text:
Why Acetone does not behave like its computational values?
I am trying to simulate the excitation state of acetone. I ran TDDFT for it both in gas phase and solvated state in water (both implicit and explicit water).
The experimental data say that acetone undergoes n->π* transition, which means it has a higher wavelength in gas phase ($\mathrm{\lambda_{max}}=$~$276~\mathrm{nm}$) and a shorter wavelength in the solvated state ($\mathrm{\lambda_{max}}=265~\mathrm{nm}$).  I started with gas phase and expected something a wavelength of about $276~\mathrm{nm}$ ($4.49~\mathrm{eV}$ ), but surprisingly I got $\mathrm{\lambda_{max}}=136.172~\mathrm{nm}$ ($9.326~\mathrm{eV}$)! I really cannot understand why there is such a big discrepancy!
Here is my GAMESS input file. What is wrong with this molecule ?
!   File created by the GAMESS Input Deck Generator Plugin for Avogadro
 $BASIS GBASIS=N311 NGAUSS=6 $END
 $CONTRL SCFTYP=RHF RUNTYP=ENERGY TDDFT=EXCITE  DFTTYP=B3LYP $END
 $CONTRL ICHARG=0 MULT=1 $END
 $TDDFT NSTATE=9 $END
 $STATPT OPTTOL=0.0005 NSTEP=99 METHOD=RFO UPHESS=MSP HSSEND=.T. $END
 $SYSTEM MWORDS=1000 PARALL=.TRUE. $END
 $SCF DIRSCF=.T. DIIS=.T. DAMP=.T. $END

 $DATA
Title
C1
O     8.0     0.00000    -1.27900     0.00300
C     6.0    -0.00000    -0.05800     0.00100
C     6.0     1.29700     0.69100    -0.00000
C     6.0    -1.29800     0.69000    -0.00000
H     1.0     1.35900     1.32900    -0.90600
H     1.0     1.35900     1.33200     0.90300
H     1.0     2.15700    -0.01300     0.00100
H     1.0    -2.15700    -0.01400     0.00100
H     1.0    -1.35900     1.32900    -0.90600
H     1.0    -1.35900     1.33200     0.90300

 $END

Is geometry optimized (at the same level of theory)? 2) Where is the output? 3) B3LYP is not the best choice for TD-DFT, basis set could also be better.

In the mean time, I suggest first to try the PBE0/6-311++G(d,p) level of theory instead of B3LYP/6-311G.

I did a few calculations: pbe0/def2tzvpp 143 nm; df-bp86/def2tzvpp 158 nm; pbe0/6-311++g(df) 147 nm. @Wildcat While I agree, that B3LYP is not the best choice, it seems like this is not a problem of only this method.

@Martin-マーチン, ok. But now we are certain that it is not the failure of B3LYP. :)

@Wildcat Usually I like pointing out the problems of B3LYP, but in this case it seems to behave normally. Even stranger CIS//BP86/def2TZVPP gives $\pi\rightarrow\pi^*$ with 130 nm.

@Martin-マーチン , I found this paper saying TDDFT with 6-311(2+,2+)G** has satisfactory results but it gives 6.98 eV !!! .  What kind of satisfactory result is it ?  http://pubs.acs.org/doi/pdf/10.1021/jp014123x

@Aug, wait, wait, wait. I think we all took your statement about $\lambda_{\mathrm{max}} = 276 , \mathrm{nm}$ for granted, but where actually did you get that? Could you quote a research paper? Take a look at this paper (PDF, 963 KB). The absorption in the region from 4 to 5 eV is almost zero (it scaled by 100 times in Fig.1).

@Wildcat that paper is really amazing and I need to dig into it a little more, but I got it from NIST: http://webbook.nist.gov/cgi/cbook.cgi?ID=C67641&Mask=400#UV-Vis-Spec   as well as: http://people.chem.ucsb.edu/kahn/kalju/chem126/public/solvat_intro.html

@Aug NIST reference does not claim that $\lambda_{\mathrm{max}}=276 , \mathrm{nm}$. Besides, clearly only a small region of wavelength is shown on the graph. Finally, in the paper quoted on NIST page it is clearly said that the peak at 2800 A is "very weak with its oscillator strength $\sim 0.0004$".

It seems that the results of the calculations are more or less fine and the OP just misinterpreted the NIST data. As I said in my comment above, NIST does not claim that $\lambda_{\mathrm{max}}=276 \, \mathrm{nm}$. Clearly only a small region of wavelength is shown on the graph and in the paper[1] referenced on the NIST page it is said (emphasis mine):

There are two diffuse ultraviolet bands; the first at 2800 Å. is very
  weak with its oscillator strength $f \sim 0.0004$ and the second at
  about 1900 Å  is moderately intense with a maximum extinction
  coefficient $\epsilon_{\mathrm{m}} \sim 1000$. McMurry has identified
  the 2800 Å. band as a forbidden $\pi^* \leftarrow n$ transition
  involving excitation of a non-bonding $\ce{O}$ electron to an
  anti-bonding $\pi$ orbital between the $\ce{C}$ and $\ce{O}$ of the
  carbonyl group.

A high-resolution photoabsorption spectrum in the energy range 3.7-10.8 eV can be found, for instance, in this recent study[2] and is more or less consistent with the result of the OP's calculations.

Noel S. Bayliss, Eion G. McRae, J. Phys. Chem., 1954, 58 (11), 1006–1011.
M. Nobre, A. Fernandes, F. Ferreira da Silva, R. Antunes, D. Almeida, V. Kokhan, S. V. Hoffmann, N. J. Mason, S. Eden, P. Limão-Vieira, Phys. Chem. Chem. Phys., 2008, 10, 550-560. (available at researchgate.net)

Interesting indeed. || I have included the complete references to the papers you cited, I find it better to cite like journals would do.

this is your output, right?
TRANSITION DIPOLE, A.U.  OSCILLATOR
                    HARTREE          EV         X       Y       Z     STRENGTH
   0  A         -193.0290234748    0.000
   1  A         -192.8724089055    4.262     0.0001  0.0000  0.0001    0.000
   2  A         -192.7831335626    6.691     0.4051 -0.0004 -0.0000    0.027
   3  A         -192.7317175333    8.090     0.0018 -0.0976  0.0001    0.002
   4  A         -192.7220472153    8.353    -0.4522 -0.0001  0.0000    0.042
   5  A         -192.7210047911    8.382    -0.0121 -0.0000 -0.0006    0.000
   6  A         -192.7176816741    8.472    -0.0001 -0.0013  0.0724    0.001
   7  A         -192.7167365427    8.498     0.0006 -0.0002  0.0075    0.000
   8  A         -192.6964881601    9.049     0.0029 -0.3152  0.0008    0.022
   9  A         -192.6862850361    9.326    -0.0022  1.1274 -0.0093    0.290

I am confused, why do you take the 9ths excitation and assume it is the experimentally measured one? The first excitation agrees reasonably quite well with the experimental data.
If you look at the exp spectrum eg here http://www.sciencedirect.com/science/article/pii/S0022407308000307
you can see that there is a 2nd absorption band below 200nm (6eV). This seems to be the 2nd excitation.
Did you look at the orbitals involved in the first excitation? I would assume it shows a n-pi* transition and then your are all good.
edit:
This paper contains some references (i cannot access the cited references therein)

http://pubs.acs.org/doi/full/10.1021/jp046334i

They report the following

The n → π* electronic transition of acetone corresponds to the excitation of an electron from the nonbonding 2py orbital on the oxygen to the antibonding π* molecular orbital on the carbonyl. For an acetone molecule in its C2 or C2v geometry, the transition is
  symmetry-forbidden; as a consequence of vibronic coupling, in the gas
  phase, a weak band at ΔE = 4.4−4.5 eV is observed, with an oscillator
  strength of f ≈ 0.0004.

Do we see that in the TD-DFT output? Mostly, yes! The oscillator strength is smaller than the printout (bad program), but if you use e.g. ORCA you get a non-zero oscillator strength. It is still much smaller than the reported value but that is somewhat expected from TD-DFT in this case (partially CT-state!) The small oscillator strength is troublesome, but the excitation energy is ok. It could indicate a ghost state (=purely artificial state), but these are not often seen in hybrid-functional TD-DFT calculations. Likely all states are physical.
The first transition is a HOMO-LUMO transition and if you bother to look at the orbital it is a nice n-pi* transition as expected.

The first excitation has a strength of 0.000 and can therefore never be $\lambda_\mathrm{max}$. It also corresponds to the HOMO LUMO excitation, which is $\sigma\rightarrow\pi^*$, i.e. orthogonal orbitals. I am sorry, but I need to downvote this answer.

@Martin-マーチン , so which one do you select as λmax ? is it just based on the largest oscillator strength or something else shoul be mentioned ?

@Aug, photoabsorption intensities are indeed directly related to TD-DFT oscillator strengths, so yes, the transition with the largest oscillator strength corresponds to the most intense line in the absorption spectrum.

@Martin-マーチン, There is something really interesting in your comment that can help me a lot: you said its HOMO-LUMO transition is σ -> π* which is definitely true, but how did you realize that? was it just based on experimental knowledge or there is something in the output file that you noticed ? This can be answer to my other question here ( I had to put a bounty on it becuase there is no satisfactory answer yet !): http://chemistry.stackexchange.com/questions/29464/uv-vis-spectroscopy-transition-reference

@Aug The output gives you the orbitals that correspond to the transition. You just have to look at them, and you can identify them. It will get much harder for open shell problems, so natural transition orbitals give you a qualitative picture.

@Martin-マーチン, some of them are rather straight forward (#1 at 4.26 is ~0 for X,Y,Z that I can guess is a σ and for #2 I can guess a π (X=0.4051). But how do you intepret #3,#4 with mixed and negative values?    I think it can be a great reference for me and people in the future if you put an answer in the link in previous comment ( I know it takes time and I appreciate that ! )

@Aug I really don't understand what you are asking with the question you put the bounty on. You put in a reference request, and that nothing I can give you. If you want to have help on how to interpret calculations, then I can be of help. Those are all straight forward, and the result is included in the output. All you need to do is look at the transitions. There is no guessing involved. I have no access to Gamess, but I guess the above is not the only output they give.

@Martin-マーチン,   I made another question here.:  http://chemistry.stackexchange.com/questions/31306/how-to-interpret-orbital-transition-in-tddft

@Martin-マーチン Obviously the 0 Transition is the ground state!

Yes obviously, it has 0.0 eV relative energy. I do not see how this would make your answer correct though.

@Martin-マーチン sorry i misread your comment. I edited the answer, maybe it helps with some of the confusion.


text:
Which unit cell do we obtain if spheres are arranged in ABABAB type in 2D and then that 2D layer is arranged in AAAAAA type in 3D?
Arranging spheres ABABAB type in 2D gives us close packing. Now, this layer if arranged in ABABAB (in 3D) gives us HCP and if arranged in ABCABC (in 3D) gives us FCC.
What do we get if we arrange it in AAAAA type (in 3D)? (we are talking about one type of spheres only)

The question is a little confusing, so I will interpret your 'ABABAB' type in 2D to mean a simple hexagonal layer (so, the same layer as used for your hcp and fcc packing). So, a stacking of a simple hexagonal layer, each layer directly above the other, will give a structure looking like right hexagonal prisms - or right triangular prisms if you look a bit more carefully.
Anyway, the resulting structure is the simple hexagonal Bravais lattice with a single atom basis. Hcp is also a simple hexagonal Bravais lattice, but with a two atom basis, one each from plane A and plane B. I do not know of any elements that crystallize in this crystal structure (but am open to corrections!).

You have interpreted it in the right way. I see that in both the cases (using the simple hexagonal 2D layer in ABAB and AAAA way in 3D) we are getting the same simple hexagonal bravais lattice.  But, the atomic arrangement is totally different in both. In chemistry books simple hexagonal bravais lattice is always means ABAB type. Nobody talks about AAAA. Maybe, it is because it does not exist in nature.   Whatever, thanks for the answer.

No, the atomic arrangement is not totally different - the underlying Bravais lattice is the simple hexagonal lattice. The unit cell that sits on that lattice is different, leading to a different point group. As for it existing in nature, it is true that I can't find an element that crystallizes in simple hexagonal, but I suspect a good synthetic chemist could come up with a unit cell likely to result in simple hexagonal...

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text:
is it possible to break the bonds of diatomic elements such as fluorine or iodine and create positive and negative ions by electron bombardment?
If I had a container in vacuum filled with $\ce{I2}$ gas and then I bombarded it with high speed electrons using an electron gun, would  be able to get both $\ce{I+}$ and $\ce{I-}$ ions or would I only get $\ce{I2+}$ plus an ejected electron?

also, if I supply I2 with low energy electrons would the iodine attract the electron due to its high electron affinity?

Yes, it is likely that one of the iodine atoms in this case will indeed capture the bombarding electron.

Sure, why not. It's just the question of your bombarding electrons having the right energy. At low energies, a bombarding electron will likely be captured by one of the iodine atoms:
$$
\ce{I2 + e- -> I- + I} \, .
$$
But somewhere around $8.62 \pm 0.06 \, \mathrm{eV}$ according to this study, you'll indeed have the ion pair formation:
$$
\ce{I2 + e- -> I- + I+ + e-} \, .
$$
There are some other processes which contribute to the ion pair formation. For details, read the paper quoted above.

With respect to the formation of $\ce{I2+}$ note that to form it a bombarding electron have to, formally speaking,

Either first break the $\ce{I-I}$ bond homolyticly and then remove two electrons from one of the $\ce{I}$ atom which requires it to carry about 30 eV of energy (the sum of the homolytic bond dissociation energy and of the first and the second ionization energies).
Or break the $\ce{I-I}$ bond heteroliticly and then remove the second electron from $\ce{I-}$ ion which requires it to carry also about 30 eV of energy (the sum of the heterolytic bond dissociation energy and of the second ionization energy).

$%edit$

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text:
In Crystal Field Theory what does Δo=10Dq mean?
While studying the Crystal Field Theory I was told $\mathrm{Dq}$ is a unit, related to the unit $\Delta_{\mathrm{O}}$ by the relation $\Delta_{\mathrm{O}} = 10\ \mathrm{Dq}$. But aren't $\Delta_{\mathrm{O}}$ and $\mathrm{Dq}$ variables, not units? The unit is of energy, such as $\mathrm{eV}$ or Joules, these are symbols to represent a particular value of energy, which could be anything. So what do we mean when we say $\Delta_{\mathrm{O}}= 10\ \mathrm{Dq}$? What exactly is $\mathrm{Dq}$ (I haven't been able to find this in any of my textbooks, they just either state this relation, or not at all) and are they actually units of energy?

Dq was derived for quantum mechanic description of the electrostatic model of crystalline fields and is connected to the radial electron density of the d-electrons, the charge of the metal and the distance of ligands and metal. The term Dq is the product of two terms D and q and is called Differential of quanta
implying the energy. The word D has been coined from D State (L=2 )

We should point out that Dq is D times q, and q is not a subscript....

Dq is differential of quanta and is the ligand field splitting parameter.

Though I don't think I can provide a full answer, I did see this in a lecture series. $\Delta_{\mathrm{O}}$ is the ligand field splitting and it is a measure for the strong or weak field that the ligands of a complex create. Depending on the ligands the energy levels of, for example octahedral iron complexes (below), are closer to each other in a low field and farther apart for a higher field. This is dependant on $\Delta_{\mathrm{O}}$. This is also an energy difference and the units are in joules or electron volts. See the picture below. You might also want to look up Tanabe-Sugano Diagrams.
I was told (sorry, no proper reference) that the expression $\Delta_{\mathrm{O}}= \pu{10Dq}$ comes from quantum mechanics. The reason that this term is simply stated instead of defined, lies in the origin of CFT. The theory was largely empirical and  thus $\Delta_{\mathrm{O}}$ was used. This difference in energy levels could also be described in quantum systems and is called $\pu{Dq}$ after its quantum number. The difference between the levels appeared to be $\pu{10Dq}$.
I hope this is not a gross oversimplification, but a rather helpful one!

The origin of the usage of $\pu{Dq}$ is discussed elsewhere and is out of scope for this question; but I can tell you that it is indeed a relative unit.
When looking at complexes without knowing their specific electronic structure and specific energy levels, the $\mathrm{Dq}$-expression is a simple way of comparing the relative electron stabilisations. The idea is that the entire field split is given as $10~\mathrm{Dq}$ and that a hypothetical zero-level is introduced such that one electron in each orbital would give an overall stabilisation of $0~\mathrm{Dq}$. So in the case of a $\mathrm{d^3}$ complex, the ‘stabilisation’ would be $-18~\mathrm{Dq}$.
The convention has a number of flaws:

As noted, the overall energy can differ greatly. No two complexes will have exactly the same value for $10~\mathrm{Dq}$.
The value of $10~\mathrm{Dq}$ even depends strongly on the geometry: octahedral complexes have a much larger value of $10~\mathrm{Dq}$ than tetrahedral ones:

$$4.44~\mathrm{Dq}(\text{oct}) \approx 10~\mathrm{Dq}(\text{tet})$$

The notation does not extend well to complexes in geometries other than octahedral or tetrahedral. For Jahn-Teller distorted or square-planar complexes, the definition of $10~\mathrm{Dq}$ becomes unclear and less useful. It is often given as the energy difference corresponding to the highest-wavelength electronic absorption.

So overall, while it is helpful to use it at a basic level, it outlives its usefulness quickly due to being, as you said, relative.


text:
How many stereoisomers are possible for hepta-1,6-diene?
I am asked to identify the number of stereoisomers that are possible for hepta-1,6-diene $(\ce{H2C=CHCH2CH2CH2CH=CH2})$.
I drew the bond-line structure as follows, and I think that there can be cis and trans isomers, but my answer is wrong. The correct answer states that since neither of the double bonds exhibit stereoisomerism, this compound does not have any stereoisomers.
Originally, I thought that single bonds can be considered too. But I think I my error is that by looking at the single bonds, we are considering the constitutional isomers of this compound, rather than the stereoisomers (which are determined by the 'fixed' double bonds). Since double bonds lack the free rotation that single bonds have, for any compound that may exhibit stereoisomerism, we can only consider the double bonds. Also, are we considering only one double bond at a time?
Is this reasoning correct? Can someone clarify and expound more on this please?

In the case of your original drawing with the terminal double bonds, there are no E or Z (trans or cis) isomers!

E and Z isomers are only possible if both ends of the $\ce{C=C}$ double bond each carry a substituent other than a hydrogen atom.

As far as the other structure is concerned, E and Z isomers look as follows:

Thank you for the explanation. I realized my bond-line structure is incorrect, and I've redrawn it accordingly. However, in another problem, a trans stereoisomer was present, despite the only substituent groups of the C=C double bond being ethyl and isopropyl. http://imgur.com/vbGcWZC Can you clarify this?

@imaginov The annotations on your image are correct. Since ethyl and isopropyl are not the same group, isomerism will be present.

The textbook says there is a trans configuration present, but I do not understand how this is true. If there are two substituent H atoms, an ethyl, and an isopropyl group, how does the compound exhibit stereoisomerism? I thought stereoisomerism excludes substituent H atoms.

@imaginov I've updated my answer for other compound.

So, E and Z isomers aren't possible if the only substituents are hydrogen atoms? But if the substituents are hydrogen atoms and say, a methyl group, E and Z isomers are possible?

There are no stereoisomers (cis/trans) of this compound because both the double bonds have three identical substituents and one different substituent. In order to have stereoisomerism across a double bond you need to have two different groups on each side of the bond.
For example but-2-ene has one methyl group and one hydrogen on each side of the double bond and so has two stereoisomers: trans-but-2-ene and cis-but-2-ene:

For double bonds with at least three different substituents the E/Z notation system is used. Z denotes substituents which are on the same side of a double bond and E denotes those on the opposite side. The four substituents are assigned a priority in accordance with the Cahn-Ingold-Prelog priority rules and then the stereochemistry is assigned based on the relative position of the two highest priority substituents. In this example, bromine has a higher priority than chlorine so the label is assigned based on the position of the bromine and the chlorine across the double bond, making this the Z isomer.

There is no Cis/Trans isomerism across the double bonds because the molecule you have there is a terminal alkene which is an alkene with double bonds at the first 2 and last 2 carbons or just at the first 2 carbons if there are no higher priority groups or just at the last 2 carbons if there are higher priority groups closer to the end with all single bonds.
If you had other substituents you could have R and S stereoisomers and even combinations of R and S at different chiral carbons.
When you look at the differences in bonding than yes you are looking at structural or constitutional isomers.
But you can use cis/trans for multiple double bonds. For example oct-2,6-diene(or 2,6-octadiene) can be:
oct-(2 cis, 6 cis)-diene
oct-(2 cis, 6 trans)-diene
oct-(2 trans, 6 trans)-diene
or
oct-(2 trans, 6 cis)-diene


text:
Does DMT-Acetate become Polar in H2O, with added NaCl?
I'm extracting DMT (N,N-dimethyltryptamine) from Acacia confusa via acid/base. I extracted the DMT with $\ce{H2O}$ and acetic acid, converting it to DMT-acetate.  Here, I used a non polar solvent, naptha, to defat the material, causing an emulsion.
I used $\ce{NaCl}$ to break the emulsion, this caused a thick layer of black tar to form on top of the aqueous solution, and a glossy black tar on the bottom of the vessel.  I need to freebase the DMT acetate; however, I'm concerned that DMT-acetate migrated into one of the two polar layers.  (Correction: Either the non polar layer on top or gunk layer at the bottom)
How can I confirm if this is so? Does $\ce{NaCl}$ cause DMT acetate  (or any acetate) to become polar in $\ce{H2O}$? Why and how? Will the DMT acetate have migrated into any of these two polar layers?

You are doing what?! Are you sure you have the proper permit to perform such operations? In A LOT of parts in the world you could get locked up for this.

I don't need one.  I am sorry to hear that you live in a place where it is a controlled substance.

Interesting. Where are you from? DMT is classified as a Schedule I drug under the UN 1971 Convention on Psychotropic Substances.

Does NaCl cause DMT acetate (or any acetate) to become polar in $\ce{H2O}$?

In aqueous acidic solution, that is in the presence of a sufficient amount of acetic acid, N,N-dimethyltryptamine is protonated. The acetate is the counter ion of the ammonium salt. The compound does not become more polar by adding sodium chloride!

Will the DMT acetate have migrated into any of these two polar layers?

Are you having two polar layers? I doubt it. You're having a top unpolar layer, the aqueous layer and some gunk.
Take the separated aqueous layer, neutralize carefully, extract with an unpolar solvent of your choice, dry over sodium sulfate, remove the solvent in vacuum. If necessary, recrystallize and determine the yield.
Please keep in mind that DMT is a controlled substance in many countries!

Thank you so much for your quick and concise answer! My apologies.  Indeed there was a top unpolar solvent saturated with plant tars hardened like a rock, the aqueous solution, and goo below. I have never seen any account of sodium chloride used in the Acetate state to enhance de-fatting.  Most suggest using water only. Is the addition of sodium chloride at this stage is worth while?  I am surprised with the quantity of plant material removed.  After shaking the material heavily, the solvent was not recoverable, it turned into an inch thick layer of air filled tar...

Klaus I very much doubt that it is legal in any country in the world. It is regulated under UN schedule. I believe only some traditional tribes have legal access to it, but they certainly do not need any help.

@Jori I know that it's Schedule 1 drug in the Single Convention on Narcotic Drugs, which, over time, has been ratified by quite a number of countries. Amazingly, the Vatican State is not among them. Anyway, I'm not a lawyer and I can't control who is in posession of a exceptional permit. Sometimes, strange things are required from students. I remember a student who was told by his professor to order morphine and use it as chiral base :D. Back in the days, students would have asked older group members for hints, nowadays they consult Chem.SE ;-)


text:
Fluorescence versus phosphorescence
Not in an atomic aspect, I learned that fluorescence is the emission of light from substances until the source of light disappears (there is no emission in the dark), on the other hand, phosphorescence, due to a blocking of electrons on their way back to normal state, is the emission of light from substances even without any source (emission in the dark).
If it is so, then why do we call glowing objects in the dark as fluorescent?

If it is so, then why do we call glowing objects in the dark as fluoroscent?

I think this is just marketing: most people have heard the term fluorescence, while phosphorescence or luminescence are largely unknown terms.
While a delayed emission from a singlet state is possible, most glow-in-the-dark items do in fact show phosphorescence, i.e. emission from a tripet state.


text:
How to make polyester casting resin?
What are the components of and how would one go about making polyester casting resin?

https://en.wikipedia.org/wiki/Resin_casting, https://en.wikipedia.org/wiki/Polyester_resin


text:
Identifying chiral centers
The compound below is fexofenadine and the only chirality center, as identified by the textbook is, the one circled in red.
I don't understand why the carbonyl carbon is not a chiral center too. Isn't it attached to a tert-butyl group? So isn't it attached to four different substituent groups? Or is it that the hydroxide group is attached to the carbon by the oxygen, thereby making a 'duplicate' substituent group of the carbon already double bonded to the oxygen?

Take a look at your carbon circled in red, it has a hydrogen on it giving it four different groups.
Now look at the C circled in blue. There is no hydrogen attached to that carbon because it already has four bonds (two to the O, one to the OH, and one to the rest of the molecule.) Therefore it only has three substituent groups attached and not four.

So an atom connected by a double bond is still considered to be a single substituent group. That was my mistake. Thanks!

Even if it were counted as two substituent groups, the carbon would be achiral because the two substituent group would be identical and you're looking for four different things attached to the carbon.

Since this is the accepted answer, I'll point out the 1,1'-Bi-2-naphthol is optically active but it doesn't have a carbon with 4 different attachments. The optical activity is the result of steric hindrance into two different forms.

If in doubt, simplify, examine both centres separately and draw.

Can you do the same for the carboxylic acid?

I see that the central carbon is only connected to three substituent groups, two of which are bonded by their oxygens. So this carbon cannot be a chiral center. http://imgur.com/oH6gX2B

No.
If you look at the carbon circled in blue, it only has three not four substituents. If you examine the three-dimensional structure around the carbon, you will find that there is a plane of reflectional symmetry through the plane of the C=O bond which means it cannot be a chiral centre.
4 different things attached to a carbon in 3D space means it will be chiral. 3 does not.

Can you please explain what you mean by "plane of reflectional symmetry through the plane of the C=O bond"? If I understand correctly, only sp3 hybridized C atoms can be chiral centers.

@imaginov Yes, you re right. Only sp3 hybridised carbons can be chiral centres but the reason is because, if they have 4 different substituents, they lack a plane of reflectional symmetry. The carbon in a C=O bond is sp2 hybridised and, because this make it flat, there is a mirror plane through corresponding to the three substituents. Therefore, no chirality as the molecule will inevitably look just like its mirror image.

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text:
Why does this compound only have one pair of enantiomers?
I've been doing some reading about enantiomers, which are apparently chiral molecules that are non-superimposable mirror images of each other. An example I found of this in some problems that I've been attempting seems to have stumped me somewhat:
Take the compound $\ce{CH2=CH-CHD2}$, where $\ce{D}$ is deuterium, or hydrogen-2, and react it with deuterium gas. A pair of enantiomers is formed, both with the condensed formula $\ce{DCH2-CHD-CHD2}$. The answers provided say that these enantiomers are as follows:

...where $\ce{A}$ is standing in for deuterium.
Why don't you get enantiomers where the orientation of the $\ce{C-D}$ bonds at both ends of the molecule varies? (ie. where those $\ce{C-D}$ bonds are also in varying planes, represented by different combinations of hashed and solid wedges on paper.) Here's an example of what I mean:

Why aren't these two structures considered to be another pair of enantiomers also?
Is this - the orientation of bonds at the end of carbon chains counting effectively for nothing - the case with all stereoisomerism in alkanes?

http://chemistry.stackexchange.com/questions/29155/can-a-meso-compund-be-rotated-in-any-sort-of-way-in-order-to-show-its-achirality

You should draw the molecule showing the $\ce{H}$ atoms as well as the D (or A) atoms.
In the hydrogenated/deuterogenated molecule, all the $\ce{C-C}$ bonds are single bonds, meaning there is free rotation about all of them.
The left (as you have drawn it) end of the molecule is $\ce{DH2C-~}$ and the right end is $\ce{~-CHD2}$.  Since each end-cap carbon has two identical substituents, there is no chirality.  Flipping the whole molecule about a left-to-right axis parallel to the plane of the screen, and then rotating the $\ce{ C-C}$ bond going to a particular endcap carbon, would transform the "chiral" representation you drew into its "enantiomer".  That wouldn't / can't happen with the truly chiral central carbon atom.

Usually it doesn't, as rotation around single bonds is generally very quick in room temperature. However, sometimes it matters, because rotation can be blocked by bulky substituents, or other parts of the molecule. Even double bonds, usually connected with E/Z isomerism, can in certain situations cause optical activity.
These are all examples of so called axial chirality. In the first case (rotation blocked by substituents) it's called atropoisomerism. Second (blocking by other parts of molecule) is present for example in helicenes. Third case (cumulated double bonds) occurs in allenes.


text:
Should chemistry students worry about quarks and such?
Even chemistry textbooks less than five years old start by describing atomic structure based only on protons, neutrons and electrons. Is an understanding of chemistry fundamentals likely to become out of date when the textbooks start to include quarks, leptons and the dozens of other sub-atomic particles?

Why would they start to do it?

There are only two types of fundamental particles in chemistry: electrons and nuclei.

There actually have been studies of chemistry based on particles other than electrons, protons and neutrons, especially using muons (which are leptons, not quarks). However, as the name "exotic atom" evokes, it's some very particular stuff.

Well in my humble opinion a true scientist always pursues to expand their knowledge about the inner workings of nature. Therefore, if you like science you should attempt to at least understand something of everything.

Muons are pretty important in solid state chemistry; look up spin polarized muon spectroscopy...

Is an understanding of chemistry fundamentals likely to become out of date when the textbooks start to include quarks, leptons and the dozens of other sub-atomic particles?

Well, the general chemisty books do already include leptons, at least electrons and usually positrons.
Other than protons, neutrons, electrons and positrons (and photons), I think that when discussing beta decay and electron capture, neutrinos and antineutrinos should be included, otherwise the nuclear reactions aren't really balanced (lepton number must be conserved).  Alternatively, the general chemistry books could omit all discussion of nuclear reactions, as being physics.
Other than that, I think the rest of the particles can wait for physics class, rather than general chemistry class.
Beyond general chemistry, other particles are sometimes discussed in more advanced chemistry books, such as quantum chemistry texts, where examples involving muons are common.
For some chemistry topics, such as isotope differences in rotational-vibrational spectra of small molecules distinction between fermions and bosons is critical.  Trying to explain the difference between fermions and bosons (and Fermi-Dirac statisitics vs Bose-Einstein statistics vs Boltzmann statistics) naturally leads to discussion of the standard model and the particles in the standard model.  So by the time a student gets to physical chemistry, around the third year of college, they should have some understanding of the standard model.

Chemistry is governed by what electrons do, and generally... what valence electrons do.  There really is no need to dive into subatomic particles.  It may help explain what keeps a nucleus held together but we generally gloss over that aspect and simply have students accept this as a fact.  Subatomic particles, of which there are many, are generally covered in physics.  We usually touch upon them in nuclear chemistry when decay processes are introduced.
The main point here is that Chemistry is essentially all about what the electrons are doing.

I mean to be honest with you, when I was in Chem 1, we never went into the details of quarks and other sub-atomic particles except once; in nuclear chemistry. Even then, you didn't have to have an idea of these subatomic particles to understand what was going on.
So, if you think about it, wouldn't they (teachers) have changed the curriculum to include, other, dozens of subatomic particles if it was needed?
Although, you still have a good point. But, overall, from my definition of a chemistry student, I wouldn't think that worrying about quarks and other subatomic particles.

The real question is not ‘Why don’t books include X?’, but ‘What would change if books included X?’.
To the best of my knowledge and understanding, what protons, electrons and neutrons are is fundamental to understanding a wide range of stuff that happens in chemistry. (Yes, neutrons play a role for deuterium exchange effects etc.)
However, there is hardly anything that would change if descriptions included lower-level particles like quarks. I can’t think of any chemical reaction or other chemical process that cannot be sufficiently explained only going back to protons, neutrons and electrons — usually, even neutrons are irrelevant. So basically, you’re suggesting increasing complexity with no added benefit to the learner. Why would one do that? Just for the sake of correctness? If anything, a note that protons and neutrons are made up of smaller particles suffices for any beginner’s level textbook.
In fact, there is quite a lot of chemistry that can be just explained with the concepts of atoms and bonds. The only reason why the sub-atomic particles are included in the first place, is because salts are introduced early and can be explained a lot better, if electrons and nuclei are known. (And it eases the explanation of ‘what exactly is a bond?’)

I'm going to take a shot at this question from a different angle, aiming for a more objective analysis. For that, I wish to bring up several concepts. Bear with me!
First of all, what do we mean when asking if something is relevant or not to Chemistry? A precise and exact definition is not trivial, but let us consider a simple argument: chemical reactions are pretty important in Chemistry! In other words, we like to see atoms swapping around positions. It turns out that we have a fair idea of how quickly these events happen. We know that bonds in molecules vibrate at certain frequencies, which is how infrared spectroscopy works. The quickest bonds to vibrate do so at around $\rm{3500\ cm^{-1}}$, which implies a frequency of about $\rm{10^{14}\ Hz}$. This establishes a useful timescale - the fastest atomic displacements relevant to bonding in a molecule happen on the order of $\rm{10^{-14}\ s = 10\ fs}$. This means even the fastest chemical reactions need at least about this much of time to happen. Any event which begins and ceases within a timescale much shorter than $\rm{1\ fs}$ means that molecules don't even have time to react to the event. Even a molecule in a reaction transition state, as short-lived as it is, would be essentially frozen during time intervals much smaller than this. There are important processes which happen faster than $\rm{1\ fs}$, such as electronic transitions and nuclear decay, but they are important in Chemistry because what matters here is not how long the transition takes, but that the resulting state is sustained for around $\rm{1\ fs}$ or more before it decays. As an interesting aside, virtual electronic states used in Raman spectroscopy exist for times as long as $\rm{0.1\ fs}$, which further shows how this scale is too quick for molecules to respond.
Now for subatomic particles. Not counting antiparticles, there are at least 17 types of fundamental particles (though 4 of those are force carriers, and the Higgs boson is something else entirely, so we'll put those aside), and over a hundred composite subatomic particles. Yet, for some reason, almost all of Chemistry only deals with three fundamental matter particles (the electron, up quark and down quark), and two composite particles (protons and neutrons, i.e. the up-up-down and up-down-down triquarks, respectively). Why is that so?
Without getting into the nitty-gritty details of particle physics, it turns out that particles tend to decay into their lightest possible relatives while obeying certain rules, such as charge conservation. Electrons are the lightest negatively-charged particle, so they are indefinitely stable (the same is true for positrons, the lightest positively-charged particle). Protons could conceivably decay into something lighter thanks to the positron, but in this case there is another type of conservation law which strongly forbids them to do so (baryon number conservation), so if they are not indefinitely stable, they are at least very long-lived ($\rm{10^{35}}$ $\rm{a}$ or more).
Neutrons are an interesting case. They're very slightly heavier than protons, by about 1%. This is enough to allow for decay, and indeed a free neutron has a mean lifetime of about $\rm{15\ min}$ before it suffers $\beta^-$ decay into a proton. If this happened to every neutron, the Universe would be filled solely with hydrogen, and life could not exist. Remarkably, the mass difference between protons and neutrons is small enough that it can be compensated by the stabilising nuclear forces inside the nucleus, and thus neutrons inside nuclei are stable with respect to decay for an amount of time comparable to the proton.
So up to now, we know that electrons, protons and neutrons are three examples of subatomic particles which are so long-lived that we can expect them to be relevant in chemical timescales. Are there any more? This is where things go awry. The next most stable subatomic particle other than the ones I just cited is the muon, a particle identical to the electron except for its larger mass, with a mean lifetime of just $\rm{2.2\ \mu s}$. This is still many of orders of magnitude more than the femtosecond benchmark established earlier, and in fact muons can display meaning chemistry, even taking part in reactions [1], [2]. However, you can imagine that processes involving muons are expected to be much, much rarer, and not have as much importance in chemistry. This is not to say that muons are irrelevant to Chemistry, just that they are of interest only in a much narrower range of circumstances relative to electrons, neutrons and protons. The problem is even worse for other subatomic particles, which often have much shorter lifetimes. Looking at some lists of particles, such as baryons, mesons and leptons, one sees just how short-lived they are; about half of them decay far quicker than $\rm{1\ fs}$ and could not even in principle partake in chemistry. There are still a couple dozen of particles which decay between $\rm{100\ ns}$ and $\rm{1\ fs}$, for which chemistry could exist at an even more limited scope, but again these are rarely relevant.
After all the information exposed above, should the average Chemistry student worry about these subatomic details? No, I don't think so. In fact if these issues were to be mentioned as more than a passing comment in when first teaching the structure of matter, I believe it would be counter-productive. That said, I do think it's a shame that an enormous majority of chemists never seem to peer a little deeper into the structure of matter, even years after graduating. It would be nice if at least knowledge of the fundamental particles were more disseminated. The Standard Model of particle physics is quite an amazing achievement, and there's more than enough beauty to appreciate even without breaking out the massive integrals!


text:
Ripping apart plastic: is this a chemical or physical change?
I was holding a piece of plastic earlier, and I ripped it in half. This caused me to start to think about what was happening at the atomic level. If the plastic is being ripped, clearly bonds are being broken and new bonds are being formed, so is this not a chemical reaction?
I realize that the plastic retains all of its chemical characteristics despite it being ripped, so I am quite confused about how to classify what is occurring.

When you rip, tear or mechanically deform a polymer (for example a piece of plastic) you are putting energy into the material.  The energy from this deformation causes the polymer chains in the vicinity of the deformation to attempt to align.  To some degree this partial alignment makes continued deformation easier.  To continue tearing the polymer apart you just need to overcome the van der Waals forces holding the chains together.
In addition to the polymer chains being pulled apart, some chemical bonds will be broken.  I don't know how many bonds are broken per chain pulled apart, but I suspect it is low.  Low or not, some free radicals (or ions) will be generated when the chemical bonds break.  If the experiment is performed under ambient conditions, the radicals or ions will quickly react with oxygen or ions in the air.  If the experiment is performed in an inert atmosphere, then the radicals will persist (dangling bonds) and can be observed in an esr spectrometer.  It has been demonstrated that cross-linked polymers, which are less prone to flow deformation and therefore more likely to undergo bond breaking, do produce larger esr signals.
So most of what is going on is molecular deformation and pulling apart of the polymer chains - but some bond breaking does occur.
It would be an interesting experiment to cut two pieces of plastic and then cut one of them into finer shreds.  Place the two samples into separate tubes and then record their esr signals.  How much more intense would the signal be in the finely cut sample compared to the bulk sample?  Polycarbonate-A, straight out of the reagent bottle, produces an esr signal.

Agreed: a physical action can cause a chemical reaction. For example, any attempt to manipulate $\ce{NI3}$ physically causes it to decompose (detonate). For that matter, dividing metals into tiny pieces changes chemical behavior, because bonds that would have been hidden in the interior are exposed at the surface. Finely divided iron, for example, becomes pyrophoric, yet bulk iron can be heated in air without rapid oxidation..
The question is one of degree: how much of a chemical change resulted from ripping the plastic. Can you perform any chemical test that would distinguish the shreds from the whole, such as solubility? For example, bulk carbon, whether in graphite or diamond, is not soluble in most liquids at room temperature (though they dissolve in some molten metals), but as fullerene particles, carbon is much more soluble.
This could be an interesting experiment: prepare some bulk samples of a plastic, and grind some others (perhaps in a chilled ball mill?) and compare some chemical characteristics.
See Properties of Nanoparticles and Size matters: why nanomaterials are different.

If you subject an aqueous  solution of polyacrylamide to a lot of mechanical shearing then the polymer chains tend to become shorter. I know that during plastics processing that polymer chains can become shorter due to the shear forces which occur during in extruders.


text:
How to interpret orbital transition in TDDFT?
Consider the following TDDFT run with GAMESS:
!   File created by the GAMESS Input Deck Generator Plugin for Avogadro
 $BASIS GBASIS=N311 NGAUSS=6 $END
 $CONTRL SCFTYP=RHF RUNTYP=ENERGY TDDFT=EXCITE  DFTTYP=B3LYP $END
 $CONTRL ICHARG=0 MULT=1 $END
 $TDDFT NSTATE=9 $END
 $STATPT OPTTOL=0.0005 NSTEP=99 METHOD=RFO UPHESS=MSP HSSEND=.T. $END
 $SYSTEM MWORDS=1000 PARALL=.TRUE. $END
 $SCF DIRSCF=.T. DIIS=.T. DAMP=.T. $END

 $DATA
Title
C1
O     8.0     0.00000    -1.27900     0.00300
C     6.0    -0.00000    -0.05800     0.00100
C     6.0     1.29700     0.69100    -0.00000
C     6.0    -1.29800     0.69000    -0.00000
H     1.0     1.35900     1.32900    -0.90600
H     1.0     1.35900     1.33200     0.90300
H     1.0     2.15700    -0.01300     0.00100
H     1.0    -2.15700    -0.01400     0.00100
H     1.0    -1.35900     1.32900    -0.90600
H     1.0    -1.35900     1.33200     0.90300
 $END

From this we obtain the following part of the output. (The complete calculation can be found here.) How can we interpret the orbital transitions, in terms of for example $\mathrm{n}\to\sigma^*$ or $\pi\to\pi^*$?

From this output it is guessing only, could you include or upload somewhere the complete in- and output files.

@Martin-マーチン,  I uploaded the output here:http://s000.tinyupload.com/?file_id=63700463386440496706

and the input is here: http://s000.tinyupload.com/?file_id=46568865464361587097

*.rtf is not the native file format is it? I cannot open this file with any of my visualisation programs. What OS are you running it on? Can you just zip the whole folder...

@Martin-マーチン,   Oops sorry, it is TextEdit under Mac OS X.  I uploaded the zip file here: http://s000.tinyupload.com/?file_id=41117960135750882611

@Martin-マーチン,  a pdf version also, just in case:   http://s000.tinyupload.com/?file_id=93272470283596227104

To visualise the orbitals of your calculation, use a program of your choice. For Gamess, there are a few options availabile. I use ChemCraft and Molden, and they work quite well. Here is a compilation of some with the former mentioned:

You can further use the summary to identify the most interesting, strongest transitions:
                          SUMMARY OF TDDFT RESULTS

   STATE             ENERGY     EXCITATION  TRANSITION DIPOLE, A.U.  OSCILLATOR
                    HARTREE          EV         X       Y       Z     STRENGTH
   0  A         -193.0290234748    0.000
   1  A         -192.8724089055    4.262     0.0001  0.0000  0.0001    0.000
   2  A         -192.7831335626    6.691     0.4051 -0.0004  0.0000    0.027
   3  A         -192.7317175333    8.090     0.0018 -0.0976  0.0001    0.002
   4  A         -192.7220472153    8.353    -0.4522 -0.0001  0.0000    0.042
   5  A         -192.7210047911    8.382    -0.0121  0.0000 -0.0006    0.000
   6  A         -192.7176816741    8.472    -0.0001 -0.0013  0.0724    0.001
   7  A         -192.7167365427    8.498     0.0006 -0.0002  0.0075    0.000
   8  A         -192.6964881601    9.049     0.0029 -0.3152  0.0008    0.022
   9  A         -192.6862850361    9.326    -0.0022  1.1274 -0.0093    0.290

For the calculated 9 states that is the last one, with an oscillator strength of 0.290. Now skip back one section to where you find:
      -------------------
      SINGLET EXCITATIONS
      -------------------

Look for excited state number 9:
 STATE #   9  ENERGY =    9.326387 EV
 OSCILLATOR STRENGTH =    0.290423
 LAMBDA DIAGNOSTIC   =    0.561 (RYDBERG/CHARGE TRANSFER CHARACTER)
 SYMMETRY OF STATE   =    A
                 EXCITATION  DE-EXCITATION
     OCC     VIR  AMPLITUDE      AMPLITUDE
      I       A     X(I->A)        Y(A->I)
     ---     ---   --------       --------
      9      17    0.137515       0.018404
     15      17    0.694034      -0.078753
     13      18   -0.085768      -0.006093
     14      19   -0.059549      -0.004510
     16      19   -0.289828       0.018031
     15      20    0.051487       0.010566
     11      21    0.053380       0.009416
     13      21   -0.060120      -0.009392
     16      22   -0.607830       0.005676
     11      24    0.052546       0.012226
     13      25   -0.039117      -0.008987
     16      26    0.112895       0.016712
     16      30   -0.041659      -0.014274
     16      33    0.052880       0.019973
     16      36   -0.031783      -0.011699
     13      40   -0.030352      -0.013004

There you find the orbital transitions. You can also have a look at the amplitudes to identify the most dominant one. In this case it is probably 15 to 17 and therefore corresponds to $\pi\to\pi^*$.
The more complicated the molecules get, the more confusing will this process be. Also the higher the level of theory, the more transitions you will need to consider. In this case Natural Transition Orbitals will certainly become very helpful, see Richard L. Martin, J. Chem. Phys., 2003, 118, 4775-4777.

Form this output you dont.
Find this in the output:
             EXCITATION  DE-EXCITATION
 OCC     VIR  AMPLITUDE      AMPLITUDE
  I       A     X(I->A)        Y(A->I)
 ---     ---   --------       --------
 10      17   -0.039945       0.001926
 16      17    0.996619      -0.032612
 16      20   -0.070681       0.003443

it shows you the involved orbitals in the transitions. This a dominant transition from the 16ths orbital to the 17ths (HOMO to LUMO). You need to visualize the MOs and decide on the nature (sigma,pi,n) of the orbitals. Since they are MOs, it can be difficult with large molecules. You can look up natural transition orbitals that aim to simplify the visualization.
How you can look at orbitals with gamess? I have no clue.

jmol is useful for visualising orbitals.  (@hokru So how do you tie your 16 -> 17 transition to the appropriate eV result in the OP's table?)

@user1945827 the section i gave is printed for every transition (1 to 9). OP's table is the summary at then end of the td-dft output. It becomes clear if you download the output file from OPs post.


text:
Investigating a factor affecting buffer capacity
I'm preparing for an experiment to investigate some factor affecting buffer capacity. At the moment, I'm looking at using a $\ce{NH_3}$ ($2\,\mathrm{M}$) $+$ $\ce{NH_4Cl}$ buffer, but I'm not sure which factor to test. My research tells me that only the concentrations of each will affect buffer capacity, and that the ratio of their concentrations will affect the $\mathrm{pH}$ of the buffer (by the Henderson-Hasselbalch equation). Are there other factors which I can test, because I'm trying to make it a bit more interesting? As far as I can tell, something like temperature won't affect buffer capacity.
As far as the actually experiment goes, what would be the best way to measure buffer capacity? I could just do a titration and record the $\mathrm{pH}$ as I go, but I'm thinking of using an indicator and measuring absorbance. I'm open to any ideas here, because I'm struggling to think of anything "out of the box"?

As far as I can tell, something like temperature won't affect buffer capacity.

In this particular case, I wouldn't be too sure about it. Isn't it conceivable that your buffer will lose ammonia upon heating?
Did you already decide about the pH indicators? For the lower end, thymol blue might be a good choice. The supplementary data
to J. Mater. Chem. B, 2015,3, 2801-2807 contains a UV/VIS spectrum of this indicator.

What do you mean by "lose ammonia"?

@deusy You are having an aqueous solution of a gas. I would expect evaporation of a part of ammonia if you heat it up (too much).


text:
How to dissolve composite In2O3 with SnO2 without affecting stainless steel?
I found this article, but it was not helpful. The $\ce{In2O3}$ is probably amorphous. $\ce{HNO3}$ and $\ce{HCl}$ do not work. Any ideas?

In case your question is about the recovery of indium from ITO, the following might be helpful:
In Recovery of indium from indium tin oxide by solvent extraction, published in Hydrometallurgy, 2011, 107, 56-61 (DOI), the abstract seems to suggest that $\ce{H2SO4}$ is more suitable than $\ce{HNO3}$. Furthermore, a solvent extraction scheme for indium, using di-(2-ethylhexyl)phosphoric acid (DEHPA) is proposed.
In Recovery of indium from used indium–tin oxide (ITO) targets, published in Hydrometallurgy, 2011, 105, 207-212 (DOI), the authors agree on using $\ce{H2SO4}$ (at 90 °C) and decrease the ymount of tin in solution by sulfide precipitation.

Thak you a lot! I will try it.

there will be definitely problem with the steel substrate, H2SO4 will attack it as well

@JaroslavKotowski My pleasure! Is there a chance to use find an autoclave with at teflon inlay? In any case, good luck with the experiments!


text:
What is the Ksp of Al(OH)3?
Top hits on google are:
This site
Aluminum hydroxide $\ce{Al(OH)_3}$ >> $    1.3×10^{–33}$
and
This site
Aluminum hydroxide $\ce{Al(OH)_3}$ >> $    3×10^{–34}$
which is different by a factor of four, approximately.
What can be the reason for this variation? Experimental conditions?
Where can I get accurate or reliable results/data of $K_{\mathrm{sp}}$ in general?

Let me add to the confusion:

https://books.google.de/books?id=imBzjH6SHi8C&pg=PA303&lpg=PA303&dq=solubility+product+constant+aluminum+hydroxide&source=bl&ots=jXoPiVihnF&sig=eSK8saszXshTXVEKGOlThUiKVr0&hl=de&sa=X&ei=Xl5TVcepDuj_ywOXmYDQBA&ved=0CG8Q6AEwCTgK#v=onepage&q=solubility%20product%20constant%20aluminum%20hydroxide&f=false
$2\times10^{-32}$
http://faculty.ncc.edu/LinkClick.aspx?fileticket=smTKdcQlPY4%3D&tabid=1903
$1.8\times10^{-33}$
http://www.austincc.edu/dlyon/gen_chem_1_summer04/Chapter%2020%20solubiliity%20product%20f08.pdf
$1.9\times 10^{-33}$
http://www.csudh.edu/oliver/chemdata/data-ksp.htm
$4.6\times10^{-33}$
http://www.periodni.com/solubility_product_constants.html
$3\times 10^{-34}$
https://users.stlcc.edu/gkrishnan/ksptable.html
$3\times 10^{-34}$

Usually I'd first have a look at the CRC Handbook of Chemistry and Physics. Unfortunately, I couldn't find the data for aluminium hydroxide.

Umm..... $3*10^{-34}$ is the answer as it appears most times.

The reason for the differences is? Experimental conditions?

@Dexter Honestly, I don't know. All the online data are from tables without further explanation of the methods and conditions used. It is rather unsatisfying!

I think the confusion comes from the fact $\ce{Al(OH)3}$ exists in several forms (amorphous, boehmite, bayerite, hydrargylite) each having a different solubility product!
For example, my source gives a value of $\mathrm{K_{sp}} = 10^{-32.7}$ [1].
Each modification also hydrolyses differently. Base dissociation constant for the amorphous form is $\mathrm{K_{sp}} \approx 10^{-9.5}$ while hydrargylite has $\mathrm{K_{sp}} \approx 10^{-14.6}$. This means at pH 12, the amorphous form will dissolve completely, while the hydrargylite only slightly [1].
What kind of $\ce{Al(OH)3}$ is  obtained depends on the way the $\ce{Al(OH)3}$  precipitates and it is very difficult to obtain just a single form. For example, precipitating the hydroxide from acidic solution with ammonia gives mostly amorphous hydroxide, while precipitation from strongly alkaline solution gives a mixture of hydrargylite and bayerite. The precipitation can be direct, homogeneous or depending on the hydrolysis. Each method gives different results that are also influenced by foreign ions in solution.
Relatively pure boehmite can be obtained by aging the precipitate [1] and then doing the measurement.
A method for the preparations for relatively pure hydrargylite (alpha form) and bayerite (beta form) are given in Brauer[2].
Aluminium hydroxide tends to form gelatinous precipitate that easily adsorbs various ions (this property is actually utilized in waste water treatment and chromatography), hence proper washing is needed to separate the pure hydroxide from other ions such as ammonium.
Hope this helps clearing out the confusion a bit. Try to look for solubility constants of the specific modifications, then apply the correct constants for your case.
Note that although the solubility products differ by a magnitude or two, the compounds is extremely insoluble in either case and unless you do extremely precise analytical work, the difference won't affect your calculations much. It's like the difference between black and red mercury sulfides, which differ by four orders magnitude, yet the less soluble form is still practically completely precipitated.
[1]: Erdey, László. Gravimetric Analysis: International Series of Monographs on Analytical Chemistry, Vol. 7. Vol. 7. Elsevier, 2013.
[2]: Brauer G. Handbook of Preparative Inorganic Chemistry. Vol. 1. Academic Press, 1965

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


text:
Why does addition polymerisation of high density poly(ethene) use low temperature and pressure?
Logically thinking, I thought that it would be high pressure and high temperature that would create high density poly(ethene)... However, that's not the case. Is it because of the catalyst? Why is the condition low pressure/temperature?

The actual density difference at high temperatures should be negligible, the name is due to the higher crystallinity and therefore density of cold HDPE.

In fact, when polymerizing ethylene using a radical initiator at high temperature and high pressure, we favor the formation of low-density polyethylene. This is because the radical at the end of the growing chain can be transferred to any carbon atom of the chain by radical rearrangement (favored at high temperature) and due to close distances among molecules (high pressure) a branch chain can grow on the principal chain, and we get a branched polyethylene.  The chains of a branched polymer cannot align together, and eventually the density of the resultant polymer is low. We have got LDPE.
On the other hand, using a specific catalyst, known as Ziegler-Natta catalyst and under mild conditions of temperature and pressure, we get linear chains of polyethylene that can be aligned together and the density of the resultant polymer is high. We have got HDPE.


text:
Is there an electrolyte that will halt the production of Hydrogen in water electrolysis?

Care must be taken in choosing an electrolyte, since an anion from the electrolyte is in competition with the hydroxide ions to give up an electron. An electrolyte anion with less standard electrode potential than hydroxide will be oxidized instead of the hydroxide, and no oxygen gas will be produced. A cation with a greater standard electrode potential than a hydrogen ion will be reduced in its stead, and no hydrogen gas will be produced.

From what I understand, that bit states that certain electrolytes will stop the production of hydrogen gas and only oxygen gas and whatever gas from the electrolyte will be produced.
So I have two questions:

What are some electrolytes that will stop the production of hydrogen gas in water electrolysis?
What kind of water will be left behind if distilled water and an aforementioned electrolyte runs through electrolysis until all of the oxygen has been exhausted from the solution?

Please correct me if I'm misunderstanding anything.

The reduction of hydrogen is what the reduction potential of every other reduction potential is based off of, so:
$$\ce{2 H+_{(aq)} + 2e^{–} ->H2_{(g)}}~~~~~\Delta E^{\mathrm{o}}=0\,\mathrm{V}$$
So if a species has a has a reduction potential $>0\,\mathrm{V}$, then it will be be reduced before the naturally occurring $\ce{H+}$ ions are (these exist in solution due to the auto-protolysis of water, $\ce{H2O -> H+ + OH-}$, thereby not producing any $\ce{H2_{(g)}}$.
If your goal is to produce $\ce{O2_{(g)}}$ from the oxidation of $\ce{OH^{-}_{(aq)}}$, the reaction for this is:
$$\ce{4OH^{−}_{(aq)} -> O2_{(g)} + 2 H2O_{(l)} + 4 e^{−}}~~~~~\Delta E^{\mathrm{o}}=-0.401\,\mathrm{V}$$
Similarly, if a species has a has a oxidation potential $>-.401\,\mathrm{V}$, then it will be be oxidized before the  $\ce{OH-}$ ions, thereby not producing any $\ce{O2_{(g)}}$.
Wikipedia has a rather comprehensive standard reduction potential table, so use that to compare the reduction potential of ions. Choose your salts wisely!
To answer your other question, if this reaction continued indefinitely, a very acidic solution would result, as for every $\ce{OH-}$ that is oxidized, a $\ce{H+}$ is left in solution. Once all of the other cation from the salt is reduced, however, the $\ce{H+}$ would then start being reduced, producing $\ce{H2_{(g)}}$.


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Planes of Symmetry and Chirality
I am trying to identify whether the following objects possess planes of symmetry or not, but my answers seem different, from the textbook solutions (Klein Organic Chemistry 2e, Page 220, Question 23). Some of them I am unsure of and would like to make sure I know the proper reasoning for. Can someone please clarify?
(b) The correct response is yes. Is the plane of symmetry I have drawn acceptable?
(c) The correct response is no, but I thought it is possible to have a plane of symmetry like the one I have drawn.
(e) The correct response is yes, which I think I understand. I think this has three planes of symmetry.
(f) The correct response is no, which I agree with. However, it doesn't seem to concur with the response for the follow-up question.
Follow-up: Which image has three planes of symmetry? The correct answer is apparently, (f). My response would have been (e).

The follow-up question\answer must be wrong. F has no plane of symmetry while E has 3 planes as you've identified.  I think C has a plane (the one you've drawn) too, unless the two metal pieces are not symmetric, it's a little hard to tell from the drawing.

For (e), assuming the holes go all the way through the brick, the third plane of symmetry would be cutting halfway deep through, not diagonally as is drawn.

I couldn't take a picture of the image in my textbook so I tried to find a similar one (mainly in terms of the angle of the picture, because I thought that matters); however, the brick has three holes, all rectangular and evenly spaced.

Let us address the elephant in the room first and foremost.  All of these objects do not have any symmetry if you wanted to get real technical about it.  However, let us just assume certain things for this useful exercise to be relevant.
b.) The teacup has a vertical plane of symmetry splitting the teacup in half along the handle axis and intersecting with the centroid of the cup and saucer.  I would say that your guess is correct though it would be useful to have drawn in a plane rather than simply a line (for all examples).
c.) At first glance the pliers have a couple planes of symmetry BUT upon further thought it may not due to the way the rotating JOINT is constructed.  See the example below:

While it is difficult to see in your picture, below it is made very clear.  Therefore I will say that this object does not contain a plane of symmetry.  It would be POSSIBLE that it could have a plane of symmetry, one that lies in the plane of your monitor screen, cutting the object in half from tip to handle end... but it would be impossible to tell without seeing the other side. Many times pliers come with an inscription on one side of the joint, thereby breaking this symmetry.
e.) The brick does indeed have multiple planes of symmetry.  One vertical plane bisects the brick along its width and one bisects the brick along its length.  In addition, one plane of symmetry lies in the plane of the brick (the plane the brick is lying on), cutting the brick in half (think of it as coming in through the side of the brick).
f.) The hand has no plane of symmetry that I can see at all.  The back of the hand doesn't look like its front... the left and right side of the hand are clearly not mirror images of each other...
The brick has three planes of symmetry.

I never thought about the inscription aspect of the pliers. Thank you for your insightful response!


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Why sodium sulphate has water absorbing capability?
At first I thought maybe sodium sulphate in contact with water produces sulphuric acid which absorbs water but I do not think it is actually a valid reason.

Sodium sulphate reacts readily with water at room temperature to form hydrates up to sodium sulphate decahydrate, $\ce{NaSO4\cdot10H2O}$. $$\ce{NaSO4 + 10H2O -> NaSO4\cdot10H2O}$$ This means that $\ce{NaSO4}$ can absorb up to 10 mol of water for every 1 mol of salt that is used, making it one of the most effective drying agents in terms of sheer capacity.
Using data from this source we can calculate the Gibbs energy change for the above reaction at different temperatures. $$\mathrm{\Delta G_{20^\circ C} = -1.33~kJ~mol^{-1}}$$ $$\mathrm{\Delta G_{30^\circ C} = 1.28~kJ~mol^{-1}}$$
Since the entropy change for the reaction is negative, we can see that the cooler the solution you are trying to dry is, the more effective $\ce{NaSO4}$ will be as a drying agent.


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Why can't a primary cell be recharged?
The explanation I have found is that primary cells have irreversible reactions, so passing current in the opposite direction cannot recharge the cell. But as far as I understood it, no reaction is really irreversible - what one calls irreversible simply has very high values of the equilibrium constant, so for all intents and purposes proceeds to completion. Nevertheless, some small - maybe undetectable practically - amount of reactant does remain and the reaction does reach equilibrium. So even in a primary cell, when it is used up, equilibrium has been achieved, so recharging should be possible - it might take long to build up the initial value of potential since the required concentration of the electrolytes will be very different from that at equilibrium, but it should be possible. Why then is it not?

In some cases, a primary cell can be recharged, e.g. Mn-alkaline cells can take a partial charge. However, there are a number of reasons primary cells cannot be fully recharged, and that secondary cells have to be specifically designed to allow recharging.

Gas evolution: once most of a cell is charged (though all material has not been converted to its original state), the electrolyte is electrolyzed (apologies for the pleonastic redundancy), usually evolving hydrogen and oxygen, causing the cell to rupture. Nickel-metal hydride cells, for example, get around that issue by incorporating a catalyst to recombine them, $\ce{2 H2 + O2 -> 2 H2O + heat}$.
Physical change in the reactants: the standard "dry cell" (Leclanché cell) is made with an outer shell of zinc. As the cell discharges, parts of the shell are dissolved, eventually leaving holes. Reversing the current flow will cause zinc to redeposit, but not necessarily filling in the weakened areas. Even lead-acid cells designed for storage have limited lifetime because of the buildup of large, insoluble crystals of lead sulfate (sulfation, aka sulphation in GB), which may break off from the lead plates, preventing reversal of the chemical reaction during charging.
Physical damage to the insulator: metal dendrite may grow during charge-discharge cycling, piercing the dielectric between cells, even in well-designed secondary cells.

Irreversible in this context isn't a thermodynamic or kinetic thing, necessarily. It simply means that applying a current in reverse doesn't result in the original reagents. It may be that there is another possible reaction at a similar redox potential that leads to a different compound being formed or that the manner in which it is deposited is not suitable, e.g. the zinc in an alkaline battery being plated back onto the anode unevenly or such that it damages the ion conducting membrane.
It's also possible that recharging may make a primary cell inoperative by some other mechanism, such as gas formation in the potassium hydroxide electrolyte of an alkaline cell deforming the case or damaging the half-cell separator. Primary cells are also typically constructed cheaply as they don't have to withstand the stresses of secondary cells, e.g. the steel casing of alkaline batteries is susceptible to corrosion.
Things like the various lithium battery chemistries make better secondary cells as they use organic solvents which reduce the problem of gas evolution and lithium can form intercalated compounds where the lithium can be evenly spread instead of growing dendrites of solid lithium which can damage the cell when recharged. That's the main difference between disposable and rechargeable lithium cells: disposable cells use solid lithium electrodes because the anode only has to dissolve once and the difficulty of replating the anode is not an issue, whereas a rechargeable cell needs to return the anode to its original state many times and an intercalated lithium compound, though more expensive, is far easier to replenish with lithium.

I have heard that technically, a dry cell CAN be recharged, but it takes a very long time to do it without rupturing the cathode - the zinc shell -, enough time that the electricity used costs more than replacing the battery, plus, when recharged, it does not have the life of a brand new battery, so it just is not worth the trouble.


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Energy of photons emitted by de-excitation of electrons
When an electron of an atom returns from an excited state to the ground state, it emits energy in the form of a photon. How does the change in energy level compare to the energy of the emitted photon?

Welcome to chemistry.SE. I edited your question to make the title and tags actually relevant. Please feel free to take the tour to find out more about the site. In future try to make your titles actually useful as this is the first thing people see when they see your question so titles like 'chemistry periotic table' don't really say anything about the question's content.

When an electron of an atom returns from an excited state to the ground state, it emits energy in the form of a photon.

That is one of several pathways for deexcitation. Often, internal conversion, the radiationless deeexcitation happens instead.

How does the change in energy level compare to the energy of the emitted photon?

It depends on the relaxed geometry of the excited state, as compared to the ground state.
Remember that the excitation happens on a femtosecond scale. That is amazingly fast, light travels a distance of just $\mathrm{0.3\,\mu m}$ in $\mathrm{1\,fs}$.
Since it takes longer than some femtoseconds to move the much heavier nuclei, the molecule in its electronically excited state still has the geometry of the ground state. This can be far from the lowest vibrational level (the relaxed geometry) of the excited state.
Before emission of a photon occurs, ectronically excited molecules relax and adopt their lowest vibrational state.
This means that the energy of the emitted photon often is (much) lower than the excitation energy. This effect is known as Stokes shift.

I wanted to add a couple of figures to elaborate on Klaus's answer. An electron "lives" in quantized energy levels, but the atom also vibrates in quantised energy levels, called vibrational states. There are many vibrational states within each main quantum number $n$ (K shell, L shell, etc.). These can be "crudely" approximated by the harmonic oscillator, or more accurately by Morse potentials. Morse potentials are depicted in the figures as the curved lines, one for each main quantum number (e.g. $n=1$ and $n=2$). The horizontal lines within the potential are the various vibrational states. Note that the first vibrational state is not at the bottom of the potential; even at 0 K there will be at least some vibrations!

Figure 1. Morse potentials for two main quantum numbers. Source: Atkins' "Physical Chemistry", 9th ed, page 505.
The absorption of radiation is depicted by an arrow going from the lowest vibrational state in the lowest electronic state, up to a high vibrational state in the next electronic state. The tiny arrows illustrate the "vibrational relaxation" explained by Klaus. This is a non-radiative process. When the electron reaches the lowest possible vibrational state in $n=2$, then an electronic transition takes place. Note that the electron does not necessarily de-excite directly to the lowest vibrational level. Similarly, note that during the absorption, the electron is not just excited to the lowest vibrational state of $n=2$, but some higher vibrational state. So we have such vibrational relaxation in both $n=1$ and $n=2$. The energy lost during vibrational relaxation leads to the emitted photon being of lower energy than the incident photon. But why is the electron excited to a high vibrational state, and not just to the lowest possible?
The transition in the figure is a vibronic transition, meaning that both a  vibrational and an electronic  transition occur. Also note that the two Morse potentials in both figures are not perfectly aligned above each other; the upper one is shifted to the right. Now look at Figure 2. The colored curves are the probability distributions of where to find the electron in each vibrational state. See how the number of nodes in each vibrational state increase by 1 for each vibrational transition? The most likely electron position shifts toward the edges for each vibrational state. At the point of convergence of the vibrational state, the electron vibrates so much that it is ejected out of its state and becomes free.
Back to why the electron does not simply jump to the lowest vibrational state in $n=2$. The Franck-Condon principle states that the most likely vibrational state in $n=2$ for the electron, is the one with which the current vibrational state overlaps with the most. The two ground state vibrational states does not overlap significantly, and so such a transition will be very weak. However, as is depicted by the blue arrow in Figure 2, a transition from the ground vibrational state in $n=1$ overlaps very well with the third vibrational ($\nu ' = 2$) state in $n=2$, and so this transition will be very strong. Finding itself $\nu ' = 2$, the electron relaxes to $\nu ' = 0$, and so undergoes de-excitation to $n=1$ ($E_0$ in the figure). This time the best overlap is also for the third vibrational state. Then the electron relaxes some more. Note that several vibronic transitions occur, but some are much more probable than others.

Figure 2. Probability densities for each vibrational state. Source: Source
So that is why the emitted wavelength is longer than the excitation wavelength; it is due to vibrational relaxation occurring due to the vibronic transition. The Franck-Condon principle gives a way to predict the vibrational transition based on the vibrational wave-function overlaps in the initial and final state.

But for the green transition in Figure 2, the energy of the emitted x-ray has exactly the difference in energy between the two orbitals. That is less than the energy of the excitation x-ray.

When an electron de-excites from a higher energy orbital to a lower energy orbital it loses energy by emitting a photon. Since energy must be conserved, the energy of this photon is equal to the amount of energy the electron has lost, or in other words the difference in energy of the two orbitals involved. The frequency of the photon can then be calculated using $E=hf$ where $h$ is Planck's constant and has a value of $\mathrm{6.626\times10^{-34}~J~s}$.


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How do I calculate the viscosity of a mixture of toluene and isopropanol?
I am trying to understand why my nanoparticles will only electrophoretically deposit in a non-polar solvent and a non-polar/polar solvent mix but not in the polar solvent. I am hoping to use zeta potential and Hamaker's equation to understand this


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2,4-dinitrophenyl hydrazine test
I found this reaction over here:

which illustrates the carbonyl test but looks fuzzy.
Assuming that the R in the product is a typo for H, I couldn't conserve the number of Nitrogen atoms.
What is the correct reaction?
Is this correct?
Is there a possibility of a methyl substitution on the ring in a variant of this test?

Somebody messed up the scheme completely!
As you have already imagined, there is no magical $\ce{NO2}$ to $\ce{CH3}$ conversion in this classical reaction ;-)

In the times before modern spectroscopy, these reactions were valuable analytical tools: the melting points of the dinitrophenylhydrazones and/or other derivatives were used to identify the carbonyl compounds.

The methyl group on the product is an error. It should be a nitro group.
Also you are right that the R group should be an H (or the other way around depending on what they intended). It looks like someone stuck two different images together but forgot to check to see if they matched up.
Other than that the reaction shown is correct. The reagent used in 2,4-dinitrophenylhydrazine, commonly know as DNPH.


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Nitrogen chain nomenclature
I know that nitrogen is versatile enough that it can form branched chains if it is all single bonds.
For example:
$\ce{H2N-N(NH2)-NH2}$
Is this compound a nitrane (nitrogen alkane)?
If so how would I name this branched nitrogen compound?

According to the IUPAC Gold book, these are known as azanes, from the information:

Saturated acyclic nitrogen hydrides having the general formula $\ce{N_{n}H_{n + 2}}$

For the example you are looking at, according to the IUPAC document P-67 Mononuclear and polynuclear noncarbon acids and their functional replacement
analogues as parents for naming organic compounds (p. 15) is named as 2-aminotriazane.

Additionally, it is probably going to explode.

This is a substance that I am definitely not keen on being around.

Yeah I can see why azanes would be explosive.


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Reactivate dried up ink of ballpoint pens
According to this question, ballpoint pens start writing intermittently when the ink is drying up. In other words, the ink is becoming more viscous and does not flow any more.
Are there ways to make it flow again?

As ink typically consists of non-polar dyes dissolved in a non-polar liquid, I would recommend using a straightened paper clip and a drop of acetone to clear out the dried ink clogging up the ink cartridge. This works by agitating the dried ink, as well as by replenishing the ink's solvent. Just be careful to not make too much of a mess!
Another method I have seen work before is heating the end of the pen under a flame, which melts the ink clogging the tip. For this method, my advice would be to not hold the pen in one place for too long, lest you risk melting the plastic.


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Enthalpy and Bond Dissociation Energies
I'm not sure if I understand the reasoning behind this particular question.
The sigma and pi bonds of $\ce{C=C}$ have a combined bond dissociation energy of 632 kJ/mol. Using this information, predict whether the following reaction is exothermic or endothermic.
$\ce{C2H4 + H2O -> CH3CH2OH}$
My attempt:
The bonds being broken are the $\ce{C=C}$ pi bond and the $\ce{H-OH}$ bond. I realize there are two bonds being formed, but I am unable to correctly identify the second one. One is $\ce{CH_3CH_2-OH}$. I think the other should be $\ce{CH_2-H}$. The textbook says this bond is $\ce{CH_2H-R}$. I think R is referring to the side group, which varies (and in this case, is $\ce{CH_2-OH}$?), but the enthalpy value for breaking the $\ce{CH_2H-R}$ is listed as 410 kJ in the solutions. According to the table in my book, the value for a $\ce{CH_3CH_2-H}$ is 410 kJ/mol. I do not understand how these two bonds are equivalent. Can someone please explain this?

Tabulation table such as the ones you reference, describe 'typical' bonds when using generic groups such as R.  You are right in your discomfort that the numbers should be somewhat different, but that would require a thermochemical measurement for every known molecule, which is difficult to achieve in practice.
Calculations such as the one you are performing are 'estimates', and while good ones, they are never as credible as an actual experiment.


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Transformation of acyl group to carboxyl group in biphenyl system with two acyl groups at para position
How do I perform this reaction?
I have prepared 4,4'-acylbiphenyl and want to make one acyl group into $\ce{-OH}$ and other into $\ce{-COOH}$ group.

Is it correct that you are aiming to synthesize 4-(4-hydroxyphenyl)benzoic acid [58574-03-1]?

The target compound is available in one step from biphenyl using potassium peroxydisulfate $(\ce{K2S2O8})$ as an oxidant, Palladium(II)-trifluoroacetate, $\ce{Pd(OCOCF3)2}$, as a catalyst, and formic acid as the carbon source for the $\ce{-COOH}$ group. The synthesis was published in
Org. Lett., 2004, 6, 2437-2439 (DOI).

EDIT
Limited resources and experience will probably rule out Stille-type coupling reactions too.
Alternatives pretty much depend on which 4,4'-disubstituted biphenyl are available.
A methoxyarene can be converted to the corresponding phenol by $\ce{BBr3}$
or the cheaper hydrobromic acid. The $\ce{-OCH3}$ group resists oxidation, is a +M substituent and is not acidic. All this might be helpful (see below).
In order to introduce the $\ce{-COOH}$ substituent, two approaches seem to be possible in an undergrade lab:

oxidation of a methyl group by permanganate
Grignard reaction of a bromoarene with carbon dioxide

Thank you sir. But I am just a UG student and I don't think these reagents will be available for me sadly.


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Do I understand why heat capacity is called 'heat *capacity*'?
The notion of heat capacity seemed a little odd to me. I thought that the word 'capacity' could be throwing me off. Do I understand the notion correctly?
I take the word 'capacity' to express, roughly, 'the maximum amount of something that a thing can take'. My first intuition is that the maximum temperature of a system equals whatever temperature corresponds to every molecule in that system moving at the limit of the speed of light. Accordingly, if 'heat' denotes a certain species of energy transfer, then the heat capacity (i.e. the maximum amount of energy that a system can receive) should be equal to the number calculated by, for each molecule in the system, taking the difference between the speed of light and the molecule's starting speed, summing those differences, and then dividing by the number of molecules in the system. Accordingly, if everything else is equal (i.e. pressure and volume), then the heat capacity of a system is a function of the starting temperature of the system and its volume. At first blush, that conception of heat capacity seems compatible with the equation
$C_x = \frac {Q}{\Delta T_x}$.

There is no maximum bound to how much energy a system (or even a single subatomic particle) can contain, in principle. Even though its speed is limited to the speed of light, kinetic energy increases infinitely as the speed barrier is approached. Whether there is a maximum temperature is a more subtle topic. Neither of these concepts are necessary to understand heat capacity, though.

@NicolauSakerNeto Okay, so if energy increases infinitely, then heat capacity doesn't seem to have anything to do with capacity as I've understood: every system has an infinite capacity to receive energy. Clearly, I misunderstand something. How should I think of heat capacity instead?

I don't think there's a particularly deep meaning for it. Heat capacity is just the proportionality constant relating how much a system's temperature increases/decreases when a certain amount of heat is introduced/extracted. Different materials will have different proportionality constants.

Expanding on Nicolau Saker Neto's comments above, the heat capacity dictates the (reversible) heat flux required to increase the system's temperature by one unit.
Definitions of the heat capacity ($C_V$ and $C_p$)
Another way to look at it is that the heat capacity dictates the response in internal energy ($U$)/enthalpy ($H$) of the system to temperature, as described below. Note that the heat capacity at constant volume ($C_V$) is not the same as the heat capacity at constant pressure ($C_p$) - and that both may be defined in several ways, the most common being:
$$ C_V = \left(\frac{\partial U}{\partial T}\right)_{V,\mathbf{n}} = \left(\frac{\partial Q}{\partial T}\right)_{V,\mathbf{n}} $$
$$ C_p = \left(\frac{\partial H}{\partial T}\right)_{p,\mathbf{n}} = \left(\frac{\partial Q}{\partial T}\right)_{p,\mathbf{n}} $$
Contributions to the heat capacity
In should be added that while translation - the kinetic energy of molecules in motion - are an important contribution to the heat capacity of a gas, both vibrations and rotations also contribute for all but the simplest molecules. For high temperatures (approx. $T \geq 3000\ \mathrm{K}) $, the excitation of outer electrons will also contribute to the heat capacity. For even higher temperatures, the substance in question will behave as plasma.
In short
There are many complicating factors to the precise understanding of the origin of contributions to the heat capacity - both at the high end and low end of the temperature range. In all cases, however, the heat capacity will give you the answer to the question:

"How much energy do I have to add to this system at the given
  conditions in order to increase the temperature with one unit?"

Further reading on Wikipedia is recommended for anyone interested.


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Is the resonance depiction of benzene typically portrayed incorrectly?
I can understand that the carbon atoms in benzene have delocalized p electrons that are shared amongst all 6 of the carbons in a cloud on either side of the ring. I also understand why this could be represented as resonance structures like so:

I can also understand representing it in this form:

What I cannot understand, however, is why only one of the first form appears so often in depictions of organic molecules. For example, this:

Since this is a "resonance structure," is this not incorrect to list only one of the structures?
One might argue that the reader understands what is being portrayed or that this is done in aims of saving time, but that doesn't sit right with me. The entire point of nomenclature, drawing a chemical, or anything scientific is that it can be interpreted by anyone in exactly the same way. I feel that if someone is deciding to draw a chemical using the resonance structures of benzene, then they should draw both resonance structures, lest it be mistaken for 1,3,5-cyclohexatriene.

On a side note, I don't think 1,3,5-cyclohexatriene exists.

Regardless, isn't it improper?

I'm confused. According to the wiki 1,3,5-cyclohexatriene is an alternative name for benzene.

I think Binary Geek meant a six-membered ring with alternating double and single bonds, i.e. non-resonanced. @JohnSnow

@JohnSnow I did mean the non-resonanced molecule as Jan has pointed out.

Starting with a short and blunt answer to the title’s question: Yes. But you likely already knew that, otherwise you wouldn’t have written the rest.

One might argue that the reader understands what is being portrayed or that this is done in aims of saving time, but that doesn’t sit right with me. The entire point of nomenclature, drawing a chemical, or anything scientific is that it can be interpreted by anyone in exactly the same way. I feel that if someone is deciding to draw a chemical using the resonance structures of benzene, then they should draw both resonance structures, lest it be mistaken for 1,3,5-cyclohexatriene.

Well, there is a lot more time saved that you seem to be imagining according to your quote.
You seem to suggest that there are only two correct resonance structures for benzene, which is incorrect. Even if you only consider the $\pi$ system, there are a lot more possible resonance structures which contribute to the final picture in a non-zero manner.

The ones I displayed in the picture are just a very small subset. In fact, solely for the $\ce{H2}$ molecule, a professor of mine claimed to have drawn some 18 (or was it 80?) different resonance Lewis structures and also claimed to be able to accurately calculate the energy of the molecule by adequately weighting all those structures. (‘Accurately’ meaning getting an energy comparable to one calculated using molecular orbital theory.)
This actually moves us towards another question: What actually is a bond? There is a lot more to a bond than simply drawing a line between two atoms and saying that ‘they are now sharing an electron pair.’ If you calculate simple-looking molecules like formaldehyde using a program like Gaussian, you are able to see a multitude of molecular orbitals — at least twelve. Two are defined as ‘core orbitals’, namely they represent the 1s orbitals of carbon and oxygen, respectively; all the others somehow contribute to bonding or antibonding interactions somewhere across the whole molecule.
So what would an accurate representation be? Drawing the atoms geometrically correctly, a cloud around them and writing a number 12 inside or next to the cloud? That way, it would be made clear that there are twelve electrons delocalised across the whole of the molecule and somehow keeping it together. But what would be gained from this type of more accurate representation? Or rather: What would be lost? You would lose the possibility to easily discuss lone pairs on the oxygen, bond order between carbon and oxygen and a lot more.
Similarly for benzene and all molecules that contain a six-membered ring of carbons with a $\pi$ system with six electrons (a.k.a. the benzene derivates you’re asking about): If you insisted on drawing ‘the two’ resonance structures, you would still be far from the truth. Furthermore, for most molecules that are not benzene but contain the same ring, the bonds are no longer equal and their bond orders are no longer exactly $1.5$ (true for any molecule that has lost the $\mathrm{D}_{6\mathrm{h}}$ symmetry). So drawing two sets of double bonds and claiming the structures to be equal would be outright wrong.
Finally, though, every chemist has learnt to get a more or less thorough understanding on the concept of a bond and on the concept of aromaticity that it can reasonably be assumed that what I presented is accessable somewhere at the back of the head if really required. Thus, every chemist should have a rather good understading that a line on paper is a line, a bond is not just a pair of electrons and a benzene ring can be drawn in multiple ways. When it really matters, we can either expect the chemist to instantly recognise the mattering due to a special case, or we can bring the mattering matter to the chemist’s attention by explicitly drawing it.
So there is a lot of time gained by simplifying drawings of molecules in the way they are and also cases where it matters can easily be illustrated by drawing more elaborate drawings which aren’t skipped in one’s attention because they are rarely encountered. And we also keep the ability to easily recognise and discuss simple concepts.

P.S: It’s not necessary for anyone to interpret these concepts the same way. In fact, it’s a good thing if an artist takes the picture of benzene and interprets it completely differently. If anything, ‘anyone’ should be restricted to the chemical — or maybe the entire scientific — community, but there I can see the interpretation safely, regardless of whether aromaticity is drawn or not.

Great answer.  It's almost like we need a tag for "historical chemical concepts/vocabulary do not always substitute for solving the Schrodinger equation; sometimes you need to just solve the equation"

The problem of drawing all the resonance in salicylic acid (or many of the molecules with a benzene ring), is that none of them convey all the resonance involved. You can easily see that you can move the 3 double bonds so that the carbons in the ring should have equal bond lengths.
What is not evident is that the carboxylic acid is planer to the ring (the p-orbitals in the carbon and 2 oxygens are aligned with the p-orbitals in the ring) and the hydrogen is traded back and forth with the 2 oxygens. With this extra resonance, the carbon in the ring near the carboxylic acid should have a different bond length than the carbon across from it.
And, the alcohol will also have a contribution to the resonance.


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In order to solve for this equation for heat capacity, do I need to treat it as a system of equations?
For a lab, we took the temperature of 15ml of tap water, $q_{cool}$, then added 20ml of tap water warmed to 80 degrees Celsius, $q_{warm}$.
We recorded a baseline temperature of 24.8, a temperature of 56 after adding the water, and a temperature of 42.9 100 seconds later.
I have the following equations:

$q=m × 4.184 J/(g*C) × ΔT$
$q_{warm} + q_{cool} + q_{calorimeter} = 0$
$q_{calorimeter} = - (q_{warm} + q_{cool} ) = C_{calorimeter} × ΔT$.

In the first formula $ 4.184 J/(g*C)$ is the specific heat of water.
To the best of my knowledge, specific heat is usually expressed in grams-Kelvin; however, the TA may have provided the info in grams-Celsius (their units are equal, so I presume they're the same thing). However, she also uses $C$ to designate heat capacity as she does in the third equation.
If, in the first equation, $C$ designates heat capacity, then I have three equations, and three unknowns: $q_{calorimeter}, q_{warm}$, and $q_{cool}$, and I'd treat it as a system of three equations and solve for them.
However, if, in the first equation, $C$ designates 'Celsius' then I could just plug in the values for $q_{warm}$ and $q_{cool}$ and solve for $q_{calorimeter}$.
Two related questions:

In the first equation, what does $C$ designate.
If it designates the heat capacity of the system, do I solve for $q$ by solving a system of equations?

In the first equation, what does C designate.

Since the final value is energy in joules, then it should be the the unit for temperature. She should have used °C to be clear.

If it designates the heat capacity of the system, do I solve for q by solving a system of equations?

No, because you don't have the correct equations. You need the amount of heat released from the hot water minus the amount of heat absorbed by the cold water to give you the heat absorbed by the calorimeter.

In the wide world of science, sadly we don't have enough letters in any alphabet for what we need them to represent. One guideline that is helpful at least when typesetting: use emphasis/italics for quantities, constants, and variables (things that represent a number) - $C_{\text{p}}$. Use normal text for units $^\circ\text{C}$.