Nomenclature for different types of ice How many types of ice exist and what is their nomenclature? I know that the ice on Earth is $I_h$, but why so?

The first thing to note about ice water is:

All ice is not crystalline and not all crystalline water is ice

There appear to be fifteen crystalline phases of ice, named 'Ice 1' to 'Ice 15' using Roman numerals $I$ to $XV$. There also appears to be three 'polyamorphic' states:.

$LDA$ : Low Density Amorphous

$HDA$ : High Density Amorphous

$VHDA$ : Very High Density Amorphous

All crystalline ice is tetrahedrically coordinated (where the oxygen atoms are in fixed positions relative to each other but the hydrogen atoms may or may not be disordered but obeying the 'ice rules'), and the polymorphs differ primarily in H-bond angles and next-nearest neighbour separations.

Ice I encompasses all ice found on naturally on earth (in a stable form).

This is usually split into two phases called $I_h$ and $I_c$.

Hexagonal ice (ice $I_h$) is 'ordinary ice', the form of all natural snow and ice on Earth. This is evidenced in the six-fold symmetry in ice crystals grown from water vapor (that is, snow flakes).

Cubic ice (ice $I_c$) is a solid form of water that can form in high clouds. Ice Ic is metastable, and irreversibly converts to ice $I_h$. Both forms contain similar layers of hexagonal rings, but differs in the way these layers are connected.

For more information on their structure and properties of ice, check out 'the Many Phases of Ice'.

Martin Chaplin maintains a fairly thorough and well referenced online resource on water, which includes a comprehensive section on the ice phases.

Finally, Google Books has the definitive 'the Physics of Ice' by Petrenko & Whitworth available online.


What is the lewis structure of NO I am a bit confused about the lewis structure of $\ce{NO}$ The answer from my worksheet is $\ce{N=O}$ with nitrogen having a free radical. But I drew the structure as N triple bonded to $\ce{O (:N#O:)}$ and oxygen has the free radical. I am confused as to why my answer is not the case

As an extension of Mithoron's comment, if you drew the structure as N triple bonded to O, you would fulfil both the octets without placing the free radical electron.

Now, you'd have to extend the octet of either nitrogen or oxygen, but you have no free orbitals to place the extra electron.

Hence, the first one is the correct Lewis structure.


How to delete everything but a specific ligand from a PDB file using Linux-compatible free software? I would like to delete everything from a PDB file except a specific ligand that I know the PDB code for. Is there a simple way to using free Linux-compatible software? If you want a specific example for your answer 1YAT and its ligand FK5 (tacrolimus) is an example I'm presently working on. I think PyMOL might have this capability using the delete command but running:

delete /1YAT/A

which should delete the macromolecule (chain A) at least but I saw no visually-obvious change. I have viewed the PyMOL Wiki article on the delete command but it was not helpful as what name meant was not obvious to me.

By free software I mean free of charge not free as in freedom. One piece of software that I used to do this with on Windows was Accelrys (or BIOVIA) Discovery Studio Visualizer, tried to run it on Arch Linux, however, as it has a Linux binary, but it failed to run. This is presumably because it is for Red Hat Enterprise Linux/SUSE Linux Enterprise which has too old of system libraries for Arch Linux compatibility.

What about a straightforward

grep -w " AAA " file.pdb > ligand.pdb

? Where you substitute AAA with the name of the residue you want.


What is the pH of CsBr? What is the pH of CsBr?

  3.8
  7
  9
  12

According to me,

$\ce{Cs+ + H2O <<=> CsOH + H+}$   (because $\ce{CsOH}$ is a strong base, the forward reaction should not occur, right?)

$\ce{Br- + H2O <<=> HBr + OH-}$ (because $\ce{HBr}$ is an acid, the forward reaction should also not occur, right?)

So the answer should be $\mathrm{pH} = 7$.

The reactions you have shown could in fact occur (in the forward direction) to a slight degree in a solution of cesium bromide in water. But I would expect them to occur to a lesser degree than the autoprotolysis of water (formation of hydronium and hydroxide ions from two water molecules), so their effect on pH would be insignificant, and the pH would effectively be neutral.


Color shifting reaction in black sealcoat When I have my driveway seal coated, it turns blue a few days after it is done and then turns back to black. I've heard it called "blue flash." What actually happens chemically to cause this?

It's not a chemical reaction, AFAIK, but due to the Tyndall effect or Rayleigh scattering from the fine particles of the asphalt-water emulsion as water evaporates and the size and/or dispersion of the asphalt particles change.

It is also possible that interference of light in an adhered water layer plays some part. This is the same effect that gives soap bubble their rainbow colors, which change as the bubbles dry -- note that just before it pops from drying, one area of a bubble  may go colorless due to thinning of the film beyond the range of visible light interference.


Stability of formyl chloride Why is formyl chloride unstable? I know it is just a combination of CO and HCl. But why it is unstable whereas other acyl chlorides are stable, like acetyl chloride?

It is unlike other acyl chlorides as formyl chloride dissolves in water. At the same time, it reacts vigorously with it, and no other acyl chlorides produce carbon monoxide and hydrochloric acid.


Can I add the molarities of multiple solutes together? If I am calculating the molarity of a solution in which more than one solute is dissolved, can I add the total number of moles of all the solutes together and divide it by the total volume of solvent to get the molarity of my solution?

It doesn't make sense to add the total number of moles then divide by volume to find molarity. By doing that you are assuming that all the solutes have the same concentration which is not true(you are finding an average concentration of the solution but it's not necessarily correct). The solution contains different solutes at different concentrations. I think what you're doing only works if you know that the concentrations of solutes in the solution are the same.


Can we chemically characterize a flame? Can we chemically characterize a flame ?

Recently, while I was lighting a candle, I came across this question; can we find out the chemical composition of a particular flame? If yes then how? In other words, can we know which elements are present in flame, and not just the salts we test using the flame tests?

If you are using a gas stove with a methane ($\ce{CH4}$) flame, an interesting things occurs if you heat $\ce{CuO}$ in the gas part of the flame. I once  created cupric oxide from heating a penny to glowing red and then shaking in air. Then, I inserted the dark colored penny into the flame zone, and it appeared to be cleaned!

Apparently, the hot cupric oxide is reduced by $\ce{CH4}$ to $\ce{Cu2O + Cu}$. As such, my take on a possible reaction liberating elemental copper, $\ce{Cu2O}$ and carbon:

$$\ce{CH4 + 3 CuO -> 2 H2O + Cu2O + C + Cu}$$

Now, the flame test for Carbon is red-orange and for Copper green/blue, so a mixture of colors could theoretically be observed in this experiment as the methane flame itself participates in element production.

Normally, however, this does not occur.


Preparation of 6-bromo-1-hexanol I am refluxing $\ce{HBr}$ with 1,6-hexanediol for 16 hours at $80\ ^\circ\mathrm{C}$ in toluene. How can I be sure that more than $90\,\%$ of the product is brominated only at one terminal? If the amount of double brominated product is in more than $10\,\%$ of the total yield, then the product mixture will be useless for me.

In an article by Katharina C. Kress, Martin Kaller, Kirill V. Axenov, Stefan Tussetschläger, and Sabine Laschat, published in
Beilstein J. Org. Chem., 2012, 8, 371-378 (DOI), this procedure is described. From the Supporting Information, I read that the authors obtained the product in 72% yield after

  column chromatography on silica gel (hexanes/EtOAc, 1:1)

I wouldn't be astonished if that's as good as it can get.


Testing for the presence of Hexamethylenetetramine / Formaldehyde I read in Russian popular articles that it is allowed to preserve herring/fish with Hexamethylenetetramine in Europe. Since Soviet times, I like to buy salty herring and marinade it with Vinegar. The food-safety article says that Hexamethylenetetramine + Vinegar = Formaldehyde, which is a serious poison.

  How can I easily test presence of Hexamethylenetetramine or Formaldehyde in food?

The salty herring that I buy is sold loosely, without packaging, so that I cannot simply read the ingredients.

I don't believe the premise that E239 (the E number designated to hexamethylene tetramine) is used to preserve fish within the EU is correct - in January, 2014, the European Food Safety Authority published an article in their journal about this preservative where it states quite clearly that the only permitted use for this additive is in Provolone cheese. There are concerns that over-consumers of this particular cheese may accrete more levels of E239 within the body than are healthy.

E239 does appear to be used in the UK for preserving fish, or at least, it is listed as one of its intended uses. Whether its actually used or not I haven't been able to determine, but given we're in the EU, technically, it shouldn't be. If I source any other research regarding this, I'll add it later, but currently, it seems it depends where you're buying your fish from in the first place, that is, the country of origin.


Make very accurate solutions without weighing I am currently facing a hen-egg problem. For a LC-MS method development I need to prepare reference solutions which are used to calibrate the method. Since the method is to be used for quantification in the pg/mL range, the reference solutions should have the most accurate concentrations possible. Of course I know how to prepare exact solutions in volumetric flasks, but here substance quantities are used which can be weighed easily and also quite exactly. In my case, only a few milligrams of the substance are available to me, which is why I cannot weigh high masses. The precision balance that I have at my disposal has a lower weighing limit of 1 mg, which is about the same as the mass that I would have to weigh. However, since the lower limit is very error-prone, I do not like this method. Unfortunately, I cannot think of an alternative solution that would allow me to create an exact solution without weighing. One could of course dissolve any amount of substance in the desired solvent and then determine the concentration by LC-MS, but this is exactly where the hen-egg problem mentioned above occurs. I cannot quantify via LC-MS until I have a suitable method and an internal standard. Are there any other ideas how to get exact concentrations using small amounts of the desired substance?

Are there any other ideas how to get exact concentrations using small amounts of the desired substance?

Every mass spectrometry lab should have microbalances to prepare samples and standards. A lab which can afford an LC-MS can also afford a semimicro balance or a microbalance. Such balances are meant to weigh 0.01 mg to 0.001 mg (with extreme caution).
Two simple options:
(i) Since you are apparently not the first user of LC-MS system in your group, ask how other senior researchers prepared standards in your lab?
(ii) Another (crude) option is that you try weighing by difference. You will tare your analytical balance to zero. By an analytical balance I mean a balance which reads to four decimal places in grams, this is the most commonly used in research labs. Place your standard container on the balance and read the mass. Carefully transfer approximate amounts to your desired container in which the standard will be prepared, and note the weight of the standard container again. The change in mass corresponds to the amount transferred. Practice it with some commonly available substance to see if this works and if your balance is sensitive enough or not. Balances age with time.


Polarimetry test for cholesterol? If I have a pure substance which may be cholesterol can I test that by polarimetry? In other words the idea would be to dissolve the sample in dichloromethane or ethanol, then test the rotation using a polarimeter.

My concern is that I read that cholesterol will not actually dissolve, only form a colloid, so that may invalidate the possibility of a test.

Taken from the website of a chemical supplier, the following solubility data can be easily obtained. Historically benzene and chloroform are commonplace for rotation measurements, however it can be any solvent in which the sample is soluble (or mixtures of solvents). You might be best to find a literature value for cholesterol (scifinder/reaxys) and then measure in the same solvent so at least you have a reference point to compare to.

One gram dissolves in 2.8 ml of ether, in 4.5 ml of chloroform, and in 1.5 ml of pyridine. The product is soluble in acetone, dioxane, ethyl acetate, benzene, petroleum ether, oils, fats, and in aqueous solutions of bile salts. Solutions should be protected from light.
Source Sigma Aldrich

Generally speaking, optical rotation isn't a method commonly used to determine whether or not you have a given compound. Methods like NMR, IR, MS, or even just TLC comparison are far more suited to this.
Optical rotation is useful if you already know you have the right thing, and want to know what your optical purity is like. Without having already confirmed that the sample is 1. clean and 2. the right thing, you'll struggle to make sense of the data since the polarimeter will simply give you a number.
Depending on where you bought your cholesterol from, and various other things, this number may be close to the literature value (maybe some confidence in it being the right thing), however even then you cant be sure its cholesterol as many other things will have similar values.


HNMR - Peak Splitting If you have $\ce{CH3-\color{\red}{CH2}-CH2Br}$

Then the $\ce{CH2}$ (in red) would have a peak splitting of $12$ because $(n+1)(m+1)$
where $n$ for the $\ce{CH3}$ is $(3+1=4)$ and m for $\ce{CH2}$ is $(2+1=3)$ so its $4\times3=12$

source:

  In the general case, a signal will be split into $(n + 1) \times (m + 1)$
  peaks for an $\ce{H}$ atom that is coupled to a set of $n~\ce{H}$ atoms with one
  coupling constant, and to a set of $m~\ce{H}$ atoms with another coupling
  constant.

How can we determine differences in the coupling constant by just looking at the structure? Otherwise how do I know when to apply $n+1$ OR $(n+1)(m+1)$?

But in the other textbook it says:

$\ce{CH3-\color{\red}{CH2}-CH2-COOH}$

The $\ce{CH2}$ (in red) has a peak splitting of $6$ because $(n+1)$ where $n=5$, $(5+1=6)$ and the $n$ is the $3~\ce{H}$ on $\ce{CH3}$ plus the $2~\ce{H}$ on $\ce{CH2}$.

Source:

Which is the correct method? Or are both right? and why?

Alternatively symmetric adjacent hydrogen groups:

$\ce{CH3–\color{\red}{CH2}–CH3}$

Is the $\ce{CH2}$ split to 4 peaks due to symmetric adjacent hydrogen environments? Due to $n+1$, where $n=3$ so $(3+1=4)$.  I read that it should be split into 7 peaks - but where are they getting 7 from - is it $n+1$, where $n=6$ so $(6+1=7)$. For symmetric adjacent groups do you allow $n$ to be the sum of all hydrogens adjacent/or not?

Both are correct to some degree.

The first book predicting 12 lines is being very theoretical.  In principle with narrow enough line widths 12 lines would be observed.  However, due to coupling constants (J) being similar from the beta protons to the alpha and gamma protons, in practice with reasonable resolution, only 6 lines are observed (the first book is actually explaining this in fine print).

(source: ucalgary.ca)

Credit: Univ. Calgary


Reaction of NaOD/D2O with semicarbazone I managed to solve the question till D but I'm stuck at the last reaction.
I tried to find its solution and the source suggest it as

Upto (D) you are correct From there it's :


How does the vapor pressure in a vessel affects the total pressure after a combustion reaction? The problem is as follows:

  A sealed container of $3\,L$ in capactity is filled with hydrogen and
  oxygen at $314\,torr$ each one at a temperature of $27^{\circ}C$. A
  spark produces the reaction between oxygen and hydrogen. Find the
  total pressure after the reaction supposing that the temperature does
  not change. (You may use the vapor pressure of water at $27^{\circ}C$
  is $27\,torr$)

The alternatives given are:

$\begin{array}{ll}
1.&157\,torr\\
2.&184\,torr\\
3.&312\,torr\\
4.&468\,torr\\
5.&494\,torr\\
\end{array}$

What I've attempted to do to solve this problem was to establish the initial conditions for the reaction:

$$O_2+2H_2\rightarrow 2H_{2}O$$

The number of moles for each gas can be found as follows: (for the sake of brevity I'm omitting units)

$n_{O_{2}}=\frac{PV}{RT}=\frac{314\times 3}{62.4\times 300}= 0.0503$

$n_{H_{2}}=\frac{PV}{RT}=\frac{314\times 3}{62.4\times 300}= 0.0503$

Both gases have the same number of moles.

From the reaction I could spot that the limiting reagent in this case is the hydrogen thus it will the one to be used for the calculation of the amount of water produced.

Since both hydrogen and water are in the same proportion the moles of water will be the same as what are the hydrogen.

$n_{H_{2}O}=0.0503$

While for the amount of oxygen which will remain will be half of the value since:

$0.0503\,mol H_{2}\times\frac{1\,mol\,O_{2}}{2\,mol\,H_{2}}=0.0252\,mol\,O_{2}$

Therefore this will be established as follows:

$\begin{array}{llll}
O_2&2H_2&\rightarrow&2H_{2}O\\
1\,mol&2\,mol&&2\,mol\\
0.0503&0.0503&&0\\
-0.0252&-0.0503&&+0.0503\\\hline
-0.0251&-0.0503&&+0.0503\\
\end{array}$

From this I went on to calculate the pressure for each:

For the water:

$P=\frac{nRT}{V}=\frac{0.0503\times 62.4 \times 300}{3}= 313.82$

For oxygen:

$P=\frac{nRT}{V}=\frac{0.0251\times 62.4 \times 300}{3}= 156.62$

But since it was given that the vapor pressure of water is $27\,torr$ made me confused on how should I use this given information. Does it mean that should I subtract from the oxygen?.

If I do that I obtain:

$156.62-27=129.62$

Then I assumed that this pressure can be added to the pressure of water:

$313.82+129.62=443.44\,torr$

But using this is way off from any of the alternatives in the given answers:

But If I add all the numbers:

$313.82+156.62+27=497.44\,torr$

Which is seems close to one of the alternatives but the number swapped. $468\,torr$

I'm not certain if this is due significant figures or an error in the problem. The source of my major confusion is exactly how should it be assesed the information regarding vapor pressure in a combustion reaction?. I'm assuming that after the combustion the water vapor formed will excert a pressure in the container so does this must be added to the pressure of the water formed resulting from the moles produced?.

Can somebody help me with this and clear out my doubts?.

First, kudos for actually showing work on a problem!

Second, I'd solve this by inspection, rather than using the Ideal Gas Law, where time is critical on a test.

As you correctly calculated, the limiting reactant is $\ce{H2}$. Therefore the pressure of $\ce{H2}$ is nil; all the $\ce{H2}$ is in the water.

Half of the $\ce{O2}$ is used, reacting with the $\ce{H2}$ to make $\ce{H2O}$, so the remaining partial pressure of $\ce{O2}$ is ~157 torr.

The volume taken up by the liquid water is ~1/2,000 that of the original gases, so the volume left in the flask is still ~3 l, though slightly less. Since the answer choices are not close together, and the decrease in volume is negligible, I'd ignore it.

The vapor pressure of water at 25° C is ~24 torr, and at 30° C it's ~32 torr, so at 27° C, it would be about halfway, by linear interpolation, say ~28 torr.

The sum of those partial pressures is close to an answer choice.

Trivia: This question seems to be citing pi with the initial value, 3[.]14 torr. If the temperature had been -8° C (ice at that temperature has a v.p. ~2 torr), the answer would have been 159 torr, so the first six digits of pi could have been used; i.e. 314 ->159.


Synthesis of 2-phenoxybutanal from but-1-yne How can I convert 1-butyne to 2-phenoxybutanal as shown above? I thought of using a hydroboration-oxidation sequence to introduce the carbonyl group, but I do not know how to add the phenoxy substituent.

Water, BH3, H2O2, NaOH

This reaction, an hydroboration, yelds butanal

Br2, H2SO4

This reaction, carried without light, is an alpha carbon bromination, and yelds 2-bromobutanal

Sodium phenoxide

This is a simple Williamson Ether Synthesis, which yelds the final product


Are the hydrogen atoms of 2-methylindane homotopic, enantiotopic, diastereotopic or unrelated? I'm very confused. According to the answer key, the circled hydrogens of 2-methylindane (2-methyl-2,3-dihydro-1H-indene) are diastereotopic. I thought they were enantiotopic.

My thought process is the following. They aren't homotopic because replacing one of them with another atom would create a chiral center. Due to them creating a chiral center, I thought it was enantiotopic.

I thought that for it to be diastereoptopic, the carbon with the methyl group off of it would have to be chiral as well, but I don't think it is chiral. The molecule is symmetrical off both sides. Does diastereotopic rely on that carbon being chiral after the new atom replaced the hydrogen? If so, then I understand.

I was under the assumption that the state of that carbon must be chiral before the hydrogen is even replaced.

The process to determine whether two protons are enantiotopic or diastereotopic is to replace each with deuterium and look at the relationship of the resulting compounds. For the indicated hydrogens in 2-methylindane, replacement of each gives a pair of diastereomeric compounds (Figure, top). The compound also contains enantiotopic protons (Figure, center). The carbon where the replacement happens doesn't necessarily become a stereocenter in this process (Figure, bottom).


Confusion regarding the mechanism of acidic/basic buffer solutions While learning about acidic and basic buffer solutions we were told that they can resist pH change if we add small amounts of acid or base by neutralizing the $\ce{H+}$/$\ce{OH-}$ ions from the added acid or base. But I am confused about the fact that if we add a small amount of base to an acidic buffer then after neutralization of the excess $\ce{OH-}$ ions from the base by the $\ce{H+}$ ions from the buffer, shouldn't the concentration of $\ce{H+}$ ions in the solution decrease, thereby making the solution acidic and reducing the $\mathrm{pH}$. (vice-versa for the addition of acid to a basic buffer)

While learning about acidic and basic buffer solutions we were told that they can resist pH change if we add small amounts of acid or base by neutralizing the $\ce{H+/OH−}$ ions from the added acid or base.

Yes, the pH changes less when you add small amounts of acid or base to a buffer, and more if you add it to pure water or an unbuffered aqueous solution.

But I am confused about the fact that if we add a small amount of base to an acidic buffer then after neutralization of the excess $\ce{OH−}$ ions from the base by the H+ ions from the buffer, shouldn't the concentration of $\ce{H+}$ ions in the solution decrease,

Yes, that is what happens. As a consequence, the weak acid/base reaction of the buffer  substances is no longer at equilibrium, and there will be a net reaction to replenish some of the $\ce{H+}$ that was lost.

thereby making the solution acidic and reducing the pH. (vice-versa for the addition of acid to a basic buffer)

This is probably a typo. When you add a small amount of base, the pH would go up, buffered or not. In the presence of buffer, it goes up a little, and in the absence of buffer, a lot.

(vice-versa for the addition of acid to a basic buffer)

Buffers stabilize the pH near the $\mathrm{p}K_\mathrm{a}$ of the conjugate weak acid in "both directions", so even if the pH of a solution is slightly acidic, it is able to buffer against addition of acid or base.


Why is an isothermal process not unique in nature? To be clear, I have got two doubts,

One, about the non-uniqueness of an isothermal process &
Two, about the uniqueness of an adiabatic process.

The first doubt

The Wikipedia page of isothermal process states that:

  For an adiabatic process, in which no heat flows into or out of the gas because its container is well insulated, Q = 0. If there is also no work done, i.e. a free expansion, there is no change in internal energy. For an ideal gas, this means that the process is also isothermal.[4] Thus, specifying that a process is isothermal is not sufficient to specify a unique process.

Now I am confused about the meaning of the last line.

  Thus, specifying that a process is isothermal is not sufficient to specify a unique process.

I am not able to understand the significance of that last statement. Does it mean that any two states, say A & B, can be connected by more than one isothermal processes?

The second doubt

Similarly, I had read in my textbook that an adiabatic process is unique in nature. That is, between any two states only one adiabatic process can exist. To say the same thing in other words,
if two thermodynamic states are connected by a reversible adiabatic process, then the same two states cant be connected by an irreversible adiabatic process

Kindly help me understand these statements with the help of some PV/VT/PT diagrams.

Any help would be deeply appreciated.

(1) An isothermal process is one during which the system temperature $T=constant$. For an ideal satisfying $pV=RT$ this means that $pV=constant$ irrespective of how you achieve it. You can go up and down, back and forth on the same isothermal, but it is still an isothermal process having the same temperature. So it is not a unique process between two fixed points.

(2) The uniqueness of the adiabatic process is true only for a simple system, that is one whose state can be described by temperature and one additional "mechanical" parameter, e.g., for a gas $p = p(T,V)$, or a magnet $M=M(T,B)$, etc. It is definitely not true for more complicated systems, for example a gas with magnetic properties $p=p(T,V,B)$. Instead what is true is that through every equilibrium point there is a unique adiabatic surface (not a curve) determined by a pair of variables say $T,V$ or $V,B$ or $T,B$ such that any curve lying in that surface is an adiabatic process. That such adiabatic surfaces exist is the basis of one of the mathematical proofs of the existence of the entropy function and this quite famous proof is due to Caratheodory.


The number of optical isomers of 3,4-dichlorobutan-2-ol The number of optical isomers possible for the given organic molecule is?
$\ce{CH_3-C^*H(OH)-CHCl-CH_2Cl}$ (the configuration is fixed at $\ce{C^*}$)

The given answer is 2.
The carbon marked with an asterisk is chiral, and the carbon to the right of it is also chiral. Doesn't that give 4 isomers? What does "configuration is fixed" mean?

The carbon marked with the asterisk cannot change its configuration, for it to be counted as a stereocentre, it must be able to. hence 2 isomers only


Why proton concentration is divided by 10⁻⁷? I am reviewing the book Biochemistry Concepts and Connections by Appling, Cahill, and Mathews and I cannot understand why they divide by the hydrogen concentration by $10^{-7}$.  Why not just leave it at the antilog(-8.1) over $\pu{1 M}$ like the other concentrations? I literally have see nothing else on the internet like this.

The textbook is precisely correct.
The equilibrium constant $K$ which the logarithm is taken of is dimensionless, and includes activities or fugacities, and not concentrations and pressures.
In practice this is achieved by using standard states which refer to the pure materials: standard concentration $c^⦵$ and standard pressure $p^⦵$.
One must be very fastidious with units when finding the equilibrium constant.
For example, the reaction

$$\ce{aA + bB <=> cC + dD}$$

equilibrium constant $K_c$ is exactly

$$K_c = \frac{([\ce{C}]/c^⦵)^c\cdot ([\ce{D}]/c^⦵)^d}{([\ce{A}]/c^⦵)^a\cdot ([\ce{B}]/c^⦵)^b}$$

For pure water in its standard state $c^⦵ = [\ce{H+}] = \pu{1e-7 M}$.
It also correlates with so-called biological standard state of $\mathrm{pH} = 7$.
You probably haven't seen it before because many authors use sloppy notations omitting mentioning standard states since they can often be cancelled out.
In this case those cannot be cancelled out, and must be written explicitly.

In fact, your own textbook contains extensive explanation [1, p. 91]:

  For chemical reactions the standard state for solutes is defined as $\pu{1 M}$; however, in living cells the concentration of $[\ce{H+}]$ is roughly $10^{-7}~\mathrm M$, much lower than the standard value of $\pu{1 M}$.
  It is therefore appropriate to define the reference concentration of $\ce{H+}$ in biochemical reactions relative to the $\ce{H+}$ concentration found in the living state (i.e., $10^{-7}~\mathrm M$), rather than the value $\pu{1 M}$ defined by the chemical standard state.
  Recall that when a solute in a dilute solution has a concentration of $\pu{1 M}$, the activity of that solute is unity.
  For the biochemical standard state we define the activity of $\ce{H+}$ to be unity when $[\ce{H+}] = 10^{-7}~\mathrm M$.

  [...]

  The mass action expression $Q$ is unitless.
  We strip the units from each concentration term in $Q$ by dividing each by its
  proper standard concentration (e.g., $\pu{1 M}$ for all solutes
  except $\ce{H+}$; $10^{-7}~\mathrm M$ for $\ce{H+}$; $\pu{1 bar}$ for gases, etc.).

Refrences

Appling, D. R.; Anthony-Cahill, S. J.; Mathews, C. K. Biochemistry: Concepts and Connections (Global Edition); Pearson: Boston, 2015. ISBN 978-1-292-11210-7.


Drawing (R)-1,1,2-trimethylcyclohexane I tried drawing out (R)-1,1,2-trimethylcyclohexane but I can't seem to figure it out. I think I"m getting hung on assigning priorities.

I've labeled the chiral carbon with an asterisk. I've also drawn out all the groups it is attached to. I gave the hydrogen the lowest priority (4) and I'm giving the methyl carbon a priority of 3. I am giving the other carbons priorities of 1 and 2 (as can be seen in the picture). However, this doesn't seem to make any sense; I get a molecule that I can neither assign S nor R to since the numbers are out of order.

Also is this really an image of (R)-1,1,2-trimethylcyclohexane?

You've correctly identified the chiral center in your molecule (the carbon with the asterisk), that is the first step.  When doing work with chiral centers, it is very important to use solid lines to denote bonds in the plane of the screen, bold wedges to denote bonds coming from the plane of the screen towards the viewer and dashed wedges to denote bonds from the plane of the screen away from the viewer.  Drawing all 4 bonds as simple solid lines, as you did, will often lead to confusion.  Look at the figures below, both enantiomers of 1,1,2-trimethylcyclohexane are shown.  The next step in assigning R/S configuration is to rank the 4 groups around the chiral carbon by molecular weight, with "1" being assigned to the highest molecular weight group, and "4" to the lowest (you also did this correctly).  Now, grab the "4" group in your hand so that groups "1" through "3" form a circular pattern between your eye and your hand.  If moving from 1->2->3 involves clockwise movement, then this is the "R" configuration.  If the movement is counterclockwise, then it is the "S" compound you are observing.

Take a look at this link for a nice picture of what I'm describing.

You cannot determine the R/S configuration of the chiral carbon in the compound you've drawn on paper at the top of your post because it is unclear which group 3 or 4 is projecting towards the viewer and which is projecting away from the viewer.  Finally, note that my "R" drawing matches the "R" drawing at the bottom of your post.


The confusion of entropy change The entropy is defined as randomness of a substance or a system.
The randomness increases as we go from solids to liquids to gases. Similarly from hard diamond to soft graphite. Since graphite is soft and slippery, the orderness would be less than diamond.

But acc. to the formula for entropy change during phase transition,
$\Delta S=Q\ \text{(at const. pressure)}/T$
$\Delta S=\Delta H/T\ \text{(at const. pressure)}$

But $\Delta H$ of reaction/process
C graphite --> C diamond is positive
Meaning $\Delta S$ is positive. But it should be negative due to less entropy of diamond.

Where is the problem?

The standard free energy of formation of graphite is $0\ \mathrm{kJ/mol}$, and the standard free energy of formation of diamond is $2.9\ \mathrm{kJ/mol}$.  So, at $25\ \mathrm{^\circ C}$ and $1\ \mathrm{bar}$, graphite is favored thermodynamically over diamond.  Let's see what pressure would be required for $\Delta G$ to be zero at $25\ \mathrm{^\circ C}$.  Since the temperature is constant, the change in $G$ is described by $\mathrm dG=v\,\mathrm dP$, where $v$ is the molar volume.  The density of graphite is $2.26\ \mathrm{g/cm^3}$ and the density of diamond is $3.51\ \mathrm{g/cm^3}$.  So the molar volumes are $5.31\times 10^{-6}\ \mathrm{m^3/mol}$ for graphite and $3.42\times 10^{-6}\ \mathrm{m^3/mol}$ for diamond.  So, at pressure $P$ (in Pa), the Gibbs free energy for graphite and diamond are:
$$G_\text{Graphite}=0+5.31\times 10^{-6}P$$
$$G_\text{Diamond}=2900+3.42\times 10^{-6}P$$
So, $\Delta G$ at pressure $P$ is $$\Delta G_\mathrm{G\rightarrow D}=2900-1.89\times 10^{-6}P$$So the change in free energy is zero at $P=1.53\times 10^{9}\ \mathrm{Pa}=15\,300\ \mathrm{bar}$

Diamonds are formed naturally at even higher pressures and higher temperatures.  But this is the approximate order of magnitude.

So at $25\ \mathrm{^\circ C}$, graphite is highly thermodynamically favored over diamond.  The actual presence of diamonds at $25\ \mathrm{^\circ C}$ must therefore be a matter of kinetics, rather than thermodynamics.


Can the hydroxyl group of 4-hydroxybenzoic acid react with phosphorus pentachloride? I learned that, a phenol will not react (or react very slowly) with $\ce{PCl5}$ due to its stabilized structure.

How about the $\ce{-OH}$ group on the benzene ring of 4-hydroxybenzoic acid? Will that $\ce{-OH}$ group react with $\ce{PCl5}$? If the answer is yes, is it anything to do with the $\ce{-COOH}$ group?

a phenol will not react (or react very slowly) with $\ce{PCl5}$

That's not true. Phenol and $\ce{PCl5}$ do react, producing mixture of compounds of formula $\ce{PCl_n(C6H5O)_{5-n}}$ (with $n > 1$ ). It is even possible to produce $\ce{(C6H5O)5P}$![1]

However, unlike in alcohols, in phenols:

the carbon–oxygen bond is much stronger, having a bond order of more than one and being formed with $\mathrm{sp^2}$ hybridized orbitals
the formation of the corresponding transition state for nucleophilic substitution would destroy the aromaticity of the benzene ring.

Thus, the formation of chlorobenzene does not proceed at a significant rate.

It is still possible to replace the phenol group by a chloride atom in some cases. For example, it is possible to prepare picrylchloride (1,3,5-trinitrochlorobenzene) by reaction of pyridine picrate and $\ce{PCl5}$, since three nitro groups greatly increase the reactivity of the phenyl fragment in nucleophilic aromatic substitution reactions.[2] However, as a general rule, benzene derivatives are not active in nucleophilic substitution except if there are a lot of strong π-electron withdrawing groups in the ring.

As for 4-hydroxybenzoic acid, my guess is that it should react as both phenol (as above) and benzoic acid (which forms benzoyl chloride).

Addition

While I was unable to find precise information on para-hydroxybenzoic acid, there is some info on ortho-hydroxybenzoic (salicylic acid). $\ce{PCl5}$ replaces the $\ce{OH}$ fragment of the carboxylic acid group with chlorine. The phosphorus oxochloride then reacts with the phenolic group, producing an $\ce{O-POCl2}$ fragment. The information, however, does not look absolutely trustworthy. Still, assuming it is correct and extrapolating it onto para-hydroxybenzoic acid, I'd say that the product should be $\ce{Cl-C(O)-C6H4-O-P(O)Cl2}$.

References

[1]: Ramirez, F.; Bigler, A. J.; Smith, C. P. Pentaphenoxyphosphorane. J. Am. Chem. Soc. 1968, 90 (13), 3507–3511. DOI: 10.1021/ja01015a038.

[2]: Boyer, R.; Spencer, E. Y.; Wright, G. F. Can. J. Chem. 1946, 24b (5), 200–203. DOI: 10.1139/cjr46b-025.


Three Phases in TFM test instead of two I am currently trying to perform the Total Fatty Matter (TFM) test on soap and for some reason I am getting three phases instead of two.
The surfactant used is SLES and other reagents include hot water, HCl, petroleum ether.
I have no idea as to what could be causing the 3 phases and my knowledge in organic chemistry is very little. Any help would be greatly appreciated. Below is the procedure:
I dissolve soap in hot water and then I transfer the cooled solution to a decantation flask. I add concentrated HCl (37% stock solution) and then petroleum ether to increase extraction ratio.
I repeat this extraction two more times (and after this I am getting 3 phases) and then I purify the organic phase twice using hot water to eliminate any HCl.
I then heat the organic phase to vaporize ether and calculate TFM %.
The surfactant used in the soap is sodium laureth sulfate (SLES), and I think that it may relate to why I have 3 phases but I can't tell why. What is the reason for this?

Soap is a fatty acid salt of an alkali metal, typically sodium or potassium. The fatty acid itself is always insoluble in water, and therefore acidification of the soap solution produces two layers, a lower aqueous layer with the alkali salt of the acid and an upper layer of fatty acid laced with aqueous material. This can be treated with an organic solvent to dissolve the fatty acid and expel the aqueous material, giving two immiscible layers.
I suspect that your soap is a combination of fatty acid soap plus SLES.
Sodium laureth sulfate is not a soap, it is a surfactant, and it is so constructed so as to be more soluble in water than sodium laurate (which is a soap) or sodium lauryl sulfonate (which is a surfactant) by addition of typically 3 -OCH$_2$CH$_2$O- (ethylene glycol) groups. This complicates the precipitation and extraction; another variable is the solvent: while petroleum ether is organic, it has poor solvent power, and long chain acids are not well dissolved. That's why you have to do the extraction many times.
I visualize your experiment to go this way:

When you add the concentrated HCl to the sodium laureth sulfate, you insolubilize the HLES (the acid form), but it is not easily separable. A variable here is whether there is excess acid and how much. Another variable is that the precipitation is incomplete.

You add petroleum ether to extract the HLES and reject water and excess acid, leaving, hopefully, mostly NaCl in the aqueous phase. You do this again, extracting more HLES.

However, HLES is somewhat soluble in water (because it contains 3 -OCH$_2$CH$_2$O- groups), and is quite a good emulsifier for organics, and pet ether has a lot of low mw molecules, so before the last of the HLES is extracted, you find that the HLES has attracted a lot of water and a lot of pet ether. But this combination cannot contain all the water, so it is immiscible with the lower aqueous layer; it contains too much water to be miscible with the pet ether, so it is a third phase. Excess acid in step 1 complicates things because acids increase the solubility of ethoxylated surfactants.

You get something like this:

So instead of dealing with a soluble/insoluble situation, you have a significant but incomplete precipitation of HLES, and what doesn't precipitate out manages to emulsify with some of the pet ether and create the third phase.
A possible way to improve your separation is to add NaCl to the mixture after adding the HCl to maximize the salting-out effect. Another adjustment is to minimize excess acid (add just enough acid to precipitate the HLES - measure pH with paper) or add NaOH to adjust pH to near neutral (pH 6-8), when you get the third phase. This minimizes acid solubilization of the HLES, but keep the aqueous layer saturated with NaCl by adding solid NaCl if necessary.
You could also substitute toluene for the pet ether, but there is somewhat more hazard. Diethyl ether would be the extractant of choice many years ago, but alas, we must emphasize safety.


What inference can be made from the fact that there is no reversal in how three components eluate? I gave an answer this question about TLC. I now realize that I am blundering my way through this question.

The problem statement doesn't specify if the compounds to be eluted are polar or non-polar. So this seems to leave a door open.

Question 1 - Could compounds be either relatively polar or relatively non-polar?

Question 2 - If an absolutely non-polar solvent like hexane were used, and some other more polar solvent, wouldn't a reversal of the order of the elution of the compounds happen?

Question 3 - Have I in fact, like a blind squirrel, finally found the nut?

EDIT 1

The problem specifies that "normal phase TLC" is being used so the stationary phase is polar.

The problem only states that the elutants have different polarity and not that any is actually non-polar like hexane.

EDIT 2

With Question 3 I was trying to be funny and ask if my two answers to the original question are right.

Question 1
Yes, of course! Something that is "relatively" non-polar would travel further with a non-polar mobile phase because it would have less attraction to the stationary phase. While a polar substance would not travel as far with the mobile phase because it would have a greater attraction to the stationary phase.
Question 2
Now, that depends. Are you also changing the stationary phase? Exactly how polar is the new polar-mobile phase? It would really depend on which phase the solutes have a greater attraction to.
Question 3
You are on your way. Quantitative HPLC is the bane of my existence. There's a million factors to consider, such as column temperature, flow rate of the solvent, particle size, polarity and charge. Usually, it takes a bit of experimentaton to reach a perfect combination that allows complete separation of whatever it is you're interested in. Usually, your best bet in identifying a compound at a particular point is to run a little bit of pure sample on its own. In the end, above all else, you need to find a combination that doesn't result in chemicals overlapping with each other.
In qualitative chromatography, as long as you can resolve all possible signals, you've done well enough.


Why is LiOH a weaker base than NaOH Why do LiOH is weak base as compared to NaOH even though it's hydration enthalpy is greater than Na?

$\ce{NaOH}$ and $\ce{LiOH}$ are both classified as ionic compounds, but as with most things in chemistry, this is always on a continuum scale, in this case between ionic and covalent bonds.  $\ce{NaOH}$ is 'more' ionic in character which means its $\ce{OH^-}$ is more frequently unaccompanied by its counter cation, $\ce{Na+}$ as compared to $\ce{LiOH}$ and its corresponding $\ce{Li+}$.

Without the cation to balance the character, $\ce{OH^-}$ will balance its charge via deprotonation, and hence it is a stronger base.


Can salicylic acid be synthesized from salicylaldehyde? You have a bottle of salicylaldehyde. And you want to make salicylic acid from salicylaldehyde. You can use all reagents and catalyst to make salicylic acid.

I think salicylic acid can be synthesized from salicylaldehyde since salicylaldehyde has a structure similar to salicylic acid. I'm thinking that it might be possible to use $\ce{KMnO4}$, but I'm afraid the hydroxyl group attached to the benzene would be oxidized too.

There are a large number of reagents out there that will transform aldehydes into the corresponding acids (benzaldehydes and benzoic acids in this case) and $\ce{KMnO4}$ is indeed one of them.

Oxidation of the phenolic hydroxide is only possible if at the same time the aromatic system is destroyed. However, it is rather hard to come up with a good product of this oxidation since a simple quinone isn’t possible. It therefore takes much harsher conditions than the oxidation of an aldehyde.

Things would change if a second hydroxy group were attached to the same aromatic system either ortho or para to the first. Then the oxidation to the corresponding quinone is much easier.


Can atomic partial charges be measured in molecules experimentally? Can we measure atomic partial charges in molecules experimentally? The charge of isolated ions can be measured, but when atoms are part of a molecule, the case is much more difficult. We do not really know where one atom starts and where another one stops (the atomic radii are diffuse). So I am inclined to say that partial charges must be based on theoretical concepts, such as the Mulliken population analysis (or a range of other methods), or be derived (estimated) from measurables such as dipole moment.

However, is it possible to probe molecules at the atomic level, using STM, TEM, or some other electron probing technique, and get the partial charges that way, perhaps based on electron-electron repulsions? Still, it seems to me, that the border between the atoms need to be properly defined in order to get the partial charges.

According to the Wikipedia page "Partial charge", several experimental techniques can be used to estimate the partial charge, e.g. XPS, NMR, EPR, UV/vis, with more. But do these techniques measure something which is a direct result of the effect of partial charge, or is some other physical observable measured, from which the partial charges are "guessed"? For example, using XPS to get the kinetic energy of the photoelectron seems like a "direct" measurements (although I know that what the detector sees is not really the kinetic energy, but something with which the kinetic energy correlates: calibration is needed). However, the idea of partial charge seems to be somewhat vague. Numerical values of partial charges depend on the definition of where one atom starts and where another ends, which is an abstract concept: this border cannot be observed or measured.

Another issue that comes to mind, is that the partial charge is not necessarily uniformly distributed around each atom, but will depend on bonds, lone pairs, interactions with other molecules, etc.

Let me put it this way. There are no atoms in molecules. There is just a continuous cloud of electron density, and as you correctly pointed out, we have no way to pinpoint where one atom ends and another starts. Or rather, we have many ways to do that, all slightly different and all inherently arbitrary. (My personal favorite is the concept of Bader charges, but that's a matter of taste.)

So yes, the very idea of partial atomic charges is unavoidably vague.


Proving using ideal gas law that molar volume at NTP is 22.4 litres The definition of NTP as given by IUPAC is

NTP – Normal Temperature and Pressure – is defined as air
at $20\ ^\circ\mathrm C$ ($\pu{293.15 K}$, $\pu{68^{\circ}F}$) and $\pu{1 atm}$ ($\pu{101.325 kN/m2}$, $101.325\ \mathrm{kPa}$, $14.7\ \mathrm{psia}$, $0\ \mathrm{psig}$, $29.92\ \mathrm{inHg}$, $407\ \mathrm{in}\ce{H2O}$, $760\ \mathrm{Torr}$).
It is also known that volume occupied by $1$ mole of gas at NTP is
$\pu{22.4 L}$.

My attempt at proving the above statement,
For $1$ mole,  $PV=RT$
Substituting $P=\pu{1 atm}$, $T=\pu{293 K}$,  $R=0.0821\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}$, we get $V=\pu{24.05 L}$ as the molar volume.
On putting $T=\pu{273 K}$ and $P=\pu{1 bar}$ (conditions of STP), we get $V=\pu{22.2 L}$
Therefore, molar volume at STP is $\pu{22.4L}$.
Why is the above calculation in NTP condition giving wrong results? How else to prove that molar volume at NTP is ${22.4\ \mathrm L}$?

The issue is that one mole of a gas at NTP should not be 22.4 L based on the ideal gas. This is only true for a gas at STP. The key difference between the two conditions is they are defined by different temperatures (STP is for 273 K while NTP is for 293 K).

They are also different in that the IUPAC now defines STP with respect to 1 bar while NTP is defined with respect to 1 atm.

See The Engineering ToolBox for more info.


Does oxygen gas form hydrogen bonds? I learned that hydrogen bonding is formed from a 'sandwich' of hydrogen and either oxygen, nitrogen, or fluorine where nitrogen, fluorine, or oxygen is the bread.

Will oxygen gas and water form hydrogen bonds? The oxygen-hydrogen sandwich could be made, but oxygen gas has no hydrogen attached. Is it necessary for both molecules to contain hydrogen for hydrogen bonding to occur?

To make it clear, the question is: do both molecules have to contain hydrogen in order for hydrogen bonding to occur?

To make it clear, the question is: do both molecules have to contain hydrogen in order for hydrogen bonding to occur?

No, as noted by Prakhar in his answer, tertiary amines are one class of compounds that can participate in hydrogen bonding (as an acceptor) when there is no hydrogen atom directly bound to the participating atom.

As an example, trimethylamine can hydrogen bond just fine with water:

More information can be found, e.g., here at ChemGuide (scroll down to "Solubility in water").


Is the phenyl group in benzophenone electron donating/withdrawing? The ketone carbonyl group is less reactive to nucleophilic attack than aldehyde since the alkyl group in ketone is electron-donating towards the carbon.

Is this the same case for benzophenone, where there is a phenyl group?

Phenyl groups are usually electron-withdrawing, but in the case of benzophenone, would it be electron-donating through resonance?

A keto or an aldo group with phenyl ring is less reactive towards nucleophillic addition then their aliphatic counterparts(acetaldehyde or acetone).With phenyl group the positive charge on carbonyl carbon is delocalised via resonance (structures  2 , 3 , 4 , 5 ) .

Resonance with phenyl ring is due to conjugation.

such conjugation is absent in aliphatic aldehydes and ketones.Therefore carbonyl compounds with phenyl rings bonded to carbonyl group are less reactive then their aliphatic counterparts.

With respect to + carbon on carbonyl group pi electrons on phenyl group are polarized towards it , in a way electron donating .


Did I accidentally create phosgene gas I accidentally mixed bleach and acetone trying to remove hair dye stains from my shower wall. The mixture turned orange when the bleach and acetone touched. Could I have created phosgene gas unintentionally? I’m really concerned for my families safety. I opened the window in the bathroom and closed the door. I poured water on the surfaces where the bleach and acetone mixture was a few times and all around the tub to make it sure there was none left. Should I be concerned or no?

I see the post was closed. I appreciate the feedback I received. Unfortunately poison control did absolutely nothing for me so that’s why I had to resort to online help. Thanks.

Sodium hypochlorite (the active ingredient in bleach) and acetone react to produce chloroform.

Chloroform can slowly decompose into phosgene in air with exposure to oxygen
$$\ce{2CHCl3 + O2 ->[h\nu] 2COCl2 + 2HCl}$$

but not in significant quantities under the conditions you describe.

You did the right thing by opening the window for ventilation and isolating the fumes to the bathroom and rinsing with copious water.

It is best not to mix bleach with anything except water and especially not household cleaning products.


Why is CO a major species only in fuel-rich combustion? In fuel-rich combustion, my understanding is that enough fuel reacts and consumes all of the oxygen present. The combustion products $\ce{CO2}$ and $\ce{H2O}$ undergo dissociation until equilibrium is established and one of these is $$\ce{CO2 -> CO + 1/2O2}.$$

From the graph below, the equilibrium mole fraction of combustion products are plotted vs. equivalence ratio. Why is there a sudden rise in $\ce{[CO]}$ as we move into fuel rich combustion? Is this due to the drop in $\ce{[O2]}$ such that $\ce{[CO]}$ must increase to compensate for equilibrium to be established?

I think it is quite simple: the higher the proportion of the compound that will burn in the mixture, the more oxygen will be required to a complete combustion. If you have a constant amount of oxygen (in the reactor), and more is requires while increasing the amount of fuel, the complete combustion will gradually decrease, and the ratio $CO/CO_2$ produced will increase as a result of that.

As it can be seen on the graphic, from the point that the equivalent ratio of fuel and oxygen is larger than one, the $CO$ starts being formed in a considerable proportion, and over a ratio of 1:3, starts being the major product.


The effect of HCL concentration on the Solubility of Hydroxyapatite Recently I conducted an experiment to investigate the effect of HCL concentration on the solubility of hydroxyapatite [Ca5(PO4)3(OH)]. This was done by taking a sample of the solution after hydroxyapatite was left to dissolve in HCL, and conducting an EDTA titration using this solution to find how many calcium ions were present. I have already established that hydroxyapatite is more soluble in solutions of HCL at higher concentrations (lower pH). However, I was wondering what the exact scientific theory behind this reaction was.

The $Ksp$ expression for hydroxyapatite is: $\ce{ Ca5(PO4)3(OH)_{(s)} -> 5Ca^{2+}_{(aq)} + 3PO4^{3-}_{(aq)} + OH-_{(aq)}}$.

When $\ce{HCl_{(aq)}}$ is added, it reacts with the dissolved $\ce{OH-}$ to form water, thereby decreasing the $\ce{OH-}$ concentration. Le Chatlier's principle predicts that more hydroxyapatite will dissolve (shift to the right).


Why is Lanthanum a "fluoride ion sensitive electrode" and hence used for measuring "flouride concentration"? My chemistry text book says that "Atomic Absorption Spectrometry (AAS) can be used to determine the concentration of metal ions, but other methods must be used for anions. One convenient method is to use an ion-selective electrode (ISE)." What is the reason? Why can't ASP be used for determining the concentration of anions?

The author then argues that "Lanthanum fluoride crystal doped with europium" is the element used for measuring the concentration of fluorine in water because "fluoride ions bind to the crystal". What is the reason behind choosing "Lanthanum doped with europium"?

What is the reason behind choosing "Lanthanum doped with europium"?

Well, you said it yourself, because "fluoride ions bind to the crystal".

Here's a great explanation from Wikipedia:

In the lanthanum fluoride electrode, the sensing element is a crystal of lanthanum fluoride LaF3, doped with europium fluoride EuF2 to create lattice vacancies. Such a crystal is an ionic conductor by virtue of the mobility of fluoride ions which jump between lattice vacancies.

Here's a rough sketch I made to help you visualizing this. This is a simplistic drawing and the actual crystalline structure of the compound is 3-dimensional.

I must stress this: Both the structure and the proportions are probably nowhere near the ones in my drawing, it is only for didactic purposes.

An electrochemical cell may be constructed using such a crystal as a membrane separating two fluoride solutions. This cell acts as a concentration cell with transference where the fluoride transport number is 1. As transference of charge through the crystal is almost exclusively due to fluoride, the electrode is highly specific to fluoride. The only ion which significantly interferes is hydroxide (OH-). Generally such "alkaline error" can be avoided by buffering the sample to a pH below 7.

We had the pH around 5.5. We also used TISAB for the preparation of water samples. The Wikipedia article on TISAB gives an excellent explanation on it's usage in fluoride ion analysis.

Why ASP cannot be used for determining the concentration of anions?

Well, I cannot give you an authoritative answer on that. I'll ask a teacher of mine this week and edit the answer with an update - if someone reading this knows it feel free to improve this answer or add a new one. In the case of fluoride it's probably because the Lanthanum electrode has pretty decent detection limits (it does). It's a lot cheaper to use it instead of AAS.


Curious about the chemistry of Carbon-14 after it's produced in the atmosphere It's often explained how $\ce{^14C}$ is formed in the upper atmosphere, a neutron hits a nitrogen atom and ejects a proton.  Source.  Since this is likely to happen to an $\ce{N2}$, I'm curious what the chemistry that follows is.

Does the energy of impact split the molecule, leaving a lone $\ce{^14C}$ and a lone $\ce{^14N}$ (and a lone hydrogen), (each highly eager to bond with something) or does it leave a $\ce{C-N}$ which might also readily bond with $\ce{O2}$ or maybe something else.    I'm mostly curious what the immediate chemical reactions are after the $\ce{^14C}$ forms cause it seems to me that it should be initially pretty reactive.

Edit & maybe a partial answer:

It occurs to me after giving it some thought that if an $\ce{N2}$ is split into a free $\ce{C}$ and a free $\ce{N}$, each would be most likely to bond to an $\ce{O2}$, or a single $\ce{O}$ from $\ce{O3}$ if the split happens near the ozone layer (which is possible given the listed height where most $\ce{^14C}$ forms, "altitudes of 9 to 15 km" - same link as above).
Initially I was thinking it might be more exotic than that, but having given this some thought I think that's what happens.   The $\ce{C}$ and the $\ce{N}$ split and mostly each binds with $\ce{O2}$, 2nd most common they bind with $\ce{O}$.  $\ce{CN}$ is probably pretty uncommon, now that I think about it.

wikipedia article cites: "After production in the upper atmosphere, the carbon-14 atoms react rapidly to form mostly (about 93%) 14CO (carbon monoxide), which subsequently oxidizes at a slower rate to form 14CO2, radioactive carbon dioxide. The gas mixes rapidly and becomes evenly distributed throughout the atmosphere (the mixing timescale in the order of weeks)."


Performing an SN1 reaction in a polar aprotic solvent Straightforward question
What would be the outcome of having an SN1 reaction carried out in a polar aprotic solvent? I have found this question approached scarcely and with unsatisfactory answers.
What I know so far
I know the SN1 reaction is carried out in a polar protic solvent since it helps stabilize the carbocation that gets formed (thanks to the lone pairs adjacent to the electronegative atoms of the solvent itself) and also the leaving group (stabilized by hydrogen bonding).

Also, many such reactions are performed as solvolysis and it is preferred - and is much more efficient - to have the solvent act as the nucleophile when this is possible. Based on this we can ascertain the fact that a polar protic solvent would be the best choice for such a reaction and its polar aprotic homologue would not do as well.

All of the reaction rates comparisons I have seen so far take into account a polar protic solvent and a nonpolar one. But what about the protic-aprotic dichotomy?
Final note
However, my question is - can we not use a polar aprotic solvent at all? I have no problem imagining it would perform in a less efficient manner. However, stabilization is not completely absent - at least it has the capacity to stabilize the carbocation. Would the sheer lack of stabilization of the leaving group (and of the nucleophile even though many SN1 reactions do not use charged nucleophiles) render this kind of solvent completely useless for this reaction? Or would it work to some extent?

The rate of an $S_N1$ reaction may be increased by increasing the polarity of the solvent of reaction.  Because, the more polar a solvent (as measured by the dielectric constant / relative permitivity, the easier to separate ions from each other; here: carbocation and its counter anion.
Contrary to polar-aprotic solvents (think DMF, DMSO) eventually leading to «naked ions», polar-protic solvents (like ethanol, methanol, water) may weakly interact with these ions; similar to a Lewis acid/base reaction, they stabilize the ions to stay in proximity to each other as an ion pair.  Eventually, it is by experiment to figure out which effect is dominating at the reaction conditions (structure of the substrate and reagents, concentration of starting material and reagents, temperature, etc.).


Structure that breaks InChI I am currently testing a piece of software that generates the InChI for a given structure. I also want to test error situations as end users will also deliver the mol files that will be used as input for the software.
I have found this description about what the current version of InChI cannot represent (Technical FAQ section 4.15 at inchi-trust.org)

InChI currently does not support the representation of:

Polymers
Complex organometallics
Markush structures
Mixtures
Conformers
Excited state and spin isomers
Non-local stereochemistry/chirality
Topological isomers
Cluster molecules
Polymorphs
Unspecific isotopic enrichment
Reactions

Also, InChI is not suitable for very large compounds; technically, InChI input may not contain more than 1023 atoms.
Please note that there are IUPAC InChI subcommittee working groups currently addressing some of these matters. Details of these efforts can be found at: http://iupac.org/web/ins/802

Unfortunately I am lacking the chemical background to generate a suitable structure as molfile that meets the described features to produce an error as an InChI cannot be generated for it.

Okay, let's tackle at least one problem here. Consider the rotation of 1,2-dichloroethane (BP86/cc-pVDZ):

These conformational changes can be further rationalised:

C and C' are the same conformation, since these are mirror images.
The same applies to B and B'.
A and C are local minima and can be referred to as conformes in the above given way
B and D are transition states - they are not stable.
If you consider the rotation A to C as a reaction, this state would mark the maximum energy needed for the conversion. I believe with reactions it is meant, that an accurate description of the transition state by InChI is not possible.

For the term non-local chirality I suggest you look at BINAP or BINOL.

In this case I have used a slightly abridged version of BINAP for educational purposes, where $\ce{Ph -> Me}$.

Other wonderful examples of non-local chirality include helicenes, which have helical chirality. Here is one example of a fancy azonia derivative of hexahelicene (parent compound and links). The structure is available at the Cambridge Crystallographic Data Centre (CCDC) Database. I removed the anion and converted it to xyz coordinates which I include in the appendix. Here are the structures, left P-chirality, right M-chirality:

On another account of non/local chirality, there is the planar chirality, as can be seen in trans-Cyclooctene (the bottom two). For further reading I suggest Neuenschwander et.al. "The Conformations of Cyclooctene: Consequences for Epoxidation Chemistry."

Markush Structures are a kind of summary structure for certain compound classes. A very simple example would be a monosubstituted biphenyl system, where the upper structure is a representation of all three below. However, it is noteworthy, that each instance of a Markush structure can be well explained by InChI.

More complicated structures are often used in patents and in other scientific publications, usually specifically referring to a very general procedure or mechanism. I found one on wikipedia commons, that could well represent a million molecules:

And another way to break the InChI are clusters. Although the definition is not very specific, the $\ce{[KO{}^{\mathit{t}}Bu]_4}$ tetramer (wikipedia, chemspider) certainly contains one. While the monomer can be described via InChI, the tetra- or Polymer should fail. It was published by Chisholm et.al. in Polyhedron and the *.cif file can be obtained from CCDB. I cleaned it up a little bit and included the *.xyz in the appendix. You can quite nicely see the $\ce{K4}$ tetrahedral cluster in the middle (hydrogens omitted for clarity).

Last but not least, I'd like to cover the matter of complicated organometallic molecules. As such I'd like to introduce the Grubbs' catalysts. Grubbs obtained the Nobel Prize in Chemistry 2005 for his research in conjunction with Yves Chauvin and Richard R. Schrock. One of the key publications was "Ring-opening metathesis polymerization (ROMP) of norbornene by a Group VIII carbene complex in protic media". The crystal structure can also be obtained at CCDC. Another example stems from "Synthesis and Applications of $\ce{RuCl2(=CHR')(PR3)2}$:  The Influence of the Alkylidene Moiety on Metathesis Activity" and is also available via CCDC. This structure can also be found in the appendix. It is generally known as a 1st generation Grubbs catalyst (Hydrogens omitted for clarity).

Appendix

The following section contains the nuclear coordinates in *.xyz(XMol) file format and angstroms as unit. (If not denoted otherwise, structures are obtained at BP86/cc-pVDZ.)

1,2-dichloroethane A

C        0.000000000      0.000000000      0.000000000
C        0.000000000      0.000000000      1.521307000
Cl       1.723641000      0.000000000     -0.593797000
H       -0.486731000     -0.904743000     -0.410753000
H       -0.486731000      0.904743000     -0.410753000
Cl      -1.723641000      0.000000000      2.115104000
H        0.486731000     -0.904743000      1.932060000
H        0.486731000      0.904743000      1.932060000

1,2-dichloroethane B

C        0.030242000      0.085780000     -0.007620000
C       -0.030242000      0.085780000      1.528927000
Cl       1.518312000     -0.758527000     -0.631286000
H       -0.837191000     -0.442465000     -0.444148000
H        0.070555000      1.115212000     -0.412109000
Cl      -1.518312000     -0.758527000      2.152593000
H        0.837191000     -0.442465000      1.965455000
H       -0.070555000      1.115212000      1.933416000

1,2-dichloroethane C

C        0.066377000      0.127741000      0.004558000
C       -0.066377000      0.127741000      1.516749000
Cl       1.083384000     -1.241726000     -0.615213000
H       -0.919786000      0.050600000     -0.492443000
H        0.568277000      1.063384000     -0.317636000
Cl      -1.083384000     -1.241726000      2.136520000
H        0.919786000      0.050600000      2.013750000
H       -0.568277000      1.063384000      1.838943000

1,2-dichloroethane D

C        0.164887000      0.141535000     -0.000240000
C       -0.164887000      0.141535000      1.521547000
Cl       0.337146000     -1.473219000     -0.795160000
H       -0.633932000      0.666019000     -0.558099000
H        1.123149000      0.665665000     -0.177529000
Cl      -0.337146000     -1.473219000      2.316468000
H        0.633932000      0.666019000      2.079406000
H       -1.123149000      0.665665000      1.698836000

R-BINAP

C       -0.086423000     -0.055821000     -0.031546000
C       -0.090885000     -0.047699000      1.480085000
C        1.148913000     -0.158515000      2.237313000
C        1.064686000     -0.139089000      3.676951000
C       -0.131097000     -0.014450000      4.357670000
C       -1.357362000      0.116010000      3.652970000
C       -1.345263000      0.107319000      2.211098000
P        2.768926000     -0.321256000      1.451350000
H        1.988600000     -0.233144000      4.265654000
H       -0.135982000     -0.008251000      5.459401000
C       -2.590782000      0.269421000      4.370415000
C       -2.623021000      0.305506000      1.559779000
C        0.437392000      1.073619000     -0.788457000
C        0.410661000      0.991852000     -2.228054000
C       -0.079670000     -0.105834000     -2.908992000
C       -0.579162000     -1.233570000     -2.204534000
C       -0.579999000     -1.219208000     -0.762676000
P        1.107602000      2.557133000     -0.001640000
H        0.790955000      1.839345000     -2.816335000
H       -0.077540000     -0.112552000     -4.010726000
C       -1.063317000     -2.378355000     -2.921905000
C       -1.047368000     -2.424579000     -0.111239000
C       -3.792005000      0.431032000      3.705083000
H       -2.558632000      0.263827000      5.471798000
H       -4.735477000      0.552772000      4.258523000
C       -3.797353000      0.457087000      2.279022000
H       -2.662047000      0.340912000      0.466043000
H       -4.750761000      0.602046000      1.748556000
C       -1.503463000     -3.517255000     -0.830391000
H       -1.036094000     -2.475958000      0.982488000
H       -1.854657000     -4.415344000     -0.299810000
C       -1.523650000     -3.499569000     -2.256520000
H       -1.052652000     -2.347534000     -4.023279000
H       -1.890198000     -4.377463000     -2.809877000
C        4.061075000     -0.509159000      2.730645000
C        2.844045000     -1.846744000      0.446017000
C        1.585856000      3.772639000     -1.280334000
C       -0.173490000      3.383258000      1.008127000
H        0.254332000      4.289560000      1.485098000
H       -1.067700000      3.703930000      0.418133000
H       -0.532498000      2.717241000      1.815764000
H        1.997887000      4.670081000     -0.777298000
H        2.377005000      3.362349000     -1.938832000
H        0.749130000      4.118465000     -1.941170000
H        5.044249000     -0.601334000      2.227532000
H        4.100503000      0.383637000      3.385840000
H        3.944572000     -1.404262000      3.395364000
H        3.842692000     -1.931312000     -0.030903000
H        2.674387000     -2.779210000      1.039384000
H        2.087489000     -1.826871000     -0.361580000

S-BINAP

C       -1.383946000     -3.448543000     -2.256996000
C       -1.065989000     -2.266144000     -2.910129000
C       -0.652376000     -1.110469000     -2.179491000
C       -0.563545000     -1.177237000     -0.738280000
C       -0.899026000     -2.413534000     -0.098700000
C       -1.299620000     -3.519323000     -0.838209000
C       -0.154302000     -0.008671000      0.004167000
C        0.166467000      1.182512000     -0.673969000
C        0.072046000      1.221328000     -2.098779000
C       -0.323252000      0.113168000     -2.831301000
C       -0.061735000     -0.093981000      1.507135000
C       -1.110430000      0.372113000      2.322674000
C       -0.989013000      0.267385000      3.742090000
C        0.132224000     -0.289631000      4.336601000
C        1.210137000     -0.777916000      3.542877000
C        1.113248000     -0.682564000      2.103804000
C        2.206553000     -1.174312000      1.320794000
C        3.329975000     -1.726929000      1.922868000
C        3.419059000     -1.820539000      3.340084000
C        2.377998000     -1.353835000      4.130094000
P       -2.590653000      1.186493000      1.502738000
C       -3.750658000     -0.293139000      1.402088000
P        0.636241000      2.692004000      0.339668000
C        2.511918000      2.555623000      0.253510000
C       -3.385046000      2.079953000      2.947654000
C        0.388676000      4.078351000     -0.898992000
H       -1.802841000      0.635802000      4.383638000
H        0.202504000     -0.360978000      5.434128000
H        0.319305000      2.150696000     -2.632672000
H       -0.386992000      0.166196000     -3.930250000
H        2.436377000     -1.419019000      5.228476000
H        4.313241000     -2.262708000      3.805535000
H        2.146133000     -1.110484000      0.224695000
H        4.157224000     -2.097887000      1.298347000
H       -0.834575000     -2.478936000      0.997042000
H       -1.552289000     -4.458551000     -0.322477000
H       -1.128190000     -2.199481000     -4.008215000
H       -1.700498000     -4.331984000     -2.832408000
H        2.950535000      3.473538000      0.698290000
H        2.842563000      1.687658000      0.858736000
H        2.884117000      2.439555000     -0.786611000
H        0.678948000      5.023670000     -0.395042000
H        0.990357000      3.982867000     -1.828352000
H       -0.686746000      4.150262000     -1.160750000
H       -4.302108000      2.573768000      2.564446000
H       -3.673500000      1.421270000      3.794842000
H       -2.700063000      2.872570000      3.312044000
H       -4.742112000      0.063129000      1.051704000
H       -3.359638000     -1.014277000      0.656570000
H       -3.871382000     -0.807607000      2.379049000

M-8a-azonia[6]helicene (from crystal structure)

N        3.565811000      0.259443000     13.724558000
C        7.445232000     -1.196589000     12.564039000
C        8.629416000     -1.762757000     12.151335000
C        8.885324000     -1.978555000     10.783139000
C        7.914184000     -1.694865000      9.864763000
C        6.665024000     -1.171130000     10.265448000
C        5.584378000     -1.088690000      9.330104000
C        4.339207000     -0.800151000      9.738213000
C        4.088732000     -0.406137000     11.083608000
C        2.763235000     -0.187914000     11.543327000
C        2.522579000      0.029096000     12.830042000
C        3.246403000      0.443720000     15.064121000
C        4.198045000      0.692252000     15.970654000
C        5.531025000      0.961394000     15.546639000
C        6.527506000      1.327523000     16.488699000
C        7.769629000      1.675468000     16.086777000
C        8.044342000      1.864594000     14.697195000
C        9.259232000      2.459858000     14.281133000
C        9.466841000      2.806590000     12.982045000
C        8.445715000      2.608977000     12.035743000
C        7.265466000      1.985829000     12.396483000
C        7.061105000      1.553020000     13.720492000
C        5.833567000      0.912900000     14.172787000
C        4.884651000      0.303088000     13.284635000
C        5.151668000     -0.288539000     12.002691000
C        6.444932000     -0.841371000     11.632053000
H        7.296301000     -1.066868000     13.466331000
H        9.304643000     -2.036748000     12.840647000
H        9.760406000     -2.388329000     10.496986000
H        8.038683000     -1.964007000      8.943383000
H        5.780686000     -1.333585000      8.395468000
H        3.604979000     -0.909262000      9.084780000
H        2.029733000     -0.315211000     10.922946000
H        1.672696000      0.060618000     13.273677000
H        2.290439000      0.327335000     15.253240000
H        3.954366000      0.824398000     16.914659000
H        6.255785000      1.333585000     17.427224000
H        8.447606000      1.891266000     16.702563000
H        9.925491000      2.630800000     14.988120000
H       10.286033000      3.200604000     12.619714000
H        8.559741000      2.909640000     11.135042000
H        6.570317000      1.891266000     11.711236000

M-8a-azonia[6]helicene (from BP86/STO-3G)

N        0.241946000      3.510759000     13.746203000
C       -1.207087000      7.492952000     12.512135000
C       -1.801472000      8.692275000     12.089683000
C       -2.102492000      8.912745000     10.707738000
C       -1.853647000      7.899258000      9.772330000
C       -1.278056000      6.643678000     10.182732000
C       -1.146616000      5.553783000      9.229249000
C       -0.770948000      4.283969000      9.646130000
C       -0.392477000      4.045979000     11.023979000
C       -0.161103000      2.688046000     11.484550000
C        0.042219000      2.428893000     12.821687000
C        0.400500000      3.183323000     15.136378000
C        0.639392000      4.176721000     16.059802000
C        0.949043000      5.527644000     15.625469000
C        1.361398000      6.519481000     16.597330000
C        1.808770000      7.766658000     16.182869000
C        1.981223000      8.066019000     14.770425000
C        2.627560000      9.287588000     14.362758000
C        2.913363000      9.530579000     13.012425000
C        2.578886000      8.546652000     12.028110000
C        1.916409000      7.363991000     12.392255000
C        1.565834000      7.097902000     13.763105000
C        0.907348000      5.851096000     14.211296000
C        0.281492000      4.885356000     13.292158000
C       -0.309154000      5.144896000     11.968134000
C       -0.894790000      6.445456000     11.575101000
H       -1.007504000      7.339193000     13.585554000
H       -2.048406000      9.472285000     12.834293000
H       -2.558542000      9.868716000     10.388465000
H       -2.123797000      8.035037000      8.708340000
H       -1.422274000      5.738729000      8.172836000
H       -0.763274000      3.430969000      8.943112000
H       -0.251900000      1.843110000     10.776492000
H        0.087281000      1.420761000     13.274365000
H        0.295196000      2.107892000     15.372408000
H        0.697253000      3.916258000     17.133249000
H        1.320711000      6.254708000     17.669775000
H        2.110372000      8.530075000     16.926174000
H        2.921715000     10.016221000     15.141242000
H        3.423497000     10.463796000     12.708003000
H        2.853573000      8.716110000     10.969950000
H        1.691114000      6.611596000     11.618453000

$\ce{KO{}^{\mathit{t}}Bu}$ tetramer (from crystal structure)

K       -1.309381000     -1.309381000     -1.309381000
O       -1.313567000      1.313567000     -1.313567000
C       -2.107232000      2.107232000     -2.107232000
C       -3.502008000      2.298114000     -1.478495000
H       -3.842748000      1.465100000     -1.465100000
H       -4.060420000      2.821364000     -2.059512000
H       -3.516240000      2.821364000     -0.619528000
O       -1.313567000     -1.313567000      1.313567000
O        1.313567000     -1.313567000     -1.313567000
K       -1.309381000      1.309381000      1.309381000
K        1.309381000      1.309381000     -1.309381000
C       -1.478495000      3.502008000     -2.298114000
C       -2.298114000      1.478495000     -3.502008000
K        1.309381000     -1.309381000      1.309381000
C       -2.107232000     -2.107232000      2.107232000
C        2.107232000     -2.107232000     -2.107232000
O        1.313567000      1.313567000      1.313567000
H       -1.465100000      3.842748000     -1.465100000
H       -2.059512000      4.060420000     -2.821364000
H       -0.619528000      3.516240000     -2.821364000
H       -1.465100000      1.465100000     -3.842748000
H       -2.821364000      2.059512000     -4.060420000
H       -2.821364000      0.619528000     -3.516240000
C       -1.478495000     -3.502008000      2.298114000
C       -3.502008000     -2.298114000      1.478495000
C       -2.298114000     -1.478495000      3.502008000
C        3.502008000     -1.478495000     -2.298114000
C        1.478495000     -2.298114000     -3.502008000
C        2.298114000     -3.502008000     -1.478495000
C        2.107232000      2.107232000      2.107232000
H       -1.465100000     -3.842748000      1.465100000
H       -2.059512000     -4.060420000      2.821364000
H       -0.619528000     -3.516240000      2.821364000
H       -3.842748000     -1.465100000      1.465100000
H       -4.060420000     -2.821364000      2.059512000
H       -3.516240000     -2.821364000      0.619528000
H       -1.465100000     -1.465100000      3.842748000
H       -2.821364000     -2.059512000      4.060420000
H       -2.821364000     -0.619528000      3.516240000
H        1.465100000     -1.465100000     -3.842748000
H        3.842748000     -1.465100000     -1.465100000
H        4.060420000     -2.059512000     -2.821364000
H        3.516240000     -0.619528000     -2.821364000
H        2.059512000     -2.821364000     -4.060420000
H        0.619528000     -2.821364000     -3.516240000
H        1.465100000     -3.842748000     -1.465100000
H        2.821364000     -4.060420000     -2.059512000
H        2.821364000     -3.516240000     -0.619528000
C        1.478495000      2.298114000      3.502008000
C        3.502008000      1.478495000      2.298114000
C        2.298114000      3.502008000      1.478495000
H        1.465100000      1.465100000      3.842748000
H        2.059512000      2.821364000      4.060420000
H        0.619528000      2.821364000      3.516240000
H        3.842748000      1.465100000      1.465100000
H        4.060420000      2.059512000      2.821364000
H        3.516240000      0.619528000      2.821364000
H        1.465100000      3.842748000      1.465100000
H        2.821364000      4.060420000      2.059512000
H        2.821364000      3.516240000      0.619528000

Grubbs I catalyst

Ru      1.923022    1.782458    4.279841
Cl      3.508375    2.054525    6.061616
P       0.620071    3.495015    5.336792
C       2.967497    2.881791    3.239789
Cl      0.101984    1.197998    2.838894
P       2.855052   -0.401312    3.741075
C       2.954465    3.343954    1.839333
Cl      3.310061    4.678711   -2.473972
C       1.880873    3.257547    0.960743
C       1.973325    3.667349   -0.352875
C       3.185197    4.168660   -0.812833
C       4.252303    4.292067    0.012884
C       4.145603    3.890748    1.341393
C      -0.637087    4.207002    4.175769
C      -1.979932    4.694567    4.692950
C      -2.932151    4.983356    3.538285
C      -2.339331    5.946324    2.517809
C      -0.957803    5.505405    2.056512
C      -0.041546    5.252113    3.235271
C       1.483967    4.856669    6.241316
C       0.518279    5.817396    6.934517
C       1.256971    6.738015    7.899778
C       2.346867    7.495106    7.161401
C       3.302557    6.547168    6.465690
C       2.565538    5.630984    5.497585
C      -0.333586    2.698072    6.719178
C      -1.014762    1.418485    6.241149
C      -1.859827    0.796208    7.358000
C      -1.043744    0.536540    8.601177
C      -0.361483    1.805588    9.075860
C       0.493782    2.427234    7.979254
C       2.628515   -0.975242    1.991593
C       3.236690   -0.059043    0.937820
C       2.687448   -0.416553   -0.435697
C       2.971677   -1.856587   -0.778868
C       2.466883   -2.799363    0.290130
C       2.999176   -2.423218    1.675026
C       4.622054   -0.540862    4.248552
C       5.523290    0.453308    3.517538
C       6.887074    0.518241    4.175936
C       7.549339   -0.842629    4.233326
C       6.649166   -1.859830    4.903939
C       5.271941   -1.927013    4.232992
C       1.995357   -1.737789    4.704830
C       2.010202   -1.452357    6.212537
C       1.292567   -2.535071    7.008639
C      -0.120796   -2.764155    6.507353
C      -0.110733   -3.109699    5.031603
C       0.567738   -2.010334    4.218268
H       3.716836    3.154747    3.670969
H       1.120535    2.917815    1.276640
H       1.189061    3.632874   -0.962082
H       5.017596    4.645026   -0.262690
H       4.792512    4.003484    1.897392
H      -0.815388    3.399122    3.615754
H      -1.902150    5.510235    5.223685
H      -2.378700    3.974732    5.228704
H      -3.726326    5.342882    3.821555
H      -3.181986    4.184074    3.125510
H      -2.984577    6.066660    1.748479
H      -2.250423    6.781889    2.909669
H      -1.054116    4.674447    1.537657
H      -0.576082    6.145292    1.467384
H       0.723372    4.965838    2.939787
H       0.125317    6.138803    3.761321
H       1.921040    4.375890    6.908583
H       1.676706    6.235725    8.636983
H      -0.145301    5.369077    7.433963
H       0.091950    6.422548    6.224250
H       0.676388    7.347621    8.295653
H       2.791319    8.070082    7.723424
H       1.896364    8.099751    6.522077
H       3.914233    7.048872    6.008409
H       3.698100    6.058287    7.055823
H       3.116098    5.021349    5.094850
H       2.158660    6.180195    4.786983
H      -1.006024    3.336226    6.935353
H      -1.559257    1.598311    5.456258
H      -0.325083    0.765200    6.003389
H      -2.608220    1.404611    7.524315
H      -2.214635    0.029761    7.044111
H      -0.418035   -0.117788    8.397718
H      -1.627822    0.182306    9.277813
H       0.176244    1.645756    9.829964
H      -1.053680    2.460437    9.333028
H       0.877440    3.225032    8.287287
H       1.235327    1.881817    7.763580
H       1.720632   -0.869158    1.853889
H       4.184273   -0.146643    0.965428
H       3.012517    0.820454    1.136093
H       1.847900   -0.200343   -0.409930
H       3.096540    0.151437   -1.100956
H       3.904789   -1.982340   -0.878422
H       2.584028   -2.105280   -1.617970
H       2.763571   -3.703670    0.105411
H       1.556168   -2.778368    0.317905
H       4.607903   -0.216537    5.166797
H       3.990062   -2.477871    1.686571
H       2.642313   -3.025155    2.339113
H       5.156005    1.289122    3.512016
H       5.590006    0.129866    2.598457
H       6.737696    0.838429    5.032942
H       7.431919    1.106030    3.739569
H       8.390108   -0.830548    4.679900
H       7.729962   -1.148361    3.309561
H       6.514020   -1.632968    5.797587
H       7.041226   -2.668142    4.887375
H       4.662063   -2.600521    4.676553
H       5.418404   -2.225128    3.326293
H       2.516573   -2.545817    4.572816
H       1.564734   -0.621934    6.319621
H       2.866067   -1.356878    6.528769
H       1.774146   -3.375710    6.968817
H       1.285356   -2.270001    7.937592
H      -0.502310   -3.449358    7.037418
H      -0.746615   -1.928105    6.670990
H       0.328488   -3.937948    4.917492
H      -1.013735   -3.235955    4.696631
H       0.558651   -2.294513    3.264385
H       0.034321   -1.214512    4.228140


How does partially thawing a frozen solution affect concentration? I have a mixture of antibiotics at a concentration of 100X and it is in the frozen state. Now some volume of the frozen mixture has been liquefied. Will the concentration of that liquefied part be 100X, or will it be less than that?

In general, no, a frozen solution that partially thaws is not the same concentration as the original solution.

The reason for this is the same reason chemists can purify a compound by crystallizing it, and one of the reasons why north polar ice cap is freshwater, despite floating in salt water.

When something (e.g. water) crystallizes (e.g. forms ice) the crystal lattice tends to reject impurities (e.g. things dissolved in the water). This means that as the solution crystallizes/freezes, the solid portion becomes depleted in the solute, and the remaining liquid portion becomes enriched in the solute. Eventually, though, if things are cold enough, even the concentrated residual liquid will freeze. But this mean that you now no longer have a uniform concentration of solvent in your frozen sample.

The reverse happens when you thaw it. Since you don't have a uniform concentration of solute, not all portions of your solid sample have the same freezing point depression. So what tends to happen is that portions of the solid which are more concentrated in solute will thaw first (as they have a lower melting point), whereas portions which are depleted in solute will thaw later. -- This is all generalization, though. What actually thaws first depends a lot on the details of where these regions are in relation to the heat source, and how thermal energy is flowing through the sample.

The up shot of this is, though, that if you freeze a stock solution, you can't partially thaw it and expect to get the appropriate concentration. You have to completely unthaw the stock solution and mix it thoroughly to make sure you have a homogenous mixture of the correct concentration.

This is why it is typically recommended to make single-use aliquots of stock solutions like antibiotics. Not only do the smaller aliquots thaw faster, but you don't put your entire stock solution into multiple freeze/thaw cycles, which can degrade some of the more delicate compounds.


How were the absolute masses of atoms calculated before the mass spectrometers? I understand that true (non-relative) masses of atoms are calculated with a mass spectrometer. using this relationship: Centripetal Force = Force due to magnetic field(B)

But I was wondering how masses of atoms were calculated before the mass spectrometers were invented.

(Please don't get this mixed up with relative atomic mass. That's not what I'm talking about.)

Absolute mass can be determined electrochemically, once the coulomb is defined in terms of numbers of electrons, as through the Millikan oil-drop experiment, to be  $6.24×10^{18}$ electrons per coulomb.

See Determination of Avogadro's Number and Absolute  Determination  of  the  Electrochemical  Equivalent and  the  Atomic  Weight  of Zinc.

One thing that caused confusion in the determination of atomic mass is natural isotopic separation, e.g. sulfur derived from condensation from vulcanism vs. from mineral compounds, so that there were slightly different experimental results depending on source.


How and why can more volatile acids be made with sulfuric acid? I have read that sulfuric acid has low volatility. Then it states that more volatile acids can be made using sufuric acid and their corresponding salts, because of the low volatility.

Can you help me understand the chemistry behind the second statement?

An acid according to the Brønsted-Lowry definition is any substance that can release $\ce{H+}$ ions. By having done so, it becomes a Brønsted-Lowry (conjugate) base which can accept $\ce{H+}$ ions. This is shown in equation $(1)$:

$$\begin{array}{ccccc}\ce{HA} & \ce{<=>} & \ce{H+} & + & \ce{A-}\\
\text{acid} & & \text{proton} & & \text{conjugate base}\end{array}\tag{1}$$

As already hinted, a Brønsted base is any substance that can accept such an incoming $\ce{H+}$ ion (reaction $(2)$)— as such, an acid will always react with a base and vice-versa; it is not possible for only an acidic reaction to happen.

$$\begin{array}{ccccc}\ce{B} & + & \ce{H+} & \ce{<=>} & \ce{HB+}\\
\text{base} & & \text{proton} & & \text{conjugate acid}\end{array}\tag{2}$$

Let me quickly quote part of the question:

  [M]ore volatile acids can be made using sulphuric acid and their corresponding salts

The corresponding salt of a more volatile acid could also be termed its conjugate base. So the reaction we are talking about includes the conjugate base of one acid and sulphuric acid as a second acid. To exemplify things, I will use hydrochloric acid as the first acid whose conjugate base is chloride. If we thus mix chloride ions (e.g. in sodium chloride) with sulphuric acid, the following equilibrium will result:

$$\ce{Cl- + H2SO4 <=> HCl + HSO4-}\tag{3}\\[1.2em]
K = \frac{[\ce{HCl}][\ce{HSO4-}]}{[\ce{Cl-}][\ce{H2SO4}]} = \frac{K_\mathrm{a} \left ( \ce{H2SO4} \right )}{K_\mathrm{a} \left ( \ce{HCl} \right )}$$

The equilibrium constant $K$ is the ratio of the acid constants of hydrochloric and sulphuric acid. Since hydrochloric acid ($\mathrm{p}K_\mathrm{a} \approx -8$) is more acidic than sulphuric acid ($\mathrm{p}K_\mathrm{a} \approx -3$), we expect the equilibrium to be shifted towards the reactant side — which would be true for a closed, pressurised system. Once we open the system, and allow for gases to be exchanged, the picture changes. Compare equation $(\text{3'})$, wherein the aggregate states of the reaction partners are mentioned.

$$\ce{Cl- (s) + H2SO4 (l) <=> HCl (g) ^ + HSO4- (s)}\tag{3'}$$

Hydrochloric acid is a gas and can diffuse away while sulphuric acid is a liquid and thus cannot diffuse away. (And cannot drip away since we are using a proper beaker of some kind.) If we constantly remove $\ce{HCl}$ from the reaction mixture by letting it diffuse away, we are exercising pressure onto the system, and per Le Châtelier’s principle the product side (one of whose species we are removing) will now be favoured.

This general principle applies to any acid, as long as it is more volatile than sulphuric acid, i.e. has a lower boiling point under the pressure used. If the second acid happens to be liquid rather than gaseous, a simple distillation will suffice to remove it from the reaction mixture.


What is the scaling for % completion time in a Michaelis–Menten scenario as a function of substrate concentration? Say I have a straightforward Michaelis-Menten reaction where an enzyme $E$ at some fixed concentration $[E]$ forms an interaction complex $ES$ with a substrate $S$ (with initial concentration $[S]_0$), and at a rate $k_{cat}$ can irreversibly convert the substrate $S$ to a product $P$.  The rate of interaction complex formation $ES$ is $k_f$ and its dissociation rate is $k_r$, such that overall, we have the reaction:

$E + S \rightleftharpoons^{k_f}_{k_r} ES \to_{k_{cat}} E + P$

Is there an explicit expression for the time $\tau_{v}$ to convert some percentage $v$ of the substrate $S$ to product $P$?  In lieu of an explicit expression for $\tau_{v}$, what scaling for this time should we expect for $\tau_{v}$ as a function $[S]$ (holding $[E]$ fixed)?

Ultimately, for $\tau_{v} \approx 1$ I suppose the steady-state approximation will begin to fail?

In short, yes.

To obtain the relation between τ and v, we need the actual time evolution of the concentration of the product by solving the differential equations (DEs) and then some way to get steady-state concentration. The DEs are non-linear. So we have to find approximate solutions. This paper develops a solution. It's not nice..

I will probably come back to add more. For the time being, I highly recommend those interested to browse through the paper.

And what do you mean by τ≈1 leading to failure of steady-state approximation please?

Edit: The tittle of the paper is "Analytical solution of coupled nonlinear rate equations : I. Michaelis-menten kinetics" by Phillipson


What will eventually happen to uncovered vegetable (seed) oil exposed to the air? If I leave some olive oil uncovered and fully exposed to indoors air for enough time (though away from UV exposure or any other environmental factors), what will happen to the oil?

I read here that some of its constituents ('lighter' molecules) will start to evaporate but that heavier ones will be left behind, which will 'not quite' evaporate.

If that is true (it may not be), would the non-evaporative part of the oil eventually harden and assume a cake-like consistency, almost like wax or hardened animal lard stored at refridgerator temperatures?

Some of the lighter constituents could evaporate. There are a lot of minor components in olive oil, and they are largely responsible for the fragrance and flavor. As these evaporate, the fragrance and flavor would change. It is unlikely that they would all evaporate though, as many of these components likely have high boiling points/low vapor pressures.

The oil itself would not evaporate, but it would not solidify either. At most you might notice a thickening of the oil.

Over long periods of time the oil may react with oxygen to form various oxidation products. This process is so slow, however, that it is definitely not a concern for olive oil you might leave out in your kitchen. The oil will start tasting and smelling stale long before significant oxidation occurs.