text:
How does the pyramidalization effect the bonding angles in a distorted octahedral environment?
This is a follow up on "Generally Octahedral - Meaning?".
Since Dissenter asked in the comments to my answer:

I'm not sure why the $\ce{F_{eq}-S-F'_{eq}}$ bond angles would be compressed, even if the $\ce{S}$ were out of the plane.

Since ron's answer already satisfies the original question about $\ce{BrSF5}$ and I did not want to clutter my answer even further, I am asking it here in a Q&A style.
How does the pyramidalization effect the bonding angles in a distorted octahedral environment?

So we are again talking about $\ce{BrSF5}$ but this post extends to all generally octahedral complexes/ molecules and even quadratic planar geometries. The main requirement is, that the molecule has a $C_{\mathrm{4v}}$ point group. (It may be extended to all non-linear $C_{n\mathrm{v}}\ (n>1)$ point groups.) First of all reconsider the molecular shape again.

Now admittedly the degree of pyramidalization is more than minimal, with less than $1^\circ$ deviation from the plane. However, this has a slight effect on the equatorial $\angle(\ce{F-S-F'})$ bond angle, purely to symmetric reasons.
Basic trigonometric considerations are enough to understand this phenomena. Consider the following scheme, where the peak of the pyramid is represented by $\mathbf{\ce{Y}}=\ce{S}$ and the base is span by $\mathbf{\ce{X_{eq}}}=\ce{F_{eq}}$, $\mathbf{\ce{X'_{eq}}}=\ce{F'_{eq}}$, $\mathbf{\ce{X''_{eq}}}=\ce{F''_{eq}}$ and for completeness $\mathbf{\ce{X'''_{eq}}}=\ce{F'''_{eq}}$.

So we have some basic quantities, that we can address, but in the end we can derive the connection of the two angles without caring about the other quantities, only knowing the point group.

$a$ is the $\mathbf{d}(\ce{F_{eq}\cdots{}F'_{eq}})$ distance
$d$ is the $\mathbf{d}(\ce{F_{eq}\cdots{}F''_{eq}})$ distance
$s$ is the $\mathbf{d}(\ce{S\cdots{}F_{eq}})$ distance
$h$ is the height of the pyramid and it refers to the out of plane shift of $\ce{S}$ compared to the $\ce{F_{eq}\cdots{}F'_{eq}\cdots{}F''_{eq}\cdots{}F'''_{eq}}$ plane
$h'$ is another virtual quantity, the angle bisector of $\angle(\ce{F_{eq}-S-F'_{eq}})$
$\alpha$ refers to the bonding angle $\angle(\ce{F_{eq}-S-F''_{eq}})$
$\beta$ refers to the bonding angle $\angle(\ce{F_{eq}-S-F'_{eq}})$

From Pythagoras's Theorem we can easily derive the following connection (square in the scheme):
\begin{align}
             && d^2 &= a^2 + a^2\\
  \therefore && d   &= \sqrt{2}\cdot a\tag{1}
\end{align}
So now we are going to use some trigonometric functions for the angles (look at the triangles):
\begin{align}
             && \sin\left(\frac{\alpha}{2}\right)
                    &= \frac{\frac12d}{s}\\
  \therefore && s   &= \frac{\frac12d}{\sin\left(\frac{\alpha}{2}\right)}\tag{2}
\end{align}
\begin{align}
             && \sin\left(\frac{\beta}{2}\right)
                    &= \frac{\frac12a}{s}\\
  \therefore && s   &= \frac{\frac12a}{\sin\left(\frac{\beta}{2}\right)}\tag{3}
\end{align}
Let's derive the connection between the angles by equalizing $(2)$ and $(3)$:
\begin{align}
             && \frac{\frac12d}{\sin\left(\frac{\alpha}{2}\right)}
                     &= \frac{\frac12a}{\sin\left(\frac{\beta}{2}\right)}\\
  \therefore && \sin\left(\frac{\beta}{2}\right)
                     &= \frac{a}{d}\sin\left(\frac{\alpha}{2}\right)\\
  \text{with (1)}\implies &&
                \sin\left(\frac{\beta}{2}\right)
                     &= \frac{1}{\sqrt{2}}\sin\left(\frac{\alpha}{2}\right)\\
  \equiv     && \sqrt{2} &= \frac{\sin\left(\frac{\alpha}{2}\right)}{\sin\left(\frac{\beta}{2}\right)}\tag{4}
\end{align}
From this you can see, that the degree of pyramidalization is codependent on the equatorial bond angles. When the pyramidalization increases, $\alpha\to0$, then the equatorial bond angle has to decrease, $\beta\to0$, i.e. a constant ratio.
In the special case of $\alpha = 180^\circ$ this simplifies to $\beta=90^\circ$. In all other cases you can derive the degree of pyramidalization from the equatorial bond angles and vice versa.
Transformation of $(4)$ yields a function that displays the equatorial bond angle in dependence of the pyramidalization, with $\alpha\in~]0^\circ;180^\circ]$ or $\alpha\in~]0;\pi]$.
$$\beta = 2\cdot\arcsin\left[\frac{1}{\sqrt{2}}\cdot\sin\left(\frac{\alpha}{2}\right)\right]$$

Addendum
The out of plane height can also be derived from Pythagoras's theorem (look at right triangle):
\begin{aligned}
                 && h^2 &= s^2 + \left(\frac{d}{2}\right)^2\\
  \therefore     && h   &= \sqrt{s^2 + \left(\frac{d}{2}\right)^2}\\
  \text{with (1)}&& h &= \sqrt{s^2 + \frac{1}{2}a^2}\\
\end{aligned}
This is usually a value that is given for domed complexes, so it is nice to know how to find it.


text:
What is the active ingredient in this "stimulant?"
It is an easily defensible argument that this is a medical question rather than a chemistry one, however, I feel that it leans ever so slightly towards chemistry, and so I posted it here.
There is a "new" OTC stimulant being pushed around that shall remain unnamed so that this isn't flagged as an advertisement or anything.  In looking at the list of ingredients, I don't see anything that would act as a stimulant anywhere in that list.  Having said that, I'm not an expert.
So, I came to the experts.  ;)
Here are the ingredients according to the bottle:

Just to clarify, the heart of this question is "Which ingredient is the stimulant?"

There appear to be multiple stimulants. 1,3,7-trimethylxanthine is a fancy name for caffeine. Synephrine is a sympathomimetic that is a common ingredient in, e.g., nasal decongestants due to the vasoconstriction produced as a side-effect of adrenergic activation. N,N-dimethyl-4-hydroxyphenylethylamine is unfamiliar to me, but appears to be a structural analog of tyramine, which is known to have mild stimulant effects (and is potentially lethal in high doses when combined with MAOIs). Beta-phenethylamine is an endogenous neurotransmitter that has no effect when orally ingested.

My suspicion is that most of the stimulant effects are the results of caffeine content, and secondarily possibly synephrine. The addition of the other ingredients strikes me as dubious at best.

@GregE. I'd say that that would make a great answer.

@GregE. Please submit an answer based on you comments so that I may mark it as correct.

If you google Schizandrol A ( 1,2,3,10,11,12-hexamethoxy-6,7-dimethyl-5,6,7,8-tetrahydrodibenzo[a,c]cycloocten-6-ol, here's a link to a picture of the structure) you find that it is an herb capable of curing just about anything.  If you poke around a little deeper, you find two repetitive themes 1) it increases focus and 2) synergistically enhances the effect of other chemicals ingested along with it.  You also note this herb has caught the attention of the bodybuilding community.  The lifters use terms like "pump and rush", "being on another planet", etc.  Apparently it can take focus to a whole new level. From what I read it looks like it's always ingested with other stuff so it's hard to separate out what's what.  My guess is that all the ingredients in the bottle come into play and the effect they have together, particularly with the Schizandrol A present, is much greater than any ingredient alone.

There actually appear to be multiple stimulants in this formulation:

1,3,7-trimethylxanthine is one systematic chemical name for caffeine.
Synephrine is a sympathomimetic drug, which is a common ingredient in various OTC medicines, where it finds use as, e.g., a nasal decongestant due to the vasoconstriction produced as a side-effect of adrenergic activation. If you examine its molecular structure, you'll see that it's closely related to phenethylamine, and many (if not most) phenethylamine derivatives and structural analogs exhibit stimulant effects. Notable examples include: amphetamine, phentermine, clenbuterol, phenelzine, and various others. These substituted phenethylamines are an important class of drugs and exert a diverse variety of actions; some are stimulants, others anorectics, others decongestants, yet others antidepressants, etc.
N,N-dimethyl-4-hydroxyphenylethylamine was formerly unfamiliar to me, but it appears to be a structural analog of tyramine, which is a naturally occurring amine present in a variety of foodstuffs. Tyramine is known to have mild stimulant effects itself, and is notorious for being potentially lethal in high doses if combined with an MAOI due to its hypertensive (blood-pressure raising) action.
Beta-phenethylamine is an endogenous neurotransmitter within the human nervous system. It exerts a stimulant effect, but it typically has no effect when orally ingested because it is rapidly metabolized before it can be absorbed. Note that this is the same compound as plain "phenethylamine" (described above).
Schizandrol A was also unfamiliar to me. There is at least one animal study indicating possible CNS effects, though how this translates to humans is unclear, and I wasn't able to find any clinical studies or reputable sources reporting on effects on human subjects.

In my estimation, the stimulant effects are probably prominently caused by the caffeine content of the formulation, with synephrine having secondary importance. Beta-phenethylamine in particular is unlikely to have any activity, due to its near total metabolism prior to reaching systemic circulation. As for N,N-dimethyl-4-hydroxyphenylethylamine and Schizandrol A, there doesn't appear to be enough literature on human effects to draw any conclusions, so their inclusion strikes me as dubious. They may or may not be significant in their action.

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text:
Henry's Law with varying temperature
I am trying to determine what the derived partial pressure would be for CO2 in the global atmosphere from historical inconstant temperatures and CO2 concentrations of the global atmosphere:

Is it possible to do this?  If so, how?

An expression for partial pressure, or some sort of useful single value? Did you mean partial pressure of CO2 or something else?

@rch Thank you for looking rch!  I don't think I was concentrating when I wrote that.  I'll edit...

The short answer is no.
The long answer is:
Henry's law works for small concentrations of ideal mixtures at equilibrium. Henry's Law constant varies with temperature according to the Van't Hoff equation:
$$
k(T)=k(T^{\circ})exp\left[-C\left(\frac{1}{T}-\frac{1}{T\circ}\right)\right]
$$
Here $C$ is a constant related to the enthalpy of solvation for each gas, and $^{\circ}$ represents the standard state, T = 298 K and 1 atm.
Using standard state data, you could get an estimate of the partial pressure at a given temperature. However, two key assumptions that are implicit to Henry's law and the Van't Hoff equation do not hold for global atmospheric gas composition:

The atmosphere and the oceans are not at equilibrium (not even close when you consider atmospheric layers and convection)
$\ce{CO2}$ in seawater is not an ideal mixture (very non-ideal since $\ce{CO2}$ reacts with water).

This is one of the many reasons that climate modeling is difficult - the oceans act as huge carbon sinks with a very complex and not-very-stable relationship between dissolved $\ce{CO2}$ content, temperature, composition of rock on the sea floor, convection currents in the atmosphere and the ocean, amount of shellfish (no kidding), and probably lots of other factors (I'm not a climate expert).
The net result is that while Henry's law can get you a quick idea of how much $\ce{CO2}$ could be dissolved at a given temperature under pseudo-equilibrium conditions (or what the corresponding partial pressure would be for a given concentration in water), it isn't enough to get you anywhere close to a good prediction of what the global atmospheric composition will be, particularly for $\ce{CO2}$ (due to its reactivity with water and incorporation into biomass).
Now, if you really just want to convert $\ce{CO2}$ concentration in ppm to a partial pressure, you can do that.
$\ce{CO2}$ in air is close enough to ideal that we can use the ideal gas law. For an ideal gas, the partial pressure is the same as whatever the pressure would be if it were in a container by itself:
$$
P_1 = \frac{n_1RT}{V}
$$
You can find V for a given mass of air, find the mass of $\ce{CO2}$ from the concentration, then convert to moles and plug it into the above.

You currently have numbers in ppm (of ${CO_2}$); ppm values for gas tend to be equivalent to ${\mu mol/mol}$ units. Can't you just use that, the the definition of partial pressure, and the atmospheric pressure and be set (though it varies by altitude and a little over time too)? Some forum has a similar homework-level question. Not sure if I'm missing something with that.
I take it you wanted to use Henry's Law to relate concentration to partial pressure. typically that's used in the realm of vapor-liquid systems; not sure how it fares in a mixture of many gases that I'd qualify the atmosphere as. See things to note: (Chemwiki)
Looking at a (pretty crude IMHO) pressure vs altitude graph of the atmosphere (Wikipedia: Atmospheric pressure): you can see the pressure even within the troposphere ranges from ~12kPa all the way to ~100kPa (sea level: ~101 kPa). The troposphere has 80% of the atmosphere's mass and you want the whole atmosphere anyways, so just take 101kPa as atmospheric pressure? I also assume atmosphere's mass is constant over the span of 400 million years - no clue on whether I can neglect losses and gains from weather, the biosphere, and space over this time span. Then you will have a bunch of carbon dioxide partial pressures over time.

On short-term constancy of the atmosphere's mass, see SE.Physics. The mean even changes by season. It's comparatively small comparing 1.5E+15 kg to the total of 5.15E+18 kg.

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text:
How does intramolecular hydrogen bonding cause the molecules to be separated from each other?
I learnt about intramolecular hydrogen bonding today, which occurs between molecules such as ortho-nitrophenol. What I was told is that in case of intramolecular bonding, the molecules separate from each other, opposite of what happens during intermolecular hydrogen bonding.
I don't understand why is this. Why would attraction within a molecule cause separation of the molecules? There is obviously some intermolecular hydrogen bonding present in this case, but I understand that this might be negligible. Still, why should the molecules spread away from each other?

Molecules with inter-molecular hydrogen bonds tend to associate with one another, while molecules with intra-molecular hydrogen bonds tend to associate with themselves.
Ortho- and para-nitrophenol are often used as examples of molecules with intra- and inter-molecular hydrogen bonds respectively (see picture).

Because of their molecular structures the ortho isomer can form an intramolecular hydrogen bond with itself, whereas the para isomer can only hydrogen bond with another molecule of p-nitophenol.  One would say that the para isomer associates with other p-nitrophenol molecules through hydrogen bonding, whereas the ortho isomer associates with itself through hydrogen bonding.  Because the para isomer associates with other molecules of p-nitrophenol through hydrogen bonding, it takes more energy to break up this association and cause the molecule to melt or boil.  On the other hand, the ortho isomer melts and boils at lower temperatures because there are fewer intermolecular forces (such as the hydrogen bonding) holding the ortho molecules together.

Would it be possible to calculate the boiling/melting point of substances in an ideal situation where no hydrogen bonding existed?

There is no simple method to do this that I am aware of.  There are software packages that attempt to estimate various physical properties of compounds. I'm not sure how accurate or available they are - I think research groups are still trying to improve\refine the packages.

What I mean to say is: is the boiling point of intramolecular hydrogen bonded substances less than that if there was no hydrogen bonding present?

If all hydrogen bonds, both inter- and intra-molecular, were removed, then (I think) generally, compounds that had inter-molecular hydrogen bonds would now boil lower and compounds that had intra-molecular hydrogen bonds would now boil higher.

Ok. That's why I was told. Why would molecules with intra-molecular bonds lower their boiling temp.?

Because they tend to "stick" to themselves instead of "sticking" to other molecules.  They behave as if they have a lower molecular weight.

That was my main question. Why would they not stick to other molecules? there would obviously be some intermolecular bonding

Yes, there is still some intermolecular attraction, but less than there would be without the H-bonding.  I suspect that the H-bonding reduces the dipolar nature of the groups involved (the NO2 and OH groups in this case) and thereby lessens intermolecular electrostatic attractive forces that dipoles can create.

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text:
What are the units for m in the average kinetic energy formula?
For the equation $E_k=\frac{1}{2}mu^2$ what does the $m$ variable represent?
I would say mass but the textbook does not say explicitly that it is anything.

$$\ce{Kinetic ~Energy ~= ~1/2 (mass~of ~the~object~ [kg]) * (velocity~of~the~object~[m/sec])^2}$$

I'm not sure if the $u$ should be $\bar{u}$ and so the $m$ should be the mass of one molecule.

(if you were keeping that under your hat for the time being, I apologize :) )

I cannot overcome a temptation to post this old anecdote:  A physicist was hospitalized after a car accident. He lies and groans: - Ahh it's so good it's half. So nice it's half... - What's about this "half" thing? - the doctor asks. - It's amazing that the kinetic energy is only half em vee squared! - says the poor guy with a smile:)

$$E_k = \frac{1}{2}mu^2$$
where,
$m$ is molar mass (unit - kg/mol), and
$u$ is root mean square speed.
According to wikipedia,

The molar mass of atoms of an element is given by the atomic mass of the element multiplied by the molar mass constant.

See this page (from wikipedia) for complete explanation.

Because this is Chemistry, I'll assume you're dealing with some form or shape of Kinetic-Molecular Theory. So, with
$$E_k = \frac{1}{2}mv^2$$ $E_k$ is the
average kinetic energy, m is the molecular mass and $v^2$
 is the average of the squares of the molecular speeds. (if you're dealing with one particle, its just the velocity of said particle)
Gases at
the same temperature have the same average kinetic energy, so, on average, molecules with a lower speed have a higher
mass. If the gas is pure, all molecules will have the same mass, but at any instant the direction of movement will be
different for each.


text:
How does Dichlorofluoromethane get into the stratosphere?
I was wondering why FCKWs can destroy the ozone layer.
It seems like these gases have a high molecular weight in comparison to $\ce{N2}$ or $\ce{O2}$.
How can these gasses come up to the outer layer of our atmosphere? In particular, how long (on average) would it take to get to such an altitude (through the diffusion of such a particle)?
I mean, they are basically several times heavier than all the other components of our atmosphere combined. Hence, it shouldn't be very energetically stable for them to float on the outer layers of space on top of other lighter particles.

The dominant factor in getting into the upper atmosphere is the atmospheric lifetime of the gas. The atmosphere is fairly well mixed (obviously so since heavy gases like carbon dioxide don't settle out). But compounds that undergo easy reactions don't survive long enough to get fully mixed. CCl2FH is one of those short lived ones unlike its fully halogenated cousins. So it won't be notable in the upper atmosphere, at least compared, for example, to CCl2F2.

This is one of the more frequently asked questions about ozone depletion. I will try to give a short summary.
First of all, it is important to know that Chlorofluorocarbons (CFCs) are extremely stable at room temperature and pressure. It is therefore only natural, that in the lower atmosphere these gases are enriched, therefore they have plenty of time getting in higher altitudes via diffusion.
The earths atmosphere is in constant motion, you can experience that on a macroscopic level via different winds. The sun heats up the earth's surface, causing all sorts of liquids to be evaporated. While evaporating water, these small molecules may form little conglomerates, taking other molecules up to higher heights. This will happen to dust, so it will also happen to CFCs. There will be local areas (or volumes, layers) of the air in different heights of the atmosphere that have different densities, concentrations and composition. As a result they will also have different air pressures. You can experience this on a plane at higher altitudes. Assuming that all reactions tend to react towards equilibrium, there will be forces acting between these areas. In a consequence there will be also upward forces acting on the CFC containing gas mixture. Slowly these molecules will move upwards to the stratosphere.
As you can see, this process is heavily dependent on many different factors and may well require years. I think it is actually very hard to find any accurate numbers for this process. My research on the subject turned up nothing about that. In most of the cases it was just stated, that these gasses are there.
Some further reading:

Why does free chlorine in the stratosphere lose its ozone-depleting potential after about 100,000 reactions?
A collection of nature articles. Some are free access.
Ozonewatch @ NASA
http://www.ozzyozone.org/ (for children, in english, french, spanish)
A hole in the sky @ Science in school (different languages available)

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text:
Determining covalent bond strengths without hybridisation

Assuming pure $\ce{2s}$ and $\ce{2p}$ orbitals of carbon are used in forming $\ce{CH4}$ molecule, which of the following statements is false? (single choice question)

Three $\ce{C-H}$ bonds will be at right angles
One $\ce{C-H}$ bond will be weaker than the other three $\ce{C-H}$ bonds
The shape of the molecule will be tetrahedral
The angle of $\ce{C-H}$ bond formed by $\ce{s-s}$ overlapping will be uncertain with respect to other three bonds.

According to me, the second option is false because $\ce{1s-2s}$ overlapping is stronger than $\ce{1s-2p}$. I just randomly found this question and I don't know the correct answer.

As I already stated, asking this question is wrong on so many levels, but let's just have a look anyway.
We are assuming a set of $3$ rigid $\ce{2p}$ and $1$ rigid $\ce{2s}$ orbitals at carbon. With rigid I mean, hybridisationnote is forbidden - which is wrong, $\text{[very]}_n$ wrong.
As a consequence, a head on overlap between the hydrogen $\ce{s}$ and the carbon $\ce{p}$ results in three bonds, each perpendicular , i.e. $90^\circ$, against each other. This is the feature of the degenerate $\ce{p}$ orbitals. (Answer 1 would be correct.)
As a conclusion from that, the molecule cannot adapt a tetrahedral geometry, because with this geometric arrangement, it is impossible to form a $S_4$ symmetry axis, which is the main feature of the $T_\ce{d}$ or $T$ point groups. Answer 3 would be wrong.
The bond between the carbon $\ce{s}$ and hydrogen $\ce{s}$ also needs to be weaker than the others, since these orbitals are totally symmetric. That means there is a good portion of the carbon $\ce{s}$ orbital already blocked by the other hydrogens. On the other hand, the orbital overlap of the carbon $\ce{p}$ and the hydrogen $\ce{s}$ is quite optimal, since the main contribution of $\ce{p}$ orbitals lie outside of the carbon $\ce{s}$ orbital. (Answer 2 would be correct.)
This $\ce{s-s}$ bond would also be quite wobbly, because neither of the orbitals is directed. Therefore the position of the last hydrogen can easily change without effecting the orbital overlap, and in first approximation the bond strength. (Answer 4 would be correct.)
As a last comment, the molecule in this conformation is not stable, probably not even as an highly excited state. The kinds of thought games actually ruin the more important thing of understanding how orbitals behave. Orbitals are not rigid. They are quite flexible in the end.

Note:
Hybridisation in the generic sense means the linear combination of atomic orbitals of one atom, that contribute to a molecular orbital.

This question may have been designed to demonstrate why we need hybridization/molecular orbital theory to explain observed geometries, symmetries, and bond strengths. I find that many times beginning students have trouble making the conceptual leap from atomic orbitals to molecular orbitals, and that walking them through examples to show why atomic orbitals can't explain bonding on their own helps them to appreciate the need for molecular orbitals. Good answer, btw!

@thomij Thanks, while I think what you say is correct, I still believe that asking these questions is wrong, because they confuse more than they help. Showing shenanigans like this is alright, but deducing things from the impossible should never been done ;) But maybe it was just ripped out of context in this case.

@thomij I agree with MArtin. While I appreciate the good will behind such questions, my experience is,too, that it often confuses more people than helps. For example molecular orbitals and such localized hybrid orbitals are (rather) different as soon as we depart from methane.

@Martin, Greg - This might be a bad example of the method (especially when taken out of context), and it might not be appropriate for chemistry.stackexchange, but it is effective when used properly. It is a common approach in teaching, science, and debate. It works because it takes advantage of natural human learning processes. If you would like to learn more, check out Socratic method, thought experiments, and reductio ad absurdum.

I found this question in a book, and the answer given is option 3. The justification provided is as follows.

In the unhybridized state of carbon, $\mathrm{2p}$ orbitals are $90^{\circ}$ to one another and each one will overlap with $\mathrm{1s}$ orbital of three hydrogen atoms, thus three $\ce{C-H}$ bonds are formed which are $90^{\circ}$ to one another. For the fourth hydrogen atom, its $\mathrm{1s}$ orbital may overlap with non-directional $\mathrm{2s}$ orbital of the carbon and this σ-bond will be stronger than $\unicode[Times]{x3c3}(\ce{C-H})$ bonds formed by $\mathrm{2p - 1s}$ overlap. In such situation $\ce{CH_4}$ molecule can never have tetrahedral geometry.

Except that $\ce{CH4}$ is most definitely tetrahedral.

@bon Come here and read the question and Martin’s answer to it ;)

Bonds are the attraction the 2s electrons of carbon for the nucleus of ANOTHER atom. These 2s electrons are shielded from the other nucleus y its 2p electrons so the attraction is lessened towards another atom making the possible bond weaker. This is another reason for hybridization; it increases the attraction of the molecular orbitals to both of the atoms and reduces electron repulsion . The question is disingenuous


text:
How to identify ions by their quantum number?
If the quantum numbers of an ion are given, how can we identify this ion by its quantum numbers?
For example :

n = 3; $l$ = 2; $m_l$ = 0; m$_s$ = + ½

What I have so far: $n$ is 3 and $l$ is 2 which corresponds to $d$, so the last shell is $3d1$, therefore we have 21 $e^-$ and the answer will be $\ce{Sc}$ ion (from the periodic table) and the answer I got is $\ce{Mn^{2+}}$.   Is this the correct approach?

Please show how you arrived at Mn+.  As I mentioned on your other question, it's critical that you put forth more effort towards these problems.

Why !! at the end of each chapter in the book there are a lot of problems .. some of them, their answers are at the end of the book .. but only the final answer is there .. So, I took one of the solved questions that I dont understand and I put it here because i have no idea how to start solving it.. consequently, how do you expect me to show some effort about something that I don't know the way it is solved ?

it is better to know : I am not scientist and my major has nothing to do with chemistry, but it is a general course that I have to take .. it is hard enough.. don't make it more ! at least write down some guidelines that I should follow to show some effort :(

Maher - start by explaining what you do understand. You must know something - even if it doesn't seem like it will help. That way we have a better idea of where to start. For example, think about the quantum numbers - what do they refer to? Is it a whole atom? All the electrons? Some of the electrons? If we know what you do understand, it is a lot easier to fill in the gaps. Otherwise, we are left with trying to give an introductory quantum chemistry lecture in the space of one answer.

you don't need to give introductory because I have it, once you give the way of solution even without explaining it I will know it directly. anyway: n is 3 and l is 2 which corresponds to d .. so the last shell is $3d^1$ .. therefore we have 21e and the answer will be Sc ion ( from periodic table) ? and ( the correct answer is $Mn^{2+}$ ) ???..

@Maher The point is that from your lectures or your text, you know something towards solving the problem (as you have demonstrated with the previous comment).  Just include that in the body of the question and there will be no problem.  It doesn't matter if you are a chemistry major or an English major, we enacted this rule to keep things fair and to keep this site from deteriorating into a homework service.

I have given you the proper framework for what we expect from those asking questions based on the information given in your comment.  Please follow this from now on.

A pattern-based approach: Remember that the values for ${m_l}$ are centered about 0. This means for a d-orbital, where ${l=2}$, the values of ${m_l}$ = ${(-2, -1, 0, +1, +2)}$.
So, for ${l=2}$ and ${m_l=0}$ and ${m_s=+1/2}$, you should arrive at ${d^3}$. The orbitals each get an electron for all the positive spins first, then are filled completely with the negative spin ones.
That means you "visit" ${m_l}$ twice (in rising order: -2, -1, 0, +1, +2, repeat).
This means the first d-electron (↑ _ _ _ _) such as ${Sc}$ has the configuration ${m_l=-2, m_s=+1/2.}$
The 3rd electron (↑ ↑ ↑ _ _) would have ${m_l=0, m_s=+1/2}$.
A 5-electron configuration (↑ ↑ ↑ ↑ ↑) would have ${m_l=+2, m_s=+1/2}$.
Now on the right-hand side of the d-group, the 6th electron (↑↓ ↑ ↑ ↑ ↑) would have ${m_l=-2, m_s=-1/2}$, and then the magnetic quantum numbers rise to +2 again at ${d10}$.
Anyways, ${Mn^{+2}}$ is ${d^{3}}$ (5 to start, then subtract 2 because of its +2 charge) which matches the quantum numbers you specified.
It's not an 8 electron configuration because of the spin specified. That'd be (↑↓ ↑↓ ↑↓ ↑ ↑).


text:
Why is it written as "joule" and "J"?
Why is the unit joule is written as "joule" and "J", I mean what does "J" mean and what does "joule" mean.

I'm voting to close this question as off-topic because this is easily answerable by just about any research.

In SI, every base units and many derived units have an official name and an official symbol. For example "meter"-"m", "second"-"s" or "joule"-"J".
There are grammatical rules how to use them, but more or less they are interchangeable. If you wonder about the capitalization: in English, the unit names are common nouns and not capitalized ("joule") even if they are derived from a person's name (Joule) or the symbol is a capital letter ("J").

The joule , symbol "$\mathrm{J}$", is a derived unit of energy, work, or amount of heat in the SI  Units. It is equal to the energy expended (or work done) in applying a force of one newton through a distance of one metre (1 newton metre or $\mathrm{N\cdot m}$), or in passing an electric current of one ampere through a resistance of one ohm for one second. It is named after the English physicist James Prescott Joule (1818–1889).
In terms firstly of base SI units and then in terms of other SI units:
$$\mathrm J  = {}\rm \frac{kg \cdot m^2}{s^2} = N \cdot m = \rm Pa \cdot m^3={}\rm W \cdot s = C \cdot V$$
where $\mathrm{kg}$ is the kilogram, $\mathrm{m}$ is the metre, $\mathrm{s}$ is the second, $\mathrm{N}$ is the newton, $\mathrm{Pa}$ is the pascal, $\mathrm{W}$ is the watt, $\mathrm{C}$ is the coulomb, and $\mathrm{V}$ is the volt.
One joule can also be defined as:

The work required to move an electric charge of one coulomb through
an electrical potential difference of one volt, or one '"coulomb
volt" ($\mathrm{C\cdot V}$). This relationship can be used to define the volt.
The work required to produce one watt of power for one second, or one
"watt second" ($\mathrm{W\cdot s}$) (compare kilowatt hour). This relationship can be
used to define the watt.

Source: Wikipedia.


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Can I somehow 'reuse' the chemicals in glow sticks?
I particulary enjoy performing chemical experiments without buying any chemicals for them.
I was always sad about the fact that one glowstick lights out, you just throw it into trash can (I like all means of recycling). So I was wondering if I could (pick one or more):

Recycle the chemicals in it to make it glow again
Use the chemicals for other interesting (and maybe even useful) experiment

Here is a generalized version of what's going on chemically.  The key steps are dioxetane formation (the highly strained 4-membered ring) and its subsequent decomposition and energy transfer to a fluorescent dye, which then decays back to its ground state by emitting light.  By using different fluorescers, you can change the wavelength (color) of the light emitted.

Each company that manufactures glowsticks has their own patent(s) and proprietary process, hence the variation from product to product.

So I was wondering if I could (pick one or more):
  1.Recycle the chemicals in it to make it glow again
  2.Use the chemicals for other interesting (and maybe even useful) experiment

1) Since the starting oxalate ester and concentrated hydrogen peroxide are consumed in the reaction, to make it glow again you would need to put in a new batch of both the oxalate ester and concentrated hydrogen peroxide.  You probably wouldn't want to open the glowstick because:

there might be some concentrated hydrogen peroxide remaining (if I were the manufacturer, I would want this cheap reagent present in a slight excess to make sure all of the ester is consumed)
many of the commonly used fluorescers are irritants, both externally and internally
phthalate esters are commonly used as solvents and they pose health risks

2) Even after all of the starting materials are used up, the fluorescer still remains in solution.  Exposing the glowstick to a black light will still cause it to fluoresce.

Sorry, but what's a black light? Otherwise, thanks for such an answer, it's really good!

Thanks!  A black light is a bulb that emits UV light instead of visible light, see  http://en.wikipedia.org/wiki/Black_light

IIRC the reaction is an oxidation.  Perhaps it could be reduced.  I believe that hair permanent may be a reducing agent.  It would be interesting to see if permanent would work but only done in a safe ventilated place.


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How to determine Langmuir adsorption isotherm constants and use the models
I am doing some adsorption tests on bio materials and would like to test the data with Langmuir and Freundlich adsorption isotherm models. Assume these values:
adsorbent(g).   initial con (µg/L)  final con.(µg/L)
0.1                   1                0.65
0.1                   5                2.1

(Sample volume is 20ml. In both cases the dose of adsorbent was the same (0.1g) and in the first initial concentration is 1ppb and, after adsorption, is 0.65ppb.  In the second initial concentration is 5ppb and, after adsorption, is 2.1ppb.)
Using these data how can I use Langmuir and Freundlich isotherms models to get the constants and check whether my adsorption data fit the model or not.  (If any other parameter is needed you can assume some reasonable value.)

For langmuir isotherm:
Let say we are using linear langmuir isotherm.
The formula is : $\frac{1}{q_e} = \frac{1}{q_m} + \frac{1}{q_mKL}$
find the value of $q_e$. Using the formula $q_e = \frac{(C_i-C_f)V}{\text{mass}}$ of adsorbent, and then plot the graph $\frac{1}{q_e}$ versus $\frac{1}{C_f}$. It would be easier by using excel. From the graph, we can calculate the slope. The linear graph will produce $y=mx+c$ equation. m=intercept=1/qe, hence qe=1/m. C=1/qm.KL. From the C value, we can actually find KL.
From the graph generated using excel, it will shown the R2 value. This value is the important indicator to determine which isotherm model fit the system. The highest R2 value will fit the system. Let say R2 value langmuir is 0.8 and R2 for freundlich is 0.9. So, we can conclude that, Langmuir model fit your system.
For Freundlich isotherm
Plot the graph log qe versus log cf. From the graph you will get linear equation, y=mx + c and the R2 value. Compare the R2 value with langmuir's R2. Highest R2 value indicate that those model fit with the system.
Cf=final concentration
Ci= initial concentration

I think your equation for the linear form of the Langmuir isotherm is wrong. Reference: https://www.hindawi.com/journals/jchem/2017/3039817/


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What does 'organic/non-organic molecule' means exactly?
I am new to the organic chemistry world.
I was delving for the exact answer to my question, and I've found some results:

In wikipedia (french), is was mentioned that an organic molecule contains Hydrogen, so that $\ce{CO2}$ is a non-organic molecule, but wait, what about $\ce{H2O}$? It's almost said a non-organic molecule and it does contain Hydrogen.
And on other websites, I've read that an organic molecule is supposed to have 'Carbon' in its components, else, it will be called 'a non-organic molecule', the result that is an obvious contradictory to the first one.

What I've deducted from all my results is, that the organic molecule must contain Hydrogen or Carbon, and I'm not pretty sure of my deduction.

Long story short, organic molecules are those which for historical reasons are studied in the department of organics.

There are many different answers to the question, "What is an organic compound".  Here are a few examples:

Pudue - "compounds that contain both carbon and hydrogen"
Britanica - compounds in which one or more atoms of carbon are covalently linked to atoms of other elements, most commonly hydrogen, oxygen, or nitrogen. The few carbon-containing compounds not classified as organic include carbides, carbonates, and cyanides
Wikipedia - any member of a large class of gaseous, liquid, or solid chemical compounds whose molecules contain carbon. For historical reasons discussed below, a few types of carbon-containing compounds such as carbides, carbonates, simple oxides of carbon (such as CO and CO2), and cyanides are considered inorganic
UC Davis- Any compound containing carbon atom(s) is classified as an organic compound

Trying to pull the best from each of these while still keeping it simple, perhaps a good definition of an organic compound would be - any covalent compound containing, at least, carbon and hydrogen.  Everything else would be inorganic.

For organic compounds, I like the "carbon skeleton" definition. There is some base carbon structure (linear, branched, cyclic) that we attach other structures to (functional groups). Hydrogens I think of as the "endcaps" of organic chemistry: if there is no interesting functional group to attach, you default to a hydrogen. This definition emphasizes the fact that chains of carbons (the skeleton) don't seem to have limits as to size. But we can also have very small carbon skeletons (e.g. $\ce{CH_4}$). So this is how I think about organic molecules.

So what would CsCO3 be by your definition?

Typo, should be CaCO3

I think it is a good idea to focus on the carbon skeleton (i.e. the C-C bond) instead of C-H, though obviously it makes necessary to define methane, methanol, carbonate, all carbides, if they belong or not to the organic matter.

@ron: an ionic compound? ;-) Seriously, you're right that you would have to carefully go through various exceptions (e.g. carbonate ion not organic, but soaps are) if you wanted a comprehensive definition. I just wanted the skeleton idea out there.

Organic\non-organic molecules - is archaic term, which created in that time, when scientists thought, organics cannot be created in lab, only in living organism.
Now, "Organic" mean "hydrocarbons", and "compound of hydrocarbons".


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Why do atomic orbitals have their unique shapes?
Is there a scientific explanation to why p orbitals are shaped like two balloons, etc. I think it has got to do with electron repulsions. Wikipedia says they are 'characterised by unique values of quantum numbers', of which I don't understand.

The pictures that we typically use to represent orbitals are really just graphs showing the 3-dimensional probability of finding an electron (that occupies the orbital being examined) in space.  $\ce{\Psi}$ is the wavefunction that describes the electron with a set of specific quantum numbers.  $\ce{\Psi^2}$, gives rise to the electron density distribution (the shape of an orbital) plotted in the graph.

I'm sorry to be a grave digger, but why does the $\psi^2$ have such unsymmetrical graphs for $l\ge1$? And if possible, could you explain the non mathematical aspect of it? Because i am not yet acquainted with wave equation's advanced math (i have only seen the raw form). From Resnick, I know that the three orbitals (in case of p orbitals)combined together are symmetrical, but why are they indivudally non spherical?

@FreezingFire  sounds different from the question above and the answer will require more than a sentence or two.  It might be best if you post this as a new question.

The orbitals are solutions (also called wave functions) to the (time-independent) Schrödinger equation. The solutions are related to Bessel functions and Legendre polynomials.
The drawing of orbitals as you often find in text books is in a sense the Copenhagen interpretation of quantum mechanics. In this interpretation, the wave functions (actually, the square of the wave function) are seen as probability distributions of where to find the electrons with 90 % probability (the cutoff might be different from 90 %).

I like to think of an atom as a 3-dimensional version of a vibrating string. If you pluck a guitar string while lightly touching the exact center point, it will vibrate in the first harmonic - make a sound one octave higher (twice the frequency). That vibrational mode is kind of like a 2p orbital in that there is zero amplitude at the exact center.  You get the same behavior in 2 dimensions--and more complex vibrational modes--with a vibrating drum head. Put some iron filings on a drum head and then use different frequencies of sound to stimulate vibration. Some of the vibrational modes look a lot like pictures of the orbitals.


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Why is the enthalpy of vaporization greater than the enthalpy of fusion?
Why is the molar enthalpy of vaporization of a substance larger than its molar enthalpy of fusion (at constant pressure); for example, in the case of ice and water.

Enthalpies of phase changes are fundamentally connected to the electrostatic potential energies between molecules. The first thing you need to know is:

There is an attractive force between all molecules at long(ish) distances, and a repelling force at short distances.

If you make a graph of potential energy vs. distance between two molecules, it will look something like this:

Here the y-axis represents electrostatic potential energy, the x-axis is radial separation (distance between the centers), and the spheres are "molecules."
Since this is a potential energy curve, you can imagine the system as if it were the surface of the earth, and gravity was the potential. In other words, the white molecule "wants" to roll down the valley until it sits next to the gray molecule. If it were any closer than just touching, it would have to climb up another very steep hill. If you try to pull them away, again you have to climb a hill (although it isn't as tall or steep). The result is that unless there is enough kinetic energy for the molecules to move apart, they tend to stick together.
Now, the potential energy function between any two types of molecules will be different, but it will always have the same basic shape. What will change is the "steepness," width, and depth of the valley (or "potential energy well"), and the slope of the infinitely long "hill" to the right of the well.
Since we are talking about relative enthalpies of fusion and vaporization for a given system, we don't have to worry about how this changes for different molecules. We just have to think about what it means to vaporize or melt something, in the context of the spatial separation or relativity of molecules, and how that relates to the shape of this surface.
First let's think about what happens when you add heat to a system of molecules (positive enthalpy change). Heat is a transfer of thermal energy between a hot substance and a cold one. It is defined by a change in temperature, which means that when you add heat to something, its temperature increases (this might be common sense, but in thermodynamics it is important to be very specific). The main thing we need to know about this is:

Temperature is a measure of the average kinetic energy of all molecules in a system

In other words, as the temperature increases, the average kinetic energy (the speed) of the molecules increases.
Let's go back to the potential energy diagram between two molecules. You know that energy is conserved, and so ignoring losses due to friction (there won't be any for molecules) the potential energy that can be gained by a particle is equal to the kinetic energy it started with. In other words, if the particle is at the bottom of the well and has no kinetic energy, it is not going anywhere:

If it literally has no kinetic energy, we are at absolute zero, and this is an ideal crystal (a solid). Real substances in the real world always have some thermal energy, so the molecules are always sort of "wiggling" around at the bottom of their potential energy wells, even in a solid material.
The question is, how much kinetic energy do you need to melt the material?

In a liquid, molecules are free to move but stay close together

This means you need enough energy to let the molecules climb up the well at least a little bit, so that they can slide around each other.
If we draw a "liquid" line approximating how much energy that would take, it might look something like this:

The red line shows the average kinetic energy needed for the particles to pull apart just a little - enough that they can "slide" around each other - but not so much that there is any significant space between them. The height of this line compared to the bottom of the well (times Avogadro's number) is the enthalpy of fusion.
What if we want to vaporize the substance?

In a gas, the molecules are free to move and are very far apart

As the kinetic energy increases, eventually there is enough that the molecules can actually fly apart (their radial separation can approach infinity). That line might look something like this:

I have drawn the line a little bit shy of the "zero" point - where the average molecule would get to infinite distance - because kinetic energies follow a statistical distribution, which means that some are higher than average, some are lower, and right around this point is where enough molecules would be able to vaporize that we would call it a phase transition. Depending on the particular substance, the line might be higher or lower.
In any case, the height of this line compared to the blue line (times Avogadro's number) is the enthalpy of vaporization.
As you can see, it's a lot higher up. The reason is that for melting, the molecules just need enough energy to "slide" around each other, while for vaporization, they need enough energy to completely escape the well. This means that the enthalpy of vaporization is always going to be higher than the enthalpy of fusion.

Sounds great, but how does it explain the latent heat? Where is the liquid at boiling point on your energy diagram? It seems to have a shallower slope at the right hand end, which would suggest that the heat of vaporisation should be less than the heat of fusion?

@alexigirl The red line on the second image represents the liquid at or near the boiling point. The heat of vaporization is the difference between the red and blue lines. The blue line is the liquid at the melting point. The heat of fusion is the difference between the blue line and some line (not pictured) close to the bottom of the well.

If the red line represents the liquid at or near the boiling point, it already has the energy represented by the difference between the red and blue lines. From the graph it would seem that this is enough energy to vaporise the liquid, but we know that this does not happen, we have to keep adding thermal energy for the liquid to vaporise (that would be crazy if once you had heated a pan of water to boiling point it all instantly vaporised). This extra energy is the latent heat of vaporisation. Maybe your red line is too high?

@alexigirl I think there still must be some confusion here. Enthalpies of phase change are differences in energy. The enthalpy of vaporization doesn't contain the enthalpy of fusion, it's the energy difference between states where the molecules are mobile but still in close proximity, and the state in which the molecules are completely unbound.

there is definitely still confusion here. You seem to be saying that the red line represents the energy of the liquid state at or near the boiling point, where molecules are mobile but still in close proximity. So the latent heat of vaporisation cannot include the energy difference between the red line and the blue line, because at both these lines the substance is liquid.

@alexigirl These diagrams represent the potential energy between a pair of molecules. For a single pair, the energy needed to melt is the difference between zero and the blue line. The energy needed to vaporize is the difference between the red and blue lines. At the boiling point, most of the pairs of atoms will have enough kinetic energy to reach the red line, which gives them enough separation that they can escape the liquid.

So why don't they all escape then? I can heat my kettle to boiling point. But if I then switch it off, most of the water (which was at 100 degrees) does not vaporise.

And if, at the boiling point, the pairs of molecules have enough energy to escape, ie they are already at the red line, why should I include the energy increase between the blue and the red line, which they already possess before they reach the boiling point, in the latent heat of vaporisation? The definition I have learnt of latent heat of vaporisation is the energy required to change the state from liquid to gas at the boiling point.

They will all escape if you wait long enough at that temperature. They'll all escape eventually if you wait long enough at lower temperatures too - that's why liquids evaporate. Your definition of latent heat is correct, but it applies to the bulk material. In this answer I was explaining how the potential energy between single pairs of molecules leads to the latent heats. There is a link in my answer to an article about Maxwell-Boltzmann distributions that might help you understand how this works. I don't think I will be able to adequately explain it in comments.

Understood. I was confusing enthalpy of vaporisation with latent heat of vaporisation.

Those are two different names for the same thing. The idea in both cases is that it's the additional energy required to change a bulk substance in the liquid state at the boiling temperature to a gas at the same temperature.

Ive looked up the article on Maxwell-Boltzmann distribution thanks. From this I understand that the molecules will have a range of speeds, with the peak frequency at the lower end of the speed range.  So I can see that quite a few of the molecules will be at or close to the lower line. But there will also be some near the top line. So why do you multiply by Avogadro's number? why not take the distribution of energies into account?

@alexigirl For enthalpy of vaporization, it's the energy required to vaporize the entire substance - in other words, all of the molecules need to move up to that line. That's why the total is given by Avogadro's number times the difference in energies. It's basically saying that every pair of molecules needs to be pulled apart. When looking at something like vapor pressure, you would take the distribution into account - this idea forms the basis of a lot of kinetic theories, from phase change thermodynamics to reaction kinetics.

Ice is less dense than water, that's why ice floats on water.  The lower density of ice means that the average distance between water molecules in ice is greater than the average distance between water molecules in the liquid state.  Because of the greater distance between water molecules in the solid compared to the liquid, molecule - molecule interactions (such as van der Waals and dipole interactions, as well as hydrogen bonding) will be less in the solid than the liquid.  So while we need to put energy into ice to disrupt the lattice structure and break the attractive interactions, this energy is offset to some degree by the even stronger attractive forces that exist in the liquid, since the attractive forces are actually greater in the liquid than the solid.
In the gas phase the molecules are far enough apart that attractive forces between molecules are minimal. Therefor, when we go from liquid to gas we must put in a lot of energy to break all of the strong attractive forces that exist in the liquid without any offset because of the lack of significant attractive forces in the gas.

Enthalpy is state function which is defined as $H = U + pV$, it includes internal energy of the system and energy to create the system volume (boundary work).
Internal energy may be decomposed in several terms: oscillation around a equilibrium centre, vibrational states (stretching and bending). Generally speaking (there might be counter example for some exotic material or at temperature close to $0\ \mathrm{K}$), internal energy increase with temperature because the partition of accessible vibrational states increases (higher states activate with temperature).
About the $pV$ term, you first must consider that it cannot be negative. Therefore $H \geq U$. Now you just have to consider that, generally, $\Delta V_\mathrm{fus} < \Delta V_\mathrm{vap}$. For a given amount of matter, volume difference when melting is small before volume difference when vaporizing. This inequality holds, even for water where $\Delta V_\mathrm{fus} < 0$.
Combining those two arguments, you can explain why enthalpy of vaporization is many order of magnitude higher than enthalpy of fusion.

The molar heat of vaporization is greater than that of molar heat of fusion due to the larger amount of energy required to break the strong attractive forces that exist between molecules of liquids than that of the attractive forces in molecules of gases.

Welcome to Chemistry.SE!  Anyone is welcome to contribute answers but the aim of this site is quality and usefulness to future users (essentially we aren't Yahoo Answers). Please take a minute to look over the help center to better understand our guidelines and question policies.

You should try to answer why the forces are stronger too otherwise the answer is really incomplete.


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Sign of the work done during a reversible compression

The work done during a reversible compression - if $V_\mathrm{i} = V_1$ (initial) and $V_\mathrm{f} = \frac{1}{6}V_1$ (final) - of an ideal gas is?

I chose $1.8 p_1 V_1$. Is it right or should it be $-1.8 p_1 V_1$?

PV work is defined as
$$w = -p\Delta V,$$
where $p$ is pressure and $\Delta V$ is $V_\mathrm{final} - V_\mathrm{initial}$.
It has this form because in chemistry, work is defined from the perspective of the system - if it is compressed, we say that we did work on the system and so the sign of work is positive.
In this example, $V_\mathrm{f}$ is less than $V_\mathrm{i}$, so $\Delta V < 0$. If $\Delta V$ is negative, then $w$ must be positive.
This means your answer should have a positive sign. However, I get a different number from $1.8$ - you should double-check that.


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Conversion of CO2 to methane
Suppose one could generate an enzyme with arbitrary conditions at the active site. What conditions would then be necessary to cause carbon from dissolved $\ce{CO2}$ in low $pH$ water to combine with the free hydrogen and make methane?

It's difficult to answer this question; what you're asking about appears to be both purely hypothetical and vague. It would be helpful, in my opinion, if you could be very specific about what you mean by "conditions [...] necessary" to effect the reaction. Also, are you interested in using the water itself as the source of hydrogen? If so, keep in mind that such a reaction requires considerable input of energy (it's typically done by electrolysis) and will not simply occur spontaneously. Enzymes are catalytic, so they will not help in overcoming thermodynamic barriers.

The definition of pH is the amount of free hydrogen in the solution relative to the OH (at least that's how I understand it, correct me if I'm wrong). Therefore water with a high pH would have lots of free hydrogen that could be used to make methane. The only energy input would be adding base to keep the pH constant as hydrogen is used to make methane.

The common definition is $pH = -log[H^{+}]$, i.e., colog of hydrogen ion concentration. Hence low $pH$, not high, is associated with high $\ce{H+}$ concentration. Also, "free hydrogen" is not a term I've seen used in reference to $\ce{H+}$, but I have seen it used to describe both neutral molecular and atomic hydrogen. Hence my assumption that you'd be using water as a source for molecular $\ce{H2}$. Also, one most certainly does need energy input. Every reaction has an associated activation energy barrier, at a bare minimum.

The problem with such a reaction is that it does not proceed easily because the activation energy ($E_a$) barrier is high. Methane can be produced from carbon dioxide and molecular hydrogen via, e.g., the Sabatier reaction, but this requires catalysis and high temperatures and pressures to overcome the $E_a$. If you want to extract molecular hydrogen from water, the standard and efficient approach is electrolysis, which yields $\ce{H2}$ and $\ce{O2}$. That reaction, however, is very thermodynamically unfavorable due to the exothermic formation of $\ce{H2O}$, so electrolysis is required.

The reaction does not occur easily. There is no known catalyst to allow reaction to occur at reasonable temperatures in one step. There ARE microorganisms using the reaction you consider as energy source, but the process is not understood well enough and clearly occurs via multiple stages, likely with staged electron transfers toward reducing carbon.

Let's take a close look at what the reaction entails. You want to convert $\ce{CO_2}$ to $\ce{CH_4}$ using $\ce{H^+}$. It is possible that the hydrogen ion will co-ordinate with $\ce{CO_2}$ to produce $\ce{[CO_2H]^+}$.
Unfortunately it is very unlikely to have a second hydrogen ion attaching to the same oxygen to produce $\ce{[CO_2H_2]^{2+}}$. Now for the impossible step of water leaving and producing $\ce{[CO]^{2+}}$. This step would require a large amount of energy and $\ce{[CO]^{2+}}$ would be extremely reactive. Since it is highly charged, you would not have any hydrogen ions approaching the moiety.
At this point, any further progress towards $\ce{CH_4}$ is fantasy. To stabilize the reaction, you need to add electrons to the carbon (which you don't have a source).

How about basic hydrogen peroxide?

@Dissenter Changing the pH and using hydrogen peroxide? What will most like be the product of formaldehyde ($CH_2O$) and hydrogen peroxide? I'm guessing a spontaneous reaction producing $H_2$ and $CO_2$.

Oh yes, we need something that'll reduce the carbon (add electrons to it). I don't know why I mentioned HOOH as that'll only oxidize the carbon.


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Why does pasta really boil over?
I was making pasta, and I noticed the pasta boiling over. I thought about it some more, and I realized I had no idea why this was happening. When the lid is on, the foam rises. When the lid is off, the foam dies back. Clearly there's some surfactant at work allowing bubbles to form, and the bubbles are mostly full of steam. If you remove the lid, they cool and condense down to a tiny size. This article is honestly just not specific enough. Why is the starch important?(It also happens with potatoes) How does a polysaccharide, a hydrophilic molecule, become a surfactant(if indeed it does)? Is it phospholipids from the cells that became pasta? Do starch molecules form a polymer and trap steam underneath like a balloon? Is it some kind of hybrid where bubbles are stabilized by viscosity? What's going on here?
This question on seasoned advice tries to tackle it but gets surface tension wrong(I think) so that makes me a little wary of it. Surfactants like soap reduce the surface tension, allowing bubbles to form by letting water molecules spread out into thin films. They also keep your alveoli in your lungs from collapsing. If starches increased the surface tension, wouldn't that reduce the likelihood of forming bubbles by increasing the energetic penalty? High surface tension materials act like mercury.
This answer is all over the internet but doesn't make any sense to me. You need less surface tension for bubbles, surely?
If it's just phospholipids acting as detergent, could I 'boil over' pasta in cold water with a whisk? (I tried this, you can't.)
I considered putting this on seasoned advice, but I'm not interested in how to prevent boil-over(or really any practical results), and the supplied answer there lacks scientific rigor.

The starch forms a loosely bonded network that traps water vapor and air into a foamy mass, which expands rapidly as it heats up.
Starch is made of glucose polymers (amylopectin is one of them, shown here):

Some of the chains are branched, some are linear, but they all have $\ce{-OH}$ groups which can form hydrogen bonds with each other.
Let's follow some starch molecules through the process and see what happens. In the beginning, the starch is dehydrated and tightly compacted - the chains are lined up in nice orderly structures with no water or air between them, maximizing the hydrogen bonds between starch polymers:

As the water heats up (or as you let the pasta soak), water molecules begin to "invade" the tightly packed polymer chains, forming their own hydrogen bonds with the starch:

Soon, the polymer chains are completely surrounded by water, and are free to move in solution (they have dissolved):

However, the water/starch solution is not completely uniform. In the middle of the pot of water, the concentration of starch is low compared to water. There are lots and lots of water molecules available to surround the starch chains and to keep them apart. Near the surface, when the water is boiling, the water molecules escape as vapor. This means that near the surface, the local concentration of starch increases. It increases so much as the water continues to boil, that the starch can collapse back in on itself and hydrogen bond to other starch molecules again. However, this time the orderly structure is broken and there is too much thermal motion to line up. Instead, they form a loosely packed network of molecules connected by hydrogen bonds and surrounding little pockets of water and air (bubbles):

This network is very weak, but it is strong enough to temporarily trap the air as it expands due to heating - thus, the bubbles puff up and a rapidly growing foam forms. Since they are very weak, it doesn't take much to disrupt them. Some oil in the water will inhibit the bubbles from breaking the surface as easily, and a wooden spoon across the top will break the network mechanically as soon as it touches it.
Many biomolecules will form these types of networks under different conditions. For example, gelatin is a protein (amino acid polymer) that will form elastic hydrogen-bonded networks in hot water. As the gelatin-water mixture cools, the gel solidifies, trapping the water inside to form what is called a sol-gel, or more specifically, a hydrogel.
Gluten in wheat is another example, although in this case the bonds are disulfide bonds. Gluten networks are stronger than hydrogen-bonded polysaccharide networks, and are responsible for the elasticity of bread (and of pasta).
DISCLAIMER:

pictures are not remotely to scale, starch is usually several hundred glucose monomers long, and the relative size of the molecules and atoms isn't shown.
there aren't nearly enough water molecules - in reality there would be too many to be able to see the polymer (1,000's).
the starch molecules aren't "twisty" enough or showing things like branching - the real network structure and conformations in solution would be much more complicated.

But, hopefully you get the idea!

This isn't really a biology question I don't think...
As a chemist I can tell you that the rough surface of the pasta does tend to make it produce boiling over.  The same can be produced with stones or wood or nearly any rough surface.  Chemists use little chips of alumina or some other mineral to prevent over boiling in fact.
If you look at a sealed soda bottle you can see the same effect.  Its the same scenario - in this case CO2 rather than steam is ready to come out of supersaturated liquid, but won't unless there is a Bubbles are forming at the plastic surface inside when you open the cap. If you see a stream of bubbles forming within the liquid they are always nucleating from a speck of dust or some other tiny bit of solid floating around.
The surface of mentos candy is particularly good at getting bubbles to form.  I'd bet that if you dropped mentos into boiling water, at least until the mentos dissolved you'd see a bit more bubbling than pasta.  I don't advise you to do this, its just a thought experiment.
But there is something you can do to prevent pasta from boiling over!  Sorry for including this, but cooks are always using too much energy when cooking and its wasteful. When the water starts boiling  you can turn down the heat and put a lid on the pot.  Usually putting the burner on the lowest setting is enough.  This lowers the amount of steam that is supercritical in the water and fewer steam bubbles will form.  This saves energy too, even though the temperature of the water is the same (boiling water is always about 100 deg C), it lowers the amount of supercritical steam in the water.

Apparently oil will also prevent boiling over, I suspect by depleting the available surfactants. Nucleation sites aren't the problem here, water will boil in my dirty scratched pots. I'm confused by the weirdly resilient temperature-sensitive foam and what it's made of, not the presence of boiling.

oil will, until it doesn't.  eventually you will get spatters of steam breaking through the surface explosively if the water is taking in heat. I imagine that the foam, if its pasta cooking, is probably semi-dissolved pasta, which will tend to act like the oil - coating the bubbles with a rough nucleating surface even as it blocks some steam from coming out.

The internet hints that only small amounts of oil are required, not enough to form a complete sealing layer of oil. I'm not certain nucleation is the controlling process here. I need to try a couple of experiments involving cold water and pasta and pasta and loads of oil.

In my experience, pasta only boils over if you put a lid on the pan! As Jeremy says, a small amount of oil will prevent boiling over, even with a lid: I don't measure but I'd guess I use a couple of teaspoons at most. I certainly wouldn't expect that little oil to form a meaningful film over the surface of a pan of boiling water.

the oil helps but eventually if there is a lot more heat going in than is escaping you will get boil over. if there is no lid on the pan it has to be a LOT more heat than with one.  if you turn down the heat of your burner, with or without a lid, oil or no, you can prevent boiling over. that's usually what the lowest burner settings on the stove approximate.

You can regard the starch particles in the boiling water as some kind of impurity, dirt. Dirt always accumulates at phase boundaries, because this saves surface energy. If there's a surface already, might als well put the dirt there, you only need half as much additional surface in your main phases then. (Hence the use of surfactants as detergents: The surface tension/energy is lowered so much that the dirt can leave the surfaces.)
So the interfaces in the bubbles are full of starch, which stabilises them (see other answer, H-Bonds etc.). They don't mind about the steel or glass surface of your pot and lid, because those are wet and full of starch, too. What kills the bubbles is cold and dry air, which they don't see before they've raised the lid and spilled all over.

I really wonder what so upsetting about my answer that people keep downvoting it after a year ... ;-)

Im guessing referring to pasta as dirt will tick some people off. Also, the generalization that impurities will collect at interfaces.

Well, that's what impurities do. I can't help it. 8-)


text:
Will heating alumina (Al2O3) to its melting point yield pure Al?
My inorganic chemistry is at a basic level, so my question is, if I heat alumina ($\ce{Al2O3}$), to its melting point around 2000 C, will I wind up with pure Al?  I think the oxygen will release from the $\ce{Al2O3}$ when the temperature approaches 2000 C.
I bought 5 kilograms from a friend who works at a refinery near my town (he works in logistics and does not have a chemical background) with 1.75$ of almost pure alumina ~99.95%, and I want to know if I melt that white powder will I obtain 2.5kg of 99.95% aluminium?

You've figured out one half of the Hall-Heroult process. Just melting alumina on its own won't liberate the oxygen. The reason is difficult to explain without getting into chemical thermodynamics - one way to think about it is that the lowest energy state is one where oxygen is bound to alumina - where the electrons from alumina have been "taken" by the oxygen. In diatomic oxygen, the electrons are "shared" (the bonds are covalent) and so the total energy is higher. This is a very brief and not-very-detailed explanation, but the end result is that oxygen and aluminum, if given the opportunity, would rather form a bond than not.
When you add thermal energy, unless it changes that balance (it doesn't, at least not until you get way past the melting point), all you are doing is making it more possible for the reaction to happen. If you kept going, you could eventually add enough thermal energy to completely dissociate the alumina, at which point (depending on how you did it) you might be able to remove the oxygen and recover the aluminum.
However - the amount of energy it would take to do that is extremely large. Instead, it turns out to be much easier to simply force oxygen to give its electrons back to aluminum using an electric potential. When you do that, you electrolyze the compound and create diatomic oxygen at the anode and aluminum at the cathode.
The Hall-Heroult process takes this another step to make this work at an even lower temperature (1000 C) by dissolving the alumina in molten cryolite. As a result, it is one of the most (maybe the most) efficient industrial process for producing aluminum, and is responsible for the widespread availability and low cost of aluminum in modern times (and also for the existence of Alcoa).
Your idea about using solar energy isn't bad - although you might be surprised at the amount of energy you need to produce aluminum - both to melt it, and to electrolyze it. One thing you might look into is using solar thermal heating (with a collector, for example) to melt the alumina/cryolite mixture, and then use photovoltaics just for the electrolysis step. That would be quite a bit cheaper than using photovoltaic electricity alone to run an electric heater.
Industrially, the cryolite mixture is kept molten by passing a large electric current through it. The electricity is usually produced by power plants, since the amount required is so large (15 kWh/kg according to Alcoa, and that's just for the electrolysis step!). Although coal and gas power isn't renewable, the advantages are a relatively low cost per unit of power (which might not stay true in the next few decades) and on-demand power generation (likely to always be a big advantage over solar).
There is also nuclear, hydroelectric, and geothermal power - all of which provide large amounts of on-demand power very inexpensively, once the plants are built.

Aluminum is the third most abundant element, however it was difficult to extract and purify prior to the Hall–Héroult process which came about in the 1880's. Despite its natural abundance, aluminum was considered a precious metal before the 1880's.  Napoleon's most important guests were given aluminum cutlery while more ordinary guests dined with silver.

@ron - another interesting fact: the Washington monument was originally capped with aluminum, because it was so valuable. There is also a legend that the roman emperor Tiberius had a man executed who had claimed to have discovered a way of refining aluminum (or what is assumed to be aluminum), because he was afraid it would make his stockpile less valuable.

Now I understand what I had missed in my mind, so the process is very expensive but just in my opinion is much more expensive using chemical agents like cryolite (AlF3,3NaF), cryolite goes at about 600-700 usd a tonne FOB. about 700 two tonnes of alumina ~= 1400 usd without electrolisys which is intensive power drain (consumable carbon anodes) water, what about Environmental Taxation and so on... At LME aluminium spot price is 1936 usd a tonne, and now I come with a second question, what really is the profit there? I mean a refinery will work with profits like 100 usd / tonne?

I don't know what the profit margins are, but I would imagine they are thin since it is a very energy intensive process and since the core technologies are out of patent protection. It will also depend strongly on commodity prices and worldwide demand - but in general I would guess that you need a large, efficient operation with lots of capability to absorb price fluctuations - which is probably why Alcoa has survived by buying up competitors. About the cryolite though - remember most of it can be recycled.

Because its position in the reactivity series of metals, aluminium cannot be extracted using carbon because it is above carbon in the reactivity series (more reactive than carbon in the series) carbon is not reactive enough to displace aluminium from its compounds such as aluminium oxide. But what about using calcium metal? It is more reactive than aluminium. [EDIT: Lowest melting point than aluminium, but what about dissolute aluminium oxide in some acids and then reactive series?]

You might be able to do a redox reaction to isolate aluminum from molten alumina, but consider the fact that melting it in the first place is the largest cost of the conventional process, and that any element which would reduce aluminum will likely be more expensive than aluminum itself.

I see that alumina is a ceramic material and is not soluble in any of the solvents. Piranha (5 parts H2SO4 (sulfuric acid) to 1 part H2O2 (hydrogen peroxide)) is well named, an unstable powerful oxidant is formed which is approximately H2SO5. Piranha is essentially dissolved oxygen atoms ready to burn any fuel. Spot price is 400 usd / tonne, cost effective solution but sounds to be very dangerous. That disolved alumina will release oxygen or will be another mixture of oxigen, sulf and hydrogen? Another solution is maybe microwave that oxigen at their resonant frequency to self-destruct.

What I see is that Hall-Heroult process uses 5V at about 50k Amps - 150k Amps which means very very high power consumption. A refinery what produces aluminium using that process will use at least 15kW of electrical energy to produce 1 kilogram of aluminium. So I am in impossibility to agree with that means a cost effective sollution...

@LXSoft It's not cheap in absolute terms, but all known alternatives are even more expensive.  Electricity happens to be extremely cheap in Iceland thanks to abundant hydroelectric and geothermal sources; because of this, in 2010 73% of all electricity production in Iceland went to aluminium refining (source:  https://arcticecon.wordpress.com/2012/02/15/aluminium-smelting-in-iceland-alcoa-rio-tinto-alcan-century-aluminum-corp/ )

Passing aluminium oxide through a hot enough flame will result in the formation of sapphires, ruby and corundum (amongst others) type gem stones also known as the 'Verneuil process'.
A lot of man made stones are made in similar way. Here is a video to explain this.

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text:
What is the relationship between pressure and stoichiometry ratios
This is the question from the textbook.
Ozone O3 can be prepared in the laboratory by passing an electrical discharge through oxygen gas. Assume that an evacuated steel vessel  with a volume of 10L is filled with 32 ATM of O2 at 25 degrees Celsius and an electrical discharge is passed through the vessel, causing some of the oxygen to be converted into ozone. As a result, the pressure inside the vessel drops to 30.64 ATM at 25 degrees Celsius .
The solutions manual subtracts 3x from the initial pressure of O2 to determine the final pressure of O2 after the electrical discharge(32-3x=final pressure of O2). X is the amount of pressure lost (32-30.63=1.36=x). My question is, how are we justified in using the stiochiometric ratios to determine the final pressure. In other words, how is (32-3x) justified when finding the final pressure of O2.
If my question is not clear. Please let me know and I'll try my best to rephrase it.

The actual reaction being discussed is $$\ce{3O2 -> 2O3}$$
We also know from the gas law that $$\ce{PV=nRT}$$
In the current experiment the only things changing are the pressure and the number of moles; the volume and temperature remaining constant.  In this case the gas law simplifies to $$\ce{\Delta P = \Delta n}$$
Therefor, we know the change in moles for the overall reaction (-3+2=-1) is $\ce{\Delta n}$ and by the above stoichiometry the change in the moles of oxygen will be three times this amount or $\ce{3\Delta n}$ or $\ce{3  \Delta P}$ or $\ce{3x}$.

Thanks for the reply but I am a bit confused. From what I understand from your answer is since P=N and N=-1, then P=-1. However, according to my solutions manual P=1.36 as stated in the question above.

In the stoichiometric equation delta P=delta n=-1; in your question delta P=-1.36

So what you're saying is that since delta P= delta N, and since P=1.36, for every mole, 1.36 atm will be reduced and by the above stoichiometric equation, since three moles of O2 are being decomposed, then 3(1.36) is the total pressure that will be reduced?

No, 3(1.36) is the number of moles of O2 consumed.

No, thats not correct, 1.36atm should decrease for every 1.36 mol since delta P= delta N.

Yes, but for every decrease of1.36 atm 3(1.36) moles of O2 have been consumed.

Glad we worked through it!


text:
Liquefying butane in a freezer
If place a can of pressurized liquid butane (such as a lighter refill canister) into a freezer to get it below its boiling point of $-1\ \mathrm{^\circ C}$,
and then release it into a container kept at a freezing temperature, will it remain in the liquid state without boiling off (until it reaches its boiling point), or will it boil off immediately?
I had a cold metal container kept at a stable temperature of $-7\ \mathrm{^\circ C}$, but when I released room temperature butane into it, it boiled off instantly.
Why did that happen, and what will happen when it is released from the can at a temperature at or below $-1\ \mathrm{^\circ C}$?

You should never store flammable liquids in a normal household refrigerator or freezer, which have various ignition sources (lights, switches, compressor motors) that could ignite vapours in the unventilated cooling compartment. Note that the flash point of butane (the lowest temperature at which it can vaporize to form an ignitable mixture in air) is approximately −60 °C.

Since the room temperature butane didn't instantly cool when it hit the container, it was still above the boiling point.
If you had a large freezer that was roughly at equilibrium at a temperature below $-1^\circ C$, and you put both the pressurized container and the empty container in there and let them reach equilibrium, then when you opened the pressurized container it would not boil and you could conceivably pour it into the other container.
However, butane has a large vapor pressure (the enthalpy of vaporization is relatively small). This means that even at low temperatures, it will evaporate very quickly. This is especially true because the atmosphere does not normally have any butane in it - so even a small partial pressure combined with convective transport in the air will cause it to evaporate even faster. If you imagine pouring some rubbing alcohol out on a table and then putting a fan next to it, it is the same idea. Butane molecules randomly have enough energy to escape the liquid, and air currents carry them away. Since there is no butane vapor nearby to establish an equilibrium with the liquid, more butane evaporates to take its place.
The net result of this is that even if you did the experiment in a freezer and made sure everything was below $-1^\circ C$ (which includes you wearing well-insulated gloves and not breathing on it), it might still evaporate so fast that it would look as if it were boiling.
From wikipedia, here is a plot of vapor pressure vs T for butane:
The boiling point is the point where the vapor pressure line exceeds atmospheric pressure. Atmospheric pressure is usually around 760 mmHg. This is a log plot for the P axis, which means one large grid step is ten times larger than the previous one. From this graph, you can see that at $-30^\circ C$, the vapor pressure is still ~200 mmHg. This means that roughly 26% of the air at equilibrium would be butane. Unless you have a very tiny freezer, that is a lot. This means that you might have to wear a gas mask and get a 55 gallon drum of butane before you could really do any "pouring" in the traditional sense.
At $-100^\circ C$ you are down to 1 mmHg vapor pressure. At this point I would say that even a small amount (like what is in a lighter) would stay liquid long enough that you could pour it.

Thanks for the detailed explanation. You definitely cleared a lot up about vapor pressure. Also, if the air was saturated with 100% butane (if that's the right word for it), would it still evaporate readily? And yes, I did the experiment wearing some work groves right next to a fan by an open window ;), although the fan probably made the results worse though!

Yes, the fan will definitely make it evaporate faster. If the air was at 100% saturation of butane, it would no longer evaporate. You would probably need a self-contained breathing unit at that point - so please don't try to do this without the right equipment.

Okay. Right now it's pretty much hypothetical to me so I won't be doing any more Butane experiments. Also another Butane question-I have a butane mini blowtorch. When I heat up a metal surface (it's Aluminium and is large so it stays relatively cool for 20 secs or so), many droplets of some liquid start "condensing" (???) on the metal surface within seconds around the general area of where the jet is aimed. The droplets are odorless (I hope they're safe....) but it is mysterious to me. Remember this happens when the metal is still cold, I haven't looked if try evaporate when it starts heating.

I'll make you a deal - you ask this as a separate quesition, and I'll answer it for you.

Sure, I'll do so.


text:
Butane blowtorch condensing liquid on heating surface
I have a small Butane blowtorch, and use it to heat an Aluminium plate. The plate is thick so takes 20 seconds or so to really start heating up.
As soon as I apply the heat, droplets of some liquid start "condensing" on the surface of the metal around the general area of the flame.
The liquid is colorless and odorless and is at room temperature, note this happens when the metal is still at room temperature as it hasn't heated up much yet.
What is this liquid exactly and why/how does it form? I suspect it could be a $C_{4}H_{10} + O_{2}$ reactant but not sure.  Thanks!

This is a guess - but I think it might be water formed as a byproduct of butane combustion precipitating on the plate:
$$
\ce{2C4H10(g) + 13O2(g) -> 8CO2(g) + 10H2O(g)}
$$
Aluminum has very high thermal conductivity, so it is possible that the plate is rapidly absorbing heat from the water vapor, causing it to condense. If this stops happening right around the point when the aluminum temperature reaches 100 C, that would be evidence to support this hypothesis.

That sounds right to me then. I updated my question with a thought that the $C_{4}H_{10} + O_{2}$ reactant could involve water and that makes sense now, I'll try with a smaller piece of Al tomorrow to test the theory. Thanks!

I added the reaction equation for you. If you try this, be sure to hold the thermometer right against the metal at the spot where the water is forming. That spot will heat up much faster than the rest of the plate. Also, make sure the thermometer is rated to around 600 C or you might break it accidentally.


text:
Instructions for handling toxic liquids
I would like to know what precautions should be taken when handling liquids like trichloroethylene, tetrachloroethylene, dichlormethane, and mercury. Of these, which is most toxic?
I have seen people handle mercury in rooms without fume hoods. I don't have access to a fume hood, so is it adequate protection if I wear a suitable mask when handling mercury?
Also, I am interested in the conditions in which mercury vapor is formed. Why is it formed when its gaseous temperature is high?

Each liquid produces some vapours. When you leave wet clothes, they eventually dry. Same here. It is true, that vapour pressure for mercury at room temperature is extremely low, but since mercury accumulates and has quite high toxicity, it is still a trouble. Same for other liquids.
Mercury must be kept in sealed container and/or under layer of water. Any works with it must occur in large pan, so no tiniest drop of mercury escaped. Escaped drops must be collected using thin copper plate, washed in concentrated nitric acid. All solutions of mercury compounds and remaining mercury should be utilized according to local procedures (there is an agency for it at my city). Metallic mercury may be transformed into salts using $\ce{FeCl3}$ concentrated solution. While not removing the mercury, this transforms it into compounds that does not produce vapours. However, in case of mercury leak a special agency must be informed to perform all necessary procedures. Consult local regulations on the cases.
Dichloromethane has regrettably low boiling point. It must be kept away from heat sources and in case of high temperatures it must be cooled or kept in sealed container. It evaporates swiftly on open air, carrying away much heat.
All the liquids should be used under fume hood. In case one is unavailable, an acceptable alternative is open air as long as there is a gentle wind. The chlorinated hidrocarbons mentioned has strong odor, strong enough to make handling them without good ventilation impossible. None of the liquids should contact skin. It is not immediately fatal, but still. In case the contact occurred, the chloroalkanes will evaporate quite quickly, though washing skin with soap and water is highly encouraged. Again, one-time breathing-in of the vapours is not fatal with this compounds, but repeated breathing-in guarantees health problems.
Good thing: none of the liquids is flammable, so flammability is not a trouble here. Unfortunately, heating them at presence of air is still highly unrecommended as well as continuous exposure to sunlight: it may produce phosgene.

DCM, produces phosgene when exposed to sunlight? Seems strange to me, cause that would raise its hazard diamond sky high.

@Jori google://DCM+phosgene . Though I agree, the most troublesome thing is open fire, but I would be cautious just in case.


text:
Formation Constants; Lewis Acidity
I still distinctly remember reading about how Lewis acidity/basicity was characterized using kinetics. The reason given was that because formation constants are so large, these reactions are not equilibrium reactions and completion reactions. So we just characterize strength using the rate of reaction/formation. This at least makes partial sense; formation constants are huge:
http://bilbo.chm.uri.edu/CHM112/tables/Kftable.htm
EDIT: I have found what I read:

Many reactions of nucleophiles are not reversible. A bond forms, a bond breaks, and that’s the end of the reaction. The problem with this from a measurement standpoint is that we often can’t determine an equilibrium constant for a reaction. And if we can’t do that, then we can’t develop a reactivity scale based on equilibria.
If we can’t measure equilibria, then what do we do? Well, we use the next best measurement available: to measure reaction rates.

1) Does the quoted text make sense?
2) Am I interpreting it correctly or out of context?
3) Reference to previously answered question: Lewis base strength and the hydrogen proton

I think I got this figured out.
Acidity and basicity are thermodynamic properties.
Nucleophilicity/electrophilicity are kinetic properties.
Lewis acids are electrophiles.
Lewis bases are nucleophiles.
We can use electrophile/nucleophile in describing Lewis acids/bases. But care should be taken to ensure we don't conflate kinetics with thermodynamics.
I think I finally got this issue resolved.

Nice job solving your own issue.


text:
Calculate the second ionization energy for the Mg from the following data?
here is a problem :

What I did " as shown in the picture below "
I rewrote the information given in terms of reactants and products ( equations )// in Part 1
Then I used the Hess law to obtain the desired result //in part 2
in part 3 // I added and cancelled the terms that appear in both sides
but I left with $O_{(g)}$ !! and the enthalpy is 1192 which is inconsistent with the correct value 1442 !!
What's wrong?

You forgot that you only need 1/2 mole of $\ce{O2}$. Since the product is $\ce{MgO}$, the third equation in step one should be:
$$
\ce{1/2O2(g) -> O(g)} \space \space \rm{249.35 ~kJ/~mol}
$$
Note that I divided the bond dissociation energy of $\ce{O2}$ by two as well.
This is a very common mistake in Born-Haber type problems - the way to catch it early is to write out the balanced reaction first, then adjust the coefficients of each step accordingly.

Thanks for showing your work.  We're given that
$$\ce{\Delta H_{f} (MgO (s))~ =~ -601$.$8 ~kJ/mol}$$
So any balanced equation we write that has MgO as the product will have this same heat of formation.  An equation that produces MgO and allows us to solve the problem is
$$\ce{Mg^{+2}(g) + O^{-2}(g) -> MgO (g)}$$
The heat of formation of $\ce{O^{-2}}$ is given by
$$\ce{\Delta H_{f} (O^{-2} (g))~=~1/2 the ~O2~ bond~ energy~ (only~ half~ because~ we~ only~ need~ 1~ oxygen~ atom) + the ~first ~ionization~ potential~ for~oxygen + the~second~ionization~ potential~ for~ oxygen}$$
or
$$\ce{\Delta H_{f} (O^{-2} (g))~=~1/2(498.7)~-~141+780~ = ~$888.4$~ kJ/mol}$$
Similarly the heat of formation of $\ce{Mg^{+2}}$ is given by
$$\ce{\Delta H_{f} (Mg^{+2} (g))~=~ \Delta H_{sublimation}Mg(s) + the ~first ~ionization~ potential~ for~Mg + the~second~ionization~ potential~ for~ Mg}$$
or
$$\ce{\Delta H_{f} (Mg^{+2} (g))~=~130 + $738.1$ + IP_2(Mg)}$$
Before solving for $\ce{IP_2(Mg)}$ there is one more adjustment to make.  Everything we've done so far has been for the gas phase except for our starting heat of formation for $\ce{MgO}$ which was for the solid, so let's adjust this heat of formation to its gas phase value.
$$\ce{\Delta H_{f} (MgO (g))~ =~ \Delta H_{f} (MgO (s)) + lattice~ energy~ MgO }$$
or
$$\ce{\Delta H_{f} (MgO (g))~ =~ $-601.8$+ 3800 ~=~$3198.2$~ kJ/mol}$$
Now, inserting all of these values into our equation we get,
$$\ce{130 + $738.1$ + IP_2(Mg) + $888.4$~=~$3198.2$}$$
or
$$\ce{IP_2(Mg)~=~$1441.7$~kJ/mol}$$