text:
Metal rod and NMR machine
I have a metal rod in my leg from previous surgery. I believe it is titanium. Would this significantly affect/be affected by an nmr machine? I noticed that a sign by it said not to bring metal objects in the room and I wanted to be sure.

You should speak to the people who run the NMR facility at your department.       Very old machines were poorly shielded, making it impossible to go near them with metal implants/pacemakers. Modern machines are highly shielded to the point where you can get very close or even touching without risk of harm.   In most NMR suites there are lines drawn on the floor/ropes tied around to show where the magnetic field of the machine extends to. But in any case. Ask the NMR guys at your institution... modifications are often made to NMR spectrometers so they might not be as the manufacturer intended.

A little bit about me - I manage a busy research NMR Facility at a leading international tertiary institution. I deal with questions like this not infrequently, and there are some very important things you need to understand about implants and magnetic fields. I have all users of our Facility sign a waiver prior to access to indicate whether they have medical or surgical implants, and for those that do - there is probably 3-4 a year - we make a specific case assessment to determine whether it is safe for them to use the instruments. For some people, we simply arrange an alternative pathway to have samples run. This may be having others run them altogether, or load up samples into an autosampler for them.
A bit about you - that you 'think' it is a titanium implant is not sufficient to allow a proper assessment in your case. Unless you can get this is writing from your doctor, then I would have to assume a surgical grade of stainless steel. And it needs to be made perfectly clear that most surgeons don't know this information. There is no such thing as completely non-magnetic stainless steels - even the highest grade austenitic steels contain minute amounts of ferrite materials. I have had many examples over the years of people believing one thing about their implant and then discovering the truth later. In the most interesting case, I had a user who had surgery and believed they had a titanium plate in their skull. Their doctor put this in writing. Only after a follow-up MRI scan some years later did they discover that they did NOT have any implant. Nothing at all. The doctor checked his records.....ah yes - they didn't need to use one after all! Some implants have additional attachments and electronics that meed to be considered above and beyond the basic 'rod' and these can both be affected by, and affect magnetic fields.
Typically, implants will rarely affect the magnet of the spectrometer to the point of creating problems. In any case, problems only arise through moving through the field - a static object is generally no problem. With MRI, this 'movement' comes from the application of field gradients. Just as passing a current through a coil generates a magnetic field, moving a metal conductor through a magnetic field creates an electric current.
Even surgical grade metals are affected by the stray fields of magnets, and depending on the age, design and size of magnet, this may be of concern to you as a spectrometer operator. Movement of metals through the field will create local eddy currents which can result in localised heating. In MRIs, a common incident is serious burns from the embedded metal in bra straps. For biomedical implants, this risk is still considerable, and consideration should be given to:

how long ago the surgery was done. Recent surgery may not have allowed adequate time for bone growth to sufficiently embed the implant, and heating can create swelling and cause problems that may need to rectified. I am aware of one such case with an implant in the wrist that required follow-up surgery to rectify intense swelling that resulted from exposure to a strong magnetic field.
where the implant is. Some areas will be more susceptible to problems, or least symptoms. Not everyone that has an implant can detect any changes. Some people do. I'd say that most people who say they experience a symptom have implants associated with joints - elbows and wrists are the most common that come to mind. Symptoms have ranged from tingling in the joint to feeling warm. How much of this is genuine, and how much psychosomatic, I can't say. I once had an older tradesman working within the stray field for a day who commented that the tattoos on his forearm were itchy by the end of the day.
what the implant is. Some implants are just not worth taking a risk with. Aneurysm clips, for instance, are simply not worth taking a risk with. Finding an alternative way for a user to access the spectrometer is much easier than cleaning up an exploding head (yes, an exaggeration, I know) and doing the paperwork. Some pacemakers are MRI approved, and most are actually perfectly ok within the small stray fields of a spectrometer. But some aren't ok, so I wouldn't take this risk without plenty of written evidence from doctors and medical suppliers.
the extent of stray field form the magnet. All magnets are different through age, design and magnitude. An older unshielded 600 poses a very high dose of exposure; a newer shielded 300 has the 5G line contained within the magnet case, and has an almost neglible exposure. However, magnetic fields are invisible, and unless you know the details about the specific magnet you are using, it is best to assume the worst case. We base all of our stray field limits on our unshielded 400, which has a much greater stray field than our 500 for instance.

Regardless of the final assessment of your condition, I would still advise against undertaking high field procedures, such as changing probes, or doing low temperature connections etc.
Ferromagnetic materials can obviously be a problem because of the large attractive forces created. Most incidents with these arise from large external metal objects being brought within the stray field; gas cylinders and trolleys etc. MRIs are usually more problematic with stray fields due to the bore size, but small objects can also pose a serious concern in strong fields. These include handheld tools, but for NMR spectrometers the biggest problems actually come from things like staples and paperclips and even iron filings that migrate across the floor over time and can cause serious problems if they get near the magnet bore.
The take-home message for you is not to take advice from a bunch of stackexchange users beyond understanding that you need to discuss your individual case with your NMR Facility Manager, your doctor, and quite possibly your supervisor if alternative arrangements need to be considered for your access. A great resource for you to be aware of is Frank Shellock's MRI Safety site.

The only way to know this is to ask your physician. You need to know exactly what the implant is made of to assess the danger. You shouldn't rely on your own potentially unreliable understanding of the material of your implant nor on the advice some random people on the internet give you. The following is for your curiousity, but you shouldn't rely on this for your own safety.
Titanium implants can be safe for MRI, see this study as an example. They do interfere with the measurement, but they typically aren't attracted strong enough by the magnet to be dangerous. Though there is also the additional danger of induced currents by magnetic fields with implanted metal objects, which leads to heating.
The magnets inside an NMR spectrometer are very strong, potentially much stronger than a typical MRI. The amount of shielding of the magnet can also vary a lot, so some spectrometers have a much wider area around them with a strong magnetic field.
The warning are really about your own safety, the presence of metal near the spectrometer is much more dangerous to you than the potential of messing up a measurement. Though a large enough metal object can damage or destroy an NMR spectrometer if it gets near enough.
Incidents with MRIs happen pretty regularly, see this post for some examples, e.g. a patient getting hit by a flat screen or a technician with scissors in they head. The danger is mostly from getting hit by flying metal objects that are brought near the magnet.


text:
Gaseous State(Dalton's Law)
Why does the mixture of gases $\ce{NH3}$ & $\ce{HCl}$ not obey Dalton's Law of partial pressure at room temperature?
Also, what other mixtures don't obey Dalton's Law of partial pressure

Your second question  could make this post too broad.

Maybe because these gases react, forming solid ammonium chloride? See here: https://www.quora.com/What-happens-when-you-mix-hydrochloric-acid-with-ammonia

Dalton's Law of partial pressures is not applicable to mixture of reacting gases.

As defined in an excerpt from this Wikipedia entry:

"In chemistry and physics, Dalton's law (also called Dalton's law of
  partial pressures) states that in a mixture of non-reacting gases, the
  total pressure exerted is equal to the sum of the partial pressures of
  the individual gases."

Where this assumption falls apart for your question is that $\ce{NH3}$ and $\ce{HCl}$ readily react with each other, both in the gas and aqueous phases.  This alone renders Dalton's Law mute.

Ammonia and hydrochloric acid react spontaneously when put together within the same apparatus. If they react, then the volume obviously changes, which will not result in the pressure that you would predict if the two gases didn't react and remained as separate entities.

According to dalton's law " total pressure of a gas is sum of partial pressure of individual gases" but he clearly mentioned gases to be "non reacting". As
$\ce{NH3 + HCl -> NH4Cl}$
  So dalton's law not followed as they react.
And also their pressure  becomes zeo

source: Stack Exchange

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

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

sort: votes


text:
Is Nitroglycerine a Nitro Compound or an Ester?
I looked up nitroglycerine in a science dictionary (Oxford Dictionary of Science, 2003), and it explicitly mentions that nitroglycerine is an ester and not a nitro compound.
However my teacher told us in class, that nitroglycerine is a nitro compound (She didn't say it wasn't an ester though …).
The definition for a nitro compound (courtesy Wikipedia) is:

Nitro compounds are organic compounds that contain one or more nitro functional groups ($\ce{−NO2}$).

Assuming that definition is correct, as well and looking at the structure of nitroglycerine, it appears (to me) that nitroglycerine is both an ester and a nitro compound.
So who's right here?

The Oxford dictionary is correct.
A nitro compound is defined by the $\ce{-NO2}$ functional group, but only if that group is attached to a carbon atom. Once an additional oxygen is added, the molecule is basically inorganic nitrate and thus belongs to a wildly different compound class.
You think that is confusing? Well, an alcohol is defined by a hydroxy group — but only if that hydroxy group is attached to a carbon, which in turn is not attached to a different oxygen, nitrogen or sulphur atom, and which is not part of an aromatic system. Even more constraints in there.

According to the Blue Book and the Glossary of class names, nitro compounds contain a $\ce{-NO2}$ group that can be attached to any atom. Thus, nitrates are nitro compounds. However, it is correct that, without further specifications, carbon-nitro compounds are usually implied. Furthermore, nitro compounds are not necessarily named by means of the prefix ‘nitro’. In particular, it is correct that nitroglycerine is systematically named as ester of nitric acid (i.e. nitrate).

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


text:
Carbometalation of alkenes
Alkynes are capable of undergoing carbometalation, e.g. with organocopper reagents:

Why don't the corresponding alkenes undergo analogous carbometalation reactions? Is β-elimination the issue?

In carbometalation, an alkyne will become an alkene before the hydrolysis occurs (in the TS with the metal).  Using the same logic, alkene will become an alkane.  Pi bonds are higher in energy and more easily accessible spatially compared to sigma bonds, pi backbonding is also is present in alkene metal complexes.  For these reasons, the latter reaction suffers from a high activation energy barrier compared to the carbometalation of alkynes.

Out of learning curiosity: is this anywhere related to chelating?

What about steric hindrance? Hydrolysis is a separate step, so you have this cuprate floating around and you're trying to add a second alkyl cuprate. It would be interesting to think about what other ligands exist on the copper but aren't represented...

Alkenes are very capable of a similar carbometallation reaction as you can see in the scheme below (ref: R. M. Sultanov et al. / Tetrahedron Letters 54 (2013) 6619–6623):

I think the reasons why such reaction with alkenes is not so common is:

It is useless (correct me if i am wrong). Hence the reason why the work in the scheme above was published in a low impact journal. Why would anyone want to make alkanes in such a complicated way instead of just getting them from crude oil for example?
With alkenes, the reaction prefers to keep incorporating alkene to form really useful dimers or polymers. If one takes a look in the step wise mechanism of olefin polymerization (e.g. Ziegler-Natta) and dimerization the transmetallation/insertion step is equivalent to the one in the question posted:

I don't think it's useless. The organomettalic intermediate could react with all kind of electrophiles not just with a proton to give alkanes.

@Marko true, but surely you could just make a Grignard in that case? I don't know. Maybe this saves a step somewhere along the way. But I can see why this isn't a particularly... groundbreaking transformation.

@Marko Yes, i am referring to the formation of short chain common alkanes only which is a bit of a waste. The organometallic intermediate is really useful. But if we are discussing only about the organometallic intermediate then carbometallation with alkenes is as common (or maybe more) as with alkynes.

@orthocresol Don't forget that by carbometallation you add an alkyl chain to the substrate, which rapidly increases complexity.


text:
Solid State - Radius Ratio rules
Why lithium salts crystallize in octahedral lattice , although radius ratio values are much below than expected ratio?

Because chemistry's "rules" are more "occasionally useful general guidelines", in spite of what secondary school chemistry teachers told you.

I second that. Also, there is no such thing as octahedral lattice. Also, there are quite a few lithium salts out there, and they crystallize in a variety of systems. If you mean halides, it would be better to say so.


text:
How does 1 mole of a compound = the molecular weight of the compound?
I am reviewing moles, and came across this sentence in my book:

One mole of a compound has a mass in grams equal the molecular weight of the compound expressed in amu and contains $6.022\times10^{23}$ molecules of that compound.

Now, I perfectly understand the second part of the sentence. 1 mole of compound has $6.022\times10^{23}$ molecules. But how do $6.022\times10^{23}$ molecules have a mass same as the molecular weight of compound?
Ex. Molecular weight of $\ce{SOCl2}$ is $119\ \mathrm{amu}$. That is the mass of 4 atoms. How do $6.022\times10^{23}$ molecules have the same mass?

The wording in the sentence is poor. For the record... One molecule of $\ce{SOCL2}$ has a mass of 119 amu, but one mole of $\ce{SOCL2}$ has a mass of 119 grams and has $6.022*10^{23}$ molecules.

In simpler terms, if you have one mole of something, then the molecular weight of that substance will be the same as the sample weight. If you weight one mole of carbonic acid, it would be 62g which would be the same as its molecular weight (in amu). The correct term should be "molecular mass". I think that was also misleading.

Well you can actually calculate it like this
$$1\ \mathrm{amu}=1.66\times10^{-27}\ \mathrm{kg}$$
Therefore
$$119\ \mathrm{amu}=119\times1.66\times10^{-27}\ \mathrm{kg}$$
But for 1 mole, multiplying by Avogadro number you get mass of one mole as $$\begin{align}
119\times1.66\times10^{-27}\times6.022\times10^{23}&=0.11899\ \mathrm{kg}\\
&=118.99\ \mathrm g\\
&\approx119\ \mathrm g
\end{align}$$

Welcome to Chemistry! Please visit this page, this page and this one on how to format your posts better with MathJax and Markdown, then [edit] this answer to include proper formatting. Thanks


text:
Is there anything that doesn't mix with water or oil?
It's well known that water and oil don't mix, and if you put them together, the oil will float on top of the water in a distinct layer. My understanding is that this is because water is polar due to oxygens high electronegativity and oil is not. Most liquids and many solids are capable or mixing or dissolving into water, and I don't know of any other nonpolar liquids, but I imagine they would mix with oil.
In electromagnetism, there are magnetic materials that will be affected by magnetism, and nonmagnetic materials that aren't, but there is also diamagnetism which opposes influence of external magnetic fields.
So is there any kind of liquid substance that is neither polar nor nonpolar, and therefore doesn't mix with either water or oil? Could you have a beaker full of 3 liquids that has 3 distinct layers?
And if that was the case, are things that dissolve into one layer only able to dissolve into that later, or is there the ability to diffuse between layers?

Everything is soluble - it's just a matter of detection limits.

metallic mercury mixes with neither... I don't know what the absolute record is but more than 3 layers is certainly possible.

It seems that eight liquid phases is the maximum known.
https://en.wikipedia.org/wiki/Multiphasic_liquid#Higher_order_multiphasic_systems

http://chemistry.stackexchange.com/questions/778/a-quick-method-for-separating-liquids http://chemistry.stackexchange.com/questions/496/why-do-biphasic-systems-of-fluorous-and-organic-solvents-form

In short, yes, many highly fluorinated liquids are miscible with neither water nor organic solvents.
If you look at my answer to this question, you can see that I made a mixture of water, hexanes, and a fluorinated solvent called HT-110.
Likewise, the HT-110 will not dissolve most compounds, apart from fluorinated ones like Teflon AF.
Many compounds will have appreciable solubility in more than one layer and if they are in contact, there can be transfer between them.
Polarity plays a large role in whether two liquids are miscible.
Basically, for two liquids to mix, it must be energetically favourable to disrupt the intermolecular forces of the individual liquids in favour of the new intermolecular forces between the components of the mixture.
For your example of water and oil, water on its own has significant hydrogen bonding, but since the oil can't participate in hydrogen bonding, there would be a lot less hydrogen bonding in a homogeneous mixture of oil and water—the oil molecules essentially block water molecules from accessing each other.
Because the mixture would greatly reduce the intermolecular forces holding the liquids together, it is more energetically favourable to remain separate.
In contrast, a mixture of water and ethanol is energetically favourable because ethanol can hydrogen bond and is similarly polar, so the disrupted water-water intermolecular interactions are replaced by similar water-ethanol interactions.
In a little more detail, a mixture is favourable when its Gibbs energy of mixing is negative:
$$\Delta G_\mathrm{mix}
    = \Delta H_\mathrm{mix}-T\Delta S_\mathrm{mix}$$
In an ideal mixture of two components A and B, A-B interactions are the same strength as A-A and B-B, so the enthalpy of mixing, $\Delta H$, is $0$.
If A and B have similar properties, A-B interactions tend to be similar and $\Delta H$ will be close to $0$.
If A-B interactions are weaker (e.g. water and oil), $\Delta H$ will be positive.
$\Delta S$ is almost always positive for common mixtures because each component has a larger volume to spread into than if the components were separate. (The exception is when A-B interactions are much stronger than A-A and B-B interactions which prevents A and B from moving randomly through the volume.)
This means that mixtures can still form if the enthalpy of mixing is positive, if the entropic contribution can overcome it.
This explains why some mixtures are possible at elevated temperatures but not room temperature.

Sugar, cocoa, salt, water: no mix.   Heat water very hot, mix.  Then add milk, heat to serving temperature, top with little marshmallows.

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


text:
Deprotection of Boc using TAF to obtained free amine group
I know Boc can easily be deprotected using strong acid like TFA and HCl, but final product is salt of TFA or HCl, my question is that, how we can avoid formation of slat and obtained free amine group.
The proposed mechanism is


text:
Is there any rule that says that oxygen should have at least one free pair of electrons in Lewis dot structure?
For example, Ozone has a double and single bond with at least one free pair of electrons in each oxygen atom. My doubt is whether ozone can have a triple bond and single bond with one oxygen atom without a free pair of electrons.

$\ce{H4O^{2+}}$ is a counter example to such a rule.
http://pubs.acs.org/doi/abs/10.1021/ja00265a031


text:
What do these labels for molecular electronic states mean?
What do these symbols mean in excited states?
$${}^2\!A_2,{}^2\!B_1,{}^4\!A''$$
I am confused with these representations, found in the abstract of this paper. I think it is kind of a representation of electronic excited states. The left superscript means spin multiplicity ($2S+1$). What do the other parts of these symbols mean, and how can I write these symbols with other molecules?

As stated in this answer, these are irrep (irreducible representation) labels for molecular symmetry point groups. In the context of chemistry, point groups are usually introduced when learning about structural symmetry (atoms and bonds). This is a broad topic with many technical points, too many for a single answer, so I won't cover the basics but hopefully the few things needed to make the connection between what you read in a book and what's present in this paper.
The key is recognizing that the structural symmetry of molecules leads to symmetry in their molecular orbitals. Take the water molecule as an example. Water's molecular structure can be described at best by the $C_{2v}$ point group, so we say it has $C_{2v}$ symmetry. Composing its molecular orbitals from atomic orbitals is actually an exercise in symmetry, as only atomic orbitals of identical symmetry may combine. Here is the LCAO-MO diagram for water, with the oxygen 1s orbital left out due to its extremely low energy; it won't mix with the hydrogen AOs.

Here is the character table for the $C_{2v}$ point group.

AOs of identical symmetry (they transform identically for any given symmetry operation) are shown in the same color. A few things are worth pointing out:

s orbitals are perfectly symmetric (a sphere), so they will always transform as the totally symmetric irrep of a molecular point group. In $C_{2v}$, that's $A_1$. Once you learn what the operations $E$, $C_{2}$, $\sigma_{v}(xz)$, and $\sigma_{v}(yz)$ do, you can convince yourself that the $\sigma$ combination of $\ce{H2}$ AOs is also completely symmetric under those operations, so it also belongs to the $A_1$ irrep. The p orbitals are figured out from the 3rd column, d from the 4th.
The 2s and $\sigma$ AOs combine to give the $3a_1$ MO. The "3" is because it's the 3rd such MO with $a_1$ symmetry. The symmetry label is in lowercase because it for a 1-electron state. Symmetry labels are in uppercase for many-electron states, such as the total wavefunction, as you see in the linked paper. This is a matter of convention, nothing more.

Neutral water is a closed-shell molecule; there aren't any unpaired electrons, so it has singlet spin multiplicity. How do you determine the symmetry of the total wavefunction? A Slater determinant can help, or even just the Hartree product part of it. Take a fictitious system with 4 electrons in 2 orbitals. The Hartree product will look something like
$$
\Psi_{\text{Hartree}} = [\phi_{1}(1)\alpha(1)][\phi_{1}(2)\beta(2)][\phi_{2}(3)\alpha(3)][\phi_{2}(4)\beta(4)]
$$
This is a product wavefunction that might be made from an electron configuration such as $1s^{2}2s^{2}$, $1s^{2}2p_{x}^{2}$, $4p_{x}^{2}4p_{z}^{2}$. Each of these MOs has a certain symmetry; the total symmetry is given by the product of the individual symmetries. For this step one needs the point group product table (again for $C_{2v}$):
$$
\begin{array}{|l|l|l|l|l|}
\hline
            & \mathbf{A_1} & \mathbf{A_2} & \mathbf{B_1} & \mathbf{B_2} \\ \hline
\mathbf{A_1} & A_1          & A_2          & B_1          & B_2          \\ \hline
\mathbf{A_2} & A_2          & A_1          & B_2          & A_1          \\ \hline
\mathbf{B_1} & B_1          & B_2          & A_1          & A_2          \\ \hline
\mathbf{B_2} & B_2          & B_1          & A_2          & A_1          \\ \hline
\end{array}
$$
If the electron configuration of the system is indeed $4p_{x}^{2}4p_{z}^{2}$, this results in a total wavefunction symmetry of
$$
B_{1} \times B_{1} \times A_{1} \times A_{1} = A_{1}
$$
The spin components don't contribute, only the spatial components. Note that in any product table, the product of any irrep with itself is always the totally symmetric representation in that group, hence the diagonal being all $A_{1}$. Product tables are also symmetric, since irreps are commutative under multiplication. Since this is a singlet, the ground-state wavefunction can be specified as $^{1}\!A_{1}$. For a closed-shell system (RHF), due to the pairs of electrons always leading to $A_{1}$ symmetry, it will always have a total state symmetry of $^{1}\!A_{1}$. If you have an open-shell system of any kind in a ground or excited state, with either an even or odd number of electrons, the total wavefunction symmetry will be determined by electrons in those orbitals alone, since the product between fully-occupied orbitals will always be of $A_{1}$. In the simplest case of one unpaired electron, the total symmetry of the ground state will be the symmetry of the SOMO.
Now, on to the paper. The most important part for our understanding of the state label assignments is Table 3:

All the systems considered have either $C_{2v}$ or $C_{s}$ symmetry. Note how going from a point group of higher symmetry ($C_{2v}$, 4 irreps) to one of lower symmetry ($C_{s}$, 2 irreps) leads to states collapsing into fewer irreps due to fewer symmetry operations that can distinguish them. For the plots (here the last row of Figure 6),

there is the isosurface of the molecular orbital (really a natural orbital), the orbital number and symmetry, and the natural orbital occupation number (NOON) for that NO in that state.
There are a number of good Dover publications (1, 2, 3) for learning about group theory in chemistry and spectroscopy that are highly recommended, since the primary peer-reviewed literature isn't usually a good resource for learning this. An excellent resource for character tables is WebQC.

To fully understand the meaning of these symbols, you need to understand molecular symmetry groups. If you know the point group symmetry of your molecule, you can construct a character table comprised of all possible symmetry operation (rotation around an axis, reflection in a plane, inversion) and the effect of these operation on a certain electronic state (+1 for unchanged, -1 for sign change for example). A row of such a table is called an irreducible representation (irrep) and is used as a label defining a certain symmetry state. The notation follows the rules set out by Mulliken in , J Chem. Phys. 23, 1997 (1955) and have the following general meaning:
A is used when the irrep is symmetric under $C_n$ or $S_n$ for the highest $n$ in the group, in addition A is used if there are no $C_n$ or $S_n$ (where $C_n$ is a rotation by 2$\pi/n$ around an $n$-fold symmetry axis and $S_n$ is also a rotation, but followed by a reflection).
B is used when the irrep is antisymmetric under $C_n$ or $S_n$ for the highest $n$ in the group.
E is a doubly degenerate state
T is a triply degenerate.
For degenerate irreps (A and B) subscripts 1 and 2 relate to the symmetric (1) or antisymmetric (-1) characters respectively, in relation to a $C_2$ axis perpendicular to the principle $C_n$ axis, or in the absence of this element, to a vertical symmetry plane ($\sigma_v$) plane. For multidimensional representations, the subscripts 1, 2, etc are added to distinguish between nonequivalent irreducible representations that are not separated under the above rules.
Single (') and double ('') primes are used when the irrep is symmetric (+1) under reflection in a horizontal mirror plane or antisymmetric in this plane (-1), provided that the molecule has no inversion symmetry.

I think these symbols represents electronic excited states according to this paper. Maybe your answer is about the first column of the character table? I think these symbols can specify the electron state.

Exactly, they specify the symmetry of the different electronic states.


text:
Reduction of copper oxide by hydrogen in the lab to copper
In the reduction of copper oxide by hydrogen in the lab to copper, why does hydrogen need to flow through the tube even after the reduction is completed and the burner is turned off until the tube is cool?

Otherwise air will flow in, and perform the reverse transformation.

Related: http://chemistry.stackexchange.com/questions/43371/why-doesnt-hydrogen-displace-metals-from-their-oxides/43377#43377

As the system cools, the gases will contract. Unless a positive flow of hydrogen (or some other inert gas) is maintained during the cooling process, this will result in room air being pulled into your tube. The oxygen present in air would then oxidize some of $\ce{Cu}$ back to $\ce{CuO}$, so long as some regions within the tube remained hot enough for this reaction to take place (at least $\pu{300^oC}$ or so).


text:
Radioactive Isotopes form Compounds? The Hershey-Chase Experiment
Our chemistry teacher taught us last year, that radioactive isotopes of elements don't form compounds with other elements "because they're unstable" and since we're high-schoolers (I'm assuming that's why) she didn't go into details.
But today in our Biology class, we were taught about the Hershey-Chase experiment (https://en.wikipedia.org/wiki/Hershey%E2%80%93Chase_experiment) , where we come across radioactive P-32 being incorporated into the phosphate group of nucleic acids and radioactive S-35 being incorporated into amino acids in proteins.
When we reached that bit, what our Chem. teacher said came back to me and it conflicted with the idea being propounded in class. Naturally I raised that issue with our Bio teacher, but she said she wasn't sure about' radioactive isotopes not forming chemical bonds', because that was precisely what was happening in the Hershey-Chase experiment.
So can anyone tell me whether:
a) My Chem. teacher was correct (perhaps within limits), and that the incorporation of radioactive isotopes in nucleic acids and proteins is a phenomenon that occurs outside of the limiting case she mentioned.
b) My Bio. teacher was right. Radioactive isotopes do form compounds.
c) None of the above (could you specify why?)

Your Chem. teacher is quite definitely wrong. Radioactive isotopes surely do form compounds; in fact, they form the same compounds as normal isotopes would. They don't differ from normal isotopes at all, except that they are radioactive. Why else isotope separation would be such a trouble?

I am radiochemsit, actually your teacher is wrong. Radioactive isotope do form chemical bond with other compound. Very simple example is  2-deoxy-2-(18F)fluoro-D-glucose is a glucose analogue used as Positron imaging agent (PET).

Multiple elements don't even have any stable isotopes and so every atom of the element is necessarily radioactive, yet they still form an abundance of compounds. A simple example would be uranium.

Radioactive isotopes have similar chemical properties to that of nonradioactive element of same atom. They actually do from chemical bond with compound. You even heard the name of radio-pharmaceutical that involved the complex of radioactive metal with ligand or the radioactive isotope is attached to the pharmaceutical through covalent bond.
The simplest example is FDG is 18F-labled glucose analogue.

Carbon-11 is another radioactive isotope that form covalent bond similar to Carbon-12 and used in Nuclear medicine as PET imaging agent. C11-Methionine is an example used in the field of Nuclear Medicine for the diagnosis of brain tumour

Iodine-131 is another sample of radioactive isotope that is used for the detection of hypothyroidism and also form bond with compound. Like in metaiodobenzylguanidine is used in neuroblastoma
.
I can gave you lots of others examples but i think these are enough for your understanding. I hope, this will be helpful.

Or, even simpler - how could anybody mine uranium ore (an oxide) if radioactive isotopes did not form compounds - all of the uranium isotopes are unstable.


text:
Why do real gases deviate from ideal gas behavior?
While studying Gaseous state in books, there was a common question which I had difficulty answering -
Why do real gases deviate from ideal behavior (3 points)?

The (3 points) part of your question is a pretty good giveaway that you copied this question from somewhere. Could you share what your thoughts on this question are?

Actually I was studying my textbook which is the NCERT for class 11 in India. In which there is this question. I added 3 points because my teacher asked the same question - Give 3 points while taking a viva examination.

Further it didn't click me that I had to write about the assumptions that were taken by Maxwell and Boltzmann while formulating the kinetic theory of gases... I was trying to relate this question to the trend of compressibility factor of real gases vs ideal gases.

Real gases deviate from ideal behaviour because their particles (atoms for inert gases or molecules) occupy some finite space and do exert interactive forces among them. Completely ideal behaviour is hypothetical because of the reasons above. At low pressure and high temperature, real gases behave approximately as ideal gases. In ideal behaviour, gas particles don't occupy space and do not have any interaction, as assumed in the kinetic theory of gases. But in reality this is not the case: we get errors by applying the ideal gas law.
That's why van der Waals corrected it by introducing suitable constants.


text:
Ethanol for wood polishing
We use ordinary spirit to polish wood, and it contains ethanol. What makes ethanol a good polishing agent for wood? What effect does it causes on wood so that it shines?

Spirit polish, as it is termed, has shellac dissolved in the ethanol. Could you post a picture of the product you use? I suspect there is more going on here.

Ethanol is an organic solvent. I would guess that it makes wood shine by:

Cleaning the surface, dissolving away organic compounds
Maybe slightly dissolving (?) a thin layer of cellulose at the surface of the wood, removing irregularities and making it more malleable, hence easier to polish

There may be more specific interactions between wood and ethanol, but I don't know about them.


text:
Inhibitors and poisons in a reaction
What are the basic differences between inhibitors and poisons which are used in a reaction
Actually both the names confuse me so please help me out.

https://en.wikipedia.org/wiki/Reaction_inhibitor#Inhibition_and_catalyst_poisoning

Inhibition is reversible, poisoning is not.

My understanding of catalyst poisons and inhibitors is that:

Inhibitors curb the catalyzing powers of a catalyst but do not react with any of the reactants and the reaction is reversible.
Poisons react with the catalyst and/or the reactants and the reaction is irreversible.

I disagree. There are too many articles about "reversible catalyst poisoning" for an answer to be so clear cut.

Couldn't the difference be that poisons react with the catalyst, possibly reversible to inhibit the reaction, and inhibitors just inhibit the reaction by doing something without reacting with the catalyst, like changing the pH, scavenging radicals or just binding in a non-covalent manner?

@SteffX, I would say that the phrase "reversible catalyst poisoning" implies that catalyst poisoning is not supposed to be reversible, so the initial definition kinda stands, but is being pushed by current advances, and maybe will change in the near future?


text:
Why doesn't beta hydride elimination happen in alkenyl/aryl halides?
Beta hydride elimination is an important reaction in organometallic chemistry:

This reaction prohibits the use of saturated alkyl halides in coupling reactions, because it can take place after oxidative insertion of the metal. Alkenyl and aryl halides don't suffer from this unwanted reaction. Why? They also contain beta hydrogens, why aren't they readily eliminated?

Do you think  it can be because the C-H and C-M bonds on the alkene can't bend enough for the M-H interaction to occur ?

Yeah, that could be it.

In aryl it's rather simple - you'd get benzyne - can't see anything that would offset it.

With sp2 carbons the C-H bond is also stronger, which might be a factor. (As an aside I think metal benzyne complexes are known but perhaps not a metal hydride complexed to a benzyne... that would probably be less stable relative to the metal aryl.)

Theres an inherent geometric restriction if you try and beta-eliminate from an olefin- the orbitals just struggle to line up.


text:
Why is Kw the same in acid, water, and alkali?
Why is Kw the same in acid, water, and alkali? Given that Kw equals to [H+][OH-], doesn't concentration of these 2 differ in acids, water and alkali? Why is Kw only affected by temperature?

Concentrations differ, of course, but their product does not. That's why it is called a constant.

@bean as temperature increases concentration of both ions increases since more disassociation will take place.

You are correct that the concentration of $\ce{[H+]}$ and $\ce{[OH-]}$ both vary based on the the composition (acid/alkaline) of the solution, but the remarkable thing is that their product does not. When $\ce{[H+]}$ goes up, $\ce{[OH-]}$ goes down, in the same proportion.
The reason for this is that the Kw represents the equilibrium constant of the following reaction, and any excess (lack) of $\ce{[H+]}$ or $\ce{[OH-]}$ will react by LeChatelier's principle to reestablish equilibrium:
$$\ce{ H2O <=> H+ + OH-}$$
We can write the standard equilibrium constant equation as follows
$$K_{eq} = \frac{\ce{[H+][OH-]}}{\ce{[H2O]}}$$
For aqueous solutions, the concentration of water is a fixed constant1, so we can multiply both sides by it, and fold it into the equilibrium constant2. Therefore we can write:
$$K_{w} = \ce{[H+][OH-]}$$
So just as standard equilibrium constants do not differ with the concentrations of chemical species in the solution, but can vary with temperature, so too does $K_{w}$ not vary based on the concentration of chemical species, but does depend on the temperature of the solution.

1 Technically, the values in the equilibrium constant equation are activities rather than concentrations (see here and ref 1 therein for a rather technical discussion), a distinction we normally ignore, as activities are proportional to concentration in dilute solutions. Regardless, the activity of water in most dilute aqueous solutions can be considered a constant, so it doesn't change the argument.
2 Note this is only true in a solution which is almost entirely water, with just a little bit of solute dissolved. If you had a water/ethanol mixture, or a very concentrated solution of a highly soluble salt, then the water activity can't be considered a fixed constant, and you'd have to start making some adjustments based on it. (This is where the distinction between "concentration" and "activity" come into play - these adjustments are more complicated than just dividing by the nominal concentration of water.)


text:
NaCl water electrolysis without membrane, how measure the free chlorine concentration?
I'm trying to made electrolysis using water and NaCl. I need at the end a concentration of 2 g/l. How can I measure the free/active chlorine? Do you know a way to "measure" the chlorine during the reaction without chlorine sensor? Using for example the measure of conductivity of the water, the electrode current and voltage and the temperature or the salt remaining?
The electrolysis is 'open' without membrane, a titan electrode is directly put in the water + NaCl.

There are not many solutions, I've tried a lot of ISE sensors (amperometric) but it seems that the hydrogen generated influences a lot the measurements, the only functional (but not cost effective!) are colorimetric measurements. I've used Swan Analytics devices.
Actually I am looking for a cheaper solution.

The best way for determining the active chlorine amount ($\ce{ClO^-}$) in a bleach is to dilute it with water, then to add Potassium iodide $\ce{KI}$, and enough HCl solution to make it acidic. It produces $\ce{I_2}$ by the equation $$\ce{ClO^- + 2 H^+ + 2 I^- -> I_2 + H_2O}$$ Then the $\ce{I_2}$ so produced is titrated by a standard solution of sodium thiosulfate $\ce{Na_2S_2O_3}$, according to : $$\ce{I_2 + 2S_2O_3^{2-} -> 2 I^- + S_4O6^2-}$$
A quicker (but not so precise) way is to add 2 mL acetone to 50 mL bleach in a insulated flask, and to measure the change of temperature due to the reaction. The reaction produces chloroform $\ce{CHCl_3}$ and acetate. After one minute, the temperature increases to a stable final value. The change of temperature is proportional to the amount of chlorine. If $\Delta T $= 10°C, the concentration of active Cl is 5.6%. Generally speaking, the concentration of active Chlorine is : 0,56% x $\Delta T$


text:
What workflow can I use on to extract cartesian coordinates of many molecules from both .spartan and .out (Gaussian) files?
I'm searching for a workflow for extracting cartesian coordinates for multiple moelcules from both a single file (e.g., several conformers from one .spartan file or one multiple molecules, each from multiple Gaussian output files). Any insight would be appreciated. I'm using Ubuntu (14- & 16.04), on a machine on which I do not have either Gaussian or Spartan installed.
Thank you kindly

Hello. The common approach is invert some minutes in codding a simple script (using bash, python, perl, for example). It can be done in less than 15 lines of code.

The only tool that I know for reading Spartan files (besides Spartan) is Jmol. With some programming or scripting, you will probably be able to save conformers to other files in a workflow - I don't think there is a pre-existing script.
There are multiple open source tools that read Gaussian output files. I would suggest either:

Open Babel - GUI, command-line and Python
cclib - using Python

For example to extract a bunch of Gaussian files to XYZ Cartesian:
obabel *.log -oxyz -m
The -m flag indicates "multiple" and the -oxyz indicates that you want XYZ-format files. More is in the manual.


text:
Calculating the chloride concentration

What are the concentrations of $\ce{Cl-}$ in the following solution? $1.9\ \mathrm{g}\ \ce{MgCl2}$ is dissolved in water to make $1\ \mathrm{L}$ total solution

$$48.31\ \mathrm{g/mol}\ \ce{MgCl2}$$
$$(1.9~\mathrm{g}\ \ce{MgCl2})(1\ \mathrm{mol}/48.31\ \mathrm{g}) = 3.932\times 10^{-2}\ \mathrm{mol}\ \ce{MgCl2}$$
$$\ce{MgCl2 -> Mg+ + Cl2-}$$
$$(3.932\times 10^{-2}\ \mathrm{mol}\ \ce{MgCl2})((1\ \mathrm{mol}\ \ce{Cl2})/(1\ \mathrm{mol}\ \ce{MgCl2})) = 3.932\times 10^{-2}\ \mathrm{mol}\ \ce{Cl2-}$$
then
$$(3.932\times 10^{-2}\ \mathrm{mol}\ \ce{Cl2-})/(1~\mathrm{L}) = 3.932\times 10^{-2}\ \mathrm{M}\ \ce{Cl2-}$$
Is that correct? Or does the $2$ go infront of the $\ce{Cl}$, so the mole ratio is $((2\ \mathrm{mol}\ \ce{Cl-})/(1\ \mathrm{mol}\ \ce{MgCl2}))$?

What's the charge on a magnesium ion? BTW please don't close this question In fact even by our previous standards this should not be closed.

The first thing we need to do is address your chemical equations.  $\ce{MgCl2}$ dissociates in water as follows:
$$\ce{MgCl2 -> Mg^{2+} + 2Cl-}$$
Now you can see that one mole of $\ce{MgCl2}$ dissociates to give two moles of $\ce{Cl-}$. So all we have to do then is calculate the number of moles of $\ce{MgCl2}$ from the given mass and its molecular weight, then multiply that by 2 to get the number of moles of $\ce{Cl-}$:
$$\mathrm{\frac{1.9\ g\ \ce{MgCl2}}{ 95.2\ \frac{g\ \ce{MgCl2}}{mol\ \ce{MgCl2}} } = 0.0200\ mol\ \ce{MgCl2}}$$
$$\mathrm{0.0200\ mol\ \ce{MgCl2}\ *\frac{2\ mol\ \ce{Cl-}}{mol\ \ce{MgCl2}} =0.0400\ mol\ \ce{Cl-}}$$
Then since the volume is 1L, the $\ce{Cl-}$ concentration is just $\mathrm{0.040 \frac{mol}{L}}$.


text:
How can I remove acids from cola?
I would like to taste Pepsi-cola without acid. I like it's flavor but don't like acidic taste. It's acidity comes from carbon dioxide, phosphoric and citric acids (see ingredients list in wiki and answers to this question). Carbon dioxide can be evaporated easily. How can I get rid of phosphoric and citric acids? Preferably without expensive lab equipment and hard-to-get reagents.
Edit: Found one recipe by Josh Velson on quora site. He advises adding limewater until the pH reaches about 12.32. Are there any other methods?

You could try to increase the pH, e.g. by adding baking powder. But avoid incorporating large quantities of the stuff.

@aventurin According to this article. Baking powder is basically just baking soda with acid added in. So probably I should add raw baking soda instead? I'll try that tomorrow.

Yes, baking soda would be better.

Yes, do that. In chemistry, you typically don't get rid of acids by removing them. Instead, you neutralize them.

@Ivan Neretin I hoped someone would name a substance that would turn the acids to something insoluble that I can filter out. That is why I said "remove".

I would not advise you to adjust the pH to 12, which is a bit too high to drink (though not toxic if you drink just one mouthful). First of all, you will need to get rid of all the carbon dioxide dissolved, which is in equilibrium with carbonic acid (vented Coke is already disgusting!). Then you would need to further neutralize it with a base but nothing containing sodium or potassium because it would add a salty taste. I would choose calcium carbonate or hydroxide. To check the pH you should have pH paper and adjust the pH to 8.

@technocrat Yeah, that's exactly the sort of thing we normally wouldn't do. First of all, there are no (or very few) acid molecules there, just cations and anions. The cations are the same in any acid: simply $\ce{H+}$. With some effort, you could turn the anions into something insoluble, but that's not what gives us acidity. You have to remove the cations. Yes, you probably could be able to turn them into something insoluble as well (with some more effort), but there's a better way: turn them into water, and skip the filtering out. And that's what our proposed recipes are all about.

Just as a note, adding baking soda will technically remove the Citric acid and phosphoric acid. It will just leave behind phosphate ions and whatever the conjugate base of citric acid is. There will be no taste from the acid since it has been broken down.

@SteffX That could be an answer.

Since all the ingredients in Pepsi are there for a reason (not all related to taste), removing them is dangerous. And it is very hard to do. If you don't like the taste, drink something else.

You can't remove the acids, because they're dissolved in the liquid. You can, however, neutralise it.
Acid is basically $\ce{H+}$ ions dissolved in the water. By reacting it with a base, you do the following reaction:
$\ce{H+ + OH- -> H2O}$
So the acid reacts to form harmless water.
The problem is that you can't just add $\ce{OH-}$, you have to add it as some salt. Also, the citric and phosphoric acid you have in the cola were added are also dissociated from a salt or molecular form:
$\ce{H3PO4 -> H+ + H2PO4−}$ for phosphoric acid, for instance.
So whatever you put in there that reacts with the $\ce{H+}$ should also react with the $\ce{H2PO4−}$ ion to make something that's not soluble.
Usually, neutralising acids is done with carbonates such as sodium carbonate or calcium carbonate (also known as soda and lime, respectively). Both of which are safe materials to handle. The problem is that by doing that, you're introducing carbonate ($\ce{CO_3^2−}$) which might alter the taste.
A better solution is to add in hydroxides, such as sodium or calcium hydroxide ($\ce{NaOH}$ or $\ce{Ca(OH)2}$). The problem is that those are much more hazardous things which can easily hurt you if you don't use them properly. I would recommend using calcium hydroxide because then the calcium will react with the phosphate anion in the cola to make insoluble calcium phosphates and with citrate to make (somewhat) insoluble calcium citrate. Then you can just filter the stuff out and have your acid-free cola.
Basically, by adding in $\ce{Ca(OH)2}$ to your cola, you react the "acid" bit to make water, and you react the "phosphoric" and "citric" bit to make a white solid precipitate in the bottom of your glass.
DISCLAIMER
This is all in theory. I highly recommend against adding chemicals to your food and drink. If you don't like the acidic taste of cola, don't drink it. Find another brand or another drink that you do like. Tea? Anyway, do not mess with your food unless you know exactly what you're doing and you know that your chemicals are food-grade and safe. Don't go and do stuff with your cola just because someone on the internet (me) told you it will work.

The implication that you cannot remove a dissolved chemical is incorrect. One can separate a liquid into all chemical constituents using chromatography, evaporate the solvents, and recombine them.

@Codebling OP explicitly requested "Preferably without expensive lab equipment and hard-to-get reagents." So, given the constraints, I stand by my statement that this cannot be done.

That's fair! I think the materials for a simple chroma setup are minimal, but the required solvents are definitely no joke. It's a good answer

Just get simple and use plain ol' grocery store baking soda. Who cares about why? If you don't add too much, you can defeat the acidic taste and after bite without affecting the Pepsi cola syrup taste. For anything from 12 oz. to 1 litre, add 1/16th teaspoon. Try it to see how you like the taste, then move the amount of baking soda up or down to make it 'perfect'.
Baking soda, bicarbonate of soda, NaHCO3, is used as a pH buffering agent. Although another name for it is sodium acid carbonate, its pH in an 0.1 mole solution is 8.3. Therefore, it is mildly basic and can be used to estimate an amount to reduce the acidity of a cola soda water. Start with a 'pinch' and the amount can be titrated up to reach a pleasing taste for your soda.

When you spoon it in and it stops bubbling wildly, you know you are getting close to neutral. Bon appetit. :-/

Bubbling has nothing to do with removing the acid. Bubbling is caused by adding nucleation surfaces which allow the CO2 to attach and bubble out.


text:
High chemical resistance, high thermal conductivity?
I'm redesigning a microfluidic circuit to include a heating element. I'm running into problems because the ETFE tubing I'm using is too thermally resistant for me to get any energy into the fluid within. My options are to either 1) replace the ETFE tubing with a polymer with similar chemical and surface properties but a higher thermal conductivity, or 2) replace just the beginning of the circuit tubing with a similarly chemically resistant, but possibly rigid, material that can transfer a lot of heat to the fluid, which would then be insulated by the ETFE on its way to its destination.
Thoughts?
Thanks!
[Edit]: Also, affordability is a slight factor. This question is industrial in nature so a certain operating budget can be assumed - but exotics are going to be hard to get.
I like the diamond comment though; I wish I could pull that off.

Well, diamond comes to mind, but that might not be so easy...

What does it need to be chemically resistant to?

Post-edit: diamond-like coatings are not hard to get commercially, and retain some of the inertness. Also it depends on the temperature and what you want resistance to.

You may use borides


text:
How to predict SN2 reaction when comparing nucleophilicty and leaving group ability
In a lab, a common experiment is to perform SN2 reactions involving a primary alkyl bromide and sodium iodide in acetone (polar aprotic solvent).
The result is that the iodide ion is a good nucleophile, so the iodide will displace the bromide from the primary alkyl bromide, and sodium bromide will precipitate from the solution.
However, without the presence of the sodium cations, would the reaction still take place?
I'm somewhat confused, because iodide is both a better leaving group and a better nucleophile than bromide, so what is the determining factor in the reaction?
Is the strength of the nucleophile more important or the ability of the leaving group be displaced?

There aren't any silver cations. NaBr will precipitate.

I meant sodium. Changed to reflect that.

In aprotic solvents bromide is a stronger nucleophile than iodide.

@CayetanoGonçalves Please, see this.

@ChemistryHelpCenter thank you. That was my next question if it was being driven by Le Châtlier's principle

This is very famous "Finkelstein Reaction" that involved the exchange of one halogen to another through SN2 mechanism. The driving force for the reaction is the solubility of salts (NaI, NaBr, KF) under equilibrium.
There are different factor that effect the equilibrium position of a reaction, firstly, nucleophilicity of anion, presence of  good leaving group, and stabilization of anion in a given solvent.
first part of your question, I don't think with the absence of sodium cation the reaction would take place, because the driving force is the precipitation of salts of NaBr and NaCl in acetone.
In the second part of your question, you are right iodide is better leaving group compared to Bromide because iodide is weaker base compared to bromoide. Apparently, iodide can't replace bromine but here the reaction is driven toward products by "mass action" due to the precipitation of the insoluble salt as mentioned above.


text:
pH Difference Our Taste Buds Can Recognise
Water at $50~^\circ\mathrm{C}$ has a $\mathrm{pH}$ of $6.63$, while the pH of water at $0~^\circ\mathrm{C}$ is $7.47$.
The temperatures are comparable to a chilled, iced drink (at $0~^\circ\mathrm{C}$) and a hot drink (at $50~^\circ\mathrm{C}$). Would these differences in acidity be noticeable? Would they be noticeable if the drink was more acidic, such as orange juice?

First, the pH of water changes with rising temperature because the total number of $\ce{H+}$ ions (well, solvated hydronium ions, really) increases, not because it has become more alkaline. The water is still neutral, having an equal number of hydroxyl ions as hydronium ions. Heat simply increases the ionization.
Second, it depends on the individual taster, but a sample of water at $\mathbf{20~^\circ C}$ with a $\mathrm{pH}$ of $6.63$ (perhaps acidified with citric acid) would probably taste sour in comparison with a sample of water at $\mathbf{20~^\circ C}$ with a $\mathrm{pH}$ of $7.47$ (perhaps with a bit of baking soda, $\ce{NaHCO3}$).
See Temperature Dependent[sic] of the pH of pure Water for a more detailed explanation.

So the scale adjusts, but the acidity doesn't? Is that what you're saying? So even with a temperature change, though the pH value will change, all liquids will taste the same in terms of acidity (or sourness)?

No. The scale doesn't adjust. pH is just a measure of the concentration of $\ce{H+}$ ions in solution. Pure water spontaneously dissociates a little bit into $\ce{H+}$ and $\ce{HO-}$, and does so more with higher temperature. So for any given temperature, there is an equal concentration of $\ce{H+}$ and $\ce{HO-}$ in pure water, and it is neutral. When the temperature rises, both concentrations rise and the water is still neutral.


text:
Why is (NH4)2[CuCl4] square planar complex but Cs2[CuCl4] is tetrahedral?
Why is $\ce{(NH4)2[CuCl4]}$ square planar complex but $\ce{Cs2[CuCl4]}$ is tetrahedral even though both have same oxidation number of copper and same ligands?

In solution, the $\ce{[CuCl4]}$2- is expected to exhibit terahedral, or nearly tetrahedral geometry.
The Jahn-Teller theorem states that degenerate orbitals cannot be unequally occupied. Molecules with unequally occupied degenerate orbitals will distort so as to render the orbitals non-degenerate, with more electrons occupying the lower energy states. Generally, this distortion leads to a reduction in symmetry. In the case of Cu (II), it is a d9 cation with a tetrahedral electronic configuration of (e)4(t2)5, so the three degenerate t2 orbitals are unequally occupied, therefore a distortion towards the square planar geometry of (eg)4(b2g)2(a1g)2(b1g)1, with no unequally occupied degenerate orbitals, is electronically favoured. Nonetheless, experimental evidence (1) indicates that this d-orbital stabilization energy is very small. Since copper is a first-row transition metal (not very big), steric effects dominate electronic effects allowing $\ce{[CuCl4]}$2- to assume a tetrahedral geometry. Most spectroscopic methods I've read involve looking at the molecule in crystal form (because that is where this effect gets interesting), but DFT studies of the $\ce{[CuCl4]}$2- hydrate in concentrated aqueous solution indicate that the anion's tetrahedral structure is more stable (2).
The issue of $\ce{(Cs)2[CuCl4]}$ vs $\ce{(NH4)2[CuCl4]}$ comes up when the complex precipitates and crystals are formed. Due to the borderline electronic effects, the $\ce{[CuCl4]}$2- anion exhibits a great degree of structural flexibility. With large cations ($\ce{Cs+}$ radius is 0.174 nm while $\ce{NH4+}$ radius is 0.143 nm) , it crystallizes as a distorted ("squashed") tetrahedron.
$\hskip1.4in$ (3)
Square-planar co-coordination occurs in such a way that you effectively get octahedra with tetragonal distortion (4). I would imagine that this phenomenon is simply a result of the smaller cation allowing the unit cell to crystallize like that. The general idea with unit cells is that the cation is never allowed to "rattle" around, so the structure will find a way to fill in the space. Given the preexisting electronic favourability of square planar geometry, $\ce{(NH4)2[CuCl4]}$ this distortion favoured by close-packing is made even more favourable.
$\hskip2.1in$ (4)
The canonical reference for $\ce{(NH4)2[CuCl4]}$ 's crystal structure is (5) if you want to take a look at the unit cell.
(1) Emeléus & Sharpe. Advances in Inorganic Chemistry and Radiochemistry, vol. 21, p. 128
(2) https://www.ncbi.nlm.nih.gov/pubmed/21462945
(3) L. Helmholz & R. F. Kruh, J. Am. Chem. Soc., 1952, 74, 1176
(4) Molecular Structure by Diffraction Methods, vol. 1, The Chemical Society, London, 1973, p. 642
(5) R. D. Willet, J. Chem. Phys., 1964, 1964, 41, 2243

thanks for pointing out my typos!
Yes, I was describing the extreme cases; the tetrahedral form in solution and the electronically favourable square planar case. In ammonium chlorocuprate's unit cell, the ion's geometry is the intermediate, non-inversion symmetric case. If anyone has access to (5), we could describe its symmetry. Sadly, this volume seems to have been misplaced in my library.

Haha, my VPN gives me access. The article doesn’t include a picture though. To quote the discussion: ‘This structure contains the first instane of a square-planar $\ce{CuC4-}$ ion. The $\ce{Cl-Cu-Cl}$ bond angle is required by symmetry to be $90^\circ$. However, the $\ce{Cu-Cl}$ bond lengths are not required to be equal. Two are $2.300 \pm 0.005~\mathrm{A}$ long, and two are $2.332 \pm 0.004~\mathrm{A}$ long. Each copper atom completes its normal distorted octahedral configuration by forming two long $\ce{Cu-Cl}$ bonds of $2.793 \pm 0.005~\mathrm{A}$ with adjacent ions and thus forming an …

… infinite two-dimensional network. The $\ce{Cu-Cl-Cu}$ bond angle is linear ($180^\circ \pm 0.2^\circ$) to within experimental error.’

is the bond length distortion due to electronic Jahn-Teller effects (as in http://chemistry.stackexchange.com/questions/43111/jahn-teller-distortions-in-square-planar-complexes), or is it just a consequence of the crystal packing? I have VPN too, but the journal isn't available in electronic format that far back for me

They unfortunately don’t go into detail. I would say that JT doesn’t sound unreasonable, but it could also be some weird packing effects. Again, an image is not included, unfortunately.


text:
Reduction of copper oxide by hydrogen in the lab
Why does the hydrogen which is passing continuously through the tube have to be burnt when it is passing out from the tube? Why can it not be allowed to just escape simply?

You don't want it to accumulate in your room and then suddenly catch fire. Gas explosions can be pretty powerful, to the point of demolishing whole buildings.

https://chemistry.stackexchange.com/questions/43371/why-doesnt-hydrogen-displace-metals-from-their-oxides/43377#43377

Burning the hydrogen gas at the exit stream of the copper oxide reduction tube is simply one way to assure that it does not accumulate to explosive levels in the lab. Although hydrogen gas is not toxic, it is flammable in air at concentrations as low as $\pu{4\%}$. As the combustion product of hydrogen and oxygen is mainly water vapor, this technique makes for a simple, safe means for disposing of the excess hydrogen from the copper oxide reduction process.


text:
How does mass affect bond stretching in UV spectroscopy?
If, for example, you have a bond on the form:
H - X, where H is hydrogen and X is some other element.
How does the mass of X affect the stretching of the bond during UV spectroscopy? Why do heavier atoms require a less 'energetic' wave to cause them to stretch?

https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator or http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc.html

UV spectroscopy is not about bond stretching at all.

@Ivan Neretin , but all bound electronic states do have vibrational modes and these will be affected by isotopic substitution. The internuclear separation of excited or ground state is not affected by isotopic substitution, only energy levels so the new spectrum will appear shifted in frequency and changed in intensity.

I didn't even see that, just assumed it was IR spec... seems like the answer below did that too. To be fair UV spec is really not discussed in terms of bond stretching.

In a very crude approximation, every bond can be considered to be a quantum harmonic oscillator. As you might know, the energies of such an oscillator are given by
$$ E_\nu = hc\omega_e \left ( \nu+\tfrac{1}{2}\right )$$
where $h$ is Planck's constant, $c$ the speed of light and $\omega_e=\sqrt{k/\mu}$. The constant $k$ is the force constant of the bond and $\mu$ is the reduced mass
$$ \mu=\frac{m_1m_2}{m_1+m_2}$$
So if you increase the mass of one of the particles, while $k$ stays roughly the same, $\omega_e$ becomes smaller and the energy spacing between the vibrational levels decreases.

Isotopic substitution effects both ground and excited state vibrational frequencies as described above. However,  the effect is very small for all but H to D substitution as can be seem from the reduced mass equation if $m_1 = m_H = 1$, then $\mu=m_2/(1+m_2)$ which is approx 1 unless $m_2$ changes drastically on isotopic substitution. Secondly, the force constant remains $exactly$ the same on isotopic substitution; it is defined in that way since $\mu$ is separated out in the equation for frequency $\omega _e$.

Actually, the fact that the force constant is the same for different isotopologues comes not from the definition but is a consequence of the Born-Oppenheimer Approximation: the nuclei are all assumed to be of infinite mass so the BO curves for H$_2$ and D$_2$ are the same to a good approximation. One can take the finit mass of the nuclei into account and derive adiabatic correction to the BO potential and even interactions between the different potentials to obtain the nonadiabatic corrections.

The $E_v$ is a consequence of the BO approx. We write down the harmonic osc. energy using a $k$ that does not depend on mass, hence I wrote 'by definition'. In fact  in the BO approx the nuclear and electronic motion is separated to allow the Schrodinger eqn. to be solved. It is not necessary to assume infinite mass, the requirements are variously stated as $(m_e/m_a)^{1/4}<<1$ and the terms left by making this approx are  of the order of $(m_e/m_a)E_e$ . Considering that relative electron to nuclear mass are typically 1:10000 this means an error of about 1 in $10^4$ to $10^5$ cm$^{-1}$.


text:
Why do unripe papayas taste less bitter after being boiled?
An unripe papaya tastes very bitter if you try to eat it. Actually, the juice that comes from it is so corrosive that it can burn your skin, causing severe pain. I've experienced this first hand when cutting raw papayas.
However, after boiling, these papayas taste quite bland. Very little of the original bitterness remains.
What chemical substances cause the original bitterness, and what happens to them when boiled?

Unripe papaya contains a high concentration of Papain enzyme, which is an enzyme that breaks down peptide bonds and therefore "digests" proteins, including potentially human tissues.  Enzymes themselves are proteins, and proteins tend to have a bitter taste.
Papain can self digest into amino acids over time, and this process could be accelerated by heating, as chemical digestion reactions will approximately double in rate for every 10K increase in temperature.  So one mechanism is that Papain may break itself down quickly when heated.  The other is that at a high enough temperature, Papain may denature.

The heat likely destroys the compounds responsible for the bitter taste. As for what the compounds are, I haven't been able to find any information other than that they have high polyphenol content, which are common astringents, but these are not susceptible to degradation by heat.

I think that a biological process named OSMOSIS occurs when papayas are placed in boiling water. What happens is that:
The interior of the papayas has a more concentrated solution than the water surrounding it. The water molecules on the exterior move inwards through the semi-permeable membrane(the papayas outer covering) and lessen the bitter effect.

Osmosis is a physical process. 2) Osmosis may reduce perceived bitterness a tad, but not to the extent as presented in the question — and only if bitterness is due to something inside the papaya’s cells.


text:
How is the configuration of an ester enolate double bond assigned according to CIP?
Two principal modern methods for enantioselective aldol reactions — the Evans aldol and the Masamune aldol — use a carboxylic-acid derived enolate rather than a ketone enolate as e.g. in the Paterson aldol reaction. For ‘traditional’ ketone enolates, a rule of thumb dictates that (E) enolates give anti-products and (Z) enolates give syn-products, typically rationalised by the Zimmermann-Traxler transition state model. Assigning (E) or (Z) to the enolate of an ethyl ketone poses no problems, as oxygen always has a higher Cahn-Ingold-Prelog priority than carbon, so it is all about the orientation of the enolate oxygen and the methyl group (see figure 1).
An Evans aldol reaction, the enolate of which is also shown in figure 1, can be analysed similarly. Oxygen has a higher priority than nitrogen, so the same basic rules apply.
A Masamune aldol reaction, however, has two oxygens attached to the enolate carbon, formally making it an O,O-ketene acetal. The CIP rules would dictate to follow the connectivities atom by atom. The Masamune auxilliary, like any ester, has a carbon attached to the oxygen. The enolate anion may be strongly attached to a certain atom (e.g. when using a boron- or titanium(IV)-mediated aldol reaction) but also may not be, or the attachment may be ambiguous (e.g. a potassium salt used for enolating in the presence of $\ce{LiCl}$).

Figure 1: The enolates in a Paterson-, Evans- and Masamune-aldol reaction and their double bond configurations where known.
To establish whether the double bond geometry is (E) or (Z), must I know if and which atom is attached to said oxygen? What if two different possibly attaching atoms (e.g. potassium and lithium) would give two different geometries? Is it possible to ignore said attachment altogether and just define the enolate oxygen as a free monoanion?
And do the CIP rules differ from common usage?

I think I read in a Warren's book that the assignation of Z/E leads to a similar problem in thioester enolates. He recommends that oxygen should be regarded as older than sulfur because it is easier to compare these very similar reactions. A change of priorities would be ambiguous. Therefore the oxygen of the carbonyl group should always have the higher priority. I will try to look it up.

@Marko Yeah, I also stubled upon a reference yesterday according to which Evans suggested the same in Asymmetric Synthesis vol. 3. Alas, I cannot access it, my library doesn’t have it.

Whilst the (E)/(Z) nomenclature works fine for enolates of ketones and aldehydes, there is, as you've discovered an issue with the assignment of enolate geometry for esters (an amides and thioesters) since depending on the nature of the ester and the metal counter-ion used, we end up with a different answer.
Whilst the CIP rules are trivial to apply to any enolate, their use is not always conducive to describing the chemistry going on, especially when we end up with (E) and (Z) enolates both giving syn-adducts (for example) merely because we've changed a counter-ion (the actual spacial configuration of the enolate is identical in both cases).

To avoid this switching of configurations, we usually choose some other way of definition such that at the beginning of any thesis or SI where enolates are used, there is usually a statement that explains the way in which the author is planning to define their enolates (i.e. define them however you wish, but make sure you define them).
This is often not in agreement with formal CIP rules for the reasons outlined above, or at least not consistently in agreement with CIP rules (things may happen to end up named correctly regardless of how we define them).
This statement defining the enolate geometry is generally a re-hash of Prof Evans who stated:

...irrespective of the formal CIP criteria, the O-atom bearing the metal (the OM residue) is given a higher priority, and the ipso-substituent X (in enolates) the lower one...

Whilst this is widely accepted, there are some issues, mainly because there are inconsistencies between groups: Evans' calls it (E) but Myers' calls it (Z) with Furstner calling it (E) but for a totally different reason than Evans'.  For this reason alone, its always worth looking at the actual schemes and defining the enolate yourself rather than trusting that enolate X gives Y.
This issue is the source of the major objection to Evans' idea about the nomenclature. The whole reason we use E/Z/R/S is because they are well defined, I can tell you that I have (R)-..... and you'll know what I'm talking about, because we both follow the same rules, and if you don't, you're wrong.
The proposed solution to Evans' solution (Ernest Eliel) is to stop messing with these well defined descriptors, and instead go back to using cis and trans for enolate geometry, as these terms are less well defined and as such abusing them in this way doesn't lead to any significant confusion.

While this is not quite fit to be an answer, considering that it hardly answers most of your questions, it is too long and perhaps too important for a comment.
Evans writes in Asymmetric Catalysis, vol. 3 (p 11):

It has now been well established that, for many metal enolates, kinetic aldol diastereoselection is strongly influenced by enolate geometry. As a prelude to the ensuing discussion, the following nomenclature is adopted with regard to metal enolate isomerism. In considering the isomeric metal enolates 5 and 6 we adopt the E/Z notational format for olefins (35) with one operational change that greatly simplifies the subsequent discussion of enolate reactions. For the C-1 enolate substituents $\ce{R_1}$ and $\ce{OM}$, the highest priority designation is always assigned to the $\ce{OM}$ group, independent of the metal. The normal priority designations for substituents at C-2 are maintained. Thus, 5 and 6 are defined as having the Z and E geometries, respectively. The rational basis for incorporating this liberalisation of the E/Z notational format is based on the fact that, in terms of conferred reactivity, the $\ce{OM}$ ligand is the dominant substituent.

Although it's hardly a satisfying explanation, it does support what NotNicolaou wrote in his answer. Such an assignment may or may not contravene the Cahn-Ingold-Prelog rules, but is simply adopted for ease of use.
Reference 35 is J. Chem. Educ. 1971, 48 (3), 163.

The quote that Marko mentioned in the comments is from Wyatt and Warren's Organic Synthesis: Strategy and Control (p 46):

There is a problem in defining enolates as E or Z. If we follow the rules strictly, a series of enolates with the same geometry, such as the four below, would change from E to Z as we changed the metal on the enolate or as we changed from ketone to ester to thiolester. When we are discussing enolates, we shall name them E or Z according to the relationship between the substituent R and the OMetal substituent. This is the generally agreed method and takes no notice whether the enolates are strictly E or Z alkenes.


text:
What would happen if I mixed positive and negative photoresists?
When exposed to light sources with certain wavelength: for positive photoresist materials, the exposed parts will be washed off by developer and the rest of the unexposed parts will remain on the substrate; while for negative photoresist materials, the exposed parts will remain on the substrate after developing, while the unexposed parts will be washed off.
So my question is, when I mix any positive and negative photoresist together, what will happen after exposure to light sources and developing? Which parts will be washed off?

Do you know what chemicals are in your materials ? Especially what the developer is composed of ? Is it just a solvent or is there some chemistry going on ? Is it the same for the positive and negative photoresist ?

I have S1813(the positive photoresist) and SU8 (the negative photoresist) in my lab. I'm a physics student so not quite sure about the chemical structures of my materials. They both have their own developers, not the same. I'm just curious if I mix them, will their positive and negative photoresist abilities cancel out...

Also posted to Physics SE.

Crosssposted to http://physics.stackexchange.com/q/277619/2451

Without knowing what the positive developer does to the negative photoresist and vice versa, I can only guess.
Assuming the developers are just washing agents, then the positive photoresist will stay on the unexposed parts, and the negative photoresist will stay on the exposed parts. So you would get an entirely covered substrate.
Why not try the experiment you are proposing in your lab, and tell us the results? That would be neat.


text:
Explain volume contraction in mixtures of alcohol and water
Could someone explain why volume contraction occurs when you mix an alcohol such as ethanol with water in relation hydrogen bonding and the dipole-dipole forces?

I think that this answer actually addresses your question pretty nicely.   It isn't purely about the ethanol-water system, though that is given as an example, but I think all the concepts apply.

Liquid water can be thought of as having an average configuration that tends to be an open, almost crystalline structure, somewhat like that for hexagonal ice crystals $\mathrm{^{[1,2,3]}}$.  This structure has been confirmed by both x-ray and neutron diffraction experiments.  Below is a depiction of the hexagonal structure of ice for reference:

Source: Boundless. “The Structure and Properties of Water.” Boundless Chemistry Boundless, 26 May. 2016. Retrieved 06 Feb. 2017 from boundless.com
This phenomenon of the ice-like structure of liquid water is also illustrated in the observation that at temperatures below $\pu{4^oC}$, the density of water decreases with decreasing temperature as it becomes more ice-like prior to actual solid crystalline ice formation.
So, if we envision this average crystal-like configuration for liquid water, then two effects come into play regarding the volume contraction observed when water is mixed with ethanol.
Firstly, some fraction of the hydrocarbon tails of the ethanol molecules is able to fit within this lattice, filling some of the voids within this water structure.
Secondly, the hydroxyl moiety of ethanol engages in hydrogen bonding with water.  As ethanol only has a single hydrogen bond donor, however, the crystal-like structure of the liquid water cannot be fully maintained, and thus it somewhat collapses.
Also, please take a look at these answers regarding the general concept of volume contraction of liquid mixtures, not specific to the case of ethanol and water.
1 http://www.diss.fu-berlin.de/diss/servlets/MCRFileNodeServlet/FUDISS_derivate_000000003174/2_Chap1.pdf;hosts=
2 http://crystal.med.upenn.edu/sharp-lab-pdfs/sharp_EncLifeSci.pdf
3 "Water structure and changes in thermal stability of the system poly(ethylene oxide)–water", J. Chem. Soc., Faraday Trans. 1, 1981,77, 2053-2077


text:
Why is CrCl3 acidic?
Why is $\ce{CrCl3}$ acidic? I know that it has a high charge density given that it is a transition metal ion, and can hence polarize neighbouring water molecules, resulting in hydrolysis and formation of $\ce{H+}$. However, I am a little confused over how this happens? Is the water molecule that it polarizes the ones that have formed a coordinate bond with it, or simply any water molecule? Also, with the high charge density, it should be attracting electrons from the water molecule. But does it mean that it attracts the 2 electrons forming the bond between $\ce{O}$ and $\ce{H}$? And if it attracts them to itself, why doesn't it form an extra coordinate bond? Why is there still only 1 coordinate bond? Another thing that I don't understand is that $\ce{OH-}$ has a lower d-orbital splitting ability than $\ce{H2O}$, doesn't that mean coordinate bonds with $\ce{H2O}$ are preferred? Why do they still polarize $\ce{H2O}$ to get $\ce{OH-}$ instead?

For starters, note that you will probably be using the hexahydrate $\ce{CrCl3 * 6 H2O}$. In fact, there is not the hexahydrate, but three distinct hexahydrate crystal structures known. These can be described as $\ce{[CrCl2(H2O)4]Cl . 2 H2O}$, $\ce{[CrCl(H2O)5]Cl2 . H2O}$ and $\ce{[Cr(H2O)6]Cl3}$. In solution, these ions will dissociate, giving you the corresponding free cation, one to three free chloride anions and potentially additional water molecules. It is safe to say that these three anions will take part in an equilibrium reaction $(1)$.
$$\ce{[CrCl2(H2O)4]+ + 2 H2O <=> [CrCl(H2O)5]^2+ + H2O + Cl- <=> [Cr(H2O)6]^3+ + 2 Cl-}\tag{1}$$
It doesn’t really matter which one is the most prevalent species, the argument is generally the same for all three. In all three cases, we have a positively charged cation to which water forms a dative bond. In neutral solution, water is known to self-ionise according to equation $(2)$
$$\ce{H2O <<=> H+ + OH-}\tag{2}$$
The displacement of a proton is facilitated by neighbouring electron withdrawing groups: They reduce the charge density making an anion more stable. And for all intents and purposes, cations are electron withdrawing, displaying a $-I$ effect (negative inductive effect). Therefore, forming a coordinate bond to a chromium(III) ion — even if the overall complex only has a $1+$ charge — will increase the tendency of a water molecule to deprotonate. Or, in equation terms, the equilibrium of equation $(3)$ is right-shifted when compared to equation $(2)$’s equilibrium. (I’ll only discuss this for the monochloridocomplex for simplicity.)
$$\ce{[CrCl(H2O)5]^2+ <=> [CrCl(OH)(H2O)4]+ + H+}\tag{3}$$
Since displacing a proton is easier, the solution is more acidic, the $\mathrm{pH}$ is lower.

$\ce{CrCl3}$ is acidic because of the following reaction (credits):
$$\ce{CrCl3 + 3H2O -> Cr(OH)3 (s) + 3HCl}$$
The $\ce{Cr(OH)3}$ is insoluble, so it precitpitates out and the acid $\ce{HCl}$ remains.
Why $\ce{Cr(OH)3}$ is insoluble is explained here.


text:
Why is CH3CH2NH3Cl acidic?
Why is $\ce{CH3CH2NH3Cl}$ acidic?
I know that it has something to do with how it is formed from a weak alkaline, but I cannot identify which acid and base it is formed from.

Its a salt of weak base and strong acid so on hydrolysis it forms acidic solution

Have you tried drawing it? What do you think ought to be the cation and anion parts when dissociating in water?

As, mentioned its a salt formed by reacting Ethyl amine with HCl,

In short
$\ce{CH3CH2NH3Cl}$ is acidic because it is formed from a strong acid ($\ce{HCl}$) and a weak base ($\ce{CH3CH2NH2}$).

Elaborated
Notice that this compound is ionic, comprising $\ce{CH3CH2NH3+}$ cation and $\ce{Cl-}$ anion.
I mentioned that $\ce{CH3CH2NH2}$ is weak, meaning that this reaction is reversible:
$$\ce{CH3CH2NH2 + H2O <=> CH3CH2NH3+ + OH-}\tag1$$
In an acidic environment, we can rewrite it as:
$$\ce{CH3CH2NH2 + H3O+ <=> CH3CH2NH3+ + H2O}\tag2$$
Therefore, when $\ce{CH3CH2NH3Cl}$ dissolves in water, there would be initially an equal amount of $\ce{CH3CH2NH3+}$ and $\ce{Cl-}$ in terms of number of moles.
Using equation $(2)$, we know that $\ce{CH3CH2NH3+}$ will react with water reaching an acidic equilibrium.
Note that there is no $\ce{OH-}$ to start with (very little in reality), so equation $(1)$ is not applicable.
Therefore, the following species would be present:

$\ce{CH3CH2NH3+}$
$\ce{CH3CH2NH2}$
$\ce{H2O}$
$\ce{H3O+}$
$\ce{Cl-}$

Now it is apparent that $\ce{H3O+}$ makes it acidic.

No mistakes. I wish there were more first answers like this! Have my upvote!


text:
Total valence electrons in a rhodium complex

How many valence electrons does the following metal complex have?

I have read that benzene donates 6 electrons to a transition metals valence electrons. Rh has 9 from itself, and H and Cl add 1 each; hence in this assignment above I thought one would end up with: 9+1+1+2+2+6=21 valence electrons.
However, the given answer is 16. How is that so?

You add the charge of the ligands, H: and Cl: both have -1. I'm getting 17 though, but if I assume the benzene has a charge of -1 it's 16. That might explain why the benzene isn't donating all 6 aromatic electrons, but rather a single bond of 2. So, 9 - 3 + 10 = 16.

What you did is neutral counting. I typically find oxidation state counting easier. In that case, realise that rhodium is rhodium(III) (three monoanions and a neutral complex) so count $6 (\ce{Rh^3+}) + 2 (\ce{H-}) + 2 (\ce{Ph-}) + 2 (\ce{Cl-}) + 2 \times 2 (2 \times \ce{PPh3}) = 16$.

This is not a benzene ligand ($\ce{C6H6}$), but rather a phenyl ligand ($\ce{C6H5}$).

In the benzene complex, all the carbons have their hydrogens. The pi electrons of the ring act as donors to the metal. On the other hand, in the phenyl complex, the ring carbon bonded to the metal does not have its hydrogen. That carbon is formally negatively charged and the lone pair on carbon acts as a donor to the metal.
Aryl ligands, just like alkyl ligands, count as 1-electron donors within the "covalent" electron-counting system (which you seem to be using, since you counted H and Cl as 1e donors as well).
Therefore the total electron count is $9 + 1 + 1 + 2 +2 + 1 = 16$.


text:
Pi bond need sigma bond to exist
My teacher was teaching about chemical bonding and she said that pi bond needs sigma bond to exist, but she didn't know why. It is a doubt which I get in my mind, can someone explain it?

C2 does have 2pi bonds but no sigma bond.

С2 is a pretty exotic thing. As for the explanation, I'd say $\pi$ bond does not "need" $\sigma$ bond; it's just that two atoms would rather form a $\sigma$ bond, if they like each other and have spare orbitals. Only if they can't form a $\sigma$ bond (for example, because it is already there), they start looking for other options.

I think it would be more correct to say that pi bonds tend to exist in conjunction with sigma bonds, rather than that they need sigma bonds to exist - as pointed out in the comments.
This can be explained rather easily by basic molecular orbital (MO) theory, which seems like an appropriate framework for the question. Let's consider the simple case of a homoatomic bond such as C=C. When you have two C atoms coming into contact so as to form MOs, you form sigma MOs (bonding and antibonding) from the set of atomic S orbitals, and pi MOs (again, bonding and antibonding) from the Px and Py atomic orbitals.
Among the (molecular) orbitals, the sigma bonding MO generally has the lowest energy, followed by the pi bonding MOs. This means that the sigma bonding orbitals will 'fill' before the pi bonding orbitals. Because electrons preferentially occupy MOs from lower to higher energy, as long as the sigma bonding MOs are lower energy than the pi bonding MOs you will see sigma bonding orbitals populated before pi bonding MOs. Thus, to have a pi bond (occupied pi bonding MOs), you generally also have a sigma bond (occupied sigma bonding MO).
As pointed out in the comments to your question, this is somewhat of a generalization - but I think it's an accurate description of the phenomenon your instructor described.


text:
Why do raisins bob up and down in carbonated water?
Yesterday I preformed a fun little experiment where I poured a glass of seltzer water into a glass, then dropped a few raisins into it.
I observed that the raisins would float at the top of the glass for a few seconds, then sink all the way to the bottom, then float back up, and repeat seemingly without end.
Why does this happen?

The raisin has nucleation sites on it that allow bubbles of $\ce{CO2}$ to form. The raisin is light enough to be lifted by the bubbles as they push their way to the surface. As the bubbles are released into the atmosphere, the raisin once again sinks until more bubbles form on it. This will continue until the soda water has lost the majority of its dissolved $\ce{CO2}$.

The rest of the story is that a raisin (without CO2 bubbles) is denser than water and would sink in uncarbonated water. Also for this experiment the difference in density between carbonated water and uncarbonated water is negligible compared to the difference in density between water and a raisin.


text:
Gaussian calculation problem - Maxcycle values for Opt and SCF
I've been trying to calculate vertical ionization potentials (IP) and electron affinities (EA) for a bunch of prostaglandin derivatives (23-25 heavy atoms). My calculation setup in Gaussian09 looks mostly like that:

# opt=(tight,maxcycle=1000)
freq=noraman cphf=noread
b3lyp/6-31g(d) geom=connectivity
integral=grid=ultrafine scf=maxcycle=1000

As you can guess, I have been struggling with converging the geometry and finding the stationary point for my molecules. Some were kind of stubborn and wouldn't converge, so in desperation I increased the Maxcycle values for Opt and SCF from a 1000 up to 5000, and I succeeded with some of them. Now, my question is if I need to recalculate all other prostaglandins that have already nicely converged with lower Opt(Maxcycle) or SCF(Maxcycle) options (I used them in different combinations like 1000, 3000 and 5000)? The thing is that I want to be able to objectivelly compare the IP and EA values between different derivatives.
I am concerned because I remember that someone told me that if I went for SCF=QC option I would need to recalculate the whole set of molecules with the same keyword option. However, as far as I know SCF=QC is like a substantial change in the setup as it uses another algorithm for calculations. Hopefully, changing the SCF(Maxcycle) value is not that dramatic, is it? How about SCF=Verytight - would it demand recalculating everything or not?
Another thing that bothers me is Opt=Tight. What if I decided to go for Opt=Verytight. Would I need then to recalculate all of the molecules with the same setting?
And lastly, what if I changed Integral(grid=ultrafine) to Integral(grid=ultrafine, acc2e=12) which is supposedly the most precise approach?

You are mixing two independent, but slightly related topics.
First is the SCF convergence, i.e. obtaining the wavefunction. This step is necessary for any calculation, without it the Gaussian crashes. For well behaved molecules, as the prostaglandines (organics) most probably are, the solution should be found fast (<30 SCF cycles). It also does not matter which converger you choose, i.e. QC or DIIS. How "tight" you decide for the SCF affects the final energy, but only the number of digits (precision). The SCF does not have any "verytight" option, according to manual, only the Conver=N.
On the other hand, the integral and grid settings do have effect on the energies, so all calculations have to be done with the same grid. I won't recommend changing the acc2e settings, the error is negligible.
Regarding the geometry convergence, you should obtain an optimized geometry in less than 100 steps for reasonable organic systems. If this does not happen, something is rotten in your setup and you should look deeper into the problem. In principle, you cannot converge the geometry to "tight" without sufficient accuracy of the SCF, namely the ultrafine grid, which you now do use.
How good the optimization was can be guessed by the frequencies which should be zero (translations and rotations). If they are significantly off, optimize the geometry more tightly. But as said above, you cannot optimize on noisy surface, where the noise is produced by the grid and SCF convergence.

Thank you @ssavec. So the bottom line is that I can switch between Opt=Tight or Opt=Verytight, and also change Opt(Maxcycle) values. The same pretty much goes for SCF settings. I should however be consistent with the Intergral(Grid) settings. One more thing, if you will. Can I change the above mentioned options doing a Frequency job for a neutral molecule and then for its ion? You know what I mean, I'd like to go high up with the parameters calculating the neutral molecule which has to converge, and loosen them up a bit for the ions that do not have to converge for vertical energy calculation.


text:
Mass change during phase transition
When you fill up a balloon about a quarter way with liquefied butane fuel and let it sit at room temperature it will turn into gas. But why does the gas weigh the same as the liquified butane?
The liquefied butane liquid weighs just about as much as water.

Assuming you fill the balloon only with liquid butane, the answer is very simple - the gas in the balloon at room temperature IS the liquid in the balloon initially. If you put some amount of liquid butane into the balloon and seal it at t=0, then allow it to warm until the liquid has evaporated, there is essentially no transport of other matter into or out of the balloon during that time. The only thing that has happened is that the initial charge of liquid has absorbed thermal energy from the surroundings and undergone a phase change from liquid to gas.
Because there is no change in the amount of material in the balloon - there are the same number of butane molecules in the gas phase as there were initially in the liquid phase - there can be no change in mass. Liquid and gas phase contain the same number of molecules, at some fixed mass per molecule.
Note that while the mass does not change, the density does.

The mass of butane will be the same but the weight might be different due to buoyancy. The mass of air displaced by the extra volume of gaseous butane might matter a little.


text:
Do both electrons in a dative bond come from the same atom?
The textbook definition of a dative bond is that it is a bond "in which the two electrons derive from the same atom" (credits).
However, I found that the two electrons of a dative bond may come from different atoms, and the two electrons of a "normal" bond may come from the same atom.
For example, consider the following reaction:
$$\ce{CH3CHBrCH3 + NaOH -> CH3CHOHCH3 + NaBr}$$
The $\ce{C-O}$ bond in $\ce{CH3CHOHCH3}$ would be an example where both electrons actually come from the $\ce{O}$ atom.
Actually, one of the electron may not have come from $\ce{O}$, but the $\ce{Na}$ atom!
What further complicates things is that all electrons are equal, so how can we tell their origins?

Your textbook’s definition is inconsistent for exactly the reasons you point out. I am going to use slightly different examples than you did but they work equally well.
A prime example for dative bonding is $\ce{H3N\bond{->}BH3}$, amminborane or ‘inorganic ethane’. The latter name because it is isoelectronic to $\ce{H3C-CH3}$, (organic) ethane. Typically, amminborane is generated by ammonia reacting with $\ce{B2H6}$ — or at least it seems very plausible to create it in that way. Ethane can be created in two different ways:

By the recombination of two methyl radicals:
$$\ce{H3C. + .CH3 -> H3C-CH3}\tag{1}$$
By the attack of a formal methide anion with a formal methanium cation:
$$\ce{Li-CH3 + H3C-OTf -> H3C-CH3 + LiOTf}\tag{2}$$

(The second version is absolutely equivalent to your $\mathrm{S_N2}$ reaction to give isopropanol.)
Going by your textbook’s rules, the first reaction might generate a normal covalent bond, but the second should give a dative bond, since both electrons of the $\ce{C-C}$ bond came from methyllithium and none from methyl triflate. And, of course, we cannot distinguish between these electrons and they are delocalised all across the molecule — that goes for both ethene and amminborane.

A much better definition of dative versus normal covalent bonds does not consider their generation but rather what happens when the bond dissociation energy is added, i.e. their dissociation. In amminborane’s case, the dissociation happens at a rather low energy and will yield ammonia and borane again, so the electrons ‘end up where they were before.’ In more technical terms, the compound dissociates heterolyticly. (Equation $(3)$)
$$\ce{H3N\bond{->}BH3 -> H3N + BH3}\tag{3}$$
Ethane, on the other hand will dissociate homolyticly upon being fed with the (much higher) $\ce{C-C}$ bond dissociation energy. It will reform the two methyl radicals used to generate it in equation $(1)$. (Equation $(4)$)
$$\ce{H3C-CH3 -> H3C. + .CH3}\tag{4}$$

This conversation has been moved to chat per the request of one of the participants.