Why does nitric acid not turn yellow everytime? I have dilutions of different samples, which are made with 0.1 M $\ce{HNO3}$. Only I left them on the lab table for a few days and one turned yellow.

So my question is why would one turn yellow and not the other ones? They all have the same concentration of sample. The matrix in all the samples are the same. The difference is the concentration of $\ce{Cl-}$ in the samples. The one with the lowest concentration of $\ce{Cl-}$ turned yellow.

Any ideas why?

Pure nitric acid is colourless, the brown colour is caused by dissolved nitrogen dioxide, $\ce{NO2}$.

Nitrogen dioxide is formed by the slow decomposition of nitric acid, according to $$\ce{4 HNO3 -> 4 NO2 + 2 H2O + O2}$$

Nitrogen dioxide is a good oxidant and it does so - supposed that it finds a suitable partner.

This would be chloride. So, the more chloride in solution the more redox chemistry furnishing colourless products.


Why is the probability of finding an electron in an orbital not 100 %? Here is what I know and please correct me if I am wrong:
An orbit can have 2 electrons only but electrons do not exist in defined orbits. An orbital is like a hollow sphere where the 2 electrons are most likely to be found. In other words the 2 electrons of an orbit are not confined to that pathway and can be found anywhere in the orbital.

Now my question is:
Why isnt the probability of finding the two electrons of an orbital $100~\%$? Most websites say the probability is $95~\%$, where else could they be and why?

There is a nonzero probability of finding an electron anywhere except for at the nodes, where the probability is 0 by definition. An orbital might be better thought of as an infinitely large cloud that is more dense in some areas than others. Integrated over all space, there is also by definition a 100% chance of finding an electron. This should make sense intuitively (if if doesn't exist somewhere then where is it and what does the integral even mean?).

Orbitals really aren't so well-defined as the nice little shapes we represent them with (think cloud), but we can make them well-defined by arbitrarily deciding how much space we wish to enclose. It is often decided for the practicality of the representation to show a space such that the probability of finding an electron in it is 95%, though mathematically it's true that there is a chance of finding it outside the space.


Can fumes from using bleach seep through plastic packaging and glass jars A bulging can leaked inside our kitchen cabinet. I googled how to handle it. I followed the instructions I found online. I mixed 1/4 bleach and 2 cups water. Applied it on all surfaces. Then applied dish soap diluted with water.

Does this produce toxic fumes? And can the toxic fumes seep through food even if they are covered and placed about 3 meters away? The cabinet still smells like bleach. What should I do about it?

Disclaimer: This answer is provided for information only. Use it at your own risk.

  I mixed 1/4 bleach and 2 cups water. Applied it on all surfaces. Then
  applied dish soap diluted with water.

This was not smart. Bleach should never be mixed with other chemicals, cleaners, or detergents. Rinsing with plain water first would have been less risky.

With bleach it is important to ventilate the area. Bleach should be thoroughly rinsed if it gets on anyone. Gloves are highly recommended for this type of cleaning.

  Does this produce toxic fumes?

Possibly.

Bleach attacks double and triple bonds in organic molecules. This action not only inactivates most disease causing organic particles, but also destroys the chromophores which give organic material their color. This reaction can sometimes liberate malodorous organic fumes which is best not to breathe.

Most inexpensive commercial bleach is a diluted form of sodium hypochlorite. Sodium hypochlorite can release chlorine gas; however, under normal usage bleach, nor its smell is likely to be harmful. The odor threshold for common bleach is 0.002 parts per million (ppm) which is well below the concentration it will cause any harm.

If on the off chance the smell is extremely irritating or if coughing occurs then get people and pets out of the area and ventilate the gas. Wait till the odor ceases to be irritating. If the odor is just annoying then it is likely safe.

  And can the toxic fumes seep through food even if they are covered and
  placed about 3 meters away?

If ventilation is good over a relatively short period of time then seepage is unlikely. If ventilation is poor over several days then seepage could turn into a problem. If the food smells or tastes either odd or like bleach, it may be best to throw it out.

  The cabinet still smells like bleach. What should I do about it?

Ventilating the cabinet till the smell goes away is best. Rinsing thoroughly with plain water, getting the water up as best as possible, and then allowing the surfaces to dry may work. Only after thoroughly rinsing and ventilating would it be worth it to clean with dilute vinegar help with the smell. Do not allow vinegar to mix with bleach!

A better disinfectant choice for just one bulged can may have been to use 70% isopropyl rubbing alcohol soaked into a dry rag. No chlorine fumes, much less likely to discolor, while still effective with germs.  This should go without saying...  Do not allow alcohol to mix with bleach!


Does the emission spectrum for a dye depend on the frequency of light used to excite the dye? Take a look at the emission spectrum for an arbitrary fluorescent dye: http://en.wikipedia.org/wiki/File:Fluorescein_spectra.jpg

Does the emission spectrum depend on exciting the dye at its peak absorbance wavelength $\lambda_{max}$?  Consider that there is going to be a spread of eigenstates around a particular excited state (http://www.olympusmicro.com/primer/java/jablonski/jabintro/jablonskijavafigure1.jpg), corresponding to particular molecular vibration or rotational modes.

The thought is that, if we excite a dye at a wavelength slightly red-shifted with respect to $\lambda_{max}$, or slightly blue-shifted with respect to $\lambda_{max}$, we'd start out at a different eigenstate clustered around the same excited state (say $S_1$).  I imagine there's a probability of decaying from each of these eigenstates in $S_1$, which gives us the spread for the emission spectrum, so if we bias the initial eigenstate, we'd achieve a different emission spectrum (presumably "statistically" red-shifted with respect to the excitation frequency).  Or is the fluorophore taking a rapid random walk through all accessible eigenstates belonging to a particular excited state?

I suppose we could also ask a similar question with respect to the photoelectric effect in terms of the energy provided to the excited electron (conditioned on having a threshold frequency to span a band gap).

Does the emission spectrum depend on exciting the dye at its peak absorbance wavelength $\lambda_{max}$?

No, it is independent of that. The excitation $S_0 \rightarrow S_1$ is vertical, bond length are not changed in the excitation process. Compare the mass of the nuclei with that of the electrons and you know why. You'll just end up in a different vibronic level of the electronically excited state.

However, before the emission takes place, this initial Franck_condon states relaxes to the vibrational ground state of $S_1$ with an equilibrated geometry and emission occurs from that (Kasha's rule).


Can a proton be ejected from an atom? Consider a neutral atom that has $n$ protons and $n$ electrons. Is it possible to remove a proton from the atom via some technique such as bombardment with another particle or applying an appropriate amount of energy to the system?

Is it possible to remove a proton from the atom via some technique
  such as bombardment with another particle or applying an appropriate
  amount of energy to the system?

Yes, this occurs in nature and can also be done in the laboratory.  In nature the process is known as radioactive decay.  Listed below are two decay processes that will remove a proton(s)

alpha decay: an atom ejects a helium nucleus (2 protons and 2 neutrons)
$$\ce{^239_94Pu -> ^4_2He + ^235_92U}$$
beta-plus decay: a positron is emitted and a proton is converted to a neutron.

In the (nuclear) lab, a process known as phototransmutation can be employed.  If a nucleus is irradiated with high-energy gamma rays (high-energy photons) it can absorb the energy and change to another element by ejecting a proton and\or neutron
$$\ce{^25_12Mg ->C[{h\nu}]\   ^1_1H + ^24_11Na}$$

You can make a rough estimate of the energy required as follows:
$$\mathrm{24.98583692\,amu~ \longrightarrow 1.00727647\,amu + 23.99096278\,amu}$$
comparing the starting and final mass, we find that 0.01240233 amu has been gained in the process.  Since 1 amu = 931.5 MeV, the gamma ray photon would have to have an energy of at least 11.6 MeV.


Trend in atomic volume across a period "Atomic volume decreases along a period, reaches a minimum at the middle, and then increases for the rest of the period"

Why does the atomic volume, along a period, initially decrease, reach a minimum and then increase?
The atomic radius decreases along a period, so I would have thought that the atomic volume would also decrease along a period.

Firstly, we need to clearly define "atomic volume" and "molar volume". Typically, one would understand "atomic volume" to be the volume of an atom of the element and it is directly related to the radius of the atom. Molar volume is a macroscopic property that is defined by Wikipedia as the ratio of the molar mass to the mass density of the substance at a certain temperature and pressure. As such, molar volume would not only be dependent on atomic radius but would also depend on the structure and bonding of the atoms.
With this understanding, it is easier to explain the trend that you have provided. The molar masses of the group 1 metals are considerably large but yet they have a relatively low mass density. As such, they are seen as spikes in the graph, with the highest molar volumes in their respective periods. Mass density generally increases from group 1 to group 13 resulting in the fall in molar volume across the period initially while for the transition metals, they stay roughly constant and take relatively large values. However, once we reach the non-metals, the trend seems to reverse. The density values for the non-metals seem to display some variability so I am not sure how the their molar volume trend can be explained. Obviously, their molar volume would be higher than that of the transition metals that come before them as those are significantly more dense.
Density data for the elements can be found here. More detailed molar volume data for the elements can also be found on another page at the same website.


how does electronic conductance of a metal change with density? The electronic conductance of a metal depends on its density.Does this mean that if we increase the density of the metal,its electronic conductance will increase?How does it happen?

You cannot change the density of a metal without also changing other properties.

Brass is a mixture of several elements. It can have different densities depending on the mixture of the elements. I haven't researched it, but the electronic conductance would likely be different for each type of brass.

Do you have a source for your statement?

  The electronic conductance of a metal depends on its density.


Does the hydrogen taken in an E1 reaction have to be antiperiplanar? I'm wondering if the hydrogen stolen during an $E_1$ reaction has to be antiperiplanar/anticoplanar like the hydrogen in an $E_2$ reaction.

Intuitively I'd say no, because the carbocation is flat so there's less steric hindrance than an E2 reaction, but I want to check.

As an example, we could use 3-methyl-2-bromobutane. Once that carbocation forms and shifts, does it matter which hydrogen is stolen?

In an E2 reaction, there is a strong preference for the hydrogen to be coplanar with the leaving group and pointed in an anti-direction to the leaving group. In an E1 reaction, the hydrogen does not have to be anti-periplanar to the leaving group because in the E1 reaction the deprotonation event happens in a separate mechanistic step, after the leaving group has left. So there's no need to define the relative orientation of the hydrogen and leaving group in the E1 reaction.

However, for the E2 reaction, remember that the anti-periplanar arrangement is preferred because it places the C-H bonding orbital anti to the C-LG antibonding orbital. In the transition state, those are the orbitals that reorganize to create the C-C pi bond.

With respect to the E1 reaction, there is no C-LG antibonding orbital, but there is an empty p-orbital on the carbocation. The C-H bond should be parallel to the p-orbital to create a similar (but strictly speaking, not anti-periplanar) alignment of orbitals.


Oxalate ion dentiticity - why is it bidentate and not tetradentate? Why is the oxalate ion a bidentate ligand? Why does it not act like a tetradentate ligand (having four oxygen atoms, with lone pairs)?

The oxalate anion cannot be tetradentate. While there are four oxygen atoms, the ion is planar and the oxygen atoms are pointing in different directions.

The best any metal can do is bond to two of them, although there are two bonding modes possible:

Perhaps you were envisioning some $\kappa^4$-bonding mode that is
"side-on". However, most of the nonbonding electrons (orbitals) are also in the plane.

This is not to say that oxalate cannot form four bonds to metal centers; it just cannot do so with only one metal center. Oxalate can act as a bridging ligand


Dihaloalkane reaction with sodamide/ammonia In the following sequence of reactions:
$$\ce{CH3-CH2-CH2-I ->[\text{Alc.}~KOH] \textbf{A} ->[Br2] \textbf{B} ->[NaNH2/NH3] \textbf{C}}$$
the product C is:

alkene
alkanol
alkyne
alkylamine

I haven't really came across a reagent $\ce{NaNH2}/\ce{NH3}$ in my whole session. Well, I have got B which is $\ce{CH3-CH(Br)-CH2-Br}.$
Then, how do I proceed next? How will my substrate  react with that reagent and what product will I achieve?

You were correct up to this point, the strong base alcoholic KOH sets off elimination instead of substitution so you will not get an alcohol.
Accordingly your products A (presumably propene) and B are correct.
The amide/ammonia is an even stronger base than alcoholic KOH and will also produce elimination.  With a vicinal dihalide like your B, there is an added twist.  Read here to find out what happens.


Major product of mononitration of 3-methylphenol From NCERT Class Chemistry Part 2:

Example 11.5
Write the structures of the major products expected from the following reactions:
(a) Mononitration of 3-methylphenol
(b) Dinitration of 3-methylphenol
(c) Mononitration of phenyl methanoate.
Solution
The combined influence of $\ce{-OH}$ and $\ce{-CH3}$ groups determine the position of the incoming group.

Since $\ce{-OH}$ and $\ce{-CH3}$ are o,p-directing groups, shouldn’t the major product be 3-methyl-2-nitrophenol? Why not?

The product you are suggesting is very unstable as compared to the ones given as answers due to steric factors. Thus it would not be formed.

  Since −OH and −CH3 are o,p-directing groups, shouldn’t the major product be 3-methyl-2-nitrophenol? Why not?

The reasoning that hydroxyl and methyl are o,p directing groups is not at supportive as to why 3-methyl-2-nitrophenol should be formed in preference to the ones given as answers. The o,p directing effect just rules out the meta product and steric factor decides among the remaining possibilities here.


How can a protein folding transition state have zero lifetime? I'm doing a module on my Biochemistry course looking at protein folding, and in a discussion of [folding] transition states I was a little confused at the thought of a zero-lifetime transition state - especially as for folding "the TS is probably an ensemble of states (a population) with a lifetime of 0".

In searching for anything with mention of protein folding TS with lifetime of 0 I found this old discussion ported to Google Groups, in which this is described as an 'inherent contradiction' to TS theory.

  ..."capital K, double dagger" is defined as a *thermodynamic equilibrium
  constant*.  So, for example, a Transition State Theory applied to a
  bimolecular reaction involves a term corresponding to the
  *thermodynamic equilibrium constant* for the transition state [AB]**.   At the same time, the transition state itself is presumed to have
  *zero lifetime*.

  In other words, the Transition State Theory is wonderfully
  self-contradictory in its very essence (it contains an *equilibrium*
  constant for something that has zero lifetime). (via)

I'm just wondering if anyone can enlighten me / agree / disagree with this, and maybe suggest some of the original literature I might go to for an understanding of the meaning to this.

One of my chemistry textbooks has a small section on Hammond's postulate but without explanation to why transient is defined as zero lifetime:

Further, it's possible I've brought myself to the wrong equations but if I understand correctly the mass balance equation (described from the source above as the reason TS is only a 'quasi-steady state') says here the dm/dt does not tend to zero, i.e. zero lifetime is overextrapolation of the theory :

  one can assume steady state for a species as long as its production
  rate and its lifetime t have both remained approximately constant for
  a time period much longer than t. When the production rate and t both
  vary but on time scales longer than t, the steady-state assumption is
  still applicable even though the concentration of the species keeps
  changing; such a situation is called quasi steady state or dynamic
  equilibrium. The way to understand steady state in this situation is
  to appreciate that the loss rate of the species is limited by its
  production rate, so that production and loss rates remain roughly
  equal at all times. Even though dm/dt never tends to zero, it is
  always small relative to the production and loss rates. (via)

Please let me know in comments if I can make this question more clear, thanks.

Update: This is a question about protein folding transition states - please can you not give me answers or comments directing me to resources covering the basics of transition state theory.

I don’t know where the person from the Google Groups you quote got the information that a transition state is "presumed to have zero lifetime", but a far as I can tell it is wrong. On the contrary, the existence of transition states is postulated in transition state theory (TST), which evidently implies that within the TST framework transition states have non-zero lifetime. Existence clearly contradicts zero lifetime: if some object has zero lifetime, what do you mean then by its existence?

Next, while the lifetime of a transition state in TST is postulated to be not zero, it is also proposed to be quite small, on the order of the time of a single molecular vibration ($10^{-13}$ seconds). Why do we have to make such a proposition? Well, transition state can be defined as the state corresponding to the highest potential energy between the reactants and the products (i.e. along a reaction coordinate) such that the probabilities of forming the reactants or products are equal. On potential energy surface (PES) then transition states correspond to the so-called saddle points (first-order saddle points, to be more precise), which, loosely speaking, can be thought of as a maximum in one direction (along the reaction coordinate) and minimum in all other directions. And structures that correspond to maxima (even in one direction) on PES) are unstable: once a system is at a maximum on PES it will almost instantly fall from it to one of the two neighbouring minima (either to reactants or to the products side).

Due to small assumed lifetime of a transition state, it is often said that "no physical or spectroscopic method is available to directly observe the structure of the transition state" (quote from Wikipedia article on TST). Although, with femtosecond IR spectroscopy one can actually observe molecular structures extremely close to the transition state structures. The practical problem is that quite often along the reaction coordinate in addition to a maximum corresponding to transition state between the reactants and the products there exist a minimum (or few) corresponding to the so-called reactive intermediates, and it is quite difficult to distinguish between transition states and reactive intermediates.

Note here that reactive intermediates correspond to minimum in all directions on PES, and are thus stable, or, to be more precise, metastable, since they are high-energy molecules which are quickly converted into more stable molecules (either another reactive intermediate or eventually the products). Reactive intermediates are also short-lived but with lifetimes appreciably longer than a molecular vibration, so they cannot usually be isolated, but they can be observed with the already mentioned femtosecond IR spectroscopy.

To summarise:

One should not in the first place think about transition states being real (and consequently detectable). A transition state is a part of a model description of a chemical reaction introduced by TST, where we just postulate their existence, and say no word on are they really exist or not. Irregardles of their real existence, transition states exist in our model description, and their existence implies non-zero lifetime.
Besides, with femtosecond IR spectroscopy it is possible to directly observe a transition states for some chemical reactions and it was actually done.


Program to convert WFN-file to Molden I have tried google but without success so here the question: does anybody know a tool that can covert WFN-file to Molden or to Molpro format (including orbital info not the coordinates only).

When a wfn-file is loaded into Multiwfn and one tries to convert it into a molden file via 100 > 2 > 6, it says Error: This function works only when input file contains basis function information.

So I guess, that as the wfn file lacks important information, it’s not easily possible to convert it into a molden file.


Why does latex tubing become white and brittle over time, especially where it's bent? Here's a picture of some latex tubing we have in the lab, bought many years ago and stored in a drawer.

What's responsible for the white corners near places where the tubing is bent? Is this dehydration, oxidation, an aging effect accelerated by stress, or something else?

Natural rubber instantly crystallises (reversibly!) under mechanical deformation. That is actually very desireable, because it gets tougher under load! High quality tyres (for trucks, not regular car tyres) today still contain quite a large amount of the world's natural rubber production for that reason.

You can try to pull at a lenght of the rubber tube, it should become pale, and go back to brownish-opaque when you release the pull.

The drawback is this also makes the material a bit brittle, if deformed beyond a certain limit. I believe this is what you see with your rubber tubing. If you unbend the kinks, the white colour should fade, but the tube is probably damaged at this point.


Bonding in diatomic C2, a carbon-carbon quadruple bond? Carbon is well known to form single, double, and triple $\ce{C-C}$ bonds in compounds. There is a recent report (2012) that carbon forms a quadruple bond in diatomic carbon, $\ce{C2}$. The excerpt below is taken from that report. The fourth bond seems pretty odd to me.

  $\ce{C2}$ and its isoelectronic molecules $\ce{CN+}$, BN and $\ce{CB-}$ (each having eight valence electrons) are bound by a quadruple bond. The bonding comprises not only one σ- and two π-bonds, but also one weak ‘inverted’ bond, which can be characterized by the interaction of electrons in two outwardly pointing sp hybrid orbitals.

  According to Shaik, the existence of the fourth bond in $\ce{C2}$ suggests that it is not really diradical...
  If $\ce{C2}$ were a diradical it would immediately form higher clusters. I think the fact that you can isolate $\ce{C2}$ tells you it has a barrier, small as it may be, to prevent that.

Molecular orbital theory for dicarbon, on the other hand, predicts a C-C double bond in $\ce{C2}$ with 2 pairs of electrons in $\pi$ bonding orbitals and a bond order of two. "The bond dissociation energies (BDE) of $\ce{B2, C2}$, and $\ce{N2}$ show increasing BDE consistent with single, double, and triple bonds." (Ref) So this model of the $\ce{C2}$ molecule seems quite reasonable.

My questions, since this is most definitely not my area of expertise:

Is dicarbon found naturally in any quantity and how stable is it? Is it easy to make in the lab? (The Wikipedia article reports it in stellar atmospheres, electric arcs, etc.)
Is there good evidence for the presence of a quadruple bond in $\ce{C2}$ that wouldn't be equally well explained by double bonding?

Okay, this is not so much of an answer as it is a summary of my own progress on this topic after giving it some thought. I don't think it's a settled debate in the community yet, so I don't feel so much ashamed about it :)
A few of the things worthy of note are:

The bond energy found by the authors for this fourth bond is $\pu{13.2 kcal/mol}$, i.e. about $\pu{55 kJ/mol}$. This is very weak for a covalent bond. You can compare it to other values here, or to the energies of the first three bonds in triple-bonded carbon, which are respectively $348, 266$, and $\pu{225 kJ/mol}$. This fourth bond is actually even weaker than the strongest of hydrogen bonds ($\ce{F\bond{...}H–F}$, at $\pu{160 kJ/mol}$). Another point of view on this article could thus be: “valence bond necessarily predicts a quadruple bond, and it was now precisely calculated and found to be quite weak.”

The findings of this article are consistent with earlier calculations using other quantum chemistry methods (e.g. the DFT calculations in ref. 48 of the Nature Chemistry paper) which have found a bond order between 3 and 4 for molecular dicarbon.

However, the existence of this quadruple bonds is somewhat at odds with the cohesive energy of gas-phase dicarbon, which according to Wikipedia is $\pu{6.32 eV}$, i.e. $\pu{609 kJ/mol}$. This latter value is much more in line with typical double bonds, reported at an average of $\pu{614 kJ/mol}$. This is still a bit of a misery to me…


Thermal stability of alkali metal hydrides and carbonates Why is it that thermal stability of alkali metal hydrides decreases down the group, but for carbonates, it increases?

I used Fajan's rule to check for ionic character but somehow this is only applicable for carbonates.

Down the group, atoms of the alkali metals increase in both atomic and ionic radii, due to the addition of electron shells. This results in the charge density of their corresponding cations decreasing down the group. Thus, as we go down the group, the cations become "softer".

Now, note that the hydride ion is "hard", having high negative charge density. While the carbonate ion is "soft", having a lower negative charge density.

Applying the "hard-soft acid base" (HSAB) principle, we would expect the "hard" ions to form more stable compounds with each other and the "soft" ions to form more stable compounds with each other.

Since the heavier alkali metal ions are softer, we would expect them to form more stable compounds with carbonate. As for the harder lighter metal ions, we would expect them to form more stable compounds with hydrides.


Why can entropy be measured in joules per kelvin? I'm a little confused as to what the measurement associated with entropy represents. It's obviously the joules of energy per kelvin, but how does this measure the disorder of a system?

The definition of entropy in 'Classical Thermodynamics' is $$\Delta S = \int\frac{\delta Q}{T}$$

The quantity on the left is the entropy change associated with a physical process. $\delta Q$ is an inexact differential, made exact by an integrating factor $\frac{1}{T}$ This notion was devolped by Kelvin and Cartheodory (I believe). This comes out of the analysis of certain kinds of differential equations called Pfaffian forms.

One notes the dimensions ("units") of this quantity, and observers them to be $\mathrm{J\ K^{-1}}$

It was only later, that the statistical definition of entropy (developed by Boltzman) came to be (and was found to be equivalent to the thermodynamic definition).

$$S = -k_B \sum_{i}p_i\ln(p_i)$$

In this framework entropy is a logarithmic measure of the number of states with significant probability of being occupied. In this expression, the the quantity is changed from dimensionless to dimensionful, by the Boltzmann Constat, which has the units, (you guesed it)  $\mathrm{J\ K^{-1}}$

The measure of "disorder" is in the term $\sum_{i}p_i\ln(p_i)$, which has no units. The units are appended to it by the Boltzmann Constant.


How to find the oxidation number of Co and Cr in [Co(NH3)6] [Cr(C2O4)3]? In the question I actually got I had to name the above compound, and I know how to name it but I don't know how to find the oxidation number of the above compound(which has to be mentioned in the IUPAC name). I actually know how to find the oxidation number in complexes but here both the ions are complex ions which makes it confusing. If someone needs the name its Hexaamminecobalt(?) trioxalatochromate(?)

$\ce{Co}$ is known to have two possible oxidation numbers in solution : +II and +III. Both make complexes with 6 $\ce{NH3}$. If $\ce{Co}$ is at +II, the charge of the complex is $2$+, as $\ce{NH3}$ is not charged.
If $\ce{Co}$ is at +III, the charge of the complex is $3$+ for the same reason.
For chromium, it is similar, as $\ce{Cr}$ may have oxidation number +II and +III. But the ligand oxalate is charged $-2$. The complex of $\ce{Cr}$ with three oxalate ions must compensate the charges of the cobalt complex (+$2$ or +$3$). So the charge of the chromium complex has to be -$2$ or -$3$. In other words, the complex ion is either $\ce{[Cr(C2O4)3]^{2-}}$ or $\ce{[Cr(C2O4)3]^{3-}}$. Let's compare the two possibilities, and the oxidation numbers of $\ce{Cr}$ in these two structures.
To get a complex $\ce{[Cr(C2O4)3]^{2-}}$, $\ce{Cr}$ oxidation number should be equal to $x$, with $x + 3(-2) = -2$. So $x$ = $+4$ or +IV. And unfortunately, this oxidation number IV does not exist for $\ce{Cr}$ in solution. This structure has to be abandoned,
On the other hand, to get a complex $\ce{[Cr(C2O4)3]^{3−}, Cr}$ must have an the oxidation number $y$ given by : $y +3(-2) = +3$. So $y$ = $+3$ = +III, and this is possible and even welcome. It corresponds to one of the requirements developed previously. As a final result, the complex ion $\ce{[Cr(C2O4)3]^{3-}}$ has a charge $3-$ that can be coupled with $\ce{Co(NH3)6^{3+}}$.
As a consequence, both Co and Cr are at oxidation number +III in the original structure $\ce{[Co(NH3)6][Cr(C2O4)3]}$


Mixing a Tagged Sample of Lead Iodide with another Iodide Salt A sample of sparingly soluble $\ce{PbI_2{(s)}}$ containing radioactive $\ce{^{133}I}$ is added to 0.10 M $\ce{KI_{(aq)}}$ and stirred overnight, what happens?

I The radioactivity of the liquid phase increases  significantly.

II The concentration of the $\ce{I^{–}}$  ion in solution increases significantly.

The answer is I. I thought II would be true because some of the water evaporates, but I guess not to a "significant" extent. So that marks II off, and leaves II. I'm fairly new to nuclear chem so I don't know much about how/why radioactivity increases, could someone please explain this to me? How could radioactivity "increase?"

You are missing a basic aspect of how a precipitate in solution behaves. The precipitate isn't just statically immersed in the solution but rather some is dissolving and some of the ions (or molecules) in solution are precipitating out. So for a salt like $\ce{PbI2}$ there will be an exchange of ions which are in the precipitate with those which are in solution.


Why "monoxide" but not "diodine"? When naming compounds, sometimes when there are two vowels in a row the second is elided: this happens for example with "mono-oxide", which becomes "monoxide" instead.
Why is this not always applied, e.g. with "diiodine"? Why aren't the repeated i's removed to make it "diodine"?

Both "monooxide" and "monoxide" are used in the literature, yet "monoxide" is being  used more often (Google Books Ngram Viewer).
Although this is an accepted elision, it is not the preferred one, and must not set a precedent for other cases when multiplicative prefix ends with the same vowel as the root word begins with, such as "diiodine".
According to the current version of Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 [1, p. 31]:

IR-2.7 ELISIONS
In general, in compositional and additive nomenclature no elisions are made when using
multiplicative preﬁxes.
Example:

tetraaqua (not tetraqua)
monooxygen (not monoxygen)
tetraarsenic hexaoxide

However, monoxide, rather than monooxide, is an allowed exception through general use.

Further, from section IR-5.2 Stoichiometric names of elements and binary compounds [1, p. 69]:

The multiplicative preﬁxes precede the names they multiply, and are joined directly to them without spaces or hyphens. The ﬁnal vowels of multiplicative prefixes should not be elided (although ‘monoxide’, rather than ‘monooxide’, is an allowed exception because of general usage).
[…]
Examples:
[…]

$\ce{NO}$ nitrogen oxide, or nitrogen monooxide, or nitrogen monoxide

References

IUPAC. Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (the “Red Book”), 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; RSC Publishing: Cambridge, UK, 2005. ISBN 978-0-85404-438-2. IUPAC website


Does magnesium sulfate in aqueous solution form magnesium (hydrogen-)carbonate when hydrogen carbonat is present? I'm using an additive of magnesium sulfate to complement my fertilizer. The additive lists the following ingredients:

I know that it does NOT contain magnesium oxide, because due to historical / commercial reasons ingredients of phosphorus, potassium or magnesium are indicated as mass percentages of the oxides with the element in question. Also the merchant confirmed that the source of magnesium is magnesium sulfate.
I noticed the precipitation of a white powder at the bottom of my reservoir. I suspect it to be magnesium carbonate. The  mains water contains the following chemicals. The dose of magnesium sulfate is within the range recommended. The precipitation occurs even if I do NOT use any fertilizer, so I think It is a reaction with chemicals in mains water supply.

Is the precipitate magnesium carbonate?
How can I prevent this precipitation? What are factors influencing it besides the availability of hydrogen carbonate?
EDIT:
User AJKOER suggested that the precipitate is calcium sulfate. I don't think that is the case because:

The solubility of calcium sulfate is given at 2.1g / liter @ 20° C  (https://en.wikipedia.org/wiki/Calcium_sulfate). This equates to about 2.1 * 0.71 ~ 1.48g sulfates per liter. (mass of SO4 /  mass of CaSO4 = 0.71)
I'm adding only about 92.1 mg / liter sulfates from magnesium sulfates
The mains water contains only 13.3mg / liter sulfates
92.1 mg / liter + 13.3mg / liter <<<< 1480 mg / liter

The white precipitate is more likely insoluble white calcium sulfate.
Per a reference on the making of fertilizer, to quote:

Calcium, magnesium, and sulfur are also important materials in plant growth. They are only included in fertilizers in small amounts...

Net ionic reaction:
$\ce{ Ca^2+ (aq) + SO4^2- (aq) -> CaSO4 (s) }$


Influence of vapor pressure on surface tension How does a liquid's vapour pressure affect its surface tension?

i.e. to say, if a liquid's vapour pressure increases, will the surface tension increase or decrease?

The SERP for this question yields only research papers to which I don't have access. This theory isn't given in my textbook either.

High surface tension is caused by strong intermolecular forces, the same are responsible for a low vapour pressure.

It is however quite a different matter to put a molecule on a surface and to completely strip it from a condensed phase and put it into the gas phase. There sure is no (inverse) proportionality or similar between vapour pressure and surface tension, and they don't "affect" each other.

A really stupid test question, imo.


Preparation of Mercurous nitrite I read that Prafulla Chandra Ray prepared Mercurous nitrite for the first time and published a paper.

Both Mercurous ions and nitrite ions are quite stable. So they should readily form the compound.

Why is it difficult to prepare it, and how to prepare it?
Can it be obtained by simply mixing $\ce {Hg2Cl2}$ and $\ce{AgNO2}$, where $\ce{AgCl}$ will be precipitated?

1.Why is it difficult to prepare it, and how to prepare it?

Prafulla Chandra Ray prepared mercurous nitrite, $\ce{Hg2(NO2)2}$ by reacting metallic mercury with cold, dilute nitrous and nitric acid. According to his paper(1):

  Mercurous nitrite is the product of the combined action of nitrous
  and nitric acids on mercury :
  $$\ce{2Hg + NO2.H + HO.NO2 -> Hg2(NO2)2 + H2O}$$
   Some of the nitrite is decomposed by the nitric acid, the quantity
  of nitrous acid thus rapidly growing, until mercurous nitrite and nitrate
  are accumulated in molecular proportions. After some, the nitrous acid remains constant in the solution. It has become the catalytic agent between the nitric acid and the mercury,which now interact in the following manner:
  $$\ce{4Hg + 4HNO3 -> Hg2(NO2)2 + Hg2(NO3)2 + 2HO}$$

It is difficult to prepare the compound because many mercury-nitrite species is generated in the solution like  $\ce{Hg(NO2)2}$, $\ce{Hg2(NO3)2.2H2O}$, $\ce{Hg2(NO3)2.OH.Hg(NO3)2}$ etc. These salts exist in the solution and it is difficult to separate the main salt i.e mercurous nitrite from these salts and isolate in pure solid salts as all salts have similar properties and they tend to impure the solid mercurous nitrite.

  2.Can it be obtained by simply mixing $\ce{Hg2Cl2}$ and $\ce{AgNO2}$, where $\ce{AgCl}$ will be precipitated?

Reaction between $\ce{HgCl2}$ and $\ce{AgNO2}$ to yield mercuric nitrite $\ce{Hg(NO2)2}$ has been reported (2) but reaction between $\ce{Hg2Cl2}$ and $\ce{AgNO2}$ to yield mercurous nitrite $\ce{Hg2(NO2)2}$ is yet to be reported.

References

http://pubs.rsc.org/en/content/articlepdf/1905/ct/ct9058700171
http://chemiday.com/en/reaction/3-1-0-5880
http://www.insa.nic.in/writereaddata/UpLoadedFiles/IJHS/Vol49_4_3_AChakravorty.pdf
http://nopr.niscair.res.in/bitstream/123456789/11011/1/IJCA%2050A(2)%20137-140.pdf
https://www.researchgate.net/publication/289285311_On_mercurous_nitrite_and_a_basic_mercurous_nitrate_derivative


Mechanism of 5‐(iodomethyl)oxolan‐2‐one synthesis from 4‐pentenoic acid and iodine in acetonitrile There is a question in my organic chemistry textbook with an answer. What it doesn't explain (and neither does my theory textbook, I checked) is the mechanism involved in the reaction. The reaction is as follows:

Can someone help me?

https://en.wikipedia.org/wiki/Iodolactonization
I+ reacts with the CC-π electron donor. A cyclic iodonium three-membered ring intermediate is formed. The carboxylate group opens the electrophilic intermediate to form the product lactone. 5-membered rings (gamma-lactones) https://en.wikipedia.org/wiki/Lactone are prefered due to optimum entropic and enthalpic parameters of activation.


Is it good practise to mix double and triple zeta basis sets? I need to compute extremely accurate single point energies using the coupled cluster method. The methods to be employed scale largely with the number of basis functions. A few basis functions may have a huge impact in the computational cost.

I planned assigning a 3ζ basis set to a central atom, and 2ζ for the surrounding atoms. I cannot compute everything at 3ζ.

Clearly, these small basis sets are intended for estimate the contribution of high excitation.

Are there any reason which prevent this mixture, or any argument supporting the usage of 2ζ for every atom?

It is perfectly fine, and actually quite common, to use big basis sets for the most important atoms (perhaps at the active site of the chemical reaction) and a smaller basis set for the surrounding hydrogen atoms (for example).

The only problem you might face would be, that this complicates the basis set extrapolation to get closer to the complete basis set limit. Such an extrapolation works best if you are, for example, extrapolating from 2Z & 3Z to infinity (not a mixed extrapolation such as 2Z/3Z & 3Z/4Z to infinity). In fact you won't be able to extrapolate because you likely won't be doing calculations with a 1Z basis set, and you won't be able to do a 3Z/4Z mix (because if you could do that, then you wouldn't need to mix 2Z into your 3Z calculation!).

I do want to point out that if you need to compute "extremely accurate single point energies" then even if you could use 3Z on the whole molecule, it would not likely even get you to 1 kcal/mol precision for energy differences. So as long as "extremely accurate" means in your context "+/- 10 kcal/mol for energy differences", you are doing everything just fine (and don't need a basis set extrapolation anyway). If you are aiming for better accuracy than that, you'll probably need a basis set extrapolation (meaning you want all atoms represented at 3Z level).

In the following reference a "mixed" basis set was used;
in this paper TZP (3Z) is used for the atoms Fe, S, and Mo while SVP (2Z) is used for C, H, O, and N:

Li, Z.; Li, J.; Dattani, N. S.; Umrigar, C. J.; Chan, G. K. The electronic complexity of the ground-state of the FeMo cofactor of nitrogenase as relevant to quantum simulations. J. Chem. Phys. 2019, 150 (2), 024302. DOI: 10.1063/1.5063376.

Since the molecule is FeMoco (an iron-sulfur complex) it is likely that the Fe and S are very important (so they have a big basis set). It is also believed that the molybdenum (Mo) is important, because why else would a biological molecule have Mo in it, so it has a big basis set too. The rest have a small basis set.


Reaction mechanism for reduction of alkyl halide to alkane Alkyl halides are reduced by zinc and dil.  $\ce{HCl}$ into alkanes
   $$\ce{ CH3Cl + H2 ->[Zn][dil\text{.} HCl]CH4 + HCl}$$

What I don't understand is the underlying mechanism behind this reaction,  is it free radical or nucleophilic substitution.  If it is the latter, then why isn't $\ce{CH3-CH3}$ a product?
I would appreciate if someone could explain the entire mechanism behind this reaction.

The action of HCl on Zn results in the formation of hydrogen.

Next, there is some surface chemistry where the diffusion of atomic hydrogen (or hydrogen radical) occurs with select metals. See, for example, this 2008 thesis "Alkaline dissolution of aluminum: surface chemistry and subsurface interfacial phenomena", by Saikat Adhikari, who cites the "ingress of atomic hydrogen into the metal".

Then, the action of a hydrogen radical on an alkyl halide:

$\ce{.H + CH3Cl -> HCl + .CH3}$  (Source on radical reactions)

$\ce{.H + .CH3 -> CH4}$

I suspect more reactions are possible resulting in minor products, for example:

$\ce{.CH3 + .CH3 -> C2H6}$

$\ce{.H + CH4 -> H2 + .CH3}$

......

The final products include HCl, CH4 and other minor products.


How to evenly ionize in mass spectrometry I know that in mass spectrometry they charge different molecules and then run them through a magnetic field and then see where they land.

Depending on this, they calculate their masses relative to each other. To do this however, they have to ionize the different particles evenly. If one molecule is charged +2 and the other +1, then the results would be thrown off.

How do they do they make sure that the charges of all the ions are the same?

I know that in mass spectrometry they charge different molecules and then run them through a magnetic field and then see where they land.

That is indeed how some mass spectrometers work, but not most of them.  Most mass spectrometers in the world do not use magnets of any type.  Instead they rely on electric fields to move the ions (i.e. charged molecules) and to see where they land.

  Depending on this, they calculate their masses relative to each other. To do this however, they have to ionize the different particles evenly. If one molecule is charged +2 and the other +1, then the results would be thrown off.

Multiple charge states definitely happen in mass spectrometry.  Mass spectrometrists like to say that name "mass spectrometry" is a misnomer, since nearly every "mass" spectrometer in the world is really a mass-to-charge ratio spectrometer.  That is, MS doesn't measure the mass of molecules, it measure the mass-to-charge ratio (or $m/z$) of an ion.

The problem of multiple charge states is especially common in the mass spectrometry of proteins, because proteins are very large molecules with many sites that can become charged.  For example, the image below shows the "mass" spectrum (i.e. $m/z$ spectrum) of horse myoglobin, which is a compound with the formula of $\ce{C769 H1212 N210 O218 S2}$.  The spectrum was obtained from this web page which further describes the multiple charge state phenomenon.

The different peaks correspond to different charge states.  The uncharged molecule has a mass of 16951 Daltons.  The peak at an $m/z$ 737.99 corresponds to a +23 charge state, and the peak at 1542.02 corresponds to a charge state of +11.  The peaks in between correspond to charge states from +22 to +12.

  How do they do they make sure that the charges of all the ions are the same?

There is no foolproof way to do this chemically or physically.  Some modes of ionization, such as electron ionization or APCI tend to favor the formation of singly-charged ions relative to electrospray ionization, but these modes of ionization are sadly ineffective on large molecules such as proteins.  And there are many exceptions to this tendency.

The mass spectral data can be manipulated computationally through a process called deconvolution in an attempt to computationally estimate the true mass spectrum at a given point.  A 1992 paper describes one computational method for spectral deconvolution.  This and/or very similar algorithms are often provided by mass spectrometer vendors in their software for browsing and analyzing MS data.


Why does thermodynamics only apply to macroscopic systems at equilibrium? A text says:

  The laws of thermodynamics deal with energy changes of macroscopic systems involving a large number of molecules rather than microscopic systems containing a few molecules. Thermodynamics is not concerned about how and at what rates these energy transformations are carried out but is based on initial and final states of system undergoing the change. Laws of thermodynamics apply only when a system is in equilibrium or it moves from one equilibrium state to another equilibrium state. Macroscopic properties like pressure and temperature fo not change with time for a system in equilibrium state.

Why do the laws of thermodynamics not apply when the system is microscopic or not at equilibrium but in intermediate stage? How can we say that thermodynamics cannot tell anything about rate and method of energy transformations?

The mathematics of equilibrium systems are far simpler than others

It isn't strictly true that thermodynamics doesn't apply to systems that are not at equilibrium only that the simple theorems and formulae of equlibrium thermodynamics only apply to systems at equilibrium.

The reason why most discussion and most teaching is about those equilibrium systems is because the mathematical results are far simpler and far easier to teach. They are also widely useful even with systems that are not exactly at equilibrium. Chemists can understand the key features of many reactions, for example, by understanding the equilibrium thermodynamics of the reaction (like the heat of formation of the products). Exceptions can be explained by understanding kinetic barriers in the reaction pathway.

But when details of non-equilibrium behaviours are required, the behaviour of many systems is so complicated there are no easy rules to teach or to talk about.

Non-equilibrium thermodynamics is actually a very large field and key results have won Nobel prizes (Prigogine won the 1977 Nobel in Chemistry for work on key aspects of it). Arguably a great deal of the complexity of the universe and life itself is the result of the complexity and structures that can arise when systems are not at equilibrium. So you could even argue that our ability to understand any thermodynamics is a result of the complexity the can arise from non-equilibrium systems. If the subject is that complex, there are not going to be many simple formulae or generalisations to teach or talk about.

And that is probably the real reason why some people think thermodynamics only applies to systems at equilibrium: anything else is too complicated to talk about so educators and textbook writers tend to avoid the subject.


What are the degrees of freedom (from Gibbs phase rule) for an azeotropic mixture? For an azeotropic binary mixture, the number of phases is 2 and number of components are 2. So, according to Gibbs phase rule, the number of degrees of freedom should be 2. But, the right answer is 1.
Can someone explain this?

Gibbs' phase rule simply counts the number of variables required to specify the system and the number of relationships between these variables (constraints). Any failure of Gibbs' phase rule is symptomatic of additional variables or constraints that have not been taken into account.

This system is generally specified by four variables, the mole fraction of one species in the liquid phase, $x$, the same in the gas phase, $y$, the pressure $P$, and the temperature $T$. Equilibrium of the liquid and gas phases in both species provides two constraints. The specification of an azeotrope, for which $x = y$, provides a third constraint, as has been mentioned in the comments. One degree of freedom remains.


Does spectrochemical series trend depend on the metal center? Can the trend of spectrochemical series vary/fluctuate according to the central metal atom (CMA) that is being used?
For example, for the CMA $\ce{Ni^2+}$ the ligand $\ce{H2O}$ will have more CFSE than $\ce{NH3}$ because $\ce{NH3}$ acts as a weak field ligand (WFL) for $\ce{M^2+},$ even worse then $\ce{H2O},$ though $\ce{NH3}$ is placed higher then $\ce{H2O}$ in the spectrochemical series.

The spectrochemical series of ligands, our instructor clarified, gives only a rough idea of the relative strengths of ligands, and may vary for different Central Metal Atoms (CMAs).
An important example that I have encountered is of $\ce{[Co(H2O)6]^+^3}$, wherein $\ce{H2O}$ acts as a Strong Field Ligand (SFL), despite its lower position in the series.
Also, JD Lee mentions in his Concise Inorganic Chemistry (5th ed., Wiley publications) that there is an increase in CFSE as we go down the group in d-block, and several instances may be found where the ligands designated as WFL begin to act as SFL for 4-d or 5-d members.
Overall, I believe there are certain deviations from the series, but it generally works well.


Structure of benzene dimer in an vibrationally excited state It is known that the benzene dimer has few stable structures, T-shaped, parallel-displaced, and sandwich structure.

Now if we excite the dimer such that it goes to a higher vibrational level (which is close to the dissociation limit but the dimer is still bound).

What happened to the structures then?
Can we still have different structures or it will be some kind of mixture?

The binding energy of the benzene dimer is really quite small. Ref. [1] reports the parallel-displaced and T-shaped dimers to have a binding energy at about 2.6 kcal/mol. This is about 900 cm$^{-1}$.

So, it is quite likely that if any vibrational mode with more energy than this is excited that upon redistribution of the energy, the dimer will fall apart.

On the other hand, excitation of vibrations less than the binding energy will likely lead to a very rapid interconversion between the T-shaped and parallel-displaced (or maybe sandwich) isomers. I have read that it is believed this happens basically no matter what.

Also, the sandwich isomer has a binding energy closer to 1.5 kcal/mol, so it is potentially possible to excite some low frequency mode which would cause the sandwich isomer to dissociate but not the T-shaped or parallel-displaced isomers.

Note that because this is such a weakly bound dimer, experimental measurement can only be made at very low temperatures. Ref. [2] describes one measurement of the conformers using supersonic jet expansion to form the dimers. This means temperatures close to milliKelvin.

References:

[1]:  Janowski, T., & Pulay, P. (2007). High accuracy benchmark calculations on the benzene dimer potential energy surface. Chemical Physics Letters, 447(1-3), 27-32.

[2]: Scherzer, W., Krätzschmar, O., Selzle, H. L., & Schlag, E. W. (1992). Structural isomers of the benzene dimer from mass selective hole-burning spectroscopy. Zeitschrift für Naturforschung A, 47(12), 1248-1252.


What is the lowest mass measurable using a 4 decimal point analytical balance? I think it is 1 mg, but I've heard people saying 0.1 mg. As 0.1 mg = 0.0001 g, which is the uncertain digit, I think such a measurement is not reliable at all.

Any opinions ? Thanks.

Keep in mind that for many analytical instruments, you cannot assume that the uncertainty is ± the last digit.
It's usually necessary to look at the specifications in the manual or on a sticker to find the measured uncertainty.
For instance, one of our balances, a Sartorius CPA225D (manual), is readable to 0.01 mg (i.e. the display can display increments of 0.01 mg), but the specifications list the repeatability as only ±0.05 mg (±1 s.d.).
This means that repeated measurements of the same mass will be (ideally) normally distributed with a standard deviation of 0.05 mg and that even though the instrument has the resolution to report in 0.01 mg increments, the uncertainty in a measurement is higher.


Are there atoms with different numbers of protons, but the same number of electrons, with different orbital occupations? Imagine there is an electron-stripped nucleus with $Z$ protons, in high vacuum and isolated. Then you add electrons, one by one, and check which orbitals are filled. Will this depend on $Z$ at all, or does it only depend on the number of electrons that are added?

In other words, will an ion have different orbits filled than the corresponding neutral atom with the same number of electrons?

In other words: are there atoms with numbers of protons $Z_1$ and $Z_2$, both with $n\leq\operatorname{min}(Z_1, Z_2)$ electrons that have different orbital fillings?

Yes, the number of protons in the nucleus does affect how the electrons are filled into orbitals. There are a number of real-world examples that usually stem from transition metals since they do not only have a rather rich oxidation state chemistry but also two energy levels (3d and 4s) which are rather close by each other.

To give you a concrete example, take $\ce{Fe^2+}$ and $\ce{Cr^{\pm 0}}$. They both have the same number of electrons. However, chromium’s ground state electronic configuration is given as $\mathrm{3d^5 4s^1}$, while iron(II) typically has $\mathrm{3d^6 4s^0}$.

The general explanation is that the higher nuclear charge in the iron nucleus attracts the electrons better than the lower nuclear charge in chromium. This difference causes the orbitals to contract. The d-orbitals which are closer to the nucleus, are contracted more and thus experience more stabilisation than the s-orbitals of the next shell.


XlogP of octanol and water itself According to Pubchem, the XLogP of octanol is 3 and that of water is -0.5. Why aren't they both 0 or 1 which seems to make more sense to me?

These values are consistent with the slight solubility of water in octanol and octanol in water.

For example, the solubility of 1-octanol in water is 0.46 g/L. Assuming the density of a saturated solution of octanol in water is approximately that of water, then the concentration in grams per liter can be converted to moles per liter:

$$\dfrac{0.46\ \mathrm{g\ octanol}}{\mathrm{L\ water}}\times \dfrac{\mathrm{mol\ octanol}}{130.23\ \mathrm{g\ octanol}}=3.53\times 10^{-3}\ \mathrm{M}$$

The molar concentration of pure 1-octanol can be calculated from its density:

$$\dfrac{824 \ \mathrm{g\ octanol}}{\mathrm{L\ octanol}}\times \dfrac{\mathrm{mol\ octanol}}{130.23\ \mathrm{g\ octanol}}=6.33\ \mathrm{M}$$

And the $\log P$ is:

$$\log P = \log \left( \dfrac{\mathrm{[octanol]_{octanol}}}{\mathrm{[octanol]_{water}}} \right) \log \left( \dfrac{6.33\ \mathrm{M}}{3.53\times 10^{-3}\ \mathrm{M}} \right)=3.25 $$

Likewise, if we knew the solubility of water in 1-octanol, we could do the same calculation.


Adiabatic processes and the Ideal Gas Law in my first year chemistry textbook it wrights that in adiabatic proccess when there is a system of canister filled with gas and a piston above(and wheights on it) if we let the piston move upward by removing some of the weights the temperature of the gas will decrease since the gas has made work.
i understand that in the opposite case when we compress a gas in adiabatic proccess the temperature will increase.

**my question is: isn't it contradicts the ideal gas law which says that the volume of the gas is proportional to the temperature and vice versa

PV=nRT

so when the volume of the gas is increasing, the temperature should increase.

also by intuition: if the volume is increased, there is less pressure , the molecules has now more space and theire kinnetic energy should increase and therefore also the temperature.

how can i look at this?

thanks
 **

No, it doesn't contradict the ideal gas law at all. When you are saying that rise in volume means increase of temperature, you are inherently assuming that Pressure remains constant. But that is not the case at all in an adiabetic expansion.

In fact, you should realise that it is the product $PV$ which is decreasing. So, Volume might increase, but pressure drops even faster than volume, that's why the product $PV$ overall decreases. If some of the weights are removed, the gas expands slowly according to the equation, $PV^{\gamma} = C$. For most of the gases $\gamma > 1  .(\gamma = \frac{C_p}{C_V})$ Thus by simple observation, you can see if Volume increases $x$ times, Pressure will decrease $x^{\gamma}$ times, which is more dominating in the product.Thus, the product $PV$ will definitely decrease with increase in volume in case of an adiabetic process.

Also, your second argument is also not acceptable, because you already realise that pressure will decrease and by Kinetic theory of ideal gases $PV = \frac{1}{3}m\ n\ v^2_{rms}$. Thus if you understand the decrease in the product $PV$, you will also understand that $v^2_{rms}$ will also decrease, and hence the kinetic energy of gas molecules also.

Also you are thinking in a much more complicated way towards the situation. Think simply. For ideal gases, internal energy ($U$) is a measure of the translational kinetic energy of gas molecules and also $U = \frac{3}{2}RT$ depicting that for ideal gases, kinetic energy is only a function of temperature only. Also, by the first law of thermodynamics, $dU = \delta q + \delta W$ ( $\delta $ here represents inexact differentials). For adiabetic process, $\delta q =0$. So, $dU = \delta W$. Now, $\delta W$ is work done on the system which is negative for an adiabetic expansion, which makes $dU$ also negative, i.e. internal energy decreases and so does temperature.


Dissolving sand (or something similar) in a bottle of water? I would like to create an illustration to demonstrate something during a talk. The idea goes that there is sediment in a bottle of water that sits at the bottom until shaken up then adding another substance to the water makes the sediment dissolve or appear to vanish.

My question is; What would make this possible? What materials or chemicals would work?

To clarify: it sounds like you want a layer of insoluble sediment, so that if you shake the bottle, the water looks cloudy. Maybe it settles back out quickly, or maybe the water stays cloudy.

Then, you add something and shake the bottle again. This time, the sediment disappears.

There are lots of possibilities here. One that comes immediately to mind is lime (calcium hydroxide), which is only slightly soluble in water. Add acid (vinegar if you're timid and don't mind adding a large quantity, hydrochloric acid if you want fast results), and it will be converted to a highly soluble salt.

Chalk (calcium carbonate) would work, too, but it would fizz (produce bubbles of gaseous carbon dioxide) when you added the acid.

If you can say a little more about the context of your demonstration, we might be able to provide better ideas. Is this actually about solubility, or is it a metaphor for something else?


Diffusion and potential gradients for two similar atoms Studying biochemistry I noticed that equilibrium is reached for K and Na separately (not taking into account electric potential), as would be the case I assume for any molecule or atom. This is of course entropy.

Eg: My understanding is, you have two types of atoms, and a membrane separating them, with different concentrations on each side. Each atom will reach equilibrium with itself across the membrane, as if the other type of atom doesn't exist (barring the influence of charge).

Why is it that, for instance, a K atom is so keen to reach equilibrium with another K atom in an area of lower concentration, and yet seemingly ignores the Na, even if the Na is in high concentration where the K is low? Both Na and K still have the same charge, and would also collide with each other I imagine? Why do identical atoms behave like this but non-identical ones with very similar properties don't?

Update #2:

My understanding was incorrect regarding K and Na having separate forces. The two atoms seem to be treated the same, except for selective permeability.

I'm still not understanding why the two atoms have separate gradient potentials, if they can be treated as the same. I made the following images to illustrate my point a little clearer.

There is a compartment with a left and right side, separated by a wall containing two selectively permeable channels that may open or close. Only Na can cross the green channel when it is open, only K can cross the orange channel when it is open.

First, the assumption: The main assumption behind all this is that, in answering this question, we can treat Na and K as being the same. They are effectively two marbles with identical physical and chemical properties, just with different colours. This is what I have come to understand based upon the answer by orthocrestol.

The cell on the left has a -70mv potential energy compared to the right compartment, caused by invisible negative ions which will always remain on the left side.

All that has been summarized by the discussed assumption is seen in figure 1.

Now in figure 2, we open the selective channel for Na. Since Na has a gradient potential, and there is also a potential difference, the positive Na flows into the left hand compartment.

In figure 3, an equilibrium is reached at an arbitrary value of -20mv. On average, no more Na will move to the left or the right, it's at equilibrium.

Figure 4, channels closed. Now this is where my understanding ends.

Apparently now, from what I gather, in figure 5 if we open the selective channel for orange (K+), since K+ has a gradient potential, it moves to the right. In the process, it takes its positive charge with it and restores more or less the original potential difference of -70mV on the left hand side.

But, if this is what happens, this must surely invalidate the marble analogy of the two atoms being the same. Because if they're the same, then whether we open the Na channel (which is already at equilibrium) or the K channel should make no difference, because once gain, they're the same atom with a different colour (except for being separately permeable via their selective channels).

The only difference that should be apparent is that if we leave both channels open, then the colours will eventually reach an equilibrium on both sides. But if an equilibrium is already reached for the charge (which it has in figures 3 and 4), I'm failing to grasp how opening then closing a different channel suddenly causes that equilibrium to change (as in figure 5 and 6). The explanation is the K potential gradient, but this is still the same marble as Na is it not? How come the potential difference changes?

Please direct me where my thinking has gone wrong, or if this is what even happens. And if it is, how is this possible? Many thanks.

This answer has been edited quite a bit. New additions are put in blockquotes.

To avoid any potential (pun not intended) confusion, let me make one thing clear before I start: Any mention of "potential" in this answer refers to an electric potential, not a chemical potential. The definition of electric potential can be found here on Wikipedia.

Let's start off simple and say that we have a beaker with two compartments. The two compartments are separated by a membrane that is ONLY permeable to $\ce{Na+}$ ions. In the left compartment, we put 1 M of $\ce{K+}$ ions and in the right compartment, we put 1 M of $\ce{Na+}$ ions (with appropriate counterions). If diffusion of $\ce{Na+}$ entirely ignored the presence of $\ce{K+}$ as you say, then we would expect that after a while, we would have 1 M $\ce{K+}$ and 0.5 M $\ce{Na+}$ on the LHS, and on the RHS we would have 0.5 M $\ce{Na+}$.
This does not happen. In actual fact, there would be a concentration gradient that favours net movement of $\ce{Na+}$ from RHS->LHS, and there would be a electric potential difference that favours net movement of $\ce{Na+}$ from LHS->RHS. We could model the concentration at any point in time with:

LHS - 1 M $\ce{K+}$ and $x$ M $\ce{Na+}$
RHS - $0.5 - x$ M $\ce{Na+}$

where $x$ is between 0 and 0.5.
When $x = 0$, there is no potential difference since both sides have the same amount of charge (overall both sides are electrically neutral). However, there is a steep concentration gradient, so $\ce{Na+}$ starts diffusing from RHS-> LHS, causing $x$ to increase. Now as $x$ increases, the concentration gradient becomes gentler and the potential difference starts to become larger. At one particular value of $x$, the concentration gradient will cancel out the potential difference and there will no longer be any net movement of $\ce{Na+}$ across the membrane. This is the point where we could say that diffusion has stopped.
Now, let me change the scenario and say that the membrane is permeable to BOTH $\ce{K+}$ and $\ce{Na+}$ ions. In that case, clearly the equilibrium concentrations will be 0.5 M $\ce{K+}$ and 0.5 M $\ce{Na+}$ for both the LHS and RHS, because under these conditions there is neither a potential difference nor a concentration gradient.

It is important to note that the diffusion of $\ce{Na+}$ is NOT directly affected by the pre-existing concentration gradient of $\ce{K+}$. It is indirectly affected only because the concentration gradient of $\ce{K+}$ leads to a potential difference between the LHS and the RHS, which in turn affects the diffusion of $\ce{Na+}$. If we were to replace $\ce{K+}$ with an electrically neutral species, such as glucose, the concentration gradient of glucose would have absolutely zero effect on the diffusion of $\ce{Na+}$.

Now we can go on to explain the action potential in the neuron, which is not something I originally wanted to go into, but since the original question has been edited I feel that I have to address this now. Even though the original question now makes no reference to the action potential, it is describing exactly those events that occur during an action potential.
The inside and the outside of the neuron is separated by a cell membrane, which is selectively permeable to ions such as $\ce{Na+}$, $\ce{K+}$, $\ce{Ca^2+}$, and $\ce{Cl-}$. Of these four, sodium and potassium ions are the most important as they are directly involved in depolarisation and repolarisation when an action potential is generated.
The distribution of ions is not symmetrical across the membrane. When at rest, the extracellular concentration of $\ce{Na+}$ is much higher than the intracellular concentration of $\ce{Na+}$; the converse is true for $\ce{K+}$. Due to the membrane having different permeabilities to $\ce{Na+}$ and $\ce{K+}$ (for details of this, one can consult a neuroscience text), the resting membrane potential is -70 mV. Conventionally, the membrane potential is described as the intracellular potential with respect to the potential of the extracellular potential; this means that the inside of the cell is more negative.
Neurons react to certain stimuli by having $\ce{Na+}$ enter the cell, via various means which are not important to the current discussion. Since positively charged ions are entering the cell, this causes the membrane potential to rise. Once the membrane potential has risen to roughly -55 mV, voltage-gated $\ce{Na+}$ channels open, making the membrane permeable to $\ce{Na+}$.
At this stage, we have to consider the two factors that influence the diffusion of $\ce{Na+}$:

Since the potential difference is still negative, it favours influx of $\ce{Na+}$.
Since the extracellular concentration of $\ce{Na+}$ is larger than the intracellular concentration, the concentration gradient also favours influx of $\ce{Na+}$.

Hence more $\ce{Na+}$ enters the cell, leading to depolarisation of the membrane until the membrane potential reaches approximately +30 mV. At this stage, the voltage-gated $\ce{Na+}$ channels close and the voltage-gated $\ce{K+}$ channels open. Again, we need to consider the two factors that influence the diffusion of $\ce{K+}$:

The potential difference, which is now positive, favours efflux of $\ce{K+}$.
Since the intracellular concentration of $\ce{K+}$ is higher than the extracellular concentration, the concentration gradient also favours efflux of $\ce{K+}$.

Therefore $\ce{K+}$ leaves the cell, leading to repolarisation. There are further events such as hyperpolarisation and the refractory period, but they are not relevant to this discussion. After all this is said and done, the concentration gradients of both $\ce{Na+}$ and $\ce{K+}$ are not the same as they were at the very start of this discussion. They are restored by the $\ce{Na+}$/$\ce{K+}$ pump.
The most important thing to takeaway from this is that the diffusion of species X across a membrane is affected by two things: 1) the concentration gradient of X; 2) the potential difference across the membrane, if and only if X is a charged species. Other concentration gradients of other species, such as Y, Z etc. do not directly affect the diffusion of species X unless they contribute to the potential difference.
Prior to my edit you might have seen that I wrote that "diffusion does not ignore similarly charged species". This might seem to be a contradiction with what I say now, that the concentration gradient of Y does not affect the concentration gradient of X. The answer is that the concentration gradient of Y can only affect that of X when both X and Y are charged species, and this is only an indirect effect, in that the concentration gradient of Y affects the electric potential difference which in turn affects the concentration gradient of X. If Y was uncharged, its concentration gradient would not affect the potential difference; if X was uncharged, the potential difference would not affect its concentration gradient.
I think the marble analogy is inherently flawed in that there is no real analogous concept to a potential difference. You could modify it to have some kind of analogous concept, but in my opinion it wouldn't make sense any more.


Cyclic compound stereoisomer How can the compound 1,2-cyclopentanediol have 3 stereoisomers? Isn't the configuration of the OH groups fixed in a ring?

By definition, stereoisomers are the compounds which have same connectivity of atoms but different spacial orientation of atoms and two stereoisomers do not have same orientation from any point of view.
There are different possibilities can a molecule of cylclopentane-1,2-diol can have:
For simplicity, I have made two hydroxyl ($\ce{OH}$) into two methyl groups.
But by definition you can cross out two structures because there exist a vantage point from which two molecules look the same (rotate molecule 180 out of plane and then rotate it slightly in plane, you get 2 pairs of identical molecules i.e 1,4 and 2,3. Hence, there are only two stereoisomers possible for cylopentane-1,2-diol.
Hope that helps.


Why is arginine's positive side chain classified as basic and not acidic? Arginine has a positive side chain.

Doesn't this mean it wants to lose a proton H+? Isn't something that wants to lose or donate a proton an acid?

Yet my book states, "arginine has a positively charged side chain, i.e it is
basic" .

Your textbook is imprecise. We consider whether a sidechain is acidic or basic based on the action of the sidechain when it is neutral.

Because arginine's sidechain is basic, it preferentially picks up a proton, and hence becomes positively charged. (In the same vein, carboxylic acid sidechains are acidic because they preferentially lose a proton, and exist as the negatively charged carboxylate anion.)


Is the deprotonation of phenol exothermic? I just wanted to confirm that deprotonation reaction of weak acids are exothermic?
I know that strong acids deprotonate exothermically (e.g. adding liquid $\ce{H2SO4}$ to water releases a lot of heat), but is this true for weak acids?

Any bond dissociation reaction is endothermic, hence the termin 'bond'. Heterolytic dissociation (into ions) is more endothermic than homolytic. Strong acids are not exception.

However, the words above are about molecules in vacuum. When the molecules are in a media, interaction with media may stabilize ions strongly. This is why salts dissolve and acids/bases dissociate.

In case of phenol, dissociation in vacuum is definitely endothermic. Dissociation in water is also endothermic, since the dissociation constant is small. (1) However, deprotonation implies that protons moves to another acceptor, and in this case it depends entirely on the acceptor. Similarly, in nonwater media dissociation of phenol also may be exothermic.

(1) Dissociation constant is tied to dissociation free energy with law
$\ln K_e = - \Delta G / RT$ . Since entropy in reaction dissociation reactions is definitely increasing, the only reason for $\Delta G$ to be unfavorable is enthalpy.


A mole of atoms is the number of atoms whose total weight equals the atomic weight? My textbook says, a mole of atoms is the number of atoms whose total weight = the atomic weight. There's something in the definition that's confusing me.

Now what is confusing me is:

Atomic weight = weight of 1 atom
Therefore, number of atoms whose total weight is atomic weight = 1.
Going by definition then, 1 mole atoms=1 atom. Which is incorrect.
Because 1 mole atoms = $\pu{6.022*10^23 atoms}$.
So from the definition apparently, weight of $\pu{6.022*10^23 atoms}$ = atomic weight(weight of 1 atom)
Which is contradictory.

Could someone clarify where am I wrong and explain the same?

PS: There are no units in the provided definition in the book.

Your textbook is inaccurate.

First of all, it should be total mass, not total weight.

The following statement is correct: A mole of atoms with an atomic weight of 12 has a total mass of 12 unified atomic mass units.

But you cannot say: A mole of atoms is the number of atoms whose total mass (not weight!) is the atomic weight. (wrong!) although it might look very similar.

The atomic weight is an alternate name for relative atomic mass. It is "The ratio of the average mass of the atom to the unified atomic mass unit." (source) It has no unit.

A mole is a unit of amount. "The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12. When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles." (source)

So your textbook sentence should end: A mole of atoms is the number of atoms whose total mass is 0.012 kg if they were all carbon-12 atoms.

The unified atomic mass unit (see definition of atomic weight) is "one twelfth of the mass of a carbon-12 atom in its ground state" (source).

So, a mole of atoms is the number of atoms whose total mass is the atomic weight (relative atomic mass) multiplied by one twelfth of the mass of an carbon-12 atom.

Or, a mole of atoms is the number of atoms whose total mass is the atomic weight (relative atomic mass) multiplied by the unified atomic mass unit.

I find these statements much harder to understand than the original definition ("a mole is as many elementary entities as there are atoms in 0.012 kilogram of carbon-12").

What might have some application: One mole of atoms with the atomic weight $x$ have a total mass of $x$ amu.


Paper making alternative Is there any paper making alternative? What is there in trees that makes it the only thing from which paper can be made? Do we even need paper anymore with all electronic devices and stylus', etc

Paper mostly consists of cellulose, form which wood is a good source.
Another interesting source is the hemp plant and it seems that hemp was indeed the major source in paper manufacturing up to the 19th century.

Nowadays, recycling of paper probably is the most reasonable way to reduce the wood use.

  Do we even need paper anymore with all electronic devices and stylus'

Yes, for different reasons:

Think in how long old books and handwritings have lasted. Compare that with the durability of magnetic (tapes, hard disks) or optical recordings (CD, DVD, Bluray, etc.)
Supposed that the ability to read doesn't get lost and a document is written in a well-defined alphabet, you do not need any additional technology to access the information. Who can read data from a 9 track tape nowadays (the technology is only fifty years old) or an "old" MFM hard drive? 8" or 5.25" floppy disks, anyone? ZipDrives, supposed they survived the "click-od-death"?
Most people do not use electronic signatures for documents. Papers are signed, sealed, etc. Consequently, written contracts, bills, etc. are considered more forge-proof than electronic documents. Storing them is therefore mandatory in business.


Facing a problem in enthalpy calculation If 1 mole of gaseous carbon atoms are converted into diamonds, calculate enthalpy change of process given that bond enthalpy of $\ce{C-C}$ bond is $\pu{400 kJ mol-1}.$

According to me, the answer should be that since C(diamond) has 4 bonds and C(gas) doesn't have any bonds, 4 bonds are being formed so enthalpy change should be

$$ΔH = 4\cdot (\pu{-400 kJ mol-1})$$

But in the correct solution it says each bond is counted twice, so answer will be

$$ΔH = \frac{4\cdot (\pu{-400 kJ mol-1})}{2}$$

I don't get why are we doing this. In other questions we don't consider the double counting of bonds. For example, another question I did had 1 mole phosphorus gas converting into $\ce{P4}$ molecule, and even though $\ce{P4}$ has 6 bonds, we did not consider what we did with diamond, so what is the reason for that?

$\ce{P4}$ forms a molecular solid, whereas diamond is a covalent-network solid (see eg definitions here).

In a molecular solid, molecules containing a discrete number of covalently bonded atoms are arranged in a lattice stabilized by weak interactions such as van der Waals forces or hydrogen bonds between the molecules. In a covalent-network solid, the "molecules" are extended with no finite stoichiometry definining a molecular size.

When estimating the enthalpy change, you consider only covalent bonds, not weak intermolecular forces. Therefore, for $\ce{P4}$ you consider only the 6 P-P bonds, whereas in diamond you consider covalent bonds to nearest neighbours, and correct for the fact that when you rupture a bond to a neighbour, that neighbour also loses a bond, therefore the division by 2.


What is the product of alpha decay of curium-226? What product is formed when curium-226 undergoes alpha decay?

Curium has atomic number 96. Since alpha decay decreases the number of protons and neutrons by 2, the product formed has atomic number 94 and atomic mass 222, thus the product is plutonium-222. However this was not an answer choice. The only answer choices were isotopes of radon, radium, and thorium. Either I don't know how to do a very simple question, or there are alternative representations of radioactive isotopes that I don't know about.

I would like to determine, if the question itself is wrong or if I am making a mistake.

The question is flawed (possibly a misprint), your reasoning is sound (given the information provided to you).
Curium-226 does not exist, according to the Web Elements page on the isotopes of curium, the 'smallest' isotope is curium-240, which undergoes $\alpha$ decay to become plutonium-236.
$$\ce{^{240}_{96}Cm -> ^{236}_{94}Pu + ^{4}_{2}\alpha}$$
The most likely candidate of what the question should have been asking about is radium-226 decaying to radon-222,
$$\ce{^{226}_{88}Ra -> ^{222}_{86}Rn + ^{4}_{2}\alpha}$$
which is part of the uranium-series decay chain:

This is the most likely scenario, as the smallest isotope of thorium is thorium-227 (WebElements) and for radium, it is radium-223 (WebElements), both have too high a mass for anything with an atomic mass of 226 to directly decay into by alpha decay (as per the question).


amu and g/mol relation Do we have that $\pu{1 g/mol} = \pu{1 amu}$ ?

Because we have, for the mass of an atom of carbon 12, call it $m(\ce{^12C})$, that

$$m(\ce{^12C}) = \pu{12 amu}$$

and furthermore

$$\pu{1 mol} \cdot m(\ce{^12C}) = \pu{12 g}$$

therefore

$$m(\ce{^12C}) = \pu{12 amu} = \pu{12 g/mol}$$

So finally we get that $\pu{1 g/mol} = \pu{1 amu}$ .

However, my chemistry teacher is telling me that those are two completely different things and that I am confused between the mass per atom and the mass per $6.022\cdot10^{23}$ atoms. I can't understand how, and this is really bugging me, so help is very appreciated.

Note that this requires the mole to be a number (or a "constant"), which may be where I'm wrong.

You are correct, but to make it a little more clear you can include the assumed "atom" in the denominator of amu:

$$
\begin{align}
m_{\ce{C}^{12}} &= \pu{12amu atom^-1} \\ \\
m_{\ce{C}^{12}} &= \pu{12g mol^-1} \\ \\
\pu{12amu atom^-1} &= \pu{12g mol^-1} \\ \\
\pu{1amu atom^-1} &= \pu{1g mol^-1}
\end{align}
$$

In other words, the ratio of amu/atom is the same as the ratio of g/mol. The definitions of amu and moles were intentionally chosen to make that happen (I'm surprised your teacher didn't explain this, actually). This allows us to easily relate masses at the atomic scale to masses at the macroscopic scale.

To check this, look at the mass of an amu when converted to grams:

$\pu{1amu}= \pu{1.6605E-24 g}$

Now divide one gram by one mole:

$\pu{1g mol^-1}= \frac{\pu{1 g}}{\pu{6.022E23 atom}} = \pu{1.6605E-24 g atom^-1}$

It's the same number! Therefore:

$\pu{1g mol^-1}= \pu{ 1 amu atom^-1}$


ʟ-Glutamine solubility in hot water What happens when you try to dissolve ʟ-glutamine in hot water (above 80 °C)? Does the heat destroy the ʟ-glutamine?

Glutamine in aqueous solutions degrades slowly when stored in room temperature. Therefore, we can expect that rate of degradation is faster at higher temperatures. It is evident that the hydrolysis product of glutamine is 5-pyrrolidone-2-carboxylic acid (not glutaric acid as shown for enzymatic degradation) and ammonia (Ref.1):

The Ref.1 states that:

The degradation kinetics of ʟ-glutamine (Gln) in aqueous solution was studied as a function of buffer concentration, $\mathrm{pH}$ and temperature. Stability tests were performed using a stability-indicating high-performance liquid chromatographic assay. The degradation product of Gln was 5-pyrrolidone-2-carboxylic acid. The reaction order for Gln in aqueous solution followed pseudo-first-order kinetics under all experimental conditions. The maximum stability of Gln was observed in the $\mathrm{pH}$ range from 5.0 to 7.5. The $\mathrm{pH}$–rate profile described by specific acid–base catalysis and hydrolysis by water molecules agreed with the experimental results. Arrhenius plots showed the temperature dependence of Gln degradation, and the apparent activation energy at $\mathrm{pH}$ 6.41 was determined to be $\pu{9.87 \times 10^4 J mol−1}$.

Factors affecting the stability of ʟ-glutamine in solution has also been studied (Ref.2). They also found that glutamine degradation resulted in the equimolar formation of ammonia and no associated formation of glutamate. At $\pu{22–24 ^\circ C}$, the degradation rate of glutamine was variable depending on the type of solution used. It was 0.23% in water $\mathrm{pH}$ 6.5; 0.22% in dextrose/water (15% $w/v$); 0.8% in mixed total parenteral nutrition (TPN) solution; and on the $\mathrm{pH}$, molarity and type of buffer used. The degradation rate was essentially unaffected by light and $\ce{O2}$.
Nonetheless, the solubility of ʟ-glutamine in water increases with temperature (Ref.3).
References:

Kanji Arii, Hideyuki Kobayashi, Toshiya Kai, Yukifumi Kokuba, "Degradation kinetics of L-glutamine in aqueous solution," European Journal of Pharmaceutical Sciences 1999, 9(1), 75-78 (https://doi.org/10.1016/S0928-0987(99)00047-0).
K. Khan, M. Elia, "Factors affecting the stability of ʟ-glutamine in solution," Clinical Nutrition 1991, 10(4), 186-192 (https://doi.org/10.1016/0261-5614(91)90037-D).
Jan P. Amend, Harold C. Helgeson, “Solubilities of the common ʟ-$\alpha$-amino acids as a function of temperature and solution $\mathrm{pH}$,” Pure and Applied Chemistry 1997, 69(5), 935-942 (https://doi.org/10.1351/pac199769050935).


What happens to the first ionization potential when a hydrogen-like atom captures a particle? This is a textbook problem from Resonance DLPD Physical Chemistry, Page #83:

  The mass of a proton is $1836$ times the mass of an electron. If a subatomic particle of mass $207$ times the mass of an electron is captured by the nucleus, what happens to the first ionization potential of H?

My answer is that it may either increase or decrease depending on the charge of the captured particle.

However, the correct answer according to my book is that the ionization potential increases.

How do I arrive at this solution?

When we solve the Schrodinger equation for the hydrogen atom we general make the simplifying assumption that the proton stays fixed and the electron moves in the potential of the fixed positive charge. So when we write, for example, the $1s$ orbital as:

$$ \psi_{1s} = \frac{2}{a_o^{3/2}} e^{-r/a_0} \tag{1} $$

the variable $r$ is the distance from the proton, and in the equation for the Bohr radius:

$$ a_0 = \frac{\hbar^2}{me^2} \tag{2} $$

The $m$ is the mass of the electron. If you're interested in some detail this is discussed on the Physics SE in Reduced mass in quantum physics (Hydrogen Atom) but taking into account the motion of the hydrogen atom turns out to be surprisingly simple. We simply define $r$ to be the distance to the centre of mass of the atom, and the mass $m$ then becomes the reduced mass of the electron-proton system:

$$ m = \frac{m_e m_p}{m_e + m_p} \tag{3} $$

If we take the limit of $m_p \to \infty$ then the reduced mass just becomes the electron mass $m_e$, but for finite $m_p$ the reduced mass is less than $m_e$.

Given all this you can now see what the question is getting at. If the proton captures a neutral particle of mass $207m_e$ the effect is to increase the mass of the proton. This increases the mass $m_p$ we have to put in to equation (3) so it increases the reduced mass in equation (2) and hence the wavefunction (1). The end result is that in the equation for the ionisation energy:

$$ I = \frac{me^4}{8 \epsilon_0 ^2 n^2h^2} \tag{4} $$

the reduced mass $m$ is increased slightly so the ionisation energy increases slightly.

A good example of this is to compare the ionisation energies of hydrogen, deuterium and tritium, where the increased nuclear mass of deuterium and tritium increase the ionisation energy by the mechanism discussed above. In fact there is an existing quesion discussing exactly this: How can difference in neutron number cause a difference in ionisation enthalpies?


Which chemical reactions occur between my tongue and the food I eat? There are various chemical reactions that can occur between materials.

Which chemical reactions occur with the tongue?
How come all reactions are reversible (nothing sticks to my tongue)?
What's special about the tongue-water interaction that makes it tasteless?

1) I would guess that all of the different interactions are possible except for ionic and covalent because those are much stronger than the rest and it would be difficult for the food molecules to disassociate from the taste receptors to be used again. Enzymes in saliva break down complex food molecules to smaller and simpler ones.

2) Started explaining in 1, molecules are not usually permanently bound to the taste receptor in order for the taste receptor to be used again (another molecule being able to bind to it). Some toxins form covalent bonds to receptors, such as curare. I'm not sure what kinds of receptors taste ones are but toxins target different specific receptors. Curare binds to nicotinic acetylcholine receptors, found in muscles.

3) We don't taste things unless a molecule binds to a taste receptor.


Naming a hexasubstituted benzene ring In naming the attached compound, is there a reason it is termed:

  3,5-dibromo-1-chloro-4-ethyl-2,6- dimethylbenzene

as opposed to:

  1,3-dibromo-5-chloro-2-ethyl-4,6-dimethyl benzene?

Shouldn't bromine be assigned the smaller numbers, given that it comes first alphabetically?

The most important simplified criteria for the numbering are:

lower locants for the group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The compound that is given has no principal characteristic group that would be expressed as a suffix. Thus, Rule (c) is not relevant in this case.
No locants are assigned to the delocalized electrons of the benzene ring. Thus, also Rule (e) is not relevant.
Usually, Rule (f) is important for benzene rings with multiple substituents. In this case, however, all six substitutable positions are completely substituted; i.e. the locant set (all considered together in a series of increasing numerical order) is always ‘1,2,3,4,5,6’ – no matter how we turn it.
Finally, Rule (g) stipulates that the lowest locant is assigned to the substituent that is cited first as a prefix in the name. Therefore, the correct name for the compound that is given in the question is 1,3-dibromo-5-chloro-2-ethyl-4,6-dimethylbenzene.


How to balance a redox equation in molecular form? I've spent a while trying to balance this equation in molecular form with no success.

$$\ce{Cu + HNO3 -> Cu(NO3)2 + 2NO + H2O}$$

I need to find the coefficient of HNO3, which is 8. However, this was my attempt:

$$\ce{Cu^0 + H + (NO3)- -> Cu^2 + 2(NO3)- + 2NO-2 + H2O }$$

$$\ce{Cu^0 -> Cu^2 + 2e- }$$
$$\ce{2x (NO3- + 2H^1 + e--> 2NO^0 + H2O)}$$

$$\ce{Cu^0 + 4H+ +2NO3- -> Cu^2+ + 4NO^0 + 2H2O + NO3^-}$$

Image of process: http://imgur.com/LysN7FZ.png

NO3 is the spectator polyatomic ion I left out during my balancing. I have no idea whether I did it right and where to go from here.
Thanks!

Judging from the picture you attached you have successfully identified the species that gets oxidised
$$\ce{Cu -> Cu^2+ + 2e-}\tag1$$
and reduced, i.e. $\ce{NO3-}$. While balancing this reaction you have made a few mistakes. You cannot leave it out as a spectator ion, since some of it is involved in the reaction.
Consider the oxidation state in nitrate and nitrogen monoxide. How many electrons have to be transferred?

   three, nitrogen goes from $+5$ to $+2$ (you have only written one)

How many water molecules will be produced? How many protons are needed?

   Two water, four protons

Write down the full half reactions

   $$\ce{\overset{+5}{N}O3- + 4H+ + 3e- -> \overset{+2}{N}O + 2H2O}\tag2$$

Now you need to find the factors to multiply $(1)$ and $(2)$ with to balance out the electrons.

   $3\times(1)$ and $2\times(2)$

This results in the following equation

   $$\require{cancel}\ce{Cu + 2NO3- + 8H+ + \cancel{6e- } -> 3Cu^2+ + \cancel{6e- } + 2NO + 4H2O}$$

How many more nitrate ions do you need to charge balance the protons? What is the final balanced equation?

   Six more nitrate ions, giving a total of eight
 $$\ce{Cu + 8HNO3 -> \underbrace{3Cu^2+ + 6NO3- }_{3Cu(NO3)2} + 2NO + 4H2O}$$


Why is a silent electric discharge used for the preparation of ozone from oxygen? I learnt that ozone is prepared by passing oxygen through a 'silent' electric discharge. Can someone please explain what is meant by a 'silent' discharge here, and why that helps? I couldn't get a satisfactory answer while Googling.

According to this Wikipedia article:

  Dielectric-barrier discharge (DBD) is the electrical discharge between two electrodes separated by an insulating dielectric barrier. Originally called silent (inaudible) discharge and also known as ozone production discharge or partial discharge, it was first reported by Ernst Werner von Siemens in 1857. On right, the schematic diagram shows a typical construction of a DBD wherein one of the two electrodes is covered with a dielectric barrier material. The lines between the dielectric and the electrode are representative of the discharge filaments, which are normally visible to the naked eye.

  DBDs can be used to generate optical radiation by the relaxation of excited species in the plasma. The main application here is the generation of UV-radiation. Those excimer ultraviolet lamps can produce light with short wavelengths which can be used to produce ozone in industrial scales. Ozone is still used extensively in industrial air and water treatment. Early 19th-century attempts at commercial nitric acid and ammonia production used DBDs as several nitrogen-oxygen compounds are generated as discharge products.


Why doesnt the overall rate of a reaction increase if you increase the concentration of a reactant in a non rate determining step? We say that the overall rate doesnt increase if we increase the concentration of a reactant in a non rate determining step. Does that mean that the speed of that non rate determining step increase but at the same time something happens to counter the increase in speed? Because it doesnt make sense to say that the speed doesnt increase with increase in concentration.

Let's think about this reaction:

$\ce{ A + B->[k_1][] C + D ->[k_2][] E }$

let's say $k_1 << k_2$ so the first reaction is much, much slower than the second step (at equal concentrations)

Now it get's a little bit tricky since the reaction rates depend on the concentration. Let's assume for now that the concentrations of $A$, $B$ and $D$ are in a way that the reaction rate of the first step is really much faster than the second step.

What happens is that $A$ and $B$ react to $C$ quite slowly, but once you got any $C$ it reacts very fast with $D$ to $E$. This means your concentration of $C$ will always be quite low, since everything that is produced will react to $E$. But this also means that the concentration of $E$ is basically increasing with the rate of the first step. The slow reaction part is limiting the overall rate.

What happens if we increase the concentration of $D$? Well, the second step will be even faster, so all the $C$ that comes out of the first step will be converted to $E$ even faster. But the first step is still slow, so your $E$ concentration will still increse approximately with the rate of the first step.

Only if we would reduce the concentration of $D$ to very low levels (much lower than $A$ and $B$) we would see a difference, in that case the second reaction will get slower than the first one (reaction rate depends on concentration) and will be the limiting step.


List in order of increasing polarity? I am doing some exam preparation and I came across this Question which I am finding challenging:

List from least polar to most polar, the following molecules:

Naphthalene, Phenol, 3-aminophenol and chlorobenzene.

I believe I know that Naphthalene is least polar as the benzene ring is very stable however, the confusion arises with chlorobenzene as the substituent Chlorine is very electronegative hence, it is strongly inductively electron-withdrawing thus is less polar than a simple benzene ring. So which is less polar, chlorobenzene or naphthalene?

Also which is most polar? 3-aminophenol or phenol? Phenol has a lone pair on oxygen in the oH group which can donate electrons to the ring, although with 3-aminophenol, both the NH2 and OH donate electrons, so wouldn't 3-aminophenol be more electron rich and hence more polar?

Please give the order of the molecules from least to most polar, explaining the order itself.

Thank you!

I would expect that the amino phenol is more polar than the phenol. This due to the fact that $\ce{NH2}$ donates electrons to the Benzene ring, making the difference in charge between the Benzene Carbon and the Oxygen stronger.

The Chlorine atom is less electro negative than oxygen (if I'm not mistaken) so that would mean: Naphatlene < Chlorobenzene < Phenol < Amino Phenol.