Why is dialysis tubing impermeable to sucrose when it should retain >12,000 Da? While writing up a lab for biology, I realized that the dialysis tubing for sale online seemed to retain large molecules (12-14,000 Da) but didn't specify an upper bound. I learned that the tubing is impermeable to Sucrose, molecular weight 342 Da, because it's too big. This doesn't make sense to me; the site says the tubing "retains >12,000 Da molecules", but Sucrose is clearly smaller than that.

So either A) Sucrose does indeed pass through dialysis tubing and I misheard/mis-learned or B) my class used tubing with much smaller pores.

There exist many types of semi-permeable membranes (the ones used for dialysis tubing), with various pore sizes. One of the very common lab experiment on the topic of osmosis is using a sucrose solution (sucrose is cheap) and small-pores membrane, such that water and small ions (typically Na+ and Cl–) can pass, but not sucrose. The one which you link to simply happens to be a variety with much wider pores.


Is Sodium Hydroxide a Bronsted-Lowry Base? I know that sodium hydroxide dissociates into sodium cation and hydroxide, so it must be an Arrhenius base. I am also aware of the rule that all Arrhenius acids and bases are considered to be Brønsted-Lowry acids and bases. However, when it comes to sodium hydroxide itself, it seems as if it does not accept proton, but hydroxide does. And someone on this StackExchange stated that it is not a Brønsted-Lowry base, but an Arrhenius base. I am confused now. Is this an exception?

All Arrhenius acids and bases are also Bronsted-Lowry bases.
According to this website,

  The Bronsted-Lowry theory doesn't go against the Arrhenius theory in any way - it just adds to it.

Remember, when we write NaOH (aq), what we really mean is Na+ (aq) and OH- (aq). Sodium hydroxide is a Bronsted-Lowry base because the hydroxide ions that are a part of sodium hydroxide accept protons.


Dissolving/denaturing bitterant residue In what seemed like a good idea at the time, I used a can of compressed air to clean out my Magic Mill flour grinder.

Picture of one: http://ic.pics.livejournal.com/ccdesan/4606041/68177/68177_320.jpg

It did a great job of cleaning out all the extra flour stuck to the bottom of the top part. However, what I didn't realize at the time was it also left bitterant residue, ruining a batch of flour I just made!

The MSDS for the compressed air (Compucessory Power Duster) just lists diflouroethane, but it says right on the can it contains a bitterant!

The bitterant is likely Denatonium (http://en.wikipedia.org/wiki/Bitterant). Is there any household chemical that will dissolve it and the other common bitterants, or better yet, break them down into non-bitter compounds?

Since probably only ppm levels of denatonium benzoate are in the compressed air, there are ppm levels or lower of denatonium in your flour grinder.  At that concentration, water should dissolve denatonium benzoate.  Is that a viable way to clean your grinder?

According to this solubility table, the best solvent for denatonium benzoate is methanol.  However I would not recommend using methanol on anything you are using for food.  Ethanol is only slightly less good.  Maybe you can wash your grinder in some cheap vodka a few times?

Here is a random thread from Google Groups where someone claims to have identified a vendor for non-bitterized air duster cans.


Does salt on the snow and ice covered roads and sidewalks make the nearby air temperature cooler? I  noticed when it snows and salt is melting the snow on the roads in the city, that the air temperature feels a lot cooler than the temperature indicated on the thermometer.
A 40 degree fahrenheit temperature feels like 30 degrees. Is it the moisture from melting snow or rain that makes the air temp feel ten degrees cooler or is it the salt brine (melted snow or ice  and salt?  Is it similar to adding salt  to the ice in an Hand powered ice cream churn to make the cream container colder to make ice cream?

As water melts due to adding a salt (AKA melting point depression), it absorbs energy from the environment and the temperature of the salt/ice/water mixture does, indeed, drop. Whether that is sufficient to cool the air above the ice perceptibly, though, is a matter of opinion.

Another effect that might make the air feel colder is increased humidity, particularly as the salt slush is splashed into the air by vehicles.

You could perform an experiment yourself to determine the air temperature above two containers, one with fresh snow and the other with snow and salt. If you do, please post your results as an answer!


What is the nature of the mysterious Kw, the ionic product of water? In pure water, [H$^+$] = 10$^{-7}$ M and [OH$^-$] = 10$^{-7}$ M. So, we conclude pH of water is 7 and so is the pOH. We define the ionic product of water as Kw = [H$^+$][OH$^-$] = 10$^{-14}$ M$^{2}$. Till this, everything is fine.

Now, without any premise, it is concluded that the product [H$^+$][OH$^-$] of all concentrations in solutions of all sorts is the same, i.e. 10$^{-14}$.

This leads to the conclusion that for all acidic, basic and neutral solutions, pH + pOH = 14.

So, why is it concluded that [H$^+$][OH$^-$] = 10$^{-14}$ M$^2$? What is the nature of Kw, the ionic product of water, that mysteriously is applicable to all solutions?

The solution product for the reaction $\ce{H2O -> H+ + OH-}$ is a constant, $10^{-14}$, which we call $K_w = [\ce{H+}][\ce{OH-}]$. To make any aqueous solution, we put a solute into water. Because the dissociation constant (which is basically an equilibrium constant) of a substance only changes with temperature, $K_w$ remains the same. Thus, for any aqueous solution at a given temperature, $K_w$ is constant.

You need to treat $K_w$ as an equilibrium constant. For example, if I have a reaction $\ce{A -> 2B + C}$, $K_{eq} = \frac{[\ce{B}]^2[\ce{C}]}{[\ce{A}]}$; let's say I measured at equilibrium $[\ce{B}] = 1$ M, $[\ce{C}] = 2$ M, and $[\ce{A}] = 0.1$ M . This means $K_{eq} = 20$. Now, if I increased the concentration of $[\ce{A}]$ to $5$ M, $K_{eq}$ wouldn't change: the concentrations would have to adapt to my change. This is the same for water: once I determine $K_w$, it is the same for all concentrations of $\ce{H+}$ and $\ce{OH-}$.


Fehling's solution turning yellow and then black? While testing for formaldehyde (an aldehyde), I added Fehling's solution to the sample and let the test tube simmer in a water bath for some time. I returned ~15 minutes later to see a black/dark grey precipitate awaiting me, which looked very metallic (Almost like Tollen's test, but darker)

I try again and this time heat the tube for only five minutes: I obtain a buff yellow coloured precipitate which, on standing, gives me the same metallic precipitate I observed earlier.

I expected that Fehling's solution will give a red precipitate of $\ce{Cu2O}$ but it looks like the formaldehyde seems to be oxidizing it all the way down to metallic copper. Also, what is the reason for a yellow precipitate to form rather than a red one?

I could not take any pictures: We're not allowed to carry phones/cameras into the lab.

In the reaction with aldehydes, Fehling's solution produces $\ce{Cu2O}$ as an orange or yellow precipitate. I don't know why the color is not always the same. The oxidation of an aldehyde produces an acid. And usually an acid cannot be more oxidized. There is one exception : methanal $\ce{HCHO}$. Because it can be oxidized in formic acid $\ce{HCOOH}$, and then into $\ce{CO2}$ . That is why Fehling's reagent produce metallic copper $\ce{Cu}$, $\ce{CO2}$ and no acid when reacting with $\ce{HCHO}$.


Reason for usage of Metal chlorides and Platinum or Nichrome wires in flame test I'm studying about the s-block elements in my course of Inorganic Chemistry and I happened to come across the following text.

  Electrons may be quite readily excited to a higher energy level, for example in the flame test. To perform this test, a sample of Metal Chloride or any salt of metal moistened with conc. HCl, is heated on a platinum or nichrome wire in a Bunsen burner flame.

Why are metal chlorides specifically used and also what is the role of the Plantinum or Nichrome wire?

Metal chlorides are usually much easier volatilized than oxides or sulfates. They melt or sublime at temperatures much lower than 800°C. So they are volatilized in any flame like Bunsen burners.

Platinum is not attacked by hot chlorides or acids, and is not oxidized in hot air. So it can be used for a great number of consecutive tests in the flame. It will never wear out in the flame.


What ion remains in the largest concentration at equilibrium: 0.1 M H₃PO₄ and 0.2 M KOH? Mix equal volumes of $0.1~\mathrm{M}$ $\ce{H3PO4}$ and $0.2~\mathrm{M}$ $\ce{KOH}$. What ion remains in the largest concentration at equilibrium?

My approach: suppose we mix $1~\mathrm{L}$ of each solution. For $\ce{H3PO4}$, $K_{a_1}=7.5\cdot10^{-3}$ and $K_{a_2}=6.2\cdot10^{-8}$, so in terms of the dissociation of $\ce{H3PO4}$, the only ions produced in non-neglible quantities are $\ce{H^+}$ and $\ce{H2PO4-}$. Furthermore the quantities of $\ce{H^+}$ and of $\ce{H2PO4-}$ are both less than one mole, since there is less than 1 mole $\ce{H3PO4}$. We have 2 moles $\ce{OH^-}$ from the $\ce{KOH}$, but since of the $\ce{OH^-}$ neutralizes the $\ce{H^+}$, in particular less than a mole $\ce{OH^-}$ is used in this neutralization reaction. Thus at least one mole $\ce{OH^-}$ remains, at most 1 mole $\ce{H2PO4-}$ remains, and all other ions remain in negligible quantities, so the answer is $\ce{OH^-}$.

Apparently, this is wrong - the answer is $\ce{HPO4^2-}$. I don't see how this is possible because we already established that the amount of $\ce{HPO4^2-}$ formed is negligible from the $K_a$ values, and since we are adding a strong base to a weak acid we know the resulting solution will be basic, so we should expect from the beginning that $\ce{OH^-}$ appears in large concentration.

I don't understand what's wrong with my reasoning?

Actually, the ion with the largest concentration is $\ce {K+}$.

But that is not what they want. Since potassium hydroxide is a strong base, then the concentration of $\ce {OH-}$ starts with 0.2 M (roughly pH of 13.3). To neutralize this, it would take 2 hydrogens from $\ce {H3PO4}$ at the specified concentration to get both concentrations near $10^{-7}$ (pH ~7). So the major anion is $\ce {HPO4^{2-}}$.

You can find the exact concentrations if you need.

Addendum:
Since $K_w = 10^{-14}$, then $\ce {[H^+][OH^{-}]=10^{-14}}$
Adding 1 L of each solution would give 0.2 moles of $\ce {OH^-}$ to react with any liberated hydrogen from the phosphoric acid. Removing one hydrogen from all the phosphoric acid gives 0.1 mole of $\ce {H+}$, which reacts with the hydroxyl, leaving 0.1 mole of $\ce {OH^-}$ (pH of 13.0). Since this is smaller than ${K_{a2}}$, more hydrogen is released from $\ce {H_2PO_4^{-}}$.

Assuming 0.2 moles of hydrogen is released, then there is an equal amount of $\ce {H+}$ and $\ce {OH^-}$ in the solution. To have $K_w = 10^{-14}$, then each of the ions needs to be near $10^{-7}$M.

Calculations:
Find $\ce {[H_2PO_4^{-}]}$:
$$\ce {H+ + HPO4^{2-} <=> H_2PO_4^{-}}$$
$K_{a1}=\frac {[H_2PO_4^{-}]}{[H^+][HPO_4^{2-}]}$ where $\ce {[H+]}=10^{-7}$M, and $\ce {[HPO_4^{2-}]}=0.05$M.
$7.5\times 10^{-3}=\frac {[H_2PO_4^{-}]}{10^{-7}\times 0.05}$
$[H_2PO_4^{-}]={10^{-7}\times 0.05}\times {7.5\times 10^{-3}}$
$[H_2PO_4^{-}]=3.75\times 10^{-10}$M

Find $\ce {[PO_4^{3-}]}$:
$$\ce {HPO4^{2-} <=> H^+ + PO_4^{3-}}$$
$K_{a3}=\frac {[H^+][PO_4^{3-}]}{[HPO_4^{2-}]}$ where $\ce {[H+]}=10^{-7}$M, and $\ce {[HPO_4^{2-}]}=0.05$M.
$4.5\times 10^{-13}=\frac {10^{-7}\times [PO_4^{3-}]}{0.05}$
$[PO_4^{3-}]=\frac {0.05\times 4.5\times 10^{-13}}{10^{-7}}$
$[PO_4^{3-}]=2.25\times 10^{-7}$


Bohr's model of atom "the most remarkable success of the Bohr's theory is that it provides a satisfactory explanation for the line spectrum of hydrogen."

But what is the problem with line spectrum of hydrogen which could not explained before the Bohr's theory was introduced and how it was explained by Bohr's theory?

Since this is pretty well covered in countless number of books and other sources I will try to just emphasise the key points.

The emission spectrum of atomic hydrogen was well known at that time, the only problem was that no one was able to explain it. :D Where does all this lines come from, i.e. what physical process gives rise to such discrete spectrum?
The wavelengths of spectral lines were known to be very well predicted by the Rydberg formula, but again no one was able to explain why this formula works, i.e. what are its physical justifications.
Bohr model explained discrete spectrum as follows: spectral lines are due to electron transitions between stationary orbits. Bohr was also able to determine the actual energy spacing between levels in hydrogen atom and explain the reason for the structure of the Rydberg formula.
However, Bohr failed to predict the lines in the emission spectrum even for helium atom. It happened simply because Bohr model is physically  wrong. Electrons do not orbit around the nucleus.


How does the size of a molecule affect acidity? Today I learned that $\ce{H2S}$ is more acidic than $\ce{H2O}$ because $\ce{H2S}$ has a larger molecule size. But $\ce{Fe^3+}$ is more acidic than $\ce{Fe^2+}$ and $\ce{Fe^3+}$ is smaller than $\ce{Fe^2+}$.
Likewise, formic acid is more acidic than acetic acid but formic acid is smaller than acetic acid.

How does this go together?

It is key to learn about the applicability of certain concepts — or why does a concept work in a certain case — along with the concepts themselves. Most notably, atom size (it should not be molecule size, but for $\ce{H2O/H2S}$ it does not matter due to hydrogen’s tinyness) is only of limited help in a few select cases to determine acidity.

When comparing the acidity of two compounds you should always actually look at the following equation and then compare the anionic products of two different acids:

$$\ce{HA + H2O <=> A- + H3O+}\tag{1}$$

Whichever generates the more stable anion is the stronger acid. Thence, a number of concepts have been developed to determine the more stable anion from a given list of alternatives — one of them is the atomic size. It is useful only when directly comparing two otherwise group-identical species — like oxygen and sulfur which are in the same group but different periods or chlorine and iodine. Basically, under otherwise identical conditions (same valency and same charge) the larger anion is able to stabilise the negative charge better. Thus, $\ce{H2S}$ is a stronger acid than $\ce{H2O}$ and $\ce{HI}$ is stronger than $\ce{HCl}$ but we cannot say anything about the difference between $\ce{H2S}$ and $\ce{HCl}$ or even $\ce{H2S}$ and $\ce{HF}$ from that.

All the other pairs you listed do not fit into that difference concept at all. The acidity of $\ce{Fe^n+}$ is not explained simply by looking at the metal ion. And the difference between acetic and formic acid is on a much more subtle scale since both are oxyanions. To reiterate: do not only learn concepts but learn what concepts are for.


Physical intuition behind negative values for wave function? So a positive and a positive wave function create a bonding orbital where the probability of finding an electron is summed while a positive and a negative create an anti-bonding orbital with a lower electron probability in the region between them leading to a repulsion. My confusion stems from not having any idea as to what a negative wave function is representing - can anyone give me some physical intuition on how this negative wave function could be correlated to something in reality?

(I tried asking this question on physics.SE but it would seem that in physics you just don't talk about or use the wave function until after you square it)

The wavefunction of a particle actually has no physical interpretation to it until an operator is applied to it such as the Hamiltonian operator, or if you square it which gives its probability of being at a certain place. So having a negative wavefunction doesn't mean anything physically. However, let's say for a particle in a box, if you solve the momentum operator for the $n^\text{th}$ stationary state, you would get two solutions: $+\hbar k$ and $-\hbar k$, where $k = n\pi/L$. The plus and minus signs mean that there is a $50$% chance the particle could be moving from the left to the right and a $50$% chance of the particle moving from right to left.

In the case of summing of two wavefunctions to get the wavefunction of a molecular orbital, a negative wavefunction doesn't mean anything at all. The reason why when a positive and negative wavefunction is added, an antibonding orbital is formed is simply because when you add a positive and negative number, you get $0$ or a really small number. This is known as destructive interference.  Therefore the new wavefunction will contain a region where the y values equal $0$. So when you square this new wavefunction of the molecular orbital you are going to get that there is very 0 probability of finding the electron in that position (known as a nodal plane), and hence it is called an antibonding orbital.However note that just because there is negative region in the wavefunction, doesn't mean that it will always add to form an anti-bonding MO. For example, if you add 2 negative wavefunction values (known as constructive interference), you get a large negative value. When you square becomes a very large positive value. Hence this means that there is a large probability of the electron being in that region and hence it will be a bonding MO.


Why is potassium ferrocyanide considered safe for consumption, when it is just one reaction away from the highly toxic potassium cyanide? According to this answer, heating potassium ferrocyanide will decompose it into potassium cyanide.

Now, potassium ferrocyanide is an approved anti-caking agent, and hence used in table salt, for example.

Table salt is often heated as part of cooking. So, isn't there a chance for trace quantities of potassium cyanide entering the body?

NB: I understand that the dosage here, if any, might not be lethal; am more concerned about trace amounts causing inefficiencies in bodily functions.

Under biological conditions it is almost impossible to release HCN

Free cyanide can be released from potassium ferrocyanide by heating or by strongly acidic conditions (and some heat). Neither of these conditions is possible in a living organism.

The reason why this is so is because the cyanide ligands are very strongly bound to the iron (in slightly more technical terms the dissociation constant is very, very small (ca 10-24)). Most of the mechanisms that generate cyanide require much more violent chemical attack: waiting for the cyanide to dissociate and then protonating it to give deadly HCN will never result in a notable amount of NCH. This means that, in biological systems, ferrocyanide is effectively inert.

This explains why the LD50 in mice, for example, is about 6g/kg which is a lot ([see the safety info here]2 for both ferri- and ferrocyanides).

It is also worth noting that food safety authorities have studied the compounds and agree that:

  The Panel concluded that ferrocyanides (E 535–538) are of no safety concern in these current authorised use and use levels.

So, while it might look like you are "one reaction away from cyanide" in practical terms this is not remotely true.


Difference between subshell and sub-subshell? I consider the s p d f orbitals as a subshell, but I have heard about the term sub-subshell. what is the difference between these two?

As you pointed out the the s, p, d, and f denoted subshells of atoms.

The term sub-subshells is very uncommon. I did find a couple of uses.

Materials Science
By G. K. Narula, K. S. Narula, V. K. Gupta

pdf file about Modern Atomic Mechanical Theory, page 24

What the authors are noting is that the 2p subshell will have three orbitals which can support two electrons each. Each of these three orbitals is being referred to as a sub-subshell.

As I said it seems a very uncommon term, and I'd suggest you avoid using it.


What is the reason for including excited states in the wave function in post-Hartree Fock methods? I'm new to computational chemistry, so perhaps this question has an obvious answer. I'm wondering what the intuitive reason for constructing Slater determinants that involve excited states is, when one tries to describe correlations, starting from the Hartree-Fock picture. I understand that the single determinant in the Hartree-Fock model only describes exchange (antisymmetry of fermionic wave functions).
I am trying to understand the logic behind why constructing Slater determinants involving excited states can give us a picture of correlation. I guess I don't understand the relation between excited states and correlations (especially thinking about it from the point of view of DFT, which is interested only in the ground state, but correlations are still important).

I think that here it is not accurate to say "starting from the Hartree-Fock picture", instead of it, I find better to say "starting from the Hartree-Fock method". When this method is carried out we get a lot of orbitals and that allow us to construct a lot of Slater determinants, even if, we can be only interested in one of them.

Given a complete set of functions $\{f_i(x)\}$, any function $g(x)$ can be thought as a linear combination
$$g(x) = \sum c_i f_i(x).$$

For a two variable function $g(x_1, x_2)$, fixing $x_2$ we have
$$g(x_1, x_2) = \sum c_i(x_2)f_i(x_1).$$
But as $\{f\}$ is a complete set:
\begin{align}
c_i(x_2) &= \sum d_{ij} f_j(x_2) &\text{and}&&
g(x_1,x_2) &= \sum d_{ij}f_i(x_1)f_j(x_2).
\end{align}

This reasoning is trivially extended to a function of $n$ variables.

For an $n$-fermions system, the wave function depends of $n$ spin-orbitals ($n$ variables). To get a mathematic valid form, we only need to antisymmetrize the generalization of the expressions above.

Take in mind that none of these determinants are the true state (they are not eigenfunctions of the true hamiltonian). The same happens with the orbitals (that are eigenvectors of the Fock operator).

So, the use of excited determinants arises naturally from a math view point. Just because we already have functions for the determinantal expansion, and we know that they are pretty good in some sense, because the are the best we can get at least for one determinant (the Hartree Fock solution).

It is not needed to think configuration interaction as a mixture of stationary quantum states (ground and excited states), as said above they truly aren't.

Direct to your question: We can think the problem just as a decomposition of the wave function, somewhat like Fourier series, where the functions involved are the Slater determinants, just because it is convenient. But, this doesn't imply a physical meaning.

Response to comment:

  Also this discussion is quite old, I would like to add that the term "picture change" is frequently used in relativistic quantum chemistry, when (effective) Hamiltonians are compared to each other. E.g., when approximating the four-component (4c) Dirac equation by an effective two-component (2c) formalism, one usually introduces further approximations. In this respect, "starting from the Hartree-Fock picture" is a correct terminology, as it implies certain approximations on the Hamiltonian. However, the better way is usually to end up with a certain picture, compared to starting from it... (TheFox)

I agree: By choosing an effective Hamiltonian one is working on a class of approximations that build up a picture for a set of analogous systems. When considering correlation we have in mind a "picture" of the system. Being that H-F is a mean field method, it belong to some picture, but I feel that when dealing with Coupled Cluster the H-F method is just a part of a bigger procedure consisting in finding some functions in order to obtain a solution for another picture. In such sense, while the H-F method itself can be understood as the method for obtaining a solution of the H-F picture, in this context is just part of method for solving a problem in another picture. I added the word here for strictness.


In a battery what would happen if you had a perfectly uniform and pure cathode? I have been watching some videos on how batteries work and they seem to talk about the cathode being the most important. I also watched another video a long time ago and while my memory is blurred it talked about impurities in the cathode being the only reason the electrical charge doesn't flow constantly. While I found this weird I also wonder how generates seem to get this unlimited supply of electrons also so it's only fair for me to assume that the battery can also.

I don't think that's accurate. Even the universe doesn't have an "unlimited supply of electrons".
Impurities in the cathode likely give rise to internal resistance inside the battery (among other things), leading to deviation from an ideal cell. I would guess that as your cathode approached purity, your battery would approach the behavior of 'ideal cells', and that discharge rates would approach those of capacitors, and efficiency would approach 100%. To get an ideal cell though, the anode and electrolyte would also have to be 'pure' and some real-world effects would have to be ignored.

Because of internal resistance, the terminal voltage of a cell that is discharging is smaller in magnitude than the open-circuit voltage and the terminal voltage of a cell that is charging exceeds the open-circuit voltage. An ideal cell has negligible internal resistance, so it would maintain a constant terminal voltage of $ \epsilon $ until exhausted, then dropping to zero. If such a cell maintained 1.5 volts and stored a charge of one coulomb then on complete discharge it would perform 1.5 joule of work. In actual cells, the internal resistance increases under discharge
Wikipedia

What do you mean by pure, exactly? Complete homogeneity and lack of any other elements? That sounds like a physics problem where you "ignore air resistence" and other real world effects for simplicity. If this is what you're after, search for things like 'Ideal Battery' and 'Ideal Electrochemical Cell'.


Each electron shell can hold up to 2n² electrons. Is there a geometric reason for this? It seems such a simple formula, that each electron shell can hold in principle two times a square number. I imagine there must be a geometrical reason for this, that's to do with the way electron orbitals "fit" around the nucleus. If this is so, what is it?

A good insight!

The number of electrons in each orbital are double the odd numbers. The s orbital can hold 2, p holds 6, d holds 10 and f has 14: double 1, 3, 5 and 7. If you add the odd numbers, you get a square number:

1 = 1

1 + 3 = 4

1 + 3 + 5 = 9

1 + 3 + 5 + 7 = 16

The doubling is due to the Pauli exclusion principle; in each orbital, an electron of a specific energy can be in one of two spin states.

It is interesting to see spectral lines split in a magnetic field (Zeeman effect) or an electrostatic field (Stark effect): one line may become 3, 5 or 7 lines. It clearly shows how orbital theory was derived.


How many hydrogen bonds can methanol make? I believe that it can form 2 hydrogen bonds, one with oxygen and the other with hydrogen but a classmate said that oxygen can form 2 hydrogen bonds? Is this true? I can't find anything on google to support or refute that claim.

Although this was an MD study (and might be said to be somewhat dated), quoting from Matsumoto and Gubbins (Ref. 1):

  Most of the molecules are in the two-bonding state, but a
  small fraction of molecules are in the one- or three-bonding
  state, as already pointed out in previous studies. 7
  ,8

According to this study most of the methanol molecules form 2 H-bonds but ~4-7% form 3.

References

Matsumoto, M., Gubbins, KE. The Journal of Chemical Physics 93(3):1981-1994 (doi 10.1063/1.459076).


Why is +M effect of OH greater than OR? In some books it is written that +M  (mesomeric effect)  effect of OH is less than that of OR . The reason they give is inductive effect of R (group) which sounds senseful . But at some places it is written +M of OR is less (which is correct). But I am not able to understand the reason which they give.

First of all, you must understand that the reactivity of OR changes as you change the alkyl group (-R).

For example, consider, the methoxy group (-OCH3), as per proton NMR spectra of anisole and phenol, the reactivity is almost similar in both the cases. In fact the reactivity is so similar that undergraduate texts just mention the fact that anisole has reactivity similar to phenol. (Source 1)

On the other hand, if you consider more bulky alkyl groups, steric hindrance might be a contributing factor to reactivity, competing with the mesomeric effect. But to a large extent, it is sufficient to say that both hydroxyl and (-OR) groups have similar reactivity.


How are octet-rule models supported by thermodynamics and quantum theory? I understand that the four quantum numbers dictate how many electrons can "fit in a shell," but I don't understand how gaining a full octet is particularly energetically favorable.

The example of the ionization of sodium seems to make sense: the atom loses a single 3s electron, and the energy of the system is lowered significantly down to the 2p energy level.

However, the ionization of chlorine isn't as obvious to me: it is indeed granted its "full octet," but filling the last available space in the 3p suborbital would not lower the energy of the system. Rather, it would raise it slightly, as the newly added electron would experience some electron/electron repulsion.

Is the main drive for this ionization "reaction" the drop in energy in the formation of the cation? Do I have to consider molecular orbital theory to answer this question for covalent interactions?

I'm mainly wondering why there's a drive to form a full octet, particularly for the halogens, chalcogens, and pnictogens.

You can't really compare the energy of $\ce{Cl-}$ to $\ce{Cl.}$, since the two states have different composition. Instead, you're interested in the difference in energy between $\ce{Cl-}$ and $\ce{Cl. + e-}$, where the electron is completely free and resting in a vacuum with no interactions with anything else. We give that electron a reference potential energy of 0. Then the question becomes, how much potential energy is lost by interacting with various atoms, and the thermodynamically favorable state is the one with the most negative potential energy.

If we consider nuclei to be point positive charges and electrons to be negative charge distributed in a much wider space around the nucleus, we find that as that free electron gets closer to the nucleus, it experiences an electrostatic interaction with the positively charged nucleus even if the atom overall is neutral. We can simplistically estimate the magnitude of the potential energy loss based on basic principles of electrostatics, where potential energy is a function of $\frac{q^+q^-}{r}$.

Where molecular orbital theory comes in is when we calculate the charge and the distance r. Simplistically, orbitals that are higher in energy on the same atom are higher in energy because they have a longer average distance from the nucleus (denominator in PE term is bigger) and because there is more electron density from other electrons between them and the nucleus, so the "effective" positive charge of the nucleus is reduced (numerator in PE is smaller). This is called "shielding".

Slater's rules are simple heuristics that provide a fairly good approximation of the effective nuclear charge experienced by an electron based on the atomic number and orbital.

Conceptually, then, the answer to your question is that adding an electron to a neutral atom like $\ce{Cl.}$ is most favorable when the electron experiences the best combination of effective nuclear charge and short average distance to the nucleus. This happens when adding the last electron to an octet because that is the element of that period with the greatest nuclear charge. Adding an electron to the next neutral element Argon is much less favorable because it has to go in the next shell (n=4) meaning that it has a larger r (larger denominator) and has much more electron density between it and the nucleus, so the effective nuclear charge is less (smaller numerator) even though the atomic number of Ar is one more than Cl.

For $\ce{Na+}$, it isn't correct to say that the system has lower energy after an electron is removed, but rather that it takes a lot less energy to remove that electron than an electron from neutral neon, which hopefully makes sense based on the discussion above.


What governs the merging of globs of inmiscible liquid? If I add a small amount of chloroform to a bunch of water, it falls to the bottom as a glob.

If I shake the container vigorously (being carefully not to spill any of the dangerous substance!), the globs of chloroform break apart into many tiny globules. These globules immediately sink to the bottom and collect together by the action of gravity the moment I stop agitating my mixture.

If I watch this pile of globules for some time, I observe that they tend to merge after several seconds, perhaps even a minute or more for some. I suspect that if I reached in with a needle and prodded the globules, they would merge faster. Also, I suspect that if I were to tabulate how long each individual globule survives before merging, I would obtain a half-life curve.

What governs the merging of these globules (I'm guessing surface tension)? Why don't they instantly merge? Why do a few little globules sometimes persist for many minutes, stuck to the side of a much larger globule?

Emulsion chemistry is actually rather complicated, but I'll try to give a conceptual description. Whenever you have a surface dividing two phases, there is an energy associated with creating a certain amount of surface, expressed as a surface energy density with dimensions of energy divided by area. This represents the energy cost of disrupting the intermolecular interactions of each phase at the interface, e.g. for your water/chloroform system, water can't hydrogen bond across the interface so it takes energy to break all the bonds required to make the surface. The larger the surface area, the more energy it takes. Logically, you can see that the lowest energy state is the one in which the interfacial area is minimized, which is why in the absence of external forces like gravity, liquids assume spherical shapes which give the smallest surface area for a given volume.

When you have many small droplets, there is a large surface area to volume ratio which is a high energy state, so it is energetically favourable for the droplets to form larger droplets. However, the reason it doesn't occur instantaneously is that an energy barrier can exist between the two states, for various reasons (electrostatic repulsion is a big one). In most cases, smaller droplets are lost first because the energy barrier to coalescing is surface area dependant. Larger droplets are more stable. Depending on the liquids, different separation mechanisms can occur with different kinetics. (coalescence, Ostwald ripening) I can't give a definite answer, but I'm guessing in your system you would not see an exponential decay-type half-life as that assumes a stochastic process where the probability is always the same. Since the stability depends on surface area, this isn't true in this case.


How do I find the thermal effect of this reaction? I have graphite turning into diamond.I need to find the thermal effect of this reaction.How would one write that?I'm not sure how to proceed here.The teacher told us to use oxygen?

Look at the difference between the enthalpies of formation. Not sure why you'd need oxygen, unless there's something I'm missing.


Dissolving silver braze from tungsten carbide I'm working out the best process to chemically strip silver brazing from tungsten carbide (WC) bits. My main objective is to recover the silver as a relatively pure metal. When I use $\ce{HNO3}$ at anywhere from 20-70%, it dissolves the solder and I end up with a red solution.  When I try to use cementation (putting copper metal in the solution) to precipitate the $\ce{Ag}$ as sponge, it isn't working.

Can any one help me figure out what's going on here?  The brazing being used should be 50-60% $\ce{Ag}$ by weight. I'm not sure yet if there is $\ce{Pd}$ or not.

Most common silver brazes are 50 to 90 % silver with copper or maybe zinc.  A complication may be the WC; it is sintered with 2 to 20 % cobalt ( usually).


Is there a list of must-have columns for LC-MS/MS used in biochemistry/clinical lab setting? Considering a wide variety of chromatographic columns used for (U)HPLC based on

phase type (normal or reverse);
column dimensions (geometry, length, inner diameter);
particle material $(\ce{SiO2},$ $\ce{ZrO2},$ $\ce{TiO2},\ldots),$ size, shape and porosity;
dopants/modifiers etc.,

as well as proprietary manufacturing techniques involved, is there a narrowed-down list of the bare-minimum column types that every analytical lab doing specifically liquid chromatography tandem mass spectrometry must have to cover the majority of the analytical routines (qualitative and quantitative analysis of short-chain fatty acids, primary metabolites, amino acids etc.)?

Background: a new LC-MS/MS lab is being created within a university's biochemistry division from the ground up both for the fundamental research and providing analytical services, so listing the most versatile and recent columns would be appreciated.
I do realize the question is borderline off-topic being somewhat opinion-based, so it would be nice if the answer were based on a trustworthy source.

This answer will change fairly frequently because technologies are always improving.  These days, I think a bare minimum list might be something like:

a 2.1 mm × 50 mm C18 column for rapid reversed-phase methods.
a 2.1 mm × 150 mm C18 column for slower, broader-coverage reversed phase methods.
a 2.1 mm × 50 mm HILIC column for rapid methods to detect polar metabolites.
a 2.1 mm × 150 mm HILIC column for slower, broader-coverage methods to detect polar metabolites.

Some further notes:

If your lab is going to use ion-pairing based methods, you will probably want separate columns for ion-pairing methods and "clean" methods without ion-pairing reagent.  Ion-pairing agents like tributylamine and/or perfluoroheptanoic acid are very difficult/impossible to remove from columns once they have been used for this purpose.
Fairly soon the best answer might be to use 1.0 mm diameter columns instead of the "standard" 2.1 mm diameter that is most common nowadays.
The most widely used HILIC column seems to the "BEH Amide" column chemistry from  Waters.  Following close behind may be ZIC-HILIC from SeQuant and/or Agilent's HILIC-Z chemistry.
For C18 chromatography, performance differences between different vendors are comparatively minor (although real).  A wide variety of chemistries are used.


House-hold chemistry experiments for a high-school student I want to experiment the things which we learn in chemistry as a twelfth grader.I want to do them at my home but I'm not supplied with enough reactants.What can I do by using such reactants.

I tried to carry out paper chromatography at home and that was the separation of a mixture of inks. I didn't have  filter paper so I used a tracing paper.
My solvent was a common organic solvent. But to my dismay,solvent front didn't rise so high. So my experiment didn't succeeded but I'm hopeful I can do other things too which'd be successful.
It was first experiment

Can someone recommend a few projects/experiments (or sources that do the same) for a school student who's looking to improve her understanding of basic chemistry?

My favorite "textbook" on home-chemistry is the Golden Book of Chemistry Experiments (Robert Brent).

It's a neat little book, and since it was published so long ago (1960s) it's not really available as a (literal) book these days. But fear not, because the PDF version can be easily found online!

A reliable place to download it from is archives.org.

Small warning:

I wouldn't recommend this to young children because quite a few experiments/projects mentioned there make use of an open flame (that's pretty much the only reason...). So be sure to conduct them in an open space far from anything inflammable.

Maybe this is the book/resource you're looking for? ;)

(A spot of light reading)

However, as the gents in the comments section have pointed out, conducting home-experiments won't (necessarily) make you a "Chemist". But I highly recommend you experiment (safety is important too) as frequent as you can, because over time it can firmly root a scientific bent of mind in you.

Intelligence, inquisitiveness and experience (and empathy) are highly prized values in the scientific world.

Besides, as you experiment at home (while you are still a high-school student) you'll soon stumble upon an extremely important idea:

No real experiment can be perfectly replicated

And you can only come to truly appreciate this fact when you experiment yourself...a lot.

No matter what experiment you do, no matter how perfect your set-up or procedure is, every time you conduct the same experiment you will never get the exact same result. Right now this "idea" may not seem like much (as a high-schooler, you might initially find this unbelievable), but its implications are profound and far-reaching.

That the fact that replicating experiments perfectly is impossible will soon lead you to another piece of insight: Probablity & Statistics rule science.

(End of light reading session)

Stemming from @andselisk's idea (in the comments under your post), you might find the following Youtube channels interesting (I'll try to add more when I get the time):

Nurdrage   <--- Highly recommended
NileRed
Cody'sLab
Doug's lab

The next two are largely DIY channels, but some of their videos include projects that count as "Chemistry":

NightHawkinLight
The King of Random

 Thus armed with this knowledge, you shall sally forth and become a chemist (of sorts...)!


Complete Combustion of Inorganic, Hydrogen Containing Covalent Compounds in the Atmosphere From what I've seen in alkanes, their complete combustion creates carbon dioxide, which has carbon in its highest possible oxidation state.  I've been wondering what complete combustion would look like in similar compounds, such as ammonia, phosphine, hydrogen sulfide, and silane.

After looking at through several search results of The combustion of <compound> and <compound>'s reaction with oxygen, I was surprised to only see that the search results for silane and phosphine generally showed that the combustion products attained their maximum oxidation state.

For example, the general result for ammonia seems to be nitrogen monoxide or nitrogen dioxide, which make sense as products for incomplete combustion, but I would think that a complete combustion would produce dinitrogen pentoxide.  Hydrogen sulfide's combustion also seems to form sulfur dioxide or disulfur dioxide rather than sulfur trioxide.

This really confused me, since I was expecting the main products of all these reactions to be fully oxidized nonmetallic oxides.  My question is, how can one predict what the fully combusted products of an inorganic, hydrogen containing compound look like?

Links for my ammonia query:

http://www.questionsolutions.com/ammonia-gas-reacts-molecular-oxygen-gas/
https://answers.yahoo.com/question/index?qid=20150615223643AArPdtu
https://answers.yahoo.com/question/index?qid=20080504104957AAgQrL3

Links for my hydrogen sulfide query:

https://chemiday.com/en/reaction/3-1-0-151

https://www.sciencedirect.com/science/article/pii/S0082078473800463

The "completeness" of combustion depends on the proportion of the reactants, and on the temperature cycle, not only the peak temperature, but the rate at which the products are quenched. For example, using ammonia and air as fuel in internal combustion engines produces mostly $\ce{H2O}$ and $\ce{N2}$, with less than 0.1% $\ce{NO_x}$ according to Kobayashi, et. al. and Zacharakis-Jutz.

The greatest oxidation state of a substance is often not thermodynamically stable. Perchlorates, chlorine dioxide , $\ce{H2O2}$ and similar compounds are somewhat unstable (i.e., BANG) and are not generally released in great quantities by combustion.


How does functionalized activated charcoal work? If I understand correctly it is possible to trap stuff inside activated charcoal pores to give it specific retention properties (by exposing it to functionalizing compounds while heated to open up the pores?).
When a solution for instance is passed through activated AC, what happens inside the pores with the functionalizing compound to trap stuff that otherwise wouldn't be trapped? Can it be said that the functionalizing compound precipitates stuff inside the pore by producing low solubility products?

No. It is not a precipitation reaction. It is physical reaction. The pores are made of the rest of the membranes of the cells. When wood is overheated in air, it burns. When this is done without too much contact with the air, the cellular water gets evaporated. The content of the cell (cytoplasm, proteins, etc.) is pyrolyzed, destroyed and mostly transformed into gases. The membrane is carbonized or charred, but remains here, forming charcoal. The rest of the cell being eliminated, the charcoal contains a huge amount of the previous cells which behave like pores. Big molecules can enter these shells, and have difficulty to get out later on.


Nanoparticle tracking analysis v.s Transmission Electron Microscope I have various bimetallic nanoparticles which have formed alloys.
I then used NTA and TEM to characterize the size of my nanoparticles.
The TEM shows that the size of nanoparticles is considerably less than the NTA.
Can anybody help explain the difference in size?

You don't mention how homogeneous your solutions are or how much "considerably less" is. As summarized in this paper, "Under ideal conditions, particle size analysis by DLS and NTA both can be accurate to 2%, but heterogeneous samples can have errors an order of magnitude larger." (See references in that paper for details).
It's also important to keep in mind that you are comparing techniques which measure samples in very different ways and under very different conditions. TEM measures individual particles in a dried (and often treated) state in a vacuum, while NTA measures an ensemble of particles dissolved in solution. In each case, the size distribution is often just as important as the so-called "size". If the size-distribution of your measurements overlap, perhaps there's no conflict after all.
Also, if your particles have some sort of coating which Au nanoparticles, for example, often do, then you might expect differences in the apparent size measured in these two different states.
Using controls is a good way to get around some (but not all) of these problems, but then the same questions need to be raised about the control particles and any difference in their behavior from the particles under study.
So, as is usually the case in well-done science: "It's complicated..."


If there is more than one branch in a hydrocarbon, which branch do you name first? I have read that branches need to be named in alphabetical order, but I do not really understand this, or do you start with the branch closest to the edge of the hydrocarbon?

Branches must be numbered in alphabetical order, along with their positions.
Suppose you have a chloro group on the first carbon, a bromo group on the second carbon and an amino group on the fourth carbon of your carbon chain (butane).
One must therefore go in alphabetical order which is amino, then bromo and then chloro.
Therefore you naming will be something like this:-

$$\ce{3-Bromo-4-chloro-butanamine}$$

The naming is in alphabetical order but the numbering starts according to the following rule:

$$\ce{RCOOH>RCOOR>RCOCOOR>RCOCl>RCONH2>RCHO>CO>OH>NH2>double bond>triple bond>halogen>NO2>ROR>alkyl}$$


Van der Waals pressure I was solving some problems and I had a doubt a problem since the resulting pressure gives me a negative value. The problem was the following:

  Could $100~\mathrm{g}$ of $\ce{N2}$ in a $200~\mathrm{mL}$ container exert a pressure of $25~\mathrm{atm}$ at $28~^\circ\mathrm{C}$ if it behave like an ideal gas? If your answer is no, what pressure would exert the gas, knowing that the van der Waals constants for $\ce{N2}$ are $3.11~\mathrm{\frac{L\,atm}{mol^2}}$ and $0.076~\mathrm{\frac{L}{mol}}$.

I have converted the 100 grams of $\ce{N2}$ to moles of $\ce{N2}$, and with that I've replaced the data in the van der Waals formula, using $a = 3.11~\mathrm{\frac{L\,atm)}{mol^2}}$ and $b = 0.076~\mathrm{\frac{L}{mol}}$. The result was a negative pressure, and I am pretty sure that's impossible (or at least that's what my chemistry teacher said).

This is what I did:

$$\begin{align}
P &= \frac{nRT}{V-nb} -\frac{an^2}{V^2}\\
  &= \frac{(3.57~\mathrm{mol})(0.082\mathrm{\frac{L\,atm}{K\,mol}})
    (301.1~\mathrm{K})}{(0.2~\mathrm{L})
   -(3.57 ~\mathrm{mol})(0.076\mathrm{\frac{L}{mol}})}
   -\frac{(3.11~\mathrm{\frac{L~atm}{mol^2}})
    (3.57~\mathrm{mol})^2}{(0.2~\mathrm{L})^2}
\end{align}$$

The result was:
$-1626.19~\mathrm{atm} - 9.9075~\mathrm{atm} = -1636.0975~\mathrm{atm}$

Let's get the obvious out of the way: The answer to that exercise is yes. Hence it is completely unnecessary to do the second part. With the parameters given in the exercise I calculated a pressure of about 440 atmospheres using the ideal gas law.

But then since we're all here to learn, let's also have a look at the second part. The good news is that I can't see anything wrong with how you used the equation. The problem is already pointed out by orthocresol in the comments, which I will quote here for safekeeping (emphasis by me):

  Ok. Firstly, the constant $a$ has units of $\mathrm{L^2~atm~mol^{-2}}$ (not $\mathrm{L~atm~mol^{-2}}$), but I am guessing it is just a typo on your part. The issue seems to arise from the values given for $a$ and $b$. Wikipedia gives the constants as $1.352$ and $0.0387$ respectively (converted to the same units as yours). Using those values I get a pressure of 997.3 atm.

I would assume that nobody caught this mistake earlier since you are not actually supposed to do that part of the question.


Why does the equilibrium constant change on scaling the stoichiometric coefficients of a reaction? I read in my textbook that if we multiply a chemical reaction by some factor(let's say $b$) its new equilibrium constant becomes $K^b$.But I don't understand why this happens..
What is the difference between the reaction $\ce{A} + \ce{B} = \ce{C}$ and $b \ce{A} + b \ce{B} = b \ce{C}$ ?
Aren't the two reactions same? So why does the equilibrium constant change?

Have a look at this answer of mine.
There you can find the derivation of the formula that defines the equilibrium constant

\begin{equation}
\log \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \log K = -\frac{\Delta G^{0}}{RT}
\end{equation}

of a general reaction

\begin{equation}
 \ce{\nu_{1}A + \nu_{2}B + ... + \nu_{$x$}C <=> \nu_{$x+1$}D + \nu_{$x+2$}E + ... +\nu_{$n$}F}
\end{equation}

and it is also shown how this definition relates to the maybe more familiar definition

\begin{equation}
 K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ .
\end{equation}

Now, the reason why the equilibrium constants of the general reaction above and the same reaction multiplied with a constant factor $\alpha$:

\begin{equation}
 \ce{\alpha\nu_{1}A + \alpha\nu_{2}B + ... + \alpha\nu_{$x$}C <=> \alpha\nu_{$x+1$}D + \alpha\nu_{$x+2$}E + ... + \alpha\nu_{$n$}F}
\end{equation}

are not the same but are related to each other via:

\begin{equation}
 K_{\alpha} = K_{1}^{\alpha}
\end{equation}

comes from the definition of $K$.

That there must be a difference is also plausible: The defining formula for $K$ relates it to the standard Gibbs free energy of reaction $\Delta G^{0}$ which in turn is defined to be the amount of Gibbs free energy that is produced per formula conversion (meaning that it is defined for an extend of reaction $\xi = 1 \, \text{mol}$).
So, it is the amount of Gibbs free energy produced by the reaction of $\nu_{1}\, \text{mol}$ of $\text{A}$ and $\nu_{2}\, \text{mol}$ of $\text{B}$ etc. to $\nu_{x+1} \, \text{mol}$ of $\text{D}$ and $\nu_{x+2}\, \text{mol}$ $\text{E}$ etc.
But if you multiply the reaction by $\alpha$ the standard Gibbs free energy of reaction is the Gibbs free energy produced by the reaction of $\alpha\nu_{1}\, \text{mol}$ of $\text{A}$ and $\alpha\nu_{2}\, \text{mol}$ of $\text{B}$ etc. to $\alpha\nu_{x+1} \, \text{mol}$ of $\text{D}$ and $\alpha\nu_{x+2}\, \text{mol}$ $\text{E}$ etc. and so there should be $\alpha$ times as much Gibbs free energy as before:

\begin{equation}
 \Delta G^{0}_{\alpha} = \alpha \Delta G^{0}_{1} \ .
\end{equation}

And this is indeed the case when you have $K_{\alpha} = K_{1}^{\alpha}$ because according to the logarithm laws

\begin{equation}
RT \log K_{1}^{\alpha} = RT \alpha \log K_{1} \ .
\end{equation}


On adding sugar to water, does the entropy increase or decrease? This question is a little confusing. According to my teacher, the entropy should decrease, but I think that sugar is crystalline first and later it dissolves, so the randomness increases and so the entropy should also increase. What is the correct answer?

While it may seem that randomness always increases when a crystal is dissolved into a liquid phase, it does not have to be that way.

Concerning sugar, the molecule has a large number of hydroxyl groups and is generally rather large when compared with the water molecule around it — much larger than your average sodium or chloride ions. Every hydroxyl will act as both a hydrogen bond donor and an acceptor, likely to two different water molecules, creating a cage of water that is rather ordered and large again when comparing to the sodium or chloride ion’s solvation cell. This will decrease the disorder (and hence entropy) of the free water so much, that it counteracts the increase in entropy of the crystal dissolving. (The effect is more pronounced with sucrose when compared to glucose due to the molecule’s size and the number of hydroxyls.)

On the other hand, the molecules in a sugar crystal are likely less ordered than the ions of a salt crystal. They will form extensive hydrogen-bonding networks, yes; but overall if you shift one row of molecules relative to its neighbour you won’t necessarily break the crystal as you won’t suddenly have ions of the same charge next to each other. And finally, the single salt crystal will break down into much smaller fragments than the sugar molecules.

So summed up, the effects are:

large solvation cells
crystal breaks down into larger (less disordered) fragments upon solvation
sugar crystals themselves are less ordered than salt crystals

Meaning that there is a lot less entropy to be gained.


Degree of dissociation of weak electrolyte at infinite dilution Why the degree of dissociation of weak electrolyte is 1 at infinite dilution and how do we get the result? Like I know that at infinite dilution the concentration of solution is 0 but what does that have to do with degree of dissociation, secondly in oswald's law of dilution we already assume that dissociation is almost negligible so we can't obtain the result from that also

There are a number of ways to explain this. In words, the entropic gain with dissociation overwhelms any attraction between counterions as the electrolyte is diluted.

Mathematically: for a m:n electrolyte undergoing dissociation according to the reaction

$$\ce{A_mB_n -> mA^{\nu_A -} + nB^{\nu_B +}}$$

if the initial concentration of electrolyte $\ce{A_mB_n}$ is $c_{T}$, define the degree of dissociation as

$$\alpha = \frac{c_{AB}}{c_{T}}$$

where $c_{AB}$ is the concentration of dissociated $\ce{A_mB_n}$. Assuming ideality and unit molarity as standard concentration, the equation for the equilibrium constant can be written as $$K=\frac{c_A^m c_B^n}{c_{AB}}$$ and solved in terms of the degree of dissociation. For a 1:1 electrolyte this leads to

$$\alpha = \frac{\beta}{2}\left[\left(1 + \frac{4}{\beta} \right)^{1/2} -1\right] \tag{1}$$

where $\beta = K/c_T$. As $c_T \rightarrow 0$ it follows from this last expression that $\alpha  \rightarrow 1$ #QED

PS The following equation (from Wikipedia's page on Ostwald's law of dilution) $$\alpha=\sqrt{\frac{K}{c_T}} \tag{2}$$ does not hold over the entire concentration range $c_T$. It can be obtained from Eq. (1) by assuming $K<<1$ and therefore $\beta<<1$. For increasingly dilute solution Eq. (2) fails and  $\alpha  \rightarrow 1$.


Home test for magnesium carbonate People that do rock climbing and weightlifting usually use magnesium carbonate on their hands to decrease slippery hands.

I would like to know how can someone that buys MgCO3 can really be sure it's magnesium carbonate and not another thing.

Is there a simple test people could do at home?

If all you want to know is the relative amounts of calcium and magnesium carnbonates disolve it in vinegar (it will fizz). Now add lye until the pH is about neutral (use pH test strips) in order to neutralize any excess acid. Obtain a test kit for determining water hardness that will give both the magnesium and calcium hardness values for a sample. Dilute with DI (or nearly DI i.e. RO) water to be within range of the kit Be sure you understand what units the kit is reporting in. Usually it is units related to the equivalence of the calcium and magnesium so that if calcium harness turns out to be 1/5 magnesium hardness that tells you the original mix has 1/5 of a mole of CaCO3 for each mole of MgCO3.


Comparing basic strength order of amines In the above question I was asked to compare the basic strength order.
So I did it like this:

The Nitrogen in I and II will not participate in resonance so the only effect it could produce is inductive (-I) .
Now the compound IV has less resonating structures than the other compounds so the the Nitrogen atom will have larger electron density.

So IV should be most basic then III > II > I because of (-I) effect.
But my book mentions (C) as the answer.

How? Any help will be appreciated

You don't have problem with assigning compound IV (1, 4, 5, 6- tetrahydropyrimidine) as the most basic compound among given four amines. I agree with you, but not the same reasons. According to Ref.1, the 1,4,5,6-tetrahydropyrimidines (THPs) are extremely basic compounds, as might be expected from their amidine structure (e.g., 1,8-Diazabicyclo[5.4.0]undec-7-ene or DBU). The parent THP, 1,4,5,6-tetrahydropyrimidine (IV) has a $\mathrm{p}K_\mathrm{a}$ of about $13$ at $\pu{20 ^\circ C}$, which is even reacted with water (Ref.1). This highly basic nature of IV can be attributed to its stabilization by resonance between two equivalent structures, V and V', when protonated:

Also, keep in mind that it can also get protonated on 3-nitrogen atom (VI; inside the blue box) as well (Ref.1).

Among the rest of other 3 amines, two are aminopyridines (I and II), and the third one is aniline (III). Since the lone pair of nitrogen in pyridines does not contributed to aromatic system (hence it is readily available for proton) while that of aniline is involved in aromatic system by resonance (not readily available), generally pyridine is more basic than aniline. Even though I and II both have extra amino group, they are expected to behave like pyridines. By this analysis, we can conclude that the least basic amine among given four amines is aniline (III).

Now, let's look at what happens when I (4-aminopyridine) is protonated:

From above illustration, you can clearly see conjugate acid of I is well stabilized by resonance including full contribution of the lone pair of 4-amino-nitrogen. If you visualize the same with II (3-aminopyridine), it is not stabilized by the extra amino grop as depicted in following diagram:

Thus, by these facts, we can conclude that II is less basic than I. Therefore, the order of basicity strength of four given amines is: IV > I > II > III.

References:

D. J. Brown, R. F. Evans, “100. Hydropyrimidines. Part I. 1, 4, 5, 6- Tetrahydropyrimidine and its derivatives,” Journal of the Chemical Society 1962, 527–533 (https://doi.org/10.1039/JR9620000527).


Cheap metal to replace platinum in hydrogen fuel cell Platinum is used as a catalyst in hydrogen fuel cells because it increases the efficiency. However, if I just want to make a fuel cell for demonstration purposes, I don't want to use a lot of money to buy platinum wires. What "common" metal (aluminium, iron, etc.) works the best for a hydrogen fuel cell?

We have tinkered with this a little. Currently, a handful of others have been experimenting with certain blends of metals, using lower costs to create equal reactions.

I have heard a lot of different ratios of Ni/Cu and Pd blends. But a good experimental blend (if your making a reasonable size batch), would comprise of a ZrO2–Pd(x)D core, with (x) varying as a slew of other metals.
 This article is regarding the direct deposition of hydrogen on palladium, probably a little more refined towards what you where searching.

While im not sure how intricate your going to get with your experimental cell, here is some good food for thought., If you can create a vessel with proper pressure safety, academic groups have been experimenting with super critical fluids, trying to test metallic blends as catalysts under states past standard classical critical conditions.


Reaction of vicinal diol in presence of benzaldehyde Could someone please help me figure out the products of the following reaction?

Here are few possibilities I've come up with:

The attacking reagent could the given vicinal diol, with its nucleophilic lone pair, attacking the electrophilic carbon of benzaldehyde. However, that doesn't seem to lead me anywhere.
As the reaction is carried out in acidic medium, one of the -OH groups (or both?) could be protonated, to become better leaving groups leading to formation of a carbocation intermediate. What after that?

Chances are it's probably a nucleophilic addition, proceeding through a tetrahedral intermediate, than a carbocationic one.

It'd be great if someone could show a proper mechanism for the reaction, and explain why it should be so. Thanks a lot!

This reaction is similar to the one where you use ethylene glycol to form a temporary protecting group for an aldehyde that you do not want to react in further steps. The answer therefore should be (b).


Is the middle ring of this organic compound aromatic? Is the middle ring with 5 atoms of (4Z)-4-[(4-methoxyphenyl)methylidene]-5-methyl-2-phenyl-2,4-dihydro-3H-pyrazol-3-one aromatic?

The carbonyl group could become a negatively charged oxygen, the tertiary amine could then form a double bond with the $\ce C$ atom of the carbonyl group. The secondary amine could contribute one electron to the $\pi$-system. And the $\ce C$ atom of the ethen-group could contribute two electrons if the other $\ce C$ atom becomes positively charged, but I'm not sure if this is likely.

You don't have to force a positive charge on the carbon between the central and left rings.  The methoxy oxygen on the left ring can donate an electron pair to take up the positive charge.  The left ring then becomes quinoid and the central ring becomes aromatic.  Such switching between quinoid and aromatic rings is common in conjugated polycyclic structures.


Reactivity order in hydrogenation reaction of but-2-yne and butene isomers Question 1:

  The reactivity order towards hydrogenation of the following compounds is:

  but-2-yne ($\ce{CH3-C\bond{3}C-CH3}$)
  cis-but-2-ene ($\ce{CH3-CH=CH-CH3}$)
  trans-but-2-ene ($\ce{CH3-CH=CH-CH3}$)
  2,3-dimethyl-but-2-ene

Question 2:

  Which of the following has the lowest heat of hydrogenation per mole:

  cis-2-butene
  trans-2-butene
  1-butene
  1,3-butadiene

In the first question, I understand that 2 > 3 (cis is more reactive than trans generally)  and 4th is the least reactive (steric hindrance). But I am not quite sure about the order of (1). I could think that alkynes are more reactive toward hydrogenation because they have more pi electron density because of which its adsorption on the surface of metal catalysts will be easier. Is that the reason?
I have also read that alkynes are more reactive because they have less number of substituents around them. Which logic is correct and why?

I am more confused in the second question since there are just so many factors to compare - more no of pi bonds (more electron density) (according to my above logic, it should increase reactivity), steric hindrance, increased reactivity of cis isomers. I can make out that a > b (cis is more reactive than trans)  and d > c (more electron density) in terms of reactivity toward hydrogenation but how do we compare the others?

Monosubstituted alkene is more reactive than disubstituted because directly attached alkyl groups stabilize the system. Therefore 1 butene is more reactive.

Cis is more reactive than trans because Cis compounds are unstable. As they have similar groups on same side, it results in stearic crowding.

Thirdly, conjugated dienes are very stable beacuse they are resonance stabilized.

Thus order will be
1 butene will be most reactive, then Cis, then trans and then 1,3 butadiene..


(+)-Bicuculline, what does (+) mean? I was looking up some ion channel blockers. What does (+) or (-) below mean?

(+)-Bicuculline

(-)-Bicuculline methiodide

Bicuculline is a chiral molecule.  That means that its mirror images are non-superimposable.  These non-superimposable images are termed "enantiomers."  When you see a "(+)" or "(-)" in front of a chemical name it tells you that only one the enantiomers is present (e.g. it is not a racemic mixture of the enantiomers).  Chiral molecules have the ability to rotate the plane of a polarized light beam that passes through it. Enantiomers will rotate the plane of polarization in exactly equal amounts (same magnitude) but in opposite directions.

Dextrorotary designated as (+), indicates clockwise rotation (to the right)

Levorotary designated as (-), indicates anti-clockwise rotation (to the left)


Why do transition elements make colored compounds? Why do transition metals element  make colored compounds both in solid form and in solution? Is it related to their electrons?

You are absolutely correct, it all about the metal's electrons and also about their d orbitals.

Transition elements are usually characterised by having d orbitals. Now when the metal is not bonded to anything else, these d orbitals are degenerate, meaning that they all have the same energy level.

However when the metal starts bonding with other ligands, this changes. Due to the different symmetries of the d orbitals and the inductive effects of the ligands on the electrons, the d orbitals split apart and become non-degenerate (have different energy levels).

This forms the basis of Crystal Field Theory. How these d orbitals split depend on the geometry  of the compound that is formed. For example if an octahedral metal complex is formed, the energy of the d orbitals will look like this:

As you can see, previously the d orbitals were of the same energy, but now 2 of the orbitals are higher in energy. Now what does this have to do with its colour?

Well, electrons are able to absorb certain frequencies of electromagnetic radiation to get promoted to higher energy orbitals. These frequencies have a certain energy which correspond to the energy difference between different orbitals. Now most substances are only able to absorb frequencies of radiation which are outside the visible light spectrum, for example they might be able to absorb radiation which has a frequency of $300$GhZ (that is infrared radiation). This means that it reflects all other types of radiation, including the full spectrum of visible light. So our eyes see a mixture of all the colours; red, green, blue, violet, etc. This is seen as white (this is why several organic compounds are white).

However transition metals are special in that the energy difference between the non-degenerate d orbitals correspond to the energy of radiation of the visible light spectrum. This means that when we look at the metal complex, we don't see the entire visible light spectrum, but only a part of it.

So for example, if the electrons in an octahedral metal complex are able to absorb green light and get promoted from the $d_{yz}$ orbital to the $d_{z^2}$ orbital, the compound will reflect all other colours except for green. Therefore by using the colour wheel, we can find the complementary colour of green which will be the colour of the compound, which is magneta.

This explains why not all transition metal complexes are colourful. For example copper sulfate is a bright blue compound, however zinc sulfate on the hand is a white compound despite being a transition metal. The reason behind this is because zinc's d orbitals are completely filled up with electrons, meaning that it is not possible for any electron to make a d-> d transition as they are all filled up. Hence you might sometimes see zinc referred as not being a transition metal.


How are t-butyl substituents named according to IUPAC nomenclature? What is the IUPAC name of this structure?

I've actually came up with two names, the first one is "1-tert-butyl-5-heptyl-2-propylcyclononane" and the other one is "1-(1,1-dimethylethyl)-5-heptyl-2-propyl-cyclononane". I'm confused which one is right or if both are correct.

According to Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book) (P-57.1.2), the preferred prefix for $\ce{-C(CH3)3}$ is "tert-butyl".

As for where it comes in the name:

  P-14.5 Alphanumerical order
  [...] Alphanumerical order is applied as follows in organic nomenclature. Nonitalic Roman letters are considered first, unless used as locants or part of a compound or composite locant [...]

so the prefix tert is not considered (except in more complicated cases, cf. P-14.5.3). Therefore, the preferred IUPAC name for the molecule in question is 1-tert-butyl-5-heptyl-2-propylcyclononane. Note that there are no parentheses around tert-butyl.

Be warned: tert-butyl has received special treatment from IUPAC on this front. Other non-systematic prefixes have fallen from favour, most notably isopropyl, sec-butyl, and isobutyl. The preferred prefixes are respectively propan-2-yl, butan-2-yl, and 2-methylpropyl.


How toxic is disilanol? Silicon has been known to form analogues of organic compounds. For example, silicon with 4 hydrogen atoms forms silane as opposed to methane. This continues up the alkane list, with disilane for ethane, and so forth.

Silanes also have with hydroxyl groups to form "silanol." Thus, disilanol is the silicon analogue of ethanol. ($\ce{H3Si-SiH2OH}$)

How toxic is disilanol?

Difficult to say. After a fair amount of Googling, I was unable to find any reliable sources on the compound you ask about, $\ce{H3Si-SiH2OH}$. From the Wikipedia article on silanols, it appears that the simple silanols are not stable, but no reference is provided.

It stands to reason that if $\ce{H3SiOH}$ is not stable, there is a good chance that $\ce{H3Si-SiH2OH}$ is also not very stable. Disiloxane $\ce{H3Si-O-SiH3}$ is much more stable, and silanols likely convert into siloxanes.

Good knowledge about the health hazards of a substance requires thorough investigation, which is not possible when the substance is not stable.

Alkyl silanols, like trimethylsilanol are stable and used for a variety of applications. Acutley, trimethylsilanol(pdf) is a volatile liquid that irritates skin, eyes, and mucus membranes (e.g. the lungs). The language on the MSDS is "may be harmful if..." This language suggests that you want to avoid contact with it but that some exposure will not cause serious harm. Note the msds for ethanol(pdf) has much scarier language.


How is it determined how many electrons are transferred in redox reactions? For example, iron and tin(II) sulfate will react to produce iron(II) sulfate and tin:

$\ce{Fe + SnSO4 -> FeSO4 + Sn}$

With the following half equations:

$\ce{Fe + SO4^{2-} -> FeSO4 + 2e-}$

and

$\ce{Sn^{2+} + 2e- -> Sn}$

but where does the "2" come from? Why can't it be:

$\ce{Fe + SO4^{5-} -> FeSO4 + 5e-}$

and

$\ce{Sn^{5+} + 5e- -> Sn}$

In short, the species you mention ($\ce{Sn^{5+}}$ and $\ce{SO_4^{5-}}$) do not exist in any chemical system at anywhere near ordinary conditions. The number of electrons shuffled in the reaction is not chosen arbitrarily, but is based on the initial and final oxidation numbers of the elements in the reaction, after the equations are balanced. The only stable compound with formula $\ce{SnSO_4}$ is made of $\ce{Sn^{2+}}$ and $\ce{SO_4^{2-}}$ ions. Hence in your reaction, $\ce{Sn}$ atoms start with oxidation number +2 and end with 0 (therefore gaining two electrons) and $\ce{Fe}$ atoms start with oxidation number 0 and end with +2 (therefore losing two electrons).

As another example of electron counting, here's the reaction between $\ce{Ba}$ and $\ce{Al^3+}$:

Oxidation: $\ce{Ba^0 -> Ba^{2+} + 2e^{-}}$

Reduction: $\ce{Al^{3+} + 3e^{-} -> Al^0}$

To balance the equations, the oxidation half-reaction must be multiplied by 3 and the reduction half-reaction must be multiplied by two. Therefore:

Oxidation: $\ce{3Ba^0 -> 3Ba^{2+} + 6e^{-}}$

Reduction: $\ce{2Al^{3+} + 6e^{-} -> 2Al^0}$

Global: $\ce{3Ba^0 + 2Al^{3+} -> 3Ba^{2+} + 2Al^0}$

In the global reaction, six electrons are involved.


Acetone-resistant clear plastic Is there a plastic that's clear, acetone resistant, and ideally sold in sheet stock that's ~1/8" or 1/4" thick?  Both acrylic and polycarbonate are out, they turn cloudy on contact with acetone.  I tried a piece of PETG and it was better, but still turned fairly cloudy after a few seconds.  I don't need continuous service, but I need it to still look decent if it gets wiped down with acetone.

According to Palstics International's Chemical Resistance Chart, the A-rated plastics (no solvent attack) towards acetone are:

ECTFE (Halar®): transparent films available
Fluorosint® PTFE: white
HDPE: transparent films available
Nylon®, Type 6/6: white
PP: clear sheets available
PPS: opaque white
PTFE: white

Among these, polypropylene appears to be the only plastic that satisfies all criteria.


protecting contact between aluminium and copper wires The quarters in which I live has both aluminum and copper household wiring. This results in corrosion of the contacts and can lead to sparking, overheating, and even fire.

Is there any way to inhibit the corrosion (if not stopping it completely), except for changing the whole wiring?

This may depend on your location and local electrical codes.  I have the same issue in my house.  In the USA, there are NEC (National Electrical Code) compliant wire nuts rated for Al to Cu connections.  These come pre-packed with a protective goop. The goop can also be purchased separately in a tube for use with ordinary wire nuts. You may have to ask at your local electrical supply store.  In my area the pre-packed wire nuts are a distinctive purple color not found on regular wire nuts (for example: Home Depot Al/Cu nuts).

For code compliance, use approved solutions only, even if you think you know more chemistry than the electricians. Having your house burn down from a wiring issue is no fun.


Clarification in the mechanism for Molisch's test for glucose Wikipedia says

  Molisch's test is a sensitive chemical test, named after Austrian botanist Hans Molisch, for the presence of carbohydrates, based on the dehydration of the carbohydrate by sulfuric acid or hydrochloric acid to produce an aldehyde, which condenses with two molecules of a phenol (usually α-naphthol, though other phenols such as resorcinol and thymol) also give colored products), resulting in a red- or purple-coloured compound.

Can somebody provide an arrow pushing diagram to illustrate the intermediate steps of these reactions? (especially for the dehydration of glucose)
Also, for the reaction with of 5-(hydroxymethyl)furfural with 2 moles of phenol, it seems that the acyl group is acting like an electrophile and naphthol is engaged in an electrophilic aromatic substitution, but then again, I am not sure how 2 moles of naphthol combined there. Any help would be appreciated.

YUSUF HASAN: 5-Hydroxymethyl furfural (5-HMF) 1 when protonated on the carbonyl oxygen becomes a reactive electrophile. Addition of phenol 2 at the reactive para-position affords 3 which deprotonates at the para-position, rearomatizing the ring to a phenol and liberating a proton. Secondary alcohol 4 is protonated on oxygen with loss of water to form stabilized cation 5. A second equivalent of phenol adds to the cation as previously described leading to the triarylmethane 6. The methane hydrogen is susceptible to oxidation by atomspheric oxygen, which is in a triplet state (unpaired electrons behaving as a free radical). Resonance stabilized radical 8 is formed along with the hydroperoxy radical 7, which can abstract the hydrogen from triarylmethane 6 forming more of radical 8 and hydrogen peroxide. Oxidation of radical 8 with oxygen gives the resonance stabilized carbocation 9 and superoxide anion, aka, superoxide radical anion 10. This species can be protonated by the the strong acid conditions of the Molisch test to form more of the hydroperoxy radical 7.


Electrophilic substitution of an indole as opposed to benzofuran and benzothiophene Regioselectivity in electrophilic aromatic substitution of benzofuran and indole

This response as to the regioselectivity of the electrophilic aromatic substitution of indole left me with a few more questions than it did answers.

In figure 3b, does the pi bond in the bottom half of the benzene on the left get pushed to the nitrogen, allowing for stabilization across the benzene ring in addition to the resonance structures of the nitrogen cation? And if so, and this is the case of a viable nitrogen cation contributing to the stabilization of a valid resonance structure, is it true that all (or most) polycyclic compounds that undergo electrophilic substitution at a location on the nitrogen-containing ring that result in the temporary resonance structure of a nitrogen cation along with stabilization across a benzene ring will be more stable than substitution at a location that would only provide the resonance across the benzene (e.g. without the possible nitrogen cation, had electrophilic substitution occurred elsewhere)?

Furthermore, does this imply that in the electrophilic aromatic substitution pattern of benzothiophene, seeing as how sulfur is less electrophilic than both oxygen and nitrogen, that we would see a similar substitution pattern at the 3-position, as opposed to the substitution at the 2-position that we would see (And that is, to my puny little brain, comprehensible) on benzofuran?

because when electrophile react on 3 position more resonance structures are posible than the 2 position and thus it show stability.Thats why we say that position 3 is more good for indol to attach electrophile.


DFT Calculations, Atomic Ionization Potentials -- Which Exchange-Correlation Functional to Use, to Preserve Koopmans Theorem? I have a program which can perform density-functional calculations for atoms, given a density functional.

Of course the simplest form of exchange potential to use is one relevant for a uniform electron gas (i.e. the original Kohn-Sham exchange, proportional to $n_e^{1/3}$). A good correlation functional is also available. However when I performed calculations for some atoms the energy of the highest occupied orbital differs greatly from the ionization potential (off by a few eVs!).

That the energy of the highest occupied orbital equals the ionisation potential of the neutral atom is known as Koopmans' Theorem. I get a feeling from the literature that DFT can be quite accurate. So what's wrong?

Interestingly when I use the Slater exchange functional, coupled with a correction suggested by Skillman (i.e. replacing the potential by $1/r$ when the overall potential drops below that value), the results improve significantly. (See the book "Atomic Structure Calculations" by Herman Skillman). Well, maybe I should follow this recipe, but it seems the procedure is quite ad-hoc.

My question is, are there any functionals which will give reasonable values for the atomic ionisation potentials? I do not want to implement methods like the optimised effective potential method since the latter is not readily generalised to a finite-temperature scenario. Thanks.

I think the most widely-used approach at the moment comes from Roi Baer and Leeor Kronik and others, e.g.

Phys. Rev. Lett. (2010) 105, art. 266802
Phys. Rev. B (2011) 84, art. 075144

Basically the idea is to tune the range-separation parameter $\gamma$ between short-range and long-range electrostatic effects in a range-separated hybrid functional. You find a match such that Koopmans' theorem holds.

The result is an "optimally-tuned" range-separated hybrid DFT functional (OT-RSH).

It's not perfect, for example, it violates size consistency: J. Chem. Phys. (2013) 138, 204115


What is the largest, noncrystalline, non-polymer inorganic molecule? A question came up in class from one of my chemistry students. He asked what the largest, non-crystal, non carbon-based, molecule is. We ruled out complex proteins, DNA, and other polymers, as well as crystalline structures and metals from the list. I didn't have an answer for him, so I looked it up but could not find a definite answer. I found two threads on Reddit, listed below. Polyoxometalates come up, but without any specifics. I know "largest" can be subjective, so please specify if the molecule is "largest" because of its physical size or because of its mass.

https://www.reddit.com/r/chemistry/comments/2phl4s/whats_the_largest_inorganic_molecule_you_know/

https://www.reddit.com/r/askscience/comments/i6v07/what_is_the_biggest_noncrystalline_inorganic/

I agree with the redditors that the answer is likely a polyoxometalate.
$\ce{Na48[H_xMo368O1032(H2O)240(SO4)48] · \text{ca.}~1000 H2O}$ aka
$\ce{\{Mo368\}}$ (core: $M \approx\pu{60.7 kDa},$ dimensions $\approx \pu{2.5 nm} × \pu{4.0 nm})$ is the primary candidate [1]:

Figure 2. Structure of 1a in crystals of 1 in polyhedral (a) and ball and stick (b) representation perpendicular to the $C_4$ axis and along the $C_4$ axis ((c) and (d); building units $\ce{\{Mo1\}}$ yellow, $\ce{\{Mo2\}}$ red, $\ce{\{Mo(Mo5)\}}$ blue with blue-turquoise pentagonal bipyramids; $\ce{O}$ atoms small red spheres, $\ce{S}$ atoms gray spheres).
Note that one capping part is turned against the other by $360/16^\circ.$
The different building blocks and the arrays of different local symmetry are assigned.

A recent review of mass-spectrometry studies of large polyoxometalates is also available [2].
References

Müller, A.; Beckmann, E.; Bögge, H.; Schmidtmann, M.; Dress, A. Inorganic Chemistry Goes Protein Size: A $\ce{Mo368}$ Nano-Hedgehog Initiating Nanochemistry by Symmetry Breaking. Angewandte Chemie International Edition 2002, 41 (7), 1162–1167. DOI: 10.1002/1521-3773(20020402)41:7<1162::AID-ANIE1162>3.0.CO;2-8.
Surman, A. J.; Robbins, P. J.; Ujma, J.; Zheng, Q.; Barran, Perdita. E.; Cronin, L. Sizing and Discovery of Nanosized Polyoxometalate Clusters by Mass Spectrometry. J. Am. Chem. Soc. 2016, 138 (11), 3824–3830. DOI: 10.1021/jacs.6b00070.


pka from molecular dynamics How can one calculate pKa values from first principle calculations, static or dynamics?

I have encountered some works using CPMD where people try to find pKa values from -(i) constrained Molecular Dynamics (MD) and (ii) metadynamics. These are somewhat more intricate ways, I believe. Is there any other simpler ways (theoretical/computational)? Can not we do it from simple MD?
Please suggest some references as well, if possible.

Thanks in advance.

Have you considered a free-energy calculation?

That is, calculate the free-energy of the dissociated species and of the neutral one, and calculate the pKa via the formula for activity.

There was this GROMACS tutorial, but it seems the link is broken.


What is the empirical formula of Haemoglobin? Everywhere I look on the internet tells me a different formula. Are there genuinely different formulas for Haemoglobin, or are some of my internet sources incorrect?
(I know there are different formulas for oxyhaemoglobin, deoxyhaemoglobin, carboxyhaemoglobin, etc but the formulas I am seeing are very varied and do not account for this... I don't think)

Unfortunately this is only half an answer. I'd be thankful if someone could chip in with the rest.

When you want structural information about proteins, the first place to look is the Protein Data Bank, a freely accessible repository of protein structures determined by techniques such as X-ray crystallography, protein NMR and neutron diffraction, among others.

The problem now is being sure about what you want to find. A basic search for human hemoglobin currently provides 190 hits, so you need to browse a bit to find the most adequate entry. To keep things simple, I looked for a structure for adult hemoglobin, with no mutations, ligands or associations with other proteins. There are some database entries which seem to be right on target, namely 2DN2 for deoxyhemoglobin and 2DN1 for oxyhemoglobin.

There is a lot of information in each entry page. The whole structure can be downloaded as a .pdb and opened in appropriate structure visualization programs, of which there are probably several free options. Depending on the program used, retrieving the empirical formula can be trivial.

Forgoing the use of molecular visualization software, there is an "Experimental Data & Validation" section in the protein page, containing a "Full Validation Report" link to a PDF file. There, you can find a section called "Entry Composition" with useful information. For example, in the deoxyhemoglobin structure there are 4777 non-hydrogen atoms. The protein is composed of two copies each of two unique subunits (the alpha and beta subunits), totalling four subunits. The alpha subunit contains 1069 non-hydrogen atoms (685 $\ce{C}$, 187 $\ce{N}$, 194 $\ce{O}$ and 3 $\ce{S}$), while the beta subunit contains 1123 non-hydrogen atoms (724 $\ce{C}$, 195 $\ce{N}$, 201 $\ce{O}$ and 3 $\ce{S}$). Additionally, each subunit also contains one molecule of protoporphyrin IX coordinated to an iron atom. One important thing to note is that the structure also contains solvent molecules, in this case a total of 221 molecules of water, which must be removed from the atom count.

The only flaw is that hydrogen atoms are not shown. This is because X-ray crystallography measurements generally cannot detect hydrogen atoms, and so they are usually added separately to the backbone of the protein, in such a way to complete all atom valencies. I don't know how this can be done other than using a program.

A last thing to keep in mind is that no matter how good the measurement or the data processing, you can very well end up a few atoms off, especially regarding hydrogen atoms. Hopefully perfect precision is not required.


Electrophilic halogenation of 4-halo-3-(tert-butyl)phenols Question

Correct answer
A,B,C
My thoughts
Correct answer should be A,B. The main role is of the steric hindrance between the halogen atom and the tert-butyl group. So why are we even considering the electronic effects of both the groups? They are common in all the cases.
Further, the answer should be A only as its only the halogen atom changing each time, and hence the steric hindrance due to it. Why are we even considering other factors, as they are common in all the cases?

The options ABC are correct answers
Reason:

A: Steric effect of halogens will definitely be taken into account since all 3 halogens are acting on the same compound hence the primary factor which will make the product formed for each case be different is the steric hindrance due to their respective sizes.
B: For steric hindrance to be taken into account both atoms need to be bulky and have equal role to play in the first option we affirmed that due to different sizes among halogens there'll be steric hindrance causing different types of products to be formed however that is only true when we have a bulky group( here, tert butyl) providing steric to the group trying to attach(the halogens) to the adjacent c atoms. Hence for us to take steric hindrance as a factor for different products the steric hindrance caused due to the tert-butyl group and halogens is considered
C: The substitution of the halogens is taking place due to the +M effect of the -OH group thus option C must be true for us to proceed even a bit and it is the first requirement and only after this would we take into consideration the steric hindrance. In other words the reaction is taking place due to the electron donating effect of the -OH group since it is donating electrons to the ring via mesomeric and not inductive. And the steric hindrance serves so as to differentiate and predict which product will be formed


How can alkali metals form covalent bonds? Alkali metals have spherical outer s orbitals unlike elements like carbon which have protruding p orbitals that can "attach" onto other orbitals to form covalent compounds. Even in cases like methane elements like carbon can form hybrid orbitals that still have a protruding p like character.

So how can covalent alkali metals form covalent bonds given that their orbitals are spherical and not protruding?

Does the s orbital distort to be more p like in covalent compounds?

Can the s orbital hybridize and steal a more protruding like character from internal orbitals or from the orbitals of another atom?

For the most part, they don't.
However, lithium sometimes can. This is due to their valence shells being relatively close to the nucleus, so they have some electronegativity and can form somewhat covalent bonds.

For more info see this.


In infrared spectroscopy, why is there no C=C reading for E alkenes? For the (E)-2-pentene, there was no C=C reading, about 1600. For the (Z)-2-pentene, there is a C=C. Why is there a reading on the Z but not on the E?

The $\ce{C=C}$ stretch is responsible for this ir peak. For an ir absorbtion to occur, the absorption must result in a change in dipole moment.  If we examine the $\ce{C=C}$ stretch in cis- and trans-2-butene we find that

image source

cis-2-butene has a dipole moment (0.33 D) and upon stretching the double bond the dipole moment will change; therefore we would expect to see an ir peak for the $\ce{C=C}$ stretch in this molecule
trans-2-butene does not have a dipole moment  (0 D) and upon stretching the double bond the dipole moment will remain zero, there is no change; therefore we would not expect to see an ir peak for the $\ce{C=C}$ stretch in this molecule

A general rule based on these dipole moment arguments is that, for symmetrically di-substituted olefins $\ce{C=C}$ stretches in trans olefins are non-existent or very weak, the absorption is much stronger in the cis isomers.

We can extend this reasoning to molecules with dipole moments, if the dipole moment is very small.  Trans-2-pentene does have a dipole moment but it is very small and it is less than the dipole moment in the cis-isomer and the dipole moment in the trans-isomer (being smaller to begin with)  will change less upon stretching the double bond than it will for the cis isomer.  Therefore we would expect a very weak absorption in trans-2-pentene, while the cis isomer should show a stronger absorption.


Why doesn't graphite dissolve in organic solvent? The graphene layers are held by weak van der Waals' force, this should be comparable to the force holding solvent particles. Why doesn't graphite dissolve in organic solvent and separate into graphene layers?

While you cannot directly make a suspension or dispersion of graphene nano-platelets from graphite, it is possible to perform very similar process, called exfoliation, on graphite oxide, as graphite oxide has functional groups decorating it that reduce the functional strength of the van der Waal's forces between layers.
The exfoliation process can create a dispersion of graphene nanoplatelets in an organic environment, however, outside energy is often (if not always) necessary to promote the exfoliation of the individual layers. Most often, this energy is provided by sonicating the solution.
A very quick search of suggests that this method has been shown to work using NMP, DMF, THF, and ethelyne glycol. (see Paredes, Villar-Rodil, Martínez-Alonso, and Tascón, 2008 for more info)


When are transition state's energy barrier "reasonable" at a certain temperature? Every computational chemist who does a little bit of mechanistic studies has asked himself this question. When you obtain your first transition state and look at the activation barrier needed to access it, when would you tell that this required energy is reasonable and can be reached at a certain temperature ?

I have heard (never found a proof yet) that a biological process can be performed under standard conditions ($\pu{37 ^\circ C}$) with energy barrier as high as $\pu{15 kcal/mol}$.

Does anyone has a guideline for this?

Your question is really about what you consider to be a reasonable rate constant and so is somewhat subjective. The fastest a bimolecular reaction can be in solution is given by how fast the reactants can collide and then react on first collision. i.e reaction is diffusion controlled. In this case as reaction is at first meeting the activation energy must be small, $\le \pu{5  kJ mol-1}$. In water the diffusion controlled rate constant is in the region of a few times $\pu{10^10 dm3 mol-1 s-1}$. Most reactions are far slower than this.

The simplest, but very approximate, approach for other reactions is to use the Arrhenius equation to estimate the rate constant. The equation is
$$ k = A \exp(-E_a/RT),$$
where $E_a$ can be taken as your activation energy and you have then to estimate $A$. This can  be done for simple reactions using statistical mechanics and calculating the partition functions, but for anything complex you are left only with guessing values based on experimental data from similar reactions. However, because of the exponential dependence of the rate constant on activation energy even a small change in this can have a dramatic influence on the rate constant, far bigger than any similar change in $A$.

If you do have a potential energy surface then you can calculate trajectories along this surface by solving the (classical) equations of motion, repeatedly using energy sampled from a Boltzmann energy distribution,  and so work out a theoretical rate constant.


Sonication to purify solid product- the crude product just disappeared I am forming Fmoc-valine-citrulline dipeptide, and below is the literature protocol I am following (from Bioconjugate Chem 2002, 13 (4), 855-869):

Formation of Fmoc-Val-Cit

Fmoc-Val-OSuc (succinimidyl ester) ($\pu{14.91 mmol}$) in DME ($\pu{40 mL}$) was added to a solution of Cit ($\pu{2.743 g}$, $\pu{1.05 equiv.}$) and NaHCO3 ($\pu{1.315 g}$, $\pu{1.05 equiv.}$) in water ($\pu{40 mL}$). THF ($\pu{20 mL}$) was added to aid solubility, and the mixture was stirred at room temperature for 16 h. Aqueous citric acid (15%, $\pu{75 mL}$) was added, and the mixture was extracted with 10% 2-propanol/ethyl acetate ($\pu{2 \times 100 mL}$). The solid product began to precipitate but remained in the organic layer. The suspension was washed with water ($\pu{2 \times 150 mL}$), and the solvents were evaporated. The resulting white solid was dried in vacuo for 5 h and then treated with ether ($\pu{80 mL}$). After brief sonication and trituration, the white solid product was collected by filtration

I confirmed the formation of the crude product through HRMS and NMR. I then proceeded to carry out sonication to purify. I thought the purpose of sonication was to initiate crystallization. However, when I sonicated the crude product (~20 mins), I observed that the product was well-dissolved in ether, rather than forming a crystal. I used a Büchner funnel to collect the product but there was nothing left on the filter paper. Would pouring more slowly over the filter paper solve the problem? And more importantly, I want to understand the purpose of sonication step.

So I think we have established the problem (thank you @user6376297) . The material still contained enough iPrOH to cause it to dissolve in ether rather than give a filtrable solid because it was not adequately dried. What you need to do with your ethereal solution of product is remove the solvent under reduced pressure then dry the resulting solid under vacuum for at least 5hrs and preferably overnight. Then briefly (~1 min) sonicate the dried solid with ether and collect by filtration.
There is a lesson here: if you have a literature procedure that delivers your desired product in good yield then you follow it - without shortcuts.


Why some materials are transparent, but other ones aren't? We have two materials; say glass and wood. Light passes through glass but it doesn't pass through wood. Is it caused by atoms and molecules arranged in glass and wood such that they interact differently with light?

A brief summary: It is caused by the electrons in the particles in the substance. When electrons are orbiting the nucleus of an atom they can move to higher and lower energy levels. They do, however, need energy in the form of photons to do this, this means that when a ray of light hits the atom, the electrons in it will absorb some of the energy from the light ray. It needs a specific amount of energy to move up to the next energy level so will only absorb light from a specific part of the spectrum. This means that in some materials most of the light can pass through it and it may only absorb light waves not of visible light.

So wood, to some frequencies of light, is see through but not to the light waves we can see. That is why we can have mobile and radio signal from inside houses but only see through glass.

(Electrons are actually much more complicated and theoretical than this suggests but this)

Try watching this:  https://www.youtube.com/watch?v=Omr0JNyDBI0