text:
Spin in meta-GGAs
In density functional theory (DFT) we know that the local density approximation (LDA) leads to an exchange-correlation energy kernel of the form:
$$E_\mathrm{xc}^\mathrm{LDA} =\int e_\mathrm{xc}\left( \rho \right) dr$$
If we want to be able to differentiate between $\alpha$ and $\beta$ spin we can introduce local spin density approximation (LSD), giving as kernel of the form:
$$E_\mathrm{xc}^\mathrm{LSD} =\int e_\mathrm{xc}\left( \rho_\alpha,\rho_\beta \right) dr$$
Further improvements can be made by introducing the gradient of the density, giving the rise to the generalized gradient approximation (GGA).
Under the spin density anzats, if we want the size of density gradient, we can define it as (this is the square of the size):
$$\gamma_{\sigma,\sigma '} = \nabla\rho_\sigma \nabla\rho_{\sigma'}$$
Giving the kernel:
$$E_\mathrm{xc}^\mathrm{GGA} =\int e_\mathrm{xc}\left( \rho_\alpha,\rho_\beta, \gamma_{\alpha, \alpha},\gamma_{\beta, \beta},\gamma_{\alpha, \beta} \right) dr$$
Note here that $\gamma_{\alpha,\beta} = \gamma_{\beta,\alpha}$ due to the definition.
Now going the next step, metaGGA can be introduced. Here we take the Laplacian of the density. Now this is where I get stuck.
How is the metaGGA extension defined, in a similar manner to how I defined the GGA? Since the Laplacian operator also only works on a single density, I guess the scalar quantity arising from the Laplacian can be given as:
$$\eta_{\sigma,\sigma'} = \nabla^2 \rho_\sigma \nabla^2 \rho_{\sigma'}$$
But I am not sure of this. Further, in the literature, the metaGGAs are almost always associated with the kinetic energy density. I know that the density is defined as:
$$\rho_\sigma = \sum_i\left| \psi_{i\sigma} \right|^2$$
I cannot figure out if the above definition directly gives rise to the kinetic energy density, when taking the Laplacian of $\rho_\sigma$. Or if the kinetic energy density is independent from the Laplacian of the density?
For the completeness of my question, I have some problems even when given the expression of the kinetic energy density, which is given as:
$$\tau_{\sigma} = \frac{1}{2}\sum_i^\mathrm{occ}\left| \nabla\psi_{i\sigma} \right|^2$$
From the literature, it would look like the kinetic energy density only depends on one spin at the time. I guess this is because $\left| \nabla \psi_\alpha \nabla \psi_\beta \right| = 0$, because the overlap of two spin functions is zero. But I am not fully sure about this argument.
Since I was unable to make the connection between the Laplacian of the density and the kinetic energy density I am not sure which of the following two kernel I expect to end up with:
$$E_\mathrm{xc}^\mathrm{metaGGA} =\int e_\mathrm{xc}\left( \rho_\alpha,\rho_\beta, \gamma_{\alpha, \alpha},\gamma_{\beta, \beta},\gamma_{\alpha, \beta}, \tau_{\alpha}, \tau_{\beta} \right) dr$$
or,
$$E_\mathrm{xc}^\mathrm{metaGGA} =\int e_\mathrm{xc}\left( \rho_\alpha,\rho_\beta, \gamma_{\alpha, \alpha},\gamma_{\beta, \beta},\gamma_{\alpha, \beta}, \eta_{\alpha, \alpha}\eta_{\beta, \beta},\eta_{\alpha, \beta}, \tau_{\alpha}, \tau_{\beta} \right) dr$$

https://www.tandfonline.com/doi/abs/10.1080/00268976.2017.1333644 This article goes into some detail about how to express the exchange functional for spin polarized LDA, GGA, and mGGA.

This might be my own ignorance on this subject, but what is the purpose of defining $\gamma_{\sigma\sigma'}$? It would seem you could write the GGA XC energy with just $\nabla \rho_{\alpha}$ and $\nabla\rho_{\beta}$.

@Tyberius It appears in the XC contrib to the Fock matrix: https://chemistry.stackexchange.com/a/81832/194

@Tyberius Yes. This is essentially the same as the depends of $\gamma_{\sigma \sigma '}$ that I wrote. Since this $\gamma$ also just depends on single index gradients.

Please note, that I made some major edits to the question.

The idea of a MGGA is not so different of a GGA. The Laplacian of the density or the kinetic energy density are employed to give a better description about the inhomogeneity of chemical systems. These parameters or "ingredients" can improve for instance the characterization of bonds. Take for instance the parameter $\alpha=(\tau - \tau^{W})/\tau^{unif}$ that is used in the SCAN
density functional [1]
where $\tau$, $\tau^{W}$ and $\tau^{unif}$ are the exact (in KS method) kinetic energy density, the Weizsaecker kinetic energy density and the kinetic energy density of the uniform electron gas, respectively. $\alpha$ can make a density functional recognize covalent, metallic and weak bonds.
Now, to the math...
First, the Laplacian operator is defined as,
$$
\nabla^{2} = \nabla \cdot \nabla = \left[ \frac{\partial}{\partial x} \hat i + \frac{\partial}{\partial y} \hat j + \frac{\partial}{\partial z}\hat k \right] \cdot \left[ \frac{\partial}{\partial x} \hat i + \frac{\partial}{\partial y} \hat j + \frac{\partial}{\partial z}\hat k \right] = \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} + \frac{\partial^{2}}{\partial z^{2}} .
$$
We see that the laplacian operator is not a vector as the gradient operator. Second, we apply the laplacian to the density of a given spin,
$$
      \nabla^{2} \rho_{\sigma}(\pmb{r}) =
      \nabla^{2} \sum_{i}^{occ} \psi_{i\sigma}^{*}(\pmb{r}) \psi_{i\sigma}(\pmb{r}) =
      2\sum_{i}^{occ} [\psi_{i\sigma}^{*}(\pmb{r}) \nabla^{2} \psi_{i\sigma}(\pmb{r}) + \nabla \psi_{i\sigma}^{*}(\pmb{r}) \cdot \nabla \psi_{i\sigma}(\pmb{r})].
\label{laprho}
$$
We found in the last term of $\nabla^{2} \rho_{\sigma}(\pmb{r})$ the kinetic energy density which we're going to discuss below. The Laplacian of the total density is defined by,
$$
      \nabla^{2}\rho(\pmb{r}) =
      \nabla^{2} \left[ \rho_{\alpha}(\pmb{r}) + \rho_{\beta}(\pmb{r}) \right] =
      \nabla^{2} \rho_{\alpha}(\pmb{r}) + \nabla^{2} \rho_{\beta}(\pmb{r}).
$$
Now, we turn to the kinetic energy density. Rearranging the terms in $\nabla^{2}\rho_{\sigma}(\pmb{r})$ and dividing by 4 yields,
$$
\frac{1}{2}\sum_{i=1}^{occ} \nabla \psi_{i\sigma}^{*}(\mathbf{r}) \cdotp \nabla \psi_{i\sigma}(\mathbf{r}) = \frac{1}{4} \nabla^{2} \rho_{\sigma}(\pmb{r}) - \frac{1}{2}\sum_{i}^{occ} \phi_{i\sigma}^{*}(\pmb{r}) \nabla^{2} \phi_{i\sigma}(\pmb{r}).
$$
Finally, we obtain the kinetic energy density,
$$
\tau_{\sigma} (\mathbf{r}) = \frac{1}{2}\sum_{i=1}^{occ} \nabla \psi_{i\sigma}^{*}(\mathbf{r}) \cdotp \nabla \psi_{i\sigma}(\mathbf{r}) = \frac{1}{2}\sum_{i=1}^{occ} |\nabla \psi_{i\sigma}(\mathbf{r})|^2.
$$
Since we have alpha and beta spin contributions in the laplacian of the total density we get,
$$
\tau(\pmb{r}) = \tau_{\alpha}(\pmb{r}) + \tau_{\beta}(\pmb{r}).
$$
With the equations above you can define MGGA's that depend on $\nabla^{2} \rho$ and/or $\tau$.


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Why are calorimeters often made of materials with low specific heats?
When building calorimeters, a material with a low specific heat is often used (such as Styrofoam, aluminum, etc.). However, why would a material with a low specific heat be preferred? My thought process is below:
Scenario: An exothermic reaction occurs in a calorimeter.
Principle: In calorimetry, we are always trying to minimize the amount of heat transfer from the calorimeter to the surroundings. The amount of heat transferred from the reaction to the calorimeter doesn't really matter because we can calculate this quantity of heat using the calorimeter's calorimeter constant, which we can derive experimentally.
Case 1: Calorimeter made of material with low specific heat
If the calorimeter had a low specific heat, it would absorb less heat, but its temperature would increase more. Differences in temperature cause heat transfer, so a larger difference in temperature between the calorimeter and the surrounding air would mean more heat transfer from the calorimeter to the surroundings.
Case 2: Calorimeter made of material with high specific heat
If the calorimeter had a high specific heat, it would absorb more heat, but its temperature would not increase much. Differences in temperature cause heat transfer, so a smaller difference in temperature between the calorimeter and the surrounding air would mean less heat transfer from the calorimeter to the surroundings.
Therefore, a calorimeter with a high specific heat would be more effective because it would minimize heat transfer between the calorimeter and the surroundings.
Where is the flaw in my logic?
Additionally, according to http://www.greenspec.co.uk/building-design/insulation-materials-thermal-properties/,

A good insulator has a higher specific heat capacity because it takes time to absorb more heat before it actually heats up (temperature rising) to transfer the heat.

Actually, a calorimeter with high specific heat would have less temperature change, which is harder to measure. Therefore, the error in the calculated enthalpy of the reaction would be higher.
Specific heat does not mean heat conductivity, as described by Fourier's Law.

Is the low thermal conductivity of Styrofoam and aluminum the main reason why they are used in calorimeters? Also, what is the relationship between specific heat and thermal conductivity?

I don't think designers care about the specific heat of the materials.  Their primary considerations are going to be standard engineering considerations, e.g. strength, lightness, cost, ease of fabrication, resistance to corrosion and wear. Aluminum scores well on all of these, so it's a very common material for use in building small devices.  Styrofoam can be used in a situation of extreme economy, because it's solid, has moderately low thermal conductivity, and is very easy to fashion into the right shape.

Aluminum has a very high thermal conductivity, while Styrofoam has a very low thermal conductivity (https://en.wikipedia.org/wiki/Thermal_conductivity#Experimental_values).
Does thermal conductivity matter when designing a calorimeter? Why or why not?

It certainly matters when designing some parts of the calorimeter, e.g. its thermal insulation.  It doesn't matter at all for designing other parts, like its case or inner lining.

So, is aluminum often used to build the case or inner lining of a calorimeter instead of the thermal insulation of the calorimeter?

There is no flaw in your logic, but there is a bigger flaw behind it. All right, maybe high specific heat would be better. It is just that the said value is less important than heat conductivity. Heat escapes to the surroundings and leaves you with no way to evaluate it. You don't want that to happen. That's when styrofoam comes into play.
Aluminum is another story. You can't make a combustion chamber out of styrofoam. It has to be a metal. Well, then why not use aluminum. Then it is not a material with low specific heat. Compared to other metals, it is higher than many.
So it goes.

I'm not actually a chemist:), but I teach chemistry at a homeschool co op and do have a science background academically. I think what you might be missing in your thought process is clarifying between heat conductivity and change in energy. I was a bit confused on this too. As others have mentioned, heat conductivity is not correlated to specific heat. When you create a calorimeter, if you are trying to make the heat transfer to the calorimeter negligible, you would use something with a low specific heat: q (energy/heat)=(m)(c)(change in temperature) where m=mass of the calorimeter and c= specific heat. So therefore, the lower the specific heat, the lower the overall heat absorption from the item put in the water.

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Why do aluminum and air have low specific heats?
According to Wikipedia,

...specific heat capacity is a function of the structure of the substance itself. In particular, it depends on the number of degrees of freedom that are available to the particles in the substance; each independent degree of freedom allows the particles to store thermal energy. The translational kinetic energy of substance particles which manifests as temperature change is only one of the many possible degrees of freedom, and thus the larger the number of degrees of freedom available to the particles of a substance other than translational kinetic energy, the larger will be the specific heat capacity for the substance. For example, rotational kinetic energy of gas molecules stores heat energy in a way that increases heat capacity, since this energy does not contribute to temperature.

However, upon reading this, one question comes to mind. After doing some research, I found that found that aluminum and air relatively low specific heats (https://www.engineeringtoolbox.com/specific-heat-capacity-d_391.html). This implies that aluminum and air have a low number of degrees of freedom. However, why is this the case? I know that air is mainly composed of $\ce{N2}$ and $\ce{O2}$ gas. Also, I know that aluminum has a crystal structure. Do these factors restrict the number of degrees of freedom? For example, I suppose that the crystal structure may restrict the rotational kinetic energy of each aluminum atom.

tl;dr They don't.
You seek to explain one peculiar isolated fact from a bigger picture. When looking at the picture as a whole, the fact is neither isolated nor peculiar. It is the same as pointing at the guy in the background of "Washington Crossing the Delaware" and asking why is his head bandaged, and why is he wearing a fur coat and a funny hat; isn't all that strange? In a today's city street, maybe; but in the context of the broader picture, not at all.
Same thing here. Heat capacity of air is not particularly low, nor is it high. In fact, it is the same as that of any other diatomic gas, even though the table might seem to suggest otherwise. How so? Well, you already know that heat capacity is explained via degrees of freedom, but what is it that has those degrees of freedom? Guess what? It is molecules, not kilograms. So let's compare the comparable. Don't look at the heat capacity per kilogram. Look at the values per one molecule (or rather per one mole, which is the same, only without nasty $10^{\pm23}$s). Then you'll be able to verify what I said before.
Now to aluminum. Its heat capacity is not all that low. It is way higher than those of titanium, iron, copper, not to mention the heavier metals. It is, however, low when compared to water. Well, that's an expected thing: most solids have lower heat capacity than most liquids or gases. In particular, like you said, this is because the particles in a solid are less free to move, hence fewer degrees of freedom.
So it goes.

I think the most important answer to your question is just that the densities of air and aluminum are low.  The relationship between the heat capacity and the availability of degrees of freedom in which energy can be stored requires you to compare equal number of degrees of freedom -- which is equal moles not equal grams.  Specific heat capacity is heat capacity per gram.  Anything with a low density will have fewer degrees of freedom per gram, by definition, and therefore a lower specific heat.
If you compare molar heat capacities, then Al is about the same as any other metal (e.g. Al 24.3 J/K-mol, Cu 24.5 J/K-mol) and not very different from air (e.g. N2 21 J/K-mol).  The real standout in molar heat capacities are substances like liquid water (75 J/K-mol).  The reason for this is that substances like liquid water have a lot of degrees of freedom that are frozen out at lower temperatures (meaning the minimum energy needed to excite them isn't available) and are becoming available at room temperature.  That means you can add in a lot of energy without changing the temperature much.  But this only happens under certain conditions.  At higher temperatures the molar heat capacity of H2O is unremarkable.

Why does water have so many more degrees of freedom than air or metals such as aluminum and copper? Do hydrogen bonds provide degrees of freedom? If so, how?

Be careful to distinguish between total number of degrees of freedom and dof that are available for energy storage without raising the temperature, which is what heat capacity measures.  All systems have the same total number of dof, 6N, where N = number of particles.  But some dof have low energy states that get "filled up" at low T, and you can't dump more energy into them without raising T.  Other dof have higher energy states that don't become accessible until higher T, and when these states "open up" it becomes possible to put a lot of energy in without raising T.

The most obvious example is at phase transitions, when a whole slew of dof "open up" and the heat capacity becomes very large until they're all "filled up."  Water in liquid state has a lot of dof that are just "opening up" so you can dump a lot of energy in without raising T.  (I'm putting phrases in "" here because to be precise about what's going on we need to get back to def'n of heat capacity C = T/N (dS/dT) and talk about how rate at which entropy increases when you dump extra energy in.)

What are the degrees of freedom that "open up" in liquid water?

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How heat makes raw food become eatable?
How can heat cooks food? My classmate asked me this but I didn't imagine how it works.

A great deal of food consists of mixtures of long-chain carbohydrates (starchy stuff) and proteins. Some of these things are hard for the human digestive system to process (try getting any calorific value from a raw potato!).
Cooking causes substantial chemical changes to those not-very-digestible components. Proteins are denatured by heat and made easier to digest; carbohydrates are broken up to some extent and smaller, easier to digest shorter chain carbohydrates are formed. In addition, the structural components in some foods are degraded making it easier to get the digestible components.


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Why are temperature and pressure represented as the number 2 in Gibbs' Phase Rule?
In our class we are learning about Gibbs' Phase Rule; $F=C-P+2$, in which $F$ stands for degree of freedom, $C$ for number of components, and $P$ for number of phases in the system. Our professor told us that the number $2$ stood for Pressure and Temperature, but I don't understand why they are treated as constants when Composition isn't.
I must be missing something obvious, but right now I can't see what it might be...

If the number of components is 1 and the number of phases is 1, then the number of degrees of freedom is 2:  you can vary pressure and temperature independently.

The underlying ah-ha moment is that the Gibbs phase rule works best in (constructing) a phase diagram. And those are usually 2-dimensional - with pressure along one axis and temperature the other. So the "2" simply reflects that 2 dimensionality. At 2 degrees of freedom, there are no constraints and the system can change either pressure or temperature and be stable without a phase transition. If you have 1 degree of freedom, then you either have a constraint on $P = k$ or $T = k$ or a "mixture" $f(P,T) = k$ (If P is changed, T must be a specific value) . At 0 degrees of freedom you are at a triple point $(P =k, T=k)$ or similar (there are a handful of named "points") - any movement in either direction means the universe is going to change the phases present or the components present.
So they are not really constants, but if the degrees of freedom are 0, then they are... "fixed" - or better yet "constrained".
The archetypical example is Ice-Water-Steam. There will be at least 3 lines in this diagram from 2 phases in equilibrium and the 1 component - and 1 point.
Out of the 3 components, lets take 2 and see what we get.
$$DoF = 1-2+2 = 1$$.
So we see that for 2 phases to be in equilibrium we have 1 constraint eating away at our DoF and we know then that there will be a fixed relationship between T and P here - we can change P and still have the equilibrium but then we have to change T as well with a specific amount. This creates a line in our T-P diagram. Since there are 3 ways to arrange this we get 3 lines (I+W) + (S+W) + (I+S)
So lets see what happens if we force all three phases to be at equilibrium:
$$DoF = 1-3+2 = 0$$.
So that is no wiggle room. There will be a point where this happens, a specific combination of P and T - where this equilibrium exists.
Now lets say I invent a new phase, and I call it Ice-9. Lets see if we can find a place where this coexists with all the others.
$$DoF = 1-4+2 = -1$$.
Nope. This is not stable. No equilibrium can exist which permits my Ice-9, regular ice, water and steam at the same time. So now we know that Kurt Vonnegut is just making stuff up, what a hack. The result can be seen here... 3 lines and a point. Don't mind the dotted stuff, that is because water is more complicated than what it is at first glance

Gibbs rule applies to systems in equillibrium. The composition is no real degree of freedom: Whatever you add (of one pure component) will partially (or completely) turn into sth else, unless it's a solid and the system contains the same solid phase at the given temperature and pressure.


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What does the osmotic pressure do in an electrolytic solution?
I have learnt that in an electrochemical cell, the osmotic pressure in the solution hinders the flow of ions from the electrode to the solution, rather some ions may accumulate on the electrode because of this osmotic pressure. As far as I know, osmotic pressure is a measurement of the tendency of the solvent molecule to flow from one solution to another.
My question is, how does this pressure cause the ions in the solution to gather at the electode?

Hint: The electrode has a negligible amount of ions (it's mostly neutral metal) and the solution has a lot of ions. Which way would the ions go?

Actually the ions cause what is called osmotic pressure, not the pressure cause the ions to flow. As positive charge accumulates on the electrode, it has a tendency to attract extra negative charge towards itself, as alike attract each other. It starts behaving as the cathode and this phenomenon starts to develop what is called osmotic pressure.


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Is there a difference between Ca++ and Ca(2+)?
In this EKG tutorial it says

Depolarization conducts slowly through the AV node, since it is carried by slow moving $\ce{Ca++}$ ions...

Is there a difference between $\ce{Ca++}$ and $\ce{Ca}^{2+}$?

Welcome to the Chemistry StackExchange. I suppose that it would be more prudent to ask such yes/no questions on chat... So please, don't refrain from using the Periodic Table, the general discussion chat for Chemistry.StackExchange.com. Although, I'll admit, I'm not sure if this question is right at home here either... Also, google clears things up sometimes :)

https://www.quora.com/Are-Ca-and-Ca2-the-same-thing has the exact answer...

They are the same.
The IUPAC Green Book writes:

$\ce{Al^{3+}}$ is commonly used in chemistry and recommended by [74]. The forms $\ce{Al^{+3}}$ and $\ce{S^{-2}}$, although widely used, are obsolete [74], as well as the old notation $\ce{Al^{+++}}$, $\ce{S^=}$, and $\ce{S^{--}}$.

The answer to this question is: No. There is no difference between the two, they are merely two different notations.

One of which is standard... $\ce{Y^X+}$ for compund Y, net charge X.


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Why nature favours lower potential energy?
It is very abstract but I was asked this question recently and unfortunately I was not able to answer it decently enough.
I've seen it coming up in various forums of physics, chemistry or biology. Most of the answers I've read give examples from everyday life or approach the matter more philosophically. I'd appreciate any clarification in this matter.

The reason you can't answer this question is because it's a "why" question with no satisfactory answer. Why does nature favor lower energy? Why does nature anything? It just does.

So there is absolutely no way to, at least partially, justify why?

Well if it didn't then we would have a perpetual source of energy. Lower states of energy are more stable.

Second Law of Thermodynamics. It is less probable that energy once distributed will again concentrate in one place. A ball will not generally get up in air on its own when kept on a table although the the heat energy in the table is enough to push it upwards.

I'm voting to close this question as off-topic because not really about chemistry.

You may want to read about the principle of least action

It doesn't.  Nature favors higher entropy, and in many situations higher entropy can be achieved by turning potential energy into heat, so it looks like the general proposition is that systems evolve to states of lower energy. In fact, that was considered as a possible general law of nature by early philosophers.
However, there are plenty of counter-examples.  For example, consider the smoke from a fire.  It rises spontaneously, so the atoms and molecules in it are spontaneously ascending to a higher state of gravitational potential energy.  Or a steam engine, in which the flow of heat from a fire can easily do work (compress springs, move objects uphill, run a generator to charge a battery, and otherwise increase the potential energy of things).  Very early philosophers rationalized these distinctions by saying that things tended to "seek their own level," with certain things seeking states of low energy and others seeking states of higher.  This isn't very satisfactory, of course, because there's no way to determine ahead of time which thing seeks which state.  One of the greatest triumphs of 19th century science was to discover the underlying principle of entropy, and that all spontaneous changes can be understood as tending to increase the entropy of the universe.
The question of why nature favors states of higher entropy is much harder to answer, because such a preference is not built into the laws of physics.  Basic physics tells us states of higher entropy are favored only when you generally begin in states of low entropy.  So far as anyone can tell, the universe generally began in a state of very low entropy, and it has been increasing ever since, and that's the only reason why we observe an evolution to state of higher entropy.  Why the universe started off in a state of low entropy, nobody knows.

I see. So preference of lower energy levels comes as a sequence of preference for higher entropy levels. Rearding the last part, could we say that it is the result of mere observation of nature?

We don't know.  The Second Law (that entropy always increases for spontaneous processes) is an observed fact, but we don't know if it's an "accident" having to do with the initial conditions of the universe (e.g. whether a universe could exist without the Second Law because it had different initial conditions) or  whether it's required by some aspect of physical law still unknown to us.

authors: Αντώνιος Κελεσίδης


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Why do we use different arguments for determining the strength of hydracids and solubility of ionic compounds?
HI is a stronger acid than HF. Why? Because when dissolved in water, the bigger iodide ion handles the negative charge way better than the small fluoride ion. So Iodide ion is a weak conjugate base making HI a strong acid.  Which one is more soluble in water, AgF or AgI? Answer is AgF. Why? It is an ionic compound and by Fajans' rules, an ionic bond shows more covalent character if it is more polarised. The smaller the anion, the less polarisation there will be, thus more ionic character, hence more solubility. But what if I said that AgI is more soluble in water, because what happens when AgI dissolves in water? It turns into Ag+ ion and I- ion. And I- handles the negative charge better than F-. What's wrong with that?

Ever heard of HSAB?

nope...but had a look in the wikipedia few moments ago.

When pondering solubility, you can't just think about what happens to the compound, you have to think about what happens to the solvent, too.  Solubility is a balance between the loss of interactions between the particles of the compound and the gain of interactions between those particles and the solvent particles.
In this case, the small highly charged F- anion will have much stronger ion-dipole interactions with water, as reflected in its much higher heat of hydration (-524 kJ/mol for F- versus -308 kJ/mol for I-).  That suggests to me that even if the lost Ag+/F- interaction (whether defined as "ionic" or "covalent") is greater than the equivalent lost Ag+/I- interaction when AgI dissolves, the much strong interaction of the F- with water more than compensates, giving AgF the greater solubility.

can you please describe the words "lost Ag+/F- interaction is greater than the lost Ag+/I- interaction". What I am thinking is maybe lost Ag+/F- interaction means the breakdown of the AgF bonds giving Ag+, F- ions. If that's correct then I don't get "the much strong interaction of the F- with water" compensates what. Because the previous line would itself mean AgF is more soluble.

Yes, that's about it.  But recall the free energy change for the process is sum of dG for each step, so dG_dissolve = dG_ion_breakup  + dG_ion_solvation.   It may well be that breaking up the Ag/F interaction in AgF has a dG that is more positive than breaking up the AgI interaction in AgI.  But if dG_ion_solvation is much more negative for Ag+/F- due to the strong new attractions between F- and water, then the overall process may be more favorable for AgF than AgI.


text:
How can I solve this synthesis?
I have this task to do:
My solution:

Is that possible? Can you please help me with the steps 2 and 3?

This is what I did, but my teacher told me I should make the diketone (compound no. 5) from the starting compound (allyl bromide) by Grignard reaction:

I think your endgame is good (from aldol through the end). There is probably another way to get to pentane-2,4-dione from allyl bromide that doesn't require a four-carbon reagent (like acetoacetate). Are there any other restrictions on the synthesis? For example, sometimes the question is posed that you are only allowed to use reagents with a certain number of carbons.

@jerepierre No, there are no 'limits'... I know, by adding acetoacetate I get one more carbon that I don't need. Unfortunately I didn't have another idea how to start the synthesis.

Treat your  hexenone with Ozone and dimethyl sulfide to give 3-oxo-butanal (2), react this with Methyl Grignard to give the secondary alcohol (3) selectively over addition to the ketone. Oxidise this to acetoacetate.

Oh of course, thank you very much! @Waylander

This begs the question, why would anyone want to make that dione? That's commercially available and cheaper from an atom economy perspective than the allyl bromide... The end game, as has been noted, is quite nice.

I think merely to demonstrate a range of transformations,and get the student used to breaking a larger molecule down to accessible fragments.

Exactly, I would't search for any practical usage in this case :)

@Waylander: Malonaldehyde (pKa=5) and acetylacetone (pKa=9). I'd estimate 2-oxobutanal with a pKa=~7). Does a methyl Grignard add or enolize?

@user55119 Its 3-oxobutanal in the scheme

@Waylander: My bad. Change 2- to 3-oxobutanal. I suspect at least some enolization.

@user55119 so would I, would also expect some reaction at the ketone. IRL I would protect the ketone before the ozonolysis, but this isn't a real scheme


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What is the difference, if there is, between this two distillation methods?
Reading about water vapor distillation, I found two different approaches:

Direct
Indirect

The direct approach has an arrangement like this:

Here the chamber contains the water and the plant. It is cooled and as the water vapor flows away it carries the low volatile compound vapor of the plant/vegetable. Both condense and the oils can be obtained.
The plant matter is in the same pot as the water.
The indirect approach is setup like this:

Here the water vapor is generated in the first container (the one furthest to the left) and when flowing through the second flask, it carries the oil vapor. I should remark: the plant matter is in a different pot than the water.
Question
Do you see any fundamental differences in these processes or something related?
It might have to do with the aggregation state of the material we need to extract, but I'm not sure.

Larger diagrams where we can actually read the text and/or clear labels would be really helpful here...

The images are fine, but I have no idea what's going on because I don't know what the substances are.

The upper picture is very small and grisly. Upon a second look, I get it, and the problem is that extracted stuff drops back in the water vessel below.

Of course not. The idea with the second setup is that you don't have your vegetables or whatever in the high-powered steam can, which walls might be considerably hotter than 100°C. Steam distillation is used to distill non-volatile, high-boiling (>100°C) compounds without heating them above 100°C. Picture 1 looks cute but is a technical nonsense imho. More AD1500 alchemy-style.

Partial pressure of what? I said the wall of the steam can can be hotter than 100°C. Unless you run it at low power, in which case you will never finish distilling your product, which has a low vapour pressure at 100°C. You need to go through a lot of water for a successful steam distillation, unless your compound actually boils below 100°C. In which case I wouldn't usually do steam distillation.

There is no fundamental difference between the two methods.
In both cases steam is formed and it is passed through the plant material.
The intention is to extract the non volatile water insoluble essential oils from the plant.
The method is called steam distillation.
On the other hand, normally it is not done in this way.
Since generally only the essential oil is of interest, the integrity of the plant does not matter, it is placed in water and the whole is heated and the water is distilled.
The students do this in the lab with cinnamon and orange peels, for instance.
It is much easier because you only have to heat one vessel.
In these two step setups, you must isolate the part containing the plant very well so that the steam does not condense, once the whole system is hot enough.
Preferably, you should also heat the part containing the plant, especially in your second example.

Thanks! Possibly the direct method is what you described, I choose that image because I suppose them almost the same...

Well, not exactly. In your first image the plant is above the water. What we normally do is to put it inside the water.

But we can call it direct method

Nice, thanks. I have tried the second option today on a lab, and it works quite good. But it is really nonsense if can be done as you suggest, I suppose...; The only advantage is that, in my opinion, it is visually nicer...

It complicates the setup but it might have sense if you need the plant for a second different extraction, maybe with solvent. When you boil the plant you get a mass that can not be handled very well. If you need to extract it after the distillation and then filter it, it might be very complicated. Then it is possibly better to use a two stage steam distillation to maintain the plant as whole as possible.

I like how people here think they know who downvoted them. :-/ And I still say method one and even more so the all in one pot variant are inefficent both speed-wise and in being dangerous to your precious product, which comes in contact with the hot (>100°C) wall of the vessel. It's a lousy technique, which might work for some separation problems, but is also slow. The one advantage is that it doesn't need the separate steam can. Poor people's steam distillation, who become even poorer when the energy bill comes in later.


text:
Evaporation of concentrated formic acid and semi-concentrated sulfuric acid at room temperature
I recently looked at a reaction where CO is synthesized using formic acid and sulfuric acid, which seems to be one of the easier to control reactions for this, and obviously ventilation etc is necessary.
However I was wondering, is ventilation only necessary because of the dangerous nature of CO, or also because of the vapors or formic and sulfuric acid? I was quite into chemistry in school and after, but I do not remember that much, so I feel like I should know this but I cannot recall anymore.
So at room temperature, if you would take a small amount of each of these acids and keep them either separately or mixed together, how fast/how much would evaporate and create dangerous acid vapor?

So the labels / other warnings cautioning of vapors are exaggerated? Or to use a concrete example, formic acid is also used by beekeepers against certain pests, and I read a report of one beekeeper losing his sense of smell because a bottle of formic acid (it didn't say which concentration) tipped over in his truck. How long would it take for this to happen or under what conditions?

Lets put it more precisely, H2SO4 has no vapor pressure to speak of in r.t. but HCOOH does. Still, corrosiveness of conc. formic and toxicity of CO are more important IMO. There is some danger of HCOOH vapor, then again it's hardly as volatile as, say, NH3.

Formic acid is pretty pungent. Though fumes are toxic, the odor should be an initial warning. See chemical data safety sheets, e.g.https://www.basf.com/documents/corp/en/sustainability/employees-and-society/employees/occupational-medicine/medical-guidelines/Formic_acid_C_BASF_medGuidelines_E036.pdf

Related: Making carbon monoxide gas from formic acid and sulfuric acid?

I might sound like a kill joy but I strongly suggest that you do not attempt to make carbon monoxide. The reaction you describe is a not one for the faint hearted. I would like to point out that aerosols of sulfuric acid are very harmful to the lungs, eyes and the rest of your body.
I think that any work done in the lab with concentrated sulfuric acid should be done in a fume hood to protect you against aerosols of the acid. The fumes from the formic acid are going to be corrosive.
In my youth we used to work on the open bench with glacial acetic acid, this is not a good thing to do as the whole lab soon becomes like a viniger factory. I fear that with formic acid being handled outside the hood the lab will soon become rather unpleasnent.

Thanks for the answer! I do not want to attempt anything, just to understand/learn and know what is theoretically possible with relatively little effort/equipment. And seeing as formic acid is used by beekeepers (my uncle had bees once, but cannot remember if he used it), it seems readily available and so it shouldn't be that difficult to handle if I was so inclined. Compared to Mithoron above, so you would say that there is actually vapor pressure from the aerosols, or are we talking about different things here?


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Why is graphite solid but water liquid?
$\ce{H2O}$ has primary covalent and secondary hydrogen bonds (hence it is liquid). Graphite has primary covalent and secondary Van der Waals bonds. Still, graphite is a solid. Why is this so?

Do you know how many atoms are there in a water molecule?

Truthfully water and graphite are so different that it makes no sense to be trying to compare them.

Water can also be solid, or a gas. This depends on so much more than just the kind of bonding.

Have you seen the structure of graphite?

Bonding types are not a good guide to the form of a compound: you also need to know the structure
The types of bonding involved in a compound tell you something about what form it will take. But not very much. The structure of the compound matters far more.
Water is a small molecule with only three covalently bonded atoms in it. They mystery is why isn't it a gas (the structurally similar but heavier hydrogen sulphide is a gas at normal conditions!). The strong hydrogen bonds explain why it is a liquid: they stick the molecules together more tightly than the much weaker inter molecular bonds in the sulphide).
Graphite isn't a small molecule: it is a nearly-infinite array of hexagonal sheets of carbon atoms. These sheets are held together by van Der Waals forces. The components of graphite are not small molecules but those large sheets which is why it is not a liquid under normal conditions (few substances made of near-infinite arrays of atoms are). But the relatively weak bonds holding the extended sheets together do have an impact on the nature of graphite: the sheets can move relative to each other easily. This makes graphite behave enough like a liquid in some spatial directions that it is sometimes used as a lubricant.
So the very different properties of water and graphite can be understood only by knowing both the sorts of bonds involved and the structure of the components making up the compounds.


text:
Why are long oxygen chains unstable?
H-O-H (water), H-O-O-H (hydrogen peroxide), H-O-O-O-H, H-O-O-O-O-H, etc.
This is analogous to hydrocarbon chains, methane, ethane, propane, butane, etc.
Note the Hydroxygen Chains satisfy 2 covalent bonds for every oxygen atom, just as every Hydrocarbon Chain satisfies 4 covalent bonds for every carbon atom. Yet the oxygen chains seem unstable. Why?
I can't imagine it has anything to do with shells deeper than p. And carbon and oxygen are very close together on the same row, so it would appear the valence shells are the same and thus governed by the same laws.
I'm also aware of those weird exceptions, Ozone and Red Oxygen. Maybe that has something to do with it, but I don't know how those bonds work either.
I also saw this related question. It just asks if they're possible. I didn't see any explanation about why they're unstable.

The top of the p block is unusual in the atoms being so small they can readily form strong multiple bonds, because (from a VB point of view) they can get close enough for good side-by-side p orbital overlap.  Hence they prefer to form multiple bonds if they can, and both N and O can do that to form diatoms.  So extended chains of either N or O atoms are generally less stable than a collection of N2 or O2 molecules.  Or if you have extra capping atoms around, like H, then a collection of N2/O2 plus some H2O/NH3, et cetera.
Carbon is the exception because it has four valence electrons, and a quadruple bond to form C2 is too high energy.  (From a VB point of view, there are no valence atomic orbitals left to overlap to form the 4th bond.)  The next smallest molecule it can form with itself with lots of double bonds is buckminsterfullerene (C60).  So without another element around, a long chain of O atoms can form a more stable collection of O2 molecules, while a long chain of C atoms can't form anything more stable at all.  (When other elements are around, all bets are off: long chains of carbon tipped with hydrogen are notoriously unstable to CO2 and H2O if there's plenty of oxygen around.)


text:
Relating properties and structures of polymers

Why would it be more favorable for a polymer designer to design copolymers with irregular chain structures, rather than design homopolymers?

Hint: Discuss intermolecular bonds
I understand that copolymerization allows for greater variation in polymers and there properties, but am unsure how exactly to relate this to intermolecular forces. Would it also depend on the intended use of the polymer?

I think that the best thing for you to start with is to compare the radical formed low density polyethylene (LDPE) and the Ziegler-Natta high density polyethylene (HDPE).
The low density polyethene has plenty of side groups which are randomly arranged through the length of the chain, this can be regarded as a copolymer of a range of different alkenes which are randomly arranged.
The high density polyethene (HDPE) is an almost perfect homopolymer of CH2 units, this polymer has far better mechanical properties, it is a polymer where the chains pack closer to each other than the chains in LDPE.
There is a version of HDPE which has properties closer to that of LDPE, this is formed by using a mixture of alkenes using the organometallic Ziegler-Natta cataylsts.


text:
Calculating the x, y and z distances between two molecules - AMBER (Molecular Dynamics)
I am interested in calculating the x,y and z distance between two molecules over a given simulation time. I know how to calculate the total (r) distance between them, but I want the distance between them along all three dimensions.
For the total distance, cpptraj can be used as:
parm prmtop-file
trajin netcdf-file
distance DIST :1-4@heavy-atoms-from-molecule-1 :5-8@heavy-atoms-from-molecule-2 out dist.out
run
quit

But like I said before, I do not want the r-distance. Is there anyway to calculate the x,y and z-distances in cpptraj? If not, what other alternatives do I have?

I'm not familiar with AMBER, but in LAMMPS there is a file containing all the atom positions and a utility to convert it to a .xyz file which is just a list of all the coordinates. You might try searching for something like an .xyz file?

I can find the coordinates via a file in AMBER called a .pdb file, but I will have to manually calculate the distances from this, which is over 20000 frames to consider.
I guess I might as well just do it.

I ended up realizing that if you run:
parm prmtop-file
trajin netcdf-file
vector v0 :1-4@heavy-atoms-from-molecule-1 :5-8@heavy-atoms-from-molecule-2 out v0.out magnitude
run
quit

You will get the total distance as well as x, y and z distances between your two molecules. The output file will contain 8 columns. The AMBER manual states that they are identified as follows:
Frame | x-Vector-Coordinate | y-Vector-Coordinate | z-Vector-Coordinate | x-Origin-Coordinate | y-Origin-Coordinate | z-Origin-Coordinate | Magnitude
The vector coordinates are your x, y and z distances. Magnitude is the total distance.
Decided to share this as someone might run into the same problem one day.


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How does the screening effect work for orbital in the same shell?
Energy of an orbital in a single electron system depends solely on n, while energy of orbital in multi electron system depends on n and l, why is that? I understand that the screening effect is at work between two different shells, but how does it work in the same shell, is there any other reason?

How is the situation with the same shell different from that with two different shells?

Screening within the same shell is generally poor, which is why effective nuclear charge rises going across the Periodic Table: each added electron screens the others poorly from the added proton.  And, yes, as you suspect, the quality of screening varies with l.
In addition to the screening effect, in many-electron atom you have electron-electron repulsion, as well as quantum exchange-correlation effects between electrons.  Since these are interactions between electrons they depend on the distance between the electrons, which in turn depends not only on the distance between each electron and the nucleus (controlled by n) but also on the angular distribution of electrons around the nucleus (controlled by l).


text:
Solutions : Henry Law
In below solution why the moles are considered half as their is nothing clearly specified in question. I mean it is not given that that mole fraction we get is for litre water.

They just wrote the multiplication by liters in an unusual way. You are going from molarity to moles, which means multiplying by .5 L or 1/2 L. @user560199

If you look at the question closely, it states in the equation, moles per litre. The reason the moles are divided by two is the fact you need to find the moles of CO2 in 500 ml and not 1000 ml.


text:
Is there a blanket term for all variants (both nuclides and ions) of a chemical element?
I'm hoping to catalogue the various nuclides and oxidation states of an element but can't find a single term that describes them all, save for "chemical element variations". Is there a specific term used for this?

I don't think so. For what?

Why would there be such a term? Nuclides and oxidation states are farther away than just unrelated. It would be like a blanket term that covers human body parts and bank accounts.

I agree they are very different, but they are nonetheless connected by the nomenclature in use, in that both would be labelled with the base element's symbol (Na+, 24Na). If a body part and a bank account are both called "Sam" then they should be in a list of all things called "Sam". That they are fundamentally different things is beside the point in my case.

Yes, for example consider selenium. You can say "selenium species".

Exactly, I always counter "nitrogen species" for instance, to account for ammonium, nitrite, nitrate and all other forms.

True how about "sulfur species", maybe this could be a new game for the chemists here. We might have to each think of an element and name several forms. For sulfur, sulfide, sulfite, thiosulfate, sulfate and peroxydisulfate.

That works (and I see that confirmed here: https://goldbook.iupac.org/html/C/CT06859.html). A little ambiguous as it seems to have some overlap but seems like it's the best option there is. Thanks!

Thank you Dan for adding the reference to the IUPAC gold book


text:
Difference between titanium and titanium dioxide electrode
I need an inert anode to perform some electrolysis.  I've tried graphite, which works to some extent, however I want to upgrade to titanium, which is pretty reasonably priced, and will perform better than the graphite.
My question is whether there is a difference between "Titanium oxide electrodes" and plain titanium.  Does a titanium dioxide electrode need to be specially treated to give it more inert characteristics, or does the oxide layer form naturally on the titanium?
I ask because I have seen this type of electrode classified as a Mixed Metal Oxide (MMO) electrode, but I have also seen some say that titanium is inert on its own.  I would like to now whether plain titanium is a viable electrode so I don't end up buying some that ends up corroding quickly during electrolysis!!
Any information in this subject would be greatly appreciated.
Thanks in advance!!

What gives you the idea that titanium is better than graphite? Did you check the literature? What reaction?

I realize that graphite could be preferable under some circumstances, however several factors point me to choose titanium or some other metal electrode.  First, graphite releases carbon particles into the solution during high amperage applications.  Second, it is more difficult to attach wires to graphite electrodes.  Third, graphite is found in rods, but I wish to increase the surface area of the electrode to increase speed of reaction.  Finally, graphite seems to generate significant amounts of heat (because of its higher resistance) at high amperage applications.

Used as an anode, a normal untreated piece of titanium metal will "anodize", forming a non-conductive titanium dioxide coating.  In my experience, the current through a large untreated titanium sheet anode drops below the milliamp level in under a minute.
But there are a variety of electrically conductive but oxidation resistant micron-scale coatings that can be formed onto titanium, including platinum (a "platinized titanium anode"); conductive oxides of common metals like Pb or Sn (lead dioxide or tin oxide); or conductive oxides of rare metals like Ir, Ru, or Ta (a "mixed metal oxide anode").  There are some decent electron microscope microstructure photos in Umicore's promotional catalog.
The advantage to using a "valve metal" like titanium or aluminum as the substrate on a coated anode is if part of the fancy conductive coating flakes off, the bare titanium just anodizes and stops conducting, rather than getting eaten away and dissolving into the electrolyte.


text:
Why do we dip noble metals (Ir, Pd) into NaOH for an oxide coating in thermal oxidation?
I've noticed that a common technique for creating a layer of oxide onto palladium or iridium is to dip it into high concentrated NaOH prior to putting it in a furnace for ~800C.
I've been reading numerous papers and it leads me to a textbook that I can't find accessible: Kauko, Y. and Knappsberg, L. (1939) Z. Elektrochem. 45, 760
The papers that I've been reading are
W. T. Grubb and L. H. King, Analytical Chemistry 1980 52 (2), 270-273
DOI: 10.1021/ac50052a013
and
https://doi.org/10.1016/0005-2795(76)90263-4
if anyone can help me solve this problem, I would be eternally grateful!


text:
Is there a substance that hardens in water?
Im trying to invent a product that starts as a liquid but when exposed to water it hardens. Is there any chemicals that can cause this reaction??

Superglue, amongst many...

Oh sure.  The most famous is calcium orthosilicate, otherwise known as Portland cement.  So what you're wanting to invent already exists in the form of concrete, which as you know is poured as a liquid, and reacts with water to produce a hard durable substance.  It's also extremely cheap, since the raw material is just a certain (common) type of rock.  So one way to go about your new invention is to think about how what you want to invent would differ from concrete.  You could start by thinking of concrete's limitations, and whether these are fundamental or could be fixed by some change in material  properties.


text:
How do you properly dispose of hexane and cyclohexane
How do you properly dispose of hexane and cyclohexane?

It is a rather short question, the problem is that it offers very little information about what the person is asking. The safe and legal disposal of "chemicals" is not a trivial matter. I will have a go at answering in the near future.

For amounts common for a small-scale lab it is safe to let it evaporate into atmosphere or burn it.  For industrial scales you could run into legal matters and should consult local law.

@permeakra - in my labs, I am prohibited from intentionally letting waste evaporate to dispose of it, regardless of quantity. Check and follow all applicable procedures and local, regional, and country laws.

Hexane and cyclohexane waste is hazardous waste, the law on the disposal of waste will vary from one part of the world to another. From a UK point of view, you need to transfer the waste to a licensed waste contractor. The producer of waste has a responsibility to make sure that they only use a licensed waste contractor rather than some fly by night criminal who would dump it in a graveyard, play park or commit some other waste related crimes.
A common method in the UK and other places to dispose of solvent waste is to use it as fuel for the production of cement, or in other places where some liquid fuel is needed to do things like start up a furnace. I know that incineration plants oftein use some liquid fuel to start a furnace before changing over to using domsetic rubbish (MSW) as the fuel.
Hexane and cyclohexane might be a bit too flammable to use in some applications, but I am sure that if care is taken then they could be used as industrial fuels.
If the waste is mainly one solvent or if the final user does not care too much about the purity of their solvent (consider how much heptane you would tolerate in your toluene) then one option is to distill the solvent waste and then to recycle the solvent. The problem with this is that if you have a waste which is 95 % hexane and 5 % involatile stuff then you will have to sacrifice some hexane to avoid the still pot becoming filled with some involatile residual which you have to then use hexane (or another solvent) to wash out.
If the solvent waste has some very horrible impurity in it, such as imagine PCBs from a process in which old transformers casings have been washed out. Then the hexane waste will have to go to a high temperture incinerator which is optimised for the destruction of the PCBs rather than energy recovery. Many incinerators are designed to recover energy from the burning process, a MSW incinerator will form some dioxin but the amount will be very low as the temperture and the chemistry of the plant is chosen such that very little is formed, what is formed is normally filtered.
On the otherhand if a plant was to be fed with things like PCBs (polychlorobiphenyls) and unwanted 2,4,5-T from the 1970s which contained dioxin then I think more effort would have to be made to destroy the existing dioxin and also prevent the formation of new dioxin. This could well reduce the amount of energy recovered per ton of waste. For example you would be able to get some reduction in the net dioxin production by adding an afterburner to the incinerator. But the afterburner will require a supply of fuel. Even if you preheat the air going into the afterburner with heat from the air leaving the afterburner, there will be some fuel needed to operate the afterburner. As a result the energy ecconomy of the site will become worse.


text:
are half of "H2SO4 pH sample problem" webpages just wrong, or am I wrong?
I did a Google search for "h2so4 pH sample problem" and went through the results, and it looks like about 50% of the webpages "explaining" how to do this problem give the wrong answer when calculating the pH of a solution of $\ce{H2SO4}$.  The incorrect sources either (a) assume that each molecule of $\ce{H2SO4}$ donates only one proton (approximately true only at high concentrations); or (b) assume that each molecule of $\ce{H2SO4}$ donates both protons (approximately true only at low concentrations).  Only about half of the sources do the problem correctly by assuming that all $\ce{H2SO4}$ molecules donate the first proton and then using the second dissociation constant to compute the degree of dissociation of the second proton.
I'm no expert, so am I missing something, or are half of the sources out there just wrong?
Googling "h2so4 pH sample problem", here are the first eight hits after removing pages that did not contain an $\ce{H2SO4}$ sample problem, and how I scored them:

http://che1.lf1.cuni.cz/html/Calculations1112_II_handout.pdf -- says "For $\ce{H2SO4}$: $\ce{[H+]}$ = 2 $\times$ $\ce{[H2SO4]}$" -- incorrect.
https://www.quora.com/How-do-I-calculate-pH-of-10-3-M-H2SO4 -- the top-voted answer also says the concentration of $\ce{[H+]}$ is twice the molarity of $\ce{[H2SO4]}$ which 0.001, which is incorrect.  (This gives the incorrect answer of pH = 2.69; the correct answer is 2.73.)
https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Acids_and_Bases/Monoprotic_Versus_Polyprotic_Acids_And_Bases/Calculating_the_pH_of_the_Solution_of_a_Polyprotic_Base%2F%2FAcid -- uses the second dissociation constant correctly and an ICE chart -- correct
https://www.youtube.com/watch?v=gd8mYfiURx0 -- "Calculating the pH value of 1 M H2SO4" -- also assumes both protons are donated -- incorrect
https://www.youtube.com/watch?v=bfdbmT27L1c -- "Calculate pH of a Strong Acid" -- also assumes both protons are donated -- incorrect
https://web2.0calc.com/questions/what-is-the-ph-of-a-0-00050m-solution-of-h2so4_1 -- uses the second dissociation constant -- correct
If the pH is zero, what would be the normality of a sulphuric acid solution? -- I'm counting this as HALF-correct, because it is assumed that each $\ce{H2SO4}$ molecule donates only 1 proton.  This is approximately true at 1 M concentration, but I think it should be stated explicitly that the second proton doesn't get donated because the concentration is so high
http://www.sparknotes.com/chemistry/acidsbases/phcalc/problems/ -- correctly explains the partial donation of the second proton, so even though they don't show the working out, it's still correct

So it looks like even for just a slightly-non-trivial problem, almost a majority of free online help resources get the answer wrong.  Is there some subtlety I'm overlooking?

And this is why I always told people to be careful about looking for help online. Personally, I've found that it's way worse that 50% for quantum mechanics problems...

Yes, you got it right. The web is dark and full of errors.

@IvanNeretin I see what you've done here :D Nice pun. BTW I'm afraid OP already asked about it earlier. This question is kinda waste of time. Question is trivial, and no idea why someone would even dig through all these pseudo "resources". Also difference between 2.69 and 2.73 as far as pH goes is negligible and approximate calculations are employed often. Approximation isn't equal to error as every value describing physical world is approximate unless it's a constant made to be precise.

@Mithoron this is not the question I asked about earlier.  Earlier I mentioned on meta that 90% of online automated pH calculators give the wrong answer when computing for H2SO4 (because the creators forgot to include special-case code for a polyprotic acid).  In this case, I'm looking at free help resources where the authors actually worked out the problem by hand, so that reduced the chance of making a mistake, but still half of all the resources were wrong.

And, one of the reasons I document these things is because I get sick of people telling newbies, "All of this information is out there for free, go and google it and stop wasting our time," when the reality is that an extremely high proportion of resources that come up on Google are wrong.

I'm voting to close this question as off-topic because it's essentially a rant how people on internet are usually wrong, not a question.

I had read your meta post yesterday and was planning to convert this into a proper Q&A, however, this isn't the way I would have done it. Your question: "am I missing something, or are half of the sources out there just wrong?" has the obvious answer that the sources are wrong. In fact, you've labeled them as being incorrect or correct yourself. I fail to see what is being asked, thus.

@GaurangTandon note my meta post was about something slightly different; the meta post was about the fact that 8 out of 9 online "pH calculator" widgets get the wrong answer when calculating for sulfuric acid; this was about the fact that 4 out of 8 manually worked out sample problems get the wrong answer.  My adding the line "Am I missing something?" was not just a cheat way of turning it into a question :) I really did want to know if there was some subtlety I had overlooked.  Christopher Grayce had a good answer: there are often inaccuracies we overlook, like the auto-ionization of water...

@GaurangTandon ... even though, in this case I didn't agree with the overall point because the auto-ionization of water doesn't cause an error at two significant figures (whereas ignoring the second proton of H2SO4 does), but it did get me to at least think about it some more.  Regardless, since you said you had an idea for converting the "pH calculator" widgets question into a "proper Q&A", how would you have done it?

...and you didn't even get into the fact that almost all explanations will ignore the autodissociation of water!
In some cases your online resources will be wrong, of course, and it would be odd to expect uniform correctness in something people donate for free on the Internet (including here BTW).  So that's worth keeping in mind.
However, it's also the case that it is not always important to treat a problem in all its subtlety.  For example, the difference you get in the calculated pH of 1M H2SO4 by treating or ignoring the second dissociation equilibrium is likely to be smaller than your measurement error in a typical undergraduate lab.  So why bother?  On the other hand, if you were running a very careful experiment with 0.01 M H2SO4 you might well want to take those other equilibria into account, and if you were doing something super sensitive with 10^-7 M H2SO4 then you'll need to worry about the autoionization of water, too.
Generally we use the least complex model that gets us the answer consistent with our measurement limitations.  That maximizes insight and reduces the obfuscation of complex calculation, which can easily make it so we fail to recognize an erroneous answer.   This runs all through science and engineering, and it's why simple models have such great staying power even when they are superseded by more sophisticated understanding.  Naturally a concomitant is that the good scientist must become very adept at identifying the least complex model he can use.  That takes deep insight into the limitations of each model, the probable sources of error, the limits of measurement, and the quality of data on hand.  Being good at this is a much more important skill in the senior scientist than mastering the details of the most complex and realistic model there is.

Well I figured that in most cases the question is asking about accuracy to two or three significant figures (that's what is used in the answers given), so you can ignore anything that's less significant than that (e.g. the auto-ionization of water), but you have to factor in anything that will make the answer incorrect at two-significant-figures accuracy (e.g., the second proton of H2SO4 at 0.001 M).

I agree if they are giving answers to more precision than is justified by the ignoring of the weak-acid equilibria, that is a significant mistake.  Goes back to the often neglected, boring, but important treatment of "appropriate precision" in calculated values.  All that sig dig stuff students hate, ha ha.


text:
Diluting two solutions together
I have two different mixtures:
1 gallon of "A mixture", which is 99% substance A and 1% water
An arbitrarily large supply of "B mixture", which consists of 4% substance B and 96% water.
How do I go about calculating how much of "B mixture" (plus extra water if needed) to add to the 1 gallon of "A mixture", to produce a final combined mixture consisting of 70% substance A, 2% substance B and the remaining 28% water?

The problem is that to get 2% (v/v) of B from a 4 % (v/v) solution of B then to make 1 litre of the final solution we would need 500 ml (0.5 L) of the 4 % stock solution of B (at 4 %).
As the stock of A is at 99 % (v/v) then when the concentration of B is 2 % then the highest concetration of A which is possible will be 49.5 %. You stated that you wanted to have a final mixture which is 70 % A and 2 % B.
I think that you would be better off getting a stronger stock solution of B, for example if you had a 20 % solution of B, then to make 1 L of a 2 % solution then you would need 100 ml of the stronger B solution.
Then if you were to have a 70 % final concetration of A then you would need 707 ml of the 99 % A solution. Then you would need to add 193 ml of diliuent (I assume water) to make the final mixture up.
I am making the assumption that the volume of the final mixture is equal to the sum of the volumes of the original liquids before mixing. It is interesting that ethanol / water mixtures have a volume which is rather less than the sum of the volume of pure ethanol and pure water.


text:
Both calcium hydroxide and ethanol have a hydroxide group, then why isn't ethanol ionic?
I am confused on what is a hydroxide exactly is. For example, ethanol has a hydroxide group so does this mean it contains hydroxide ion? Does this make it ionic? I know it is a covalent molecule but why isn't it ionic?
Calcium hydroxide has $\ce{OH-}$ but how is this different than the $\ce{OH}$ group in ethanol? Is it covalent due to the fact it does not dissociate into ions when in aqueous solution? I know I'm missing something here.

You must compare ethanol with water and sodium hydroxide with sodium ethoxide

@RaoulKessels what do you mean?

Sodium or calcium hydroxide has $\ce{OH^-}$ and sodium or calcium ethoxide has $\ce{EtO^-}$, while water is $\ce{H-OH}$ and ethanol $\ce{Et-OH}$. You can not cross compare them

@RaoulKessels So more or less one has ions and the other does not? So The OH in calcium hydroxide is already an ion where as in water it is not? Kinda confused with that.

Yes, and the same with ethanol

@AnuragBaundwal Since you are editing the question, can you format the OH- properly, please?

No, it is not ionic.
Your doubts are somehow legitimate if you are not aware of the concept of electronegativity. As a rule of thumb if two atoms have similar electronegativity is harder to have ionic bonds. You can use Pauling scale of electronegativity to have some insights on it.

A difference of electronegativity of over 1.7 is likely to be ionic,
and a difference of less than 1.7 is likely to be covalent.
Source: Atkins, Peter; Loretta Jones (1997). Chemistry: Molecules, Matter and Change. New York: W. H. Freeman & Co. pp. 294–295. ISBN 0-7167-3107-X.

This means you will often deal with ionic compounds made of one ion with a strong electronegativity and one with a weak electronegativity. In fact, ions are composed of anions and cations. With a lot of approximations, we can calculate the electronegativity of a group averaging the electronegativities of the components for $\ce{-OH}~\dfrac{3.44 + 2.82}2 = 3.13$.
In your example, $\ce{C2H5OH}$ ethanol the bond is with $\ce{C}$ ( electronegativity 2.55)  so the difference is 0.58 hence covalent. If the bond is with $\ce{Ca}$ (1.00) the difference would be 2.12 which is greater than 1.7 hence is categorized as ionic.
Of course is not always easy to determine if it's ionic or not so in complex cases we can say that it has an ionic character o covalent character.


text:
What is meant by work done "expanding against atmosphere"?
In the book, Chemistry : A molecular approach, I came across this line:

Over here what does the author mean by work done by expanding against atmosphere?

Imagine someone was pushing you against a wall. If you spend some energy overcoming that push, wouldn't you say that you did some work (expenditure of energy) against that pressure?
With a similar line of reasoning, the atmosphere is exerting a pressure of one atmosphere trying to compress your gas down. That's a huge pressure (nearly $10^5$ pascals!) and the gas in question has expanded against it. The wording makes sense.
In general, whenever gases are produced as a result of some reaction (in this case combustion), they are initially at a higher pressure than atmospheric pressure. This isn't an equilibrium situation. So, they expand, and lose their pressure until it finally becomes equal to that of the atmosphere.

It just means that when you burn natural gas, the reaction increases the volume of the matter involved in the reaction, so the system expands.  The author's wording is odd when they said it expands "against the atmosphere"; that just means the atmosphere is always there, exerting pressure, but the system expands anyway.


text:
Why is the equilibrium constant of water with its vapour equivalent to its vapour pressure?
Can someone conceptually explain why the equilibrium constant is equivalent to the vapour pressure of water for the above equilibrium reaction?
For example if $K = x$, then it is said that vapour pressure $= x$ (in bars). However I seem to cannot understand why.

The equation is pretty much a definition of the equilibrium between liquid water and water vapour. That is what creates a vapour pressure.  So, of course, the vapour pressure is tightly related to the equilibrium constant.

Lets start by writing the definition of the equilibrium constant, for some general reaction:
$$\sum_i^{n_\text{reac}} a_iA_i = \sum_j^{n_\text{prod}}b_jB_j$$
The above is a general reaction, where $a_i$ and $b_j$ is the stoichiometry and $A_i$ and $B_i$ is the molecules. For this reaction we can write the equilibrium constant:
$$K=\frac{\prod_j^{n_\text{prod}}\frac{B_j^{b_j}}{U_j^{\circ}}}{\prod_i^{n_\text{reac}}\frac{A_i^{a_i}}{U_i^{\circ}}}$$
Here $\prod$ is the symbol that denotes we take the product over a range. The $U_i^\circ$ are a standard unit, to ensure that our equilibrium constant is well defined.
Now if we look at your reaction:
$$\mathrm{H_2O\left( l \right)} \rightleftharpoons \mathrm{H_2O\left( g \right)}$$
We can from the above equations identify that; $a_1 = 1$, $b_1=1$, $A_1=\left[ \mathrm{H_2O} \right]$ and $B_1=p_\mathrm{H_2O}$.
We can thus write our equilibrium constant as:
$$K=\frac{\frac{p_\mathrm{H_2O}}{1\text{ bar}}}{\frac{\left[ \mathrm{H_2O} \right]}{1\text{ M}}}$$
As you might note, I have also inserted the $U$s. For pressure $U=1\text{ bar}$ and for concentrations $U=1\text{ M}$. Now we need one last ingredient. You might have learned that the activity of liquids is $1$, i.e. the means that we have to set $\frac{\left[ \mathrm{H_2O} \right]}{1\text{ M}}=1$, this is due to the definition of activity. If interested I would suggest to take a look at this. As our final result we can see that:
$$K=\frac{p_\mathrm{H_2O}}{1\text{ bar}}$$
I.e. we have that the equilibrium constant equals the partial pressure of water in standard units.

Hi Erik, why is $ B_1 = [P_{H_2 O}] $?

It is not $\left[ p_\text{H_2O} \right]$, but $ p_\text{H_2O}$. I have now corrected this in the answer. Sorry for the confusion.
It is the pressure instead of the concentration because it is on gas form. Often when things are on gas form, it is the partial pressure that is interesting instead of the concentration (Concentration would be a very small number).


text:
Would the change in enthalpy change of chemical reaction in space be considered as change in internel enrgy?
The formula of enthalpy change is
change in H = U + P*(change in V)
In space since there is no air, thus atmospheric pressure = 0 , P= 0 , Pv = 0 ,
So will H = U  ?

Yes. Then again, every compound has a vapor pressure, so in the equilibrium (given $P=0$) they all will up and evaporate, leaving you with a pretty dull environment.

Well, in principle yes, for example if you had a blob of liquid floating in empty space and wanted  to consider the whole blob as a thermodynamic system.
But it may be worth mentioning that wouldn't be a typical use of thermodynamics in astrophysics.  Usually you'd consider your thermodynamic system to be some small part of a larger system and the larger system would be a thermal and volume reservoir, so T and P would both be nonzero.
For example, it's common to consider the thermodynamics of stars, but what you do is consider a small packet of gas to be your system, and the rest of the star to be surroundings that establish a constant T and P for your packet.


text:
Why the migrator aptitude are different about different books?
In Solomons 8th
H> phenyl >3 order alkyl > 2order alkyl >1 order alkyl > methyl
In Wade 8th or Bruice 5th, they also means
H > 3 order >2 order ≈phenyl > 1 order
why the description are different about phenyl?

see https://chemistry.stackexchange.com/questions/10833/migratory-aptitude-in-pinacol-pinacolone-rearrangement/16470#16470


text:
Last step of nicotine extraction using DCM
After using hydro-distillation method with distilled water and NaOH, separating the extract with dichloromethane, distillate it at 40 °C/104 °F, I'm left with some yellow pale oil.
I'm just wondering if traces of DCM could still be detected in the final product or if I need more steps to purify the extracted alkaloids ?

DCM is really pretty volatile and what you have done should shift it effectively. Have you a vacuum dessicator you could leave it in for a few hours? Failing that, just letting it sit at room temp open to the atmosphere will get rid of the last traces.

If you are worried only with DCM, then you should be fine since it is quite volatile. To be sure you have no DCM left in your nicotine leave it for a few days in a closed recipient filled with some dried silica gel or calcium chloride (here in my country those are easily found, even for domestic use), and if you are in a proper laboratory just leave it in the desiccator with the vacuum on for a few hours.
However, nicotine is hard to isolate since other alkaloids from tobacco are also extracted in the most commonly used methods. I don't know which extraction methodology you used but just for comparison, here are two papers with some information for isolating nicotine:
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3263647/
https://www.ncbi.nlm.nih.gov/pubmed/19135499


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What is a reasonable mechanism for the following reaction?

Hello all! I'm stuck on this one question and am not sure how to finish the mechanism. What I did was attack the alkyne first with the N farthest to the left. I'm not sure what do to after that though. Any help would be greatly appreciated! Thank you!

Michael attack by the ring N, as you describe then cyclisation onto the protonated CN by the exocyclic NH2

I’m trying to figure out how or what to attack after the ring-nitrogen attack of the alkyne. When I attack the resulting product from the first step with the exocyclic N I get a very strained cyclic structure with two consecutive alkenes within the 6 membered heterocycle. Does that sound like the right way to do this mechanism?

That doesn't sound right.  The neutral intermediate has an acrylonitrile unit added to the ring N, and =NH exo to the ring.

I think I got it...

attack alkyne with cyclic N and protonate nitrile
push LP on nitrile-N back into conj. system and protonate one of the alkenes, getting the the acrylonitrile intermediate you mentioned
attack nitrile-C, forming a six membered ring
delocalize middle N LP to the left most N

Don't forget the contribution of the ring O to step 3


text:
How to apply Henry's law down the sea?
Background: Henry's law states that:

At a constant temperature, the amount of a given gas that dissolves in
  a given type and volume of liquid is directly proportional to the
  partial pressure of that gas in equilibrium with that liquid.

And mathematically:
$$S=k\times p_{\text{gas}}$$
where $S$ is solubility, $k$ is Henry's constant and $p$ is the partial pressure of the gas.
Question: I want to calculate the oxygen solubility down the sea. Is partial pressure the atmospheric one? If it is so, we have same solubility on all sea deepness, which is contradictory. What am I missing?

Edit
I know this law can be thought as depending on this equilibrium:
$$\ce{A {(g)} <=> A {(s)}}$$

What does it even mean to calculate or measure solubility of something down the sea?
Why, there is a precise and well-defined meaning, and it is hinged on equilibrium between the gas and the solution, which supposedly should be established at the said conditions. Our imaginary gas bubble, like everything else, must withstand the immense pressure of all those layers of water up above.
So no, the partial pressure is not just the atmospheric one.
Deep abyss of the seas is undersaturated in all gases, including oxygen, and that by a wide margin.

Do you mean deep on the sea there is no equilibrium with atmospheric gases? (I don't know why) so this law is just valid for a small surface layer of the sea?

To be in equilibrium, two things have to be in touch to begin with. The gas of the atmosphere is not in touch with the deep waters, hence no luck.

I don't think so, we can suppose that there is a dynamic equilibrium between the oxygen down the water and on the top...; Suppose the oil water system, there you have no touch between liquid and water vapor, and a equilibrium, isn't it?

OK, well, skip the "in touch" part. (Indeed, I have mentioned the oil/water system in a closed vessel on an earlier occasion.) The counterparts in equilibrium have to be in the same conditions, which they are not, thanks to the gravity potential. In a laboratory setup yon can neglect it; in 100m down the sea you can't.

Just in case, do you know any expression to estimate solubility in the example I ask? Taking int account gravity maybe

Taking into account gravity is the same as taking into account pressure, just like I said.

Yes but there are no gases there, that's what stuck me...

Oh, Im sorry, what I wanted to ask is if there is a way to estimate oxygen concentration on deep sea, but i'll ask on a diff question


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Diffusion experiment with dialysis membrane bag
Recently I have performed a diffusion experiment with a dialysis membrane containing methylene blue. A schematic picture of the experiment:

I want to calculate the flux (amount diffused/area*time) from the samples I have taken. However I am stuck with how to calculate the surface area of the dialysis bag so I can use Fick's law to estimate the flux. The dialysis bag is cylindrical, however when you put the 2 pins above to close it, it is not completely cylindrical anymore as you can see in the picture. Anyone that could help me out with this? If in need of membrane specifications I can post them.

Why don't you measure the area with equivalent square pieces of paper or some other material? You might not be able to calculate analitically..

You can double-check the area after you're done:

Poke a small hole in the membrane and drain it completely.
Stretch gently by pulling the clips (don't let them slip) until the area between them is rectangular.
Multiply that area (L x W) by two, because the membrane has two sides around the now-flattened middle.

There will always be some small error, because the wet membrane can be stretched somewhat, but that should be insignificant compared to other sources of inaccuracy, such as edge-effects where the clips narrow the membrane, impeding flow.


text:
How to calculate the equilibrium constant of acetic acid-hydroxide ion equilibrium?
I am trying to calculate the equilibrium constant of the following equilibrium:
$$\ce{CH3COOH (aq) + OH- (aq) <=> CH3COO- (aq) + H2O (l)}$$
I am aware that this reaction essentially goes to completion (the position of the equilibrium lies far to the right). However, I still would like to calculate the equilibrium constant.
I know that $K_\text{eq} = K_\mathrm{a}(\ce{CH3COOH}) \times K_\mathrm{b}(\ce{OH-)} / K_\mathrm{w}$ (I have derived this from the definitions $K_\mathrm{a}$ and $K_\mathrm{b}$). I know that $K_\mathrm{a}(\ce{CH3COOH}) = 1.8\times10^{-5}$ and that $K_\mathrm{w} = 10^{-14}$. However, I am struggling to determine what $K_\mathrm{b}(\ce{OH-})$ is.
I learnt that $K_\mathrm{b}$ is called the "base ionization constant." However, $\ce{OH-}$ is a base, but it is already an ion! So, how can something that is already an ion, ionize? According to this webpage, $\ce{OH-}$ has a $K_\mathrm{b}$ of 1.0. However, what is the equilibrium that led to this number? Is it $\ce{OH- + H2O <=> H2O + OH-}$?
Moreover, if the $K_\mathrm{b}$ of $\ce{OH-}$ is 1.0, is $\ce{OH-}$ considered to be a weak base? Generally, strong bases have a $K_\mathrm{b}$ that is much greater than 1.0. On the other hand, if bases that contain hydroxide ions, such as sodium hydroxide and potassium hydroxide, are strong bases, shouldn't the hydroxide ion itself also be a strong base?

Interesting question! Though in the third step you've written $K_\text{eq} = K_\mathrm{a}(\ce{CH3COOH}) \times K_\mathrm{b}(\ce{OH-)} / K_\mathrm{w}$ and said that "I have derived this from the definitions $K_\mathrm{a}$ and $K_\mathrm{b}$" That means you know the definition of $K_\mathrm{b}$ (base dissociation constnat) for $\ce{OH-}$ ion, don't you? If not, then how did you insert it into that formula in the first place?

Good point. I shouldn't have made that assumption. I have used the equation Ka(HA) * Kb(B) = Keq * Kw in the past for acid base reactions of the form
HA + B <=> BH+ + A-, and I assumed that it would work for the acetic acid - hydroxide ion equilibrium.

Firstly, to derive this equilibrium constant ($K_\text{eq}$), you don't need to bother about $K_\mathrm{b}$ of $\ce{OH-}$. There is a a far simpler way of doing it.
So from the equation, $\ce{CH3COOH(aq) + OH-(aq) -> CH3COO-(aq) + H2O(l)}$, you can easily write, $$K_{\text{eq}} = \frac{\ce{[CH_3COO^-]}}{\ce{[CH_3COOH][OH-]}} = \frac{\ce{[CH_3COO^-]}[\ce{H^+}]}{\ce{[CH_3COOH]}} \times \frac{1}{\ce{[H^+][OH^-]}}= \frac{K_{\mathrm{a}}(\mathrm{CH_3COOH})}{K_\mathrm{w}}$$ So, here no $K_\mathrm{b}$ of $\ce{OH^-}$ arises and therefore we don't have to bother about it.
Secondly, if we want to find out $K_\mathrm{b}$ of $\ce{OH^-}$, At $\pu{25^o C}$, $\mathrm{p}K_\mathrm{w} = 14$, and from the equation of dissociation of water, $\ce{H2O(l) <=> H+(aq) + OH-(aq)}$, we can consider water as a weak acid, and $K_\mathrm{a} = \ce{[H^+][OH^-]} = K_\mathrm{w}$, so, $\mathrm{p}K_\mathrm{a}= \mathrm{p}K_\mathrm{w}$, and from the relation $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = \mathrm{p}K_\mathrm{w}$ (note that in the relation the definitions are that $\mathrm{p}K_\mathrm{a}$ is the acid dissociation constant of a weak acid and $\mathrm{p}K_\mathrm{b}$ is the base dissociation constant of its conjugate base). The conjugate base of water is $\ce{OH-}$, so $\mathrm{p}K_\mathrm{b}=\mathrm{p}K_\mathrm{w} - \mathrm{p}K_\mathrm{a} = 0$ as, both are same for water. So, $K_\mathrm{w}= 10^{-\mathrm{p}K_\mathrm{a}} = 10^0 =1.$  Now, you can see why $K_\mathrm{b}$ of $\ce{OH-}$ is 1 at $\pu{25^o C}$
Thirdly, yes, $\ce{OH^-}$ can't be said as a strong base. It's a moderately strong base. And, the concept that $\ce{NaOH}$ or $\ce{KOH}$ being a strong base should imply $\ce{OH^-}$ should be strong is a totally wrong concept. There is no logic behind it. Those are ionic compounds and completely dissociates and, those are Arrhenius bases. But to become a base, $\ce{OH^-}$ should accept a proton, and hence its Brønsted base, and there can't be a comparison as such. If you have a weak Brønsted acid, its conjugate base must be strong and vice versa. But here the cases are completely different, and there can't be any direct implication.

Are NaOH (aq) and KOH (aq) also Bronsted-Lowry bases, since technically OH- (aq) is a "part" of NaOH (aq)?

Not at all. If they have to behave as Bronsted-Lowry Base, the whole molecule should undergo reaction like $$\ce{NaOH + H^+ <=> NaOH2^+}$$ Does such equilibrium ever exist ???

True, that equilibrium does not exist.
But NaOH (aq) is not a molecule, right? Isn't it just a bunch of Na+ ions and OH- ions floating around in water?

That's the point. Now, you see the definition of Bronsted Bases... Is it allowed to first dissociate into smaller molecules and then get protonated ??


text:
Gas concentrations along the gradient from sea level to deep-sea
Studying scuba diving-related phenomenas I thought about how to estimate gases concentration down the sea.
Of course, solubility of gases at, for example, $5$ atm of pressure, can be estimated with Henry's law, i.e
$$S=k\,P_g$$
But this is not the case down the sea, cause there is no gas in equilibrium with sea-water. There should be a way to relate solubility on the deep with the surface pressure, but can't find how.
So, is there any way to estimate gases concentration deep into the sea?

Gas concentrations down the sea can't be estimated; instead, they must be measured. Equilibrium is of no help here, just like I said.

Yes, and your opinion is valuable @IvanNeretin, but I don't see there is a reason for that. There is no eq but maybe it can be estimated.

It can't. It is below the equilibrium pressure (and well below it, in fact), but that's about it. You can't tell any more than that from the first principles. Can there be a sea of water not containing any gas at all? Why, sure. Can there be a sea nearly saturated with oxygen? Why, sure. Where is the real, actual sea? Someplace between.

@IvanNeretin There should be something like an exponential decrease...(I'm just joking). If there isn't any valuable answer or opinion I'll remove it, but let me wait.

Says who? Why an exponential decrease? In fact, there actually is a decrease, and probably it can be well approximated by a downward exponent, but the reason for this (if any) lies far away from chemistry.


text:
What is the theoretical quantum ground state energy of helium hydride positive ion?
What is the theoretical quantum ground state energy of helium hydride positive ion?
I am speaking of the usual assumptions such as the approximate modeling of the two nuclei here - one being a proton, the other being 2 protons plus 2 neutrons - as two infinitely massive particles alone in the universe at some equilibrium distance, in the non-relativistic simple case alone considering no spin dependent terms etc.
So actually only the charge of the nuclei is relevant in this approximation unless the vibrational energy - perhaps modeled in first approximation as a simple harmonic oscillator with some force constant - is also taken into account - which would be the positive 'zero' point value of harmonic oscillator.
I would assume that it would have to be lower (that is greater bonding magnitude or absolute value of the negative energy) than that of the neutral helium atom, but I am not sure. But one place on internet gives $\pu{3600 cm^-1}$ as a fundamental frequency (but of what?) which - if I am not mistaken comes - to $\pu{1.64E-2}$ Hartrees or atomic units - so that cannot be correct or the value i am looking for as the helium atom is about -2.904 Hartrees or atomic units. I suppose it would be almost impossible to practically devise any experiment to arrive at any experimental value? Though, if so, what is that value?

Just to make sure, are you referring to $\ce{HeH+}$? https://en.wikipedia.org/wiki/Helium_hydride_ion

You could get a quick estimate of the ground state energy with paper and pencil by using Heitler-London method, since all you need to do is change the charge on the second nucleus.
I would assume the vibrational energy absorption energy quoted referred to the He-H bond vibration.


text:
organic spectroscopy: n-phenethylacetamide base peak ion
This is the n-phenethylacetamide mass spectrum

His base peak ion is at 104 m/z. I have long been thinking about what its structure might be, but with a m/z ratio of 104 I can not think of anything plausible. Can you help me?
Thanks in advance

What ionisation method was used?

@Waylander With so much fragmentation it will have been electronic impact

Source Temperature: 200 °C, Sample Temperature: 150 °C, RESERVOIR: 75 eV

@Waylander Done! Long time since I did MS :-|

Assuming electronic impact, the most electronegative atom will loose an electron and remain positively charged.
Amides typically undergo a McLafferty rearrangement.
In this case, the rearrangement can only be to one side, because the other chain is too short.
A neutral fragment of acetamide is lost (in its enolic form) and a positively charged styrene ion remains:
$\ce{C6H5-CH=CH2^+}$: $77+13+14=104$
The rearrangement is as follows:

(Sorry, I do not have a proper chemical drawing program with me; the arrows have been drawn on the touchscreen.)

That fits nicely

Well: so, is a styrene ion?


text:
Acidity and alkalinity versus inorganic carbon concentration
Alkalinity represents the amoount of carbonates in water. When I compare concentration of inorganic carbon measured by TOC-analyzer, with concentration of inorganic carbon calculated from alkalinity,they are in pretty good accordance with each other. But, when there is also acidity present, then the better correlation is achieved with adding the amount of acidity to alkalinity value before multiplying with 12.01. pH is around neutral (6.8)
What is the basis of this? Why this adding gives better estimation of inorganic carbon concentration?

Alkalinity is more than just a measure of carbonates. It includes anything that can resist acidification. What else is in your solution?


text:
Volume of composition of air
My teacher stated:

At standard conditions, 1 liter of air at 21% oxygen possesses $\pu{0.21 L}$ of oxygen. Since at STP, 1 mole of gas occupies $\pu{22.4 L}$, simply divide $0.21/22.4$, to arrive at 0.0094 moles of oxygen.

I have a question regarding this. What if I trap $\pu{100 ml}$ of air in a vessel and maintain $\pu{273 K}$ and pressure $\pu{1 atm}$? So, in that case, what does it mean that oxygen has volume of $\pu{21 ml}$? Won’t the volume of oxygen and nitrogen be same inside vessel? Or am I wrong somehow?

Note that, apparently, your exercise still uses the old definition of STP $(T=273.15\ \mathrm K$, $p = 1\ \mathrm{atm})$, which was used until 1982. The molar volume of an ideal gas at the current definition of STP $(T=273.15\ \mathrm K$, $p = 1\ \mathrm{bar})$ actually is $V_\mathrm m=22.710947(13)\ \mathrm{l\ mol^{-1}}$.

@Loong interesting comment would you happen to have a link to any primary sources?

If you took the 100 mL sample of air, and took out everything but the oxygen, then the amount of oxygen trapped would fill 21 mL at STP.
One way to think of it is, even though the oxygen would expand to fill the container, its pressure would need to reduce according to Boyle's law: P1V1 = P2V2. This is where the concept of Partial Pressure comes from. But if you then compressed that oxygen so that P = 1 atm (since you specified initial pressure to be 1 atm), the oxygen would take up 21 mL of volume.

So I calculated some values. If I want to maintain a volume of air as 100 ml at pressure 1 atm and temperature 273 K  , I should have 0.00353 mole of nitrogen molecules and 0.00094 mole of oxygen molecules . And If I keep that same mole of Oxygen molecules in another vessel (whose volume can be changed) at 1 atm pressure and 273 K , the volume came out 21 ml . So correct . Thanks again.

@ShawnDypx That's because this definition of volume composition is just a fancy way of saying mole fraction. Sigh.

@AnuragBaundwal yes. Indeed.

@ShawnDypx And this is not even the "volume composition" — this definition is like saying that light can't escape a neutron star, if the neutron star is somehow compressed to the density of back hole — it's wrong because it's not a neutron star anymore and light can in fact escape an actual neutron star. They've all bent and twisted the definition, making it invalid. This "definition" of "volume composition" is only more confusing (and incorrect) compared to more fraction, if anything. Sigh.

Yeah , they should really say mole fraction rather than volume composition. It can be confusing .

Air is 21% oxygen by moles, not by volume. Remember that gases occupy the whole of the container in which they are kept and not just a fraction of its volume. This is because the gas molecules are constantly moving about. Even if you tried to confine oxygen (along with air) in 21% volume of a container, the oxygen molecules would move about randomly and eventually occupy the whole container. The other gases can't stop this.
Clarification: Air is actually composed of approx. 21% dioxygen by moles, and not by mass or volume. A simple calculation of the molar mass of air assuming this composition and comparing it with the actual data will verify this. Sorry for the confusion.

Air has an oxygen content of about 21% by volume. This is wrong, hence the down-vote.

@Martin-マーチン isn't by moles and by volume the same for gases?

The correct interpretation of the ideal gas laws here is that at a constant pressure the volume occupied by a gas is proportional to the number of moles of a gas. So if we take partial pressures in the atmosphere the volume % is the same as the molar %. Removing everything but oxygen from a vessel is the wrong comparison, the right comparison is removing everything else but keeping the pressure the same in which case the volume % will equal the molar %.

Thnx @matt_black, that helped me. (I wasn't claiming that the answer is good.)


text:
In a copper wire, how many electrons does each atom give away?
As an ion, copper can give off 1, 2, 3 or 4 electrons. But it has 1 s electron in the last shell and 10 d electrons. So as a metal, how many of those are delocalised and free to move around, and how many are staying with the atom?

All electrons that go out on one end of the wire, are replaced with the same quantity at the other end, so the net loss is 0

@RaoulKessels Sure, but I'm interested in the amount of electrons that can freely move inside the wire.

$I=\frac{q}{t}$ and the charge of one electron is $1.6\times10^{-19}$C

This is a number which can be measured via the Hall effect. This reference give the Hall coefficient as $-5.4\times10^{-11}\,\mathrm{m^3/C}$ for a number density of charge carriers as
$$n_\mathrm e=\frac1{\left(-5.4\times10^{-11}\,\mathrm{m^3/C}\right)\left(-1.602\times10^{-19}\,\mathrm C\right)}=1.16\times10^{29}/\mathrm m^3$$
The number density of copper ions is
$$n_\ce{Cu}=8920\,\frac{\mathrm{kg}}{\mathrm{m^3}}\times\frac{1000\,\mathrm g}{\mathrm{kg}}\times\frac{1\,\mathrm{mol}}{63.546\,\mathrm g}\times\frac{6.022\times10^{23}}{\mathrm{mol}}=8.45\times10^{28}/\mathrm m^3$$
So that works out to about $1.37$ charge carriers per ion.

A good first guess is that there's a gap between the 3d and 4s bands in the electronic structure of solid Cu, and since the 3d band is filled, and the 4s band half-filled, that means only the 4s electron can be considered nearly free.  (Recall Cu = [Ar] 3d10 4s1.)


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Units for density in nanorod image
I am counting the number of nanorods contained in a sample (see picture below). The length scale is 1 µm. What units does the density have? Is it µm-2?

If you are counting them "per square micrometer".

What units does the density have?

If you are considering volume density, the units of the density would be per cubic micrometer, or μm-3. Else if you are considering surface (areal) density, the units of the density would be per square micrometer, or μm-2.
This makes sense because the length scale is already defined at one micrometer, so the other two units will just be the multiplication of the length unit with itself.
If you're measuring - say - the weight of the nanorods instead, you would also need to prepend the appropriate weight (as defined by the scale).