text:
Diagonalization of Hessian H= PkP$^{-1}$: what do the elements of P matrix physically mean?
This is quoted from Computational Chemistry by Errol J. Lewars' chapter 2's Stationary Points & Normal-Mode
Vibrations: ZPE:

\begin{align}\mathbf H &=\begin{bmatrix}
    \dfrac{\partial ^2 E}{\partial q_1 \partial q_1} & \dfrac{\partial ^2 E}{\partial q_1 \partial q_2} & \dfrac{\partial ^2 E}{\partial q_1 \partial q_3} & \dots  & \dfrac{\partial ^2 E}{\partial q_1 \partial q_9} \\
    \dfrac{\partial ^2 E}{\partial q_2 \partial q_1} & \dfrac{\partial ^2 E}{\partial q_2 \partial q_2} & \dfrac{\partial ^2 E}{\partial q_2 \partial q_3} & \dots  & \dfrac{\partial ^2 E}{\partial q_2 \partial q_9}\\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    \dfrac{\partial ^2 E}{\partial q_9 \partial q_1} & \dfrac{\partial ^2 E}{\partial q_9 \partial q_2} & \dfrac{\partial ^2 E}{\partial q_9 \partial q_3} & \dots  & \dfrac{\partial ^2 E}{\partial q_9 \partial q_9}
\end{bmatrix}\\ \\ &=\underset{\mathbf P}{\begin{bmatrix}
    q_{11} & q_{12} & q_{13} & \dots  & q_{19} \\
    q_{21} & q_{22} & q_{23} & \dots  & q_{29} \\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    q_{91} & q_{92} & q_{93} & \dots  & q_{99}
\end{bmatrix}} \underset{\mathbf k}{\begin{bmatrix}
    k_{1} & 0 & 0 & \dots  & 0 \\
    0 & k_{2} & 0 & \dots  & 0 \\
    \vdots & \vdots & \vdots & \ddots & \vdots \\
    0 & 0 & 0 & \dots  & k_{9}
\end{bmatrix}}\mathbf P^{-1}\end{align}
The 9 × 9 Hessian for a tri-atomic molecule
  (three Cartesian coordinates for each atom) is decomposed by diagonalization into a P
  matrix whose columns are “direction vectors” for the vibrations whose force constants
  are given by the $\bf k$ matrix. Actually, columns 1, 2 and 3 of $\bf P$ and the corresponding $k_1,k_2,k_3$ of $\bf k$ refer to translational motion of the molecule (motion of the whole
  molecule from one place to another in space); these three “force constants” are nearly
  zero. Columns 4, 5 and 6 of $\bf P$ and the corresponding $k_4,k_5,k_6$ of $\bf k$ refer to
  rotational motion about the three principal axes of rotation, and are also nearly zero.
  Columns 7, 8 and 9 of $\bf P$ and the corresponding $k_7,k_8,k_9$ of $\bf k$ are the direction
  vectors and force constants, respectively, for the normal-mode vibrations: and
  refer to vibrational modes 1, 2 and 3, while the 7th, 8th, and 9th columns of $\bf P$ are
  composed of the x, y and z components of vectors for motion of the three atoms in mode
  1 (column 7), mode 2 (column 8), and mode 3 (column 9).

I couldn't conceive what actually $\bf P$ implies; it is, of course, the eigenvector matrix of the Hessian $\bf H $ but what does it represent?
Lewars wrote that columns of $\bf P$ are actually direction vectors; now what did he mean by direction vector?
I thought $q$ represents the geometric coordinate; I got $k$ is the force-constant; but what are those series of $q$ s in $\mathbf P\;?$
Are the $q$ s in $\bf P$ coordinates of the atoms in molecules? If so, aren't they always changing since they are vibrating, rotating...?
Can anyone please explain this to me what those $q$ s actually represent? What did Lewars want to mean by direction vectors?

After a long time, I got a downvote; thank you- I was missing it. But I would be happier if Mr. Downvoter attempts to waste a few minute to tell where I committed a sin so that I may try to undo that. Please : (

Are the $q$s in $\mathbf{P}$ coordinates of the atoms in molecules?

Close, but not quite. The $q_{i,j}$ represent the displacements from the equilibrium geometry of the appropriate atom/coordinate $i$, for each mode j. Thus, for example, $q_{5,9}$ corresponds to the displacement from equilibrium of atom $2$ along axis $2$ (typically the $y$-axis) when the molecule is undergoing vibration in mode $9$.

If so, aren't they always changing since they are vibrating, rotating...?

There are no changes due to translation or rotation. It is customary to "project out" the translational and rotational motion from the Hessian and the geometry.  Qualitatively, this means that we simply ignore the translation and rotation and consider the molecule as "fixed" in its equilibrium configuration (allowing for vibrational 'wiggling'), with that configuration always oriented in the same way relative to the coordinate system.
As for the vibrational motion, though: yes, the $q_{i,j}$ values do technically also 'wiggle' as the molecule vibrates. However, the magnitudes of the perturbations $\delta q_{i,j}$ are in many cases very small, and can be neglected. In those cases where they cannot, the harmonic oscillator approximation to the vibrational motion breaks down and is no longer a useful model for vibration along that mode.
See my answers here and here for some links to literature on situations where this occurs (emphasis on the 'internal rotation' topics, though some of the 'anharmonic' and 'Coriolis' references might be somewhat relevant). Here is a Chem.SE question about nitrogen inversion that might be of interest.  More generally, a Google search for "low-frequency vibrational modes" should turn up lots of interesting reading.

First one should understand what normal coordinates and (mass-weighted) cartesian coordinates are: They are both a means to specify where in space each single atom of a molecule is located. In cartesian coordinates you do this simply by specifying the x,y and z coordinates of each single atom. Normal coordinates are a little trickier to understand, since they include always several atoms.
A simple example might illustrate this: Suppose you have two masses which are connected by a spring and can move only in one direction (the direction of the spring). Also assume that their mass is $m=1$ so we don't have to distinguish between the usual and mass-weighted cartesian coordinates. Then there are two ways to specify the position of the two masses in space: You can specify the x coordinates of the two masses (cartesian coordinates) or you can specify the center of mass and the distance between the two masses (normal modes).
To be more concrete, the dynamical matrix (Hessian of the potential in mass-weighted coordinates) of the above problem is
\begin{equation}
\mathbf H = \begin{pmatrix}
k & -k \\
-k & k
\end{pmatrix}
\end{equation}
where k is the force constant of the spring.
Diagonalization then yields the force constants 0 and 2k and the matrix
\begin{equation}
\mathbf P = \frac{1}{2} \begin{pmatrix}
1 & -1 \\
1 & 1
\end{pmatrix} \end{equation}
The matrix $\mathbf P$ now tells us how to compute the values of the normal coordinates $Q_i$ from the values of the cartesian coordinates:
\begin{equation}
Q_i = \sum_j q_i P_{ij}
\end{equation}
From this we see that the first normal coordinate is proportional to the center of mass and the second one proportional to the distance of the two masses. Note that my example used usual cartesian coordinates as $q_i$ while in real problems you usually use displacement coordinates. So you can see that the matrix $\mathbf P$ just let's you express the values for one set of coordinates in terms of the values for the other.
You can of course also do it the other way round and ask: what is the position/displacement of some atom, when I know that the normal coordinates have the values $Q_i$. Then you can calculate your cartesian coordinates via
\begin{equation}
q_i = \sum_j (P^{-1})_{ji} Q_j = \sum_j P_{ij} Q_j
\end{equation}


text:
What is the Lewis Structure for CH3CONHCH3?
Chemical formula: $\ce{CH3CONHCH3}$
How could I draw the Lewis structure so that oxygen and nitrogen have a full octet? Currently, I have $\ce{CH3-C=O-NH-CH3}$.

The oxygen is doubly bonded to the carbon, and not bonded to the nitrogen.

The part you get wrong is just like what Lighthart says, the oxygen is double bonded to carbon and not with nitrogen. Therefore, the nitrogen will be to bonded with the carbon. Second, you have a mistake when you write the first one as $\ce{CH_3}$. This is because it's carbon which bonded with the other carbon and not hydrogen. So the proper lewis structure will be like the picture below.

source: Stack Exchange

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sort: votes


text:
How does complex ion differ from ligand complex?
As I understand, in terms of "colouring", ligand complex cause a transition metal cation centre's d-shell to split, allowing the electrons in that region to absorb specific frequencies of light, and thus resulting in colouring, does the same apply for complex ions such as chromate? Does the oxygen atoms/ions around chromate cause d-shell splitting and thus "colouring" effect?

ligand complex? Where'd you get such name?

Chemistry class, from the international baccalaureate, what I mean by that is a positively charge metal centre surrounded by datively bonding neutral molecules or anions. Such as hydrated copper ion.

And doesn't chromate fit to your definition?

IIRC, chromate has colour due to metal-oxide charge transfer in the pi-cloud due to photoexcitation of electrons.

Can you elaborate "metal-oxide charge transfer " or direct me to a source?

Wrong way round, it is called ligand-metal charge transfer, the answer here describes it http://chemistry.stackexchange.com/questions/39829/how-can-the-intense-color-of-potassium-permanganate-be-explained-with-molecular/39848

@orthoseal What a coincidence that I just quoted from that question, too =D

To quickly pre-answer the question at the end of the body: Yes, the oxygen atoms around the chromium(VI) cause d-orbital splitting. However, chromium(VI) is a $\mathrm{d^0}$ system and thus the colour is not caused by d-electron transitions. Rather, as I highlighted in my permanganate answer, the cause of the colour is a ligand to metal charge transfer transition. For details, see there.

The question in the title basically refers to the difference between a complex such as $\ce{[CoCl4]-}$ and an ion such as $\ce{SO4^2-}$ or $\ce{CrO4^2-}$. These can be considered two extremes of a transition much like the extremes unpolar covalent bond and ionic bond when used to describe bonding.
All coordination complexes are nothing more than dative bonds, which in turn are nothing more than a special subset of weak covalent bonds. From a qualitative bonding point of view, there is really no difference between either compounds: For both you get orbital mixing, a bonding and an antibonding set of orbitals and an overall stabilisation of the system. And if you totally ignore energy differences and the like, you will find that each of the three compounds written up there have more or less the same MO scheme.
The difference from a theoretic point of view is the bond strength and the energy difference between the central atom’s and the ligands’ orbitals. For a complex ion, the orbital energies of the unbonded atoms are very similar to each other resulting in a large energy difference upon forming a bond and therefore a large difference between bonding and antibonding orbitals. For typical ligand (or coordination) complexes, the ligands have much lower orbital energies than the central metal and thus the energy gained in the bonding process is comparatively small. On this scale between $\ce{[CoCl4]-}$ and $\ce{SO4^2-}$, chromate(VI) would be somewhere just to the sulphate side of the axis’ centre.
What does this difference mean from a practical point of view? Well, one typical reaction of coordination complexes is the ligand exchange reaction: Add a better ligand into the solution and more often than not it will exchange with a worse ligand. Sometimes this results in a different geometry. $\ce{[CoCl4]-}$ can be synthesised by preparing an aquaeous cobalt(III) solution which contains $\ce{[Co(H2O)6]^3+}$ ions and adding excess chloride ions — the complex will form spontaneously. The reverse reaction is possible by adding silver nitrate which will precipitate the chloride ions, reforming the hexaaqua complex. Ligand exchange is simple.
The sulphate ion, as the extreme opposite example, has such strong $\ce{S-O}$ bonds that they will not ‘simply dissociate by offering a better ligand.’ (In fact, I don’t know if a better ligand even exists.) It is impossible to displace the oxygens to attach other ligands — and most notably, there is no good oxophilic compound known that could simply remove (i.e. precipitate) those oxygens. Note, however, that sulphuryl chloride $\ce{SO2Cl2}$ will readily react with water to form sulphuric acid (diprotonated sulphate) and hydrogen chloride, so one type of ‘ligand exchange’ is still possible.
Chromate(VI), as written earlier, is closer to sulphate than to the cobaltate complex. It will — much like permanganate — not take part in displacement reactions, the metal–oxygen bonds are pretty stable. Again note that an analogous chloride complex exists with chlorochromate $\ce{CrClO3-}$.
Note that not always will the ligand exchange I highlighted above occur easily. Ammonia should form a more stable complex with iron(III) than water. However, when adding ammonia to an iron(III) solution, iron hydroxides will precipitate rather than an ammin complex being formed. So the practical point of view is a really rough approximation that only works sometimes.

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text:
Calculating the molar mass of a gas given state function values
Question:

A gas has a volume $255\ \mathrm{mL}$, a pressure $733\ \mathrm{mmHg}$, a temperature $548\ \mathrm{^\circ C}$, and a mass $1.22~\mathrm{g}$. What is the molar mass?

My attempt:
$$pV = nRT$$
Convert $733\ \mathrm{mmHg}$ to $\mathrm{atm} = 733~\mathrm{mmHg} \cdot 1\ \mathrm{atm}/760~\mathrm{mmHg} =   0.9645\ \mathrm{atm}$
Convert $\mathrm{^\circ C}$ to $\mathrm{K}$: $548\ \mathrm{^\circ C} + 273.15\ \mathrm{K} = 821.15~\mathrm{K}$
Convert $\mathrm{mL}$ to $\mathrm{L}$: $255~\mathrm{mL} \times \frac{1~\mathrm{L}}{1000~\mathrm{mL}} = 0.255~\mathrm{L}$
$$0.9645\ \mathrm{atm} \times 0.255~\mathrm{L}= n \times \frac{0.8206~\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}}  821.15~\mathrm{K}$$
$$\frac{0.2459}{673.84} = n \times \frac{673.84}{673.84}$$
$$0.000364\ \mathrm{mol} = n$$
$$M = \frac{m}{n} = \frac{1.22~\mathrm{g}}{0.000364~\mathrm{mol}} =   3351.648$$

The mistake you made was the value of $R$. If pressure and volume are taken in $\mathrm{atm}$ and $\mathrm{L}$, then the value of $R$ is $0.0821$. (Not $0.82..$).
Now, $$n=\frac{0.9645\times 0.255}{0.0821\times 821.15}=0.0036$$
$$M=\frac{1.22}{0.0036}=334.412\mathrm{g}$$


text:
Why does microwaving napalm make it nonflammable?
In this video, microwaving napalm made it nonflammable. What caused this to happen?
I assume it was some sort of chemical reaction. Could someone explain? I understand some chemistry, but a not-too-technical answer would be appreciated.

In a very similar video I found, some gentlemen microwave gasoline, the flammable component of napalm, along with aluminium foil. The aluminium sparks when exposed to the microwaves, igniting the gasoline vapors in the microwave. Once all the vapor in the microwave is combusted, the microwave must then once again fill with gasoline vapor. When it once again comes into contact with the ignition source, the vapor combusts again.
Since the gentleman in the video you used did not use an ignition source in the microwave, the gasoline in the napalm vaporized, but did not ignite. The vapor filled the microwave, and what was left was the much less flammable gelatin.

I'd agree with most but the last phrase should be "and what was left was the much less flammable gelatin." I'm sure that the gelatin part would burn when a sufficient ignition source was used.

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text:
Does NH4 anion exist?
$\ce{BF3}$ can react with $\ce{F-}$ to give $\ce{BF4-}$ where we have for $\ce{B}$ in this molecule the 2s orbital and the three 2p orbitals  filled totally (hybrid orbitals).
In case of $\ce{NH3}$ we have the 2s orbitals and the three 2p orbitals of $\ce{N}$ filled;
So i am asking this: can the hydride ion $\ce{H-}$ make a dative bond with the 3s orbital of $\ce{N}$ (which is vacant) to form $\ce{NH4-}$? If not, why?

The simple answer is that the 3s orbitals of nitrogen are too high up in energy. You could theoretically have such a species exactly as how you described. But the chances are that it would have so much energy that it would simply fly apart.

You mean the dative bond will not occur since the overlapping of the 3s orbital of NN with the 1s orbital of HH needs high energy?@orthocresol

Why the comparison with boron and nitrogen? or ammonia and boranes?

The 3s orbitals are far too high in energy to participate in chemical bonding interactions. Lower in energy are the empty σ* antibonding orbitals of the N–H bonds, which can still be energetically relevant in further reaction chemistry. As a thought experiment, it might be possible for a hydride anion to donate electron density into an empty N–H σ* orbital in a sort of Lewis acid-Lewis base interaction. In fact, this is a hallmark of so-called `hypervalent' bonding of heavier main group elements. Rather than using d-orbitals (as is often incorrectly taught to intro chem students), these acid-base interactions are how the phosphorus(V) center of PF$_5$ bonds to the fifth fluoride ion, forming what is known as a three-center four-electron bond.
Why isn't this type of bonding seen with NH$_3$ and H$^–$ to form NH${_4}^-$? These `hypervalent'-type interactions involving 3c-4e bonds depend on the outer atoms having a high-enough electronegativity to stabilize a positive charge at the central atom. In the case of the hypothetical NH${_4}^-$ anion, this just isn't the case. Hydrogen is in fact far less electronegative than nitrogen.

So we might expect an interaction through the hydrogen end of the ammonia dipole. As illustrated in another answer, that interaction is known in a more limited form.

Among isolable species, arguably the closest thing there is to an ammonium anion can be seen in ammine magnesium borohydride, $\ce{Mg(BH4)2•2NH3}$[1]. In this compound hydridic (negatively charged) hydrogen is dihydrogen-bonded to the protic hydrogen in the ammonia. In the molecular-orbital rendering of hydrogen bonding, the electrons from the boron-hydrogen bond overlap with the formally vacant antibonding orbitals from the ammonia (both of these being orbitals polarized towards hydrogen) to delocalize the bonding into the "intermolecular" region. No contribution from outer orbitals on the nitrogen are involved as this is a hydrogen-hydrogen interaction.
As with other cases of hydrogen bonding, the electron source (borohydride hydrogen) remains bonded primarily bonded within its own "molecule" (meaning to boron); the dihydrogen bond is only a secondary interaction. Were the boron-hydrogen bond to be broken, for instance by by thermal decomposition, the hydridic and protic hydrogens would simply combine to form $\ce{H2}$ (hydrogen evolution on heating is the intended result in the referenced study).
Transient ammonia-hydride species have been identified using photoelectron spectroscopy. Snodgrass et al[2] identify $\ce{H^-(NH3)}$ and $\ce{H^-(NH3)2}$, which are considered hydride ions solvated by ammonia molecules and thus would involve protic-hydridic interactions similar to that in the complex borohydride salt described above. They also found a peak attributed to a tetrahedral $\ce{NH4^-}$ isomer in which all the hydrogens are positive; an ordinary ammonium ion accompanied by two Rydberg electrons. Diaz-Tonoco and Ortixz[3] give ab initio calculations supporting the Rydberg configuration and extend the concept to higher $\ce{N_mH_{3m+1}}^-$ species.
Reference

Grigorii Soloveichik, Jae-Hyuk Her, Peter W. Stephens, Yan Gao, Job Rijssenbeek, Matt Andrus, and J.-C. Zhao (2008). "Ammine Magnesium Borohydride Complex as a New Material for Hydrogen Storage: Structure and Properties of Mg(BH4)2·2NH3". Inorg. Chem. 47, 10, 4290–4298.
https://doi.org/10.1021/ic7023633

Joseph T. Snodgrass,   James V. Coe,   Carl B. Freidhoff,   Kevin M. McHugh  and  Kit H. Bowen (1988). "Photodetachment spectroscopy of cluster anions. Photoelectron spectroscopy of H–(NH3)1, H–(NH3)2 and the tetrahedral isomer of NH4-". Faraday Discuss. Chem. Soc., 86, 241-256, https://doi.org/10.1039/DC9888600241.

Manuel Díaz-Tinoco, J. V. Ortiz (3019). "Double Rydberg anions with solvated ammonium kernels: Electron binding energies and Dyson orbitals ". _J. Chem. Phys. 151, 054301. https://doi.org/10.1063/1.5113614.

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text:
Dissolve Silver Oxide
I was looking at how to dissolve silver oxide (in a solvent). I know that monovalent silver oxide will not dissolve in water, but are there any solutions in which monovalent silver oxide will dissolve? I was looking whether they would do so in alkaline solutions, but I could not find anything.

Well, $\ce{Ag2O}$ is just a basic oxide. As such, it would dissolve in suitable acids ($\ce{HNO3}$ would do), but I guess that's not quite what you want. Well, some metal oxides ($\ce{ZnO}$, for instance) are amphoteric and thus would dissolve in $\ce{KOH}$ as well, via formation of hydroxo complexes. Sadly, this is not the case with $\ce{Ag2O}$. Then we must resort to some specific complex which would be stable in alkaline media.
Indeed, there is such complex: $\ce{Ag(NH3)2+}$, the active component of Tollens' reagent, which is prepared by interaction of $\ce{Ag2O}$ with aqueous ammonia, like I said. If you are actually going to do it, read the page carefully, including the "Safety" section, or at least don't make that solution by gallons and don't leave it unattended for years. Also, note that it is normally made using freshly prepared $\ce{Ag2O}$; dry reagent off the shelf would likely dissolve as well, only much slower.

Not every acid would dissolve it, as HCl e.g. would give a precipitate.

Regarding safety of Tollen's reagent, see this:-http://chemistry.stackexchange.com/questions/43441/is-tollens-test-safe

If one needs to work with organic solvents, silver(I) oxide can be dissolved in trifluoroacetic acid (TFA) producing silver(I) trifluoroacetate $\ce{AgOCOCF3}$, which is a versatile reactant in organic synthesis (compact overview: [1]); also a precursor or chloride precipitant in the synthesis of metal complexes (see, e.g. [2]).
References

Wistrand, L.-G.; Seapy, D. G. Silver(I) Trifluoroacetate. In Encyclopedia of Reagents for Organic Synthesis; Wiley; Chichester, UK, 2009. DOI:10.1002/047084289X.rs031.pub2.
Kuehn, C. C.; Taube, H. Ammineruthenium complexes of hydrogen sulfide and related sulfur ligands. J. Am. Chem. Soc. 1976, 98 (3), 689–702. DOI: 10.1021/ja00419a010.


text:
How can I make my yeast experiment more chemistry-oriented?
For a school chemistry design lab I tried to find the optimal sugar concentration for fastest yeast fermentation. I put the same amount of yeast and sugar water solutions with different concentrations of sugar together and measured the carbon dioxide produced using a water displacement apparatus. My teacher approved the topic, which is why I went along with it, but our teachers are often too lenient with approving topics, and this will be moderated so I can't just squeak through.
My main problem is that I feel it is too biology-oriented. There is very little chemistry in it. What can I do (preferably with yeast because I still have quite a bit and it wouldn't seem like a complete change of topic to my teacher)

If there really is little hope, if anyone could suggest a different area or experiment I could look into that would be great.

So you don't like your topic with yeast for being too biology-oriented, hence you want to switch  to another topic, also with yeast? Just how would that be less biology-oriented than your current topic? Well, I guess it would if you'd kill all the yeast beforehand (say, to measure nitrogen in them by Kjeldahl method - very chemical indeed), but that's cruel and unnecessary, and not that interesting after all.

@Ivan Neretin Hi there, yeah, I see what you mean. It's hard to evade the biology-ness of yeast after all. It's generally okay for our experiments to have a mix of two sciences, but I feel that this one is really too bio-ish. If I could just have some chemistry elements I think it'd be fine. I was thinking about using yeast as a catalyst in hydrogen peroxide decomposition, but I don't know what exactly I'd be looking for (optimization?)

There's really no need for me to continue with yeast, I just thought that'd be nice so that I don't have to ask my teacher for materials. I'm thinking more to find another experiment that won't require any special chemicals (just stuff like NaOH or something that the school will have on hand as our draft is due in around two weeks.

A too high sugar content will give your yeast a killing osmotic pressure. Not enough other nutrients (as @Ryan hinted) will set a limit on producing capability.

@GyroGearloose do you know where I might be able to find some papers that talk about at what point the sugar solution becomes hypertonic? I've managed to find a few that say 200g/L but it's really hard to find papers that are available to high school students.

Embrace some biochemistry.  Research how to measure the enzyme activity of catalase in yeast that is exposed to hydrogen peroxide.  It breaks down hydrogen peroxide and makes water and O2 gas.  Try testing different conditions like pH or temperature.  Maybe too advanced, or it might be more interesting to you.

It sounds like a perfect chemistry experiment to me. It sounds good enough to even be one of my AP chemistry labs. As an objective for the lab, you could try to determine the amount of CO$_2$ gas produced per amount of sugar you use, (with all other ingredients in excess). I'm pretty sure the equation for fermentation is C$_6$H$_{12}$O$_6$ → 2 C$_2$H$_5$OH + 2 CO$_2$ (although note that common granulated sugar has the equation C$_{12}$H$_{22}$O$_{11}$ and decays into two glucose molecules when combined with water).Why not test a certain amount of sugar (like .1 moles) and see if you yield the theoretical .2 moles of CO2. You could use the ideal gas laws (PV=nRT) to determine the moles of CO2 product you yielded. Compare what you yielded to the theoretical amount you should've yielded. If you yield less, then you could include in your report/conclusion including why you think your yield was short.
The lab would involve gas laws, chemical equations, and percent yield, all of which are chemistry topics/principles. What is your current level of chemistry experience? If you haven't learned of anything I mentioned, I can help explain it better and point you in the right direction if you'd like.

Hi Ryan, thank you for this answer. I like the idea of keeping the experiment more or less the same and having the chemistry part be in the analysis. I am familiar with percent yield and gas laws, but thank you very much for your offer.


text:
Why are acids usually depicted in fiction as green?
In cartoons, etc., acid is generally shown as vividly green, as in this scene from the Simpsons:

We even have the colour "acid green".
But when I used to use acids in chemistry class, they weren't bright poisonous green. They were mostly colourless, as, for example, this picture of sulfuric acid from Wikipedia:

So where does this idea of green acid come from? Are there acids which are actually green? Or perhaps acids are green when highly concentrated?

my guess it comes from slang 'acid' for LSD, which was a source of psychedelic art, making a lot of use of bright 'acidic' colors.

BTW, acid green color is not as much a color of acid - it is a color used to depict any toxic or dangerous substance. Green is a common color-code for 'poisoned' state of a character. Not sure where it came from.

According to one source, green is associated with 'dangerous' from the association of (radioactive and dangerous) radium paint used to make green-colored glow-in-the-dark watch faces. Perhaps 'dangerous' = 'glowing green' is applied to acid as well. Also, transparent wouldn't render too well in a cartoon :)

I guess it was simply the one basic, bright color left: blue is for pure water, red is for blood which leaves green for acid.

@Thawn, what colour would the grass be? ! :)

I thought it was because of the movie Soylent Green ;-) https://en.wikipedia.org/wiki/Soylent_Green

The green is the color of bile which usually accompanies gastric acid.  Acid  in its original context referred to a bitter taste.  Bitter tasting compounds, such as vinegar, is what led to the modern chemical definition.  The chemical definition is less likely to have influenced the populace than the experience associated the somewhat bitter, acidic, taste of gastric acid accompanied by bile.

Good answer. However, I think you mean sour or acrid taste not bitter...

Bitter is correct in this answer's context. In the early days of chemistry several acids, for example tannic and hydrocyanic acid, were discovered by extracting plant material which tasted bad, or the "bitter principle."  Bitter, along with sweet, was regarded as one of only two types of tastes. This thinking, started by Aristotle in 350 BC, did not change until the early twentieth century when the number of accepted taste types expanded to four with a division between sour and bitter.   Thankfully discussing taste in non-food related chemistry is an abandoned practice.

Interesting, thanks for the historical perspective :)

Actually, I'd rather say the green-yellowish color is attributed to both toxic chemicals and radioactive waste in general and not only to the super acids. It might as well be that the corrosive acidic stuff seen in cartoons, movies and video games is green due to the cliché introduced by the comic book authors, namely during the Golden Age inbetween late 1930s and early 1950s. The first issue of Batman from 1940 introducing Joker already contains a pool of green waste (BTW it's acidic as we shall see later):

I cannot speak for the exact motives of the comic books' artists, but at that time period there already were several chemicals that could affect the choice of the "acid-green" color for the years to come; I tried to sort them from the most fictional to the most plausible and fitting the "green acid" term:

Copper-doped zinc sulfide ($\ce{ZnS:Cu^2+}$). Polycrystalline samples are capable of green luminescence. Mentioned by Batman as the chemical responsible for the green color of the acid pool in Batman Vol 2, 14, 2013:

Batman: I know its components by heart. The whole list.
Eleven percent sodium hydroxide. Thirty-four percent sulfuric acid. Five percent chromium solution. Zinc sulfide, doped with copper, which gives it its green glow.

I'm purposely leaving no comments on co-existing of $\ce{NaOH}$ and $\ce{H2SO4}$ in the solution, as well as absence of hydrolysis of $\ce{ZnS}$. Bruce Wayne is rich enough to know better.
Fluorescein. The most famous dye tracer known for over hundred years. Fluorescein was used in 1877 to prove that the Danube and Rhine rivers are connected by underground channels. These days is also used for entertainment purposes, e.g. to turn rivers green on St. Patrick's Day. Fluorescein is not acidic (it fact, requires mildly basic medium) and not corrosive or particularly hazardous, but the result of dissolving of the few grams of powder in a ton of water turning it green looks impressive, which could've served as inspiration for artist of that time period.
Scheele's Green $\ce{CuHAsO3}$. Used as a green pigment in XIX century. An acidic copper arsenite salt is both toxic and carcinogenic.
Absinthe. A highly alcoholic green beverage due to the presence of chlorophyll. Not really acidic, but, when consumed excessively, a violent hangover takes place along with xanthopsia (a vision of the surrounding reality in a yellow-green color), not to mention uncontrollable rage, nausea and psychological disorders. Favored by certain painters, probably some comic book artists were also influenced by the aftermath this drink causes.
Used chromic acid $\ce{H2SO4 / CrO3}$. Chromic acid a mixture of concentrated sulfuric acid and potassium dichromate commonly used for cleaning heavily polluted glassware a few decades ago. Initial color is brown-yellowish and is attributed to the formation of chromium trioxide $\ce{CrO3}$. Very strong oxidizing agent and a corrosive substance. As the result of oxidizing organic pollutants, chromium(VI) is reduced to chromium(III) having green color, so that the used chromic mixture gradually changes its color to green, although remaining acidic (and carcinogenic, too). These days is replaced by piranha solution which contains hydrogen peroxide instead of chromium as a more environment-friendly oxidizer.
Chlorine trifluoride $\ce{ClF3}$. A famous "N-Stoff", synthesized on industrial scale in the Nazi Germany from 1938 till 1945. Pale-green liquid below 12 °C or when pressurized. Corrosive on its own, and violently hydrolyses to hydrofluoric $\ce{HF}$ and hydrochloric $\ce{HCl}$ acids. Has been proposed as a rocket propellant, poison gas, incendiary and corrosive agent to penetrate metal alloys and concrete. Probably comes as close as it gets to the definition of the "green acid of choice for a super-villain".

Finally, it's also worth noticing that the green-yellowish color of approx. 550 nm $\color{#80FF00}{\Large\bullet}$ is the most noticeable color to the human eye under usual ambient daylight conditions, helping to quickly adjust viewer's focus to what's more important or to just overwhelm.

One chemistry related answer could be that chlorine gas has a greenish hue after which it was named(wikipedia):

Sir Humphry Davy in 1810, who named it from Ancient Greek: χλωρóς (khlôros) "pale green

Also, chlorine gas is closely related to hydrochloric and hypochlorous acid:

Solutions of chlorine in water contain chlorine (Cl2), hydrochloric acid (which is arguably the most commonly known acid), and hypochlorous acid:
$\ce{Cl2 + H2O <=> HCl + HClO}$

Finally, in WWI, chlorine gas was used as a chemical weapon, that caused acid burns to skin, eyes and lung:

By April, German chemists had tested a method of releasing chlorine gas from pressurised cylinders and thousands of French Algerian troops were smothered in a ghostly green cloud of chlorine at the second Battle of Ypres. With no protection, many died from the agonies of suffocation

Maybe it could be good answer, but right now it looks like speculation.

@Mithoron Well, this question is unlikely to get a definitive answer because it basically asks for the intentions of comic artists.


text:
How to extract water from a solution in formamide
I'd like to determine the water content in a sample without buying an expensive machine (like Karl Fischer titrator). So I'd like to measure the water volumetrically. I've chosen formamide to dissolve the sample because of its high polarity.
I've dissolved a sample (approx. 20 g; composed of sucrose, glycerol and glucose syrup, water) in 50 ml formamide and like now to separate the water fraction again.
1) Let's assume there is 0,5 ml water dissolved in the formamide fraction, is it possible to add 30 ml solvent B and get so a fraction of 30,5 ml (30 ml + 0,5 ml)?
2) Instead of formamide, do you know another solvent A which dissolves the sample including the water (and the watercontent of approx. 20% in the glucose syrup)) and a solvent B which extracts the water phase from the solvent A?

You can use Dean-Stark azeotropic removal of water with toluene or benzene

Can't do you a brine wash? Prepare a saturated aqueous solution of salt, then do a liquid-liquid partition and collect the aqueous layer. The volumes of brine and water will not be additive, but the masses will be. It's not a perfect separation, but certainly among the easiest options.

Maybe try silica gel?

@K_P Thank you K_P for your useful advice! Is it possible to heat my solution of formamide up to 105°C without distilling also the glycerol? (I think glycerol (because it's an alcohol) will make an azeotrop with water and therefore I doubt if this method is suitable)

@NicolauSakerNeto Water is solvable in formamide. I'll try, if I get a phase separation. Among sucrose, glycerol, glucose syrup and water, which ingredients will go into the aqueous phase?

@G-man I'm not sure if silica gel also binds some part of formamide and of  glycerine.

I don't think you will have any problem with glycerol since it has a boiling point of $\ce{260 ^{\circ} C  }$ Also I wouldn't use formamide at all. Just mix your sample with enough toluene with good stirring and let it reflux using the Dean-Stark and collect and measure the amount of water. Formamide usually contains some water so this can affect your calculations depending what accuracy you need

I'd like to determine the water content in a sample without buying an expensive machine (like [a] Karl Fischer titrator).

Rather than try to extract the water and measure it volumetrically, I think you could try to measure its activity, or else perform your own Karl-Fischer titration, which does not necessarily require any equipment.
Water activity measurement
Practically this means hygrometers.  That may sounds like another expensive machine, but E-bay tells me that capacitive hygrometers are available for under $20.  If your sample consists of sucrose, glucose, and glycerol, those substances are unlikely to interfere with capacitive hygrometry measurements of water activity.
You might even get very crude accuracy by incubating a mirror at a known (chilled) temperature, and then waving it above your water/sucrose/glucose/glycerol sample, seeing if dew forms, and then repeating the process at several different temperatures.
Karl Fischer titration on your own
As an alternative, couldn't you just set up your own Karl Fischer titration?  Commercial titrators often use coulometric detection for better accuracy, but setting up a volumetric titration is pretty easy.  You need:

Methanol (or some other alcohol)
Imidazole (or some other amine base)
Sulfur dioxide
Iodine

Imidazole and sulfur dioxide should be combined in a 1:1 molar ratio in an excess of methanol.  Iodine should be dissolved in methanol.
A 2012 paper has a nice description of how to do KF with visual end-point determination.  You just need to buy a KF titration solution, or make your own:

Commercially available [...] Test Kit from [...] was used for reference moisture determination. All the equipments needed were provided with the kit. It consisted of two-component ethanol-based reagents: [a] working medium containing [a mixture of] imidazole, sulphur dioxide and diethanolamine, and [separately a titrant solution] containing iodine. According to manufacturer's instructions for use, 20 mL of [working medium] was added to the titration vessel, which was then tightly sealed with a septum. The titrant was delivered through the septum with a syringe to react with the present water, until a color change from colorless to yellow occurred. A known amount of sample was added by a syringe to the titration vessel via septum, and a newly filled titration syringe was used for a new titration until the color change from colorless to yellow occurred. Consumption volume of the titrant was read off the titration syringe. The water content, in percent by volume, was calculated from the consumption of the titrant and the sample volume.

If you've diluted with formamide

I've dissolved a sample (approx. 20 g; composed of sucrose, glycerol and glucose syrup, water) in 50 ml formamide and like now to separate the water fraction again.

The nice part about both water activity measurement and Karl Fischer titration is that they are both sensitive and accurate enough to still work on your sample even after dilution or dissolution in formamide.

Thank you Curt F. for your ideas! (We already use a device which gives a water activity of let's say 60%. Unfortunately it's not possible to calculate back to the water content.) There are three questions: 1) How can I bring in SO2 in methanol most easily? 2) If I don't know the water content exactly, will it works if I use twice the molar amount of SO2/imidazol as stochiometrical needed? 3) Is it possible to see the colorchange from colorless to yellow?

I'm not sure about your question (1).  The answers to (2) and (3) are both "yes".

The answer to #1 depends a lot on the equipment you have.  I suppose it's possible to use sodium sulfite as the $\cf{SO2}$ source and then ion exchange the resulting solution with an imidazolium-based ion exchange resin.  Bubbling in $\cf{SO2}$ gas to a solution of imidazole free base also probably works.


text:
Is this the correct formula for dopamine?
Google says that $C8H11NO2$ is the correct formula for dopamine. Is this so?

Without meaning to be rude, you could work this out, given that you know how to use google. Search for the molecular structure. To help you along, the CAS number for dopamine is 51-61-6.

Do you have a particular reason to suspect that google is wrong?

I'm voting to close this question as off-topic because it is trivial to answer with a search engine.

Take a look at the structure, you can easily work it out for such a small molecule.
Lets start with the really obvious ones:
Oxygens: 2
Nitrogens: 1
You then have to recognise that in a skeletal formula, the carbons are simply represented by 'corners' (i.e. where two lines meet):
Carbons: 8
Finally, you need to work out how many hydrogens are present. Hydrogens attached to heteroatoms (Oxygen and nitrogen) are shown, hydrogens attached to carbon are not, and hence you have to rely on your chemistry knowledge to work out how many should be attached to each carbon:
Hydrogens: 11
In short, yes, the formula is right for dopamine.


text:
Infrared Spectroscopy of Hexachlorocyclotriphosphazene
When assigning the IR spectrum of hexachlorocyclotriphosphazene there are two prominent peaks (one due to the nitrogen-phosphorus bond and the other due to the phosphorus-chlorine bond).
There is one peak at 1179cm⁻¹ and another at 594cm⁻¹. I think that the peak at 1179cm⁻¹ is due to the phosphorus-nitrogen bond and the peak at 594cm⁻¹ is the phosphorus-chlorine bond.
My reasoning for this is that the phosphorus-nitrogen bond is shorter and therefore stronger thus it will have the higher stretching frequency.

Is my reasoning correct?

I highly doubt you can see P-N bond IR. It should be a very weak and complex set of bands. Very possibly both peaks are due to PCl2 fragment. Look in alkane IR spectroscopy to see how several bands appear from CH2 bridge.

Or you can find a guy with access to package with analytic second derivitives (like Gaussian) and ask him to run IR spectrum calculation on the molecules. Granted, the methods sometimes (quite often) fails miserably, but It should give an idea at least.


text:
Non-standard source of lead?
Is there a way to get lead cheaply from local sources? I can buy ingots off eBay and there is a non-ferrous metal supplier I know who sells it for $12 for 5 pounds, but the shipping is nuts.
I considered salvaging lead from a battery, but apparently that is not viable cause the battery is contaminated with a lot of lead oxides, lead carbonates and antimony.
I also considered fishing weights, but I am concerned about alloyed contaminants and also it is getting difficult to find lead fishing weights, especially large ones.
I am interested mostly in "technical grade" just for weighting stuff down (like apparatus stands), but also having a small amount available that is reagent quality.

If you are at, say, a university - ask around! You might be surprised what folks have on hand. Otherwise, your cost problem with all lead products is the shipping - it is heavy, after all. Perhaps find a local metal supplier and see what they have and can offer.  Since you are concerned about contaminants, that affects all 'non-standard' sources.

@JonCuster Right, that is something I have considered, sort of Craigslist type of local pickup option. I am asking the question because I thought somebody might know of a common product (not a battery) that I could just pull lead out of for free.

@ShakaBoom can you pick up from here? http://www.nuclead.com/leadproducts.html

@DavePhD That's a great find! Thanks for that vendor, I didn't even know they existed. West Bridgewater is a bit of a drive, but not too bad.


text:
pH-dependency of the acidic dissociation of carbonic acid
I know the product of $\ce{CO2 +H2O}$ yields $\ce{H2CO3}$. This is a reversible equation so $\ce{H2CO3}$ goes back to $\ce{H2O + CO2}$. However, my question is: Is there a certain pH which will dissociate $\ce{H2CO3}$ into $\ce{H+}$ and $\ce{CO3^2-}$ ions? If so, lets say you started with $2\ \mathrm{mol}$ of $\ce{H2O}$ and $1\ \mathrm{mol}$ of $\ce{CO2}$, how many moles of $\ce{H+}$ and $\ce{CO3^2-}$ ions will form?

Hi and welcome to chemistry.stackexchange.com. Since you already took the tour, all that remains is to guide you towards the [help] where you can find useful hints, e.g. on how to format your posts with MathJax — check out what I did while editing your post ;)

Related: Does H2CO3 exist in solution?

At what temperature and pressure? Solubility of CO2 depends on them strongly.

at 25 degrees and 1 atm


text:
Why do SN1 reactions happen slowly?

'The first step where the carbonium ion forms is slow and because it contains only the halogen, the general speed of the reaction will be slow.'

I don't understand  why should SN1 reactions happen slowly.
ps. Please be specific about the question, I am aware of other characteristics of SN1 reactions.

It appears that the first sentence is a quote. Where is the quote from? A book? Why do you think SN1 is slow? It has something to do with one of those characteristics of SN1.

Yes it is written in my Albanian book , I don't understand that either.

Perhaps try a different organic chemistry textbook? The questions posted  are answered by any UK/US introductory organic textbook (Clayden, McMurry, March etc.).

From what I understand from the quote, it's not saying that $\ce{S_N1}$ reactions occur slowly, but rather that if the first step of the reaction occurs slowly, it'll cause the overall reaction to occur slowly.
So, the  $\ce{S_N1}$ reaction contains the two steps, the first one where the carbonium ion forms as your book says, and the second step where the substitution occurs. The first step is the slowest step of these two steps, whereas the second one is quite fast. The first step is thus called the rate determining step, meaning that it "controls" the overall rate of the reaction. If it is slow, overall the reaction is slow. If it is fast, then the overall reaction will be fast. At least that's what I understand from it.


text:
Do SN1 reactions have dependency on the nucleophile?

'The rate of the reaction of tertiary butyl bromide with the hydroxy ion is dependent on the concentration of the tertiary butyl bromide, but is independent of the hydroxy ion.'

This means that the rate of SN1 is independent from the nucleophile. Why does that happen?

'tertiary butilic bromure' do you perhaps have an image/structure to accompany your quote? As is, the question doesn't entirely make sense. You might also like to edit the post to remove un-necessary capitalisation.

I'm guessing that "tertiary butilic bromure" is "tertiary butyl bromide", but please correct me if that's wrong.

see http://chemistry.stackexchange.com/questions/42566/does-a-reaction-have-to-have-a-rate-determining-step - Your 3 question stem from what is rate determining step of this reaction.


text:
Why do SN1 reactions prefer tertiary carbons?
I can understand that $\mathrm{S_N2}$ reactions prefer primary carbons, because of the steric hindrance that would be an obstacle.
But what is the exact reason that $\mathrm{S_N1}$ reactions prefer tertiary carbons ?

http://chemistry.stackexchange.com/questions/6791/how-does-the-inductive-effect-explain-the-reactivity-of-alcohols-with-halogen-a

The intermediate in an SN1 reaction is a carbocation. The stability of that intermediate determines how favorable that particular reaction is, so a more stable intermediate means a more favorable reaction.
Carbocations are quite unstable on their own, but the inductive effect of nearby C-C bonds provide some electron density to balance out the positive charge, making the carbocation intermediate more stable. Tertiary carbons have the largest number of adjacent C-C bonds, the largest inductive effect, the most stable carbocation intermediate, and are thus favored in SN1.

Because this is about reaction rate, you should refer to transition state energies for the most correct response.

Some answers are already addressing the key issues, but I see some important missing elements. While cation stability is important, we're talking about reaction rates, so we should be using kinetic, rather than thermodynamic arguments. Here, I will invoke the Hammond postulate to convert a thermodynamic argument into a kinetic one.
First, we establish that the $S_{\mathrm{N}}1$ mechanism proceeds via an intermediate cation. This intermediate is, importantly, higher in energy than the reactants.
Second, the first step of the $S_{\mathrm{N}}1$ reaction is the slow step and therefore (given the first fact), also the rate determining step.
By the Hammond postulate, in an endothermic reaction, the transition state will be structurally and energetically similar to the reaction product. In this specific case, the transition state will resemble the cation. This means that we can expect the transition state for the rate determining step to be strongly affected by factors that can stabilize the intermediate cation.
Generally, we speak about hyperconjugative effects as the factors that most influence the stability of the cation. Specifically, with increasing substitution on the cation, adjacent $\ce{C-H}$ bonds are able to donate some of their electron density into the empty orbital of the cation, helping to stabilize it.
Therefore, $S_{\mathrm{N}}1$ reactions will be fastest for species that generate a tertiary cation, because those reactions go through cation-like transitions states that are lower in energy.


text:
Are elemental compounds composed of different isotopes polarised?
Are elemental compounds composed of different isotopes polarised? I'd expect any such effect would only really be noticeable for very light elements such as hydrogen.
I'm not sure which way the charge would be distributed and whether a deuterium protium pair would be more like deuterium protide or protium deuteride.

yes! $\ce{C^1H^2H^3HC{^1H}_3}$ would be optically active.

When taken isotopes into account, properties of compounds change. The polarisation of the compound included. Your hunch is correct, this probably only matters for very light compounds, or more precise compounds where the mass difference between the isotopes is large. The most extreme case is probably $\ce{HT}$, which has roughly double the mass of $\ce{H2}$.
I found an article by M. Trefler and H. P. Gush on the electric dipole moment of $\ce{HD}$ which explains the presence of the dipole moment as follows:[1]

The center of positive charge oscillates along the axis of the molecule during a molecular vibration because the excursions of the proton are greater than those of the deuteron; since the electrons do not follow exactly the motion of the positive charge, a weak, vibrating dipole moment results.

There are more articles of the dipole moment of $\ce{HD}$, I simply did not have the time to go through all of them. I also think the above already makes the point you were interested in.[2]
A very thorough theory is presented by W. R. Thorson et. al.,[3] which essentially explains the dipole moment of HD as follows:

[...] ; it arises directly from asymmetry of the electronic Hamiltonian, provided that we cast. that Hamiltonian in an appropriate set of canonical coordinates. The model of this Hamiltonian as having a pair of nuclei with slightly different effective charges is particularly pleasing, since we can link that directly to the orbital-size and binding-energy effects which we know must be associated with the difference in electronic reduced masses for D and for H.

M. Trefler and H. P. Gush found the dipole moment to be very small, i.e. $\mathbf{p}=(5.85\pm0.17)\times 10^{-4}~\mathrm{D}$. Hence to speak of deuterium protide or protium deuteride is a stretch.
Since the effect is quite small in $\ce{HD}$ already, one can probably assume it will not have a significant effect on the polarisation in other than symmetric diatomic molecules.
However, with isotopes a few notable applications can be achieved. One of the prime examples is “chiral acetate”, (R)-deuterotritioacetic acid $\ce{HDTC-COOH}$, which helped elucidate biosynthetic transformations. For a quite comprehensive, educational article see [4].

References

M. Trefler and H. P. Gush; Phys. Rev. Lett. 1968, 20, 703.
W. R. Thorson, J. H. Choi, and S. K. Knudson; Phys. Rev. A 1985, 31, 22.
(a) S. M. Blinder; J. Chem. Phys. 1960, 32, 105. (b) S. M. Blinder: The dipole moment of HD. http://hdl.handle.net/1811/7809 (c) J. Bradley Nelson and G. C. Tabisz;
Phys. Rev. Lett. 1982, 48, 1393. (d) G. Herzberg; Nature 1950, 166, 563. (e) etc. see google scholar
Addison Ault; J. Chem. Educ. 2003, 80 (3), 333.


text:
Speed up polymerisation of formaldehyde
Is there a way to speed up the formation of para-formaldehyde in a formalin solution?

In pure enough formaldehyde, that is, with low methanol content, para-formaldehyde will start forming as a precipitate, but very slowly. I have tried finding some answers by simply googling, but all I get is how to make formaldehyde out of para-formaldehyde...

According to this patent, alkaline catalyst of composition sodium and potassium hydroxides, carbonates, formates, and acetates can be used to speed up formation of paraformaldehyde from formalin solution (aq. formaldehyde).

This invention relates to a new and improved process for the
  production of formaldehyde polymers. More particularly it relates to a
  process whereby high strength aqueous formaldehyde solutions can be
  concentrated to paraformaldehyde at a greatly accelerated rate.

Moreover, the patent describes the process of this reaction in details. Have a good read.


text:
why are cyclic dienes much more reactive?
Cyclic dienes are very reactive, for example cyclopentadiene in room temperature, it even dimerizes.
What is the factor that indicates such reactivity of cyclic dienes ?

They are fixed in the s-cis conformation, which is required for the Diels-Alder reaction. Whereas acyclic dienes can flip back and forth between the s-cis (reactive) and s-trans (unreactive) conformations

http://chemistry.stackexchange.com/questions/5289/a-few-questions-about-the-conditions-of-the-diels-alder-reactions


text:
How does ice melt?
I know that when water freezes, it forms a crystalline structure created by the hydrogen bonds between polar water molecules. Also, I know that during a phase change, the potential energy of the molecules increases to the point that it can break this structure.
However, what happens to this potential energy? Does it perhaps turn into kinetic energy as soon as the bonds are broken or maybe alter the structure of the atoms? Also, why does ice need to get to $0^{\circ}$ Celsius to melt when it can just use the energy to go through a phase change immediately?
If the energy does, in fact, turn into kinetic energy as the bonds are broken, then wouldn't temperature, the measure of average kinetic energy, increase dramatically as soon as the phase change is completed?

This may be somewhat long but I Hope to be useful..
There are only two major types of energy, namely, kinetic and potential energy. So the answer for your first question**"Does it perhaps turn into kinetic energy as soon as the bonds are broken or maybe alter the structure of the atoms?"** is that the first sentence is correct, and no change in the structure of atoms(the individual atom) occurs nor in the molecules, but only in the crystal lattice structure which is held by inter-molecular hydrogen bonds as you stated.
For the second question, Temperature is a measure of the kinetic energy of the molecule which is expressed in solids by vibration motion, so raising the temperature of, say, an ice cube from -3 to 0 means that the molecules in the crystal lattice vibrate by a larger frequency.
Immediately after reaching 0 Celsius at atmospheric pressure the kinetic energy is not sufficient to make the water molecules to move relatively freely as those of liquid because of the potential energy expressed by hydrogen bonds in the crystal lattice which hold the molecules together (cohesive forces).
To melt the ice cube you need more energy (latent heat of fusion) than that gained by heating to 0 Celsius, but it is not sensible(cannot be measured by thermometer) since it is a form of Potential energy.This latent heat is only to break inter-molecular hydrogen bonds.That is why freezing and fusion point are the same point on temperature scale.Only After complete melting the energy acquired can shift to raise the kinetic energy or the temperature of liquid water.

Sorry for the mistake, I corrected here the wrong sentence about assisting cohesion by pressure after reaching the melting point. In fact this happens before the melting point but after that the only barrier to melting is the chemical bonding .

If the hydrogen bonds are broken, then why does liquid water still have hydrogen bonds?

They are not completely broken, only those responsible for the characteristic crystal lattice. http://www.chem.uwec.edu/Chem150_S07/elaborations/unit3/unit3-a-states/water-ice.png


text:
When naming hydrocarbons, when do you number the branch or double bond?
I have realized that not all chains and double bonds are numbered. I have noticed that up until 3 carbons in a chain, the double bond is not numbered, but I am not sure about the chain.

Numbering is only necessary where the un-numbered name is ambiguous. I'm not really sure what your question is.

When there are 2 carbons in a chain, the double bond will be at the number 'one' position. Therefore it's not always mentioned in the naming.
For example:  Ethene
It can be written as 1-ethene or eth-1-ene. Since the position of the double bond can be at no other places than the first, the position in the naming you saw, wasn't mentioned.
When 3 or more carbons are present, the double bond can be between any 2 of the carbons, ie. either the first position or the second. (but-1-ene/butene or but-2-ene)
Position of double bonds and that of branches, must always be mentioned while naming organic compounds.

Prop-1-ene and prop-2-ene are the same.

Maybe you meant 4 Carbon -enes: (but-1-ene and but-2-ene)

Ahh yes sorry. I meant butene.


text:
If there is more than one branch in a hydrocarbon, which branch do you name first?
I have read that branches need to be named in alphabetical order, but I do not really understand this, or do you start with the branch closest to the edge of the hydrocarbon?

Related: Nomenclature of organic compounds containing complex side chains and Nomenclature of an organic compound.

Branches must be numbered in alphabetical order, along with their positions.
Suppose you have a chloro group on the first carbon, a bromo group on the second carbon and an amino group on the fourth carbon of your carbon chain (butane).
One must therefore go in alphabetical order which is amino, then bromo and then chloro.
Therefore you naming will be something like this:-
$$\ce{3-Bromo-4-chloro-butanamine}$$
The naming is in alphabetical order but the numbering starts according to the following rule:
$$\ce{RCOOH>RCOOR>RCOCOOR>RCOCl>RCONH2>RCHO>CO>OH>NH2>double bond>triple bond>halogen>NO2>ROR>alkyl}$$


text:
Why is Hessian Matrix called 'force-constant' matrix?
Hessian Matrix is a square matrix containing the elements as the second-order partial derivatives of energy-function of a molecule; the derivative is done with respect to geometric coordinates of the molecule.
However, I couldn't get why Hessian matrix is referred to as force-constant matrix; as is done in the book I'm following viz. Computational Chemistry: Introduction to the Theory and Applications of Molecular and Quantum Mechanics by Errol G. Lewars and in many other sources.
If the energy function is a quadratic function which is mostly the case for a molecule following 1D PES; the function which a Hookean spring follows for small displacement, then the second derivative does indeed give the value of force constant, that is $$\begin{align}E- E_0&= k(q-q_0)^2\\ \implies \frac{\mathrm{d}^2 E}{\mathrm{d}q^2}&= 2k\;.\end{align}$$
However, for molecules following multidimensional PES, most of the time, the energy function is not simple quadratic; how can the second order derivative give the force constant? Second order derivatives vary from point to point unlike the simple case above.
So, can anyone tell why Hessian matrix is called force-constant matrix?

I didn't know what to tag it: since I got the issue related to computational chemistry, I tagged it so. I wanted to tag it as [tag:soft-question]; but it doesn't exist here:(

Excuse me if this comment is more of a question than a comment, my background is math rather than chem, and, in math, Hessian matrix has a far more abstract definition than it is for chem, guessing by your question. So for the outsiders, how is your Hessian matrix defined? (Link preferred above explanation).

@GyroGearloose http://www.bioinfo.rpi.edu/bystrc/courses/biol4550/lecture20.pdf (please see slide 20-29)

Thanks, so "your" Hessian is about the energy of a molecule by displacement of the atoms, right?

The Hessian is calculated in the equilibrium geometry when used for force constants and such. The fact that the reality is non quadratic, it does not influence the fact that we use quadratic models (or else it wouldn't be constant, would it be) - most approximations still use the harmonic approach for the force fields.

You are right that multidimensional PES is not simple quadratic, but close to the PES well minima it can be approximated as harmonic oscillator. Also if you really want to include the real PES you can include anharmonicity in your calculation. In that case you have to include an anharonic constant $x_e\;.$ Your PES can be written as
$$E=\left(n+\frac{1}{2}\right)h\nu-x_e\left(n+\frac{1}{2}\right)^2h\nu$$
Usually in computational chemistry the energy can be expanded into a Taylor series
$$E=E_0+\frac{\mathrm{d}E}{\mathrm{d}x}(x-x_0)+\frac{1}{2}\frac{\mathrm{d}^2E}{\mathrm{d}x^2}(x-x_0)^2+\textrm{higher order terms}\;.$$
And you can usually neglect those first order term and higher order terms for most of the practical problems. And then your PES will follow the trajectory of a harmonic oscillator.

Thanks for the answer.... could you please tell how you came up with the formula above the expansion?

Please see the attached link:
http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/06._One_Dimensional_Harmonic_Oscillator/Harmonic_Oscillator/Anharmonic_Oscillator

@mamun The Taylor expansion is missing a  1/2

yes, you are right. I am changing it.


text:
Increasing Covalent character increases Colour intensity
I observed some trends as follows:-

$\ce{PbCl2}$ white and $\ce{PbI2}$ yellow
$\ce{SnCl2}$ white, $\ce{SnCl4}$ red and $\ce{SnI4}$ black
$\ce{AgCl}$ white, $\ce{AgI}$, $\ce{AgBr}$ and $\ce{Ag2CO3}$ yellow

In all these cases as the size of anion increases ( polarisability increases)  or charge on cation increases (polarising power increases) colour exhibited becomes darker.. How does increasing covalent character affect this?  Or is there  any other reason?

In silver compounds colour is mostly due to charge transfer.

tin(IV) chloride is a colorless liquid and tin(IV) iodide is orange solid.

In molecular orbital theory, as the covalent character increases the homo-lumo gap decreases, and the excitation happens more easily. Thus the colour intensity increases.


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Role of the counter/auxiliary electrode in cyclic voltammetry
I am doing a school project on cyclic voltammetry and am currently studying the topic. I have learnt that the reaction mostly takes place around the working electrode where the current is measured together with the counter/auxiliary electrode. My question is, how does the counter/auxiliary electrode pass current without oxidising or reducing the analyte?
For example, I were to increase the potential in a solution of $\ce{Fe^3+}$ ions. The working electrode would be the cathode and would reduce the ions, giving them electrons and forming $\ce{Fe^2+}$ ions. In this case, what would happen around the counter electrode? How would the electrode be able to pass current to balance it? Would it oxidise the newly formed $\ce{Fe^2+}$ ions? Wouldn't this result in a never ending process? If someone has already asked this question, I apologize. Please link the page, thank you.

In order for the solution to remain electrically neutral something has to be oxidized and something else has to be reduced.

For cyclic voltammetry basics, read this article.

You are right that the counter electrode must balance the redox reactions happening at the working electrode.
For every electron that goes to reduce an $\ce{Fe^3+}$ ion at the working electrode, an electron must be collected from an oxidation reaction at the counter electrode.
What that reaction actually is depends on a lot of things.
First, we must consider what is actually in the solution to oxidize at the counter electrode.
In an $\ce{Fe^3+}$ solution, unless the working electrode is very close to the counter electrode, the newly formed $\ce{Fe^2+}$ ions will not diffuse efficiently to the counter electrode, so something else must balance the bulk of the current.
In many aqueous solutions, this might be the water itself being oxidized, unless there is a more easily oxidized substance in the solution (a supporting electrolyte, for instance).
If the counter electrode material is chosen poorly, it may even be the counter electrode itself oxidizing.
In practice, electrochemists often design cells such that the reactions at the counter electrode are unimportant, perhaps by placing the counter electrode in a separate vessel connected with a salt bridge, or at the least assuring that potential counter electrode products will not interfere with the reactions at the working electrode.


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Fermentation of different types of sugar
I'm searching for the difference in the fermentation of muscovado sugar, honey and sugar cane. I have used a specific microbiota to to ferment these three types of sugar and have to explain why I've got different results regarding acetic acid and ethanol production in the final beverage, as well as different amounts of sucrose, glucose and fructose in same.

Micro-organisms eat the sugars then excrete all sorts of compounds. Acetic acid and/or ethanol is commonly excreted by anaerobic organisms.  Anaerobic organisms can derive energy from multiple reaction paths, even within the same micro-organism (consider the case of facultative anaerobes.)
Since the same organism can use different paths to derive it is quite natural to assume what comes out is naturally affected by what goes in.

authors: ϻ αɳαɳƚԋ2129


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How many valence are in each molecule?
I have Benzene, Dewar Benzene, Prismane, and 1,3,5 cyclohexatriene (which looks just like benzene but with different lengths between the $\ce{C-C}$ bonds).
They are all $\ce{C_6H_6}$ chemical formula and I need to find out how many valence are in each molecule. I looked up what valence is but I'm confused if valence is the same thing as valence electrons or if valence is the amount of bonds the molecules can have. I think the amount of valence $\ce{H}$ can have is 1 but and $\ce{C}$ can have 4 so is the amount of valence 30?
i'm pretty sure the amount of valence electrons in all of them are 6.

What do you mean by " all of them "?

@M.Ibrahim for benzene, dewar benzene, prismane, and 1,3,5-cyclohexatriene

Which valence is equal to six ? carbons, hydrogens or both ?

@M.Ibrahim that's kind of my question, i dont know how to find out how many valence are in each element.

I think that it is the same in all compounds of carbon and hydrogen, C=4 and H=1, whenever the molecule is neutral and stable . How are you sure that the valence is 6 ?

@M.Ibrahim no i'm not sure the valence is 6. would the valence then be 30 if each C=4 and H=1?

Why not ?, 24 carbon electron +6 hydrogen electrons =30

A molecule doesn't have a valence (usually spelled valency). Valency is a property of atoms.

I assume that the answer is that they are all C6H6 compounds (sorry for the terrible image!):

As others have pointed out in the comments, it doesn't really make sense to talk about the valency of a molecule. You could say they all have the same elemental formula, but then so do around 200 other molecules (in theory).


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Sensitive test for lead in water?
Some of my friends are in Flint and I'm horrified by their situation. I would like to learn how to test for lead independently.
After all if you can't trust the state sending them water to test is not sufficient. The testing procedures I have read about give too many type II errors. A procedure with a bias toward type I errors would be better since then I could pay a professional lab to see if it was real.
For tests that only detect high levels of lead, could they be made more sensitive by evaporating the water then testing a larger sample?

The situation in Flint doesn't need more gasoline on the fire. A water sample needs to go to a laboratory with appropriate credentials to do water analysis. This is absolutely not a situation for some novice chemist.

As far as I understand the problem is not the low quality water getting into your house (bad enough on its own), but possibly lead pipes within your house. I wonder if there is a way to inject some detecting agent at the location where the municipal water enters,  something that attaches to lead pipes, and check how much of it is still there when the water gets out of the tape in your kitchen.

I used to work in an environmental water testing laboratory.  The regulatory standard method for testing for lead in water uses an Atomic Absorption (AA) spectrophotometer.
It seems that the issue with Flint's water appears to be that the new source of water is corrosive, the water has not been treated for corrosion, and pipes in Flint are commonly made out of lead.  So if your pipes at home are made of lead and are corroding, you may have a problem that the lab at the water treatment plant was not able to test for.
If you are curious about checking your water at home, you can obtain a water testing kit from a local hardware store.  These kits come with test strips that allow you to check your water for things like nitrates, pH, alkalinity, hardness etc.  Make sure your kit allows you to test for lead.

I would doubt that a home water testing kit could detect lead at ppb levels. It might include a test to detect lead solder on copper water pipes since there is always excess solder on the outside of the joints.

Apparently, they can.  From the first lead test kit specification that pops up on Amazon: "Lead kit can detect dissolved lead at levels below the EPA Action level of 15 parts per billion (ppb)"

Could you post a link to the product? I tried to find the result you mentioned, but I guess Amazon is customizing search results, because I couldn't find it at all. Best I could find was a kit which claimed to detect at 5 ppm (1000x higher, and probably not too helpful).

Ok, I don't personally think it is kosher to post direct links to sites advertising products, but here is an example of a lead test kit that meets the specifications above.

Testing for trace minerals is quite tricky and the procedures can be elaborate if done by chemical means. Note that the lead will generally be in the form of a soluable compound like lead acetate, or lead carbonate or whatever. In many cases other chemicals, especially chlorine, if they are present can interfere with obtaining accurate results. If you go to NIST's web site, you can probably find the standard procedures (which is what the labs do). Doing these procedures probably requires a well-equipped lab, and also probably takes a considerable amount of skill and practice. You can't just read a few instructions and get it right off the bat if you are not an expert chemist.
There are two alternatives:
XRF machines. They cost $30,000 and up. Check with a manufacturer to find out their senstivity.
Spectrographic analysis. If you can obtain access to a high-quality spectrometer, you can easily do a test. It may require some filtering. Just tell the physicist who runs the spectrometer that you want to test for trace amounts of the compounds you are interested in (lead carbonate, etc). This kind of instrument can identify very tiny amounts of any element or compound reliably.

I was an XRF specialist. XRF isn't going to be able to measure Pb in ppb levels in a water sample. ICP could do this easily, but it is a very expensive instrument.

@MaxW I think you evaporate the water first before using the XRF. Only the solid residues of the water are tested.

I did XRF for years - both WDXRF and EDXRF. Evaporating to solid residues won't work. XRF just isn't the way to do this analysis. This is exactly why some novice chemist shouldn't be trying to do lead water analysis. This needs to be done by a lab certified to do such analyses using a standard technique.

It’s possible to purchase an instantaneous lead water test kit in Home Depot, Lowes, Ace, or another home improvement, retailer. Test kits may also be bought from internet retailers, such as Amazon or eBay. To know more how to test lead in the water you can also read this article for better understanding.

A spectrophotometric method would be the cheapest available method, however, the lead would have to be dissolved using a complex ion formation that has a distinct color or absorption pattern that would be is difficult to figure out. There is a commercial lead test that does this, from the Hach Company, but the extraction method requires specific reactants and procedures that I am not versed on.


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What is the correct Lewis structure for calcium phosphate?
In calcium phosphate, $\ce{Ca3(PO4)2}$, since the calcium and phosphate share an ionic bond, and the phosphorus and oxygen share a covalent bond, should the Lewis structure be like the following?

While I was checking other web pages I noticed there was another Lewis structure for the same molecule:

Are they both possible Lewis structures, or is the first structure the only "correct" version of the molecule?

The first one is definitely more realistic.

Note that calcium phosphate is ionic, not covalent.

Calcium phosphate crystallises in a number of different modifications of which β-$\ce{Ca3(PO4)2}$ is the stable room temperature and ambient pressure modification. It has a rather complex structure not easily described, consisting of chains and rings with phosphate and calcium ions in specific patterns. It is orthorhombic, but $c \gg b \approx a$.
Any attempt to simplify its structure by drawing it in pseudocovalent depictions like your lower scheme is doomed to fail. Better just use the upper scheme; unless you are expected to know a certain salt’s crystal structure that one is safe.

Just for reference: M. Yashima et. al, J. Solid State Chem. 2003, 175 (2), 272-277. DOI: 10.1016/S0022-4596(03)00279-2. Unfortunately no ESI, and no deposition in ccdc.

authors: Георги Гущеров


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Stability of CN molecule

The $\ce{CN}$ molecule has been observed spectroscopically in comets where pressure has been estimated to be as low as $10^{-10}\ \mathrm{mmHg}$. Draw a Lewis structure for $\ce{CN}$. Suggest an explanation as to why this species can persist in a comet but is unstable under normal terrestrial conditions?

Having drawn the Lewis structure, the molecule is a free radical, hence it is very unstable. Would the species be unstable under terrestrial conditions due to a higher temperature (hence pressure) there? Or more specifically, how does a change in pressure effect a free radical molecule?

Temperature is unrelated to pressure. As for the radical, it is stable by itself, but very reactive (=unstable) when it meets another radical (or just about anything else, for that matter), which happens more often when there is more stuff around.

@IvanNeretin I'm not a native english speaker, but reactive not means unstable. For example the complex of cobalt with EDTA is very very stable in water but if you have cobalt in water as a complex like $[Co(H_2O)_6]^{3+}$ it is absolutely unstable but to totally inert, then if you add a bit of EDTA you will never get the complexe of Co with EDTA if you don't heat up your solution...

Neither am I. Sure, these two words are not synonyms, but in this particular context they can be used this way. OK, I should have said "extremely reactive to everything around (including itself), and hence unstable".

Some molecules are inherently unstable, CN, CN- or even CN+ aren't.

Is there a direct relationship of the stability of a molecule to the pressure it is undergoing? @IvanNeretin

$\ce{NC-CN <=> 2CN}$
Which side is favored at low pressure?

Would the right side be favoured at low pressure because there is less force exerted on the molecules to drive them together, hence will remain separate?

@mnmakrets The right side will be favored, but not for the reason you have given.  The concentration (and partial pressure) of CN will be squared in the equilibrium expression. So for example if K=1, then the system will be at equilibrium when, for example, p(CN) = p(NCCN) = 1. But if the pressure is decrease such that p(CN) = p(NCCN) = 0.01, then the system will not be at equilibrium and will need to shift far to the right to achieve equilibrium.  Read more about pressure effects on equilibrium generally.


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Relative rates of two acyclic dienes in the Diels-Alder reaction
As for factors that affect the rate of the Diels Alder reaction, having an s-cis conformation is primary.
But in a case as shown below, what makes the second diene react faster compared to the third?
The only reason I can think of is that the second diene has substituents pointing outward so they cause less steric hindrance as it reacts with a dienophile thus has a lower activation barrier. Are there any other underlying reasons apart from hindrance?

You are correct that the having the diene in an s-cis conformation is extremely important.  Only from this conformation can the Diels-Alder occur. If there is a high concentration of the s-cis conformer the Diels-Alder will likely occur rapidly; if there is a low concentration the reaction will occur more slowly.
In cyclopentadiene, the diene component has no choice, it is locked in an s-cis conformation by the ring.  However with acyclic 1,3-dienes an equilibrium exists between the s-cis and s-trans conformations.
In 1,3-butadiene the amount of the s-cis conformation is relatively small (~2%) because of the steric interaction between the "internal" hydrogens at the ends of the chain in this conformation. In accord with this expectation, 1,3-butadiene is not as reactive as cyclopentadiene in the Diels-Alder reaction.

If we replace one of these "internal" hydrogens with a methyl group (penta-1,cis-3-diene), the steric interaction between the "internal" methyl and hydrogen in the s-cis conformation is even greater, further destabilizing the s-cis conformation and reducing the amount present at equilibrium.  On the other hand, replacing one of the "exterior" hydrogens with a methyl group (penta-1,trans-3-diene) doesn't cause any increased steric interference in the s-cis conformation and the amount of the s-cis conformer is about the same as 1,3-butadiene. Again, in accord with expectations, penta-1,trans-3-diene reacts at a rate similar to 1,3-butadiene, whereas penta-1,cis-3-diene reacts slower.

(image source)
Using this information we can make predictions for your compounds. We would expect cyclopentadiene to react the fastest because it is held in an s-cis conformation. In the case of cis, cis-2,4-hexadiene we now have an internal methyl-methyl interaction, which severely destabilizes the s-cis conformation, so there is very little present at equilibrium and this compound reacts the slowest in the Diels-Alder reaction. trans, trans-2,4-hexadiene only has an "internal" hydrogen-hydrogen interaction when in the s-cis conformation, therefore it should react about as fast as 1,3-butadiene - in between the extreme rates of the other two compounds.

It is indeed steric hindrance. It looks a lot less severe in the picture than it actually is. Remember that each end of the line is not just a carbon atom but a methyl group $\ce{CH3}$. That means that there are three further undrawn atoms bonded to the final atom of each side of the chain.
Now making the simplistic assumption that $\ce{C-H}$ bonds are similar in length to $\ce{C-C}$ bonds and just boldly drawing the hydrogens in a $120^\circ$ angle from those final carbons we notice that one hydrogen each would land on the carbon of the opposite end of the chain — very bad. If we twisted and turned a little, we could get around that, but all of the hydrogens would still be outstandingly close to each other. Therefore, (Z,Z)-hexa-2,4-diene will have a really hard time adopting the s-cis configuration.
The middle molecule, (E,E)-hexa-2,4-diene only has two hydrogens in the middle, which would hardly even notice each other.


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Can I use fluorescence microscopy with a solution of hydrogen peroxide?
I want to use microspheres which have been dyed with a fluorescent dye for fluorescence microscopy in a solution that has a 1-5% concentration of hydrogen peroxide. Is this possible or will the hydrogen peroxide always bleach any particles I put in?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

The Hydrogen peroxide will definitively accelerate bleaching of your fluorescent microspheres. In fact, we routinely add glucose oxidase, glucose and catalase to our single molecule fluorescence experiments in order to get rid of oxygen and hydrogen peroxide in our solution and thus reduce photobleaching. However, Fluorescence microspheres are very photostable and whether or not they work for your application depends on several factors:

how long the microspheres have been in the solution before you start imaging (the sooner you start the better, because of less reaction time between dye and hydrogen peroxide).
How long you need to image (the shorter the better, because of less photobleaching)
How large the microspheres can be (the larger the better because large microsphere contain more dye that is less accessible to the hydrogen peroxide in solution).

Basically, I suggest you try (and let us know how it went :) )

It worked fine and the spheres I used did not get bleached by the hydrogen peroxide. I think they were dyed all through, rather than surface coated. I left them in the hydrogen peroxide solution for 4 hours and I could still see them just fine.


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How to dilute a stock solution when the resulting solution has a smaller volume?

Describe how you would dilute $2.0~\mathrm{L}$ of a $0.85~\mathrm{mol/L}$ magnesium hydroxide stock solution to make $200.0~\mathrm{mL}$ of $0.30~\mathrm{mol/L}$ magnesium hydroxide.

I know the $$c_1 V_1 = c_2 V_2$$ formula and that magnesium hydroxide's molar mass is $58.32~\mathrm{g/mol}$. However, I am confused because $V_2$ is lower than $V_1$.

Maybe the question was posed this way to make you think practically.
Approach this problem as if you were in a lab and you actually had a bottle of $2.0~\mathrm{L}$ of a $0.85~\mathrm{mol/L}$ magnesium hydroxide stock solution, and you wanted to make a new solution of $200.0~\mathrm{mL}$ of $0.30~\mathrm{mol/L}$ magnesium hydroxide.
You do not need to dilute the entire stock, but just a portion of the stock solution.  Describe what volume of stock solution you would take out, and how you would dilute it, using the $$c_1 V_1 = c_2 V_2$$ formula, to reach $200.0~\mathrm{mL}$ of $0.30~\mathrm{mol/L}$ of magnesium hydroxide.


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Why is the enthalpy of solution for NaCl 3.88 and not 4.1 kJ/mol?
Based on Hess' law, the molar enthalpy of solution is equal to the sum of the enthalpies of formation of products minus reactants. In the case of dissolving table salt ($\ce{NaCl}$) in water, those reactants are $\ce{NaCl (s), Na+ (aq)}$, and $\ce{Cl- (aq)}$, with enthalpies for formation of $-411.2$, $-239.7$, and $-167.4\ \mathrm{kJ/mol}$, respectively.
So, based on Hess' law, the enthalpy of solution should be $4.1\ \mathrm{kJ/mol}$. However, the actual enthalpy of solution in this case is $3.88\ \mathrm{kJ/mol}$ according to multiple sources. The discrepancy appears to be that the enthalpies of formation for $\ce{Na+}$ and $\ce{Cl-}$ don't exactly equal the combined enthalpy of formation for $\ce{NaCl (aq)}$.
I am at loss for why there is this small offset. Is it due to only partial dissociation of $\ce{NaCl (aq)}$ while in solution? And, if so, wouldn't that have a larger effect? Or is the effect slight because there are some inter-molecular forces at play that only minimally affect the total enthalpy of solution? It could be that my numbers are wrong, but a cursory search on the internet matches the values for $\ce{Na+}$ and $\ce{Cl-}$ in the textbook I have on hand.

Your approach is correct. The small discrepancy of the results when using parameter values from different sources could be caused by inconsistent standard states (temperature and pressure), or simply by measurement uncertainty (in this case, the resulting difference would not be considered significant).
By way of comparison, I found the following parameter values for a temperature of $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$ and a pressure of $p=1\ \mathrm{bar}=100\ \mathrm{kPa}$.

Standard molar enthalpy of formation $\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{NaCl(s)})=-411.2\ \mathrm{kJ\ mol^{-1}}$ [1]
Standard molar enthalpy of formation $\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{Na+(aq)})=-240.1\ \mathrm{kJ\ mol^{-1}}$ [2]
Standard molar enthalpy of formation $\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{Cl-(aq)})=-167.2\ \mathrm{kJ\ mol^{-1}}$ [2]
Standard molar enthalpy of solution $\Delta_{\mathrm{sol}}H_{\mathrm m}^\circ(\ce{NaCl(s)})=3.88\ \mathrm{kJ\ mol^{-1}}$ [3]

Calculating the standard molar enthalpy of solution from the given values for the standard molar enthalpy of formation yields
$$\begin{align}
\Delta_{\mathrm{sol}}H_{\mathrm m}^\circ(\ce{NaCl(s)})&=\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{Na+(aq)})+\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{Cl-(aq)})-\Delta_{\mathrm f}H_{\mathrm m}^\circ(\ce{NaCl(s)})\\[6pt]
&=-240.1\ \mathrm{kJ\ mol^{-1}}-167.2\ \mathrm{kJ\ mol^{-1}}+411.2\ \mathrm{kJ\ mol^{-1}}\\[6pt]
&=3.9\ \mathrm{kJ\ mol^{-1}}
\end{align}$$
Considering the number of significant digits, there is no significant difference between the calculated result of $\Delta_{\mathrm{sol}}H_{\mathrm m}^\circ(\ce{NaCl})=3.9\ \mathrm{kJ\ mol^{-1}}$ and the corresponding literature value of $\Delta_{\mathrm{sol}}H_{\mathrm m}^\circ(\ce{NaCl})=3.88\ \mathrm{kJ\ mol^{-1}}$.

References
[1] “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.
[2] “Thermodynamic Properties of Aqueous Ions”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.
[3] “Enthalpy of Solution of Electrolytes”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.


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Why did my deep eutectic solvent solidify?
I'm finally getting around to beginning to test this deep eutectic electrolyte.  I mixed choline chloride with urea in a 1:2 mole ratio in a glass container.  It formed a sludge.  I sealed the container and then came back two days later to begin electrolysis and the mixture had formed a hard solid!
The whole time the mixture was at room temperature, and except for the mixing stage it was sealed.
So what am I missing?  Is this a sign my reagents are bad?  Or do these eutectics have a limited mixed life?
Update: Following Beerhunter's suggestion I ran the reagents in a blender until they formed a very liquid (low viscosity), homogenous mixture.  But one day later the DES is crystallizing again at room temperature:

You have to keep mixing the components until the liquid forms. If you have a sludge, you will have mixed something, but not everything intimately, so you won't have the correct component ratio. The DES is not going to form spontaneously,  so leaving it will not have helped.
We have had DES's remain as liquids for 6 months and counting.

Are there any good indicators that it is properly mixed?  Like do you know what sort of viscosity should be attained with these particular eutectics?  Can water or some other solvent, and/or heat, be introduced to improve the mixing?

We haven't measured the viscosity, but if you tip it at 45 degrees, it takes over a second to level itself. If it's mixed properly, it will become a liquid. We use automation, but old school is mortar and pestle. The literature frequently makes reference to using water to dissolve the solids, then evaporating the water (under vacuum may be better?) to leave the DES. DES properties can still be exhibited with small amounts of water, but obviously this composition change alters the properties vs the pure DES


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Thermodynamic and kinetic products - carbocation stability

The professor wants to know which is the thermodynamic product and which is the kinetic product.
He says there are only two products - the two on the right that result from the resonance-stabilized complex. I can see a third product - the product on the far left with the question mark over an arrow.
The argument for the far right product being the kinetic product is that it is a tertiary carbocation and therefore the energy barrier to formation is probably lower than the energy barrier to formation of the secondary carbocation.
Alright, sure. However, could one also argue that the far left product is the kinetic product? It is after all the result of a tertiary carbocation. And it is the result of a "pure" tertiary carbocation - not a tertiary carbocation "diluted" by resonance; the far right product results from a resonance-stabilized complex that has both secondary and tertiary carbocation character. This was something I thought of at first. But then I realized I was probably taking resonance contributors too literally and that the resonance-stabilized complex would probably be more favored intermediate because the positive charge is "delocalized".
So what's your take?

Your last line is pretty much it. The simple way of looking at it is that delocalisation always leads to stabilisation and therefore, a tertiary allylic carbocation is more stable than a tertiary non-allylic carbocation. If delocalisation was unstable, the molecule would not do it. It is a bit counter-intuitive that resonance mixes in some character of a higher-energy state and it leads to stabilisation, but that's what it is.

@orthrocresol can you expand on your comment? I'm sure it'd make a great answer!

Related http://chemistry.stackexchange.com/questions/59964

Background
First off, let's discuss what we mean by "kinetic" and "thermodynamic" products.  The kinetic product is the first formed product, the one that forms the fastest, the one with the largest rate constant. The rate constant, k, can be described using the Arrhenius equation
$$\mathrm{k=Ae^{\frac{-Ea}{RT}}}$$
where A is the pre-exponential factor, a measure of the probability of a successful collision. Ea is the activation energy, the energy required to surmount the barrier to reaction, R is the gas constant and T is the temperature in Kelvin.
Usually, in a series of related compounds like we have here, the A factor (which is a very large number to begin with, so would need to change by a large amount to alter the course of a reaction) is assumed to remain relatively unchanged.  Consequently the activation energy is viewed as the controlling factor in terms of which reaction pathway is favored.
The thermodynamic product is the product with the lowest energy of formation, the most stable product.
In a reaction where several products can be formed the thermodynamic product could be the same as the kinetic product, or it could be different.
Two examples to further clarify kinetic and thermodynamic products can be found in this earlier answer describing the Diels-Alder reaction between cyclopentadiene and maleic anhydride. The endo isomer is formed first - it is the kinetic product, but if the reaction is run under conditions permitting equilibration, then ultimately the more stable exo product will be predominate. Another example can be found in the sulfonation of naphthalene where naphthalene-1-sulfonic acid is formed initially (the kinetic product), but upon heating will equilibrate and rearrange to the more stable 2-isomer.
A key point here is that in order for the kinetic product to rearrange to the thermodynamic product, the reaction must be reversible under the reaction conditions.
Back to your question

could one also argue that the far left product is the kinetic product?

Technically, according to your diagram, the answer is "no". There are no equilibrium arrows in your reaction diagram, therefore if the far left product were the kinetic product it would be observed.
But maybe you inadvertently left out the equilibrium arrows, what would the answer be if you included them? All 3 possible products are rather analogous, so for starters we would expect similar A factors in the 3 pathways and look to differences in the activation energies to decide which is the preferred pathway (kinetic product). Your product on the left requires a tertiary carbocation as its intermediate, the other two products require a resonance-stabilized allylic carbocation which is further stabilized by a methyl substituent.  The latter carbocation is lower in energy (resonance stabilization plus inductive and hyperconjugative stabilization by the methyl group) and so the kinetic product will derive from it.
Of the two allyl resonance structures you've drawn, the one on the right will carry a bit more importance in terms of describing the actual allyl carbocation due to the methyl stabilization described above. Hence I would expect the kinetic product to involve attack at the methylated allyl carbon leading to your far right product being the kinetic product.
Attack at the other allyl terminus produces a product with a methyl group on the double bond. We know from heats of formation that alkylation of a double bond stabilizes the system (because electrons are moving from higher [$\ce{sp^3}$] to lower [$\ce{sp^2}$] energy). Hence, your middle product is a possible  thermodynamic product.  Your molecule on the far-left also contains a tri-substituted double bond, it may also appear as a thermodynamic product.
Whether or not equilibration between the kinetic and thermodynamic products occurs depends on the reaction conditions.  Assuming equilibration is possible, then short reaction times at low temperatures will favor formation of the kinetic product, while higher temperatures and longer reaction times will facilitate transformation of the kinetic product into the thermodynamic product.

are you saying that kinetic products are in equilibrium with something?

What I am saying is...if, in any reaction, the kinetic product is transformed into the thermodynamic product, then the kinetic product must be in equilibrium with the intermediate that partitions into the thermodynamic and kinetic products.  The pathway to the thermodynamic product will have a higher activation energy, but for every billion or so kinetic product molecules that return to this intermediate, a few will have enough energy to follow the reaction coordinate leading to the thermodynamic product.  The thermodynamic product, being lower in energy, is less likely to return to

this intermediate. Hence, given time and energy, the thermodynamic product will accumulate at the expense of the kinetic product.

The product on the far left cannot form in any reasonable amounts. Consider the HOMO of s-cis-butadiene (which is a close enough model system to your dimethylcyclohexadiene): Those contributions that extend into space furthest are those on the terminal carbons.

(Image credit: ChemTube3D.com by Nick Greeves. Ignore the right half of the image, it is not relevant to this part of the question.)
Therefore, it is most likely for an electrophile to attack one of these two carbons. Then also, the attack on the terminal end will give an allylic cation while attacking the middle carbon will result in an unstabilised (albeit tertiary) cation — an allylic cation is much more stabilised and therefore the other cation is unlikely to form at all, even if we don’t consider the lesser contributions of the central carbons to the HOMO.
The next step is to identify a thermodynamic product. This is clearly the one with the higher-substituted double bond; your middle product. Thankfully, there is little discussion as to what is a thermodynamic product.
Determining a kinetic product is not easy at all. There are numerous factors to consider:

ease of attack
stability of the intermediate
favourable orbital overlap (this is probably most important)
…

The thing about our orbitals here is that we only have one allylic cation where both ends of the allylic system share approximately $+\frac{1}{2}$ charge and being very similar. Thus, we don’t truly know whether it is the slightly larger contribution of the tertiary carbon to the allylic system’s LUMO that is more important or whether it is the more accessable (for a large atom such as bromide) secondary carbon with less steric hindrance. I for my part would not allow myself to judge these two effects and would rely on experimental evidence for one or the other. However, I am inclined to believe that the kinetic product is that with less steric stress (that the steric argument outweighs the LUMO argument) and that thus kinetic and thermodynamic products are identical.

For the image you linked, it is absolutely necessary to know, that in the right column, the metal ion is not in the same plane as the butadiene; it is $\nu^4$ coordinated. I know that this is not about the point you are making, but it got me thinking and it might get some others thinking, too.

Didn’t you mean $\unicode[Times]{x3b7}^4$ coordinated?

Yes, I did... I was typing that on the phone and some things just slip away ;)