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Why does graphite have a high melting point even though its intermolecular forces are weak?
From what I've read online, when a substance undergoes a state change, the intermolecular forces are broken. The covalent bonds are not broken (http://www.bbc.co.uk/schools/gcsebitesize/science/add_ocr_pre_2011/chemicals/airmolecularrev2.shtml)
My textbook states that layers of graphite can slide over one another easily and can break of one another as they have weak intermolecular forces.
I thought that if an object has weak intermolecular forces, its melting point is lower. Why does graphite have such a high melting point if its intermolecular forces are weak?
My textbook also states that the strong covalent bonds are broken during state changes, isn't this incorrect?

Graphite does not consist of molecules to begin with.

Graphite has got a structure similar to books stacked on top of each other. Multiple layers on top of each other and each layer going by the name graphene.
Atoms in each individual layer is covalently bonded, which is quite strong. Remember covalent bond is the one that holds diamond together, which is one of the hardest substances. Atoms in the individual layers of graphite are strongly held with only three of the four potential bonding sites satisfied. The fourth electron is free to migrate in the plane, making graphite electrically conductive. However, the different layers are held together by weak van der Waal forces, which enables them to slide on top of each other, making graphite a good lubricant.
Now melting is essentially turning a highly ordered state of molecules to a disorderly one. That comes at an expense of energy. In this case, since the constituent molecules of graphite are held together by a strong covalent force, a high amount of energy is needed to weaken that bond. That explains graphite's high melting point.

Thanks a lot. Maybe my teacher has taught me wrong, but doesn't melting involve breaking down of intermolecular forces? I read that when something simple like water boils, the intermolecular forces break down, and the covalent bonds do not. Is this correct?

This answer starts off well, but I'm afraid it reaches the correct conclusion by an incorrect argument. The covalent bonds within each sheet are not really relevant to the melting point. If the covalent bonds broke or changed, the compound wouldn't melt, it would decompose.

@NicolauSakerNeto but graphite is already in elemental form. What would it decompose to?

@ChristopherU'Ren Yes. Melting involves breaking of intermolecular forces. We know water molecules in ice are held together by hydrogen bonds, which is the intermolecular force in this case. However in graphite, the molecules are carbon atoms themselves. These "molecules" are held together by covalent bonds, which plays the role of intermolecular bond in here. Melting/sublimating graphite involves breaking these bonds.

@Gimelist After some reflection, I realise the melting of covalent network solids must involve the breakage of covalent bonds. This is in stark contrast with molecular materials, where breaking covalent bonds is necessarily some chemical transformation. In some sense, the melting of a covalent network solid is a kind of decomposition, except that the original structure is recovered during freezing.

@ChristopherU'Ren Unless you're doing graduate studies, the standard teaching method is to simplify everything and pretty much lie most of the time (but not in a big way). Water in liquid form is held together by a combination of atmospheric pressure and inter-molecular forces. To boil it, you need to disrupt either of the two (or both). Both decreasing pressure and increasing temperature will make individual molecules more likely to escape as gas (and less likely to condense back again - don't forget water constantly evaporates and condenses back, as do all liquids).

(cont.) In this view, Abirbhav is right; the strong covalent bonds of graphite are fundamental to its high melting point. However, this is only true because the (non-covalent) intermolecular interactions are so strong in the first place that it actually becomes more viable to break the molecules down during melting rather than making them slip against eachother whole.

@ChristopherU'Ren But water isn't graphite. The sheets are in a sense "one molecule" - we're talking about "intramolecular bonds" there. It's similar with metals - again, it's not just a mix of individual molecules, metals form large crystals that are essentially huge molecules. To melt them, you need to break the far stronger bonds between the atoms. And even with water, don't confuse liquid water and ice. Ice again forms crystals, where the molecules stick together rather strongly than in liquid water, which has the intermolecular bonds continually break and reform.

@NicolauSakerNeto This is always true when dealing with elemental substances, of course - it doesn't matter if it's graphite, iron or solid oxygen (though oxygen will quickly form dioxygen molecules as it melts - it's a lower energy configuration). Still, the first thing that breaks are the "crystals" - it's just that the strength difference between the "edge" and "centre" in a graphite sheet is far smaller than the difference between the bonds between water molecules in ice and the bonds between individual atoms in the water molecules. If there was no difference, we'd only have monocrystals.

@Luaan I understand what you're getting at, but I feel there is a qualitative difference. A covalent network solid can have bonds which are both highly directional and localised. This differs from metals and ionic solids, for example, where bonding is extended over a large volume and is less directional.

It's not usual to consider graphite as a material composed of "molecules" in the typical sense, though it could be viewed as a kind of polymer with two-dimensional macromolecules. Regardless, it is fruitful to analyse the individual sheets in graphite as the limit of increasingly large polycyclic aromatic hydrocarbons (PAHs). The sequence goes: benzene ($\ce{C6H6}$) → coronene ("superbenzene", $\ce{C24H12}$) → $\ce{C54H18}$ → ...
The main kind of intermolecular interaction relevant to this sequence of compounds is pi-stacking. For the smallest example in the sequence, benzene, it appears that the strength of this intermolecular interaction is only about $\mathrm{10\ kJ\ mol^{-1}}$. This represents only a modest attraction; a hydrogen bond can easily be twice as strong even though it involves less atoms. In this sense, when "normalised" by the number of atoms taking part, pi-stacking is indeed a comparatively weak intermolecular interaction.
Furthermore, $\mathrm{10\ kJ\ mol^{-1}}$ is comparable to the average thermal energy of particles in ambient conditions ($\mathrm{k_BT_{amb}=2.5\ kJ\ mol^{-1}}$), so it doesn't take too much effort to pull apart benzene molecules. Indeed, benzene melts at $\mathrm{5.5\ ^oC}$ and boils at $\mathrm{80\ ^oC}$ under one atmosphere.
However, the next compound in the sequence, coronene, already melts at $\mathrm{437\ ^oC}$ and boils at $\mathrm{525\ ^oC}$. Larger PAHs would almost certainly have even greater values, eventually reaching the limit of graphite, which melts around $\mathrm{4000\ ^oC}$ under pressure. The type of intermolecular interaction hasn't changed, so why are these molecules suddenly so difficult to pull apart? The answer comes from realising that, though individual sections of each molecule interact weakly, the sum of many weak intermolecular interactions over an entire molecule leads to a very strong intermolecular interaction overall.
Crudely speaking, imagine that each individual aromatic ring contributes $\mathrm{10\ kJ\ mol^{-1}}$ worth of intermolecular attraction. Coronene contains 7 fused aromatic rings, which would lead to a total interaction of $\mathrm{70\ kJ\ mol^{-1}}$ between two molecules. As the molecules get larger, this value increases further and further. Eventually the total intermolecular interaction between two very large PAH molecules becomes enormous. For the substance to liquefy, it is only necessary to "break" a fraction of these intermolecular interactions (breaking all of them is turning the material into a gas), but even a small fraction eventually represents a huge amount of energy, so melting only happens at very high temperatures.
It is interesting to note how often chemists make the mistake of neglecting weak long-distance interactions (e.g. van der Waals), especially in the presence of stronger ones. For example, van der Waals interactions are fundamental to the stability of alkyl-substituted hexaphenylethane derivatives. In proteins, weak interactions are often overlooked in favour of hydrogen bonding ($\alpha$-helices and $\beta$-sheets), though they can be decisive in determining the correct conformation of an enzyme or how a protein interact with medicinal compounds.
As a last slight tangent, I just want to point out that in undergraduate-level chemistry, often one finds statements such as "boiling points for covalent compounds increase with their molecular weight. Now it should be evident this is not strictly true. It merely happens that compounds with higher molecular weights tend to be larger and allow a greater amount of intermolecular interactions per molecule, thus leading to higher boiling points.

If we take graphene layers as the limiting case of PAH's then the graphenes layers couldn't slip (we know graphite is a good lubricant). How can your explanation be compatible with this fact? Also, the layers are not stacked above each other but are slighlty off.

@Anton I'm not sure I understand your lack of slipping argument. Even though the sum of the intermolecular forces between two macroscopic graphene layers is very large, if you were to slip the sheets against each other by the width of a few atoms, then almost all the atoms from both sheets are still in contact (only the edges are dangling - in fact, they would curl on themselves), so this state is only very slightly higher in energy. You can apply this same argument over and over. This requires very much less energy than necessary to liquefy - think defect propagation in solids.

So they are slippery because we can move them a little bit each time. So if we try to separate them vertically which is equivalent to break all the interactions at the same time it would be difficult right (even though the final states are of similary energy)?

@Anton Yes, the sheets are slippery because they can move past each other with almost no energy change (especially considering a single unsupported sheet can easily roll itself up) and also importantly almost no energy barrier. Peeling the sheets apart by pulling perpendicularly to the surface is comparatively much harder (this would result in a very different final state with considerably higher energy if the sheets are pulled apart in a vacuum). The scotch tape method to make graphene is more complicated, because you have to consider the interactions between graphene and glue.


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Does sodium borohydride reduce C=N group?

Q.15 the major product of the following reaction is:

I have read that $\ce{NaBH4}$ reduces only aldehyde, ketone and acid chlorides. In some cases it may reduce double bonds conjugated with carbonyl groups. So shouldn't the answer here be option 1?

Sodium borohydride readily reduces isolated C=N bonds, read up on reductive amination.

@Waylander Can you provide some reference in the literature?

try this http://commonorganicchemistry.com/Rxn_Pages/Reductive_Amination/Reductive%20Amination_NaBH4.htm and this  https://www.sciencedirect.com/science/article/pii/S1001841709003167

Waylander's comments answer the title question as well as I could.  Sodium borohydride, and even modified versions where some of the hydride ligands are replaced by others like $\ce{NaBH3CN}$, can reduce imines.
But, if we have only a limited amount of borohydride we will likely see the carboxyl group primarily reduced, as it has a more polar pi bond and thus it is a more reactive electrophile.  So you may in principle get mostly 1 with limited sodium borohydride, or 4 if you have enough sodium borohydride to attack both the carboxyl and less reactive imine functions.
The carbon-carbon bond here, not conjugated to either electronegatove function, is interactive to borohydride and thus products 2 and 3 are not reached.

If the question is written exactly as I had uploaded in the question (nowhere excess or anything is mentioned) do you think I should be penalized for writing 1 and not 4 ? Note the question has specifically one correct answer and there are no other details than that I have given in my original question... please reply

Then either 1or 4 might be formed.  I really can't decide because I do not have ESP and cannot figure whether they mean one or possibly two moles of borohydride reacting.  Always ask precise questions!

Thanks for your reply. The question is exactly as written. No other details given. Then what do you think? Please share your opinions, it matters a lot

The examination commitee has decided that 4 is correct and decided to penalize option 1 . However they are ready to get their answers challenged if proper citations are provided. Please help me if you think that $1$ should not be penalized for the question written with no other specific details. Please sir. (This question is from an undergraduate entrance examination for high school students)

Then they apparently decided to assume excess borohulydride.  I have no more to say.

I really would not attempt to reduce an imine in the presence of a carbonyl group. My gut feeling is that the difference in reactivity between the two is not enough that you could differentiate between the two by limiting the amount of borohydride.

@Waylander okay... if you do not attempt to reduce imine in presence of carbonyl then you would agree that 1 is valid as an answer... wouldn't you?

No, I think you would get a mixture of 1 and 3 with a limited amount of borohydride. This why 4 is the most sensible answer for the major product.

How is there 3?  Double bond is n9t conjugated with carbonylgroup.

Oscar, I think you wrote carboxyl instead of carbonyl in your answer

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Aqua complex scarcity for transition metal ions of second and third row
I've read that for the second and third row of the transition metal ions, almost no [M(OH$_2$)$_6$]$^{2+}$ or [M(OH$_2$)$_6$]$^{3+}$ complexes are known, while they're very common for the first row.
My question is: why is that so? I think it has to do with the fact that elements in the second and third row in oxidation state +II and +III tend to be readily oxidized further, but does somebody have a deeper and clearer explanation?

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adiabatic expansion vs Joule-Thomson Effect

Source: The1stLawofThermodynamicsLesson4
I am looking for difference between adiabatic expansion and Joule-Thomson effect. Looking at the setup demonstrating Joule-Thomson effect, I find no essential difference between the set-up for the two.
What is the feature of the setup for Joule-Thomson effect that makes the process isenthalpic, as oppose to adiabatic expansion? (I am not looking for proof initial H= final H)
What is the use of the throttle/ porous plug? If it is the throttle that makes the process isenthalpic, what is the physical reason behind it?
From another source,

The gas is first compressed isothermally, then expanded isothermally
  (may be at different temperature as initial temperature)

I am not sure what it means. Does it mean that its temperature jumps abruptly from initial temperature to final temperature? It has only two temperature throughout the process? How is this possible? I expect the temperature to change continuously, because the volume of the gas changes throughout the process.

In the Joule-Thompson experiment, there is major viscous dissipation of mechanical energy (in this case pressure) to internal energy occurring within the pores of the plug.  This is analogous to Poiseuille flow of a viscous fluid in a tube.  If the tube is adiabatic, the fluid flowing through a tube experiences a temperature rise as a result of the viscous "heating."  The work to push the fluid through the plug is totally dissipated and converted to increased internal energy of the gas.  The reason that viscous effects are so important in flow through a plug is that the pores are very small.  So flowing the gas through the plug is analogous to flowing the gas through an array of very tiny tubes in parallel.
Ordinarily when a gas expands adiabatically, it cools.  But, for flow of an ideal gas through a porous plug, the adiabatic cooling is exactly offset by the viscous heating.  So there is no temperature change.  And, as a result, there is no enthalpy change.

Interesting explanation.  Does this also explain why Joule-Thompson expansion is not reversible, while adiabatic is?  Because there is nothing to "cool" the gas while it is being pushed through porous plug toward higher pressure?

Yes.  But I think you meant "toward lower pressure."  In fact, it is also the explanation of entropy generation in fluids experiencing irreversible mechanical deformations.

You are right, there is no Joule-Thomson compression.  It is impossible for the gas at a lower pressure to go through porous plug to the higher pressure.  So the process is completely impossible, and the Joule-Thompson expansion is inherently irreversible.  [I was a bit confused because some web pages talk about Joule-Thomson expansion and compression]


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Why enthalpy is defined at a constant pressure?
How is absolute enthalpy and change in enthalpy defined? What is the clear difference between enthalpy and heat? Is it the constancy of pressure what makes the enthalpy a state function?
My teacher told me in very simple words that when heat goes inside a system, it is called enthalpy. Is this correct? Would constant pressure make the volume constant in a thermodynamic system?

How are absolute enthalpy and enthalpy change defined?

Enthalpy $H$ is defined as in your question. The most basic definition of enthalpy change when going from state 1 to state 2 is the enthalpy of state 2 minus the enthalpy of state 1:
$$\Delta H = H_2 - H_1$$

What is the clear difference between enthalpy and heat?

Enthalpy is a state function and heat is not. Heat is energy transferred from a hotter system to a cooler system (not to be confused with thermal energy, that part of internal energy due to the kinetic energy of particles randomly moving, i.e. what makes a sample have a certain temperature).
A simple experiment to experience the difference: If you rub your hands together, they warm up. If you wait a while until they are back at the original temperature, they will be in the same state (no change in enthalpy). At the same time, there was a heat exchange with the environment, increasing its temperature a bit.
In essence, some mechanical energy was converted into thermal energy. There was some heat transfer $q > 0$, but the enthalpy did not change ($\Delta H = 0$) in this thermodynamic cycle. So enthalpy change and heat are different concepts, and there are cases where their values are different.

Is it the constancy of pressure what makes the enthalpy a state function?

No, it is because of how enthalpy is defined. The internal energy is a function of state, and P and V are a description of the state. This means that by the definition of enthalpy, the same state will have the same enthalpy. (If you want to know why internal energy is a function of state: it is related to the first law of thermodynamics, but more fundamentally - it is what all measurements so far have shown).
There is a reason why looking at enthalpy instead of internal energy is often simpler, see below, related to how much of what we experience is at constant pressure (usually atmospheric pressure).

My teacher told me in very simple words that when heat goes inside a system, it is called enthalpy. Is this correct?

This is a shortcut for the following: If you run a process in a way that the only forms of energy transferred between the system and the surroundings are heat and pV-work, the change in enthalpy will be equal to the heat transferred. (In my example above, there was mechanical work; that would break the equality of $\Delta H$ and $q$, as would electrical work done by or on the system, or electromagnetic radiation going in or out.) Also, the system has to be closed (no matter entering or leaving the system).
pV-work is a term for the work of expanding the system's volume by $\Delta V$ into the surrounding. When this happens, you work against the pressure of the surrounding. Enthalpy is defined in a clever way so that it does not change due to pV-work at constant pressure:
$$ H = U + PV$$
$$ $$
$$ \Delta H = H_2 - H_1$$
$$ = U_2 + PV_2 - (U_1 + PV_1)$$
$$ = \Delta U + P\Delta V$$
It turns out the change in internal energy due to pV-work (at constant pressure) is $-P\Delta V$ (it takes work to push against the surrounding, done by the system, so the system loses energy), which cancels out against the $P\Delta V$ term in the calculation above. Thus there is no change in enthalpy due to pV-work at constant pressure.

Would constant pressure make the volume constant in a thermodynamic system?

No. Take an ideal gas (in a closed system) whose temperature rises for some reason. If you keep the pressure constant (by having the gas inside a floppy mylar baloon), the volume will increase. Enthalpy is a useful state function for systems at constant pressure because it does not change just because the volume changes.

Why is this answer so long and complicated (I added this question)?

At my college, enthalpy is introduced as a "fact" in our first chemistry course. A semester later, we learn about first and second law of thermodynamics, and we discuss the tools of thermodynamics a year or two later. So it is not taught from first principles, and you have to learn some stuff without a full picture of the why and how. What I summarized here would be taught in three different courses at my college (along with other material, of course). If you have not been introduced to all of these topics yet, this answer will contain some concepts unfamiliar to you.

sir bundle of thanks! i got this...i learnt imp thing from you i-e i have to skip over some hard concepts. i will control my curiosity of how and why!. Because due to this reason i give much time to a single concept and as a result i find no time for the other chapters and low marks in paper!

You are very welcome. Good luck with your learning!

@KarstenTheis Can you please see my comment on the answer by Chet and provide your insight?

Enthalpy is a physical property of the material being processed (and does not depend on the details of any specific process), while heat is a form of energy that is transferred to or from the material that is being processed, so it is not a property of the material, but it is a feature of the specific process being applied to the material.  So enthalpy and heat are two fundamentally different entities.
Enthalpy is a property defined in terms of the internal energy, pressure, and volume of a material:
$$H=U+PV$$It does not apply only to constant pressure.  However, if a process is carried out on a material at constant pressure, the change in enthalpy of the material is equal to the amount of heat added to the material during the process.
Finally, with regard to your question as to whether constant pressure makes the volume constant, the answer is No.  Even if the pressure is constant, if the temperature increases, the volume increases in proportion (according to the ideal gas law).

How can we define $PV$ in the above equation as it is not work done?

@Apurvium It has nothing to do with work done.  It is simply pressure times volume.

It is a form of energy but how can we define this energy? What is this pressure times volume?

@Apurvium You seem to be trying to assign some sort of physical significance to enthalpy.  Enthalpy is just a convenient function that arise in the solution to many kinds of thermodynamic problems.  It is just shorthand for U + PV

Yes... My confusion is, enthalpy is equal to addition of internal energy + some other energy. I am trying to find the name of that "other" energy.

https://physics.stackexchange.com/questions/155966/what-is-the-interpretation-of-pressure-times-volume-as-energy

@Apurvium I don't understand why you waste your valuable time trying to interpret this.  My advice is to you is to spend your time more profitably by solving some textbook problems involving enthalpy.  You'll get the idea of how it works.

@Apurvium Perhaps $PV$ gives us a measure of how much volume a system is sustaining against how much external pressure. After all $w=-P_{ext}dV$


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Correlation energy in quantum chemistry
In density functional theory (DFT) there is classic Coulomb energy and non-classic correlation energy. But what is difference between them? In books these energies are explained very similarly. Both contributions are explained by the fact that electrons are negatively charged particles.

I would say its not the correlation energy which is nonclassical, since it is just a consequence of the motion of the electrons being correlated. What is nonclassical is the exchange energy (which is often lumped together with the correlation in DFT). The exchange energy is nonclassical because it is Coulomb like, but only arises due to the fact that electrons are fermions (i.e. that their wavefunction is antisymmetric with respect to exchange of particle labels). @Stanislav

But hence why we cannot consider that correlated motion via only Coulomb interaction? Coulomb repulsion doesn't really include that correlated motion?

The correlation is due to e-e repulsion. However the Coulomb energy is generally used to refer to the energy based on the average effect of the electrons repelling each other.

All mentioned terms -- classical Coulomb, exchange and correlation energy -- have the very same physical origin: the electron-electron Coulomb repulsion.
The split into these contributions is only an artifact arising from Hartree-Fock theory.
Here the electronic wave function is approximated by a Slater determinant (the anti-symmetrized product of orbitals). This results in a mean-field description of the electron-electron interaction, and the error with respect to the exact solution is the correlation energy.
Plugging the Slater determinant into the Schrödinger equation eventually leads to an expression for the Hartree-Fock energy:
\begin{equation}
  E_\mathrm{HF} = \sum_i^{N_\mathrm{elec}} \langle i|\hat h|i\rangle + \sum_{i>j}^{N_\mathrm{elec}} [ii|jj] - [ij|ji]
\end{equation}
The three terms in that equation use a short hand notation. Explicitly written those are the one-electron integrals
\begin{equation}
  \langle i|\hat h|i\rangle = \int\chi_i^*(\boldsymbol x_1)\hat h\chi_i(\boldsymbol x_1) \mathrm{d}\boldsymbol x_1
\end{equation}
which cover the kinetic energy and electron-nuclear Coulomb interaction.
The other two terms originate from the electron-electron interaction.
The first one is called Coulomb integral, and gives the contribution of the classical Coulomb energy
\begin{equation}
    [ii|jj] = \int\chi_i^*(\boldsymbol x_1)\chi_i(\boldsymbol x_1) \frac{1}{r_\mathrm{12}}\chi_j^*(\boldsymbol x_2)\chi_j(\boldsymbol x_2) \mathrm{d}\boldsymbol{x_1}\mathrm{d}\boldsymbol{x_2}
\end{equation}
The only reason for its name is, because with $\chi_i^*(\boldsymbol x_1)\chi_i(\boldsymbol x_1) = \rho_i(\boldsymbol x_1)$ and $\chi_j^*(\boldsymbol x_2)\chi_j(\boldsymbol x_2) = \rho_j(\boldsymbol x_2)$ one obtains the formula for the coulomb interaction of two classical charge distributions $\rho_i(\boldsymbol x_1)$ and $\rho_j(\boldsymbol x_2)$. However, we are doing quantum mechanics here, so do not over interpret that analogy.
The last term is the exchange integrals
\begin{equation}
    [ij|ji] = \int\chi_i^*(\boldsymbol x_1)\chi_j(\boldsymbol x_1) \frac{1}{r_\mathrm{12}}\chi_j^*(\boldsymbol x_2)\chi_i(\boldsymbol x_2) \mathrm{d}\boldsymbol{x_1}\mathrm{d}\boldsymbol{x_2}
\end{equation}
and simply named like that because it is essentially the Coulomb term but two indices have been exchanged.
In DFT these three terms show up, because it borrows some concepts from HF and but tries to add the missing electron correlation.
Interpretation of Electron Correlation
I always struggle with the "correlated motion" thing to explain what electron correlation is. I don't think it helps anyone to get an intuitive understand what this means. I do like however the following pictures from the book by Helgaker, Jorgensen and Olsen.

They consider the electron density in $\ce{H2}$ in HF (mean-field) approximation in the top plots and the exact solution below.
To the left we have the 1-electron density which is obtained by integration (=averaging) of the 2-electron density the in middle and right plots.
The positions of the two Hydrogen atoms are obvious by the peaks in the 1-electron density.
The 2-electron density of the HF solution has 4 peaks:
The 2 at ($r_1\approx 1$, $r_2\approx -1$) and ($r_1\approx 1$, $r_2\approx -1$) are the probability of finding both electron at different atoms. The other 2 at ($r_1\approx 1$, $r_2\approx 1$) and ($r_1\approx -1$, $r_2\approx -1$) are the probability of finding both electrons at the same atom. The latter 2 peaks are missing in the exact solution, because the electrons try to avoid such situations.
HF however is unable to consider this correlation. Note that this effect is not visible in the 1-electron densities. HF only works based the 1-electron wave functions (orbitals) and constructs the 2-electron wave function by simple multiplication the 1-electron wave functions.


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Slope of Graph of Wave Number vs inverse square of the principal quantum number(1/n^2) in Hydrogen Emission spectrum
What is the final expression we get for wave number in a hydrogen emission spectrum? Let us say for example from 6 to an orbit $n$. Considering that it is an emission spectrum, I think it would be right to say $n<6$. And wave number is a positive quantity so I thought it would be $R_h(\frac{1}{n^2}-\frac{1}{36})$ and slope of the graph between wave number and $\frac{1}{n^2}$ would be $R_h$. But I would like to ask if it is right though. I recently wrote an exam who says the slope is $-R_h$, where $R_h$ is Rydberg's constant.
Thanks!

https://chemistry.stackexchange.com/questions/108066/slope-of-graph-of-wave-number-vs-inverse-square-of-the-principal-quantum-number

The slope is $R_h$ in emission. It is -$R_h$ in absorption.

Firstly, the question doesn't seem logical as n can take only 5 discrete values here, i.e. 1,2,3,4,5. So the graph will have just 5 isolated points. Still, if you do have to plot a continuous linear graph through these points, the slope of the line will be positive and equal to +Rh. So you are right.

There are no restrictions on how high $n$ can be. It can also be infinite and it's totally fine. Also, plotting continuous line across discrete data points is just wrong (you can estimate the slope though).

Thanks Ankit. So it should increase with increase in (1/n^2) then. Thought so too. The key is probably wrong then, thanks!

@andselisk, the question talks about de-excitation from the  6th shell, so I stated that n can take values only from 1 to 5. Thus, 5 discrete points.


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Which Carbocation is most stable among the three?

According to me the answer should be (C)  as that Carbocation will be stablized by resonance with Chlorine.
I have read that resonance is the biggest deciding factor about the stability of a carbocation
But some argue that (A) would be most stable as it is farthest away from a highly electronegative Chlorine atom.
Please help

Are you sure B can be discharged out discussion?

i do not agree with you because Cl is considered a deactivating(electron withdrawing) groupand the carbon will be very posistive which this itself causes instablitiy.

If you're not "supposed to" think about chloronium in such case then such question is pretty much waste of time.

You are correct as resonance is most powerful factor for stabilisation of carbonation So C is most stable and after that due to hyperconjugation B would be more stable than A. Order for governing stability is
$\text{resonance}>\text{hyperconjugation}>\text{inductive}$

What about overlap strength? I mean, I know that for fluorine, overlap of lone pairs from 2p of fluorine to empty p of carbocation is quite stabilising. Can the same be said for overlap of empty p of C and 3p of Cl?


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Changing the CO2 concentration (adding or subtracting) inside a sealed vessel?
I am looking for a method of modifying the CO2 concentration inside a sealed container that is both precise and repeatable (C02 needs adjustment every time the container is sealed, also need to be able to make additional adjustments to counteract potential slow leaks.
Is there a way I could do this electrically? Voltage across a sample of some precursor? Coating a heating wire with something that could be gradually burnt off? I read an online article from 2016 about a type of battery that actually removed CO2 from the atmosphere and used it to generate a current (and in the process reduced the concentration) - could not tell from the article if all the parts of the battery were solid - if it were then perhaps a scaled down cell?
Very much appreciate any suggestions so I have some options.

Without a lot more detail (how big is the container, what pressure do you need, over what timescale, what else is in the container, what is the container made out of, ...) this is really hard to answer.

size would level between envelope and suitcase size, - I'm working out a system to detect intrusion into a package or luggage - the idea I had was to modify the CO2 level inside once it was sealed so that upon opening the CO2 levels would rush back to normal ambient levels and a CO2 sensor would detect the intrusion.

I'm not prepared to give you a full answer, but maybe you should look into the membrane technologies. Existing carbon dioxide capture membrane prototypes are made of wide range of materials, including organic ones that work at fairly low temperatures (e.g. DOI 10.1021/ie070794r). As for the "battery," molten carbonate fuel cells can be used for $\ce{CO2}$ transport (e.g. DOI 10.1021/ie302725a). Anyway, it might be easier to pump the package with pure nitrogen, either from the gas bottle or from the generator, and detect the intrusion by measuring the oxygen partial pressure.

@voffch - it turns out that CO2 sensors are more robust and cheaper than O2 sensors. For the OP, particularly since the container is not 'sealed', you likely will have some issues with shipping companies by having power, compressed gases, etc. inside the volume. There are other ways to see if something has been opened or not in transit.

Just to clarify the objective is a small battery powered device that drops inside the bag - no gas bottles - any reactants for creating the additional CO2 (only need enough to make a small relative percentage difference) need to start off in a solid phase and only a small battery will be available.

Reading the paper "The O2-assisted Al/CO2 electrochemical cell:
A system for CO2 capture/conversion and electric
power generation" and they seem to use a room temperature electrolyte.

Sodium Carbonate decomposes into Sodium Oxide and CO2 starting at 500 centegrade, while Carbonic acid decomposes to H2O and CO2 starting at only 50 centegrade. a pellet with embedded heating wire could make for a compact generator.


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Why HI isn't fully miscible with cold water when HF is?
HF is more soluble in water than HI, due to greater hydrogen bonding in HF, accepted, but on the other hand HF is a weak acid so it's dissociation will be meagre when compared to HI's dissociation, so why is HF and not HI more soluble, am I misinterpreting the concept of solubility?

If you know what solubility is, I wonder how you'd suppose to misinterpret it. Also your assumptions about HF are wrong  https://chemistry.stackexchange.com/questions/34818/why-is-hf-the-least-acidic-hydrogen-halide

I still could not understand , what I was able to gather from that link is , HF does dissociate but due to high electronegativity diff. thr H+ is not freed resulting in low acid strength , am I right?

This is not off-topic, the author gives a very clear attempt to understand the underlying principles.

@NutanPrakash Pretty much this. BTW your question may be better worded as why HI isn't fully miscible with cold water when HF is.

Hydrogen iodide is not listed as miscible with water because it is a gas. Miscibility is a property generally ascribed to pairs of liquids. Hydrogen fluoride is a liquid, and so it can be described as miscible with water. HF is a weak acid, but its high solubility can be explained by the high polarity of the H-F bond and the ability of HF to be both a hydrogen bond donor and accepter.
Hydrogen iodide is a gas at room temperature. It's boiling point is $-35\ ^\circ\text{C}$. Just because it is not miscible does not mean that it is not highly soluble. The solubility of HI in water is 245 g per 100 mL. In other words, a saturated solution has more HI by mass than water. HI is a strong acid, and the solubility of the gas is due to the nearly complete ionization of HI. Since HI is a gas, I presume that the solubility limit occurs when the vapor pressure of HI above the solution is high enough that vaporization begins to compete with ionization.
Physical data on HI from Wikipedia

that means solubility can be attributed to both - high polarity and complete ionization in particular cases.And I chose HI just because I wanted to compare solubilities of species which dissolve due to complete ionization - like strong acids and HF due to hydrogen bonding .

HF is a liquid if your lab is relatively cold (bp ~19 °C). Anyway, the main point is that solubility and ionisation are not strongly related.


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Angular momentum of an electron in an atom
The Orbital Angular momentum of an electron in an atom is given by $\hbar \sqrt{l(l+1)}$ and the Spin Angular Momentum is $\hbar \sqrt{s(s+1)}$. So what is the resultant Angular momentum of an electron in an orbit in an atom?
I came across a question which asked the value of Angular momentum of an electron in its ground state in an atom. Does this refer to resultant of both kinds of Angular momentum, or am I misinterpreting the question?
Also, what exactly does ground state refer to in an atom? Is it the 1s orbital?

Electrons do not orbit the atom. This is confusing, but try to just accept it: they have angular momentum, but are not "orbiting" as in moving in an orbit around anything. They have spin angular momentum, but are not spinning.

I would argue that individual electrons do not have well defined angular momentum in an atom, only the many-electron state can be assigned to a well defined angular momentum. Also, not all states are angular momentum eigenstates.

You should start by looking for LS coupling (Russell-Saunders coupling) in a undergraduate level spectroscopy or phys. chem. textbook. The angular momenta are added in a vectorial like manner (vector model of angular momentum).  There is a standard method by which to find the total angular momentum quantum number, usually called J. (The Hyperphysics site has some good pics.)

@porphyrin, Thanks for the reference. At present, this seems to be somewhat out of my scope, since I'm only in Grade 12, but I'll try my best to understand.

The question does not state what kind of atom, but I assume this is about the Hydrogen atom. In this case you can combine orbital and spin angular momentum according to Russell–Saunders coupling.
Since we are talking about the Hydrogen atom, the ground state is essentially just the $1s$ orbital. In general however, for many-electron systems, electronic states and orbitals are different things!

So what would be meant by 'ground state' in any general atom, in view of the quantum mechanical model of an atom? Isn't the 1s orbital the one with the lowest energy even in a multi electronic atom?

Yes $1s$ is the lowest orbital, but it is not a state! Within Hartree-Fock, a state is a Slater-Determinant, which is build up from many orbitals and therefore corresponds to an electron configuration. But this is really oversimplifying things.

Maybe better put it this way: An orbital is a one-electron wave function, two electrons if you are talking spatial orbitals, but never more! An electronic state on the other hand is a many-electron wave function, meaning it can be more than just one or two electrons. The state is what you can physically measure, an orbital is just a mathematical tool.

For completeness, I would suggest to mention the alternative for heavier atoms, j-j-coupling.

In case of a single electron, LS and jj coupling are equivalent.


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Balancing an equation
so I might be just way to tired for this, but I somehow have a stupid mistake here when trying to balance an equation.
$$Zn(s)+MnO_2(s)+NH_4^{+}(aq)=Zn^{2+}(q)+Mn_2O_3(s)+NH_3(aq)+H_2O(l)\tag{1}$$
Here's what I do:
$$A\cdot Zn(s)+B\cdot MnO_2(s)+C\cdot NH_4^{+}(aq)=D\cdot Zn^{2+}(q)+E\cdot Mn_2O_3(s)+F\cdot NH_3(aq)+G\cdot H_2O(l)\tag{2}$$
We get
$Z_n: \quad A=D$
$Mn:\quad B=2E$
$O:\quad 2B=3E+G$
$N:\quad C=F$
$H:\quad 4C=3F+2G$
$z:\quad  C=2D$
We see that
$$Z_n \ \& \ N \ \& \ z: \quad A=D=\frac{1}{2}C=\frac{1}{2}F\tag{3}$$
and
$$O \ \& \ Mn:\quad 4E=3E+G \quad \Rightarrow \quad E=G\tag{4}$$
so it follows
$$Mn:\quad B=2E=2G\tag{5}$$
we also see
$$N \ \& \ H: \quad 4F=3F+2G \quad \Rightarrow \quad F=2G\tag{6}$$
so we get the relationship:
$$A=D=\frac{1}{2}C=\frac{1}{2}F=G=E=frac{1}{2}B\tag{7}$$
in correct order:
$$A=\frac{1}{2}B=\frac{1}{2}C=D=E=\frac{1}{2}F=G\tag{8}$$
Now the actual solution is:
$$Zn(s)+2MnO_2(s)+2NH_4^{+}(aq)=Zn^{2+}(q)+Mn_2O_3(s)+2NH_3(aq)+H_2O(l) \tag{9}$$
so I should have gotten:
$$A=2B=2C=D=E=2F=G\tag{8}$$
I don't see what's wrong. :/

No, your solution is right because $A = 1$ and $B = 2$, not the other way around. However, you tried to solve a seven variable system of equations without the help of linear algebra, which just made the process more error prone. This problem is even easier if you know about oxidation states.

I could have made n matrix and solved that, sure. But in general it's not worht the effort for my basic chemistry here. I also just noticed that if I choose $A=1$ then $1=0.5B \Rightarrow B=2$. Thanks

Your solution is right, and the answer given is the correct answer. What you have written in equation 8 is incorrect. The numbers you have multiplied in each variable should actually be in the denominator, since they are the ratios.


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Why enzyme-catalyzed reactions mostly go one way?
I listened to the lecture about enzymes. Professor downplayed the question in the subject. He said that enzymes are able to work both ways, but practically speaking reactions with overwhelming probability flow in only one direction. He said that a variety of factors cause this, and there is nothing interesting there beyond this.
I am not clear what the reasons are. There is the reaction rate constant. Does this mean that rate constants are very different in value depending on direction? Or the reason is that reaction product molecules are always quickly consumed by the next step, and as a result the concentration of the product is always low?

Some enzyme-catalyzed reactions always go in the same direction in a live cell (e.g. reactions of citric acid cycle, RNA polymerase) and others go in either direction depending on circumstances (for example, some enzymes of glycolysis are also involved in the reverse pathway that makes glucose).
As for reasons why a reaction might go in one direction only, you are on the right track. You mentioned two reasons you could imaging for a preference in one direction:

Does this mean that rate constants are very different in value depending on direction?

If a reaction is far from equilibrium (i.e. proceeds mainly in one direction), the forward rate and the reverse rate will be very different from each other. This could be because the rate constants a very different from each other (with similar concentrations of reactants and products) or because the concentrations of reactants and products are very different from each other, or both. So the safe thing to say is that the rates differ by a lot (rather than the rate constants).

Or the reason is that reaction product molecules are always quickly consumed by the next step, and as a result the concentration of the product is always low?

This case also occurs. In RNA synthesis, for example, the reaction of a NTP (i.e. a triphosphate) with the growing RNA occurs near equilibrium, yielding longer RNA and diphosphate (pyrophosphate). In a second step, pyrophosphate hydrolyzes quickly, removing one of the products and making the first step virtually irreversible once the second step has occurred.
In the end, whether the reaction goes in both directions or just in one is not dependent on the enzyme: the enzyme is not capable of changing the position of an equilibrium. Whether the reaction goes forward or backward depends on the equilibrium constant of the net reaction (which the enzyme is not a part of) and the concentrations of the reactants and products.
You could envision a scenario where all enzyme-catalyzed reactions happen to be far from equilibrium under physiological conditions, but this is not the case, as the counterexamples mentioned above show.
Enzymes that catalyze reactions that are far from equilibrium are often regulated (turned on and off in various ways) to regulate entire pathways. If a reaction is near equilibrium and is used in different directions by distinct pathways, regulating the enzyme catalyzing the reaction would affect both pathways, so usually these enzymes are not regulated.


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Which etching process is better, photochemical or electrochemical?
My boss and I are experimenting with several different ways to micro machine .01" and .005" stainless steel. When I look up how it's done professionally, I see an overwhelming majority of companies use photochemical etching (acid etching). I have also seen electrochemical etching as an option, but not as a "precision" method. So far, both methods have achieved sub-millimeter precision and electroetching currently takes a slight lead over photochem, but neither have achieved the incredible accuracy that I've seen online that appears to have resolution finer than the human eye can distinguish.
Electrochem etching has also proven to be much simpler and safer since we don't have to be involved with potentially dangerous chemicals. My suspicion was that acid etching would be more precise since it is so widely used, but our results are not supporting that hypothesis. I am wondering what advantages does acid etching have over electroetching that makes it so much more popular? I'm curious to know more so we can further refine our etching process.

While I have no expertise in this subject I did a web search for "advantages acid etching has over electroetching" which led to the following which I copy verbatim, specifically the part that seems relevant to your comments:

Electro-etching offers a great alternative to etching with acid, but
  as with everything, it also has some disadvantages:
• A feature of electro-etching is that it will bite isolated lines deeper than those that are close together, or cross hatched lines.
  Also it tends to bite harder in the external areas than at the center
  of plates (edge-effect).
  In addition, electrolytic etching bites harder on the edges of wide
  lines and edges of resists of large areas of open bite. This is more
  noticeable in prolonged bites. However, all these drawbacks can be
  compensated for by using a cathode grid in the electro-etching unit
  instead of a cathode plate.


text:
why do I get chlorine gas production during electroetching?
I was electroetching stainless steel plates with table salt as electrolyte and the power supply was providing 5V, 6A when I placed the anode and cathode about .25 inches apart. I started smelling a swimming pool chlorine smell and when I pulled out the anode, it had a yellow film on it which smelled strongly of chlorine also, which I am guessing is some sort of chlorine precipitate. I thought that distinct smell came from chloramines though, not a chlorine gas, so I don't know exactly what compound I was smelling.
However when I returned the plates to .5 inches apart and heated the solution from about 15C to 35C, I was drawing 5V, 8A since the heat allows the reaction to happen more vigorously. However, I no longer smelled any chlorine even though there was a higher power consumption. What is going on here? Clearly the distance between the two plates affects this, but can someone give a chemical explanation as to why?

Smaller distance means less inner resistance, ergo smaller voltage difference within the solution, and higher voltage difference at the electrodes (if both experiments are run at 5V total). I assume (or guess) you just passed the overvoltage for $\ce{2Cl- -> Cl2 + 2 e-}$. Also the change in temperature possibly lowers the overvoltage of chlorine production.


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Why does bond breaking in ATP release energy?
We always come across that formation of bonds releases energy, and bond breaking requires energy, but in the case where ATP converts into ADP or AMP (bond breaking), energy is released. Why?
Further, how bond formation releases energy?

In most chemical reactions where bonds are broken, other bonds are formed. Take your example of the hydrolysis of ATP. A bond between two phosphate groups breaks, but one of the phosphate groups forms a new bond with the oxygen of water. Whether energy is released or taken up (whether the reaction is exothermic or endothermic) depends on the sum of the energies associated with breaking and making bonds in that reaction.

How does bond formation release energy?

In the simplest case in the form of kinetic energy, heating up the reaction mixture. In the case of ATP hydrolysis in biological processes, the energy is sometimes converted into mechanical energy (muscle contraction), used to run pumps (transmission of signals in the nervous system), or other processes that would not go forward on their own.

I love this question!
I teach Chemistry at various levels and this concept around ATP hydrolysis causes more issues for my students than any other. Often, this is the first time that a student meets a concrete example of bonding (in a Biology class) and they so often walk away with the wrong idea about the processes of bond forming and breaking.
Breaking a bond, in isolation, never releases energy. Bonding is a stable state compared to the unbonded species, where opposite charges are closer together when bonded compared to unbonded and the whole system is at a lower (electrical) potential energy. The bond broken in the hydrolysis of ATP is no different. It is a fairly weak bond, but still requires energy to be broken.
The reason there is energy released in the process is because the products formed (ADP and hydrogenphosphate/phosphate) have stronger covalent bonds (plus intermolecular forces with the surrounding solution and dissolved ions) than the starting materials. This is the case for any exothermic process. As you break the P-O bond in ATP a new P-O bond is formed in the hydrogenphosphate, but you also need to look at the interactions of the starting materials compared to the products with the solution. We should also note that the water that attacks the phosphate group in the hydrolysis reaction will then  need to be deprotonated and the hydrogenphosphate ion formed will partially dissociate to phosphate, so there's a lot going on!
Also, it is worth noting that when people say "energy is released in ATP hydrolysis" they are normally referring to Gibbs Free Energy, which also includes the contribution made by the system entropy change (times temperature) as well as the enthalpy change (determined by bond and other electrostatic interaction strength). In the case of ATP hydrolysis, under most conditions, we also have an increase in the entropy of the system and this drives the process to be even more exergonic (favorable, can be used to drive other processes) than the enthalpy alone would suggest.
Please understand: the chemistry involved here is actually very complex and the total usable energy made available depends on many factors beyond the structures of the starting materials and the products. To truly understand ATP hydrolysis requires knowledge of all species' concentrations (as this affects the driving force) including various dissolved ionic species that aren't normally included in the simple reaction equation.
To answer your last part, bond formation from isolated species always releases energy as opposite charges are getting closer together and potential energy is decreasing.


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what is the exact meaning of oxidation state?
can you mention the concept of oxidation state without its simple definition i-e it is the apparent charge +ve or -ve.....
I found these in wikipedia but didn't understand."Conceptually, the oxidation state, which may be positive, negative or zero, is the hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic, with no covalent component. This is never exactly true for real bonds.". what does it mean? please clear with example if possible.

The gist is that no bond is 100% ionic or covalent. Obviously some are mostly ionic or covalent, but not 100% of either extreme.

In the "oxidation state" theory, we assume that the "oxidation state" is always an integer.  What the sentence in Wikipedia means is just that the real charge is usually not an integer.

Oxidation state is a formal number ascribed to atoms, to tell if it has more or less electrons than in its natural state.
For single atoms it is fairly straightforward, as electrons can only be removed or added in discrete units. Therefore, an atom or ion has an oxidation state that equals it electric charge.
But electrons are not small particles, orbiting the molecule. In many ways an electron is better described as a cloud of negative charge hovering around the molecule and chemical bonds.
When different atoms are connected in a molecule, this cloud is not evenly distributed between the atoms. Example: In Carbonmonoxide (CO) the Oxygen attracts most of the electron-cloud. But not all, if so the molecule would part into ions.
In reality, the oxidation state of the atoms in CO should be calculated with decimals instead of whole numbers, but for practical reasons we ascribe the whole of an electron in a bond to one of the atoms.
In Carbonmonoxide the atoms are bonded by a double-bond (4 electrons) and if all of these “belongs” to oxygen it will have two more than the basic atom and thus the oxidation state is -2.

Oxidation states are whole numbers assigned to individual atoms that are part of molecules or ions. They serve to describe oxidation and reduction reactions. An increase in oxidation state is associated with oxidation (loss of electrons) while a decrease is associated with reduction (gain of electrons).
There are multiple sets of rules how to assign oxidation states to atoms (see for example section on oxidation reduction reactions here or what to do for organic molecules here). Sometimes, there is disagreement how exactly to assign oxidation states (it is a conceptual number, you can't measure it), but in all cases, the sum of the oxidation states of all atoms in a molecule or ion or complex has to equal the net charge of that species. This means you have to (artificially) assign each electron to an individual atom based on specific rules or based on electronegativity scales.
To give examples of oxidation states, the carbon in methane has an oxidation state of minus four, while the carbon in carbon dioxide has  an oxidation state of plus four. Hydrogen is mostly assigned an oxidation state of plus one, and oxygen mostly an oxidation state of minus two.
When methane is burned (i.e. reacts with elemental oxygen), it is oxidized to form carbon dioxide in an oxidation process. At the same time, the elemental oxygen is reduced to form water and parts of the carbon dioxide. Together, this is called a redox reaction, and there is a transfer of electrons (electrons are available from the oxidation half reaction, and are required for the reduction half reaction).


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Copper Mirror Reaction
I would like to create a copper mirror on a plate of sandblasted glass.
I know that a reaction exists:
$$\ce{CuSO4 + N2H2 -> Cu + N2 + H2SO4}$$
I know that such a plate of copper is very weak. Therefore, I would like to use sandblasted or laser etched glass.
My question(s) are:

Would laser etched glass be sufficient to retain the deposition? I.e. if I plate a complete glass slide, part of which is etched, and part of which is smooth – can I safely wipe off the smooth part from deposited copper, while the etched part provides a good adhesion to the deposited layer?
Is there a way to replace the hydrazine? I read that fructose could work – but is not reliable. I also know that $\ce{Zn}$ and $\ce{P}$ also work as reducing agents. But they work at an elevated temperature. Is it possible that we can do this at room temperature with an easily available, less poisonous or explosive (possibly from the drug store or the hardware store) agent?

Zn works at room temperature just fine. The resulting Cu won't stick to glass, though.

You will find a lot more recipes for silvering a glass surface. Once you have a thin silver layer, just electrodeposit the copper on top?


text:
Is it possible to prevent saponification by modifying free fatty acids?
Not big on chemistry so this question might be a bit out there. It is my understanding that a soap molecule is made up of a free fatty acid and lets say a sodium ion. Correct? Would it be possible to modify the free fatty acid (with a solid or liquid substance) in such a way that when it comes in contact with the sodium ion it wouldn't be able to form a soap molecule?
Thanks

What really matters is the counterion that the sodium is associated with. If the counterion is something (like hydroxide OH-) that will remove the acid proton from the free fatty acid then you get the sodium salt of the fatty acid formed - this is the soap. If the counterion is something that does not remove the acid proton (chloride for instance) then you don't get the soap.

That's not what is called saponification.

Waylander are you saying if chloride is added to a free fatty acid it won't form a soap molecule even if later on both come in contact with sodium (Na+)?

NaCl and a fatty acid gives a salty fatty acid. Why not Google "saponification"?

@okonolapota you have to have a base (an anion that grabs protons) to get the salt, if there is no base present then all you have is a mixture of free acid plus salt

You can do a lot of chemistry with the fatty acid. But to what purpose? You only say what you don't want.

In principle You can replace the Sodium ion with any other cation. But all You get is another kind of soap: Potassium gives a smoother substance, Calcium/magnesium gives the soap-stone you see in washing machines, etc.
Another possibility is to esterify the fatty acid with an alcohol. This is widely used in the industry to produce non-ionic surfactants, and these do not react with cations. But off course, then it really isn’t a fatty acid anymore.

That's what I'm looking for, something that wouldn't be a fatty acid anymore and thus wouldn't be able to become a soap molecule. Is there anything else besides alcohol? What about precipitating the fatty acid? I read that free fatty acids can be precipitated as alkali soaps. What does this mean?

What about xylitol?

In chemistry an alcohol is basicly any organic molecule that contains an -OH group: ethanol, isopropanol etc, but also xylitol, glucose and glycerol are alcohols. There is a group of detergents called Alkyl plyglycosides (APGs) that are esters of fatty acids and (poly) glycose-units.

So basically by adding an alcohol to a fatty acid you get a surface active ingredient?

Very basically: Yes. But in reality it is more complicated. With simple alcohols just mixing it could yield a significant amount of ester (Fisher reaction), A catalyst, such as a strong (sulfuric)acid, is however needed as well as elevated temperature. If water is present (and water is formed during the reaction) it will hydrolyze the ester and thus reducing the yield.

Saponification is the process of creating the free fatty acids from the triglycerides that carry fatty acids around in the body. Triglycerides are glycerol esters of fatty acids and are the major constituent of bodily fat in animals.
Soap is created by breaking down the triglyceride esters releasing the free fatty acids and is usually accomplished by reacting the fat with sodium or potassium hydroxide. The resulting compounds–the sodium or potassium salts of the free fatty acids–are what we call soap.
Soaps act as surfactants which can solubilise dirt and fat in water. Their primary action as surfactants results from the long chain fatty acids having a polar end that likes water and a non-polar end that likes the fatty and water-insoluble elements in the dirt. It is the fatty acid not the counter-ion that dominates this effect. Swapping the counter-ion (sodium or potassium, etc.) for something else make a small difference to the character of the soap but not a large difference to the surfactant action as that is dominated by the free fatty acid.
So swapping the counterion isn't the point of saponification: the point is to get the salt of the fatty acid free from the triglyceride. And the resulting salts are ionic compounds where the counter ion of the acid salt is not hugely important to the function as a soap.

Thanks for explaining what is saponification but it did not help with my question.

@okonolapota Given what saponification actually is your question "Is it possible to prevent saponification by modifying free fatty acids?" doesn't really make any sense. Free fatty acid salts are soaps. Esters of free fatty acids are fats not soaps. Changing the metal counter ions only makes a marginal difference to their soapiness. Different fatty acids make different soaps. Some non fatty acid compounds make good surfactants. The fats that fatty acids are derived from are not soaps. What is the question?

I said that the question "may be a little out there" implying that what I asked wasn't even possible. Still got a few interesting answers. If you have a FFA without the sodium/potassium ion it isn't a soap it is just a FFA right? Let's say you have a situation where a free fatty acid which will eventually bind with a sodium ion - this is something you cannot control, it just will happen. My question was if it were possible to modify the FFA with a substance (one answer was alcohol) so that when it eventually does comes into contact with the sodium ion it won't be able to form soap.


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Electrolysis in water with baking soda
Many demonstrations of electrolysis show the use of baking soda and a 9 volt battery for electrolysis of water. As a science experiment, I tested different voltages by adding more batteries. In my experiment, I used distilled water. After letting the electrolysis happen for 5 minutes, the 18 volt test had yellow discoloration in the water. Because I couldn't find any places online that mention this or explain what it is, I thought to ask. Perhaps it is not a common thing?
[Edit] Answer: Most likely cause was degradation of poor electrode material with too much electricity.

Electrode material is more important than you seem to assume.

Note A: The electrodes (the exposed bits) I am using are stainless steel alligator clips.
Note B: Kosher salt is NOT a suitable medium, it uses yellow prussiate of soda as an anti caking agent. YPoS (Ferrocyanide) is stable until mixed with Hydrogen and Ample electricity, after which it turns into hydrogen cyanide.

Stainless or not, pretty much any metal anode will dissolve.

Thank you, I hadn't considered the electrodes as an input for contamination. Less worried now.

18 volts are also pure overkill. Above 3 - 4 V or so, you just produce more heat with higher voltage.


text:
How to quantify the difference between two xyz files for the same molecule?
Sometimes I encounter xyz files with slightly different coordinates, for example, before and after optimization.
What is the easiest way to compare how different are they?
In the first approximation, I would define the difference as a minimum of the sum of squares of differences between individual coordinates, where the minimum is taken over translations/rotations of one of the representations.

RMSD, or maybe TM-score.

Try https://github.com/charnley/rmsd

The answer is also dependent on what you want to know / to compare. It is a different situation when you are interested in if two optimized geometries are the same, or if you try to compare different comforters.

What you describe is to calculate the RMSD after superposition. It is not necessary to do a minimization, there is a classic algorithm to get the best superposition directly: Kabsch algorithm. Sometimes, only a subset of coordinates will be superimposed, either because coordinates are missing in one of the structure, or because some parts of the structure show larger differences and interfere with superimposing the similar parts of the structures.
In some cases, where the structures are very different from one another but there are domains of conserved structures, you could also compute differences of pairwise distances and plot them as a distance difference matrix. Distance differences within the conserved domains will be small, while distance differences from domain to domain will be large.
When you do a superposition and RMSD calculation for the first time, it is nice to be able to check the superposition. Many structure viewers have the capability to visualize the superposition while also calculating the RMSD, for example the "compare" command in jmol.

As Karsten Theis said there is the Kabsch algorithm for this purpose, but this algorithm claime to make pairs atoms in two structures (for minimization RMSD between structures via rotation and translation). However, sometimes this is not easy. Though if your order of atoms doesn't change it will not a problem, otherwise there is ICP ICP algorithm or global optima ICP, but both algorithms consider atoms as points and hence have a lots disadvantages, because global minima between two point clouds is not global minima between two molecules in common.

As already mentioned in the comments there exists a Python implementation of the RMSD calculation which can be found at https://github.com/charnley/rmsd.
If you want to instead get detailed overview over what changed, maybe my Python script at https://github.com/Krzmbrzl/xyz-diff might be suitable for you. It will effectively calculate a diff-matrix and output that (so you see the diff for every individual coordinate). Note that no prior aligning is performed, so the output will be the raw diff.
Example output:
O         +0.00000000   +0.00000000   -0.01143253
O         +0.00000000   -0.00000000   +0.01143253
Cu -> Co  -0.01804938   +0.00829356   +0.00000000
Cu        +0.01804938   -0.00829356   +0.00000000
N         -0.04042133   +0.03424848   +0.00000000
N         +0.04042133   -0.03424848   +0.00000000
N         -0.02848896   -0.00191986   +0.01590888
N         -0.02848896   -0.00191986   -0.01590888
N         +0.02848896   +0.00191986   +0.01590888
N         +0.02848896   +0.00191986   -0.01590888
H         -0.04673929   +0.02155138   +0.00000000
H         +0.04673929   -0.02155138   +0.00000000
H         -0.04417332   +0.03825913   +0.00254647
H         -0.04417332   +0.03825913   -0.00254647
H         +0.04417332   -0.03825913   +0.00254647
H         +0.04417332   -0.03825913   -0.00254647
H         -0.02376989   -0.00091570   +0.01582489
H         -0.02376989   -0.00091570   -0.01582489
H         +0.02376989   +0.00091570   +0.01582489
H         +0.02376989   +0.00091570   -0.01582489
H         -0.02343687   -0.00428525   +0.01206063
H         -0.02343687   -0.00428525   -0.01206063
H         +0.02343687   +0.00428525   +0.01206063
H         +0.02343687   +0.00428525   -0.01206063
H         -0.03131781   -0.00348489   +0.01817259
H         -0.03131781   -0.00348489   -0.01817259
H         +0.03131781   +0.00348489   +0.01817259
H         +0.03131781   +0.00348489   -0.01817259

(in a terminal the output is color-coded, so large diffs can more easily be spotted)


text:
What is the role of ammonium chloride in the workup of a Grignard reaction?
In the following Grignard reaction, why is aqueous ammonium chloride used to get to the products? I don't see how it participated in or modified the reaction from what is normally expected.

This question has been edited to be more obviously on-topic.

In scientific writing, brevity is highly valued. I do not see the point of padding a post when the question is perfectly clear and answerable. It is a waste of time for the editor as well as for future readers. To make myself abundantly clear, I am not criticising the edit, but rather the situation which necessitates this edit.

Ammonium chloride ($\ce{NH4Cl}$) is the work-up reagent that quenches the magnesium alkoxide product of the Grignard addition. It is the reagent of choice as it is a proton source without being acidic; acidic conditions could result in protonation of the tertiary alcohol product and elimination to the alkene. It also ensures that all inorganic salts of Mg will extract into the aqueous phase.

Oh I actually never knew that! In that case, other weak acids can also be used right? For example, what about weak organic acids like ethanoic acid?

Yes, coud use that or something like sodium dihydrogen phosphate.

@Waylander I don't understand it is a proton donor yet not acidic?

I would not favor weak organic acids. In undissociated form they could get into the organic medium making extraction of pure priduct more difficult. An appropriate inorganic salt would avoid this.

@OscarLanzi  It depends on what you use. You are correct about acetic, but citric and tartaric acid can be useful as they complex metal ion residues.


text:
Meaning of red heat
I have two reactions below.

What is the difference between red heat and normal heat? Is it related to iron as a catalyst when it is red hot?

Red heat is normal heat, quite a bit hotter than 473K. It is not related to iron in any way.

All black bodies (at least the ones that don't decompose when hot) emit essentially the same radiation. The colour of that is a good indicator of the temperature. Red heat is the temperature where things glow red whether they are glass or steel or some other inert substance. And that is pretty hot (as the answer says 500-600 °C).

I removed second question that was homework question. read carefully.

We know that an object when heated upto a certain temperature emits radiation in form of light. This is called thermal radiation. The color of the radiation depends on the temperature and/or physical proerties of the object. The radiation obeys Wien's law. So, when said "red-hot", it means that the object is heated upto a certain temperature such that the object glows red.
After a quick Google search, I came across a reference that said "red-heat" is observed somewhere between 500-1000°C and is standardized somewhere at 555°C although this range varies according to property of object.
For a particular metal, range of red heat is:
\begin{array}{c|c}
\mathbf{Color} & \mathbf{Temperature(°C)} \\\hline
\text{Faint red } & \text{500}\\
\text{Blood Red} & \text{580} \\
\text{Dark Cherry } & \text{635}\\
\text{Medium Cherry } & \text{690}\\
\text{Cherry} & \text{745}\\
\text{Bright Cherry} & \text{790}\end{array}

What do you mean by rubber? Rubber most certainly does not survive even the lowest of the mentioned temperatures.

@IvanNeretin Normal rubber melt at lower temperature(180 C) but rubber tires do melt at even higher temperature(600  C). But for the benefit of doubt, I am removing the data for rubber but no to say at least that the data came from rubber industry and thus I interpreted the object to be rubber.

Thats the reason i asked red hot. How can crystals become red hot? Its why I thought of iron as a catalyst

Anything can become red hot, unless it decomposes before that.


text:
Attack of 3-methylbutan-1-ol on benzene in acidic medium

That's how I proceeded with the question. However the answer is tert-butylbenzene. I can't figure out why do I have an extra carbon atom.
SOURCE
- JEE Advanced 1994

Yes, your mechanism seems quite correct. I believe it must be a printing error

I agree. No way can you get t-Bu Benzene from this sytem.

I feel like dehydration of the alcohol is going to be much more likely under the forcing conditions whereby you might actually generate that primary cation...

Benzene, the nucleophile, attacks the alcohol - as the cation - which is the electrophile. ;)

The original question doesn't have an extra carbon.


text:
Why are hydrogen-bonded compounds (such as NH3) excellent solvents for most lab reagents?
Hydrogen bonds are said to be one of the strongest molecular interactions in nature. Then why can most substances that interact through hydrogen bondig such as water and $\ce{NH3}$ able to dissolve a large variety of both ionic and covalent laboratory reagents? Does it have something to do with the structure of $\ce{NH3}$ or something else?

Hydrogen bonds are the weakest interactions that still make directed bonds. Water is not "hydrogen-bonded", it makes hydrogen bonds to itself and to other substances, in 99% of the cases these are exactly those substances that mix well with or are soluble in water.

Who says they're "excellent"? Most compounds aren't particularly soluble in them or react.

@Mithoron Thank you for the point. On research, found that most existing compounds (by numbers) aren't soluble in solvents that interact through hydrogen-bonding. Edited the question to "laboratory reagents" from "compounds".

Recall that hydrogen bonding interactions result from the polarized nature of bonds between hydrogen and nitrogen, oxygen, and fluorine. This polarized bond results in a hydrogen atom with very low electron density—a property that allows the atom to interact very strongly with negatively charged particles.

Thus, anions and electronegative atoms in polarized molecules have very strong electrostatic interactions with the low-electron-density hydrogen atom in $\ce{NH3}$ or $\ce{H2O}$, for example. Cations and positively charged elements in polarized molecules can also be stabilized by the electronegative atoms in molecules exhibiting hydrogen bonding ($\ce{N}$ and $\ce{O}$ in this case).

These strong interparticular interactions ultimately result in polarized or ionic solutes generally being very soluble in solvents with hydrogen bonding such as $\ce{NH3}$ or $\ce{H2O}$.


text:
Phase abbreviations for non-aqueous solutions
When writing a chemical equation, how do you indicate that a chemical is dissolved in a non-aqueous solvent?
For example, it would be inappropriate to write
X (aq)
when X is dissolved in benzene, since aq stands for "aqueous," implying water soluble. A previous answer provided the potential abbreviation (sol), but I wonder if that is actually correct/commonly used.
How would you indicate that X is dissolved in a non-aqueous solvent?

In general I would specify X in benzene soln, there is no generally accepted notation for non-aqueous solution

Even if not all that common, (sol) stands a good chance of being understood correctly.

This is a matter of convention of course. I wonder what iupac has to say?

According to IUPAC, general notation for any solution would be $\ce{X(sln)}$ [1, p. 54]:

(vi) States of aggregation
The following one-, two- or three-letter symbols are used to represent the states of aggregation of chemical species [...] The letters are appended to the formula symbol in parentheses, and should be printed in Roman (upright) type without a full stop (period).
$$
\begin{array}{ll}
&\ldots\\
&\text{s} &\text{solid}\\
&\text{sln} &\text{solution}\\
&\text{vit} &\text{vitreous substance}\\
&\ldots\\
\end{array}
$$

Similarly, when denoting relation of a given quantity/symbol, subscript $\mathrm{sln}$ notation is used, e.g. $Δ_\mathrm{sln}G^\circ(T,p^\circ)$ [2, pp. 241, 276]:

4.2 Subscripts and superscripts
4.2.1 Subscripts
$$
\begin{array}{ll}
&\ldots\\
&\text{r} &\text{reference, reduced}\\
&\text{sln} &\text{solution}\\
&\text{s} &\text{saline solution}\\
&\ldots\\
\end{array}
$$

References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7. (PDF)
Gamsjäger, H.; Lorimer, J. W.; Scharlin, P.; Shaw, D. G. Glossary of Terms Related to Solubility (IUPAC Recommendations 2008). Pure and Applied Chemistry 2008, 80 (2), 233–276. DOI: 10.1351/pac200880020233. (Free Access)


text:
How do electrostatic analysers act as kinetic energy analysers?
Consider a mass spectrometer. The ions produced are accelerated across a potential V.
Now, the kinetic energy of an ion with charge q is given by qV. When it enters the analyser with electric field strength E (orthogonal to the trajectory), the radius of the trajectory is given by
r = 2K/qE, where K is the kinetic energy.
Substituting with qV, we get
r = 2V/E
The mass and charge terms disappear! So how exactly is this a kinetic energy analyser? Because as per this equation, all ions, irrespective of charge, mass and kinetic energy must have the same trajectory. Where is the flaw in my reasoning?

I think this question may be more appropriate at physics SE. In any case, off the cuff it seems your curvature calculation ignores that the velocity under E is changing ie Fperp = qE so a = dv/dt = (q/m) E while vpar is constant.

The electric field in an analyser is orthogonal to the trajectory.

@user1089 Related:http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/maspec.html#c2

I know, that's why I used labels perp and par. Label perp refers to motion orthogonal to V. The prep dimension is the resolving dimension. Note the q/m factor

@Try Hard I meant that the electric field is orthogonal throught out. It's a  radially symmetric field.

@but what about particles with same charge (it's not uncommon for most ions produced during mass spectroscopy to have a single positive charge). Under the same potential, wouldn't they have the same KE. Or am I forgetting to consider their initial velocities at the time of ionisation? Thanks, by the way, for everything so far.

@Try Hard makes sense

Nothing wrong with your equations, afaik. So perhaps there is something with the original motivation of the equation. Where did you find the description of the system?

'Mass Spectrometry' by Hoffman and Stroobant.

The point is that all ions produced in the ion source will not have a kinetic energy outside of the acceleration region equal to $q V$. When produced the ions will most probably be produced at (slightly) different positions in the acceleration field, with a small $\Delta V$ potential deviation from the average acceleration potential, and with a thermal kinetic energy distribution ($E_{kin,thermal}$).
In these conditions, the real kinetic energy of the ions as they exit the ion source will be slightly different:
$$E_{kin,real} = q V + q \Delta V + E_{kin,thermal}$$
and the purpose of the electrostatic analyser is to separate the ions depending on their actual kinetic energy and thus compensate for the distribution of kinetic energies of the ions exiting the ion source.

Thanks! I had suspected this to be the case, but since neither my textbook nor my professor had mentioned this, I was a little unsure.


text:
Using TBAI (tetra-butylammonium iodide) as a catalyst. Ideas on the mechanism or the driving force?
for my etherification of an alcoxide with alkylbromide I used TBAI as a catalyst. The catalyst transfers the alkylbromide into an alkyliodide which is more reactive.
I was wondering, what is the driving force for the Iodide to bond to the carbonchain instead of the bromide? Is there a mechanism to explain that?

Tell us more about the conditions you used

Classic conditions as usally described in literature. NaH in THF at room teperature, then add the alcohol (cooling because is quite exothermic) and TBAI, then add the alkylbromide. Stir overnight at RT.

iodide replacing bromide is usually driven by solubilities, the iodide is the more soluble is organic media

I was wondering, what is the driving force for the Iodide to bond to the carbonchain instead of the bromide?

It's not so much a driving force as much as it is a dynamic equilibrium of sN2/sN1 reactions. Halides can exchange each other in equilibrium if one is not removed thus processes that remove the unwanted product such as the Finkelstein reaction are use to achieve complete substitution (think Le'Chatlier's Principle).
$$\ce{R-Br +I- <=> R-I + Br-}\qquad\quad\ \ \  \text{(Equillibrium)}$$
$$\ce{R-Br +I- +Na+ <=> R-I + NaBr (s)}\qquad \text{(Finkelstein Reaction)}$$

Is there a mechanism to explain that?

Given the information you provided, Your reaction is proceeding as follows:
$$\ce{R-Br + I- -> R-I + Br-}\tag{1a}$$
$$\ce{R-I + R'-O- -> R-O-R' +I-}\tag{1b}$$
Where as the reaction:
$$\ce{R-Br + R'-O- -> R-O-R' +Br-}\tag{2}$$
cannot proceed or does not proceed at a fast enough rate and thus the catalyst can improve the reaction rate by first substituting the less reactive halide. The Wikipedia article on Williamson ether synthesis agrees with this:

However, if an unreactive alkylating agent is used (e.g. an alkyl chloride) then the rate of reaction can be greatly improved by the addition of a catalytic quantity of a soluble iodide salt (which undergoes halide exchange with the chloride to yield a much more reactive iodide, a variant of the Finkelstein reaction).

You don't form the iodide from all of the bromide, and then the iodide reacts.
The iodide is formed from the bromide and then consumed to form the bromide via the reverse reaction. The bromide is more stable, so there's more of it and probably very little of the iodide sitting around, especially if you use the TBAI in catalytic quantities.
This reaction likely is under conditions that could reasonably described as Curtin-Hammett. The iodide and bromide are in equilibrium and the more reactive species is the one that reacts, regardless of their relative concentrations.
This provides a faster overall pathway to the product by performing substitution on the iodide, but that doesn't mean that the iodide is more preferred overall.


text:
A liquid that forms a surface membrane that prevents evaporation but allows oxygen to pass through
I'm a biologist. I have a water solution with living cells in a tiny well (10 um). I need to keep the solution from evaporating for several days in a warm incubator. I'm looking for a liquid to place on the surface of the solution that will create a surface membrane/film that prevents or slows evaporation but continues to allow oxygen exchange. I welcome any other suggestions for achieving the same result, prevent/slow evaporation while allowing oxygen exchange.
Here are some things that we've tried and some constraints:

Increasing the humidity in the incubator is not sufficient (we've tried)
Lowering the temperature in the incubator is not an option (we want the cells to keep doing what they do)
Sealing the well is not an option (we need the solution to remain oxygenated)
What can be added to the solution is limited to substance we are sure won't react with the cells

Maybe a fluorosurfactant will work. Fluorocarbons provide high oxygen solubility but are very dense, whereas the surfactant should coat the surface. When you say you tried to increase of the humidity in the chamber what exactly did you attempt? If the atmosphere is saturated then any loss from your well means there is a sink somewhere else where water is condensing. Question is what is that sink. Also, there are humidity controlled incubators on the market.

An alternative is to increase the oxygen content in the chamber (a fire risk though) while narrowing the exposure of the water surface (for instance with a lid with a hole). But it becomes a balancing act because carbon dioxide may accumulate in the well, depending on how dense and active your cells are.

How about keep the solution in a closed container at $\pu{1 atm}$ $\ce{O2}$ pressure?

A ten µm droplet that is supposed to survive for days? Look for a poymer membrane with sufficient oxygen permeability and place it on top.The absolute amount of oxygen needed can't be too high in such a small system, I guess.

You should consider an non-ionic surfactant such as diethanolamine cocoamide or an ethoxylated alcohol.

Thank you for the responses and suggestions. It turns out that there exists a product for exactly this purpose. "Light Mineral Oil is a sterile light mineral oil intended for use as an overlay when culturing in reduced volumes of media to prevent evaporation, and to protect the media from changes in osmolality and pH."
https://en.wikipedia.org/wiki/Mineral_oil#Cell_culture

I don't think you or your cells want any strange low molecular weight substance in your experiment.
You can buy polymethylpentene (PMP) foil, this stuff has a rather low density compared to PE and other polyolefines, and therefore a high permeability for gases (10x higher than PE for oxygen, and PE is already not exactly tight against gases, x20 higher permeability than PVC for example). 50µm films are readily available, you find them online in a minute.

1) Octadecanol, a solid, waxy material, not soluble in water, has been scattered on reservoirs (in Australia, and probably other places) to inhibit evaporation of water. It's biodegradable, so it only lasts a few days.
Octadecanol has been emulsified with equal amounts of soap and diluted for application onto fresh concrete to reduce evaporation, but the extra water in the diluted spray may be part of its success.
2) Melting point capillary tubes are small in diameter (~1 mm), but long enough to reduce the rate of evaporation of a microdrop in the bottom. Water would diffuse out (not really evaporate as if in the open) while O2 will diffuse in. Immersing the melting point tube in a large beaker of water so its water level is almost at the top of the melting point tube could keep the humidity high at the mouth and seriously impede evaporation. The melting point tube would tend to float, so it will have to be weighted down. And maybe put a watch glass over the beaker to form a somewhat enclosed pocket in the incubator. Oxygen will still get in easily, but the humidity will have an actual high level rather than one which is reduced by leaks, insufficient evaporation from the water source, etc.
If this works somewhat, you can draw hot glass (or warm PE tubes) to very small diameters and very long lengths. And I can guess that there will be a lot of breakage!

Seal the well, add some phytoplankton and store with plenty of light. Or seal in a larger volume (1 ml) and hope the cells don't run out of oxygen.