Must lactose-free milk be ultra-pasteurized? In this question I asked why lactose-free milk lasted such a long time. The answer was because it is ultra-pasteurized. This leads naturally to another question: does lactose-free milk have to be ultra-pasteurized?My initial guess is that it does because:Since the lactase added to make the product lactose free breaks down lactose into twice as many sugar molecules, any living bacteria have twice as much nutrients to thrive in.Since the disaccharide lactose has been broken down into the simple monosaccharides glucose and galactose, the bacteria do not need to produce their own lactase to use the sugar; thus any non-lactase producing bacteria can thrive in the milk.And finally, I have been adding lactase to regularly pasteurized milk to produce my own lactose-free milk. However, this milk almost always goes bad a couple of days before the Best-By date stamped on the container.This all leads me to think that lactose-free milk must be ultra-pasteurized in order for it to have any reasonable shelf life.

You can now buy (at least in Switzerland where I live) lactose-free milk which is not ultra high-temperature pasteurized. They are kept chilled in the supermarket alongside "normal" milk. So I would say based on this, no lactose free milk does not have to be ultra pasteurized. Though I will note, they hardly tasted sweet at all and they are quite expensive, nearly double the price of ordinary milk.


Iodine color change in different complexes Why does the color of iodine differ when it forms a charge transfer complex with benzene (pi donor) - pink and ethanol (electron pair donor) - brown? Does the HOMO-LUMO energy differ, if so how?

The formation of charge transfer complexes between iodine and aromatic hydrocarbons according to

\[\ce{Ar + I2 <=> Ar\cdot I2}\] is a long known phenomenon and has been described by Benesi and Hildebrand in A Spectrophotometric Investigation of the Interaction of Iodine with Aromatic Hydrocarbons. The following data for the absorption maxima of iodine in the VIS and UV range (wavelength in nm) in different solvents are taken from that article.

\begin{array}{lrrr}
\mathbf{solvent} & \lambda_{max}(\textrm{VIS}) & \lambda_{max}(\textrm{UV})\\
\hline
\textrm{benzene} & 500 & 297\\
\textrm{toluene} & 497 & 306\\
\textrm{p-xylene} & 495 & 315\\
\textrm{mesitylene} & 490 & 333\\
\hline
\textrm{trifluormethylbenzene} & 512\\
\textrm{n-heptane} & 520 & \\
\end{array}

Note that the effect is stronger in the UV than in the VIS range and the shift correlates with the (combined) $+I$ effects of the methyl groups, while a $\ce{CF3}$ substituent disfavours then formation of a charge transfer complex.

Ethanol, on the other hand, is a protic and polar solvent, nice to stabilize charged species.

One might argue that under these conditions, other species than in the case of aromatic solvents exist - the solvation of triiodide $\ce{I3-}$
 is certainly possible here.

In addition to the coordination of iodine by ethanol - we cannot exclude interaction with several molecules of ethanol:

\[\ce{n\ EtOH +I2 <=> I2\cdot(EtOH)_{n}}\]

it is conceivable that disproportionation of iodine and solvation of the resulting charged species might play a role here:

$$\ce{ I2\cdot(EtOH)_{n} + I2 <=>[\ce{EtOH}] I+\cdot(EtOH)_{n} + I3-} $$


Williamson ether synthesis Williamson ether synthesis is conducted using vanillin, $\ce{KOH}$, $\ce{EtOH}$, water and benzyl chloride.

Why is water needed for the synthesis?
Why must the solution be refluxed for an hour to gain the desired product?
Why is water added after reflux to create a cloudy solution?

1.

Consider the phases for potassium hydroxide and benzyl chloride at room temperature.
Why might you not want to use ethanol as the solvent to form the benzyl alkoxide of vanillin in?
(As an aside, think about why you might not want to use water over, say, acetonitrile as a solvent for this reaction.)

2.

What governs the rate of the Williamson ether synthesis?
Why would you want to heat a chemical system you want to have a reaction in?

3.

What solvents would the product ether dissolve in?
Which by-product(s) would dissolve in water but not ethanol?
How would you remove the product ether from the solvent mixture?

I'm trying not to give away the answers directly but these bullet points should help you work out each point.


What gas was produced in this orbiting space-balloon from 33 pounds of benzoic acid and anthraquinone? Why these two chemicals? As pointed out in this answer, on page 4 of NASA Technical Note TND-1366 The Orbital Behavior of the Echo I Satellite and its Rocket Casing During the First 500 Days (June, 1962) says:

PHYSICAL CHARACTERISTICS

Echo I has an effective cross-sectional area of 7,854 square feet. The launch weight of 157 pounds decreased to 124 pounds with the loss of 33 pounds of benzoic acid and anthraquinone which were used to maintain inflation for the first few  weeks in orbit. Thus, the initial ratio of the weight to mean drag area for Echo I was 0.020 pound per square foot and reduced to 0.016 pound per square foot after several weeks in orbit.

Echo-1 was about 100 feet in diameter, so that change in weight of 0.004 pound per square foot of drag area corresponds to the 35 pounds mentioned.
Questions:

What kind gas would have been produced by this mixture, and what fraction of the initial mass would have been converted to gas? Could it have been all of it as suggested? (the idea is that after the gas leaked out the balloon would have become rigid enough to retain its shape while in orbit.
Why both benzoic acid and anthraquinone? Did they react? Did they sublimate only? Is there any chemistry-related reason why both substances were used?

below x2: cropped and full size from Echo, NASA's First Communications Satellite

There are three materials to investigate: the two solids in question have been described as having fast and slow rates of sublimation. The balloon was sent up with 4.5 kg of benzoic acid (bp 250C) and 10.6 kg of anthraquinone (bp 380C). https://en.wikipedia.org/wiki/Project_Echo
The third material is the film: 12.7 micron biaxially oriented polyethylene terephthalate metallized with 0.2 micron aluminum. The two sublimatible solids were used to rapidly inflate the balloon and then to keep it inflated in case of micrometeorite strikes.
Echo-2 was made with a thinner PET film (9 microns) sandwiched between two thicker aluminum films (4.5 microns each), and the balloon was inflated to a pressure that slightly deformed the aluminum sandwich and kept the whole structure more rigid, so continual sublimation was not required.
The PET films are so thin! By comparison, a human hair is typically 100 microns in diameter. Perhaps the choice of sublimatible materials was made for non-reactivity (with the film) rather than with any reaction to form a gas. But the rationale is still not clear - it certainly couldn't have anything to do with cost.


Finding the number of orbitals on a central atom In $\ce{BeCl2}$ the number of orbitals on central atom, i.e. on beryllium, are 2. In $\ce{BF3}$, the number of orbitals on central atom , i.e. on boron, are 3.

Similarly in $\ce{NH3}$ there are 4, on $\ce{PCl5}$, $\ce{SF4}$, and $\ce{ClF3}$ there are 5. But how?

According to the question, it is totally not clear that whether you want to know filled orbitals or total orbital (filled + vacant). Well, but after seeing the examples you gave, it can be concluded that you are talking only about filled orbitals. As I'm not clear and assuming that you are only talking about filled orbitals, so if you are talking about total, leave a comment.

As each orbital contains 2 electrons, therefore it is easy to find number of filled orbitals.

The number of filled orbitals in central atom will be equal to the sum of number of bonds and number of lone pairs present on central atom.

Example

As in case of $\ce{BeCl2}$, $\ce{Be}$ makes 2 bonds (both with $\ce{Cl}$) and have no lone pair, therefore,

$\text{No. of filled orbitals} = 2$

Let us take another example of $\ce{NH3}$.

In case of $\ce{NH3}$, $\ce{N}$ makes 3 bonds (each with $\ce{H}$) and have 1 lone pair, therefore,

$\text{No. of filled orbitals} = 4$

Take another example of $\ce{ClF3}$

In this case, the central atom $\ce{Cl}$ makes 3 bonds with $\ce{F}$ by sharing its one electron in each bond and have 2 lone pairs on it. (As $\ce{Cl}$ have 7 electrons, 3 shared in bonding and 4 left = 2 lone pairs.) Therefore,

$\text{No. of filled orbitals} = 5$

Hope, now everything is clear to you!


How does the acidity of soft drinks not affect certain drugs, such as opiates? What about drugs and stomach acid? Here's a very random introduction. I heard the term, "dirty Sprite" (it's some rap album/slang term) on a YouTube advertisement. I Googled "dirty Sprite", and it's apparently a recreational drug: codeine syrup in Sprite lemon-lime soda. People allegedly get high off of it, meaning the opiates in codeine is metabolized properly. My question is, how does harsh acidity of the Sprite (pH of 3.4-3.8 if you trust Google) not interfere with the codeine structure and mitigate its metabolism?

Then, this turned into a meta philosophical question: how does the acidity of the stomach, which is more acidic than Sprite, not interfere with not only codeine, but other substances?

Many drugs are not very soluble in water what is a common property for a lot of organic substances, including Codeine. Therefor it is very common to turn them into their hydrochlorid which is simply $\text{Drug}\ce{.HCl}$. [4] E.g., Crystal Meth got it's name from the crystals that the hydrochloride forms.

$\hskip1.7in$
Source: Wikimedia Commons

Other examples are, e.g., Cocaine, Cetirizine, hydroxyzine, and many more.

As your substance is either the hydrochloride or would turn into the hydrochloride in stomach there should not be any destructive influence on the structure of the drug. In fact it even promotes the absorption.

On the other hand, the better solubility of the corresponding hydrochloride is a relatively big problem for those drugs that should not get absorbed in the stomach. They need to be coated to "survive" the stomach acidity to reach the intestine where they should get absorbed.


Bleaching of chlorophyll stains using hydrogen peroxide I have been doing some experiments concerning bleaching of chlorophyll stains from cotton clothing. Currently I have found that $\ce{H2O2}$ is fairly effective, but still leaves visible stains. Would a crude Horseradish root extract containing peroxidase get me better results?

The reason I'm asking about the crude extract and not the purified enzyme is because I don't have access to a laboratory or more complex materials and my work also aims to produce simple methods of chlorophyll degradation that could be replicated at home.

Horseradish peroxidase is not hydrogen peroxide, $\ce{H2O2}$, but rather an enzyme that breaks $\ce{H2O2}$ into $\ce{H2O}$ and $\ce{O2}$. In general, enzymes ending in -ase are lytic enzymes, catalyzing the breakdown of a similar-sounding substance. So, if you add ground horseradish to hydrogen peroxide, bubbles of oxygen are released. BTW, horseradish roots differ in peroxidase potency, so test beforehand with a particular batch before doing this as a demonstration... sometimes it fizzles out, or not.

Another example of lytic enzyme, amylase in the mouth, breaks apart amylose in starch, so a starchy cracker tastes sweeter as it melts in your mouth.

As @Buttonwood suggests, alcohols, ether and oils are more effective solvents than water.


Do molecules with polar bond, but with no dipole moment experiences a greater effect from the london dispersion forces? Do molecules that have polar bonds and a structure that makes the polarity of those bonds cancel each other out, experience a greater effect from the London dispersion force, than molecules with a similar size, bond length, shape, and mass, but with non-polar bonds?

Yes, an example of a molecule would be CO$_2$, which has no dipole moment but does have two polar bonds. Although CO$_2$ has no dipole moment, its charge distribution gives rise to a quadrupole moment. This quadrupole moment provides a handle for the molecule to interact with other particles (with other quadrupoles, but also with dipoles, induced dipoles or electric charges). As you might know, dipole-dipole interaction scales as $r^{-3}$ where $r$ is the distance between the particles and Van der Waals interaction (induced dipole-induced dipole interaction) scales as $r^{-6}$. It can be shown that quadrupole-quadrupole interaction scales as $r^{-5}$ and therefore represents a stronger interaction than the induced dipole-induced dipole interaction.


Extraction of starch from soup I have a solution containing many different organic compounds (originally from leek and potato soup) and I would like to extract the starch from it. However, starch is insoluble in water. How could I go about extracting the starch?

You can't simply filter it and take the feed because all the vegetables fiber remain in the filter with your starch. If you want to make a starch with a decent level of purity you should start directly with the potatoes: grating the potatoes, putting them in hot water (preferably salted: this make more effective the separation from proteins) for a bit of time. Then take the filtered solution (not the residual) make the water evaporate now what you get is the starch with the potato protein, amino acids and minerals (what is called fruit water). There is this video if you want to see.

If you want to have a more pure starch also you can try to precipitate the proteins warming the solution and adding an acid. If you are at home acetic acid could be enough.
Eventually you can cool down the solution this make starch precipitate so you can remove the water with a filter and take the precipitate.

If you really want to isolate starch there are more sophisticated methods see here and here for references.


Whats the difference between reversible process in thermodynamics and reversible chemical reactions? We know that in thermodynamics, reversible process never occur since entropy should be conserved. However, in chemistry we do have a lot of reversed chemical reactions that forms reactants from products, previously formed by a direct reaction. So, what is the misunderstanding here?

There is no misunderstanding at all.

The second law of thermodynamics can be roughly expressed as: in a given process the entropy of the universe will always increase (defining universe as system + surroundings).

Reversible processes

For reversible processes the entropy of the universe remains constant. Most natural occurring processes are irreversibles.

A reversible process is a process where you can change its direction by inducing infinitesimal changes to the system. Throughout the entire process the system is at equilibrium, so from one state A to a state B you will not observe macroscopic changes.

The system goes from state A to state B in the following way: A + dA = B where B is infinitesimally different from A. These changes are applied over properties of the system like temperature and pressure by modifying its surroundings.

Reversible chemical reaction

In the case of a reversible chemical reaction it means that if you combine the products of that reaction you’ll obtain at least some of the reactants again. Usually these reactions require the same amount of energy that was consumed or released to obtain the products (This means that you must take into account the enthalpy of the forward reaction).

A typical example are lead batteries where lead reacts with sulfuric acid to form lead sulfate and $\ce{H+}$ ions while the battery is discharging and then you apply a direct current to revert the reaction and obtain lead, sulfuric acid and lead oxide again.

Even when a reaction is thermodynamically possible you must take into account the kinetics parameters. For example: it is thermodynamically possible to turn graphite into diamond but kinetically at standard conditions (1 atm, 25$^\circ$C) it will happen in millions of years.


Why can there not be more than one sigma bond in a set of bonds? A question on an exam asked why there is exactly one sigma bond in double and triple covalent bonds. I looked in my text and online after the exam, but couldn't find an anawer to the question.

Why can there not be more than one sigma bond in a set of covalent bonds?

Why can there not be more than one sigma bond in a set of bonds?

There can be, even in simple carbon compounds. Bent bonds, tau bonds or banana bonds; whatever you might like to call them were proposed by Linus Pauling; Erich Hückel proposed the alternative $\sigma - \pi$ bonding formalism. Hückel's description is the one commonly seen in introductory texts, but both methods produce equivalent descriptions of the electron distribution in a molecule.

In order to better understand the bent bond model let's first consider its application to cyclopropane and then move to ethylene.

In cyclopropane it has been found that significant electron density lies off the internuclear axis, rather than along the axis.

(image source)

Further, the $\ce{H-C-H}$ angle in cyclopropane has been measured and found to be 114 degrees.  From this, and using Coulson's Theorem $$\ce{1+\lambda^2cos(114)=0}$$ where $\ce{\lambda^2}$ represents the hybridization index of the bond, the $\ce{C-H}$ bonds in cyclopropane can be deduced to be $\ce{sp^{2.46}}$ hybridized.  Now, using the equation $$\ce{\frac{2}{1+\lambda^_{C-H}^2}+\frac{2}{1+\lambda_{C-C}^2}=1}$$ (which says that summing the "s" character in all bonds at a given carbon must total to 1) we find that $\ce{\lambda_{c-c}^2~=~}$$\mathrm{3.74}$, or the $\ce{C-C}$ bond is $\ce{sp^{3.74}}$ hybridized.

Pictorially, the bonds look as follows.  They are bent (hence the strain in cyclopropane) and concentrate their electron density off of the internuclear axis as experimentally observed.

We can apply these same concepts to the description of ethylene. Using the known $\ce{H-C-H}$ bond angle of 117 degrees, Coulson's theorem and assuming that we have one p-orbital on each carbon (Hückel's $\sigma - \pi$ formalism), we would conclude that the carbon orbitals involved in the $\ce{C-H}$ bond are $\ce{sp^{2.2}}$ hybridized and the one carbon orbital involved in the $\ce{C-C}$ sigma bond is $\ce{sp^{1.7}}$ hybridized (plus there is the unhybridized p-orbital). This is the "$\ce{sp^2}$" description we see in most textbooks.

Alternately, if we only change our assumption of one pi bond and one sigma bond between the two carbon atoms to two equivalent sigma bonds (Pauling's bent bond formalism), we would find that the carbon orbitals involved in the $\ce{C-H}$ bond are again $\ce{sp^{2.2}}$ hybridized, but the two equivalent $\ce{C-C}$ sigma bonds are $\ce{sp^{4.3}}$ hybridized. We have constructed a two-membered ring cycloalkane analogue!

Hybridization is just a mathematical construct, another model to help us understand and describe molecular bonding.  As shown in the ethylene example above, we can mix carbon's atomic s- and p-orbitals in different ways to describe ethylene. However, the two different ways we mixed these orbitals

one s-orbital plus two p-orbitals plus one unmixed p-orbital (Hückel's  $\sigma - \pi$), or
two s-orbitals plus 2 p-orbitals (Pauling's bent bond)

must lead to equivalent electronic descriptions of ethylene. As noted in the Wikipedia article (first link at the beginning of this answer),

  There is still some debate as to which of the two representations is
  better,[10] although both models are mathematically equivalent. In a
  1996 review, Kenneth B. Wiberg concluded that "although a conclusive
  statement cannot be made on the basis of the currently available
  information, it seems likely that we can continue to consider the σ/π
  and bent-bond descriptions of ethylene to be equivalent.2 Ian
  Fleming goes further in a 2010 textbook, noting that "the overall
  distribution of electrons [...] is exactly the same" in the two
  models.[11]

The bent bond model has the advantage of explaining both cyclopropane (and other strained molecules) as well as olefins.  The strain in cyclopropane and ethylene (e.g. heats of hydrogenation) also make intuitive sense with the bent bond model where the term "bent bond" conjures up an image of strain.

So, bent bonds are a mix of both sigma and pi properties and multiple bent bonds  can be used to describe the bonding between adjacent atoms.


Is UV-Vis absorption polarization dependent ? Why? Consider molecule of interest be dissolved in a solvent (and this molecule has absorption in the visible region). We are using a broadband visible light source say tungsten lamp.  This light passed through a polarizer and then split into two beams, one passing through sample cell, and other through reference cell having solvent.

Would there be difference in the electronic absorption spectra when measured keeping the polarizer parallel or perpendicular with respect to the incident polarization ?  Also, no external electric field is present and we have randomly oriented molecules.

(Consider the transmission of optics and detector are corrected for this polarization difference)

With the experimental setup you describe, you would not observe any intentional polarization dependence, because the light passes only through a beam splitter, not a polarizer.

There can be unintentional polarization due to the nature of optical materials and manufacturing, caused by

  the diffraction grating,
  the polarization contribution of mirror reflections, and
  the diffraction grating slits.

With that out of the way, UV/Vis absorption is polarization dependent. In theory, polarization can occur in any spectroscopic technique where the polarization of the incident EM radiation can be controlled. To quote the Wikipedia article,

  In general, any kind of anisotropy in the media results in some sort of change in polarization. Such an anisotropy can be either inherent to the media (e.g., in the case of a crystal substance), or imposed externally (e.g., in the presence of magnetic field in plasma).

Assuming that no other external fields are applied other than your probe beam and your bulk phase has no anisotropy, any differential absorption of polarized light must come from the molecules themselves, namely those that are "optically active". Linear dichroism and particularly circular dichroism are used to describe the differential absorption of linearly and circularly polarized light, respectively. Electronic (UV/Vis) CD is widely applied in the determination of protein secondary structure and the absolute configuration of stereocenters in chiral molecules. An ECD spectrum is usually presented as a difference spectrum between the left- and right-polarized spectra, which themselves are no different from a typical UV/Vis spectrum.

The parent dichroism article is a good starting point for learning about these spectroscopies.


Which is more likely to form: silver chloride or silver phosphate? If silver nitrate is added to an aqueous solution (such as tap water) that contains both chloride ions and phosphate ions, which precipitate is more likely to form: silver chloride, silver phosphate or both should form?
And how do you know?

This is NOT a homework question. I'm 32 and learning chemistry on my own as a hobby and doing experiments using a Chemistry Kit for kids.
The reason I'm asking the question is because I did the experiment on the tap water at home and I didn't get any silver phosphate precipitate, but I did get a milky white appearance which I'm assuming is silver chloride (or another halide). I'm wondering if I got no silver phosphate precipitates because all the silver ions were "used up" by the chloride ions or simply because the concentration of phosphate ions is either zero or too small.

I tried to figure out if silver chloride formation takes precedence (so to speak) over forming silver phosphates using enthalpies of formation, but I got lost with that a bit.

The key concept here is solubility products which are a form of chemical equilibria. Once you have understood that, you will realise that it depends on the ions’ concentrations and relative solubilities which one will precipitate first.

In general, you have to consider the following principal equilibria:

$$\ce{Ag3PO4 v + 3 Cl- (aq) <=>[$k_1$] 3 Ag+ (aq) + 3 Cl- (aq) + PO4^3- <=>[$k_2$] 3 AgCl v + PO4^3-}\tag{1}$$

and futher ones that derive from the acidity equilibrium of the phosphate anions as well as the formation of silver hydroxide at higher $\mathrm{pH}$ values. I will choose to ignore these.

Equilibrium $k_1$ is actually only the precipitation of silver phosphate; we might instead write it as:

$$\ce{3 Ag+ (aq) + PO4^3- (aq) <=>[$k_1$] Ag3PO4 v}\tag{2}$$

The equilibrium constant is thus:

$$K_1 = \frac{a(\ce{Ag3PO4})}{a(\ce{Ag+})^3 \times a(\ce{PO4^3-})}\tag{3}$$

We are dealing with activities here; the activity of a solid is, by definition, always $1$.
This equilibrium is usually given in the inverse form in which the dissolved ions are the product and the solid salt is the reactant; in that case $K_1{}^{-1}$ is what you might know as the solubility product.

The same discussion can be made for the other equilibrium:

$$\begin{gather}\ce{Ag+ (aq) + Cl- (aq) <=> AgCl v}\tag{4}\\[1em]
K_2 = \frac{a(\ce{AgCl})}{a(\ce{Ag+}) \times a(\ce{Cl-})}\tag{5}\end{gather}$$

Again, the more common form is $K_2{}^{-1}$ which is also known as the solubility constant.

Knowing the values of both $K$ as well as the initial concentrations of all ions will allow you to calculate in which direction the reaction goes although the calculation may be tedious.


Visualizing atoms in XYZ file with colors representing partial charges I have a file containing the atomic coordinates (in XYZ format) of a structure, and I also have a list of the partial atomic charges for each atom. I'd like to create an image of the structure where the colors of the atoms correspond to the computed charges. What are some good ways for doing this?

As far as I know, no software can do this right away but with some work, this should be possible.

Use CYLview to create a povray file, change the atoms to the desired color and render it with Povray. The problem here is that the coordinates of the xyz file and the povray are not the same. But as far as I can tell the numbering stays the same. So if the 3rd atom in your xyz is this:

N       -1.227679000      0.558365000      0.000000000

the 3rd sphere in the povray file would look like that:

sphere {<-1.9762436, 0.9491377, 0.0000000> 0.2812500
   pigment{color rgbt <0.4,0.4,1.0,0.00>}
   finish{F_normal}

All you need to do is change the color here. It should also be pretty easy to automate this color change with some lines of code.

So you can go from this

to for example this

by simply changing the color value shown above to

pigment{color rgbt <0.2,0.8,1.0,0.00>}


Which are the radicals containing phosphorus in ATP? The ATP molecule contains 3 $\ce P$ atoms linked through esteric bonds. The IUPAC name of triphosphate adenosine is "[[(2R,3S,4R,5R)-5-(6-aminopurin-9-yl)-3,4-dihydroxyoxolan-2-yl]methoxy-hydroxyphosphoryl] phosphono hydrogen phosphate" as PubChem states. This name presents 3 groups containing phosphorus in the end: hidroxyphosphoryl, phosphono and hydrogen phosphate. I'm not sure about how are these groups placed in the molecule because I can't seem to find the hydroxyphosphoryl. To ilustrate what I mean, I've drawn it. This is how I found the other groups, but I'm unsure it's the right way to look at the end of the molecule. In blue: methoxy, In green: hydrogen phosphate, In red: phosphono. The other black part does not match the hydroxyphosphoryl group.

You have almost answered yourself completely.

'Phosphoryl’ in current IUPAC Nomenclature of organic chemistry, preferred names (2013):

  P-67.1.4.1.1.2 Fundamental acyl groups for substituent derived from mononuclear noncarbon oxoacids

  (…)

  $\ce{-P(O)<}$
  phosphoryl

Example compound with “hydroxyphosphoryl” group:

$\ce{C2H5-P(O)(OH)-CH2COOH}$
[ethyl(hydroxy)phosphoryl]acetic acid

(software generated, in the book there is an example without hydroxy parens: O-({[(2R)-2,3-bis(octadecanoyloxy)propoxy]hydroxyphosphoryl})-L-serine)

‘Phosphono´:

  P-67.1.4.1.1.1 Retained names for substituent groups derived from mononuclear noncarbon oxoacids

  (…)

  $\ce{-P(O)(OH)2}$
  phosphono (preselected prefix)

(There's no “phosphonyl” if someone wonders, there's however different group with similar name, phosphonoyl $\ce{-PH(O)\bond{-}}$.)

Example:

$\ce{(HO)2P(O)-CH2COOH}$
phosphonoacetic acid

And, I think that the correct layout for the name provided should be rather:
{[(2R,3S,4R,5R)-5-(6-aminopurin-9-yl)-3,4-dihydroxyoxolan-2-yl]methoxy}(hydroxy)phosphoryl phosphono hydrogen phosphate

This name is somewhat overcomplicated. ‘Triphosphate’-based name could be used, see

  P-67 Mononuclear and polynuclear noncarbon acids and their functional replacement analogues as functional parents for naming organic compounds

  (…)

  P-67.2 Di- and polynuclear noncarbon oxoacids

  (…)

  P-67.2.1 Preselected names

  (…)

  $\ce{(HO)2P(O)-O-P(O)(OH)-O-P(O)(OH)2}$
  triphosporic acid (preselected name)

and

  P-106 Nucleotides

  (…)

  P-106.2 Nucleotide diphosphates and triphosphates

  Diphosphate, triphosphate, etc. esters of nucleosides are named by citing a phrase such as diphosphate, after the name of the nucleoside. The presence of hydrogen atoms on the diphosphate, triphosphate, etc. component of the molecule is indicated by the words ‘hydrogen’, ‘dihydrogen’, etc. Parentheses are used to avoid ambiguity.

  Examples:

  uridine 5’-(tetrahydrogen triphosphate)

(but this is natural compounds section, which is more permissive)

Seemingly apparent reason why ‘triphosphate’ is currently not used in software generated systematic names is that there's an ambiguity in binding the residue to the first (last), or the middle phosphorus' oxygen (positional isomers), and no numbering system for non-carbonic acids is defined.

However it is defined (P-67.2.2.1), with esters examples (P-67.2.5.2). The numbering is from-to central atoms including the oxygen connectors, so the terminal phosphorus atoms get numbers 1, 3, 5; so (generalized) names for the two isomeric alkyl triphosphates (alkyl on terminal or middle phosphate unit, respectively), are

O1-alkyl tetrahydrogen triphosphate
O3-alkyl tetrahydrogen triphosphate

which might be confusing, especially in the latter case. So this can be the main reason, why these are not used.

Let's use it anyway (with a minor improvement – indicated hydrogen):
O1-[(2R,3S,4R,5R)-5-(6-amino-9H-purin-9-yl)-3,4-dihydroxyoxolan-2-yl] tetrahydrogen triphosphate.

It should be OK to omit ‘tetrahydrogen’, as phosphates are ionized in physiological conditions.

(Not speaking of yet better names like adenosine 5’-triphosphate or ATP.)


Why does substance that sublime not come to liquid state? A substance that sublimates doesn't come to a liquid state at room temperature. But when we melt the substance it melts! Why does it skip the liquid state at room temperature?

From a thermodynamic point of view there are two competing possibilities for condensation of a vapor:

(1) vapor $\ce{->}$ liquid, with some negative $\Delta H$

(2) vapor $\ce{->}$ solid, with an absolutely larger negative $\Delta H$

In the absence of large quantum mechanical effects (see here), the second process with a more negative enthalpy change must become favorable at a low enough temperature above absolute zero; the actual temperature where this happens  depends on the material.  So when the condensation is done at a low enough temperature, it goes directly from vapor to solid, and the reverse process of evaporation then also misses the liquid.


What is the "smell" of solid sulfur A solid piece of sulfur has a characteristic smell.

Is the smell of solid sulfur the result of the release of compounds like sulfur dioxide or hydrogen sulfide from a reaction with air or moisture, or is it from elemental sulfur particles that are being released by sublimation?

If it is from pure Sulfur particles (at least in part), what form is it released into the air, ie $\ce{S8}$ or $\ce{S2}$ or something else?

The primary candidates for volatilization of odorous sulfur gases are $\ce{SO2}$ and $\ce{H2S}$.  Unless the elemental sulfur has been physically ground and dispersed into the air it is not likely the direct source of the smell.

According to the International Volcanic Health Hazard Network, who of course study gaseous volcanic emissions with respect to safety issues:

  "Sulfur dioxide ($\ce{SO2}$) is a colourless gas with a
  characteristic and irritating smell. This odour is perceptible at
  different levels depending on the individual's sensitivity, but is
  generally perceived between 0.3-1.4 ppm and is easily noticeable at 3
  ppm."

Regarding $\ce{H2S}$, according to the US Occupational Safety and Health Administration (OSHA), $\ce{H2S}$ has a perceptible smell at between 0.01 and 1.5 ppm.

The two gases have different smells, $\ce{H2S}$ being that of rotten eggs and that of $\ce{SO2}$ being described as "pungent and irritating".  So that would be one way to differentiate which gas you are smelling.

The most likely candidate is $\ce{SO2}$ as our air is generally oxidizing, and to me elemental sulfur smells more like $\ce{SO2}$ than rotten eggs.


Building a solid-liquid interface I am trying to build a solid-liquid interface to be simulate using molecular dynamics. Since, I would be using ab initio molecular dynamics (cpmd or cp2k), I need the minimum cell dimensions. Please suggest some ways to fix the cell dimensions.

The issues that I am concerned about are given in for example in the science paper (http://science.sciencemag.org/content/303/5658/658.full). It is said that there should be a minimum size of the system (specially in the direction perpendicular to the slab) to attain trustworthy results. I am not able to find how and why? For example, if I take 3-10 layers of atoms for making the surface how to determine which one would yield the trustworthy results?

First, when it comes to periodic boundary conditions, you should make sure that your system does not interact with itself through the boundaries.
A way to check it is to see if the electron density is low in vacuum (in case your system has some void separating surfaces).

Energies should also be checked to be converged with respect to cutoff energies, $k$ point selection, etc..
This means that you have attained the "maximum performance" you could; e.g. a well-converged result means that improving parameters do not significantly changes results.

Furthermore, convergence with respect to slab thickness is also important: as you increase slab thickness, the properties of the inner layer should approach the properties of the bulk.
How fast this happens depends on your system.

A good reference for all this is this one.


Why is water instead of hydroxide ions used to balance oxygen atoms in redox half equations? When balancing half equations of redox reactions, $\ce{H2O}$ is used to balance any oxygen atoms? For example, this half equation

$$\ce{NO3- → NH4+}$$

is balanced into:

$$\ce{NO3- + 8 e- + 10 H+ → NH4+ + 3 H2O}$$

Why is $\ce{OH-}$ not used instead of $\ce{H2O}$? E.g. the balanced equation becomes

$$\ce{NO3- + 8 e- + 7 H+ → NH4+ + 3 OH-}$$

It's not that we really have a choice in this case between water or hydroxide or pick one or another arbitrarily to fit the stoichiometry of the whole redox reaction. The redox reaction is obviously taking place in quite a strong acidic medium (makes sense, e.g. if $\ce{Zn}$ is used as reducing agent) as you supply protons $\ce{H+}$ to the nitrate according to your reaction.

In fact, 10 protons for each reduced nitrate group is needed; and even if you propose hydroxide ion formation, you'll still need 7 protons (second reaction). Either way, there is no chance for $\ce{OH-}$ to coexist with $\ce{H+}$ in acidic medium and equilibrium will be shifted towards water formation instead.

Note that it doesn't mean that reduction of nitrate is forbidden if hydroxide ions are formed. For example, you can use another reducing agent such as $\ce{Al}$ and do perform the reduction in basic media instead. In this case you are going to end up with hydroxide ions all right:

$$
\begin{align}
\ce{Al + 6 OH- &→ [Al(OH)6]^3- + 3 e-} &|\cdot 8 \tag{ox}\\
\ce{NO3- + 8 e- + 6 H2O &→ NH3 + 9 OH-}&|\cdot 3 \tag{red}\\
\hline
\ce{8 Al + 3 NO3- + 21 OH- + 18 H2O &→ 8 [Al(OH)6]^3- + 3 NH3} \tag{redox}
\end{align}
$$

Just as with the formation of human personality, environment also plays gigantic role in chemistry.


Which reactant to use to calculate yield? A chemist mixes 24.0 g $\ce{H2}$ with 8.0 g $\ce{N2}$. Assuming the reaction goes to completion, which reactant should she use to calculate the yield?

$$\ce{3H2 + N2 -> 2NH3}$$
I believe it should be nitrogen. With 24 g of $\ce{H2}$, we have 11.9 mol of $\ce{H2}$. With 8 g of $\ce{N2}$, we have 0.3 mol of $\ce{N2}$. So, when I express the moles of $\ce{N2}$ in terms of $\ce{H2}$, I see that I only need 0.9 moles of $\ce{H2}$ – which means that $\ce{N2}$ is the limiting reacting and then used to calculate the theoretical yield, right?

You have the right thought process - you need to find which is the limiting reagent.
Your calculation is correct, $N_2$ is limiting, so it can be used to calculate yield. Looks like we weren't really needed!


Why is sodium potassium tartarate used in Fehling's solution? Why is this salt used specifically in Fehling's solution, instead of a simple disodium or dipotassium salt of tartaric acid?

Fehlings soluition is alkaline. Mixing cupric ions with an alkaline solution would result in formation of cupric hydroxide of very low water solubility. By adding tartrate ions, the cupric ions will form a complex with the cupric ions which will keep them dissolved in the alkaline solution. The type of cation of the tartaric acid is of no importance other than for solubility reasons.

The same principle is used in Benedicts´s solution, where tartaric acid is replaced by citric acid or a salt thereoff.


What is the mass of 1 Liter of ethylene at 2000 bar pressure? I was wondering the weight of 1 liter of ethylene at 2000 bar and 310 Kelvin?

I have tried to just multiply the weight at 1 bar with 2000 but I don't think i got the correct answer because the answer I got was just below 2.4 kg per liter which seems way too high.

I calculated it with constant temperature.

If we treat it as an ideal gas, then
\begin{aligned}
  PV &= n\mathcal{R}T \\ \mathcal{R} &=0.08314~\mathrm{L~bar~K^{-1}~mol^{-1}}\\
  n &=\frac{PV}{\mathcal{R}T}=\frac{2000\cdot1}{0.08314\cdot310}=77.6~\mathrm{mol}\\
  M(\ce{ethylene}) &=28~\mathrm{g~mol^{-1}}\\
  \text{mass of ethylene} &= (28\cdot77.6) = 2.17~\mathrm{kg}
\end{aligned}

Edit:
 Using the link[a] in the comment I find that the density of ethylene at $310~\mathrm{K}$ and $1800~\mathrm{bar}$ is $\varrho = 20.385~\mathrm{mol/L}$, and $\varrho = 21.065~\mathrm{mol/L}$ at $2200~\mathrm{bar}$.  Averaging these two numbers gives an estimated density of $\varrho = 20.725~\mathrm{mol/L}$ at $310~\mathrm{K}$ and $2000~\mathrm{bar}$.  So 1 liter of ethylene under these conditions would contain
\begin{aligned}
  \varrho\cdot M_{\ce{C2H4}}\cdot V & ~= m_{\ce{C2H4}}\\
  20.725~\mathrm{\frac{mol}{L}} \cdot 28~\mathrm{\frac{g}{mol}}\cdot 1~\mathrm{L} & ~\approx 580~\mathrm{g}.\\
\end{aligned}
  Is that more in line with your expectations?

[a] Majid Jahangiri, Richard T Jacobsen, Richard B. Stewart and Robert D. McCarty: Thermodynamic Properties of Ethylene from the Freezing Line to 450 K at Pressures to 260 MPa. J. Phys. Chem. Ref. Data 1986, 15 (2), 593. (DOI:10.1063/1.555753) (mirror at www.nist.gov)


Why does atomic charge not balance in nuclear reactions? In the discovery of neutron, the bombardment of beryllium with alpha particles is well known.

$$\ce{^9_4Be + ^4_2He^2+ -> ^12_6C + ^0_1n}$$

I don't understand how the electrons add up. As alpha particles do not have electrons, the total electrons should be 4 accounting for the electrons of beryllium, but after bombardment, there is a deficiency of 2 electrons in carbon atom. Shouldn't the $\ce{^12_6C}$ atom be positively charged? Like:

$$\ce{^9_4Be + ^4_2He^2+ -> ^12_6C^2+ + ^0_1n}$$

There are two reasons why nuclear reactions don't consider (or often account for) charge.

The first is that charge is usually irrelevant to the reactions. If we are looking at changes in nuclei we can for all practical purposes ignore the electrons that would be associated with the nuclei under normal conditions. They are just irrelevant to the nuclear reaction taking place.

The second is that they do account for charge, but only in nuclear processes. Some radioactivity (isolated neutrons decay or any nucleus undergoing beta emission) can lead to the decay of neutrons into a proton and electron and some other minor particles. The electron clouds that would normally surround the nucleus are still irrelevant to these processes but the equation for the nuclear reaction should account for the emitted particles. When writing these equations only the nuclei (and the emitted particles) are accounted for and any net change of charge is ignored as a description of the nuclei involved implicitly describes the net nuclear charge (in the same way that many organic chemical structures don't explicitly draw every hydrogen as chemists can fill them in knowing that the total valence around carbon should add to 4).


Temperature Dependence of The Conductivity of Electrolytic Solutions I have learned in my chemistry text books that the conductivity of electrolytic solutions increase with the increasing of temperature because the ions of electrolytic solution move faster by getting more thermal energy. But a question arose in my mind if there exist any mathematical equation which can describe the conductivity of electrolytic solutions as a function of temperature. I searched about it on internet but I got nothing satisfactory. Does any equation exist which can model the electrolytic conductivity as a function of temperature?

One useful approximate relationship that can be used to predict the temperature dependence of the conductivity can be obtained by relating the electrical mobility to the viscosity.
For a $z_+:z_-$ electrolyte the limiting molar conductivity can be expressed as
$$\Lambda^\circ _m =  F^2\left(\frac{z_+^2\nu_+}{f_+} + \frac{z_-^2\nu_-}{f_-}  \right)$$
where the $f_i$ are frictional coefficients  for each ion, z is the ionic charge, $\nu$ is the stoichiometric coefficient, and F is the Faraday constant (for instance, for $\ce{Al2O3}$, $z_+=3$,$\nu_+=2$,$z_-=2$,and $\nu_-=3$.
The frictional coefficients can be related to the viscosity, for instance using Stokes' relation $f=6\pi \eta r$ for a sphere being driven through a uniform viscous medium with a low Reynold's number. From the temperature dependence of the viscosity, and assuming the ion radii are independent of T, you can derive a relation for the T dependence of the limiting conductivity:
$$\Lambda^\circ _m = \frac{F^2}{6\pi\eta (T)}\left(\frac{z_+^2\nu_+}{r_+} + \frac{z_-^2\nu_-}{r_-} \right)$$
The equations so far apply to dilute solutions where the viscosity is independent of the concentration of electrolyte and ion-ion interactions can be neglected.
Note that even simpler theories predict a somewhat complex dependence of the molar conductivity on temperature. One can however attempt to fit an expression inspired by those above, of the sort
$$\Lambda _m = A \times \eta (T)^{-1} $$
where A is assumed independent of T. This might work over a narrow T range and could be used to assess the importance of ion-ion interactions.


Corrosion of a galvanic couple made with silver and gold If a bangle is made out of gold and silver, connected with each-other would there be corrosion happening?
If so, can it be explained using the galvanic series?

Also do those metals undergo oxidation under normal conditions?

As you state, though silver is not very reactive, silver jewelry would corrode more rapidly when in contact with gold and a bit of sweat as electrolyte. Though the difference in electronegativity of pure gold and pure silver is 0.15V, and is considered acceptably corrosion resistant, commercial jewelry contains copper and other metals in the alloys. There is clearly evident corrosion in photos of copper/silver jewelry. According to the Victoria and Albert Museum staff, "All metals, with the exception of 24 carat gold, suffer the effects of corrosion."

However, in a bangle, physical abrasion is likely to cause more damage and loss of material than galvanic corrosion. Wear the jewelry and enjoy it, but it will not last forever.


How to draw glucose using a Fischer projection? I asked how to draw glucose as a Fischer projection from its given chair conformation but all I got was an unsatisfactory ad hoc explanation involving rotation of my neck muscles to somehow see where the hydroxyl groups were. How does one actually draw a Fischer projection of a sugar molecule from its chair conformation?

Here is a web page that shows how to interpret the position from the structure.
The UP groups go to the left side of the backbone and the DOWN groups go to the right side.


Why does methyl not shift in the mechanism of isoborneol to camphene? After the formation of carbocation, I was confused why would the reaction not go in favor of forming bornylene. I am still not sure why but I think that $\mathrm{sp^2}$ hybridized carbon will cause a lot of strain in the ring. Another thing confuses me, why does methyl not shift in place of the ring itself breaking? Is this anywhere similar to Bredt's rule? Please help me with these two doubts.

This is a class of carbocation 1,2-rearrangement reactions, and historically named Wagner–Meerwein rearrangement after two chemists who discovered the reaction (see following diagram):

Although it can be explained considering as an rearrangement of a classical carbocation, there are a tremendous amount of work has been done considering 2-norbornene cation as non-classical carbocation (see comment of  orthocresol elsewhere). However, my interned task is to explain why it is not possible to have 1,2-methide shift instead of alkyl shift (Wagner–Meerwein rearrangement) considering classical carbocation.

First, why the reaction not go in favor of forming bornylene? Since the reaction is in acid medium, fast equlibrium of protonation (to give initial 2-norbonyl carbocation derivative) and deprotonation (to give bornylene derivative) could be expected until most thermodynamically stable product is formed. This can be expected from a $2^\circ$-carbocation, if there is a possibility of rearrangement to a more stable $3^\circ$-carbocation.

Now, the initial 2-norbonyl carbocation derivative is $2^\circ$-carbocation with a possibility of rearranging to a more stable $3^\circ$-carbocation by either 1,2-methide shift or alkyl shift (which is the one eventually happens). If you look at initial carbocation without the bridge-head, the bottom cyclohexyl part is in boat confirmation, which is locked in by the rigid bridge-head carbon. Now that restricts the 1,2-methide migration, which would made energetically unfavored triangularpyramidal $3^\circ$-carbocation. Thus, the readily available alkyl migration is energetically favored and fast. The rearanged crbocation would readily dehyrogenate to give camphene (2,2-dimethyl-3-methylidenebicyclo[2.2.1]heptane).

Also keep in mind that not only it is not favored by the steric resons, but also 1,2-hydride shift would generate a $3^\circ$-carbocation, which would have violate Bredt's rule.

Note: Nonetheless, a recent publication has discussed this scenario and 1,3-hydride migration to give 7,7-dimethyl-2-methylidenebicyclo[2.2.1]heptane as elimination product (Ref.1).

References:

Huidong  Zheng, Yi Lin, Mengchan Wang, Jie Liu, Dan Wu, Jingjing Chen, Guanghua Yin, S. Ted Oyama, Suying Zhao, “The influence of solvent polarity on the dehydrogenation of isoborneol over a $\ce{Cu/ZnO/Al2O3}$  catalyst,” Catalysis Today 2019, 323, 44-53 (https://doi.org/10.1016/j.cattod.2018.10.021).


Relationship between surface area of electrode and reaction rate of hydrogen in fuel cells I am looking at the effects of increasing the surface area of which the flowing hydrogen fuel is in contact with the anode in a fuel cell.

At the anode the hydrogen is catalytically split into protons and electrons. The catalyst used is most commonly platinum. So, for a fixed flow rate of hydrogen, increasing the contact surface area with the anode will increase the number of hydrogen atoms in contact with the catalyst, and the duration that they are in contact with it. Therefore the number of oxidization half cell reactions will increase. But what is the mathematical relationship that governs the number of reactions that will occur? Is it just a linear relationship, i.e doubling the surface area doubles the number of reactions, or is it more complex?

If it is impossible to get an exact mathematical relationship, what is a good approximation on how these variables will scale with each other? Also, how do chemist usually predict the number of reactions/reaction rate, and could this be applied to my example?

Any galvanic cell produces its equilibrium voltage at open-circuit.  Let's say your cell is at +3 V before you allow any current to flow through it.  As soon as you throw the switch, you allow that 3 V to push a current through the cell and some external load that you connect to it. There are several resistances within the cell that act to reduce the cell voltage from the initial 3 V.  The electrolyte resistance, concentration dependent resistances at the electrodes and electron transfer resistances all contribute to the total voltage drop in the cell, and hence, effectively limit the amount of current it can generate.  Typically, one of these resistances is much larger than any of the others depending on the exact configuration and operating conditions of the cell.  It is the electron transfer resistance that you are concerned with in your question.  The traditional way of expressing the voltage drop due to electron transfer resistance is the Tafel equation:

$$\eta _{a}=a+b\, log\, i$$

where $a$ and $b$ are empirical constants, $i$ is the current density and $\eta _{a}$ is what is called the activation overpotential.  We see from this expression that the amount of voltage drop in the cell due to electron transfer increases linearly with the logarithm of the current density.  It is desirable then to minimize $\eta _{a}$ for a battery or flow cell application so that we have the maximum amount of voltage available for pushing current through the load.  A very good way to do this, as you pointed out, is to simply increase the electrode area.  By doing so, we decrease the current density which decreases $\eta _{a}$.

So, the exact mathematical relationship you seek would be the empirical Tafel expression given above, or even better, the Butler-Volmer equation which is really just a more fundamental way of expressing the Tafel relationship.  All you should need to do is locate some references on platinum polarization in aqueous solutions.  There just happen to be tons of papers on this subject.  Here is one to get you started:

  Sheng W. et al, Journal of The Electrochemical Society, 157 11
  B1529-B1536 2010


Arrange these compounds: CO2, CH3OH, RbF, CH3Br in order of increasing boiling points Arrange these compounds: $\ce{CO2}$, $\ce{CH3OH}$, $\ce{RbF}$, $\ce{CH3Br}$ in order of increasing boiling points.

I think I should consider the forces between them, that is:

$\ce{CO2}$: dispersion forces
$\ce{RbF}$: dispersion and  ionic forces
$\ce{CH3OH}$: Dipole-dipole interactions, Hydrogen bonding and dispersion forces
$\ce{CH3Br}$: Dipole-dipole interactions and dispersion forces

It is obvious that $\ce{CO2}$ is the smallest one and because $\ce{CH3OH}$ stronger than $\ce{CH3Br}$ it will have higher boiling point

But how to arrange the rest? Or how to compare ionic forces with Dipole?

You know $\ce{CO_2}$ is gaseous at room temperature, so let's put that at the bottom. Methanol  forms hydrogen bonds, so that will be above bromomethane which does not. At last we have rubidium fluoride which is a salt. Salts generally have a very high boiling point (> 1000 °C, much higher than molecular structures) because of the ionic (electrostatic) interaction between the ions, so that one will be at the top. Ionic forces can be seen as extreme dipoles in a certain way, there is a grey area when electronegativity becomes large enough, that it can be seen either as a molecular structure or ionic structure. Consulting online information about the boiling points of these compounds (i.e. just check Wikipedia or some MSDS site) confirms the theory.


Stability comparison of benzylic carbocation and benzene Which is more stable - benzylic carbocation or benzene - and why?

Benzylic carbocation may have more resonance structure, but these structures are destroying its aromaticity as well. So, how do we compare its stability with benzene?

First refer to https://en.m.wikipedia.org/wiki/Chemical_stability to clean up the situation.

There is not much in comparing the stability of individual different molecules which are not among the reactants or the product of a specific reaction or in contact to a specific and same environment.

The question arises from a common confusion for we are used to compare the "stability" of different limiting forms (resonance structures ) of a hybrid of resonance.

Or, differently, but still analyzing entities with the same constitutional formula, the stability of various conformers.

An analysis in the terms meant by the question could be done to compare the stability of different carbocations, for instance, and see if there could be mesomeric effects able to delocalise the charge. In such a sense the presence of the phenyl ring has a stabilising effect as compared to, e.g., a linear alkyl chain.

This state, but anyway answer the question, the number of resonance structure does not even count as for all those of the benzyl cation are to be supposed much less stable  than those of benzene, the carbonation is obviously charged so this will be the main difference anyway.

You won't ask what is more stable between methane and ethyl carbocation. The fact that in benzyl cation there is a phenyl ring can only tell you that its formation under a specific mechanism will be favoured as compared to the case of ethane.


How to decide whether to pick 4-bromo-2-methoxyphenol or 4-bromo-6-methoxyphenol as the name for the compound? Why is the following compound named as 4-bromo-2-methoxyphenol and not 4-bromo-6-methoxyphenol?
Shouldn't the substituents be numbered according to their alphabetical order?

When choosing which name to use when the only difference is the numbering, we choose the combination that gives the lowest possible numbers, 2 and 4 in this case.
So the name will be 4-bromo-2-methoxyphenol. The substituents are given in alphabetical order.


Are there any molecular modeling programs that can build materials made up of three or more elements into a cylindrical shape? I would like to build a material similar to what is in the image below.

I have tried programs such as Material Studio, Quantum ATK, and VNL, but most of them can only do materials with 2 elements.

Various custom software exists to create nanotubes.

The following appears suitable:
http://www.jcrystal.com/products/wincnt/

A Wolfram Demonstrations application may also be of some use:
https://demonstrations.wolfram.com/NanotubeBuilder/

Furthermore, you may have further considerations like how to generate molecular mechanics parameters. For example, you may be interested in how to generate the nanotubes to be compatible with AMBER, or GROMACS. The following may be of use if that is the case:
https://github.com/bio-phys/cnt-gaff
http://www.gromacs.org/Documentation/How-tos/Carbon_Nanotube

I should also mention that VMD and Avogadro, which are more easy to use, can create carbon nanotubes. VMD can create boron-nitride nanotubes
http://www.ks.uiuc.edu/Research/vmd/plugins/nanotube/


What happens to the complex part of the wave function when doing LCAO? I understand that in general the wave function can take complex numbers, however when talking about combining atomic orbitals to form molecular orbitals we talk about the phase of the orbital being positive or negative to create constructive or destructive interference. It seems as if we have lost the aspect of the wave function taking on complex phase, what happened to it?

I understand that in general the wave function can take complex numbers [...]

It is not that it can, rather it should: by the very postulates of quantum mechanics wave function is a complex-valued function. So, both the spin orbitals (that are one-electron wave functions) and the Slater determinant built out of them are in principle complex-valued functions.

However, in solving the HF equations by the SCF procedure it is quite typical to impose some constraints on the spin orbitals, for instance, restrict them to be real-valued functions rather then complex-valued ones. As with any other constraint this restriction might lead to what is called SCF instabilities, i.e. situations when relaxing the constraints lead to a different variational solution of lower energy.

To elaborate a bit more let us mention few other well-known (as well as lesser-known) constraints for the SCF procedure:

The infamous restricted Hartree-Fock (RHF) model with the requirement that spin orbitals come in pairs: two spin orbitals corresponding to two different pure spin states are constructed from the same spatial orbital:
$$
\begin{aligned}
  \psi_{2i-1}(1) &= \phi_{i}(1) \alpha(1) \, , \\
  \psi_{2i}(1)   &= \phi_{i}(1) \beta(1)  \, .
\end{aligned}
$$
In the unrestricted Hartree-Fock (UHF) model the requirement above is relaxed and we exclusively use spatial orbitals from one set to construct $\alpha$ spin orbitals and spatial orbitals from another set to construct $\beta$ spin orbitals:
$$
\begin{aligned}
  \psi_{2i-1}(1) &= \phi_{i}^{\alpha}(1) \alpha(1) \, , \\
  \psi_{2i}(1)   &= \phi_{i}^{\beta}(1) \beta(1)   \, .
\end{aligned}
$$
A bit less known is the fact that the unrestricted Hartree-Fock method is not so unrestricted. In fact, there is a constraint here: each and every spin orbital describes an electron in a pure spin state, either $\alpha$ or $\beta$, while in general (in accordance with the postulates of quantum mechanics) electron can be in superposition of this states:
$$
  \psi_{i}(1) = \phi_{i}^{\alpha}(1) \alpha(1) + \phi_{i}^{\beta}(1) \beta(1)
$$
This most general setting is known as the general Hartree-Fock (GHF) method.

And all these three methods (RHF, UHF, GHF) exist in two variants: the real one, in which spin orbitals additionally required to be real-valued functions, and the more general complex one. All this give rise to six variants of the HF method with many possible instabilities between them discussed it some details in the seminal paper by Schlegel & McDouall1:

There is also a very similar earlier paper by Seeger & Pople.2 which suddenly is not freely available.

Its important to realize that any constraint of the mentioned above can only raise the electronic energy: with a great deal of certainty, we may expect that if any constraint is relaxed, the variational procedure will result in a lower energy due to greater variational freedom. Thus, for instance, for the most usual different real variants of the Hartree-Fock method we have
$$
    E_\mathrm{e}(\mathrm{RGHF}) \leq E_\mathrm{e}(\mathrm{RUHF}) \leq E_\mathrm{e}(\mathrm{RRHF}) \, .
$$
The same, of course, is true for the corresponding complex variants of these methods,
$$
    E_\mathrm{e}(\mathrm{CGHF}) \leq E_\mathrm{e}(\mathrm{CUHF}) \leq E_\mathrm{e}(\mathrm{CRHF}) \, .
$$
And for any of the three formalism
$$
    E_\mathrm{e}(\mathrm{C}x\mathrm{HF}) \leq E_\mathrm{e}(\mathrm{R}x\mathrm{HF}) \, , \quad \text{where} \quad x = \mathrm{G}, \mathrm{U}, \mathrm{R} \, .
$$

RHF/UHF/GHF and correct spin symmetry

The possible rise of electronic energy that accompanies introduction of more and more constraints in the GHF -> UHF -> RHF sequence seems to be quite contrary to the goal of the variation method. However, UHF and RHF constraints are nothing but symmetry constraints: they arise by requiring an approximate electronic wave function to have the same spin symmetry as the exact non-relativistic one, i.e. to be an eigenfunction of spin operators, the total spin-squared operator $\hat{S}^2$ and the $z$-component of the total spin operator $\hat{S}_{z}$.

With respect to $\hat{S}_{z}$ operator it is well known that any Slater determinant built out of spin orbitals corresponding to pure spin states is an eigenfunction of $\hat{S}_{z}$. Thus, RHF and UHF wave functions are eigenfunctions of $\hat{S}_{z}$, but GHF wave function is not.
The situation with $\hat{S}^{2}$ is a bit more complicated, but all in all only in RHF (but not in UHF or in GHF) formalism it is possible to construct an approximate electronic wave function which is an eigenfunction of $\hat{S}^{2}$.3

To conclude, as it was pointed out by Löwdin, we always face with a dilemma here: should we seek a solution that is a true variational minimum or should we seek a solution with the correct spin symmetry?

1) H. B. Schlegel and J. J. W. McDouall, Do You Have SCF Stability and Convergence Problems? in Computational Advances in Organic Chemistry: Molecular Structure and Reactivity, Springer Netherlands, 1991, pp. 167-185. DOI: 10.1007/978-94-011-3262-6_2. Free PDF from wayne.edu.

2) Seeger, R., & Pople, J. A., Self‐consistent molecular orbital methods. XVIII. Constraints and stability in Hartree–Fock theory. The Journal of Chemical Physics, 66(7), 1977, 3045-3050. DOI: 10.1063/1.434318.

3) This can be done even for an open-shell electron configuration, although it would require a linear combination of Slater determinants with constant coefficients, none of which alone is an eigenfunction of $\hat{S}^{2}$.


What is spatially and chemically heterogeneous materials? Follow this paper,

Structural analysis of humins formed in the Brønsted acid catalyzed dehydration of fructose

They explained "humins are spatially and chemically heterogeneous materials."

I found "spatial" means "about space" but I think this meaning doesn't correlate with this context"

In summary,
What is spatially and chemically heterogeneous materials? and How did they know humins are spatially and chemically heterogeneous materials?

Spatially heterogeneous means that the composition varies as a function of position within the material (varies "in space").

Chemically heterogeneous means that the chemical composition just varies in general. In this case, I assume they desire to emphasize that the composition varies among species of widely differing molecular weights—from large polymers down to small molecules.

So, it appears to me that "spatially and chemically heterogeneous" was used to emphasize that the chemical composition varies significantly, in particular as compared to prior assumptions; and that this compositional variability itself varies as a function of position within a given volume of the material.

In general, "spatially and chemically heterogeneous" would be rather a redundant term: a material can't really be spatially heterogeneous without also being chemically heterogeneous: if a substance is chemically pure, there is no possibility of any spatial distribution of compositional variations. So, I think the particular history of this material, which apparently had been assumed to consist only of polymeric material, is what led the authors to use the phrase.


Weak acids compared to strong ones in pH > 7 solution In the book of Molecular Cell Biology (Big Alberts) by Alberts, at page 46, it is stated that

Acids—especially weak acids—will give up their protons more readily if the concentration of > H3O+ in solution is low and will tend to receive them back if the concentration in solution is high.

Does this mean that, in a solution with high pH, the weaker acids will behave like a strong acids in water, in the sense that they will be more inclided to give up their protons compared to strong acids?

When the ratio of conjugate base concenteration to conjugate acid concenteration is more than 1000 , we  say that the  dissociation of acid is complete.
According to Henderson–Hasselbalch equation the ratio of conjugate base concenteration to conjugate acid concenteration gets more than 1000 , when PH-PKa is more than 3.
For strong acids, this ratio is more than 1000 in PH>2 ((  actually it is true for most strong acids )) , but for weak acids , there is no specefic PH . Actually, higher PH , higher dissociation ratio.
So ,when the PH gets higher, the dissiciation ratio gets greater and more poroton will be given up.
Actually, when PH-PKa gets more than 3 for weak acids , they behave like strong acids and allmost all of the acid will dissociate.


Why doesn't calcium carbonate dissolve in water even though it is an ionic compound? I studied the solubility of compounds in water. I have found that calcium carbonate doesn't dissolve in water. The teacher stated that the ionic compounds dissolve in water except some carbonates. What would be a clear elaboration of this phenomenon?

As someone said here, this:

  The teacher stated that the ionic compounds dissolve in water except
  some carbonates.

Is indeed an oversimplification. First of all, the distinction between an "ionic compound" to other compounds isn't too defined. What your teacher probably said, or didn't say but wanted to, is that some ionic compounds easily dissolve in water. Salt (halite - NaCl) is the best example.

Calcium carbonate, in nature, also commonly dissolves. It's just not as immediate as dissolution of the more soluble ionic compounds. You are probably familiar with this phenomenon:

This forms because calcium carbonate dissolves. The rock is limestone, which is usually composed of pure calcium carbonate. Acidic water greatly enhances the solubility of calcium carbonate, and it doesn't even need to be highly acidic. Rain or river water that come into contact with the atmosphere absorb the $\ce{CO2}$ as $$\ce{H2O + CO2 <=> H2CO3},$$ which then facilitates calcium carbonate dissolution with $$\ce{CaCO3 + H2CO3 <=> Ca^2+ + 2HCO3-}.$$


HCH bond angle in cyclopropane and cyclobutane I understand why the $\ce{CCC}$ bond angles in cyclobutane and cyclopropane are less than ideal $109.5^\circ$ due to the geometric restrictions they encounter in a cyclic structure. However, it is not clear to me how the $\ce{HCH}$ angles (which turn out to be greater than ideal) are affected by this constraint.

Computational investigations of geometry, aromaticity and bond energies at HF/6-31G** theory level [1] revealed the following deformed  geometrical ($\alpha$) and interpath angles ($\beta$ -- the angle between two bond paths) for $\ce{H-C-H}$ bond, which are slightly deviating from the strain-free $109.5^\circ$ due to varying Baeyer strains:

\begin{array}{lr}
\hline
\text{molecule} & \alpha,~^\circ & \beta,~^\circ \\
\hline
\text{cyclopropane} & 114.1 & 117.0 \\
\text{planar cyclobutane} & 107.9 & 109.7 \\
\text{puckered cyclobutane} & 108.3 & 110.0 \\
\hline
\end{array}

P. S. Unfortunately, this data cannot be taken from single-crystal x-ray experiment as this method doesn't allow precise determination of H-positions, and I didn't find a neutron diffraction data on solid cycloalkanes mentioned.

Reference

Cremer, D.; Gauss, J. J. Am. Chem. Soc. 1986, 108 (24), 7467–7477 DOI: 10.1021/ja00284a004.


Do we consider the spatial orientation of a molecular orbital while assigning plane of symmetry? I am following the video lectures by Neeraj Saini sir and in this video$^{\dagger}$ he said that $\ce{CO2}$ has infinite planes of symmetry.
But I think that if we take into account the orientation of $π$ bonds (which I don't know if we have to or not) then $\ce {CO2}$ molecule must  have $2$ planes of symmetry, as per the following argument:
A $\ce {CO2}$ molecule has similar $\pi$ bond orientation as that of an allene.
Now if we consider the orientation of pi bonds then $\ce {CO2}$ must have only 2 planes of symmetry.
So, to check whether infinite planes of symmetry statement is correct or not, I have the following question:

Do we consider the spatial orientation of molecular orbitals while assigning plane of symmetry? And if not then why not?

$\dagger$ Note that he uses Hindi language for the purpose of conversation.

The reason that orbitals are not considered in the symmetry of a molecule is that the orbitals cannot decrease the symmetry of the molecule. If we look at the canonical delocalized molecular orbitals of any molecule, everyone of them is symmetric or antisymmetric with respect to a symmetry operation of the molecule, with those symmetry operations determined only by the positions of the atoms, not the electron density.
Looking specifically at carbon dioxide, the simplistic picture of $\ce{CO2}$ as being like allene with lone pairs instead of the C-H bonds is not an accurate view of the electron density. If one looks at the delocalized pi-type orbitals, both span all three atoms in either the xz or yz planes (using the standard convention of the z axis being aligned with the molecular axis). Thus, both orbitals are symmetric with respect to the xy plane of symmetry of the molecule (which passes through the C).
Similarly, the lone pairs do not occupy separate sp2 orbitals that have specific alignment in a plane perpendicular to the p orbital involved in pi bonding. Instead, the create a combined electron density that can be deconvoluted into an sp-type orbital along the z-axis and partial contributions from the px and py orbitals, such that the total has no bias along the x or y axis.
If we were to look at the total electron density of $\ce{CO2}$, we would therefore see only a pill shape that does not vary as one rotates around the z-axis.
To visualize this, I recommend looking at all of the calculated delocalized molecular orbitals of $\ce{CO2}$. You can do this on the molcalc.org website if you do not have other software for that purpose.


How do I eliminate the disagreeable odor of soy milk? I've was making tofu from soy-milk, but noticed this disagreeable odor produced during the process.

I conducted a string of internet searches in an attempt to find a method to eliminate this odor.

I learnt that the enzyme lipoxygenase is primarily responsible for this odor, and is activated upon grinding soy beans in water. Additionally, it appears that lipoxygenase is inactivated in the presence of  D-glucose and the enzyme glucose oxidase.

Obtaining D-glucose is fairly simple, but I'm at odds as to where I can find glucose oxidase.

Further investigation suggests that commonly available honey contains glucose oxidase.

Armed with this knowledge, I want to know:

Can I add a measured quantity of dilute honey, along with some D-glucose, to the soy milk and inactivate the lipoxygenase enzyme this way?

One more query...
I found somewhere that glucose and glucose oxidase in water, makes hydrogen peroxide and this H2O2 inhibit the lipoxygenase enzyme... So my query is.... What, if add hydrogen peroxide directly, in a measured quantity, in water and grind the soybeans with this solution.

A few things first:

Enzymes are proteins (usually) that promote or accelerate a biochemical reaction. The enzyme lipoxygenase in particular, promotes the dioxygenation of some lipids called Polyunsaturated Fatty Acids (PUFAs); for this, the enzyme requires molecular oxygen (which is present in the air, and gets dissolved in the soy milk) . You can think of this as, "it makes the soy milk go bad". The oxidation of the PUFAs in the soy milk doesn't need lipoxygenase...but having lipoxygenase around makes things faster ;  if the oxidation of over 90% of the PUFAs in a bowl of soy milk takes (say) 10 days without the lipoxygenase...then it would take (say) a day with lipoxygenase present . As to what specific substance(s) is responsible for the smell, I don't really know yet.

The enzyme glucose oxidase promotes the breakdown of D-glucose. Once again, it too requires molecular oxygen.

Now the basis of that patent you cited (I haven't actually read the thing) is apparently along the lines of:

  If you dump in glucose and (a small amount of) glucose oxidase into the soy milk, the reaction that follows (oxidation of glucose) should hopefully decrease the molecular oxygen content on the soy milk. This should decrease the activity (not complete inactivation though) of lipoxygenase and hence should reduce the intensity of the odor (not complete removal of the odor)

Now you suggest using dilute honey in place of pure glucose oxidase (since the former's easy to get). But that would barely work.

The composition of honey depends on multiple factors; type of bee, region, season, etc. But from looking up the composition of various kinds of honey (courtesy: Google), apparently glucose oxidase constitutes a fraction of honey. So diluting honey here would make things worse (lower concentration of glucose oxidase). Obviously, adding concentrated honey is also out of question...since that would alter the flavor of the tofu altogether.

As @airhuff mentioned in the comments below: Use of honey is still an option...but a poor option, nonetheless. Why you ask? Because there are no simple means for one to go about quantifying the glucose oxidase content in some random honey-sample. The process(es) are time-consuming, pricey and hence, (probably) impractical for your needs. Additionally, even if it contains the "right" amount of glucose oxidase, I fear that even dilute honey would still alter the taste of the tofu (there's a reason why honey is the popular epitome for "sweetness").

What can you do now?

1) You can buy glucose oxidase online at Sigma-Aldrich. But if your tofu-making session is a one-time thing (i.e- You don't run a tofu-making factory) then this is a bit impractical.

2) Minimize contact with air! The primary reason the soy milk reeks is because the lipoxygenase is breaking down PUFAs using molecular oxygen. So if you want to limit the enzyme's activity, you must keep contact between the soy milk and air to a minimum.

Some ways you can do this:

a) As soon as your done grinding the soy beans with water, transfer it to an air-tight container (fill it all the way to the brim) until you need it.

b) When you're transferring the soy milk from one vessel to another, be sure to avoid air-bubbles. This answer by @Buttonwood may prove useful in that regard.

On a side note:

I'm genuinely pleased to see that you've gone to great lengths to find a solution to your problem before you approached us at Chemistry.SE. For someone who probably doesn't have much of a background in Chemistry (In the original post, you mentioned you were a "commerce student"), this post came as a pleasant surprise. I myself am only student, and I'm relatively new to Chemistry.SE (been here for under a year)... and you'd be surprised to know how so very few users (from a "non-Chemistry" background such as yours) proactively attempt to resolve their problems before resorting to Chem.SE; so kudos to you! We look forward to your future contributions on Chem.SE. Cheers!


If all chemical processes are equilibria at least to some extent, where does the activation energy come from? If all chemical processes are equilibria at least to some extent, then where does the energy to overcome the activation energy come from?

Is a process with a very high activation energy (i.e. the ionization of sodium vapor into sodium ion) still considered an equilibrium process at least to some extent?

If so, where might the energy to furnish the forward reaction come from naturally - i.e. with no human input of energy?

I don't think the statement that all processes are in equilibrium to some extent is true. By definition, dynamic equilibrium means that the rate of change in one direction is equal to the rate of change in the opposite direction.

This is certainly not true for all processes, otherwise there would not be a process - nothing would appear to change.

On the other hand, for any given reaction equation that is known to occur, we can say:

for a reaction at equilibrium, neither the concentration of reactants nor the concentration of products will be zero

I think that may be what you meant in your question?

If that is the case, then let's look at your example:

  Is a process with a very high activation energy (i.e. the ionization of sodium vapor into sodium ion) still considered an equilibrium process at least to some extent?

It's not an equilibrium process. However, the reaction:

$\ce{Na(g)->Na(g)+ + e-}$

could be at equilibrium. In other words, the ratio of reactants to products could be such that the Gibbs free energy is at a minimum for this temperature and pressure. A more mathematical way to say that is:

$Q = K_{eq}$

where $Q=\frac{[Na^+]}{[Na]}$ and $K_{eq}$ is the equilibrium constant under these conditions. (Brackets mean "partial pressure" - not concentration here.)

Under normal lab conditions (1 atm, 298 K), this is not thermodynamically favorable. The reaction quotient heavily favors the reactants, or another way to say it is $K_{eq}$ is really close to zero. So close you may not be able to measure it. The activation energy is extremely high (equal to the ionization energy), and so any electrons that are removed would have to get that energy from somewhere - which was your next question:

  If so, where might the energy to furnish the forward reaction come from naturally - i.e. with no human input of energy?

In an idealized, isolated system, you only have one source of energy for things like that: kinetic. Thermal motion of the atoms could conceivably put enough energy into a single atom that an electron could be "kicked off" and result in an ion. However - the odds of that happening are extremely low, given that the distribution of speeds would have a vanishingly small, although still non-zero, probability of occurring at the magnitude needed to overcome the activation energy barrier. Since the reaction is thermodynamically unfavorable, the probability that the electron would jump right back on is very high - much higher than it would be for leaving in the first place.

Now when you have $6.022 \cdot 10^{23}$ opportunities each femtosecond, very low odds can still work out. But the total amount that would be ionized at any one time is still very, very low. Possibly so low that you would have to wait for an essentially infinite amount of time to see a measurable amount of it.

In a real system, a much larger fraction would be ionized (and this may or may not be reflected in $K_{eq}$, depending on how it is measured). The reason is that there is always ionizing radiation present on earth, and there will be more or less depending on where you are located. Statistical thermodynamics doesn't take this into account, so any prediction of the relative amounts of reactants and products would not be correct. This is kind of a unique example in that respect - most equilibrium reactions, even with very slow kinetics, deal with systems where the thermal energy of the system has a much bigger effect than radiation.


Is an SN1 reaction an elementary reaction/one step reaction? I need some help. From what I understand, elementary reactions are reactions which only has one transition state. Another definition that I have come across is that elementary reactions obey the rate law.

I believe that $S_N1$ are NOT elementary reactions because they proceed through more than one transition state. However I am not so sure about the other definition (the rate law one). Could someone explain to me what exactly an elementary reaction (or a one step reaction) is? And can someone explain to me if an $S_N1$ reaction is an elementary reaction or not.

From http://goldbook.iupac.org/E02035.html:

  A reaction for which no reaction intermediates have been detected or need to be postulated in order to describe the chemical reaction on a molecular scale. An elementary reaction is assumed to occur in a single step and to pass through a single transition state.

So you're right that elementary reactions occur in a single step/have a single transition state. I'm not sure what you mean by "obey the rate law", but necessarily, a reaction that has only a single transition state is first order in each reactant. Because there is only a single step, the concentration of every molecule involved in that step effects the rate.

In practice, because the probability of three molecules colliding together in a way that can cause a reaction is very low, all elementary reactions involve one or two molecules. A single molecule reaction is just a dissociation, e.g. $A_2 \rightarrow A + A$, with a first order rate in A and an overall order of one. A two molecule reaction could be something like $A+B\rightarrow AB$, with still a first order rate in each of the reactants for an overall order of two.

In your question, you've identified that SN1 reactions have multiple transition states, so SN1 cannot be an elementary reaction. In fact, the 1 in SN1 means that the overall order of the reaction is one, but since it involves two reactants, the order would be two for a single step reaction. Naturally, an SN2 reaction has an overall order of two and is an elementary reaction, proceeding through a single transition state where a bond between the incoming nucleophile and the carbon centre is formed at the same time as the leaving group's bond is broken.


Why is carbon dioxide gas used in soda? Why is carbon dioxide used in soda? What makes it preferable to other gases for this purpose?

You are approaching this question from the wrong angle. One does not simply ‘add a gas to a drink’ for no reason. If one wanted to add any gas, one would probably use boring old nitrogen since it is the most abundant in the atmosphere (See the Guiness example as pointed out by Ivan in a comment).

Carbon dioxide is present in fizzy drinks such as soda to give them their fizz. The reason lies in the equilibrium between $\ce{CO2 + H2O}$, $\ce{H2CO3}$, $\ce{HCO3-}$ and $\ce{CO3^2-}$. No other gas would display this behaviour and give a slight acidic taste along with the tingle on the tongue.

Note that oxygen is sometimes added to drinks to improve the assumed value. It doesn’t really do anything but sells for a higher price. I don’t think that would be the case for nitrogen- or argon-enriched water (but I am likely underestimating the powers of marketing here). Many other gases would simply make no sense to add at all because they are either toxic ($\ce{Cl2, H2S}$) or disgusting ($\ce{H2S}$) or simply way too expensive ($\ce{Xe}$).


Shape and polarity of the iodate ion I am trying to figure out the shape of the iodate ion or $\ce{IO3^-}$. No matter how many times I draw it, I keep getting iodine in the centre, and 3 oxygen atoms surrounding it, 2 with single bonds, and 1 with a double bond. This would make it a trigonal planar shape. It would mean that it is not a "bent" shape.

I went to double check on the internet, and I am now confused. On Wikipedia it says it is in fact polar and a pyramidal shape.

Why does iodine have a pair of lone electrons in its shell? Why would it not combine with one of the oxygen atoms to make a double bond?

Looking at the formal charges on the atoms, each oxygen has a charge of -2. With the added electron, the iodine has a formal charge of +5. This leaves 2 electrons in the valance shell for the lone pair. So there are 4 items (3 oxygens and a lone pair) to place around the iodine atom, giving it a tetrahedral shape. Since each oxygen has the same charge of -2, there must be 2 bonds to each oxygen. The extra electron is placed on the molecule, not the iodine atom. Since you have a tetrahedral shape with a different corner (lone pair), it has a dipole (like ammonia).


How to disperse salicylic acid in sodium lauryl sulfate solution? Anti-dandruff shampoo contains salicylic acid as an active ingredient, but when salicylic acid powder comes in contact with a base solution such as sodium laurylsulfate and water (1:3 wt/wt) caking of the particle occurs, forming a non-dispersible aggregate.

I know from the study of pharmaceutics that decreasing the contact angle between particles and solution leads to a decrease in surface tension and thereby dispersion of the particles.

How can one disperse the salicylic acid particles specifically in this base solution?

I don't have a lot of experience with salicylic acid as I'm just starting to work with it myself at the moment. However, I find that it disperses better if it is added to the solution at a high temperature (before surfactants) and then homogenised (once thickening agents, if any, have been added).

The other thing which may help is by dissolving in denatured alcohol, although I haven't personally tried this myself. Although it's not a shampoo, I have recently formulated a facial cleanser and as described above I firstly added the thickening agents to the water and then salicylic acid, heated to approximately $70~\mathrm{^\circ C}$ and then homogenised.


The reasoning behind predominance diagrams Consider a one dimensional predominance diagram for a simple acid-base reaction:

$$\ce{AH <=> A^- + H^+}$$

With equilibrium constant:

$$K_a = \frac{[A^-]_{eq}[H^+]_{eq}}{[AH]_{eq}}$$

If I understand correctly, $[H^+] > K_a$, then $[AH]>[A^-]$. At first I thought I understood why this was so: the $[H^+]$ cancels out on both sides of the inequality, and then you bring $[AH]$ to the other side. Only now I realized that you can't actually cancel $[H^+]$ because on one side of the inequality it's actually $[H^+]_{eq}$, not the current amount of protons but the amount at equilibrium. So now I'm lost. What justifies this rule for determining predominance?

I think you might be making a fundamental mistake when you differentiate between the "current amount of protons" and the "amount at equilibrium." All dissociation constants are based on equilibrium concentrations, and predominance diagrams refer to the concentrations present only after equilibrium has already been established.

Consider the concrete example of the dissociation of ammonium: $\ce{NH4+ <=> NH3 + H+}$. The equilibrium constant expression for this chemical process is:

$$
K_\text{a} = \frac{\ce{[NH3][H+]}}{\ce{[NH4+]}}
$$

The $\mathrm{p}K_\text{a}$ of this process has been experimentally determined to be approximately $9.25$. Therefore, the following equation should hold:

$$
-\log(K_\text{a}) = -\log\frac{\ce{[NH3][H+]}}{\ce{[NH4+]}} = \mathrm{p}K_\text{a} = 9.25
$$

If a given solution has a measured $\mathrm{pH}$ precisely equal to the $\mathrm{p}K_\text{a}$ of the above process, then what we have is the following (keeping in mind that $\ce{[H+]} = 10^{-\mathrm{pH}}$):

$$
\log\frac{[\ce{NH3}]\cdot 10^{-9.25}}{[\ce{NH4+}]} = -9.25
$$

Taking the anti-log of both sides of the equation:

$$
\frac{[\ce{NH3}]\cdot 10^{-9.25}}{[\ce{NH4+}]} = 10^{-9.25}
$$

Then dividing both sides of the equation by $10^{-9.25}$, we find:

$$
\frac{[\ce{NH3}]}{[\ce{NH4+}]} = 1
$$

That is to say, the concentrations of ammonium and ammonia in the solution are perfectly equal when the $\mathrm{pH}$ and $\mathrm{p}K_\text{a}$ are equal. If the $\mathrm{pH}$ is higher (that is to say, more basic), then you'll find that your conjugate base (in this case, $\ce{NH3}$) will predominate by factor of $10^{\mathrm{pH} - \mathrm{p}K_\text{a}}$. Conversely, if the $\mathrm{p}K_\text{a}$ is higher, then the conjugate acid ($\ce{NH4+}$) will predominate by a factor of $10^{\mathrm{p}K_\text{a} - \mathrm{pH}}$.

If you prefer to use an inequality and some generic conjugate acid/base pair $\ce{HA/A- }$, then the logic goes as follows.

If $\mathrm{pH} > \mathrm{p}K_\text{a}$:

$$
\mathrm{pH} > \mathrm{p}K_\text{a}\\
-\log\ce{[H+]} > -\log\frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\
\ce{[H+]} < \frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\
1 < \frac{\ce{[A- ]}}{\ce{[HA]}}\\
\ce{[HA]} < \ce{[A- ]}
$$

If $\mathrm{pH} = \mathrm{p}K_\text{a}$:

$$
\mathrm{pH} = \mathrm{p}K_\text{a}\\
-\log\ce{[H+]} = -\log{\frac{\ce{[H+][A- ]}}{\ce{[HA]}}}\\
\ce{[H+]} = \frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\
1 = \frac{\ce{[A- ]}}{\ce{[HA]}}\\
\ce{[HA]} = \ce{[A- ]}
$$

If $\mathrm{pH} < \mathrm{p}K_\text{a}$:

$$
\mathrm{pH} < \mathrm{p}K_\text{a}\\
-\log\ce{[H+]} < -\log\frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\
\ce{[H+]} > \frac{\ce{[H+][A- ]}}{\ce{[HA]}}\\
1 > \frac{\ce{[A- ]}}{\ce{[HA]}}\\
\ce{[HA]} > \ce{[A- ]}
$$


Why V can't have the valence 7 by taking an electron from the 3p orbital and putting it into 3d? Does that happen because 3p already has a bond?

Vanadium has a stable valency of 5.
Electrons cannot be transferred from $3p$ orbital to $3d$ orbital as the $3p$ orbital is completely filled and has attained stability.
It requires high energy so as to transfer electrons from $3p$ orbital to $3d$ orbital.
Hence,Vanadium has a valency of 5 only and not of 7 or 6.


How is (SO4)^2- Charged if S and O all are Neutral Themselves? As far as I understand $\ce{SO4^2-}$ structure,  sulfur makes two coordinate bonds with two oxygen atoms and two double bonds are formed with oxygen after sulfur expands it's valance shell to 6. At this point calculating formal charge of those two coordinate bonded oxygen gives total formal charge of −2 and the formal charge on $\ce S$ and the double bonded $\ce O$ is zero. Thus the minus 2 charge on $\ce{SO4^2-}$ can be shown but initially all the atoms were neutral. While making bonds electrons were shared only.

How did excess 2 negative charge pop up?

Please help, I am certainly making some mistakes

The ion is not one S atom and 4 O atoms together. The ion on the whole is having two negative charges, Thus it is somewhat like 1Sulphur + 4 Oxygen+2 electrons. It is not as though there was one sulphur and four neutral oxygen atoms, and they became negatively charged during bond formation. The bond must have formed in some reaction, and in that reaction the charge would have been conserved.


How does the equipartition theorem apply to vibrational modes? Specifically, how, based on the idea that energy will be shared equally among all degrees of freedom, does one come to the conclusion that each vibrational mode of a molecule will occupy $kT$ units of energy?

When learning statistical thermodynamics, classes often spend a lot of time talking about the Maxwell-Boltzmann distribution and even deriving that the expectation value for the energy of a monoatomic ideal gas is $\frac{3}{2}kT$ where $\frac12 kT$ comes from energy distributed to each cartesian coordinate plane.

Now, I find that I learn a whole lot from mathematical derivations of these types of things, and from simply working through the math and understanding its physical meaning.

So, I want to know, both on a conceptual level and through some sort of mathematical explanation, why does each vibrational mode of a molecule contribute $kT$ to the overall energy of the molecule when those modes are occupied?

My only guess is that one degree of freedom is kinetic energy (associated with the moving atoms) and the other is potential energy (analogy with a spring works here) but, if that is true, much clarity would be gained from a mathematical demonstration of this.

As you probably already know, the equipartition theorem states that each quadratic
degree of freedom will, on average, possess an energy $\frac{1}{2}kT$, where a ‘quadratic degree of freedom’ is one for which the energy depends on the square of some property.
Now lets consider the kinetic and potential energies associated with translational, rotational and vibrational energy:

Translational degrees of freedom:  $~E = \frac{1}{2}mv^2$

Rotational degrees of freedom:  $~E = \frac{1}{2}Iω^2$

Vibrational degrees of freedom: $~E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$

As you can see that for translational and rotational degrees of freedom, there is only one quadratic. Hence each translational and rotational state will contain $\frac{1}{2}kT$ energy.
However for vibrational states, the energy contains two quadratic contributions. Therefore each vibrational state will contain $kT$ energy.
Now lets consider the reasoning behind this. You are right when you say that the vibrational degree of freedom depends on kinetic and potential energy and hence has 2 quadratic contributions. In the equation for the energy of the vibrational state, $\frac{1}{2}mv^2$ represents the kinetic energy and $\frac{1}{2}kx^2$ represents the potential energy.
A molecule in the vibrational state acts like a harmonic oscillator. So let consider $H_2$. You can imagine that the bond between the 2 hydrogen atoms is like a spring which contracts and expands in length.

The average bond length can be thought as the equilibrium length of the bond. When the bond the contracts, the displacement from this equilibrium length is known as 'x'. Now using Hooke's Law, the potential energy of the system is $\frac{1}{2}kx^2$ where k is the spring constant while the kinetic energy is $\frac{1}{2}mv^2$. Hence the total energy of the system is given by: $~E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$


Benzaldehyde and Cyanide - if you can smell one can you smell the other? Cyanide and benzaldehyde both have an almond-like smell.

Some people cannot smell cyanide (specifically, HCN). Can they also not smell benzaldehyde?

In other words, do they both trigger the same receptors?

Answer is, they are most probably not correlated. Smell requires molecules to bind to receptors, which they do by structure and functional group. CN is a very small linear molecule that does not even share its heteroatoms with benzaldehyde, an aromatic compound several times cyanide's size. Ergo, it's improbable they share the same scent receptors.