text:
Warning during geometry optimization via Gaussian
I have some recurrent problem with my DFT calculations [functional: hseh1pbe; basis:6-31+g(d,p), object: porphyrin derivative]. During the MO setting I get:
**** Warning!!: The largest alpha MO coefficient is  ***D+03
**** Warning!!: The smallest alpha delta epsilon is  ***D-01
I did some first-step research over the internet: the only considerable suggestion I found was to decrease the basis since some MOs got linear-dependent. I already decreased it from 6-311++g(d,p) so I assume that it didn't help much. Moreover, simplifying the basis set would mean increase in inaccuracy, right?
Did someone meet and solve this issue before?

For HD & DFT calculations a near-linear-dependency in a basis is usually not a problem, so you can ignore it. For post-HF treatments this might be an issue. Details.

Thanks for the comment. The link you gave it is what I mentioned. I still feel a bit awkward about those warnings.

First, are you sure you actually need diffuse functions (*)? Secondly, try Ahlrichs basis sets instead of Pople ones, say, Def2SVP instead of 6-31g(d,p) or Def2TZVP instead of 6-311g(d,p).

As for diffuse functions, no, I am not sure. I would think that it is good to include them for higher accuracy.
It's nice that you mentioned about Ahlrich's basis sets: I initially started with DGTZVP (in Gaussian) but the same problem occurred so then I started to minimize the basis set.


text:
What are the height and width of the large and small nodes of the sp3 hybridized orbitals of carbon and silicon?
I will make an effort to interpret answers based on wave functions, but I am a senior pursuing a bachelor's degree in Biology, so I don't have the background in quantum physics for a purely mathematical answer to make intuitive sense.
I would appreciate any partial answers which cover only carbon or only silicon, as well as partial answers that give the height and width of only the small node or only the large node of the sp3 orbital. I am asking about the width of the large and small nodes at their widest points, not at their center of electron density.
I am aware the sp3 hybrid orbitals only form when these atoms are bonded to four other atoms, so there are no empirical observations showing the height of sp3 orbitals from isolated atoms (for this question, assume a tetrahedral lattice composed purely of carbon or silicon). That said, any estimates based on halving bond lengths should account for orbital overlap in these bonds so that the height of the orbital from each bonding atom is not underestimated.
Pieces of information that I would find very useful as an addendum to answering my main question are estimates of the electric potential energy, kinetic energy, and total energy (constant) of an electron as it travels in the largest ellipse contained within the sp3 orbital of either a carbon or silicon atom in a lattice – including the large and small nodes, and assuming this ellipse does not move from one bonded atom to another.
Thank you!
EDIT to update: Having taken organic chemistry I and II, I am also aware that orbitals are simply the 95% confidence interval for finding an electron. I am also aware that the common understanding of electron movement in orbitals is that they can move between states of equal energy without occupying all of the states in between. In my question about an electron moving in an elliptical path, I am asking for answers to assume that the electron is following such a path, and that this path is a planar ellipse which corresponds to a cross-section of the outer edge of the 95% confidence interval. Smaller elliptical paths would have lower energy, and I'm asking this because it is the highest energy path an electron would take in the orbital if it were following a planar elliptical path.

Please realize that those pretty pictures in the books are not real. Think of them as some kind of contour plot to give you the general idea of what the electron distribution looks like. To get a better (but more difficult picture) you can start with the Laplace spherical harmonics, and then hybridize them to sp3. You will then have to determine what you mean by 'width' by defining a differential probability distribution.

You just can't assume that electron is running along elliptical path. True, there was once so-called Bohr-Sommerfeld model which managed (by a combination of serendipity and sheer luck) to get some things right. But that model would not let you describe hybridization, or even atoms with more than one electron, let alone molecules. Also, sp3 hybridization does not require 4 other atoms, sometimes 3 are enough, and you may "see" a "naked" sp3 orbital. But it does not have anything like "height" or "width".

So, after I get past the shock and awe that you are asking us to describe a fictional object,
this page has some pictures which will be suitable for you to construct visualization: ].
I eyeball it as about 4 times larger in all three dimensions for the 2s/2p level.  The radial distribution function for 3s/3p has its highest density about 2.2 times further away from nucleus as the 2s/2p level
]

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text:
Zinc oxide reduction in ethanol
I'm trying to prepare a ZnO thin film deposited on a glass substrate. The precursor solution prepared by sol gel process through dissolving 1 M zinc acetate dihydrate $\ce{Zn(CH_3COO)_2.2H_2O}$ in ethanol with ethanolamine as stabilizer in 1:1 mol ratio with $\ce{Zn(CH_3COO)_2.2H_2O}$. I aged the solution for about 1-2 weeks in a closed glass container. But, after I actually deposited and annealed at 600 °C, the film's XRD measurement seems to show ZnO as well as zinc peak, with zinc somewhat showing stronger peak.
However I also have the one with precursor aged for only 1 day but annealed at 800 °C, this one show peak characteristic that of ZnO with no noticeable Zinc peak. Other samples I have are aged for 3-4 days added with 1-5at% Nickel doping and annealed at 600 °C, they also show strong Zinc peak along with ZnO peak.
In the ones with Ni doping, Zn peak relative to ZnO also seems to get stronger as Ni concentration increasing. It seems like nickel (or its derived compound) acts as catalyst in the reduction of ZnO. The ones with 3at% and 5at% nickel concentration also have solution with green-bluish turbid solution with white precipitates (this probably zinc) unlike 0at% and 1at%, which yields clear solution. This might support that nickel act catalyst.
So, I guess there must be more rate of reduction reaction of ZnO during aging process, or maybe some other reaction that produces Zinc. Maybe something happens during annealing as well, but that seems unlikely since ZnO seems to be stable at 800 °C.
The question:

Looking at the precursor solution, what possible reactions that would produce Zinc from ZnO, and how likely they are?
Is it possible that the reaction of producing Zinc happens during annealing? How likely they are? How much is the rate, can they be neglected?
Do Nickel doping contributes to the production of Zn? If yes, what mechanisms that seem most likely to affect the production of Zn?

https://chemistry.stackexchange.com/questions/47894/zinc-oxide-reduction-in-ethanol

As for your first question, I have personally had better luck only aging 24-48 hours (actually unsealed, just covered with Kimwipe to prevent contamination, but allowing for some of the solution to evaporate).  We're getting good $\ce{ZnO}$ XRD peaks on this process, and are currently developing an SOP/Optimization for $\ce{ZnO}$ by sol-gel technique. I'll let you know when we publish.
When you say you're getting $\ce{ZnO}$ and Zinc peaks, do you mean you're seeing XRD peaks matching Zinc as crystalline metal?  To your question precisely: "What possible reactions would produce $\ce{Zn}$ from $\ce{ZnO}$, and how likely are they?" Remember that your sol-gel is not necessarily $\ce{ZnO}$ yet.  The reaction that produces $\ce{ZnO}$ begins with the dissociation of $\ce{Zn}^{2+}$ and $\ce2{CH_3COO}^{-}$ ions in ethanol, followed by a hydrolysis. As to how you're getting the extra electron to precipitate metal, I don't know. What we have seen has been the precipitation of $\ce{Zn(CH_3COO)_2}$ after about 96-120 hours after originally mixing.  (We're pretty sure it's acetate because it smells like vinegar when we wash it out of the flasks). There is a very good article going over the chemistry and technique of $\ce{ZnO}$ sol-gel: article

+1 to @ReedThomas's question-The redox reaction between Zinc Oxide & Ethanol is this- $\ce{ZnO + C2H5OH -> Zn(OH)2 + C2H4}$

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text:
Thermodynamic versus kinetic reactions 2
I want to ask more following the previous Question. (Thermodynamic versus kinetic reactions)

in this reaction below, is the first product 'kinetic product'? and second one is 'thermodynamic product' ?
upper pathway is acid/base reaction but doesn't seem reversible, thus looks like far from thermodynamics. but I learned acid/base reaction is about thermodynamics. what did I get wrong?

I'd appreciate any comments^_^

Note that it is possible that the thermodynamic and kinetic products are the same.

even when two different products can come up like a case above?

Yes, even so. An example is radical bromination. Radical bromination favors formation of the most stable radical (and by Hammond's postulate) this radical forms fastest.

if one product is kinetic and thermodynamic product at the same time, why would there any kind of other product?

@NKYu Because a mixture of products always has a lower Gibb's energy than a pure product.

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text:
Formation of nitrate esters

When nitric acid is protonated by sulfuric acid, why is the hydroxyl group protonated? There are two other possible oxygens with partial negative charge that would seem to be better proton acceptors.

Because that pathway leads to a reaction. Reaction mechanisms describe the sequence of steps required to reach a certain product. They don't describe every possible step that can occur.

but shouldn't pathway resulted in product be likely to happen?

I think I understood it. Thank you^^ I should've thought more


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Is organic carbon the same thing as organic compounds?
Are they the same thing or is there some minute difference? I have looked all over the web and you would think this would be a common question but I can't find any answer. If they are different then what is the difference?
Thanks

Organic carbon is the carbon contained in organic compounds.

@IvanNeretin the context of this is for an emissions inventory so I don't think that is quite right

Context matters! Plants, animals and atmospheric CO2 are enriched in C-14 compared to e.g. coal deposits. "Organic carbon" might be interpreted as carbon-containing compounds with this particular isotope ratio.

Organic compounds typically refers to particular, identified compounds and/or compound classes.
Organic carbon, ususally expressed as total organic carbon (TOC), is a relevant term in waste (water) management.
In this context the total carbon (TC) is expressed as the sum of total inorganic carbon (TIC) and total organic carbon (TOC).
TIC and TOC are often measured as the amount of $\ce{CO2}$ liberated from a sample.

UPDATE
While Ivan Netetin's remark on the broader context is certainly true, terms like TOC and COD (in German CSB), the chemical oxygen demand, play an important legal role in the question whether and under which conditions waste water may be submitted to the public sewer systems. This involves water directly from procdution processes, as as the elution of material from from landfills.
In Germany, this is specified in the Abwasserverordnung - AbwV. Here, §6(3) states

Ein nach dieser Verordnung einzuhaltender oder in der wasserrechtlichen Zulassung festgesetzter Wert für den Chemischen Sauerstoffbedarf (CSB) gilt unter Beachtung von Absatz 1 auch als eingehalten, wenn der vierfache Wert des gesamten organisch gebundenen Kohlenstoffs (TOC), bestimmt in Milligramm je Liter, diesen Wert nicht überschreitet.

Translated to non-legalese English, this means: Business-specific limits are considered to be met (= it is legal to submit the water to the sewer) if the fourfold TOC does not exceed the COD given by law.

Well, it's not necessarily water, but also air and soil. Also, it's not necessarily waste. Organic (as well as inorganic) carbon is found pretty much everywhere, carbon cycle is a huge thing, and human activity is but a small part of it.


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Determine reaction rate from reaction time?
These are the results of my experiment (solution A and B were reacted and the reaction time was measured).
Data Table Edit:

How would I determine the rate of reaction (so I can find the rate law) from the reaction time?
Solution A: $\pu{4.3 g}$ of potassium iodate per $\pu{1 L}$ of water
Solution B: $\pu{2 g}$ of sodium bisulphite, $\rm5~mL~1.0~M~\ce{H2SO4}$, 4 g soluble starch per $\pu{1 L}$ of water
I found the concentrations of $\ce{[KIO3]}$ by dividing mass of $\ce{KIO3}$ by molar mass of $\ce{KIO3} (\rm214.001\frac{~g}{mol})$ and then dividing that by volume of solution A. I also divided the mass of $\ce{NaHSO3}$ by molar mass of $\ce{NaHSO3} (\rm104.061\frac{g}{mol}$) and then dividing that by volume of solution B to find the concentrations of $\ce{[NaHSO3]}$.
I then tried to find the reaction rate by $$\frac{1}{3}\times\frac{\ce{[NaHSO3]}}{t}$$ and $$\frac{\ce{[KIO3]}}{t}$$ as the reaction is $$\ce{IO3- + 3 HSO3- -> I- + 3 HSO4-}$$
Not sure if I'm on the right track or not.
Source:

Reaction Rate was found by $[\ce{NaHSO3}]$/Reaction Time and $[\ce{KIO3}]$/Reaction Time.
Rate order was found by log(Reaction Rate)/log([$\ce{NaHSO3}$]) and log(Reaction Rate)/log([$\ce{KIO3}$]).
Would appreciate it if anyone could tell me if I did this correctly.

This is an example of an iodine clock reaction. If you really want to solve it, here is a hint: 1. Start with your stock solutions A and B and calculate the molar concentrations. 2. Calculate the initial concentrations of iodate and bisulfite in each of your experiments. 3. Group your data. 4. Plot your data.

Sorry for the late reply. I didn't have access to Internet. The reaction is IO3− + 3 HSO3− → I− + 3 HSO4−, so I tried to find the rate of reaction by [NaHSO3]/t*1/3 and [KIO3]/t.

The reactant orders in iodine clock reactions generally turn out to be first order.  Your calculated orders are suggestive of first order dependences, but there is a systematic error evident in the data that possibly stems from the experimental design.  In addition, I don't believe that the manner in which you calculated the orders is correct.  The approach in this experiment is to use the method of initial rates.  Briefly, a differential rate law (which gives instantaneous rate as a function of concentrations) can be written as
rate $=k[\ce{A}]^n [\ce{B}]^m$ for a reaction $\ce{A + B -> products}$.  To find the order of reactant A, for example, one varies the initial concentration of A and keeps the concentration of B constant. This generates a pseudo-rate law in which the concentration for B is treated as a constant(k'):  rate = $kk'[\ce{A}]^n$.  This relation can be linearized as follows:
$$
\begin{align}
\log \text{rate} &= \log kk'[\ce{A}]^n\\
\log \text{rate} &= \log [\ce{A}]^n + \log kk'\\
\log \text{rate} &= n \log[\ce{A}] + \log kk' \tag{I}
\end{align}
$$
Trials are conducted in which [A] is varied and the initial rate is determined.  A plot of log rate vs. log[A] should be linear and its slope should equal n, the order of reactant A. If, on the other hand, we solve EQ I for n we get
$$
\begin{align}
n &= \frac{\log \text{rate} - \log kk'}{\log[\ce{A}]} \\
n &= \frac{\log \frac{\text{rate}}{kk'}}{\log[\ce{A}]}
\end{align}
$$
Thus calculating n as $\log \text{rate}/\log[\ce{A}]$ (your approach) is not fully correct because the pseudo-rate constant $kk'$ is not accounted for.  I plotted your $[\ce{HSO3-}]$ data in the form of $\log \text{rate}$ vs. $\log [\ce{HSO3-}]$ and obtained a value for n (the slope) of 0.84 with $r = 0.997$ (note: the rates must be treated as negative values because reactant is being consumed).  I advise you to attempt such a plot for the data where $\ce{[IO3-]}$ was varied.  Regarding the systematic error, you should think about the fact that you determined average rates experimentally, meaning reactant concentration was actually decreasing during the timed intervals and, therefore, the rates you calculated reflected an average reactant concentration, not an initial concentration.


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Exhaust gas purification
I'm more asking this because I'm not sure how far science has come.
Pretty much, I've been thinking about the car exhaust system lately. The two main components that lead to hazards from cars are $\ce{CO}$ and $\ce{NO_{$x$}}$. Carbon monoxide is hazardous because it thrives to become $\ce{CO2}$ and robs living beings of the oxygen in their bloodstream to become carbon dioxide. And the nitrogen oxides are hazardous only as that compound, but nitrogen and oxygen are quite harmless as $\ce{N2}$ and $\ce{O2}$.
Would it be possible to separate the oxygen from $\ce{NO_{$x$}}$ and somehow give that to the $\ce{CO}$, and any left over oxygen simply becoming $\ce{O2}$?
This all might be a stretch, but I have an active imagination and like seeking out answers to these questions.

@AndrewL, there are ways you can split a compound into elements, but the process requires energy through the roof to be supplied to the compound(or maybe use a catalyst). Also even if we succeed in splitting a compound, it will react within microseconds with either the same atom or the nearest feasible option to decrease it's energy, because that's why the compounds exist in first place. If you want to split $\ce{CO2}$ you would want to make sure that when the atoms of each element react back, they don't form the same compound (maybe by introducing a more feasable choice, I don't know what)

Oh, I got an example $\ce{ 2NaOH + CO2 -> Na2CO3 + H2O}$, here the Carbon dioxide reacts to form, non hazardous componds in general. $\ce{Na2CO3}$ is solid when distilled out from the mixture

"Carbon monoxide is hazardous because it thrives to become CO2 and robs living beings of the oxygen in their bloodstream to become carbon dioxide." No. $\ce{CO}$ blocks $\ce{O2}$ transport because it binds to hemoglobin more strongly than $\ce{O2}$ and in addition hinders the release of $\ce{O2}$ from hemoglobin at the remaining binding sites.

@cbeleites beat me to it.  CO is harmful for reasons that have nothing whatsoever to do with CO2.

This pretty much what happens (among other processes) in a three-way catalyst of a car:
$$\ce{2CO + 2 NO ->[\mathrm{cat}] 2CO2 +N2}$$

for reference : https://en.wikipedia.org/wiki/Catalytic_converter

And a simpler example: https://en.wikipedia.org/wiki/Electrolysis_of_water

@Nick Electrolysis and catalysis are completely different...

@J..., I was referring to a simpler example of decomposition. The title of the question has changed from "can you split a compound into elements", though, making my example less relevant.


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Energetics of Bond formation
Back to chemistry after 2 months and got stuck in the beginning! That's why I don't like it. Anyways, come to the question please.
Following are the words from my textbook (Don't ask the name ;-)

$$\ce {Na -> Na^+ + e^-}\quad\Delta{H}=+495 \frac{KJ}{mol}$$
$$\ce {Cl + e^- -> Cl^-}\quad\Delta{H}=-348 \frac{KJ}{mol}$$
$$\ce {Na^+ + Cl^- -> NaCl}\quad\Delta{H}=-786 \frac{KJ}{mol}$$
We conclude that it is essential for the formation of ions and ionic bond  that the sum of energies released in second and third reactions must be larger than what is required in the first step.
The great attractive forces (bonds) reduce the energy of the system greatly as it emits 786 KJ of energy.

a. Why should it happen? i.e. sum must be greater et cetera.
b. How does attractive forces tend to reduce the energy?

For second question I think the answer is that that bonds reduce the momentum of ions by not letting them move freely and therefore heat energy is reduced and released. But for question a I have absolutely no idea.

Ok your question is a little ambiguous but I'll try.
a. Why should it happen? - If you look at Na, it has a lone electron in its outer shell and that essentially unbalances the atom, so it prefers to lose it and become an ion, this is almost spontaneous and hence requires less energy. When $\ce{Na+}$ joins up with $\ce{Cl-}$ they neutralise the electrostatic charge introduced by them becoming ions, releasing more energy. More energy because they have gone from an energy state A (medium energy) to B (higher energy) to become an ion. Then when they come together they both, each atom, falls to a lower energy state C than state A even and hence the change (NB) from B to C is greater than the change from A to B. In your book the change from my A to B is the change of an electron ( first two equations in your book) and then my change from B to C is the Na + Cl part.
b. How do attractive forces tend to reduce energy? - This is an important one to note the point of the text book, which is pretty badly written if you ask me, is referring to the energy of the system and not the energy of attractive forces. When it says system it means the newly bonded together $\ce{NaCl}$. The energy is reduced because of the neutralisation mentioned above. Positive and negative prefer to be neutral and hence, are attracted to one another. But before that, you must realise that being charged or an 'ion' is less stable and therefore in a state of higher energy than a neutral system. So it is a matter of stability as well.
If you want quantum explanations please say but I get the feeling you'd prefer the simpler option.
I hope this helps, even if only a bit.

Why do ions have greater energy as compared to their neutral state? Why would $Na$ have less energy than its ion $Na^+$. And in fact ions have their octet complete so they must be stable which implies they should have less energy. No?


text:
Net ionic equations for an aqueous solution
Given the reaction of $\ce{Ni^{2+}}$ with aqueous ammonia to give $\ce{Ni(OH)_2}$ Would this be the correct reaction?
$$\ce{Ni^{2+} (aq) + 2NH3 (aq) + 2H2O(l) -> Ni(OH)2 (s) + 2 NH4+ (aq)}$$

You might just as well start with $\ce{Ni^2+}$ and $\ce{OH-}$.

Or use this: $$\ce{Ni^{2+}{(aq)} + 2NH4^+OH^- -> Ni(OH)2{(s)} + 2NH4^+}$$

Your answer is probably correct. There are some extra details one might wish to consider. If a short yes/no answer was enough, you must not read any further.
First, let us remind ourselves that in aqueous solutions metal ions coordinate with water molecules. Octahedral complexes are most common.
$$\ce{Ni^2+(aq) + 6 H2O(l) <=> [Ni(OH2)6]^{2+} (aq)}$$
Then $\ce{NH4OH}$, better described as a hydrate $\ce{NH3.H2O}$, is added.
$$\ce{NH3 (aq) + H2O (l)<=>NH3.H2O (aq)<=>NH4+ (aq) +OH-(aq)}$$
Now either $\ce{NH3}$ or $\ce{OH-}$ steals a proton from $\ce{[Ni(OH2)6]^{2+}}$. Note that the resulting complexes are equivalent.
$$\ce{[Ni(OH2)6]^{2+}(aq)+NH3.H2O(aq) <=> [Ni(OH)(OH2)5]+(aq) + NH4+(aq) +H2O(l)}$$
$$\ce{[Ni(OH2)6]^{2+}(aq) + OH-(aq)<=> [Ni(OH)(OH2)5]+(aq) + H2O(l)}$$
If another proton is taken, we reach the required precipitate $\ce{Ni(OH)2}$. (Due to the equivalency mentioned, I will write only one option.)
$$\ce{[Ni(OH)(OH2)5]+ (aq) + OH-(aq)<=> [Ni(OH)2(OH2)4](s) + H2O (l) \tag 1}$$
Often, however, the $\ce{Ni(OH)2}$ precipitate is not visible. The whole point of this excercise was to give background to the explanation that follows. When $\ce{NH3.H2O}$ is in excess, the following processes occur. The precipitate could directly react, in which case we might end up with
$$\ce{[Ni(OH)2(OH2)4](s) + 6 NH3.H2O (aq)<=> [Ni(NH3)6](OH)2 (aq) + 10H2O(l)}$$
Generally, the preferred explanation is through Le Chatelier's principle. $\ce{[Ni(OH2)6]^{2+}}$ ions react via $(2)$
$$\begin{align}
\ce{[Ni(OH2)6]^{2+} + 4 NH3.H2O (aq)} & \ce{<=> [Ni(NH3)4(OH2)2]^{2+}(aq) + 8H2O(l)}\\
\ce{[Ni(OH2)6]^{2+} + 6 NH3.H2O(aq) } & \ce{<=> [Ni(NH3)6]^{2+}(aq) + 10 H2O(l)}\\
\ce{[Ni(NH3)6]^2+(aq) +2 OH-(aq) } & \ce{<=> [Ni(NH3)6](OH)2(aq)}
\end{align}$$
Since the $\ce{[Ni(OH2)6]^{2+}}$ ions are used up, the equilibrium of $(1)$ tilts to the left, thus the precipitate disappears.
Once more, this was simply to explain why showing $\ce{Ni(OH)2}$ as a precipitate might not correspond with reality when there is (excess) dissolved ammonia.

I'm not sure where $(2)$ is.

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text:
Why is HCl a stronger acid than Acetic Acid?
Drawing the conjugate bases,  Oxygen is more electronegative than Chlorine so negative charge on Oxygen is more stable than Chlorine,  Thus My argument that Acetic Acid is more acidic.
Also the O¯ ion is in resonance with another Oxygen so that should also add to the acidic nature.
Where am I going wrong?

I can't dig into my textbook right now, but I believe the reasoning is that water solvates chloride ions extremely well - so a massive amount can dissolve very easily. Acetate ions are stable, but don't interact as well with water. I'll try and get a proper answer written later this evening, if I haven't forgotten (:

Also acetic acid tends to dimerise, thus reducing the effective number of acetate ions

The chloride anion and the acetate anion are such different species that the "why" in the question is hard to explain without blowing smoke. The place to start is to recognize that the Pauling electronegativity scale is for free atoms in space, not  ions in aqueous solution. // Obviously in aqueous solutions you can lookup pKa's to order acid strengths.

A simpler question, would be why is water a weaker acid than $\ce{HCl}$?  If we were to look at that issue, we might clear some of the mess up.
Strongly reactive compounds produce weakly reactive products.
In the case of acid base theory, chloride ion is a much weaker base than hydroxide.  Why is this so?  The hydroxide is much smaller, and what is referred to as 'harder'.  The guts of the issue is that in order for a base to deprotonate anything it has to donate two electrons to a bonding region near a proton (and in doing such render the orbital manifest, and the bond existent).  In oxygen based anions, those two electrons are more likely to be in a confined region, and hence more likely to donate to the putative $\ce{HO-H}$ bond that is forming, while $\ce{Cl^-}$, being larger has a harder time getting two electrons into the bonding space required to form a $\ce{Cl-H}$ bond.  So hydroxide is a stronger base.  This means that when donating a proton, it is likely to just yank it back again from whatever it just gave it up to, as compared to $\ce{Cl^-}$.  $\ce{HCl}$ gives up the proton and is more likely to not recapture it.
Groovy?  Back to acetate.  The resonance in acetate creates some diffusion of the negative charge and makes it harder to form said $\ce{AcO-H}$ bond, and as such it is a weaker base than hydroxide.  However, the smaller oxygen atoms are still better and forming bonds than $\ce{Cl^-}$ and as such it is a stronger base than chloride, and therefore its conjugate acid is less acidic.
Strong acids: weak conjugate bases.
Strong bases: weak conjugate acids.

Yep. Chloride ions are incredibly terrible bases, which makes their conjugate acid (HCl) very powerful.

"Strong acids: weak conjugate bases. Strong bases: weak conjugate acids." This is really more of fact that could be derived from pKa values than an "explanation." In aqueous solution: $$\mathrm{pKa + pKb = 14}$$

Acid strength depends upon ease of ionization of the 'acidic hydrogen' forming hydronium ions by proton transfer. (Bronsted-Lowry Theory).  HCl is one of the 'strong acids' and as such ionizes 100% whereas Acetic Acid (HOAc <=> H + OAc) is a 'weak acid' and ionizes less than 100% (~1.3% at 25C). For a 0.10M HCl(aq) solution => 0.10M in hydronium ions, but a 0.10M HOAc(aq) solution => only 0.0013M in hydronium ions at 25C. Structurally, the electronegativity of the chloride ion is much less than that of the acetate ion which means the covalent bond of the ionizable hydrogen in HCl is much weaker and more easily ionized in aqueous solution than the ionizable hydrogen to oxygen bond in HOAc.

Acid strength is determined by the stability of the conjugate base. Here are the factors that influence stability of the conjugate base:

Ion size (when going down the rows of the periodic table)
Electronegativity (when going in the same row of the periodic table across)
Resonance
Inductive Effects

If you look at the conjugate bases, in acetic acid, the negative charge falls on the oxygen and in HCl, the negative charge falls on the chlorine. Because of the fact that oxygen and chlorine are on different rows, we use ion size as the most significant factor for determining CB stability. The size of the chloride ion is greater than the size of the oxygen ion and therefore the CB of HCl can distribute the charge more evenly and is thus more stable. Moreover, the methyl group in the acetic acid is also electron-donating, meaning that it actually reduces how much the charge gets spread around on the molecule’s conjugate base. However, this is not really a significant factor.


text:
Why is tetramethylsilane (TMS) used as an internal standard in NMR spectroscopy?
I understand that TMS has a large proton number, giving a strong peak in the NMR spectrum. But apart from that is there any other reason why just TMS is used and not any other molecule?

In theory, you could use something like cyclohexane and get a similar result, but you have a much greater chance of overlap. Having a TMS singlet also allows you to shim your sample off the shape of the peak if you have no other strong, non-overlapping singlets. Most people just shim off the lock signal, but there are times that's not perfect.

TMS has 12 protons which are all equivalent and four carbons, which are also all equivalent. This means that it gives a single, strong signal in the spectrum, which turns out to be outside the range of most other signals, especially from organic compounds.
Although the chemical shift scales are still zeroed at the TMS peak, most spectra are now calibrated against the residual solvent peak. Typically, deuterated solvents such as $\ce{CDCl3}$ and $\ce{DMSO}$-$\ce{d6}$ are used and these contain a very small amount of undeuterated, or partially deuterated, solvent which produces a peak in hydrogen NMR. In carbon NMR the solvent peak is recognised by its splitting pattern, which is a triplet for $\ce{CDCl3}$ and a heptet for $\ce{DMSO}$-$\ce{d6}$ and the fact that it always comes at exactly the same place.

DMSO is a quintet in proton NMR, while chloroform is a singlet.  When you said heptet, I lost the carbon context, then I realized this comment should be added.

Other than for chloroform-d, it is not undeuterated solvent that is referenced, but rather the perdeuterated material. DMSO-d6 contains a small amount of DMSO-d5, which is why the 1H signal appears as it does. Undeuterated DMSO appears as a singlet. DMF is about the only common mulitply-deuterated solvent that usually contains completely undeuterated material

@long Yes you are right. I didn't think about that.

TMS is notoriously nontoxic too.

The newer IUPAC referencing standards actually do NOT reference the spectrum off the residual solvent peak.  Rather, all referencing is done relative to the 2H frequency of the solvent.  Modern instruments all use the deuterium signal from the solvent to lock the instrument (against magnetic field drift) and so this lock signal is now used for referencing all other chemical shifts (1H, 13C, 31P, 19F, etc.)  This is what "referencing by solvent" in modern NMR software is actually doing.

TMS was first proposed as a reliable internal chemical shift reference in 1958 by Tiers. Back in them good ol' days, 1H NMR was called proton nuclear spin resonance, or nsr, and the tau scale was  used for reporting chemical shifts (10ppm in the delta scale was set to 0 and positive values were read to the right. So TMS came at t10.0), and CCl4 was the one of the most common solvents used for running dilute solution samples.
TMS was proposed as a reference at the time becasue it offered the following advantages:

the chemical shifts are largely temperature and concentration independent
it offered a single phase method for referencing (That is, it was an internal standard. Most common methods at the time used an external reference, such as a coaxial insert; a method which is still acceptably used today)
The TMS peak lies outside the 'usual spectral region and is readily identifiable'
It is largely inert, and is not likely to react with most samples
It can be used for almost amy solvent (except H2O, D2O)

Today, TMS is still accepted as the single universal reference standard for NMR. For over a decade, the IUPAC recommendations have stated that, other than for 13C, all nuclei are referenced to the frequency of the 1H signal of 1% TMS solution in CDCl3. Yes, your 31P and 77Se peaks should actually be referenced to 1H of TMS. This unified reference scale is very easy to implement on modern instruments that incorporate a digital 2H lock (ask your friendly NMR custodian if they use it!). It is in fact arguably more accurate that referencing to internal solvent residues which are much more likely susceptible to concentration and temperature factors. Internal solvent residues are called secondary references, and are less accurate than the primary reference. However, they are accurate enough for the very vast majority of work done in the laboratory.
It has long been shown that the chemical shift of TMS is, in fact, both temperature and concentration dependent. But not much. There are actual corrections that should be applied to measured chemical shifts at extreme temperatures, but these are very small and rarely seen in the literature.
TMS is not water soluble, and in aqueous samples DSS is the recommended reference standard. It contains signals in the 1H spectrum at 3.1, 1.9 and 0.8ppm, all at a fraction of the intensity of the main reference signal, and these can interfere with analysis of other spectral regions. A deuterated version is widely used in the literature. An alternative, TSP, is sometimes used, although is more pH dependent. Other reference standards can also be used, however, and as long as the reference method and sample conditions are reported and verified, then all is good. For 1H NMR, compounds such as acetonitrile, methylsulfone, benzene, 1,4-dioxane and dichloromethane are all examples of internal standards that are sometimes used, both for chemical shift referencing and concentration measurements, as they all largely satisfy that important list of reference characteristics above. They all have very simple 1H spectra, soluble in common deuterated solvents and fairly unreactive. Plenty of other compounds with multiple peaks can be used as well, such as toluene and dmf.
So, to specifically answer your question, TMS is the adopted primary standard for 1H chemical shift referencing for the reasons stated above. Any other material can be used as a secondary reference, either internal or external, provided its chemical shift can be accurately determined against the chemical shift for TMS. Results reported should include a clear description of how chemical shifts were determined. For example, 1H chemical shifts of all reported compounds were measured relative to the methyl peak of n-dodecane, determined as 0.88ppm. The disadvantages of using other materials is that it needs to be shown that the chemical shift of these signals is independent of sample conditions (concentration, temperature etc), and that they will often have peaks overlapping regions of interest.

Another reason that TMS was chosen as a good reference is that it's relatively volatile and, thus, can be easily removed from the sample later (back in the "old days" when sensitivity was very low, it was important to recover the material in the NMR tube for further analysis or further reactions).

Also, although TMS is technically the origin of the new IUPAC referencing standard, the new standard is designed to remove the need for putting TMS in your sample.  Every deuterated solvent that is in common use has had its deuterium chemical shift referenced against TMS.  Modern NMR software contains tables of these calibrated values, so that all NMR (1H, 13C, 31P, 195Pt, etc.) are now referenced directly off of the deuterium signal.  I haven't used Bruker software for a long time, but the Varian/Agilent software automatically references off the 2H signal and ignores the TMS peak.

@S.Burt - Both good comments. Yes, automatic referencing is alluded to in the answer above. And while, by definition, it is a secondary reference, it is incredibly easy to implement, especially when running obscure nuclei with rather hazardous reference materials, such as 199Hg or 205Tl. It does, however, require that the instrument is set up correctly.


text:
Ion/Counter ion layers in the colloid precipitate of silver chloride

You use a precipitation method to determine [Ag+] in an unknown sample using Cl- ions. Reaction is conducted in presence of HNO3 in excess (relative to Ag+) Which statement is correct concerning the colloid precipitate at the start of the rxn (rxn not yet complete).
A) The ion in Primary Absorption layer is Cl-
B) The Colloid Particle is (-) charge
C) Ag+ is the major ion in Counter-ion Layer
D) Most Cl- ions added are in the precipitate form.

So I think the correct answer is D through process of Elimination and not chemistry knowledge. I know B is incorrect, B/C the precipitate must be AgCl, or possibly AgNO3 since there's Nitric acid here. Both of these are neutrally charged.
I'm pretty sure that Primary absorption layer is the "opposite" of the counter ion layer. So whichever ion goes into the Primary, the opposite ion goes in the counter. Therefore I eliminated A and C, since one cannot be correct without the other being correct as well. (although my first instinct was to put Ag+ in the primary ion layer).
I still have doubts about D. Since the reaction is not complete, and we are adding Cl- to a solution of Ag+, its safe to assume that any Cl- added will react. But HNO3 is a strong acid which dissociates completely, that means there's an excess of NO3- ions floating around. I'm not given any precipitation constants. Can I safely assume Ag+ will bind with Cl- before making AgNO3?
Also could you please confirm or correct my assumptions about Primary/counter-ion layer? I can't find any info it besides vague definitions.

It isn't absolutely clear from the wording, but it seems you're adding $\ce{Cl^{-}}$ to a $\ce{Ag^{+}}$ solution.
As some of the $\ce{Cl^{-}}$ is added, $\ce{Ag^{+}}$ is still in excess in the solution so virtually all of the $\ce{Cl^{-}}$ precipitates. $\ce{Ag^{+}}$ will be the primary adsorbent on the precipitate, and the colloidal particles will have a positive charge.
Since the colloidal particles have a positive charge, the ion in the counter-ion layer around the particle needs to be negatively charged. There isn't any  $\ce{Cl^{-}}$ in solution so it has to be $\ce{NO3^{-}}$ as the anion.
Thus the correct answer is (D).

Thanks for clearing this question up. So what happens if this or similar reaction is run in, for example, water. The Ag+ will be in the primary ion layer. and there is no Cl- in solution because it all precipitates out. Will the counter ion layer simply not exist? or maybe it fills with OH-?

(1) I would assume that the original reaction was in water since you didn't specify otherwise.  (2) You have to be careful when you say "all" the $\ce{Cl^-}$ precipitates. $\ce{AgCl}$ does have a Ksp. (3) With $\ce{Ag^+}$ in solution there must be some anion. You can't have a solution with just cations. So that anion would be in the counter layer.


text:
What is the criterion for filling up of electrons in molecular orbitals of increasing energy in simple diatomic species?
So far I have found different ways in different books.
For e.g.
One says

"In MO electronic configuration of diatomic species up to 14 electrons, $\sigma2p_z$ orbital (considering $z$ as the internuclear axis) is greater in energy than $\pi2p_y$ and $\pi2p_x$ orbitals. For species with more than 14 electrons, $\sigma2p_z$ orbital is lower in energy than $\pi2p_y$ and $\pi2p_x$ orbitals."

Whereas the other book used 16 as the no. of electrons instead of 14. This leads to different ways of filling electrons in molecules with 15 electrons such as $O_2^{+}, NO$ etc. I want to know which one is correct.

What is known for sure (and is mentioned in all the books I've seen so far) is that for homonuclear diatomics the "$\pi \rightarrow \sigma$ switch" happens when going from $\ce{N2}$ to $\ce{O2}$: for 14-electron case $\pi_{2p_{x,y}}$ orbitals are lower in energy then $\sigma_{2p_{z}}$, for 16-electon one it is other way around. The usual "explanation" is that moving from $\ce{Li2}$ to $\ce{F2}$ the successive decrease of the $s-p$ interaction between the $2s$ orbital of one atom and the $2p_z$ orbital of another atom decreases the energy of the $\sigma_{2p_{z}}$. But you have to understand that this is merely a manifestation of a known fact: we know how orbitals of these molecules are arranged from the molecular orbital theory calculations and use the notion of $s-p$ interaction to rationalize this knowledge.
The problem is that no information about 15-electron case can be inferred from the above mentioned knowledge about molecular orbitals of homonuclear diatomics from $\ce{Li2}$ to $\ce{F2}$. A priori we could not tell will $\sigma_{2p_{z}}$ orbital be lower in energy than $\pi_{2p_{x,y}}$ orbitals or higher then them for a 15-electron case. Besides, things coul be different for different 15-electron cases, such as $\ce{N2-}$, $\ce{O2+}$, $\ce{NO}$, etc.: for some $\sigma_{2p_{z}}$ orbital might be lower than $\pi_{2p_{x,y}}$ orbitals, for others - higher. There is no simple rule to use, one have to resort to molecular orbital theory calculations to tell the truth.

See also the related question on MO diagram for $\ce{N2-}$. The answer here is basically the same.

In fact, this question might be even a duplicate of that one...

Nicely explained. However, could please you tell me (a bit unrelated) if molecular orbitals formed by overlapping of pz orbital of one atom and s orbital of another is possible? If yes, give an example too.

I had this doubt in my mind for a while now but didnt want to start another question-answer thread on it here.

@Phill2, it depends on what the nuclear axis is. If it assumed to be $z$, then surely $\mathrm{p}_{z}$ and $\mathrm{s}$ orbitals can overlap to form a $\sigma$ bond.


text:
Analysis of metal carbonyl bonding in Fe(CO)5 using molecular orbitals
I am really struggling with this question we have been set in class:
Metal-carbonyl bonding: Conceptionally, one can build up $\ce{[Fe(CO)5]}$ from the $\ce{Fe(CO)4}$ fragment plus free $\ce{CO}$ in order to analyse the bonding between the two moieties. Consider first the HOMO and LUMO of free $\ce{CO}$: Describe both MOs with respect to their $\ce{C-O}$ bonding character (bonding/anti-bonding). Note the type of bond that would be formed if $\ce{CO}$ bonded to another moiety through the HOMO and LUMO respectively.
Predict how the $\ce{C-O}$ bond itself would be affected
(i) if electrons were donated into the LUMO (e.g., from a metal centre);
(ii) if electrons were withdrawn from the HOMO (e.g., by a metal centre).
Next consider the $\ce{Fe(CO)4}$ complex fragment: Does its LUMO possess significant amplitude at the location where the additional $\ce{CO}$ will bind? Does the LUMO’S shape and symmetry (by visual inspection) match the symmetry of the HOMO of $\ce{CO}$? Conversely, does the HOMO of the complex fragment match the LUMO of $\ce{CO}$?
Based on these considerations, describe the bonding and orbital interactions between the $\ce{Fe(CO)4}$  fragment and $\ce{CO}$. Predict how binding to the metal centre would affect the structure and vibrational frequency of $\ce{CO}$.
Compare your prediction with the structural and vibrational data obtained for $\ce{CO}$. $\ce{[Fe(CO)5]}$ and $\ce{[Fe(CO)4]^{2-}}$and rationalise the observed trends.
I know that the HOMO of $\ce{CO}$ is 3σ which is primarily carbon in character and the LUMO is $2π^*$ which is also more carbon in character.
Symmetry is one of my worst topics so if someone could give me a hand on this and could even suggest some further reading material it would be awesome :)

I love symmetry and I can probably help you. Just to check how much you understood though, can you show a molecular orbital diagram of how you derived the HOMO and LUMO of CO? If you don't know how to do that, I will have to make my answer a lot longer...


text:
how to identify reaction medium by looking at the equation ( for h2o2)?
$\ce{Ag2O + H2O2 -> 2Ag + H2O + O2}$
My book says the medium is alkaline as $\ce{H2O2}$ gives $\ce{O2}$ on reduction only in alkaline medium only but $\ce{O2}$ is produced in acidic medium too. Is there really a way of recognizing the medium by looking at reaction equations??

Is there really a way of recognizing the medium by looking at reaction equations??

Yes and no...
For the particular reaction, you have to know some chemistry. In acid solutions hydrogen peroxide acts as an oxidizing agent: $$\ce{H2O2 + 2H^+ +2e^- -> 2H2O}$$
In neutral and basic solutions hydrogen peroxide acts as an reducing agent: $$\ce{H2O2 -> 2H^+ +2e^- + O2(g)}$$
The silver oxide is being reduced so the hydrogen peroxide must be acting as a reducing agent.
The real answer is that the balanced reaction isn't really enough information to duplicate the reaction. What temperature? How long for the reaction to take place? Since reaction would use excess hydrogen peroxide, how much was in excess and so on...
PS - Sometimes you have to consult the oracle to know the laboratory conditions. I asked Carnac the magnificent for more information.

This reaction was done by a red-headed girl. She was wearing a white lab coat over a red blouse and a blue skirt. She did have safety glasses which is a really sexy look over her hazel eyes in the chem lab. She was also wearing white leather sneakers.

My book say h2o2 can act as reducing and oxidising agent in both alkaline and acidic media.

Silver oxide is being reduced. So the hydrogen peroxide must be the reducing agent. In order to reduce hydrogen peroxide in acid you need to use a strong oxidizing agent which can't be silver oxide since it is already oxidized.

Now it's getting clearer. Here's how I should analyze. I look at the equation, see that h2o2 is reducing agent. As h2o2 would like to act as reducing agent in alkaline medium or in acidic medium if it's reacting with a more powerful oxidizing agent, but ag2o isn't a powerful oxidizing group h2o2 must be in an alkaline medium. Wow. Great reasoning. But can you suggest how would I analyze if h2o2 is the oxidizing agent. I just can't think anymore, my brain is overheated right now.

If you were using the reaction in a publication then you'd explicitly specify all the reaction conditions so that a subsequent researcher could duplicate your work. So for example the time, temperature and pH for the reaction would be given in text not just the overall balanced reaction. Real science is about leaving footprints, not creating puzzles - like teachers do to test students... ;-)

Yup. I think I was going in too much detail. Time to stop.


text:
Do absorbances of different compounds at one wavelength sum up?
Question on my practice exam. The answers to the questions were not released, otherwise I could work backwards to figure out how to solve it. As it stands, though, I can only guess the method to solve this question.
The extinction coefficients of the 4 natural nucleoside-5'monophosphates @ 260nm
(Cavaluzzi,MJ,Nucleic Acids Res., 2004, 32,e13).
*pdA = 1.506 E4 L/mol/cm
*pdG = 1.218 E4 --
*pdC = 7.10 E3 --
*pdT = 8.56 E3 --

Find the Extinction Coefficient of a single stranded, 5'-phosphorylated Dickerson Dodecamer sequence d(pCGCGAATTCGC)
If I saw this question on a test, my first instinct would be to add up the values for pdC+pdG+pdC etc...(following the sequence of letters). Or maybe average them all.
We did learn the Beer-Lambert Law used in absorption spectrum, where e was extinction coeffecient: $$A=E(extinction)*b*c$$  but I can see no way to use this equation, since i'm only given a wavelength and a whole bunch of coefficients.
Does anyone know how to solve this? Is there an equation I'm missing?
Thanks in advance.

It would seem that you'd just add them up. Presumably the extinction coefficients are the molar extinction coefficients for the lone precursors. One mole of the stand would contain multiples of each precursor. // You're not asked to solve for the absorbance at all. So you don't need the Beer-Lambert Law.

Does it make sense to add them? I don't understand the concept of extinction coefficient enough to confidently make that decision. What exactly IS an extinction coefficient? Absorbance I can visualize, path length = size of the cuvette, Molarity no problem. extinction coefficient?

[…] I can see no way to use this equation, since i'm only given a wavelength and a whole bunch of coefficients.

We have all the data that we need.
You have already mentioned the Lambert-Beer law.
For the solution of a single compound S in a cuvette of width d, the absorbance $E_\lambda$ at a particular wavelength $\lambda$ is proportional to the concentration $c$ of S.
\[E_\lambda = \log_{10}{\left(\frac{I_0}{I} \right)} = \epsilon_\lambda\cdot c \cdot d \]
A higher concentration of S will result in higher absorbance.
Now imagine that the light-absorbing molecules S are linked together by units that do not absorb the light at the wavelength observed.
Let's compare two equimolar solutions of
$$\ce{-linker-S-linker-S-linker}$$
and
$$\ce{-linker-S-linker-S-linker-S-linker-S-linker-S-linker}$$
The second will show a higher absorbance, since the concentration of S is higher!
Now let's have a look at equimolar solutions of different compounds S1 and S2 and their absorbance.
We have outlined above that the absorbance $E_\lambda$ is proportional to the concentration of the light-absorbing species.
But apparently, different compound to not all absorb the same at a particular wavelength!
Exactly this is expressed with the molar absorption coefficient $\epsilon_\lambda$. The higher $\epsilon_\lambda$, the stronger the attenuation of incoming light. Note that $\epsilon_\lambda$ is

always given for a particular wavelength
a compound-specific constant

Let's tie the pieces together!

When light-absorbing units are linked together by 'blind' linkers, the effective concentration of the absorbing units counts.
Different compounds can absorb at the same wavelength, but due to their specific $\epsilon_\lambda$, they will do this to different amounts.

Consequently, we can sum up the combined absorbances in the following way:
$$E_\lambda =  \left(\epsilon_A \cdot c_A + \epsilon_C \cdot c_C + \epsilon_G \cdot c_G + \epsilon_T \cdot c_T \right)\cdot d$$

@JessL. My pleasure :)

@JessL. Welcome to Chemistry.SE!

There seems to be a typo in the first equation, should it not be $\log_{10}(I_0/I)$, and decadic log as this is the normal way of reporting extinction coefficients?

@porphyrin Yes! Thanks a lot fot spotting the typo!


text:
Potassium Alum Crystals
This is a question regarding the crystallization of $\ce{KAl(SO4)2 . 12 H2O}$. I have an idea of what to do (I'll explain later in the explanation), but I would like to hear other suggestions encase the following ideas do not work out.
I placed a specific amount of the salt into hot distilled water and began stirring, later setting it out to grow my seed crystals. I have been waiting for the past 3 days to receive any potential seed crystals, however, I have not received even one. I noticed that over the 4 days a significant amount of water has evaporated, and at the bottom of the beaker has accumulated a white powder I'm assuming is the $\ce{KAl(SO4)2 . 12 H2O}$. I fail to understand why it has not crystallized into a small singular crystal?
My idea is to scratch the bottom of the beaker (it is new) to provide a location for the crystal to begin it's growth. Another idea pertains to adding more solute ($\ce{KAl(SO4)2}$) into the solution. Honestly, I don't know why I would add more solute if I am seeing the powder fall out of the solution. This should mean it is already saturated. Finally, I am deciding about adding some of the undissolved salt from the packet, into the solution. Hopefully this would provide a nice "infant crystal" (I am aware that there is no such thing an "infant crystal", I simply want to refrain from referring to "seed crystal" since these are not my seed crystals.) for my ions (dissolved solute) to latch onto and form my seed crystals.
Any suggestions would be appreciated.
Just to mention, my final technique drives through a more mathematical approach. I found the molar solubility from the $\mathrm{p}K_\mathrm{sp}$ of the crystal ($\mathrm{p}K_\mathrm{sp}$ of salt from aqion.de) I did my calculations to see a total of 0.3 grams of $\ce{KAl(SO4)2}$ per $1~\mathrm{mL}$ of water to obtain a saturated solution. Aesthetically, this doesn't make any sense to me. I feel the website I used may not have provided accurate experiments to reveal the $\mathrm{p}K_\mathrm{sp}$. If you can link me to any other websites or locations I would be more than grateful.

What's so aesthetically unpleasing about the solubility figure, whatever that might be?

google is your friend https://www.google.com/search?q=Potassium+Alum+Crystals&ie=utf-8&oe=utf-8#q=Potassium+Alum+Crystals+growing

@MaxW lmgtfy and similar references to Google are generally not all too welcome here … just sayin’.

There's a video on this page: https://www.thoughtco.com/is-alum-safe-to-eat-use-607367

Typically, you will want to make sure that your solution is saturated. Don’t add a specific, certain amount; rather, add the solid to hot distilled water until it actually won’t dissolve any more. Then quickly filtrate the hot solution through a hot funnel (don’t burn yourself though, please). Let this filtered solution cool down and you should obtain acceptable seeding crystals in your beaker.
By using this supersaturation technique, you can be sure that

you have a saturated solution so crystallisation can occur immediately and
since you cool down the solution rather quickly and since solubility of aluminium potassium sulphate increases with increasing temperature, it becomes very supersaturated very quickly.

Once you have a seeding crystal, e.g. by this method, you can use the hanging technique mentioned by Carlos with the remaining solution (which is still supersaturated). If you do so, you can use an almost infinitely thin string; a human hair typically works perfectly and is even less visible. Another option is to filtrate the solution again so that the bottom of the flask is clean and place your crystals there to grow. You will need to turn them around every now and again to allow for even growth.
And you can make your aluminium potassium sulphate crystals even nicer by cocrystallising it with chromium potassium sulphate. They crystallise in exactly the same structure but the chromium(III) ion is coloured (aluminium(III) is not).

I have done this experiment a few times in the past, and nice crystals have always been produced.
One thing you might consider doing is to drop a single tiny grain of alum (as small as a sand grain) into your saturated solution, so that the alum crystallizes around it. Don't add "some" alum crystals to the solution; add only a tiny one.
The nicest crystals are produced by means of this technique. The only problem here is that the thread becomes "part" of the crystal, and you may not like the final result, aesthetically speaking. And if the thread has a dark colour (like the one shown in the photo) you may like the final result even less. So if you decide to apply this technique be sure to use a white or even a transparent thread (for example, a thin nylon filament). The "seed" crystal doesn't need to be as big as the one in the photo -- a smaller one will be good enough.


text:
How to calculate the electrical potential and mols of electrons transferred from combined half cell equations?
Given 2 half-cell rxn's and their potentials, balance the full cell and solve for ECell:
$$\ce{Fe^{3+} + e^{-} ->Fe^{2+}}; E=0.68V$$
$$\ce{Cr_2O_7^{2-}+14H^+ +6e^- -> 2Cr^{3-} + 7H_2O};E=1.33V$$
So I know that both these half-cells are Gaining electrons (Reduction), and in order to solve for the full cell I need to flip one. I flip Fe Cell to oxidation b/c it has a lower potential and adjust for number of electrons moving. To Calculate ECell: $E(Cr)-E(Fe)$ despite adjustment of mols.
So my full Cell is
$$\ce{6Fe^{2+}-> 6Fe^{3+} + 6e^{-}}$$
$$+$$
$$\ce{Cr_2O_7^{2-} +14H^+ +6e^- -> 2Cr^{3-} +7H_2O}$$
My question is, why is the formula to solve ECell:  $E(Cr)-E(Fe)$ and NOT $E(Cr)-6*E(Fe)$?
I'm pretty sure that more electrons generally = more electricity, so why is it when I multiply the amount of electrons, I don't multiply my half cell E?

In the top two reactions both these half-cells are gaining electrons electrons which is reduction.

TL;DR
This basically boils down to the fact that $\mathrm{E_{cell}}$ is an intensive property, not an extensive property.
Long Version
Lets define some terms first:
Intensive Property:
A property that doesn't change when the size of the sample changes. For example, density and concentration are examples of intensive properties
Extensive Property:
A property that changes when the size of the sample changes. For example, mass and volume are examples of extensive properties
$\mathrm{E_{cell}}$ is the energy in joules of 1 coulomb of charge (1 volt = 1 joule per coulomb). Hence it is an intensive property. Therefore the number of coulombs of charge or number of electrons doesn't effect the value of $\mathrm{E_{cell}}$.
Another way of thinking of it is that by doubling the amount of moles, the  amount of coulombs and joules also double, hence $\mathrm{E_{cell}}$ remains the same.

Thanks for answering. So the electrical potential stays the same despite the amount of mols of electrons flowing because of how the units work out. I recall a little bit of physics that electrical potential can be equated to the slope of a hill. So adding more of the hill doesn't nescessarily increase the steepness, but overall there will be more velocity when the ball hits the bottom the the hill. Does the analogy hold in this case? would I be correct to say that 100 mols of ReDox solution has a higher "electric power" than 1 mol of the same ReDox solution?

Yes, that is a valid and also a great analogy that also applies to the case of volts. If we were to plot a graph of electrical energy vs number of moles of electrons, the gradient would represent the voltage. Hence increasing the number of moles won't effect the voltage, but gives it greater electrical potential energy.


text:
Autoprotolysis Constant of Water
My chemistry textbook states that $[\ce{H3O+}][\ce{OH-}] = K_w$ in all aqueous solutions, which we know of as $K_w$. However, I was wondering why this was true. The textbook states that it is true because $K_w$ is an equilibrium constant, but why are we allowed to assume that the other ions take no part in this reaction?
To clarify further, consider the following reaction
$$\ce{HCl + H2O (l) <=> H3O+ + OH- + Cl- }$$
Of course this isn't balanced, but why doesn't the hydrogen ion on the left affect the equilibrium constant?

Your textbook is correct, the $K_\mathrm{w}$ expression is as follows:
$$K_\mathrm{w} = [\ce{H3O+}][\ce{OH-}]=1.0\times10^{-14}$$
Since for every $\ce{OH-}$ there must be one $\ce{H+}$ ($\ce{H+}$ in water is synonomous with $\ce{H3O+}$) the mol ratio is 1:1, if moles are equal the molarities are equal. I'm going to substitute $x$ in for both
$$K_\mathrm{w}=[x][x]=[x]^2=1.0\times10^{-14}; x= 1.0\times10^{-7}.$$
So the molarity of $[\ce{OH-}]$ and $[\ce{H3O+}]$ are both $1.0\times10^{-7}$ That's a really really low amount, small enough to ignore.
However, a lot of reactions (especially in Organic Chemistry) you'll see a negative ion, for example, $\ce{Cl-}$ pulling the $\ce{H+}$ off of a water. So when $[\ce{OH-}]$ or $[\ce{H3O+}]$ are needed, water can be torn apart to make it.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

Thanks Martin. I'll be sure to learn Latex before posting again. Sorry you had to go through and correct my post.

You are welcome. I am happy when my edits are appreciated. For me it is no problem, and I am just trying to help. Don't worry too much about it, learn from the edits others make on your posts and you'll do just fine. If you want to play a little, you can use the sandbox on meta.

!!Bing!! !!Bing!! !!Bing!! Congratulations - You're thinking like chemist!!
Your textbook has stated an oversimplification which isn't really true. Let's define $\ce{K_{w}}$ based on concentrations as: $$\ce{K_{w} = [H^+][OH^{-}]}$$
(1) The value of $\ce{K_{w}}$ changes with temperature.
(2) $\ce{K_{w}}$ changes with pressure.
(3) $\ce{K_{w}}$ depends on the total ionic strength of the solution, which includes spectator ions.
See Wikipedia article "Self-ionization of water" section "Dependence on temperature, pressure and ionic strength."
The "real" relationship at any temperature and pressure is given in the goldbook as noted by Martin: $$\mathrm{K^{*}_{w} = 1.00 \times 10^{-14} = a(H^+) a(OH^-)}$$
where $\mathrm{a(H^+)}$ and  $\mathrm{a(OH^-)}$ are the activities of $\ce{H^+}$ and $\ce{OH^-}$. In dilute solution the activity of an ion can be approximated by its concentration. A "dilute solution" is dilute enough if a linear relationship holds between concentration and activity. In general this would be 0.1 N or less.

Wait sorry, I made a false conclusion off of what the chemistry textbook said. It says $[H_3O^+][OH^-] = K_w$ always, but thank you for taking note of this error.

The self-ionization constant of the reaction $\ce{HA + B <=> A- + HB+}$ is certainly defined as $K_\mathrm{w} = a(\ce{A-})\cdot a(\ce{HB+})$ at any given temperature or pressure. See goldbook). OP makes no statement about the book contradicting that. It just simplifies that instead of activities the concentrations are used, and that it is true for any aqueous solution (which is a stretch).

Trying to be a smartass I tripped over my own feet. I think I have it fixed correctly now.

Hm, I'm sorry but now you contradict yourself. Your points 1-3 are absolutely correct. The value of $K_\mathrm{w}$ is only $1.0\times10^{-14}$ for standard conditions. The definition, i.e. product of the activities is generally valid for any pure phase of water. (And maybe you'd try to ping me, so that I can remove my vote)

@Martin - マーチン - Don't know if this will ping you are not. Did you miss that I defined $\mathrm{K_w}$ as depending on absolute concentrations, and $\mathrm{K^{*}_{w}}$ as depending on activities?

It did ping. You write:  The "real" relationship at any temperature and pressure is given in the goldbook as noted by Martin:
$\mathrm{K^{*}_{w} = 1.00 \times 10^{-14} = a(H^+) a(OH^-)}$. This is incorrect, as $1.0\times10^{-14}$ is the value for $25~^\circ\mathrm{C}$.


text:
Why can't we find the density and molarity given the masses of solute/solvent and molar masses?
From 2013 ACS Local Exam:
We have a $\ce{KNO3}$ solution prepared in water, and we know the masses of solute and solvent as well as the molar masses. I know that we can easily find molality, but how come density and molarity cannot be determined?
Since the solvent is water, can't the mass be converted to volume to find both values?

Just how are you going to convert mass to volume, if you don't know density?

As a side note in dilute solutions you could assume about 1.00 kg per L. There are lookup tables of course which you didn't have access. http://chemistry.mdma.ch/hiveboard/rhodium/pdf/chemical-data/prop_aq.pdf#page=17

You know the volume of the original KNO3 and water for the pure substances, but there is a change in total volume when they are mixed to form a solution.


text:
What are the oxidizing and reducing agents in the reaction?

What are the oxidizing and reducing agents in the reaction?
  $$\ce{Pb(s) + PbO2(s) + 2H^+ + 2HSO_4^- -> 2PbSO4(s) + 2H2O(l)}$$

I thought that, since the solid lead is going from $0$ to $2+$ it is being oxidized, and in the half reaction it works with the $\ce{HSO4^-}$, so the $\ce{HSO4^-}$ would be the oxidizing agent.

Sulfur can have different oxidation states, but it is $\ce{SO4^{2-}}$ on both sides of the equation so it isn't being oxidized or reduced.

You are correct, the Pb(s) is oxidized, which makes it the reducing agent. However, there is no change in the oxidation state of the H or the SO4−, so it is not the oxidizing agent. Rather the PbO2(s) is the oxidizing agent, as the Pb goes from +4 in the PbO2(s) to +2 in the PbSO4(s).


text:
reconciling free energy equations
I am struggling with a very basic concept and I think the issue lies in the definition of standard states.
When calculating standard state free energy change via Gibbs equation $$\Delta G^ \circ = \Delta H^\circ  – T \Delta S ^\circ$$  one simply plugs in the standard state entropy and enthalpy, and the corresponding temperature, assuming that $\Delta H^\circ$ and $\Delta S^\circ$ are relatively temperature independent. Since these are standard state values, all species are at $1~ \mathrm{M}$ (in solution). But if all species are at $1~ \mathrm{M}$, then using the equation $\Delta G^ \circ = -RT ~ \ln K$ yields $\Delta G^ \circ =  0$. What am I missing?

K is the equilibrium constant; the concentrations in it are the equilibrium concentrations, and those are not 1.

Any assumption that standard state entropy and enthalpy are relatively constant is doomed to failure. There is a reason that there are publications on, say, the accepted Gibbs free energy functions vs temperature for the elements (A.T. Dinsdale, 'SGTE Data for Pure Elements', CALPHAD 15(4) 317-425 (1991)).

Ivan, I agree that the equilibrium concentrations will not be 1 M. But if dGo is the standard state free energy of formation, and the standard state is the most stable configuration with concentrations at 1 M, then how do I reconcile this definition with the equilibrium constant not simply being K = 1 (because dGo = -RT ln K, so for dGo we have concentrations at 1 M, thus K = 1 by this reasoning)?

'if dGo is the standard state free energy of formation' - I believe that dG0 is not quite that. There is a difference between dG0 for formation and dG0 for reaction. dG0(reaction) = dGo(formation products) - dG0(formation reactants). Also I don't believe it's accurate to say that the standard state is the most stable one. It's just a convenient comparison point. Most of the time, dG0(rxn) != 0 as the equilibrium concentration will not be 1, so at 1 concentration the reaction will want to go forward or backward.


text:
What is the chemistry behind heat produced when blood is bleached?
When I discard blood (20-40 mL) in a beaker containing bleach (400mL), I can feel that heat is produced in the beaker. What is the chemistry behind it? Will heat be produced due to Lysis of haem protein in RBCs? Or any other mechanism?

It is simply because of oxidation. Bleach is an oxidizing agent.

This reaction produces heat for the same reason as burning methane in oxygen. There's a lot of chemical energy stored in the reactants, and comparatively little in the products.
The hypochlorite is oxidising the organic content of the blood - not just breaking proteins but then breaking down the amino acids too, as well as anything else that happens to be in the sample.


text:
How to choose the correct volumes when preparing a dilution series?
Original Question

A solution of dye, Amido Black 10B dye ($10~\mathrm{mM}$, $6.165~\mathrm{g\over L}$), as-provided absorbs light much too strongly for measurement of an absorbance value. It is therefore necessary to dilute it before proceeding with an experiment to construct a standard curve of absorbance vs. concentration or amount of dye. The stock $10~\mathrm{mM}$ Amido Black solution needs to be diluted $200\times$ using purified (RO) water. To achieve the dilution, it is recommended that serial dilutions be used.
How would you dilute the Amido Black 10B $200\times$ to a total volume of $3~\mathrm{mL}$? Assume you only have pipettes that cannot dispense less than $20~\mu\mathrm L$.

Comments
We've already been through this question together in class but despite asking a lot of questions I still didn't understand what my tutor was trying to explain. I already have the answers, but I want to know how to figure it out for myself. My tutor's way was this:

Adding $50~\mu\mathrm L$ of dye into $50~\mu\mathrm L~\ce{H2O}$ gives a $1$ in $2$ dilution.
Take $30~\mu\mathrm L$ of the new solution and add $2970~\mu\mathrm L~\ce{H2O}$, giving a further $1$ in $100$ dilution.

I do not understand where she got $50~\mu\mathrm L$ of dye and $50~\mu\mathrm L$ of $\ce{H2O}$ if it's $1\!:\!2$. I know she split $1\!:\!200$ into $1\!:\!2$ first because the pipettes can't dispense less than $20~\mu\mathrm L$.  I also don't understand where she got the $30~\mu\mathrm L$ from in the second step.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. This appears to be a homework question, please share your thoughts and attempts towards the solution. I am sure you can figure out how much the 200 fold dilution of your mother solution is. Just figure out how much water you would need to dilute it in one go, include it in the post, and we'll reopen it to cover the rest.

It is impossible to answer this since you don't specify what pipettes and volumetric flasks you have to use. It would seem reasonable to assume that you have more than enough stock solution to prepare the 3ml of dilute solution.

To your questions:

I do not understand where she got $50~\mu\mathrm L$ of dye and $50~\mu\mathrm L$ of $\ce{H2O}$ if it's $1\!:\!2$.

It's always problematic expressing dilutions as ratios, because it's ambiguous as to what the numbers are referring to.  Here, the "$1$" is referring to the volume of the solution to be diluted, and the "$2$" is referring to the final volume after dilution.  Thus, the $50~\mu\mathrm L$ ("$1\times$") is being diluted to $100~\mu\mathrm L$ ("$2\times$"), and in order to do that you need to add $100~\mu\mathrm L-50~\mu\mathrm L=50~\mu\mathrm L$ of the RO $\ce{H2O}$.  This could just as validly be expressed as a $1\!:\!1$ dilution; thus the ambiguity.

I also don't understand where she got the $30~\mu\mathrm L$ from in the second step.

Per the same $\mathit{Initial~Volume : Final~Volume}$ ratio convention, a $1\!:\!100$ dilution to a final volume of $3~\mathrm{mL}=3000~\mu\mathrm L$ will require starting with $3000\div 100 = 30~\mu\mathrm L$ of the intermediate solution.

More generally:
I prefer to express dilutions using an "n-fold" notation, e.g.:

Dilute $50~\mu\mathrm L$ of dye solution $2$-fold with RO $\ce{H2O}$
Dilute $30~\mu\mathrm L$ of the solution from step ($1$) $100$-fold with RO $\ce{H2O}$

Though there is still room for ambiguity in this phrasing, to me diluting $n$-fold indicates much more clearly that the final volume $V_f$ should be $n\times$ the initial volume $V_i$, and thus that the volume of diluent needed is:
$$
V_d = V_f-V_i = nV_i-V_i = V_i\left(n-1\right)
$$

The whole problem is contrived to test you knowledge of dilutions.
(1) Assuming that you had a 15 μL pipette and a 3ml volumetric flask then you'd just take 15 μL from the stock solution and add that to 3ml volumetric flask and fill with water to the mark. Done...
(2) The solution using a 50 μL pipette and then a 30 μL pipette is sort of stupid. You don't need two different pipette sizes. You could mix 30 μL of stock solution with 30 μL of water to get a solution with half the concentration. Taking 30 μL of the 1/2 concentration solution and diluting to 3ml would give you another 100 fold reduction in concentration. So overall the two steps would reduce concentration to 1/200th of the original.


text:
Addition of anodes and cathodes in electrolysis
I am doing a research on the production of hypochlorite through an electrochemical cell using a brine feed. I know that using using an undivided cell would allow for the chlorine atoms that was oxidised at the anode to react with the $\ce{OH-}$ ions in the solution. However, I was just wondering what the addition of more anodes and electrodes would do to the product?

You might find NOTE ON THE ELECTROLYTIC PREPARATION OF
DILUTE SODIUM HYPOCHLORITE SOLUTIONS
(DAKIN'S SOLUTION), J. Biol. Chem., 1919, * 37*, 519-523 helpful.


text:
Separate toluene from acetone and isopropyl alcohol
I found a cellulose thinner that is made of toluene 60-80%, acetone 10-20%, and 2-propanol 5-10%. As acetone and 2-propanol are water soluble, but toluene isn't, would it be possible to just add a lot of water and then remove it, and then end up with just toluene?

This should work, though not as perfectly as it might seem.

It may work, but fractional distillation would be better.

Multiple washes with smaller quantities of water will be more efficient than "just adding a lot of water" in a single wash. The toluene will be "purified" further with each successive wash. Some toluene will be lost with each wash also to the degree of it's solubility in water at the temperature of the extraction (about 0.03% soluble in water at room temperature). As stated in a comment however, distillation would be a superior method for recovering the pure toluene.


text:
Voltage sensitive compounds that undergo a conformational change?
In nature, the Ca/Na/K channels which consist of groups protein are often voltage sensitive and undergo conformational changes in response to voltages on the order of mV.
Are there any known classes of other simpler organic or inorganic compounds which also undergo rapid conformational changes in response to applied voltages?

http://pubs.rsc.org/en/content/articlelanding/2007/cs/b604484k#!divAbstract you might be interested in these. No time to summarise now - someone else feel free to use it.

bon. Thank you for the general site information.

orthocresol. Thank you for the interesting link. Had not considered the molecular machine aspect.


text:
How accurate would be the following conclusion on 'Critical Temperature'?

"Boiling does not occur when liquid is heated in a closed vessel. On heating continuously vapour pressure increases. At first a boundary is visible between the liquid and vapour phases because liquid is more denser than vapour. As temperature increases more and more molecules go to vapour phase and density of vapours rises. At the same time liquid becomes less dense. It expands because molecules move apart. When the density of liquid and vapours becomes the same; the clear boundary between the two phases disappears. This temperature is called critical temperature"

That is EXACTLY what my textbook says. And that is where I am having doubts. It's because - a gas might still be found in its gaseous phase below the critical temperature if the pressure is considerably below its critical pressure ($P_c$) and it has a volume sufficiently larger than its critical volume ($V_c$).

Just consider any point in the Pressure-Volume isotherms of $\ce{CO2}$ given above, which is in the region of gas-liquid equilibrium (within the shaded dome shaped curve) but has a volume greater than the critical one. For e.g a point that lies

Within the shaded region.  and
On a line passing through B and parallel to the pressure axis.

Let that point represent the state of liquid inside the vessel.
From there if we increase the temperature, we would arrive at a point on the boundary of the dome shaped curve, (viz. B) but that point would be on a lower isotherm than the critical isotherm. And it looks like B is also a point where the two densities becomes equal.(isn't it?)
Based on that premise, we can't say that the temperature arrived at in the above (heating liquid in a clossed vessel) situation would necessarily be the Critical temperature.
What I want to ask is if my reasoning is correct or am I missing anything?
My question can be rephrased in the following manner also
Is it true that critical temperature is the only temperature at which the density of the gaseous phase of a gas becomes equal to its liquid density?

Providing that there is enough liquid in the vessel to begin with, your book is correct.

"Providing" of course is his quibble, so he is right. However if the liquid were evaporating then you'd see the volume of liquid getting less and less as the temperature is raised until it was all gone. The book is proposing that a significant volume of liquid just all of a sudden disappears. For example a container half full of liquid just below Tc.

What i am saying is that i found the quoted paragraph in my textbook and i am not sure if the information provided is sufficient to conclude that it would surely be the critical temperature whenever the phases become indistinguishable, because gas below the critical temperature is also not always in the liquid phase and requires sufficient pressure to reach the equilibrium phase.
Note: Right now i am not in a position to upload a diagram to make my point clearer to others. I would try my best to describe it with words.

Could you please insert the citation to the textbook you use for proper attribution and transparency.

Here. But, frankly it is not just my book that says so. I've found similar paragraphs in several places too. Which makes me believe that i might have missed some points.

@Phill2 it is not about the contents of the passage or the generality of such a statement. You quite verbatim from a certain source and proper scientific conduct is to reference this. We use a model here that also requires attribution if anyone copies content, hence it is only fair to require proper attribution from the users.

Your question refers to points "in the shaded region".  This region does not represent a stable state; in that region dP / dV (at constant T) is positive; this violates principles of thermodynamic stability.  Points in the shaded region will result in phase separation into two phases, marked "B" and "C" in the text.  Since density is reciprocal volume (as long as n the number of molecules total is constant), your book is correct, at least as far as I can see.

In other words, points on the line "BC" will represent a two-phase system, with variable proportions of the two phases, but with phase densities given by point "C" for the liquid and "B" for the vapor phase.

Ok, I added a point Z to the diagram which I've added below, and I'm going to quote you text in italics below the image...

Just consider any point in the Pressure-Volume isotherms of $\ce{CO2}$ given above, which is in the region of gas-liquid equilibrium (within the shaded dome shaped curve) but has a volume greater than the critical one. For e.g a point that lies:

Within the shaded region. and
On a line passing through B and parallel to the pressure axis.

Let that point represent the state of liquid inside the vessel.
Question 1 - My point Z meets your two conditions doesn't it?
... From there if we increase the temperature, we would arrive at a point on the boundary of the dome shaped curve, (viz. B) but that point would be on a lower isotherm than the critical isotherm.
Assuming you agree with point Z, I agree with the idea above. The next two sentences is where you're wrong.
And it looks like B is also a point where the two densities becomes equal.(isn't it?) Based on that premise, we can't say that the temperature arrived at in the above (heating liquid in a closed vessel) situation would necessarily be the Critical temperature.
Let's start at point A, and follow the purple isotherm. At point A all the carbon dioxide is gas. At point B all of the carbon dioxide is still gas. But as you start to goto point C from point B the carbon dioxide vapor becomes supersaturated and liquid carbon dioxide forms. Now when you get to C there is so little volume that all the carbon dioxide is liquid. Thus only in the blue shaded area under the curve do you have two phases gas and liquid.
Liquid carbon dioxide is denser than gaseous carbon dioxide. So the gas will be on top and the liquid below.

To your question: yes, Z is the point..
Regarding your answer: I am getting it now. But could you tell me what is supersaturated? Is it same as simply saturated? Assuming it is, if we reach point B, we get a vessel full of saturated vapour. Phase distinction is still not lost at that temp. because we can get liquid if we want by reducing the volume. Now, above B, our vessel is full of unsaturated vapour for each corresponding higher temp. untill we reach the critical isotherm and that is when the two phases literally merge, and their densities become equal. Am I correct now?

Supersaturated in this case means that there is too much carbon dioxide in the gas phase and some of the carbon dioxide has to go into the liquid state to reach equilibrium between the gas phase and liquid phase. (Like fog forming when temperature drops outside.) // The gas density and the liquid density never will be equal. At point E carbon dioxide is one phase, a "supercritical fluid", which is different than a gas or a liquid.  https://en.wikipedia.org/wiki/Supercritical_fluid

Also, if any gas above its critical temperature and pressure is called supercritical fluid, what is it called when its above critical temp. but below critical pressure?

A "liquid" see phase diagram for CO2 https://en.wikipedia.org/wiki/Supercritical_fluid#Phase_diagram

Its gas not liquid.. :)

@Phill2 - Mea culpa - yes you are absolutely correct. It is a gas.

Lets pick up any point in the gas-liquid equilibrium. That would be any point on an isotherm below critical temperature but the point should be in phase equilibrium. Now when we are increasing the temperature while keeping the volume constant, we are increasing pressure. Now as we increase pressure we are moving the point to the isotherm of a little higher temperature. Now if we keep on increasing the temperature, we will reach on a point on the critical isotherm. The point will be below the critical pressure, not the critical temperature. Further heating would make it supercritical.

Below the critical pressure, not temperature? Could you elaborate.. There's a dome shaped region in the isotherm that represents the gas liquid equilibrium phase and that has its peak touched by the critical isotherm.  Now, isnt it true that the boundary of that dome on the right side of its peak represents the points where the line  between gas and liquid phase disapear? Note: Right now i am not in a position to upload a diagram to make my point clearer to others. I  would try my best to describe it with words.

I am unsure about some details on what you exposed. For any two points have the same density in the diagram above, they have to be located in the same vertical line.  The right hand side of the dome line correspond to the volume of pure gas in hypotetical equilibrium with the liquid (both have the same chemical potential). Notice that the total amount of matter is fixed. Left side of the curve correspond to liquid volume. Below critic temperature, liquid is denser than vapour. As temperarure increase they densities get closer up the critic point that they turn to be the same.

Bear with me on this one. We chose a point in the shaded region and just vertically below the point B. Now keeping the volume constant we start heating. As random motion of particles rises, we get an increase in pressure. Pressure keeps increasing until it reaches the point B. Here the liquid is in equilibrium with gas. The distinction between liquid and gas phase is there. But if we keep heating further the equilibrium mixture turned completely into gas and on further heating pressure will reach upto a point on the dark blue isotherm. All points on an isotherm i.e. the dark blue line are on the same temperature, hence our fluid reached the critical temperature. Distinction is lost when we get above that dark blue line. The shaded region is not an isotherm. It represents the region where gas and liquid are in equilibrium and the boundaries of the shaded region represent the point where our substance just started phase change and got into equilibrium. Phases are distinct. Isotherms are the lines on the graph and the critical isotherm ( dark blue line, damn it how many times do I have to say that) is the deciding factor. Above it our substance behave as a supercritical fluid ( has the properties of both liquid i.e. solubility and gases i.e. mobility).

The "shaded region" is an unstable state and represents a region where vapor an liquid are in equilibrium.
Your question refers to points "in the shaded region". This region does not represent a stable state; in that region $\left(\frac{\partial P}{\partial V}\right)_T$ is positive; this violates principles of thermodynamic stability. Points in the shaded region will result in phase separation into two phases, marked "B" and "C" in the text. Since density is reciprocal volume (as long as n the number of molecules total is constant), your book is correct, at least as far as I can see.
In other words, points on the line "BC" will represent a two-phase system, with variable proportions of the two phases, but with phase densities given by point "C" for the liquid and "B" for the vapor phase.
The Point "Z"
Other answers mention a new point "Z".  This is also in the shaded region.  Therefore it represents a region where phase separation into liquid and vapor occur.
The impossibility of liquid and vapor densities being equal...
The liquid density is given by the x-coordinate of "C"-like points, i.e. points on the left side of the edge of the shaded region.
The vapor density is given by the x-coordinate of "B"-like points, i.e. points on the right side of the edge of the shaded region.
Since the right side of the shaded region always has a higher x-coordinate (in this case, x represents volume), the vapor density will always be greater than the liquid density.
As temperature rises, the difference in densities of vapor and liquid gets smaller and smaller, as we ascend the right side (vapor phase) and left side (liquid phase) of the shaded region.
...except at the critical point
When we reach the critical point, the difference between liquid and vapor density disappears.  Because the shaded region is convex, there is no possibility of this occurring anywhere else.


text:
Denatured Ethyl Alcohol Purification
I have denatured ethyl alcohol which was obtained this formula;
In 100 L ethyl alcohol there is 0.8 gr denatonium benzoate and 78 gr tert-Butyl alcohol.
Is there any way to obtain absolute ethyl alcohol from that mixture.

What applications are you conceiving needed absolute ethanol for? Most things I can think of (besides drinking and spectroscopy) should tolerate those two impurities.

I'd never try to "undenature" ethyl alcohol to drink it. // Denaturants are chosen specifically to be hard to remove. The denatonium benzoate could probably be removed effectively by distillation which wouldn't remove the tert-Butyl alcohol which is poisonous for animals (including humans).

I suggest purchasing Everclear (95% $\ce{EtOH}$) and using calcium chloride to dry it.

To remove tert-butanol you could add hydrochloric acid to selectively produce tert-butyl chloride, which would be easier to separate.

@A.K. after adding hydrochloric acid is it possible to remove mixture from tert-butyl chloride with distillation and also denatonium benzoate? and also how is it work other method with everclear?

Please note that the following is not meant to be used as an instruction!
According to an article on Detection of Denatonium Benzoate (Bitrex) Remnants in Noncommercial Alcoholic Beverages by Raman Spectroscopy, published in J. Forensic Sci. 2014, 59, 1358-1363
(DOI), the manufacturers of illegal alcoholic beverages often use sodium hypochlorite ($\ce{NaOCl}$, household bleach) to remove the bitter taste of tax-free technical ethanol.
In the study above, the authors used legally available legal alcoholic beverages (vodka, cognac, etc.), and denatured/de-denatured them under controlled conditions:

The separate batch of purchased alcoholic beverages was
  denatured with 20 mg/L of denatonium benzoate, after which 1.33 mL/L sodium hypochlorite was added.

Spectroscopical measurement of the samples thus obtained clearly showed degradation products of the bitterant.
While hypochlorite treatment might be suitable to remove the bitter taste, this doesn't imply that the result is healthy or recommended to consume.

The hypochlorite treatment wouldn't remove the tert-Butyl alcohol which is poisonous.

@MaxW "Poisonous" is a bit exaggerated, isn't it? LD50 (oral) is about 2.7g/kg in rats. It is H332 (harmful if inhaled), not H330 ot H331.

actually I dont have problem with bitter taste. Other ingredients apart from ethanol brokes my sensitive calculation and almost impossible to find pure ethanol because of idiot regulation for tax here. There is only way to make own ethanol via fermantation. But takes a lot of time and so tiring for me. That is why I am asking is there a way remove other chemicals from the denatured alcohol.

Does not address the specific question, but rather a comment on the original:
Use Benzene-Water-Ethanol ternary azeotrope (bp 64.9, 74.1% Benzene, 7.4% Water) to dehydrate Everclear.  You'll need to add 10 times as much benzene as water, (therefore roughly half the total amount of liquid you are dehydrating).   Adding excess benzene leads to the Benzene Ethanol azeotrope (bp 67.8, 67.6% benzene), so err on the side of slightly more than half the volume of alcohol.
Both of these azeotropes are well less than the bp 78,2 for ethanol, so with some good technical skill you will be able to distill this.  You will lose roughly 15% of the alcohol in the azeotropes.
This is more laborious than using CaCl2.


text:
Solublity values interpretation
"Solubility:    Soluble in water (>25 mg/ml), and DMSO (50 mM)."
It happens to be talking about a drug but that's irrelevant to my question.  Does this solubility statement reasonably indicate that I can dissolve 10mg/ml in water and expect it to stay in solution?  if possible, what does the dmso bit mean?  (maybe mM is a typo of mL?)

Yes at least 25 mg should dissolve per ml of water. Dimethyl sulfoxide (DMSO) is a common organic solvent.

@MaxW You should post that as an answer.

wow, great answers everybody, thanks!  I'm asking this as a guest, so I don't seem to see an "Accept Answer" button, which I would click if I could.  all my q's have been answered in prefect detail.  thank you, people!

Yes.
The compound is soluble in water to at least 25 mg/ml - which is (obviously) the same as g/l. Presumably they didn't test at higher concentration.
Alternatively, it is soluble to 50 mM, i.e. mmol/l, in pure DMSO.
Why they would use different units is beyond me,  but I would guess this has to do with the units conventionally used in the two experiments they've run.

and mM means milli-molar


text:
Find the molarity of the filtrate
I can't figure out this stoichiometry question!
10.0 mL of 1.15M copper(II) nitrate reacted with 20.0 mL of .85M ammonium hydroxide.  The solution is filtered. Determine the molarity of the filtrate.
I figured out that the net ionic equation is Cu + 2OH -> Cu(OH)2, so the 2NH4NO3 dissolves, and the Cu(OH)2 probably doesn't pass through the filter, but I don't know where to go from there. Thanks in advance!

The question is poorly worded since it doesn't specify for which chemical species the molarity is desired. Copper(II) would seem to be the species of interest.

$$\ce{Cu(NO3)2}+2\ce{NH4OH} =>\ce{Cu(OH)2}(s)+2\ce{NH4NO3} $$
Essentially, we're trying to find the Molarity of $\ce{NH4NO3}$ since the copper filters out(as you stated)
First solve for limiting reagent. $$10mL\ce{Cu(NO3)2}*\frac{1L}{1000mL}*\frac{1.15mol}{1L}=0.0155mol\ce{Cu(NO3)2}$$
$$20mL\ce{NH4OH}*\frac{1L}{1000mL}*\frac{0.85mol}{1L}=0.0170mol\ce{NH4OH}*\frac{1}{2}=0.0085$$

Why do we divide mols of $\ce{NH4OH}$ by 2? Remember that for every molecule of $\ce{Cu(NO3)2}$, there must be 2 molecules of $\ce{NH4OH}$. $$\frac{2\ce{NH4OH}}{\ce{Cu(NO3)2}}$$ for this reason we need to divide the amount of $\ce{NH4OH}$ by 2 to compensate for stoicheometry.
So actually, $\ce{NH4OH}$ our limiting reagent.
Since you didn't give my any solubility constants, I'm going to assume 100% of the Copper became  $\ce{Cu(OH)2}(s)$ and all of the ammonium became $\ce{NH4NO3}$. In the real world, perciptates never form at 100%.

0.0085 mols of $\ce{Cu(OH)2}$ is produced. You want $\ce{NH4NO3}$  Can you use stoichometric ratios to proceed from there? Molarity - (mol/L). Make sure you add up the two volumes to find the Total volume.

You assumption that all the copper was converted to copper hydroxide is wrong.

Yeah I know. but usually in HS chemistry when students are learning stoichometry, they generally ignore Solubility constants and whatnot. I'll edit my post a little.

I still can't figure it out. I found that .83 grams of Cu(OH)2 was produced using stoichiometry, but I don't think this would help. I don't know what to do with the Total volume of 30 mL. I also figured out that .003 moles of excess reactant remains after the reaction, so that might affect my answer. The answer key says  the answer is .10M

Whoops. That's embarrassing- looks like I didn't help much. In that case use the numbers you got.  0.003mol/30mL= .003mol/.03L . =0.10 mol/L = molarity of excess reactant in solution.

Thank you! I think I was confused about what molarity meant (since we learned about it a while ago), but now it makes sense.

:) Glad I could help, even it's just a bit. ;)

Dah... the answer is wrong. Assume all ammonia is present as species $\ce{NH4^+OH^{-}}$. So there is 0.0085 mols of $\ce{Cu(OH)2}$ produced, but there is 0.155 - 0.0085 = 0.070 moles/L of Cu left in solution.

That's what I thought! Maybe there was more to the question that he didn't post.


text:
Real Gas Equation
When a real gas has a pressure lower than the expected value from ideal gas (because of the intermolecular force of attraction) , we have -
$$P_{\text{ideal}}= P_{\text{real}} + \frac{an^2}{V^2}$$ where a is the Van der Waal's constant for intermolecular forces of attraction between a gas.
Similarly, a real gas has a lower volume than the calculated volume in ideal gas. Here, when we say volume , we mean the empty space around molecules or atoms.
So, for volume, we have -
$$ V_{\text{real}}=V_{\text{ideal}}-nb$$
where b is the Van der Waal's constant for the volume of a molecule or atom.
Having said this, when we plug in these values into the ideal gas equation to get the real gas equation.
We put $\rightarrow$
$$ \left(P+\frac{an^2}{V^2}\right)(V-nb)=nRT $$
But shouldn't this be wrong? Because for pressure, we put $P_{\text{ideal}}$ but for Volume we put $V_{\text{real}}$. Shouldn't we put both $P_{\text{ideal}}$, $V_{\text{ideal}}$ so as to satisfy ideal gas equation?
Why do we not do so?
Please give a theoretical explanation along with a mathematical one if possible.

You have the volumes wrong. The ideal gas law is: $$\mathrm{ P_{ideal}V_{ideal} = nRT}$$
So: $$ V_{\text{ideal}}=V_{\text{real}}-nb$$ when compared to Van der Waal's equation. Thus the "ideal" volume is less than the "real" volume because the molecules take up space. At least that is the rationale.
I think the "truth" is closer to the fact that the extra terms allow more constants to be fit to a particular gas. There are a number of "real gas" equations basede on various assumptions about how a "real" gas would behave. Which equation works best for a particular gas is a matter of the temperature and pressure range. So you have to try each and just see which one works the best.

So you're saying that Ideal gases have lesser volume than real gases?  The real gases have considerably large sized molecules and thus the net volume for them to move is lessened. Define "Volume" of a gas. Is it the space left empty ? or the space occupied by the particles?

I was regurgitating the assumptions behind Van der Waal's equation. An ideal gas is supposed to contain gas molecules which are points in space with no volume, not molecules of finite volume. So an "ideal gas" is supposed to occupy 22.4 liters at STP.  But the "real gas" molecules occupy some of that volume, so the "empty" space is less than 22.4 liters.  Look at a quick dimensional analysis. In the correction the factor $n$ is the number of moles of gas and $b$ is the volume per mole of the molecules.

22.4 liters is 22.4 liters. But if you put a mole of real gas let's say the molecules occupy 0.2 liter of the volume. Then the empty space is only 22.2 liters, but it should be 22.4. So you let the pressure drop and the gas expands to 22.6 liters. Now we have 0.2 liters for the molecules and 22.4 liters of empty space.