What is meant by electrons moving to lower potential energy? I am currently learning about chemical bonds in chemistry, and came across somebody who wrote this (source):

"Chemical bonds certainly "contain" potential energy, and the atoms want to move to a lower potential energy (become more stable).

When methane,
CH
4
, forms, the valence electrons end up in more stable (lower energy) C-H bonds.

These bonds are fairly strong, so methane is relatively inert.

However, if you add energy to the methane in the form of a flame or a spark in the presence of oxygen, some of the molecules will have enough energy to overcome an activation energy barrier.

Some of the
C-H
 bonds will break.

The electrons can then enter an even lower energy state by forming
C=O
 and
O-H
bonds rather than staying as
C-H
 and
O=O
 bonds."

What is meant by this? What does it mean when electrons enter a lower energy state, do they move to an even lower ground state?

What is meant by electrons moving to lower potential energy?

In a chemical reaction, the electronic states in the reactants are different from the electronic states of the products because the atoms are arranged differently into molecules. Lower potential energy of the electrons is associated with stronger bonds. When the products have stronger bonds than the reactants, energy is released in the reaction (exothermic). Conversely, when the bonds of the reactants are stronger than the bonds of the products, energy is absorbed in the reaction (endothermic).

  do they move to an even lower ground state?

Yes. For example, the ground state in the hydrogen molecule $\ce{H2}$ is lower than in the hydrogen atom $\ce{H.}$ (because there is more room for the electrons as they are interacting with two nuclei instead of one). If you want to break the molecule into atoms, you have to provide energy to move the atoms apart (i.e. break the bond) because separating them will result in a higher ground state. For molecules reacting to other molecules, the energy balance depends on the relative strength of the bonds (and the number of bonds).

Follow-up questions

  [OP's comment]: Could you just elaborate on the part where you say "because there is more room for the electrons as they are interacting with two nuclei instead of one"? Why does the ground state get lower because of that?

This might be jumping ahead a bit too much, but electrons don't get closer and closer to the nuclei (even when that would increase interactions) because "they don't like to be pinned down" (Heisenberg's uncertainty principle). If an electron can keep the same average distance to the nuclei but occupy a large space, it will lower its energy. This is reflected in the color of molecules with conjugated double bonds (carotene in carrots, for example), and also in a model called particle in a box. One analogy is strings of different length in music, where longer strings produce a lower pitch (lower energy) and short strings produce a higher pitch (higher energy).

  [OP's comment]: I read this online. It talks about how the bonds formed in the products after exothermic reactions have lower potential energy. Is it the electrons which have lower potential energy or the new molecule? This answer states that it's the electrons which have lower potential energy, which one is more correct?

Typically, adjectives for bonds are strong or weak. If you are actually measuring an energy, it is the energy of the entire molecule. The reason energies are different is because the electrons have different energies (nothing happens in the nuclei). So saying electrons have lower potential energy makes the most sense to me. Distinct bonds don't really exist anyway, we just use them for making sense of the molecular properties.

  [OP's comment]: So are bond strength and energy of the entire molecule both dependent on the electron energy?

Yes, much of chemistry is about the electrons. Chemical bonds and intermolecular interactions both are a reflection of "what the electrons are doing", and chemical reactions go along with "electrons doing something new", so knowing the electronic states of the molecules carries enormous predictive power.


The naming of benzene Why is benzene not called as a cyclo triene? Wikipedia says benzene is a IUPAC name, but isn't it a common name?

It would be wrong to say cyclotriene, rather cyclohexatriene would be an appropriate name. And it would also be a highly systematic IUPAC name for said compound, following the rules of e.g. the cyclopentadienyl anion, the cycloheptatrienyl cation, cyclooctatetrene and more.‡

Let’s look at benzene’s history. It was discovered as early as 1825 by Michael Faraday who called it bicarburet of hydrogen. In 1833, Eilhard Mitscherlich isolated it and named it benzin. In 1836 it was isolated by Auguste Laurent who called it phène. The name benzene also dates back to the first half of the 19th century and, like benzin, derives from the gum benzoin. Note that this already predates the founding of IUPAC by about 80 years. The first IUPAC nomenclature was not established until 1921.

At the time when it was discovered, chemists had no clue as to what structure it had. It wasn’t until 1865 when Friedrich Kekulé published his ground-breaking paper[1] on the structure of benzene which in the end turned out to be correct (-ish). At that time, nomenclature of chemicals was still the Wild West, since there was hardly any way for structure elucidation and thus most chemicals acquired trivial names — some of which survive up to date (acetic acid being a prime example).

When IUPAC finally derived comprehensive nomenclatures for organic (and inorganic) compounds, and when structure elucidation was finally good enough to prove structures, the names benzene, benzin and possibly also phène had been in use for many decades, much like many other trivial names. Had IUPAC decided to obsolete them all altogether, they could have just as well trashed their own recommendations, because nobody would have followed them — even today, many chemists are too traditional to take up newer nomenclature schemes even if they have been around for two decades.

Probably to help gain acceptance, IUPAC has always allowed a certain set of retained names to persist under official nomenclature. While the list has gotten smaller over the years, it is extremely unlikely that benzene, phenol, phenyl etc. will ever get removed from it, since calling those compounds that way is so prevalent — and short.

Similar things happened to carboxylic acid. Hardly anybody says methanoic acid or butanoic diacid in place of formic acid and succinic acid. And sometimes, trivial names even become part of the official nomenclature, as was the case with fullerene because it provided a much shorter way to label a certain class of compounds.

So the short answer to your question: IUPAC has retained a common name. The reasons are tradition, prevalence and simplicity.

Note:

‡: Yes, as a chemist I know that the structure of benzene is incorrectly represented by drawing three pure double bonds and three pure single bonds, that in reality the molecule has a very high $D_\mathrm{6h}$ symmetry rendering all bonds equal and the electrons in the π system are delocalised all across the molecule. However, that is not what the question is asking about, and not something IUPAC nomenclature cares about.

Reference:

[1]: F. A. Kekulé Bull. Soc. Chim. Paris 1865, 3, 98.


Does crystalline glass exist? As far as I know, glass is a non-crystalline amorphous solid. Does glass exist in crystalline forms? Or is it the definition of glass that it's non-crystalline, in which case my question is based off of a false assumption?

As Ivan explained, a glass is by definition amorphous. However, you can take a quartz crystal (silicon dioxide, which is the main ingredient of glass) if it is of high purity and optical quality, and machine it into a transparent object such as a window pane.


Misconception? "That gas is heavier so it sinks." I've heard people say many times that one gas is heavier than the other so the heavier gas sinks. Now this must certainly be true to a certain extent (I think of gas used in WWI that sunk into trenches). However, this idea of the weight of a gas being a driving force seems unusual because, in my experience, entropy is the dominating motive for particles in the gas phase. A good example of this is that gas phase clusters are studied extensively in the literature due to how unusual they are.

So, why would some gases "sink" and give up their entropic gain for what doesn't seem to be much of an energy minimization? Could the answer be simply that this concentration of heavier gases is quite short lived and they soon float away to their entropy filled wonderland?

I would also give as an example the fact that CO2 certainly gets up high in the atmosphere quite well and that is more than twice the "weight" of other particles in the air. So what's different about that than about the examples given previously?

Without significant mixing, diffusion takes a long time to mix gases.  Our understanding of entropy tells us that we will indeed finish with mixed layers, but that doesn't give us a time frame for that mixing, only an outcome.

Given a slow but steady production of a dense gas, layers absolutely will form due to density differences.  There are plenty of places where this can easily be seen, from your WWI example to the ability to pour carbon dioxide or hexane vapor into a container.  They're not stable over the long term, but they do exist for substantial periods of time, days in some cases.

If confined by an open container, huge areas can experience this effect, as at Lake Nyos.  Heck, even small temperature based density differences drive atmospheric layering that has major impacts on weather and pollution.

It's not an enormous effect, but you may be underestimating the strength of the forces involved here.  The same pressure differential that forces a helium balloon upward acts on an open volume of helium as well, and more strongly due to the missing mass of the balloon.  A body of a gas like $\ce{NO2}$ has about as much force pushing it downward (in air) than an equal volume of helium has pushing it upward.  Without outside mixing to cause it to disperse, it will happily form a short term layer until diffusion carries it away.


Endothermicity of Reaction Regarding this question, how can one solve this without access to thermodynamic data?

I suppose we should keep the following information in mind:

1) Double bonds are not double the strength of single bonds. Same goes for triple bonds. Reason has to do with orbital overlap with regard to sigma and pi bonds.

2) In order of increasing stability: $\ce{O_2, O_3, O}$. We know this because diatomic oxygen is what usually exists in large quantities at ground level; ozone is a pollutant because it is quite reactive and can thus harm tissues; we don't usually hear of monoatomic oxygen.

Therefore:

a) Cannot be the answer. Going from ozone to diatomic oxygen is exothermic. At least part of this reaction is exothermic.

b) Definitely not the answer. Going from ozone to diatomic oxygen exclusively.

c) This is endothermic, but perhaps not as endothermic as creating monoatomic oxygen.

d) Looks pretty endothermic; we are splitting oxygen.

e) Definitely not the answer, this is fundamentally exothermic.

How can we definitely pick out the right answer? What reasoning would you employ?

I believe that you would have to know at least these things to start with:

All enthalpies of bond dissociation are endothermic
Complete dissociation of triple bonds is more endothermic than double bonds, which is more endothermic than single bonds (when compared for the same elements).
The difference between going from a triple bond to a double bond is smaller than double to single, and that is smaller than going from single to complete dissociation (you referenced this in your question)

We can summmarize these facts with the following figure:

Here D(O-O) represents the bond dissociation enthalpy of an oxygen-oxygen single bond. The numbers inside of circles represent "steps" between each successively stronger bonds, and are there just for easier reference.

With that in mind, let's look at the options again, and count how many of each type of bond are broken and formed:

$ a) \space O_3(g) \rightarrow O_2(g) + O(g) $

Ozone has a resonance structure, which means that we need to consider the bonds as being somewhere between a single and double bond in strength. We could imagine them being about halfway up step 2.

Broken: 2 x single/double resonance bonds

Formed: 1 x double bond

So we are looking at an overall reaction that would be slightly more endothermic than the breaking of a single bond. To make that easier to see, think about taking these energetic "steps" using the figure above:

(+1.5, +1.5) - breaking the two resonance bonds

( -1, -2) - forming the double bond

$ b) \space  2O_3(g) \rightarrow 3O_2(g) $

Here we are breaking two sets of resonance bonds, but forming 3 double bonds. So our steps look like:

(+1.5, +1.5, +1.5, +1.5) - breaking six resonance bonds

(-1, -2, -1, -2, -1, -2) - forming three double bonds

There are three -1, -2 steps, each of which is larger than a +1.5 step. However, they are not much larger, and the total of the difference may or may not be greater than the fourth 1.5 step. This means we can't say for sure whether it is exo- or endo-thermic, but we can at least say it would be less endothermic than breaking a single bond.

$ c) \space  3O_2(g) \rightarrow 2O_3(g) $

Now we are doing the opposite of (b). We still don't know the sign, but we know the magnitude is the same as (b), which was less than (a), so either way it can't be (b) or (c).

$ d) \space  O_2(g) \rightarrow 2O(g) $

Here we are breaking a double bond, and forming nothing. This is the clear winner so far, with steps +1, +2.

$ e) \space  O_3(g) + O(g) \rightarrow 2O_2(g) $

In the final case, we break two resonance bonds, and form two double bonds. Our steps are:

+1.5, +1.5, -1, -2, -1, -2

Since the formation of each double bond releases more energy than the dissociation of each resonance bond, this is clearly exothermic, as you stated in your question.

Taking all of this together, we have to choose (d) as the correct answer.


Why CCl4 does not dissolve in water while SiCl4 does? Why $\ce{CCl4}$ does not dissolve in water while $\ce{SiCl4}$ does?
Does it have to do with d-orbitals? If yes, then how are they related?

$\ce{SiCl_4}$ does not quite dissolve in water; rather, it reacts with water. So does $\ce{CCl_4}$, albeit extremely slowly, so for most purposes one may safely assume it doesn't. The difference is mostly due to the atomic size of C and Si. Smaller С atom is completely blocked off by four bulky chlorine atoms, so the water molecule can't reach it. Larger Si is easier to attack.

Another factor is that silicon is a less electronegative element than carbon, so that the Si-Cl bonds will have more polar character.

Whether or not the d-orbitals of silicon contribute significantly to the transition state, is (IMHO) a question of secondary importance.


What did I do wrong in my temperature change calculations? I am doing an online assignment where I am being asked to find temperature change but there is something wrong in my calculations as I am getting the answer wrong. Can someone help me figure out where I went wrong in my calculations?
Assignment Instructions:
Solve for the unknown variable using the specific heat formula. Input your answer (without units) and using the correct number of sig figs.
Question:
If 31,500J of heat is used to warm 750g of water, what is the temperature change?  The specific heat of water is 4.184J/goC.
My Answer: 10
Since I am being asked to find the change in temperature in this question, I split the original equation up. First, I did (750) * (4.184) and got 3138. Then, I did (31,500) / (3138) and got 10.0382409178. The lowest number of significant figures is two, so I rounded 10.0382409178 to 10. Why is the computer counting my question as wrong?
I used the equation temperature change = (heat) / (mass) * (specific heat). The variables I used are q (heat), m (mass), and c (specific heat).

You're not really wrong: the computer program is apparently expecting a float and not an integer, which is really stupid because the question is about temperature change and not the vagaries of significant figures.
Instead of you learning about heat capacity and the relation to temperature change, you're instead learning about significant figures first, and the chemistry/physics part second.
I find this sort of thing patently awful and a frustrating way for you or anyone else to "learn" anything about the principles.
You have to report your answer to 2 significant figures:
750 has 2 (the limiting number); 31500 has 3; and 4.184 has 4.
$$Q = m\times c\times \Delta T$$
or
$$\Delta T = {Q\over m\times c}$$
so
$$\Delta T = {31,500\ {\rm J}\over 750\ {\rm g} \times 4.184\ {\rm J}\cdot {\rm g}^{-1}\cdot {\rm ^{\circ}C}^{-1}}$$
which, to 2 significant figures, is:
$$\Delta T = 10.\ ^{\circ}{\rm C}$$
Notice I have included a decimal point, so the now ridiculous answer that should be accepted is 10. and not 10
Meanwhile, you've lost a lot of time wondering where your calculation went wrong because some programmer doesn't know how to cast. Or maybe you're really supposed to report numbers with a terminal decimal point, and I'll challenge anyone to find any respected publication where that happens:

"The temperature change, in degrees Celsius, is 10.."

Note that 10.0 has three significant figures, so that's also "wrong."
And that, my friend, is why "learning" from a computer is a lazy way out for the "teacher."


Why is pyridinium chlorochromate selective compared to chromate, dichromate etc.? One can use PCC (pyridinium chlorochromate) to oxidise an alcohol selectively up one level - to an aldehyde/ketone, without further oxidation to a carboxylic acid.
Why is pyridinium important to this selectivity?

I understand that usually PCC is dissolved in dichloromethane and not in anything like THF, and this prevents the formation of a hydrate, which can then act "like" an alcohol in it's oxidation to a carboxylic acid, but I can't figure out why it's important that pyridinium is used.

Though pyridinium chlrochromate (PCC) is a salt, the pyridnium cation is easily dissolved by a wide variety of organic solvents. As such, the chlorochromate anion then becomes dissolved in the organic solvent, and can oxidize $1º$ and $2º$ alcohols present in solution to carbonyls. Oxidation of carbonyls to carboxylic does not occur when using PCC for the very reason you said yourself—no water is present to hydrate the carbonyl species to its geminal diol to allow for further reaction.

Other benefits of PCC include that it is not particularly hygroscopic, is commercially available, and can be stored for some time, though other methods (Swern oxidaiton, Moffatt oxidation, Dess-Martin Periodinane) are generally preferred as the reaction workup for PCC can be tedious and the chromium(IV) species produced is toxic.


Do we use activities or concentrations in the rate expression? The equilibrium constant of a reaction is a dimensionless quantity due to the fact that we use activities instead of concentrations(in moles/liter).
But, in the rate expression for a reaction, do we still use activities?

If this is so, then isn't the unit of the rate constant always (moles/litre-sec), since these are the dimensions of the rate of a reaction ?

The rate expression is always described in terms of molar concentrations. Units of the rate constant will change based upon the overall order of the reaction. This happens such that the rate is always given in M/s.


How to calculate the amount of substance of sodium that are in 10 mL of 30% sodium chloride? How much of sodium is there in 10 mL of 30% sodium chloride?

$30\%$ of $\ce{10 ml}$ is $\pu{3 g}$ $\ce{NaCl}$. There is $\pu{3000 mg}$ of $\ce{NaCl}$ and would that be divided by the molecular weight of $\ce{NaCl}$ or of just $\ce{Na}$?
I divided $\pu{3000 mg}$ of $\ce{NaCl}$ by $\pu{58.44 mg/mol}$ and got $\pu{51.44 mol}$ $\ce{NaCl}$.
Since there is one mole of $\ce{Na}$ for every mole of $\ce{NaCl}$ there would be $\pu{51.44 mol}$ of \ce{Na}.
My teacher divided $\pu{3000 mg}$ (he didn't express what he had $\pu{3000 mg}$ of) by $\pu{22.99 mg/mol}$ of $\ce{Na}$ to get $\pu{130.5 mol}$ of $\ce{Na}$.  I have found other similar problems on the internet and feel I am correct.  My teachers only response was to read the question.  If I am wrong please help me understand why.

You are on the right track and with a few assumptions I'd say your calculation is correct. The one of your teacher is not.
Let's back up a moment and understand what is going on.

First of all we should make sure, that we are asking the right question. As you have given it, it seems incomplete. It should probably rather be:

  What amount of substance of sodium (ions) can be found in 10 mL of a 30% aqueous solution of sodium chloride?

We have been given a mass fraction, which we need to convert to the amount of substance concentration first. The general formula for this is
$$w_i = \frac{\rho_i}{\rho} = \frac{c_i M_i}{\rho},$$
where $\rho$ is the density of the solution, $\rho_i$ is the mass concentration of component $i$, $M_i$ is the molar mass of component $i$, and $c_i$ is the amount of substance concentration of $i$.

In your calculation you made the assumption, that the density of a 30% sodium chloride solution is $\rho = 1~\mathrm{g/mL}$, which is not correct. You can look up a couple of values for brine solutions on wikipedia.
For the purpose of this example, let's assume $\rho(\ce{NaCl~30\%}) \approx 1.2~\mathrm{g/mL}$.

The molar mass of sodium chloride is $M(\ce{NaCl}) = 58.4~\mathrm{g/mol}$. Be careful with the units, as you used the incorrect ones.

The amount concentration is given through the formula
$$c_i = \frac{n_i}{V},$$
where $n_i$ is the amount of substance (not number of moles!) and $V$ is the volume of the solution.

You correctly identified that there is a one-to-one correlation between sodium ions and sodium chloride as expressed by
$$\ce{NaCl (s) ->C[H2O] Na+ (aq) + Cl- (aq)}.$$
Therefore we can write
$$n(\ce{Na+}) = n(\ce{NaCl}).$$

Putting everything together we arrive at the following formula:
\begin{align}
n(\ce{Na+}) &= \frac{w_{\ce{NaCl}}\cdot \rho(\ce{NaCl~30\%})}{M(\ce{NaCl})} \cdot V\\
            &= \frac{0.3\cdot 1.2~\mathrm{g/mL}}{58.4~\mathrm{g/mol}}\cdot 10~\mathrm{mL}\\
            &= 0.061~\mathrm{mol}
\end{align}

Basically you thought process was correct under the assumption of the inaccurate density. And it was off by $10^{-3}$ because you (and your teacher) used the wrong units.


Does "covalent" AgCl (aq) exist and if so, how to quantify its concentration? I am reading Ionic Equilibrium by James N. Butler.  Using silver(I) chloride in excess aqueous chloride as an example it speaks of several forms:  $\ce{AgCl (s)}$ i.e. the ionic crystalline salt, $\ce{Ag+(aq)}$ and $\ce{Cl^{-}(aq)}$ the aqueous ions from dissociation, and finally $\ce{AgCl(aq)}$, which Butler purports to be "dissolved covalent silver chloride molecules as distinct from solid salt or dissolved ions."

Does this "molecule" of $\ce{AgCl}$ really exist in solution?  If so, is there a way to quantify the concentrations of this covalent species?  Equilibrium solubility product constants seem to give the concentration of the aqueous ions only.

Some amount of $[\ce{AgCl(H2O)}]$ particles should exist in solution from pure thermodynamic point of view. See, $\ce{Ag+}$ ion can coordinate two ligands, and in solution both $\ce{Cl}$ and $\ce{H2O}$ are present. However, it should be really hard to detect trace amounts of this particle, as $\ce{AgCl}$ is almost insoluble. In case one can get concentrated salt solutions, similar particles may exist and hints for their existence may be got from conductivity measurements and osmotic pressure. Sometimes a direct study using NMR is possible, which can give some structural information if used properly (see 2D-NMR methods). As I remember, $\ce{Cr(CH_3COO)2}$ hydrate is slightly soluble in water but undergoes little to no dissociation.


Measure of partial pressure on blood Here is the issue of blood gases partial pressure.
On medical reports is common to see partial pressure of gas on blood, usually oxygen, but there is no gas in equilibrium with blood, so it is confusing. What they mean with partial pressure of a gas on blood is, according to this answer:

the partial pressure which would prevail in a gas phase in equilibrium
with the liquid [blood] at the same temperature.

I suppose they use a partially filled vessel with blood and a manometer or something like that, but I would like to know the way they measure it.
Question
How do they measure $p(\mathrm{O}_2)$?

In principle you could measure it by getting a sample of blood and putting it in contact with gaseous O2 at various pressures.   Whatever external pressure of O2 results in the amount of O2 dissolved in the blood neither increasing nor decreasing would equal p(O2) in the blood.

As for how it's actually measured, I believe the usual answer these days is spectroscopically.  That is, you shine light through a sample of blood and look for the absorption at certain frequencies which are typical of hemoglobin with O2 bound to it.  The amount of absorption is proportional to the concentration of oxygenated hemoglobin, and your software has some internal chart that converts that to partial pressure.

As for why it's measured that way, in part it's probably just tradition, and medicine changes very slowly and conservatively, for obvious reasons (you're experimenting with health and life).  In part, though, it may be that it directly relates to something important: the partial pressure of O2 in the atmosphere, which is about 160 mmHg.  That's the concentration of O2 in the lungs.  What fraction of that reaches the blood at various points in the circulation is a natural indicator of how well oxygen is being taken up, transported, and delivered.  That is, you can directly relate this particular measure of O2 concentration to the form of concentration usually used to describe a mixture of gases, in this case the air you breathe, which is the origin and destination of the blood gases.


In a battery, what happens to LiCoO2 when the lithium leaves? It becomes CoO2? What chemical compound exists at the cathode of a lithium-ion battery when the lithium is not there?

The cathode is usually described as $\ce{LiCoO2}$, so does it become $\ce{CoO2}$?

$\ce{CoO2}$ puts cobalt in an oxidation state of +4, which is possible but rare and unstable. I think $\ce{CoO2}$ doesn't even have a Wikipedia page.

The general textbook formula for the Li ion battery is
$\ce{Li_{1-a}Ni_{1-x-y}Mn_{x}Co_{y}O2}$
, and the simplest/earliest type has $y=1$, i.e. cobalt-only. These "mixed (lithium transition metal) oxides" have a layered structure (see e.g. wikipedia), where you can relatively easily electrochemically remove (and later reinsert) a part $a$ of the Li atoms, without destroying the whole structure. You cannot increase $a$ to much more than 0.1 - 0.3 or so, depending on the exact formulation/grain size/quality/temperature/etc., before the structure does break down.
Even the "uncharged" $a=0$ mixed Li-cobalt(III) oxide gives off oxygen (!) if heated above 180°C. (It is produced by tempering at ~800°C in a pure oxygen atmosphere.) With $a>0$, this happens earlier already, and basically always leads to a runaway destruction of your whole battery setup. With higher contents of Ni and Mn, this danger becomes less pronounced. I´m not sure the Co-only variant was ever commercialised. (?)
Simply speaking, a cobalt-rich battery blows up when overcharged, and in ones with more nickel and manganese the layered structure still breaks down, and you get somewhat stable Ni/Mn(III/IV) species (and still quite a lot of overheating/general destruction). In any case this happens long, long before you reach the ideal $\ce{CoO2}$ (or generally $\ce{MO2}$) stochiometry.
If you look closely (e.g. via XRD) at the crystal structure during charging, you seemingly find that there are a number of intermediate deformed structures, and even more mixed/disordered states in between. And that of course becomes even more complicated (or washed out) with a (more or less) random arrangement of different transition metals in the structure, which will depend also on the specific synthesis route.


HPLC Columns For Compounds Not Soluble In Non- Or Low Polar Solvents Can someone recommend HPLC columns for compounds which are not soluble in no- or low polar solvents (hexane, isopropanol, toluene, and dichloromethane)?

Try searching Google Scholar and write HPLC separation and the names of your compounds. You will find plenty of publications and of course column names. If your compounds are new, you should test the solubility in MeOH, ACN, and 50:50 MeOH/H2O and 50:50 ACN/H2O. Do they dissolve in such solvents systems? If they do, you would use C-18 RPLC columns. There are literally hundreds of columns with similar C-18 chemistry.

If your compound is highly polar (carboxylates, amines, phosphates etc.), even bare silica column can do the separation. This technique is called HILIC. I have written several articles on HILIC. Try Google Scholar for reviews on HILIC.


How long does it take to run an MS/MS experiment I am not planning to do these types of experiments but I would like to know the order of magnitude of the time it takes to run an LC-MS/MS experiment. Let's assume that I have the sample prep (protein) finished and I just need to load the sample into the LC. How long would I need to wait before I would get MS/MS data?
I know that for a typical HPLC experiment, the sample takes around 20 minutes to go through the column.

The MSMS part is the detection of the LC in this case (nb. you don't need to do LC-MSMS, you can do direct infusion quite happily if you don't need to separate anything), so the actual experiment wouldn't need to take much longer than any other LC (as CHM says).

However, calibration + set-up of these things, and extensive cleaning afterwards, can make it a day's job to run just a few samples (even direct infusion). If you don't have much information on the system already, you may also find that you need to do a 'sighting' run, to work out which ions you need to examine more closely (MSMS). And then there is data processing.....

All in all: quick, if you know exactly what you want, and someone is running everything for you. If you want to argue that it is an easy job, you can, but be aware that this is not always gong to be the case.


Does the position of a functional group in an organic compound have any effect on the compound's heat of combustion? Is there a link between where the functional group is located in an organic compound and the compound's heat of combustion? It has been mentioned before that it is difficult to accurately get the enthalpy of a reaction experimentally; thus, enthalpies of formation are used instead when calculating it.This is why I looked up the enthalpy of formation of liquid at STP data for two compounds, 1-Heptanol and 2-Heptanol, which are $\pu{403.4 kJ mol^{-1}}$ and $\pu{416.9 kJ mol^{-1}}$ respectively.

The exhaustive combustion of an organic compound consisting of carbon and hydrogen only will yield carbon dioxide and water.  Thus, if the heat of formation for isomers differ from each other, then the heat of combustion will differ, too.
This entry in the English Wikipedia states, for example
$\pu{−146.9 kJ mol^{-1}}$ for n-pentane, $\pu{−154.4 kJ mol^{-1}}$ about isopentane (methylbutane), and $\pu{−167.8 kJ mol^{-1}}$ for neopentane / dimethylpropane; all of same number of atoms $\ce{C5H12}$ (pentane), but with different connection of methyl groups to a backbone in common.  The very table continues for other branched isomers of alkanes, too.

Possibly there is a misunderstanding between the computation of heat of formations from paper (or, in a computer program), on one hand, and the experimental determination of these properties, too.
For the former, the approach typically taught in schools to determine the enthalpy of formation is

to imagine the reaction of $\ce{C_xH_y}$ with $\ce{O2}$ to yield $\ce{CO2}$ and $\ce{H2O}$
to look up the standard heat of formation of $\ce{CO2}$ and $\ce{H2O}$, hopefully with a discern between water as gas and water as liquid
to apply Hess's law to compute $\Delta_fH^\circ$ for $\ce{C_xH_y}$ with the convention that $\Delta_fH^\circ$ for elements in their most stable form (here, oxygen) is set zero

This approach however is an approximation which does not account how the atoms are connected with each other and their neighbours, or bond order; as if e.g., a carbon in acetylene does not differ from the carbon in methane.
If one leaves the imaginary reaction to $\ce{CO2}$ and $\ce{H2O}$, there are empirical increment systems based on experimentation, set to improve this approximation when computing this on paper.$^*$  With the advent of quantum chemistry, it is up to you to select a theory suitable enough to compute this in silico, too (example by MOPAC).
Among the classic approaches to determine experimentally the heat of formation via the heat of reaction (again, Hess' law is applied) is bomb calorimetry.  Here, exhaustive combustion a known quantity of your compound takes place in an atmosphere of pure high-pressure oxygen.  It is one of the experiments students may encounter early when studying physical chemistry.
$^*$  Publications like 1993ChemRev2419, 2011JPhysChemA10576, or 2012JPhysChemA7196 review / present extensions, or are basis for more recent work, too (cited by below the landing page of the publication) in addition to the increment system above mentioned Wikipedia entry currently presents.  Of course, one equally may  adjust existing / establish a new increment systems, e.g., as combination of own work (bomb calorimetry) and using data already recorded by others and presented in primary references (articles) and tables (e.g., CRC Handbook of Chemistry and Physics, NIST Webbook Chemistry)


HPLC Lab Report: Why is weight/purity % sometimes over 100%? I help with the supply chain of a small supplement company, sometimes the HPLC lab reports we get back have a weight % over 100%. What does this mean, how does this kind of thing happen? Obviously the sample can't be over 100% pure.

By far the most likely reason for reported purities of over 100% when using HPLC is the "analytical uncertainty of the method", but let me illustrate exactly how this is likely to arrive.

A standard procedure for estimating purity is:

Run an exhaustive series of known standards with (assumed) exactly known purities, or weight percents, on the HPLC.
Plot this calibration data with peak area on one axis, and the known weight % on the other axis.
Fit a line to this data.  The slope of this line is called the "response factor" or sometimes just the "response".
Run some unknown samples and measure the peak area for the unknowns.
Using the response factor, calculate a weight percent for the unknown sample.

If there are analytical uncertainties in the calibration standards, then the response factor may be randomly off by small amounts from the "true" value of the response factor.  Similarly, if there are uncertainties in the peak area of the unknown sample, even using the "true" response factor may give a value that is slightly off.

Statistically, these errors should be random, so that sometimes when the "true" purity is 100.000 % the measured value will be 100.6%, and sometimes it will be 99.4%.

Chemically, what is the source of the errors?  Well, maybe the autosampler that injected the unknown sample was supposed to inject 2.0000 μL of sample for analysis, but due to instrument imperfections actually injected 2.012 μL instead.  Maybe there was a tiny power surge in the AC lines going to the UV detector that was measuring the peak right when the compound of interest eluted, leading to an ever-so-slightly greater signal at that moment...

The list of possible causes is nearly endless, but none of these things are really cause for concern, unless you really need to know the purity to three or more decimal places.  From the report you provided, you can safely assume that the uncertainty in their analytical method is at least 0.6%, but not much else.


Why is negative charge more stable on sulphur than on oxygen? Consider this problem:

Which is more acidic, $\ce{MeOH}$ or $\ce{MeSH}$?

According to my teacher:

To determine which is more acidic, figure out which forms the more stable conjugate base. In this case, $\ce{MeS-}$ is more stable than $\ce{MeO-}$. In general stability reduces in the order: $$\ce{RS- > RP- > RO- > RN- > RC-}$$

Why is the compound with $\ce{S-}$ or $\ce{P-}$ more stable than that with $\ce{O-}$? I know that oxygen is more electronegative than either sulphur or phosphorus. Shouldn't the descending order start with $\ce{RO-}$?

There is more to it than meets the electronegative eye.
Get into the habit of transferring attributes into energy.  In this case the relevant energies are the electron affinity of the base money ($\ce{RO}$ vs $\ce{RS}$) versus the bond energy of the moiety to hydrogen.  A strong anionic Bronsted-Lowry base will have a low value for the first versus a high value for the latter.
Surprise!  Sulfur has a higher electron affinity than oxygen!  Add the fact that the oxygen-hydrogen bond is stronger than the hydrogen-sulfur bond and by both measures, $\ce{RO^-}$ appears as the stronger base.
The higher electron affinity of sulfur and even selenium, compared with oxygen, has another curious effect.  We know about the aromaticity of certain ring ketones in which the electronegative oxygen atom imparts a positive charge to a three- or seven-membered carbon ring.  It turns out that despite what an elctronegativity table might suggest, the higher electron affinity of sulfur and selenium versus oxygen causes this characteristic to carry over to the corresponding $\ce{C=S}$ and $\ce{C=Se}$ compounds.


What is quasi-aromaticity? During my study of aromaticity, I came through this topic of quasi-aromaticity. Can you please elaborate what Quasi aromatic compounds are and how are they designated?

In general, we can describe a quasi aromatic compound as a compound, which is ionic in nature with a counter ion, and the $\pi$ electrons in such compounds follow Huckel's rule ($4n+2$).

In other words, quasi aromatic compounds are those in which the charges present on the molecule are a part of aromaticity of the compound. A few examples of such compounds are depicted in the diagram:

However, it has a deeper and broader meaning. For a more in-depth explanation, please read the given references.

References:

T. M. Krygowski, B. Bankiewicz, Z. Czarnocki, M. Palusiak, “Quasi-aromaticity—what does it mean?” Tetrahedron 2015, 71(30), 4895–4908 (https://doi.org/10.1016/j.tet.2015.05.074).
E. Kleinpeter, A. Koch, “Characterization and quantification of quasi-aromaticity by spatial magnetic properties (TSNMRS),” Tetrahedron 2015, 71(33), 5275–5284 (https://doi.org/10.1016/j.tet.2015.06.019).


Equivant mass of double salt What is the equivalent mass of double salt like dolomite? And what if the double salt is hydrated?

Is a double salt taken as a single unit while equating the gram equivalent in any reaction?
For example, in reaction

$$\ce{CaCO3.MgCO3 + 4 HCl -> CaCl2 + MgCl2 + 2 CO2}$$

is

$$\frac{\text{wt of dolomite}}{\text{eq wt of dolomite}} = \frac{\text{wt of}~\ce{CO2}}{\text{eq wt of}~\ce{CO2}}?$$

$\ce{CaCO3.MgCO3}$ reacts with $\ce{4 HCl}$, therefore $m_\mathrm{eq}=m/4$.
$\ce{CaCO3}$ reacts with $\ce{2 HCl}$, therefore $m_\mathrm{eq}=m/2$.
$\ce{MgCO3}$ reacts with $\ce{2 HCl}$, therefore $m_\mathrm{eq}=m/2$.
$\ce{Na2CO3.10 H2O}$ reacts with $\ce{2 HCl}$, therefore $m_\mathrm{eq}=m/4$.
$\ce{1 CO2}$ of carbonate origin is released by $\ce{2 HCl}$, therefore $m_\mathrm{eq}=m/2$.


Most activated position for Electrophilic addition in substituted biphenyls I know the first structure will go fastest electrophilic addition due to +mesomeric effect but I was wondering what would be the product(major) if a electrophile attacks

Will the electrophile attack both ring or the ring near to $\ce{OH}$ group? Also the negative charge due to resonance will resonate on both the rings then how will I decide which ring to attack to give me major product?

According to me, the ortho positions of the ring near the $\ce{OH}$ group will have a negative charge - as well as the para position - but the para position is hindered. So, the electrophile will attack the ortho positions.

If this the correct explanation of the question, then I have one more question as well, that: when negative charge delocalises to the para position, it can delocalise to the other ring as well. Then, the electrophile could attack in the second ring also, where the negative charge would be generated by resonance. So, will the electrophile attack-the first ring or the second ring?

Dipole moment is what matters in substituted biphenyls.

The dipole moment in the first case is from left ring to right ring due to -OH groups predominant +M effect to the ring.(the alkene being an EWG(electron withdrawing group)), further enhances this effect. So effectively, electron density is more in the right hand side (I know it is counter-intuitive but it is how it is).
So EAS takes place preferably/majorly in the right side ring.

In the second case, NO2 shows -M effect and this dominates that of the alkenyl group. So dipole moment is from right to left ring and EAS takes place preferably/majorly in the left side ring.

Note: If there is any group(which is symmetric, or else we need to consider electronic effect due to assymetric grp also) between the C-C bond linking the two rings together, then the answer is as it "intuitively seems/as you expect". Try to rationalize this yourself.


Electroplating diamond grit I want to do electroplating with diamond grit and make profiling, grinding and cutting disks. (see the attached photos)

I want to know the process to make the machine or the company whom is selling machinery to do it.

UPDATE :
My question is not about how to shape metal like this, I know how to do it.
My question is that, how can I electroplate diamond grit on the surface of these profiling wheels?

Maybe these links would help:
US3046204A - Method for making diamond tools - Google Patents
https://patents.google.com › patent

The Basics: Electroplating & Vacuum Brazing - Diamond ...
https://www.dpcanada.com › articles › pl...

US4079552A - Diamond bonding process - Google Patents
https://www.google.com › patents

https://www.jstage.jst.go.jp/article/matertrans/43/10/43_10_2604/_article/-char/ja/

https://www.sciencedirect.com/science/article/pii/S0263436808001649

https://patents.google.com/patent/US6286498B1/en


Rate of O exchange with H2O We have to find the order of rate of exchange of O with H2O of the following compounds

$\ce{CCl3CHO}$
$\ce{CH3CHO}$
$\ce{CH3COCH3}$
$\ce{CF3CHO}$

According to me, higher rates of exchange would be when electron density is be less on the carbonyl carbon

But how we will compare 1. & 2.? In first compound there is hydrogen bonding (chloral hydrate)

When you talk about the rate of exchange of oxygen with water it might be instructive to realise that actually any aldehyde or ketone in the presence of water is in equilibrium with the hydrate, as shown below.

Source: UC Davis

There are a few ways we know that this is indeed an equilibrium process, but the main method that was used to fully elucidate the mechanism was by using isotopically labelled water containing 18O (the most abundant isotope of oxygen is 16O).
By doing this the scientists could observe the 16O in the starting aldehyde being slowly replaced with 18O from the enriched water.
Given that an equilibrium is involved, there are actually two pertinent questions...

What is the position of the equilibrium?
How fast do we reach the equilibrium?

The table below might be of some use to you. On the left, you can see a variety of carbonyl compounds, and on the right, the equilibrium constants for the hydration.
Two of the compounds you listed (acetaldehyde and trifluoroacetaldehyde) are listed in the table, and you can see that the acetaldehyde is significantly less hydrated at equilibrium than the trifluoroacetaldehyde.

Source: J. Am. Chem. Soc. 1983, 105, 868 (American Chemical Society)

From the table, some trends can be picked out (excluding formaldehyde which is almost exclusively hydrated). There is here a relationship between kinetics and thermodynamics (the more favourable a compound is to form the hydrate, the faster it will form it)

Aldehydes are far more likely to be hydrated than ketones. This is largely though not exclusively a steric effect.

Electron withdawing/electronegative substituents favour hydration. On a very simple level this is due to the fact that they make the carbonyl more electron poor, thus promoting the addition of water via its lone pair.

The order should therefore be 4>1>>2>3   (fluorine is more electron withdrawing than chlorine)


How does a catalyst accelerate both forward/reverse reactions simultaneously? A catalyst is known to speed up both forward/backward reactions of a reversible reaction. But how does this work, because the mechanism for the forward and backward reactions could be different right? I mean the adsorbed species for the forward and backward reactions when conducted independently will be different right? Is that because all the steps are exactly reversible?

The idea is that the activation energies is lowered for both the forward and backward reactions.

Though the activation energy for the backward reaction is higher than the activation energy for the forward reaction, it is nevertheless lowered.

To answer your second question, reactions aren't always reversible, but a lot of the time they are, and are subject to the principle of microscopic reversibility. That means when reactions are performed as a series of steps, at equilibrium, each individual reaction occurs in such a way that the forward and reverse rates of reaction are equal, and these equilibriums can be affected to drive the reaction one way or the other.


Electrochromic Materials, electrode setup I've read a lot about electrochromism. This article on electrochromic materials has also been additionally helpful.
I was wondering, for these electrochromic materials, can you simply attach a small electrode to either side of the film like in setup 2, then apply a small voltage? Will the electrochromism only occur at the edges in this case?
Does anyone know how material dependent this effect is? I was hoping to use a polymer film in particular.

I will answer your last question first, this is very material dependent. Not all materials will be electrochromic. It is a very specific effect where the material has different oxidation states that have different absorbance profiles(colors if you will). The most common ones are conducting polymers (https://en.wikipedia.org/wiki/Conductive_polymer) since these materials have different oxidation states that have vastly different electronic structure.

Two electrode electrochemistry can get pretty hard to understand. Two compare your setups 1 and 2, the main differences are the electrode area and the electrode distance. Other than that, the two are the same. So, yes, electrochromism will happen in setup2 as well. However, it is hard to predict exactly where it will happen and to what extent. It all depends on the material you select, it's resistance and doping characteristics. In either setup, it will start at the electrode and work it's way inward, obviously, setup 1 is more advantageous with the larger electrode area and shorter inter-electrode distance.


Why is sodium methanethiolate still a better nucleophile than sodium methoxide when the solvent is ethanol I'm looking at a problem where sodium methanethiolate and sodium methoxide, both in equal amounts, are being reacted with $\ce{CH3I}$ and the solvent is ethanol.

I understand that sodium methanethiolate is a better nucleophile than sodium methoxide when the two are reacted in equal amounts with something like methyl iodide in a polar aprotic solvent. But why is sodium methanethiolate still a better nucleophile than sodium methoxide when the solvent is polar protic, like ethanol?

I thought the ethanol would reduce the nucleophilicity of both, but would do so at by different amounts, given the different electronegativity of sulfur and oxygen. If ethanol reduces the nucleophilicity of both at the same rate, why even use ethanol as a solvent in this case?

Thiolates are generally more nucleophilic than alcoholates. So whenever the choice happens to be between a nucleophilic sulphur and a nucleophilic oxygen, the sulphur will win.

Both the methanolate and the methanethiolate will show reduced nucleophilicity due to partial protonation, though. And of course you have introduced ethanolate as an additional nucleophile, too (which should be the least reactive, though).


What is the name of the formula ΔE = q + w? What is the name of this formula? Google could not provide me with an answer.

$$\Delta E=q+w$$

This is a statement of the first law of thermodynamics. The first law says that the energy of an isolated system is conserved. An isolated system is one in which neither mass nor energy is exchanged with the surroundings. This equation shows the two means by which a closed system (meaning a system in which mass is conserved but energy is not) can exchange energy with its surroundings - by heat, or by work.

$$
\Delta E=q+w
$$

All energy transfers can be described as either heat flow or as work done on or by a system, and therefore the total amount of heat energy that goes in or out ($q$), plus the total amount of work energy that goes in or out ($w$), must be equal to the total internal energy change of the system ($\Delta E$).

This equation is particularly important in chemical thermodynamics because we have no way of measuring the total internal energy of a real system directly - all that we can measure is the amount of heat and work that goes in or out of the system, and therefore the change in internal energy.


Do SN2 reactions always invert absolute configuration? I know that an $\mathrm{S_N2}$ reaction proceeds through complete inversion of configuration, but is it necessary that the absolute configuration (R/S) of the chiral centre change after an $\mathrm{S_N2}$ attack ?

For the ease of thinking about this, let's imagine $\mathrm{S_N2}$ attack of $\ce{CN-}$ on bromochlorofluoroiodomethane, where iodide is being displaced.

Absolute stereochemistry is retained in this case because the highest priority substituent is becoming the lowest priority substituent. If you were to find the absolute stereochemistry of $\ce{2 ->3->4}$ in (R)-bromochlorofluoroiodomethane, which becomes $\ce{1->2->3}$ after attack, you would see that it is S and after attack becomes R. You can see that here:


Dissociation reaction notation This equation has been described as an equilibrium in an article, how would you explain it to someone with not a lot of understanding of it.

The equation is:
$\ce{CH3COOH <-> CH3COO- + H+}$

Although the description is of an equilibrium, it can be clearly seen that it is a resonance arrow. So how would you describe it?

No, no it's not a resonance arrow my friend, it has to be an arrow for equilibrium because it is a reaction equation of $\ce{CH3COOH}$ in its aqueous solution. It will dissociate feebly into its constituent ions because its a really weak acid. Thus establishing equilibrium quite a lot first.

Moreover in a resonance arrow we describe the same compound with different structures (which are all correct) and that is why we establish a resonance hybrid to denote all such possible resonating diagrams in a single diagram fulfilling the properties of the compound.

Nevertheless a resonance arrow can't anyway turn into a dissociation reaction arrow.


What's the simplest way to etch Nickel and Chromium sputtered layers? I have to etch a very thin film (50 nm) of Nickel ($\ce{Ni}$) on a 100 nm $\ce{SiO2}$/$\ce{Si}$ substrate. Additionally, there is a ~5 nm adhesion layer of $\ce{Cr}$ under the $\ce{Ni}$ that I'd also like to etch.

So I found this page which has a fairly long list of etches for $\ce{Ni}$ and $\ce{Cr}$.

Ni:

1 : 1 : 1 - HNO3 : HAc : Acetone
1 : 1 - HF : HNO3
30% FeCl3
3 : 1 : 5 : 1 - HNO3 : H2SO4 : HAc : H2O 85 C 10 microns/min
3 : 7 - HNO3 : H2O
1 : 1 - HNO3 : HAc
10% g/ml Ce(NH4)2(NO3)6 : H2O
HF, concentrated – slow etchant
H3PO4 – slow etchants
HNO3 – rapid etchant
HF : HNO3 – etch rate determined by ratio, the greater the amount of HF the slower the reaction
4 : 1 - HCl : HNO3 – increase HNO3 concentration increases etch rate
30% FeCl3
5g : 1ml : 150ml - 2NH4NO3.Ce(NO3)3.4(H2O) : HNO3 : H2O – decreasing HNO3 amount increases the etch rate
3 : 3 : 1 : 1 - H3PO4 : HNO3 : CH3COOH : H2O ~15min/micron @ RT with air exposure every 15 seconds

Cr:

  2 : 3 : 12  KMnO4 : NaOH : H2O
  3 : 1   H2O : H2O2
  concentrated and dilute     HCl
  3 : 1   HCl : H2O2
  2 : 1   FeCl : HCl
      Cyantek CR-7s (Perchloric based)    7 min/micron (24A/s new)
  1 : 1   HCl : glycerine     12min/micron after depassivation
  1 : 3   [50gNaOH+100mlH2O] : [30g K3Fe(CN)6+100mlH2O]   1hr/micron

I'd really like to avoid HF if I can (it terrifies me!), but I have pretty much everything else at my disposal.

So far, everything I've done has been at room temperature.

First I tried concentrated $\ce{FeCl3}$. No bubbling or signs of anything after ~20min.

Then I tried weak $\ce{H3PO4}$. Nothing.

Today I tried strong (98%) $\ce{H3PO4}$, nothing.

Then I tried concentrated fuming $\ce{HNO3}$. Nothing after ~1hr.

Then I tried about 1:1 $\ce{HNO3}$:$\ce{HCl}$, nothing.

Am I missing something here? Do the solutions need to be heated up or something?

Nickel can be etched using diluted (1:1 V) $HCl$. 50 nm will be dissolved within seconds.

The etching solution for the chromium layer contains $\ce{Ce(NH4)2(NO3)6}$+$\ce{CH3COOH}$+$\ce{H2O}$
in ratio 165g : 45 ml (con.) :100 ml DI water. The chromium layer is dissolved within 5 minutes of etching.


Oxidation State Of Propene I was just trying to understand the oxidation states of organic compounds. A more electronegative atom than carbon increases it's oxidation state by 1 and less electronegaitve atom decreases it's oxidation state by 1. Consider the molecule  propene $\ce{CH3-CH=CH2}$  the oxidation state of left carbon is -3, that of the middle one is -1 and that of the right one is -2? So, what is the oxidation state of carbon in   propene? Same complexity arises in 2-propyne or any such molecule.

Reconsider if a more electronegative element bound to carbon and bond to carbon will -- formally -- take the electrons, or donate the electrons in shared the bond with carbon, and how this is reflected in the oxidation state.  In each instance, watch out the signs, please.

After getting familiar with the basics, look up references focussing on organic chemistry and redox chemistry.  Albeit organic chemists tend to prefer moving electrons by hooks and arrows, there are references like this (hint: it includes an answer you may compare with the one you should find yourself, too) or even video classes like this.

Eventually, "left" and "right" will advantageously become obsolete when you gain enough credit to upload structural schemes on SE, too.


How to tell/control whether sodium ethoxide will react by a substitution or elimination pathway? Why is the behavior of sodium ethoxide inconsistent?

In a reaction with 1-bromobutane, the anionic part of sodium ethoxide substitutes for $\ce{Br}$:

$$ \ce{CH3CH2CH2CH2Br + C2H5ONa -> CH3CH2CH2CH2-O-CH2CH3}$$

Conversely, in a reaction with 2-bromobutane, the anionic part eliminates $\ce{Br}$:

$$\ce{CH3CH2CHBrCH3 + C2H5ONa -> CH3CH2CH=CH2}$$

I simply don't see how changing the position of $\ce{Br}$ changes the type of reaction. I believe sodium ethoxide, being bulky, must always do elimination, consistent with Rule #3 here.

Given the above: can sodium ethoxide be forced to one type of reaction with a given reactant, by enforcing some reaction conditions?

The above reactions belong to Williamson Synthesis of Ethers, probably the best of the alternatives for prepartion of ethers!

The Williamson Ether synthesis reactions follow SN2 mechanism. Since the SN2 mechanism proceeds through a single step where the nucleophile performs a “backside attack” on the alkyl halide, the only thing stopping this, is steric hindrance.

Methyl and primary alkyl halides are excellent substrates for this synthesis, as they provide no/very less hindrance for the approaching nucleophile.

Since alkoxides are strong bases, competition with elimination (E2) becomes a concern once the alkyl halide becomes more sterically hindered. This is exactly why, the first reaction that you’ve mentioned, proceeds by SN2 but in the second both elimination(major) and substitution (minor) takes place.

Also note that, when tertiary alkyl halides are used, there is no hope for substitution because of the bulky groups present around carbon. Hence, only elimination takes place in this case.

So the reactions are,

With primary alkyl halide,
$$ \ce{CH3CH2CH2CH2Br + C2H5ONa -> CH3CH2CH2CH2-O-CH2CH3}$$
With secondary alkyl halide,
$$\ce{CH3CH2CHBrCH3 + C2H5ONa -> CH3CH2CH=CH2(major)}$$

Moving onto conditions to favour a particular type of reaction:

One way to attempt to get the SN2 to be favoured over elimination is to use a polar aprotic solvent (such as acetonitrile or DMSO) that will increase the nucleophilicity of the alkoxide.
Increasing the temperature favours elimination over substitution.

Source for the above conditions: chem.ucalgary

  I believe sodium ethoxide being bulky should always do elimination

In most cases, rather than the incoming nucleophile, the hindrance in the the alkyl halide plays a more important role.

The rule #3 mentioned in the link you stated clearly says that the elimination takes place for a tertiary carbon in presence of a strong base like $\ce{CH3CH2-O-Na}$.


Calculation of proportion of zwitterion from this please. This is the question. I tried to post it in the title section but it was rejected because it was considered lengthy. Question: Valine is one of the 20 natural amino acids and has an acid -COOH (pka1=2.6) and an amine-NH2(pka2=9.6). If we consider a valine solution with pH=10.6, what is the proportion (%) of the zwitterion form of the solution? A) 2.4 B) 3.6 C) 5.7 D) 9.1 E) 10.2.
Sorry if this might be too elementary but I don't know how to solve this question. I need help, thanks.

(I shouldn't answer this because it is a homework question with no effort, but since someone posted a wrong answer I'm answering anyway)

Only the pK of 9.6 is relevant because the pH is 10.6; therefore, think of the situation as having just two species, zwitterion (acid) and negative (conjugate base).

Because pH is one unit above the pK, the ratio of acid to conjugate base is 1 to 10.

Ratio of acid to total is 1 to 11.

1/11 = 0.0909


Why don't sponges foam up an hour later? When I add dish soap to a sponge it is able to foam up. But if I leave the sponge alone for an hour, it doesn't foam up, not even if it is wetted.

Why? The sponge should still contain dish soap an hour later.

Dish soap traditionally contained foaming agents such as sodium laureth sulfate which is chemically stable over long periods of time. But more recently environmentally conscious soap companies are turning to foaming agents that have a shorter half-life. These chemicals tend to break down sooner - perhaps just by being exposed to air. This helps reduce foaming in the settling basins of wasterwater treatment plants, or storm drains that enter the ocean and it takes less water to rinse away the soap.

Read the ingredients of the soap you are using and research to see which are the foaming agents, and what the stable half-life is.


Why doesn't NF₃ undergo hydrolysis? I was reading Jan's answer to the following question-:
Why can NCl3 be hydrolyzed but CCl4 cannot?.
Since nitrogen is less electronegative than fluorine in $\ce{NF3}$, the electrophilic site is the Nitrogen atom. That means the water molecule should attack nitrogen atom in an $\mathrm{S_N2}$ manner. The stearic crowding around the atom is also less. Then why is the reaction not taking place? Is it only because fluoride ion is a poor leaving group due to its high basic strength?

Nitrogen trifluoride is extremely inert to hydrolysis, attack by base or acids or oxidation because of the fact that there is no d-orbitals in Nitrogen trifluoride and fluorine is far more electronegative than nitrogen. However, in the presence of base or acids, nitrogen trifluoride can however undergo hydrolysis because there is now excess of nucleophile which can attack nitrogen trifluoride.

There is a old paper which discuss the hydrolysis of nitrogen trifluoride in presence of acids and base.

  In the presence of aqueous base, however, slow hydrolysis occurs at
  100°C yielding nitrite and fluoride.

  $$\ce{NF3 + 4 OH- ->[100℃] NO2-  + 3F- + 2H2O}$$

It further says:

  This behavior differs sharply from that of nitrogen trichloride,
  which is known to give ammonia and hypochlorite under similar con-
  ditions. The latter products are readily explained in terms of
  nucleophilic attack directed at the chlorine atoms, a mechanism which
  appears reasonable in view of the fact that the electronegativities
  of $\ce{N}$ and $\ce{Cl}$ are very nearly the same and that the
  halogen may easily  expand its valence shell. Obviously, these
  considerations cannot be  applied to the nitrogen trifluoride molecule
  since fluorine is considerably more electronegative than nitrogen and
  it has no available  d orbitals. Although the nitrogen atom also has
  no free orbitals,  the relatively low electron density would at least
  offer less resistance to the approach of a nucleophile .

  The proposed intermediate $\ce{HONF2}$ would be expected to be
  unstable with  respect to the loss of $\ce{HF}$, as is apparently the
  case with the unknown  perfluoro alcohols .

  Two other reaction modes worth considering include the formation of
  an intermediate activated complex with water.


The oxidation state of iodine in iodoxybenzoic acid I am slightly puzzled by this seemingly trivial question. The idea behind assigning oxidation states is simply to treat every single bond made by the atoms as ionic bonds. That is to say, both the electrons of a covalent bond in a molecule would go to the more electronegative atom. For example, in the case of the $\ce {O-I}$ bond, the bond electrons would go to the oxygen atom, leaving the oxygen atom with a negative charge while the iodine atom incurs a positive charge. Based on this, we would say that the oxygen atom doubly bonded to the $\ce {I}$ atom has an oxidation state of -2.

However, I am slightly confused as to why various sources assign the oxidation state of $\ce {I}$ in IBX as +5, rather than +3. On the Pauling scale of electronegativity, it is shown that $\ce { I (2.66)}$ is indeed slightly more electronegative than $\ce {C (2.55)}$. Based on this premise, we would reasonably assign the $\ce {I}$ atom in IBX with a +3 oxidation state as the pair of electrons in $\ce {C-I}$ bond would go to $\ce {I}$ rather than to $\ce {C}$.

Image taken from here

This is a written reply from Professor Zhdankin*, author of Hypervalent Iodine Chemistry:

Indeed, many literature sources classify pentavalent iodine compounds as iodine in the oxidation state +5. It makes sense because the electronegativity of iodine is almost the same as carbon, or even less than carbon. According to different studies, electronegativity of iodine varies from 2.21 [A. L. Allred and E.G. Rochow, J. Inorg. Nucl. Chem., 5, 264 (1958).] to 2.66 [A. L. Allred , J. Inorg. Nucl. Chem., 17, 215 (1961)]. The original Pauling's value of iodine electronegativity is 2.5, exactly the same as carbon. This terminology reflects the similarity of iodine in IBX to iodine in $\ce {I2O5}$ or $\ce {HIO3}$ by chemical and structural properties.
Personally, I prefer not to use the "oxidation state" terminology regarding organoiodine compounds. Pentavalent iodine is a perfectly correct classification for IBX.

*He has given his permission for his reply to be shared here.


Is there a convention as to what order to write reactants and products in a reaction? When doing homework, it can be time-consuming to check the answers in a redox reaction due to different order of the reactants and products.

For example, while $$\ce{2MnO4- + 5Fe^2+ + 16H+ -> 2Mn^2+ + 5Fe^3+ + 8H2O}$$ and $$\ce{5Fe^2+ + 16H+ + 2MnO4- -> 5Fe^3+ + 8H2O + 2Mn^2+ }$$ are equivalent, they are not visually equivalent and to mark them as correct/incorrect takes a lot of time.

Is there a convention of what order to write these reactants/products in? If so, what is this convention?

There is no convention.

There might be some cases where one choice simply makes more sense to people. For example I have yet to see anybody write

$$\ce{NaCl(s) <=>[H2O] Cl-(aq) + Na+(aq)},$$

but there is no convention dictating that you have to write $\ce{Na+}$ before $\ce{Cl-}$.


how to display diffusion to a young audience? I need to display some science topic pertaining to the atomic bombs, or atomic history in some way to children of approximately middle school level. So I've decided that something interesting to cover would be gas diffusion (as used to isolate U-235 for the atomic bombs). I tried to see if there were any non-toxic, easy to obtain colored gases, but I couldn't find any. So the next thing I thought of is liquids.

Now I'm wondering if I can do something similar with liquids but I'm not sure how liquids would behave. Could I set up an effusion device, fill it with water and red food dye on one side, and water with blue food dye on the other and open a hole between the two and have them slowly diffuse? Will the diffusion of these two side be related to the molar mass or concentration of the dyes? Or would it follow some other law?

As you mention most colored gases (Cl2, Br2, NO2, I2) are toxic and being a gas difficult to handle. Therefore not suitable for such demonstration. Maybe, you could try two small colored smoke bombs in a transparent container although the colour is not really a gas (more like fine particulate) in this case.

I would use liquids with the most common diffusion demonstration being a drop of blue ink in a glass of water. However, we need something fancier than that so how about this: Prepare a solution of red dye in water and a solution of blue dye and sugar or salt in water as well. In a 120 mL volumetric cylinder (needs to be large and for people to be able to see) add the sugar or salt blue solution and then on top of that layer with the red water solution by adding it slowly and carefully. Due to the larger density of the blue solution the two solutions will not mix instantly. You should end up with a bottom blue solution and a top red solution. Diffusion will start where the two liquids come in contact giving you a purple band in the middle which will expand over time with the speed being depending on the concentration of the bottom solution in sugar or salt (you ll need to tune this according to your presentation). You could even show how diffusion becomes faster with temperature by heating with a hairdryer but the effect will be masked a bit by convection.

This is how we grow crystals of compounds in the lab as well: Dissolve your compound in a heavy solvent such as DCM and layer it with a light solvent the compound doesnt dissolve in such as hexane. As the hexane slowly diffuses in the DCM solution the solubility slowly drops resulting in slow formation of high quality crystals suitable for X-Ray diffraction studies.


How to predict the products of a organic chemical reaction Given a set of reactants, what aspects should I consider when predicting possible products?

You need to build a working knowledge of known reaction mechanisms that include the functional groups of your reactants. Put another way, (1) identify the functional groups present in your reactants and (2) check known reaction mechanisms that proceed by reaction of these functional groups.

Really, the only way to be good at this is to study and memorize organic reactions. There's a number of resources to help with this (e.g. online flashcards, and I really like Chemistry by Design, an app that you can use on a smartphone).


Are the bonding orbitals in methane equivalent - photoelectron spectrum The low energy portion (the part dealing with the $\ce{2s}$ and $\ce{2p}$ electrons) of the photoelectron (PE) spectrum of methane is reproduced below.

(image source)

The reaction being examined is the following one-photon process
$$\ce{CH4 + ~h\nu~ -> [CH4]^{+} + ~e^-}$$

The fact that 2 peaks in roughly 1:3 intensity are observed is often used to argue against the $\ce{sp^3}$ hybridization model for methane. Simply put, how can a model with 4 equivalent $\ce{sp^3}$ bonds produce 2 peaks in the PE spectrum.  The argument then goes on to point out that, according to MO theory, methane has one triply degenerate bonding MO and a second single bonding MO and these two nonequivalent orbitals would explain the 2 peaks in the PE spectrum with their 1:3 ratio (see p. 827 here for an example of such an MO-based explanation).

The PE spectrum of methane shows 2 transitions with 1:3 intensity.  How does one know that it is the methane ground state that is responsible for producing two peaks with a 1:3 ratio in the PE spectrum?  Why couldn't one argue that it is $\ce{[CH4]^+}$, where we clearly have 3 $\ce{C-H}$ bonds containing a pair of electrons and one $\ce{C-H}$ bond with one electron - two different types of $\ce{C-H}$ bonds in a 1:3 ratio - that is responsible for the two peaks in the PE spectrum?

Can the PE spectrum of methane really be used as evidence to argue against the hybridization model?

EDIT:

Poking around on the internet, I came across this Wikipedia article on Orbital Hybridization - Photoelectron Spectroscopy.

It says,

  "There is a popular misconception that the concept of hybrid orbitals
  incorrectly predicts the ultraviolet photoelectron spectra of many
  molecules. While this is true if Koopmans' theorem is applied to
  localized hybrids, quantum mechanics requires that the (in this case
  ionized) wavefunction obey the symmetry of the molecule which implies
  resonance in valence bond theory. For example, in methane, the ionized
  states (CH4+) can be constructed out of four resonance structures
  attributing the ejected electron to each of the four sp3 orbitals. A
  linear combination of these four structures, conserving the number of
  structures, leads to a triply degenerate T2 state and a A1 state.[16]
  The difference in energy between each ionized state and the ground
  state would be an ionization energy, which yields two values in
  agreement with experiment." (emphasis mine)

The bold face portion seems to support my contention (if I'm understanding it correctly) that it is the energy differences between the ground state and the excited state that is responsible for the spectrum.  Therefore, orbital splitting and orbital degeneracies in the excited state are likely an explanation for the observed spectrum.

This question goes along the line of what does it mean when it is said that an sp3 orbital has 25% s-character. It also intrigued me so I have tried to find answer, which would not break my hybridised orbital view.

What you clearly see from the spectrum is that there are two bands, corresponding to different energy levels in the molecule and their intensity is 3:1. You can say, it is $3\times 2p : 1 \times 2s$ and you are done. Not at all. The most important thing to deal with is the $T_{d}$ symmetry of the molecule. That means, all four $\ce{C-H}$ bonds are equivalent.

But it is this symmetry which has the answer. Leaving out the $1s$ orbital of carbon, the occupied orbitals must have the following symmetries:
$1\times A_1$ and $3\times T_2$. And this is what you see in the spectrum.

Now how it comes with the hybridisation? Upon simple HF/STO-3G calculation, the orbital populations are following:

$A_1$: 0.62 C 2s + 4x 0.18 H 1s  ... fully symmetric, great
$T_2$: 0.57 C pz + 0.3 1H 1s + 0.3 2H 1s - 0.3 3H 1s - 0.3 4H 1s
$T_2$: 0.57 C py + 0.3 1H 1s - 0.3 2H 1s - 0.3 3H 1s + 0.3 4H 1s
$T_2$: 0.57 C px + 0.3 1H 1s - 0.3 2H 1s + 0.3 3H 1s - 0.3 4H 1s

Well, as if you would copy the symmetry table ;) The downside of this picture is that all the bonds are completely delocalised (as you would expect from canonical MOs). Although observable through PES, they do not allow you to draw line between any C and H atom.

Where do we get the hybridised orbitals? If we want to localize the orbitals, whatever scheme you choose, you just form linear combinations of the existing MOs so that you have least number of atoms involved in any bond. One of such methods is NBO and as a result you obtain:

C - H1  ( 52.55%)   0.7249* C    s( 25.00%)p 3.00( 75.00%)
        ( 47.45%)   0.6888* H  1 s(100.00%)
C - H2  ( 52.55%)   0.7249* C    s( 25.00%)p 3.00( 75.00%)
        ( 47.45%)   0.6888* H  2 s(100.00%)
C - H3  ( 52.55%)   0.7249* C    s( 25.00%)p 3.00( 75.00%)
        ( 47.45%)   0.6888* H  3 s(100.00%)
C - H4  ( 52.55%)   0.7249* C    s( 25.00%)p 3.00( 75.00%)
        ( 47.45%)   0.6888* H  4 s(100.00%)

All with 2 electron occupation and energy -0.63479 Eh.

So where is the truth? ... $\Psi$ is the answer and you can massage it to your will, depending what you want to see. You want localized bonds? You will get them. You want orbitals with the proper symmetry required by spectroscopy? As you wish.

Does the hybridisation concept have sense? In localized bond picture yes. Is it observable? No, and was never meant to be.


Why do we multiply two wavefunctions in valence bond theory instead of adding them? The combined wavefunction of a hydrogen molecule in valence bond theory is the product of two wavefunctions, while in molecular orbital theory it is the sum of two wavefunctions.

Actually, the MO theory wavefunction can also be written as a product. While the individual molecular orbitals are just weighted sums of the atomic orbitals, the wavefunction can, to a first approximation, be written as a Slater determinant of the occupied orbitals.

For your example of the hydrogen molecule, this would mean taking the antisymmetrized product of the occupied $\sigma$ spin orbitals. (Note, if you haven't encountered this before, spin orbitals are a spatial and spin description of a single electron, rather than the commonly encountered spatial orbitals which describe two electrons of opposite spin.


Active masses of solids and liquids In the calculation for equilibrium constant, we take the active mass of solids and liquids as unity. Intuition tells that more amount of a solid reactant should result in more product being formed, but according to the formula for the equilibrium constant it has no role in determining that.
Is it unity in all possible cases? How can we assume it to be one and have it not affect the equilibrium calculation?

You might view the relevant principle as follows: components whose activity changes during a reaction tend to put a stop to a reaction before they are completely consumed. Components with constant activity (independent of the amount of other reactants), on the other hand, can be completely consumed (exhausted): they can be limiting reagents, provided they are less abundant (stoichiometrically) than the other reagents. If a reagent with constant activity is exhausted before equilibrium is reached the reaction stops - it is then a limiting reagent. Reagents whose activities change during a reaction (because their concentration changes, for instance) are in principle not exhausted this way because their activity increases too much for the reaction to proceed any further. Reactions involving only reagents with variable activity tend to go to an equilibrium in which none of the species is completely exhausted. If reagents with constant activity are present in excess (such that their initial amount exceeds that at equilibrium), then they will not limit the extent of reaction. Note also that if the increase in activity with dilution of a reagent with variable activity is negligible, its concentration at equilibrium may be so low that it can be considered exhausted (and limiting).

A good example of the relevant principle is a mixture of two reactive solids or two immiscible liquids of constant activity that react at the interface between them. In either case, if the reaction is favorable ($Q\gt K_{eq}$) then it will proceed until one or both of the substances has been completely consumed (both being completely consumed if present in stoichiometric proportion). On the other hand, when a solid dissolves into a solvent, the activity of the solid remains constant and it will continue to dissolve provided the activity of the dissolved species does not increase above that of the solid. If it doesn't, then the solid is completely dissolved (exhausted).

If a reaction is already at equilibrium, adding more of a solid won't make a difference (provided there are no surface effects). But if you are not at equilibrium, it will if the amount present is limiting. A good example again is solubilization of a solid in a solvent. Added solid will continue dissolving until you reach the saturation point. Past this point the concentration of the compound in solution does not change any further. In addition, with saturation the activity of the compound in the solid and solution phases becomes equal.


Human gastric lipase Does anyone know if the human gastic lipase has a quaternary structure? If not, why not? I can not seem to find this information anywhere.

There is a crystal structure of human gastric lipase (PDB ID 1hlg). The accompanying paper does not mention any quaternary structure and shows the protein as single chain. Automated analysis of the coordinates support that. The enzyme is glycosylated, so the molar mass does not match the one expected from the sequence. The apparent molar mass determined experimentally is about 50,000 g/mol, corresponding to a monomer.

However, the annotation in the structure (probably erroneously) say that the biological assembly consists of the two chains in the asymmetric unit. Also, there is a report that the size of the molecule changes with pH (90,000 g/mol at pH 6 vs 45000 g/mol at pH 3).

The same group that solved the human lipase structure later solved the dog lipase structure (in an open conformation with a bound ligand) where the annotation in the PDB deposition says it is a monomer. I doubt that there are interspecies differences between dog and human in terms of quaternary structure, so that would be another piece of evidence supporting that the human enzyme is a monomer in its native form.


Can cis-trans isomers interconvert? Can cis-trans isomers interconvert via tunneling ?

Electron tunneling
Tunneling refers to elementary particles moving "through" a potential barrier in the absence of sufficient energy to go "over the barrier". This is roughly equivalent to saying that a particle can occupy two locations and traverse between them even though all paths between those locations have a zero probability of the particle being there.
Some orbitals such as the p-orbital describe a probability density with areas of non-zero probability ("lobes") separated by space with zero probability ("nodes").
When electrons are transferred from one electron carrier to another, the transfer can occur through bonds or through space; the through-space transfer is either based on orbital overlap or on tunneling.
Proton tunneling
Proton tunneling is much slower because the proton is about 2000-times heavier than the electron. Tunneling has been used to explain reactions involving protons that are faster than expected from inspecting the potential energy landscape. This proton tunneling also explains proton transfer in hydrogen bonds.
For formic acid and acetic acid, proton tunneling explains cis-trans isomerism at ultra-low temperatures. Acetic acid, at room temperature, has a cis and a trans conformation.

There is a rotational barrier (which one may explain via resonance involving the carbonyl double bond) which is lower than thermal energy at room temperature [Ref 1]. For related systems like the peptide bond, cis and trans forms are isomers because the rotation barrier is higher.
At ultra-low temperature in a nobel gas matrix, the trans form is dominant and does not convert into the cis form unless you excited the molecule at the right wavelength. This means that the cis and the trans form are isomers with individual properties. Under these conditions, it is possible to enrich the cis form, and to observe the conversion to the trans form in the dark [Ref 2]. The temperature-dependence of the kinetics (and kinetic isotope effects) suggest that this cis-to-trans reaction occurs via tunneling.
Hydride tunneling in enzymes
Many redox reactions involve hydride transfer, e.g. from alcohol to NAD+ in the alcohol dehydrogenase catalyzed reaction. For enzymes, it is typically not possible to measure reactions at ultra-low temperatures; instead, a key experiment is to study the primary and secondary kinetic isotope effect [Ref 3]. From studies of multiple enzymes that catalyze hydride transfer it is evident that tunneling does occur, and the extent of tunneling is modulated by mutations in the active site. The review makes the argument that tunneling is easier to study for hydrogen (i.e. proton, hydrogen radical and hydride) transfer than for electron transfer because in the former case, isotopes (with a different deBroglie wavelength) are available as a probe. While the examples in the review are all redox reactions rather than E/Z isomerization reactions, one could envision the latter based on hydride tunneling as well.
"Double" tunneling
If there are two atoms connected to a double bond (such as in the molecule $\ce{H2C=CDH}$ mentioned in the comments), isomerization through tunneling would be much harder because the hydrogen and the deuterium have to somehow switch position. To avoid a clash, they would have to tunnel simultaneously, and could not both take a path along the saddle point of the potential energy landscape. Therefor, a cis-trans isomerization of $\ce{H2C=CDH}$ seems much less feasible, and there does not seem to be any experimental evidence for it.
References

Barriers to rotation adjacent to double bonds. 3. The carbon-oxygen barrier in formic acid, methyl formate, acetic acid, and methyl acetate. The origin of ester and amide resonance. Kenneth B. Wiberg and Keith E. Laidig. Journal of the American Chemical Society 1987 109 (20), 5935-5943. DOI: 10.1021/ja00254a006
Rotational isomerism of acetic acid isolated in rare-gas matrices: Effect of medium and isotopic substitution on IR-induced isomerization quantum yield and cis→trans tunneling rate. E. M. S. Macoas, L. Khriachtchev, M. Pettersson, R. Fausto, M. Rasanen
J. Chem. Phys. 121, 1331 (2004); DOI:10.1063/1.1760733
Tunneling and Dynamics in Enzymatic Hydride Transfer.
Zachary D. Nagel and Judith P. Klinman, Chem. Rev. 2006, 106, 3095-3118, DOI:10.1021/cr050301x


Isothermal vs adiabatic compression of an ideal gas Could someone please explain these specific questions physically (especially the first question!)? I understand the corresponding mathematical proofs, but not the physical reasoning? I have looked online and can only find answers to what I believe correspond with my second question.

Why is isothermal work larger than adiabatic work for the compression of an ideal gas for the same change in pressure?
Why is isothermal work less than adiabatic work for the compression of
an ideal gas for the same change in volume?

There are a number of ways to look at this. One is to look at the equations. Another is to plot the pressure p as a function of volume V. The area under the pressure curve on the plot is the total work done on the system.

It is very easy to answer your second question by inspection of this figure, but perhaps not the first question without actually evaluating integrals (areas).

Yet another way to answer this, but perhaps hand-wavy, is to think of what happens as you compress the ideal gas either adiabatically or isothermally. In the adiabatic compression no energy is lost to the surroundings as heat. All of the energy transferred to the gas as compression work is retained by the gas. But the only way to increase the internal energy of an ideal gas is to raise its temperature. Therefore the effect of the adiabatic work is to increase the temperature of the gas. The increase in temperature is accompanied by an increase in pressure above that due to Boyle's law. This follows from
$$p=\frac{nRT}{V}$$
In the adiabatic process p increases with the higher temperature above the level it would have if the heat had been lost to keep the temperature constant, as it would in an isothermal process. In that way the adiabatic compression is a more efficient way of increasing the pressure. This means less work is required to arrive at the same final p. By the same logic, if you want to achieve a determined final volume, you will have to work harder in an adiabatic setting because the pressure you are working against increases more rapidly as the gas gets hotter.


Products formed when solid CuCN is dissolved in aqueous KCN solution Copper cyanide forms a complex coordination compound, when dissolved in a solution containing excess ($\ce{CN^-}$) ions.

My question is whether it forms $\ce{[Cu(CN)_4]^{3-}}$ only or both $\ce{[Cu(CN)_3]^{2-}}$ and $\ce{[Cu(CN)_4]^{3-}}$ complexes?

Wikipedia says that both complexes are formed, but I am not sure if its a reliable source.

No, both $\ce{[Cu(CN)3]^2-}$  and $\ce{[Cu(CN)4]^3-}$ species are formed. According to Wikipedia article of copper cyanide:

  ... but rapidly dissolves in solutions containing $\ce{CN−}$ to form
  $\ce{[Cu(CN)3]^2-}$ and $\ce{[Cu(CN)4]^3-}$, which exhibit trigonal
  planar and tetrahedral coordination geometry, respectively. These
  complexes contrast with those of silver and gold cyanides, which form
  $\ce{[M(CN)2]−}$ ions in solution.The coordination polymer
  $\ce{KCu(CN)2}$ contains $\ce{[Cu(CN)2]−}$ units, which link together
  forming helical anionic chains.

See this paper for more details.


Reverse Phase Chromatography Please help me figuring out what is going on with my RP-HPLC protocol.

I am purifying a small peptide out of E.coli Culture, heat stable enterotoxin, its amino acid sequence is:

N-T-F-Y-C-C-E-L-C-C-N-P-A-C-A-G-C-Y.

I am collecting culture supernatant and using amberlite XAD2 to concentrate and desalting the peptide in 80% and 99% methanol fractions. These fractions were subjected to flash evaporation. Then I am using on MCI gel CHP20P (glass column) for further concentration and removing unwanted proteins (polar and less hydrophobic ones) using stepwise elution with 20%, 40%, 60%, 80% and 100% methanol fractions. Each fraction is subjected to flash evaporation. A sample (1 ml) from each fraction (60, 80 and 100%) was then loaded to RP-HPLC using xbridge C8 preparative column (19x25 mm,10 micron, 130 angstrom) on a gradient system (solvent A: water/ solvent B: 80%methanol) initial-5 minutes 100 A, 5-10 minute 30% B, 10-65 minute 100% B, 65-80 minute 100% B at flow rate 2 ml/minute. The chromatogram showed a single large peak (3 AUF) at 25 minutes, B solvent around 45-50%.

I am wondering that the retention time is lower than that was reported before, 35 minute at 55-60% B (I followed almost the same protocol 5 years ago except using MCIgel methacrylate resin and Vaydc C8 column instead of mci gel CHP20P and xbridge C8 column). It is also very frustrating that I loaded 0.5 ml of the collected peak on Waters analytic C8 coulumn, 5 micron particle size and it  elutes very early at 5-6 minutes (almost in the mobile phase A: water). My frustration comes from that I am using highly hydrophobic resin particle (MCI gel CHP20P) to capture hydrophobic molecules and loaded these fractions (60%, 80% and 100% methanol MCI gel) on C8 column and eluted so early.

I am sorry for this long message but I rely on your understanding. Any answer would be highly appreciated. Thank you in advance

I am just wondering how it comes that 60-100% methanol fractions from MCI gel CHP20P eluted very early on C8 column?!

The gel you are using seems to be a porous divinyl benzene / styrene resin.  The aromatic groups of the resin may interact more strongly with aromatic side chains on your molecule (e.g. Y residues) than does the aliphatic C8 groups of the reversed-phase HPLC column you are using.

"Hydrophobic" vs. "hydrophilic" is a very useful axis for understanding solute interactions with stationary phases, but unfortunately not everything can be collapsed into this single-variable scale.

Another possibility is that residual solvents or very large amounts of co-solutes are interfering with the RP-HPLC method you are using, causing premature elution.  As an extreme example of what I mean, imagine if you injected a few μL of chloroform along with your sample in the RP-HPLC method.  The chloroform would prevent column binding and cause premature elution.  Now, it's unlikely your samples have chloroform, but they could have large amounts of methanol or other "solvent-like" impurities.

These are just general points; unfortunately I can't be sure that either is the precise source of your particular problem, without more (and more clearly presented) information on your former method from five years ago, your method today, and the differences in observed vs. expected results.


is it possible to break the bonds of diatomic elements such as fluorine or iodine and create positive and negative ions by electron bombardment? If I had a container in vacuum filled with $\ce{I2}$ gas and then I bombarded it with high speed electrons using an electron gun, would  be able to get both $\ce{I+}$ and $\ce{I-}$ ions or would I only get $\ce{I2+}$ plus an ejected electron?

Sure, why not. It's just the question of your bombarding electrons having the right energy. At low energies, a bombarding electron will likely be captured by one of the iodine atoms:
$$
\ce{I2 + e- -> I- + I} \, .
$$
But somewhere around $8.62 \pm 0.06 \, \mathrm{eV}$ according to this study, you'll indeed have the ion pair formation:
$$
\ce{I2 + e- -> I- + I+ + e-} \, .
$$
There are some other processes which contribute to the ion pair formation. For details, read the paper quoted above.

With respect to the formation of $\ce{I2+}$ note that to form it a bombarding electron have to, formally speaking,

Either first break the $\ce{I-I}$ bond homolyticly and then remove two electrons from one of the $\ce{I}$ atom which requires it to carry about 30 eV of energy (the sum of the homolytic bond dissociation energy and of the first and the second ionization energies).
Or break the $\ce{I-I}$ bond heteroliticly and then remove the second electron from $\ce{I-}$ ion which requires it to carry also about 30 eV of energy (the sum of the heterolytic bond dissociation energy and of the second ionization energy).

$%edit$


What are the products of the hydrolysis of ICl3? I have learnt that interhalogens on hydrolysis give a halic acid derived from the smaller atom and an oxoacid derived from the larger atom. For example hydrolysis of $\ce{BrF5}$ would give me $\ce{HF}$ and $\ce{HBrO3}$.

Similarly the hydrolysis of $\ce{ICl3}$ should give me $\ce{HCl}$ and $\ce{HIO2}$ but my textbook says that $\ce{HIO2}$ is too unstable to be isolated. What then would be the products of the reaction?

$$\ce{3ICl3 + 6H2O → 9HCl + HI + 2HIO3}$$

  Iodine(III) chloride react with water to produce hydrogen chloride,
  hydrogen iodide and iodic acid. The reaction is called Hydrolysis of
  iodine trichloride.(source)

The reaction is same as @xavier_fakerat noted in the comment section.


What is the Lewis structure of Ozonide ion? What will be the Lewis structure of the ozonide ion, including formal charges?

Do all oxygen atoms have filled octate?

From a cursory e-book search:

  Ozonide ion is paramagnetic and is shown to have magnetic
  susceptibility. Bond length is 135 Å and bond angle is 108°. The
  structure is:


Find the velocity constant K given the speed This reaction is of first order in [A]. Consider that

  What is the velocity constant $K$ here?

So let's see... The velocity equation is

$$V = K \cdot [A]^n\cdot[B]^m$$

Assuming that it is bimolecular... But... we don't know if it is bimolecular! All they tell me is that there is some reaction with a molecule $A$ involved.

Anyway, they also tell me that the reaction is of first order... on [A]. It is my understanding that this is called a partial order, and the global order is not necessarily the same, as it would depend on the involvement of any other molecule (which we don't know about). Therefore, it is not necessarily true that the equation of velocity is

$$V = K \cdot [A]^n\cdot[B]^m$$

So how do I even approach this kind of problem?

I can't even calculate the general velocity of the reaction, because I don't know the coefficients of the reagents. Heck, I don't even know how many reagents and products are there!

(I know this is an extremely simple exercise and the answer is also quite basic, but I just fail to find an example of exactly this scenario anywhere).

You are really confusing yourself. The problem is very simple. First of all, remove the fact that the rate of reaction depends on all reactants from your mind. It is not the case. Sometimes, even though multiple reactants are involved, only some of them actually matter when we need to find the rate. Let me explain it without chemistry.

Consider a factory where a car is made. The first level makes the engine in 1 hour. The second level makes the tyres in 1 hour. The third level makes the rest of the parts in 1 hour. The fourth level puts them all together in 1 day. Now, even though all the 4 levels are involved in the process, it is the slowest level (level IV) that determines the rate of the process because the rest of the levels can keep sending the parts, but the finished car won't exit the factory unless level IV puts them all together. This is the case with chemical reactions too! The slowest step decides the order and rate of the reaction.

Now, back to the problem. Since they have given that the reaction follows first order kinetics in $[A]$, the rate law can be written as $R=k[A]$. After a little bit of integration and other stuff, we have the equation, $k=\frac{ln(2)}{t _{1/2}}$ , where $t _{1/2}$ is the half-life, i.e. time taken for concentration (actually, it's no. of molecules) of reactant to become half its original value. In this problem, $t _{1/2}$=10.0 seconds. Substitute and find the answer.


How are the boiling points of Tungsten and other metals determined and proved? The boiling point of Tungsten is 10,030 degrees Fahrenheit. How was this determined and proved? And more generally how are the boiling points of metals determined and proved?

Is it really so simple as "heat it up as high as you can and see if (and at what temperature) it evaporates"?

I feel like there is a more sophisticated process, but I can't really guess at what that might be, especially if the boiling point was determined years and years (100+?) ago.

The Internet is really good at telling me the boiling point of a lot of (or all, I haven't exhaustively checked) elements and alloys, unfortunately it seems it's not very good at telling me how these seemingly arbitrary temperatures were determined.

According to the 1913 paper THE VAPOR PRESSURE OF METALLIC TUNGSTEN, The Physical Review, vol. II, pp. 329-342:

The vapor pressure (p) of a pure substance is related to its evaporation rate (m) in a vacuum by the following relationship:
$$ m= \sqrt{\frac{M}{2\pi RT}}p$$

Evaporation rates were measured at temperatures in the range from 2400K to 3100K, heating by electric resistance of a filament, and determining the evaporation rate by weighing the filament before and after the evaporation.
The temperature was determined by observing the color and intensity of the light emitted by the filament. (Paper says they will publish a separate paper about the details of the temperature determination technique).
Vapor pressure was extrapolated to a boiling point of 5110 K using the Clausius-Clapeyron relationship.


Search materials by HOMO, LUMO and work function I need to search organic semiconductors for organic photovoltaics (OPV) by their HOMO, LUMO, work function at 300K (as I understand, WF cannot be derived from HOMO and LUMO because semiconductor can be differently doped and Fermi level depends on concentration of dopants) and deposition methods (such as CVD, PVD, spin-coating, electrodeposition, etc)..

Is there some free site where I can either download a machine-readable table of needed data or make a search based on that criteria? What is it?

Indeed, Prof. Alán Aspuru-Guzik has placed his "clean energy project" database on the web in a searchable form.

At the moment, it covers ~2.3 million compounds.

  Use of the data requires citation of the appropriate literature sources. The data in this database is released under the Creative Commons Attribution ShareAlike license and therefore any academic publications associated with the CEPDB should be appropriately cited.

As you point out, the database only covers HOMO and LUMO orbital eigenvalues. While this may not be the most useful for experimental applications (i.e., you really want ionization potentials and first excited state energies or electron affinities) it should be a helpful "first screen" to find interesting lead compounds.


What is this Lab equipment? I want to know the name of this equipment with a sealed graduated tube and the purpose it's being used.

I went through all the results of Google, but I was unable to find this exact equipment. I believe someone here should know this very well.

This lab equipment is called 'Fermentation Tube'.

If you want to see how an experiment is carried out with this equipment, please watch the video below.

Mark Garcia: Lab Protocol - Fermentation in Yeast (Unit 9 Respiration); YouTube