lie on the principal axis
Now let's consider the $2s$ orbital or carbon atom 1. It should be fairly obvious that it transforms back to itself under all the operations of the point group (unchanged under identity $E$, any of the $C_3$ rotations, any of the $\sigma_v$ reflections), and so it transforms as $A_1$. It should also become very obvious that this is also true for the $2s$ orbitals of carbon atom 2 and nitrogen atom 3, even though these atoms are not symmetry-related, their atomic orbitals can have the same symmetry under the point group. This will be important when we come to construct molecular orbitals.
$p$ orbitals that lie on the principal axis
For p orbitals, we need to impose an axis system. Conventionally, the principal axis is taken as the $z$ direction. You have to follow these conventions when doing symmetry analysis.
Let's consider the $2p_z$ orbital or carbon atom 1, which lies along the principal axis. This again transforms as $A_1$.
The $2p_x$ and $2p_y$ orbitals are a little tricky. They remain identical under the identity operator $E$, giving a character of 2. For the reflection operator $\sigma_v$, one of the three $\sigma_v$ planes would keep one identical while inverting the other one (giving zero), and the other two $\sigma_v$ planes give characters of zero, so the overall character for $3 \sigma_v$ is zero. The $C_{3v}$ is the difficult one, where you have to put vectors on the axis, rotate them, work out the geometry with trigonometry, form the $2 \times 2 $ matrix and work out the trace (i.e. the character) is -1. Or you can use the fact that the character tables show (x,y) transform as the irreducible representation $E$.
Again, the same symmetry labels apply for all the atomic orbitals for all atoms that lie on the principal axis.
Set of hydrogen $1s$ orbitals that do not lie on the principal axis
For these hydrogen $1s$ orbitals we examine explicitly how they behave under all operations of the point group.

$E$: 3  (all three $1s$ remain unchanged from themselves)
$C_v$: 0 (all three change and do not remain at the same position)
$\sigma_v$: 1 (for any one operation, only 1 of the orbitals remain the same, the rest change)

This is a reducible representation. You need to reduce it to find the symmetry labels for the hydrogen $1s$ orbitals.
Symmetry labels for all valence atomic orbitals
For the whole molecule of acetonitrile, the valence atomic orbitals give the irreducible representation $4 A_1 \oplus 4 E$. From the discussion above you should be able to deduce the exact symmetry label for any atomic orbital on any atom of the molecule.
Molecular orbitals
You know that you need to generate as many molecular orbitals as atomic orbitals, and only the atomic orbitals of the same symmetry can mix. Molecular orbitals use lower case letters, so you expect two types of symmetry labels based on the irreducible representations of the VAOs. You should be able to say how many molecular orbitals of each type you expect, remembering that the irreducible representation $E$ is two-dimensional and indicates a degenerate pair of VAOs.
Symmetry analysis cannot give you the energy of these molecular orbitals, however, so you cannot know how they are ordered relative to one another.

I have no problem with working acetonitrile as $C_{3v}$. Clearly if you treat the methyl as fixed, that would be correct. However, in most cases, when bonds have free rotation (e.g., terminal $\ce{-CH3}$ or $\ce{-OH}$ you really should approximate as a spheroid) IMHO.

Reality, of course, is somewhere in between. Depending on the spectroscopic technique, you'll either observe behavior approximating $C_{3v}$ or $C_{\infty v}$.

I'm afraid I disagree. I believe the normal modes you find using a $C_{3v}$ and $C_{\infty v}$ would have different symmetry inherently, as $C_{3v}$ has a $E$ irreducible representation while $C_{\infty v}$ has $E_1, E_2 ... E_n$, so a simple approximation would not give you normal modes with the same IR. I haven't gone through the whole normal mode analysis, so I can be wrong here. But that is my gut feeling! Of course, what you actually observe as normal modes may have accidental degeneracy as well, but that is a possibility with any symmetry analysis.

For vibrational spectroscopy, if you used $C_{\infty v}$ instead, you will miss out on all the asymmetric modes. http://webbook.nist.gov/cgi/cbook.cgi?ID=C75058&Mask=800#Electronic-Spec

@selkie222 You might want to add that the 3 Hydrogen s orbitals can be used to build an orbital that forms a basis for $A_1$ and a pair of linear combinations that form a basis for the representation of E. With this we have in total a representation that is a sum of $5A_1 \oplus 5E$ which agrees with the total number of valence orbitals which is 15.

It depends a bit on how you handle the $\ce{CH3}$ group. Usually, you assume free rotation, so it's effectively a sphere. In that case, $\ce{CH3CN}$ is linear and asymmetric. Since the two ends are different, that places it in the $C_{\infty v}$ point group.
Typically, though, we don't assign molecular orbitals using $C_{\infty v}$ or $D_{\infty h}$ representations. It's hard to do the math when using an infinite rotation axis, so we usually use $C_{2v}$ and $D_{2h}$ instead.
That is, you'll want to use the irreducible representations from $C_{2v}$ for acetonitrile.

"$\ce {CH3}$ is linear and asymmetric."  How can it be linear ?

In acetonitrile if you treat the methyl as spherical, then you have three linear spheres in the molecule, but of course it's asymmetric because CH3 and N are different

@GeoffHutchison Do you know papers where this is done ? I remember some discussions with a Professor from the theoretical chemistry department who was quite adamant that this should not be done and would be interested if there a publications that use this approach.

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Please explain the mechanism of Cannizzaro's reaction
I did not completely understand the mechanism of Cannizzaro's reaction, especially the rate determining step i.e. the hydride transfer step, and the effect of electron withdrawing substituents on this step.
Like, I was taught that the intermediate which donates hydride, would donate hydride more readily when electron donating substituents were attached to carbon atom with the outgoing hydrogen atom, as they would increase electron density on that carbon, and therefore create a "pushing" effect on the outgoing hydride. But in all the questions I have attempted, the answer says that electron WITHDRAWING substituents facilitate hydride transfer.
Please tell how the electron withdrawing substituents affect the hydride transfer, and why. Also, please explain how to determine the products of a crossed Cannizzaro's reaction. Do we determine them using the concepts of the RDS,or otherwise?


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Can we repair a dead body?
The following passage has been extracted from the book "Organic Chemistry-10th edition by Solomon and Frhyle":

"The growth of living things from microbes to elephants rests on organic reactions, and organic reactions provide the energy that drives our muscles and our thought processes"

Is the above statement true that, organic reactions provide energy that drives our thought processes? Is there any other process by which we are given energy to drive our thought processes?
If you all agree with the statement, then when a person is called "dead" (or after a while being called dead), I hope the reactions will cease (because he can't perform "thought processes"), isn't it?
If these all points are agreed, is their a possibility of repairing a dead body by restarting the reactions immediately?
I agree that, it is not as easy as I said, there is a complexity, but I believe we all can do some research "together" and try to do a "little" contribution to save ourselves.
So, can we repair a dead body?

Sorry, I didn't find Biochemistry Stack exchange, so I posted here. If this is not the right place, please inform me about where I should post this question.

There's Biology.SE where that might be on topic, especially concerning the definition of "death". Also, restarting (high-level) cognitive processes might not induce (low-level) cellular repairs.

@tschoppi: Is this not on topic here?

As long as your question only concerns the chemical reactions that provide this energy and alternative sources, this is on topic. But life and death really is more in the realm of biology and ultimately philosophy, so I'd try to cut out as much as possible from that here.

I'm voting to close this question as off-topic because it's rather a biology question.

No, you cannot. There is no way to reset the transient pulses in the neurons that carry the essence of thought processes. The information is already gone.
You will find the neural synapses all in the discharged state. All thought process, all memory, all of what makes a person is gone.

This answer is very short and has no explanation. Perhaps you could add some more detail about why this is the case and also comment on the main point of the question which is about the reactions that drive our though processes.

Could you be a little more elaborate, please? Like I dunno, by saying how you decide that the info is gone, by bringing references or simply by explaining more.


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How to study Sodium hypochlorite stability with materials

first of all I'm not not sure to be in the right stack exchange site, but I'm a material engineering student and this year I have a project: choose a material for a container filled with Sodium hypochlorite, my university has given us CES Edupack, that contains a lot of infomartion about mechanical and physical properties, but nothing about material stability.
Since we haven't done any lecture about materials stability/decomposition, I'm here to ask if there is someone that could enlight me about how to proceed to evaluate the stability of a material.The only thing that I came across while thinking about it is to avoid materials(polymers) that has got active groups with Sodium hypochlorite, is it a good approch?
EDIT
Maybe my question is a little dispersive so I make it clear here: my main concert is more general: how do I study the stability of sodium hypochlorite inside the container? What should I check to be sure that the container doesn't "melt"?

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site.

Sodium Hypochlorite  is soluble in water and reacts with all organic solvents. Sodium hypochlorite is extensively used in house-holds as a bleaching agent and it is a chemical compound with formula $\ce{NaClO}$.
Household sodium hypochlorite bleach was introduced to Americans in 1909 and sold in steel containers, then in glass bottles. In the early 1960s, the introduction of the plastic jug brought a cheaper, lighter, and nonbreakable packaging alternative. It reduced transportation costs and protected the safety of workers involved in its shipping and handling. Additionally, the thick plastic did not permit ultraviolet light to reach the bleach, which improved its chemical stability and effectiveness. In recent years, how-ever, plastic containers have become an environmental concern because of the time it takes the material to decompose in a landfill. Many companies that depend on plastic packaging, including bleach manufacturers, have begun to reduce the amount of plastic in their packaging or to use recycled plastics. In the early 1990s, Clorox introduced post-consumer resins (PCR) in its packaging. The newer bottles are a blend of virgin high-density polyethylene (HDPE) and 25% recycled plastic, primarily from clear milk jug-type bottles.
Read more: http://www.madehow.com/Volume-2/Bleach.html

Thanks for the reply, but my main concert is more genera: how do I study the stability of sodium hypochlorite inside the container? What should I check to be sure that the container doesn't decompone?

To keep sodium hypochlorite solution stable, protect it from light,including direct sunrays and store it separated from acids.

To be sure that the container doesn't decompose, I propose to monitor the composition of hypochlorite solution using gas phase chromatography GPC or HPLC.


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Is a carbon-fluorine bond stronger than a carbon-chlorine bond?
In hydrolysis/nucleophilic substitution of haloalkanes, the bond enthalpy indicates the rate of hydrolysis (e.g. the $\ce {C-I}$ bond is weaker than the $\ce {C-Cl}$ bond so in hydrolysis the alkane with the $\ce {C-I}$ bond will break more quickly and react with the $\ce {OH-}$ ions)
However, in free radical substitution reactions, if you have a molecule like $\ce {CCl2F2}$ , then is the $\ce {C-F}$ bond stronger or weaker than the $\ce {C-Cl}$ bond (so which will require less energy to break)?

For free radical reactions, the most important parameter in assessing bond strength is bond dissociation enthalpy (BDE). Typical values for $\ce {C-F}$ bonds are around $\mathrm {100\ kcal/mol}$, while for $\ce {C-Cl}$ bonds are around $\mathrm {80\ kcal/mol}$; $\ce {C-Cl}$ bonds are therefore weaker.
This can be rationalized by considering the poorer orbital overlap between the $\ce {Cl}$ bonding orbital (3 s/p character, more diffuse than 2s/p) and the $\ce {C}$ $\mathrm{sp^3}$ hybrid (predominantly $\text{2p}$ character) than that between the $\ce {F}$ bonding orbital (2 s/p character) and the $\ce {C}\ \mathrm{sp^3}$ hybrid.
Some data on typical BDE values in $\ce {F/Cl}$ compounds can be found here.

But why would polar bonds have a higher bond enthalpy?

This is often the case; bonds with poorer orbital overlap are typically weaker even with a better electronegativity match.

Rather than looking at different bond strengths, when determining which bond reacts favourably one should look at reactants and products.
In nucleophilic substitutions or eliminations, this means considering the more stable halide anion. The larger the halide, the more diffuse and thus the better stabilised the negative charge is.
In radical reactions, the stability of the product is also more important. Here again, the lower an element is in the periodic table, the less unstable the radical is.

I would like to offer an alternative explanation. The answer provided by J. LS does not provide the main reason for why the $\ce {C-F}$ bond is stronger than the $\ce {C-Cl}$ bond. Yes, it is true that fluorine's $\ce {2p}$ orbital is less diffuse than chlorine's $\ce {3p}$ orbital. However, it is also important to note that orbital energies need to be accounted out in determining the effectiveness of the covalent interaction between two orbitals. In order for two orbitals to interact strongly, they must be of similar sizes and energies. The Pauling electronegativity values of fluorine and chlorine are $\ce {3.98}$ and $\ce {3.16}$ respectively while the value for carbon is $\ce {2.55}$. Clearly, the orbital energies of chlorine and carbon would be more similar than the orbital energies of fluorine and carbon. Thus, we cannot conclude that the covalent interaction between carbon's and fluorine's atomic orbitals is stronger than that between carbon's and chlorine's atomic orbitals.
What then can we use to rationalise the bond strengths?
We can actually explain this using the idea of bond polarity and ionic contributions to the covalent bond. Lemal (2004) reports that the partial charge on carbon in the molecule $\ce {CF4}$ is $\ce {+0.76}$. This illustrates that partial charges are actually very significant when looking at the $\ce {C-F}$ bond. Thus, it is due to this ionic character of the $\ce {C-F}$ bond, that is, the attraction between the partial positive charge on carbon and the partial negative charge on fluorine, that results in the $\ce {C-F}$ bond being significantly stronger than the $\ce {C-Cl}$ bond.
Reference
Lemal, D. M. (2004). Perspective on Fluorocarbon Chemistry. The Journal of Organic Chemistry,69(1), 1-11. doi:10.1021/jo0302556

authors: ورده الياسمين


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Organic synthesis problems
I am an A-level student going into university and I am looking for a source of organic synthesis problems and answers that is freely available on the internet. The difficulty level I am looking for is something like first year undergraduate level. Any links would be greatly appreciated.

I believe there is some site where you can practice synthesizing a compound (theoretically of course) by selecting reactants and starting materials. Very nice idea, but unfortunately I cannot find it anymore... anyone?

Search for the latest IChO tests and preparatory problems. You can find quite a few interesting problems.
Here are some tests:
http://www.iuventa.sk/files/documents/5_icho/past%20competition%20tasks/volume3-icho41-45.pdf
http://icho2013.chem.msu.ru/materials/Preparatory_problems_IChO_2013.pdf (problems 18-21)
http://icho2014.hus.edu.vn/Preparatory-Problems/0001/01/Preparatory-Problems-224.html (problems 18-26)

Janice G. Smith's textbook "Organic Chemistry", while unfortunately expensive, has a good interactive workbook online which is useful even without the companion text.
You can find it here.

That website is not working for me. Does it work for anyone else?

@yolo123 Select the chapter, then open the corresponding multiple choice quiz.

I would recommend doing textbook problems e.g. those of David Klein and Clayden. In addition, it might be useful to refer to MIT open courseware's organic problems. I have included their links:
https://ocw.mit.edu/courses/chemistry/5-12-organic-chemistry-i-spring-2003/
https://ocw.mit.edu/courses/chemistry/5-12-organic-chemistry-i-spring-2005/
These problems might be old, but I still found them useful... As for which year to do, I don't think it matters too much

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What is the reason behind the numbering of nitrogen bases?
I am studying nitrogen bases and their structures and roles in nucleotide, polynucleotides and nucleic acids formation. In the structures of the nitrogen bases like Adenine, Guanine, Cytosine and Thymine, the numbering is baffling me. Why are the atoms in these molecules numbered so? Can someone please specify the rules for numbering these four molecules? I am a biologist and I'm not so good at chemistry.

A minor correction: I think cytosine has a single bond to nitrogen atom of the amino group at the 4-position.

This customary numbering is not systematic. https://doi.org/10.1351/goldbook.P04953

ACDLabs maintains a nice summary of the IUPAC nomenclature rules with examples:
Heterocyclics (pyrimidine): http://www.acdlabs.com/iupac/nomenclature/79/r79_702.htm
Fused heterocycles (purine): http://www.acdlabs.com/iupac/nomenclature/79/r79_960.htm
Briefly, apply the following rules:

Number your rings so that the nitrogens end up with the lowest number combination. Thus pyrimidines have (1,3).
Number your rings so that other functional group have lowest possible numbers. Hence pyrimidine is numbered to the exocyclic carbonyl groups are (2,4).
In a fused system, numbering should prefer (in this order): ring with more nitrogens, rings with other heteroatoms, larger rings, nitrogen atom closer to ring junction.
Number toward the farther ring junction from where you started. Do not number junction positions.

Number 3 is fine and all, but purine seems to have a numbering scheme that does not match these rules, and it is supported by IUPAC. Likely, this represents a historical numbering pattern that predates the IUPAC rules. Everyone was using it and so it was adopted. You will likely just need to memorize it (or look it up when you need to).

Rule 4 clearly has some exceptions, one of which deals with non-carbon atoms (i.e. N) present at fused junctions.


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How can I relate the reactivity series to electronegativity and ionization energy?
I am trying to figure out how the reactivity series comes about. My understanding is that elements with a higher electronegativity will be more reactive than elements with a lower electronegativity, and that elements with a low ionization energy will be more reactive than elements with a high ionization energy.
Here is a chart of electronegativity (from Wikipedia):

This shows electronegativity decreasing (as reactivity increases) down Group 1 and it also decreasing (as reactivity decreases) down Group 7.
Here is a chart of ionization energy:

(source: genchem at www.dartmouth.edu)
This shows ionization energy decreasing (as reactivity increases) down Group 1 and it also decreasing (as reactivity decreases) down Group 7.
Neither electronegativity nor ionization energy can predict both reactivity trends, so what is going on?

Two thoughts: 1. More reactive with what? 2. The alkali metals are very power reducing agents - their reaction with water gets more explosive as you go down the group (to lower electronegativities).

@BenNorris I guess this question is referring to the metal reactivity series. The WP page says that: "It is used to summarize information about the reactions of metals with acids and water, double displacement reactions and the extraction of metals from their ores."

With less electronegativity, there is also less ionization energy. This means for metals, it is easier to lose the electrons. When it is easier to lose the electrons, the metal becomes more reactive because it uses less energy to fulfill its goal; to gain a full outer shell.
For nonmetals, the main goal is to gain electrons and not lose them. Thus, an increase in electronegativity and ionization energy means an increase in reactivity because the nonmetals want to hold on to their electrons and gain more vigorously to achieve the full outer shell.

Reactivity is not only about the tendency of an element becoming an ion( as reflected in ionisation energy and electronegativity). Reactivity takes into account the stability of the produced ion. Often reactions that take place in an aqueous solution takes into account the stability of the ion in water.

Following is from Wikipedia

Electronegativity, symbolized as χ, is the tendency for an atom of a given chemical element to attract shared electrons (or electron density) when forming a chemical bond.1 An atom's electronegativity is affected by both its atomic number and the distance at which its valence electrons reside from the charged nucleus.

Whereas an element's reactivity is based more on how ready it is to gain or lose electrons. Here is a visual explanation of the situation.

As the atoms move from left to right, the electronegativity of these atoms are decreasing as the distance between the electrons and the positively charged nucleus is would be greater, giving it lower ability to attract electrons. However, as you said the reactivity series says potassium is more reactive than sodium. Let me explain...
The potassium would have a lower tendency of attracting electrons compared to sodium, it would hence more readily lose electrons to form potassium ions (K+) to form ionic bonds with other elements. The reaction would tend to occur more readily hence the reactivity of potassium is higher than sodium.
There are also other cases in the reactivity series that are explained in a similar way. Eg. Sodium and Magnesium. (For this you need to consider only the net charge of the ions formed as the charge would be considered neutral once electrons and protons are equal). The magnesium ion has charge of 2+ while Na has a charge of 1+, this would mean that the there is stronger attraction between the valence electrons of Mg than the valence electrons of Na. The magnesium will lose the electrons less readily and so it will be placed lower than sodium in the reactivity series.
The reason why it is different for the non metallic elements is that they need to gain electrons in order to react and form bonds. This would mean that a higher electronegativity would give it a higher ability to gain electrons and hence react more readily, placing them higher in the reactivity series, unlike the metallic elements that would go through a different scenario.
In summary, the electronegativity of an atom affects the reactivity whether it needs to gain or lose electrons to react. Hope this answers your question!


text:
Balloon from float to sink?
Everytime I walk by the balloon stall, I know that they use helium to make the balloons float in the air. But my problem is why when I leave the balloon aside, the balloon sinks to the bottom. Is there anything happen to the  helium inside?

some of the helium leaks out

maybe I don't understand your question properly, but helium can slowly pass through the wall of the balloon (depending upon the material used), or through an imperfectly sealed closure such as a knot.

Maybe the wall.

see: http://www.karaloon.com/en/home/quality-and-safety/balloonsheliumfloating-time.html

Helium gas is made from tiny atoms which effuse through the pores in the balloon, usually rubber has more pores than a foil balloon so will sink faster

When the balloon is fully inflated with helium its density (mass/volume) is less than that of the air around it, allowing it to float. Over time, some of the helium will "leak" out of the balloon. As the helium is released the mass of the balloon decreases slightly, but the volume decreases at a much faster rate, resulting in a balloon with a density greater than that of the air around it, which is why it falls to the ground.


text:
What do different forms of ice look like?
Everyone is familiar with the very common, hexagonal ice. Bit there are many other forms of ice - what do they look like? If you had big blocks of ice in all the different known phases next to a block of hexagonal ice, what obvious differences would you notice?

As you know, ice can exist in at least 15 or so solid phases. Of these, the first phase, $Ice$ $I$ (ice-one) is the only phase found naturally on earth, mostly in the form of $Ice$ $I_h$ hexagonal crystal as seen in glaciers and snowflakes.

The visible appearance of ice (eg: transparency and clarity) depend also upon the level of purity of the crystals formed. For example, air can get trapped within inclusions of the ice, making it translucent or opaque white, as shown in the image below.

The cubic form $Ice$ $I_c$ is metastable so is rarely found, generally in high clouds in the upper atmosphere, where it is formed at ambient pressure by the condensation of water vapour at very low temperatures (-80°C).
Most other forms of crystalline ice (designated as $Ice$ $II$ to $XV$ are metastable, being produced in the laboratory by the application of high pressures (above 200MPa) and/or low temperatures (below 72K), resulting of denser packing of the water molecules.
For example, $Ice$ $II$ (see image below) is formed by compressing 'normal' $Ice$ $Ih$ to 200MPa at temperature of 198K.

The SEM image above is of $Ice$ $I$ decorated grain boundaries in fine-grained $Ice$ $II$. The $Ice$ $I$ stands in raised relief relative to the $Ice$ $II$ due to volumetric expansion to the lower-density phase.
Scientists believe this form of ice can be found on the interior of some large icy moons orbiting Jupiter.
The macroscopic appearance of the 'exotic' forms of ice depends upon many variables such as crystal grain size. The laboratory production of these 'novel' forms of ice to ensure proper formation of fine crytallite grains (less than 10 microns) is time consuming, requiring careful control of temperature and cycling of very high pressures.
For example, $Ice$ $IV$ is formed by heating high-density amorphous ice at a slow (0.4K/min) rate from 145K and at a constant pressure of 0.81GPa. It forms a rhombohedral crystal (see image (a) below).

Similarly, $Ice$ $V$ is formed from liquid water at 500MPa by lowering its temperature to 253K. It has a more complex lattice structure and forms monoclinic crystals (see image (b) above).
"Square snow" or $Ice$ $VI$ is formed from liquid water by compressing it at 1.1GPa (10,850atm) and lowering its temperature to 270K. Its unit cell forms tetragonal crystals with all water molecules hydrogen bonded to four others, resulting in the formation of dendritic crystals (approx 45 microns across, see image below).

For more images from the laboratory, see: http://web.mit.edu/wbdurham/www1/short%20htmls/more%20ice%20photos.html
References:
Water structure and science, Ice phases, known ices, Martin Chaplin.
Grain size-sensitive creep in Ice II, Kubo, Durham, Stern & Kirby, Science, Vol 311 (2006).
Visual observations of crystal morphologies and melting of Ice IV at pressures up to 1GPa, Chou and Haselton, Review of High Pressure Science Technology Vol 7 (1998).
Dynamic pressure-induced dendritic and shock growth of ice IV, Lee, Evans & Yoo, Proceedings of the National Academy of Sciences, Vol 104, No. 22.

Strictly speaking, Ice Ih is no longer alone on Earth.  We have Ice VII in diamonds.

I added a mention of Ice VII naturally occurring in diamonds. Now Ice I is the only naturally occurring hase "in large quantities" on Earth. Feel free to roll back.

Was I rolled back?


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Inductive effect in Organic Chemistry

Why does $\ce{O-}$ show +I effect?
On the other hand why does $\ce{OH-}$ show -I effect?
both oxygens are electron rich and are negatively charged. my teacher says that charge on the species is a headache for the species ,so according to me both the oxygens should show +I effect, that is donating electrons tendency. Plus they have lone pairs too.
How can we extend the similar explanation for the halides? they have 3 lone pairs so they should also not take up more electrons.
If carbon is double bonded to oxygen then is there any inductive effect taking place over there?

Organic is confusing me. Help would be appreciated.

Lone pairs can sometimes be delocalized. An acceptable rule of thumb is that if an atom possesses lone pairs and is connected to an sp2 atom, then the two atoms participate in resonance. So consider phenol. Oxygen has lone pairs. It's connected to an sp2 carbon. And so yes, the oxygen can participate in resonance with the sp2 carbon it's connected to, and also the other sp2 carbons in the aromatic ring. We can ascertain the fact that every atom is phenol is sp2 by noting that phenol is planar.

In general, a negative formal charge makes delocalization a (more) favorable move as this decreases the electron (charge) density on one atom. Spreading out electron density between or among multiple atoms generally lessens reactivity.
Now, your question is why would oxygen exhibit a +I or electron-donating effect despite its high electronegativity. Well, it can participate in resonance donation as explained above; see phenol and its conjugate base, the phenoxide ion, for examples. Examining both in terms of electrophilic aromatic substitution also reveals that the oxygen atom in both is strongly electron donating. True, oxygen can withdraw electron density via sigma bonds (inductive effect) but doing so is unfavorable especially in the phenoxide ion in which the oxygen already bears a negative formal charge. As your teacher explained, bearing charge isn't something that is very favorable.
And in general, you're going to find that resonance effects outweigh inductive effects. Hence the reason why an sp2 oxygen connected to another sp2 atom generally exhibits a +I effect. Consider dimethyl malonate and methyl acetoacetate . Which is more acidic? From inductive effects alone we'd say dimethyl malonate because the oxygen is highly electronegative as opposed to an sp3 carbon and should withdraw more electron density via the inductive effect. From a resonance-standpoint we'd say the opposite; the oxygen in the ester donates electron density via resonance more than a methyl group can inductively donate electron density. And you'd have to realize that the resonance-standpoint is more important in evaluating the acidities of the two weak acids.
A common exception to resonance>inductive effect in organic chemistry is an sp2 atom connected to a halogen. Yes, the halogen has lone pairs and these can be delocalized into the adjacent p-orbital. But the halogens (except possibly fluorine) are much bigger than many other atoms, and therefore the magnitude of resonance donation is limited. So the great electronegativity of the halogens trumps their magnitude resonance donation, and we can see this in the context of electrophilic aromatic substitution again; halogens are ortho-para directors (delocalize the lone pairs in the Lewis structures to see why) but are also mild deactivators.


text:
Reactions in water radiolysis - producing hydroxyl radicals
There’s an excellent question on Stack Exchange regarding the radiolysis of water here but I have one simple outstanding question that perhaps someone might now; in radiobiology, a common DNA damaging free radial is caused by the radiolysis of water. This reaction is something like
$$\ce{H2O + $h\nu$ -> H2O+ + e- }$$
$$\ce{H2O+ + H2O ->  H3O+ + OH. }$$
which produces a hydroxyl radical which can prove fatal to cells. I follow the logic here, but given the massive energy of the ionizing radiation, how come a cleaving reaction like the following isn’t common?
$$\ce{H2O + $h\nu$ -> H. + OH. }$$
I’ve seen the energy quoted to break water this way at around $\pu{493.4 kJ/mol}$, which is way below the energy ($h\nu$) of an x-ray photon, even after compton scattering. Is there a reason why this reaction isn’t valid or common?

H (atomic hydrogen radical) is in fact formed.  It is 9% of the radial species formed by radiolysis.
See Reactivity of the Hydroxyl Radical in Aqueous Solution and references cited therein for further information.

Thanks for the link - I'm more curious why the hydroxyl radical isn't formed by direct cleaving the H20 into a H and OH radical? Am I missing something obvious?

@DRG, that reaction is equation 5 of the reference and it says on page 6 that it is proven to occur.

So it does, thank you - is there any discussion on the relative probability of the various possible reactions?

@DRG try looking at Yield and Reactivity of Electrons and H Atoms in Irradiated Aqueous Solutions Chem. Phys. 37, 1865 (1962) http://scitation.aip.org/content/aip/journal/jcp/37/8/10.1063/1.1733378

For ionizing radiation, various mechanisms for interaction with matter are possible. Depending on the energy, the most important types for gamma radiation are photoelectric effect, Compton scattering, and pair production. Charged particles, such as alpha and beta particles, mainly interact through coulomb forces between their charge and the negative charge of the orbital electrons of the absorber atoms.
All these processes lead to partial or complete transfer of the radiation energy to electron energy; thus, they lead to electron excitation or ionization:
$$\begin{align}
\ce{H2O &->[\gamma][\quad] H2O^{$*$} }\\
\ce{H2O &->[\gamma][\quad] H2O+ + e- }
\end{align}$$
The interaction is very fast. The excitation or ionization event occurs within a time scale of $< 10^{-16}\ \mathrm{s}$. This process is too fast for immediate chemical reactions. By way of comparison, the frequencies of molecular vibrations range from less than $10^{12}\ \mathrm{s^{-1}}$ to approximately $10^{14}\ \mathrm{s^{-1}}$.
Accordingly, the first chemical reactions follow slightly later, within a time scale of $10^{-14}\ \mathrm{s}$ to $10^{-13}\ \mathrm{s}$:
$$\begin{align}
\ce{H2O^+ + H2O &-> *OH + H3O+ }\\
\ce{H2O^{$*$} &-> H. +\, .OH }\\
\ce{H2O^{$*$} &-> H2 + O. }
\end{align}$$
The radiolysis products proceed to diffuse, and further chemical reactions occur.


text:
Why is the +3 oxidation state of Cerium more stable than +4?
For the 4f inner transition metals, +3 is the most common oxidation state (OS). Why is the +3 OS of cerium considered more stable than +4, at which it attains noble gas configuration?

The question is a problem, it contains a fact which can not be trusted. It has the assumption that one oxidation state is more stable than another oxidation state.
The problem is that the relative stability of oxidation states depends on the chemical environment of an element. If we consider cobalt as an example then it is clear that cobalt(II) in a simple aqueous phase (dilute perchloric acid) will be more stable than cobalt(III) as the cobalt(III) is able to oxidize water and then convert itself into cobalt(II). There the cobalt will be in the form of [Co(H2O)6]^2+
Howver cobalt(II) in an aqueous cyanide solution is in the form of [Co(CN)6]4- which is a very strong reducing agent, this will reduce water to hydrogen. Thus in cyanide media cobalt(III) is the more stable oxidation state.
If we consider solid oxides then the cobalt(III) oxidation state is much more stable, in general the oxide ligand is a harder (in terms of Pearson's HSAB theroy) ligand than a water. The hard ligands tend to increase the stability of metals in the higher oxidation states. I know that in a lithium ion battery that there is an equilibrium between LiCoO2 and CoO2 at one of the electrodes. Here the oxide ligands are making the higher oxidation states of the cobalt more stable than they would be in aqueous media.
In the same way lead(IV) is far more stable in the form of solid metal oxides than it is in aqueous solution. Consider for a moment the mixed oxidation state oxide known as "red lead" which is Pb3O4 and the lead dioxide found on one of the plates of a lead acid battery (used normally as a "car battery").
Now back to the element which the question was about, the problem with the question is that it did not state what environment the cerium was in. if the cerium is in the solid state as an oxide then cerium dioxide (ceria) CeO2 will be very nice and stable. The other lanthanides do not form dioxides with the same ease. The reason why I think that cerium forms the dioxide with such ease that it forms a +4 ion with  a noble gas configeration.
In aqueous media cerium(IV) is possible to form but I am sure that it is less stable, it is a strong oxidant.
Equally sometimes in the aqueous media there are times where the higher oxidation state is more stable. If we consider plutonium. Just the other day I exposed a little plutonium to ozone, this converts it from Pu(IV) to Pu(VI). The +6 oxidation state is sufficently stable in aqueous media (nitric acid) for me to do wet chemistry with it. But if we consider the metal oxides then plutonium trioxide is not something which can be formed with ease. The normal oxide for plutonium is the dioxide. There have been atttempts to make plutonium trioxide but when I last checked the literature it was not clear if anyone has ever managed to make a sample of it.

Ce has electronic configuration [Xe]$\ce {4f^1}\ce{5d^1}\ce{6s^2}$. Electronic configuration of the Ce$^{3+}$ is [Xe]$\ce{4f^1}$ while that of Ce$^{4+}$ is [Xe]$\ce{4f^0}\ce{5d^0}\ce{6s^0}$ or simply [Xe]. It is evident from this that Ce$^{4+}$ has a noble gas configuration which means it should be more stable than Ce$^{3+}$ in which there is one left over f electron.
But a 4f electron is the one that is significantly deeper inside the atom $i.e.$ it is considerably closer to the nucleus compared to a 5d or a 6s electron. Even if you attain $+4$ for Cerium, it would require a large amount of energy to extract it from this much inside. However, you would already have achieved the $+3$ state before that in which you have already removed 5d and 6s electrons.
In other words, we cannot compensate the tremendous surge in energy of Ce$^{4+}$ with the stability achieved by attaining noble gas configuration.

When applying MathJax, it is better to apply it to everything maths related and not use it as a markup to. For example $\ce{Ce^4+}$ for $\ce{Ce^4+}$. You'll find more information on [meta]. As a happy accident, you can use mhchem also for electron configurations: $\ce{[Xe] 4f^1 5d^1 6s^2}$


text:
What is the reaction mechanism of a 5-endo-trig cyclization?
I am trying to solve this mechanism for a disfavoured 5-endo-trig cyclisation to generate the disfavoured product (4) from starting material (1). The favoured 5-exo-trig is given by (2), which I have got a mechanism for. My efforts take me to intermediate (3) as shown below, but I'm not too sure on how to generate the desired product without reagent reduction.

Your mechanism to the 5-endo-trig product (4) is very close.  Notice however that when you do the initial cyclization, the intermediate formed has a positive charge on nitrogen along with the negative charge on oxygen (as you showed).  This cyclized intermediate then just needs to move a proton from the nitrogen to oxygen to form the neutral enol which then converts to the (ester) carbonyl - which is your desired product.  No other reagents are required.

Additionally, the OP's second step is not possible, the pi bond cannot move into the ring since the destination carbon already have four bonds


text:
Disproportionation Reactions of the Halogens: Writing balanced net ionic equations for four reactions. See details
I first conducted the following reactions, each in separate test tubes:

$\ce{NaBr}$ + $\ce{H2SO4}$ + $\ce{NaBrO3}$
$\ce{KI}$ + $\ce{H2SO4}$ + $\ce{KIO3}$

...with water and dichloroethane added to each as well.
(In the experiment, I put a small quantity of sodium bromide in one test tube and a similar quantity of potassium iodide in the other test tube. I then added 1 mL of water, 3 drops of 9M sulfuric acid, and 2 mL of dichloroethane to each. I then added a small quantity of sodium bromate to the first test tube and a similar quantity of potassium iodate to the other.)
Then, I removed the aqueous layer from each test tube and added $\ce{NaOH}$ to the dichloroethane layer in each.

(#1) $\ce{NaOH}$ added to dichloroethane
(#2) $\ce{NaOH}$ added to dichloroethane

So far, I have the following equations written:

$\ce{NaBrO3 + 5NaBr + 3H2SO4 -> 3Na2SO4 + 3H2O + 3Br2}$
$\ce{KIO3 + 5KI + 3H2SO4 -> 3K2SO4 + 3H2O + 3I2}$
$\ce{3Br2 + 6OH^{-} -> 5Br^{-} + BrO3^{-} + 3H2O}$
$\ce{3I2 + 6OH -> 5I^{-} + IO3^{-} + 3H2O}$

Please check my work. If without error, I just need to convert these to equations to net ionic.

(answer to original question)
No, you are not on the right track.
If bromate is a starting material, it should not appear as a product in reaction 1. Instead, consider how bromide and bromate can react with each other.
If iodate is a starting material, it should not appear as a product in reaction 2. Instead, consider how iodide and iodate can react with each other.
For reactions 3 and 4, a hint would be "disproportionation".

Okay, so my initial reaction (#1) would be NaBrO3 + 5 NaBr + 3 H2SO4 --> 3 Na2SO4 + 3 H2O + 3 Br2. Then, adding NaOH to the dichloroethane layer, I would have 3 Br2 + 6 OH(-) --> 5 Br(-) + BrO3(-) + H2O. Correct?

@JaxBragg looks much better, the 3 Br2 + 6 OH- reaction could use balancing with respect to O though


text:
How to synthesize cyclopentane from cyclohexane?
Does this route work to synthesize cyclopentane from cyclohexane?

Free-radical halogenation.
Elimination with strong base to form cyclohexene.
Ozonolysis to make di-aldehyde.
Oxidation to make di-carboxylic acid.
Esterification with ethanol, acid, and heat.
Alpha-proton elimination with ethoxide ion.
Intramolecular attack of other carbonyl carbon; loss of ethoxide leaving group.
Acid, high heat, water to return to carboxylic acid. Expulsion of ethanol.
Heat for decarboxylication. Loss of carbon dioxide.
Not illustrated: removal of carbonyl using Wolff-Kishner and Clemmensen.

I think a few modifications I could make are:

Formation of methyl ester using diazomethane. This would not involve heat and acid. But it would involve something that is potentially explosive.
Alternately, make an acid chloride and then go down to the ester if I wanted to avoid the heat involved with Fischer esterification.

Whether this route works or not isn't the point. There is absolutely no good reason to make one readily available hydrocarbon from another via complex chemistry.

Why not just use the Aldol condensation on the dialdehyde produced from ozonolysis and save some steps?

What does the heat, copper, and quinolone do?

Just typical conditions for a decarboxylation, quinoline is a high boiling solvent.

I know this is an old thread that only just became back to life so it's probably no longer worth it, but I'd think you could shorten these routes a bit by using some ring-contraction chemistry (e.g. Favorskii):

Refs:
1. M. T. Musser, Cyclohexanol and cyclohexanone. Ullmann's Encycloedia of industrial chemistry, 2000.
2. Synth. Comm. 2007, 37, 4149.
3. JACS 1960, 82, 4307.
4. J. Chem. Soc., Chem. Commun. 1983, 939.

Your route looks good until the last step. After decarboxylation, the intermediate will be cyclopentanone. So all that will be left to do is reduce the ketone to an alkane (you probably know a couple of options).

Oops, forgot about that. Wolff-Kishner or Clemmensen should work.

I would go with a different route, I would aim to convert the cyclohexane into cyclohexanone maybe via cyclohexanol. I would be looking for some bacteria which can do a selective oxidation or I would aim for using some CH activation chemistry which Derek Barton worked on.
An alternative is to brominate cyclohexane to cyclohexyl bromide and then react with 2-nitropropane to form cyclohexanone under basic conditons. Here is a diagram showing the core of my route.

I would want to react the cyclohexanone with nitric acid using vanadium as a cataylst to form adipic acid. I would convert this into the diethyl ester. I would then want to condense it with itself using a base such as sodium ethoxide to form a ketoester.
I would then react this with aqueous base and then aqueous acid to get cyclopentanone. I would then reduce this using either zinc amalgum in hydrochloric acid or hydrazine under basic conditions.
Another method which is a tad on the brutal side is to make the adipic acid and then to heat it up with a mixture of manganese carbonate and barium hydroxide to form the cyclopentanone. See Slomp,G. et al. - Journal of Organic Chemistry, 1960, vol. 25, p. 514 - 518 for the use of this type of reaction. This relates to the synthesis of phenylacetone from phenyl acetic acid and acetic acid using thorium dioxide. I have worked once with phenyl acetic acid and it has a disgusting smell, I recall it was like old socks.


text:
Titration: Standardizing NaOH using PHI
Q: A student standardizes sodium hydroxide by weighing out 1.207 grams of potassium hydrogen iodate, dissolving the compound in water and titrating to a suitable endpoint:
If the volume to reach the endpoint is 32.80 mL, what is the concentration of sodium hydroxide?

The molar mass of PHI is $389.75 \text{g/mol}$. if 1.207g is added to the titrand beaker, then there are:
$$\frac{1.207\text{g}}{389.75\text{g/mol}} = 0.0031\text{mol of KHI in the titrand beaker}$$
NaOH reacts with KHI in a 1:1 molar ratio, and 32.8mL of NaOH was added to reach the endpoint. This means that 0.0031mol of NaOH was added. So, the concentration of NaOH is:
$$\frac{0.0031\text{mol}}{0.0328\text{L}} = 0.0945\text{M}$$
This is coming up as a wrong answer in my book. Can anyone explain where I'm going wrong?

I get the same answer you do. What is the "correct" answer supposed to be?

I got the same answer too. Is the book wrong?


text:
How do I calculate the mass percent in the original mixture?
I have the question:

"A mixture of $\ce{KClO3}$ and $\ce{KCl}$ with a mass of 0.950g was heated to produce $\ce{O2}$. After heating, the mass of residue was 0.700g. Assuming all the $\ce{KClO3}$ decomposed to $\ce{KCl}$ and $\ce{O2}$, calculate the mass percent of $\ce{KClO3}$ in the original mixture."

I took the amount of $\ce{O2}$, converted it to moles $\ce{O2}$, then divided it to moles of just O, then divived by 3, bcause for every 3 O (for the 3 in the $\ce{KClO3}$) there is a mole of $\ce{KClO3}$ so I had moles of $\ce{KClO3}$. Then I converted that to grams of $\ce{KClO3}$, but thaat ended up as more then the original mixture's mass.
Am I even thinking about this the right way?

Looks close. Your math looks sound, but the reasoning is messy. $\ce{O2}$ is the only oxygen here.
$$\ce{2KClO3 -> 2KCl + 3O2}$$
You might have skipped writing out that reaction — it helps a lot.
If $0.700~\mathrm{g}$ of residue remained, that means $0.950\ \mathrm{g} - 0.700\ \mathrm{g} = 0.250\ \mathrm{g}$ of $\ce{O2}$ left the container (which means that $0.250\ \mathrm{g}$ of $\ce{O2}$ was made!)
We can calculate the number of moles, similar to what you did.
$0.250\ \mathrm{g} \times \frac{1\ \mathrm{mol}}{32\ \mathrm{g}} = 0.00781\ \mathrm{mol}\ \ce{O2}$
However, we have no reason to divide by 2 — $\ce{O2}$ is the only way (well, at least for this purpose) oxygen can exist in nature. Treat it as "oxygen".
Let's apply a mole ratio, now. There were $\ce{2 KClO3}$ molecules used for every $\ce{3 O2}$ molecules we got.
$0.00781\ \mathrm{mol}\ \ce{O2} \times \frac{2~\mathrm{mol}\ \ce{KClO3}}{3~\mathrm{mol}~\ce{O2}} = 0.00521\ \mathrm{mol}\ \ce{KClO3}$ which we had at the beginning. Note that we can do this because it says all of it reacted completely. If only $95~\%$ of it reacted, we would multiply by $0.95$, for example.
Find the grams of $\ce{KClO3}$ expended in reaction.
$0.00521\ \mathrm{mol} \cdot \frac{122.55\ \mathrm{g}}{1\ \mathrm{mol}} = x~\mathrm{g}\ \ce{KClO3}$
Your percent is pretty easy from there.
$\frac{x\ \mathrm{g}}{0.950\ \mathrm{g}} \times 100~\% = y~\%$
Where I'm sure you can figure out $x$ and $y$. (Remember to find $x$ to 3 significant digits).
Leave a comment if I wasn't clear on something!


text:
How do I approach drawing all of the possible constitutional isomers?
Does anyone have any tips on how to approach these types of questions?
A specific example would be:

Draw all constitutional isomers of $\ce{C2H4O}$

I can get the easier ones like vinyl alcohol, but it's hard to get ethylene oxide and it takes a long time for me to get acetaldehdye. And how would I know if I got all of the possible constitutional isomers?

I approach questions like this by considering all possibilities of arrangements of the non-hydrogen atoms. Obviously, there's C-C-O, C-O-C, and then the cyclic ethylene oxide.
Then I make sure I've matched bond orders to get the hydrogen count.
So I get vinyl alcohol, acetaldehyde, and ethylene oxide. H2C-O-CH2 would only work for a biradical, which is probably not expected.
The systematic fashion then is to arrange the "heavy" atoms, first in all linear forms, then all cyclic forms, then all two-ring forms, etc. Of course with large formulas, the number of isomers goes up exponentially.


text:
What conditions can cause a material to exist as a gas?
So my question is, how can something exist a gas. Is it when the pressure exerted by the gas is greater than the atmospheric pressure, then it is a gas?
And also how do manometers work, how can you measure the pressure of a gas if the atmospheric pressure is greater?

(source: duckduckgo.com)

Taking a look at the example above, I understand that the pressure
exerted by the gas can be greater than the atmospheric pressure,
because only then you have a gas. But what I don't understand is why a
substance exists as a gas when the atmospheric pressure is greater?
Shouldn't the atmospheric pressure crush the substance to a solid or
liquid?

I know this is a lot, but please help.

The question of how a manometer works could probably be fit to be its own question.

And when you say, how can something be a gas, do you mean become a gas, or exist as a gas?

@JohnSnow I mean that exist as a gas.

@AwesomeFlame123 and you mean exist as a gas under what conditions?

@Dissenter I want to know what conditions. What conditions can something exist as a gas. I also think that something cannot exist as a gas if the atmospheric pressure is greater than the gasses exerted pressure.

The atmosphere consists of molecules with much space between them; it does not constrain evaporation until the vapor pressure approaches atmospheric pressure. In a constrained volume such as a piston if the applied force is greater than the pressure the vapor will not form.

Generally, when a substance is at > 0 K (0 Kelvin or -273°C) the atoms/molecules making up that substance possess kinetic energy, which means that each atom/molecule are [almost] constantly moving.  If that atom/molecule is in a vacuum, it would move in [almost] a straight line.  However, in the ordinary matter that we encounter everyday, atoms/molecules exist in macroscopic phases which are composed of a humongous number of particles.  Which means that each particle experiences frequent collisions with neighbouring particles, making the trajectory of each particle to appear zig-zag like.
The higher the temperature in the vicinity of an atom/molecule, the higher the kinetic energy of the particle, and the faster that particle's motion becomes.
Also, depending on the properties of the atoms/molecules in the substance, each atom/molecule experiences electrostatic forces (either attractive or repulsive) exerted by neighboring atoms/molecules.
At the interface between a liquid and a gas (e.g., water in a tumbler sitting on a table in a room at sea level), the surface of the liquid is constantly being bombarded by gaseous particles that have much higher kinetic energies than the molecules of the water in the tumbler.  This bombardment imparts enough kinetic energy to some of the water molecules at the surface.  Some of the bombarded molecules gain enough kinetic energy to enable them to leave the water surface to become gaseous water molecules.  Some of the bombarding particles (oxygen molecules, nitrogen molecules, water vapour molecules, etc.) from the air lose enough kinetic energy so that they become stuck into the liquid surface.  At equilibrium (maximum humidity saturation of the air in the room), the number of water molecules leaving the liquid surface is equal to the number of molecules entering the liquid surface.

It is the net result of factors 2, 3 and 4 that determines the phase of a substance.  For example, substances like water, which are composed mainly of polar molecules (molecules possessing an asymmetrical distribution of charge around it, making it behave like a bipolar/tripolar/multipolar magnet), have relatively high melting and boiling points.  Below the melting point, the molecules exist in a [3D] crystalline lattice.  Each molecule are in constant motion but are at the same time held together by electrostatic forces (hydrogen bonds).  Since in a solid, the kinetic energy or velocity of each molecule is not high/fast enough, those particles cannot overcome/escape from the electrostatic attraction exerted by their immediate neighbours.  Therefore, the molecules stay stuck in the crystal lattice, exhibiting their kinetic energy as vibrational motion.
When the melting point is reached, the kinetic energy of the molecules starts to overcome the electrostatic attraction from neighbouring particles, allowing them to move away from their neighbours.  This is the liquid state, in which the kinetic energy of the particles are enough to prevent the particles from being stuck in one place, allowing them to move about, making the phase fluid; however, the kinetic energies are not enough to allow the particles to move farther away from each other.  In the liquid state, the distances between neighbouring particles are almost the same as that in a solid.
For a liquid to become a gas, two factors must be overcome: (a) the intermoleculer forces in the liquid, and (b) the pressure above the liquid surface.  At the boiling point, the kinetic energies of the particles become high enough to overcome (a) and (b) to enable the particles to move farther away from each other, and we get a gaseous phase.
Factor (a) is overcome, for example, by heating the liquid in the container.
You can appreciate the effect of factor (b) when you go mountain-climbing (5000 feet or higher).  Take a thermometer with you to the mountain top.  You will observe that water at high altitudes boil at temperatures lower than 212°F (100°C).
Another way to experience the effect of factor (b) is through the following experiment:  Get a 50-mL (1 to 2-oz) syringe, remove the needle, draw water to about 25% capacity of the syringe barrel, plug the inlet with your index finger, then with your other hand, pull the plunger until it is almost at its maximum position.  You should see the water in the syringe boil without the application of heat.
This seems to be a simple question but a full answer can be appreciated only with some knowledge of statistical thermodynamics.
BTW, the pressure of a gas is the macroscopic result of a humongous number of particles impinging upon a surface (the diaphragm of a barometer, the surface of a manometer liquid, the walls of a container, etc.).

You can also give the example of a pressure cooker to show the opposite, that water will boil at temperatures higher than 100 degrees.

First statement hopefully a typo! At zero K molecules have No, Zero random kinetic energy. 4th paragraph The gas particles from the atmosphere do not impart energy to the liquid unless the pressure compresses the liquid. At constant pressure the vapor pressure is a function of the temperature of the liquid. the Energy distribution among the molecules determines the evaporation in an equilibrium. In an energy flux the temperature and energy input are important.

At absolute zero molecules have no random velocities and zero kinetic energy [Unless there is a QM zeropoint heat content That is yet to be discovered] Molecules are in their lowest quantum mechanical energy states. As energy is added the energy is distributed among the QM states and the modes of motion. the molecules gain heat content and have random velocities that are distributed by distribution functions; at room T this is the Boltzmann distribution. This means that above a certain low temperature every substance has a number of molecules with sufficient velocity and kinetic energy to escape the forces that hold it to the solid or liquid, the intermolecular forces. The evaporated molecules establish a vapor pressure. As the temperature is increased more and more molecules have sufficient energy and eventually, for small molecules at least, the kinetic energy of the particles prevent the formation of liquid or solid [at reasonably low pressures at least] This temperature is the critical temperature of the gas. Every substance possibly has a vapor pressure depending on the temperature, its chemical properties, and its stability.

"QM zeropoint heat content That is yet to be discovered" - you think why He isn't solid if not under high pressure?


text:
Inductive vs resonance effects and the acidity of phenol
A lot of people rationalize the acidity of phenol by saying that resonance is responsible for much of phenol's acidity as opposed to aliphatic alcohols.

However, this image suggests that in fact the inductive effect is responsible for a great deal of phenol's additional acidity. As we can see in the picture immediately above, it seems that the sp$^2$ carbon neighboring to the oxygen is pulling most of the weight here; that the resonance contributors with charges on carbons are minor contributors. The additional resonance contributors for phenol seem to be doing almost nothing here.
So, who's right? Resonance $\gt$ inductive effect in phenol or the other way around?
EDIT: The argument I have summarized above is presented here at this website entitled "I Judge People By Their Grammar and Knowledge of Phenol"

While the chemistry in the paper was good, as The Chem Blog has noted,
the authors’ lack of attention to detail was borderline disrespectful.
I expected more from one of the most storied labs in synthetic organic
chemistry.
[...]
It turns out that an inductive effect—not a resonance effect—is the
predominant reason for the increased acidity of phenol relative to
aliphatic alcohols.
[...]
For those interested, these data come from Evans’ Chem 206 lecture
notes (Lecture 20, restricted access), where the point is hammered
home in glorious detail. Professor Evans’ PowerPoint slides should be
framed and displayed in the Smithsonian.

Addressing jerepierre's point
$\ce{Phenol + acetone~enolate <=> phenoxide + acetone ~enol}$
So, from the $\mathrm{p}K_\mathrm{a}$ values we know that the above equilibrium lies mostly to the right (according to jerepierre). Given this, we can conclude that the reactants are less stable in the acid-base sense than the products; acetone enolate is less stable than phenoxide.
Phenol is also less stable than acetone enol.
But I don't see how this contradict's the blog article. It does contradict ron's point about being unable to use the provided $\mathrm{p}K_\mathrm{a}$ values to compare the relative stabilities of the acids.

Are your 3 pKa values all measured in the gas phase (I can't imagine that someone measured the pKa of propen-2-ol in solution)?

You seem to forget that the $\ce{C=C}$ double bond in the enol is able to provide resonance stabilization as well and I would think that in both cases - for phenol and the enol - the resonance stabilization has the major effect on the acidity not the hybridization.

@Philipp True, but that makes for what? Two contributing resonance structures for the enol? A carbocation and an oxide ion? Phenoxide however has four ... the oxide ion and three carbocations. It seems that the two remaining carbocation contributing structures give rise to less than an order of magnitude of pKa difference. The presence of an sp2 carbon itself gives rise to much, much more than an order of magnitude of pKa difference.

@ron - why wouldn't one be able to measure the pKa of an unstable enol in solution? Also, how does one get pKa in the gas phase? Doesn't Ka and therefore pKa imply the presence of water?

http://blog.chembark.com/2007/03/10/i-judge-people-by-their-grammar-and-knowledge-of-phenol/

@Dissenter The number of resonance structures is not always the key to stability. The effect of providing any possibility for delocalization might give the major contribution (I don't know) and the possibility to spread the charge over an even larger "area" in Phenol might simply not make that much of a difference. Coulomb's law goes as $1/r^{2}$ so the difference between having the electrons very close together and moderately far apart is much more pronounced than between going from having electrons moderately far apart and very far apart. But my argument might be wrong.

@Philipp I understand; it just seems that in this case the resonance contributors are more or less similar. The main difference I see is that for one set there is a ring and for another set there is not a ring.

@Dissenter Yes, and the ring is only slightly better at providing stabilization. And one would have expected the ring to be better at this than the enol from the resonance structures, so that is alright. The only unexpected thing is that the ring doesn't seem to make that much of a difference but that might be alright if my line of thought is remotely correct. I also think that expecting to predict pKa values in any quantitative manner from simple mesomeric-structure arguments is a little bit to much to ask from such a simple model. It is only good for painting the broad strokes.

@Dissenter, perhaps you can find an example where the inductive effect of an sp2 hybridized carbon is present, but resonance is not possible because it's impossible to align the lone pair with the pi-system.

The acidity of $\ce{A-H}$ is a measure of where the following equilibrium lies.
$$\ce{A-H <=> A^- + H+}$$
This means that we are comparing the relative stabilities of the products and reactants ($\ce{\Delta G=-RTlnK}$).
In the present example (assuming your $\mathrm{p}K_\mathrm{a}$ for propen-2-ol is really being measured in water, note too that in water the relative concentration of this enol of acetone is ~ $10^{-8}$, how can you measure the $\mathrm{p}K_\mathrm{a}$ of something so dilute?) we are comparing the relative stability between [phenol and phenoxide] to the relative stability between [propen-2-ol and the corresponding anion].  Just because the two systems have the same difference in relative stabilities does not mean that phenol and propen-2-ol have the same stability.  In other words, just because the two systems have similar $\mathrm{p}K_\mathrm{a}$'s this cannot be interpreted to mean that the additional resonance structures in phenol and phenoxide are not significant contributors to both their stability and description.
EDIT: Jerepierre's comment as worked up by Dissenter (Addressing jerepierre's point)
in his edit Dissenter wrote,

Given this, we can conclude that the reactants are less stable in the acid-base sense than the products

I agree
Dissenter further posits,

acetone enolate is less stable than phenoxide. Phenol is also less stable than acetone enol.

That's where I disagree.  We know that the reactants are less stable than the products, but I don't see how we know which reactant(s) is less stable than which of the products.
Let's say the reactants are $10\,\mathrm{kcal/mol}$ (made up number) less stable than the products.  Is phenol $5\,\mathrm{kcal/mol}$ less stable than phenoxide and acetone enolate $5\,\mathrm{kcal/mol}$ less stable than acetone enol?  Or is it $10$ and $0$ or $0$ and $10$, or $15$ and $-5$?  How can you break down the overall reactant/product difference of $10\,\mathrm{kcal/mol}$ further?

So in your opinion the author of the blog article is wrong in his reasoning? I was this close to actually showing the article to my professors because they, like most, stick with resonance in explaining the acidity of phenol.

Yes, in my opinion.  Do you see any flaws in my argument?  Do you see the flaw I'm pointing out in his?

@Dissenter I disagree. pKa can be used as a measure of which conjugate base is more stable. We can easily calculate the the equilibrium constant of any acid base reaction if we know the pKa's of the acids. In this case, if acetone enolate is reacted with phenol, an equilibrium will be established with a ~10:1 ratio of phenolate to acetone enolate (setting aside protonation at the alpha-carbon). Doesn't that tell us that phenolate is more stable than acetone enolate? Essentially we are using the proton as the basis of comparison.

@jerepierre  What is it that you are disagreeing with?

Your argument is that pKa tells us the relative stability of the acid and (the conjugate base + proton), but can't be used to say anything about the relative stabilities of phenol and propen-2-ol. However, the pKa's do tell us about the stabilities of the conjugate bases of those species.

@jerepierre  I don't follow, could you expand your comment into an answer?

@ron i have expanded on what jerepierre is saying in my post.

" 'Acetone enolate is less stable than phenoxide. Phenol is also less stable than acetone enol.' That's where I disagree." Do you disagree with the first statement, the second statement, or both?

Perhaps what we are missing here is that these pKa values have to be on the same scale (e.g., the water scale) so that they can be compared or tell us anything meaningful.

@jerepierre  rather than disagree, I'd say I don't know if both of the statements are true or false.

@jerepierre  I agree, that's my first comment under Dissenter's question.

http://isites.harvard.edu/fs/docs/icb.topic93502.files/Lectures_and_Handouts/20-Acidity_Concepts.pdf - page 7 has some more information.

That blog post you are referencing voices very strong opinions with which one can agree or disagree. I personally love one of the comments under the blog saying

I judge people by the units they use. Mr. Evans used kcal, so he lost.

But of course, that doesn’t discredit the valuable contributions Mr Evans made especially to synthetic organic chemistry.
However, there is a flaw in the argument Mr Evans and the blog writer uses. Both are comparing the $\mathrm{p}K_\mathrm{a}$ of phenol with that of prop-1-en-2-ol. But the standard way of measuring $\mathrm{p}K_\mathrm{a}$s (see equation) is in aquaeous solution, and prop-1-en-2-ol is not stable in aqaeous solution and will rearrange to acetone! Ron mentioned that the relative concentration of prop-1-en-2-ol in aquaeous solution is $10^{-8}$ which, for a $1\,\mathrm{M}$ solution, is lower then the concentration of $\ce{H3O+}$ by autoprotonation of water.
$$\ce{HA + H2O <=> A- + H3O+}$$
As such, I wish to see exact measurement details before I accept both values as equivalent, and therefore I reject the comparison of $\mathrm{p}K_\mathrm{a}$ values made like this.
Instead, I would propose measuring the $\mathrm{p}K_\mathrm{a}$ values of these two related species which are guaranteed to be stable in the drawn form.

Let us continue this argument when these values are presented.

This question was answered well enough 5-7 years ago, but there is something still lacking. The resonance diagrams drawn in brackets suggest that resonance is responsible for increasing the acidity of phenol; I find the wording misleading. Better would be that the resonance of the phenyl ring is responsible for the stabilization of phenolate anion. IMHO, drawing diagrams with two charges and switching them around the molecule amounts to creative imagination - sometimes useful, but only as a last resort.
Let's first picture what's going on with phenol, or more accurately, with the $pi$ electrons of benzene, the ultimate resonance molecule (Ref 1). Resonance, the amount of energy of hydrogenation (or combustion) less than calculated for a compound with the same number of C-H and C-C and C=C bonds, occurs because three localized C=C double bonds (which will have energy at the dotted line) interact and become more delocalized and stable: the totally interacting combination is significantly more stable than the three original double bonds (it contains 2 electrons; it is one molecular bond). Then because of symmetry, there are two more combinations of the three original orbitals, each with a node perpendicular to the ring, and degenerate (having the same energy). Each of these bonding combinations is a bond that is more stable than one of the original C=C double bonds.

The diagram in brackets in the original question suggests that a $pi$ lone pair from the more electronegative oxygen plops onto the phenyl ring, leaving oxygen with a positive charge. Not likely. However, a base can capture the phenolic proton, leaving oxygen with negative charge, more than it usually has. Too much, in fact. How does it adjust? By sharing the lone pair in an orbital perpendicular with the ring; remember, the oxygen is sp$^2$ hybridized, with 2 lone pairs now in the plane of the ring and a third perpendicular. The lone pair in an atomic orbital on oxygen interacts with one molecular orbital of the ring. That's my simplification. If we talk about molecular orbitals, we should include all the orbitals - but for simplicity, I'm going to discuss only a bond between the one oxygen atomic orbital and one of the phenyl ring molecular orbitals.
Which phenyl orbital should we consider? The three bonding orbitals are filled; in my simplification, they are tied up, out of consideration. However, there are three anti-bonding $pi$ orbitals, two of which have low enough energies to be somewhat compatible with the lone pair of oxygen - they could combine to give a spread-out molecular orbital, including the oxygen and the phenyl ring, reducing charge concentration on oxygen while increasing bonding somewhere else. Which orbital would be most effective?
Take the carbon at 3 o'clock to be the one connected to oxygen. The orbital labeled $pi_5$ has a node at that carbon and hence does not see the oxygen. However, $pi_4$ has significant density there (and so does $pi_6$, but it is too high energy to be concerned with) and so can unite with the oxygen AO to form a phenolic MO with electron density spread over the ring (phenol is then activated toward electrophilic attack).
This explains how phenol is more acidic than cyclohexanol.
There is another explanation of phenolic acidity, and I bring it up only to point out that I believe it is an oversimplification at least, and perhaps very inadequate (Ref 2):

It just smears everything together and gives no insight, and not even the creativity of drawing lines and arrows.
Connecting vinyl alcohol with phenol and cyclohexanol seems to be possible because its pK$_a$ was given as 10.9. Literature favors a different number: 19-20 (Ref 3). With this larger pK$_a$, there seems to be no close connection to discuss.
At room temperature, acetaldehyde (H$_3$CCH=O) is more stable than vinyl alcohol (H$_2$C=CHOH) by 42.7 kJ/mol (Ref 4). Vinyl alcohol can be stabilized by utilizing kinetic favorability of the deuterium kinetic isotope effect (k$_{H+}$/k$_{D+}$ = 4.75, k$_{H2O}$/k$_{D2O}$ = 12). The tautomerization process is significantly inhibited at ambient temperatures, and the half-life of the enol form can be increased to t$_{1/2}$ = 42 minutes. The difficulty of dealing with vinylic alcohols is a further disconnect from phenol and cyclohexane.
For reference, the empirical resonance energy of benzene is 143.1 kJ/mol (Ref 5). The stabilization of phenolate will be considerably smaller than this, and present only for the ion. If phenol were stabilized as much as the phenolate ion is, there would be no increased acidity compared to an alcohol. Comparing the resonance stabilization of benzene with ordinary bond energies shows a fairly small energy: a C-C bond has about 347 kJ/mol strength; a C=C bond about 614 kJ/mol, so a $pi$ bond has about 267 kJ/mole strength. The resonance energy of benzene is worth about a half of a $pi$ bond (Ref 6).
Ref 1. https://www.masterorganicchemistry.com/2017/05/05/the-pi-molecular-orbitals-of-benzene/
Ref 2. http://www.chemhume.co.uk/A2CHEM/Unit%201/Notes/1A%20Benzene%20and%20phenol.pdf
Ref 3. https://www.organicchemistrytutor.com/topic/enolization-keto-enol-tautomerism/
Ref 4. https://en.wikipedia.org/wiki/Vinyl_alcohol
Ref 5. https://en.wikipedia.org/wiki/Resonance_(chemistry)
Ref 6. https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Bond_Energies


text:
Is the phenoxide ion aromatic?

Is the phenoxide ion aromatic? I read that technically it isn't aromatic because it violates some arcane rule, but I can't find any references in the literature.
I don't see why it couldn't be aromatic, since it's

flat/planar
has a system of conjugated p-orbitals.
has 4n+2 pi electrons: 2 from the oxygen lone pair that is delocalized and n = 0.
is cyclic.

I agree absolutely with Max, and while I was still running the calculations, he already provided the answer.
The goldbook provides a nice definition for aromatic compounds:

In the traditional sense, 'having a chemistry typified by benzene'.
A cyclically conjugated molecular entity with a stability (due to delocalization ) significantly greater than that of a hypothetical localized structure (e.g. Kekulé structure ) is said to possess aromatic character. If the structure is of higher energy (less stable) than such a hypothetical classical structure, the molecular entity is 'antiaromatic'. The most widely used method for determining aromaticity is the observation of diatropicity in the 1H NMR spectrum.
The terms aromatic and antiaromatic have been extended to describe the stabilization or destabilization of transition states of pericyclic reactions The hypothetical reference structure is here less clearly defined, and use of the term is based on application of the Hückel (4n + 2) rule and on consideration of the topology of orbital overlap in the transition state. Reactions of molecules in the ground state involving antiaromatic transition states proceed, if at all, much less easily than those involving aromatic transition states.

Therefore not all aromatic compounds have to strictly full-fill the Hückel (4n + 2) rule, but all that obey that rule are aromatic. Phenols certainly behave like benzene and are therefore aromatic, please note that aromaticity is an experimentally observed feature.
You can see the aromaticity from the $\pi$ type orbitals, that are in both cases quite similar and extensively delocalised. The calculations were done at the DF-BP86/def2-SVP level of theory.
Benzene exhibits a very high symmetry $D_\mathrm{6h}$, therefore the last two of the shown orbitals are degenerate.

The restriction is not present in the phenolate, as the symmetry is only $C_\mathrm{2v}$.

Also due to the missing symmetry constraint the neutral phenolic radical, $\ce{H5C6O.}$, and the phenolic cation, $\ce{[H5C6O]^{\oplus}}$ would be aromatic.

Even when the molecule is present in a neutral form, as a phenol, the oxygen will still be close to a $\ce{sp^2}$ shape with the $\ce{p}_z$ orbital partially overlapping the $\ce{p}_z$ orbital of the carbon bound to it. The proton NMR of phenol clearly shows characteristic aromatic peaks, demonstrating that the ring is aromatic.
Hückel's rule states that a fully conjugated ring will be aromatic if it has $4n+2$ electrons, and antiaromatic if it has $4n$ electrons, where $n$ is an integer $\geq 0$. The rule is a generalization that comes from how the bonding and nonbonding orbitals are filled as electrons are added to an isolated conjugated ring.
In analyzing the resonance of the phenol, the Hückel rule applies only to the phenyl ring. It gives no understanding of how the hydroxide atom affects the system because the oxygen atom is not within the ring.
The hydroxide will affect the electron distribution around the ring. This animation shows it nicely with resonance structures:
(source)
The resonance pictures make it easy to see, but it is important to keep in mind that molecules do not actually have resonance structures. A more accurate way of seeing it would be:
The $\ce{p}_z$of $\ce{O}$ is delocalized into $\pi^{*}$ of $\ce{C1}$, weakining $\ce{C1-C2}$ and $\ce{C1-C6}$ bonds, increasing the density on $\ce{C2}$ and $\ce{C6}$ by kicking the electrons in the bonding orbitals closer to them. The $\ce{p}_z$ electrons of $\ce{C2/C6}$ move into the $\pi^*$ of $\ce{C3/C5}$, which in turn results in an increased density on $\ce{C4}$.
This is still an MO simplification, and to get the accurate picture you would simply have to do the calculations.
It is still possible for an aromatic ring current to flow through the ring in the presence of a magnetic field, as we can see from NMR, but perhaps it feels more like a roller coaster now.

As phenoxide ion has 8 $\pi$ electrons due to presence of lone Pair on oxygen, it counts in $π$ electrons so it is antiaromatic but not aromatic

When doing the electron counting for a ring it is normal not to include the electrons which are on atoms outside the ring.

Please provide strong evidences to counter the claims already given by the two highly voted answers.


text:
How significant is air decomposition on formation of nitrogen oxides?
How significant is $\ce{NO$_x$}$ formation from $\ce{O2}$ (reactant) and $\ce{N2}$ decomposition from simply heating the two as opposed to from $\ce{CO2}$, $\ce{H2O}$, (and $\ce{N2}$) dissociation? In other words, which of the two pathways for $\ce{NO$_x$}$ formation is more significant (I imagine this will vary with equivalence ratio):

$\ce{O}$ and $\ce{N}$ radical from $\ce{N2}$ and $\ce{O2}$ dissociation
$\ce{O}$ radical from $\ce{CO2}$ and $\ce{H2O}$ dissociation with $\ce{N}$ radical

My feeling is that the second becomes more significant as equivalence ratio increases (more fuel means more $\ce{O2}$ is consumed in combustion).

I have updated your post with chemistry markup. If you want to know more, please have a look here and here. Please do not use markup in the title field, see here for details.

In situations where NOx originiates form $\ce{N2}$, such as lightning or a high temperature furnace, the Zeldovich mechanism dominates:
$$\ce{O2 -> O +O}$$
$$\ce{O + N2 -> NO +N}$$
$$\ce{O2 + N-> NO +O}$$
If a combusting fuel actual contains nitrogen, other mechanisms can be involved.

Generally speaking, would the proportion of NOx formed from $N_2$ and $O_2$ above (air decomposition) decrease as we increase the equivalence ratio (oxygen will favour reaction with fuel as opposed to decomposition)?


text:
Why does pasta become sticky after being cooked?
This question was posed by my biochemistry professor at the end of a class and then never answered.
What I've gathered is that:

It has something to do with the denaturing of a protein, since we were on the subject of denaturing that lecture.
It most likely involves gluten, because that's the main protein in wheat. And, from personal experience of making glue, gluten is very sticky.
The answer might involve hydrophilic/hydrophobic interactions, because those interactions are disturbed during the denaturing process.

Any ideas?

My intuition is that most "stickiness" is caused by hydrogen bonding.

After cooking, pasta has a lot of free surface starch ready to bond with things. This is why we rinse well after cooking to remove extra starch and cool down the pasta to prevent it from producing more surface starch.

Does this tie in at all with proteins and denaturation?

Pasta contains starch (carbohydrate). It has nothing to do with proteins.


text:
Explanation for behavior of acetic acid in aquaponic/hydroponic systems?
Because aquaponics is hydroponics with fish & bacteria, options for adjusting pH are limited.
Primarily people use phosphoric acid, hydrochloric acid, and nitric acid (heavily diluted).  A less popular but easy to acquire material is vinegar.
I've tried using it in a very alkaline solution (100 ppm calcium carbonates, 8.4 pH) and noticed that the pH would drop by a small amount before returning to it's original state.  Anecdotally, other aquaponic hobbyists have noticed a similar effect that is not replicated by the other acids. What is happening that the pH would lower temporarily and then rise back up over several hours?

The source of the acid shouldn't matter much in adjusting the pH by a small amount.  What is the source of the calcium carbonate?
If it is dissolved calcium carbonate only, as you might find in hard water, then this is strange.  The solution is buffered by the combination of dissolved carbon dioxide and carbonate minerals, so its pH might be difficult to change, but once it has been changed it should stay where you put it, more or less.
If you have solid calcium carbonate (cement, limestone gravel, some pottery (notably not terracotta) or marble chips, perhaps), then what is happening makes a lot more sense.  You add acetic acid, lowering the pH and consuming some of the carbonate from the solution.  There is a large reservoir of carbonate available though, in the form of the carbonate minerals, and that eventually redissolves and the solution becomes more basic with time.

In my recent case, yes there was an abundance of carbonates as measured by a KH test.  But other's have also repeated similar observations of pH bounce back using vinegar(acetic acid).  Looking at the full equations for HCL, phosphoric, and nitric acid they create compounds that are more readily accessible to plant/animals.  With Calcium acetate, it looks like it's inaccessible to plant/animals, so perhaps the pH drop is just the creation of CO2 from acetic acid + calcium carbonate.  After the CO2 is outgassed the pH would go back up?

Marking this as the answer, will have to hunt down a biologist to see if they have any ideas/thoughts to explain the pH bounce back from a biological standpoint.

@David An aquarium site/forum, particularly one for freshwater planted aquariums, like The Planted Tank, might be able to offer you some insight.

its some kind of carbonate bounce- if you fight the ph war long enough, you'll eventually start getting some precipitates (cloudiness, or formed on vessel). Once the ph drops enough, these precipitates dissolve and it starts the cycle over again.


text:
What will happen to the entropy and free energy of the gasses when the partition is removed?

Consider a container of volume $ 5.0$  L that is divided into two compartments of
    equal size. In the left compartment there is nitrogen at $1.0$ $atm$ and $25 °C$; in the
    right compartment there is hydrogen at the same temperature and pressure. What
    will happen when the partition is removed?

$A) $The entropy decreases, and the free energy decreases.
  $B)$ The entropy increases, and the free energy decreases.
  $C) $The entropy increases, and the free energy increases.
  $D) $The entropy decreases, and the free energy increases.

Logic tells that upon removing the partition, randomness increases and hence entropy increases. I am confused about free energy. First law of thermodynamics has to be applied , I think. But I can't seem to get the right direction.

A Spontaneous process is characterized by an increase in the total entropy (for both system and surroundings).
Spontaneous processes are characterized by a decrease in free energy (analogous to the decrease in gravitational potential energy occurring for a ball rolling downhill).

We know that free energy of a process is given by:
$$ \Delta G_\mathrm{mix}  = \Delta H_\mathrm{mix}  - T \Delta S_\mathrm{mix} $$
The molar entropy of mixing is given by:
$$\Delta S_\mathrm{mix} = -R(x_\ce{A}\ln(x_\ce{A}) + x_\ce{B}\ln(x_\ce{B}))$$
For an ideal gas: $\Delta H_\mathrm{mix} = 0$ giving free energy as:
$$\Delta G_\mathrm{mix}  = -T\Delta S_\mathrm{mix} = RT(x_\ce{A}\ln(x_\ce{A}) + x_\ce{B}\ln(x_\ce{B}))$$
since $x_\ce{A}$ and $x_\ce{B}$ are mole fractions ($x_i \le 1$), the natural log will always yield a negative number.  Therefore the entropy will be positive and the free energy will be negative.


text:
What is the bonding picture in Peroxomonosulfate?
In $\ce{SO5^{2-}}$, can somebody tell me which orbitals are used to form the pi bonds in this anion.
Please tell the hybridization.
And the lewis structure.
Well I tried and the answers I got was sp3 & the pi bonds are formed using d orbitals.I just need to know whether I'm correct

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Here is already the Lewis structure

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I see, sorry I'm rather new on the website, I just wnted give a little help :x

The peroxomonosulfate is a hypercoordinated molecule, in which the sulfur appears to exceed Lewis' rule of two, better known as the octet rule. In former times it was believed that d orbitals would contribute to bonding. This is not the case.
Bonding is better represented by

three equivalent $\ce{S-O}$ single bonds
sulfur having a formal charge of +2
these three oxygens having a formal charge of -1
one $\ce{S-O}$ single bond that leads to the peroxyl oxygen
this oxygen has a formal charge of zero
one $\ce{O-O}$ single bond
and the terminal oxygen having a formal charge of -1

This bonding picture is much more in accordance with quantum chemical calculations and an NBO analysis on the DF-BP86/def2-SVP level of theory. The involvement of d orbitals is minimal (<1%).

It is noteworthy, that this molecule pretty much has the same bonding around the sulphur as in sulphate. It can therefore similarly be described. Have a look at my rant against hypervalency or the bonding picture of the $\ce{PO4^{3-}}$ anion, which is isoelectronic to sulphate.


text:
Form the oxides of the element Z=39
Its electronic configuration is $4d^15s^2$ so in normal state it has valence=1 and in energetic state, valence=2 (because our teacher said we count only the electrons in the outer shell firstly) and valence =3. First question: is it true what my teacher said? It was for Z=22 example, where she said it has the valence 2 and  4). Second question when we form the oxide we should use only the valence in the energetic states and not the one in the normal state?

No one looks at the periodic table through the lens of Z.  This assignment is ... dumb.

Z=38 is Sr, isn't it, and the electronic configuration is 4d0 5s2.

I am not sure that I understand your question, but in terms of establishing the valency in this case the key point is that in d block elements the energy level of the outer s electrons are very similar to that of the d electrons in the shell below, so the valency in this case will be 3.