text:
Real space term for PPPM vs PME
I was under the impression that differences between the various long range electrostatic algorithms, ewald, PME, PPPM, to name a few, lied solely in their treatment of the reciprocal space term. Specifically, I thought all methods looped over a particles neighbor and summed q(i)*q(j)erfc(alphar)/r for the real space calculation. However, this is my impression, which I have yet to validate with others. I became confused when I recently read "Ewald summation techniques in perspective: a survey" whose equation 22 seems to have suggested a different way to calculate the short range potential. Could anyone else chime in here, and either tell me where I am wrong, or I am correct?
Thanks!

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There are a number of ways of looking at Ewald's summation, but one is that instead of trying to solve the original problem where the potential is due to a set of point charges you, instead, solve the system for a set of smeared charges and add a correction. Now there are limitations on what are good ways to do the smearing, but there are still quite a lot of possible choices. The most common is to replace the original delta function charge with a Gaussian, the width of which is determined by alpha in your question. This leads to the error function and exponentials you are used to. However it's not the only possible choice. In equation 22 in the paper you refer to a different way of smearing the charges has been chosen, a linear ramp centred on the particle. This is described in equation 21. As the smearing is different the individual terms in the Ewald summation have different forms when compared to Gaussian smearing, but it's no less valid a way of proceeding.
If you want to learn more about this look into Fourier Transforms and the Convolution theorem. A bit of reading on that and the Ewald Summation in all its forms becomes pretty straightforward.

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text:
What is the structure of nitrite?
I know that $\ce{NO2}$ has 3 effective pairs,  so it is likely to have trigonal planar structure. In trigonal planar structure we usually see three atoms bonded to central atom at 120°. But in $\ce{NO2-}$, I observe that though it has three effective pairs,  but still, it is not having the desired structure,  with only 2 O atoms bonded to N atom. Can anyone explain the situation?

Nitrogen is actually surrounded by 3 effective pairs. But 2 of them are bonding and the third one is a lone pair. According to VSEPR theory, the structure should be  trigonal planar.

The angle is not exactly $120^\circ$, as the electrostatic  repulsion between non-bonding pair and bonding pair is higher than the repulsion between two bonding pair. So, the angle of $\ce{O-N-O}$ is slightly smaller than $120^\circ$.
The distribution of the three electron effective pairs is trigonal planar. But the molecule structure is bent (like letter V).

source: Stack Exchange

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text:
What makes many water soluble fertilizers blue?
I've noticed that water soluble fertilizers tend to be blue. What chemicals or chemical reactions cause this color here? Is this necessary or for aesthetic purposes? I haven't seen this in organic fertilizers. I figured chemists might be able to examine the ingredients of Miracle Gro or something and tell me. I'm guessing it has something to do with copper.

What chemicals or chemical reactions cause this color here? Is this necessary or for aesthetic purposes?

I have some Miracle Gro on hand, the water soluble kind: The blue stuff is copper sulfate, or ${\rm CuSO}_{4}$. Here is the MSDS for Miracle Gro Water Soluble All Purpose Plant Food, which lists the chemicals in the mixture.

Is this necessary or for aesthetic purposes?

It's necessary in these formulations for purposes detailed (in part) here and here. In short, it is included in these formulations because plants need copper for biosynthesis; and because it acts as a fungicide, presumably leading to a nicer ecosystem in one's lawn/garden.

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text:
How can I use sunlight to heat a RBF and reflux a reaction?
This may seem like a off topic question but I really need help on it. I am currently working on a polymerization reaction in which I'm breaking down plastic polymers into smaller molecules. However, since this is a green chemistry project, I decided, with the suggestion of other chemists, that I should a method that doesn't require a lot of electrical energy to heat it. One idea was an enzyme, but I decided I want to use solar energy to reflux the RBF. I have a few ideas and please comment on them and also provide your own suggestions:

First is using a parabolic mirror below the flask to focus the heat towards it.
Second idea is painting the RBF black and hoping that is sufficient to heat it to reflux.

How about buying solar panels and connecting them to a nice electrical heater?

@Jori It sounds like a nice idea but converting sunlight to electrical energy and then to heat isn't actually that efficient (mainly inefficiency of the solar panel). On the other hand, parabolic mirrors and/or black surfaces can be pretty effective. You are probably looking for something along the lines of this.

If you get good sunlight, the idea provided by @bon can be done with very simple materials, i.e. some cardboard and the shiny side of a piece of aluminum foil.

Thanks everyone. I love the solar heater idea. I talked to my professor and we will try to build our own next week. If it doesn't heat enough, then I will buy this solar cooker. ITs suppose to reach up to 230 C


text:
Calculating Fluoride in toothpaste

Question
  A linear regression line was obtained; $$\log(\ce{[NaF]})~(\mathrm{mol~L^{-1}})~ \mathrm{vs~EMF}:
y = 59.8x +600.8$$
  $0.7540~\mathrm{g}$ of toothpaste was diluted in a$50~\mathrm{mL}$ and tested to obtain a EMF value of $390$. what is the ppm of $\ce{F-}$ in $198~\mathrm{g}$ of toothpaste?

I plugged in $390$ into the eqn to get $-3.525$, taking the antilog to obtain $0.0002985$, and times it by the volume $0.05~\mathrm{L}$ to get the moles of $\ce{NaF}$.
Due to stoichiometry, this also equals $1~\mathrm{mol}$ of $\ce{F-}$. I multiplied by $18.998~\mathrm{g~mol^{-1}}$ to obtain $0.000283527~\mathrm{mol}$ of $\ce{F-}$. I then multiplied by the factor: $\frac{198}{0.750}$ to obtain $0.074454015~\mathrm{g}$ of $\ce{F-}$ in the $198~\mathrm{g}$ of toothpaste.
I'm stuck here, how can I calculate the ppm?
The expected value of fluoride in toothpaste is $1000~\mathrm{ppm}$

I'm having difficulty getting your expected answer.  I follow your method through moles of fluoride, but at that point, your units get messed up.  Multiplying by the molar mass gives you 0.0002835 grams of fluoride.  That value divided by the original mass of the sample, 0.7540 grams of toothpaste, multiplied by one million, should give you ppm, regardless of sample size.  I get 376 ppm.  Parts per million is a concentration, so if you figure it out once for any quantity of toothpaste, it should apply to all quantities.

thanks for your support, i guess it has to deal with the regression line, it must be wrong? i get the same answer.

@Spencer-Trinh I'd guess you're past caring at this point, but I'm curious, was this a word problem or a lab experiment?

@JasonPatterson I think you already pretty much gave the answer. Would you mind writing your comment up and posting it as an answer?

Working with your numbers gives me the same result through moles of $\ce{NaF}$.  However, at this point you can multiply by the molar mass of fluoride, 18.998 g/mol, to get grams of fluoride, not moles, specifically, 0.0002835 g $\ce{NaF}$.
Parts per million is a mass-based concentration measure.  It's similar in concept to percent mass, but instead of parts per hundred, it uses parts per million.  Since we know the original mass of this sample, we can calculate the concentration in ppm based on that value.  It will be the same concentration as in the larger sample.
I get the following:
$\frac{0.0002835 \textrm{ g } \ce{F-}}{0.7540 \textrm{ g toothpaste}}\times 1000000 = 376.0\textrm{ ppm }\ce{F-}$


text:
Why does acetate aldol with Evans auxiliary give almost no diasteroselectivity?
The following acetate aldol reaction gives no diasteroselectivity:

Because I don't know what the basis of above statement is (I heard it in a lecture and read it on some internet web pages), I expect the two products are formed as a racemic mixture.
For the related propionate aldol we can draw two (considering R of aldehyde to always be equatorial) possible transition states:

The question is why we don't get good selectivity if we compare the acetate aldol with the same transition states for the propionate aldol?
As a side-note: in practice chlorine is used instead of Me to obtain good selectivity and the chlorine will be eliminated finally by use of Zn/AcOH.

Quick reaction :) But still it's amide not ester.

The Evans auxiliary is an imide.

Yeah, is it hydrolyzed later?  Btw your question is very narrow - it lowers chances for good answer

With $\ce{LiOH/H2O2/NaHSO3}$ it can be hydrolized to a $\beta$-hydroxy-carboxylic acid.

Do you have any sources for the transition states and the reactions itself? What are we talking about when we say failed selectivity, ist it racemic or only slight discrimination?

Aldol reactions with Evans aldol can be described with this Zimmerman-Traxler transition state (I have no resources than those of the lecture). I expect people mean racemic when they say "no selectivity", otherwise they would need to say "bad selectivity" if they mean slight discrimination.

You might find something useful here

@ron I haven't found an explanation in your link. Unfortunately I've not enough time now to look into all the "Other Useful References" listed there.

I bet if you build a cyclohexane-chair model you'll find that when the methyl group is present it strongly favors the cyclohexane conformation with the methyl equatorial (there won't be much of the other chair where the methyl would be axial).  When you change Me to H you lose that conformational preference and both stereoisomers can now be formed, one from each of the 2 possible chairs.

I don't think so, because (in my knowledge) it's the isopropyle groupe of the imidazolidinon (conformation determined by dipole minimization) and not the methyl which primatily determines from which side the aldehyde can attack. But I agree with you exchange of Me by H would give (very slight?) discrimination. May the question is to which extent the i-Pr and to which the Me dictates from which side the attack can happen.

1. The acetate aldol reaction
Before looking at selective variants, its worth pointing out what an acetate aldol is.
An acetate aldol (below), is the simplest kind of aldol in which the β-hydroxyketone product has no methyl group at the α-position to the carbonyl. This is in contrast to the (more usual?) propionate aldol reaction in which the α-position contains a methyl group (as is common in polyketides).

Source: Modern methods in stereoselective aldol reactions, Wiley.

Early studies into acetate aldols quickly established that it was far harder to set a single stereo centre than to introduce two stereocentres concomitantly. This was due to the cyclic transition states involved (Zimmerman-Traxler model) in which the methyl group bound to the enolate (in the propionate aldol) exhibits a level of control which isn't present in the acetate aldol.
2. The Evans' auxiliary and its applications in the aldol reaction
Whilst many aldol reactions exhibit diastereoselectivity, they show no inherent enantioselectivity. Several approaches were developed to fix this including chiral auxiliaries and chiral reagents. One of the most 'famous' of these was Dave Evans' oxazolidinone auxiliaries.
These are used to gain enantioselectivity, but can also be used to override substrate control to get the desired product.

Source: Dave Evans' CHEM206 Lecture notes, Harvard

With propionate aldols, this works excellently (see above), giving d.r.'s well in excess of 20:1 (about the maximum NMR can detect), with a 300:1 d.r. being quoted by Evans'. With acetate aldols however, the selectivity drops down to 1:1.
Let us consider the TS for the propionate aldol.

Source: Dave Evans' CHEM206 Lecture notes, Harvard

Evans' himself discusses the two diastereomeric TS structures. The favoured on top we now rationalise by opposing the dipoles between the auxiliary and enolate.
An alternate reason for why the disfavoured TS was disfavoured was provided by computational work, in which it appears that the methyl group in the propionate system clashes with the auxiliary.

Source: Evans' and Houk.

By disfavouring one TS (Steric clash) and favouring another (opposing dipoles), we gain good diastereoselectivity.
When the α-methyl group is taken away (as it is in the acetate aldol), we lose the unfavourable interaction (no more steric clash between the enolate and the auxiliary) leaving us only with the favourable dipole-dipole minimisation.
There is perhaps also another argument here which is that without the α-methyl group the enolate is considerably smaller, meaning that the TS doesn't even need to be quite so organised in order for the reaction to proceed.
3. Selective acetate aldols
As a closing point, switching from boron --> tin, and using a slightly modified auxiliary does allow for selectivity to be achieved.

Source: Unknown

This works because the sulfur and tin remain coordinated throughout the reaction rather than with the Evans' chemistry where the oxygen points into space to allow the dipoles to oppose. This results in a much more ordered TS.

The dipole-dipole interaction will remain the same for both acetate and propanoate enolates, and this would lead to some stereoselectivity in both cases, but the steric interaction is diminished for acetate, hence there is no stereoselectivity. This makes me conclude that the steric factor is the only important factor here, am I right? By the way, great question and answer.


text:
Boiling/melting points of cyclohexane and cycloheptane
I’m looking at a chart in my book, and while cycloheptane has a higher boiling point than cyclohexane like I’d expect, cycloheptane has a melting point of −12 °C, while cyclohexane has a melting point of 6.6 °C. Why does it have a lower melting point only?

There are not many reputable sources on the matter (as I could find) since it seems that cycloheptane isn't the most popular chemical around (poor guy) but I will present you with the best explanation/argument/insight into the situation as I can.
Note that it would be solely my own interpretation, and a case of explaining results that we see, rather than formulating an explanation to encompass natural phenomena without observing the results prior.
The key ingredient that I noticed while looking through charts and whatnot was bonding which is connected to sterics. More on that later.
Melting and Boiling Points of Cycloalkanes
Above is a chart of the melting and boiling points of cycloalkanes. As expected, boiling point increases linearly. That is lovely! Thus, we can explain boiling point with natural intuition, similar to regular alkanes. More is more! If only life was so simple.
Now let's take a look at melting points. HOLY! Cyclobutane? Why you so high? (it is actually an error by the book. Nonetheless, it is higher by a little than cyclopentane and that is the point) And wow, is cyclohexane a monster, past the melting point of our dearest water. Something is wrong with this universe. It must be the martians.
But no. It is not the martians. At least I do not believe so. The answer, I believe, is sterics.
Sterics? you ask. Are you trying to say that cyclobutane bonds more structured than cyclopentane and cyclohexane bonds more structured than cyclohexane?
Precisely.
Although you may not expect much difference from cyclobutane and cyclopentane, cyclobutane is very close to a right angle (perfect square) and this would allow tight packing. Cyclopentane, on the other hand, is normally as an "envelope" which allows less close packing.
Similarly, cyclohexane has  very close cystalline packing which may account for its low melting point.
Then, after its crystalline structure melts away, our intuition gives way to the boiling point expectancies.
Again, this is not definite. If possible, I would like to juxtapose the solid structures of each cycloalkane, but unfortunately data seems scarce, especially for cycloheptane, which is of concern.


text:
How many electrons can have the quantum number set n=5, l=3, ml=-1?
How many electrons can have the quantum number set $n=6,\ l=3,\ m_l=-1$?
Also, please explain why. I know that n describes the number of shells in an atom but what do n, l, and ml have to do with electrons?

Unfortunately, as $n$, $l$ and $m_l$ are directly related to the electrons, even if we answer this question, your confusion might remain. Would you mind adding more info about your thoughts about the problem and the concepts in it?

possible duplicate of What do the quantum numbers actually signify?

This post answers the question of the of what the other quantum numbers mean:
What do the quantum numbers actually signify?
As for how many electrons can have $n=5,\ l=3,\ m_l=-1$...
The Pauli Exclusion Principle states that two identical fermions (an eletron is a type of fermion) cannot occupy the same quantum state.

In other words, two electrons (in the same atom) cannot have identical quantum numbers.

For each $n$, the values of $l=0,1,2,...,n$
For each $l$, the values of $m_l=-l,...,-1,0,+1,...,+l$

For each $m_l$ there are only two allowed values of $m_s$, namely $m_s = + \frac{1}{2}$ and $m_s = -\frac{1}{2}$. Thus, each set of $\{n,l,m_l\}$ can only describe two electrons. In your case, we have two electrons with $n=5,\ l=3,\ m_l=-1$:

$\{n,l,m_l,m_s\}=\{5,3,-1,+\frac{1}{2}\}$
$\{n,l,m_l,m_s\}=\{5,3,-1,-\frac{1}{2}\}$


text:
Shapes of ionic compound
Can anyone explain the shapes of ionic compounds?  What does it mean? What does the coordinate number mean? I know that in $\ce{NaCl}$, one $\ce{Na}$ atom is surrounded by 6 $\ce{Cl}$ atoms, and vice versa, but can it be explained with the help of coordinate numbers?

Another more complicated case that can occur is when the ions interact with molecules or when there are resonance hybrids between ionic bonds and covalent bonds. Than one needs both VESPR or MO and methods for dealing with ionic compounds.

The question is ambiguous. Do you mean crystals of 'inorganic' compounds, say NaCl or k$_4$Fe(CN)$_6$ etc or do you mean the structure of ions in solution or these ions in a crystal e.g. Fe(CN)$_6^{4-}$.  What about ions of organo-metallic compounds, say Ruthenium tris(2,2'-bipyridyl) chloride.

Only for solids that have regular and repeating pattern of constituent particles we define shape and geometry.
So, on the basis of order of arrangement of constituent particles solids are classified into crystalline and amorphous.The former has regular pattern of arrangement of particles which repeat itself periodically over the entire crystal (as shown below is the two dimensional structure of quartz)

Amorphous solids consist of particles of irregular shape(as shown below is 2d structure of quarts glass.

If the three dimensional arrangement of constituent particles of crystalline solid is represented diagramatically in which each particle is depicted as a point, the arrangement is called crystal lattice
Unit cell is the smallest portion of crystal lattice which when repeated in all directions generate the entire crystal (see below). There are only 14 possible crystal lattices called Bravais Lattices( after a French mathematician who first described them.

On the basis of unit cell parameters (axial distances and axial angles)
we get the following seven primitive unit cell which defines the shape of solid.

COORDINATION NUMBER
this term is used in crystallography in a somewhat different sense than it is in coordination compounds.
The number of nearest neighbors (or touching particles) that a particle has in a crystal is called its coordination number. All atoms in bravais lattice have the same coordination number.
In a simple or primitive unit cell, around any given atom there would be six equally spaced nearest neighbours.Thus the coordination no=6.

Any other particle in crystal lattice will be at a greater distance from the particle under consideration. This also apply to your NaCl crystal.


text:
R32 Space Group
Which are the elements of the Twofold Screw Axis in this image?
The "+" and "-" elements that intercept the red line are the elements of the Twofold Screw Axis 2/3?
And if so, is the height of the red line 2/3 plus or 2/3 minus?
But why the height of the red line is 2/3?
I have the same problem with the other colored lines. Are the elements of the green line 1/3 minus or plus?

I would strongly recommend to consult the references I mentioned in this answer earlier: R32 Space Group (Crystallography)
In particular have a look at the meaning of the individual symbols and how space groups are represented in these kinds of diagrams. Again, the International Tables of Crystallography, Volume A, has plenty of explanation on this in its first part.
I would also recommend not to start out with the rhombohedral groups, as they are probably the trickiest of them all due to the slightly uncommon choice of centring. Try to start with an orthorhombic space group, for instance, at least until you do not have troubles with the symbols any longer.
As for an answer: your question unfortunately does not make a lot of sense to me. I'll go with a hunch here: Could it be that you are mistaking the "+" and "-" signs for something they are not, like having anything to do with height (which they do not)?
They describe the orientation of the motif - think "upwards-/downwardsfacing".
Hence (screw-)axes obviously do not have a "sign", but change the sign of the motif they operate on.

authors: ນາງ ວົງປະພັນ ສຸຂະວົງ


text:
Mixing two containers of each water and ethanol to varying degrees of purity
In the following question, the water/wine mixing problem was recast in chemical terms. However, partial molar volumes can make this a much more challenging problem than it appears on the surface.
https://chemistry.stackexchange.com/questions/33579/wine-water-mixing-problem
For example, consider a container of $\pu{10 L}$ of ethanol and another container of $\pu{10 L}$ of water. To make a point, let's transfer $\pu{5 L}$ of ethanol to the water container and then transfer $\pu{5 L}$ of the water/ethanol solution back to the ethanol container. What is the purity in each container at the end?
We need to consider that mixtures of water and ethanol have less volume than the two do separately:

For example, after the first transfer, the water container contains $\pu{10.000 kg}$ of water and $\pu{3.945 kg}$ of ethanol for a total mass of $\pu{13.945 kg}$. However, we need to compute mole fraction to determine the volume of the solution:
$$\begin{array}{c|c|c}
\ & \text{water} & \text{ethanol}\\ \hline
\text{mass in kg} & 10.000 & 3.945\\ \hline
\text{moles} & 555 & 85.6\\ \hline
\text{mole fraction} & 0.866 & 0.134
\end{array}$$
From our graph, it looks like we have an excess volume of $\pu{-0.75 mL/mol}$. Since we have total moles $=640.6$, we have an excess volume of $\pu{-0.75 mL/mol}\times \pu{640.6 mol}\times \pu{1 \times 10^{-3} L/mL}= \pu{-0.480 L}$. Thus the actual volume is not $\pu{15 L}$ but more like $\pu{14.5 L}$.
If we transfer $\pu{5 L}$ back, we are left with less than $\pu{10 L}$ in the water container. We will also have less than $\pu{10 L}$ in the ethanol container!

What volume $V_1$, as a fraction of the original volume $V_0$, needs to be transferred in order for both containers to have the same final purity?

Interesting question. Do you want the purity as mole fraction or $%v/v$ or $%w/w$?

Well, mol faction can't be 0.5 since original containers have $\pu{555 mol}$ water and $\pu{171.2 mol}$ ethanol. Thus it should be either $%(v/v)$ or  $%(w/w)$. Would you please clarify which way you want?

@MathewMahindaratne I was really hoping you would ‘take your shot’ on this one! Please post something so I can award the bounty to you and not just have it vanish away, softly and suddenly, like the hunter who spies a snark! (Apologies to Lewis Carroll.)

@MathewMahindaratne Unless Ben Norris says otherwise, just do as you think is most sensible!

@Ed V: I'll try my best to solve it. :-)

$$\begin{array}{c|c|c}
\ & \text{water} & \text{ethanol}\\ \hline
\text{Volume in concern, in L} & 10.00 & 5.00 \\ \hline
\text{Density in kg/L} & 1.000 & 0.789 \\ \hline
\text{Molar mass in g/mol} & 18.015 & 46.069\\ \hline
\text{mass in kg} & 10.000 & 3.945\\ \hline
\text{moles} & 555 & 85.63\\ \hline
\text{mole fraction} & \frac{555}{555+85.63} = 0.866 & \frac{85.6}{555+85.63} = 0.134
\end{array}$$
From the given graph, it looks like we have an excess volume of $\pu{-0.75 mL/mol}$ (at $\chi_\text{EtOH}=0.134$). Since we have total amount of water and ethanol, $555+85.63=\pu{640.63 mol}$, we can calculate the excess volume of the container:
$$\pu{-0.75 mL/mol} \times \pu{640.63 mol} \times \pu{1 \times 10^{-3} L/mL}= \pu{-0.480 L}$$
Thus the actual volume is not $\pu{15 L}$, but $\pu{(10.00+5.00-0.480) L}=\pu{14.52 L}$
Suppose, we transfer $V_1$ of water/ethanol mixture back to the container-2 containing the remaining of $\pu{5.00 L}$ of ethanol. That $V_1$ contains:
$$\frac{\pu{555 mol}\text{ water}}{\pu{14.52 L}}\times V_1 \ \pu{L} = 38.22V_1 \ \pu{mol}\text{ water}$$
and,
$$\frac{\pu{85.63 mol}\text{ EtOH}}{\pu{14.52 L}}\times V_1 \ \pu{L} = 5.90V_1 \ \pu{mol}\text{ EtOH}$$
Thus container-2 has $38.22V_1 \ \pu{mol}$ of water and $\left(5.90V_1 + \pu{85.63 mol}\right)$ of ethanol after the addition of $V_1 \ \pu{L}$ of water/ethanol mixture from container-1.
Suppose we have made both containers have the same purity (by $\%(w/w)$) by this action. Let's calculate the masses of water and ethanol ($m_{w2}$ and $m_{et2}$, respectively) in container-2:
$$m_{w2} = 38.22V_1 \ \pu{mol}\text{ water}\times \frac{\pu{18.015 g}\text{ water}}{\pu{1.0 mol}\text{ water}}= 688.5 V_1 \ \pu{g}\text{ water}$$
Similarly,
$$m_{et2} = \left(5.90V_1 + 85.63\right) \ \pu{mol}\text{ EtOH}\times \frac{\pu{46.069 g}\text{ EtOH}}{\pu{1.0 mol}\text{ EtOH}}= \left(271.8V_1 + 3945\right) \ \pu{g} \text{EtOH}$$
Now, we'll calculate the masses of water and ethanol ($m_{w1}$ and $m_{et1}$, respectively) in container-1. Amounts of water and ethanol in this container is:
$$\frac{\pu{555 mol}\text{ water}}{\pu{14.52 L}}\times (14.52 - V_1) \ \pu{L} = \left(555 - \frac{555V_1}{14.52}\right) \ \pu{mol}\text{ water}$$
and,
$$\frac{\pu{85.63 mol}\text{ EtOH}}{\pu{14.52 L}}\times (14.52 - V_1) \ \pu{L} = \left(85.63 - \frac{85.63V_1}{14.52}\right)  \ \pu{mol}\text{ EtOH}$$
Thus,
$$m_{w1} = \left(555 - \frac{555V_1}{14.52}\right) \ \pu{mol}\text{ water}\times \frac{\pu{18.015 g}\text{ water}}{\pu{1.0 mol}\text{ water}}= \left(555 - \frac{555V_1}{14.52}\right)\times   \pu{18.015 g}\text{ water}$$
Similarly,
$$m_{et1} = \left(85.63 - \frac{85.63V_1}{14.52}\right)  \ \pu{mol}\text{ EtOH}\times \frac{\pu{46.069 g}\text{ EtOH}}{\pu{1.0 mol}\text{ EtOH}}= \left(85.63 - \frac{85.63V_1}{14.52}\right)\times   \pu{46.069 g}\text{ EtOH}$$
Theoretically, the condition for same purity by $w/w$ is:
$$\frac{m_{w1}}{m_{w1}+m_{et1}} = \frac{m_{w2}}{m_{w2}+m_{et2}} $$
or
$$\frac{m_{et1}}{m_{w1}+m_{et1}} = \frac{m_{et2}}{m_{w2}+m_{et2}} $$

It took you less than two hours to get an answer for a question that sat for more than 4 years (!) without even a comment! Wow!   Thank you!

I will now award the bounty! Boy, this is fun! Thanks again!

@ Ed V: I apologize to take this long to response. I was rushing to go home last night. Well, I also calculated the first situation (5 L ethanol to water and back 5 L of water/ethanol). For first transfer, answer is ther (14.52 L). When second transfer has been done, new volume of original ethanol container is 9.65 L (water to ethanol mole fraction: 0.624:0.376). After transfer, volume of original water container is 9.52 L. Lost about 0.8 L!

No problem at all! Even if it should happen that your answer is not totally correct, and I am not implying that, the important thing is you stepped up and screwed this problem to the wall like so much sheetrock! That is worth 50 points anytime!

I think we don't have to calculate masses in container-1 using $V_1$. I'd correct it sometimes today. I have a meeting in few min. :-)

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text:
How to reverse damage done by nail polish remover without acetone to a laptop keyboard?
Today I was trying to clean my laptop with a nail polish remover (ingredients: ethyl acetate, peg-40, hydrogenated castor oil, panthenol, parfum, d-limonene, geraniol, ci 26100, ci 60725). Not only it didn't get clean, but some white thing remained on the keys.
I tried boiling water and a piece of cloth to make it warm and easier to remove without any success.
What can be done to reverse the damage?

Short answer: ethyl acetate is almost as bad as acetone when it comes to plastics. Try rubbing alcohol.

Quite possible the damage has been done by dissolving a layer of plastics.

The solvent appears not only to have removed paint on the keys but also to have caused solvent crazing in the plastic itself. Regrettably, there is probably not much that can be done to actually remove the crazing, which may have penetrated too deeply to remove by buffing, but you might be able to hide it with a substance such as an oil that has the same index of refraction as the plastic.
If the appearance is unacceptable, you might replace the keyboard, use a plug-in keyboard, or get a keyboard cover, such as the Kuzy Russian Keyboard Cover, specific to the Mac Pro. Individual key stickers would probably not adhere well.

Thinkpad (TP) keyboards have some sort of anti-microbial cover, which looks like a rough porous coating under the microscope. This developed surface prevents skin oils from sticking to the keys; but it can also be easily destroyed with solvents. And there is no way to restore its feature without replacing the keyboard, which is probably the ideal solution. TPs are designed with high modularity in mind, and there are thousands affordable options on how to get a new/refurbished/second-hand keyboard. Installation is also a piece of cake (Info from a biased TP X220t user:) ).


text:
How is spiro[4.4]nonatetraene antiaromatic
How is spiro[4.4]nonatetraene antiaromatic? I know that spiro compounds exhibit homoaromaticity, but how do they exhibit antiaromaticity?

Spiro compounds are also fun! Do you have a reference suggesting that this compound or one like it is antiaromatic?

Ian Flemin's "Molecular Orbitals and Organic Chemical Reactions: Student Edition", page 57, problem 5 asks to show that the spirononatetraene 1.31 is spiro-antiaromatic.

http://pubs.acs.org/doi/pdf/10.1021/ja00803a021

The orbital interaction between the 2 butadiene units in spiro[4.4]nonatetraene is referred to as spiroconjugation.  Note that the 2 rings attached through the common or spiro carbon are perpendicular to one another.  Each cyclopentadiene ring contains a 1,3-butadiene unit.  In the figure below, I've drawn in the molecular orbitals for the highest occupied molecular orbital ($\Psi_2$) of 1,3-butadiene.

It is a bit easier to analyze the situation using a Newman projection.  In the Newman projection on the right I only show the 4 carbons (and their P-orbitals) that are directly attached to the central spiro carbon, since these atoms and orbitals are the only ones close enough to interact between one ring and the other.  Notice how all of the through space interactions between nearby P-orbitals is stabilizing (indicated by arrows).  That is, all of the overlaps are construcive (bonding}, either + + or - -.  This indicates that $\Psi_2$ has the correct symmetry to interact with itself.  Look at the molecular orbitals (MO) for 1,3-butadiene and convince yourself that $\Psi_1$ and $\Psi_3$ do not have the appropriate symmetry to interact with $\Psi_2$ (e.g. there are no + + overlaps between the carbons attached to the spiro carbon, only + - overlaps that cancel out; and energetically, $\Psi_4$ is to far away to interact.  When the two $\Psi_2$ MOs interact, as usual, new MOs are formed; one is stabilized and one is destabilized and the destabilized orbital is always slightly more destabilized than the stabilized orbital is stabilized.  The net result in this case is that spiroconjugation produces a (small) net destabilization and the molecule is termed antiaromatic in a spiroconjugation sense.
On the other hand, consider spiro[2.4]heptatriene. Now the only interaction of consequence (e.g. where the interactions are of the ++ or -- type rather than +-; said differently, only $\Psi_2$ of the diene and $\ce{\pi_^{*}}$ of the ethylene have the appropriate symmetry to interact) occurs between the HOMO of the diene ($\Psi_2$) and the LUMO of the ethylene in the 3-membered ring.

(image source)
In this case, as the orbital mixing diagram indicates, there is a net stabilization of the pi electrons that results.  Spiro[2.4]heptatriene is said to be aromatic.
Spiroconjugated compounds containing 4n+2 pi electrons will turn out to be stabilized (aromatic), while those with 4n pi electrons will be destabilized (antiaromatic).  Remember, these effects are small, but they can be seen in UV spectral shifts and even in solvolysis reaction rates of related compounds.  If you google spiroconjugation or spiroaromaticity you can find links to the original articles that appeared in the late 60's, as well as some more recent information on the subject.

Why does the ovelap in spiro[4.4]nonatetraene cause a destabilization, but the similar overlap in spiro[2.4]heptatriene cause a stabilization? Is it because spiro[4.4]nonatetraene fills the MOs, and spiro[2.4]heptatriene doesn't?

Yes, in spiro[4.4]nonatetraene two identical MOs ($\Psi_2$), both containing electrons, interact and for the reasons given above, a net destabilization results.  In spiro[2.4]heptatriene an unfilled and a filled MO interact and a net stabilization of the filled MO results.


text:
Effect of Catalysis on Reaction Rate in elementary Reaction
I understand that there are many different catalytic mechanisms that suggest different ways of accounting for the effect of a catalyst on reaction rate. My question is:
Given a termolecular elementary (it is known it is an elementary reaction) reaction of the form:
$$\ce{A + B + C <=> D + E + C}$$
Does the rate of reaction simplify down (according to collision theory) to:
$$r = [\ce{C}]\bigr(k_\mathrm{fwd}[\ce{A}][\ce{B}] - k_\mathrm{rev}[\ce{D}][\ce{E}]\bigr)$$
Edit: To clarify my question, I have a gas phase elementary reaction of the form:
$$\ce{A + B + C -> D + E + C}$$
I have another of the form:
$$\ce{D + E + C -> A + B + C}$$
Whether they are exactly the same mechanism is irrelevant. My question relates to how quantitatively does a collision partner's concentration in a gas mixture affect reaction rate. i.e. is forward rate defined by (assuming that I know the reaction acts as an elementary reaction in that there is no intermediate):
$$r_\mathrm{fwd} = k_\mathrm{fwd}[\ce{A}][\ce{B}][\ce{C}]$$
and reverse rate defined by (reverse is also elementary reaction - not concerned if same mechanism just also acts in the rate law as a single collision event):
$$r_\mathrm{rev} = k_\mathrm{rev}[\ce{D}][\ce{E}][\ce{C}]$$
Therefore, the total rate, not assuming they're the same reaction just two elementary reactions that go in opposite directions and we've called on forward and one reverse, the total rate forward, taking into account our other reaction, is:
$$r = r_\mathrm{fwd} - r_\mathrm{rev}$$
$$r = k_\mathrm{fwd}[\ce{A}][\ce{B}][\ce{C}] - k_\mathrm{rev}[\ce{D}][\ce{E}][\ce{C}]$$
$$r = [C]\left(k_\mathrm{fwd}[\ce{A}][\ce{B}] - k_\mathrm{rev}[\ce{D}][\ce{E}]\right)$$
Therefore quantitatively a collision partner does not change the overall rates direction but rather enhances it in both directions. Is this correct?

You are correct.  What you call a collision partner is a specific example of the general concept of catalysis.  "C" is a catalyst in your scheme.  Catalysts don't change thermodynamic equilibria, just the speed at which it is approached (i.e. kinetics).  Accelerating the forward rate but not the reverse rate of an equilibrium reaction would alter the equilibrium point, and is not possible for any true catalyst.

Thank you very much for all your help, if you wouldn't mind just putting that exact paragraph into an answer I'll be sure to accept it for future viewers.

I have downvoted your question for several reasons. The main reason is, that you are still describing an equilibrium as an elementary reaction. Additionally I believe your thinking is still not correct, but I won't and cannot go into detail here. Instead of altering your question to the point where my answer becomes irrelevant you could have asked a new question with a clarification.

I understand that there were errors in my question, but I feel as though even without the clarification whilst correct your answer failed to address my issue. Whilst you could have pointed out this issue, I feel as though the main query was clear enough. Also since your answer was not actually answer put rather pointing out an error in the question, I feel it may have been inappropriate to put it in the answers and maybe would have been better in the comments. I do appreciate that my question wasn't the best and understand the down-vote and I thank you for the time put into your answer :)

Also I called it a set of elementary reactions to get across the point that I knew the effect of the concentration of other reactants in both directions. The equilibrium sign was just for convenience.

You are correct. What you call a collision partner is a specific example of the general concept of catalysis. "C" is a catalyst in your scheme. Catalysts don't change thermodynamic equilibria, just the speed at which it is approached (i.e. kinetics). Accelerating the forward rate but not the reverse rate of an equilibrium reaction would alter the equilibrium point, and is not possible for any true catalyst.

If it is in equilibrium, you are already looking at it from a macroscopic scale. It can therefore not be an elementary reaction. For example, $$\ce{A + B -> C + D}$$ is an elementary reaction, the reverse $$\ce{C + D -> A + B}$$ is a different elementary reaction. The rates of both reactions
\begin{align}
r&=k_\mathrm{fwd}[\ce{A}][\ce{B}]\\
r&=k_\mathrm{rev}[\ce{C}][\ce{D}]\\
\end{align}
would determine the location of the equilibrium
\begin{align}
\ce{A + B &<=> C + D}& K&=\frac{k_\mathrm{fwd}}{k_\mathrm{rev}}.
\end{align}

For a true termolecular reaction, i.e. three species react in one concerted transition state,
$$\ce{A + B + C -> D + E + F},$$
the rates are always given as
\begin{align}
r&=k_\mathrm{fwd}[\ce{A}][\ce{B}][\ce{C}]\\
r&=k_\mathrm{rev}[\ce{D}][\ce{E}][\ce{F}].\\
\end{align}
You cannot simplify a rate law of an elementary reaction.$%edit$

But if F = C, then you could just to algebra to simplify the rate law to be equivalent to what user2353082 wrote in the question.  Surely algebraic manipulation is still allowed, even for a rate law of an elementary reaction?

@Curt No you cannot. An elementary reaction is only dependent on itself, it has only one direction, i.e. one transition state, therefore there is only one possible rate law. If F = C, then these are still two completely different elementary reactions. You can determine the the location of the equilibrium (and its rate law) in the suggested way, but this is no longer an elementary reaction.

Your answer so far is very good, but is answering something the question didn't ask, which is, "what is the strict definition of elementary reaction".  I think it's possible that the OP meant to say "Consider this system of two coupled elementary reaction_s_..." and really wants a confirmation of their algebra that the final rate law is correct.  The algebra is correct.  As an aside, I'm not really sure what use the elementary reaction concept is, if it is immune to algebra and does not know about microscopic reversibility either.

@CurtF. You are correct, this was my question. I have tried to include some edited clarification.  So are you saying that my thinking is indeed correct?

@CurtF. The question clearly did as for an termolecular elementary reaction. I clarified, that there is only one possible rate law for this kind of reaction and that this thinking in this case is wrong.  If the OP means something different than what he asks, this is not my fault.

@user2353082 I am very sorry for the confusion. I will however not change anything in that answer, even if you downvoted it. I still do think, that it is not correct.


text:
Why are there peaks in electronegativities in d-block elements?
Looking at the Pauling electronegativities in the Periodic Table (below, from ChemWiki):

Asides from the overall trend of increasing electronegativity across and up the Periodic Table (towards fluorine), there seems to be a minor peak in the first 3 periods of the $d$-block elements, this peak becomes more pronounced as you go down the $d$-block, with the most noticable peaks in Periods 5 and 6 - seemingly the reverse of the overall trend. This is shown in the graph below (Image source):

In the ChemWiki page, there is also a reference to the Allred-Rochow Electronegativity Scale, which also has peaks occurring most prominently in Periods 4 and 6:

Image source
The Wikipedia page Electronegativity details several methods of calculation, which brings into question how well-defined the electronegativity values are; however, there still seems to be the peaks occurring, particularly in Periods 5 and 6, and to a lesser extent in Period 4.
What is the reason for these electrongativity peaks in the $d$-block?

Perhaps a question of more immediate concern before tackling the title question is, just how well-defined are the electronegativities for the transition metals?

@NicolauSakerNeto that could be part of an answer

at your link they say nearly at the end (below slater's rules), that at the time allred and rochow constructed this scale the values for the covalent radius were wrong. they gave also a corrected table

Good points - I have added these details into the revised question

You should know that closeness to filling a shell and atomic size are the reasons for the overall electronegativity trend. This also applies to the subshells, how close is an element to filling, or half-filling, a subshell? How large is the atom?
The atomic size trend for the d subshell is opposite to that of the overall trend. In period 4 the 3d subshell is deep in the core orbitals and the electron density is high, whereas in the lower periods the d orbital is more diffuse and can more easily accommodate more electrons, leading to a better energy payoff for completing the subshell. For the elements that take s electrons to fill the d subshell, this energy payoff is even greater from then filling the s subshell as they are very diffuse.

The peaks in electronegativity for elements may be named as groups as next:
V Cr Mn has a peak and then Mn drops down. Iron is the key as shown next.
Zr Nb Mo has a peak and then a a plateau. Zirconium is the key as shown next.
My research is displayed :
Periodic Table of Elements : Shape of Each Nucleus affecting electronegativity

http://pyramidalcube.blogspot.com/2017/08/periodic-table-of-shapes-of-nuclei.html
The shape of the nucleus affects the properties of the element. Notice how
Manganese is the last in a series from the carbon core cube-2, then iron starts the cube-3 shape. That transition across the Table causes the abrupt change in electronegativity.
Alan Folmsbee

Do you have a reference for this beyond a blog entry that you wrote yourself (which subsequently links to another blog)?

Dear Jonska, Neon has a shape like in my Table. https://phys.org/news/2012-07-atomic-nucleus-fissile-liquid-molecule.html#nRlv and also Radon has a shape like in my table: http://ns.ph.liv.ac.uk/~lg/papers/StudiesOfNuclearPearShapesUsingAcceleratedRadioactiveBeams_Archive.pdf and also Barium has a shape like in the Table https://phys.org/news/2016-06-heavy-barium-nuclei-pear.html . I see a minus five insult from your welcoming committee. Progress is here, enjoy.

I did not downvote, but as I mentioned, it was probably due to having insufficient references . It may help others if you incorporated brief summaries of some of that research into your answer rather than only the chart.


text:
Search for the last step of an altered Raschig process reaction mechanism
The Raschig process (according to the german wikipedia) is divided in the following steps:

$\ce{2NaOH + Cl2 -> NaOCl + NaCl + H2O}$
$\ce{NaOCl + H2O -> HOCl + NaOH}$
$\ce{HClO + NH3 -> NH2Cl + H2O}$
$\ce{NH2Cl + NH3 -> N2H4 + HCl}$
$\ce{N2H4 + 2NH2Cl -> N2 + 2NH4Cl}$

But what happens now if one changes $\ce{NH3}$ into $\ce{R-NH2}$, with $\ce{R}=$?
Starting from step 3 of the Raschig process I came up to the reaction of the produced hydrazine derivate to give an azo compound.

$\ce{HClO + R-NH2 -> R-NHCl + H2O}$
$\ce{R-NHCl + R-NH2 -> R-NH-NH-R + HCl}$
$\ce{R-NH-NH-R + x~R-NHCl -> R-N=N-R~+~?}$

The last reaction step bothers me a little ... is it simply
$$\ce{R-NH-NH-R + 2R-NHCl -> R-N=N-R + 2R-NH2Cl}$$
but without the cleavage?

Aliphatic diazocompounds (say, $\ce{CH3-N=N-CH3}$) are very unstable and tend to fragment into dinitrogen molecule and two alkyl radicals after a funny look.

You are right at this. I added the information on my R- to the question. And that compound is stable as it was synthesised - not only be me.

It can't be
$$\ce{R-NH-NH-R + 2R-NHCl -> R-N=N-R + 2R-NH2Cl}$$
The reason is that while the atoms are balanced, the electrons aren't. Going from $\ce{R-NH-NH-R}$ to $\ce{R-N=N-R}$ involves the loss of two protons and two electrons, but the $\ce{R-NHCl}$ are really only picking up two protons. (Nitrogens with four substituents are positively charged.)
When thinking of mechanisms, it may help to explicitly split the final listed reaction into two steps.
5a. $\ce{H-NH-NH-H + NH2Cl -> H-N=N-H + NH4Cl}$
5b. $\ce{H-N=N-H + NH2Cl -> N2 + NH4Cl}$
That is, each chloramine is involved in pulling one dihydrogen (two protons and two electrons) off of the diazo. - The addition of a "hydride" (a proton and two electrons) to the nitrogen kicks the chlorine off as a negatively charged ion, and the remaining proton protonates the lone pair on the nitrogen.
The analogous last step for the $\ce{R-NH2}$ case might be:
$$\ce{R-NH-NH-R + R-NHCl -> R-N=N-R + R-NH3Cl}$$
This is equivalent to reducing the $\ce{N-Cl}$ covalent bond to $\ce{R-NH2 + HCl}$ followed by an acid/base reaction.


text:
Are the elements lanthanum and actinium considered to be in the d block or the f block of the periodic table?
Many periodic tables place lanthanum and actinium in the f-block of elements, for example, this periodic table from Los Alamos National Laboratory.
However, this table from the Royal Society of Chemsitry places lanthanum and actinium in the d-block.
Which is the most appropriate placement of these two elements on the periodic table? What is the most recent scientific consensus regarding this matter?

It's really better if you reword this. I think I get what you mean, but atoms aren't in subshells.

I understand this question now, there is a bit of contention about where La and Ac actually should be placed - I recall reading an article recently, I will see if I can find it.

I edited your question to make it more clearly about what I think you mean - the placement of these two elements in the periodic table. If it is about their electron configurations, I think there is less of a debate.

https://lavelle.chem.ucla.edu/wp-content/supporting-files/Publications/Lavelle_LaAc_in_d_Block_JCE2008p1482.pdf

As per IUPAC, lanthanum and actinium are f-block elements. source:
https://iupac.org/what-we-do/periodic-table-of-elements/

@DhankharNaman Your answer has been converted to a comment as it does not add much beyond what is already included in other answers.

If IUPAC says these elements are in the $f$ block, I'll roll with it.
The real lesson here is that the boundaries between "blocks" of the Periodic Table, like the boundary between "strong" and "weak" acids or bases or even between what is a stable compound and what isn't, is not sharp. Some other examples of a rough, spotty, changeable real world:

Most simple magnesium compounds, even the best known hydride and boride compounds, are primarily ionic, but when they have covalent character the magnesium often bonds tetrahedrally as if using $3p$ as well as $3s$ valence orbitals. See for instance the coordinated structure given here for methylmagnesium chloride in THF. Beryllium shows this effect even more prominently in its wider variety of covalent compounds.

Calcium could be called a $d$-block element when it bonds with its $3d$ orbitals in this calcium(I) compound[1],[2] (and yes it is +1, showing multiple oxidation states like a transition element).

Cerium, the second element in the lantanide series, does some straddling of its own between $d$ and $f$ blocks as its valence in the metal is changeable between 3 (with a core-like $5f$ electron) and 4 (with this electron engaged in the metallic bonding). See Johanssen et al. 3 and the WP article citing this reference.

Among the actinides there is also thorium, which appears to involve only $7s$ and $6d$ valence electrons in the gas phase but brings in $5f$ orbitals in the metal (see this answer and the references therein).

References

Sven Krieck, Helmar Görls, Lian Yu‡, Markus Reiher, and Matthias Westerhausen (2009).
"Stable “Inverse” Sandwich Complex with Unprecedented Organocalcium(I): Crystal Structures of [(thf)2Mg(Br)-C6H2-2,4,6-Ph3] and [(thf)3Ca{μ-C6H3-1,3,5-Ph3}Ca(thf)3]". J. Am. Chem. Soc. 131, 8, 2977–2985.
https://doi.org/10.1021/ja808524y.

Krieck, Sven; Görls, Helmar; Westerhausen, Matthias (2010). "Mechanistic Elucidation of the Formation of the Inverse Ca(I) Sandwich Complex [(thf)3Ca(μ-C6H3-1,3,5-Ph3)Ca(thf)3] and Stability of Aryl-Substituted Phenylcalcium Complexes". Journal of the American Chemical Society. 132 (35): 12492–12501. https://doi.org/10.1021/ja105534w. PMID 20718434.

Johansson, Börje; Luo, Wei; Li, Sa; Ahuja, Rajeev (17 September 2014). "Cerium; Crystal Structure and Position in The Periodic Table". Scientific Reports. 4: 6398. Bibcode:2014NatSR...4E6398J. https://doi.org/10.1038/srep06398. PMC 4165975. PMID 25227991.

Lanthanum and Actinium are generally both accepted as f-block elements. You will notice in the Royal Society's periodic table, La and Ac are both color coded to match the other f-block elements. They do this to show the beginning of both the Lanthanide and Actinide series, while maintaining a reasonable symmetry for their periodic table (that is, without "holes" in their table, similar to the first link you provided).

Then what are your counter-arguments against this?

I suppose I was conforming to the more generally accepted placement of La and Ac (Lavelle's "general chemistry books"). Realize the topic of his paper is to suggest why both La and Ac should be considered f-block and not d-block elements, the latter of which is the general view--my view. Albeit my argument is less technical, I support the general acceptance of Ac and La being members of the f-block; and view/explain the differences in the above periodic tables with my generally accepted view.

Correction: Lavelle argues La and Ac should be d-block, not f-block.

Lanthanum and actinium are usually regarded as d-block elements (Myers, Oldham & Tocci 2004, p. 130) and generally counted as lanthanides and actinides (the rest of which occupy the f-block). Some periodic tables, like the one from LANL emphasise chemical similarities, so lanthanum and actinium are shown at the foot of the table, along with the f-block elements. That's fine, but leaves a hole under yttrium so it isn't clear which two elements go there. Other tables, like the one from the Royal Society of Chemistry, presumably rely on electron configurations to work our which elements fit under yttrium, and use colour shading to emphasise similarities in chemical behaviour.
Myers RT, Oldham KB & Tocci S 2004, Holt Chemistry, Holt, Rinehart and Winston, Orlando

From the IUPAC red book:

The groups of elements in the periodic table (see inside front cover) are numbered from 1 to 18. The elements (except hydrogen) of groups 1, 2 and 13–18 are designated as main group elements and, except in group 18, the first two elements of each main group are termed typical elements. Optionally, the letters s, p, d and f may be used to distinguish different blocks of elements. For example, the elements of groups 3–12 are the d-block elements. These elements are also commonly referred to as the transition elements, though the elements of group 12 are not always included; the f-block elements are sometimes referred to as the inner transition elements. If appropriate for a particular purpose, the various groups
may be named from the first element in each, for example elements of the boron group (B, Al, Ga, In, Tl), elements of the titanium group (Ti, Zr, Hf, Rf), etc.
The following collective names for like elements are IUPAC-approved: alkali metals
(Li, Na, K, Rb, Cs, Fr), alkaline earth metals (Be, Mg, Ca, Sr, Ba, Ra), pnictogens$^8$ (N, P, As, Sb, Bi), chalcogens (O, S, Se, Te, Po), halogens (F, Cl, Br, I, At), noble gases (He, Ne, Ar, Kr, Xe, Rn), lanthanoids (La, Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb, Lu), rare earth metals (Sc, Y and the lanthanoids) and actinoids (Ac, Th, Pa, U, Np, Pu, Am, Cm, Bk, Cf, Es, Fm, Md, No, Lr).

So to decide which block La and Ac belong to, you would have to know which group they are in (group 3 or no group). In the periodic table shown in the same document, there is only an asterisk in period 6 and 7 group 3. So according to the red book, La and Ac (and Lu and Lr) are in the f-block.
If a periodic table shows elements in group 3, this would imply they are in the d-block. This is true even if they are color-coded in the same way as the lanthanoids and actinoids.
The IUPAC red book also makes a statement about the terms lanthanide and actinide:

Although lanthanoid means ‘like lanthanum’ and so should not include lanthanum,
lanthanum has become included by common usage. Similarly, actinoid. The ending ‘ide’ normally indicates a negative ion, and therefore lanthanoid and actinoid are preferred to lanthanide and actinide.

The placement of La and Ac in the periodic table is debatable, and maybe one day IUPAC will change its mind.

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


text:
Questions regarding mol files
I have some questions regarding generation of mol files-

For bond block the numbering can be done at random or it has to follow some order
Also in the atom block the order in which atoms occur can it be random or again it has to follow any order
Is order of numbering done in bond block and atom block related
Is there any free software which converts mol files to structure

Here is a sample structure with mol file-

  Sample

 22 23  0  0  0  0  0  0  0  0999 V2000
   -2.5962    2.2535    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -2.5962    0.7512    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -1.2958    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    0.0000    0.7512    0.0000 N   0  0  0  0  0  0  0  0  0  0  0  0
    0.0000    2.2535    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -1.2958    3.0047    0.0000 N   0  0  0  0  0  0  0  0  0  0  0  0
   -4.6479    1.7042    0.0000 F   0  0  0  0  0  0  0  0  0  0  0  0
    1.3005    3.0047    0.0000 O   0  0  0  0  0  0  0  0  0  0  0  0
   -1.2958    4.5070    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -1.2958   -1.5023    0.0000 O   0  0  0  0  0  0  0  0  0  0  0  0
    1.3005    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    1.3005   -1.5023    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    2.6009   -2.2535    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    3.8967   -1.5023    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    3.8967    0.0000    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    2.6009    0.7512    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
    5.1972   -2.2535    0.0000 Cl  0  0  0  0  0  0  0  0  0  0  0  0
    2.6009   -3.7559    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -3.8967    3.0047    0.0000 C   0  0  0  0  0  0  0  0  0  0  0  0
   -5.1972    3.7559    0.0000 F   0  0  0  0  0  0  0  0  0  0  0  0
   -3.1455    4.3052    0.0000 F   0  0  0  0  0  0  0  0  0  0  0  0
    3.9014   -4.5070    0.0000 O   0  0  0  0  0  0  0  0  0  0  0  0
  1  2  2  0  0  0  0
  2  3  1  0  0  0  0
  3  4  1  0  0  0  0
  4  5  1  0  0  0  0
  5  6  1  0  0  0  0
  1  6  1  0  0  0  0
  1 19  1  0  0  0  0
  5  8  2  0  0  0  0
  6  9  1  0  0  0  0
  3 10  2  0  0  0  0
  4 11  1  0  0  0  0
 11 12  1  0  0  0  0
 12 13  2  0  0  0  0
 13 14  1  0  0  0  0
 14 15  2  0  0  0  0
 15 16  1  0  0  0  0
 11 16  2  0  0  0  0
 14 17  1  0  0  0  0
 13 18  1  0  0  0  0
 19  7  1  0  0  0  0
 19 20  1  0  0  0  0
 19 21  1  0  0  0  0
 18 22  2  0  0  0  0
M  END

Can you explain, how the answer to your previous question does not cover your current question. For example, I read there, that the hydrogen atoms are stored implicitly.

@Martin-マーチン Yeah sorry for that.i will edit and remove that part of my question..But still others questions were not covered in my last post

For bond block the numbering can be done at random or it has to follow some order

Bonds don't need to be in any particular order, although usually bonds are created first for atom 1, then atom 2, etc. However, the first bond could be between atom 101 and atom 100 if you wish.

Also in the atom block the order in which atoms occur can it be random or again it has to follow any order

Atoms can be in any order. Some programs use "canonical ordering" so that the same molecule will always have the same atomic order, but this is rarely used because it can be time-consuming.

Is order of numbering done in bond block and atom block related

As indicated above, both the atom and bond blocks could be in random order. The relation, obviously, is that for a given molecular structure, there will be bonds between bonded atom pairs.

Is there any free software which converts mol files to structure

There are many, but you don't describe whether you want 3D (presumably) or 2D structures.
Personally, I suggest Avogadro which can create 3D coordinates for files.
If you're generating a large batch of 3D structures, you will likely want to use Open Babel or RDKit or something similar. Both are open source and free.

Thanks a lot.I still have a question. In the mol file I have provided in that consider     first six lines of bond block.1 is connected with 2,2with 3,3-4,4-5,1-6.So in this they have started from 1 and then continued and returned to 1 after 4-5.Can I just finish up 1 first then go to 2 and so on.I mean can I write this way- 1-2, 1-6, 1-19, 2-3. So,I finished up 1  and then started 2. Can I do this way?

You can enter the bonds in any order, since they will be read separately. Yes, you could write all the bonds of atom 1 and then 2, etc. or any way you see fit.

Sorry but I have another question.Can x,y,z coordinates in atom blocks  be random if I have to write mol file by hand without any software?Because looking at the picture how is it possible to tell precise coordinates of the atoms?

No, they cannot be random. In this case, it's a 2D structure, so the Z coordinate should be uniformly 0.0, and the Mol file marked as such. The X and Y coordinates should be mapped out by the image, since 2D structures have meaningless bond lengths - they're for display purposes much like the image. But if you want image -> structure, I suggest OSRA which is open source and already does this.


text:
Singlet/triplet oxygen cycloadditions
Why can singlet oxygen participate in cycloaddition reactions while triplet oxygen, on the other hand, can't?

Normal dioxygen ($\ce{O2}$) exists as a ground state, triplet biradical. This is an example of a molecule that, thermodynamically should be quite reactive, yet is kinetically unreactive - once again, a case of kinetics vs. thermodynamics.
Because spin must be conserved in a reaction, if ground state, triplet $\ce{O2}$ were to react with something, it would have to produce a triplet product. For most compounds, the triplet state is considerably higher in energy than the singlet state.  Hence reaction with ground state, triplet $\ce{O2}$ would produce a higher energy, triplet product. Since ground state, triplet dioxygen doesn't have any excess energy, the activation energy required to produce a high energy triplet product just isn't there - hence, no reaction.
On the other hand, singlet $\ce{O2}$ is a spin-paired excited state with plenty of excess energy (the first singlet lies some 95 kJ/mol above the ground state triplet).  Basically, singlet $\ce{O2}$ behaves like any other electrophilic (due to the electronegativity of oxygen), reactive (due to its excess energy), compound with a double bond. It readily undergoes the Diels-Alder, ene and even 2+2 cycloaddition reactions - again, just like any other reactive olefin.


text:
What causes that foul taste with combination of toothpaste and orange juice?
Ugh, I drank orange juice too soon after brushing my teeth and my mouth feels awful!
What causes that foul taste with this combination of toothpaste and orange juice? Is it a reaction between the oil of wintergreen in the toothpaste and the citric acid in the OJ? Is it something else? What's the explanation for this?
As a second part to this question, is there a way to stop this reaction from happening, aside from not drinking orange juice after brushing?

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

Why do juice taste awful?

Awful taste is due to sodium laureth sulfate, also known as sodium lauryl ether sulfate (SLES), or sodium lauryl sulfate (SLS)- depending on which toothpaste you use.
SLES and SLS are surfactants(wetting agent). Both chemicals are added in toothpaste to create foam and make the paste easier to spread around your mouth.
Both chemicals they suppress the receptors on our taste buds that perceive sweetness, inhibiting our ability to pick up the sweet notes of food and drink. Also break up the phospholipids on our tongue.  These fatty molecules inhibit our receptors for bitterness and keep bitter tastes from overwhelming us, but when they're broken down by the surfactants in toothpaste, bitter tastes get enhanced.
Thus, anything you eat will be less sweeter and more bitter!
Source: mentalfloss

How to prevent awful taste after brushing?

After a bit research I found few ways to avoid awful taste:

Avoid brushing. (not recommended)
Wait for about 30 minutes.
Look for SLS-free toothpaste.
eat a little bit of cheese first.(suggested by repurposer)
hot water will do the trick.

Nice answer, but could you elaborate a bit about the second part of the question (how to avoid this)? I know it'd be something like don't eat anything when brushing teeth, but I'd like to see a scientific approach to avoid that.

You answered my life-long question! BTW I don't think there is an easy solution to this...

So the answer to the second part is, eat a little bit of cheese first

@M.A.Ramezani I have added second part. I hope you will try them ;-)

Nicely answered, @Freddy... Granted this was a hypothetical scenario, even though I have gone through it at least once or twice in my life, so well done!

@repurposer Would that be because that reintroduces phospholipids into the mouth?

@Eliseod'Annunzio Yes! As mentioned phospholipids, I immediately thought of sommoliers, naturally. And, I tried it out, it works beautifully.


text:
Equilibrium of more than two inter reacting species
We know that equilibrium in a chemical system is attained when forward and backward reaction rates are equal. What if the reaction mixtures involve more than one reaction?
For example, consider three inter-reacting species A, B and C. The rate constants for the forward and backward reactions of $\ce{A -> B}$ are $k_1$ and $k'_1$, that of $\ce{B -> C}$ are $k_2$ and $k'_2$ and that of $\ce{C -> A}$ are $k_3$ and $k'_3$. What is the relation between the six rate constants so that the system remains in equilibrium? Assume first order kinetics for all the reactions. Thanks in advance.

See also http://chemistry.stackexchange.com/questions/15511/can-a-multi-species-system-oscillate-around-equilibrium

At equilibrium $~r_1~=r_2~=r_3~=~0$.
So, $$k_1[A]-k_{-1}[B]=0$$
$$k_2[B]-k_{-2}[C]=0$$
$$k_3[C]-k_{-3}[A]=0$$
So, The criteria for equilibrium is
1) $$k_1[A]=k_{-1}[B]$$
2) $$k_2[B]=k_{-2}[C]$$
3) $$k_3[C]=k_{-3}[A]$$
Now defining equilibrium constant as $K_i=\frac{k_i}{k_{-i}}$
We can derive a formula
$$K_1 \times K_2 \times K_3~=~1$$

The previous answer was not done correctly.
These are the balances on the three species:
$$\frac{dA}{dt}=-(k_1+k_{-3})A+k_{-1}B+k_3C$$
$$\frac{dB}{dt}=k_1A-(k_{-1}+k_2)B+k_{-2}C$$
$$\frac{C}{dt}=k_{-3}A+k_2B-(k_3+k_{-2})C$$
If the system is at equilibrium, the three time derivatives must be equal to zero.  So one must have:
$$-(k_1+k_{-3})A+k_{-1}B+k_3C=0$$
$$k_1A-(k_{-1}+k_2)B+k_{-2}C=0$$
$$k_{-3}A+k_2B-(k_3+k_{-2})C=0$$
These relationships represent 3 simultaneous homogeneous linear algebraic equations in three unknowns, A, B, and C.  The condition for these equations to be satisfied for all values of A, B, and C is that the determinant must be equal zero.
Chet

Your mathematical explanation is for steady state solution not for equilibrium. It's more applicable for flug flow reactor catalytic system. At equilibrium, you must have equal forward and reverse rate.

mamun comment is incorrect.  At equilibrium of a system involving multiple species and reactions, each individual reaction does not have equal forward and reverse rate.  The condition for equilibrium of such a system  is that the free energy of the  reaction mixture is minimum.  That condition is achieved when the relationships presented in my previous post are satisfied.

That's a common mistakes to mix up the concept of equilibrium and steady state. I would suggest you to please read the following link:                                  1)  https://en.wikipedia.org/wiki/Steady_state_(chemistry)                                      2)https://www.researchgate.net/post/Can_anyone_explain_briefly_the_difference_between_the_equilibrium_and_steady_state_system_in_chemistry

3)http://chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/Chemical_Equilibria/Principles_of_Chemical_Equilibria/Principles_of_Chemical_Equilibrium/Characteristics_Of_The_Equilibrium_State  4)   https://en.wikipedia.org/wiki/Chemical_equilibrium                                                                  -Thanks

In a batch reactor, how do you determine when a system at steady state is not at equilibrium, given that the concentrations of all the species are not changing with time?  If the system is not at equilibrium, then it should be moving spontaneously toward equilibrium, and the concentrations should be changing.

Ah man! Equilibrium concept is only applicable for closed system. Now if you use a close batch reactor with no input and output, the system will slowly approach equilibrium and the net rate will reach zero. But you don't want that because that's not economical. To disturb the equilibrium, you are withdrawing the product and giving more reactant input. So, the system is at steady state and the driving force towards equilibrium is the highest.

In Chemical Engineering we study thermodynamics to know the boundary and then we study heat and mass transfer to utilize those boundaries as much as possible.  - Thanks!

Hi Mamun, I just realized that your 3 equations are a linear combination of my three equations, and visa versa, so we must both be correct.  I'm sure that if I evaluate the determinant of my 3 homogeneous linear algebraic equations, it will lead to your final result.  So I unjustifiably criticized you in my original post.  Sorry about that.  Incidentally, I am a chemical engineer also (1967 PhD) with over 50 years of experience, so I fully understand what you are saying.


text:
Carbohydrate configurations
I want to study several carbohydrates that includes but not limited to the following molecules:

Glucose
Galactose
Mannose
Fructose
Tagatose
Sorbose

I know that for any carbohydrate, possible list of carbohydrates isomers are

single chain (L and D isomer)
$\alpha$ and $\beta$ pyranose or six member ring structures
$\alpha$ and $\beta$ furanose or five member ring structures.

Now, for ring structures there are several rotamers, mainly gg, gt and tg (g=gauche, t=trans) and also there are several anomers but I am not sure about them. Is there anything I'm missing?
For accurate thorough computational study, I need to identify all the isomers. Is there any single book or document that can guide me to screen the full potential energy surface?
Thanks for your kind help.

There are 32 isomers of C6H12O6, of which you have named 6. The most common one is D-glucose (dextrose). I don't think all of the 16 L-saccharides are found in nature.
α and β refer to the connection between rings, whether the ketal (or aldehyde) oxygen is axial or equatorial.

  Structures made with ChemDraw.

Worth noting that $\alpha$ and $\beta$ conformations will interconvert with each other in solution, in the case of single, unsubstituted rings (i.e. hemiacetals equilibrate, acetals do not).  So $\alpha$-D-glucopyranose and $\beta$-D-glucopyranose will interconvert in solution, as will any $\alpha$ and $\beta$ pair in the awesome chart above.  But e.g. cellulose is a polymer of $\beta$-D-glucopyranose monomers, and in the polymer, there is no significant interconversion to $\alpha$.

I'm using this two figures almost everyday and they are actually very handy to remove any confusion regarding structures. Thank you! Can you please give me something similar for open chain structures? Thanks!

You can get ChemDraw for free and have all the structures you need available from the templates. http://www.acdlabs.com/resources/freeware/


text:
Is it possible for a group to substitute itself?
When I first learned about substitution reactions, I started wondering something that I never ended up figuring out:
Suppose you have a strong nucleophile which is also a good leaving group, such as I$^{-}$. Can such a group substitute itself using the SN1 mechanism? If so, would that mean if such a group were bonded to a chiral carbon, an enantiomerically pure solution could spontaneously racemize?

Very short answer: Yes. Possibly also via SN2.

Has spontaneous racemization been observed?

see this and this

The reaction you refer to has been done and racemisation observed, however I don't have the reference

Yes - a nucleophile can replace itself in a substitution reaction. Under the SN1 mechanism, once the carbocation is formed, it becomes prone to attack from any negatively charged species, including the one that left it in the first place. Likewise, in an SN2 reaction, there is the possibility of a nucleophile replacing itself. Given enough time, I would expect alkyl halides to become racemic in solutions that contain the halide anion; however, substitution reactions remain stereo-selective. This is because when you are preforming a substitution reaction, you normally use a better leaving group than the nucleophile that you are adding. Because of this, the nucleophile in solution will preferentially replace the initial leaving group over its resulting product. This is because the rate laws for both SN1 and SN2 reactions depend on the concentration of the alkyl group - implying that an alkyl group with a better leaving group will react faster than one with a weaker leaving group. In SN1, the rate determining step is the leaving group dissociating from the carbocation, which happens quicker with a better leaving group. In an SN2 reaction, the rate determining step involves both the nucleophile attacking and the leaving group leaving. This - once again - means that the reaction will proceed quicker with a better leaving group. Given enough time, your product should racemize; however, your substitution reaction should be done before this can have a major effect on your yield.

Iodine is quite a good nucleophile for saturated carbon atoms because of its softness. a good nucleophile is a poor leaving group - this statement only really applies to reactions with hard electrophiles like carbonyl groups.

@NielsKornerup Can you clarify what you mean when you say, "Because of this, the nucleophile in solution will preferentially replace the initial leaving group over its resulting product"?

I modified my answer. Hopefully this helps!


text:
Reverse Phase Chromatography
Please help me figuring out what is going on with my RP-HPLC protocol.
I am purifying a small peptide out of E.coli Culture, heat stable enterotoxin, its amino acid sequence is:
N-T-F-Y-C-C-E-L-C-C-N-P-A-C-A-G-C-Y.
I am collecting culture supernatant and using amberlite XAD2 to concentrate and desalting the peptide in 80% and 99% methanol fractions. These fractions were subjected to flash evaporation. Then I am using on MCI gel CHP20P (glass column) for further concentration and removing unwanted proteins (polar and less hydrophobic ones) using stepwise elution with 20%, 40%, 60%, 80% and 100% methanol fractions. Each fraction is subjected to flash evaporation. A sample (1 ml) from each fraction (60, 80 and 100%) was then loaded to RP-HPLC using xbridge C8 preparative column (19x25 mm,10 micron, 130 angstrom) on a gradient system (solvent A: water/ solvent B: 80%methanol) initial-5 minutes 100 A, 5-10 minute 30% B, 10-65 minute 100% B, 65-80 minute 100% B at flow rate 2 ml/minute. The chromatogram showed a single large peak (3 AUF) at 25 minutes, B solvent around 45-50%.
I am wondering that the retention time is lower than that was reported before, 35 minute at 55-60% B (I followed almost the same protocol 5 years ago except using MCIgel methacrylate resin and Vaydc C8 column instead of mci gel CHP20P and xbridge C8 column). It is also very frustrating that I loaded 0.5 ml of the collected peak on Waters analytic C8 coulumn, 5 micron particle size and it  elutes very early at 5-6 minutes (almost in the mobile phase A: water). My frustration comes from that I am using highly hydrophobic resin particle (MCI gel CHP20P) to capture hydrophobic molecules and loaded these fractions (60%, 80% and 100% methanol MCI gel) on C8 column and eluted so early.
I am sorry for this long message but I rely on your understanding. Any answer would be highly appreciated. Thank you in advance

"I am sorry for this long message" -- it would be much easier to read if it were in paragraphs

What evidence is there that the peaks you are seeing have anything to do with your target peptide?  Do you have an activity assay for that peptide?   That peptide has a lot of cysteines.  Is its chromatographic behavior affected by inter- or intramolecular disulfide formation?  How are you assessing the function of all the columns you are using?  Do you have a standard mix for column capacity and separation factor estimation?

Thank you for your response. I am using in vivo suckling mouse model to test its biological activity. I am just wondering how it comes that 60-100% methanol fractions from MCI gel CHP20P eluted very early on C8 column?!

Is there any reason why you're growing such a small peptide in culture instead of producing it through solid phase peptide synthesis? You'd have a lot less crap to remove and could probably use a simpler reverse phase purification strategy. And do you have access to LC-MS? You could directly test the mass of each peak and confirm it matches the peptide rather than rely on a suckling mouse model.

As far as I know that cystiene-rich peptide is a problematic on SPPS. Do you know trusted company for synthesis this peptide efficiently. Thank you

@Nasr-EldinAref We do all of our own peptide synthesis, so we never had to think about hiring a company to do it. the cysteines could cause problems, but protecting groups exist to help facilitate SPPS with cysteines. Does the peptide have specific disulfide bonds you need to maintain? This thesis might have helpful information on cysteine protecting groups and peptide synthesis.

would you please email me your contact information on aref71@hotmail.com. Thank you

@Nasr-EldinAref This paper describes a synthetic approach to make your peptide with proper disulfide linkages.

I am just wondering how it comes that 60-100% methanol fractions from MCI gel CHP20P eluted very early on C8 column?!

The gel you are using seems to be a porous divinyl benzene / styrene resin.  The aromatic groups of the resin may interact more strongly with aromatic side chains on your molecule (e.g. Y residues) than does the aliphatic C8 groups of the reversed-phase HPLC column you are using.
"Hydrophobic" vs. "hydrophilic" is a very useful axis for understanding solute interactions with stationary phases, but unfortunately not everything can be collapsed into this single-variable scale.
Another possibility is that residual solvents or very large amounts of co-solutes are interfering with the RP-HPLC method you are using, causing premature elution.  As an extreme example of what I mean, imagine if you injected a few μL of chloroform along with your sample in the RP-HPLC method.  The chloroform would prevent column binding and cause premature elution.  Now, it's unlikely your samples have chloroform, but they could have large amounts of methanol or other "solvent-like" impurities.
These are just general points; unfortunately I can't be sure that either is the precise source of your particular problem, without more (and more clearly presented) information on your former method from five years ago, your method today, and the differences in observed vs. expected results.

Actually, I have methanol in my loaded samples with varying concentration. I have 3 fractions from MCI gel CHP20P namely: 60%, 80% and 100% methanol. The last 2 fractions still have some methanol after flash evaporation (they are not turn solid upon freezing) but the first fraction did not. I have only one large peak (AUF=2+) on preparative  C8 from the three fractions at the same retention time.

My previous protocol had the same steps, however I used MCI gel methacrylate based resin instead of divinyl benzene / styrene resin (MCI gel CHP20P). Thus actually I used AmberliteXAD2 followed by MCI gel methacrylate resin, then Vydac preparative C8 (300 Angstrom) column. The chromatogram showed some peaks at first then my target peak at RT 36 minutes B % 55.  Thank you for your help

Another piece ofinformation from my previous protocol is that my target peptide was present mostly in the 60% MCI gel methacylate based resin fraction.


text:
Why is an alcohol proton so shielded?
Considering how electronegative oxygen is, why is the chemical shift of alcohol protons (1H-NMR) not so high? The chemical shifts of the protons on 3-propanol are about 1.20 for the carbon-1 hydrogens, about 2.2 for the alcohol proton, and about 4.01 for the carbon-2 hydrogens. Why on earth would the electrons on the carbon-2 hydrogens feel more pull from the oxygen than those on the alcohol proton?

Oh, I was just reading about this. This excerpt points out the factor contributing to the chemical shift observed in alcohols

The extent of hydrogen bonding varies with concentration, temperature, and solvent, and changes in the degree of hydrogen bonding can cause substantial shift changes. This is very evident in the nmr spectrum of ethanol taken at different concentrations in CC14 (Figure 9-29). The hydroxyl resonance will be seen to move upfield at the lower concentrations. This is the result of decreasing association by hydrogen bonding through equilibria...

Source
So basically the reason (or a reason) for alcohols moving upfield due to hydrogen bonding, which results in decreasing association.
I looked at your link and there are a few notable things. They mention *variable, highlighting that it is not necessarily constant based on different conditions, which connects to hydrogen bonding. For carbon-2, we have the normal ~1.7 for methine protons + 2.5 from oxygen of alcohol, coming to be a pretty close value. So no, it's not "why on earth would the electrons on the carbon-2" but what on earth is going on with that alcohol?

Should I understand this as: the protons involved in hydrogen bonding are partly being shielded by the electrons of the oxygen that is "attacking" it?

Perhaps. I looked around but there isn't really that definitive of an explanation about the effects of hydrogen bonding in NMR. The key thing is probably to note that at very low concentrations, the chemical shift would be more upfield. At higher concentrations, the chemical shift is downfield due to increased hydrogen bonding.  For purposes of identification, deuterium is often used to make the signal magically "disappear". To better understand, I would direct you to this link about hydrogen bonding and NMR: http://www.biozentrum.unibas.ch/~grzesiek/TEACHING/MATERIAL/2005/hbond_all.ppt.pdf


text:
Why is the IUPAC name of glycerol expressed in the way that it is?
I have come across the name trihydroxy propane for glycerol, which I think should've been Propan-1,2,3-triol.
Why is it named this way? Is there any  other rule of nomenclature working?

It should be propane-1,2,3-triol instead of Propan-1,2,3-triol. Because If there are are more then 1 hydroxyl groups, e is not supposed to be removed.(That's what my text-book says)

Glycerol and trihydroxypropane are common names. The IUPAC name is propane-1,2,3-triol. The common name glycerol comes from the root glyco- which means sweet. Glucose, glycogen, and glycerin (another name for glycerol) have the same etymology.
Trihydroxypropane makes sense as a common name because there are only 3 locations for the hydroxyl groups on the 3 carbon chain, so the name is basically unambiguous - multiple hydroxyl groups on the same carbon are not stable (see geminal diols).

https://en.wikipedia.org/wiki/Methanediol

Thanks for pointing that out. The geminal diol form is indeed favored for methanediol (as well as a few other compounds) but generally, the carbonyl form is predominant, so it still makes trihydroxypropane a logical common name.

Trivial names have stuck around for so long despite their lack of systemization because that's what most people call them including biologists. Think of glyceraldehyde,aniline, pyruvate, toluene etc. There is no choice but to leave these named artifacts as acceptable by IUPAC.

IUPAC nomenclature is intended to be as un-ambiguous as possible. Trihydroxy propane is not unambiguous? Where are the three hydroxy groups. Does it have a geminal diol?
The "1,2,3-" part of the IUPAC name specifies exactly where each -OH is.


text:
Temperature Dependent Specific Heat of Nitrogen
Calculate the change in enthalpy as 1kg of nitrogen is heated from 1000K to 1500K, assuming the nitrogen is an ideal gas at a constant pressure. The tempreature dependent specific heat of nitrogen is Cp = 39.06 - 512.79T^1.5 + 1072.7T^2 - 820.4T^3 where Cp is in kJ/kg-mol and T is in K
Answer is 697.27 kJ
This is the first time i've ever encountered something like "Temperature Dependent Specific Heat"
So I know
H = m*C_p*(T2-T1);
Cp = 39.06 - 512.79T^1.5 + 1072.7T^2 - 820.4T^3

So I try to make something like C_p1:
C_p1 = 39.06 - 512.79T^1.5 + 1072.7T^2 - 820.4T^3; where T = 1000
C_p1 = -8.1934*10^11

C_p2 = 39.06 - 512.79T^1.5 + 1072.7T^2 - 820.4T^3; where T = 1500
C_p2 = -2.766466215*10^12

Cp2 - Cp1 = -3.8585 * 10^12

So yeah... Its not looking too good. Am i missing a formula?
The function looks like its easy to differentiate or integrate. Should I be doing something involving calculus?

The function looks like its easy to differentiate or integrate. Should I be doing something involving calculus?

Yes, exactly!  The equation you should be using is the differential form of the enthalpy equation you are already using.
$$dH = m~C_p(T)~dT$$
$$\int_{H_1}^{H_2} dH = m \int_{T_1}^{T_2} C_p(T)~dT  $$
As you can verify, if heat capacity is independent of temperature, i.e. a constant, the equation reduces to
$$H_2 - H_1 = m~C_p~\left (T_2 - T_1 \right ) \implies \Delta H = m~C_p\Delta T$$
This is the equation you tried to use originally.  However, since we derived this equation assuming a heat capacity that is independent of temperature, and your question explicitly involves dependence on the temperature, we should not use the already integrated equation.  Instead you'll have to do the integration yourself by substitution of temperature-dependent equation.

I think im getting close to the answer; when i do try to integrate it assuming i  convert kg-mol to kg's i get an answer thats 10^14

I can get close to the answer by using T/1000 instead of T (this is commonly how temperature figures into what you have here, the Shomate equation) after doing the integration, but only by fudging the first term - that said, I feel that something is off in the coefficients you have given us. Can you confirm/deny? For reference, see values here from NIST.

I see 3 problems:

All of the exponents should be negative.
Temperature (in Kelvin) should be divided by 100.
Resultant $C_p$ will be in $\pu{kJ kmol-1 K-1}$. Have to divide by molecular weight to get $\pu{kJ kg-1 K-1}$.


text:
Unknown Mass in Dalton's Law of Partial Pressures
An air with mass of $0.454\ \mathrm{kg}$ and an unknown mass of $\ce{CO2}$ occupy an 85 liters tank at $2068.44\ \mathrm{kPa}$. If the partial pressure of the $\ce{CO2}$ is $344.74\ \mathrm{kPa}$, determine its mass.
Answer is $0.138\ \mathrm{kg}$
Ok lets setup Dalton's Law:
$$\begin{align}
p_\text{total} &= p_\text{air} + p_{\ce{CO2}} \\
2068.44 &= \left(mRT/V\right)_\text{air} + 344.74 \\
m_\text{air} &= 0.454\ \mathrm{kg} \\
R_\text{air} &= 0.287\ \mathrm{kJ/(kg\ K)} \\
T_\text{air}/V_\text{air} &= 18847.73126
\end{align}$$
Ok so I need to get something out of:
$$\begin{align}
T_\text{air}/V_\text{air} &= 18847.73126 \\
V_\text{total} &= V_\text{air} + V_{\ce{CO2}} = 0.085\ \mathrm{m^3}
\end{align}$$
Ok so im basically stuck there. only 2 equations for 3 unknowns.
Any hint? or am I totally approaching this the wrong way?

How about 344.74$kPa*V=nRT$?

This gas mixture is stored in the same container. So they should have the same volume - which is the volume of the container. Unlike liquids or solid state, the gases have ample of free space. So the fact their volumes overlap isn't an issue. So don't you think, V_total = V_air + V_co2 should be, V_total = V_air = V_co2 ?

@bonCodigo If $V_{total} = V_{air} + V_{co2}$ and $V_{total} = V_{air} = V_{co2}$, then the only solution is $0m^3$. If $V_{air} = 1m^3$ and $V_{co2} = 1m^3$, then $V_{total} =2m^3$, but the third equation says it should be $1m^3$. So, no, $V_{CO2} \neq 85L$.

@LDC3 I did not mean Vtotal=Vair+Vco2 but Vco2 = Vair = 85L. Partial pressures ratio is proportional to the mole fraction. I seem to be crashing here... care to give a heads up?

@bonCodigo Even if $V_{air} = V_{CO2} = 42.5L$, since $V_{total}$ needs to be 85L, it won't give you the correct answer.

Assuming that the total pressure $p_\text{total}=2068.44\ \mathrm{kPa}$ is equal to the sum of the partial pressures of the individual gases in accordance with Dalton’s law, the partial pressure of air $p_\text{air}$ can be calculated as follows:
$$\begin{align}
p_\text{total}&=p_\text{air}+p_{\ce{CO2}}\\[6pt]
p_\text{air}&=p_\text{total}-p_{\ce{CO2}}\\[6pt]
&=2068.44\ \mathrm{kPa}-344.74\ \mathrm{kPa}\\[6pt]
&=1723.70\ \mathrm{kPa}
\end{align}$$
Since molar mass $M$ is defined as
$$M=\frac mn$$
where $m$ is mass and $n$ is amount of substance, the ideal gas law
$$p\cdot V = n\cdot R\cdot T$$
where $V$ is volume, $R$ is the molar gas constant, and $T$ is temperature, can be rewritten as
$$\begin{align}
p\cdot V&=\frac{m\cdot R\cdot T}M\\[6pt]
\frac V{R\cdot T}&=\frac m{M\cdot p}
\end{align}$$
This equation applies to air
$$\frac V{R\cdot T}=\frac {m_\text{air}}{M_\text{air}\cdot p_\text{air}}$$
where $m_\text{air}=0.454\ \mathrm{kg}$ is the given mass of air and $M_\text{air}=28.97\ \mathrm{g\ mol^{-1}}$ is the average molar mass of air, as well as to $\ce{CO2}$
$$\frac V{R\cdot T}=\frac {m_{\ce{CO2}}}{M_{\ce{CO2}}\cdot p_{\ce{CO2}}}$$
where $m_{\ce{CO2}}$ is the unknown mass of $\ce{CO2}$ and $M_{\ce{CO2}}=44.01\ \mathrm{g\ mol^{-1}}$ is the molar mass of $\ce{CO2}$.
Therefore,
$$\begin{align}
\frac {m_{\ce{CO2}}}{M_{\ce{CO2}}\cdot p_{\ce{CO2}}}&=\frac {m_\text{air}}{M_\text{air}\cdot p_\text{air}}\\[6pt]
m_{\ce{CO2}}&=\frac {M_{\ce{CO2}}\cdot p_{\ce{CO2}}\cdot m_\text{air}}{M_\text{air}\cdot p_\text{air}}\\[6pt]
&=\frac {44.01\ \mathrm{g\ mol^{-1}}\times344.74\ \mathrm{kPa}\times0.454\ \mathrm{kg}}{28.97\ \mathrm{g\ mol^{-1}}\times1723.70\ \mathrm{kPa}}\\[6pt]
&=0.138\ \mathrm{kg}
\end{align}$$
Note that the given value for the volume $V=85\ \mathrm l$ is not required for the calculation of $m_\text{air}$. Thus, the uncertainty of the volume $V=85\ \mathrm l$ (with only two significant digits) does not affect the uncertainty of the result.


text:
Negative and Zero Enthalpy?
I've been investigating chemical kinetics and stumbled across the CHEMKIN program manuals and the JANAF tables. In the original CHEMKIN database they list 'standard enthalpy'. I've checked and this is not in fact enthalpy of formation as this is a separately listed value. What I'm confused about is that I cannot find any reference to molar enthalpy any where else except in the the introduction to the JANAF tables and the CHEMKIN database. What's more it lists the enthalpy of $\ce{H2O}$ and $\ce{H2O2}$ as zero and other substances have negative enthalpy. Questions:
Is specific molar enthalpy in comparison to water or some other substance?
If it is still a relative measurement of enthalpy what advantage does it provide over heat of formation in practical use?
How can the two substances both have the same enthalpy even if it is relative if one has an extra atom and an extra bond?
Am I completely wrong and the whole thing is something else entirely?

Just browsing... Sounds like a reference state problem. In most cases, the reference point is set to ideal gas standard conditions so that the enthalpy description (i.e. enthalpy departure calculations) are consistent with respect to chemical reactions. However, if chemical reactions are not a concern, then we could equally define the reference point to be at some temperature $T_{ref}$ in e.g. the liquid phase (at, let's say, the normal boiling point), or we could simply define the reference to be zero if we are being lazy...

The key point is that we are always moving from some known state to a state of interest, and if we choose the known state well we can save some time by reducing the required steps in the calculation route (strategy).

At a glance, the values listed in the table in the CHEMKIN database are standard enthalpies of formation at 298 K, $\Delta _fH^○(T)$, and standard entropies at 298 K. Why? If only because CHEMKIN's $H^○(298)$ for the elements in their stable states are presented as equal to zero.
You may have just mixed up different enthalpies. Along with the heats of formation, thermodynamic tables (including NIST JANAF) often present the enthalpy increments (sometimes called simply enthalpies), $H^○(T)-H^○(0)$ or $H^○(T)-H^○(298)$. These are not to be confused with the enthalpies of formation $\Delta _fH^○(T)$.
By definition, the standard enthalpy of formation (at a certain temperature) of a complex substance AB is equal to the enthalpy of a chemical process A + B = AB, where A and B are elements in their most stable modifications (at a certain temperature). By convention, $\Delta _fH^○(T)$ for the elements in their most stable modifications is zero.
The enthalpy increment $H^○(T)-H^○(0)$ is an integral of the heat capacity at constant pressure, $\int\limits_0^T C_p(T)dT$. It is no less important than the heat of formation. For example, knowing $\Delta _fH^○(298)$ for our hypothetical substance AB and all the enthalpies $H^○(T)-H^○(298)$ for A, B and AB, we can calculate $\Delta _fH^○(T)$, an enthalpy of formation at different temperature:
$\Delta _fH^○(T) = \Delta _fH^○(298) + [H^○(T)-H^○(298)]_{AB} - [H^○(T)-H^○(298)]_{A} - [H^○(T)-H^○(298)]_{B}$
Note that similar formulae can also be found in the introduction to NIST JANAF (at least to the 4th ed. of it).


text:
What are the magnetic quantum numbers for the three real p orbitals?
The form of the p orbitals that we are familiar with are the $\mathrm{p}_x$, $\mathrm{p}_y$, and $\mathrm{p}_z$ orbitals:

(source: ChemTube 3D)
I also know that the p subshells have the quantum number $l = 1$, meaning that the magnetic quantum number can take the three values $m_l = -1, 0, +1$. Which orbital has which value of $m_l$?

The answer is... it is not so simple. Some quantum mechanics follow, but the TL;DR version is that while $m_l=0$ corresponds to $p_z$, the orbitals for $m_l=+1$ and $m_l=-1$ lie in the $xy$-plane, but not on the axes. The reason for this outcome is that the wavefunctions are usually formulated in spherical coordinates to make the maths easier, but graphs in the Cartesian coordinates make more intuitive sense for humans. The $p_x$ and $p_y$ orbitals are constructed via a linear combination approach from radial and angular wavefunctions and converted into $xyz$. Thus, it is not possible to directly correlate the values of $m_l=\pm1$ with specific orbitals. The notion that we can do so is sometimes presented in introductory courses to make a complex mathematical model just a little bit simpler and more intuitive.
From Physical Chemistry by Atkins and DePaula, the three wavefunctions for $n=2$ and $l=1$ are as follows.
$$\begin{align}
&\Psi_{2,1,0}&&=r\cos{\theta}f(r)\\
&\Psi_{2,1,+1}&&=-\dfrac{r}{\sqrt{2}}\sin{\theta}\mathrm{e}^{\mathrm{i}\phi}f(r)\\
&\Psi_{2,1,-1}&&=\dfrac{r}{\sqrt{2}}\sin{\theta}\mathrm{e}^{-\mathrm{i}\phi}f(r)\\
&\end{align}$$
The notation is $\Psi_{n,l,m_l}$, $r$ is the radius, $\theta$ is the angle with respect to the $z$-axis and $\phi$ is the angle with respect to the $xz$-plane.
$$f(r)=\sqrt{\dfrac{Z^5}{32\pi a_0^5}}\mathrm{e}^{-Zr/2a_0}$$
in which $Z$ is the atomic number (or probably better nuclear charge) and $a_0$ is the Bohr radius.
In switching from spherical to Cartesian coordinates, we make the substitution $z=r\cos{\theta}$, so:
$$\Psi_{2,1,0}=zf(r)$$
This is $\Psi_{2p_z}$ since the value of $\Psi$ is dependent on $z$: when $z=0;\ \Psi=0$, which is expected since $z=0$ describes the $xy$-plane.
The other two wavefunctions are unhelpfully degenerate in the $xy$-plane. An equivalent statement is that these two orbitals do not lie on the $x$- and $y$-axes, but rather bisect them. Thus it is typical to take linear combinations of them to make the equation look prettier. Linear combinations are allowed by the maths of quantum mechanics. If any set of wavefunctions is a solution to the Schrödinger equation, then any set of linear combinations of these wavefunctions must also be a solution. We can do this because orbitals and the wavefunctions that describe them are not real physical objects. They constitute a mathematical model.
In the equations below, we're going to make use of some trigonometry, notably Euler's formula:
$$\mathrm{e}^{\mathrm{i}\phi}=\cos{\phi}+\mathrm{i}\sin{\phi}$$
$$\sin{\phi} = \frac{\mathrm{e}^{\mathrm{i}\phi}-\mathrm{e}^{-\mathrm{i}\phi}}{2\mathrm{i}}$$
$$\cos{\phi} = \frac{\mathrm{e}^{\mathrm{i}\phi}+\mathrm{e}^{-\mathrm{i}\phi}}{2}$$
We're also going to use $x=\sin{\theta}\cos{\phi}$ and $y=\sin{\theta}\sin{\phi}$.
$$\begin{align}
\Psi_{2p_x}=\frac{1}{\sqrt{2}}\left(\Psi_{2,1,+1}-\Psi_{2,1,-1}\right)=\frac{1}{2}\left(\mathrm{e}^{\mathrm{i}\phi}+\mathrm{e}^{-\mathrm{i}\phi} \right)r\sin{\theta}f(r)=r\sin{\theta}\cos{\phi}f(r)=xf(r) \\
\Psi_{2p_y}=\frac{\mathrm{i}}{\sqrt{2}}\left(\Psi_{2,1,+1}+\Psi_{2,1,-1}\right)=\frac{1}{2\mathrm{i}}\left(\mathrm{e}^{\mathrm{i}\phi}-\mathrm{e}^{-\mathrm{i}\phi} \right)r\sin{\theta}f(r)=r\sin{\theta}\sin{\phi}f(r)=yf(r)\\
\end{align}$$

So, while $m_l=0$ corresponds to $\Psi_{p_z}$, $m_l=+1$ and $m_l=-1$ cannot be directly assigned to $\Psi_{p_x}$ and $\Psi_{p_y}$. Rather $m_l=\pm1$ corresponds to $\{\Psi_{p_x},\Psi_{p_y} \}$. Put another way, I suppose we could say that $m_l=+1$ might correspond to $\Psi_{p_{x+y}}$ and $m_l=-1$ might correspond to $\Psi_{p_{x-y}}$.

Why did we assign the value of $m=0$ to the $p_z$ orbital,  is  it conventional  or it was solved using equations (i.e. In the equation,  was  it like this that putting $m=0$, we get $\Psi=z \cdot f(r) $?)

The angular wavefujctions are functions of two angular variables relative to the axes. They are defined relative to the z axis. However, the assignment of the three axes is arbitrary and all possible assignments are degenerate.

Is there any definition of z-axis quantum mechanics  (I hope it's called so), since axes are arbitrary, and how we would define that in a sample of, say Hydrogen atoms, where zillions of molecules are randomly oriented.

@BenNorris Do you mind elaborating on the case for $d$ orbitals and $f$ orbitals as well? Thanks!